diff --git "a/stack-exchange/math_overflow/shard_15.txt" "b/stack-exchange/math_overflow/shard_15.txt" deleted file mode 100644--- "a/stack-exchange/math_overflow/shard_15.txt" +++ /dev/null @@ -1,30139 +0,0 @@ -TITLE: Torsion subgroups of hyperbolic groups are finite? -QUESTION [14 upvotes]: Is it true that torsion subgroups of hyperbolic groups are finite? -I have a vague memory that this is true, perhaps due to Ol'shanskii, but have been struggling to find a reference. -(By a torsion subgroup I mean a subgroup $T\leq H$ such that every $t\in T$ has finite order. Although there are finitely many conjugacy classes of finite order elements in $H$, this does not immediately imply that $T$ is finite.) - -REPLY [2 votes]: Alternatively, one can refer to Lemma 3.4 from Osin's article Acylindrically hyperbolic groups. -Proposition: If a group acts on a hyperbolic space uniformly properly$^*$ with unbounded orbits, then it contains a loxodromic element. -This implies directly that torsion subgroups of hyperbolic groups are finite. The result is attributed to Ivanov and Olshanskii, the proof given by Osin being extracted from Lemma 17 in Hyperbolic groups and their quotients of bounded exponents. -$^*$An action $G \curvearrowright (X,d)$ is uniformly proper if, for every $R \geq 0$, there exists $N \geq 0$ such that $\# \{g \in G \mid d(x,gx) \leq R \} \leq N$ for every $x \in X$.<|endoftext|> -TITLE: Endomorphisms of elliptic curves with CM; can we have an order? -QUESTION [6 upvotes]: Let $E$ be an elliptic curve over $\mathbb C$ with CM by ring of integers $O_K$ of an imaginary quadratic number field $K$. Let $O$ be an order of $O_K$. -Is there a number field $L$ such that $E$ has a model $E_L$ over $L$ with $\mathrm{End}_L(E_L) = O$? -Of course, if $O = O_K$, the answer is positive. But what if $O$ has a non-trivial conductor? -Concretely: Let $E : y^2= x^3 +x$ over $\mathbb Q$, and let $f\geq 2$ be an integer. Is there a number field $L$ such that $\mathrm{End}_L(E_L) = \mathbb Z[fi]$? - -REPLY [13 votes]: No. If $\alpha$ is an endomorphism over $\bar L$ such that $n\alpha$ -is defined over $L$, then so is $\alpha$. Proof: suppose some element -of Gal$(\bar L / L)$ takes $\alpha$ to an endomorphism $\beta$. -We shall prove that $\beta=\alpha$. Indeed by hypothesis $n\beta = n\alpha$. -But then $n(\beta-\alpha) = 0$, so the endomorphism $\beta-\alpha$ -takes $E$ to the $n$-torsion group $E[n]$, which is finite. -Hence $\beta-\alpha$ is the zero endomorphism; that is, $\beta=\alpha$. -Therefore $\alpha$ is Galois-invariant, -and thus defined over $L$, as claimed. $\Box$ -This proof works more generally with the elliptic curve $E$ -replaced by any abelian variety.<|endoftext|> -TITLE: Binomial again, and again -QUESTION [37 upvotes]: Let $\lceil a\rceil=$ the smallest integer $\geq a$, otherwise known as the ceiling function. When the arguments are real, interpret $\binom{a}b$ using the Euler's gamma function, $\Gamma$. -Recently, I posted a problem on MO about "sin-omials". Here is another one in the same spirit. - -QUESTION. Numerical evidence suggest that - $$\left\lceil \int_0^n\binom{n}x dx\right\rceil=\sum_{k=0}^n\binom{n}k.$$ - If true, how do we prove this? - -REMARK. The RHS evaluates to $2^n$; I only want to stress parallelism between the integral and sum. - -REPLY [5 votes]: The $2^n$ result in Noam Elkies answer generalizes to $n=s$ for $real(s)>-1$, as he notes in MSE. In this range, the generalized binomial coefficient $ \binom{s}{\alpha}$ is a Fourier band-limited fct. in $\alpha$, or, equivalently a sinc fct. interpolation of the sequence $\binom{s}{n}$. See my notes "Fractional calculus and interpolation of generalized binomial coefficients" for simple derivations and my contribution to the MO-Q Generalizing a problem to make it easier for the integral of the sinc fct. over infinte limits, also simply derived. -The Wikipedia article on the classic Gibbs phenomena explores the extrema of $Si(x)$, easily visualized as a convolution of the sinc with the rectangle function, and, therefore, bounds on the residual of the ceiling. -The sinc function interpolation is an important illustration of the Nyquist-Shannon sampling theorem and a special case of a generalized Chu-Vandermonde identity using Euler's reflection formula for the gamma/factorial function. (I remember long ago reading somewhere that Ramanujan noted this. Anyone know a ref? This would be simple check for his Master Theorem/Formula.)<|endoftext|> -TITLE: Preservation results in abstract logics -QUESTION [7 upvotes]: In retrospect the original version of this question was impossibly bloated. Here's a better version: -There are many results about when first-order sentences are preserved by algebraic operations on model classes; for example, in first-order logic the sentences preserved by taking substructures are those semantically equivalent to (= having the same models as) universal sentences. -I'm broadly interested in preservation results beyond first-order logic. In particular, I'd like to know if preservation under taking (arbitrary) Cartesian products nicely characterizable in second-order logic (preservation under taking substructures is actually simpler for $\mathsf{SOL}$ than $\mathsf{FOL}$ since the former can directly quantify over substructures). -To pose this precisely: - -Is there a computable set of $\mathsf{SOL}$-sentences $X$ such that the $\mathsf{SOL}$-sentences preserved by Cartesian products ("productive sentences") are exactly those semantically equivalent to elements of $X$? - -Note that the productivity of a given $\mathsf{SOL}$-sentence is not set-theoretically absolute. However, neither is semantic equivalence between even individual $\mathsf{SOL}$-sentences, so this doesn't immediately give a negative answer to the question. -I'm also interested in what happens if we weaken "semantically equivalent to elements of $X$" to "semantically equivalent to possibly infinite conjunctions of elements of $X$." - -REPLY [2 votes]: EDIT: Now understanding "semantically equivalent", I change my answer (my previous answer, to a different question, is further below)...I'm working in ZFC. -Yes, there is such a computable set $X$; that is, $X$ is a set of productive second order sentences, and for every productive second order $\mathcal{L}$-sentence $\psi$, there is a semantically equivalent sentence $\varphi\in X$ (having the same models). Here we can take "productive" either to mean w.r.t finite products or set-sized products, and we fix countably many constant/function/relation symbols that we deal with. -Proof: The plan is to (i) give a recursive procedure to pass from sentences $\varphi$ to sentences $\psi_\varphi$ such that $M\models\psi_\varphi$ iff $M$ is a product $\Pi_{i\in I}M_i$ such that $I\neq\emptyset$ and each $M_i\models\varphi$, -and (ii) then use $X=\{\psi_\varphi\bigm|\varphi\text{ is an second order }\mathcal{L}\text{-sentence}\}$. -First let's consider the set-product version, so $I$ can be whatever set above. -Note that if we can do (i), then $X$ as in (ii) works: Each $\psi_\varphi$ -is easily productive, and if $\varphi$ is productive then $\varphi$ is semantically equivalent to $\psi_\varphi\in X$. (If $M\models\varphi$ then $M\models\psi_\varphi$ (take the product of only $M_0=M$, with $I=\{0\}$), and if $M\models\psi_\varphi$ then because $\varphi$ is productive, $M\models\varphi$.) -So it suffices to do (i). A slightly annoying thing to deal with here are empty models and models of cardinality 1, because they contribute less to the cardinality of a product. Let $M$ be the model in question. Then $\psi_\varphi$ says: if $M\models\neg\varphi$ then $M$ is isomorphic to a product $\Pi_{i\in I}M_i\times\Pi_{i\in J}M_i$ where $I\cup J$ has cardinality $>1$ and each $M_i\models\varphi$, and for each $i\in I$, $M_i$ has cardinality 1, and for each $i\in J$, $M_i$ has cardinality $\neq 1$. Note that we can now ignore the case that some $M_i$ is empty, because this just makes the product empty (hence isomorphic to $M_i$). -Note that we can identify (in SOL) whether $M$ is infinite. -Case 1: $M$ is infinite. -Note that in order to be a product as above, $M$ must have cardinality $\Pi_{i\in J}\lambda_i$ where $\lambda_i=\mathrm{card}(M_i)$. Since $M$ is infinite, and $\mathrm{card}(M_i)>1$ for $i\in J$, we have $0<|J|<2^{|J|}\leq\mathrm{card}(M)$, so we can find a subset of $J'\subseteq M$ of cardinality $|J|$. -Similarly, each $\lambda_i\leq\mathrm{card}(M)$, so we can find a subset $N'_i\subseteq M$ with $\mathrm{card}(N'_i)=\lambda_i$, and a structure $M'_i$ on $N'_i$ such that $M'_i\equiv M_i$ (note these are all structures in the same finite language that $\varphi$ uses). Fixing bijections $\pi:J'\to J$ and $\sigma_i:N'_i\to M_i$, -let $M'_i$ be the structure on $N'_i$ isomorphic to $M_i$ via $\sigma_i$. -We can find some (finitely many) relations describing these things; e.g. -we have a binary relation $R$ where $R(x,y)$ iff $x\in J'$ and $y\in N'_{\pi(x)}$. -Suppose for the moment that $I=\emptyset$; we just want to assert that $M$ -is isomorphic to the product $\Pi_{i\in J}M_i$. For this, we want to assert -(using SOL) that there is a binary relation $R$ (on the universe of $M$) -and some other relations $R^P$ etc for the symbols $P$ etc used in $\varphi$, -coding a set $J'$ and models $M'_i$ as above, each $M'_i$ models $\varphi$, -and $M$ is isomorphic to their product. So it just remains to assert the isomorphism. But this is just done by coding an isomorphism through another relation: Code an isomorphism -$\tau:M\to\Pi_{i\in J}M_i$ via the ternary relation $R_\tau$ where -$R_\tau(a,x,y)$ iff $a\in M$ and $x\in J'$ and $R(x,y)$ and $\tau(a)(\pi(x))=\sigma_i(y)$. -(That is, $\tau(a)$ is an element of the product, i.e. a function -$\tau(a):J\to\bigcup_{i\in J}M_i$ with $\tau(a)(i)\in M_i$ for each $i\in J$. -And $\pi:J'\to J$ is the bijection above. So $\tau(a)(\pi(x))=\tau(a)(i)\in M_i$, -so this is of the form $\sigma_i(y)$ for some $y\in N'_i$.) -Thus, we just have to assert that there is a binary relation $R$ on $M$, and -various relations $R^{P_1},\ldots,R^{P_n}$ for the symbols $P_1,\ldots,P_n$ in $\varphi$, -such that $R,R^{P_1},\ldots,R^{P_n}$ code a collection of models, indexed -by $\mathrm{dom}(R)$, each modelling $\varphi$, and there is a ternary relation $R'$ on $M$, such that $R'$ codes an isomorphism between $M$ and the product -of the models just mentioned, as coded above. -This deals with the $I=\emptyset$ case. Now suppose $I\neq\emptyset$. -The slight irritation here is that these factors of the product don't add to the cardinality of $M$, so it might be that $I$ has cardinality bigger than $M$. However, note that this case is irrelevant, and in fact, we can assume that $I$ is finite. This is because there are only finitely many non-isomorphic models with 1 element and in the symbols used in $\varphi$, and the if $M_i,M_{i'}$ are isomorphic 1-element models, then note that their product is still isomorphic to the same model; likewise for infinite products of the same 1-element model. So we may assume $I$ is finite. That being the case, it is easy to modify the preceding paragraph to incorporate the full product $\Pi_{i\in I}M_i\times\Pi_{i\in J}M_i$. -Case 2: $M$ is finite. -We deal with $J$ just as before. For $I$, in case $M$ has too small cardinality, -we can just hard code all of the (finitely many) possible products -of 1-element models, in the language of $\varphi$, into a disjunction -of possibilities. That is, first compute a list $A_1,\ldots,A_k$ of all such products (up to isomorphism). Then when writing $\psi_\varphi$, just say "There is $i\leq k$ such that $M$ is isomorphic to $A_i\times\Pi_{i\in J}M_i$ (where $J,M_i$ are as before)". -This completes the description of $\psi_\varphi$, and note that $\varphi\mapsto\psi_\varphi$ is recursive. -That was the case for arbitrary products. For finite products it is the same, except that we must also assert that $I,J$ are both finite. But since we have SOL, this is no problem. - -REMARK: It's important above that "false" is productive, so that we can (and must) have sentences in $X$ which have no models. Suppose one modified the question and demanded that all the sentences in $X$ be true. Then (1) it is no longer possible. Similarly, (2) there is no computable set $X$ of true SOL sentences such that for each true SOL sentence $\varphi$, there is some $\psi\in X$ which is semantically equivalent to $\varphi$. For (2): Let $\delta$ be the least ordinal such that $V_\delta\equiv_{\Sigma_2}V$. Then $V_\delta$ is also the least set containing models of all consistent SOL sentences. But this means that using $X$, we can define $\delta$ in a $\Sigma_2$ way, and since $X$ is computable, therefore $\{\delta\}$ is $\Sigma_2$, which contradicts the definition of $\delta$. Note that this shows that in fact no such $X$ can be $\Sigma_1$-definable. For (1) it is similar, because the true productive SOL sentences are still cofinal in the true SOL sentences. (Given $\varphi$ true productive, $\psi_\varphi$ (as above) is "above" $\varphi$.) - -Original answer (actually to another question): I think the answer to the first question is no, but I'm not totally sure what "semantically equivalent" should mean. -As a warm-up, first consider the version of the question for finite products (I'm not sure whether you meant finite or arbitrary products). -And avoiding the question of "semantically equivalent", -let's restrict attention to the second order language $\mathcal{L}=(0,1,+,\times)$ of arithmetic (without $<$), and show that the set $X$ of productive sentences of $\mathcal{L}$ is itself not computable. -(Also, I'm taking the product $P$ of two $\mathcal{L}$-models $M,N$ -to have $c^P=(c^M,c^N)$ for constants $c=0,1$, and $+/\times$ are likewise defined component-wise. Likewise for arbitrary products.) -To show $X$ (the finite product version) is not computable, we just identify a computable sequence -$\left<\psi_n\right>_{n\in\mathbb{N}}$ of sentences -such that $\psi_n$ is finitely productive iff $n\in 0$-jump. -Observe first that there is a finitely productive sentence $\psi$ of second order $\mathcal{L}$ which asserts "The model is a product of finitely many, but at least 1, copies of $\mathbb{N}$". -For this, say (i) The sub-model generated by $(0,1,+)$, extended with the restriction -of $\times$, is isomorphic to $\mathbb{N}$, and (ii) there is some $n\in\mathbb{N}$ and an isomorphism between the $n$-fold product of $\mathbb{N}$ and the full model. -Now for $n\in\mathbb{N}$ let $\psi_n$ be the sentence $\psi$+"if the model is not isomorphic to $\mathbb{N}$ then $n\in 0$-jump", -where $0$-jump is defined using the copy of $\mathbb{N}$ found as in (i) above -(note that because it models $\psi$, we can indeed find a copy of $\mathbb{N}$ in this way to start with). -Then $\psi_n$ is finitely productive iff $n\in 0$-jump: If $n\in 0$-jump -this is clear; if $n\notin 0$-jump then note that $\mathbb{N}\models\psi_n$ -but $\mathbb{N}^2\models\neg\psi_n$ (it's easy to see that $\mathbb{N}$ and $\mathbb{N}^2$ are not isomorphic). - -Edit: generalization for arbitrary products. The new thing here is to appropriately adapt $\psi$. But for this, I want to work instead -with the language $\mathcal{L}'=(0,1,+,\times,\leq)$. -I am taking products of relations to be defined using the "for all" quantifier, i.e. in our case, for the product $P$ -of a sequence $\left_{i\in I}$ of models, -$f\leq^Pg$ iff $f(i)\leq^{M_i}g(i)$ for all $i\in I$. -I'll write $<^*$ for the strict part of $\leq^P$ below -(note this need not be the same as the product of the strict parts of respective $\leq$s). -So we want a sentence $\psi$ which says "the model is the product -of set-many, but at least 1-many, copies of $\mathbb{N}$". -Let $P$ be the product of $I$-many copies of $\mathbb{N}$. -Given $j\in I$, let $f_j\in P$ be the function $f_j:I\to\mathbb{N}$ -where $f_j(j)=1$ and $f_j(i)=0$ for $i\neq j$. -Note that the set $A=\{f_j\bigm|j\in I\}$ is -definable over $P$: given $f\in P$, we have -$$ f\in A\iff [f\neq 0^P\text{ and for all }g\text{, if }g<^*f -\text{ then }g=0^P],$$ -where $<^*$ is the strict part of $\leq^P$. Then from $A$, -we can uniformly identify the models $\mathbb{N}_f$, for $f\in A$, -generated by $\{f\}$ by adjoining $0^P$ and closing under $+$, interpreting $1$ as $f$ and restricting $\times,\leq$ from $P$. And then $P$ is isomorphic to $\Pi_{f\in A}\mathbb{N}_f$. -So to assert that a given model $M$ (in the language above) is some product of set-many (at least 1) copies of $\mathbb{N}$, just define the set $A^M$ in the same manner, assert that every $f\in A^M$ generates (by adjoining $0$, closing under $+$, and interpreting $1$ as $f$) a model $\mathbb{N}_f$ which together with the restriction of $\times^M,\leq^M$, is isomorphic to $\mathbb{N}$, -and $\mathbb{N}_f\cap\mathbb{N}_g=\{0^M\}$ when $f\neq g\in A$, -and that $M$ is isomorphic to the product $\Pi_{f\in A}\mathbb{N}_f$. To say the latter, -just say that there is a binary relation $R\subseteq M^2$ which codes an isomorphism $\pi:M\to\Pi_{f\in A}\mathbb{N}_f$, via $R(x,y)$ iff $y\in\mathrm{rg}(\pi(x))$. -It is easy enough to write down the properties that $R$ needs in order to code an isomorphism in this way. -This $\psi$ is (set-)productive, and has the right meaning. Now define $\psi_n$ as before, and we get the same result as before.<|endoftext|> -TITLE: What is the negative cyclic homology of a smooth projective variety? -QUESTION [7 upvotes]: Let X be a smooth and projective variety. Hochschild homology and cohomology have a very simple definition in terms of Ext groups of the diagonal of X. The Hochschild-Kostant-Rosenberg (HKR) theorem then tells us that these coincide with sums of cohomology of differential forms or polyvector fields (one reference for this is the papers by Caldararu). - -Is there a simple definition of negative cyclic homology for X? -Is there an HKR-type theorem which relates this to more understandable gadgets? -Same question for ordinary cyclic and periodic cyclic homology (and cohomology)? - -REPLY [12 votes]: There are conceptually simple definitions, but they require a more symmetric definition of Hochschild homology. The Hochschild homology of $X/k$ (with coefficients in $\mathcal O_X$) is the homology of the complex -$$ \mathbf R\Gamma(X\times X, \mathbf R\Delta_*\mathcal O_X\otimes^{\mathbf L}\mathbf R\Delta_*\mathcal O_X)). $$ -Equivalently, in the language of derived algebraic geometry, this is the complex $\mathcal O(LX)$ where $LX=X\times^{h}_{X\times X}X$. The circle comes in because $S^1$ is the homotopy pushout $*\coprod^h_{*\amalg *}*$, so, formally, $LX=X^{S^1}$ is the free loop space of $X$ (it is a derived scheme whose underlying scheme is $X$). Now $S^1$ is also a group and it acts on itself, hence it acts on $LX$ and on the complex $\mathcal O(LX)$ computing $HH_*(X)$. -The complexes computing the cyclic, negative cyclic, and periodic cyclic homology of $X$ are respectively the homotopy orbits $\mathcal O(LX)_{hS^1}$, homotopy fixed points $\mathcal O(LX)^{hS^1}$, and Tate fixed points $\mathcal O(LX)^{tS^1}$ of this circle action. (For $G$ a compact Lie group acting on a spectrum or chain complex $E$, there is a norm map $(\Sigma^{\mathfrak g}E)_{hG}\to E^{hG}$ whose homotopy cofiber is by definition $E^{tG}$. In our case this cofiber sequence induces the usual long exact sequence relating these three homology theories.) -I'm not sure what the cyclic cohomology of a $k$-scheme is. At least if $X$ is affine, it is computed by the dual of the complex $\mathcal O(LX)_{hS^1}$, i.e., by the complex of $S^1$-invariant maps $\mathcal O(LX) \to k$. -ETA: Hochschild cohomology of $X$ with coefficients in $\omega_X$ is computed by the complex $\omega(LX)$, which has an $S^1$-action. So perhaps one gets reasonable "cohomology" versions of the above theories by replacing $\mathcal O(LX)$ by $\omega(LX)$. -To relate this to the traditional definitions, one shows that there is an equivalence between $S^1$-equivariant chain complexes and mixed complexes, and that the $S^1$-action on $\mathcal O(LX)$ corresponds to Connes' operator $B$. -Here's yet another way to understand the $S^1$-action on Hochschild chains which applies to the noncommutative setting as well. The complex $\mathcal O(LX)$ can be identified with the Euler characteristic (=trace of the identity) of $D_{qcoh}(X)$ in the symmetric monoidal $\infty$-category of presentable dg-categories. It is a general fact that the Euler characteristic of any object comes with an $S^1$-action. From the point of view of the cobordism hypothesis, this $S^1$ is now the framed diffeomorphism group of the circle, which is the Euler characteristic of the universal dualizable object in $\operatorname{Bord}_1^{fr}$. -If $\mathbb Q\subset k$ and $X$ is smooth and affine, the relations with Kähler differentials are given by: -\begin{align*} -HC_n(X) &= \Omega^n(X)/B^n(X) \oplus \bigoplus_{i\geq 1} H^{n-2i}_{dR}(X),\\ -HC_n^-(X) &= Z^n(X) \times \prod_{i\geq 1} H^{n+2i}_{dR}(X),\\ -HC_n^{per}(X)& = \prod_{i\in\mathbb Z} H^{n+2i}_{dR}(X). -\end{align*} -(Reference: Loday's book, 3.4.12, 5.1.12). The formula for $HC^{per}_n(X)$ remains valid if $X$ is not affine (because it satisfies Mayer-Vietoris), but the first two become more complicated... -Update. Here are the global formulas, I hope I got the indices right: -\begin{align*} -HC_n(X) &= \bigoplus_{-\dim(X)\leq i\leq n} H^{n-2i}_{Zar}(X, s_{\leq n-i}\Omega^*_X),\\ -HC_n^-(X) &= \prod_{0\leq i\leq \dim(X)-n} H^{n+2i}_{Zar}(X, s_{\geq n+i}\Omega^*_X). -\end{align*} -Here $H_{Zar}$ is Zariski hypercohomology, $s_{\leq k}$ and $s_{\geq k}$ denote the stupid truncations, and $\Omega^*_X$ is the de Rham complex.<|endoftext|> -TITLE: Аrе thеsе integrals known? -QUESTION [6 upvotes]: While studying some dark matter related stuff, I came across to the following interesting identities: -$$\int\limits_0^\infty\sqrt{\frac{y}{xp}}\,e^{-y}\left(K(p)-E(p)\right)dy= -\frac{\pi x}{4} \left[I_0\left(\frac{x}{2}\right)K_0\left(\frac{x}{2}\right)-I_1\left(\frac{x}{2}\right)K_1\left(\frac{x}{2}\right)\right],$$ -and -$$\int\limits_0^\infty\sqrt{\frac{y}{xp}}\,\frac{K(p)-E(p)}{y^2}\,dy= -\frac{\pi}{2x}.\tag{1}$$ -Here $K$ and $E$ are complete elliptic integrals of the first and second kind, -$I_0$,$I_1$ are modified Bessel functions of the first kind, $K_0$,$K_1$ are modified Bessel functions of the second kind, and -$$p=\frac{x^2+y^2-\sqrt{(x^2-y^2)^2}}{2xy}=\left\{\begin{array}{c}\frac{y}{x},\,\,\mathrm{if}\,\,x\ge y, \\ \frac{x}{y},\,\,\mathrm{if}\,\,x\lt y. -\end{array}\right . $$ -are these identities known? I got them only indirectly by calculating a gravitational field strengths by two different methods. Is there any simple method to prove them? -Using $$\frac{y}{x}\left[\frac{K(m)}{x+y}+\frac{E(m)}{x-y}\right]=2\,\frac{\partial}{\partial y}\left [\sqrt{\frac{y}{xp}}\,\left(K(p)-E(p)\right)\right],$$ where $$m=\frac{2\sqrt{xy}}{x+y},$$ -these integrals can be rewritten as follows -$$\int\limits_0^\infty e^{-y}y\left[\frac{K(m)}{x+y}+\frac{E(m)}{x-y}\right]dy=\frac{\pi x^2}{2}\left[I_0\left(\frac{x}{2}\right)K_0\left(\frac{x}{2}\right)-I_1\left(\frac{x}{2}\right)K_1\left(\frac{x}{2}\right)\right],$$ and -$$\int\limits_0^\infty \left[\frac{K(m)}{x+y}+\frac{E(m)}{x-y}\right]dy=\pi.$$ -Eq. (1) can be rewritten in the (perhaps more interesting) form: -$$\int\limits_0^1\frac{k+1}{k^2}\left[K(k)-E(k)\right ]dk=\frac{\pi}{2}.$$ - -REPLY [2 votes]: I have not found the exact reference for your integral identity but you may like to check this out in the meantime. -Also, explore this table of integrals.<|endoftext|> -TITLE: Set theoretic forcing, large cardinals and probabilistic methods -QUESTION [7 upvotes]: This question is motivated by the work of Ajtai "The complexity of the pigeonhole principle" and similar works. In this paper, the author proves that $PHP_n$, the pigeonhole principle for $n,$ does not have polynomial-size constant-depth Frege proofs. The method of proof is an arithmetical analogue of forcing (of a kind already used by Paris and Wilkie), plus a probabilistic argument to handle the relevant combinatorics. -Now my questions are the following. - -Question 1. Are there similar works, which connect set theoretic forcing with probabilistic arguments in an essential way? -Question 2. Are there works, which connect large cardinals and probabilistic arguments? - -REPLY [2 votes]: Today I saw the following paper in which probabilistic arguments are used in a forcing argument: -Halfway New Cardinal Characteristics. -See the proof of 3.4. The paper is written by Jörg Brendle, Lorenz Halbeisen, Lukas Daniel Klausner, Marc Lischka and Saharon Shelah.<|endoftext|> -TITLE: Problem on triangles -QUESTION [6 upvotes]: Let $T\subset \mathbb{R}^2$ be any triangle and $T^t$ a deformation of $T$. Call $l_1,l_2,l_3$ the squares of the lengths of the sides of $T$ and $l_1^t,l_2^t,l_3^t$ the squares of the lengths of the corresponding sides of $T^t$. Let $\phi_t:T\rightarrow T^t$ be the affine map which sends sides to corresponding sides. -This is the problem I'd like to solve: -Find triangles $T$ such that the following equality is verified for any deformation $T^t$: -$$Max\{\frac{l_1}{l_1^t},\frac{l_2}{l_2^t},\frac{l_3}{l_3^t}\}=Lip(\phi_t)^2$$ -where $Lip(\phi_t)$ is the Lipschitz constant of $\phi_t$. - -REPLY [4 votes]: The equality is never satisfied. -We send $T$ to $T^t$ by an affine map. This sends the unit circle to an ellipse. The bilipschitz constant is either the half of the major axis or the reciprocal of the half of the minor axis. When we look at how the sides of a triangle are distorted, we "measure the ellipse" in three directions only. There is no guarantee that one of these directions corresponds to the maximum or minimum distortion. -More concretely, for every triangle you can keep the length of one side and move the opposite vertex in a direction perpendicular to that side. The maximum distortion is along this perpendicular and is bigger than the distortion along any of the sides. -On the other hand, one can ask for an estimate of the bilipschitz constant in terms of the distortions along the sides of a triangle...<|endoftext|> -TITLE: Zeroes of a not quite holomorphic (but random if helpful) function -QUESTION [5 upvotes]: I’m interested in the zeroes of the complex function -$f(z,\bar{z}) = p(z) + \frac{1}{log(|z|)} q(z)$ -where both $p$ and $q$ are polynomials of the complex variable $z$ (and are therefore holomorphic). I’m interested even in restricted cases, such as when both $p$ and $q$ are of low degree. I would also be very happy if anything could be said about the case where $p$ and $q$ are Kac-polynomials of high degree. -While I don’t know the literature well, I am aware that there are beautiful results relating to the clustering of zeroes on the unit circle in the context of random polynomials --- the simplest being Kac-polynomials. Presumably the factor of $\frac{1}{log(|z|)}$ spoils the usage of these results though. -I am also vaguely aware of Rouche’s theorem. However, since $f$ is not holomorphic, this result can not be used directly. -I would like to know how to count the zeroes of this function. Treating the coefficients as independent and identically distributed random variables would be fine too if that would be useful. To be honest I have no idea where to even start looking, so any and all suggestions are most welcome. - -REPLY [14 votes]: There is no general theory which applies here. However your problem can be restated -as a problem about zeros of harmonic maps, or about fixed points of anti-holomorphic maps, if you rewrite your equation as -$$\overline{z}=\exp(-2q(z)/p(z))/z.$$ -Of course I am aware that this last equation has more solutions than -the original one (in fact infinitely many in most cases) but still it is useful. -This type of problems (solutions of equations $\overline{z}=R(z)$, where $R$ -is meromorphic) have been considered in -MR2431564 Khavinson, Dmitry; Neumann, Genevra, From the fundamental theorem of algebra to astrophysics: a "harmonious'' path. Notices Amer. Math. Soc. 55 (2008), no. 6, 666–675. -MR2676458 -Bergweiler, Walter; Eremenko, Alexandre, -On the number of solutions of a transcendental equation arising in the theory of gravitational lensing. (English summary) -Comput. Methods Funct. Theory 10 (2010), no. 1, 303–324. -arXiv:1507.01704 Walter Bergweiler and Alexandre Eremenko, -Green's function and anti-holomorphic dynamics on a torus. -The first paper treats the equations -$$p(z)|z|+q(z)=0,\quad\mbox{and}\quad p(z)|z|^2+q(z)=0.$$ -The methods developed in these papers also apply to the present problem. -EDIT. Using these methods, one can obtain that the number of solutions -is at most $$3\max\{ m,n\}+2m,$$ -where $m,n$ are degrees of $p,q$. This is sometimes exact: your equation with $p(z)=1,\; q(z)=1-z^n$ has $3n$ solutions. -Also when $m=n$, consider the equation -$$n\log|z|=3\log 2\frac{z^n-1}{z^n+1}.$$ -It is easy to see that for $n=1$ it has 5 real roots (1/2,1,2 plus two negative). Replacing $z$ by $z^n$ we obtain an equation with $5n$ roots. These examples are due to Walter Bergweiler. Using similar examples we could show -that the estimate is exact for $m=0$, for $n\leq m$ and for $n=2m$. -EDIT2. Concerning the lower estimates, consider the function -$$g(z)=\log|z|-q(z)/p(z).$$ -It is easy to see that it defines a continuous (actually smooth) map -of the Riemann sphere. This map has a topological degree, which is easily seen to be $\deg(q/p)=\max\{ m,n\}$. Therefore we have at least -$\max\{ m,n\}$ solutions. This is exact for all $m,n$. -Here is a sketch of the proof. (Last updated on 11/24).<|endoftext|> -TITLE: Conceptually, what does unitization do? -QUESTION [7 upvotes]: Let $(\mathcal A,||\cdot||)$ be a normed algebra (with or without a unit). The unitization of $\mathcal A$ is the space $\mathcal A_+:=\mathcal A\oplus \Bbb C$ where the multiplication operation $\cdot$ and norm $|||\cdot |||$ are defined by -$$\begin{align} -(a,\lambda)\cdot(b,\mu) &:=(ab+\mu a+\lambda b,\lambda\mu) \\ -|||(a,\lambda)||| &:= ||a||+|\lambda| -\end{align}$$ -for any $a,b\in\mathcal A$ and $\mu,\lambda\in \Bbb C$. It can be easily verified that $(0,1)$ is a unit in $\mathcal A_+$ with norm $1$, and that $\mathcal A$ isometrically embeds into $\mathcal A_+$. This construction seems to be a very fundamental tool used to augment a unit into a space without one -, it came up in my first class of the chapter on Banach Algebra. -Our professor tried to convince us that unitization is done in the same spirit as completion of metric spaces. However, this construction troubles me for 2 main reasons. -Firstly, nothing stop me from unitizing a unital normed algebra. This gives me an essentially different space than the one I started with. This is not the case for the completion $\hat X$ of a Banach space $X$, in which we have $\hat X \cong X$. To add salt to the wound, the original unit in $\mathcal A$ is no longer a unit in $\mathcal A_+$. -Secondly, unlike the process of completion, the adjoined unit seems quite artificial. In metric space completion we only fill in "holes" but it seems like in unitization we artificially add genuine new direction into our space. - -I know that my professor's analogy shouldn't be taken literally but in what sense does unitization resemble metric completion? Conceptually, what is unitization? - -It also occurred to be that I'll learn to appreciate unitization as I encounter more and more results in this field. Nevertheless, I would like to have an intuitive understanding of unitization. -For example, take $(\mathcal A,+,\cdot,||\cdot||)$ to be $(L^1(\Bbb R),+,*,||\cdot||_{1})$, where $*$ is the convolution. This is a non-unital algebra since the Dirac distribution does not belong to $L^1$. In this case what does $\mathcal A_+$ look like? - -REPLY [2 votes]: When $A$ is a C*-algebra, the norm - $$\Vert (a,\lambda)\Vert := \Vert a\Vert + |\lambda|$$ -does not make $A_+$ a C*-algebra because the axiom $\Vert a^*a\Vert = \Vert a \Vert^2$ will no longer hold. So one usually replaces this norm by - $$\Vert (a,\lambda)\Vert := \sup_{b\in A,\ \Vert b\Vert \leq 1}\Vert ab+\lambda b\Vert.$$ -In case $A$ is already unital, with unit $1_A$, the above definition will give $\Vert (1_A,-1)\Vert = 0$, so this process only works when $A$ is not unital, meaning neccessarily without a unit. -One may thus interpret this as saying that unital C*-algebras do not like to be unitized :-)<|endoftext|> -TITLE: Greatest lower bound for subordination -QUESTION [7 upvotes]: Consider the set $X$ of all analytic functions $f$ in the unit disk $U$ satisfying -$f(0)=0, f'(0)\neq 0$. We say that $f\prec g$ if there exists -$\phi\in X$ which maps $U$ into itself, and $f=g\circ\phi.$ -This is called subordination. There are hundreds of papers with "subordination" in the title, but I could not find the answer to the following question: - -Does the greatest lower bound with respect to $\prec$ exist for arbitrary pair of functions in $X$ ? - -Precisely: Let $f,g$ be elements of $X$. Does there exist $h\in X$ such that -$h\prec f,\; h\prec g$, and for every $p$ such that $p\prec f,\; p\prec g$, we have -$p\prec h$? -I am especially interested in the subclass $X_0\subset X$ of locally univalent functions, $f'(z)\neq 0$ for all $z\in U.$ -Remark. -For univalent (injective) functions, the g.l.b, corresponds to the intersection -of the image domains. That's why I conjecture that g. l. b. must always exist. -EDIT. For meromorphic functions, the answer is negative. Let $f$ be a conformal map of the unit disk onto the disk $|z-1|<2\;$, $f(0)=0$ and $g$ be a conformal map onto the disk $\{ |z-1|>1/2 \}\cup\{\infty\},\;$ $g(0)=0$. As the intersection of the two images is a ring, there is evidently no g.l.b. -For holomorphic functions it remains open. - -REPLY [2 votes]: It turns out that the greatest lower bound may not exist. The counterexample -can be constructed using the example of Milnor -explained in the paper by Poenaru, Extension des immersions en codimension 1, Séminaire Bourbaki, 10 (1966-1968), Exp. No. 342. Milnor's example was brought -to my attention in a comment by Misha to my other question -Uniqueness theorem for conformal maping<|endoftext|> -TITLE: Binomial coefficients and "missing primes" -QUESTION [17 upvotes]: Define the following two subsets of prime numbers -$$\Pi_n:=\{p: \text{$p$ is prime, $p\leq n$}\}$$ -and -$$B_n:=\{p: \text{$p$ is prime, $p$ divides $\binom{n}k$ for some $0< k< n$}\}.$$ -Denote their respective cardinalities by $\pi(n):=\vert\Pi_n\vert$ and $\,b(n):=\vert B_n\vert$. - -QUESTIONS. Experimental evidence suggest that -(a) $\pi(n)-b(n)=0$ or $1$. -(b) If $\pi(n)-b(n)=1$, then the missing prime is the largest prime divisor of $n+1$. -Are these true statements? - -REPLY [15 votes]: Suppose that $p $ is a prime satisfying $p\le n$ and $p\nmid \binom{n}{k}$ for all $k=1,\ldots,n-1.$ Then from Kummer's theorem it follows that the base-$p$ representation of $n$ is $n=(a,p-1,\ldots,p-1)_p$ with $1\le a\le p-1$. So $n=(a+1)p^r-1$, where $r$ is the number of base-$p$ digits in $n$. Then $n+1 = (a+1)p^r$ with $a+1 \leq p$, so any prime factor of $n+1$ besides $p$ is a prime factor of $a+1$, so it has to be less than $p$. -Here is a second solution, using Lucas's theorem: writing $n = n_0+n_1p+\cdots +n_rp^r$, where $0 \leq n_i \leq p-1$ and $n_r \geq 1$, -each $k$ in $\{0,1,\ldots,n\}$ has base-$p$ representation $k_0+k_1p+\cdots +k_rp^r$ where $0 \leq k_i \leq p-1$ (maybe $k_r = 0$ if $k$ is much smaller than $n$). Then Lucas' theorem says -$$ -\binom{n}{k} \equiv \binom{n_0}{k_0}\binom{n_1}{k_1}\cdots \binom{n_r}{k_r} \bmod p. -$$ -For digits $n_i$ and $k_i$, which are in $\{0,1,\ldots,p-1\}$, we have $\binom{n_i}{k_i} \equiv 0 \bmod p$ if and only if $n_i < k_i$ (otherwise the factors in $\binom{n_i}{k_i} = \frac{n_i(n_i-1)\cdots(n_i-(k_i-1))}{k_i!}$ are all nonzero modulo $p$). Thus we have $\binom{n}{k} \not\equiv 0 \bmod p$ for all $k$ from $0$ to $n$ (let's allow $k=0$ and $k=n$ since those cases are not divisible by $p$) if and only if every $k \leq n$ has $k_i \leq n_i$ for $i < r$ and $k_r \leq n_r$, and that's equivalent to $n_i = p-1$ for $i < r$ (if some $n_i < p-1$ then at $k = (p-1)p^i$ we have $\binom{n}{k} \equiv 0 \bmod p$). Such $n$ have the form $(p-1)+(p-1)p\cdots+(p-1)p^{r-1} + n_rp^r = (n_r+1)p^r-1$. If $r = 0$ then $n = n_r \leq p-1$. If $r \geq 1$ then $p \mid (n+1)$, and the first solution shows easily in this case why $p$ has to be the largest prime factor of $n+1$. In summary, every prime that is $\leq n$ divides some $\binom{n}{k}$ for $1 \leq k \leq n-1$ except when $n = mp^r-1$ where $r \geq 1$, $p$ is prime, and $2 \leq m \leq p$, in which case the prime $p$ divides no $\binom{n}{k}$. - -REPLY [13 votes]: I like the following argument. -Let $p \leq n$ be a missing prime. Then $p \leq n/2$ as otherwise $p $ divides $\binom{n}{n+1-p}$. Also $p$ divides $n+1$ as $p$ does not divide $\binom{n}{k}$ for $0\leq k \leq p$. Similarly $p^b$ -must divide $n+1$ for all $b$ up to the highest power of $p \lt n$, otherwise there is a $k\lt p^b$ (or $k \leq n/2$) which reveals $p$. So $n+1 = cp^b$ for some $c\lt p$. So both conditions hold, and the latter can be strengthened to apply only with $n=cp^b - 1$. -Gerhard "Likes Writing Low Power Arguments" Paseman, 2016.11.17.<|endoftext|> -TITLE: Algebraic characterization of convergence spaces -QUESTION [8 upvotes]: Is there an algebraic characterization of convergence spaces similar to Barr's characterization of topological spaces as lax algebras for the ultrafilter monad? I'm also curious about the same question for uniform and Cauchy convergence spaces. - -REPLY [9 votes]: This is sort of an extended comment. -If you're interested in this sort of thing, you should be aware that there's a whole cottage industry centered around generalizations of Barr's characterization which apply to uniform spaces, approach spaces, and other "topological" categories. See, for example, Clementino, Hofmann, and Tholen. The basic setup is that you have a monad on the category of Sets, you extend this to (a generalization of) the category of relations, and then you define lax algebras for the monad. A lot of work goes into seeing which aspects of topology can be generalized to this setting. I can't resist mentioning that this story fits into a common abstract setup with the notion of a multicategory -- see Cruttwell and Shulman. -Note that the axioms for a convergence space are all phrased in terms of convergence to a single point $x$. So in a Barr-like axiomatization of convergence spaces, one expects there will be no analog of Barr's lax associativity axiom, which relates convergence at many different points to convergence at a single other point. In fact, if you simply drop the lax associativity axiom from Barr's characterization, (so you just have a relation from ultrafilters to the original set which is lax-unital) you get pseudotopoogical spaces, which can be identified as a certain subcategory of convergence spaces. -To get convergence spaces you would presumably have to work with the filter monad rather than the ultrafilter monad. Be aware that there is more than one reasonable way to extend the filter monad from $\mathsf{Set}$ to $\mathsf{Rel}$ -- see Seal, which discusses two important ones, and Schubert and Seal, which discusses them more generally.<|endoftext|> -TITLE: Maximality and non-Hausdorffness -QUESTION [6 upvotes]: We say that a non-$T_2$ topology $\tau$ on a set $X$ is maximal non-$T_2$ if every topology $\tau'$ strictly containing $\tau$ is $T_2$. - -Is every non-$T_2$ topology contained in a maximal non-$T_2$ topology? - -REPLY [7 votes]: Yes, every non-Hausdorff topology is contained in a maximal non-Hausdorff topology. -To see this, let's start with a different question: What do the maximal non-Hausdorff topologies on an infinite set look like? -To answer this auxiliary question, let $Y = X \cup \{p,q\}$ be an infinite set, $p,q \notin X$. Suppose $\mathcal U$ is an ultrafilter on $X$ and consider the topology on $Y$ defined by - -Every point of $X$ is isolated. -Sets of the form $\{p\} \cup A$, where $A \in \mathcal U$, form a neighborhood basis for $p$. -Sets of the form $\{q\} \cup A$, where $A \in \mathcal U$, form a neighborhood basis for $q$. - -Let us call this topology $\tau$. It is easy to see that $\tau$ is a non-Hausdorff topology on $Y$. -I claim that any proper refinement of $\tau$ is Hausdorff. To see this, let us consider topologies of the form $\langle \tau \cup \{B\} \rangle$ where $B \subseteq Y$ (recall that $\langle \tau \cup \{B\} \rangle$ is defined to be the smallest topology that refines $\tau$ in which $B$ is open; equivalently, it is the topology having $\tau \cup \{B\}$ as a sub-basis). It is not too hard to check that any proper refinement of $\tau$ having this form is Hausdorff. [Here is a short proof. If $p,q \notin B$ or if $X \cap B \in \mathcal U$, then $\langle \tau \cup \{B\} \rangle = \tau$. Thus we may assume that at least one of $p$ and $q$ is in $B$, and that $B \cap X \notin \mathcal U$. If only one of $p$ or $q$ is in $B$, $\langle \tau \cup \{B\} \rangle$ is Hausdorff (in fact, a maximal non-discrete topology). If both are in $B$, $\langle \tau \cup \{B\} \rangle$ is the discrete topology.] -Now suppose $\tau'$ is any topology properly refining $\tau$ (we do not assume it has the form $\langle \tau \cup \{B\} \rangle$ for some $B \subseteq Y$). Since $\tau'$ is a proper refinement, there is some $B \in \tau' - \tau$. By the previous paragraph, $\langle \tau \cup \{B\} \rangle$ is Hausdorff. Any refinement of a Hausdorff topology is Hausdorff, so $\tau'$ is Hausdorff. Therefore the topology described above really is a maximal non-Hausdorff topology. -Now that we know what (at least some of) the maximal non-Hausdorff topologies look like, we can answer Dominic's question. Let $\sigma$ be any non-Hausdorff topology on an infinite set $Y$ (the question is trivial for finite sets). Let $p,q$ be two points in $Y$ that cannot be separated with open sets. Let $X = Y - \{p,q\}$ and observe that -$$\mathcal F = \{U \cap V \cap X : p \in \mathrm{Int}(U) \ \text{ and } \ q \in \mathrm{Int}(V)\}$$ -is a filter on $X$. Let $\mathcal U$ be any ultrafilter on $X$ extending $\mathcal F$. Then the topology described above is a maximal non-Hausdorff topology refining $\sigma$. -Incidentally, this also shows that every maximal non-Hausdorff topology has the form of the topology $\tau$ described above.<|endoftext|> -TITLE: Does it make sense to talk about expansion in irregular graphs? -QUESTION [8 upvotes]: Usually, one defines an expander graph to be a regular graph satisfying one of the following properties: -Either the edge-expansion is large, or -the spectral gap is large, or -the mixing time is at most logarithmic in the number of vertices, or -it satisfies a mixing-lemma type property. -It doesn't really matter much which property we require, since up to a constant they are all equivalent. This is also the reason why the concept of expander graphs is so powerful. -My question is: What happens in graphs that are far from regular? One can look at each of these properties separately, but is there a sense in which they are equivalent? If not, does it make sense to talk about expander graphs in this setting? - -REPLY [3 votes]: All the inequalities relating the three "definitions" of expander graphs are known in fully explicit forms, so that one can see what happens in first approximation for not completely regular graph (the versions I know only involve the minimal and maximal degrees, which might be too coarse for what you have in mind?) But whether one definition is or is not "better" might then depend on applications, and in different contexts, one of the three aspects might be more suitable (and close to the traditional expanders), while the others have different behavior. -(For instance, Valiant has proposed computation algorithms based on graphs that he feels might be realistic models for the way the brain works, and he mentions explicitly that expansion is a useful property, but if I understand right, his graphs have typically a degree that is approximately the square-root of the number of vertices, which is much more than a classical expander).<|endoftext|> -TITLE: Why symplectic geometry gives Poisson geometry -QUESTION [20 upvotes]: One way I've learned to understand Poisson geometry is to consider it as symplectic geometry with no open conditions - i.e. no condition of nondegeneracy. This idea can be applied to many other theories, including Kahler geometry, Riemannian geometry, Riemann-flat geometry, and hyperKahler geometry. -Of those, only symplectic geometry seems to lead to a deep and interesting theory. Is there a specific reason for that? -EDIT: I guess this comprises a couple of questions. -Question 1: Is there a specific reason other "degeneracy-generalization" theories don't seem to be studied much? Something that makes symplectic nicer than others? -Question 2: Is there some nice geometric or algebraic reason why symplectic geometry should lead to a dimension-independent Poisson geometry, and most other "nondegenerate" theories don't without losing some information? In particular, why no other integrable theories (at least among those checked) give dimension-independent equations? -Or generally: what is it about symplectic geometry that makes Poisson geometry "nice" that other geometries lack? - -I've tried to work out "Poisson-Riemann" and "Poisson-Kahler" algebras, and none of them seem to work out as nicely as symplectic geometry. For example, "Poisson-Kahler algebras" can be made, but the relevant equations depend on the dimension in an obvious way (that is, there's an easy way given the dimension to write the equations). "Poisson-Riemann algebras" can be made in multiple ways; the easiest way has no good generalization of the curvature tensor, and the hardest way doesn't seem to work at all. "Flat Poisson-Riemann algebras" seem to make sense, but the equations seem to depend on dimension in a non-obvious way. Is this one reason that Poisson geometry is studied, and other aren't? -One part of the problem with other theories is that they lack a natural way to do a Lie bracket of vector fields coming from functions. As such, there are tensors that can be made using the Lie bracket that can't be generated by the "Riemann bracket" or "Kahler bracket" (generalizing the Poisson bracket). This contrasts with the fact that in Poisson geometry, the Lie bracket on vector fields from functions effectively is the Poisson bracket. Might this be related? - -Some additional detail and calculations to demonstrate a couple of points from the above: -For a "Poisson-Riemann algebra" from a Riemannian metric, the easiest way to do it is to define a bracket $\lfloor f, g\rfloor = \langle \nabla f, \nabla g\rangle$. This gives no relations beyond symmetry and the usual tensor relation $\lfloor fg, h\rfloor = f \lfloor g, h\rfloor + g \lfloor f, h\rfloor$. This, similar to the usual symplectic geometry, is a dimension-independent definition, but doesn't include the curvature tensor. -If $f_1, f_2, f_3, f_4$ are functions and $X_1, X_2, X_3, X_4$ are their gradients, then $R(X_1, X_2, X_3, X_4)$ can be calculated: -$R(X_1, X_2, X_3, X_4) = \lfloor f_1, \lfloor f_3, \lfloor f_2, f_4\rfloor \rfloor \rfloor - \lfloor f_1, \lfloor f_4, \lfloor f_2, f_3\rfloor \rfloor \rfloor - \lfloor f_2, \lfloor f_3, \lfloor f_1, f_4\rfloor \rfloor \rfloor + \lfloor f_2, \lfloor f_4, \lfloor f_1, f_3\rfloor \rfloor \rfloor + \lfloor f_3, \lfloor f_1, \lfloor f_2, f_4\rfloor \rfloor \rfloor - \lfloor f_4, \lfloor f_1, \lfloor f_2, f_3\rfloor \rfloor \rfloor - \lfloor f_3, \lfloor f_2, \lfloor f_1, f_4\rfloor \rfloor \rfloor + \lfloor f_4, \lfloor f_2, \lfloor f_1, f_3\rfloor \rfloor \rfloor - \lfloor \lfloor f_1, f_3\rfloor, \lfloor f_2, f_4\rfloor\rfloor + \lfloor \lfloor f_1, f_4\rfloor, \lfloor f_2, f_3\rfloor\rfloor + 2 \langle [X_1, X_2], [X_3, X_4]\rangle - \langle [X_1, X_3], [X_4, X_2]\rangle - \langle [X_1, X_4], [X_2, X_3]\rangle$ -The last three terms are key; they are the only terms that can't be directly expressed in terms of $\lfloor ,\rfloor$. -We could extend the algebra to include the curvature with a tensor $R(f_1, f_2, f_3, f_4)$, but we'd then need some relation so that $R$ is uniquely determined if $\lfloor ,\rfloor$ is nondegenerate. As far as I've calculated, there is no dimension-independent way to do this. -If we instead turn to making "Flat Poisson-Riemann algebras", then I think we shouldn't need any extra tensors (as extra tensors lead to local invariants, and flat Riemann manifolds have no local invariants). This should be the closest to being analogous to Poisson geometry, as both should be "Poisson generalizations" of integrable systems. But the relevant equations, unlike in the Poisson geometry setting, seem to depend on dimension (although I'm still working on the exact equations). -The situation is easier for "Poisson-Kahler algebras". A Kahler form gives rise to a (neither symmetric nor antisymmetric) bracket $\lceil f, g\rceil = \lfloor f, g\rfloor + i \{f, g\}$. The conditions on a $2n$-dimensional Kahler manifold are then: -(Pointwise condition): $\sum_{\sigma \in S_n} (-1)^{sgn(\sigma)} \prod_{i = 1}^n \lceil f_i, g_{\sigma(i)} \rceil = 0$ -If the structure is nondegenerate, this is equivalent to the division of the space into $n$-dimensional holomorphic and antiholomorphic parts. -(Integrability of the complex structure): Two equations, which can be gotten by each other by "switching sides" in the bracket. -$\sum_{\sigma \in S_n} (-1)^{sgn(\sigma)} (\lceil f_1, \lceil f'_1, g_{\sigma(1)}\rceil \rceil - \lceil f'_1, \lceil f_1, g_{\sigma(1)}\rceil \rceil) \prod_{i = 2}^n \lceil f_i, g_{\sigma(i)}\rceil = 0$ -In the case that the structure is nondegenerate, the two equations say that the bundles are integrable - the above equation says that the Lie bracket of two holomorphic vector fields is holomorphic, and its counterpart says the same for antiholomorphic. -(Integrability of the symplectic structure): $\lceil f, \lceil g, h \rceil \rceil - \lceil \lceil f, g\rceil, h\rceil + \lceil \lceil f, h\rceil, g\rceil - \lceil f, \lceil h, g\rceil + \lceil h, \lceil f, g \rceil \rceil - \lceil \lceil h, f\rceil, g\rceil + \lceil \lceil h, g\rceil, f\rceil - \lceil h, \lceil g, f\rceil + \lceil g, \lceil h, f \rceil \rceil - \lceil \lceil g, h\rceil, f\rceil + \lceil \lceil g, f\rceil, h\rceil - \lceil g, \lceil f, h\rceil = 0$ -As far as I can tell, "Poisson-Kahler algebras" still have the problem of having no curvature tensor, but even without it the dimension-dependence seems clear. - -REPLY [3 votes]: Look up sub-Riemannian geometry. This is a quite developed theory now, but here the Riemannian metric is defined only along a subbundle of the tangent bundle which is assumed to be completely non-integrable (so that geodesics reach evewhere). You take the inverse of the Riemannian metric and assume that it may be degenerate. -For your curvature expression, see section 2 of - -Mario Micheli, Peter W. Michor, David Mumford: Sobolev Metrics on Diffeomorphism Groups and the Derived Geometry of Spaces of Submanifolds. Izvestiya: Mathematics 77:3 (2013), 541-570. Izvestiya RAN: Ser. Mat. 77:3, 109-136. (pdf) - - -There is a nice relation between this curvature formula and O'Neill's formula for the behavior of curvatures under a Riemannian submersion: exactly the last 3 terms in your formula have further terms when compared to pullbacks; these are the O'Neill terms.<|endoftext|> -TITLE: Does $2^n-n$ have infinitely often a prime divisor greater than $n$? -QUESTION [11 upvotes]: I think the question in the title is clear. -Let $n\in \mathbb{N}$. It is a nice exercise to show that every prime number divides infinitely many terms of the sequence $2^n-n$. (For example take $n=(p-1)^{2m} , m\in \mathbb{N}$) -I would like to show that there are infinitely many primes for which $2^n\equiv n\pmod{p}$ is actually satisfied for some $0N^{1/3}$. Note that in this case $p^\ell$ is huge compared to any fixed power of $N$. Thus we can try the same argument with arbitrary $m -TITLE: Finite-dimensional approximations of the shift operator -QUESTION [8 upvotes]: On the standard space $l^2$ let us consider the left shift operator -$$ -L(c_1,c_2,c_3,\ldots)=(c_2,c_3,c_4,\ldots). -$$ -It is well known that the spectrum of $L$ is the whole unit disk in the complex plane. I would like to approximate $L$ by some sequence of finite-dimensional operators $L_n$. A naive way to do this is to set $L_n$ as follows -$$ -L_n=\left(\begin{array}{ccccc} -0 & 1 & 0 & \ldots & 0 \\ -0 & 0 & 1 & \ldots & 0 \\ -0 & 0 & 0 & \ddots & 0 \\ -0 & 0 & 0 & \ldots & 1 \\ -0 & 0 & 0 & \ldots & 0 -\end{array}\right) -$$ -However the spectrum of $L_n$ consists only of $0$. Could one suggest more reasonable finite-dimensional approximation sequence $L_n$ such that spectrum of operator $L_n$ gradually fills the unit disk? References are welcome. - -REPLY [6 votes]: If you replace it with the cyclic shift operator, you get a circulant matrix (the same as your $L_n$ except that the bottom-left entry is $1$). The eigenvalues of that matrix are the $n$th roots of unity. So as $n$ grows, the spectrum fills the unit circle (it does not fill the unit disk, though). -Your $L_n$ is a highly non-normal matrix; the circulant version is normal. If you want to understand this better, read Chapter 7 of Trefethen & Embree's Spectra and Pseudospectra, which deals specifically with your example. - -REPLY [3 votes]: I think that the numerical range is an appropriate tool for your question. Your naive approximations $L_n$ of the shift operator are nilpotent. For such matrices $M$ (nilpotent of size $n$), the numerical range ${\cal H}(M)$ is a disk $D(0;r_n)$ with radius -$$r_n=\|M\|\cos\frac\pi{n+1}\,$$ -where $\|M\|$ is the standard operator norm. In your situation, $\|L_n\|=1$, so that -$$r_n=\cos\frac\pi{n+1}\rightarrow1^-.$$ -I suspect that for reasonnable operators $L$, the finite dimensional approximations $L_n=P_n^*LP_n$ ($P_n$ the orthogonal projection on an increasing sequence of subspaces) has the property that the union ${\cal H}(L_n)$, which is a non-decreasing sequence for inclusion, contains the spectrum of $L$, exactly as ${\cal H}(M)$ contains the spectrum of $M$. This would be true if $L\mapsto {\cal H}(L)$ is lcs for a rather weak topology on operators.<|endoftext|> -TITLE: Is X+X finitely representable in X? -QUESTION [5 upvotes]: I wonder if the following assertion in true: -Conjecture. Let $X,Y,Z$ be infinite-dimensional Banach spaces such that both $Y$ and $Z$ are crudely finitely representable (c.f.r. for short) in $X$. Then $Y\oplus Z$ is c.f.r. in $X$. -Remark some equivalent formulations of the above conjecture. -(A) For every infinite-dimensional Banach space $X$, $X\oplus X$ is c.f.r. in $X$. -(B) For every infinite-dimensional Banach space $X$, $X\oplus X$ is isomorphic to a subspace of an ultrapower of $X$. -I guess this question could be connected with those of whether a Banach space isomorphic to its square (solved in the negative by Figiel and later improved by Gowers). But perhaps it is much simpler. -Thanks in advance. - -REPLY [7 votes]: The conjecture which you stated is false. A counterexample is contained in the proof of Figiel [Studia Math. 42 (1972), 295–306]. He actually proves that squares of finite-dimensional subspaces of the space he constructs are not uniformly embeddable into the space itself. -I am unaware of a simpler counterexample for infinite-dimensional spaces.<|endoftext|> -TITLE: Gamma-free definition of binomial coefficients -QUESTION [8 upvotes]: In an answer to T. Amdeberhan's recent question, Noam D. Elkies gives, as a by-product, an elementary computation of the classical integral -$$\int_{-\infty}^{+\infty}{\sin x\over x}dx=\pi,$$ -and suggests a trigonometric definition of binomial coefficients ${n\choose x}$ for non-negative integer $n$ and real or complex $x$, as: -$${n\choose x}:=\frac{n!}{\pi} \, \frac{\sin \pi x}{(n-x)(n-1-x) \cdots (1-x) x}\ , -$$ -which is consistent with the definition via Gamma function, $\frac{n!}{\Gamma(1+x)\Gamma(1+n-x)} -$, using the functional equation and the symmetry relation $\Gamma(x) \Gamma(1-x)=\pi/\sin\pi x$. (We may even hide the sine function and $\pi$, using the infinite product for $\sin(\pi x)$, which gives a plain infinite product -$${n\choose x}=\prod_{k=1}^n \Big(1+{x\over k}\Big)\prod_{k>n} \Big(1-{x^2\over k^2}\Big),$$ -but this formula seems somehow less handy to work with). -I'd like to elaborate this idea, mainly for didactic purposes, but not only, and see what can be deduced from the trigonometric definition without resorting to the Gamma function. - -It is clear that it actually reduces to the classical definition for $x\in\mathbb{Z}$, taking a limit as $x\to k$. -The symmetry property ${\big({n\atop x}\big)}={n\choose n-x \ }$ and the recursive relations ${\big({n\atop x}\big)}={n\over x}{n-1\choose x-1 \ }$ and ${n-1\choose x}+{n-1\choose x-1\ }={n\choose x}$, just follow from elementary identities. -Many identities seem to have a continuous counterpart, like Amdeberhan's -$$\int_{\mathbb{R}}{n\choose x}a^xb^{n-x}dx=\sum_{k\in\mathbb{Z}} {n\choose k}a^kb^{n-k}=(a+b)^n.$$ - -While these properties motivate and justify as quite natural the choice of the above definition for the binomials ${n\choose x \ }$ for real $x$, it remains the want of a necessity for this choice, that is, a uniqueness result analogous to the Artin-Bohr-Mollerup characterization of the Gamma function. -So here is my question: - -Is there a Gamma-free characterization of the extended binomials $\textstyle{n\choose x}$? In other - words, what property, together with one or more of the above relations, - would lead to the above trigonometric definition? - -REPLY [3 votes]: Let $\ \mu\ $ be a measure in $\ \mathbb C.\ $ Then define $\ \mathbb P_\mu(z)$: -$$ \mathbb P_\mu(z)\ :=\ e^{ \int_\mu {\log(z-t)\,dt} } $$ -We may conveniently define also the respective Newton coefficient: -$$ \binom z{\mu,n}\ :=\ \frac{\mathbb P_\mu(z)}{\mathbb P_\mu(n)} $$ -when $ \mathbb P_\mu(n) \ne 0\ $ and is well-defined (for arbitrary $\ n\in\mathbb C$) -- this coefficient is more like a generalized polynomial. - -EXAMPLE   Let $\ \mu\ $ be an atomic measure where $\ \mu(\{k\}) := 1\ $ for every $\ k=0\ \ldots\ n\!-\!1\ $ (this time $\ n\ $ is a natural number), and there are no other atoms. Then $\ \binom z{\mu,n}\ $ is equal to a respective basic integer polynomial or Newton polynomial (coefficient) $\ \binom zn.$ - -This my definition is just a draft of a note. - - -EDIT   Let me add another definition along similar lines (a measure). My note is a draft since--for instance--I am not discussing here the question of the set of arguments of the introduced functions; also, for the second definition below, one should also discuss the question of the measure staying in some sense away from $\ 0\in\mathbb C\ $ (e.g. let $\ \mu(B(0; r)) = 0\ $ for certain $\ r>0\ $ (but much less can and should be assumed). - -  -Let $\ \mu\ $ be a measure in $\ \mathbb C.\ $ Then define $\ \mathbf p_\mu(z)$: -$$ \mathbf p_\mu(z)\ :=\ e^{ \int_\mu {\log(\frac zt-1)\,dt} } $$ -We may conveniently define also the respective Newton coefficient: -$$ \binom z{\mu;n}\ :=\ \frac{\mathbf p_\mu(z)}{\mathbf p_\mu(n)} $$ -when $ \mathbf p_\mu(n) \ne 0\ $ and well-defined (for arbitrary $\ n\in\mathbb C$). - -WARNING This time the Newton coefficient notation features ";" instead of ",".<|endoftext|> -TITLE: Definition of cusped manifold? -QUESTION [6 upvotes]: There is much talk about hyperbolic cusped 3-manifolds, but almost no definition of what a cusped manifold is. -One definition I found was that it is a result of a parabolic transformation on H^n, fixing the infinity point (?). -Is there a more intuitive definition for the 3-dimensional case. What about a general n-dimensional case. Is there a definition of a cusp? - -REPLY [11 votes]: Cusped manifolds are noncompact complete hyperbolic manifolds with finite Riemannian volume. -More precisely, a cusped hyperbolic n-manifold is a Riemannian manifold (without boundary) of constant negative curvature, which is metrically complete and has finite Riemannian volume, but is not compact. If you prefer to look at group actions, then you can see it as the quotient of n-hyperbolic space by a group of isometries acting properly discontinuously, freely, and with a fundamental domain of finite volume but no compact fundamental domain. -This definition does not depend on the dimension. The name "cusped" comes from the fact that these manifolds have the following structure: say $M$ is one such, then it retracts onto a compact submanifold $M'$ which has a boundary consisting of flat manifolds $T_1, \ldots, T_h$. The rest of the manifold consists of so-called "cusps", which are warped products $T_i \times [1, +\infty[$ with the metric $(dx^2 + dt^2)/t^2$ (where $dx$ is a flat metric on $T_i$). The uniformisation of a cusp is as follows: take the upper-half plane model for hyperbolic space, so that it decomposes as $\mathbb H^n = E^{n-1} \times ]0, +\infty[$ where $E^{n-1}$ is (n-1)-Euclidean space. Let $\Lambda$ be a $n-1$-dimensional torsion-free crystallographic group (a subgroup of Euclidean isometries of $E^{n-1}$ with a compact fundamental tile, for example $\mathbb Z^{n-1}$ acting by translations). It acts (by parabolic isometries) on $\mathbb H^n$ preserving the subsets $E^{n-1}\times\{t\}$; then the quotient $\Lambda \backslash E^{n-1} \times[1,+\infty[$ is a cusp. -In 3 dimensions, by Perelman's solution the Thurston geometrisation conjecture there is a topological characterisation of hyperbolic cusped manifolds (I am writing from memory here, there might be imprecisions): they are exactly the interior of irreducible compact manifolds with a nonempty boundary consisting of tori, which contain no other essential tori up to homotopy, and no embedded annuli connecting two essential curves on distinct boundary components (In this setting irreducible means that every sphere bounds a ball and every disc bounds a half-ball). (EDIT: added the condition on annuli and precised that only the interior is hyperbolic above)<|endoftext|> -TITLE: Is there a theorem whose only known proof uses "$A$ or not $A$" for undecidable $A$ -QUESTION [16 upvotes]: Is there a known theorem $T$ in $ZF+DC$ (or $ZF$ or $ZFC$) such that the only proof we know of $T$ is by using the LEM applied to $A$ ( "$A$ or not $A$" ), where $A$ is independent of $ZF+DC$ ? - -REPLY [2 votes]: When I was in grad school I learned about the following fact which was actually intended as a joke. -Theorem: There are irrational numbers $x$ and $y$ such that $x^y$ is rational. -Proof. Look at $\sqrt2^{\sqrt2}$. Is it rational? If so, take $x=y=\sqrt2$, and we are done. Otherwise, take $x=\sqrt2^{\sqrt2}$ and $y={\sqrt2}$, and we are again done since - $$x^y = \left(\sqrt2^{\sqrt2}\right)^{\sqrt2} = \sqrt2^{\left(\sqrt2\times\sqrt2\right)} = \sqrt2^2 = 2.$$ -QED -I have no idea if the rationality of $\sqrt2^{\sqrt2}$ is decidable or not (most likely it is), so this is not quite an answer.<|endoftext|> -TITLE: The stationary tower and supercompactness -QUESTION [5 upvotes]: I'm currently reading through Larson's book on the stationary tower and a point confused me. -Let $\delta$ be Woodin and let $j:V\to M\subseteq V[G]$ be an elementary embedding associated with the (full) stationary tower $\mathbb P_{<\delta}$. Then $V[G]\models {^{<\delta}}M\subseteq M$. This seems close to $\operatorname{crit}j$ being $\theta$-supercompact for every $\theta<\delta$, in $V[G]$, with the exception that the generic extension might contain more subsets of $\operatorname{crit}j$, so that the measure on $\operatorname{crit}j$ isn't total. -Now say there is a proper class of Woodins, and let $j:V\to V[G]$ be an elementary embedding associated with the stationary tower $\mathbb P_\infty$. Again, is $\operatorname{crit}j$ now 'close' to being supercompact in $V[G]$? -Is this the same thing as $\operatorname{crit}j$ being generically supercompact? Since we can choose $\operatorname{crit}j$ to be any inaccessible, does this then mean that a proper class of Woodins implies a proper class of generically supercompacts? - -REPLY [2 votes]: Apparently this has a name. To any large cardinal notion, there is a virtual variant, in which the elementary embeddings in question lie in some generic extension. Reformulating my scenario, I pointed out that a Woodin $\delta$ consistency-wise implies a virtually $\theta$-supercompact for every $\theta<\delta$. Using the countable tower instead, as Yair pointed out, we also get the result that a Woodin consistency-wise implies that $\omega_1$ is virtually almost huge. -But as is pointed out in Gitman's slides, all virtual large cardinals lie consistency-wise below $0^\sharp$, so this observation (that Woodins lie above virtuals) is pretty moot.<|endoftext|> -TITLE: Understanding Homology Operations and how to compute them -QUESTION [11 upvotes]: I stumbled upon this Lemma: - -Let $X$ be a spectrum and $H_p(X;\Omega_q^{Spin})\Rightarrow MSpin_{p+q}(X)$ the Atiyah-Hirzebruch spectral: - -The differential $d_2\colon H_p(X;\Omega_1^{Spin})\to H_{p-2}(X;\Omega_2^{Spin})$ is the dual of $Sq^2\colon H^{p-2}(X;\mathbb{Z}_2)\to H^p(X;\mathbb{Z}_2)$ -The differential $d_2\colon H_p(X;\Omega_0^{Spin})\to H_{p-2}(X;\Omega_1^{Spin})$ is reduction mod $2$ (denoted with $r$) composed with the dual of $Sq^2$ - - -And the proof is based on the following observation: $d_2$ is a stable homology operation and thus induced from elements in $$[H\mathbb{Z}/2, \Sigma^2 H\mathbb{Z}/2]\cong \mathbb{Z}/2\langle Sq^2\rangle$$ or $$[H\mathbb{Z}, \Sigma^2 H\mathbb{Z}/2]\cong \mathbb{Z}/2\langle Sq^2\circ r\rangle$$ -where $HR$ is the Spectrum representing singular cohomology with coefficient $R$. -In this paper, def 3.1 suggest that an Homology Operation is a natural transformation between homology functors, but for example Switzer introduces the Homology Cooperations in a complete different way (they might be different but it seems that in some cases they re the dual of the cohomology operations). So my questions are: -1) First of all, what's the relationship between homology operations and the homology cooperations described in Switzer's book for example. -2) Why are we looking at the stable Cohomology operations and then dualise it? -3) later in the paper the following reasoning is done: the edge homomorphism for the AHSS for stable homotopy is a stable homology operation form stable homotopy to singular homology, i.e. an element of $[S^0,H\mathbb{Z}]\cong \mathbb{Z}\langle h \rangle$ where $h$ is the Hurewicz homomorphism. Therefore (after testing with certain space) we can conclude that the edge homomorphism is given by the Hurewicz map. Where can I see some reference for the fact that $[S^0,H\mathbb{Z}]\cong \mathbb{Z}\langle h \rangle$? why aren't we dualise anything here as done above? (I know that homology is not necessarily the dual of cohomology, but I want to understand why we dualise above and not here, since map of spectra indices stable Cohomology operations. - -REPLY [10 votes]: I will work stably: everything in sight will be a spectrum. - -It is well known and classical that cohomology operations correspond to map of spectra: that is if $E,F$ are two spectra, a natural transformation $E^*X→F^{*+n}X$ correspond by Yoneda to a map of spectra $E→\Sigma^nF$. -On the other hand if you have a map of spectra $E→\Sigma^nF$, you also get a homology operation $E_*X→F_{*-n}X$. This is because homology can be written as -$$ E_*X = \pi_*(E\wedge X),\qquad F_{*-n}X=\pi_*(\Sigma^nF\wedge X)\,.$$ -So understanding cohomology operations will give you also a supply of homology operations. This is not a dualization as you said, it is just a way to get a homology operation out of a map of spectra: no duality required. Moreover we can prove that all homology operations are of this form. -In fact any such operation is determined by its restriction to finite spectra (since every spectrum is a filtered homotopy colimit of finite spectra), and using the canonical equivalence $\mathrm{Sp}_{fin}\cong\mathrm{Sp}_{fin}^{op}$, such a natural transformation is the same thing as a natural transformation -$$E^{-*}X=E_*DX→F_{*+n}DX=F^{-*-n}X$$ -when $X$ varies among finite spectra and $D$ is the Spanier-Whitehead dual. But we know that these operations are induced by maps $E→\Sigma^nF$. -Basically, when you restrict to finite spectra homology and cohomology are pretty much the same thing: you just need a Spanier-Whitehead dual to pass from one to the other. -We can say more about the second differential in the AHSS: if we look at the construction of the AHSS that uses the Postnikov tower for the spectrum $E$ instead than the one using the cellular filtration for $X$, we immediately see that the differential $d_2$ is the map of spectra given by the k-invariants of $E$. That is, we can look at the AHSS as the spectral sequence associated to the exact couple -$$\require{AMScd} -\begin{CD} -\dots @>>> \pi_*(P_{n+1}E\wedge X) @>>> \pi_*(P_nE\wedge X) @>>> \pi_*(P_{n-1}E\wedge X) @>>> \cdots\\ -{} @VVV {} @VVV {} @VVV {}\\ -\dots @. \pi_*(H\pi_{n+1}E \wedge X)=H_*(X;\pi_{n+1}E) @. \pi_*(H\pi_nE\wedge X)=H_*(X;\pi_nE) @. \pi_*(H\pi_{n-1}E\wedge X)=H_*(X;\pi_{n-1}E) @.\cdots -\end{CD}$$ -That is the $d_2:H_*(X;\pi_nE)\to H_{*-1}(X;\pi_{n+1}E)$ is induced by the map of spectra -$$H\pi_nE\to \Sigma P_{n+1}E\to \Sigma H\pi_{n+1}E\,.$$ -So we can leverage our understanding of the cohomology operations to get at the $d_2$ of the AHSS. - -Now let us tackle homology cooperations. These are the homotopy classes of $E\wedge E$. I really do not have too much to say about them: they show up because they are important for the Adams spectral sequence (in fact they basically run the show there). They are also the "dual" of the cohomology operations. If $E$ is an $E_\infty$-ring spectrum we can use the universal coefficient spectral sequence to go -$$\mathrm{Ext}_{E_*}(\pi_*(E\wedge E),E_*) \Rightarrow \pi_*F_E(E\wedge E,E)=\pi_*F(E,E)$$ -where $F_E$ and $F$ are the internal hom in $E$-modules and spectra respectively. For example when $E=Hk$ for $k$ a field (the main case people treat I think) this spectral sequence degenerates telling you that the $k$-linear dual of $\pi_*(Hk\wedge Hk)$ is the ring of cohomology operations. As you can see these have no direct relation to homology operations and they arise in a different setting altogether. - -Finally, the Hurewicz homomorphism. This is, by definition, the map on homology theories defined by the map $h:\mathbb{S}\to H\mathbb{Z}$ sending 1 to 1. In fact, if $f:\Sigma^n\mathbb{S}\to X$ is a class in $\pi_nX$ we are sending it to the image of the fundamental homology class of $\Sigma^n\mathbb{S}$. But the fundamental homology class is just the map -$$\Sigma^nh=(h\wedge 1):\Sigma^n\mathbb{S}\to H\mathbb{Z}\wedge\Sigma^n\mathbb{S}\,.$$ -So we are sending $f$ to $(1_{H\mathbb{Z}}\wedge f)\circ (h \wedge 1) = (h\wedge f)$, that is we are doing the homology operation represented by $h$.<|endoftext|> -TITLE: What is the automorphism group of the tensor square of the Leech lattice? -QUESTION [6 upvotes]: The tensor square of the Leech lattice is an even unimodular lattice of dimension 576 which, unless I am very mistaken, has no roots. Its automorphism group contains a group of shape $2 \cdot \mathrm{Co}_1^{\times 2} : 2$, but I expect it is larger than that. I would like to understand what I can about this group, e.g. how it is built from simple groups, what its maximal subs are, things like that. Has anything about this automorphism group been calculated? - -REPLY [3 votes]: This post provides details for the answer outlined by Noam Elkies in the comments. It is Community Wiki, so anyone can edit it to improve it. -Answer: Let $\Lambda$ denote the Leech lattice. The full automorphism group of the lattice $\Lambda^{\otimes 2}$ is the group $2\cdot \mathrm{Co}_1^{\times 2}:2 = (\mathrm{Co}_0^{\times 2}:2) / 2$, where $2\cdot \mathrm{Co}_1 = \mathrm{Co}_0 = \mathrm{Aut}(\Lambda)$. -(The action is: the wreath product $\mathrm{Aut}(\Lambda)^{\times 2} : 2$ obviously acts on $\Lambda^{\otimes 2}$, but the diagonal central $\mathbb Z/2$ acts trivially. As in the Atlas, $2$ denotes the group of order $2$, $:$ denotes semidirect product, and $\cdot$ denotes an extension that does not split.) -By Kitaoka, Scalar extensions of quadratic lattices II, minimal vectors in $\Lambda^{\otimes 2}$ are all of the form $v \otimes w$ for $v,w \in \Lambda$ minimal. -By inspecting the shapes of minimal vectors in $\Lambda$, one sees that if $v,v'\in \Lambda$ are minimal, then $|\langle v,v'\rangle| \leq 4$, with equality only if $v = \pm v'$. It follows that $\langle v\otimes w,v'\otimes w'\rangle = 8$ iff either $v = \pm v'$ and $\langle w,w'\rangle = \pm 2$ or $\langle v,v'\rangle = \pm 2$ and $w = \pm w'$. -Suppose now that $v_1\otimes w_1$, $v_2 \otimes w_2$, and $v_3 \otimes w_3$ are minimal vectors in $\Lambda^{\otimes 2}$ which pairwise pair to $8$. Then by pigeonhole at least two of the $v$s or at least two of the $w$s are equal up to sign, and so all of the $v$s or all of the $w$s are equal up to sign. (The sign of $v$ is not determined by the tensor product $v\otimes w$, of course.) If all the $w$s are equal, then all the $v$s pair pairwise to $+2$. Let us call such a triple of $v$s a "trio". -Choose a basis $e_1,\dots,e_{24}$ for $\Lambda$ consisting entirely of minimal vectors with the following property: the graph whose vertices are trios and whose edges are when trios intersect in sets of two is connected. It is easy to find such bases. -Let $\phi\in \mathrm{Aut}(\Lambda^{\otimes 2})$. Consider the set of 24 vectors $\phi(e_1\otimes e_1), \phi(e_2\otimes e_1),\dots,\phi(e_{24}\otimes e_1)$. These are all minimal vectors in $\Lambda^{\otimes 2}$ with many triples that pair pairwise to $8$ (one for each trio), and so up to perhaps multiplying $\phi$ by the "switch" $v\otimes w \mapsto w \otimes v$, we have $\phi(e_i \otimes e_1) = v_i \otimes w_1$. Since $\phi$ is metric-preserving, so is $\varphi_L : e_i \mapsto v_i$. -(Note that the sign of $w$, and hence of the $v_i$s, is ambiguous, but that is all.) -Compare now the set $\phi(e_1\otimes e_1),\dots,\phi(e_1\otimes e_{24})$. Either $\phi(e_1\otimes e_j) = v_1 \otimes w_j$ for some vectors or $v_j' \otimes w_1$ (since we already know that $\phi(e_1\otimes e_1) = v_1\otimes w_1$). But in the latter case the set of 47 linearly independent vectors $e_{24}\otimes e_1,\dots, e_1\otimes e_1,\dots,e_1,\otimes e_{24}$ is mapped under $\phi$ to the 24-dimensional space $\Lambda \otimes w_1$, and so is ruled out. Let $\varphi_R : e_j \mapsto w_j$. -Similarly, $\phi(e_2 \otimes e_j) = v_2 \otimes w_j'$, where at first all we know is that $w_1' = w_1$. But considering $\phi(e_i\otimes e_2)$ shows that $w_2 = w_2'$. -All together we find that $\phi(e_i \otimes e_j) = \varphi_L(e_i) \otimes \varphi_R(e_j)$ for $(\varphi_L,\varphi_R) \in \mathrm{Co}_0^{\times 2}$. -This description is redundant because we can switch the signs of both $\varphi_L,\varphi_R$ simultaneously without changing $\phi$. Also we allowed up to one multiplication by the "switch" earlier. All together this shows that $\phi \in 2 \cdot \mathrm{Co}_1^{\times 2} : 2$ as claimed.<|endoftext|> -TITLE: Eigenvalues of a matrix with entries involving combinatorics -QUESTION [14 upvotes]: Let $F(n, l, i, j)$ be the cardinality of the set -\begin{eqnarray*} -\{(k_1, \cdots, k_n)\in\mathbb{Z}^{\oplus n}|0\leq k_r\leq l-1\text{ for }1\leq r\leq n\text{, }k_1+\cdots+k_n=lj-i\}. -\end{eqnarray*} -Define an $n\times n$ matrix $M(l, n)$ by -\begin{eqnarray*} -M_{ij}(l, n)=(-1)^{i+j}F(n, l, i, j). -\end{eqnarray*} -In fact $M(l, n)$ is related to the Adams operations on $U(n)$, and I can show using algebraic topology that the eigenvalues are $1, l, l^2, \cdots, l^{n-1}$. Note that the last column vector of $M(l, n)$ is $(0, 0, \cdots, 0, 1)$ and so it is an eigenvector corresponding to the eigenvalue 1. When $n=2$, $M(l, 2)$ is $\begin{pmatrix}l&0\\ 1-l& 1\end{pmatrix}$. -Question: Are there more elementary ways to show this? -Added: For a fixed $n$ and different $l$, the matrices $M(l, n)$ commute with each other (more precisely, $M(l_1, n)M(l_2, n)=M(l_1l_2, n)$) and thus they are simultaneously diagonalisable and their eigenvectors do not depend on $l$. See the answer to this question for a set of eigenvectors of $M(l, n)$. A side question is how one can obtain the eigenvectors using elementary methods. - -REPLY [2 votes]: As I mentioned in the question, the matrix $M(l, n)$ is related to the Adams operations on $U(n)$. Let me elaborate further on this. -Recall that for a finite CW complex $X$, $K^{-1}(X)$ is $[X, U(\infty)]$, the group of homotopy classes of maps from $X$ to $U(\infty)$. Let $G$ be a compact Lie group and $\delta: R(G)\to K^{-1}(G)$ be the map which sends a $G$ representation $\rho$ to the homotopy class of it viewed the map $G\stackrel{\rho}{\rightarrow}U(n)\hookrightarrow U(\infty)$. If $G=U(n)$, then by a theorem of L. Hodgkin, its $K$-theory is the exterior algebra -\begin{eqnarray} -K^*(U(n))=\bigwedge\nolimits^*_\mathbb{Z}(\delta(\sigma_n), \delta(\wedge^2\sigma_n), \cdots, \delta(\wedge^n\sigma_n)), -\end{eqnarray} -where $\sigma_n$ is the standard representation of $U(n)$. The Adams operation $\psi^l$ on the primitive $\mathbb{Z}$-module of this exterior algebra has $lM(l, n)$ as the matrix representation with respect to the basis $\delta(\sigma_n), \cdots, \delta(\wedge^n\sigma_n)$. The definition of Adams operations implies that they commute with each other, so do $M(l, n)$ for different positive $l$ and fixed $n$, and their eigenvectors are independent of $l$ due to simultaneous diagonazability. -Using the definition of Chern character and Adams operation, one can easily get the following statement: - -Let $X$ be a finite CW-complex and $\alpha\in K^{-1}(X)$. Then we have - \begin{eqnarray} -\text{ch}(\psi^l(\alpha))=\sum_il^i\text{ch}_{2i-1}(\alpha) -\end{eqnarray} - where $\text{ch}_k(\alpha)$ is the degree $k$ term of $\text{ch}(\alpha)$. - -Returning to the case $X=U(n)$, we know that $H^*(U(n), \mathbb{Q})$ is also an exterior algebra on $n$ primitive generators of degrees $1, 3, 5, \cdots, 2n-1$. So if $\alpha$ is a (rational) primitive element of $K^*(U(n))\otimes\mathbb{Q}$ which is also an eigenvector of $\psi^l$, then there exists $1\leq j\leq n$ such that $\text{ch}_{2i-1}(\alpha)$ is zero for $i\neq j$ but nonzero for $i=j$, and the corresponding eigenvalue is $l^j$. So the eigenvalues of $lM(l, n)$ are $l, l^2, \cdots, l^n$. This recent paper has more details of the above discussion.<|endoftext|> -TITLE: On the partner of the Emma Lehmer quintic -QUESTION [7 upvotes]: Given, -$$x^5+10cx^3+10dx^2+5ex+f = 0$$ -If there is an ordering of its roots such that, -$$\small x_1 x_2 + x_2 x_3 + x_3 x_4 + x_4 x_5 + x_5 x_1 - (x_1 x_3 + x_3 x_5 + x_5 x_2 + x_2 x_4 + x_4 x_1) = 0\tag1$$ -then its coefficients are related by the quadratic in $f$, -$$(c^3 + d^2 - c e) \big((5 c^2 - e)^2 + 16 c d^2\big) = (c^2 d + d e - c f)^2 -$$ -This implies that such quintics come in pairs, having the same $c,d,e$ but differing only in $f$. An example would be the solvable Emma Lehmer quintic, -$$\small \color{blue}{y^5 + n^2y^4 - (2n^3 + 6n^2 + 10n + 10)y^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)y^2 + (n^3 + 4n^2 + 10n + 10)y} +1=0$$ -and its partner (also solvable) -$$\small \color{blue}{y^5 + n^2y^4 - (2n^3 + 6n^2 + 10n + 10)y^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)y^2 + (n^3 + 4n^2 + 10n + 10)y} - (n^3 + 5n^2 + 10n + 20)(n^3 + 5n^2 + 10n + 5)/25=0$$ - -Q: Is there a subset of $n$ such that the partner also has Galois order 5? - -REPLY [7 votes]: [Edited to give more details] -The subset of such $n \in \bf Q$ is empty: as long as the "partner" quintic, -call it $Q$, is irreducible, its Galois group is cyclic of order $5$ over -${\bf Q}(\sqrt{5})$, but dihedral of order $10$ over $\bf Q$. -Let the roots of $Q$ be $x_1,x_2,x_3,x_4,x_5$, ordered so that -there is a Galois automorphism $\sigma$ taking each $x_i$ to $x_{i+1}$ -(with $x_6 \equiv x_1$). Then if $Q$ is irreducible over some field $K$ -then its Galois group is cyclic if and only if -$$ -\Delta := (x_1 - x_2) (x_2 - x_3) (x_3 - x_4) (x_4 - x_5) (x_5 - x_1), -$$ $$ -\Delta' := (x_1 - x_3) (x_3 - x_5) (x_5 - x_2) (x_2 - x_4) (x_4 - x_1) -$$ -are in $K$ (because $\Delta,\Delta'$ are invariant under $\sigma$, but -taken to $-\Delta,-\Delta'$ by the involution $x_i \leftrightarrow x_{6-i}$). -Now $\Delta \Delta' = \pm \cal D$, where -${\cal D}^2$ is the discriminant of $Q$; we compute -${\cal D} = D_3 D'_3 D_4^2 / 5^3$, where -$$ -D_3 = 3 n^3 + 20 n^2 + 50 n + 50, -\quad -D'_3 = 4 n^3 + 10 n^2 + 25 n + 25, -$$ $$ -D_4 = n^4 + 5 n^3 + 15 n^2 + 25 n + 25. -$$ -Moreover, $\Delta$ and $\Delta'$ are integral over ${\bf Q}[n]$, -and even over ${\bf Z}[\frac15][n]$ with bounded denominator; -so there are not many possibilities for the factorization -${\cal D} = \Delta \Delta'$, and we can determine the correct one -by specializing $n$ and computing the $x_i$ numerically. -We find that $\Delta$ and $\Delta'$ are -$\pm 5^{-3/2} D_3 D_4$ and $\pm 5^{-3/2} D_3' D_4$ -(or vice versa if we change $x_1,x_2,x_3,x_4,x_5$ to $x_1,x_3,x_5,x_2,x_4$). -These are always rational over ${\bf Q}(\sqrt 5)$, but never over $\bf Q$ -(for rational $n$). Therefore the Galois group is always cyclic over -${\bf Q}(\sqrt 5)$ but dihedral over $\bf Q$. $\Box$<|endoftext|> -TITLE: When is the exterior algebra a Hopf algebra? -QUESTION [6 upvotes]: I have several questions on the exterior algebra of a vector space: - -Q1:When has the exterior algebra A (viewed just as an algebra, not considered as a graded algebra) of an $n$-dimensional vector space over a field $K$ the structure of a Hopf algebra? (depending on n and K) - -Note that it is not always a Hopf algebra, for example in the easiest case the exterior algebra is $K[x]/(x²)$ and this should be a Hopf algebra iff the characteristic of the field is 2. - -Q2: Is there a finite dimensional, nonprojective module M over A with $Ext_A^{1}(M,M)=0$? - -This question has the answer no when the exterior algebra is a Hopf algebra and thus this question is related to Q1. -(Q2 is also open in the graded case and has a positive solution in a special case, see the last chapter of https://arxiv.org/pdf/1701.01149.pdf ) - -Q3:Can one classify all periodic modules over this algebra? - -In general the exterior is a wild algebra for more than 2 variables and it is hopeless to give a classification of all indecomposable modules, but maybe there is an interesting classification of special modules such as indecomposable periodic modules (a module is periodic in case $\Omega^n(M) \cong M$ for some $n$). - -REPLY [3 votes]: Another proof: The exterior algebra is Koszul, it's Koszul dual is the symmetric algebra, that is commutative, but NOT super commutative, unless Char=2. Assuming $char \neq 2$, the exterior algebra cannot be Hopf because for $H$ a Hopf algebra, $Ext_H(k,k)$ is a subalgebra of Hochschild cohomology of H, hence, it should be super commutative.<|endoftext|> -TITLE: A question about billiards -QUESTION [10 upvotes]: This is a question in a rather well investigated subject of which I know very little and I have a hard time "translating" the general results available. Let me also say that I got interested in this question after a conversation with a friend about placements of cell phone towers. (It is too long a story to include here.) -Here's the problem. $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\bsP}{\boldsymbol{P}}$ $\newcommand{\bR}{\mathbb{R}}$ Fix a positive integer $n$ and let $\bsP_n$ denote the regular $n$-gon centered at the origin and inscribed in the unit circle. -The billiard in the title refers to the motion of a unit speed photon inside $\bsP_n$ assuming the edges of $\bsP_n$ are perfect mirrors. For every $x$ in the interior of $\bsP_n$, any unit vector identified with a $\theta\in S^1$ and any $T>0$ we denote by $N_T(x,\theta)$ the number of "bounces-off-mirrors" during the time interval $[0,T]$ of a photon that starts its motion at $x$ in the direction $\theta$. The question involves the behavior of $N_T(x,\theta)$ for large $T$. -I believe that there exists a measurable function $\beta_n:\bsP_n\times S^1\to [0,\infty)$ such that for almost all $(x,\theta)\in\bsP_n\times S^1$ we have -$$\lim_{T\to\infty}\frac{1}{T}N_T(x,\theta)=\beta_n(x,\theta).$$ -The questions involve the proportionality constants $\beta_n(x,\theta)$. - -Has anyone computed explicitly the function $\beta_n(x,\theta)$? -Has anyone computed explicitly the average - -$$\bar{\beta}_n(x):=\frac{1}{2\pi} \int_0^{2\pi}\beta_n(x,\theta) d\theta ?$$ -(I am inclined to believe that $\bar{\beta}_n(x)$ is constant.) -Above, the cases $n=3,6$ are of particular interest in the original cell phone tower problem. -Finally, has anyone investigated the behavior of $\beta_n(x,\theta)$ as $n\to\infty$? -Thanks. - -REPLY [11 votes]: The answer is given by the general theory of rational billiard flows (i.e., those on polygons whose angles are rational multiples of $\pi$). On such a polygon $Q$ the tangent vectors to any given orbit are parallel to a finite set of unit vectors, so that the orbits with initial direction $\theta$ lie on an invariant surface $M_\theta$ which consists of a finite number of copies of $Q$, one for each potential direction of an orbit with initial direction $\theta$. According to a theorem of Kerckhoff, Masur and Smillie (http://www.ams.org/mathscinet-getitem?mr=855297 or the announcement http://www.ams.org/mathscinet-getitem?mr=799797), for almost all $\theta$ the flow on $M_\theta$ is uniquely ergodic. It implies that for a.e. $\theta$ the projection to $Q$ of every orbit with initial direction $\theta$ is uniformly distributed in $Q$ (with respect to the Lebesgue measure). -Therefore, $1/\beta_n(x,\theta)$ coincides, for a.e. $\theta$ and any $x\in Q$, with the average length of segments in billiard orbits, i.e., by the ergodic theorem, with the average of the function $L(x,\theta)$ (the distance between the points where the line issued from $x$ in the direction $\theta$ intersects the boundary of $Q$) with respect to the Lebesgue measure on $M_\theta$, which only depends on $\theta$ and should be quite easy to caclulate explicitly in the case of regular polygons.<|endoftext|> -TITLE: On a set of sets intersecting in $1$ point -QUESTION [5 upvotes]: Let $\kappa$ be an infinite cardinal. Suppose $E \subseteq {\cal P}(\kappa)$ has the following property: for $e_1\neq e_2\in E$ we have $|e_1\cap e_2|= 1$, and suppose $|E| = \kappa$. -Does this imply that at least one of the following statements is true? - -there is $e\in E$ with $|e|=\kappa$; -there is $\alpha \in \kappa$ such that $|\{e\in E: \alpha \in e\}| = \kappa$. - -REPLY [10 votes]: Consider two sets $e,f\in E$, assume that $\max(|f|,|e|)=:\mu<\kappa$, $\{x\}:=e\cap f$. Take arbitrary element $y\in f\setminus x$, it is contained in at most $\mu$ sets from $E$. Indeed, they all have a common element with $e$, and all those common elements are different. So, the elements of $f\setminus x$ are contained totally in at most $\mu\times \mu<\kappa$ sets, thus $x$ is contained in $\kappa$ sets.<|endoftext|> -TITLE: Leopoldt's conjecture for totally real cubic and $S_3$-extensions -QUESTION [5 upvotes]: For an introduction to Leopoldt's conjecture, see part 3, Chapter X of "Cohomology of Number Fields", which is freely available here. -Write Leo($K$,$p$) for Leopoldt's conjecture for the number field $K$ at the prime $p$. If $K$ is an abelian extension of either $\mathbb{Q}$ or an imaginary quadratic field, then we know Leo($K$,$p$) for all primes $p$. (This is due to Ax and Brumer - see reference above for more details.) -Now let $K$ be a cubic extension of $\mathbb{Q}$. I am interested in Leo($K$,3). -If $K/\mathbb{Q}$ is Galois then we are done. If $K$ has mixed signature, then its Galois closure $L$ is a totally complex $S_3$ extension of $\mathbb{Q}$ and thus is a cyclic extension of an imaginary quadratic field. Thus we know Leo($L$,3) and this implies Leo($K$,3) (Leopoldt's conjecture is inherited by subfields). -My question is: are there any cases for which Leo($K$,3) is known when $K$ is a totally real non-Galois cubic extension of $\mathbb{Q}$? By the above reasoning, it would suffice to find a totally real $S_3$ extension $F/\mathbb{Q}$ for which Leo($F$,3) is known. - -REPLY [7 votes]: Out of curiosity, I thought I'd follow up on znt's suggestion. -Let $K = \mathbf{Q}(\alpha)$ where $\alpha^3 - 13\alpha + 7=0$. Then $K$ is totally real and non-Galois, and the prime 3 is totally inert in $K$. The unit group of $K$ is generated by $(u_1, u_2) = (\alpha^2 + 3\alpha - 2, \alpha^2 - 4\alpha + 2)$. -There are 3 embeddings of $K$ into the unramified cubic extension $L$ of $\mathbf{Q}_3$. Sage happens to like the polynomial $y^3 + 2y + 1$ as a defining polynomial for $L$; if $c$ is a root of this, then for $i = 0, 1, 2$ there's an embedding $\sigma_i$ for which $\sigma_i(\alpha) = c + i \bmod 3$. -Now one finds that the matrix with $i,j$ entry $\log \sigma_i(u_j)$ for $1 \le i, j \le 2$ has determinant -3^3 + 3^7 + 2*3^8 + 3^9 + 3^10 + 2*3^12 + 2*3^15 + 3^17 + 3^18 + 2*3^19 + 3^20 + O(3^21) - -which is not zero. Hence Leopoldt's conjecture holds for this extension at 3.<|endoftext|> -TITLE: are these polynomials or rationals functions? -QUESTION [15 upvotes]: Let $x$ be a variable. Define the following family of sequences (reminiscent of Lucas polynomials) according to the rule: $P_0(x):=0, P_1(x):=1$ and for $n\geq2$ by -$$P_n(x)=xP_{n-1}(x)-P_{n-2}(x).$$ -Notice that $P_n(2)=n$ for every $n\in\mathbb{Z}_{\geq0}$. Here are a few examples: -$$P_2=x, \qquad P_3=x^2-1, \qquad P_4=x^3-2x, \qquad P_5=x^4-3x^2+1.$$ - -QUESTION 1. Empirical evidence suggest that, for each fixed integers $n, k\geq1$, - $$Q_{n,k}(x):=\frac{P_1(x)^{2k-1}+P_2(x)^{2k-1}+\cdots+P_n(x)^{2k-1}}{P_1(x)+P_2(x)+\cdots+P_n(x)} \tag1$$ - is a polynomial in $x$. -This is trivial for $k=1$. Is it true for other odd powers $2k-1$? - -REMARKS. -(1) Specialized values $2k-1=3$ or $5$, etc are still interesting to me. -(2) Even the case of special valuations for $x\in\mathbb{Z}$ are appealing as well, which means (1) becomes a claim on integrality of sequences. - -QUESTION 2. Encouraged by the success with QUESTION 1, how about this? - $$R_n(x)=\prod_{j=1}^n\frac{P_j(x)^{2k-1}+\cdots+P_n(x)^{2k-1}}{P_j(x)} \tag2$$ - is a polynomial in $x$. - -REPLY [5 votes]: Question 2. -Since $P_n(2\cos t)=\sin(nt)/\sin t$, we get that the roots of $P_n$ are $2\cos(\pi k/n)$, -$k=1,\dots,n-1$. That is, a number $\kappa=2\cos(\pi a/b)$, $0 -TITLE: Is $\overline{\Delta}$ a linearly independent set? -QUESTION [6 upvotes]: Let $G$ be a connected reductive group defined over a field $F$. Let $T$ be a maximal torus of $G$ which is defined over $F$, $A_0$ the maximal $F$-split subtorus of $T$, and $X_0$ the cotorsion free subgroup of $X = X(T)$ corresponding to $A_0$. If $L$ is a finite Galois extension of $F$ over which $T$ splits, then $X_0$ can be described as -$$X_0 = \{ \chi \in X : \sum\limits_{\sigma \in \textrm{Gal}(L/F)} \chi^{\sigma} = 0 \}$$ -Now, let $\Phi$ be the set of roots of $T$ in $G$, and $\Delta$ a base for the root system $(\langle \Phi \rangle_{\mathbb{Q}}, \Phi)$, where $\langle \Phi \rangle_{\mathbb{Q}}$ denotes the vector subspace of $\mathbb{Q} \otimes_{\mathbb{Z}} X$ spanned by $\Phi$ . -The base $\Delta$ comes from a total ordering $<$ of $\langle \Phi \rangle_{\mathbb{Q}}$, or even of $X$. Suppose that $<$ is compatible with $X_0$ in the sense that if $\chi, \chi' \in X$ with $\chi + X_0 = \chi' + X_0$ with $\chi \not\in X_0$ and positive, then $\chi'$ is also positive. Then $<$ induces a total ordering on $X/X_0$. -Suppose $\Delta$ is coming from such a total ordering. Let $\overline{\Delta}$ be the image of $\Delta$ in $X/X_0$, if necessary removing the zero element. -It may not be the case that the reduction of $\Delta - \Delta \cap X_0$ modulo $X_0$ is injective. Nevertheless, let $\overline{\Delta} = \{\gamma_1, ... , \gamma_t \}$ for distinct $\gamma_i \in X/X_0$. These are positive elements in the total ordering on $X/X_0$. Is it the case that the $\gamma_i$ are linearly independent? -I have tried to prove this without success. I was thinking that $\overline{\Delta}$ should be a base for a possibly nonreduced root system. The elements of the image of $\Phi$ in $X/X_0$, removing zero, are integer-linear combinations of $\overline{\Delta}$ which are either all positive or all negative, but the linear independence does not seem obvious. - -REPLY [4 votes]: Indeed. -As explained in my answer to your previous question (https://mathoverflow.net/q/255159), the map $X\to X/X_0$ is the restriction $X(T)\to X(A_0)$. In your present question, you gave those character groups compatible orders as in "Borel Tits, Groupes réductifs". Denote $_{F}\Delta$ (respectively $\Delta$) the base of the root system $\Phi (G,A_0)$ (respectively $\Phi$) corresponding to this order. Then $\overline{\Delta} =~ _{F}\Delta $ (by "Borel Tits, Groupes réductifs, 6.8"). Being a base of a root system, they are in particular linearly independent. -In summary, as you guessed, the image of $\Phi$ in $X/X_0$ (removing zero) is itself a root system (namely the root system $\Phi (G,A_0)$, also called the relative root system), and $\overline{\Delta} =~ _{F}\Delta $ is a base of it.<|endoftext|> -TITLE: Structures in the plot of the “squareness” of numbers -QUESTION [26 upvotes]: (This is based on an earlier MSE posting, -"Structures in the plot of the “squareness” of numbers.") - -My main question is to explain the structural features of this plot: - - - -This is a plot of what I call the squareness ratio $r(n)$ of a natural number $n$ (or simply the "squareness"). -The squareness $r(n)$ is the largest ratio $\le 1$ that can be obtained -by partitioning the factors of $n$ into two parts and forming the ratio of their products. -A perfect square has squareness $1$. -A prime $p$ has squareness $1/p$. -In a sense, the squareness measures how close is $n$ to a perfect square. -The squareness ratios for the first ten number $n=1,\ldots,10$ are -$$1,\frac{1}{2},\frac{1}{3} - ,1,\frac{1}{5},\frac{2}{3},\frac{1}{7},\frac{1}{2},1,\frac - {2}{5} \;.$$ -A more substantive example is $n=12600=2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 7$, -for which the largest ratio is -$$\frac{3 \cdot 5 \cdot 7}{2^3 \cdot 3 \cdot 5}=\frac{7}{8}=0.875 \;.$$ -One more example: $n = 2^2 \cdot 3^2 \cdot 5^2 \cdot 7 \cdot 11 = 69300$, -$$r(n) = \frac{2^2 \cdot 3^2 \cdot 7}{5^2 \cdot 11}=\frac{252}{275} \approx 0.92 \;.$$ -The only feature of this plot—the rays through the origin—that is -evident to me -is explained by the fact that, for a prime $p$, -for $n = k p$ and $k=1,\ldots,p$, the squareness ratio is $k/p$, and so those -ratios lie on a line through the origin of slope $\frac{1}{p^2}$. -MSE user PattuX emphasized that similar rays occur for particular composite $n$. -Several other features could use explanation: -(1) The discernable density change at $r=\frac{1}{2}$. -(2) The (apparent) hyperbolas. -(3) The interesting structure near $r=1$, both negative (hole-)curves and positive (dot-)curves: - - - - -Detail: $35 K \le n \le 60K$ (approximately), near $r=1$. - - -I welcome explanations for (1), (2), (3), and other apparent features of the plot. -This is to satisfy curiosity; it is far from my expertise. - -Added(1): As per Gerhard Paseman's request, the plot with only odd $n$ ratios: - - - -        - -Squareness ratio $r(n)$ for odd $n$ only; even $n$ not plotted. - - -Added(2): The landscape is rather different for larger $n$ -(in accordance with Lucia's insights): - - - -        - -Squareness ratio $r(n)$ for $900{,}000 \le n \le 1{,}000{,}000$. - -REPLY [4 votes]: It is true that in the long run most points are near the bottom but there are the observed patterns in the portion shown. Here are some comments in a slightly stream of consciousness order: -At the top left the increasing curve right below the line $y=1$ must be $(k(k-1),1-1/k)$ which is certain points pretty close to $y=1-1/\sqrt{x}.$ In the plot it appears piecewise constant but that is an artifact of the plotting. -Below there we see the similar curves $(k(k-c),1-c/k).$ In the long run these become indistinguishable from $y=1$ -To be sure, plot the point using something like Maple which lets you probe for coordinates. Some work of this type could explain other features. -To my eye the top line for the (approximately) $35K \le n \le 60K$ plot seems to have $97$ points. There should be about $57$ squares $(k^2,1)$ in that range so I guess the $n=k(k-1)$ are indistinguishable but the $n=k(k-2)$ are visually lower. -Coming down and back from each point $(k^2,1)$ we have a curve $(k^2-j^2,\frac{k-j}{k+j})$ This is especially clear in the plot ending at $n=10^6.$ each of these break eventually when the numerator and denominator have many small factors. For example I think the sequence going down from (104^2,1) has $y$ values $103/105,102/106,\cdots,89/119$ however the next point in that sequence $(88\cdot 120,\frac{88}{120})$ can get lifted up to replace $\frac{88}{120}$ with $\frac{96}{110}.$ I feel that more could be said about how long it takes before the sequence can break the first time than I have figured out. After such a break the pattern might pick up again as in $\frac{87}{121}$ next is $\frac{86}{122},\frac{85}{123},\cdots$ -These curves, which do go on for a while are approximately parallel and interspersed are the curves $((k+j+1)(k-j),\frac{k-j}{k+j+1}.$ These are all the points "really near" $y=1$ (a bold statement that needs justification.) In between , then, are white curves. I count $100$ of these curves in the plot which runs from $949^2$ to $1000^2$ which is what one would expect. -For $n=2k^2$ we can get the point $(n,\frac12).$ It may be possible to have a higher point. In a dramatic case $(2\cdot 35^2,\frac{49}{50}).$ However for over $60$ values $k \le 100$ we do get $(2k^2,\frac{1}{2}).$ I didn't examine larger $k.$ -For any point $(x,y)$ we have potential points $(cx,\frac{1}{cy})$ for small $c.$ It might be possible to have a higher $y$ but if one runs over a curve then a good fraction of the points might transform that way creating a transformed curve. I imagine the curves going up from $y=1/2$ for $900K \le n \le 1000K$ are (partial) transforms of the lines coming down from $y=1$ for $450K \le n \le 500K.$ These seem to have an almost reflection below $y=1/2$ making sideways parabolas. I don't have an explanation but suspect getting the actual coordinates would reveal much. -Similarly for any point $(x,y)$ we have potential points $(xz^2,y).$ Again, it might be possible to improve on this in some cases but that would be more challenging for $y$ closer to $1.$ Likewise if $z$ has large divisors relative to $x.$ -I'll stop with the generalization that if we have $(x_1,y_1)$ and $(x_2,y_2)$ then we have the potential points $(x_1x_2,\frac{y_1}{y_2}).$ Running over two curves , both with large $y$ values we can get parametric families such as $$\left((k^2-u^2)((k+1)^2-v^2),\frac{(k\pm u)(k+1\pm v)}{(k\mp u)(k+1\mp v)}\right).$$ -I can imagine that clusters of points would result, perhaps with explainable holes. -Holding, say $(x_1,y_1)$ constant might give transforms of curves. It would be easy to probe this with some plots. For example the curve above $(104^2-j^2,\frac{104-j}{104+j})$ combined with the point $(56,7/8)$ gives $y=\frac{7(104+j))}{8(104-j})$ for $j=1\cdots 6$ getting to $\frac{770}{784}$ but then switches to $y=\frac{8(104-j))}{7(104+j})$ starting with $776/777.$ Again, some of those points might be possible to improve.<|endoftext|> -TITLE: Alternatives to "Sketches of an Elephant" Volume 3 -QUESTION [11 upvotes]: The third volume of Peter Johnstone's massive compendium of topos theory, "Sketches of an Elephant", is yet to be published. The volume is supposed to discuss cohomology and mathematical universes in the context of topos theory. While we wait for the publication of this volume, are there any alternative (hopefully just as comprehensive) references for the aforementioned subjects? - -REPLY [5 votes]: Johnstone's 1977 book 'Topos Theory' is a very good source.<|endoftext|> -TITLE: Is there a choice-free proof that a Euclidean domain is a UFD? -QUESTION [18 upvotes]: I asked this question about a week ago on math.SE, without any answers. My motivation is pedagogical, but maybe the question comes closer to research-level than I thought. -The proof (at least the proof I know) that a principal ideal domain is a unique factorization domain uses the axiom of choice in multiple ways, and the usual way to show that a Euclidean domain is a UFD is to show that it's a PID (which is easy and constructive). -Is there a direct proof, not using the axiom of choice, that a Euclidean domain is a UFD? - -REPLY [14 votes]: There are two parts to showing a Euclidean domain or a PID are UFDs: (i) existence of an irreducible factorization for every nonzero nonunit and (ii) essential uniqueness of the irreducible factorization (any two use the same number of irreducible factors and the irreducibles that occur in both factorizations can be matched termwise up to multiplication by a unit). -To prove (ii), the key point is that every irreducible element is a prime element, and to prove that you need to be able to write $px + ay = 1$ for any irreducible $p$ and element $a$ where $p \nmid a$; the nondivisibility implies (since $p$ is irreducible) that the only common factors of $p$ and $a$ are units, so Euclid's algorithm in a Euclidean domain lets you algorithmically solve $px + ay = 1$ for some $x$ and $y$. In a PID you'd instead observe that the ideal $(p,a)$ has to be $(1)$. -To prove (i) is a major distinction between Euclidean domains and PIDs. You can do this for Euclidean domains in a much more concrete way than for PIDs. I compare approaches for each as Theorems 4.2 and 4.3 at http://www.math.uconn.edu/~kconrad/blurbs/ringtheory/euclideanrk.pdf. You need to read Sections 2 and 3 first to see why I mean about being able to assume the "$d$-inequality" holds: a Euclidean domain does not have to have its "norm" function $d$ be totally multiplicative or satisfy $d(a) \leq d(ab)$, but you can always adjust the "norm" function to fit that inequality all the time if it doesn't at the start. (Some books make this inequality part of the definition of a Euclidean domain and some do not.) Of course in $\mathbf Z$ and $F[x]$ that inequality is true, so you save some time when proving those rings are UFDs compared to a general Euclidean domain. -The bottom line is that you definitely do not need to introduce the machinery of PIDs in order to prove rings like $\mathbf Z$ or $F[x]$ have unique factorization. After all, unique factorization in those types of rings as well as in $\mathbf Z[i]$ was known (say, to Gauss) long before there was a concept of PID. I remember being surprised when I first saw how PIDs are proved to be UFDs, since I knew the case of Euclidean domains already and the proof of part (i) for PIDs was rather more abstract than I expected.<|endoftext|> -TITLE: Picard groups and birational morphisms -QUESTION [6 upvotes]: Let $f:X\rightarrow Y$ be a birational morphism of projective varieties. Assume that $Pic(X)$ is a free abelian group generated by $n$ divisors $D_1,...,D_n$. -Under which hypothesis on $X$ and $Y$ is it true that then $Pic(Y)$ is the free abelian group generated by the divisors $D_i$ not contracted by $f$ ? - -REPLY [2 votes]: OK, apparently this ended up too long for a comment.... (and to see the connection, read the comments above) - -It's actually very easy to prove that if $Y$ is locally $\mathbb Q$-factorial (even a little weaker than locally factorial!), then for any point $y\in Y$ in the image of the exceptional set of any projective birational morphism $f:X\to Y$ that exceptional set has to contain a divisor whose image contains $y$, which is a bit weaker statement than that the exceptional set is of pure codimension $1$ (which I bet is what Debarre says in the reference given by Jason and it is also proved in Shafarevich's book), but it already proves Jason's comment (modulo assuming projective) that small (projective) morphisms produce not locally factorial singularities. -The proof (which is kind of fitting your setup) is as follows: Let $H$ be an effective Cartier divisor on $X$ which is not numerically trivial on a curve contracted to $y\in Y$ and consider $f_*H$; if $Y$ is locally $\mathbb Q$-factorial, then (perhaps after restricting to an open subset) some multiple of $f_*H$ is a Cartier divisor. Replacing $H$ with the same multiple we may assume that $f_*H$ is Cartier. Then $f^*f_*H$ is (numerically) trivial on any curve contracted by $f$ and hence by the choice of $H$, they cannot be equal. Therefore $f^*f_*H-H\neq 0$ is an (effective) exceptional divisor whose image contains $y$. -This proof actually shows that if $f$ is a small projective morphism (small means that the exceptional set has codimension at least $2$), then some of your generators on $X$ will likely not end up in the Picard group of $Y$.<|endoftext|> -TITLE: Characterization of exact groups via the existence of amenable actions on unital C*-algebras, part 2 -QUESTION [5 upvotes]: In this recent MOF question I asked whether exact groups could be characterized via the existence of amenable actions on unital C*-algebras. The answer, provided by Caleb Eckhardt in a comment, was that an amenable action on a unital C*-algebra leads to an amenable action on the compact spectrum of the center of said algebra, and hence exactness follows from a well known result. -The present question is pretty similar to my previous question, but I am now employing a different amenability condition, as follows: -Definition. An action $\theta$ of a discrete group $G$ on a C*-algebra $A$ is said to satisfy the approximation property if there exists a net - $\{ a_i \}_{i\in I}$ - of finitely supported functions - $$ - a_i: G \to A, - $$ - which is bounded in the sense that - $$ - \sup _{i\in I} \Big\Vert {\sum_{g\in G } a_i(g)^* a_i(g)} \Big\Vert < \infty , - $$ - and such that - $$ - \lim _{i \rightarrow \infty } \sum_{h\in G } a_i(gh)^* b\,\theta_ g( a_i(h)) = b,\quad \forall b\in A . - $$ -See the discussion before [1, Definition 20.11]. -This condition is weaker than Anantharaman-Delaroche's usual condition of amenability (definition 4.3.1 in Brown and Ozawa) in that it does not require the $a_i$ to take values in the center of $A$. Nevertheless it is enough for most purposes and in particular it implies that the crossed product is nuclear provided $A$ is nuclear [1, Proposition 25.10]. -Thus, here is the new version of my question: - -Question: Suppose that a discrete group $G$ admits an action on a unital - C*-algebra $A$, satisfying the above approximation property. Is $G$ - necessarily exact? - -Notice that, precisely because the $a_i$ are allowed to take values outside the center of $A$, it is not immediately clear that the above condition passes to the center, and hence Caleb's answer might no longer work in this case. -Reference: -[1] -Partial Dynamical Systems Fell Bundles and -Applications - -REPLY [3 votes]: The answer to the question is positive: see Remark 6.6 in the paper -https://arxiv.org/pdf/1904.06771.pdf -The approximation property implies amenability in the sense of Claire Anantharaman Delaroche, this has been proved in -https://arxiv.org/pdf/1907.03803.pdf -for discrete groups, see also -https://arxiv.org/pdf/1904.06771.pdf -Indeed, it has been stablished recently that the approximation property is equivalent to amenability even in the more general context of locally compact groups by Ozawa-Suzuki in -https://arxiv.org/abs/2011.03420 -and that the above question in this context also has a positive answer (see Corollary 3.6).<|endoftext|> -TITLE: Two statistics on the permutation group -QUESTION [8 upvotes]: Let $\mathfrak{S}_n$ be the permutation group on an $n$-element set. For each fixed $k\in\mathbb{N}$, consider the two sets -$$A_n(k)=\{\sigma\in\mathfrak{S}_n\vert\,\, \text{$\exists i,\,\, 1\leq i\leq n\,$ such that $\,\sigma(i)-i=k$}\}$$ -and -$$B_n(k)=\{\sigma\in\mathfrak{S}_n\vert\,\, \text{$\exists i,\,\, 1\leq i\leq n\,$ such that $\,\sigma(i+1)-\sigma(i)=k$}\}.$$ - -QUESTION. I believe the following is true, for each $n$ and $k$: - $$\# A_n(k)=\# B_n(k).$$ - Is there a combinatorial proof of this? If it is known, then can you provide references? - -REPLY [13 votes]: A simple variant of the "transformation fondamentale" of Rényi and -of Foata-Schützenberger does the trick. Write a permutation $\sigma$ in -disjoint cycle form, with the smallest element of each cycle first, -and the cycles arranged in decreasing order of the smallest element, -e.g., $(7,8)(5,6,9)(3)(1,4,2)$. Erase the parentheses to get another -permutation $\hat{\sigma}$, written as a word, e.g., $785693142$. This gives -a bijection $\mathfrak{S}_n\to\mathfrak{S}_n$ with the property that -$\sigma(i)-i=k>0$ if and only if -$\hat{\sigma}(j+1)-\hat{\sigma}(j)=k$, where $\hat{\sigma}(j)=i$.<|endoftext|> -TITLE: Is the space of Levi-Civita connections convex -QUESTION [12 upvotes]: More precisely, suppose we a given two metrics $g_0$ and $g_1$ on a manifold $M$. Let $\nabla_0$ and $\nabla_1$ be the corresponding Levi-Civita connections. Set $\nabla_t:=(1-t)\nabla_0+t\nabla_1$. Then $\nabla_t$ is a torsion free connection. Does there exist a continuous family of metrics $g_t$ such that $\nabla_t$ is the Levi-Civita connection of $g_t$? - -REPLY [3 votes]: The answer of Professor Bryant shows, that the space of Levi-Civita connections is not convex. On the other hand, there are natural convex subspaces, at least in dimension 2, namely the space of Levi-Civita connections for metrics of a fixed conformal class. To prove this, let us assume we are in the oriented case, so we are dealing with Riemann surfaces. Let $K\to\Sigma$ denote the canonical bundle, i.e., its sections are the complex-valued complex linear 1-forms on $\Sigma$. A conformal metric on $\Sigma$ in its conformal class defines a unique connection $\nabla$ on $K$ whose $(0,1)$-part is just the exterior derivative, which is the natural holomorphic structure on $K$. If $g=e^{2\lambda}g_0$ their connection 1-forms differ by $$\nabla-\nabla^0=-2\partial\lambda.$$ For $\tilde g=e^{2\tilde\lambda}g_0$ we obtain $$\tilde\nabla-\nabla^0=-2\partial\tilde\lambda,$$ and $$t\tilde\nabla+(1-t)\nabla=\nabla^0-2(t\partial\tilde\lambda+(1-t)\partial\lambda)$$ -which is the Levi-Civita connection for $e^{2(t\tilde\lambda+(1-t)\lambda)}g_0.$<|endoftext|> -TITLE: Morava K-theories for dummies? -QUESTION [29 upvotes]: Professor Urs Würgler passed away one year ago, and his wife engraved his tombstone with "the formula he was the most proud of" : -$B(n)_*(X)\cong P(n)_*(K(n))\square_{\Sigma_n}K(n)_*(X)$ -However she doesn't understand it, and she asked me if I can. I can't. -But I discovered it is very close to the one in Theorem 3.1 p. 121 of -Urs Würgler "Morava K-theories - a survey" -2006, in Lecture Notes in Mathematics Vol. 1474 DOI:10.1007/BFb0084741 -So I tried to understand the wikipedia page on Morava K-theory but it is way above my level (PhD in dynamics and control) -Can anybody try to explain what the formula is about in plain english, or it is definitely too abstract to express in human language ? - -REPLY [39 votes]: This is a result in algebraic topology, where we study the structure of topological spaces $X$. One early way to do this is to calculate a thing called $H_*(X)$, the ordinary homology of $X$. Later people discovered various "extraordinary homologies", which give more precise information. There are very many different extraordinary homologies, including those called $P(n)_*(X)$, $B(n)_*(X)$ and $K(n)_*(X)$. Of these, $P(0)$ (also called $BP$, or Brown-Peterson homology) is the most powerful, but it is often very hard to calculate. At the other end of the scale, $K(0)$ is the weakest and easiest to calculate. In general $K(n)$ is reasonably easy. Roughly speaking, the information in $P(n+1)$ is the information in $P(n)$ minus the information in $K(n)$, so all the $P(n)$'s are often hard to calculate. -From the definitions, the obvious guess would be that $B(n)$ is only a little easier than $P(n)$. However, this turns out to be wrong: $B(n)$ contains exactly the same information as $K(n)$ (and so is much easier than $P(n)$). If you know $K(n)_*(X)$ then Würgler's theorem allows you to calculate $B(n)_*(X)$, and a different but easier theorem lets you go in the opposite direction.<|endoftext|> -TITLE: A topology on $\Bbb R$ where the compact sets are precisely the countable sets -QUESTION [13 upvotes]: QUESTION. -In there a topology on $\Bbb R$ where the compact subsets are precisely the countable subsets? - -I am trying to create a counterexample to a certain claim, and I found that what I need is a topology of this kind. I thought hard about this and did quite a lot of searching, but could not find something relevant. Thank you in advance. -Note: "countable" includes "finite". - -REPLY [39 votes]: There is no such topology. -Suppose there were. Then $\mathbb R$ itself is not compact, so there is an open cover $\mathcal U$ of $\mathbb R$ with no finite subcover. Using recursion, we can construct a non-compact countable subset of $\mathbb R$. -To begin, let $x_0 \in \mathbb R$ and let $U_0$ be any member of $\mathcal U$ containing $x_0$. Since $\{U_0\}$ does not cover $\mathbb R$, we may choose $x_1 \in \mathbb R - U_0$, and then we may also choose some $U_1 \in \mathcal U$ containing $x_1$. Continuing in this way, suppose $x_0, \dots, x_n$ are already chosen, and so are sets $U_0, \dots, U_n$ from $\mathcal U$, with $x_i \in U_i$ for each $i$. Since $\{U_0, \dots, U_n\}$ does not cover $\mathbb R$, we may choose $x_{n+1} \in \mathbb R - \bigcup_{i \leq n}U_i$, and then we may also choose $U_{n+1}$ from $\mathcal U$ containing $x_{n+1}$. -In the end, we get a countable set $\{x_0,x_1,x_2,\dots\}$ and a countable open cover $\{U_0,U_1,U_2,\dots\}$ of this set. But it is clear from our choices of the $x_i$ and the $U_i$ that this open cover has no finite subcover.<|endoftext|> -TITLE: Second Stiefel-Whitney class is a square -QUESTION [7 upvotes]: I'm interested in examples of manifolds which are orientable and such that the second Stiefel-Whitney class is a square. (Of course the second Stiefel-Whitney class should be non-zero.) -An easy example is the real projective space $RP^n$ in the case $n \equiv 1 \ (\operatorname{mod} 4)$. But unfortunately, I don't know any more examples. I know that for $2$- and $3$-manifolds have the nice property $\omega_2 = \omega_1^2$, but it implies that orientable $2$- and $3$-manifolds have always vanishing second Stiefel-Whitney class. -If we know $\omega_2 = x^2$, we can apply $\operatorname{Sq^1}$ to it to get $\operatorname{Sq^1}(\omega_2) = 0$. On the other hand, using Wu's formula and since we assume $\omega_1 = 0$, we get $\operatorname{Sq^1}(\omega_2) = \omega_3$. So this means that I'm especially looking for orientable manifolds with vanishing third Stiefel-Whitney class ... what does this condition mean geometrically? - -REPLY [14 votes]: At least there are quite a lot of such manifolds: up to multiplying by powers of 2, any oriented bordism class contains such a manifold. -Proof: Let $f: X \to BSO$ be the universal map such that $w_2$ pulled back to $X$ is a square of a class $x$. I.e. $X$ is the fiber of a map $w_2 - x^2: BSO \times K(\mathbb{F}_2,1) \to K(\mathbb{F}_2,2)$. Then there is a Thom spectrum $Mf$ and a map of bordism groups $\pi_d(Mf) \to \pi_d(MSO)$ which becomes an isomorphism after inverting 2. If $d \geq 4$ and $M$ is some oriented manifold, then there exists an $n \gg 0$ such that $2^n [M] \in \pi_d(MSO)$ lifts to $\pi_d(Mf)$, i.e. there is an oriented bordism from the disjoint union of $2^n$ copies of $M$ to a manifold $N$ such that $\tau N: N \to BSO$ admits a lift to a map $N \to X$. If $d \geq 5$ (and maybe 4?), there is no obstruction to doing surgery on $N$ in order to make the map $N \to X$ 2-connected, and hence $H^2(X;\mathbb{F}_2) \to H^2(N;\mathbb{F}_2)$ is injective so $w_2$ stays a non-zero square in $N$. $\square$ -If you like, we can partially control the homotopy type of the manifold. For example, if $d \geq 4$ and we write $n = \lfloor d/2\rfloor$ we could use surgery to make the map $N \to X$ $n$-connected. More generally, given an $n$-dimensional finite $CW$-complex $X'$ there is only the "obvious" homotopical obstruction to $X'$ being an $n$-skeleton for a manifold with the requested property: if there exists a map $f: X' \to BSO$ with $f^*(w_2) \in H^2(X';\mathbb{F}_2)$ a non-zero square, then there exists an $n$-connected map from $X'$ to a manifold with the requested property.<|endoftext|> -TITLE: Weyl group elements fixing a set of simple roots -QUESTION [5 upvotes]: Suppose I have a root system $\Phi$ (of a semisimple Lie algebra) with a set of simple roots $\Delta$. I am interested in describing Weyl group elements $w$ preserving a given subset $\Delta'$ in the sense that $w(\Delta')=\Delta'$. -Denote by $\Delta''$ the largest subset of $\Delta$ such that $\Delta'\cap\Delta''=\emptyset$ and all roots in $\Delta'$ are orthogonal to roots in $\Delta''$. Obviously, any product of simple reflections corresponding to roots of $\Delta''$ fixes $\Delta'$. For example, one may hope that this gives me all such $w$ if $\Delta'$ is a connected subset of $\Delta$. -As far as I understand, my question is equivalent to asking which Weyl group elements fix a given Levi subalgebra. - -REPLY [4 votes]: I'm not sure whether there is an efficient way to answer your question (or a written reference), but it's possible to analyze the situation case-by-case. -It's probably best to start with an irreducible root system $\Phi$ (reduced in the Bourbaki sense), which can be asociated with a simple Lie algebra over $\mathbb{C}$ but can just as well be studied abstractly. Fix a simple system $\Delta$ and the corresponding Weyl group $W$ generated by the simple reflections $s_\alpha$ (with $\alpha \in \Delta$). Now choose a (proper) subset $\Delta' \subset \Delta$ with a corresponding root subsystem $\Phi'$ (not necessarily irreducible) and Weyl subgroup $W'$. -It's important to keep in mind that the automorphism group Aut($\Phi$) is the semidirect product of the normal subgroup $W$ (which acts simply transitively on simple systems in $\Phi$) and the possibly trivial group $\Gamma$ of graph automorphisms. An example is given by Nathan Reading for type $A_3$, where the nontrivial graph automorphism has order 2. Similarly, Aut($\Phi'$) is the product of such automorphism groups (and a permutation group on irreducible components if there is more than one), which might or might not be realized within $W$ and might or might not involve graph automorphisms. So a certain amount of case-by-case description could be needed. -[EDIT] With this set-up, suppose $1 \neq w \in W$ stabilizes $\Delta'$. To summarize the possibilities for $w$ based on the above description of Aut($\Phi'$): (1) $w \notin W'$ by the simple transitivity of $W'$ on bases such as $\Delta'$. (2) If there exist simple roots orthogonal to all $\alpha \in \Delta'$ (i.e. $\Delta'' \neq \emptyset$), then $w$ can be any nontrivial element of $W''$, the subgroup of $W$ generated by $\Delta''$. (3) As in the comments, $w$ might involve both reflections $s_\alpha$ for $\alpha \in \Delta'$ and elements outside $W'$ such as the longest element $w_0$. (Note that any reduced expression for $w_0$ involves all simple roots. Also, $w_0$ takes $\Delta$ to $-\Delta$ but might also involve a Dynkin diagram automorphism. In either case, if $w_0 \alpha = -\alpha$ in the case $\Delta' =\{\alpha\}$, then $s_\alpha w_0$ fixes $\Delta'$. Similarly for larger, or disconnected, $\Delta'$ as suggested by Arkandias.) As indicated above, case-by-case work should determine all such possibilities.<|endoftext|> -TITLE: Feynman diagrams and periods of motives -QUESTION [9 upvotes]: A recent article in the online science magazine Quanta, Strange Numbers Found in Particle Collisions, -discusses experimental evidence of a connection between Feynman integrals and periods of motives. Although Quanta usually does a great job with expository articles, in this case so much simplification was necessary for the lay reader that I was left wanting more; a lot more. -Can anyone add more detail to the article, or point me towards another, more in depth exposition? - -REPLY [9 votes]: If I understand correctly, Quantum Field Theory was successful in so far as it was predictive of experimental results. Somehow these badly divergent integrals, when combined correctly product an answer aligning with the result you see in the particle accelerator. -Having gone through a quantum field theory class, I was warned not to worry about being too rigorous as long as I can get the correct result. And the course moved so quickly there wasn't enough time to reflect whether I had seen these tools in my math courses. And I left the course confused and dissatisfied. -There are many types of math and many types of physics. So, I think a great question to ask to is which math and which physics are being related in a given paper. -A few names are mentioned again and again in the paper. Francis Brown, so here is one if his: - -Feynman Amplitudes and Cosmic Galois group - -Certainly there is earlier literature. Dirk Kreimer is mentioned so I pull one out of a hat: - -Renormalization and Mellin transforms - -None of the Hopf algebra structures or Tannakian categories discussed here ever appear in a QFT course. Real QFT homework consists of pages and pages of integerals followed by more integrals and you are never told which matematical theory can formalize this. - -Peskin Schroeder Quantum Field Theory -Zuber Quantum Field Theory - -Even worse, when you try to formalize these computations, you drown in rigor and entirely lose the spirit of the original computation. - -Using the Feynman Rules we can put together diagrams which contribute to the scattering cross sections of various quantum fields theories. In particular $\phi^3$ and $\phi^4$ theory I am seeing a lot. It doesn't matter because they all use the same diagrams. Here's an integral: -$$ \left[ \prod_{k=1}^L \int \frac{d^4 p_k}{p_k^2 (p_k + p)^2}\right]\prod_{m=0}^L \frac{1}{(p_{m+1}-p_m)^2} $$ -and later on in that paper we obtain the value of a slightly different diagram: -$$ \int [\dots] \, d^4p_k = p^2 \left( \frac{\pi}{p}\right)^{2L} \binom{2L}{L} \zeta(2L-1) $$ -for the ladder with $L$ rungs. And this is nothing short of PHENOMENAL. -I believe part of the miracle is, that although we could write down the contributing integrals, we had no idea what the $\int$ evaluated to, even in these simple cases. In class, I thought we had covered this but I guess not. -And I left out the domain of integration - which loosely involves conservation of momentum - but there may be other factors as well. -And there are many facets to these integrals that we are just beginning to find out. - -Summation of an infinite series of ladder diagrams -Multiple zeta values and periods of moduli spaces $\mathfrak{M}_{0,n}$ - -Here are the sources I have looked up. And I recognize that moduli space, it is the moduli space of $n$ marked points on the sphere. Or as I like to think of them as polyhedra (hopefully that is accurate). -The second diagram counts how many Feynman diagrams - which is like a zillion. - -Having explained a tiny bit why there is a question at all, we learn that Feynman diagrams evaluate to multiple zeta values or other periods. As I learn a bit more, maybe I can explain why the first examples of motives. QFT students doing their homework. -Here is theorem 0.1 in Brown - page 1 - -For any Feynman graph $G$ with generic kinematics $q,m$, there is a canonical way to associate to a convergent integral - $$ I_G(q,m) = \int \bigg[ \frac{P(\alpha_E) \Omega_G }{\Psi_G^A \Xi_G(q,m)^B } \bigg] $$ - -an object $\text{mot}_G$ in $\mathcal{H}(S)$, where $S$ is a zariski open [set] in the space of kinematics -a de Rham class $[\omega_G]$ in the generic fiber of $(\text{mot}_G)_{dR} $ -a Betti class $[\sigma_G]$ in a certain (Euclidean) fiber of $(\text{mot}_G)^\hacek_B$ - -such that the [Feynman diagram] integral is the period - $$ \sigma_G\big( c(\omega_G) \big) = I_G(q,m)$$ - -Unless your Deligne or Kontsevich you probably have no idea what that means. I read it was: in the process of doing Feynman integrals we have done a lot of other things: -a motive is a thing that appears in many cohomology theories - -the space you're integrating over is part of Betti singular cohomology (this is the one from topology) -the integrand is part of the deRham cohomology (this is the one from calculus) -By some sort of Poincaré duality there's a way to match the Betti and deRham cohomolog to get a number - -I believe Brown's kinematics is what happens when you evaluate the Feynman integral over momentum space $dp$ rather than position space $dx$. -He also does something rather strange getting these cohomology theories over the fractions $\mathbb{Q}$ rather than $\mathbb{C}$. And the space of kinematic data (e.g. from convervation of momentum) he says define a scheme rather than a variety. Complicating things further, but separating it from what a day-to-day high energy physics graduate student might call a "Feynman diagram" -At least now we see what the motives are and why they are appearing. If you are an expert in motives I direct you to the relevant papers and textbooks. Lastly... a great discussion on Motives (in French) - -Le Groupe Fondamental de $\mathbb{P}^1 \backslash \{ 0, 1, \infty\}$ -From Calculus to Cohomology - -the last one one of my favorite resources.<|endoftext|> -TITLE: heat kernel on closed manifolds - error in Chavel's book? -QUESTION [5 upvotes]: first of all, I am not sure if this question fits here. I asked this question on math.stackexchange also but didn't get an answer so far. -In Isaac Chavel's book Eigenvalues in Riemannian Geometry, Chapter VI, pages 151-154, the heat kernel for compact manifolds is constructed. -I am hoping for someone that is familiar with the consctrucion of the heat kernel in Chavel's book. -On page 154, the final formula for the heat kernel $p$ on the closed Riemannian manifold $M$ reads -$p=H_k+((L_xH_k)\ast F)\hspace{25em}(A)$ -where $H_k$ is a parametrix for the heat operator $L:=\Delta-\partial_t$ on $M$ and $F=\sum_{l=1}^\infty (L_xH_k)^{\ast l}$. -Shouldn't the correct formula be -$p=H_k+ (H_k\ast F)\hspace{26em}(B)$ -? -Formula $(B)$ would also correspond to the ansatz he made in equation $(42)$ on page 153. I read up other books and they all seem to use $(B)$. Additionally, $(A)$ doesn't make much sense for me. However, this still bugs me and I wanted to ask if I make an obvious mistake here, e.g. are $(A)$ and $(B)$ actually the same? -In summary, I want to know if $(A)$ is a typo or intended and would really appreciate any help. -Thanks in advance. - -REPLY [4 votes]: Yes, there is indeed a mistake. Chavels Lemma 2 on page 153 tells you that -$$L(H_k * F) = (LH_k)*F - F,$$ -so if you define $F = \sum_{l=1}^\infty (LH_k)^{*l}$ and $p= H_k + H_k * F$, then -$$ L p = LH_k + (L H_k)*F - L F = LH_k + \sum_{l=2}^\infty (LH_k)^{*l} - \sum_{l=1}^\infty (LH_k)^{*l} = 0,$$ -where all sums converge and differentiation under the sum is allowed by all the estimates on $LH_k$. Also, if you look at formula (42), you see that the definition $p= H_k + H_k * F$ is indeed the one he meant to make.<|endoftext|> -TITLE: Does finiteness of hodge numbers imply properness? -QUESTION [8 upvotes]: If an algebraic variety $X$ is proper, then the (naive) hodge numbers $h^{p,q}:= dim\, H^p(X, \Omega^q)$ are finite. - -To what extent is the converse true? - -E.g., finiteness of $h^{0,0}$ tells you that at least the variety is not (affine and not proper). -I am happy to assume X is smooth. - -REPLY [17 votes]: Let $L$ be a degree $0$, nontorsion line bundle on an elliptic curve $E$, viewed as an affine scheme $X$ over $E$ with $\pi: X \to E$ the structure morphism. I claim all the Hodge numbers of $X$ are finite. -We have a canonical isomorphism $$\omega_X = \pi^* (\omega_E \otimes L^{-1})$$ and short exact sequence $$0 \to \pi^* \omega_E \to \Omega^1_X \to \pi^* L^{-1}\to 0$$ that, together with the non-canonical isomorphism $\omega_E=\mathcal O_E$, reduce us to showing $H^p(X, \mathcal O_X)$ and $H^p(X, \pi^* L^{-1})$ are finite. -Because $\pi$ is affine, we have $$H^p(X, \mathcal O_X) = H^p (E, \pi_* \mathcal O_X)= H^p (E, \sum_{n=0}^\infty L^{-\otimes n}) = \sum_{n=0}^{\infty} H^p(E, L^{- \otimes n}) = H^p(E,\mathcal O_X)$$ because as $L$ is nontorsion of degree $0$, all its powers have vanishing cohomology in every degree. Similarly $$H^p(X, \pi^* L^{-1}) = H^p (E, \pi_* \pi^* L^{-1}) = H^p (E, \sum_{n=0}^{\infty} L^{-\otimes (n+1)}) = \sum_{n=0}^\infty H^p(E, L^{- \otimes (n+1)}) = 0$$ by the same logic. - -In general, let $X$ be a variety with compactification $\overline{X}$ whose boundary is a simple normal crossings union of smooth divisors $D_1,\dots D_k$. If I calculated correctly, a sufficient condition for finiteness of Hodge numbers is that finitely many of the cohomology groups indexed by numbers $p,q$ and tuples $m_1,\dots,m_k$ -$$H^p \left( \bigcap_{ i | m_i>0} D_i, \Omega^q_X \otimes N_1^{\otimes m_1} \otimes N_2^{\otimes m_2} \otimes \dots \otimes N_k^{\otimes m_k}\right)$$ -are nonvanishing. -These should be the cohomology groups of the associated graded components of the pole order filtration of $\Omega^q_X$, viewed as a quasicoherent sheaf on $\overline{X}$. -This condition explains my example when applied to the obvious compactification as a $\mathbb P^1$ bundle. However, it is not necessary, as can be seen by simply choosing another compactification. -This vanishing might be an interesting condition to study. There are some explicit constructions generalizing my example (for $A$ an abelian variety and $X$ any variety, $L_1$ a very general degree $0$ line bundle on $A$ and $L_2$ any line bundle on $X$, a smooth divisor $X \times A$ with normal bundle $L_1 \otimes L_2$ does the job.) -The only necessary conditions I found were a bunch of vanishing characteristic class integrals (I think the integral of any nonconstant monomial in $c_1(D_1),\dots,c_1(D_k)$ times the Chern character of $\Omega^q$ times the Todd class must vanish) which don't really say much about the structure of the $D_i$.<|endoftext|> -TITLE: Weaker version of locally extended residual finiteness -QUESTION [5 upvotes]: Definition (Property A). We will say that group $G$ has property A if for every finitely generated subgroup $H$ of $G$ there exists a group morphism $\rho : G \to F$ to a finite group such that - $\rho(H) \neq \rho(G)$. - -Of course, property A implies residual finiteness but seems to be a stronger property. (Edited later: As pointed out by YCor this is, of course, false! ) -Property A is related to the concept of LERF groups but seems to be a weaker property. Recall that a group $G$ is called LERF if for every finitely generated subgroup $H$ of $G$ and every $g \in G \setminus H$ there exists a finite index normal subgroup $K$ containing $H$ but not containing $g$. - -Question 1. Has property A already appeared in the literature ? Is it actually weaker then locally extended residual finiteness ? - -My second question concerns finitely generated subgroups of matrices groups. - -Question 2. Does there exist a finitely generated subgroup $G$ of $GL_n(\mathbb C)$ (for some $n$) which does not have property $A$ ? - -REPLY [10 votes]: A restatement is that for every f.g. subgroup $H\neq G$ of $G$ there exists a normal finite index subgroup $N$ of $G$ such that $HN\neq G$. So this is also equivalent to the condition that every proper finitely generated subgroup of $G$ is contained in a proper finite index subgroup. This appears in the literature as the "engulfing property". Its hereditary version (groups in which every subgroup of finite index has the engulfing property) is considered here (esp. Section 3.2) as "Property LPF". Many examples are described in this link. -Your claim that this property "of course implies" residual finiteness (RF) is not correct. Indeed, (same link), f.g. virtually nilpotent-by-polycyclic satisfy the even stronger property PF that every proper subgroup is contained in a proper finite index subgroup. This includes non-RF groups. -To answer your first question, the converse also fails. For instance, by Margulis-Soifer + Weisfeiler, the RF group $\mathrm{SL}_k(\mathbf{Z})$ does not have Property LPF for $k\ge 3$ (hence some - possibly all- of its finite index subgroup fails to have the engulfing property). This also answers your second question. -As you observe, the engulfing property (and actually LPF) is implied by LERF, as discussed in the above link. The converse indeed fails, for instance the Baumslag-Solitar group $\mathbf{Z}[1/n]\rtimes_n\mathbf{Z}$ is LPF but not LERF for $n\ge 2$. This answers the last part of your first question. - -Edit: I've been wondering about the difference between the engulfing property EP and property LPF (which means that every finite index subgroup has the engulfing property). I'm unable to produce an example with EP and without LPF (i.e. showing that EP does not pass to finite index subgroups), still I believe it might exist. If I'm not wrong, all papers proving EP for some class C of groups (usually related to geometric topology in some way) actually prove LPF, typically because the concerned classes C are stable under taking finite index subgroups.<|endoftext|> -TITLE: Is there a presentation to the kernel of the prime-to-$p$ fundamental short exact sequence of curves over finite fields? -QUESTION [11 upvotes]: Let $X$ be $\mathbb{P}^1_{\mathbb{F}_q}\smallsetminus \{a_1,...,a_r\}$, where $a_1,...,a_r$ are some $\mathbb{F}_q$-rational points. Let $\bar X :=X_{\bar{\mathbb{F}}_q}$. There is a short exact sequence $$1\rightarrow \pi_1(\bar X)\rightarrow \pi_1(X)\rightarrow \operatorname{Gal}(\mathbb{F}_q)\rightarrow 1.$$ -One can then apply the prime-to-$p$ functor, which takes profinite groups to their maximal prime-to-$p$ quotients. The prime-to-$p$ functor is exact on the right, but not on the left. Therefore one gets: -$$1\rightarrow N\rightarrow \pi_1'(\bar X)\rightarrow \pi_1'(X)\rightarrow \operatorname{Gal}'(\mathbb{F}_q)\rightarrow 1,$$ -where $'$ denotes "prime-to-$p$". -By SGA1, $\pi_1'(\bar X)$ is isomorphic to the prime-to-$p$ quotient of the profinite completion of $\langle \alpha_1,...,\alpha_r|\alpha_1\cdots \alpha_r=1\rangle$. -My question is: does $N$ have any well known and reasonable presentation? Can it easily be expressed in terms of the $\alpha_i$'s? - -REPLY [7 votes]: Well, $N$ is generated by the commutators of elements of $\pi_1'(\overline{X})$ with elements of the maximal pro-$p$ subgroup of $\operatorname{Gal}(\mathbb F_q)$. This is just because any such element of $\operatorname{Gal}(\mathbb F_q)$ lies in the kernel of the map to the maximal prime-to-$p$ quotient of $\pi_1(X)$, and so its commutator with any element of $\pi_1(\overline{X})$ also lies in that kernel. Vice versa, modulo this subgroup, the action of the prime-to-$p$ quotient of $\operatorname{Gal}(\mathbb F_q)$ on $\pi_1(\overline{X})$ is well-defined, and so their semidirect product is clearly $\pi_1(X)'$. However this is not very explicit. -No, I think it is not really possible to give a reasonable presentation. -First let's show why the same problem is hard if we instead take $X$ to be an elliptic curve. Then the prime-to-$p$ fundamental group of $X$ is simply $\prod_{\ell\neq p} \mathbb Z_{\ell}^2$. On this, $\operatorname{Frob}_q$ acts with eigenvalues $\alpha,\beta$ that are the two roots of an polynomial with integer coefficients $x^2 - a x+ q$. $N$ contains the eigenspace of $\alpha$ if and only if the multiplicative order of $\alpha$ mod $\ell$ is not prime to $p$, and similarly for the eigenspace of $\beta$. This is because any more of $\operatorname{Frob}_q$ whose order is a power of $p$ is an element of the maximal pro-$p$ subgroup of $\operatorname{Gal}(\mathbb F_q)$, and its commutators are generated by all its eigenvectors with nontrivial eigenvalue. -This information does not really have a presentation in terms of generators and relations. It depends on the arithmetic of $\alpha,\beta$, and the number field they lie in. -Now let's show how this problem can be reduced to your problem. -Consider the case where $r=4$, $a_1=0,a_2=1,a_3=\infty, a_4=\lambda$ for some $\lambda \in \mathbb F_q$, $\lambda \neq 0,1$. Then the subgroup of $\pi_1'(\overline{X})$ generated by all products of two elements in the set $\{\alpha_1,\alpha_2,\alpha_3,\alpha_4\}$ is the fundamental group of a degree $2$ cover of $x$ - it's the elliptic curve with equation $y^2=x(x-1)(x-\lambda)$ minus the points $x=0,x=1,x=\lambda,x=\infty$. Because the local monodromy generators at those four points are $\alpha_1^2,\alpha_2^2,\alpha_3^2,\alpha_4^2$, the quotient of that subgroup by the normal subgroup generated by $\alpha_1^2,\alpha_2^2,\alpha_3^2,\alpha_4^2$ is the fundamental group of this elliptic curve. -Given an explicit presentation of $N$, we can calculate a presentation for the intersection of $N$ with this index $2$ subgroup, and hence calculate the image of $N$ in this quotient. So the elliptic curve case can't be any harder than this case. -Using more tricks (Belyi's theorem) it is possible to reduce the problem for essentially any curve to the $r=3$ case. So I don't see this having a very easy solution. - -However I think it is possible to calculate the image of $N$ in the maximal pro-$\ell$ quotient of $\pi_1(\overline{X})$. It is the whole pro-$\ell$ quotient if the multiplicative order of $q$ mod $\ell$ is divisible by $p$ and is trivial otherwise. -The reason is that $\operatorname{Frob}_q$ acts by multiplication by $q$ on the abelianization of the maximal pro-$\ell$ quotient (as you can see explicitly by looking at its action on the Galois groups of Kummer extensions). So in particular it acts as multiplication by $q$ mod $\ell$. Hence its commutator with elements of this subgroup is trivial mod $\ell$. so $N$ is a normal subgroup of a pro-$\ell$ group whose projection to the mod $\ell$ abelianization is the whole mod $\ell$ abelianization, and hence $N$ is the whole group. -On the other hand, if the order of $\operatorname{Frob}_q$ is prime to $p$, then every element of $\operatorname{Gal}(\mathbb F_q)$ of order a power of $p$ acts trivially on the mod-$\ell$ abelianization of this pro-$\ell$ group. Hence, by the Hall-Burnside theorem, it acts trivially on the whole pro-$\ell$ group. So there are no nontrivial commutators.<|endoftext|> -TITLE: A conjecture about algebraic values of $(-q;\,-q)_\infty/(q;\,q)_\infty$ -QUESTION [14 upvotes]: Recall that $(a;\,q)_\infty$ is the $q$-Pochhammer symbol: -$$(a;\,q)_\infty=\prod_{n=0}^\infty(1-a \, q^n).\tag1$$ -Its important special case $(q;\,q)_\infty=\prod_{n=1}^\infty(1-q^n)$ is sometimes called the Euler function. It appears in Euler's pentagonal number theorem, and its reciprocal $(q;\,q)_\infty^{-1}$ is the generating function for the partition numbers. It is also related to Jacobi theta functions and Ramanujan theta functions. -Let -$$f(x) = \frac{(-q;\,-q)_\infty}{(q;\,q)_\infty},\quad\text{where}\,\,q=e^{-\pi\sqrt x}.\tag2$$ -In the OEIS entry A080054 there is an empirical observation by Simon Plouffe that apparently $f(1)=\sqrt[8]2$. - -Empirical (Simon Plouffe, Feb. 20, 2011): - $$\sum_{n=0}^{\infty}e^{-\pi n}a(n) =\sqrt[8]2.$$ - -I did some numerical experiments related to this observation, and the outcomes suggest a fascinating stronger conjecture: -Conjecture: For every $p\in\mathbb Q,\,p>0$, the value $f(p)$ -is an algebraic number. -For example, it appears that -$$f(3/5) = \sqrt[8]{2} \cdot \sqrt[4]{9 \sqrt{5}+5 \sqrt{15}-11 \sqrt{3}-19},$$ -and $f(13/7)$ is an algebraic number of degree $96$ whose minimal polynomial is -$$x^{96}-647442063456 \, x^{88}+16702438371168 \, x^{80}-529345497357824 - \, x^{72}+4159684203040512 \, x^{64}-12099397290541056 - \, x^{56}+16408771708010496 \, x^{48}-10607690933600256 - \, x^{40}+2651923007078400 \, x^{32}-367001600 \, x^{24}+257949696 - \, x^{16}-100663296 \, x^8+16777216$$ -and an isolating rational interval is $(37/36,\,6/5)$. -Is this conjecture new? Is it known to be true? If not, can you suggest any ideas how to (dis-)prove it? - -REPLY [2 votes]: This is not meant an answer, instead I wish to list some computational values for $f(x)$. Not sure if these are known. -$$f(1)=\sqrt[8]2\,,$$ -$$f(2)=\sqrt[16]2\,\sqrt[4]{\cos\frac{\pi}8}\,, \qquad -f(4)=\sqrt[16]2\,\sqrt{\cos\frac{\pi}8}\,,$$ -$$f(3)=\sqrt[4]{\sec\frac{\pi}{12}}\,, \qquad -f(1/3)=\sqrt[4]{\csc\frac{\pi}{12}}\,,$$ -$$f(1/2)=\sqrt[4]{1+\sqrt2}\,, \qquad -f(1/4)=\sqrt{1+\sqrt2}\,\,.$$<|endoftext|> -TITLE: Minimizing $x_1^2+x_2^2+x_3^2+x_1x_2+x_2x_3+x_3x_1$ -QUESTION [5 upvotes]: Look at the expression -$$ -f(x_1,x_2,x_3) = x_1^2+x_2^2+x_3^2+x_1x_2+x_2x_3+x_3x_1. -$$ -The numbers $x_1,x_2,x_3$ are non-negative, and I assume that $x_1+x_2+x_3=3$. This is a sum of squares and "cyclic correlations" of consecutive variables. Then you can check that $f$ is minimized for the values $x_1=x_2=x_3=1$. -Now look at -$$ -g(x_1,x_2,x_3,x_4) = x_1^2+x_2^2+x_3^2+x_4^2+x_1x_2+x_2x_3+x_3x_4+x_4x_1, -$$ -under the assumption that $x_1+x_2+x_3+x_4=4$. Again, this is minimized by $x_1 = x_2 = x_3 = x_4=1$. -A similar thing happens if I add "second-order cyclic correlations": Let -$$ -h(x_1,x_2,x_3,x_4) = x_1^2+x_2^2+x_3^2+x_4^2+x_1x_2+x_2x_3+x_3x_4+x_4x_1+x_1x_3+x_2x_4+x_3x_1+x_4x_2, -$$ -again under the assumption that $x_1+x_2+x_3+x_4=4$. This is also minimized for the values $x_1=x_2=x_3=x_4=1$. -Is there a simple explanation for this? Is there a simple argument showing that the same will happen for, say, 12 variables and correlations of order up to 3? - -REPLY [2 votes]: Minimizing $f$ and $h$ subject to the given constraints are quadratic programs of the form -$$\begin{array}{ll} \text{minimize} & \frac 12 \mathrm x^{\top} \mathrm A \,\mathrm x\\ \text{subject to} & 1_n^{\top} \mathrm x = n\\ & \mathrm x \geq\mathrm 0_n\end{array}$$ -where -$$\mathrm A := \mathrm I_n + 1_n 1_n^{\top}$$ -Let us temporarily ignore the non-negativity constraints. The Lagrangian is -$$\mathcal{L} (\mathrm x, \lambda) := \frac 12 \, \mathrm x^{\top} \mathrm A \,\mathrm x - \lambda (1_n^{\top} \mathrm x - n)$$ -Taking the partial derivatives and finding where they vanish, we obtain -$$\mathrm A \mathrm x = \lambda 1_n \qquad \qquad \qquad 1_n^{\top} \mathrm x = n$$ -Hence, we conclude that the minimizers are -$$\bar{\mathrm x} := \left( \frac{n}{1_n^{\top} \mathrm A^{-1} 1_n} \right) \mathrm A^{-1} 1_n \qquad \qquad \qquad \bar{\lambda} := \frac{n}{1_n^{\top} \mathrm A^{-1} 1_n}$$ -Using Sherman-Morrison, the inverse of $\mathrm A$ is -$$\mathrm A^{-1} = (\mathrm I_n + 1_n 1_n^{\top})^{-1} = \mathrm I_n - \left(\frac{1}{1+n}\right) 1_n 1_n^{\top}$$ -Hence, -$$\mathrm A^{-1} 1_n = \frac{1}{1+n} 1_n$$ -and -$$1_n^{\top} \mathrm A^{-1} 1_n = \frac{n}{1+n}$$ -Thus, the minimizer is -$$\bar{\mathrm x} := \left( \frac{n}{1_n^{\top} \mathrm A^{-1} 1_n} \right) \mathrm A^{-1} 1_n = \left( \frac{n}{\frac{n}{1+n}} \right) \frac{1}{1+n} 1_n = \color{blue}{1_n}$$ -which satisfies the non-negativity constraints.<|endoftext|> -TITLE: D-modules over algebraic curves VS differential Galois theory -QUESTION [17 upvotes]: Disclaimer: I know very little about both of the fields in question. -My question is pretty simple: - -What's the relation between differential Galois theory and D-modules - over algebraic curves? - -Differential galois theory can't subsume D-modules obviously since the latter contains also information about behavior at singularities. So, in particular, is differential Galois theory a "natural subset" of D-module theory over curves? If so then in what precise sense? If not why? Still are there any methods/algorithms/ideas from differential Galois theory which are useful for studying D-modules over curves? -Finally, are there any sources discussing this? - -REPLY [8 votes]: Consider what happens if you take a $D$-module on an algebraic curve (with field of fractions $K$) and remove all the information on the singularities. You can achieve this by tensoring over the structure sheaf with $K$, obtaining a module for the ring of differential operators on $K$. This ring is generated over $K$ by differentiation along a single meromorphic vector field (since it's generated by differentiation along all vector fields. So a module over it is just a $K$-vector space with a semilinear action of this differentiation. This will usually be finite-dimensional (I think always for holonomic $D$-modules). -To pass to the Galois theory, we pick a specific vector field and view it as a derivation $D$ on $K$, so we have a finite-dimensional vector space with an action of $D$. -From a finite-dimensional vector space with an action of $D$ one can make a differential field extension using Picard-Vessiot theory. Take a ring generated by independent transcendentals corresponding to basis of this vector space, with the $D$ action given by the $D$ action on the vector space, mod out by a maximal differential ideal, and take the field of fractions. -Any field extension generated by solutions of ODEs arises this way, because we can construct from an order $n$ ODE the vector space generated by a formal solution and its first $n-1$ derivatives and take the corresponding ring, which maps to the field, and the kernel is a differential ideal. - -I think this object, a vector space with an action of $D$ is one of the simplest objects one could study in the theory of ODEs, I guess other than an ODE itslf. To some extent, in differential Galois theory and D-module theory, we would take these objects and study them in different ways - in D-modules, one obviously passes from vector spaces to the richer $\mathcal O_X$-modules, which we can study using commutative algebra, and also allow more than one differential operator to act at the same time, creating more interesting algebra, while in differential Galois theory, we pass to studying the differential field extensions and their automorphism groups, often including, as Avi notes, higher degree differential polynomials. -However there is a specific area where they remain close together. When we study $D$-modules whose underlying $\mathcal O_X$-module is locally free, the space of analytic solutions is a representation of $\pi_1(X)$. On the other hand the space of solutions in a differential field large enough to contain all the solutions is a representation of the differential Galois group. These representations can be identified, with the image of $\pi_1$ inside $GL_n$ a subgroup of the differential Galois group - this is simply because analytic continuation around a loop in $X$ always acts as an automorphism of the field of analytic solutions. In good cases (e.g. regular singularities, by the Riemann-Hilbert correspondence), the Zariski closure of $\pi_1$ (the "monodromy group") is precisely equal to the differential Galois group, but not always - as in the case of $e^x$, which has no monodromy but a nontrivial differential Galois group. -So some aspects of the theory of $D$-modules, specifically their comparison to local systems / sheaves and the Riemann-Hilbert correspondence, are closely related to the representation theory of the relevant differential Galois group.<|endoftext|> -TITLE: Something interesting about the quintic $x^5 + x^4 - 4 x^3 - 3 x^2 + 3 x + 1=0$ and its cousins -QUESTION [6 upvotes]: (Update): -Courtesy of Myerson's and Elkies' answers, we find a second simple cyclic quintic for $\cos\frac{\pi}{p}$ with $p=10m+1$ as, -$$F(z)=z^5 - 10 p z^3 + 20 n^2 p z^2 - 5 p (3 n^4 - 25 n^2 - 625) z + 4 n^2 p(n^4 - 25 n^2 - 125)=0$$ -where $p=n^4 + 25 n^2 + 125$. Its discriminant is -$$D=2^{12}5^{20}(n^2+7)^2n^4(n^4 + 25 n^2 + 125)^4$$ -Finding a simple parametric cyclic quintic was one of the aims of this post. - -(Original post): -We have -$$x^5 + x^4 - 4 x^3 - 3 x^2 + 3 x + 1=0,\quad\quad x =\sum_{k=1}^{2}\,\exp\Bigl(\tfrac{2\pi\, i\, 10^k}{11}\Bigr)$$ -and so on for prime $p=10m+1$. Let $A$ be this class of quintics with $x =\sum_{k=1}^{2m}\,\exp\Bigl(\tfrac{2\pi\, i\, n^k}{p}\Bigr).$ I was trying to find a pattern to the coefficients of this infinite family, perhaps something similar to the Diophantine equation $u^2+27v^2=4N$ for the cubic case. -First, we can depress these (get rid of the $x^{n-1}$ term ) by letting $x=\frac{y-1}{5}$ to get the form, -$$y^5+ay^3+by^2+cy+d=0$$ -Call the depressed form of $A$ as $B$. - -Questions: - - -Is it true that for $B$, there is always an ordering of its roots such that$$\small y_1 y_2 + y_2 y_3 + y_3 y_4 + y_4 y_5 + y_5 y_1 - (y_1 y_3 + y_3 y_5 + y_5 y_2 + y_2 y_4 + y_4 y_1) = 0$$ -Do its coefficients $a,b,c,d$ always obey the Diophantine relations, -$$a^3 + 10 b^2 - 20 a c= 2z_1^2$$ $$5 (a^2 - 4 c)^2 + 32 a b^2 = z_2^2$$ $$(a^3 + 10 b^2 - 20 a c)\,\big(5 (a^2 - 4 c)^2 + 32 a b^2\big) = 2z_1^2z_2^2 = 2(a^2 b + 20 b c - 100 a d)^2$$ for integer $z_i$? - -I tested the first forty such quintics and they answer the two questions in the affirmative. But is it true for all prime $p=10m+1$? - -REPLY [3 votes]: Question 1: Yes; in fact -$$ -\sum_{n\bmod 5} y_n y_{n+1} = \! \sum_{n\bmod 5} y_n y_{n+2} = -p/5 -$$ -for the "depressed" $y_n$, using an order consistent with the -action of the cyclic Galois group. Likewise "for other values of 5". -This follows from the formula $|\tau|^2 = p$ for the absolute value of -a nontrivial Gauss sum $\tau$, together with the observation that -the quintic Gauss sums are the values at $m \neq 0$ of -the discrete Fourier transform $\hat y$ of $\vec y$: -$$ -\tau_m = \sum_{n \bmod 5} e^{2\pi i mn/5} y_n -$$ -(and $\tau_0 = 0$ thanks to the "depression"). Now the $y_n$ are real, -so $\bar\tau$ is the Fourier transform of the map $n \mapsto y_{-n}$; -thus the autocorrelation function -$$ -c_j = \sum_{n \bmod 5} y_n y_{n+j} -$$ -is the convolution of $\vec y$ with $n \mapsto y_{-n}$, -whence the Fourier transform $\hat c$ is $\tau \bar \tau = |\tau|^2$. -But that's $0$ at $m=0$, and $p$ otherwise; therefore $c_0 = 4p/5$ -and $c_j = -p/5$ for other $j$. In particular $c_1 = c_2$, QED. -P.S. I was using $x = y - \frac15$, not your $x = \frac{y-1}{5}$, -so most formulas above will have to be scaled by suitable powers of $5$ -to apply to your $y$'s.<|endoftext|> -TITLE: What is the use of Grothendieck universes in category theory? -QUESTION [11 upvotes]: First of all, I have to mention that I'm truly sorry if this question would seem inappropriate for this site for some people. Still, I think it is better to ask here rather on math.stackexchange. -I recently grasped basic notions of Grothendieck universes for category theory. We most use the axioms of Zermelo and Fraenkel, Axiom of Choice and an additional axiom saying that any set is contained within some universe. -What is more, I understood that the purpose of universes, intuitively, is that we can "do set theory" within the universe. That is, we can construct new sets out of old ones using the axioms of Zermelo and Fraenkel, and still stay within universe. But my understanding may be false, please, correct me, if I'm wrong. Many of the sources I consulted are either very technical or very brief, so my understanding of the matter is somewhat vague. -Still, as of now, I don't think I understand what Grothendieck universes and the axiom (the one about every set being in a universe) brings to the table of category theory. -Classically, people wanted to talk, for example, about all categories of sets with (or without) structure (e.g. $\mathrm{Grp}, \ \mathrm{Ring}, \ \mathrm{R-Mod}$ or even $\mathrm{Set}$). So, if we use classes, the collections of objects in the aformentioned categories would be proper classes. -Now, if we use universes, we fix a Grothendieck universe $\mathbb{U}$ and assume that collections of objects and morphisms in any category are actually sets. Then we call the members of $\mathbb{U} \ \ \mathbb{U}$-sets, the subsets of $\mathbb{U} \ \ \mathbb{U}$-classes, and we call a $\mathbb{U}$-class proper if it is not a member of of $\mathbb{U}$. -After that, we call a category $\mathrm{C}$ a $\mathbb{U}$-small if the set of objects and the set of morphisms of $\mathrm{C}$ are $\mathbb{U}$-sets. And we call $\mathrm{C}$ a locally small if the set of objects and the set of morphisms of $\mathrm{C}$ are $\mathbb{U}$-classes, and $\forall X,Y \in \mathrm{C}: \ \mathrm{Hom_C} (X,Y)$ is a $\mathbb{U}$-set. -But I have to wonder what is the motivation? I realise that in the end it is supposed to ease the foundations of category theory on set theory by erasing the need on distinguishing between sets, classes, conglomerates and whatever the "higher" notions are called (sorry, I haven't heard of them). But how is this supposed to make up for the fact that we don't study "the whole" category (such as the category of all sets, for example) anymore? And what is the motivation behind the notion of Grothendieck universe in respect of category theory? So, we can "do set theory" in a universe, and what it does for us as far as category theory goes? -To sum up: my question is what exactly the use of Grothendieck universes gives us so we don't need to study categories with proper classes of objects and morphisms, or proper conglomerates of objects and morphisms, or other collections of objects and morphisms that are not sets? If we assume that any category has only a set of objects and morphisms, but this time use the help of Grothendieck universes, why this works? Why can we forsake collections of elements that are too big to be a set in favor of sets and only sets if we fix a universe $\mathbb{U}$ and study whether sets of objects and morphisms of a category belong to $\mathbb{U}$? What is the "magic" of a Grothendieck universe? -P.S. I consulted several sources before asking this question, they were helpful to me in other aspects, but in the end they didn't resolve my confusing on this specific matter. In particular, I consulted Borceux'. Mac Lane's and Kashiwara's (coathored with P.Schapira) books on category theory, the articles by D.Murfet and M.Shulman. I know there is supposed to be an extensive exposition in SGA 4, but, unfortunately, I can't read French. - -REPLY [5 votes]: Universes provide a convenient framework for stacking many levels of largeness. But a class in NBG cannot contain any other class. The most basic application which shows that universes are more convenient is that the category of functors $[\mathcal{C},\mathcal{D}]$ between two $U$-categories $\mathcal{C},\mathcal{D}$ is again a $V$-category if $U \in V$ are two nested universes. You cannot even write down this construction in NBG (unless you use inaccessible cardinals, which then is more or less equivalent to the universe approach).<|endoftext|> -TITLE: Which triangulated categories are subcategories of compact objects "somewhere"? -QUESTION [5 upvotes]: Let $T$ be a small triangulated category. Under which conditions there exists a triangulated category $B$ closed with respect to (small) coproducts such that $T$ fully embedds into the subcategory of compact objects of $B$ (I don't need this embedding to be an equivalence; yet if some $B$ of this sort exists then one can always "shrink" it so that its subcategory of compact objects will become the Karoubi envelope of $T$)? -I would conjecture that such a category $B$ exists for any $T$; yet does there exist any way to prove this? What is the largest class of small triangulated categories for which this statement is known? -I suspect that the proof should be "known" whenever some "enhancement" exists for $T$; yet any comments or references would be very welcome! - -REPLY [5 votes]: I do not know of an answer for a general triangulated category (non-topological triangulated categories are very unusual), but as soon as you ask for some more structure the thesis follows very quickly. -Let us suppose that $T$ is the homotopy category of some stable ∞-category $C$ (that is, that $T$ is a topological triangulated category in the sense of Schwede (arXiv:1201.0899) ). Then the idempotent completion (sometimes called the Karoubi envelope) of $C$ is the category of compact object for the stable ∞-category $Ind(C)$ by lemma 5.4.2.4 in Lurie's Higher Topos Theory, so the idempotent completion of $T$ is the category of compact objects of the homotopy category of $Ind(C)$. -The fact that $Ind(C)$ is stable (and so its homotopy category is triangulated) is proposition 1.1.3.6 in Lurie's Higher Algebra.<|endoftext|> -TITLE: Meta-undecidability -QUESTION [11 upvotes]: Could there be an undecidable statement $S$ in ${\sf ZFC}$ of which one will never be able to prove its undecidability for principal reasons (ie we will never know that $S$ is undecidable)? -If this is a silly question, I apologise; feel free to vote to close. I'll remove it quickly. - -REPLY [6 votes]: Here is a general statement: - -If $X$ is the set of $\Pi_1$ statements of arithmetic (i.e., statements $P$ of the form $\forall n\in\mathbb{N}\,(Q(n))$ where $Q$ is arithmetic with bounded quantifiers), there exists no Turing machine $U$ which, given an element $P$ of $X$, (1) always halts, (2) returns "yes" if $P$ is a theorem of Peano arithmetic, and (3) returns "no" if $\neg P$ is a theorem of Peano arithmetic (or equivalently, if $\neg P$ is true, because a $\Sigma_1$ arithemetical statement like $\exists n\in\mathbb{N}\,(\neg Q(n))$, if true, is trivially provable). - -(In other words, there is no program $U$ to decide which $\Pi_1$ statements of arithmetic are theorems and which are false, even if the program is allowed to answer whatever it wants — but it must still terminate — for a statement which is true but unprovable in Peano arithmetic.) -To see why this is, consider two Turing machines $T_1$ and $T_2$ which enumerate sets $S_1$ and $S_2$ which are disjoint (and provably so in Peano arithmetic) and recursively inseparable. Then for an integer $n$, the statement "$T_2$ never generates $n$" is $\Pi_1$. Now if we had a machine $U$ as above, we could run it on this statement, if it answers "no" then by (1) and (2) we know $n\not S_1$ (because if $T_1$ generates $n$ then Peano proves that $T_2$ does not), and if it answers "yes" then by (1) and (3) we know $n\not S_2$; so we are able to recursively separate $S_1$ and $S_2$, a contradiction. -Now what does this have to do with your question? We can construct a machine which, given a $\Pi_1$ arithmetical statement $P$, searches in parallel for: (A) a counterexample to $P$, (B) a proof of $P$ in ZFC, (C) a proof of "$P$ does not follow from Peano arithmetic" in ZFC and (D) a proof of "$P$ is consistent with Peano arithmetic" in ZFC"; if it finds (A) first, it stops and answers "no", if it finds (B) first, it stops and answers "yes", if it finds (C) first, it stops and answers "no", and if it finds (D) first, it stops and answers "yes". Assuming ZFC is arithmetically sound, this machine will answer "yes" for every theorem of Peano arithmetic (because the (B) search will terminate but (A) and (C) cannot), and "no" if $P$ is false (because the (A) search will terminate but (B) and (D) cannot). By the statement above, there must be $P$ for which it does not terminate, i.e., none of the three searches finds anything: this is a $\Pi_1$ arithmetical statement which (A) is true, (B) is not provable in ZFC, (C) for which ZFC cannot prove that it is not provable, nor even that it is not provable in Peano arithmetic and (D) for which ZFC also cannot prove that it is not refutable, nor even that it is not refutable in Peano arithmetic. So this is meta-undecidable in quite a strong sense; and it is clear, by adding further searches to the machine, that one can make this even stronger. -Furthermore, this can all be made completely explicit (by choosing effectively inseparable sets $S_1$ and $S_2$).<|endoftext|> -TITLE: A question about subsets of plane -QUESTION [27 upvotes]: Is there a subset $X$ of plane with two points $x, y$ such that each one of $X \setminus \{x\}$, $X \setminus \{y\}$ is isometric to $X$? I tried hard to construct a counterexample but failed. -Sorry if this is too easy for mathoverflow. - -Edit: The question was initially asked on MathSE on Nov. 22 '16, with a reference to the book by Paul Sally, Fundamentals of Mathematical Analysis, Pure and Applied Undergraduate Texts 20. Amer. Math. Soc. 2013 (Amazon link). It appears as Problem 3.2 p.77, stated precisely as - -Suppose $A$ is a subset of $\mathbb{R}^2$. Show that $A$ can contain at most one point $p$ such that $A$ is isometric to $A\smallsetminus\{p\}$ with the usual metric. - -REPLY [22 votes]: (Initial post November 24, 2016, edited November 27, 2016) This does not exist. -The proof that $X$ doesn't exist is a bit elaborate and makes use of ends of coset spaces. I will prove: - -(a) Let $\Gamma$ be a finitely generated subgroup of the group of isometries of the plane. If $\Gamma$ is not virtually infinite cyclic, then every $\Gamma$-commensurated subset $X$ of the plane is $\Gamma$-transfixed (see terminology below). -(b) For $\Gamma$ virtually infinite cyclic (that is, with an infinite cyclic finite index subgroup), a subset of the plane as prescribed does not exist. - -Terminology (borrowed from here): if a group $\Gamma$ acts on a set $W$ (here, the plane), a subset $X\subset W$ is $\Gamma$-commensurated if $X\Delta\gamma X$ is finite for all $\gamma\in\Gamma$, and $\Gamma$-transfixed if there exists a $\Gamma$-invariant subset $X'$ such that $X'\Delta X$ is finite (clearly this implies that it is $\Gamma$-commensurated and furthermore that the cardinal of $X\Delta\gamma X$ is bounded independently of $\Gamma$). -(The link with ends of coset spaces is that for a finitely generated group $\Gamma$ with subgroup $\Lambda$, the Schreier graph has $\ge 2$ ends if and only if $\Gamma/\Lambda$ has a non-$\Gamma$-transfixed $\Gamma$-commensurated subset. Here restricting to transitive actions would be inconvenient, so I'm using the above language.) - -First (a) and (b) imply the non-existence of $X$ as required. -We start from the classical fact that in a Euclidean space, any isometry between any two subsets has an isometric extension to the whole plane (see appendix C.2 in the Bekka-Harpe-Valette book, or check directly). So the question can be reformulated as follows: - -(*) Does there exist a subgroup $\Gamma$ of the group of isometries of the Euclidean plane, generated by two isometries $\alpha,\beta$, such that, for some subset $X$ of the plane and $x,y\in X$ with $x\neq y$, we have $\alpha(X)=X\smallsetminus\{x\}$ and $\beta(X)=X\smallsetminus\{y\}$? - -Indeed, take (a), (b) for granted, and assume by contradiction that $\Gamma$ exists. Then both generators commensurate $X$ and hence $\Gamma$ commensurates $X$. Since the first of this isometries maps $X$ to $X\smallsetminus\{x\}$, its $n$-th power maps $X$ to the complement of $n$ points in $X$: in particular the cardinal of $X\Delta\gamma X$ is unbounded and $X$ cannot be transfixed. So we get a contradiction, unless $\Gamma$ is virtually infinite cyclic, which is ruled out by (b). -Finally I'll justify that such "paradoxical" decompositions can occur without any reference to free subgroups: - -(b') In the isometry group of the 4-dimensional Euclidean space, there exists $\Gamma$ and $X$ as in (*), with $\alpha,\beta$ generating a free abelian group of rank 2. - - -Let us first prove (a). The proof of the easier (b) is below. -Let $T$ be the group of translations in $\Gamma$. We distinguish two cases (the main one being the third). -1) $T$ is cyclic (either trivial or infinite). -Assuming that $X$ is not transfixed, since $\Gamma$ is finitely generated, by (Prop. 4B2 here), there exists a $\Gamma$-orbit $\Omega$ such that $\Omega\cap X$ is non-transfixed. Fixing a point $\xi$ in $\Omega$, write $\Omega=\Gamma/\Lambda$; that $\Omega\cap X$ is not transfixed implies that the number $e(\Gamma,\Lambda)$ of ends of the coset space $\Gamma/\Lambda$ is $\ge 2$ (by definition $e(\Gamma,\Lambda)>1$ means that $\Gamma/\Lambda$ has a $\Gamma$-commensurated subset that is not transfixed, or equivalently that is neither finite nor cofinite). -Then $\Gamma$ is a polycyclic group. Houghton (1982) proved that for a polycyclic group $\Gamma$, a subgroup $\Lambda$ satisfies $e(\Gamma,\Lambda)>1$ iff $\Lambda$ has Hirsch length 1 less than $\Gamma$ and has normalizer of finite index in $\Gamma$. Hence here, let $P$ be this normalizer and $P_+$ the set of motions in $P$ (of index at most 2 in $P$). Let $Y$ be the set of points in the plane fixed by $\Lambda$. Then $\xi\in Y$ and $Y$ is $P$-invariant. -Then $Y$ is an affine subspace. If $Y=\{\xi\}$, then $\xi$ is fixed by $P_+$, which is thus a group of rotations. Since $\Gamma$ is infinite, so is $P_+$, and so is some finite index subgroup $Q$ of $P_+$ that is normal in $\Gamma$; then $\xi$ is the unique fixed point of $Q$, and hence is fixed by $\Gamma$, a contradiction since $\Omega=\{\xi\}$ was supposed to contained a non-transfixed subset. If $Y$ is the plane, then $P_+$ is trivial, so $\Gamma$ is finite, a contradiction. -If $Y$ is a line, then $P_+$ acts by translations parallel to this line. Since $\Gamma$ is not virtually cyclic, $P_+$ is a finitely generated abelian group of $\mathbf{Q}$-rank $\ge 2$ and thus is 1-ended; it acts freely on the plane and hence transfixes every subset it commensurates, again a contradiction. -(Edit: see below for a more direct approach to Case (1)) -2) $T$ is not cyclic (hence either it contains a free abelian group of rank 2, or a rank 1 infinitely generated abelian group of rank 1). A result of Scott-Sonneborn (Coll. Math., 1963), later rediscovered by Oxley (Math Z, 1972) implies that $T$ is 1-ended, in the sense that every subset of $T$ commensurated by $T$-translations is finite or cofinite [*]. Hence the intersection of $X$ with every $T$-orbit is either finite or cofinite. Actually this intersection is empty or the whole orbit for all but finitely many orbits [**]. Hence we can find a subset $X'$ with $X'\Delta X$ finite, that is $T$-invariant. -Since $T$ is normalized by $\Gamma$, $\Gamma$ permutes the $T$-cosets. Hence $X'\Delta\gamma X'$, for $\gamma\in\Gamma$, is a union of $T$-cosets, and thus either infinite or empty. Since $X'$ is commensurated by the $\Gamma$-action, it is thus empty and $X'$ is $\Gamma$-invariant, so $X$ is $\Gamma$-transfixed. -[*] From Scott-Sonneborn's paper, use Theorem 5, or from Oxley's paper, use Proposition 3.3 in rank $\ge 2$ and Proposition 3.11 in the infinitely generated rank 1 case. -[**] Suppose the contrary. Consider a countable infinite disjoint union $T\times I$ of copies of $T$ and for all $i$ a subset $F_i$ of $T$ (nonempty finite or a proper cofinite subset) such that the disjoint union $\bigcup F_i\times\{i\}$ is commensurated by $T$. Passing to complements, we can suppose that all $F_i$ are nonempty finite. Fix an infinite cyclic subgroup $Z$ of $T$. We can find elements $t_i\in T$ such that the $F_i+t_i\subset T$ are pairwise disjoint, and disjoint of $Z$. Then $\bigcup F_i+t_i$ is commensurated, infinite, disjoint from $Z$ and hence with infinite complement. This contradicts that $T$ is 1-ended. -Edit: here's a more direct approach to Case (1) in the setting of the question: -(1a) If $\Gamma$ has no nontrivial translation, then it's embeddable into $\mathrm{O}(2)$ hence virtually abelian. Then $\alpha^n$ is an element of infinite order and not a translation, hence a rotation of infinite order. Hence its fixed point is unique, and hence is fixed by some finite index subgroup of $\Gamma$, and hence by uniqueness by $\Gamma$. So the action fixes a point, say $O$. By removing entire $\Gamma$-orbits in $X$, we can suppose that $X$ is reduced to the (possibly equal) orbits of $x$ and $y$. One boils down to $\|x\|=\|y\|$ by possibly performing a similarity fixing $0$ to the orbit of $y$ (mapping this orbit to a subset disjoint from the orbit of $x$). Eventually $X$ lies in a cercle centered at $0$, and we can use the abelian case as in (2) unless $\Gamma$ is virtually cyclic. -(1b) the group of translations in $\Gamma$ is infinite cyclic: then it is normal, and hence centralizes a subgroup of index 2; since the centralizer of a nontrivial planar translation is reduced to the group of rotations, we deduce that $\Gamma$ has a subgroup of index 2 consisting of translations so $\Gamma$ is virtually cyclic. - -Let us check (b). Then the "index character" $\tau:\Gamma\to\mathbf{Z}$ mapping $\gamma$ to $\#(X\smallsetminus\gamma X)-\#(\gamma X\smallsetminus X)$ is a homomorphism (this is a general fact). Here $\tau$ maps both $\alpha,\beta$ to 1; it is thus a surjective homomorphism and since $\Gamma$ is virtually cyclic, its kernel is finite, i.e. $\Gamma$ is finite-by-(infinite cyclic). -It is then easy to see that either $\Gamma$ fixes a unique point $\xi$ (which we can suppose to be 0), or fixes no point and preserves a line. -In the first case, $\Gamma$ acts freely on the complement of $\{0\}$. Being generated by two elements $\alpha,\beta$ of infinite order, it acts by rotations and hence is abelian. So we can get a contradiction exactly as in my comment for two translations (we can perform $y\mapsto\alpha^{-1}y\mapsto\beta^{-1}\alpha^{-1}y\mapsto\beta^{-1}y$). -In the second case, it preserves a line; since it is generated by two elements of infinite order, these act as nontrivial translations on this line and we deduce again that $\Gamma$ is abelian (action on this line, and hence on this coordinate is by translation, and action on the orthogonal coordinate is just by possible change of sign). We conclude again with the same trick. - -Proof of (b'): for convenience, we use complex coordinates and work in $\mathbf{C}^2$. Fix an element $\zeta$ in the unit circle of infinite multiplicative order. Define the diagonal $\mathbf{C}$-linear isometries $\alpha(z,z')=(\zeta z,z')$ and $\beta(z,z')=(z,\zeta z')$. Define -$$X=\{(\zeta^n,0):n\ge 0\}\cup\{(0,\zeta^n):n\ge 0\},\quad x=(1,0),\;y=(0,1).$$ -It is immediate that they satisfy the required hypotheses of (*). - -Edit (August 25, 2018) -This (the non-existence of a planar subset as in the OP's question) was originally proved by E.G. Straus, On a problem of W. Sierpinski on the congruence of sets, Fundam. Math. 44 (1957), p. 75-81; available without restriction (at this time) here (Biblioteka Nauki); see also the Math Review link. The proof seems completely by hand and takes 4 pages.<|endoftext|> -TITLE: Integral of the entrywise square of the exponential of a matrix -QUESTION [5 upvotes]: Note: I posted my question on math.stackexchange but got no answer. That is why I am asking it here. -Let $A$ be a $n\times n$ square matrix such that the real part of all eigenvalues are negative. For each $i,j$, let $\exp(At)_{ij}$ be the element $(i,j)$ of the matrix. It is well known that: -$$ \int_0^\infty \exp(At)_{ij}dt = -(A^{-1})_{ij}$$ -Is it possible to simplify a similar expression where each element is squared: -$$ \int_0^\infty (\exp(At)_{ij})^2 dt = ??$$ -I am wondering if it is possible to simplify the above expression. If it helps, I can assume that $A$ is diagonalizable. Note that unless for one-dimensional matrices, $(\exp(At)_{ij})^2\ne\exp(2At)_{ij}$. - -REPLY [2 votes]: Given a Hurwitz matrix $\mathrm A \in \mathbb R^{n \times n}$, let -$$\Phi (t) := \exp(\mathrm A t)$$ -be the state transition matrix, and let its $(i,j)$-th entry be denoted by -$$\varphi_{ij} (t) := \mathrm e_i^{\top} \Phi (t) \, \mathrm e_j$$ -Hence, -$$\begin{array}{rl } \displaystyle\int_0^{\infty} \left( \varphi_{ij} (t) \right)^2 \, \mathrm d t &= \displaystyle\int_0^{\infty} \left( \mathrm e_i^{\top} \Phi (t) \, \mathrm e_j \right)^2 \, \mathrm d t\\\\ &= \displaystyle\int_0^{\infty} \mathrm e_i^{\top} \Phi (t) \, \mathrm e_j \mathrm e_j^{\top} \Phi^{\top} (t) \, \mathrm e_i \, \mathrm d t\\\\ &= \mathrm e_i^{\top} \underbrace{\left( \displaystyle\int_0^{\infty} \Phi (t) \, \mathrm e_j \mathrm e_j^{\top} \Phi^{\top} (t) \, \mathrm d t \right)}_{=: \mathrm W_c} \mathrm e_i = \mathrm e_i^{\top} \mathrm W_c \mathrm e_i\end{array}$$ -where $\mathrm W_c$ is the controllability Gramian of the pair $(\mathrm A, \mathrm e_j)$ and is the solution to the following controllability Lyapunov equation -$$\boxed{\mathrm A \mathrm W_c + \mathrm W_c \mathrm A^{\top} + \mathrm e_j \mathrm e_j^{\top} = \mathrm O_n}$$ -Thus, the $n$ columns of the integral of the entrywise product -$$\int_0^{\infty} \left( \Phi (t) \circ \Phi (t) \right) \mathrm d t$$ -are the diagonals of $\mathrm W_c^{(1)}, \mathrm W_c^{(2)}, \dots, \mathrm W_c^{(n)}$, where $\mathrm W_c^{(1)}, \mathrm W_c^{(2)}, \dots, \mathrm W_c^{(n)}$ are the solutions to the following $n$ controllability Lyapunov equations -$$\begin{array}{cl} \mathrm A \mathrm W_c^{(1)} + \mathrm W_c^{(1)} \mathrm A^{\top} + \mathrm e_1 \mathrm e_1^{\top} &= \mathrm O_n\\ \mathrm A \mathrm W_c^{(2)} + \mathrm W_c^{(2)} \mathrm A^{\top} + \mathrm e_2 \mathrm e_2^{\top} &= \mathrm O_n\\ \vdots & \\ \mathrm A \mathrm W_c^{(n)} + \mathrm W_c^{(n)} \mathrm A^{\top} + \mathrm e_n \mathrm e_n^{\top} &= \mathrm O_n\end{array}$$<|endoftext|> -TITLE: Algebraic independence of $a^a$ and $b^b$ for algebraic irrationals $a,b$ -QUESTION [5 upvotes]: Let $a,b$ be algebraic irrationals. -Are there conjectures or unconditional results about the algebraic -independence of $a^a$ and $b^b$? -Probably Schanuel's conjecture is related, -maybe only $\log{a},\log{b}$ and/or $a,b$ are linearly -independent over the rationals. - -REPLY [12 votes]: They are not always independent. There are two problematic cases: the trivial case $a=b$, and the trinomial case when $c_1x^n + c_2x^m +c_3 =0$ for rational $c_1,c_2,c_3$, $a= x^n$, and $ b= x^m$, so that $ (a^a)^{c_1 /n} (b^b)^{c_2/m} x^{c_3}=1$, giving an algebraic relation between $a^a$ and $b^b$. -In every other case they are independent conditional on Schanuel's conjecture. -First we check that $\log a,\log b, a \log a, b\log b$ are linearly independent over $\mathbb Q$. If $r_1,r_2,r_3,r_4$ are rationals, not all zero, with $r_1 \log a - r_2 \log b + r_3 a \log a - r_4 b \log b = 0$ then $\frac{r_1 + r_3 a}{r_2 + r_4 b}$ is an algebraic number with $a^{ \frac{r_1 + r_3 a}{r_2 + r_4 b} } = b$, violating Gelfond-Schneider unless $\frac{r_1 + r_3 a}{r_2 + r_4 b} $ happens to be a rational $m/n$. In this case, if $m=n$ then $a=b$ and we are in the trivial case. Otherwise, we have $a^m = b^n$ so we can find $x$ with $a=x^n$ and $b=x^m$, but then $$\frac{r_1+r_3 x^n}{r_2+r_4 x^m} = \frac{n}{m}$$ so $$ (m r_3) x^n - (n r_4) x^m + (mr_1 - nr_3)=0$$ and we are in the trinomial case. -Now apply Schanuel's conjecture. The transcendence degree of $$\overline{\mathbb Q}(\log a,\log b, a\log a,b\log b, a,b,a^a,b^b) = \overline{\mathbb Q}(\log a,\log b, a^a,b^b)$$ is at least $4$, so those elements are all independent transcendentals, so in particular $a^a$ and $b^b$ are independent transcendentals.<|endoftext|> -TITLE: How to constructively/combinatorially prove Schur-Weyl duality? -QUESTION [44 upvotes]: How is Schur-Weyl duality (specifically, the fact that the actions of the group ring $\mathbb{K}\left[ S_{n}\right] $ and the monoid ring - $\mathbb{K}\left[ \left( \operatorname*{End}V,\cdot\right) \right] $ on the tensor power $V^{\otimes n}$ are each other's centralizers) for a field $\mathbb{K}$ of characteristic $0$ proven constructively? - -Let me now define the notions and explain what I mean by "constructively" and -what I want to avoid. -Notations -Let $\mathbb{K}$ be a field of characteristic $0$. Fix $n\in\mathbb{N}$, and -let $S_{n}$ be the symmetric group of the set $\left\{ 1,2,\ldots,n\right\} -$. -Let $V$ be a finite-dimensional $\mathbb{K}$-vector space. The symmetric group -$S_{n}$ acts on the $n$-th tensor power $V^{\otimes n}$ by permuting the tensorands: -$\sigma\left( v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right) -=v_{\sigma^{-1}\left( 1\right) }\otimes v_{\sigma^{-1}\left( 2\right) -}\otimes\cdots\otimes v_{\sigma^{-1}\left( n\right) }$ for all $\sigma\in -S_{n}$ and $v_{1},v_{2},\ldots,v_{n}\in V$. -Thus, the group ring $\mathbb{K}\left[ S_{n}\right] $ acts on $V^{\otimes -n}$ as well (by linearity). This makes $V^{\otimes n}$ into a $\mathbb{K} -\left[ S_{n}\right] $-module. -On the other hand, the monoid $\left( \operatorname*{End}V,\cdot\right) $ -acts on $V^{\otimes n}$ as follows: -$M\left( v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right) =Mv_{1}\otimes -Mv_{2}\otimes\cdots\otimes Mv_{n}$ for all $M\in\operatorname*{End}V$ and -$v_{1},v_{2},\ldots,v_{n}\in V$. -Thus, the monoid ring $\mathbb{K}\left[ \left( \operatorname*{End} -V,\cdot\right) \right] $ acts on $V^{\otimes n}$ as well. This makes -$V^{\otimes n}$ into a $\mathbb{K}\left[ \left( \operatorname*{End} -V,\cdot\right) \right] $-module. -(Many authors tend to restrict this module to a $\mathbb{K}\left[ -\operatorname*{GL}V\right] $-module, but this doesn't feel particularly -natural to me. Either way, these things behave pretty much interchangeably.) -Schur-Weyl duality makes the following two claims: - -(a) Each endomorphism of the $\mathbb{K}\left[ S_{n}\right] $-module - $V^{\otimes n}$ is the action of some element of $\mathbb{K}\left[ \left( -\operatorname*{End}V,\cdot\right) \right] $. -(b) Each endomorphism of the $\mathbb{K}\left[ \left( -\operatorname*{End}V,\cdot\right) \right] $-module $V^{\otimes n}$ is the - action of some element of $\mathbb{K}\left[ S_{n}\right] $. - -In general, "some element" is not uniquely determined, as none of the two -module structures is faithful. The $\mathbb{K}\left[ S_{n}\right] $-module -structure is faithful when $n\leq\dim V$; the $\mathbb{K}\left[ \left( -\operatorname*{End}V,\cdot\right) \right] $-module structure is probably -never faithful. The quotients that do act faithfully can be described, but -this is a different story. -How is this usually proven? -For a theorem that appears in every other book on representation theory, -Schur-Weyl duality seems to have a shortage of actually distinct proofs. The -argument (as given, e.g., in §4.18 and §4.19 of Pavel Etingof et al, -Introduction to representation theory, arXiv:0901.0827v5) proceeds, -roughly, as follows: [EDIT: The proof outlined in the following is neither the simplest nor the slickest version of the standard argument. The Etingof-et-al text does it in a much clearer way, by factoring out some of the semisimple-modules arguments into a general lemma. As pointed out by commenters, David Speyer and Mark Wildon (in MO question #90094) have further elementarized the argument, but their versions are still not as lightweight as I'd like them to be (e.g., they still use Schur's lemma, requiring proof of absolute irreducibility).] - -First prove part (a) using fairly elementary methods. (Outline: Let $f$ -be an endomorphism of the $\mathbb{K}\left[ S_{n}\right] $-module -$V^{\otimes n}$. Write $f$ as a $\mathbb{K}$-linear combination of -endomorphisms of the form $f_{1}\otimes f_{2}\otimes\cdots\otimes f_{n}$, -where each $f_{i}$ is in $\operatorname*{End}V$. Since $f$ is $\mathbb{K} -\left[ S_{n}\right] $-equivariant, we can symmetrize it, so that $f$ also -becomes a $\mathbb{K}$-linear combination of endomorphisms of the form -$\dfrac{1}{n!}\sum_{\sigma\in S_{n}}f_{\sigma\left( 1\right) }\otimes -f_{\sigma\left( 2\right) }\otimes\cdots\otimes f_{\sigma\left( n\right) } -$, where each $f_{i}$ is in $\operatorname*{End}V$. It remains to show that -each endomorphism of the latter form is the action of some element of -$\mathbb{K}\left[ \left( \operatorname*{End}V,\cdot\right) \right] $. This -is done using some polarization identity, e.g., $\sum_{\sigma\in S_{n} -}f_{\sigma\left( 1\right) }\otimes f_{\sigma\left( 2\right) }\otimes -\cdots\otimes f_{\sigma\left( n\right) }=\sum_{I\subseteq\left\{ -1,2,\ldots,n\right\} }\left( -1\right) ^{n-\left\vert I\right\vert }\left( -\sum_{i\in I}f_{i}\right) ^{\otimes n}$.) -Let $B$ be the $\mathbb{K}$-algebra $\operatorname*{End} -\nolimits_{\mathbb{K}\left[ S_{n}\right] }\left( V^{\otimes n}\right) $. -Then, $B$ is a quotient of $\mathbb{K}\left[ \left( \operatorname*{End} -V,\cdot\right) \right] $ because of part (a). Thus, $\operatorname*{End} -\nolimits_{\mathbb{K}\left[ \left( \operatorname*{End}V,\cdot\right) -\right] }\left( V^{\otimes n}\right) =\operatorname*{End}\nolimits_{B} -\left( V^{\otimes n}\right) $. -Recall that $\mathbb{K}\left[ S_{n}\right] $ is a semisimple algebra (by -Maschke's theorem), and thus $V^{\otimes n}$ decomposes as $V^{\otimes -n}=\bigoplus_{\lambda\in\Lambda}V_{\lambda}\otimes L_{\lambda}$ for some -finite set $\Lambda$, some nonzero $\mathbb{K}$-vector spaces $V_{\lambda}$ -and some pairwise non-isomorphic simple $\mathbb{K}\left[ S_{n}\right] -$-modules $L_{\lambda}$. Conclude that $\operatorname*{End} -\nolimits_{\mathbb{K}\left[ S_{n}\right] }\left( V^{\otimes n}\right) -\cong\prod_{\lambda\in\Lambda}\operatorname*{End}\left( V_{\lambda}\right) -$. This is a particularly tricky step, since several things are happening at -once here: First of all, we need to know that $\operatorname*{End} -\nolimits_{\mathbb{K}\left[ S_{n}\right] }\left( L_{\lambda}\right) -\cong\mathbb{K}$, which would be a consequence of Schur's lemma if we assumed -that $\mathbb{K}$ is algebraically closed, but as we don't, requires some -knowledge of the representation theory of $S_{n}$ (namely, of the fact that -the simple $\mathbb{K}\left[ S_{n}\right] $-modules are the Specht modules, -and are absolutely simple). But even knowing that, we need to know that the -endomorphism ring of a direct sum of irreducible $\mathbb{K}\left[ -S_{n}\right] $-modules decomposes as a direct product according to the -isotypic components, and on each component is a matrix ring. This is standard -theory of semisimple algebras, but also requires a nontrivial amount of work. -Now, $B=\operatorname*{End}\nolimits_{\mathbb{K}\left[ S_{n}\right] -}\left( V^{\otimes n}\right) \cong\prod_{\lambda\in\Lambda} -\operatorname*{End}\left( V_{\lambda}\right) $, so the $V_{\lambda}$ for -$\lambda\in\Lambda$ are the simple $B$-modules. Hence, the decomposition -$V^{\otimes n}=\bigoplus_{\lambda\in\Lambda}V_{\lambda}\otimes L_{\lambda}$ -can be viewed as a decomposition of the $B$-module $V^{\otimes n}$ into -simples. Hence, the endomorphisms of the $B$-module $V^{\otimes n}$ are direct -sums of the form $\bigoplus_{\lambda\in\Lambda}\operatorname*{id} -\nolimits_{V_{\lambda}}\otimes f_{\lambda}$, where each $f_{\lambda}$ lies in -$\operatorname*{End}\left( L_{\lambda}\right) $. (This, again, requires some -basic semisimple module theory.) It is now straightforward to show that all -such isomorphisms are actions of elements of $\mathbb{K}\left[ S_{n}\right] -$ (indeed, the $L_{\lambda}$ are pairwise non-isomorphic simple $\mathbb{K} -\left[ S_{n}\right] $-modules, and thus $\prod_{\lambda\in\Lambda -}\operatorname*{End}\left( L_{\lambda}\right) $ is a quotient of -$\mathbb{K}\left[ S_{n}\right] $). Due to $\operatorname*{End} -\nolimits_{\mathbb{K}\left[ \left( \operatorname*{End}V,\cdot\right) -\right] }\left( V^{\otimes n}\right) =\operatorname*{End}\nolimits_{B} -\left( V^{\otimes n}\right) $, this yields part (b). - -What do I want? -I am fairly happy with the proof of part (a) given above, but the proof of -part (b) is exactly the kind of argument I shun. It is implicit, -non-constructive and relies on half a semester's worth of representation -theory. Probably my biggest problem with it is aesthetical -- I see (b) as -a combinatorial problem (at least a lot of its invariant-theoretical -applications are combinatorial in nature), but the proof is combing this cat -completely against the grain (if not pulling it along its tail). But asking -for a combinatorial or explicit proof is not a well-defined problem, whereas -asking for a constructive one, at least, is clear-cut. -That said, I suspect that a constructive proof can be obtained by some -straightforward manipulations of the above argument. The representation theory -of $S_{n}$ can be done constructively (see, e.g., Adriano Garsia's notes on -Young's seminormal form), and most semisimple-algebra arguments can -probably be emulated by plain linear algebra (albeit losing what little -intuitive meaning they carry). I would much prefer something that avoids this -and either significantly simplifies the representation theory or replaces it -by something completely different. -What has been done? -My hopes for a better proof have a reason: Schur-Weyl duality actually works -in far greater generality than the above proof. Theorem 1 in Steven Doty's -Schur-Weyl duality in positive characteristic (arXiv:math/0610591v3) -claims that both (a) and (b) hold for any infinite field $\mathbb{K}$, -no matter what the characteristic is! The proof in that paper, however, goes -way over my head (it isn't self-contained either, so the 17 pages are not an -upper bound). Another paper that might contain answers is Roger W. Carter and -George Lusztig, On the Modular Representations of the General Linear and -Symmetric Groups, but that one looks even less approachable. -Of course, I would love to see a proof that works for any infinite field -$\mathbb{K}$, or maybe even more generally for any commutative ring -$\mathbb{K}$, assuming that we replace the endomorphisms of the $\mathbb{K} -\left[ \left( \operatorname*{End}V,\cdot\right) \right] $-module -$V^{\otimes n}$ by a more reasonable notion of $\operatorname*{GL} -$-equivariant maps (namely, endomorphisms of $V^{\otimes n}$ that commute with -the action of a "generic $n\times n$-matrix" adjoined freely to the base -ring). But I would be happy enough to see just vanilla Schur-Weyl duality -proven in a neat way. -One step that can be done easily is a proof of part (b) in the case when -$\dim V\geq n$. Namely, in this case, we can argue as follows: Let $\left( -e_{1},e_{2},\ldots,e_{d}\right) $ be the standard basis of $V$; thus, $d=\dim -V\geq n$. Let $F$ be an endomorphism of the $\mathbb{K}\left[ \left( -\operatorname*{End}V,\cdot\right) \right] $-module $V^{\otimes n}$. Let -$\eta=F\left( e_{1}\otimes e_{2}\otimes\cdots\otimes e_{n}\right) $. For -every $n$ vectors $v_{1},v_{2},\ldots,v_{n}\in V$, we can find a linear map -$M\in\operatorname*{End}V$ satisfying $v_{i}=Me_{i}$ for all $i\in\left\{ -1,2,\ldots,n\right\} $, and thus we have -$F\left( v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right) =F\left( -Me_{1}\otimes Me_{2}\otimes\cdots\otimes Me_{n}\right) $ -$=\left( M\otimes M\otimes\cdots\otimes M\right) \underbrace{F\left( -e_{1}\otimes e_{2}\otimes\cdots\otimes e_{n}\right) }_{=\eta}$ (since $F$ is -$\mathbb{K}\left[ \left( \operatorname*{End}V,\cdot\right) \right] $-equivariant) -$=\left( M\otimes M\otimes\cdots\otimes M\right) \eta$. -Thus, the value of $\eta$ uniquely determines the endomorphism $F$. -Furthermore, we can write $\eta$ as a $\mathbb{K}$-linear combination of pure -tensors of the form $e_{i_{1}}\otimes e_{i_{2}}\otimes\cdots\otimes e_{i_{n}}$ -and show that, for each such pure tensor that actually occurs in this linear -combination (with nonzero coefficient), the $n$-tuple $\left( i_{1} -,i_{2},\ldots,i_{n}\right) $ must be a permutation of $\left( 1,2,\ldots -,n\right) $. (To prove this, we assume the contrary; i.e., assume that the -$n$-tuple $\left( i_{1},i_{2},\ldots,i_{n}\right) $ is not a permutation of -$\left( 1,2,\ldots,n\right) $, but the tensor $e_{i_{1}}\otimes e_{i_{2} -}\otimes\cdots\otimes e_{i_{n}}$ does occur in $\eta$. Thus, either one of the -numbers $i_{1},i_{2},\ldots,i_{n}$ is $>n$, or two of these numbers are equal. -In the first case, pick an $M\in\operatorname*{End}V$ that sends the -corresponding $e_{i_{k}}$ to $0$; in the second, pick an $M\in -\operatorname*{End}V$ that multiplies the corresponding $e_{i_{k}}$ by a -generic $\lambda$. Either way, again use the $\mathbb{K}\left[ \left( -\operatorname*{End}V,\cdot\right) \right] $-equivariance of $F$ to obtain -something absurd. Sorry for the lack of details.) The result is that $\eta$ -is a $\mathbb{K}$-linear combination of various permutations of -$e_1 \otimes e_2 \otimes \cdots \otimes e_n$; and therefore, $F$ (being -determined by this $\eta$) is the action of some element of -$\mathbb{K}\left[S_n\right]$. -However, this argument completely breaks down when $\dim V -TITLE: Stabilization of representation of the symmetric group -QUESTION [13 upvotes]: In their seminal work on "representation stability" (https://arxiv.org/abs/1008.1368), Church and Farb deal with a stabilization procedure for representations (up to isomorphism) (over $\mathbf C$) of the symmetric group. It's a stabilization in the sense that it produces a representation of $\mathfrak S_{n+1}$ out of a representation of $\mathfrak S_n$. -The procedure is defined for irreducible representations as follows. -Irreducible representations are in bijective correspondence with Young diagrams and their procedure consists in sending the representation associated to a Young diagram on the irreducible representation associated to the Young diagram with one more box on the first line (the biggest one). -In general, a representation is isomorphic to a sum of irreducibles and its image is the sum of the images of each summand. -My question is: does this stabilization procedure correspond to an intrinsic procedure on $\mathfrak S_n$ representations? Or equivalently, can you describe it only in terms of representations (without knowing the diagrams of the irreducible summands and/or in a functorial way)? - -REPLY [5 votes]: Here is one idea that is implicit in the $\rm FI$ module approach: -If you have a fixed $S_{n_0}$ representation $W$, then the sequence of representations $n \mapsto {\rm Ind}_{S_{n_0} \times S_{n - n_0}}^{S_n} W$, for $n \geq n_0$ is representation stable: by the Pieri rule, the decomposition into irreducibles eventually only differs from $n$ to $n+1$ by adding a box to each top row. -And, if a sequence of $S_n$ representations $n \mapsto V_n$ is representation stable, it can be written as quotient of families of $S_n$ representations that are induced from a finite level. More precisely, there exist finitely many $S_{n_i}$ representations $W_i$ and $S_{m_j}$ representations $U_j$ and and a family of maps $$f(n): \bigoplus_j {\rm Ind}_{S_{n_j} \times S_{n - n_j}}^{S_n} U_j \to \bigoplus_i {\rm Ind}_{S_{n_i} \times S_{n - n_i}}^{S_n} W_i $$ -such that $V_n$ is the cokernel of $f(n)$ for $n\gg 0$. You can see this from the Pieri rule as well. -So this doesn't give an answer to what "adding a box" means intrinsically for passing from $S_n$ to $S_{n+1}$ representations. Instead, it suggests that eventually differing by "adding a box" should be equivalent to being a quotient of representations that are induced from some finite set of $S_{n_i}$ such that the complement of $n_i$ acts trivially. This isn't true in the setup I've given so far, because there should be some restriction on the maps $f(n)$ that makes them uniform in $n$. This is what $\rm FI$ modules formalize: the domain and target of the maps $f(n)$ are exactly "free" or projective $\rm FI$ modules, and the uniformity condition is that the $f(n)$ be a map of $\rm FI$ modules.<|endoftext|> -TITLE: Is Cauchy induction used for proofs other than for AM–GM? -QUESTION [23 upvotes]: The proof by Cauchy induction of the arithmetic/geometric-mean inequality is well known. I am looking for a further theorem whose proof is much neater by this method than otherwise. - -REPLY [2 votes]: Another inequality can be proved by Cauchy induction :$$\prod_{i=1}^n(x_0+x_i)\geqslant\left(x_0+\prod_{i=1}^nx_i^{1/n}\right)^n\quad\text{for all}\quad n=1,2,...\;\text{and}\; x_0,...,x_n\geqslant0.$$ Details can be found here. I don't see a different easy way to do it. While this involves a geometric mean, it doesn't look much like the AM–GM inequality, because there is a product on both sides. Perhaps, though, it could be derived from AM–GM. -Edit:$\quad$As I have later learned, it can indeed be derived from AM–GM. First, write this standard inequality with variables $x_1,...,x_n>0$. Now replace each $x_i\,$ ($i=1,...,n$) respectively by $x_0^2/(x_0+x_i)$. In parallel, replace each $x_i$ respectively by $x_0x_i/(x_0+x_i)$. Add the two consequent inequalities, and the result easily simplifies to the inequality above. (The extension to $x_1,...,x_n\geqslant0$ is trivial.)<|endoftext|> -TITLE: Is there an analogue of CW-complexes built from $K(\mathbb Z, n)$ instead of $S^n$? -QUESTION [6 upvotes]: The question is motivated by Eckmann-Hilton duality and certain flaws of the homotopy category of CW-complexes. Unfortunately, I do not know the formalism of model categories, so excuse me if it is a basic fact concerning them. -First consider a circle $S^1$ with a fixed point. All spaces are supposed to be connected and have a fixed point, and I will suppress it in notations. Then there are its $(n-1)$-suspension $S^n:=\Sigma^{n-1} S^1$ and its $(n-1)$-delooping $K(\mathbb Z, n):=\Omega^{-(n-1)}S^1$ (I believe that one can construct it geometrically). From them we can build $n$-th homotopy groups $\pi_n(X):=[S^n, X]$ and $n$-th (integral) cohomology groups $H^n(X):=[X, K(\mathbb Z, n)]$, where $[A, B]$ means a set of homotopy classes of maps from $A$ to $B$, and give them a natural group structure, see Fomenko, Fuchs Homotopical topology, $\S$1.4. -As suspension $\Sigma$ and looping $\Omega$ are adjoint, one has $H^i(S^n) \simeq \mathbb Z$ for $i=n$ and $0$ otherwise, and $\pi_i(K(\mathbb Z, n)) \simeq \mathbb Z$ for $i=n$ and $0$ otherwise (we ignore $\pi_0$ and $H^0$). Moreover, in the category of CW-complexes $K(\mathbb Z, n)$ is defined by this condition up to homotopical equivalence, but $S^n$ is not -- there is Poincaré homology sphere $\widehat{S^3}$. -A quick idea is that it happens because CW-complexes are constructed from $S^n$, so one should work in the category of spaces constructed from $K(\mathbb Z, n)$, with something like Postnikov towers instead of CW-complexes' skeletons. There are certain conditions on CW-complexes, the main one is that the quotients of skeletons are bouquets of spheres; of course, one may ask that the homotopy fibers of Postnikov tower are products of $K(\mathbb Z, n)$, that is that the homotopy groups are torsion-free, but the resulting category looks very small, without spheres. -So a question is whether there is a nice category that is "Eckmann-Hilton dual" to CW-complexes? A perfect case would be if one can dualize any statement about CW-complexes and get a true statement about objects in this category, and if it contains spheres and $K(\mathbb Z, n)$, so that we have homotopy and cohomology groups. It is wrong for CW-complexes themselves: for example, the pullback of a fibration by a cofibration is always a fibration, but the pushforward of a cofibration by a fibration is not always a cofibration. -Update. According to the article May The Dual Whitehead Theorems from Dan Ramras' comment, another candidate to "Echmann-Hilton dual" category to CW-complexes is simple spaces, that is whose $\pi_1(X)$ acts trivially on $\pi_n(X)$ (in particular, it acts by conjugation on itself, so is abelian). Simple CW-complexes are those for which Postnikov towers are usually defined. - -REPLY [2 votes]: Repeating my comment above as answer: -You may find Peter May's article "The Dual Whitehead Theorems" interesting in this context. It's at math.uchicago.edu/~may/PAPERS/47.pdf<|endoftext|> -TITLE: The philosophy behind local rings -QUESTION [26 upvotes]: This question has been bugging me for a while and I can't seem to make sense of it on a clear conceptual level. -The theory of local rings is given by taking the theory of rings and adding the axioms -\begin{eqnarray} -(0=1) \vdash \bot \\ -x + y = 1 \vdash \exists z : (xz = 1) \vee \exists z : (yz = 1) -\end{eqnarray} -This theory is geometric and thus has a classifying topos $S[\text{Local Rings}]$. With a bit of work one can show that this topos is actually given by the Zariski topos, i.e. given as the topos of Sheaves on the Zariski site $CRing_{fp}^{op}$ of finitely presented rings with coverings given by partitions of unity. -As an algebraic geometer with an acquaintance of the functor of points POV this is already a nice fact, as this basically means that scheme theory is a straight up consequence of the notion of a local ring (schemes can be identified with those sheaves that can be covered by representables). -But this is not where the story ends. Write $S[\text{Rings}]$ for the classifying topos of the theory of rings. A ring $R$ is equivalent to a geometric morphism $\text{Sets} \rightarrow S[\text{Rings}]$ (that's just how classifying toposes are defined after all). We also have a geometric morphism $S[\text{Local Rings}] \rightarrow S[\text{Rings}]$. It turns out that the topos theoretic pullback along those two geometric morphism is just the topos of Sheaves on $\text{Spec}(R)$. (This was mentioned in this mathoverflow post by Peter Arndt) (EDIT: As was pointed out by Simon Henry below, this should be taken with a grain of salt until it is clear in what sense this pullback is meant.) -$\require{AMScd}$ -\begin{CD} - Sh(Spec(R)) @>>> \text{Sets}\\ - @V V V @VV V\\ - S[\text{Local Rings}] @>>> S[\text{Rings}] -\end{CD} -hence the construction of the spectrum of a ring follows directly from looking at the fibers of the geometric morphism $S[\text{Local Rings}] \rightarrow S[\text{Rings}]$. The slogan is "the spectrum of a ring is the universal way of making the ring into a local ring" (the corresponding local ring is given by the structure sheaf, which is a local ring internal to $Sh(Spec(R))$ and given by the geometric morphism $Sh(Spec(R)) \rightarrow S[\text{Local Rings}]$). -Moreover, locally ringed spaces as special cases of locally ringed topoi are considered by some writers as the right notion a "geometric" space - for example Jacob Lurie generalizes basic notions of geometry to higher geometry using locally ringed spaces. There's a natural notion of what it means to be locally isomorphic to a given locally ringed space and the concept of smooth manifold, complex analytic manifold and also scheme derive naturally from that. The category of locally ringed topoi is simply the overcategory (or slice category) of $\text{Topos}$ over $S[\text{Local Rings}]$ - a nice fact is that this automatically gives the right notion of a smooth map between smooth manifolds, holomorphic map between complex analytic manifolds, etc. -Some other remarks from the side of constructivist mathematics are that linear algebra in a constructive setting works best over local rings (as there are different notions of field in constructive mathematics). -So to come to my question: What is going on here? Let me be specific: I can grasp the concept of a group as a syntactic description of what we mean by the symmetries of an object - what is the underlying essence of a local ring? What part of our human intuition do they describe naturally? -A little remark: There is a partial answer to be found in the following: A local ring has a canonical apartness relation $x \# y$, where $x \# y$ iff $x-y$ is invertible. In fact a ring object in the category of sets equipped with apartness relations is the same thing as a local ring. I am however not yet satisfied with this. - -REPLY [8 votes]: As already mentioned in the comments, local rings are supposed to abstract the idea of germs of functions. One can make this precise as follows. -A ring of germs is defined as a homomorphism $$\mathrm{ev}_p : R \to K$$ between commutative rings, interpreted as the evaluation of certain functions at some point $p$, such that the following properties hold: - -$K$ is a field. -$\mathrm{ev}_p$ is surjective -If $f \in R$ has the property that $\mathrm{ev}_p(f)$ is invertible, then $f$ is invertible. - -A morphism of ring of germs is just a commutative diagram. -It is easy to see that the category of rings of germs is equivalent to the category of local rings. Here, a local ring is defined as a commutative ring which has exactly one maximal ideal (namely, the kernel of $\mathrm{ev}_p$). Thus, the equivalence to the first-order definition mentioned by Georg needs the axiom of choice. I believe that we can also give a definition of rings of germs which is directly equivalent to local rings in the first-order definition.<|endoftext|> -TITLE: Reference for symplectic structures on schemes? -QUESTION [13 upvotes]: My original goal was to read the PTVV paper Shifted Symplectic Structures https://arxiv.org/pdf/1111.3209v4.pdf. I was quickly humbled! -Being told the theory ought to generalize symplectic structures on algebraic varieties and schemes I was unable to find a clear reference for these structures. I could get a hold of a paper or talk here and there giving some definition, but nothing futher. - -REPLY [18 votes]: Dear past life Jacob, -Find a specific geometric problem that this stuff solves, which you think is interesting, and then you will find yourself magically learning it. For you, this problem was extending Donaldson-Thomas theory to Calabi-Yau 4-folds. Even in the Calabi-Yau 3-fold case, shifted symplectic structures show Behrend's notion of symmetric obstruction theory comes from a secret $-1$-shifted symplectic structure (this amazed you). -First read Behrend's paper (https://arxiv.org/abs/math/0507523) introducing symmetric obstruction theories, then possibly Joyce's paper on derived critical loci and DT3 invariants, and the Joyce-Borisov/Cao-Leung papers (https://arxiv.org/pdf/1504.00690.pdf /https://arxiv.org/pdf/1407.7659.pdf) on DT4 invariants. At that point you will not be able to sleep until you read PTVV.<|endoftext|> -TITLE: Set of integral curves of a vector field -QUESTION [6 upvotes]: Let $V \colon [0,1]\times \mathbb R^d \to \mathbb R^d$ be a Borel vector field which is globally bounded, $V \in L^\infty$. -I am looking for a reference for the following result (which I suppose it is true and already been proved somewhere). - -The set of integral curves of $V$, i.e. -$\displaystyle \mathcal C :=\big\{ \gamma \in C([0,1];\mathbb R^d): \gamma(s)-\gamma(t) - \int_t^s V(\tau,\gamma(\tau)) d\tau = 0, \quad \forall 0\le t \le s \le 1\big\}$ -is a Borel set in the space of continuous path $C([0,1];\mathbb R^d)$ endowed with the topology inherited from the $\sup$ norm. - -[The integral curves are defined for all times being the vector field bounded] -It is immediate to show that for continuous vector fields the statement holds (the set $\mathcal C$ is closed) but I do not find anything for the general case. - -REPLY [5 votes]: Actually this follows plainly as a consequence of an important classical result in descriptive set theory, namely, the set of all Borel real-valued functions on a metric space is the smallest class containing the continuous functions, and closed under point-wise convergence, that is, the Baire class. (And, of course, nothing changes, if you consider $\mathbb{R}^d$-valued maps instead, or also, maps valued in the unit ball of $\mathbb{R}^d$). -A suitable reference is e.g. A. S. Kechris' Classical Descriptive Set Theory, ch. 11, or this paper by R. W. Hansell available on-line, that contains all ingredients. (Alternatively, you may replace "Borel vector field" with "Baire vector field" in the statement). -For any bounded Borel map $V:[0,1]\times\mathbb{R}^d\to\mathbb{R}^d$ we can consider the operator $F_V$ on the Banach space $E:=C^0([0,1],\mathbb{R}^d), \|\cdot\|_\infty$ defined by $F_V( \gamma)(t)=\int_0^tV(\tau,\gamma(\tau))d\tau$ for any $\gamma\in E$ and $t\in[0,1]$. Two simple facts easily follow by bounded convergence: - -If $V$ is continuous on $[0,1]\times\mathbb{R}^d$, then $F_V$ is continuous on $E$. Indeed, if $\gamma_n\to\gamma$ in $E$ -$$\big\|F_V(\gamma )-F_V(\gamma_n)\big\|_\infty\le\int_0^1\big\|V(\tau,\gamma(\tau))-V(\tau,\gamma_n(\tau))\big\| d\tau=o(1).$$ -If $V_n$ converges point-wise to $V$ on $[0,1]\times\mathbb{R}^d$, then $F_{V_n}$ converges point-wise to $F_V$ on $E$. Indeed, if $V_n\to V$ point-wise, then for any $\gamma\in E$ -$$\big\|F_V(\gamma )-F_{V_n}(\gamma)\big\|_\infty\le\int_0^1\big\|V(\tau,\gamma(\tau))-V_n(\tau,\gamma(\tau))\big\| d\tau=o(1).$$ - -As a consequence, the set of all bounded Borel maps $V:[0,1]\times\mathbb{R}^d\to\mathbb{R}^d$ such that $F_V$ is a Borel map on $E$ contains all bounded continuous functions and it is closed under point-wise convergence. Therefore it is the whole class of bounded Borel maps. -To conclude, let's denote ${\mathbb 1}$ and $\mathrm{ev_0}$ the identity map on $E$, respectively, the evaluation at $0$ (seen as a bounded, linear rank-one operator on $E$). The set -$$\mathcal{C}=\big\{\gamma\in E : \gamma(t)-F_V(\gamma)(t)-\gamma(0)=0,\; \forall\, t\in[0,1] \big\},$$ -which is the zero-set of the Borel map ${\mathbb 1}-F_V-\mathrm{ev_0}$, is therefore a Borel set. -Rmk: From the above argument also follows by induction that, more precisely, if $V$ is in the Baire class $\alpha$ for a countable ordinal $\alpha$, then so is $F_A$ on $E$, so that $\mathcal{C}$ is in the corresponding Borel set hierarchy.<|endoftext|> -TITLE: Torsion of submanifolds -QUESTION [5 upvotes]: Studying curves in the Euclidean three dimensional space, one usually defines the curvature and the torsion of a curve. If I am not missunderstanding the thing, I guess that a curve has zero torision if and only if it is contained in a plane. -The curvature is generalised by the second fundamental form of a submanifold. -What it is the generalisation of the torsion for a sub manifold $N$ of a Riemannian manifold $M$? I would like to preserve the property that the torsion vanishes if and only if the submanifold is contaned in a totally geodesic hypersurface. -My guess, if $N$ is a curve, would be something like $\nabla_vII(v,v)-\nabla_{II(v,v)}v-[v,II(v,v)]$, where $\nabla$ is the Levi-Civita connection, $II$ is the seocond fundamental form, and $v$ is a tangent vector to $N$. This guess is insipired by the other use of the word torsion in Riemanninan geometry, and this is likely to be missleading, in view of the question Relating curvature and torsion of a connection to those of a curve -Probably this material is standard, but the references I found are either about torision of a curve in the Euclidean space (and usually they are very computational), or torsion of a connection. - -REPLY [3 votes]: CLARIFICATION: The connection $\nabla$ used below is the Levi-Civita connection for the Riemannian metric on $M$ and not $N$. -First, assume that $N$ is a curve -and recall how to construct the Frenet frame of a curve. -Let $e_1 \in T_*N$ be a unit tangent vector along the -curve. If $\nabla_{e_1}e_1 \ne 0$, then, since $e_1\cdot\nabla_{e_1}e_1 -= 0$, there isa a unique unit vector $e_2$ and function -$\kappa_1 > 0$ such that $e_2\cdot e_1 = 0$ -and $\nabla_{e_1}e_1 = \kappa_{1} e_2$. -If, in turn, $\nabla_{e_1}e_2 \ne 0$, then, $e_2\cdot \nabla_{e_1}e_2 = -0$ and $e_1\cdot \nabla_{e_1}e_2 = -e_2\cdot\nabla_{e_1}e_1 = --\kappa_{1}$. Therefore, there exists a unique unit vector $e_3$ -and function $\kappa_2 > 0$ such -that $e_3\cdot e_1 = e_3\cdot e_2 = 0$ and -$$ -\nabla_{e_1}e_2 = -\kappa_{1} e_1 + \kappa_{2} e_3. -$$ -If, at each stage, $\nabla_{e_1}e_k \ne 0$, then this leads to a -uniquely defined orthonormal frame of vectors $e_1, \dots, e_n \in -T_*M$ along $N$, known as the Frenet frame and corresponding functions -$\kappa_1, \dots, \kappa_{n-1}$ such that -$$ -\nabla_{e_1}e_k = -\kappa_{k-1}e_{k-1} + \kappa_{k}e_{k+1}. -$$ -If $M$ is flat, then the vanishing of $\kappa_{j}, -\dots, \kappa_{n-1}$ implies that $N$ lies in an affine subspace of -dimension $j$. The function $\kappa_1$ is usually called the curvature -of $N$, and the functions $\kappa_2, \dots, \kappa_{n-1}$ considered as -torsion functions. -This can be generalized to higher dimensional submanifolds -as described by Liviu. Here is a sketch of one way to do it: -If $n_1 = \dim N$, let $e_1, \dots, e_{n_1}$ be an -orthonormal frame of vectors tangent to $N$. Suppose at each point -in $N$, the vectors $e_1, \dots, e_{n_1}, -\nabla_{e_j}e_i$, where $1 \le i, j \le n$, span an $n_2$-dimensional -subspace, where $n < n_2$. Extend the orthonormal frame to a basis $e_1, \dots, -e_{n_2}$ of this larger subspace. The generalization of $\kappa_1$ for a -curve to a higher dimensional submanifold is the second fundamental -form -$H_{ij} = H^\mu_{ij}e_\mu$, where $H^\mu_{ij} = -e_\mu\cdot\nabla_{e_i}e_j$ and $n+1 \le \mu \le n_2$. Now suppose that $e_i, \nabla_{e_j}e_i$, -$1 \le i,j \le n_2$ span an $n_3$-dimensional subspace of -$T_*M$. Again, extend the orthonormal frame to one that spans this -subspace. -Then one can define a torsion tensor $T_{\mu,\nu} = T_{\mu\nu}^\eta -e_\eta$, where $n+1 \le \mu,\nu \le n_2$ and $n_2+1\le \eta \le n_3$. -This process can be continued. If the resulting frame does not span $T_*M$, then -one might call $N$ degenerate. If $M$ is flat, this implies that $N$ -lies in an affine subspace. Otherwise, one gets a higher dimensional -version of a Frenet frame. The frame is not unique, but the -the nested sequence of osculating subspaces obtained in this way are.<|endoftext|> -TITLE: Limit of a sequence of locally presentable categories -QUESTION [6 upvotes]: Let $\dotsc \to \mathcal{C}_2 \xrightarrow{F_1} \mathcal{C}_1 \xrightarrow{F_0}\mathcal{C}_0$ be a sequence of cocontinuous functors between locally presentable categories. Consider the limit $\mathcal{C}=\lim_n \mathcal{C}_n$ inside the $2$-category of categories. It consists of sequences $X_n \in \mathcal{C}_n$ of objects equipped with isomorphisms $F_n(X_{n+1}) \to X_n$. -1) Why is $\mathcal{C}$ again locally presentable? (Of course, it is cocomplete. The only problem is to show that there is a strong generating set of presentable objects.) -2) Can we say anything about the explicit description of equalizers in $\mathcal{C}$? (Of course, they are not computed pointwise since the $F_n$ are not continuous.) Probably it will be some kind of coreflection of the pointwise kernel, but in that case I would like to learn more about that coreflection. I am not only interested in existence results. -I am aware of Bird's thesis on limits in $2$-categories of locally presentable categories, but since this apparently only exists as a poorly scanned copy, it is very hard to find the corresponding results. Actually I would like to prefer a self-contained answer if possible. - -REPLY [2 votes]: Here's a proof that if $\kappa$ is an uncountable regular cardinal, then the 2-category $\mathsf{Pres}_\kappa$ of locally $\kappa$-presentable categories are closed in $\mathsf{Cat}$ under PIE limits. Here the 1-morphisms are cocontinuous functors which preserve the $\kappa$-presentable objects, or equivalently have $\kappa$-accessible right adjoints. But since $\mathsf{Pres}_\kappa$ is a non-full sub-2-category, the limits in $\mathsf{Cat}$ do not necessarily continue to be limits in $\mathsf{Cat}$. But from the description below it will be clear that the inclusion $\mathsf{Pres}_\kappa \to \mathsf{Cat}$ reflects inserters, equifiers, and $\kappa$-small products. -Products -It's easy to check that locally $\kappa$-presentable categories are closed under small products. For colimits can be computed pointwise, and a generating set is given by objects whose components are $\kappa$-presentable, and initial except on a $\kappa$-small set of coordinates. If we were working with general $\kappa$-accessible categories, this would be more complicated -- I'm not sure, we might have to raise the degree of accessibility. This does not require $\kappa$ to be uncountable. But note that if the product is of size $\kappa$ or more, then generally it will not actually be a product in the 2-category $\mathsf{Pres}_\kappa$ because a functor all of whose projections preserve $\kappa$-presentable objects will generally not itself preserve $\kappa$-presentable objects. -Inserters -Now consider the inserter $Ins(F,G)$ between functors $F,G: \mathcal{C} \to \mathcal{D}$. Colimits are computed as in $\mathcal{C}$, and it's easy to see that the objects whose underlying $\mathcal{C}$-object is $\kappa$-presentable are $\kappa$-presentable. So it suffices to show that the colimit closure of these objects is all of $Ins(F,G)$ -- by Theorem 2.5.1 in Makkai-Paré, or by Lemma 3.7 in these notes by Mike Shulman and the usual strong-generator characterization of locally $\kappa$-presentable categories. Consider an object $F(c) \overset{\gamma}{\to} G(c)$ of $Ins(F,G)$. Then $c$ has some presentability rank $\lambda$. If $\lambda \leq \kappa$ we are done, otherwise $\lambda = \mu^+$ for some uncountable $\mu$, and $c$ is a $\mu$-sized colimit of $\kappa$-presentable objects (see Adámek and Rosický Remark 1.30). Let $K: I \to \mathcal{C}$ be such a diagram. There are two cases: either $\mu$ is regular, or $\mu$ is singular. -If $\mu$ is regular. -In this case, we can replace $I$ with a $\mu$-sized poset with $\mu$-small colimits and $K$ with a functor that preserves these colimits and takes values in the $\mu$-presentable objects of $\mathcal{C}$. To see this, first replace $I$ with its free completion under $\mu$-small colimits (extending $K$ by $\mu$-small-cocontinuity); then an easy induction allows one to choose a cofinal poset closed under $\mu$-small colimits. Construct a cofinal chain $(i_\alpha)_{\alpha < \mu}$ in $I$ and a natural family of morphisms $\gamma_\alpha: F K i_\alpha \to G K i_\alpha$ as follows. Enumerate the objects of $I$ as $(x_\alpha)_{\alpha < \mu}$. Let $i_\alpha^0 = \sup(\{i_\beta \mid \beta < \alpha \} \cup \{x_\alpha\})$. Because the $Fi$'s are $\mu$-presentable and $I$ is $\mu$-filtered, we may inductively choose objects $i_\alpha^n$ and morphisms $i_\alpha^n \to i_\alpha^{n+1}$ and $\gamma_\alpha^n : F K i_\alpha^n \to G K i_\alpha^{n+1}$ such that the following diagram commutes for all $\beta < \alpha$: -$\require{AMScd} -\begin{CD} -FK i_\beta @>>> FKi_\alpha^0 @>>> \cdots @>>> FKi_\alpha^n @>>> FKi_\alpha @>>> Fc \\ -@V{\gamma_\beta}VV @V{\gamma_\alpha^0}VV @VVV @V{\gamma_\alpha^n}VV @V{\gamma_\alpha}VV @V{\gamma}VV \\ -GK i_\beta @>>> GKi_\alpha^1 @>>> \cdots @>>> GKi_\alpha^{n+1} @>>> GKi_\alpha @>>> Gc -\end{CD}$ -In the diagram, we have set $i_\alpha = \varinjlim_{n<\omega} i_\alpha^n$ and observed that becuse this colimit is preserved by $K$, $F$, and $G$, we can define $\gamma_\alpha : FKi_\alpha \to GKi_\alpha$ to be the colimit of the $\gamma_\alpha^n$'s, and this is indeed how we define $i_\alpha$ and $\gamma_\alpha$. This is where we need $\kappa$ to be uncountable -- otherwise we cannot take the colimit of this chain when $\mu = \kappa$. -Now it is easy to see that the chain $(i_\alpha)_{\alpha < \mu}$ is cofinal in $I$, and the arrows $(FKi_\alpha \overset{\gamma_\alpha}{\to} GKi_\alpha)_{\alpha < \mu}$ define a functor into $Ins(F,G)$ taking values in the $\mu$-presentable objects whose colimit is $Fc \overset{\gamma}{\to} Gc$. By induction, the $\mu$-presentable objects are in the colimit closure of the $\kappa$-presentable objects (this is where we use that $\mu$ is regular -- otherwise $\mu$-presentability is the same thing as $\lambda$-presentability!). So $Fc \overset{\gamma}{\to} Gc$ is in this colimit closure too. -If $\mu$ is singular -The basic idea is to do what we just did for a cofinal sequence of smaller regular cardinals, and take a colimit. (Interestingly, this argument seems to work for any limit cardinal $\mu$). -Choose a sequence of regular cardinals $(\mu_\alpha)$ satisfying $\kappa \leq \sum_{\beta < \alpha} \mu_\beta < \mu_\alpha < \mu$ with supremum $\mu$. Similarly to the preliminaries before, we can replace $I$ with a $\mu$-sized poset equipped with an exhaustive filtration $I_0 \subset I_1 \subset \dots \subset I$, $\cup_\alpha I_\alpha = I$ satisfying the conditions that $|I_\alpha| = \mu_\alpha$ and $I_\alpha$ is closed under $\mu_\alpha$-small colimits, which are preserved by the inclusion $I_\alpha \to I$. And we may assume that $K$ preserves $\mu$-small colimits and that $Ki$ is $\mu_\alpha$-presentable for $i \in I_\alpha$ . We construct a chain $(i_\alpha)$ with $i_\alpha \in I_\alpha$ and a natural family of morphisms $\gamma_\alpha : Fi_\alpha \to Gi_\alpha$ by taking $i_\alpha^0 = \sup (\cup_{\beta < \alpha} I_\beta)$ and performing an iterative construction as before to choose $i_\alpha$ and $\gamma_\alpha$. It is harmless to assume that $i_\alpha \in I_\alpha$ -- otherwise $i_\alpha$ first appears in $I_{\alpha'}$ for some $\alpha'>\alpha$, and we simply modify our choice of $I_\beta$ for $\alpha \leq \beta < \alpha'$ by adding in $i_\alpha$ and closing under $\mu_\beta$-filtered colimits in $I_{\alpha'}$. As before, we have defined a chain in $Ins(F,G)$ whose colimit is $Fc \overset{\gamma}{\to} Gc$. Moreover for each $\alpha$, the object $Ki_\alpha$ is $\mu_\alpha$-presentable, and $\mu_\alpha$ is a regular cardinal strictly less than $\lambda$, so by the inductive hypothesis, $Fc \overset{\gamma}{\to} Gc$ is in the colimit closure of the $\kappa$-presentable objects as desired. -Equifiers -I think that a similar argument to the inserter case will show that locally $\kappa$-presentable categories are closed under equifiers for uncountable $\kappa$.<|endoftext|> -TITLE: Where does the name "R-matrix" come from? -QUESTION [17 upvotes]: In quantum integrability and related topics a lot of not-so imaginative terminology is used. One may hear people talk about "Q-operators", "R-matrices", "S-matrices", "T-operators", as well as "L-operators". In fact, this has further led to even less poetic names such as "RTT-relations" and "RLL-relations". -My question is if anyone knows what the origin of this terminology is: where and when were they first used? I am particularly interested in this for the name "R-matrix", and why the letter "R" was chosen. - -The physical roots of quantum integrability lie in several different areas, which has led to quite a few examples of terminology that (only) makes sense in one of these contexts. An example is "auxiliary" vs "quantum" (or "physical") space: this is sensible nomenclature for spin chains in the quantum-inverse scattering method, but arbitrary from the vertex-model viewpoint. In any case, having these different possible origins makes it a bit harder to trace back terminology. -After a bit of research my best guess for an explanation of the origins of the above terminology is as follows. I'd be glad for any corrections or comments. -The S-matrix was probably first; it comes from QFT, and the "S" stands for "scattering". In quantum-integrable cases many-particle scattering can be factorized into two-particle scattering processes, and the consistency condition for this factorized scattering is a version of the Yang-Baxter equation. -Perhaps, then, the R-matrix's name (or notation $R$) was chosen to resemble "S-matrix" (or $S$). Note that Yang writes $Y$ for the R-matrix. Baxter usually uses $w$ (or $W$) for the vertex weights, i.e. the entries of the R-matrix: he seems to prefer the star-triangle relation, i.e. the Yang-Baxter equation in components. According to Perk and Au-Yang the notation "R" comes from the Leningrad school. -In can further imagine that Baxter might have denoted his operator by $Q$ to continue this (anti)alphabetic sequence. -Given these names it's not strange that other quantities may get (nick)names that are formed in analogy. This has happened to the operators originating in the classical inverse-scattering method: the Lax operator, sensibly denoted by $L$, is also called the "L-operator" and the monodromy matrix, $T$, the "T-operator". (Maybe the notation $T$ for the latter comes from its relation to the transfer matrix, for which $t$ is a reasonable notation -- although Baxter writes $V$ for the latter.) -Finally the names "RTT-" and "RLL-relations" just represent the form of those equations (expressing the quasitriangularity of the R-matrix for a representation) given the preceding notation. - -REPLY [9 votes]: (See update at the bottom) -I could not resist: in the meanwhile I have searched the literature further. I think the following is an interesting addition to my best guess from the OP, which is why I have chosen to share it, even if it isn't the definitive answer to my question. -As I (almost correctly) mentioned, Perk and Au-Yang suggest the Leningrad group was first to use the name "R-matrix". The first publication of the Leningrad group where the symbol "R" appears in this context seems to be the founding paper of the quantum-inverse scattering method, - -Sklyanin, Takhtadzhyan and Faddeev, Theor Math Phys (1979) 40: 688, - -with the Russian original submitted on April 17, 1979. In this paper it is subsequently referred to simply as "the matrix R". To the best of my knowledge the first time where the name "R-matrix" is used in this context appears to be (see the bottom of p20) - -Takhtadzhan and Faddeev, Russ Math Surv 34 (1979) 11, - -received by the editors on June 1, 1979. The authors were certainly familiar with the term "S-matrix", so it is not strange that the matrix R would quickly have gotten a similar name. - -Note. In fact the references of the above mention two earlier papers by the Leningrad group on this topic: - -Sklyanin and Faddeev, "Quantum-mechanical approach to completely integrable models of field theory", Dokl Akad Nauk SSSR 243 (1978) 1439 = Soviet Physics Dokl 23 (1978) 902, -Faddeev, "Completely integrable quantum models of field theory", preprint Leningr Otd Mat Inst, P-2-79, Leningrad (1979); in: Problems of Quantum Field Theory, Dubna (1979) 249 (Russian). - -These might also mention the R-matrix, which would push the earliest occurrence back a bit futher, but I cannot find those references online. -It would be great if anyone could provide a link to a digital version of either of these papers! - -Interestingly, tracing back the references of the aforementioned paper by Sklyanin, Takhtadzhyan and Faddeev shows that the notation "R" was in fact already used in Baxter's rather earlier work on the eight-vertex model's partition function, - -Baxter, Ann Phys (1972) 70: 193 - -for the operator containing the local Boltzmann weights at the vertices. I can't find earlier sources where "R" is used in this context; in particular it is not used in the references of Baxter's paper. Thus it seems reasonable to assume that Sklyanin et al adopted Baxter’s notation for the R-matrix. -It would still be nice to know if Baxter’s paper is indeed the first time the notation "R" was used for what is now known as the R-matrix, and, if so, whether there was any specific reason why Baxter chose to use that symbol in his paper. (The rather different context, of lattice models in statistical physics, suggests that Baxter did not choose R in analogy with S-matrix.) - -Update (2020.XII.28). At the start of this year I had the opportunity to ask Baxter in person, at the conference Baxter2020 in Canberra in honour of his eightieth birthday. When I told Baxter the above and asked him if he had any reason why he chose the letter R he looked somewhat surprised at the question, shrugged and said with a smile "It was probably the next letter that came to mind!" -I also asked Baxter about the choice for Q for his Q operator, and the answer was similar. -At the conference I did, however, find out the origin of some similarly imaginative terminology for three systems of functional equations by talking with Klümper, Kuniba and Pearce: -The "Y system" was named so in - -Ravanini, Tateo and Valleriani, Int. J. Mod. Phys. A8 (1993) 1707, arXiv:hep-th/9207040 - -in honour of Yang and Yang, who invented the thermodynamic Bethe ansatz that leads to these equations. -The "T system" first appeared in work of Klümper and Pearce, where t was used since it is a linear combination of transfer matrices T. Kuniba then introduced "T system", where - -By T we meant Transfer matrices, but it can either be thought as Toda or Tau. - -The "Q system", finally, - -are named so in [1] after the notation $Q^{(a)}_m$ due to [81, 92], which was adopted to mean "quantum character" [304]. - -Here [1] refers to Kuniba, Nakanishi and Suzuki (1994), [81, 92] to Kirillov and Reshetikhin (1990) and Kirillov (1989), and [304] to Kirillov (2008), see the following for detailed references. The preceding quotes are taken from - -Kuniba, Nakanishi and Suzuki, J Phys A44 (2011) 103001, arXiv:1010.1344<|endoftext|> -TITLE: A relation between a binomial sum and a trigonometric integral -QUESTION [6 upvotes]: May not be a research-level problem for an expert, but non-trivial for a non-expert: why do we have - $$ \sum_{k=0}^n \frac{(-1)^k}{2k+1} \binom{n}{k} = \frac12 \int_0^\pi (\sin x)^{2n+1} dx $$ -and what is the asymptotic / good lower bound for this as $n$ grows? Thanks! - -REPLY [11 votes]: As I hinted in my comment, the $(-1)^k \binom{n}{k}...$ should set off alarm bells. There is for example a basic reciprocity law, that for functions $f, g: \mathbb{N} \to \mathbb{R}$, we have -$$g(n) = \sum_{k=0}^n (-1)^k \binom{n}{k} f(k) \qquad \text{iff} \qquad f(n) = \sum_{k=0}^n (-1)^k \binom{n}{k} g(k)$$ -(see Concrete Mathematics by Graham, Knuth, and Patashnik, Trick 3 on page 192; it's an easy exercise using e.g. exponential generating functions). -Using this, the claimed identity is equivalent to -$$\frac{1}{2n+1} = \sum_{k=0}^n \binom{n}{k} \frac1{2} \int_0^\pi (-1)^k (\sin x)^{2k+1}\; dx$$ -which follows easily since the right side is (by the binomial theorem) -$$\frac1{2} \int_0^\pi \sin(x) (1 - \sin^2 x)^n\; dx = \frac1{2} \int_0^\pi (\cos x)^{2n} \sin x\; dx$$ -followed by a straightforward integration. - -As noted by Ira Gessel, the closed-form evaluation of $\sum_{k=0}^n (-1)^k \binom{n}{k} \frac1{2k+1}$ can be achieved by any number of methods. In the spirit of finite difference calculus (alluded to in my comment under the question), where the $n^{th}$ falling power -$$x^{\underline{n}} = x(x-1)\ldots (x-n+1)$$ -plays the role of $x^n$ in the ordinary continuous calculus, we have $x^{\underline{m+n}} = x^{\underline{m}} (x - m)^{\underline{n}}$, giving the definition $x^{\underline{-m}} = \frac1{(x+1)(x+2)\ldots (x+m)}$, and the difference formula $\Delta x^{\underline{k}} = k x^{\underline{k-1}}$ (the Pascal triangle identity in disguise) holds for all integers $k$. By means of this one quickly computes -$$\Delta^n \frac1{2x + 1} = \frac{(-1)^n n! 2^n}{(2x+1)(2x + 3)\ldots (2(x+n) + 1)}$$ -and then my comment which says $(-1)^n (\Delta^n f)(0) = \sum_{k=0}^n (-1)^k \binom{n}{k} f(k)$ leads to the evaluation -$$\sum_{k=0}^n (-1)^k \binom{n}{k} \frac1{2k+1} = \frac{n! 2^n}{1 \cdot 3 \cdot \ldots \cdot (2n+1)} = \frac{(n!)^2 2^{2n}}{(2n+1)!}$$ -as Ira Gessel and T. Amdeberhan were saying.<|endoftext|> -TITLE: On existence of a certain irreducible character of $SL(5, q)$ -QUESTION [6 upvotes]: Let $q=p^f$ be a prime power such that $q \equiv 1 \pmod 5$. According to the list of irreducible (complex) character degrees of $SL(5, q)$ in Frank Luebeck's homepage (here), $SL(5, q)$ has 20 irreducible characters of degree $1/5(q^2+1)(q^2+q+1)(q+1)^2(q-1)^4$. Using Lusztig's parametrization of irreducible complex characters of groups of Lie type, I want to prove the existence of such an irreducible character for $SL(5, q)$ when $q$ is sufficiently large: Assume that $\chi \in \mathrm{Irr}(SL(5, q))$ is of degree $1/5(q^2+1)(q^2+q+1)(q+1)^2(q-1)^4$. By Lusztig's result, $\chi$ must belong to a Lusztig's rational serie $\mathcal{E}(SL(5, q), s)$, where $s$ is a semisimple element of $PGL(5, q)$. Moreover, the degree of $\chi$ is given by $\chi(1)=\frac{|SL(5, q)|_{p'}}{|C_{PGL(5, q)}(s)|_{p'}}\psi_s(\chi)(1)$, where $\psi_s(\chi)(1)$ is the degree of a unipotent character of $C_{PGL(5, q)}(s)$ (see Theorem 13.23 and Remark 13.24 of Digne and Michel's book). It is known that if $p$ is sufficiently large (I think $p>5$ would be sufficient), then all the $p'$-degree irreducible characters of $SL(5, q)$ are preciesly the so-called semisimple characters. Therefore, $\chi=\chi_s$ is a semisimple character and $\chi(1)=\frac{|SL(5, q)|_{p'}}{|C_{PGL(5, q)}(s)|_{p'}}$. Thus, the existence of such $\chi$ depends on the existence of a semisimple element $s \in PGL(5, q)$ such that $|C_{PGL(5, q)}(s)|_{p'}=5(1+q+q^2+q^3+q^4)$. Question: Is there any semisimple element $s$ of $PGL(5, q)$ satisfying $|C_{PGL(5, q)}(s)|_{p'}=5(1+q+q^2+q^3+q^4)$? Some Thoughts: We deduce from above argument that we must have $|C_{PGL(5, q)}(s)|_{p'}=5(1+q+q^2+q^3+q^4)$. So if $s_1$ is a preimage of $s$ in $GL(5, q)$, the characteristic polynomial of $s_1$ is an irreducible polynomial of degree $5$ in $\mathbb{F}_q[x]$. Hence we have $C_{GL(5, q)}(s_1)\cong GL(1, q^5)$ and $C_{PGL(5, q)}(s)\cong C_{GL(5, q)}(s_1)/Z(GL(5, q))$. But this implies that $|C_{PGL(5, q)}(s)|_{p'}=1+q+q^2+q^3+q^4$ and so $SL(5, q)$ could not contain an irreducible character of degree$1/5(q^2+1)(q^2+q+1)(q+1)^2(q-1)^4$! How can I explain this? I would be grateful if you could hint me how to find a proper semisimple element corresponds to $\chi$. - -REPLY [9 votes]: [The author or this question made me aware of this thread, so I send the answer here.] -The description in the question is almost correct, except when it comes to the centralizer of the semisimple element $s$ in $PGL_5(q)$. For $q \equiv 1 \pmod 5$ the characters of $SL_5(q)$ of degree $1/5(q^2+1)(q^2+q+1)(q+1)^2(q−1)^4$ are in Lusztig series corresponding to certain elements $s$ of order $5$ in the dual group $PGL_5(q)$. Preimages $s_1 \in GL_5(q)$ of such $s$ can be found as elements in the cyclic Coxeter torus (a maximal torus of order $q^5-1$) which have an eigenvalue $x$ of order $5(q-1)$. Its other eigenvalues are $x^q, x^{q^2}, x^{q^3}, x^{q^4}$ (the successive quotients of these eigenvalues are a fixed $5$-th root of unity). -Such $s_1 \in GL_5(q)$ are regular and have a torus of order $q^5-1$ as centralizer. Its image $s \in PGL_5(q)$ has a non-connected centralizer generated by a maximal torus and a $5$-cycle in the Weyl group. -To $s_1$ corresponds one irreducible character of $GL_5(q)$ (which is isomorphic to its dual group) and to $s$ corresponds the restriction of this character to $SL_5(q)$ which splits into $5$ irreducible constituents of same degree, these are the characters in question. Since there are $4$ primitive roots of $5$ we get altogether $4 \cdot 5 = 20$ characters of that degree.<|endoftext|> -TITLE: Mathematical applications of quantum field theory -QUESTION [62 upvotes]: I understand that quantum field theories are interesting as physics; however, there is also a large community of mathematicians who are interested in them. For someone who is not at all interested in physics, what are some compelling mathematical applications of this work? I've search for such things on the internet, but all I find are speculation and philosophy, neither of which interest me very much. I prefer concrete theorems about concrete mathematical objects (eg in topology, algebraic geometry, number theory, etc). The only counterexample to "not finding stuff" I have seen concerns gauge theory and its applications to geometry and topology (especially in dimension 4). Since this is so well-documented, I'd prefer to exclude it from this discussion. - -REPLY [2 votes]: Gromov-Witten theory and DT theory have a lot of these sorts of results. For a sample, you may have a look at: -https://arxiv.org/abs/math/0312059 -https://arxiv.org/pdf/math/0405204.pdf -https://arxiv.org/abs/1404.6698<|endoftext|> -TITLE: Why are there spikes in the number of exotic $n$-spheres for $n \equiv 3 \!\pmod 4$? -QUESTION [11 upvotes]: Looking at Kervaire and Milnor's1 classification of exotic spheres, for each $n$ there are a finite number of distinct (up to diffeomorphism) smooth structures you can impose on the $n$-sphere (OEIS sequence A001676). The first few terms of this sequence, denoted $[\Theta_n]$, are -$$ -\begin{array}{c|cc} -n & 1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&\dotsb\\\hline -[\Theta_n] & 1&1&1&\geq1&1&1&28&2&8&6&992&1&3&2&16256&\dotsb\\ -\end{array} -\;.$$ -There are obvious spikes in the value of $[\Theta_n]$ for $n \equiv 3 \!\pmod 4$. Is there a nice reason why this happens? The details of the linked paper are a bit over my head. -Wikipedia gives a brief explanation: $\Theta_n$ is the group of $h$-cobordism classes of the homotopy $n$-sphere, and there's a correspondence between the elements of this group and the number of smooth structures of the $n$-sphere. Each group $\Theta_n$ contains a cyclic subgroup $bP_{n+1}$ of the "$n$-spheres that bound parallelizable manifolds." And then for some reason the size of $bP_{n+1}$ can be very large for $n \equiv 3 \!\pmod 4$ and somehow relates to Bernoulli numbers. - - -Kervaire and Milnor, Groups of homotopy spheres: I, Ann. Math. (2) 77 (1963), 504-537 - -REPLY [3 votes]: I wish I can give a more detailed explanation here. -The group $\Theta_n$ fits into an exact sequence $0 \rightarrow bP_{n+1} \rightarrow \Theta_n \rightarrow Coker J$, which is essentially Theorem 4.1 in Kervaire & Milnor. $bP_{n+1}$ is the subgroup of homotopy n-spheres(under connected sum) which are bounded by some n+1-manifolds. $CokerJ$ is the cokernel of the stable J-homomorphism in that dimension. The last map will have cokernel $\mathbb{Z}/2$ if $n = 2$ mod $4$ and Kervaire invariant in that dimension is nontrivial. Otherwise it is surjective. $bP_{n+1}$ is trivial when $n+1$ is odd. And is $\mathbb{Z/2}$ or trivial, when $n+1 = 2$ mod $4$, depending on whether we have a trivial Kervaire invariant in this dimension or not. The invariant is defined in the last section of the paper. If you google Kervaire invaraint you can see a lot of stuff about this. -The case you are interested in is when $n+1$ is a multiple of $4$. The formula I mentioned in comment is given in the paper after Corollary 7.6 with a promise of a proof in a non-exist following paper. However a proof can be found in various places. I learned a proof from Kosinski's book Differential Manifold(Theorem IX.8.7 in the book), which is a great book on everything you need for understanding Kervaire & Milnor. The number you see is so big basically because such coefficients appear in the Hirzebruch signature theorem, which you can use to compute signature of a manifold by Pontrjagin classes.<|endoftext|> -TITLE: Which C*-algebras are complemented in their bidual? -QUESTION [6 upvotes]: Every von Neumann algebra is 1-complemented in its bidual, and so is every injective C*-algebra. Also, if $C_0(X)$ is infinite-dimensional and separable then it is not complemented in its bidual, and $\mathcal{K}(\mathcal{H})$ is not complemented in its bidual when $\mathcal{H}$ is infinite-dimensional. -How much more than this is known? Is every infinite-dimensional separable C*-algebra not complemented in its bidual? Is anything known about the nonseparable case? - -REPLY [5 votes]: It seems that only for C*-algebras that are 1-complemented in the second dual we may say something definite (they are AW*-algebras as observed by Hannes). -A separable C*-algebra $A$ that is complemented in $A^{**}$ must be finite-dimensional, as by a result of Pfitzner, it is a Grothendieck space. It is an easy exercise on the Eberlein–Šmulian theorem that a separable Grothendieck space is reflexive. Reflexive C*-algebras are finite-dimensional. -Beyond this case we cannot say much more even in the commutative case. For example, kernels of non-weak*-continuous characters on $\ell_\infty$ are complemented in their biduals (being Banach-space isomorphic to $\ell_\infty$). The question of for which (locally) compact space $K$ the space $C_0(K)$ is isomorphic to a dual space (such spaces are complemented in their second duals) is discussed in the forthcoming book by Dales, Dashiell, Lau and Strauss. Even in this case, the current state of affairs it far from a complete characterisation. Indeed, it is still open whether injective Banach spaces are isomorphic to 1-injectives ones (that is, to commutative AW*-algebras) and a commutative C*-algebra complemented in the second dual is injective.<|endoftext|> -TITLE: Irreducible $S_n$-modules and $S_n$-actions on projective spaces -QUESTION [8 upvotes]: Let $V$ be an $(N+1)$-dimensional vector space with an action of the symmetric group $S_n$, such that $V$ is an irreducible $S_n$ -module. -Let $\{p_1,...,p_h\}\in \mathbb{P}(V)$ be $h\geq N+2$ points such that $S_n$ acts transitively on $\{p_1,...,p_h\}$. -Is it true then that there exist $N+2$ points $p_{i_1},...,p_{i_{N+2}}\in \{p_1,...,p_h\}$ that are in linear general position in $\mathbb{P}(V)$ ? - -REPLY [5 votes]: EDIT: Actually, I made this way too difficult. -Points in projective space are in general linear position if the corresponding lines are the multiples of a basis of $V$. Thus, the question is just if the lines corresponding to $p_1,\dots, p_h$ span the vector space $V$ (since in this case, some subset will correspond to a basis). The span of these lines is invariant and non-zero, and so must be all of $V$ by its irreducibility.<|endoftext|> -TITLE: Definition of $\textrm{GSpin}_{2n}$ and its root datum -QUESTION [5 upvotes]: I'm trying to get my hands on the general spin group $G = \textrm{GSpin}_{2n}$. It have seen it mentioned as a connected, reductive group whose derived group is $\textrm{Spin}_{2n}$, which is the unique semisimple, simply connected group having the same root system as $\textrm{SO}_{2n}$. -The general spin group is for example mentioned in a paper here https://math.okstate.edu/people/asgari/Files/gspin.pdf . -I have read that it is difficult to view $\textrm{Spin}_{2n}$ as a closed subgroup of some general linear group. According to (9.16) in Linear Algebraic Groups and Finite Groups of Lie Type, the smallest $m$ for which $\textrm{Spin}_{2n}$ can be embedded in $\textrm{GL}_{m}$ is $2^{[\frac{2n-1}{2}]}$. -As for $G$, I am not really sure how that should be defined. Since its derived group is $\textrm{Spin}_{2n}$, it should be quotient of $\textrm{Spin}_{2n} \times S$ by a finite normal subgroup, where $S$ is some torus. -So my questions are: - -1 . How should the general spin group be defined? -2 . What is the most straightforward way to compute the roots, coroots, and root datum of the general spin group as well as its derived group? - -For the group $\textrm{SO}_{2n}$, I did (2) by finding a maximal torus and computing the Lie algebra. I imagine there must be a different approach when one is not working with an explicit embedding into $\textrm{GL}_{m}$. - -REPLY [2 votes]: If you'd like to be completely explicit, the Atlas of Lie Groups and Representations software can do this. See www.liegroups.org. Comments are in braces { }. -atlas> {start with simply connected group Spin(8)xGL(1), -of type D4.T1, then mod out by specified element of order 2 -so G=Spin(8)xGL(1)/Z_2 -Z_2 given by the vector [1/2,1/2,1/2]: element of center -of Spin(8)xGL(1)=Z/2 x Z/2 x C^x, i.e. the element (-1,-1,-1)} -atlas> set rd=root_datum(Lie_type ("D4.T1"),[1/2,1/2,1/2]) -Variable rd: RootDatum -atlas> rd -Value: simply connected root datum of Lie type 'D4.T1' -{derived group is simply connected} -atlas> set roots=simple_roots (rd) -{simple roots as columns of a matrix} -Variable roots: mat -{The software chooses a basis so that X^*(T)=Z^5, X_*(T)=Z^5, -with the dot product pairing, and then gives roots and corooots -as integral vectors} -atlas> set coroots=simple_coroots (rd) {simple coroots as columns} -Variable coroots: mat -atlas> roots -Value: {roots are columns, 4 vectors of size 5} -| 2, -1, 0, 0 | -| 3, 0, -1, -1 | -| 1, 0, 0, 0 | -| -2, 0, 0, 2 | -| 0, 0, 0, 0 | - -atlas> coroots -Value: {coroots are columns} -| 1, -2, 1, 1 | -| 0, 1, -2, 0 | -| 0, 0, 2, 0 | -| 0, 0, -1, 1 | -| 0, 0, -1, 0 | - -atlas> ^roots*coroots {compute the pairing of roots and coroots} -Value: -| 2, -1, 0, 0 | -| -1, 2, -1, -1 | -| 0, -1, 2, 0 | -| 0, -1, 0, 2 | - -atlas> ^roots*coroots=Cartan_matrix(rd) -Value: true {this confirms that these are valid roots and -coroots of the root system}<|endoftext|> -TITLE: What is the name of this combinatorial object and place to read about it? -QUESTION [27 upvotes]: The title is admittedly noninformative but I could not figure out how to squeeze into it the description of the object I am interested in. Judge by yourself. -I have an alphabet on $d$ symbols. I want to build a rectangular array with $pd$ rows and $qd$ columns filled with them in such a way that each row contains $p$ copies of each symbol, each column contains $q$ copies of each symbol, and the following number $n$ is minimized. -For each pair $i,j$ of columns, let $n_{ij}$ be the number of rows where these columns contain identical symbols. Then $n$ is the largest of the $n_{ij}$ for all $i\ne j$. -I cannot even figure out what is the smallest possible value of $n$ for given $d,p,q$. -I tried to search for such structures. Not that I did not find anything. On the contrary, I found too many very similar investigations around gadgets called block designs, association schemes, Steiner triples and whatnot. Unfortunately I was unable to find the thing I need - I do not even give it a name since I am sure it already has a well-established name but I cannot figure out what it is. - -REPLY [37 votes]: What you're asking for is a code with two non-usual restrictions. It's a code over alphabet size $d$. You want each code word to have length $pd$ and you want to have $qd$ total code words. -The parameter you call $n$ is equal to $pd - \Delta$, where $\Delta$ is called the distance of your code. You want to minimize $n,$ which is the same as maximizing the distance $\Delta.$ -So far, your set-up is exactly the same as the usual set-up in coding theory, and this problem is extremely well-studied. You add the two additional restrictions that - -If we project onto any given coordinate, then each symbol should appear equally often (this is not really much of a restriction since we could take linear codes when say $d$ is a prime power); and -Each code word needs to use each symbol the same number of times (this is much more of a restriction, I'd say). - -In either event, if you want to get the best possible bounds, consider say the Hoffman bound (an upper bound on codes not needing to satisfy your additional 1 and 2). Then consider a suitable random code, which you need to make sure has property 2 by construction. Then it will almost have property 1, so make sure to save a few code words at the end so that you can fix this. Then hopefully this random code will match the Hoffman bound decently well. -Another thought is to look at say Reed-Solomon codes. -The case $d=2$ is very, very well-studied (binary codes), and I wouldn't be surprised if your problem is known in that setting. -Added: -In fact, it seems this has been studied. Condition 2 is called "balanced codes" (or more generally constant weight codes). They have applications to bar codes and such, apparently. See here.<|endoftext|> -TITLE: Is there a crossing-free planar embedding of the 2-skeleton of the 6-simplex? -QUESTION [5 upvotes]: A pseudo-disk arrangement is a collection of planar bodies whose boundaries are Jordan curves that pairwise intersect at most twice. -I would like to know if given seven points in the plane whether it is possible to find a pseudo-disk arrangement with $\binom 73$ disks, such that for each triple of the seven points there is one disk containing exactly them from the set of seven points. -I'm also interested in what kind of theorems there are to conclude the non-existence of such embeddings. -I've just realized that I don't even know the answer if we replace seven with five. -For five points we can find a pseudo-ball arrangement is $3d$; take a tetrahedron with a point inside, and slightly perturb the ten triangles they span. -Can we do more points in $3d$? -In $4d$ any number of points are OK if we aim for triples (just like pairs would be OK in $3d$). -Please note that I've updated the question a couple of times to make the problem more clear, some of the comments and Jeff's answer are for older versions. - -REPLY [3 votes]: I think that there is no pseudo-disk arrangement even with $\binom{5}{3}$ pseudo-disks for the 5 point case; I sketch a proof below. -Suppose for contradiction that there is such an arrangement. -Let $a_1,\dots, a_5$ be the points. For each closed Jordan curve $c_{ijk}$ that surrounds $a_i,a_j$ and $a_k$ we define a Jordan arc $d_{lm}$ that connects the remaining two points $a_l$ and $a_m$ in the unbounded region of $c_{ijk}$. We should define the $d$-curves in a way that there are no touchings between them or with the $c$-curves, and also we should make sure that the number of curve incidences in the whole picture is finite. -I claim that a pair of $d$-curves that don't share an endpoint must intersect an even number of times Wlog. consider the curves $d_{12}$ and $d_{45}$. The closed Jordan curves $c_{123}$ and $c_{345}$ together define four regions in the plane, one unbounded, one containing $a_1$ and $a_2$, one containing $a_3$, and one containing $a_4$ and $a_5$. Let $I$ be the union of the three bounded regions. Notice that the boundary of $I$ is made up of two curves, one is an arc of $c_{123}$, while the other is an arc of $c_{345}$. -Since any intersection point $q \in d_{12} \cap d_{45}$ must lie outside $I$, both $d_{12}$ and $d_{45}$ has at least one arc outside $I$. Consider the space $\mathbb{R}^2/I$. In this space, the curves $d_{12}$ and $d_{45}$ become collections of closed Jordan curves that contain the point $p$ that is the picture of $I$ in the contraction. Since one boundary arc of $I$ is only intersected by $d_{12}$ and the other only by $d_{45}$, the curve endigns around $p$ are `separable', i.e., if we label each curve ending by $a$ for the curve $d_{12}$ and by $b$ for the curve $d_{45}$, then the cyclic order of the labels around $p$ is $a^{2k}b^{2l}$ for some positive integers $k,l$. -Let $g_1,\dots,g_k$ be the closed Jordan curves of $d_{12}/I$ and let $g'_1,\dots,g'_l$ be the closed Jordan curves of $d_{45}/I$. In order to show that $|d_{12} \cap d_{45}|$ is even, it is sufficient to show that $|g_i \cap g'_j|$ is even for any pair $i,j$. We have seen that $g_i$ and $g'_j$ are two closed Jordan curves that touch at the singular point $p$. It follows that they must have an even number of crossings. -The $d$-curves give a planar drawing of the 5-clique, where the curves corresponding to any pair of non-incident edges intersect an even number of times. This contradicts the strong Hanani-Tutte theorem.<|endoftext|> -TITLE: What is the rank of the Mordell equation $y^2 = x^3 - 2$? -QUESTION [12 upvotes]: The mordell equation $E$ defined by $y^2 = x^3 - 2$ over $\mathbb{Q}$ is known to have only one non-trivial integer solution $P = (3,5)$ from here. However, the rank of Mordell-Weil group $E(\mathbb{Q})$ is not zero because formulas known since the time of Bachet show that $P$ has infinite order. -How can we compute the rank of $E$ without using MAGMA or the BSD conjecture? I'd be happy to see an argument by 3-descent. -The torsion subgroup of $E(\mathbb{Q})$ is known to be trivial. - -REPLY [12 votes]: You could have a look at this paper: - -M. Stoll, On the arithmetic of the curves $y^2 = x^\ell + A$, II; -J. Number Theory 93, 183-206 (2002). -Corollary 2.1 says that for $A = -2$, one gets a rank bound of -$1$ plus twice the $3$-rank of the class group of ${\mathbb Q}(\sqrt{6})$. -So it only remains to verify that the class number of this field is not -divisible by $3$ (in fact, it is $1$). -The method is $(1-\zeta_3)$-descent, which is pretty close to a $3$-isogeny -descent. Apart from the class number determination, no computation is -necessary; it all follows from theoretical considerations.<|endoftext|> -TITLE: Ping-pong progress through a quincunx -QUESTION [10 upvotes]: A quincunx or -Galton board consists of -staggered pegs from which ping-pong balls bounce and eventually display -a binomial / normal distribution in catch-bins. I am wondering if the -downward progress of each ball has been studied. -Assume just one ball falls, initially with some nonzero horizontal velocity -and zero vertical velocity. Gravity increases its vertical velocity over time; -assume no friction. -Perhaps the ball radius and the peg radii are identical. -The peg is immobile. Ball-peg collision is elastic. -The ball's velocity vector is reflected from the common ball-peg tangent -at a collision (like a lightray from a mirror). - -                - - -                - -(First bounce in animation below.) - - -Over time, the ball makes (sometimes slow) downward progress. -E.g., in the example below, traversing the $10$ rows of pegs would take -about $4.5$ time units in free-fall in the absence of collisions, -but the $25$ collisions retard its progress considerably: -it takes about $19$ time units to escape below. - -      - - -      - -(Repeats once; reload to replay animation. Browser dependent.) - - - -Q1. Is there a simple argument to show that the ball - actually does make downward progress, in that its expected - vertical position at time $t_2 > t_1$ is - below its expected position at $t_1$? - Gravity is pulling the free paths downward, but collisions can shoot - the ball upward. -Q2. Any insight into quantitative bounds, say, on the expected number of collisions per unit time, as a function of the ball/peg radii? - Have analogous lattice reflections been studied? - -REPLY [10 votes]: A comprehensive study of the Galton board has been published by Arthur Lue and Howard Brenner, Phase flow and statistical structure of Galton-board systems (1993). Without any friction the influence of gravity vanishes in the long-time limit, because the kinetic energy overwhelms it. In that limit the Galton board reduces to a Lorentz gas, with a chaotic dynamics that has been studied extensively. With (even a little) friction the dynamics is richer, with attractors and repellers in phase space. -The dynamics in the Galton board without friction has been analyzed in The Galton Board: Limit theorems and recursion, by Chernov and Dolgopyat (2009). (See also this 2007 publication.) The displacement of the particle $x(t)$ grows as $t^{2/3}$, sublinearly because of the backscattering by the pegs. Conservation of energy then implies that the velocity grows as $t^{1/3}$. -As an answer to Q1, the motion in the Galton board without friction is recurrent: With probability one -$$\lim_{t\rightarrow\infty}{\rm inf}\, x(t)\leq x(0).$$ -Because of the diminishing effect of gravity in the long-time limit, the particle slows down and effectively returns to a vicinity of its initial state infinitely many times. As Chernov & Dolgopyat write: "We all know that a ball thrown into a real Galton board always rolls down and ends up on the floor. But on the idealized board, rather paradoxically, the ball will almost surely bounce all the way back up!"<|endoftext|> -TITLE: Equivalences for models of spectra -QUESTION [8 upvotes]: I'm sorry if this is well-known. I'm new to stable homotopy theory. -There are many models of spectra, and an answer to this MO post says most of them are equivalent, even equivalent while taking the symmetric monoidal structure (smash product) into account. -My question is motivated by wanting to know how these equivalences are established. There's a clean development of the stable $\infty$-category of spectra and the smash product thereof in Lurie's Higher Algebra, and both are characterized by universal properties, but it seems anachronistic to say that checking these universal properties is the way to establish the equivalence of different models. -Finally, even if one is able to establish that the different model categories of spectra with smash product (symmetric, orthogonal, what have have you) are equivalent, it seems like this doesn't easily construct a functor from one model to another--for instance, it's not obvious how to take two arbitrary Omega spectra (more or less the model in Higher Algebra) and spit out a pair of symmetric spectra while verifying that their smash products in either model are equivalent. -Is there a written account of: - -Historical efforts that failed, and in particular, attempted models for spectra that are not equivalent to modern models, -A list of equivalent models, -How we know the models are equivalent, and -When known, functors between the models? - -REPLY [7 votes]: The paper you're looking for is "The stable homotopy category is rigid" by Stefan Schwede. It shows that an equivalence on the homotopy category level implies a Quillen equivalence on the model category level. -For monoidal results, check out "Monoidal Uniqueness of Stable Homotopy Theory" by Brooke Shipley. This paper has exactly the same universal property for the model category of spectra that you mention from Lurie, but of course many years before his work. I think it's safe to say these papers of Schwede and Shipley were what Lurie had in mind when he wrote the $\infty$-category version.<|endoftext|> -TITLE: Isomorphic schemes over DVR -QUESTION [10 upvotes]: Let $S, S'$ be flat schemes over a DVR. Their generic fibers are isomorphic and their special fibers are isomorphic as well. -Does that imply $S$ and $S'$ isomorphic? If not, what can go wrong? -Thanks in advance! - -REPLY [3 votes]: If we are all pitching in our favorite counterexamples, consider a smooth, projective morphism $\pi:S\to \text{Spec}(R)$ over a DVR whose generic fiber is the Hirzebruch surface $\mathbb{P}^1\times \mathbb{P}^1$ and whose special fiber is the Hirzebruch surfaces $\Sigma_2 = \mathbb{P}_{\mathbb{P}^1}(\mathcal{O}(-1)\oplus \mathcal{O}(+1))$. For such a family, there is an integer invariant; roughly the order of vanishing at the closed point of the DVR of the "modulus" of the moduli space of Hirzebruch surfaces. More precisely, it is the length of the torsion sheaf $R^1\pi_* \textit{Hom}_{\mathcal{O}_S}(\Omega_\pi,\mathcal{O}_S)$. This integer invariant can take on any positive integer value. Thus, there exist families $S$, $S'$ as above such that the invariant is $1$, respectively $2$.<|endoftext|> -TITLE: Why are Green functions involved in intersection theory? -QUESTION [40 upvotes]: I've been learning Arakelov geometry on surfaces for a while. Formally I've understood how things work, but I'm still missing a big picture. - -Summary: -Let $X$ be an arithmetic surface over $\operatorname{Spec } O_K$ where $K$ is a number field (we put on $X$ the good properties: regular, projective,...). Moreover suppose that $\{X_\sigma\}$ are the "archimedean fibers" of $X$, where each $\sigma$ is a field embedding $\sigma:K\to\mathbb C$ (up to conjugation). Each $X_\sigma$ is a smooth Riemann surface. We accomplished to "compactify" our arithmetic surface $X$, so now we want a reasonable intersection theory. -An Arakelov divisor $\widehat D$ is a formal sum -$$\widehat D=D+\sum_\sigma g_\sigma\sigma$$ -where $D$ is a usual divisor of $X$ and $g_\sigma$ is a Green function on $X_\sigma$. - -Question: -Why on the archimedian fibers $X_\sigma$ do we need Green functions? A Green function on a Riemann surface is simply a function $g:X_\sigma\to\mathbb R$ which is $C^\infty$ on all but finitely points and at the "non-smooth" points is locally represented by -$$a\log|z|^2+\text{smooth function}$$ -in other words a Green function is a decent real valued smooth function which "explodes at infinity" in a finite set of points. Simply I don't understand why in order to define a reasonable intersection theory we need an object with such properties. The key point should be that Green functions satisfy a Poincare-Lelong formula which can be expressed in terms of currents as: -$$dd^c[g]=[\operatorname{div}^G(g)]+[dd^c(g)]$$ -but still I don't get the meaning of this formula. -Somewhere I've read something like: - -Green function are useful in order to "measure distances" on a Riemann - surface. (This is not a quote, but simply what I remember). - -I really don't have any idea on the meaning of this sentence. -Addendum: -Moreover if one wants to calculate the intersection between two Arakelov divisors, at the archimedian places one has to take the $\ast$-product between Green functions. I don't understand the geometry of this product, probably because I don't understand well the geometry behind distribution and currents on Riemann surfaces. - -REPLY [3 votes]: Just as comment to previous answers: Concrete example for Green function for $X=\mathbb P^d_\mathbb Z= \text{Proj}\mathbb Z[x_0,...,x_d]$ to undrestand the meaning of "measure distances" -For an integer $p\geq 0$ let $Z^p(\mathcal X)$ be the group of cycles $Z$ in $\mathcal X$ of codimension -$p$. Any cycle $Z =\sum n_iZ_i$, where $n_i \in \mathbb Z$, is a formal sum of irreducible cycles $Z_i$, -i.e. irreducible closed subschemes of $\mathcal X$. An irreducible cycle $Z \in Z^p(\mathcal X)$ defines a real current $\delta_{Z(\mathbb C)}\in D^{(p,p)}(\mathcal X)$ by integration -along the smooth part of $Z(\mathbb C)$. More explicitly, the current $Z(\mathbb C)$ is defined by -$$\delta_{Z(\mathbb C)}(\omega)=\int_{Z(\mathbb C)_{reg}}\vartheta^*(\omega)$$ -where $\vartheta:Z(\mathbb C)_{reg}\to Z(\mathbb C)$ is a desingularization of $Z(\mathbb C)$ along the set of singular points of $Z(\mathbb C)$. Such desingularization exists due to theorem of Hironaka. -Now we define Green current. -Suppose $Z\in Z^p(\mathcal X)$. A Green's current for $Z$ is a current $g_Z \in D^{(p-1,p-1)}(\mathcal X)$ such that -$$dd^cg_Z+\delta_{Z(\mathbb C)}=[\omega_{g_Z}]$$ for a smooth form $\omega_{g_Z}\in A^{(p,p)}(\mathcal X)$. -Note that for any cycle $Z$ there exists a Green's current for $Z$. Consider the arithmetic variety $X=\mathbb P^d_\mathbb Z= \text{Proj}\mathbb Z[x_0,...,x_d]$ and the cycle -$$Z=\{=...==0\}$$ of codimension p,where for $0=a_0^{(i)}x_0+...+a_d^{(i)}x_d\in \mathbb Z[x_0,...,x_d]$.The Levine form $g_Z$ associated to $Z$ is defined by -$$g_Z(x)=-\log \left(\frac{\sum_{i=0}^{p-1}||^2}{\sum_{i=0}^{d}|x_i|^2}\right).\left(\sum_{j=0}^{p-1}\left(dd^c\log\left(\sum_{i=0}^{p-1}||^2\right)\right)^j\wedge \omega_{FS}^{p-1-j}\right)$$ where $\omega_{FS}=dd^c\left(|x_0|^2+...+|x_d|^2\right)$ is the Fubini-Study form on $\mathbb P^d_\mathbb C$. The Levine form $g_Z$ associated to $Z$ satises -$dd^cg_Z+\delta_{Z(\mathbb C)}=[\omega_{FS}]$.<|endoftext|> -TITLE: A quantitative version of Hensel's Lemma -QUESTION [13 upvotes]: I've been reading some papers on Igusa zeta functions, and they seem to be implicitly using a "quantitative version" of Hensel's Lemma, which also asserts the number of lifts of a $\mathbb{Z}/p\mathbb{Z}$-point to a $\mathbb{Z}/p^k\mathbb{Z}$-point. I'm looking for something like the following: - -Let $X$ be a smooth irreducible separated scheme of finite type of relative dimension $n$ over the ring of $p$-adic integers $\mathbb{Z}_p$. Then for any $k>0$ do we have - $$\# X(\mathbb{Z}/p^k\mathbb{Z}) = p^{n(k-1)}\# X(\mathbb{Z}/p\mathbb{Z}) \quad ?$$ - -I'm looking for either a proof or a reference where I can find a proof. - -REPLY [10 votes]: Let me try a very explicit proof. Consider the problem of taking a solution modulo $p^k$ and lifting to solutions modulo $p^{k+1}$. For this, we may assume that $X \subset \mathbb{A}^m$ is smooth and affine of dimension $n$, given by polynomials $f_1, \dotsc, f_r \in \mathbb{Z}_p[x_1, \dotsc, x_m]$. Let $\mathbf{a} \in \mathbb{Z}_p^m$ satisfy $f_1(\mathbf{a}) \equiv \dotsb \equiv f_r(\mathbf{a}) \equiv 0 \pmod{p^k}$. -To find points modulo $p^{k+1}$ that lift $\mathbf{a}$, you put $\mathbf{x} = \mathbf{a} + p^k \mathbf{y}$ and look for vectors $\mathbf{y}$ modulo $p$ giving solutions to $f_i(\mathbf{x}) \equiv 0 \pmod{p^{k+1}}$. Taylor expansion gives - -$(f_1(\mathbf{x}), \dotsc, f_r(\mathbf{x})) = (f_1(\mathbf{a}, \dotsc, f_r(\mathbf{a})) + p^k \mathbf{J}(\mathbf{a}) \mathbf{y} + O(p^{k+1})$ - -where $\mathbf{J}$ is the Jacobian matrix $(\partial(f_i)/\partial x_j)$. Dividing by $p^k$ and reducing modulo $p$ gives an inhomogeneous linear equation over $\mathbf{F}_p$: - -$\mathbf{J}(\mathbf{a}) \mathbf{y} \equiv -p^{-k} (f_1(\mathbf{a}, \dotsc, f_r(\mathbf{a})) \pmod{p}$ . - -That $X$ is smooth over $\mathbb{Z}_p$ at $\mathbf{a}$ implies that $\mathbf{J}(\mathbf{a})$, when reduced modulo $p$, has rank $m-n$. The space of solutions $\mathbf{y}$ is non-empty, by Hensel's Lemma, so is a linear space of dimension $n$ over $\mathbb{F}_p$. In this way you see that every solution modulo $p^k$ lifts to precisely $p^n$ solutions modulo $p^{k+1}$, and you get your formula by induction.<|endoftext|> -TITLE: Why do combinatorial abstractions of geometric objects behave so well? -QUESTION [94 upvotes]: This question is inspired by a talk of June Huh from the recent "Current Developments in Mathematics" conference: http://www.math.harvard.edu/cdm/. -Here are two examples of the kind of combinatorial abstractions of geometric objects referred to in the title of this question: - -Coxeter groups. These are abstractions of Weyl groups. Weyl groups have geometry coming from Lie theory: they are finite reflection groups associated to a crystallographic root system. Weyl groups (or perhaps finite reflection groups, or including Weyl groups associated to affine lie algebras, etc.) are then the "realizable" Coxeter groups. -Matroids. These are abstractions of collections of vectors in some vector space. The matroids coming from collections of vectors in some vector space (over some field, say) are again the "realizable" matroids. - -Here is what I mean by "behave so well": -Often it happens that we can associate some interesting polynomial invariant to the combinatorial object in question. Some examples are: - -The Kazhdan-Lusztig (KL) polynomial associated to a Coxeter system. -The characteristic polynomial associated to a matroid. -The recent KL polynomial associated to a matroid (see https://arxiv.org/abs/1412.7408). - -And these polynomials have surprising and deep properties (positivity or unimodality/log-concavity of coefficients) that are not at all obvious from their definitions. A recurring theme is that these properties can be established in the "realizable" cases by appealing to algebraic geometry, specifically, to some suitable cohomology theory. However, the properties continue to hold for the general, nonrealizable objects for which there is no underlying geometry. The proofs of the general result are usually more "elementary" in so far as they avoid any algebraic geometry; but chronologically they come after the realizable results. -For instance, the coefficients of KL polynomials associated to a Coxeter system are positive. This was a famous conjecture of Kazhdan-Lusztig, proved a few years ago by Elias and Williamson (https://arxiv.org/abs/1212.0791). However, positivity was known for realizable Coxeter groups much earlier by interpreting the polynomials as Poincaré polynomials for the intersection cohomology of certain Schubert varieties. -Similarly, it is conjectured that the KL polynomial of a matroid has positive coefficients (see https://arxiv.org/abs/1611.07474); and this conjecture is known to be true when the matroid is realizable, again by interpreting the coefficients as dimensions of intersection cohomology spaces on certain varieties. -Or for the characteristic polynomial of a matroid: we know that the coefficients of this polynomial are log-concave, as was recently proved in the remarkable work of Adiprasito-Huh-Katz (https://arxiv.org/abs/1511.02888). Again, this result was preceded by the same result for the realizable case, due to Huh-Katz (https://arxiv.org/abs/1104.2519), interpreting the coefficients as intersection numbers for some toric variety. -So we come to my question: -Why do combinatorial abstractions of geometric objects behave so well, even in the absence of any underlying geometry? -EDIT: At around the 50 minute mark of his plenary talk at ICM 2018 (on Youtube here: https://www.youtube.com/watch?v=-3q6C558yog), Geordie Williamson asks a roughly similar question, and suggests that it may be a "mystery for the 21st century." -EDIT 2: As mentioned in the answers of Gil Kalai and Karim Adiprasito, another good example of "combinatorial abstraction of geometric object" is the notion of simplicial sphere, where the realizable case is a boundary of a polytope. Here the realizable case is connected to algebraic geometry via the theory of toric varieties, and as always this connection enables one to prove deep positivity results (e.g. the g-theorem of Stanley); whereas again the same results for the nonrealizable case are apparently much harder and the subject of intense, current research. -EDIT 3: I'm including a very relevant passage from a preprint of Braden-Huh-Matherne-Proudfoot-Wang (https://arxiv.org/abs/2010.06088). - -Remark 1.13 It is reasonable to ask to what extent these three nonnegativity results can be unified. [The three results here are the nonnegativity of the coefficients of the KL polynomial of an arbitrary Coxeter group, the $g$-polynomial of an arbitrary polytope, and the KL polynomial of an arbitrary matroid.] In the geometric setting (Weyl groups, rational polytopes, realizable matroids), it is possible to write down a general theorem that has each of these results as a special case. However, the problem of finding algebraic or combinatorial replacements for the intersection cohomology groups of stratified algebraic varieties is not one for which we have a general solution. Each of the three theories described above involves numerous details that are unique to that specific case. The one insight that we can take away is that, while the hard Lefschetz theorem is typically the main statement needed for applications, it is always necessary to prove Poincaré duality, the hard Lefschetz theorem, and the Hodge–Riemann relations together as a single package. - -REPLY [8 votes]: I'm not exactly addressing your question about combinatorial abstractions of geometric objects, but you seem to be taking Lie theory as a given natural geometric arena. -On the contrary, the development of Lie theory itself is an awesome abstraction from even more concrete geometric notions. Some early avatars are given by the identification of $\mathfrak{so}(3)$ with $\mathbb{R}^{3}$ equipped with the cross product, or the Heisenberg Lie algebra popping out of considerations in the early days of quantum mechanics, both examples having their origins in physics. In this case, I would wager that the abstraction to the general definition of a Lie algebra works so well because the way you prove anything about these concrete examples is by using their apparent algebraic properties, which is exactly what is being codified in passing to an abstract Lie algebra. -Furthermore, I'm not so sure I'd say that abstractions of geometric objects necessarily behave so well. Using my above example again, the theory of general Lie algebras is kind of a mess (we'll never classify nilpotent Lie algebras for example), but it's an extremely rich mess that has various alleyways which are amenable to a deep analysis and classification scheme (e.g semi-simple Lie algebras). -In my opinion, the idea that abstractions of geometric objects don't have an underlying geometric companion is an ode to a romantic sense that there is some mystical quality about certain geometric objects. While I would count myself as a mystic in this sense, maybe at the end of the day the reason things "behave well" is because the abstractions aren't really any less geometric than the original objects of study, as others have mentioned above.<|endoftext|> -TITLE: Integral of power of binomials equal to sum of power of binomials? -QUESTION [14 upvotes]: Inspired by this MO question about integrating binomial coefficients and the answers, I was wondering whether integrating powers of binomial coefficients also relates to the respective sums. And indeed I have strong numerical evidence that -$$\int_{-\infty}^{\infty} \binom{n}{x}^2 dx =\sum_{k=0}^n\binom{n}{k}^2$$. (The latter expression is of course $\binom{2n}{n}$.) -So I'm wondering for which $l$ the following identity holds: -$$\int_{-\infty}^{\infty} \binom{n}{x}^l dx =\sum_{k=0}^n\binom{n}{k}^l$$. -(And furthermore one could conjecture that there are similar examples, where sum over binomials is identical to integral of the same expression over real numbers.) -EDIT: -Regarding the last sentence here an example: I conjecture (and have numerical evidence) that -$$\int_{-\infty}^{\infty} x\binom{n}{x} dx =\sum_{k=0}^n k\binom{n}{k},$$ the latter expression being of course $2^{n-1}n$. -EDIT2: -And for Vandermonde's identity it seems also (by numerical evidence) to work analogously: -$$ \int_{-\infty}^{\infty} \binom{m}{x} \binom{n}{r-x} dx = \sum_{k=0}^r \binom{m}{k} \binom{n}{r-k},$$ the latter expression being of course $\binom{m+n}{r}$. -I dare to conjecture that one can still find more examples. - -REPLY [7 votes]: The generalization looks like this -$$ -\int_{-\infty}^{\infty} \binom{n}{\alpha x}^l dx =\sum_{k=-\infty}^\infty\binom{n}{\alpha k}^l,\quad 0<\alpha\le 2/l,~l\in\mathbb{N}\tag{1} -$$ -where $n$ need not be an integer. The general theorem is given for example in the paper Surprising sinc sums and integrals. -Below I give the general outline of the proof which is based on the well known fact that the following function is band limited (its Fourier transform has limited spectrum) -$$g(x)=\binom{n}{x}=\frac{1}{2\pi}\int_{-\pi}^{\pi} (1+e^{i \omega})^n e^{- ix\omega} d\omega$$ -One can see that Fourier transform is limited to frequencies $|\omega|<\pi$. Whenever spectrum of a function $f(x)$ is limited to frequencies $|\omega|<2\pi$ one expects that -$$ -\int_{-\infty}^{\infty} f(x) dx =\sum_{k=-\infty}^\infty f(k). -$$ -Now the Fourier transform of $g(\alpha x)^l$ has a spectrum limited in the band $|\omega|<\pi\alpha l$. This is easy to see calculating Fourier transform -$$ -\int_{-\infty}^{\infty}g(\alpha x)^le^{-ikx}dx -$$ -with the help of $\int_{-\infty}^{\infty}e^{-ikx}dx=2\pi\delta(k)$, where $\delta$ is delta function. The general theorem from the paper cited above now states that -$$ -\int_{-\infty}^{\infty} g(\alpha x)^l dx =\sum_{k=-\infty}^\infty g(\alpha k)^l,\quad 0<\pi\alpha l\le 2\pi,~l\in\mathbb{N}, -$$ -which is equivalent to (1). -All this analysis also explains that when $\alpha=1$ the proposed identity holds for $l=1,2$ but not for larger $l\in\mathbb{N}$.<|endoftext|> -TITLE: Classification of quasitriangular Hopf algebras -QUESTION [6 upvotes]: The classification of hopf algebras is a big and open problem, containing various subproblems (such as: the classification of groups, of Lie algebras, the study of special classes such as (co)commutative, (co)semisimple, pointed Hopf algebras etc). -One of the first results, known since the sixties, had to do with the classification of the cocommutative Hopf algebras: - -Let $k$ an algebraically closed field, of characteristic zero, and let $H$ a cocommutative Hopf algebra (over $k$). If we denote by $G(H)$ the group of its grouplike elements and $P(H)$ the Lie algebra of its primitive elements, we have the following isomorphism of Hopf algebras - $$ -H\cong U(P(H))\sharp kG(H) -$$ - where $\sharp$ stands for the smash product Hopf algebra. The above isomorphism is given explicitly by - $$ -x_{i_{1}} \otimes -x_{i_{2}} \otimes \ldots \otimes x_{i_{n}} \sharp g \mapsto -x_{i_{1}}x_{i_{2}}\ldots x_{i_{n}}g -$$ - where, $i_{1} \leqslant \ldots \leqslant i_{n} \in I$, $n \in \mathbb{N}$, $I$ is a totally ordered set and: $\{x_{i}\}_{i \in I}$ are primitive elements of $Η$ forming a $k-$basis of $P(H)$. - (In the RHS of the last map, $x_{i_{1}}x_{i_{2}}\ldots x_{i_{n}}g$ stands for the product of the elements $x_{i_{1}} , x_{i_{2}} , \ldots , x_{i_{n}} , g$ inside $Η$). - -The above result is frequently referred to, in the literature, as "the Cartier-Konstant-Milnor-Moore theorem". (although it seems that quite a lot of people have contributed to it). -Now, my question is: Given that the notion of quasitriangularity extends the notion of cocommutativity (in the sense that cocommutative hopf algebras are trivially quasitriangular through the $R$-matrix $R=1\otimes 1$ and thus quasitriangular are a class of Hopf algebras extending the class of cocommutative hopf algebras) are there similar results to the above, generalizing the Cartier-Konstant-Milnor-Moore theorem for quasitriangular Hopf algebras over an algebraically closed field of characteristic zero? - -REPLY [5 votes]: The best I know of are some classification results for triangular Hopf algebras, which would be a subcase. These are found in several papers by Etingof and/or Gelaki. See this paper and its references, for example. Theorem 2.2.2.4 therein is a result of Kostant that generalizes your quoted result, I'll note. I don't think even the triangular case has been completely determined, though I won't profess to be certain. -There are also some classification results on pointed quasitriangular Hopf algebras, such as this paper on minimal such ones generated by skew primitives by Masuoka. -Update (5/2/17): For some reason, an update to an old (1999) preprint of Etingof and Gelaki has been posted to the arxiv: The Classification of Triangular Semisimple and Cosemisimple Hopf Algebras Over an Algebraically Closed Field. Not sure if that was some automatic system recompiling of the article or legit update at this point. Either way, it's a paper worth looking at for the triangular case.<|endoftext|> -TITLE: What is the smallest uniquely hamiltonian graph with minimum degree at least 3? -QUESTION [15 upvotes]: I would like to know more about uniquely hamiltonian graphs with minimum vertex degree at least 3, and in particular what is the smallest one. -(Recall that a graph is hamiltonian if it has a cycle passing through each vertex exactly once, and is uniquely hamiltonian if there is only one such cycle.) -Here's the smallest one that I currently know. - -Does anyone know if a smaller one (fewer vertices) has been published? - -REPLY [9 votes]: The system encouraged me to answer my own question, although it feels a bit strange to do so. -Anyway, after a bit of thinking and a (more substantial) bit of computing, I can now safely conclude that this 18-vertex 28-edge graph is the smallest uniquely-hamiltonian graph with minimum degree 3, and there are no others of this order (number of vertices) and size (number of edges).<|endoftext|> -TITLE: On the exact reference of a cute Diophantine problem -QUESTION [9 upvotes]: The problem asks to prove that the Diophantine equation $x^{3}+y^{3} = (x+y)^{2}+(xy)^{2}$ does not have any solutions in natural numbers $x, y$. -I believe that this problem appeared in the section Задачи наших читателей of the Soviet magazine Квант somewhere between the first issue of 1980 and the last one of 1989. Since I don't know much Russian, I haven't been able to locate it by surfing the archives of the magazine that are available online: to add insult to injury, it seems to me that the section in question of the magazine was not a regular one. I would like to provide the exact reference for this problem in a certain document which I am preparing and that's the main reason that has compelled me to ask you this: -Did anybody here remember seeing this cute problem in Квант once? If so, would you be so kind as to provide me with a hint that allows one to find out what the actual issue wherein it appeared was? -Please, let me thank you in advance for your attentive consideration of this query of mine. - -REPLY [11 votes]: It appeared in issue 8 of 1984 at the page 34.You can download this issue from here: http://kvant.mccme.ru/oblozhka_djvu3.htm<|endoftext|> -TITLE: Ray class groups through binary quadratic forms -QUESTION [5 upvotes]: (Cross-posted from https://math.stackexchange.com/questions/2029407/ray-class-groups-through-binary-quadratic-forms) -If $d$ is the discriminant of a quadratic number field, then the primitive classes of binary quadratic forms of discriminant $d$ form a group isomorphic to the narrow ideal class group of $\mathbb{Q}(\sqrt{d})$. The classes of forms are equivalence classes under an action by $\mathrm{SL}_2(\mathbb{Z})$. My question is, is there a collection of subgroups of $\mathrm{SL_2}(\mathbb{Z})$ for which the sets of classes of forms of discriminant $d$ under action by these groups are in bijection with the narrow ray class groups of $\mathbb{Q}(\sqrt{d})$? -(I would hope that Gauss's composition operation, which is already well-defined at the level of classes under the action of $\begin{bmatrix} 1 & \mathbb{Z} \\ 0 & 1 \end{bmatrix}$, would provide a group operation that makes these bijections into isomorphisms of ray class groups.) - -REPLY [4 votes]: I guess that the answer is no. The binary forms with discriminant -$\Delta \cdot f^2$ describe ring class fields modulo $f$ (see Cox's book), so by taking the limit as $f \to \infty$ something like the idelic version of the full ring class group will result, and this is what I expect you will get using your weak equivalence. For proving that you cannot get anything beyond ring class fields you probably should look at G. Bruckner, Charakterisierung der galoisschen Zahlkörper, deren zerlegte Primzahlen durch binäre quadratische Formen gegeben sind, Math. Nachr. 32 (1966), 317-326, where the proof in the finite case is given.<|endoftext|> -TITLE: Geometry of the cut locus -QUESTION [15 upvotes]: Let $(M^n,g)$ be a smooth complete Riemannian manifold. Let $p\in M$ be a point. Recall that the cut locus of $p$ is the set of vectors $v$ in the tangent space $T_pM$ such that $\exp(t v)$ is a minimizing geodesic for any $t\in [0,1]$, but not for $t\in [0,1+\varepsilon )$ for any $\varepsilon >0$. -Question. Does the cut locus have Lebesgue measure 0 in $T_pM$? If yes, does it have Hausdorff dimension at most $n-1$? -If the above questions have negative answers, one may ask the same questions about the exponential image of the cut locus. - -REPLY [18 votes]: The key fact is that the cut time $t_c : UM \to \mathbb{R}$, defined on the unit tangent bundle $UM$ of a complete, $n$-dimensional Riemannian manifold, is locally Lipschitz continuous around all $v \in UM$ such that $t_c(v) < +\infty$. Hence the tangential cut locus at $p \in M$, that is -$$ -\tilde{C}_p = \{t_c(v)v\mid v \in UM,\quad t_c(v) < +\infty\} \subset T_q M, -$$ -either is empty, or it has Hausdorff dimension exactly $n-1$ (being the graph of a locally Lipschitz function). The exponential map being smooth, it cannot increase the Hausdorff dimension, hence $\dim(C_p) = \dim(\exp_p(\tilde{C}_p)) \leq n-1$. -All of this is proved here by Itoh and Tanaka.<|endoftext|> -TITLE: Determinant of the "real part" of a matrix -QUESTION [9 upvotes]: Let $A$ be an $n\times n$ complex matrix, and write $A=X+iY$, where $X$ and $Y$ are real $n\times n$ matricies. Suppose that for every square submatrix $S$ of $A$, $|\mathrm{det}(S)|\leq 1$ (i.e., all minors of $A$ are complex numbers with modulus $\leq 1$). This includes the assumption that $|\mathrm{det}(A)|\leq 1$. -Question: Is $|\mathrm{det}(X)|\leq 1$? -Note: It is easy to see $|\mathrm{det}(X)|\leq n!$ by a simple induction (since every component of $A$ has modulus $\leq 1$--and therefore the same is true for $X$). However, computer simulations make me wonder if the above question might be true. I'd be up for hearing about any sort of bound which is better than $n!$. - -REPLY [12 votes]: Bound 1 does not hold already for $n=2$. Take a matrix $A=\pmatrix{e^{ia}&e^{ib}\\e^{ic}&e^{id}}$, it satisfies your conditions if $|a+d-b-c|\leqslant \pi/3$. On the other hand, $X=\pmatrix{\cos a&\cos b\\\cos c&\cos d}$ and $$\det X=\cos a\cos d-\cos b\cos c=\frac12\left(\cos(a+d)+\cos(a-d)-\cos(b+c)-\cos(b-c)\right)\\= -\frac12\left(2\sin\frac{b+c-a-d}2\sin\frac{a+d+b+c}2+\cos(a-d)-\cos(b-c)\right), -$$ -thus if $a-d=0$, $b-c=\pi$, $a+d+b+c=\pi$, $b+c-a-d=\pi/3$ (I am lazy to solve this explicitly), this expression is equal to $3/2>1$. - -REPLY [6 votes]: As a complement to Fedor's answer, let me note the following special case, where the desired claim does hold. - -Prop. Let $A$ be a complex matrix and $A=X+iY$ be its Cartesian decomposition, i.e., $X=\frac12(A+A^*)$ and $Y=\frac{1}{2i}(A-A^*)$. If $X>0$ (i.e., it is positive definite), then in fact $|\det A| \ge \det X$.<|endoftext|> -TITLE: Pure first order logic formulations of Markov's principle -QUESTION [7 upvotes]: Markov's principle is a statement of constructive arithmetic that allows classical reasoning on formulas of the shape $\exists x P$ when $P$ is a recursive predicate: -$\neg \neg \exists x P \to \exists x P$ -Its formulation is well known in the context of arithmetic, and it is well known that adding it to Heyting Arithmetic gives rise to a constructive system: when $A \lor B$ is provable, either $A$ is provable or $B$ is provable; when $\exists x \, A(x)$ is provable, there is a $t$ such that $A(t)$ is provable. -However I think it is not clear how one could formulate it in the context of pure intuitionistic first order logic (if it does make sense at all). -Various sources ([1], [2]) state it as "$\neg \neg \exists x P \to \exists x P$ for $P$ $\forall, \to$-free". Another tempting formulation would be $\forall x (P \lor \neg P) \to \neg \neg \exists x P \to \exists x P$ for any propositional $P$. As far as I can see, these two formulations are not comparable. Both these axioms, though, share the property that if we add them to pure intuitionistic first order logic we still obtain a constructive system (this is proved in [1] for the first axiom; I couldn't find references for the second axiom, but a proof can be obtained with a very similar argument). -Is there a more general analog of Markov's rule for first-order logic which preserves the disjunction property and other proof-theoretic properties of constructive systems? Or alternatively, is there some other source justifying the choice of the formalization used in [1,2] for Markov's principle? -[1] H. Herbelin, An intuitionistic logic that proves Markov's principle https://hal.inria.fr/inria-00481815/ -[2] U. Berger, A Computational Interpretation of Open Induction http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=1319627 - -REPLY [4 votes]: This is not an answer to the question, but rather a comment which benefits from proper formatting. -For me personally, Markov's principle is specifically associated with the natural numbers and not with some other set. By that I mean that I don't think of -$$ \neg\neg \exists x \in X{:}\ P(x) \quad\Longrightarrow\quad \exists x \in X{:}\ P(x) $$ -(with $P$ restricted in some form, as you detailed) as a reasonable constructive principle, if $X$ is some arbitrary set. -Recall that there is a concrete reason why some schools of constructive mathematics accept Markov's principle: consider the algorithmic interpretation. The property $P$ is such that we can test in some finite way, for any natural number $n$, whether $P(n)$ holds or not. The assumption that $\neg\neg\exists n \in \mathbb{N}{:}\ P(n)$ means that there is some number $n$ such that $P(n)$ holds, but we do not know such a number (we are not given a witness of the existential statement). But we can find such a number on our own, thereby providing a witness of the statement $\exists n \in \mathbb{N}{:}\ P(n)$, by checking $P(0)$, $P(1)$, $P(2)$, and so on. This algorithm terminates (but we couldn't provide an upper bound for its runtime). -For an arbitrary set $X$ in place of $\mathbb{N}$, no analogous approach is possible. Therefore I can't quite fathom what you're driving at. (Markov's principle certainly works for some relatives of $\mathbb{N}$, for instance $\mathbb{N}^2$, which we can too search in a sequential manner.)<|endoftext|> -TITLE: Applications of the prime avoidance lemma -QUESTION [6 upvotes]: I was wondering if the prime avoidance lemma is very useful or just a nice result. So far I know just only one application: let $R$ be a commutative noetherian ring and $I$ be a proper ideal of $R$. If $I$ consists only of zero divisors of $R$, then $I$ is contained in some associated prime ideal of $(0)$. -So my question is: are there other applications of the prime avoidance lemma in commutative algebra? Thanks in advance for your answers. - -REPLY [2 votes]: I just stumbled upon the 4-year-old question. Let me add another application to the list. -Claim. Let $A$ be a commutative Noetherian ring. If primes $\mathfrak p, \mathfrak r$ satisfy $\mathfrak p < \mathfrak r$, then the interval $(\mathfrak p, \mathfrak r)$ contains infinitely many primes or zero primes. -Chat. Here I am working in the ordered set of primes, so $\mathfrak p<\mathfrak r$ means $\mathfrak p\subsetneq \mathfrak r$. Here $(\mathfrak p, \mathfrak r)$ denotes the set of primes strictly intermediate to $\mathfrak p$ and $\mathfrak r$. Finally, this question about the structure of intervals in the ordered set of primes is asked in order to understand Spec($A$). -Reasoning for Claim. Suppose otherwise that $(\mathfrak p, \mathfrak r)$ -is finite and nonempty. Shrink this interval to a minimal such one, and we end up with an interval containing some $\mathfrak q\neq \mathfrak p, \mathfrak r$ such that $\mathfrak p\prec \mathfrak q\prec \mathfrak r$ and $(\mathfrak p,\mathfrak r)$ is finite. (The notation $\mathfrak p\prec \mathfrak q$ indicates covering, which means that $\mathfrak p$ is included in $\mathfrak p$ and $(\mathfrak p,\mathfrak q)$ is empty.) When we have shrunk to a minimal such interval, no two distinct primes strictly between $\mathfrak p$ and $\mathfrak r$ will be comparable. -Factor by $\mathfrak p$ to reduce to the case $\mathfrak p=(0)$. Then localize at $\mathfrak r$ to reduce to the case where $\mathfrak r$ is maximal. Now we have $(0)\prec \mathfrak q\prec \mathfrak r$ and that $\mathfrak r$ is maximal. In this reduced case we are trying to show that it is impossible for a Noetherian integral domain of Krull dimension $2$ to have only finitely many primes. This is where Prime Avoidance comes in. Assume that $\mathfrak q_1, \ldots, \mathfrak q_n$ is a complete list of the primes strictly between $(0)$ and $\mathfrak r$ (i.e., height-$1$ primes). -By Prime Avoidance, it is impossible to have $\mathfrak r$ contained in $\cup_{i=1}^n \mathfrak q_i$, so choose $a\in \mathfrak r - \cup_{i=1}^n \mathfrak q_i$. By the Krull Principal Ideal Theorem, any prime minimal over $(a)$ must have height $1$, so must be one of the $\mathfrak q_i$'s. But we specifically chose $a$ so that it belongs to none of them, so we are done. $\Box$ -The Noetherianness hypothesis is used to allow us to invoke the Krull Principal Ideal Theorem.<|endoftext|> -TITLE: How should a "working mathematician" think about sets? (ZFC, category theory, urelements) -QUESTION [51 upvotes]: Note that "a working mathematician" is probably not the best choice of words, it's supposed to mean "someone who needs the theory for applications rather than for its own sake". Think about it as a homage to Mac Lane's classic. I'm in no way implying that set theory is not "real mathematics" (whatever that expression might mean, though I've heard some people say it, and I don't respect this point of view that something abstract is "not real mathematics") and I have a great respect for that field of study. -However, I'm personally not interested in set theory and its logic for their own sake (as of now). For a while I have treated them naively, and it was fine as I haven't needed anything beyond introductory chapters in compheresnive books on algebra, analysis or topology. But recently I decided to understand the foundations of category theory based on Grothendieck universes and inaccessible cardinals. So, I went to read some sources on set theory. And was really confused at first about such definitions as of a "transitive set", which implicitly assume that all elements of all sets are sets. Then I read more about it and discovered that in $\mathrm{ZFC}$ - -everything is a set! - -It seemed absurd to me at first. After consulting several sources, I realized that ZFC was meant to be a (or even the) foundation for mathematics, rather than simply a theory which gives us a framework to work with sets, so at that time people thought that every mathematical object can be defined in term of sets. It didn't seem as unreasonable as before anymore, but still... -It still doesn't feel right for me. I understand that at the time when Zermelo and Fraenkel were developing axiomatic set theory, it was reasonable to think that every conceivable mathematical object is set. But it was a long time ago; is it still this way - especially concerning category theory? -If we work in $\mathrm{ZFC}$ (+ $\mathrm{UA}$) we have to assume that every object in any category is a actually as set. And the same should go for morphisms. Because, given a category $\mathrm{C}$, $\operatorname{ob} \mathrm{C}$ and $\operatorname{mor} \mathrm{C}$ are sets, so their elements, namely, objects and morphisms of $\mathrm{C}$, should also be sets. -The question is: is the assumption that there are no urelements, that is, that every conceivable mathematical object can be modeled in term of sets, reasonable, as of the second decade of the 21st century? Is there an area of mathematics where we need urelements? Can this way of thinking be a burden in some mathematical fields? (Actually, it's three questions, sorry. But they are related) -P.S. I hope this question is not too "elementary" for this site. But as I understand there are quite a lot of working mathematicians who don't think much about foundations. So, even if this question is not useful for them, it can at least be interesting for them. - -REPLY [2 votes]: Within traditional set theory (and without abandoning it for category theory) one can make a compelling case that not everything is a set, or more precisely that the assumption that everything is a set is both limiting and counterproductive when dealing with fruitful frameworks that are conservative extensions of the traditional set theory. -Thus, Edward Nelson's Internal Set Theory is a way of working with infinitesimals within the ordinary real line, modulo foundational adjustments that involve introducing a richer language into set theory. Namely, one works not merely with an $\in$-language with with a $(\in,\mathbf{st})$-language. Here $\mathbf{st}$ is a one-place predicate "standard", where $\mathbf{st}(x)$ reads "$x$ is standard". To emphasize, Nelson's theory is a conservative extension of ZFC, unlike some of the other frameworks discussed in this space. -The point is that the collection of standard $x$'s is typically not a set, even when $x$ are ordinary (integer or real) numbers. Thus the new predicate violates the axiom of separation. A blanket assumption that "everything is a set" would make Nelson's approach incomprehensible.<|endoftext|> -TITLE: Human brains considered as directed graphs -QUESTION [6 upvotes]: I assume that human brains can be considered as directed graphs with neurons as nodes and synapses as edges. I explicitly don't want to consider the weights, the dynamics of neural activity (based on the weights), and the adjustment of weights (learning) - just brains as static unweighted finite directed graphs. -Sensor neurons may be those having in-degree 0, actor neurons may be those having out-degree 0. (0 meaning "essentially 0".) -Considering human brains as finite directed graphs, for each question concerning finite directed graphs there should be an answer with respect to human brains. -Such questions might be: - -How long is the shortest path from a sensor to an actor neuron? -How long is the longest (direct) path from a sensor to an actor neuron? -What is the (global/local) layer structure (on different levels of granularity)? -What is the (global/local) cycle structure (on different levels of granularity)? - -I find it hard to get answers to such questions considering human brains as directed graphs, because neuro-scientists don't think in terms of graphs, but for example in terms of signal paths and neuro-anatomy. But then - for them - "anything goes", and "everything is connected to everything" - which is not very helpful. - -I would be very glad for any reference treating (formally) human - brains as directed graphs. - -REPLY [4 votes]: There is empirical evidence that the connectivity in the brain has the characteristics of a directed small-world network. -Small-world directed networks in the human brain: Multivariate Granger causality analysis of resting-state fMRI (Wei Lao et al., 2010): - -Small-world organization is known to be a robust and consistent - network architecture, and is a hallmark of the structurally and - functionally connected human brain. However, it remains unknown if the - same organization is present in directed influence brain networks - whose connectivity is inferred by the transfer of information from one - node to another. Here, we aimed to reveal the network architecture of - the directed influence brain network using multivariate Granger - causality analysis and graph theory on resting-state fMRI recordings. - We found that some regions acted as pivotal hubs, either being - influenced by or influencing other regions, and thus could be - considered as information convergence regions. In addition, we - observed that an exponentially truncated power law fits the - topological distribution for the degree of total incoming and outgoing - connectivity. Furthermore, we also found that this directed network - has a modular structure. More importantly, according to our data, we - suggest that the human brain directed influence network could have a - prominent small-world topological property. - -More recent studies of the human brain as a directed graph are summarised in section 7.3 of this review article.<|endoftext|> -TITLE: Belief in consistency of extremely large cardinals -QUESTION [10 upvotes]: One of the most common justifications for the consistency of large cardinals is the development of a coherent inner model theory for many large cardinal axioms. While the strength of this argument can be debated (has any axiom ever been shown to be inconsistent via the attempt to develop an inner model theory?), it appears to be pretty popular, e.g. it appears in Maddy's Believing the Axioms. -However, it is my understanding that assuming Woodin's HOD Conjecture, his Universality Theorem implies that given an extendible cardinal, there exists a weak extender model that reflects essentially all of the large cardinals that are currently studied. -If inner model theory can be extended so far in a single leap, can it still be used to justify the consistency of any particular large cardinal axiom stronger than a supercompact? If not, what is the replacement? -It's also my understanding that assuming the $\Omega$ Conjecture and a proper class of Woodin cardinals, the same class of large cardinal axioms can be ordered in consistency strength by the Borel degree of the initial segment of universally Baire sets $A$ for which they establish determinacy properties of $L(A, \mathbb{R})$. Would calibrating the place of a large cardinal axiom on this hierarchy replace other arguments for consistency? - -REPLY [7 votes]: There is another path that can be used to justify the existence of very large cardinals. Consider, for example, the abstract to Magidor's paper, "On the Role of Supercompact and Extendible Cardinals in Logic" (Israel Journal of Mathematics, Vol. 10, 1971, pp. 147-157) : - -It is proved that the existence of supercompact cardinals is equivalent to a certain Lowenheim-Skolem Theorem for second order logic, whereas the existence of [an] extendible cardinal is equivalent to a certain compactness theorem for that logic. It is also proved that a certain axiom schema related to model theory implies the existence of many extendible cardinals. - -Here the theorems: - -Theorem 1. There exists a supercompact cardinal iff there is a $\mu_0$ such that for all $R({\beta})$, $\beta$$\ge$$\mu_o$there is an $\alpha$$\lt$$\beta$ such that $<$$R({\alpha})$, $\epsilon$$>$ can be elementarily embedded in $<$$R({\beta})$, $\epsilon$$>$. The least such $\mu_0$ is the first supercompact cardinal. -Theorem 2. The first supercompact cardinal is the first $\mu_0$ such that for every structure $A$ = $<$$M$, $R_1$,...,$R_n$ $>$ $|$$M$$|$$\ge$$\mu_0$ and every $\Pi^{1}_1$ sentence $\varphi$, such that $A$$\vDash$$\varphi$, there exists a substructure $A^{'}$ = $<$ $M^{'}$, $R_1$$|$$M$,..., $R_n$$|$$M$$>$ of $A$ with $|$$M^{'}$$|$$\lt$$|$$M$$|$ and $A^{'}$$\vDash$$\varphi$ [this is the "Lowenheim-Skolem Theorem" Magidor refers to in his abstract--my comment]. -Definition. Logic is called $\kappa$-compact iff for every set of formulae $A$ in this logic, if every subset of $A$ of cardinality $\lt$$\kappa$ has a model, then $A$ has a model. The $\mathrm L^{n}_{\kappa}$ logic is like the $n$-th order logic, except that we allow conjunction and disjunction of less than $\kappa$ formulae. The usual second order logic is of course $L^{2}_{\omega}$. -Theorem 4. $\kappa$ is extendible iff $L^{2}_{\kappa}$ is $\kappa$-compact. $\kappa$ is the first extendible iff it is the first $\alpha$ such that second-order logic [$L^{2}_{\alpha}$--my comment] is $\alpha$-compact. -...We shall now show that a certain axiom schema is a very strong axiom of infinity, namely, it implies the existence of many supercompact cardinals. The axiom schema is: -($V$) If $\varphi$(x) defines a proper class of structures in the same language, then there exist two members of the class that one can be elementary embedded in the second...This axiom schema is called Vopenka's principle. -Let $V^{'}$ be the following axiom schema: ($V^{'}$) if the formula $\tau$(x) defines a closed unbounded set of ordinals, then there is an extendible cardinal in this class (i.e. "the class of all extendable cardinals is 'stationary' "). -Theorem 3. ($V$) implies ($V^{'}$) but is not equivalent to it. - -Note that the forward implication '$\Rightarrow$' of Theorems 1, 2, and 4 show that if the required large cardinal exists, the logic in question will have the required property. This suggests to me the following 'formalist' justification of the existence of large cardinals: - -If you want second-order logic to have certain model-theoretic properties, hypothesize the appropriate large cardinals. - -This also suggests the following research program: - -Discover what model-theoretic properties of n'th-order logic the existence of large cardinals imply (from $I0$ on down the large cardinal hierarchy), if any. - -Similarly, one can hypothesize the existence of certain Lowenheim-Skolem-Tarski numbers, as V$\ddot a$$\ddot a$n$\ddot a$nen shows in his paper, "Sort Logic and Foundations of Mathematics" ("Sort Logic...is a many-sorted extension of second-order logic."): - -The canonical hierarchy $\Delta_n$ ($n$$\lt$$\omega$) inside sort logic climbs up the large cardinal hierarchy by reference to Hanf-, Lowenheim-, and [Lowenheim-Skolem-Tarski]-numbers, reaching all the way up to Vopenka's Principle [pg. 185].<|endoftext|> -TITLE: Eigenvectors of a matrix with entries involving combinatorics -QUESTION [7 upvotes]: In the question Eigenvalues of a matrix with entries involving combinatorics No_way asked about eigenvectors of $n\times n$ matrix $M$ with entries \begin{eqnarray*} -M_{ij}=(-1)^{i+j}F(n, l, i, j), -\end{eqnarray*} -where $F(n,l,i,j)$ is the cardinality of the set -\begin{eqnarray*} -\{(k_1, \cdots, k_n)\in\mathbb{Z}^{n}|0\leq k_r\leq l-1\text{ for }1\leq r\leq n\text{, }k_1+\cdots+k_n=lj-i\}. -\end{eqnarray*} -These eigenvalues are known to be $1, l, l^2, \cdots, l^{n-1}$. -Let's remove signs and consider the matrix $M$ with $M_{ij}=F(n, l, i, j)$. According to my numerical experiments eigenvectors do not depend on $l$ for $l\ge 2$ and they are polynomials. - -Q1: Why do eigenvectors not depend on $l$? - -For $l=2$ we have $M_{ij}=\binom n{2j-i},$ and first examples are (eigenvectors of $M$ are rows of $V$) -$$n=2,\qquad M=\left( -\begin{array}{cc} - 2 & 0 \\ - 1 & 1 \\ -\end{array} -\right),\qquad V=\left( -\begin{array}{cc} - 1 & 1 \\ - 0 & 1 \\ -\end{array} -\right);$$ -$$n=3,\qquad M=\left( -\begin{array}{ccc} - 3 & 1 & 0 \\ - 1 & 3 & 0 \\ - 0 & 3 & 1 \\ -\end{array} -\right),\qquad V=\left( -\begin{array}{ccc} - 1 & 1 & 1 \\ - -1 & 1 & 3 \\ - 0 & 0 & 1 \\ -\end{array} -\right);$$ -$$n=4,\qquad M=\left( -\begin{array}{cccc} - 4 & 4 & 0 & 0 \\ - 1 & 6 & 1 & 0 \\ - 0 & 4 & 4 & 0 \\ - 0 & 1 & 6 & 1 \\ -\end{array} -\right),\qquad V=\left( -\begin{array}{cccc} - 1 & 1 & 1 & 1 \\ - -1 & 0 & 1 & 2 \\ - 2 & -1 & 2 & 11 \\ - 0 & 0 & 0 & 1 \\ -\end{array} -\right);$$ -$$n=5,\qquad M=\left( -\begin{array}{ccccc} - 5 & 10 & 1 & 0 & 0 \\ - 1 & 10 & 5 & 0 & 0 \\ - 0 & 5 & 10 & 1 & 0 \\ - 0 & 1 & 10 & 5 & 0 \\ - 0 & 0 & 5 & 10 & 1 \\ -\end{array} -\right),\qquad V=\left( -\begin{array}{ccccc} - 1 & 1 & 1 & 1 & 1 \\ - -3 & -1 & 1 & 3 & 5 \\ - 11 & -1 & -1 & 11 & 35 \\ - -3 & 1 & -1 & 3 & 25 \\ - 0 & 0 & 0 & 0 & 1 \\ -\end{array} -\right).$$ -Denote by $v_m=(v_m(1),\ldots,v_m(n))$ rows of $V$ ($0\le m\le n-1$). They defined up to multiplicative constant and $v_m(k)=\mu_m P_m(k)$ where $P_m(x)$ are some special polynomials of degree $m$. In particular for $m=0,1,2,3,4$ we have $$P_0(x)=1,\quad P_1(x)=2x-n,\quad P_2(x)=3x^2-3nx+\frac{n(3n-1)}{4},$$ -$$P_3(x)=4x^3-6nx^2+n(3n-1)x-\frac{n^2(n-1)}{2},$$ -$$P_4(x)=5x^4-10nx^3+\frac{5n(3n-1)}{2}x^2-\frac{5n^2(n-1)}{2}x+\frac{n(15n^3-30n^2+5n+2)}{48}.$$ - -Q2: What is the generating function for these polynomials? - -REPLY [5 votes]: In fact for a fixed $n$, the matrices $M(l, n)$ for $l>0$ commute with each other and thus are simultaneously diagonalisable. For your second question, if $\{p_j(y)\}$ is a sequence of polynomials satisfying -\begin{eqnarray} -\left(\frac{t}{\sinh t}\right)^y=\sum_{j=0}^\infty p_j(y)t^{2j}. -\end{eqnarray} -then the $i$-th entry of an eigenvector corresponding to the eigenvalue $l^{n-k-1}$ is -\begin{eqnarray} -(-1)^{i-1}\sum_{j=0}^{\lfloor\frac{k}{2}\rfloor}\frac{p_j(n)}{(k-2j)!}(n-2i)^{k-2j}. -\end{eqnarray} -I believe from here we can work out what the generating function of your $P_m(x)$ is.<|endoftext|> -TITLE: Why is the notion of algorithm a primitive one in Brouwer's intuitionism? -QUESTION [10 upvotes]: I've seen several times people mentioning that the notion of an algorithm / a computation is taken as a primitive notion in L. E. J. Brouwer's intuitionism. For instance, in Varieties of Constructive Mathematics (1987, p.1), Bridges and Richman write that: - -In Bishop's constructive mathematics (BISH), and in Brouwer's intuitionism (INT), the notion of an algorithm, or finite routine, is taken as primitive. Russian constructivism (BUSS), on the other hand, operates within a fixed programming language, and an algorithm is a sequence of symbols in that language. - -To the best of my knowledge, Brouwer's intuitionistic mathematics is fundamentally a languageless creation of the mind which is erected on the two acts of intuitionism: (i) the first act gives us the construction that corresponds to the natural numbers; (ii) the second act provides us with the notion of free choice sequents. To put it other way, there is no explicit mention of algorithms. -On the other hand, it seems that this common (?mis?)conception, namely, that the notion of an algorithm is a primitive notion in Brouwer's intuitionisim, came with the influence of the BHK interpretation, where the notion of a proof (which, by the Curry-Horward correspondence, corresponds to the notion of a program) plays the central role in this informal interpretation of intuitionistic logic. However, as I said, this is intuitionistic logic and not Brouwer's intuitionisim - and I do think it helps to keep both apart. -Can we justify this claim that the notion of an algorithm is a primitive notion in Brouwer's intuitionism? - -REPLY [4 votes]: Since the work of Church and Turing (say around 1936), the notion of algorithm is definitely not considered primitive in intuitionism. But Brouwer started intuitionistic mathematics more than 2 decades before (1907, 1912), and in the absence of a commonly accepted notion of algorithm, he formulated his idea (somewhat) like this: -A sequence of natural numbers arises in the course of time, like a walk through the (infinite) tree $\mathbb{N}^*$ of finite sequences, choosing at each 'node' a next continuation. -The result is an infinite branch, or an 'arrow' as Brouwer called it. At any node (also right at the start) one may specify a definite finite law which completely governs all next choices (so that there really is no choice left at all). If one does so right at the start, the arrow is called a 'sharp arrow' or a 'lawlike sequence'. -Since the 1930's definition of recursion, it has become standard to identify 'sharp arrow' and 'lawlike sequence' with recursive sequences, I believe.<|endoftext|> -TITLE: Intuition on Kronecker Product of a Transition Matrix -QUESTION [5 upvotes]: Let $T$ be a $N\times N$ transition matrix for a markov chain with $N$ states. Thus $T_{ij}$ is the probability of transition from state $i$ to state $j$ (and thus rows summing to one). Now consider the matrix $$T_k=T\otimes T$$. Its easy to see that rows of $T_k$ sum to one and each entry is non-negative. Thus, $T_k$ is a transition matrix for some markov chain which has $N^2$ states. What is the relation between markov chains corresponding to $T$ and $T_k$. I am not able to visualize this. - -REPLY [3 votes]: I think you just have a pair of independent chains. The probability of making a transition from (i,j) to (k,l) is $p_{ik}p_{jl}$, where the first component is the state of the first chain and the second of the second. The transitions out of the (i,j) state are found in the N(i-1) + j th row.<|endoftext|> -TITLE: two's and three's survive in gcd of Lagrange -QUESTION [24 upvotes]: Lagrange's four square_theorem states that every positive integer $N$ can be written as a sum of four squares of integers. At present, let's focus only on positive integer summands; that is, $N=a_1^2+a_2^2+a_3^2+a_4^2$ with $a_i\in\mathbb{N}$. -If now mix-up sums and products, something rather curious happens: the greatest common divisor (gcd) becomes low-primed. I wish to know why. - -CLAIM. If $n\geq26$ is an integer, then - $$\text{GCD}\,\left\{a_1a_2a_3a_4\,: \, 4n+1=a_1^2+a_2^2+a_3^2+a_4^2 \,\,\, \text{and $a_i\in\mathbb{Z}^{+}$}\right\}=2^b3^c$$ - for some $b+c>0$. - -For example, \begin{align} 4(26)+1&=7^2+6^2+4^2+2^2=8^2+4^2+4^2+3^2 \\ -&=8^2+6^2+2^2+1^2=9^2+4^2+2^2+2^2. -\end{align} -Therefore, the GCD equals $2^43^1$. - -UPDATED (later November 29, 2016). - -I now have a more specific claim for the specific powers of $2$ and $3$, showing that even these numbers are low-powered. - -CONJECTURE. Denote $a=a(n)$ and $b=b(n)$ the exponents $2^b3^c$ from above. Then, - $$b(n)=\begin{cases} 4 \qquad \text{if $n$ is even} \\ -3 \qquad \text{if $n$ is odd}; \end{cases}$$ - $$c(n)=\begin{cases} 0 \qquad \text{if $n=3k$} \\ -2 \qquad \text{if $n=3k+1$} \\ -1 \qquad \text{if $n=3k+2$}. \end{cases}$$ - -"It's easy for number theory to be hard." - anonymous. - -REPLY [21 votes]: The claim is certainly true for $n$ sufficiently large, and "sufficiently large" could be specified explicitly with more care. -We follow the suggestion of Fedor Petrov, and rely on the results of Brüdern & Fouvry (J. reine angew. Math. 454 (1994), 59-96) and of Heath-Brown & Tolev (J. reine angew. Math. 558 (2003), 159-224). These yield, for any prime $p\geq 5$, that if we count the representations $$4n+1=a_1^2+a_2^2+a_3^2+a_4^2$$ smoothly (with positive integers $a_i$), then the proportion of representations such that $p\mid a_1$ is at most $1/(p-1)+o(1)$ as $n\to\infty$, where $o(1)$ is uniform in $p$ (note that we can restrict to $p\ll n^{1/2}$). See especially Theorem 3 and Lemmata 6-7 in Brüdern & Fouvry, and the displays (340), (346), (349), (352) in Heath-Brown & Tolev. It follows that for $p\geq 7$ the proportion of representations such that $p\mid a_1a_2a_3a_4$ is at most $2/3+o(1)$, hence for $n$ sufficiently large (independent of $p$) there is a representation such that $p\nmid a_1a_2a_3a_4$. For $p=5$ we need to be more careful, because $4/(p-1)$ equals $1$ in this case, and subtract the (positive) proportion of representations such that two of the $a_i$'s are divisible by $p$. This is also covered by the mentioned works as they discuss general divisibility constraints $d_i\mid a_i$.<|endoftext|> -TITLE: Does unique factorisation hold for quiver algebras? -QUESTION [8 upvotes]: Given a finite dimensional quiver algebra A=KQ/I. It can be (possibly) written as $A= B_1 \otimes_k B_2 ... \otimes_k B_r$ and the $B_i$ can not be decomposed into smaller algebras. Is this factorisation unique? (all quivers are assumed to be connected and have at least one arrow) - -REPLY [12 votes]: In Nüsken, M. "Unique tensor factorization of algebras", Math Ann. (1999) 315-341 this is proved for $K$ of characteristic zero. As far as I know, it's still open in positive characteristic, although the same author proved some partial results in a later paper.<|endoftext|> -TITLE: May $p^3$ divide $(a+b)^p-a^p-b^p$? -QUESTION [9 upvotes]: Do there exist positive integers $a,b$ and a prime $p>\max(a,b)$ such that $p^3$ divides $(a+b)^p-a^p-b^p$? -The reader of Kvant magazine A. T. Kurgansky asked to prove that such $a,b,p$ do not exist, see here. -But discussion here -On the exact reference of a cute Diophantine problem -suggests that it should be very hard to prove. Maybe, a counterexample may be bound? Roughly speaking, a probability of this event is about $1/p^2$, for each $p$ we have about $p$ events (even for $b=1$, I was previously wrong that it may be supposed without loss of generality, thanks for Noam Elkies for noting this) and so as $\sum 1/p=\infty$, we may expect them (and even infinitely many!) But this series converges very slowly, so the minimal example may be large. - -REPLY [20 votes]: We explain the pattern observed by Joe Silverman, deducing -the existence of infinitely many solutions, some of which even have -$p^5 | (a+b)^p - a^p - b^p$. -Lemma. If $n \equiv 1 \bmod 3$ then the homogeneous polynomial -$P_n(a,b) := (a+b)^n - a^n - b^n \in {\bf Z}[a,b]$ has a factor $(a^2+ab+b^2)^2$. -Proof : Either -i) evaluate $P_n(x,1)$ and its derivative at a cube root of unity $\rho$, or -ii) use the $S_3$ symmetry of $-P_n(a,b) = a^n + b^n + c^n$ where $a+b+c=0$: -double roots at $(a:b:c) = (1:\rho:\rho^2)$ and $(1:\rho^2:\rho)$ -are the only ways to get the total number of roots to be $1 \bmod 3$. $\Box$ -[These ${\bf Q}(\rho)$-rational points on the $n$-th Fermat curve -are well known.] -Corollary. If $p$ is a prime congruent to $1 \bmod 3$, -and $a,b$ are integers such that $p^k | a^2+ab+b^2$, then -$p^{2k+1} | P_n(a,b)$. -Proof : Observe that $P_p \in p {\bf Z}[a,b]$ and use the Lemma. -Now $k=1$, and thus $2k+1=3$, is easy to obtain: -choose $a$ arbitrarily, and let $b\equiv ra \bmod p$ where $r$ is a cube root of unity -mod $p$. We can even get a factor of $p^5$ by choosing $a,b$ such that -$a^2 + ab + b^2 = p^2$ (that is, so that $a,b,p$ are sides of -a triangle with a $120^\circ$ angle); for example, -$$ -(3+5)^7 - 3^7 - 5^7 = 120 \cdot 7^5. -$$<|endoftext|> -TITLE: What is the structure associated to almost-everywhere convergence? -QUESTION [11 upvotes]: Let $M(X)$ be the vector space (actually it's an algebra) of all equivalent classes of measurable functions $X\to \mathbb{C}$ (where $X$ is a measured space) modulo equality almost-everywhere. -One can define a notion of sequential convergence on $M(X)$. A sequence $(f_n)$ of $M(X)$ converges to $f$ iff (by definition) there is a sequence of functions $g_n$ and a function $g$ such that $g_n\to g$ pointwise almost everywhere and where $g_n$ is in the class of $f_n$ and $g$ is in the class of $f$. This notion of convergence is compatible in an obvious way with the algebraic structure of $M(X)$. -From what I understood of what I read, $M(X)$ cannot be given a topology such that the notion of convergence associated with this topology coincide with the previous notion of convergence almost everywhere. -My question is about a positive result : what is the structure of almost everywhere convergence ? In other words, what is $M(X)$ ? Is it a "convergence vector space" (whatever that would mean) ? (From this, it is not clear to me that this is the case) -Has this structure been explored, for example, could we talk about completeness in this setting, how about classical topological results such as Hahn-Banach ? Are they still true in some form for this "convergence space" structure ? - -REPLY [17 votes]: Yes, this defines a "convergence vector space". In fact, it's probably the original motivating example for the generalization. In Fréchet's thesis he discussed L-spaces, which are essentially sequential convergence spaces: a set equipped with families of convergent sequences at each point satisfying the axioms that the constant sequence converges and subsequences of convergent sequences converge. The axioms of a convergence structure are intended to generalize this by axiomatizing nonsequential convergence. However, given an L-space you can define a convergence space generated by the tail filters $\{ x_n : n \geq m \}$ of the convergent sequences. Fréchet also discussed the special case of E-spaces, which we call metric spaces, and this special case became much more popular. -With messier notation than the definitions in your link, you could define a convergence structure in terms of families of convergent nets such that the constant net converges and subnets of convergent nets converge (given the appropriate notation of subnet that corresponds to a subfilter), but since there is no set of all nets on a space it's easier to just define convergent filters. -Much of functional analysis generalizes to convergence vector structures in some way. Perhaps the most salient new feature of the theory is an improved duality theory, with continuous convergence providing a dual object corresponding to aw$^*$ (almost weak$^*$) topology on locally convex spaces, which often fails to even be a topology, nevermind a linear topology. -The Hahn-Banach Theorem doesn't generalize outright, as it often fails quite badly for topological vector spaces, existence of spaces with no nontrivial continuous linear functionals. However, the question of when the Hahn-Banach Theorem holds for a convergence vector space is an interesting one, even for the duals of locally convex spaces, where the convergence vector space dual of $E$ having the Hahn-Banach extension property is equivalent to $E$ being fully complete (or B-complete) in the ordinary sense of locally convex spaces. This is connected to the generalizations of the Closed Graph Theorem for locally convex spaces, and gives better proofs of those results. -The best book on the subject is Convergence Structures and Applications to Functional Analysis by Beattie and Butzmann. It's a pretty intriguing theory, but unfortunately many of the consequences for the theory of locally convex spaces had already been discovered in different forms with less clear statements and proofs.<|endoftext|> -TITLE: When is $X_{hG} \to X/G$ a weak equivalence for $X$ a free $G$-space, $G$ compact Lie? -QUESTION [9 upvotes]: Two questions (more details below): - -Let $G$ be a compact Lie group and $X$ a $G$-space such that all stabilizer -subgroups are conjugate to a fixed $H \leq G$. Denote by $\pi: X -\to X/G$ the quotient map. Under which conditions on $X$ is $\pi$ a Serre fibration? -Let $G$ be as above and $F$ a free $G$-space. Under which conditions -is the canonical map $F_{hG} \to F/G$ from the homotopy quotient to the -quotient a weak equivalence? - -All spaces are assumed to be CGWH spaces. - -Relation between 1. and 2.: -I am most interested in 2.. However, if $F$ is a free $G$-space, $EG$ denotes -Bar construction, such that $(F \times EG)/G$ models $F_{hG}$, and both $F$ and -$F \times EG$ are as in 1., i.e., the respective quotient maps are Serre -fibrations, then it is not hard to see that $F_{hG} \to F/G$ is a weak -equivalence, using the long exact sequence of homotopy groups. -So, any condition for 1. that holds for $EG$ and is stable under products -provides a condition for 2. - -Results known to me: - -A sufficient condition for 1. is being a completely regular Hausdorff space -(also known as a Tychonoff space) by a result in Bredon's "Introduction to -Compact Transformation groups." He shows that the quotient map is a fiber -bundle in this case. However, the Tychonoff property does not seem to be -preserved by CGWH products, so it will not hold for many constructions which -one would like to have results as in 1. or 2. for. -A sufficient condition for 2. should be that $F$ is a (retract of a) free -$G$-CW-complex by using model category arguments. However, this also a rather -severe condition that can be hard or impossible to check in practice. - -Thus, I would like to know if there any other known results, preferably with a -reference, that improve the sufficient conditions outlined above. Any results for more arbitrary -topological groups will also be appreciated. - -REPLY [3 votes]: Assertions 1. and 2. hold for spaces that are (compactly generated and) Hausdorff (we still have to require $G$ to be compact Lie). This is Theorems A.10 and A.8 of my recent preprint https://arxiv.org/abs/1612.04267v3. An application, where the "results known to me" from above are not sufficient, can be found there as well. -(It's hard to find out what happens for spaces that are CGWH but not Hausdorff. This is related to the following question: Compact Lie group action on non-Hausdorff (but CGWH) space with Hausdorff quotient)<|endoftext|> -TITLE: Strongly non-Ramsey order type in polarized partition problems -QUESTION [5 upvotes]: It is known (a theorem of Komjáth and Hajnal) that it is consistent (by adding a Cohen real to a universe where there is no Suslin line) that there exists an order type $\theta$ such that for any other order type $\eta$, there exists a bad coloring $f: [\eta]^2 \to \omega$ such that for any order preserving embedding $g: \theta \to \eta$, $f''[g''\theta]^2=\omega$, i.e. $\eta \not \rightarrow [\theta]_\omega^2$, in particular $\eta \not \rightarrow (\theta)_\omega^2$ (there are more results but enough for now). -I'm wondering if there are references about similar phenomenon for polarized partition relation, more concretely: do there exist order types $\theta_0, \theta_1$ and cardinal $\gamma$ such that for any order types $\eta_0, \eta_1$, there exists $f: \eta_0\times \eta_1 \to \gamma$ such that for all order-preserving embeddings $g_0: \theta_0 \to \eta_0, g_1: \theta_1\to \eta_1$, $f''[g_0''\theta_0\times g_1''\theta_1]=\omega$ (or just not constant), i.e. ${\eta_0 \choose \eta_1} \not \rightarrow {\theta_0\choose \theta_1}_\gamma^{1,1}$. Any pointers would be appreciated. Thanks in advance. - -REPLY [2 votes]: We can't find such pair of bad order types, aka there indeed is some Erdös-Rado phenomenon in the polarized partitions with respect to linear orderings. Indeed given $\gamma$ number of colors and $\theta$, $\psi$ we can find large saturated dense linear orders $\alpha, \beta$ such that for any $f: \alpha\times \beta \to \gamma$, there exists $X\subset \alpha, Y\subset \beta$ such that $tp(X)=tp(\alpha), tp(Y)=tp(\beta)$ such that $f''X\times Y$ is constant. -To elaborate: Given $\gamma$ the number of colors and order types $\theta, \psi$, given let $\lambda$ be regular a lot larger than $|\theta|, |\psi|, \gamma$, let $\beta$ be a $\lambda$-saturated dense linear order. Let $\lambda'$ be a lot larger than $2^{|\beta|}, \gamma$ and let $\alpha$ be $\lambda'$-saturated dense linear ordering, we claim that ${\alpha\choose \beta}\to {\theta\choose \psi}^{1,1}_{\gamma}$. Given a coloring $f: \alpha \times \beta\to \gamma$, for each $i\in \alpha$, consider $f_i: \beta\to \gamma$. - -There exists a monochromatic under $f_i$ sub order of $\beta$ that is $\lambda$-saturated. - Why? Suppose not, then induct on $i\in \gamma$ to build a cut. Stage 0, there exists $\eta_0,\eta_1<\lambda$ and $A_0=\{a_j: j<\eta_0\} -TITLE: Abstract treatment of multivariate calculus relevant for optimization -QUESTION [8 upvotes]: After studying the basics of (convex) optimization, I've become convinced there's sometimes a conceptual benefit in thinking of quantities like gradients etc. in a coordinate-free way, and keeping track of the spaces they naturally come from (e.g., while oftentimes we pick a basis $\mathbb{R}^n$ and think of the gradient as a vector also in $\mathbb{R}^n$, one can define gradients abstractly and see they naturally come from the dual $V^*$; another example would be optimizational duality). -I'm looking for a text that emphasizes such abstract points of view on the aspects of multivariate analysis relevant to (convex) optimization (no upper limit to how abstract (e.g. categories are fine)... I guess). Suggestions that are not obviously related to (convex) optimization are also welcome. - -REPLY [3 votes]: I can even top Dirk's reference: - -Constantin Zălinescu, MR1921556, Convex analysis in general vector spaces, World Scientific Publishing Co., Inc., River Edge, NJ, 2002. - -(Although I draw the line at metric spaces...)<|endoftext|> -TITLE: Integer-valued power sums -QUESTION [12 upvotes]: Suppose I have a positive number $d \in \mathbb{R}$ and a sequence of numbers $a_n \in [0,d]$ for $n \in \mathbb{N}$ with the following properties -$$ -\sum_{i=1}^{\infty} a_i^r \in \mathbb{Z} -$$ -for all $r \in \mathbb{N}$ and -$$ -\sum_{i=1}^{\infty} a_i \leq d \ . -$$ - - -Does it follow that only finitely many of the $a_i$ are non-zero? - - -Note that it does not follow that $a_i \in \mathbb{Z}$ as the sequence $a_1 = 2 + \sqrt{2}$, $a_2 = 2 - \sqrt{2}$, $a_k = 0$ for $k > 2$ with $d=4$ shows. This sequence also shows that the power sums can be unbounded. - -REPLY [5 votes]: OK. I've finally got the time to correct my deleted answer. If anyone doesn't like complex analysis, this answer only uses real analysis. -Let $f(x)=\prod_{n=1}^\infty (1+a_nx)=\sum_{n=0}^\infty e_nx^n$, where $e_n$ is the elementary symmetric polynomial in $(a_n)$. Then as observed in the comments, $e_n\in\mathbb{Z}/n!$, so if infinitely many $a_n>0$, then $e_n\ge1/n!$, which implies $f(x)\ge e^x$, or $\log f(x)\ge x$ for $x>0$. On the other hand, we can pick $N$ such that $\sum_{n=N}^\infty a_n<1/2$. Then $\sum_{n=N}^\infty \log(1+a_nx) -TITLE: Automorphism of restriction of scalars -QUESTION [6 upvotes]: Let $E/F$ be a finite field extension. Let $G_E$ be a reductive linear algebraic group defined over $E$ and let $G=\mathrm{R}_{E/F}G_E$ be the Weil restriction of scalars. Then $G$ is a linear algebraic group defined over $F$ and $\mathrm{R}_{E/F}$ is a functor from the category of linear algebraic groups defined over $E$ to the category of linear algebraic groups defined over $F$. -Consider $\mathrm{Aut}_E(G_E)$, the group of automorphisms of $G_E$ as an algebraic group over $E$. The group $\mathrm{Aut}_F(G)$ is defined similarly. The functor provides a homomorphism -$$ -\mathrm{R}_{E/F}:\mathrm{Aut}_E(G_E)\to\mathrm{Aut}_F(G). -$$ -My question is, what is the quotient -$$ -\Gamma=\mathrm{Aut}_F(G)/\mathrm{R}_{E/F}(\mathrm{Aut}_E(G_E))? -$$ -If $G_E=\mathrm{GL}(n)$, then I would like to know in particular that $\Gamma$ is the Galois group $\mathrm{Gal}(E/F)$. To see concretely how the Galois group acts, we can construct the restriction of scalars by choosing an $F$-basis for $E$ that provides a map from $G_E(E)$ to $\mathrm{GL}_N(F)$ where $N=n[E:F]$. The image of this map defines an algebraic group over $F$ which represents the restriction of scalars and the Galois group action on $G_E(E)$ can be transferred to an action on the restriction of scalars. - -REPLY [9 votes]: You surely meant to assume the finite extension of fields $E/F$ is separable (otherwise ${\rm{R}}_{E/F}(H)$ is never reductive for a smooth connected affine $E$-group $H \ne 1$). Also, this is one of those cases where more generality clarifies the situation. The reason I say this is that by limiting yourself only to Weil restriction from a field over $F$ rather than more generally from a finite etale $F$-algebra (i.e., a product of several fields over $F$, all finite and separable over $F$), you prevent yourself from accessing the technique of scalar extension on $F$ to simplify computations. -So let's pose the situation more broadly as follows. Let $F$ be a field, $E$ and $E'$ nonzero finite etale $F$-algebras, and $G$ and $G'$ smooth affine groups over $E$ and $E'$ respectively such that their fibers over the factor fields of $E$ and $E'$ are connected reductive and non-trivial. We want to describe $F$-isomorphisms $f: {\rm{R}}_{E/F}(G) \simeq {\rm{R}}_{E'/F}(G')$ in terms of more concrete data, perhaps under some hypotheses on the groups (and in particular to address the case when $E = E'$ is a field and $G = G' \ne 1$). -Given -an $F$-algebra isomorphism $\alpha:E \simeq E'$ and a group isomorphism $\varphi:G \simeq G'$ over $\alpha$ there is an evident associated $F$-group isomorphism ${\rm{R}}_{\alpha/F}(\varphi): {\rm{R}}_{E/F}(G) \simeq -{\rm{R}}_{E'/F}(G')$. So we can ask the (easy) question of whether the pair $(\alpha, \varphi)$ is uniquely determined by -the $F$-isomorphism ${\rm{R}}_{\alpha/F}(\varphi)$, and the more serious question of whether every $f$ arises from such a pair. The first key point, and the reason for recasting the setup in terms of the generality of (nonzero) finite etale $F$-algebras rather than only fields (of finite degree and separable) over $F$ is that if $F'/F$ is any extension field (such as a separable closure, or simply a big enough finite Galois extension, or whatever) then -$${\rm{R}}_{E/F}(X) \otimes_F F' \simeq {\rm{R}}_{(E \otimes_F F')/F'}(X \otimes_E (E \otimes_F F'))$$ -for any affine $E$-scheme $X$ of finite type. Note that if $E$ were a field then $E \otimes_F F'$ is typically not a field, but that since $E$ is finite etale over $F$ at least $E \otimes_F F'$ is finite etale over $F'$. More specifically, a nonzero finite etale $F$-algebra $E$ always has the form $E = \prod E_i$ for fields $E_i$ that are finite separable over $F$, any scheme $X$ over ${\rm{Spec}}(E) = \coprod {\rm{Spec}}(E_i)$ always has the form $X = \coprod X_i$ where $X_i$ is the $E_i$-fiber of $X$, and for $X$ affine and finite type over $E$ we have -$${\rm{R}}_{E/F}(X) \simeq \prod {\rm{R}}_{E_i/F}(X_i);$$ -this makes contact back with the more familiar world of Weil restriction through field extensions, but the language of finite etale $F$-algebras is much more efficient for bookkeeping purposes with the notation when we want to bring in ground field extensions (as we will do below!). -Let's now come back to the question of whether an $F$-isomorphism -$f: {\rm{R}}_{E/F}(G) \simeq {\rm{R}}_{E'/F}(G')$ having the form ${\rm{R}}_{\alpha/F}(\varphi)$ uniquely determines the $F$-algebra isomorphism $\alpha$ and the group isomorphism $\varphi$ over $\alpha$. -Since we're only asking about uniqueness of such a pair, not existence, it suffices to check after making a ground field extension! And we were careful to set up our entire framework in a manner that is not destroyed by such an operation, so for the purpose of proving uniqueness we can apply ground field extension from $F$ to $F_s$ so that $F$ is separably closed and hence all finite etale $F$-algebras are split! -More to the point, now (for the purpose of proving uniqueness) we can assume as $F$-algebras that $E = F^I$ and $E' = F^J$ for non-empty finite sets $I$ and $J$, so $\alpha$ is nothing other than a bijection of sets $\tau:I \simeq J$ (with $\alpha((x_i)) = (x_{\tau^{-1}(j)})$). Likewise, now $G = \coprod G_i$ and $G' = \coprod G'_j$ with $G_i$ and $G'_j$ -all non-trivial, and -$\varphi$ is nothing other than a collection of $F$-isomorphisms -$\varphi_i:G_i \simeq G'_{\tau(i)}$ for $i \in I$. Since ${\rm{R}}_{E/F}(G) = \prod G_i$ and ${\rm{R}}_{E'/F}(G') = \prod G'_j$, -the uniqueness question comes down to the concrete problem of whether the product isomorphism of $F$-groups -$$\prod_{i \in I} G_i \simeq \prod_{j \in J} G'_j$$ -defined by $(g_i) \mapsto (\varphi_{\tau^{-1}(j)}(g_{\tau^{-1}(j)}))$ uniquely determines both $\tau$ and the collection of $\varphi_i$'s. -Since all $G_i$ and $G'_j$ are nontrivial, the uniqueness of both is immediate by chasing the image of $G_i$ under the product isomorphism. -That settles the uniqueness aspect, so now we can turn to the much more interesting existence aspect. Obviously this cannot be affirmative in general (just imagine that $G$ is itself a Weil restriction, and consider that a composition of Weil restrictions is a Weil restriction for the composite extension). So the good result is as follows: -Theorem. Assume that the connected reductive fibers of $G$ and $G'$ over the respective factor fields of $E$ and $E'$ are semisimple, absolutely simple, and simply connected. Then any $F$-group isomorphism -${\rm{R}}_{E/F}(G) \simeq {\rm{R}}_{E'/F}(G')$ has the form ${\rm{R}}_{\alpha/F}(\varphi)$ for a $($uniquely determined$)$ pair -$(\alpha, \varphi)$ as above. The same holds with "adjoint type" in place of "simply connected". -The "absolutely simple" hypothesis rules out the obstruction caused by internal Weil restrictions as noted above, the "semisimple" condition is needed prior to making any meaningful/useful notion of "absolutely simple", and the "simply connected" (or alternatively "adjoint type") hypothesis ensures that the absolute root datum (not just root system!) is a direct product of those for its irreducible components. -For a proof of the above Theorem (including a reference to its historical antecedent in the older language of Weil restriction only through field extensions, at the cost of slightly heavier notation), see Proposition A.5.14 in the book Pseudo-reductive Groups. -Now let's finally come back to the original question concerning automorphisms of $H := {\rm{R}}_{E/F}({\rm{GL}}_n)$ for a finite separable extension of fields $E/F$ and $n \ge 1$. For $n = 1$ it is really hopeless to say anything in general (since passing to the Cartier dual Galois lattice ${\rm{Ind}}_{\Gamma_F}^{\Gamma_E}(\mathbf{Z})$ shows that (if $E \ne F$) there are a lot of automorphisms, as we can see by first considering the situation after scalar extension to $\mathbf{Q}$). So let's assume $n \ge 2$. Then the derived group $\mathscr{D}(H)$ coincides with ${\rm{R}}_{E/F}({\rm{SL}}_n)$, and we have -$$H \simeq (\mathscr{D}(H) \times Z_H)/\mu = ({\rm{R}}_{E/F}({\rm{SL}}_n) \times {\rm{R}}_{E/F}({\rm{GL}}_1))/{\rm{R}}_{E/F}(\mu_n)$$ -where $Z_H$ is the (scheme-theoretic) center of $H$ and $\mu = \mathscr{D}(H) \cap Z_H -= Z_{\mathscr{D}(H)} \simeq {\rm{R}}_{E/F}(\mu_n)$. -Thus, any $F$-automorphism $f$ of $H$ is exactly the data of an $F$-automorphism $f_1$ of $\mathscr{D}(H) = {\rm{R}}_{E/F}({\rm{SL}}_n)$ and an $F$-automorphism $f_2$ of $Z_H = {\rm{R}}_{E/F}({\rm{GL}}_1)$ that coincide on their overlap ${\rm{R}}_{E/F}(\mu_n)$. The preceding stuff shows that $f_1$ is uniquely of the form ${\rm{R}}_{\alpha/F}(\varphi)$ for an $F$-algebra automorphism $\alpha$ of $E$ and a group automorphism $\varphi$ of ${\rm{SL}}_n$ over $\alpha$. Our knowledge of the automorphism group of ${\rm{SL}}_n$ over fields shows that the effect of $\varphi$ on $\mu_n$ is (up to the effect of $\alpha$) either trivial or inversion, and both options extend to automorphisms of ${\rm{GL}}_1$ over $\alpha$! Thus, whatever $f_1$ we might want to consider, a compatible $f_2$ can always be found. -Consequently, we see that the obstruction to an affirmative answer to your question is exactly the group of $F$-automorphisms of the induced torus ${\rm{R}}_{E/F}({\rm{GL}}_1)$ restricting to the identity on its $n$-torsion subgroup ${\rm{R}}_{E/F}(\mu_n)$. Since $n > 1$, whenever $E \ne F$ we can make infinitely many such extra automorphisms. I conjecture that you posed the question for ${\rm{GL}}_n$ under the mistaken belief that it would make the problem easier (whereas now you see that the central torus actually makes the situation harder), and that the positive answer for ${\rm{SL}}_n$ (and more generally as in the Theorem above) is the sort of thing with which you would be satisfied. So I'll leave it to you to mull over the extra junk automorphisms arising from the central torus if you really do need that information (which feels doubtful).<|endoftext|> -TITLE: Does $(\nabla \times F) \cdot F= (\nabla \times F) \cdot \nabla f$ have a solution? -QUESTION [7 upvotes]: A grad student asked me this question during office hours, and I couldn't for the life of me come up with a proof or counterexample: -For a given $F:\mathbb{R}^3 \to \mathbb{R}^3$, does $(\nabla \times F) \cdot F= (\nabla \times F) \cdot \nabla f$ in $R^3$ always have a solution in $f$? - -REPLY [28 votes]: There exist $F$ for which there is no global solution $f$ to the above equation. Here is how you can construct an example: -First, regard $F$ as a vector field on $\mathbb{R}^3$ and consider its dual $1$-form $\phi$. The left hand side of your equation can then be written as the Hodge dual of $\phi\wedge\mathrm{d}\phi$ and the right hand side can be written as the Hodge dual of $\mathrm{d}f\wedge\mathrm{d}\phi$, so your equation becomes -$$ -\phi\wedge\mathrm{d}\phi = \mathrm{d}f\wedge\mathrm{d}\phi = \mathrm{d}\bigl(f\,\mathrm{d}\phi\bigr).\tag1 -$$ -Now suppose that $\phi$ has compact support and that the integral of $\phi\wedge\mathrm{d}\phi$ over $\mathbb{R}^3$ is nonzero. (See below for a construction of such a $\phi$.) Then integrating the ends of (1) over $\mathbb{R}^3$ and using Stokes' Theorem will yield a contradiction. -Now, to construct such a $\phi$, let $x,y,z$ be standard coordinates on $\mathbb{R}^3$, and let $h\not\equiv0$ be a smooth function with compact support on $\mathbb{R}^3$. Set -$$ -\phi = h\,(\mathrm{d}z - y\,\mathrm{d}x).\tag2 -$$ -Then computation shows that $\phi\wedge\mathrm{d}\phi = h^2\,\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z$. Hence, its integral over $\mathbb{R}^3$ is positive, as desired.<|endoftext|> -TITLE: When do surjective morphisms induce injective maps on global sections of coherent sheaves? -QUESTION [5 upvotes]: This question is a follow-up to this question which I asked on MSE. -Let $f: X \rightarrow Y$ be a surjective morphism of schemes, and $\mathscr{F}$ a coherent sheaf on $Y$. Are there conditions we can put on $X$ and $Y$ that imply that the global pullback map $H^0(Y, \mathscr{F}) \rightarrow H^0(X, f^* \mathscr{F})$ is injective? -If $f$ is flat or $\mathscr{F}$ is locally free, this map is always injective. In addition, this answer to the linked question on MSE indicates a proof that, at least when the schemes are noetherian and we either assume that $Y$ is regular or that $X$ and $Y$ are both normal, the same is true when $f$ is any finite map (of course, by miracle flatness, this is only interesting when $X$ is not Cohen-Macaulay or $Y$ is not regular). In the comments and edits to this question, we found counterexamples when $X \rightarrow Y$ is the normalization of a curve and when $X$ is smooth but not connected. Both of these are, to some degree, related to the phenomenon of $f$ "breaking apart infinitesimal neighborhoods". (Both of the modules considered were infinitesimal thickenings of points). -Can this sort of behavior occur when $X, Y$ are connected smooth varieties (ideally over $\mathbb{C}$)? It seems to me like the natural place to find a counterexample is when $f$ is a birational morphism and $\mathscr{F}$ has support on the image of the exceptional locus, but I have not been able to construct one along these lines. - -REPLY [6 votes]: Begin with $Y$ equal to the affine plane, $\text{Spec}\ R$, for $R=k[s,t]$. Let $\overline{f}:\overline{X}\to Y$ be the blowing up of $Y$ at the ideal $\mathfrak{m} = \langle s,t \rangle$. Define $I\subset \mathfrak{m}$ to be the ideal $\langle s,t^2 \rangle$. Define $\mathcal{G}$ to be $\widetilde{R/\mathfrak{m}}$, define $\mathcal{F}$ to be $\widetilde{R/I}$, and define $p:\mathcal{F}\to \mathcal{G}$ to be the natural surjection. The pullback $\overline{f}^*\mathcal{G}$ is the structure sheaf of the exceptional divisor $E$. The pullback $$\overline{f}^*p:\overline{f}^*\mathcal{F}\to \overline{f}^*\mathcal{G},$$ is an isomorphism except at a single point $q$ on $E$. -Define $X$ to be the open complement of $\{q\}$ in $\overline{X}$. Define $f:X\to Y$ to be the restriction of $\overline{f}$ to $X$. Since we remove a single closed point from the positive dimensional fiber $E$ of $\overline{f}$, the morphism $f$ is still surjective. On $X$, the natural homomorphism $$f^*p:f^*\mathcal{F}\to f^*\mathcal{G},$$ is an isomorphism. Therefore, the pullback map $H^0(Y,\mathcal{F}) \to H^0(X,f^*\mathcal{F})$ factors through the map $$H^0(p):H^0(Y,\mathcal{F})\to H^0(Y,\mathcal{G}).$$ This map has a one-dimensional kernel.<|endoftext|> -TITLE: What is the descriptive complexity of a set added by Cohen forcing? -QUESTION [9 upvotes]: I want to think of ZFC as not fully determining the powerset of the naturals, because you can add subsets with forcing and otherwise have a lot of control over the cardinality of the powerset of the naturals. But that suggests the question: what subsets does ZFC determine? There is also the related question: how complicated are the sets added by forcing? -Since Turing machines are absolute, models of ZFC should have the same hyperarithmetical sets. So $\Delta_1^1$ is absolute. -Can Cohen forcing add a $\Pi_1^1$ set? If not, what is the descriptive complexity of sets added by Cohen forcing? -This isn't my field so I apologize if this question is ill formed or naive. References would also be appreciated. -Edit: I'm specifically interested in where a set/real $c$ added by forcing fits into the lightface hierarchy. - -REPLY [8 votes]: Jensen proved in this paper that, beginning with $V = L$, it is possible to add a real $a$ by forcing such that $a$ is $\Delta^1_3$ in $L[a]$. -This result is the best possible, in the sense that one can never add $\Sigma^1_2$ or $\Pi^1_2$ reals by forcing. This follows from Shoenfield's Absoluteness Theorem (mentioned already by Asaf in the comments). In fact, Shoenfield's theorem implies that all $\Sigma^1_2$ and $\Pi^1_2$ reals are constructible; so they can't be added by forcing because they're already in the ground model. (For a proof, see Theorem 25.20 and Corollary 25.21 in Jech's book). -I do not know whether anyone has improved on Jensen's result to show that a $\Delta^1_3$ real can be added to any ground model by forcing. -Jensen's forcing is a bit difficult to understand. For an easier-to-understand example of a non-constructible $\Delta^1_3$ real, there is $0^\sharp$. Even if $0^\sharp$ does exist (which is not provable in ZFC -- it is a large cardinal axiom), it cannot be added by any set-sized notion of forcing. So $0^\sharp$ does not answer your question, but it seems related, so I thought I'd share.<|endoftext|> -TITLE: Simultaneous near-best approximation with respect to two norms -QUESTION [5 upvotes]: Suppose that $M$ is a closed infinite dimensional subspace of $L_4(0,1)$ which is also a closed subspace of $L_1(0,1)$. Hence $M$ is isomorphic to $\ell_2$ as a subspace of $L_p(0,1)$ for $1\leq p\leq 4$. Note also that $L_4(0,1)\subset L_1(0,1)$ and $\|f\|_1\leq \|f\|_4$ for each $f\in L_4(0,1)$. -Is it possible to find a constant $C$ such that for each $f\in L_4(0,1)$ we can find $g\in M$ satisfying -$\|f-g\|_q \leq C\inf_{h\in M}\|f-h\|_q$ for both $q=1$ and $q=4$? - -REPLY [3 votes]: It seems that the answer is yes. -Denote by $M_q$ the space $M$ equipped with the $L_q$-norm; by the closednedd condition, it is a Banach space for both $q=1$ and $q=4$. The identical mapping $M_4\to M_1$ is bounded; by the bounded inverse theorem, so is its inverse. Thus there exists $\mu$ such that $\|g\|_4\leq \mu\|g\|_1$ for all $g\in M$; surely $\mu\geq 1$. We claim that $C=2+2\mu$ works. -To show this, take any $f\in L_4(0,1)$ and denote $\alpha_q=\rho_q(f,M)=\inf_{h\in M}\|f-h\|_q$. Choose $g_1,g_4\in M$ such that $\|f-g_q\|_q\leq 2\alpha_q$ and set $g=(\alpha_4g_1+\alpha_1g_4)/(\alpha_1+\alpha_4)$. We claim that $g$ fits the goal. -Firstly, notice that -$$ - \|g_1-g_4\|_1\leq \|g_1-f\|_1+\|f-g_4\|_1\leq 2\alpha_1+2\alpha_4, -$$ -so $\|g-g_1\|_1=\frac{\alpha_1}{\alpha_1+\alpha_4}\|g_4-g_1\|_1\leq 2\alpha_1$. On the other hand, since $\|g_1-g_4\|_4\leq \mu\|g_1-g_4\|_1\leq 2\mu(\alpha_1+\alpha_4)$ we have $\|g-g_4\|_4\leq 2\mu\alpha_4$. Therefore, -$$ - \|f-g\|_1\leq \|f-g_1\|_1+\|g_1-g\|_1\leq 4\alpha_1\leq C\alpha_1 -$$ -and -$$ - \|f-g\|_4\leq \|f-g_4\|_4+\|g_4-g\|_4\leq (2+2\mu)\alpha_4=C\alpha_4, -$$ -as required.<|endoftext|> -TITLE: Is there easy proof for triangle-free two-coloring of planar graphs? -QUESTION [14 upvotes]: By merging two-two color classes, the Four Color Theorem implies that every planar graph can be two-colored such that each color class induces a triangle-free graph. -Is there a simpler proof for this fact? - -REPLY [7 votes]: Thomassen does indeed prove the vertex version, but in a different paper. In fact this paper proves the stronger statement that you can get a coloring without monochromatic triangles from any 2-list coloring. -A bit of googling revealed that originally the claim of the OP was proved in - -Burstein M.I., "The bicolorability of planar hypergraphs", - Sakharth. SSR Mecn. Akad. Moambe 78 (1975), no. 2, 293–296. (MR0396314) - -This paper proves that for any planar triangulation, and any assignment of colors to the outer triangle which is not monochromatic one can extend the coloring to the entire triangulation while avoiding monochromatic triangles.<|endoftext|> -TITLE: Significance of half-sum of positive roots belonging to root lattice? -QUESTION [13 upvotes]: Let $\Phi$ be a (crystallographic) root system and $\Phi^{+}$ a choice of positive roots, with $\Delta$ the corresponding choice of simple roots. So the root lattice of $\Phi$ is just $\mathbb{Z}\Delta$ ($=\mathbb{Z}\Phi^{+}=\mathbb{Z}\Phi$). -Then $\rho := \frac{1}{2}\sum_{\alpha \in \Phi^{+}}\alpha$ (the "half-sum of positive roots", a.k.a. the "Weyl vector") is a distinguished vector associated to $\Phi$ with great significance in representation theory; see e.g. the following Mathoverflow question: What is significant about the half-sum of positive roots?. -A priori, $2\rho \in \mathbb{Z}\Delta$, but in fact it may happen that $\rho \in \mathbb{Z}\Delta$. -Question: is there any representation-theoretic significance to whether we have $\rho \in \mathbb{Z}\Delta$? -With collaborators we have discovered some combinatorial phenomenon that is apparently related to whether $\rho \in \mathbb{Z}\Delta$, and we would like to relate this phenomenon more to representation theory. -It is not hard to work out exactly when $\rho \in \mathbb{Z}\Delta$: - -Type $A_{n}$: Have $\rho \in \mathbb{Z}\Delta$ iff $n$ is even. -Type $B_n$: Always have $\rho \notin \mathbb{Z}\Delta$. -Type $C_n$: Have $\rho \in \mathbb{Z}\Delta$ iff $n \equiv 0,3 \mod 4$. -Type $D_n$: Have $\rho \in \mathbb{Z}\Delta$ iff $n \equiv 0,1 \mod 4$. -Type $E_6$: $\rho \in \mathbb{Z}\Delta$. -Type $E_7$: $\rho \notin \mathbb{Z}\Delta$. -Type $E_8$: $\rho \in \mathbb{Z}\Delta$. -Type $F_4$: $\rho \in \mathbb{Z}\Delta$. -Type $G_2$: $\rho \in \mathbb{Z}\Delta$. - -But I see no particular rhyme or reason to these root systems. - -REPLY [3 votes]: There is no single answer to this question. Here is one observation. -The element $z_G=e^{2\pi i\rho^\vee}$, where $\rho^\vee$ is one-half the sum of the positive co-roots, is a canonical (independent of the choice of positive co-roots) element of $G$, fixed by every automorphism of $G$. If $G$ is simply connected $z_G=1$ if and only if $\rho^\vee$ is in the co-root lattice. For whatever reason this $z_G$ tends to come up. For a slightly more precise list of your cases see Bourbaki, Lie Groups and Lie Algebras, Chapters 7-9, Chapter IX, Section 4, Exercise 13. -For example the Frobenius-Schur indicator of a self-dual finite dimensional representation $V_\lambda$ (telling whether the invariant form is orthogonal or symplectic) is $e^{2\pi i\langle\lambda,\rho^\vee\rangle}=\lambda(z_\rho)$. So $\rho^\vee$ is in the co-root lattice if and only if every such representation of the simply connected group is orthogonal. See [loc. cit. Chapter 9, Section 7] and -Math Overflow: Is there a formula for the Frobenius-Schur indicator of a rep of a Lie group?<|endoftext|> -TITLE: Five cubes, Hadamard and Shklyarskiy -QUESTION [7 upvotes]: Here is my(=bad) translation of from the paper about Shklyarskiy by Golovina: - -... in 1937/38 Dodik presented to school students a complete proof of Abel's theorem about equations of degree 5. He made geometric proof based on a note in the book of Hadamard, "From the existence of five cubes described in the exercise 1051, one can get nonexistence of solution in radicals to the general polynomial equations of degree five". The five cubes are the cubes inscribed in the dodecahedron... - -Do you know any written proof based on this idea? - -REPLY [4 votes]: He might be referring to Felix Klein's Lectures on the Icosahedron and the Solution of Equations of the Fifth Degree -Hadamard's Leçons de la Géométrie Élémentaire is available online. -Vol1 (Plane Geometry) and Vol 2 (Spatial Geometry) -There is a discussion of regular solids, but there's no discussion of algebra or Galois Theory. -These old-fasioned approaches to Galois Theory are really making a comeback. Here are books that I like - -Eugene Dickson Introduction to the theory of algebraic equations -Eugene Dickson Linear groups, with an exposition of the Galois field theory - -Here "Galois Field" is what we would call a "finite field" $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$ but notice that these can have group actions on solutions to equations. I think Felix Klein's discussion is the way to go since he will talk about modular forms and Galois theory. -I not 100% sure he show's the irreducibility of the quintic. The modern streamlined approach using Abstract Algebra is nicely written out by Emil Artin - -Emil Artin Galois Theory - -This is where all the Galois Theory textbooks come from. Klein's approach is very unique. - -In general the icosahedron has a large symmetry group. It is isomorphic to the alternating group $A_5 \simeq \mathrm{PSL}(2, \mathbb{F}_5)$. - -This group could also be represented as fractional linear gransformations with elements in $\mathbb{F}_5$: -$$ \left\{ \frac{az+b}{cz+d} : ad-bc = 1\right\} $$ -and this has an action on the hyperbolic plane $\mathbb{H} = \{ \mathrm{Im}(z) > 0 \}$ - -This is in Doyle-McMullen but $A_5$ also acts on the roots by permuting them. -$$ p(x) = (x - x_0)(x - x_1)(x - x_2)(x - x_3)(x - x_4) = 0 $$ -The five cubes then via Galois Theory correspond to invariants, homogenous polynomals that could be made with the $x_i$. Klein proposes a geometric solution using algebraic geometry. -The algebraic manipulations are quite elaborate. For starts: -\begin{eqnarray*} -p_1 &=& x_0 + \epsilon^1 x_1 + \epsilon^2 x_2 + \epsilon^3 x_3 + \epsilon^4 x_4 \\ -p_2 &=& x_0 + \epsilon^2 x_1 + \epsilon^4 x_2 + \epsilon^1 x_3 + \epsilon^3 x_4 \\ -p_3 &=& x_0 + \epsilon^3 x_1 + \epsilon^1 x_2 + \epsilon^4 x_3 + \epsilon^2 x_4 \\ -p_4 &=& x_0 + \epsilon^4 x_1 + \epsilon^3 x_2 + \epsilon^2 x_3 + \epsilon^1 x_4 \\ -\end{eqnarray*} -and how these act under particular transformations of the complex plane: -\begin{eqnarray*} S: z &\mapsto &\epsilon z\\ - T: z &\mapsto & \frac{-(\epsilon^1 - \epsilon^4) z + (\epsilon^2 - \epsilon^3)}{(\epsilon^2 - \epsilon^3) z + (\epsilon^1 - \epsilon^4)} \end{eqnarray*} -and these two should generate an action of $A_5$ on $\mathbb{H}$. And there's fair bit more which I can't summarize. There are two parameters: -$$ \lambda = - \frac{p_1}{p_2} = \frac{p_3}{p_4} \text{ and } \mu = - \frac{p_2}{p_4} = \frac{p_1}{p_3} $$ -and we can see all these paramters should behave nicely under permutations of the numbers 1-2-3-4-5 -Of course, there will be no solution by radicals. - -See also this blog of Oliver Nash<|endoftext|> -TITLE: How do you get the spectral curve from a Calabi-Yau? -QUESTION [8 upvotes]: In N=2 Quantum Field Theories and Their BPS Quivers by Alim, Cecotti, Córdova, Espahbodi, Rastogi and Vafa the authors give a recipe which constructs from a pair $(C,\phi)$ consisting of a Riemann surface $C$ and a meromorphic quadratic differential $\phi$ on $C$ a (non-compact) Calabi-Yau 3-fold $X$. -Essentially, $X$ is cut out from the total space of the bundle $E=T^*C\oplus T^*C\oplus T^*C$ by the equation $uv=y^2-\phi(z)$ where $z$ is a local coordinate on $C$ and $(u,v,y)$ are coordinates on the fibre $E_z$. An important feature is that $X$ contains the spectral curve of $(C,\phi)$, i.e., the curve $\Sigma\subset T^*C$ which is a branched double cover of $C$ branched at the critical points of $\phi$. -From poking around physics literature that I don't understand, I get the sense that this construction is in some sense post hoc and the natural construction is to go the other way. That is, one should start with (some particular type of) CY $X$ and extract from it the pair $(C,\phi)$. Hence my question: - -Is there a natural recipe which takes a CY $X$ and returns the pair $(C,\phi)$, inverse to the construction above? - -Any insights or references to the literature would be greatly appreciated. - -REPLY [6 votes]: In general there is no way to extract a spectral curve from a Calabi-Yau threefold. -In the study of strings on Calabi-Yaus, one object of interest is the periods, i.e. integrals of the holomorphic top form. We are particularly interested in the behavior of these period as the complex structure of the Calabi-Yau is varied. -One is interested in these questions for honest, compact Calabi-Yaus but non-compact examples are often studied for a variety of reasons, most notably because many questions have explicit answers. -Typically, the non-compact examples in question are local singularities that have been smoothed by complex structure deformation. Importantly, the singularities of interest are restricted to be those that could in principle arise at a finite distance in the moduli space of a compact Calabi-Yau (see this paper). (Practically speaking this often bounds the degrees appearing in equations cutting out the singularity.) -The constructions discussed in the referenced paper are merely a very simple class of examples. Due to the nature of that setup, the problem reduces to studying the Jacobian of a curve (we did NOT invent the construction! It is twenty years old.). The construction can be generalized to build non-compact Calabi-Yaus from n-fold branched covers (embedding general Hitchin systems into the problem). -However many more complicated examples of non-compact Calabi-Yaus exist where the period geometry does not reduce (in any known way) to the Jacobian of a curve. A key phrase to look for in the the literature is "geometric engineering." For some recent classification results, see e.g. this paper, which investigates singularities defined by a complete intersection of two hypersurfaces.<|endoftext|> -TITLE: Can an AW*-algebra be recovered from its lattice of projections? -QUESTION [5 upvotes]: Can an AW*-algebra be recovered (up to Jordan isomorphism) from its lattice of projections? This is possible in the commutative/Boolean case. - -REPLY [7 votes]: Yes, this follows from Theorem 4.2 in Dye's Theorem and Gleason's Theorem for AW*-algebras by Jan Hamhalter. -Link: https://arxiv.org/abs/1408.4597 -Here the statement is: given an AW*-algebra $A$ without type I$_2$ summands, and an AW*-algebra $B$, then every orthocomplement-preserving order morphism $\varphi:P(A)\to P(B)$ can be extended to a Jordan *-homomorphism $A\to B$, i.e., the complexification of a Jordan homomorphism from the self-adjoint part of $A$ to the self-adjoint part of $B$. -You can get rid of the type I$_2$ condition by splitting $A$ in a type I part and a type II/III part and use that the projections of a type I AW*-algebra completely determine type I AW*-algebras up to *-isomorphism, which is an observation of M. Ozawa. The type II/III part follows from Hamhalter's Theorem. Details can be found in Corollary 9.2.9 of http://repository.ubn.ru.nl/handle/2066/158429<|endoftext|> -TITLE: Formula for $U(N)$ integration wanted -QUESTION [11 upvotes]: Before you jump on the "duplicate" buttom, let me say that I do not want to hear about Weingarten calculus and I do not want to see a character of the symmetric group. -What I would like is a formula for the (normalized) Haar measure integral -$$ -\int_{U(N)} g_{i_1 j_1}\cdots g_{i_n j_n} {\bar{g}}_{k_1 l_1}\cdots {\bar{g}}_{k_n l_n}\ d\mu(g) -$$ -of the form -$$ -\left.\mathcal{D}\ g_{i_1 j_1}\cdots g_{i_n j_n} {\bar{g}}_{k_1 l_1}\cdots {\bar{g}}_{k_n l_n}\right|_{g=\bar{g}=0} -$$ -where $\mathcal{D}$ is an explicit constant coefficient differential operator of infinite order. -Of course, in this formula the $g$'s and $\bar{g}$'s are treated are $2N^2$ completely unrelated formal variables. -As an example of what I would like, in the case of $SU(N)$ and $\bar{g}$-free monomials -$$ -\mathcal{D}=\sum_{n=0}^{\infty} \frac{0!1!\cdots (N-1)!}{n!(n+1)!\cdots(n+N-1)!}\ ({\rm det}(\partial g))^n -$$ -works. -As per the "Additional remark" in my second answer to this MO question, I had a vague recollection of seeing a math-physics paper with -such a formula, but maybe my memory is faulty. So I think it's better to ask the experts. - -REPLY [6 votes]: To expand on my comments, this paper https://arxiv.org/pdf/hep-th/9209083v2.pdf by Shatashvili deals -with ``correlation functions'' of Haar unitary matrices of the form -$$ -\int_{U(N)}^{} d\mu(U) e_{}^{tr(UAU_{}^{-1}B)} -U_{i_1j_1}^{}\bar U_{k_1\ell_1}^{}\ldots -U_{i_mj_m}^{}\bar U_{k_m\ell_m}, -$$ -and provides a certain combinatorial formula for these. Then setting $A=0$ would probably -recover what you're asking about. -The same correlation functions (and an alternative formula for them) are also discussed in -https://arxiv.org/pdf/hep-th/0502041.pdf<|endoftext|> -TITLE: Status of Hodge conjecture over subrings of $\mathbb{C}$ -QUESTION [6 upvotes]: Does there exist some subring $k \subset \mathbb{C}$ such that the following assertion holds? - -($k$-Hodge conjecture) For each nonsingular algebraic variety $X$ over $\mathbb{C}$, and each $q = 0, 1, \ldots, \dim_{\mathbb{C}}(X)$, each class $\mathfrak{z} \in H^{2q}(X; k) \cap H^{q,q}(X)$ is a $k$-linear combination of classes of algebraic cycles. - -Atiyah-Hirzebruch proved in 1961 that $k \neq \mathbb{Z}$. As commented by Ben Wieland below, the argument of Atiyah-Hirzebruch also shows that $k \neq \mathbb{Z}_{(p)}$ for any prime $p$. -Remark: If the millenium problem is true, then we may take $k = \mathbb{Q}$. -(The above post has been slightly edited from the original). - -REPLY [10 votes]: The $k$-Hodge conjecture is false for any $k$ containing an irrational real $\alpha$. -Indeed we may clearly assume $\alpha$ is negative. Consider the elliptic curve $E = \mathbb C / \langle 1, \sqrt{\alpha} \rangle$. Let $e_1$ and $e_2$ be a basis of $H^1(E,\mathbb Z)$ such that the map to $H^{1,0}(E)$ sends $e_1$ to $1$ and $e_2$ to $\sqrt{\alpha}$. Then in $H^1(E,k) \otimes H^1(E,k) \subset H^2(E \times E, k)$, consider the class $$\alpha( e_1 \otimes e_1) - (e_2 \otimes e_2)$$ -In the natural projection to $H^{2,0}$, it is sent to $\alpha(1) - (\sqrt{\alpha})^2=0$. In the natural projection to $H^{2,0}$, which is the complex conjugate of that, it is sent to $\alpha (1) - (-\sqrt{\alpha})^2=0$. So it lies in $H^{1,1}(E\times E) $. -But because $E$ is not CM (because $\alpha$ is irrational $\sqrt{\alpha}$ is not a quadratic irrational), the only algebraic cycle that contributes to $H^1(E) \otimes H^1(E)$ is the diagonal, which has class $e_1 \otimes e_2 - e_2 \otimes e_1$, and this class is not a multiple of it. - -On the other hand, assuming the $\mathbb Q$-Hodge conjecture, we obtain the $k$-Hodge conjecture for any imaginary quadratic field $k$. Indeed, for $k= \mathbb Q(\sqrt{-D})$, $H^{2q}(X,\mathbb Q(\sqrt{-D}))= H^{2q}(X,\mathbb Q) + \sqrt{-D}H^{2q}(X,\mathbb Q)$. Because complex conjugation exchanges $H^{q,p}$ and $H^{p,q}$, it fixes $H^{q,q}$, so if a class in $H^{2q}(X,\mathbb Q(\sqrt{-D}))$ lands in $H^{q,q}$, both its real and imaginary parts do as well, so under the Hodge conjecture they are $\mathbb Q$-linear combinations of algebraic cycles, and the original is clearly a $\mathbb Q(\sqrt{-D})$-linear combination of algebraic cycles. -I'm not sure about number fields of degree at least $3$ that contain no real irrationals.<|endoftext|> -TITLE: Inner product on $V_{-\rho}$? -QUESTION [6 upvotes]: Prologue. To $M^n$ a compact real manifold with frame bundle $F$ (a principal $GL_n$ bundle), we associate a line bundle using the representation $M\mapsto \sqrt{|\det M|}$, the bundle of half-densities. Then the space of sections of this has a canonical inner product (multiply then integrate), invariant under all diffeomorphisms of $M$. If $M$ is oriented then one can use the representation $M\mapsto \sqrt{\det M}$, the "half-form" bundle. -Present day. -I was thinking about "parabolic category $\mathcal O$" and realized that the most parabolic is the one containing (only) the Verma module $L(-\rho)$ with highest weight $-\rho$. The corresponding Borel-Weil line bundle is the -half-form bundle, so its space of sections carries a canonical inner product... except there are no holomorphic ones. Or indeed any sheaf cohomology. - -What special structures do the very special representation $L(-\rho)$ bear? Is there any shadow of the inner product that the finite-dimensional irrep would have, if it existed? - -EDIT: "If it existed" was a lousy way to talk about that finite-dimensional irrep. As Ben points out, Borel-Weil-Bott makes clear that it does exist but is $0$. - -REPLY [2 votes]: I'm not totally sure what will satisfy you on this account. Of course you pointedly restricted yourself to the reals, but some complex manifolds do have holomorphic half-form bundles, and the flag manifold is one of them. Of course, the half-form line bundle $\omega_X^{1/2}$ "should" have sections $V(-\rho)$ (actually, this should be the cohomology in all degrees) but in order to make sense of this you have to declare that $V(-\rho)=0$. It's still true that multiplying two sections of this bundle gives a (now holomorphic) top form, but there just are no global sections. -One manifestation of this that is of interest to geometric representation theorists is that the ring and sheaf $D_{\omega^{1/2}}$ of holomorphic differential operators twisted in half-forms on any complex manifold (which make sense even if half-forms are not a well-defined line bundle) is self-opposite (the sheaf never is for other twists, sometimes the ring is various global reasons). This makes it often the most natural twist to work with, and it plays a important role in D-module theory.<|endoftext|> -TITLE: An orbit of symmetric polynomials -QUESTION [15 upvotes]: Consider the ring of polynomials $R:=\mathbb{Z}[x_1,x_2,x_3]$. Define the operators $E, I:R\rightarrow R$ by $Ef(x_1,x_2,x_3)=f(x_1-1,x_2,x_3)$ and the identity $If=f$. -Let $\mathcal{L}:R\rightarrow R$ be the operator given by -$$\mathcal{L}f=[(x_1+x_2)(x_1+x_3)E-x_1^2I]f.$$ -Let $1$ stand for the constant function $f(x_1,x_2,x_3)=1$ and $\mathcal{L}^2f=\mathcal{L}(\mathcal{L}f)$, etc. - -CLAIM. Experiments suggest that $\mathcal{L}^n1$ is always a symmetric polynomial in $R$. Any proof? - -EDIT. This has found a resolution (see Pietro Majer's answer). -For example, $\mathcal{L} 1=e_2$ and $\mathcal{L}^21=e_2^2-e_1e_2+e_3$ where $e_1=x_1+x_2+x_3, e_2=x_1x_2+x_1x_3+x_2x_3, e_3=x_1x_2x_3$ are the standard elementary symmetric polynomials. - -QUESTIONS. (EDIT) These did not find a definitive answer (apart from Brendan McKay's evidence and argument). -(1) Are there other orbits of symmetric polynomials under $\mathcal{L}$? -(2) Are there other non-trivial operators with similar property over $R$? -(3) What about over rings of many more variables? - -REPLY [7 votes]: Noting Pietro's finite check, I can report that all symmetric polynomials $p(x_1,x_2,x_3)$ up to degree 18 inclusive such that $\mathcal{L}p$ is symmetric are linear combinations of $1,\mathcal{L}1,\mathcal{L}^21,\ldots\,$. -This strongly suggests that there are no others. -An observation that might lead to an elementary proof is that, up to degree 18, all polynomials $p$ such that both $p$ and $\mathcal{L}p$ are symmetric are uniquely determined by the coefficients of the powers of $e_2$ (i.e. the terms in the representation in the base $\{e_1,e_2,e_3\}$ which have the form $c e_2^k$).<|endoftext|> -TITLE: Special fiber of the Néron model of the generalized Jacobian of a singular curve -QUESTION [6 upvotes]: Let $C$ be a curve over $\mathbf{Q}_p$ (or a finite extension) whose minimal regular model $\mathcal{C}$ over $\mathbf{Z}_p$ has a "nice" special fiber (maybe singular, with at most ordinary double singularities). Let $S \subset \mathcal{C}(\mathbf{Z}_p)$ be a finite set of nonsingular points. Let $J$ be the generalized Jacobian of $C$ relative to $S$. Do we have a description of the special fiber of the Néron model of $J$ in terms of the special fiber of $\mathcal{C}$? -In the case of the "absolute" Jacobian, Raynaud gave a description of the toric part in terms of the dual graph of the special fiber of $\mathcal{C}$. But intuitively, taking the relative Jacobian means gluing the points of $S$ together, so the the special fiber of $\mathcal{C}$ would now have multiple ordinary points so I don't know if we just have to add a "multiple" loop to the dual graph or what? -I'm interested in the case where $C$ is a classical modular curve with split multiplicative reduction at $p$ and $S$ is the set of cusps. - -REPLY [6 votes]: I'm afraid this should really be a comment not an answer, since I am giving a reference rather than a description. But a full description is a bit long. -The description you want is indeed known, and is very nicely written up in the later chapters of the book `Néron models' by Bosch et al. For example, theorem 9.5.4 relates the Néron model to a certain quotient of the relative Picard. Theorem 9.6.1 tells you what the group of connected components is. The identity component itself is also discussed in 9.5.4. The assumption that the singularities be e.g. nodal is not needed, the results are very general. The main thing you want is something like the geometric multiplicities of the components of the special fibre having gcd = 1, which is for example satisfied in the nodal case, or more generally if there exists an étale quasi-section. -I'm afraid I don't quite follow your paragraph about loops, but I hope you can resolve things from the reference - please ask if not. Certainly there are close connections between the combinatorics of the graph and the component group in the nodal case; this can be extracted quite easily from the references given above, though perhaps there is somewhere else it is written more simply. Roughly, the toric rank is the rank of the homology of the graph (see e.g. example 9.2.8 of [loc.cit.]). The component group is the `critical group' of the graph. This goes by many names, you can extract a definition from lemma 9.5.9. I like to think of it in terms of electrical resistance in a circuit, but the people who referee my papers seem to disagree ;-). -In the special case of modular curves, you might also for example be interested in the paper `On Néron models, divisors and modular curves' by -Bas Edixhoven, which makes some parts more explicit. -Thanks to Owen Biesel for pointing this question out to me. -Added later -Now to try to allow for this generalised jacobian on the generic fibre. Some parts of the above description go through OK, but for others I get a bit stuck, as we shall see. -First, I find it helpful to relate the generalised jacobian to the relative Picard of another curve. You have the curve $\mathcal C/Z_p$, and a finite set $S$ of smooth sections. Let $\mathcal C_S$ be the curve obtained by `glueing these sections together' - more formally this should be constructed as a pushout, see for example [Ferrand, CONDUCTEUR, DESCENTE ET PINCEMENT] for a nice general theory. This $\mathcal C_S/Z_p$ has non-smooth locus consisting of exactly one section (with some multiplicity) and a bunch of points lying over thickenings of the closed point in $Z_p$ (the non-smooth locus is always finite unramified in the ordinary double point case). In particular, the generic fibre is irreducible, with exactly one node. Write $P$ for the total-degree-zero part of the relative Picard of $C_S/Z_p$. Then the generic fibre of $P$ is just the degree zero part of the relative Pic of the generic fibre of $C_S$, in particular it is a smooth connected group scheme over $Q_p$. In fact, it is exactly the generalised jacobian in the sense of Serre --- this can be deduced from chapter V of [Serre, Algebraic groups and class fields]. -OK, so we have this group scheme P, whose generic fibre is the generalised jacobian. I want to relate $P$ to the N\'eron model. Note that $P$ is separated over $Z_p$ if and only if the special fibre of $C_S$ (or equivalently of $C$ is irreducible. So in general it is not separated, so is certainly not the N\'eron model. However, this is quite easy to fix - one just takes the largest separated quotient (equivalently, divide out by the closure of the unit section), to obtain a separated group scheme, let's call it $Q$. Is this $Q$ now the N\'eron model? Note that it is not obvious, since $C_S$ is not regular, so the usual argument about extending line bundles fails. -In general I do not know the answer. However, in the nodal case (where $S$ has cardinality 2) the answer is indeed yes, $Q$ is the N\'eron model *added: if one makes some suitable blowups of he special fibre first. *. This follows from a recent result of Giulio Orecchia, [Semi-factorial nodal curves and Néron models of jacobians, https://arxiv.org/abs/1602.03700 ]. The point is that all the edges in the dual graph have thickness 1 or infinity (since you started with a regular model $\mathcal C$), so Orecchia's `circuit coprime' condition is automatically satisfied once one makes some blowups to avoid degenerate loops. Note that of course $Q$ is actually a N\'eron lft-model, since it need not be of finite type. -The intersection of the closure of the unit section with the fibrewise connected component of identity will be trivial, so the identity component of the N\'eron model $Q$ will just be Pic^0 of the special fibre of $\mathcal C_S$, which has a nice explicit description as discussed above. -Even in the nodal case, this leaves the question of what the component group is. It should be possible to read this off from the dual graph of the special fibre together with a labelling of the edges by the thickness of the singularity there. Some variation of the constructions in [BLR, Neron models] should suffice for this, but I have never thought about how to do it. -It seems to me that the more difficult point is to relate this largest separated quotient $Q$ of $P$ to the N\'eron model. This is what is done by Orecchia in the case where $S$ has two elements. It seems quite possible his methods would generalise to this case, but I am not sure - maybe you should ask him...<|endoftext|> -TITLE: If a quotient ring is a projective module then the ideal is principal -QUESTION [5 upvotes]: I've been studying projective modules in Rotman, as well as the topic of localization. Now on the Wikipedia article about them, there's an example of a locally free module that is not projective. -The module in question is $R/I$ where $R$ is a direct product of countably infinite copies of $\mathbb{F}_2$ and $I$ is a direct sum of countably infinite copies of $\mathbb{F}_2$. -I understand why it is locally free, but in order to explain why it is not projective they mention the following theorem: If $I$ is an ideal of a commutative ring $R$ such that $R/I$ is a projective $R$-module, then $I$ is a principal ideal. -I'm not sure how to prove this (more general) theorem. I feel like I'm missing an obvious map to show that $I$ not principal implies $R/I$ not projective. - -REPLY [9 votes]: Let $\pi:R\to R/I$ be the natural projection map. This is an $R$-module homomorphism (as well as a ring homomorphism). If $R/I$ is projective, then this map splits. Call such a splitting $\varphi:R/I\to R$. So we have $R= \varphi(R/I)\oplus \ker(\pi)$ (as internal direct sums of $R$-modules). But $\ker(\pi)=I$, and this shows that $I$ is cyclic. [In particular, $I$ will be generated by the idempotent of $R$ which corresponds to the direct sum decomposition.]<|endoftext|> -TITLE: Hankel determinants of binomial coefficients -QUESTION [16 upvotes]: For $\{h_{n}\}_{n=0}^{\infty}$ a real sequence, denote by $H_{n}$ the $n\times n$ Hankel matrix of the form -$$ -H_{n}:=\begin{pmatrix} - h_{0} & h_{1} & \dots & h_{n-1}\\ - h_{1} & h_{2} & \dots & h_{n}\\ - \vdots & \vdots & \ddots & \vdots\\ - h_{n-1} & h_{n} & \dots & h_{2n-2} - \end{pmatrix}. -$$ -Just by a mathematical curiosity, I am interested in formulas for $\det H_{n}$ where -$$ -h_{n}=\binom{pn}{n} -$$ -where $p\geq2$ is an integer. -If $p=2$, the Hankel determinant (also called the Hankel transform) of central binomial coefficients is well-known and is not very difficult to evaluate. It reads -$$ -\det H_{n}=2^{n-1}. -$$ -For $p=3$, the formula (and its derivation) is more complicated, though it is also known that -$$ -\det H_{n}=3^{n-1}\left(\prod_{i=0}^{n-1}\frac{(3i+1)(6i)!(2i)!}{(4i)!(4i+1)!}\right), -$$ -see, for example, this paper for the proof. -Nevertheless, if $p\geq4$, I have not found any result about the corresponding Hankel determinant, nor I was able to even guess a form of the possible identity for $p=4$ while experimenting with Mathematica. -My questions are as follows: - -Is there something known about $\det H_{n}$ for $p\geq4$? -Is anybody able to at least guess a formula for the case $p=4$? -Is there any reason which might indicate that an explicit formula for $\det H_{n}$, with $p\geq4$, (like the one for $p=3$) need not exist? - -Remark: I would like to remark that I know Krattenthaler's survey on advanced determinant calculus but I didn't find an answer therein. - -REPLY [3 votes]: This is not an answer. -Although there might be no "closed formula" for the case $p=4$, I wish to add the following to the case $p=2$ for which the determinants evaluate to $2^{n-1}$. -Suppose we decide to compute the determinant of the $n\times n$ Hankel matrix with left-most corner is at any position; in detail, consider -the matrix $H_n(a)$ whose $(i,j)$ entry is given by -$$\binom{2(i+j+a)}{i+j+a} \qquad \text{for} \qquad 0\leq i,j\leq n-1$$ -then we have -$$\det H_n(a)=\prod_{j=0}^{n-1}\binom{2a+2j}{a+j}\binom{a+n+j-1}{2j}^{-1}.$$ -In particular, the original matrix $H_n=H_n(0)$ and we get a peculiar identity -$$\prod_{j=0}^{n-1}\binom{2j}j\binom{n+j-1}{2j}^{-1}=2^{n-1}.$$<|endoftext|> -TITLE: Package for the Closest Vector Problem (CVP)? -QUESTION [5 upvotes]: Let $A$ be a positive definite, real $n \times n$ matrix. This defines a norm on $\mathbb{R}^n$. Now I have a given point $p \in \mathbb{R}^n$ and I want to find the lattice point $x \in \mathbb{Z}^n$ that is closest to $p$ with respect to this norm. This is commonly known as the Closest Vector Problem (CVP) and seems to be very important. So I guess algorithms to solve this should be implemented somewhere. However, I was not able to find something. -Is this there a package for solving this problem for some common mathematical software like Maple, Mathematica, Sage, etc? -Note that I am not interested in some approximation using LLL or so, I want really the (or one) closest vector! - -REPLY [3 votes]: Sage provides function closest_vector() in the module free_module_integer.<|endoftext|> -TITLE: In how many ways can a given planar graph be mapped into the plane? -QUESTION [7 upvotes]: I feel sure that this question must have been addressed in the literature, but I can't seem to find it - I may be looking in the wrong place. -A graph is planar if it can be drawn on the plane such that no two edges intersect in an interior point. In general, there are infinitely many ways to draw a planar graph, of course. However, we can say that two such mappings are equivalent if one can be obtained from the other via continuous transformations (or isotopies), in such a way that no two edges cross along the way. -Now, the number $\pi(G)$ of equivalence classes is a finite number depending on the graph $G$. I would like to get good bounds on this number, as a function of the graph, ie the number of vertices, edges, the vertex degrees, or other properties of the graph. Do results of this type exist in the literature on planar graphs? -As an example, consider the star graph on the vertex set $0,1,...,n$, where the vertex $0$ is connected to all of the other vertices, and there are no other edges. Any drawing induces a cyclic ordering of the outer vertices, and two drawings are equivalent iff the induced cyclic ordering is the same. Therefore, in this case $\pi(G)= (n-1)!$. - -REPLY [7 votes]: It is a theorem of Whitney, that a $3$-connected planar graph has two planar embeddings (one being the other flipped over). If a graph is two-connected, then you can flip over some, but not all of the three-connected components (so you get $2^n,$ where $n$ is the number of such). Finally, if the graph is $1$-connected, you can monkey around just as you have, so the decomposition into 2- and 3- connected component should give you a (messy) formula.<|endoftext|> -TITLE: Does the existence of a non-principal measure on ω imply that of a non Lebesgue measurable set? -QUESTION [14 upvotes]: A non-principal [probability] measure on a set X is a function $\mu$ defined on all subsets of $X$, with values in $[0,1]$, which is finitely additive, satisfies $\mu(X)=1$, and vanishes on singletons. -Can one prove in ZF + DC that the existence of such a measure on $\bf N$ (or on $\omega$), the set of natural integers, implies that of a non-Lebesgue measurable subset of $\bf R$? -Comments: 1) Sierpinski proved this in 1938 in the special case that $\mu$ takes values in $\{0,1\}$ (non-principal, or non-free ultrafilter). It suffices to take $X=\{A\subset{\bf N}\mid\mu(A)=0\}\subset\{0,1\}^{\bf N}$, equipped with the standard Bernoulli measure $\lambda$ (so that $\{0,1\}^{\bf N},\lambda)\approx([0,1],{\rm Leb}$). If $X$ were $\lambda$-measurable, one would have $\mu(X)={1\over2}$ since $X^c=\{A^c\mid A\in X\}$. But $X$ is a queue event, thus by Kolmogorov, $\mu(X)=0$ or $1$, contradiction. -2) In the 1998 book by Howard and Rubin, Consequences of the axiom of choice, it is stated that the existence of a non principal measure on $\bf N$ implies that of a non-Lebesgue measurable set [Form 222 implies Form 93], and that it can be found in articles by Pincus in 1972 and by Foreman-Wehrung in 1991. However, I could not find this implication in these articles. -3) The most famous non-principal measures on countable sets are invariant means on amenable groups, starting with $\bf Z$ [mean is another name for a finitely additive probability measure defined on all subsets]. Question: does the existence of a non-principal measure on $\bf N$ (or $\bf Z$) imply (in ZF + DC) that of an invariant mean on $\bf Z$? - -REPLY [9 votes]: This is stated as an open conjecture in Pincus' 1974 paper The Strength of the Hahn-Banach Theorem, which is a pretty good sign that he didn't prove it in a paper published in 1972. That same paper contains the first proof of the analogous question for the Baire Property. -As far as I am aware, this problem is still open, even assuming the stronger assumption of a Banach limit (such a measure that is translation-invariant and whose integral gives the limit for convergent sequences).<|endoftext|> -TITLE: Geometry of convex sets in Riemannian manifolds -QUESTION [8 upvotes]: Let $M$ be a smooth Riemannian manifold without boundary. Let $X\subset M$ be a closed subset which is a smooth submanifold with boundary, $\dim X=\dim M$. Assume that $X$ is locally convex, i.e. any point of $X$ has a neighborhood $U$ such that for any two points from $X\cap U$ there exists a unique shortest geodesic in $M$ connecting them, and this geodesic is contained in $X\cap U$. -Question. Let $p\in \partial X$. Let $\gamma$ be a geodesic in $M$ such that $\gamma(0)=p$ and $\gamma'(0)$ is tangent to $\partial X$. Is it true that there exists an open (in $M$) neighborhood $V$ of $p$ such that $\gamma \cap V$ does not intersect the interior of $X$? -Remark. If $M$ is a Euclidean space then the statement is true and it can be deduced e.g. from a version of the Hahn-Banach theorem. -A reference would be helpful. - -REPLY [8 votes]: Yes it is true. -Consider the signed diststance function $f=\mathrm{dist}_{\partial X}$. -Note that it is semiconcave in a neigborhood of $X$; -moreover, $f''\le C\cdot |f|$ for some constant $C$ defined locally. -For your $\gamma$, we have $(f\circ\gamma)'=0$ and -$$(f\circ\gamma)''\le C\cdot |f\circ\gamma|.$$ -The latter implies $f\circ\gamma(t)\le 0$ for all sufficiently small $t$.<|endoftext|> -TITLE: Printing omission in Mumford's "Lectures on Curves on an Algebraic Surface" -QUESTION [7 upvotes]: On page 174 of both copies I can access of Mumford's book "Lectures on Curves on an Algebraic Surface" there is a printing omission (like this one). - -LEMMA 2: Let $A$ be a complete $p$-adic ring where $p$ is not a zero-divisor, such that $A/p$ is perfect. Then there is a 1-1 correspondence between members of $A$ and sequences $(\xi_0,\xi_1,\ldots)$ of elements of $A/p$, given by... - -Who can supply the missing text? - -REPLY [10 votes]: In the Russian edition it is $$(\xi_0,\xi_1,\xi_2,\dots)\leftrightarrow f(\xi_0)+pf(\xi_1)+p^2f(\xi_2)+\dots$$ where $f$ is the Teichmuller map.<|endoftext|> -TITLE: Non-isomorphic Heisenberg groups over rings -QUESTION [10 upvotes]: Suppose $R_1,R_2$ are finite unital commutative rings. Consider Heisenberg groups $H_3(R_1)$ and $H_3(R_2)$ (upper unitriangular marticies $3 \times 3$). - -Proposition. If $R_1 \not\cong R_2$ (as rings) then $H_3(R_1) \not\cong H_3(R_2)$ (as multiplicative groups). - -Is that true? It seem that if $R_1$ and $R_2$ have not isomorphic additive groups then $H_3(R_1) \not\cong H_3(R_2)$, since they have not isomorphic centers. But what about general case, or is that too broad? - -REPLY [11 votes]: Theorem 1.13 of this paper (The model theory of unitriangular groups, Ann. Pure Appl. Logic -68(3), 1994, 225-261) by O. Belegradek says that the answer is positive even for infinite commutative rings. -However, Proposition 1.9 in the same paper asserts that this does not extend to the non-commutative (associative unital case). The counterexamples have the form $R_1=K\times K$, $R_2=K\times K^{\mathrm{op}}$, where $K$ is indecomposable and not isomorphic to $K^{\mathrm{op}}$.<|endoftext|> -TITLE: Probability of finding one sub-object vs a million disjoint ones -QUESTION [5 upvotes]: Let $W$ be a set of words of length $n$ on the three letters $a$, $b$ and $\ast$. Say that an element $w$ in $W$ ``matches" a word $w'$ of length $n$ on the letters $a$ and $b$ if each $\ast$ in $w$ can be replaced with an $a$ or a $b$ in such a way to give $w$'. Say that two words in $W$ are disjoint if, for all $i=1,\ldots,n$, the $i$-th letter of at least one of the two words is a $\ast$. -(So, for example, if $n=4$ then $\ast a\ast b$ matches $babb$ but not $bbab$. The word $\ast\ast ab$ is disjoint from $a\ast\ast\ast$ but not $\ast a b\ast$ or $\ast\ast ab$.) -Suppose that, given a word $w'$ of length $n$ on the letters $a$ and $b$ chosen uniformly at random, with probability at least $1/20$ there is a set of a million disjoint words $w$ in $W$ that match $w'$. -Is there some $\delta>0$ independent of $n$ and $W$ such that, given a word $w'$ of length $n$ on the letters $a$ and $b$ chosen uniformly at random, with probability at least $1/20+\delta$ there is at least one $w$ in $W$ that match $w'$? -My motivation for asking this question comes from various geometric settings (Erdős–Rényi random graphs, ...) where one might wish to know that finding one sub-object of interest is definitively more likely than finding a million disjoint sub-objects of interest, even when the class of sub-object of interest is so poorly understood that you can't use anything about this class to analyze the question. -If all the words in W are disjoint from each other, then there is a positive answer (and moreover you can take $1/20+\delta$ to be very close to $1$). - -REPLY [7 votes]: As pointed out to me by Noga Alon, this was known as the Van den Berg - Kesten conjecture, which was solved by David Reimer: -If p is the probability of matching at least one word in W, then the probability of matching at least two disjoint words in W is at most p^2. -http://journals.cambridge.org/article_S0963548399004113 -(The probability of matching a million disjoint words is at most p^{1,000,000}.)<|endoftext|> -TITLE: p-adic expansion for elements in algebraic closure of p-adic numbers -QUESTION [5 upvotes]: In the following I will describe a proposal for the p-adic expansion of the elements of the algebraic closure $\overline{\mathbb{Q}_p}$ of $\mathbb{Q}_p$. My question is if this "conjecture" has been proved or disproved before. -Consider the following result proved in "Local Fields" Berlin. (1980) by Serre, J. P.: - -Theorem: If $R$ is a perfect field of characteristic $p>0$, then there exists a unique ring $W(R)$ with characteristic $0$ with discrete valuation $v$ such that the residue field is $R$, $v(p)=1\in\mathbb{Z}$, and $W(R)$ is complete with respect to $v$. Also, the field of fractions $F(W(R))$ is the only field with characteristic $0$ with discrete valuation $v$ such that the residue field is $R$, $v(p)=1\in\mathbb{Z}$, and $F(W(R))$ is complete with respect to $v$. - -In this context, the valuation ring of $F(W(R))$ is $W(R)$ the ring of Witt vectors with coefficients in $R$. For example, if $R=\mathbb{F}_p$, then $W(R)=\mathbb{Z}_p$ and $F(W(R))=\mathbb{Q}_p$. -In the article "Maximally complete fields." Enseign. Math.(2) 39.1-2 (1993): 87-106, @Bjorn Poonen describes the unique maximally complete immediate extension field of $\overline{\mathbb{Q}_p}$ as follows: -let $W(\overline{\mathbb{F}_p})((t^\mathbb{Q}))$ be the ring of all series $\sum_{g}\alpha_gt^g$ where $g\in\mathbb{Q}$, $\alpha_g\in W(\overline{\mathbb{F}_p})$, and $\{g:\alpha_g\neq0\}$ is well-ordered. -A series $\sum_{g}\alpha_gt^g\in W(\overline{\mathbb{F}_p})((t^\mathbb{Q}))$ -is null if for all $g\in\mathbb{Q}$, $\sum_{n\in\mathbb{Z}}\alpha_{g+n}p^n=0$ in $F(W(\overline{\mathbb{F}_p})).$ -The set $N$ of all null series is an ideal of $W(\overline{\mathbb{F}_p})((t^\mathbb{Q}))$ that contains the polynomial $t-p$. The quotient $$L=W(\overline{\mathbb{F}_p})((t^\mathbb{Q}))/N$$ is a field of characteristic $0$, valuation ring $\mathbb{Q}$ and residue class field $\overline{\mathbb{F}_p}$. -Also $L$ is algebraically closed, and it is the maximally complete immediate extension of $\overline{\mathbb{Q}_p}$. Furthermore, if $S\subset W(\overline{\mathbb{F}_p})$ is a set of representatives of the classes of $\overline{\mathbb{F}_p}$, then each series $\sum_{g}\alpha_gt^g$ in $W(\overline{\mathbb{F}_p})((t^\mathbb{Q}))$ is null equivalent to a unique series of the form $\sum_{g}\beta_gt^g$ for $\beta_g\in S$. In other words, we have the following expansion for the elements of $L$: -$$L=\bigg\{\sum_{g\in\mathbb{Q}}\beta_gp^g:\beta_g\in S, \{g:\beta_g\neq0\}\mbox{ is well-ordered}\bigg\}.$$ -Notice the analogy with the $p$-adic expansion for elements of $\mathbb{Q}_p$: -$$\mathbb{Q}_p=\mathbb{Z}_p((t^\mathbb{Z}))/N=W(\mathbb{F}_p)((t^\mathbb{Z}))/N=\bigg\{\sum_{n=m}^\infty\beta_np^n:m\in\mathbb{Z},\beta_n=0,1,\dots,p-1 \bigg\}.$$ -Now let's consider the following chain of valued field extensions: -$$\mathbb{Q}_p((x))\subset -\overline{\mathbb{Q}_p}((x))\subset \overline{\overline{\mathbb{Q}_p}((x))}= -\bigcup_{n=1}^\infty \overline{\mathbb{Q}_p}((x^\frac{1}{n}))\subset\overline{\mathbb{Q}_p}((t^\mathbb{Q}))$$ -Also consider the following chain of rings: -$$\mathbb{Z}_p((x))\subset -W(\overline{\mathbb{F}_p})((x))\subset -\bigcup_{n=1}^\infty W(\overline{\mathbb{F}_p})((x^\frac{1}{n}))\subset W(\overline{\mathbb{F}_p})((t^\mathbb{Q}))$$ -By taking the quotient of these rings by their ideals of null series, we obtain the following chain of field extensions: -$$\mathbb{Q}_p= -\mathbb{Z}_p((x))/N\subset -W(\overline{\mathbb{F}_p})((x))/N\subset -\bigcup_{n=1}^\infty W(\overline{\mathbb{F}_p})((x^\frac{1}{n}))/N\subset W(\overline{\mathbb{F}_p})((t^\mathbb{Q}))/N=L$$ -where there is an abuse of notation for the ideals $N$ (The N's of different quotients are different). - -Conjecture 1: The relation - $$\overline{\overline{\mathbb{Q}_p}((x))}= \bigcup_{n=1}^\infty - \overline{\mathbb{Q}_p}((x^\frac{1}{n}))$$ implies the relation - $$\overline{\mathbb{Q}_p}= \bigcup_{n=1}^\infty - W(\overline{\mathbb{F}_p})((x^\frac{1}{n}))/N= - \bigg\{\sum_{n=m}^\infty\beta_np^\frac{n}{N}:m,N\in\mathbb{Z},N\geq1,\beta_n\in - S \bigg\},$$ where $S\subset W(\overline{\mathbb{F}_p})$ is a set of - representatives of the classes of $\overline{\mathbb{F}_p}$. - -Edit: As it is noticed in the comments below, the conjecture 1 is false, since $\sqrt{-1}\in\overline{\mathbb{Q}_2}$ cannot be represented in the proposed form. Also notice that $W(\overline{\mathbb{F}_p})\neq\{x\in\overline{\mathbb{Q}_p}:|x|_p\leq1\}$ since the former has a discrete valuation while latter has a dense valuation. Also I just noticed the following result, which follows immediately by the theorem stated above and the properties of the maximally complete field $L$ described by Bjorn Poonen in its paper also mentioned above. The notations are defined above. - -Proposition: If $R$ is a perfect field of characteristic $p>0$, then $W(R)((x))/N$ is the only field of characteristic $0$, with discrete valuation $v$ such that the residue class field is $R$, $v(p)=1\in\mathbb{Z},$ and $W(R)((x))/N$ is complete with respect $v$. Hence $F(W(R))=W(R)((x))/N$ is maximally complete. - -REPLY [7 votes]: The following article discusses p-adic expansions in $\overline{\mathbb{Q}}_p$ -and of $\mathbb{C}_p$. -Algebraic $p$-adic expansions, -David Lampert, -Journal of Number Theory 23 (1986), 279–284.<|endoftext|> -TITLE: Combinatorial Identity with Connection Coefficients and Falling Factorial $\langle i x\rangle_n$ -QUESTION [7 upvotes]: Let $j, k ,n$ be nonnegative integers such that $0 \leq j, k \leq n \leq k +j $. Pick integer $m$ such that $0 \leq m \leq k + j - n$. -Let $\langle x \rangle_m$ denote the falling factorial $x(x-1)\ldots (x-m+1)$. -I've stumbled across the need to prove the following equality: -$$\sum_{i=0}^n \frac{\binom{k}{i}\binom{j}{i}}{\binom{n}{i}}(-1)^i\langle ix\rangle_m = \left\{ \begin{matrix} \frac{k! j!}{n!} (-x)^m ~~~~~~~~\text{when }m = k+j - n\\ 0 ~~~~~~~~~~~~~~~~~\text{when }m < k + j - n \end{matrix} \right.$$ -An equivalent formulation is: -$$\sum_{i=0}^n \frac{(-1)^i\binom{k}{i}\binom{j}{i} \binom{i x}{m}}{\binom{n}{i}}= \left\{ \begin{matrix} \frac{k! j!}{n!m!} (-x)^m ~~~~~~~~\text{when }m = k+j - n\\ 0 ~~~~~~~~~~~~~~~~~\text{when }m < k + j - n \end{matrix} \right.$$ -Any references to potentially related material would be greatly appreciated. -${\scriptsize \textbf{Edited to fix sign}}$ - -REPLY [10 votes]: It is very probable that what is written below is the simplification of Darij's argument. I use the notation $x^{\underline{n}}=x(x-1)\dots(x-n+1)$ [as in Knuth's books] for the falling factorial, and $[t^n] f(t)$ for the coefficient of $t^n$ in the polynomial $f$. -For a polynomial $f(t)$ of degree at most $j$ we have $$\sum_{i=0}^j (-1)^{j-i}\frac{f(i)}{i!(j-i)!}=[t^j]f(t),$$ -this follows from the Lagrange interpolation of $f$ in the points $\{0,1,\dots,j\}$. Apply this to the polynomial $f(t)=(tx)^{\underline m}(n-t)^{\underline {n-k}}$. In RHS we have 0 if $mq$ are zero -(because the factor $\dbinom{q}{u}$ makes them vanish). Hence, we can remove -them from the sum without changing the value of the sum. We thus obtain -$\sum\limits_{u=0}^{Q}\left( -1\right) ^{u}\dbinom{q}{u}\dbinom{r-u}{s} -=\sum\limits_{u=0}^{q}\left( -1\right) ^{u}\dbinom{q}{u}\dbinom{r-u}{s}= -\begin{cases} -1, & \text{if }s=q;\\ -0, & \text{if }si$, because the factor -$\dbinom{i}{u}$ makes these addends vanish). -Now, fix $u\in\left\{ p,p+1,\ldots,i\right\} $. Then, $u\leq i\leq n$. -Hence, $n-u\in\mathbb{N}$ and $i-u\in\left\{ 0,1,\ldots,n-u\right\} $. -Hence, the symmetry of Pascal's triangle (i.e., the fact that every -$N\in\mathbb{N}$ and $M\in\left\{ 0,1,\ldots,N\right\} $ satisfy $\dbinom -{N}{M}=\dbinom{N}{N-M}$) yields -$\dbinom{n-u}{i-u}=\dbinom{n-u}{\left( n-u\right) -\left( i-u\right) -}=\dbinom{n-u}{n-i}$. -But Lemma 3 (applied to $n$ and $u$ instead of $m$ and $a$) yields -$\dbinom{n}{i}\dbinom{i}{u}=\dbinom{n}{u}\underbrace{\dbinom{n-u}{i-u} -}_{=\dbinom{n-u}{n-i}}=\dbinom{n}{u}\dbinom{n-u}{n-i}$. -In other words, -(4) $\dfrac{\dbinom{i}{u}}{\dbinom{n}{u}}=\dfrac{\dbinom{n-u}{n-i} -}{\dbinom{n}{i}}$. -(The denominators here are nonzero since $u\leq i\leq n$.) -Now, forget that we fixed $u$. We thus have proven (4) for each -$u\in\left\{ p,p+1,\ldots,i\right\} $. Hence, (3) rewrites as -$\sum\limits_{u=0}^{n}\dfrac{\left( -1\right) ^{u}\dbinom{i}{u}\dbinom{j} -{u}}{\dbinom{n}{u}}P\left( u\right) $ -$=\dbinom{j}{p}\sum\limits_{u=p}^{i}\left( -1\right) ^{u}\dfrac{\dbinom -{n-u}{n-i}}{\dbinom{n}{i}}\dbinom{j-p}{u-p}$ -$=\dbinom{j}{p}\dbinom{n}{i}^{-1}\sum\limits_{u=p}^{i}\left( -1\right) -^{u}\dbinom{n-u}{n-i}\dbinom{j-p}{u-p}$ -$=\dbinom{j}{p}\dbinom{n}{i}^{-1}\sum\limits_{u=0}^{i-p}\underbrace{\left( --1\right) ^{u+p}}_{=\left( -1\right) ^{p}\left( -1\right) ^{u} -}\underbrace{\dbinom{n-\left( u+p\right) }{n-i}}_{=\dbinom{\left( -n-p\right) -u}{n-i}}\underbrace{\dbinom{j-p}{\left( u+p\right) -p} -}_{=\dbinom{j-p}{u}}$ -(here, we have substituted $u+p$ for $u$ in the sum) -$=\dbinom{j}{p}\dbinom{n}{i}^{-1}\left( -1\right) ^{p}\sum\limits_{u=0} -^{i-p}\left( -1\right) ^{u}\dbinom{\left( n-p\right) -u}{n-i}\dbinom -{j-p}{u}$ -(5) $=\dbinom{j}{p}\dbinom{n}{i}^{-1}\left( -1\right) ^{p} -\sum\limits_{u=0}^{i-p}\left( -1\right) ^{u}\dbinom{j-p}{u}\dbinom{\left( -n-p\right) -u}{n-i}$. -Recall that $j\geq p$. Hence, $j-p\in\mathbb{N}$. Furthermore, $\underbrace{i} -_{\geq j}-p\geq\underbrace{j}_{\geq p}-p\geq0$ and thus $i-p\in\mathbb{N}$. -Finally, $n-i\in\left\{ 0,1,\ldots,j-p\right\} $ (since $n-\underbrace{i} -_{\leq n}\geq n-n=0$ and $n-i\leq j-p$ (since -$p\leq i+j-n$)). Hence, we -can apply Lemma 6 to $q=j-p$, $Q=i-p$, $r=n-p$ and $s=n-i$. We thus obtain -$\sum\limits_{u=0}^{i-p}\left( -1\right) ^{u}\dbinom{j-p}{u}\dbinom{\left( -n-p\right) -u}{n-i}= -\begin{cases} -1, & \text{if }n-i=j-p;\\ -0, & \text{if }n-i -TITLE: Shapiro's lemma in the language of group extensions -QUESTION [15 upvotes]: I am trying to understand Shapiro's lemma for $H^2$ in the concrete language of extensions of finite groups. -Let $H$ be a subgroup of a finite group $G$, and let $A$ be an $H$-module. Let ${\rm Ind}_G^H(A)$ be the induced module (see Serre, Galois Cohomology, Ch. I, 2.5). Shapiro's lemma says that the inclusion $H\hookrightarrow G$ and the map $\pi\colon {\rm Ind}_G^H(A)\to A$ of evaluation at $1$ give an isomorphism $H^2(G,{\rm Ind}_G^H(A))\simeq H^2(H,A)$ (of course, this is true in any degree). -Now let us view elements of $H^2$ as equivalence classes of extensions. -An extension $0\to {\rm Ind}_G^H(A)\to \hat G\xrightarrow{f}G\to 1$ should correspond under this isomorphism to the extension $0\to A\to f^{-1}(H)/{\rm Ker}(\pi)\to H\to1$. -My question is how to realize the inverse map explicitely: -That is, starting from an extension $0\to A\to \hat H\to H\to1$, what is the extension of $G$ by ${\rm Ind}_G^H(A)$ corresponding to it under Shapiro's lemma? - -REPLY [4 votes]: Shapiro's Lemma boils down to the following isomorphism for a subgroup of finite index: Let us write $G=\bigcup_{i=1}^t g_iH$ for left coset representatives of $H$ in $G$. For a $G$-module $P$ we have the following isomorphism $$\Psi:Hom_H(P,A)\cong Hom_G(P,Ind_H^G(A))$$ -$$ \Psi(f)(p) = \sum_{i}g_i\otimes f(g_i^{-1}p).$$ -If we think now of the extension $$1\to A\to \hat{H}\to H\to 1$$ as given by a two cocycle $\beta:H\times H\to A$, we can proceed in the following way: -We have the Bar Resolutions $B_H$ and $B_G$ of the trivial module $\mathbb{Z}$ over $H$ and $G$ respectively. The two cocycle $\beta$ can then be considered as an element in $Hom_H((B_H)_2,A)$. Since $B_G$ can also be considered as a resolution for $\mathbb{Z}$ over $H$, we will have a lifting of the identity map $B_G\to B_H$ as a map of $H$-modules. This will already gives us a two cocycle in $Hom_H((B_G)_2,A)$. Now use the above isomorphism to get a two cocycle in $Hom_G((B_G)_2,Ind_H^G(A))$. This will give you the desired cocycle and therefore the desired extension. -The main computational difficulty here (besides calculating the isomorphism $\Psi$) is in lifting the identity map $\mathbb{Z}\to\mathbb{Z}$ to $B_G\to B_H$. In degree zero this is rather simple: you choose right coset representatives $g_i'$ and then map $hg_i'[]\in (B_G)_0$ to $h[]\in (B_H)_0$. In degrees one and two it will be more complicated. In many concrete cases you can calculate explicitly projective resolutions for $H$ and for $G$. Also, in case $H$ is normal in $G$ you can write things more concretely, by choosing a two cocycle for $G/H$ with values in $H$ representing the extension $$1\to H\to G\to G/H\to 1$$ (even if $H$ is not abelian). -In the general case I do not think that there is a neater way of writing this down. Of course, I will be more than happy to be proven wrong about this.<|endoftext|> -TITLE: What are braided vertex algebras? -QUESTION [5 upvotes]: The notion of vertex algebra, like any reasonable algebraic notion, makes sense inside any (sufficiently linear) symmetric monoidal category. The standard pictures of the operator product, however, suggest that symmetry is more than is needed: one should be able to formulate the axioms of "vertex algebra" internal to any (sufficiently linear) braided monoidal category. Has this been done in the literature? Are there standard examples? -For vertex operator algebras, I expect you need your braided monoidal category to be balanced. - -REPLY [7 votes]: For the case of vector spaces graded by an abelian group (with braiding determined by an abelian 3-cocycle following Joyal-Street), this was done by Dong and Lepowsky in their 1993 book "Generalized Vertex Algebras and Relative Vertex Operators". The object has the name "abelian intertwining algebra", and standard examples come from applying the lattice vertex algebra construction to rational lattices. A greater supply of examples was given in van Ekeren, Möller, Scheithauer by an orbifold construction in 2015. -As far as I can tell, the general definition has not appeared explicitly in the literature. Some variants have appeared under the name "quantum vertex algebra", though. In particular, I remember working out that Borcherds's definition of (A,H,S)-vertex algebra in his "Quantum Vertex Algebras" paper can be modified to fit the braided setting without too much trouble.<|endoftext|> -TITLE: Mitchell, Steel. FSIT. Lemma 2.8: Is $k$-solidity actually needed? -QUESTION [5 upvotes]: Consider the following result (which is Lemma 2.8 in Mitchell and Steel's paper on Fine Structure and Iteration Trees): -Lemma 2.8 Let $\pi \colon \mathcal{H} \to \mathcal{M}$ be generalized $r \Sigma_{k}$ - elementary, where $\mathcal M$ is a ppm (not of type III) and $1 - \leq k < \omega$. Suppose that $\rho_{k}^{\mathcal{M}} \subseteq - \mathcal{H}$ and $\pi \restriction \rho_{k}^{\mathcal{M}} = - \operatorname{id}$. Suppose also that $\pi(r)$ is the $k$th standard parameter of - $(\mathcal{M}, \pi(q))$ and that $\pi(r)$ is $k$-solid and - $k$-universal over $(\mathcal{M}, \pi(q))$. Then - -$\rho_{k}^{\mathcal{H}} = \rho_{k}^{\mathcal{M}}$, -$r$ is the $k$th standard parameter of $(\mathcal{H},q)$ and -$r$ is $k$-universal over $(\mathcal{H}, q)$. - -The proofs of items $1.$ and $2.$ are included in this paper and in -both cases it seems that the $k$-solidity of $\pi(r)$ over -$(\mathcal{M}, \pi(q))$ isn't actually needed. Hence I decided to -prove item $3.$ to see how $k$-solidity comes into play. However, if -my argument is correct, it doesn't rely on $k$-solidity either. -Here is my proof of item $3.$: -Proof (of $3.$). Let $A \in \mathcal{H}$ be such that $A \subseteq - \rho_{k}^{\mathcal{H}}$. Then $\pi(A) \cap \rho_{k}^{\mathcal{M}} - \in \mathcal M$. By the $k$-universality of $\pi(r)$ over - $(\mathcal{M}, \pi(q))$ there is hence some generalized Skolem term - $\tau \in S_{\kappa}$ and some $\vec{\alpha} \in ^{< \omega} - \rho_{k}^{\mathcal{M}}$ s.t. - $$ - \pi(A) \cap \rho_{k}^{\mathcal{M}} = - \tau^{\mathcal{M}}[\vec{\alpha}, \pi(r), \pi(q)] \cap \rho_{k}^{\mathcal{M}}. - $$ - Let $B := \tau^{\mathcal{H}}[\vec{\alpha},r,q] \cap \rho_{k}^{\mathcal{H}}$. Combining the fact - that $\pi$ is generalized $r \Sigma_{k}$-elementary, $\rho_{k}^{\mathcal{H}}= - \rho_{k}^{\mathcal{M}} \subseteq \mathcal{H}$ and $\pi \restriction - \rho_{k}^{\mathcal M} = \operatorname{id}$ we have that - \begin{align*} - \mathcal{H} \models A \cap \rho_{k}^{\mathcal{H}} = B &\iff \mathcal{H} \models A \cap \rho_{k}^{\mathcal{H}} = \tau^{\mathcal - H}[\vec{\alpha}, - r,q] \cap - \rho_{k}^{\mathcal{H}} - \\ - &\iff - \mathcal{M} - \models - \pi(A) \cap - \rho_{k}^{\mathcal{M}} - = - \tau^{\mathcal{M}}[\vec{\alpha},\pi(r),\pi(q)] - \cap \rho_{k}^{\mathcal{M}}. - \end{align*} - Since the last line is true, it follows that - $A \cap \rho_{k}^{\mathcal{H}} - =\tau^{\mathcal{H}}[\vec{\alpha},r,q] \cap \rho_{k}^{\mathcal{H}}$. Thus $r$ is indeed - $k$-universal over $(\mathcal{H}, q)$. Q.E.D. -Question: Did I miss something and this result actually relies on the $k$-solidity of $\pi(r)$ over $(\mathcal{M},\pi(q))$ or can this assumption be dropped? -If $k$-solidity is needed, I'd like to understand where exactly in the proof it is used and ideally I'd like to see an example in which Lemma 2.8 fails without $k$-solidity. - -PS: I am aware that this question isn't exactly 'ongoing research' and I strongly considered posting it over at MSE. However, since the group of people able to answer this question is more likely to be encountered here and since a somewhat similar question has been asked and well-received here, I decided to go with mathoverflow. - -REPLY [3 votes]: In my question I already verified that (assuming 1. and 2.) item 3. is provable without assuming that $\pi(r)$ is $k$-solid over $(\mathcal{M},q)$. To see that we can actually drop $k$-solidity in this lemma, it hence suffices to see that 1. and 2. also don't require that $k$-solidity of $\pi(r)$. -If $\rho_{k}^{\mathcal{M}} = \operatorname{Ord}^{\mathcal{M}}$, then $\pi = \operatorname{id}$ - and the lemma trivially holds. Thus assume that - $\rho_{k}^{\mathcal{M}} < \operatorname{Ord}^{\mathcal{M}}$. - -Let $\alpha \leq \rho_{k}^{\mathcal M}$. Since -$\pi \restriction \rho_{\kappa}^{\mathcal{M}} = \operatorname{id}$ -and since $\pi$ is generalized $r \Sigma_{k}$-elementary, we have - up -to a slight abuse of notation - -\begin{align*} - \operatorname{Th}_{k}^{\mathcal{H}}(\alpha \cup \{ s \}) - &= \{(\phi, \vec{a}, s) \mid \vec{a} \in ^{< \omega}{\alpha} - \wedge \mathcal{H} \models \phi[\vec{a}, s] \} \\ - &= \{(\phi, \vec{a}, s) \mid \vec{a} \in ^{< \omega}{\alpha} - \wedge \mathcal{M} \models \phi[\vec{a}, \pi(s)] \} -\end{align*} -In particular, for any $\phi \in r \Sigma_{k}$ and any $\vec{a} \in ^{< - \omega}{\alpha}$, we have -$$ - (\phi, \vec{a}, s) \in - \operatorname{Th}_{k}^{\mathcal{H}}(\alpha \cup \{s \}) \iff - (\phi, \vec{a}, \pi(s)) \in - \operatorname{Th}_{k}^{\mathcal{M}}(\alpha \cup \{ \pi(s) \}). -$$ -By enlarging $\alpha$, if necessary, we may assume that $\alpha$ -is primitive recursively closed and hence uniformly code -\begin{align*} - \{(\phi, \vec{a}) \mid (\phi, \vec{a}, s) \in - \operatorname{Th}_{k}^{\mathcal{H}}(\alpha \cup \{ s \}) \} \\ - = \{(\phi, \vec{a}) \mid (\phi, \vec{a}, \pi(s)) \in - \operatorname{Th}_{k}^{\mathcal{M}}(\alpha \cup \{ \pi(s) \}) \} -\end{align*} -as a subset $A \subseteq \alpha$. Since -$\alpha < \rho_{k}^{\mathcal{M}}$, we have -$\operatorname{Th}_{k}^{\mathcal{M}}(\alpha \cup \{ \pi(s) \}) \in -\mathcal{M}$ and hence $A \in \mathcal{M}$. By the strong -acceptability of $\mathcal{M}$ - observing that -$\rho_{k}^{\mathcal{M}}$ is an $\mathcal{M}$-cardinal - this -yields -$$ - A \in \mathcal{J}_{\rho_{k}^{\mathcal{M}}}^{\mathcal M} = \left( - H_{\rho_{k}^{\mathcal{M}}} \right)^{\mathcal{M}} \overset{\pi - \restriction \rho_{k}^{\mathcal{M}} = \operatorname{id}}{=} \left( - H_{\rho_{k}^{\mathcal{M}}}\right)^{\mathcal{H}} \subseteq \mathcal{H}. -$$ -Therefore -$$ - \operatorname{Th}_{k}^{\mathcal{H}}(\alpha \cup \{ s \}) = \{ - (\phi, \vec{a}, s) \mid \langle \phi, \vec{a} \rangle \in A \} - \in \mathcal{H} -$$ -and $\rho_{k}^{\mathcal{M}} \leq \rho_{k}^{\mathcal{H}}$. -On the other hand, suppose that -$\rho_{k}^{\mathcal{M}} < \rho_{k}^{\mathcal{H}}$. Then -$\operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{M}} \cup -\{(r,q)\}) \in \mathcal{H}$. Let -$A \subseteq \rho_{k}^{\mathcal{M}}$, $A \in \mathcal{H}$ be a -uniform code for -$$ - \{ (\phi, \vec{a}) \mid (\phi, \vec{a}, (r,q)) \in - \operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{M}} \cup - \{(r,q)\})\}. -$$ -Then $A = \pi(A) \cap \rho_{k}^{\mathcal{M}} \in \mathcal{M}$ -witnesses (as above) that -$\operatorname{Th}_{k}^{\mathcal{M}}(\rho_{k}^{\mathcal{M}} \cup -\{ \pi(r), \pi(q)\}) \in \mathcal{M}$. This contradicts the fact -that $\pi(r)$ is the $k$th standard parameter of -$(\mathcal{M},\pi(q))$! -The proof above also shows that -$\operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{H}} \cup -\{ (r,q)\}) \not \in \mathcal{H}$. Hence it suffices to show that -for all $s <_{\operatorname{lex}} r$ -$$ - \operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{H}} \cup - \{ (s,q)\}) \in \mathcal{H}. -$$ -So, fix $s <_{\operatorname{lex}} r$. Then -$\pi(s)<_{\operatorname{lex}} \pi(r)$ and hence -$$ - \operatorname{Th}_{k}^{\mathcal{M}}(\rho_{k}^{\mathcal{H}} \cup - \{ (\pi(s),\pi(q))\}) \in \mathcal{M}. -$$ -Let $A \subseteq \rho_{k}^{\mathcal{M}}$, $A \in \mathcal{M}$ be -the code of this fact as above. Since $\pi(r)$ is $k$-universal -over $(M, \pi(q))$ there is some $\tau \in \operatorname{Sk}_{k}$ -and $\vec{a} \in ^{< \omega}{\rho_{k}^{\mathcal{M}}}$ such that -$A = \tau^{\mathcal{M}}[\vec{a}, \pi(r), \pi(q)] \cap -\rho_{k}^{\mathcal{M}}$. Let -$$ - B = \tau^{\mathcal{H}}[\vec{a}, \pi(r), \pi(q)] \cap - \rho_{k}^{\mathcal{M}}. -$$ -Since $\pi$ is generalized $r \Sigma_{k}$-elementary we have, -for all $\xi < \rho_{k}^{\mathcal{M}} = \rho_{k}^{\mathcal{H}}$ -$$ - \mathcal{M} \models \xi \in - \tau^{\mathcal{M}}[\vec{a},\pi(r),\pi(q)] \iff \mathcal{H} - \models \xi \in \tau^{\mathcal{H}}[\vec{a},r,q]. -$$ -Thus $B = A \in \mathcal{H}$ witnesses that -$\operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{H}} \cup -\{(s,q)\}) \in \mathcal{H}$ and hence the -$<_{\operatorname{lex}}$-minimality of $r$.<|endoftext|> -TITLE: Splitting field of a root vector $x_{\alpha}: \mathbf{G}_a \rightarrow U_{\alpha}$ -QUESTION [5 upvotes]: Let $G$ be a quasi-split connected reductive group over a perfect field $F$, with Borel subgroup and maximal torus $B \supseteq T$ defined over $F$. Assume everything splits over a Galois extension $L/F$, with $\Gamma = \textrm{Gal}(L/F)$. The choice of $B$ gives us a set of simple roots $\Delta \subseteq \Phi(G,T)$, and for each $\alpha \in \Delta$, there exists an isomorphism of algebraic groups $x_{\alpha}$ of $\mathbf{G}_a$ onto a closed, one dimensional subgroup $U_{\alpha}$ of $R_u(B)$, such that -$$t x_{\alpha}(a)t^{-1} = x_{\alpha}(\alpha(t)a)$$ -for all $t \in T, a \in \mathbf{G}_a$. Such a map $x_{\alpha}$ is called a root vector for $\alpha$. -If $x_{\alpha}'$ is another isomorphism onto $U_{\alpha}$ with the same property, then there is a unique $0 \neq c \in \mathbf{G}_a$ such that $x_{\alpha}(a) = x'_{\alpha}(ca)$. -$\Gamma$ acts on $\Delta$ by $\alpha^{\sigma} = \sigma \alpha \sigma^{-1}$. If $x_{\alpha}$ is a root vector for $\alpha$, then $x_{\alpha}^{\sigma} := \sigma x_{\alpha} \sigma^{-1}$ is one for $\alpha^{\sigma}$. -What I would like to do is the following: - -Goal: Choose for each $\alpha \in \Delta$ a root vector $x_{\alpha}$ such that $x_{\alpha}^{\sigma} = x_{\alpha^{\sigma}}$ for all $\alpha \in \Delta, \sigma \in \Gamma$. - -This does not seem possible, since for a fixed $\alpha \in \Delta$, and $\sigma \neq \tau \in \Gamma$ we can have $\alpha^{\sigma} = \alpha^{\tau}$, and there is no guarantee that $x_{\alpha}^{\sigma} = x_{\alpha}^{\tau}$ for any choice of root vector $x_{\alpha}$ for $\alpha$. -The next best thing seems to be to, for each orbit of $\Delta$ under the action of $\Gamma$, choose an $\alpha$ in the orbit, and replace $\Gamma$ by some quotient $\textrm{Gal}(F_{\alpha}/F)$ for some appropriate Galois extension $F_{\alpha}$ of $F$. In Eisenstein Series and Automorphic L-Functions by F. Shahidi, such a field $F_{\alpha}$ is mentioned, but not defined, and it is called the splitting field of $x_{\alpha}$ over $F$. -My question is, how should this field $F_{\alpha}$ be defined, and how close can we get to achieving the Goal mentioned above? - -REPLY [5 votes]: This is explained by the Borel-Tits relative structure theory for connected reductive groups over arbitrary fields. In particular, there is no need to assume $F$ is perfect. The explanation below is long when written out, but the underlying principles are rather natural and clean. -Moreover, we will find a collection of root vectors satisfying the stronger requested Galois-equivariance property as near the start of the question. I will use Roman letters to denote roots, so "$a$" rather than "$\alpha$". -Consider a relative root $a$ in a chosen basis for the relative root system (it corresponds to a Galois orbit in a basis of the absolute root system such that the members of the orbit have nontrivial restriction to a maximal split torus, as we review below). The $a$-root space may have massive dimension as an $F$-vector space, but we will show that it is naturally a line over a finite separable extension $F_a$ of $F$. The construction of this enhanced linear structure over such an extension will make clear the sense in which $F_a/F$ and that $F_a$-linear structure refining the given $F$-linear structure is canonical for non-multipliable $a$ (and nearly canonical -for multipliable $a$). This will use in a crucial way the quasi-split hypothesis. -First, we fix what seems to be a misconception: the correct definition of $F_a$ will show that it is generally not a Galois extension of $F$. Rather, it is a finite separable extension of $F$, and the Galois orbit of absolute roots in question is not naturally indexed by the Galois group for $F_a$ over $F$ (there is no natural base point for the orbit, and no such Galois group when $F_a/F$ is not Galois) but rather is naturally indexed by the set of $F$-embeddings of -$F_a$ into a fixed separable closure $F_s$ (sanity check: ${\rm{Gal}}(F_s/F)$ does -act naturally and transitively on that set, regardless of whether or not $F_a/F$ is Galois!). -We are going to postpone the quasi-split hypothesis for as long as possible, to identify most clearly where exactly this assumption is truly essential. For an arbitrary connected reductive $F$-group $G$ (not yet assumed to be quasi-split, to clarify ideas), let $S$ be a maximal split $F$-torus. The Borel-Tits relative structure theory over arbitrary fields ensures that: -(i) $S$ is contained in some minimal parabolic $F$-subgroup $P$ of $G$, -(ii) all minimal parabolic $F$-subgroups of $G$ are $G(F)$-conjugate to each other (so by (i), if $G$ is quasi-split then $S$ is contained -in a Borel $F$-subgroup), -(iii) all pairs $(S, P)$ with $S \subset P$ are $G(F)$-conjugate to each other, and $S$ is non-central if and only if there exists a proper -parabolic $F$-subgroup, -(iv) $S' := (S \cap \mathscr{D}(G))^0_{\rm{red}}$ is a maximal split $F$-torus in $\mathscr{D}(G)$, the map $\Phi(G,S) \rightarrow {\rm{X}}(S')$ via restriction is an injection into ${\rm{X}}(S') - \{0\}$ -and constitutes a (possibly non-reduced!) root system spanning -${\rm{X}}(S')_{\mathbf{Q}}$, -(v) the positive systems of roots in $\Phi(G,S)$ are in bijective correspondence with the set of minimal parabolic $F$-subgroups $P \subset G$ containing $S$ via $P \mapsto \Phi(P,S)$ (the set of non-trivial $S$-weight occurring in ${\rm{Lie}}(P)$). -Pick a minimal parabolic $F$-subgroup $P \supset S$, and a maximal $F$-torus $T \subset P$ containing $S$ (so $T$ is also maximal as an $F$-torus in $G$). Let ${}_F\Delta$ be the basis of $\Phi(G,S)$ corresponding to the positive system of roots $\Phi(P, S)$. By the split theory applied over $F_s$, we can pick a Borel $F_s$-subgroup $B \subset G_{F_s}$ contained in $P_{F_s}$ and containing $T_{F_s}$. (We are -ultimately interested in the case that $G$ is quasi-split, so $B = P_{F_s}$, but we do not make that assumption just yet.) -Let $\Delta$ be the basis of the absolute root system $\Phi = \Phi(G_{F_s}, T_{F_s})$ corresponding to the positive system of roots -$\Phi(B, T_{F_s})$, and -let $\Delta_0 \subset \Delta$ be the set of absolute roots whose restriction to $S_{F_s}$ is trivial. The restriction map -${\rm{X}}(T_{F_s}) \rightarrow {\rm{X}}(S_{F_s}) = {\rm{X}}(S)$ clearly carries $\Delta - \Delta_0$ into $\Phi(G_{F_s}, T_{F_s})$, and since -$B \subset P_{F_s}$ it is immediate from the definitions that $\Delta - \Delta_0$ is even carried into $\Phi(B, T_{F_s})$; with some -more work one shows that it lands inside ${}_F\Delta$. -The Galois group $\Gamma_F = {\rm{Gal}}(F_s/F)$ acts naturally (on the left) on the geometric character lattice $X = {\rm{X}}(T_{F_s})$, and as such it certainly preserves $\Phi$ (since $T$ is an $F$-torus and the absolute root system encodes the weight spaces for non-trivial weights for the $F_s$-scalar extension of the $F$-linear $T$-action on ${\rm{Lie}}(G)$ over $F$. For each $\gamma \in \Gamma_F$, $\gamma(\Delta)$ is a basis for $\Phi$. It is exactly in the quasi-split case that we can find a Borel $F$-subgroup, and then pick $T$ inside that, so then the corresponding $\Delta$ would be $\Gamma_F$-stable. But we want to do better for now, so we confront the possibility that $\gamma(\Delta) \ne \Delta$ for some $\gamma$, perhaps no matter what $T$ and $\Delta$ we may try. This is handled by Tits' so-called $\ast$-action of $\Gamma_F$ on $\Delta$, as we now review. -Recall that the Weyl group of any root system acts simply transitively on the set of bases for the root system (since the set of bases is in natural bijective correspondence with the set of Weyl chambers). But for the finite etale $F$-group $W(G,T) = N_G(T)/T$ we know from the structure theory of split reductive groups (such as over $F_s$) that the natural map -$$W(G,T)(F_s) = N_{G(F_s)}(T_{F_s})/T(F_s) \rightarrow W(\Phi)$$ -is an isomorphism, so there is a unique $w_{\gamma} \in W(G,T)(F_s)$ such that $w_{\gamma}.\gamma(\Delta) = \Delta$. One checks from the uniqueness aspect that $w_{\gamma'} \gamma'(w_{\gamma}) = w_{\gamma'\gamma}$ -(using the natural $\Gamma_F$-action on $W(G,T)(F_s)$), and that $w_{\gamma} = 1$ for $\gamma$ in some open subgroup of $\Gamma_F$, so -$$(\gamma, a) \mapsto \gamma \ast a := w_{\gamma}(\gamma(a))$$ -is a continuous left action of $\Gamma_F$ on $\Delta$. In the quasi-split case, if we had chosen $T$ inside a Borel $F$-subgroup $B$ then for the corresponding $\Delta$ we would have $w_{\gamma} = 1$ for all $\gamma$, so this $\ast$-action would agree with the effect of the natural $\Gamma_F$-action on $\Phi$ that preserves the positive system of roots $\Phi(B_{F_s}, T_{F_s})$ and hence preserves its set $\Delta$ of simple elements. That establishes the contact with the quasi-split case, and in what follows we always work with the $\ast$-action to avoid a quasi-split hypothesis. -Here finally is the real substance of the matter. From the definitions one can check that the natural map -$$\pi: \Delta - \Delta_0 \rightarrow {}_F\Delta$$ -introduced above is invariant under the $\ast$-action on $\Gamma_F$, so $\pi$ carries each $\Gamma_F$-orbit to a single point in ${}_F\Delta$. -(Really we have a map $\pi:\Delta \rightarrow {}_F \Delta \cup \{0\}$, -and $\Delta_0$ is the fiber over $\{0\}$ by definition.) -Much more serious (and deeper) is the fact that $\pi$ is surjective -with fibers consisting of precisely the $\Gamma_F$-orbits. So to each simple positive relative root $a$ (i.e., element of ${}_F\Delta$) there is a $\Gamma_F$-orbit of simple positive absolute roots $a'$ (i.e., elements of $\Delta$) such that $a'|_{S_{F_s}} = a$ via the identification of ${\rm{X}}(S_{F_s})$ with ${\rm{X}}(S)$. -We can start cooking with gas. For each $a \in \Phi(G,S)$ (such as an element of ${}_F\Delta$, but can be more general for a moment) we have the associated codimension-1 $F$-subtorus $S_a = (\ker a)^0_{\rm{red}} \subset S$, and $Z_G(S_a)$ is a connected reductive $F$-group containing $T$ and $S$ in which $S_a$ is central but $S$ is not since ${\rm{Lie}}(Z_G(S_a)) = {\rm{Lie}}(G)^{S_a}$ contains the nonzero $a$-weight space for $S$ in ${\rm{Lie}}(G)$. In case $G$ is quasi-split, -so $P$ as above is a Borel $F$-subgroup, so is $P \cap Z_G(S_a)$ -(as Borel subgroups meet centralizers of their subtori in Borel subgroups -for general smooth connected affine groups over fields). Thus, -if $G$ is quasi-split then so is $Z_G(S_a)$ and hence its derived group. -The $F$-group $G_a := \mathscr{D}(Z_G(S_a))$ is connected semisimple containing the 1-dimensional isogeny complement -$S'_a := (S \cap G_a)^0_{\rm{red}}$ to $S_a$ in $S$ as a maximal split $F$-torus, and all absolute roots restricting to $a_{F_s}$ on $S_{F_s}$ -have their weight spaces in ${\rm{Lie}}(G_{F_s}) = {\rm{Lie}}(G)_{F_s}$ supported inside ${\rm{Lie}}(G_a)_{F_s}$! Also, -$T \cap G_a$ is a maximal $F$-torus in $G_a$ (since $G_a$ is normal in -$Z_G(S_a)$) and is an isogeny complement in $T$ to the maximal central $F$-torus in $Z_G(S_a)$. The absolute weight spaces and root groups for $(Z_G(S_a), T)$ coincide with those for $(G_a, T \cap G_a)$ since -passage to the derived group never affects such notions. -Thus, for any questions about the root groups or root spaces over $F_s$ associated to absolute roots extending $a_{F_s}$, we lose nothing by passing to $(G_a, S'_a, T \cap G_a)$ -(also preserving the quasi-split property if $G$ is quasi-split, -as seen above) and we gain the property of $G$ being connected semisimple with $F$-rank equal to 1. We apply the preceding with $a \in {}_F\Delta$ and are interested in the root groups and root spaces attached to the $\Gamma_F$-orbit $\pi^{-1}(a) \subset \Delta$. This brings us to the case of $F$-rank 1 with the singleton $\{a\}$ a basis for the rank-1 (possibly non-reduced!) relative root system. (One has to make to small argument involving the insensitivity of the relative root system upon passage to the derived group, applied to $Z_G(S_a)$ and its derived group, to ensure that $a|_{S'_a}$ really is a basis for the relative root system for $(G_a, S'_a)$. This argument is easy, so we omit it here.) -Passage to the simply connected central cover has no effect on the relative root system, nor on the quasi-split property (if it holds), nor on the absolute root system, nor on the absolute root groups or their root spaces or even the relative root spaces, since the Lie algebra of the (possibly inseparable) central isogeny is contained in the Lie algebra of any maximal $F$-torus. Thus, for the purpose of the question posed we may and do assume $G$ is not only connected semisimple with $F$-rank equal to 1 but is also simply connected. -Now, at last, the mythic (and canonical!) "field of definition" $F_a$ in the question (which will be finite separable over $F$) emerges from the shadows. This rests on the following fundamental result (the proof of which is part of the same circle of ideas that prove the Theorem stated in my answer to Automorphism of restriction of scalars): -Theorem. If $G$ is a nontrivial connected semisimple group over a field $F$ and it is simply connected then $G \simeq {\rm{R}}_{F'/F}(G')$ for a nonzero finite etale $F$-algebra $F'$ and a smooth affine $F'$-group $G'$ whose fiber over each factor field of $F'$ is connected semisimple, simply connected, and absolutely simple. Moreover, the pair $(F'/F, G')$ is unique up to unique isomorphism in the sense that if $(F''/F, G'')$ is a second such pair then any $F$-isomorphism -${\rm{R}}_{F''/F}(G'') \simeq {\rm{R}}_{F'/F}(G')$ arises from a unique pair $(\alpha, \varphi)$ consisting of an $F$-algebra isomorphism $\alpha:F' \simeq F''$ and a group isomorphism $\varphi:G'\simeq G''$ over $\alpha$. -This result is basically just an exercise in Galois descent and the fact that the root datum of a split simply connected group is uniquely the direct product (or direct sum?) of root data for its irreducible components. A reference for a proof is the same as for the Theorem I stated in my answer at the link above, namely Proposition A.5.14 in the book Pseudo-reducitve Groups. -Coming back to the case of interest, we use the preceding Theorem to describe our connected semisimple $G$ that is simply connected with $F$-rank 1 as ${\rm{R}}_{F'/F}(G')$ for some nonzero finite etale $F$-algebra $F'$ and $F'$-group $G'$ as in the Theorem. Letting $\{F'_i\}$ be the factor fields of $F'$ and $G'_i$ the $F'_i$-fiber of $G'$, we have -$$G = {\rm{R}}_{F'/F}(G') \simeq \prod_i {\rm{R}}_{F'_i/F'}(G'_i)$$ -where each $G'_i$ is a connected semisimple $F'_i$-group that is absolutely simple and simply connected. Maximal split $F$-tori in such a product have the unique form $\prod S_i$ -where $S_i$ is the maximal $F$-split torus in ${\rm{R}}_{F'_i/F}(S'_i)$ for a maximal split $F'_i$-torus $S'_i$ in $G'_i$ (see Proposition A.5.15(2) in Pseudo-reductive Groups for a proof in a somewhat broader setting); note that $\dim S_i = \dim S'_i$. Thus, exactly one of the factors ${\rm{R}}_{F'_i/F}(G'_i)$ has $F$-rank 1 (with the corresponding $G'_i$ of $F'_i$-rank 1) and the others are $F$-anisotropic. -Likewise, $T = {\rm{R}}_{F'/F}(T')$ for a unique maximal $F'$-torus $T' \subset G'$ and moreover $T' \supset S'$. We have $T' = \coprod T'_i$ for a unique maximal $F'_i$-torus $T'_i \subset G'_i$, and $S'_i \subset T'_i$. Thus, $T = \prod {\rm{R}}_{F'_i/F}(T'_i)$. Note in particular -that all $T_{F_s}$-weight spaces in ${\rm{Lie}}(G_{F_s})$ that extend $a_{F_s}$ are supported inside the $F_s$-fiber of the unique -$F$-isotropic factor ${\rm{R}}_{F'_i/F}(G'_i)$, and that the corresponding -root groups over $F_s$ are also contained in the $F_s$-fiber of that direct factor. -By Galois descent one also shows rather generally that if -$k'$ is a nonzero finite etale algebra over a field $k$ and $H'$ is a smooth affine $k'$-group with connected fibers over the factor fields of $k'$ then $Q' \mapsto {\rm{R}}_{k'/k}(Q')$ is a bijective correspondence between the set of parabolic $k'$-subgroups of $H'$ and the set of parabolic $k$-subgroups of ${\rm{R}}_{k'/k}(H')$, inclusion-preserving in -both directions. Thus, $P = {\rm{R}}_{F'/F}(P')$ -for a unique minimal parabolic $F'$-subgroup $P' \subset G'$ -that necessarily contains $T'$, and $P'$ is a Borel $F'$-subgroup if and only if $P$ is a Borel $F$-subgroup. Note -that $P' = \coprod P'_i$ for a parabolic $F'_i$-subgroup $P'_i \subset G'_i$ that is clearly minimal and contains $S'_i$ for each $i$. -Hence, for our purposes we can focus entirely on the factor with $F$-rank 1, which means that we can assume $G = {\rm{R}}_{F'/F}(G')$ where -$F'/F$ is a finite separable extension field and $G'$ is a connected semisimple $F'$-group that is absolutely simple and simply connected with $F'$-rank 1. We have $S \subset {\rm{R}}_{F'/F}(S')$ -for a unique $F$-split torus $S' \subset G'$ with dimension 1, -and it is easy to check by inspection of Lie algebras (and using the good behavior of Lie algebras with respect to Weil restriction) that $\Phi(G',S') = \Phi(G,S)$ (equal to $\{\pm a\}$ or $\{\pm a, \pm 2a\}$). -Finally, the payoff: we claim that the sought-after field $F_a$ -is equal to $F'$ or to a quadratic Galois extension of $F'$. The reason is due to the quasi-split hypothesis that we shall now use. The point is that although it is hopeless to describe the possibilities for the absolutely simple $G'$ over $F'$ -with $F'$-rank 1 in general, in the quasi-split case -the possibilities are very explicit: -Proposition. If $H$ is a connected semisimple group over a field $k$ and it is absolutely simple, simply connected, and quasi-split with $k$-rank equal to $1$ then $H$ is either $k$-isomorphic to ${\rm{SL}}_2$ or $H$ is the quasi-split special unitary group ${\rm{SU}}_3(K/k)$ associated to a quadratic Galois extension $K/k$. In the latter case, $K/k$ is determined by $G$ up to $k$-isomorphism. -Proof. If the absolute rank $n$ of $H$ is equal to 1 then the 1-dimensional maximal split tori are in fact maximal $k$-tori, so $H$ would be split and thus the conclusion clear. We therefore may assume $n \ge 2$. Let $H_0$ denote the split $k$-form of $H$, so $H$ is quasi-split form on $H_0$. The diagram $\Delta$ of $H_0$ is connected by absolute simplicity, and has $n$ vertices. -The automorphism group of the based root datum for $H_0$ coincides with the automorphism group of $\Delta$ since $H_0$ is simply connected. Thus, the quasi-split $k$-forms of $H_0$ are classified up to isomorphism by the pointed set ${\rm{H}}^1(k, {\rm{Aut}}(\Delta))$ of conjugacy classes of continuous homomorphism ${\rm{Gal}}(k_s/k) \rightarrow {\rm{Aut}}(\Delta)$; the dictionary goes via inner-twisting by pinned automorphisms. This pointed set is trivial unless $\Delta$ admits non-trivial automorphisms, so by the classification of reduced and irreducible root systems the only possibilities we need to consider for $\Delta$ are A$_n$ ($n \ge 2$), D$_n$ $(n \ge 4)$, and E$_6$. -The maximal $k$-tori in the Borel subgroups are the Weil restriction of ${\rm{GL}}_1$ from the finite etale $k$-algebra $k'$ associated to the ${\rm{Gal}}(k_s/k)$-action on $\Delta$: we associated to each Galois-orbit of vertices the subfield of $k_s$ associated to the open subgroup fixing a choice of vertex in the orbit. In particular, if there are at least 2 Galois orbits then the $k$-rank of the associated quasi-split form is at least 2 since ${\rm{R}}_{K/k}({\rm{GL}}_1)$ contains ${\rm{GL}}_1$ as a $k$-subgroup for any finite separable extension field $K/k$. But we assumed $H$ has $k$-rank equal to 1, so we only need to consider cases when ${\rm{Gal}}(k_s/k)$ acts transitively on the set of vertices of $\Delta$. This can only happen when ${\rm{Aut}}(\Delta)$ acts transitively on $\Delta$. Inspection of the diagrams for -A$_n$ ($n \ge 2$), D$_n$ ($n \ge 4$), and E$_6$ shows that the latter only happens for the A$_2$ diagram. The quasi-split groups of type A with absolute rank $n \ge 2$ are exactly the special unitary groups ${\rm{SU}}_{n+1}(K/k)$ associated to quadratic Galois extensions $K/k$ (and these $k$-groups have $k$-rank equal to $n-1$). The extension $K/k$ is unique up to isomorphism since it is the splitting field for the Galois action on the absolute diagram. -QED Proposition -We conclude that either (i) $G = {\rm{R}}_{F'/F}({\rm{SL}}_2)$ -or (ii) $G = {\rm{R}}_{F'/F}({\rm{SU}}_3(F''/F'))$ for a quadratic Galois extension $F''/F'$. In the first case we will show $F'/F$ satisfies the desired properties to deserve being called $F_a$, and in the second case likewise for $F''/F$. We emphasize that $F'$ is an abstract finite separable extension of $F$, not built inside $F_s$! And of course $F'$ will generally not be Galois over $F$, as was noted near the start. The preceding method does give intrinsic meaning to $F'/F$ in terms of $a$ (even without reference to $P$). A slightly annoying feature is that we have not pinned down $F''/F'$ canonically, but only up to $F'$-isomorphism. -Let's first consider case (i). We have $G = {\rm{R}}_{F'/F}(G')$ -where $G' \simeq {\rm{SL}}_2$, but we do not choose such a latter isomorphism (as we want to keep everything canonical). The key point is that the geometric unipotent radical of the Borel $F'$-subgroup $P' \subset G'$ determined by $a$ descends to a unipotent smooth connected normal $F'$-subgroup $U' \subset P'$ that is split of dimension 1. This $U'$ is normalized by $S'$, with $S'$ acting through $a$ on the 1-dimensional Lie algebra $V'$ of $U'$ over $F'$. There is a unique $S'$-equivariant $F'$-isomorphism $f:U' \simeq \underline{V}'$ to the vector group over $F'$ associated to $V'$ such that ${\rm{Lie}}(f)$ is the identity map on $V'$. -Thus, -the analogous $F$-unipotent radical $U = {\rm{R}}_{F'/F}(U')$ for ${\rm{R}}_{F'/F}(P') = P'$ is identified with the vector group -$\underline{V}$ associated to the $F$-vector space $V$ underlying $V'$, -and this is compatible with the action of $S \subset {\rm{R}}_{F'/F}(S') = T$. The $F_s$-fiber $U_{F_s}$ is the unipotent radical of the Borel $F_s$-subgroup $P_{F_s}$. The compatibility with Weil restriction with extension of the ground field and the canonical isomorphism of $F_s$-algebras $F' \otimes_F F_s \simeq \prod_{\sigma} F_s$ -via $y' \otimes t \mapsto (\sigma(y')t)$ (with $\sigma$ varying -through the set $\Sigma$ of $F$-embeddings of $F'$ into $F_s$) identifies $T_{F_s}$ with $\prod_{\sigma} (S' \otimes_{F',\sigma} F_s)$ and $U_{F_s}$ with $\prod_{\sigma} (\underline{V}' \otimes_{F',\sigma} F_s)$. -Thus, if we pick an $F'$-basis of $V'$ then this defines -an $F'$-isomorphism $x':\underline{V}' \simeq \mathbf{G}_{\rm{a}}$ -so that $U_{F_s}$ is identified with $\mathbf{G}_{\rm{a}}^{\Sigma}$ -making the $\mathbf{G}_{\rm{a}}$-factors exactly the $T_{F_s}$-root groups. This defines a canonical identification of -$\Sigma$ with the set of absolute roots $a'$ extending $a_{F_s}$; -another way to make that bijection is to identify $S'$ with $S_{F'}$ via the map $S_{F'} \rightarrow S'$ corresponding to the inclusion $S \hookrightarrow {\rm{R}}_{F'/F}(S')$ and then identify ${\rm{R}}_{F'/F}(S')_{F_s}$ with $\prod_{\sigma} (S' \otimes_{F',\sigma} F_s)$, noting that each absolute root factors through the projection to exactly one of those factors $S' \otimes_{F',\sigma} F_s$. -By design, for any $a'$ and any $\gamma \in {\rm{Gal}}(F_s/F)$, $x_{\gamma(a')}$ is the scalar extension of $x_{a'}$ through the $F$-automorphism $\gamma:F_s \simeq F_s$. Voila, so $F'$ has the desired property for $F_a$ (and we see moreover that the true choice involved in picking those root vectors is a single choice, namely a basis of the canonical $F'$-line $V' = {\rm{Lie}}(U')$. That takes care of case (i)! -Next, we turn to case (ii): $G \simeq {\rm{R}}_{F'/F}({\rm{SU}}_3(F''/F'))$ for a quadratic Galois extension $F''/F'$ (unique up to $F'$-isomorphism). In this case the relative root system $\Phi(G,S)$ is BC$_1$ (see Prop. 17.1.6(iii) with $n=3$ in Springer's book Linear Algebraic Groups); explicity, the $F'$-unipotent radical $U'$ of the Borel $F$-subgroup $P'$ is a Heisenberg group, with $S'$ acting through $2a$ on the Lie algebra of the 1-dimensional $F'$-split center and through $a$ on the 2-dimensional $F'$-split vector group quotient $U'/Z_{U'}$. In particular, $a$ is a basis for ${\rm{X}}(S')$, which is to say that $a:S' \rightarrow {\rm{GL}}_1$ (corresponding to $a: S \rightarrow {\rm{GL}}_1$) is an $F'$-isomorphism. -The root groups for the two absolute roots of $(G', T')$ corresponding to root lines in ${\rm{Lie}}(U'/Z_{U'})_{F'_s}$ give a direct product decomposition of $(U'/Z_{U'})_{F'_s}$ via the scalar extension of -the $F'$-homomorphism $U' \rightarrow U'/Z_{U'}$. -By explicit inspection of the construction of ${\rm{SU}}_3(F''/F')$, we can identify $T'$ with ${\rm{R}}_{F''/F'}(S'_{F''}) \simeq {\rm{R}}_{F''/F'}({\rm{GL}}_1)$ (latter isomorphism via $a$!) respecting the inclusion of $S'$ into each such that the linear $T'$-action on the 2-dimensional $F'$-vector space ${\rm{Lie}}(U'/Z_{U'})$ arises from a (necessarily unique) $F''$-vector space structure. Thus, -for the $F''$-line $L := {\rm{Lie}}(U'/Z_{U'})$ there is a unique -$T'$-equivariant isomorphism $U'/Z_{U'} \simeq {\rm{R}}_{F''/F'}(\underline{L})$ inducing the identity on Lie algebras, so an $F''$-basis of $L$ plays the same role as an $F'$-basis of $V'$ in case (i) to define root vectors of the desired type (thereby justifying that $F''$ deserves to be called $F_a$ (with the annoyance that $F''$ is only unique up to $F'$-isomorphism and not shown to be truly canonical over $F$).<|endoftext|> -TITLE: Jack polynomials and the Witt algebra -QUESTION [5 upvotes]: The symmetric Jack polynomials $J_n^{\alpha}(x_1,x_2,..,x_{n+1})$, a special subset of the symmetric Jack functions presented in Stanley's paper in equation a) on page 80, can be represented by the action of the operator $x^{1+\alpha}d/dx$, which specializes to reps of the Witt algebra, or centerless Virasoro algebra, for integer exponents. -Can someone provide some Web-accessible references on the relation of any subsets of the general Jack polynomials to the Witt algebra, particularly if couched in terms of Witt diff ops? - -REPLY [2 votes]: There are two very famous instances of Jack polynomials in relationship to the Virasoro algebra (there are some others, but they very often seem to be related to one of these): -Katsuhisa Mimachi and Yasuhiko Yamada. Singular vectors of the Virasoro algebra in terms of Jack symmetric polynomials. Comm. Math. Phys. -Volume 174, Number 2 (1995), 447-455. -B. Feigin, M. Jimbo, T. Miwa, E. Mukhin. A differential ideal of symmetric polynomials spanned by Jack polynomials at rβ = -(r=1)/(k+1). International Mathematics Research Notices, Volume 2002, Issue 23, 2002, Pages 1223–1237. -Perhaps you can explain a little bit more about what you want/expect to get further references!<|endoftext|> -TITLE: The expected square of the determinant of a random row stochastic matrix -QUESTION [13 upvotes]: In this -question Anthony Quas asks about the expected absolute value of -the determinant of an $n\times n$ row stochastic matrix $A$, where -the rows are independently selected from the uniform distribution on -the unit $(n-1)$-dimensional simplex $x_1+\cdots+x_n=1$, $x_i\geq -0$. I can show by a messy computation that the integral over all such -matrices of $(\det A)^2$ is $1/(n+1)!^{n-1}$. Is there a -noncomputational reason for such a simple value? - -REPLY [9 votes]: Hidden in the comment by Victor Kleptsyn is a really nice argument. Since nobody upvoted that comment yet (I just did) it's probably worth expanding it. -From a probabilistic approach it's more natural to show the equivalent statement that the expectation of $(\det A)^2$ equals $\frac{(n+1)!}{(n(n+1))^n}$. The trick is to realize the uniform measure on the simplex as the distribution of -$$ (X_1,\dots,X_n) := \frac{(Y_1,\dots,Y_n)}{Y_1 + \cdots +Y_n} $$ -where $(Y_i)$ are i.i.d. exponential random variables (say, of parameter 1). All you need to know is that $\mathbf{E} [Y_i] =1$ and $\mathbf{E} [Y_i^2] =2$. Moreover in this representation $(X_i)$ are independent from $S:=Y_1+\cdots+Y_n$. It follows that we have the following identity in distribution -$$ B = D A $$ -where $B$ is a matrix with i.i.d. exponential entries and $D$ a diagonal matrix whose entries are independent copies of $S$, with $A$ and $D$ moreover independent. Therefore -$$ \mathbf{E} [\det B^2] = \mathbf{E} [\det D^2] \cdot \mathbf{E} [\det A^2] .$$ -It is very easy to check that $\mathbf{E} [\det D^2] = (n(n+1))^n$, and $\mathbf{E} [\det B^2]$ can be expanded as -$$ \sum_{\sigma,\tau \in \mathfrak{S}_n} \varepsilon(\sigma) \varepsilon(\tau) -2^{Fix(\sigma \tau^{-1})} = n! \sum_{\pi \in \mathfrak{S}_n} \varepsilon(\pi) -2^{Fix(\pi)} = n! \det \begin{pmatrix} 2 & 1 & \cdots & 1 \\ 1 & 2 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \vdots \\ 1 & \cdots & \cdots & 2 \end{pmatrix} = (n+1)!$$ -(Here $Fix$ denotes the number of fixed points.)<|endoftext|> -TITLE: Most manifolds are hyperbolic? -QUESTION [28 upvotes]: I heard the claim as in the title for a long time, but can not find the precise reference for this claim, what's the reference with proof for this claim? Thanks for the help. -To be more precise, is there a canonical topology structure on the space $\Omega$ of all compact $n$-dim smooth manifolds, such that for any compact smooth $n$-dim manifold $M^n$, any neighborhood $U$ of $M^n$ in $\Omega$, there is a $n$-dim smooth manifold $N^n\in U$ such that $N^n$ admits a Riemannian metric with curvature equal to $-1$. Everything in my mind is just Riemannian hyperbolic, no complex structure involved. - -REPLY [10 votes]: Some answers here explain why in small dimensions being hyperbolic is generic. Maybe surprisingly, in some sense being hyperbolic is actually rare in higher dimensions (while in other senses it is generic, as Igor Belegradek explains in his answer). -In "Counting hyperbolic manifolds" Burger-Gelnader-Lubotzky-Mozes proved that for $d\geq 4$ there is a constant $c(d)$ such that for $V$ large enough, the number of $d$-dimensional hyperbolic manifolds of volume at most $V$ is bounded by $V^{c(d)V}$. -Note that by Mostow Rigidity, a given manifold carries at most one hyperbolic structure, and if it does, this manifold is completely determinded by its fundamental group. What BGLM do is actually counting possible fundamental groups. -On the contrary, in dimension 3 there exist infinitely many different (pairwise non-homeomorphic) compact hyperbolic manifolds of uniformly bounded volume (examples could be obtained by Dehn filling a knot complement). -By taking a product with a $(d-3)$-dimensional torus, one sees that also changing "hyperbolic" to "non-positively curved" (sectional curvature in $[-1,0]$) gives infinitely many compact manifolds of uniformly bounded volume. -Maybe I should (and maybe I shouldn't) note that one can generalize the above counting result to the realm of all pinched negatively curved manifolds of dimension $\geq 5$. We do this in a forthcoming paper with Gelander and Sauer. For this one should relay on Farrel-Jones Theorem instead of Mostow Rigidity. - -REPLY [4 votes]: Although this does not answer your question, there are partial answers in dimension 3. -For example, if you construct an orientable 3-manifold via a random Heegaard splitting (constructing the gluing map as a product of Dehn twists, this is where the "random" part comes in) then most 3-manifolds are hyperbolic. -https://arxiv.org/abs/0809.4881 -On a contrary note, I think the answer to your question will depend heavily on how you consider your manifolds to be generated. Some processes are biased towards things like hyperbolic manifolds. Others have drastically different biases. In that regard I don't think there is any one answer to your question.<|endoftext|> -TITLE: factorization of polynomials wrt the major index stat -QUESTION [8 upvotes]: Let $\mathfrak{S}_n$ be the permutation group on $\{1,\dots,n\}$. Given $\pi=\pi_1\pi_2\dots\pi_n\in\mathfrak{S}_n$, its major index statistic is denoted maj$(\pi)$. Define the polynomials -$$Q_{n,k}(x):=\sum_{\pi\in\mathfrak{S}_n}x^{\text{maj}(\pi)+\pi_n+\pi_{n-1}+\cdots+\pi_{n-k}}.$$ -EDIT. -maj can be replaced by the inversion number inv without affecting $Q$. This is true due to the work of Foata and Schutzenberger on the equi-distribution of the two statistics (as multisets). However, Fedor's question made me reflect again. Is it still true that -$\{\text{maj$(\pi)+\pi_n+\dots+\pi_{n-k}$}: \pi\in\mathfrak{S}_n\}$ and -$\{\text{inv$(\pi)+\pi_n+\dots+\pi_{n-k}$}: \pi\in\mathfrak{S}_n\}$ are equi-distributed? If yes, then Fedor's answer is complete. - -CLAIM. $\,\,$ Experiment supports that, for each $n$ and $k$, we have - $$Q_{n,k}(x) -=\binom{n}{k+1} x^{(k+1)n-\binom{k+1}2}\prod_{i=0}^k\left(1+x+\cdots+x^i\right)\prod_{j=0}^{n-k-2}\left(1+x+\cdots+x^j\right).$$ - Any proof? - -The above exploration was motivated by this paper. - -REPLY [5 votes]: For inv instead of maj this is rather clear. I claim that when we fix a $(k+1)$-element subset $K$ of $\{1,2,\ldots,n\}$, the polynomial -$$\sum_{\pi}x^{\operatorname{inv}(\pi)+\pi_n+\pi_{n-1}+\cdots+\pi_{n-k}} ,$$ -where the sum is running over the $\pi\in\mathfrak{S}_n$ with $\{\pi_{n-k},\dots,\pi_n\} = K$, -equals -$$x^{(k+1)n-\binom{k+1}2}\prod_{i=0}^k\left(1+x+\cdots+x^i\right)\prod_{j=0}^{n-k-2}\left(1+x+\cdots+x^j\right).$$ -Indeed, let us count the number of inversions $(\pi_m<\pi_j)$ for which $j\pi_m|$. When we sum up by all $m=n-k,\dots,n$, we get $n(k+1)-\sum_{m\geqslant n-k} \pi_m-\binom{k+1}2$. So, $\operatorname{inv}(\pi)+\sum_{m\geqslant n-k} \pi_m$ equals $n(k+1)-\binom{k+1}2$ plus the number of inversions inbetween first $n-k-1$ elements plus the number of inversions inbetween last $k+1$ elements. These are independent and have known generating polynomials $\prod_{i=1}^s(1+x+\dots+x^{i-1})$, for $s=n-k-1$ or $s=k+1$ respectively. Thus the result. -I doubt that for maj it is true, please recheck. It should not be divisible by such a high power of $x$, since it may appear that maj is small, say, maj$=n-k-1$, and $\{\pi_{n-k},\dots,\pi_n\}=\{1,2,\dots,k+1\}$.<|endoftext|> -TITLE: Can the algebraic geometry of schemes be developed internally in topoi? -QUESTION [8 upvotes]: Using the internal logic of a topos it's often possible to derive newer theorems about sheaves from earlier ones about simpler objects, assuming that you can prove the earlier ones constructively. In addition, Barr's Theorem allows one to directly reuse classical results when they have a geometric statement relative to a geometric theory. -How far can you take this? Can scheme-theoretic algebraic geometry can be developed internally in the appropriate topoi, or are there some notions that have inherently extrinsic definitions and proofs? - -REPLY [9 votes]: The notes of Ingo Blechschmidt, Using the internal language of toposes in algebraic geometry cover this. See also his lecture at Topos à l'IHÉS of the same name. -His work very much extends Hakim's thesis.<|endoftext|> -TITLE: counting monomials and integrality -QUESTION [7 upvotes]: For $n\in\mathbb{Z}^{+}$, consider the polynomials -$$P_n(x)=\prod_{k=0}^{n-1}(x^n-x^k).$$ - -QUESTION. Is it possible to find a closed formula for the number of monomials in $P_n(x)$, after expansion? - -Those interested in such enumeration may like to read this. -Here is an assertion which might be possible to prove. - -CLAIM. For any $m\in\mathbb{Z}$, the following are all integers: - $$\frac1{n!}P_n(m).$$ - -UPDATE. If it helps, I've found a stronger divisibility when $m$ is a prime $p$. -$$\frac{P_n(p)}{n!\,\cdot p^{\eta_p(n)}}=\frac{p^{\binom{n}2}\prod_{k=1}^n(p^k-1)}{n!\,\cdot p^{\eta_p(n)}}$$ -where $\eta_p(n)=\nu_p\left((p\lfloor n/p\rfloor)!\right)$ and $\nu_p(m)$ is the $p$-adic valuation of an integer $m$. - -REPLY [4 votes]: The claim is correct. We begin with a standard lemma. -Lemma: Let $p$ be an odd prime, and $x$ an integer coprime to $p$. Let $k$ be a positive integer and let $o$ be the multiplicative order of $x$ modulo $p$. Then $$v_p(x^k-1) = \begin{cases} v_p(x^o-1) + v_p(\frac{k}{o}) & o \mid k \\ 0 & o \nmid k \end{cases}.$$ -We prove the claim by comparing the $p$-adic valuation of $P_n(m)$ and $n!$ for every prime $p \le n$ (since all prime divisors of $n!$ are $\le n$). We skip the primes in the set $\{ q: q \mid m \} \cup \{2\}$ (they may be treated separately). The lemma implies that your claim is equivalent to the following inequalities: -$$(*) \forall p\le n (\text{odd, coprime to }m): v_p((\frac{n}{o})!) + \lfloor \frac{n}{o} \rfloor v_p(m^o-1) \ge v_p(n!),$$ -$$\text{where }o \text{ is the multiplicative order of m modulo p},$$ -and slightly different inequalities for $p=2$ and $p \mid m$. -We have: -$$v_p((\frac{n}{o})!) + \lfloor \frac{n}{o} \rfloor v_p(m^o-1) \ge v_p((\frac{n}{o})!) + \lfloor \frac{n}{o} \rfloor = \sum_{i \ge 1} \lfloor \frac{n}{op^i} \rfloor + \lfloor \frac{n}{o} \rfloor$$ -$$= \sum_{i \ge 1} \lfloor \frac{pn}{op^i} \rfloor \ge \sum_{i \ge 1} \lfloor \frac{n}{p^i} \rfloor = v_p(n!), $$ -where the last inequality follows from the fact that $o -TITLE: How to understand the explicit formula for zeta function? -QUESTION [6 upvotes]: The explicit formula for the zeta function, e.g. -$$\psi_0(x) = \dfrac{1}{2\pi i}\int_{\sigma-i \infty}^{\sigma-i \infty}\left(-\dfrac{\zeta'(s)}{\zeta(s)}\right)\dfrac{x^s}{s}ds=x-\sum_\rho\frac{x^\rho}{\rho} - \log(2\pi) -\dfrac{1}{2}\log(1-x^{-2})$$ -(where $\psi_0$ is the normalized Chebyshev function, $\sigma > 1$) relating zeroes of the $\zeta$ function and primes is certainly of much importance. -How does one understand this beyond what's normally given by a standard textbook on analytic number theory? -According to Connes - -The noncommutative space of adele classes of a global field provides a framework to interpret the explicit formulas of Riemann-Weil in number theory as a trace formula. - -Can someone elaborate on this? How do we interpret this as a trace formula? - -REPLY [6 votes]: The analogy between zeta functions and trace formulas goes at least to Selberg, when he proved his famous trace formula for hyperbolic surfaces and the result turned out to resemble Weil's generalization of Riemann's explicit formula quite a bit. -Riemann-Weil explicit formula: -\begin{equation*} -\begin{split} -\sum_\gamma h(\gamma)=\frac{1}{2\pi}\int_{-\infty}^{+\infty} h(r) \frac{\Gamma'}{\Gamma}\left(\frac{1}{4}+\frac{1}{2}ir \right)dr &+h\left(\frac{i}{2}\right)+h\left(-\frac{i}{2}\right)\\ -&-g(0)\ln\pi-2\sum_{n=1}^\infty\frac{\Lambda(n)}{\sqrt{n}}g(\ln n) -\end{split} -\end{equation*} -Selberg trace formula: -\begin{equation*} -\begin{split} -\sum_{n=0}^\infty h(r_n)=\frac{\mu(F)}{4\pi}\int_{-\infty}^{+\infty} rh(r) \tanh (\pi r)dr &+\sum_n \Lambda(n) g(\ln N(n)) -\end{split} -\end{equation*} -There's a big literature on this type of question, and the interplay of number theory, spectral analysis, mathematical physics... I recomend section 3 of Lagarias' survey "The Riemann Hypothesis: Arithmetic and Geometry" for references. -Connes approach started with - -"Formule de trace en géométrie non-commutative et hypothèse de Riemann" (1996) - -And was completed (as far as I know) in - -"Trace formula in noncommutative geometry and the zeros of the Riemann zeta function" (1998) - -The main result says that given a global field $K$ and a character $\alpha=\prod_v \alpha$ of the space of adele classes $A/K$, and any adecuate test function $h$, we have: -$$\underbrace{\widehat{h}(0)+\widehat{h}(1)-\sum \widehat{h}(\mathcal{X},\rho)}_{\text{spectral side}}=\underbrace{\sum_v \int_{K_v^*}' \frac{h(u^{-1})}{|1-u|}d^*u}_{\text{arithmetic side}}$$<|endoftext|> -TITLE: Is the $H$-space structure on $S^7$ associative up to homotopy? -QUESTION [23 upvotes]: Endow $S^7$ with a structure of an $H$-space induced from multiplication in octonions $\mathbb{O}=\mathbb{R}^8$. It is not associative as octonion multiplication is not associative. -Is it associative up to homotopy, i.e. are maps $m(m(-,-),-):S^7\times S^7\times S^7\to S^7$ and $m(-,m(-,-)):S^7\times S^7\times S^7\to S^7 $ homotopic? - -REPLY [6 votes]: This response is to the question raised above whether S^7_(2), the 2-localization -of the 7-sphere, can admit a homotopy associative multiplication. The answer is no. This fact was established by I.M. James. A later proof by Daciberg Goncalves, "Mod 2 homotopy associative -H-spaces" in Geometric Applications of Homotopy Theory I 198-216, Edited by M.G.Barratt and M.E. Mahowald, Lecture Notes in Mathematics 657, Springer-Verlag (1978), revealed the universal nature of the obstruction. The proof involves a secondary cohomology operation and has wider implications. In the same volume, -John Hubbock gave a K-theory proof. The calculation of the secondary operation is also presented in the book "Secondary Cohomology Operations" Graduate Studies in Mathematics 49 AMS, by the undersigned. The details are on pages 192-3, and 217-8.<|endoftext|> -TITLE: Terrible tilers for covering the plane -QUESTION [27 upvotes]: Let $C$ be a convex shape in the plane. -Your task is to cover the plane with copies of $C$, each under any rigid motion. -My question is essentially: What is the worst $C$, the shape that forces the most -wasteful overlap? -To be more precise, assume $C$ has unit area. -Let $n_C(A)$ be the fewest copies of $C$ (under any rigid motions) that suffice to cover a disk -of area $A$. I seek the $C$ that maximizes the "waste": -$$ w(C) = \lim_{A \to \infty} n_C(A) / A \;.$$ -So if $C$ is a perfect tiler of the plane, then $\lim_{A \to \infty} n_C(A) = A$ -(because $C$ has area $1$) -and $w(C)=1$, i.e., no waste. -Consider a regular pentagon $P$, which cannot tile the plane. Here is one way -to cover the plane with regular pentagons: - -          - - -If I've calculated correctly, this arrangement shows that $w(P) \le 1.510$. -So one could cover an area $A=100$ with about $151$ unit-area regular pentagons, -a $51$% waste. -I doubt this is the best way to cover the plane with copies of $P$ (Q3 below), but it is one -way. -Three questions. - -Q1. Is it known that the disk is the worst shape $C$ to cover the plane? - My understanding is that L.F.Tóth's paper[1], which I have not accessed, - establishes this for lattice tilings/coverings. Is it known for arbitrary coverings? - -Q1 Answered. Thanks to several, and especially Yoav Kallus, for pointing me -in the right direction. Q1 remains an open problem. In [2,p.15], what I call -the waste of a convex body $C$ is called $\theta(C)$. It is about $1.209$ for a disk. The best upperbound is $\theta(C) \le 1.228$ due to Dan Ismailescu, -based on finding special tiling "p-hexagons" in $C$. A p-hexagon has two -opposite, parallel edges of the same length. - -Q2. Since every triangle, and every quadrilateral, tiles the plane, - the first interesting polygonal shape is pentagons. - What is the most wasteful pentagon? -Q3. More specifically, what is the waste $w(P)$ for the regular pentagon? - - -[1] -L. Fejes Tóth, "Lagerungen in der Ebene auf der Kugel und im Raum." -Die Grundlehren der mathematischen Wissenschaften Vol. 65. Springer-Verlag, 1972. doi:10.1007/978-3-642-65234-9 -[2] -Brass, Peter, William OJ Moser, and János Pach. Research Problems in Discrete Geometry. Springer Science & Business Media, 2006. - -REPLY [15 votes]: Here is a better (possibly best) way of covering the plane with congruent regular pentagons: - -   - -The density of this covering is ${\sqrt5}/2 = 1.1180...$. -This covering is generated by the maximum-area $p$-hexagon contained in $P$, and is the thinnest among all coverings with $P$ that are of the double-lattice type. I conjecture that this density is minimum among all coverings with $P$, not just double-lattice ones. -For the notion of double-lattice see: Kuperberg, G.; Kuperberg, W. (1990), Double-lattice packings of convex bodies in the plane, Discrete and Computational Geometry, 5 (4): 389–397, MR 1043721 -This covering is obtained by a continuous transition starting with the best packing with regular pentagons, see -http://www.auburn.edu/~kuperwl/pent_movie.mp4<|endoftext|> -TITLE: Is Lagrange's Theorem equivalent to AC? -QUESTION [34 upvotes]: Lagrange's Theorem is most often stated for finite groups, but it has a natural formation for infinite groups too: if $G$ is a group and $H$ a subgroup of $G$, then $|G| = |G:H| \times |H|$. -If we assume AC, the we get a fairly straightforward proof (pick a representative of each coset, and then let your map $(G : H) \times H \to G$ be $(gH, h) \mapsto gh$). -However, somebody told me someone told them that the converse is true too, i.e. Lagrange's Theorem gives us AC. (Indeed, the wikipedia page for LT offhandedly mentions they're equivalent without a reference). Does anybody have a proof (or alternatively a proof AC is stronger than LT)? - -REPLY [8 votes]: Lagrange's Theorem does not follow from ZF. In fact, the following weaker statement also does not follow: -$\sf LT^{-}$: Let $H$ be a subgroup of $G$. Then $|H|$ divides $|G|$, i.e. there exists a set $A$ s.t. $|H| \times |A| = |G|$. -(The name LT$^{-}$ is not standard, but in the vein of the name LT$^{+}$.) -I stumbled upon the paper The Construction of Groups in models of set theory that fail the Axiom of Choice (Hickman, 1976) where he constructs a model of ZF with an amorphous group (a group whose carrier set is amorphous, i.e. an infinite set that isn't the disjoint union of two infinite sets). -He then goes on to prove several properties about amorphous groups, including the following: - -Suppose $G$ is an amorphous group, and $H \leq G$ a finite non-trivial subgroup. (Such a subgroup always exists: every element in $G$ has finite order as $G$ has no $\aleph_0$ subset.) -Suppose that there was a set $A$ and a bijection $f \colon H \times A \to G$. As $H$ is finite, $A$ must be infinite. But then for any $h_1 \neq h_2 \in H$, $f(\{h_1\} \times A)$ and $f(\{h_2\} \times A)$ are infinite disjoint subsets of $G$, contradicting $G$ being amorphous. - -So we need some 'choice' (i.e. a statement along the lines of 'no amorphous sets exist' or 'every infinite set is Dedekind-infinite') at the very least to have LT. How much is still unclear.<|endoftext|> -TITLE: naive de Rham cohomology fails for singular varieties -QUESTION [21 upvotes]: Let $X$ be a variety over a field $k$ of characteristic zero. If $X$ is smooth, algebraic de Rham cohomology defined as -$$ -H^n_{dR}(X / k)=\mathbb{H}^n(X, \Omega^\bullet_{X/k})\qquad (\star) -$$ is a good cohomological theory, for instance, is isomorphic tensor $\mathbb{C}$ to the singular cohomology of $X(\mathbb{C})$ once one chooses and embedding $k \hookrightarrow \mathbb{C}$. -It is "well known" that this fails for singular $X$ unless one considers more sophisticated definitions of de Rham cohomology, but I had difficulties finding a counterexample of, say, a singular variety $X$ such that the dimension of $(\star)$ is different from the dimension of singular cohomology. After a while, I found a paper by Arapura-Kang where the example of the plane curve -$$ -x^5+y^5+x^2y^2=0 -$$ is presented. The proof is not so easy. Does anybody know a simpler example? - -REPLY [23 votes]: The nonexactness of the naïve analytic de Rham complex for complex analytic spaces is discussed in the following papers. -H.J. Reiffen, Das Lemma von Poincaré fȕr holomorphs Differential-formen auf komplexen Räumen, Math Z. 101 (1967). -M. Sebastiani, Preuve d'une Conjecture de Brieskorn, Manuscripta math. 2 (1970). -K. Saito, Quasihomogene isolierte Singularitäten von Hyperflächen, Invent. math. 14 (1971). -There are many isolated hypersurface singularities which are not quasihomogeneous (in the sense that there are no local analytic coordinates in which the hypersurface is locally defined by a quasihomogeneous polynomial) and for any such one can globalize it to a projective complex hypersurface with one singular point, and by the above in this case the Euler characteristic of the singular cohomology differs from the Euler characteristic of the naïve de Rham cohomology.<|endoftext|> -TITLE: Binomial ID $\sum_{k=m}^p(-1)^{k+m}\binom{k}{m}\binom{n+p+1}{n+k+1}=\binom{n+p-m}{n}$ -QUESTION [5 upvotes]: Could I get some help with proving this identity? -$$\sum_{k=m}^p(-1)^{k+m}\binom{k}{m}\binom{n+p+1}{n+k+1}=\binom{n+p-m}{n}.$$ -It has been checked in Matlab for various small $n,m$ and $p$. I have a proof for $m=0$ that involves Pascal's rule to split it into two sums that mostly cancel, but this does not work for all $m$. I have looked at induction on various letters (and looked at simultaneous induction on multiple letters) without any success. -I would appreciate any help, algebraic or combinatoric. - -REPLY [7 votes]: It's a generating-function exercise. We have -$$ -\sum_{k=m}^\infty (-1)^{k+m} {k \choose m} x^{k-m} = (1+x)^{-(m+1)}, -$$ -and (with $j=p-k$) -$$ -\sum_{j=0}^\infty {n+p+1 \choose n+(p-j)+1} x^j -= \sum_{j=0}^\infty {n+p+1 \choose j} x^j = (1+x)^{n+p+1}. -$$ -The desired sum is the $x^{p-m}$ coefficient of the product -$(1+x)^{n+p-m}$ of these two power series, which is -${n+p-m \choose p-m} = {n+p-m \choose n}$, QED. - -REPLY [7 votes]: I wish to explain a modern (high tech) method called the Wilf-Zeilberger (WZ) technique which might help you (and anyone interested) with the present question and many others you encounter in the future. This will save you time from hunting the literature and comparing notes with the milliard hypergeometric formulas. -Zeilberger developed an accompanying algorithm package which is by now part of the symbolic softwares, Maple and Mathematica. In case you have access to Maple, you may download and run a freely available online copy of it from here. -Start with the discrete function -$$F(p,k)=(-1)^{k+m}\binom{k}{m}\binom{n+p+1}{n+k+1}\binom{n+p-m}{n}^{-1}.$$ -The above-mentioned algorithm furnished the companion function -$$G(p,k)=-\frac{F(p,k)(k-m)(n+k+1)}{(p+1-k)(n+p+1-m)}.$$ -Check (preferable using a symbolic software) that -$$F(p+1,k)-F(p,k)=G(p,k+1)-G(p,k).\tag1$$ -Convention: $\binom{a}b=0$ if $b>a$ or $b<0$. Now, sum both sides of (1) over all integers $k$, i.e. $\sum_{-\infty}^{\infty}$. Notice that the RHS of (1) vanishes because $\sum_{k\in\mathbb{Z}}G(p,k+1)=\sum_{k\in\mathbb{Z}}G(p,k)$. If we denote $f(p):=\sum_{k=m}^pF(p,k)=\sum_{\mathbb{Z}}F(p,k)$ then the LHS of (1) implies -$$f(p+1)=f(p).$$ -But, if $p=0$ then $f(0)=1$ and hence $f(p)=1$ identically. That means (after rewriting) -$$\sum_{k=m}^p(-1)^{k+m}\binom{k}{m}\binom{n+p+1}{n+k+1}=\binom{n+p-m}{n}$$ -as desired. - -REPLY [5 votes]: Such identities are often reduced to the Chu--Vandermonde's identity $\sum_{i+j=\ell} \binom{x}i\binom{y}j=\binom{x+y}\ell$ by using reflection formulae $\binom{x}k=\binom{x}{x-k}$, $\binom{x}k=(-1)^k\binom{k-x-1}k$. -In your case you may write -$$ -\sum_{k=m}^p(-1)^{k+m}\binom{k}{m}\binom{n+p+1}{n+k+1}= -\sum_{k=m}^p \binom{-m-1}{k-m}\binom{n+p+1}{p-k}= -\binom{n+p-m}{p-m} -$$ -as you need, so it is Chu--Vandermonde for $x=-m-1$, $y=n+p+1$, $\ell=p-m$.<|endoftext|> -TITLE: Church-Farb on the cohomology of pure braid groups and character polynomials, intuition behind proof of result? -QUESTION [14 upvotes]: Fix $n \ge 2$. Let $V_n$ be the $\binom{n}{2}$-dimensional vector space (over $\mathbb{C}$) generated by a set of vectors $\{w_{ij} : 1 \le i < j \le n\}$. Let $\bigwedge^* V_n$ be the exterior algebra on $V_n$. Note that $\bigwedge^* V_n$ is graded as an algebra (e.g. as a vector space) by $\bigwedge^* V_n = \bigoplus_{i = 0}^\infty \bigwedge^i V_n$, where of course $\bigwedge^i V_n = 0$ for $i > n$. Now let $\mathcal{P}_n$ be the quotient of the algebra $\bigwedge^* V_n$ by the ideal generated by all elements $R_{jkl}$, where $j$, $k$, $l$ are distinct and$$R_{jkl} := w_{jk} \wedge w_{kl} + w_{kl} \wedge w_{lj} + w_{lj} \wedge w_{jk}.$$Thus $\mathcal{P}_n$ is just $\bigwedge^* V_n$ where in addition we have the "Jacobi identity" $R_{jkl} = 0$ on all triples of generators. Note that the grading on $\bigwedge^* V_n$ induces a grading$$\mathcal{P}_n = \bigoplus_{i = 0}^n \mathcal{P}_n^i.$$It is a famous theorem of Arnol'd from 1969 that the algebra $\mathcal{P}_n$ is isomorphic as an algebraic to the cohomology of the pure braid group on $n$ strands, or equivalent the space of $n$-tuples of distinct points in the plane, with the $i$th cohomology group isomorphic to $\mathcal{P}_n^i$. -The symmetric group $S_n$ acts on $\mathcal{P}_n^1 = \wedge^1 V_n = V_n$ via$$\sigma \cdot w_{ij} = w_{\sigma(i) \sigma(j)}.$$This action induces an action of $S_n$ on each $\mathcal{P}_n^i$. -Consider the following result of Church and Farb, established in their paper "Representation theory and homological stability" here. - -Theorem (Church-Farb). For each $i$ there is a (unique) character polynomial $P(X_1, \ldots, X_r)$ in the cycle counting functions $X_i$ so that$$\chi_{\mathcal{P}_n^i} = P(X_1, \ldots, X_r) \text{ for all }n \ge 0.$$ - -Question. Is it possible anybody could give me their intuition behind the proof of this result? - -REPLY [21 votes]: This turns out to be a completely general phenomenon for configuration spaces on any open manifold, though we did not know this at the time; it came in the later paper "FI-modules and stability for representations of symmetric groups" by Church, Ellenberg, Farb (all references are to that paper). -Consider the following two categories: -FI = the category of finite sets and injections -FI# = the category of finite sets and "partially-defined injections" (Def 4.1.1) -The category FI arises naturally here because it parametrizes configuration spaces: having fixed a space M, associated to the finite set S is the configuration space ConfS(M) := Inj(S,M), and an injection S -> T defines a restriction ConfS(M) <- ConfT(M). Therefore the cohomology groups Hk(ConfS(M)) define an FI-module, meaning a functor from FI to abelian groups. -The category FI# is a slightly larger one that captures some additional structure that exists on (the cohomology of) configuration spaces of open manifolds. The maps ConfS(M) <- ConfT(M) "forget" points (they would go from Confn(M) <- ConfN(M) with n <= N), but on an open manifold you can also add points near the boundary, or "at infinity". This is only well-defined up to homotopy, so you don't quite get an FI#-space; but on cohomology this does extend the FI-module structure on Hk(ConfS(M)) to an FI#-module. (Prop 6.4.2) -The result you state comes from a general structure theorem for FI#-modules (Theorem 4.1.5). This theorem is very general (it holds over Z, without any finiteness conditions) but for your purposes we can use this corollary, Theorem 4.1.7(vii): - -Theorem (Church-Ellenberg-Farb). If V is an FI#-module over Q with each Vn finite-dimensional, then there is a (unique) character polynomial $P(X_1, \ldots, X_r)$ in the cycle counting functions $X_i$ so that$$\chi_{V_n} = P(X_1, \ldots, X_r) \text{ for all }n \ge 0.$$ - -In particular, the result you asked about holds for the cohomology of configurations on any open manifold. -There is also a similar theorem for FI-modules, as long as they are finitely generated (Theorem 3.3.4): - -Theorem (Church-Ellenberg-Farb). If V is a finitely generated FI-module over Q, then there is a (unique) character polynomial $P(X_1, \ldots, X_r)$ in the cycle counting functions $X_i$ and some $N_0\in \mathbb{N}$ so that $$\chi_{V_n} = P(X_1, \ldots, X_r) \text{ for all }n \ge N_0.$$ - -This applies to configurations on any manifold, closed or open (Theorem 6.2.1). I emphasize that the existence of this $N_0$ comes from a Noetherian theorem, which in general makes it completely nonconstructive. (The Noetherian theorem was proved by Church-Ellenberg-Farb-Nagpal; however for this application you only need it over Q, where it had been proved earlier by Snowden and Church-Ellenberg-Farb, which is all you need here.) However in the special case of configuration spaces, we could do some tricks to bound it; in particular when M has dim > 2, we find (Theorem 1.8) that the character polynomial for Hk(Confn(M)) has degree at most k and holds for $n\geq N_0=2k$.<|endoftext|> -TITLE: How many elementary embeddings can there be? -QUESTION [7 upvotes]: If $T$ is a complete first-order theory and $\kappa$ is a cardinal, let $\mathrm{Mod}_\kappa(T)$ be (a skeleton of) the category of $\kappa$-small models of $T$ (i.e. of cardinality $<\kappa$), with elementary embeddings as morphisms. What are the possible cardinalities of (the set of morphisms of) $\mathrm{Mod}_\kappa(T)$? And more specifically: what are the possible cardinalities of the hom-sets $\mathrm{Hom}_{\mathrm{Mod}_\kappa(T)}(M,N)$ for various $M,N \in \mathrm{Mod}_\kappa(T)$. -From a categorical perspective, this is a natural variant on classification theory and the question of the number of nonisomorphic models, which is about the cardinality of the set of objects of $\mathrm{Mod}_\kappa(T)$. It also provides a variation on Vaught's conjecture to consider. -For example, suppose the language is countable and $\kappa = \aleph_1$ and there is an infinite model. There's an obvious upper bound on $|\mathrm{Mod}_\kappa(T)|$ of $2^{\aleph_0}$, which is attained in all the examples I can think of (but I'm not actually a model theorist!). There's also a topology on the homsets given by pointwise convergence, with respect to which composition is continuous, which is a metric topology in the case of countable models, so it's tempting to try to construct perfect sets of embeddings. - -REPLY [4 votes]: It is a fact (following from the Ehrenfeucht–Mostowski theorem) that for every complete theory $T$ and for every $\lambda \geq |T|$, there is $M \models T$ with $|M| = \lambda$ and $M$ having $2^\lambda$-many automorphisms (assuming $T$ has infinite models). So if I'm understanding your question correctly then what you denote as $\mbox{Mod}_\kappa(T)$ always has cardinality $2^{<\kappa}$, at least for $\kappa > |T|$. -PS: (simplified example) To partially address your second question: for every cardinal $\lambda$ (possibly $\lambda$ finite or $0$) there are $M \equiv N$ structures in a countable language such that there are exactly $\lambda$ elementary embeddings from $M$ to $N$. Let $\mathcal{L}$ be the language with infinitely many constant symbols $(c_m: m < \omega)$ and let $T$ say that they are all distinct. For $\lambda > 0$, let $M$ be the model with exactly one unsorted element and let $N$ be the model with exactly $\lambda$ unsorted elements; then there are $\lambda$ elementary embeddings from $M$ to $N$. For $\lambda = 0$ let $M$ have 1 unsorted element and let $N$ have no unsorted elements. -PPS: On the other hand if $M, N$ are countable then the possibilities are exactly $0, 1, 2, \ldots, \aleph_0, 2^{\aleph_0}$. From the above examples we have seen these are all possible, on the other hand the space of elementary embeddings from $M$ to $N$ is a Polish space (completely metrizable space) so either is countable or has a perfect subset. (The subtleties of Vaught's conjecture are that we are looking at models up to isomorphism, so an equivalence relation on a Polish space.) - -REPLY [2 votes]: While far from being a solution of the general problem, let me confirm that if $T$ is any theory in a countable language that has an infinite model, then $|\mathrm{Mod}_{\aleph_1}(T)|=2^{\aleph_0}$. -We may assume $T$ is complete. Let $S_n(T)$ be the space of complete $n$-types of $T$ (dual to the Lindenbaum algebra of $T$ in $n$ variables). Being a second-countable Boolean space, each $S_n(T)$ is either countable, or it has cardinality $2^{\aleph_0}$. There are two cases to consider. -Case 1: $|S_n(T)|=2^{\aleph_0}$ for some $n\in\omega$. -Then $T$ has $2^{\aleph_0}$ nonisomorphic countable models (as each $n$-type is realized in a countable model, but one such model can only realize countably many types). Thus, $\mathrm{Mod}_{\aleph_1}(T)$ even has $2^{\aleph_0}$ distinct objects. -Case 2: $|S_n(T)|\le\aleph_0$ for all $n\in\omega$ ($T$ is small in model-theoretic terminology). -Then $T$ has a countable saturated model $A$. It suffices to show that there are $2^{\aleph_0}$ elementary embeddings $A\to A$. (They could even be made automorphisms by a minor modification of the argument.) -Let us enumerate $A=\{a_n:n\in\omega\}$, and let $S$ be the tree of all sequences $\langle b_i:i -TITLE: When is a flow geodesic and how to construct the connection from it -QUESTION [15 upvotes]: Let $(M,\Gamma)$ be a $C^\infty$ $n$ dimensional real manifold with a linear connection $\Gamma$ on it. I know the following: -If $\gamma:[t_0,t_1]\rightarrow M$ is a smooth curve and is a geodesic, then in local coordinates we have $$ \ddot{\gamma}^\mu(t)+\Gamma^\mu_{\alpha\beta}(\gamma(t))\dot{\gamma}^\alpha(t)\dot{\gamma}^\beta(t)=0. $$ By defining $\dot{\gamma}(t)=v(t)$, this is a first-order system of ODEs: $$ \frac{dv^\mu}{dt}=-\Gamma^\mu_{\alpha\beta}v^\alpha v^\beta \\ \frac{d\gamma^\mu}{dt}=v^\mu. $$ Since this is an ODE for the variables $(x^1,...,x^n,v^1,...,v^n)$, this is a differential equation on the tangent bundle of $M$, $TM$, moreover, this is a first-order, homogenous differential equation on $TM$, so it is represented by a vector field, $G\in\Gamma(TTM)$, which I'll call the geodesic flow. -I also know that this vector field can be defined invariantly if we take $\gamma_{(p,v)}(t)=\exp_p(tv)$, eg. the maximal geodesic with initial point $\gamma_{(p,v)}(0)=p$ and $\dot{\gamma}_{(p,v)}(0)=v$ and we take $\bar{\gamma}_{p,v}(t)=(\gamma_{(p,v)}(t),\dot{\gamma}_{(p,v)}(t))$ the natural lift of the curve to $TM$, then $$G_{(p,v)}=\left.\frac{d}{dt}\bar{\gamma}_{(p,v)}\right|_{t=0}.$$ In local coordinates $(x,v)$ this obviously has the form $$ G_{(p,v)}=v^\mu\left.\frac{\partial}{\partial x^\mu}\right|_{p}-\Gamma^\mu_{\alpha\beta}(p)v^\alpha v^\beta\left.\frac{\partial}{\partial v^\mu}\right|_{v}. $$ -I don't have any textbooks that deal with this any further however, so I have two questions. -Question 1: Given a smooth vector field $G\in\Gamma(TTM)$, what is the criteria for $G$ to be the geodesic flow of a linear connection? I can see from the local coordinate formula that the $x^\mu$ components contain only $v^\mu$ and the $v^\mu$ components are a quadratic form of the $v$ coordinates, but I am looking for invariant characterization. It is also clear to me that $G$ must not be vertical anywhere, but I don't think this is enough. -Question 2: If we are given $G$, how can I recover the torsionless connection? I mean, if I write up the local coordinate form, I can read off the connection coefficients and define the covariant derivative as $\partial_\mu V^\nu+\Gamma^\nu_{\mu\sigma}V^\sigma$ but I am once again looking for invariant characterization. A limit or $t$-derivative or something like that which reproduces the covariant derivative on the base if I know $G$. - -REPLY [7 votes]: Here is a geometric way that turns out to be equivalent to Robert's answer (i.e., to the Klein-Grifone-Foulon approach to connections associated to a second order ordinary differential equation of a manifold). -Let $\phi_t : TM\setminus 0 \rightarrow TM\setminus 0$ be the (local) flow of the second order equation, and let $D\phi_t : T(TM\setminus 0) \rightarrow T(TM\setminus 0)$ be its differential, which is also a flow. -In order to define a complement to the vertical space $V_{v_x}$ at the point $v_x \in TM\setminus 0$, consider the three subspaces -$L(0) = V_{v_x}$, $L(t) := D\phi_{-t}V_{\phi_t(v_x)}$, and $L(2t) := D\phi_{-2t}V_{\phi_{2t}(v_x)}.$ -These are three $n = \dim(M)$ dimensional subspaces in the $2n$-dimensional vector space $T_{v_x}(TM \setminus 0)$ and they are pairwise transversal if $t$ is close to zero. Now consider the harmonic conjugate (i.e., as in standard projective geometry) of this triplet: the image of $L(t)$ under the reflection that is minus the indentity on $L(0)$ and the identity on $L(2t)$. As $t$ tends to zero this harmonic conjugate gives you the complement to the vertical space (i.e., the horizonal subspace of the connection).<|endoftext|> -TITLE: Some intuition on the $SL_n$-module $V_{[1,1,...,1]}$ -QUESTION [7 upvotes]: (This question highly overlaps with this and also this.) -The irreducible ${\sf SL}_{n-1}$-module $V_{[1,1,\ldots,1]}$ is the one providing the minimal projective embedding $\mathbb{P}(V_{[1,1,\ldots,1]})$ for the complete flag variety $\mathrm{Fl}(\mathbb{C}^n)$. -However one does not need representation theory in order to realise that $\mathrm{Fl}(\mathbb{C}^n)$ is projective. First, embed -$$ -\mathrm{Fl}(\mathbb{C}^n)\subset\mathrm{Gr}(1,\mathbb{C}^n)\times\mathrm{Gr}(2,\mathbb{C}^n)\times\cdots\times\mathrm{Gr}(n-1,\mathbb{C}^n)\, , -$$ -then regard each Grassmannian $\mathrm{Gr}(i,\mathbb{C}^n)$ as a projective variety in $\mathbb{P}\bigwedge^i\mathbb{C}^n$, and finally use the Segre embedding: -$$ -\mathrm{Fl}(\mathbb{C}^n)\subset\mathbb{P}\left(\bigotimes_{i=0}^n\bigwedge^i\mathbb{C}^n \right)\, . -$$ - -QUESTION. Is there any "evident map" defined on $\bigotimes_{i=0}^n\bigwedge^i\mathbb{C}^n$ whose kernel is $V_{[1,1,\ldots,1]}$? - -By "evident" I mean definable in terms of elementary operations between tensors, like skew-symmetrisation. For instance, $V_{[1,1,\ldots,1]}$ lies in the common kernel of all the skew-symmetrisations -$$ -\bigotimes_{i=0}^n\bigwedge^i\mathbb{C}^n \stackrel{s_{ab}}{\longrightarrow} \bigwedge^{a+b}\mathbb{C}^n\otimes\bigotimes_{i\neq a,b}\bigwedge^i\mathbb{C}^n \, , -$$ -where -$$ -s_{ab}(\cdots\otimes\omega_a\otimes\cdots\otimes\omega_b\otimes\cdots):= (\omega_a\wedge\omega_b)\otimes\cdots\, , -$$ -and I suspect that $V_{[1,1,\ldots,1]}$ is exactly equal to $\bigcap_{a,b}\ker s_{ab}$, though I cannot prove it! - -QUESTION (reformulated). Is there an "evident way" of regarding $V_{[1,1,\ldots,1]}$ as a submodule of $\bigotimes_{i=0}^n\bigwedge^i\mathbb{C}^n$? - -REPLY [2 votes]: You can understand this using skew Howe duality. You can regard the tensor product $\bigotimes_{i=1}^{n}\bigwedge{}^{\!i} \mathbb{C}^n$ as a subrepresentation of $\bigwedge{}^{\!\binom{n+1}{2}}(\mathbb{C}^n\otimes \mathbb{C}^n)$ sending -$$v_a\otimes (v_{b_1}\wedge v_{b_2})\otimes \cdots \mapsto (v_a\otimes v_1)\wedge (v_{b_1}\otimes v_2)\wedge (v_{b_2}\otimes v_2)\wedge \cdots $$ -Here $\mathfrak{sl}_n$ is acting on the left copy of $\mathbb{C}^n$ and trivially on the right one. Which means that we have a commuting copy of $\mathfrak{sl}_n$ acting on the right copy; through some usual manipulations, we see that the images of $U(\mathfrak{sl}_n)$ acting via the left and right actions are maximal mutually commuting subalgebras (anything commuting with one lies in the other). In fact, as a module over $\mathfrak{sl}_n\times \mathfrak{sl}_n$, the module $\bigwedge{}^{\!\binom{n+1}{2}}(\mathbb{C}^n\otimes \mathbb{C}^n)$ breaks up as a sum $V_{\lambda}\otimes V_{\lambda^t}$ of the rep for a partition with $\binom{n+1}{2}$ boxes which fits inside an $n\times n$ square (you can check this by finding the common highest weight vectors). The copy of $\bigotimes_{i=1}^{n}\bigwedge{}^{\!i} \mathbb{C}^n$ we've embedded is exactly the lowest weight vectors of weight $-\omega_1-\cdots -\omega_n$ for the right action, so it's the common kernels of all the maps corresponding to lowering operators. These aren't the full skew symmetrizations, but rather the partial ones, where one maps $\bigwedge{}^{\!k}\mathbb{C}^n\otimes \bigwedge{}^{\!k+1}\mathbb{C}^n\to -\bigwedge{}^{\!k-1}\mathbb{C}^n\otimes \bigwedge{}^{\!k+2}\mathbb{C}^n$ -by -$$ (v_{a_1}\wedge \cdots \wedge v_{a_k})\otimes (v_{b_1}\wedge \cdots \wedge v_{b_{k+1}})\mapsto\\ \sum_{i=1}^k (-1)^{k-1}(v_{a_1}\wedge \cdots v_{a_{i-1}}\wedge v_{a_{i+1}}\wedge \cdots\wedge v_{a_k})\otimes (v_{a_i}\wedge v_{b_1}\wedge \cdots \wedge v_{b_{k+1}})$$ in each consecutive pair of terms.<|endoftext|> -TITLE: Relative version of Whitney Immersion Theorem -QUESTION [6 upvotes]: I asked the following question on StackExchange but received no response: -Let $M$ be a smooth $n$-dimensional compact manifold with boundary. Let $U$ be an open neighborhood of $\partial M$. Assume that we have a fixed immersion $\bar f : U \to \mathbb{R}^{2n-1}$. Can we always find an immersion $f : M \to \mathbb{R}^{2n-1}$ such that $f |_{\partial M} = \bar f |_{\partial M}$? -I believe this should work in $\mathbb{R}^{2n}$. Since $\bar f$ is an immersion, $d\bar f$ is a bundle monomorphism $TU \to U \times \mathbb{R}^{2n}$ covering $\bar f$. We can think of this as a section of the associated bundle over $U$ whose fibers are the Stiefel manifold $V_{n}(\mathbb{R}^{2n})$. Since $\pi_{n-1}\left(V_{n}(\mathbb{R}^{2n})\right) = 0$, we can extend the section from $\partial M$ to $M$. Now we can apply Smale-Hirsch to get the desired immersion $f : M \to \mathbb{R}^{2n}$. -Does the statement hold in $\mathbb{R}^{2n-1}$? -EDIT: To incorporate Oscar's comment, let's assume that we can perturb $\bar f$, if necessary. We can also assume that $U$ is connected. - -REPLY [3 votes]: No, you can't generally do this even with the added assumptions. The bundle $\tau = TS^{n-1} \to S^{n-1}$ is non-trivial for $n-1 \neq 1,3,7$, but it is stable after (one) trivialisation. Hence $\tau \oplus TD^n\vert_{S^{n-1}}$ is a trivial bundle over $S^{n-1}$. -By Smale--Hirsch this means there is an immersion $i: S^{n-1} \times [0,\epsilon] \hookrightarrow \mathbb{R}^{2n-1}$ whose ($(n-1)$-dimensional) normal bundle is isomorphic to (the pullback of) $\tau$. If $i$ extended to an immersion of $D^n$ then the normal bundle would extend to $D^n$ and hence be trivial, but it is not.<|endoftext|> -TITLE: Is there a theory of decomposition into indecomposables? What's the relation to idempotents? -QUESTION [6 upvotes]: Call a nonzero object of a pointed category simple if it has no proper quotients, and indecomposable if it's not the product of two objects (dual to connected). -Idempotents seem to pop up in many discussions about indecomposability (and I was told they're also relevant to simplicity). I was wondering - is there a nice theory conceptually explaining the relation between idempotents, indecomposability, and simplicity? -Very tentatively I'm hoping for something in the setting of semi-abelian categories because they admit a Jordan-Holder theorem, but this may be a fake and irrelevant justification. - -REPLY [5 votes]: Semi-abelian feels like a red herring here, since you don't even have a direct sum necessarily. What would indecomposable even mean in that context? -In an abelian category, any direct sum decomposition $A\oplus B$ is equipped with a projection map to $A$, and an inclusion of $A$ (using the usual universal properties). The composition of these gives an idempotent endomorphism of $A\oplus B$ from which you can reconstruct the decomposition, and you can easily get a decomposition from an idempotent by taking the image plus the kernel. In particular, in an abelian category, an object is indecomposable if and only if its endomorphism algebra has no idempotents other than 0 or 1. -In a more general additive category, this isn't necessarily true (consider free modules over a ring that has non-free projectives) since the image and kernel of idempotent may not be well behaved. You can fix this by passing to the Karoubian envelope (or idempotent completion) where you formally add an image to every idempotent endomorphism. Generally, indecomposable objects will not be very well behaved outside the Karoubian context.<|endoftext|> -TITLE: On a weaker version of homotopy equivalence between topological spaces -QUESTION [8 upvotes]: Consider two topological spaces $X$ and $Y$. -The notion of homotopy equivalence between $X$ and $Y$ is defined as a pair of continuous maps $f:X\to Y$ and $g:Y\to X$ such that $f\circ g$ and $g\circ f$ are homotopic to the identities. -But what if we weaken the definition a little bit, by considering the following "weaker" homotopy equivalence between $X$ and $Y$ : it is defined as a pair of pairs of continuous maps, $(f,g)$ and $(f',g')$, where $f,f':X\to Y$, and $g,g':Y\to X$ such that $g\circ f$ is homotopic to $Id_X$ and $f'\circ g'$ is homotopic to $Id_Y$. -I'm asking this question because I don't see why intuitively we would require $f'$ and $g'$ to be actually equal to $f$ and $g$. -Would this definition give an interesting notion of homotopy equivalence, or would this just coincide with the usual definition ? Or would this be simply too weak to do anything with it ? - -REPLY [11 votes]: In any category, we can say that a biretraction between objects $X$ and $Y$ is a system of maps $(f,g,f',g')$ with $gf=1_X$ and $f'g'=1_Y$, and we can say that two objects are biretractible if there is a biretraction between them. Biretractions can be composed and reversed in an obvious way, so biretractibility is an equivalence relation. -We say that a biretraction "is an isomorphism" if $f$, $g$, $f'$ and $g'$ are isomorphisms, or equivalently $fg=1_Y$ and $g'f'=1_X$. -In any category with good finiteness properties, it will work out that every biretraction is an isomorphism. For example, in the category of finite-dimensional vector spaces over a field this follows easily from consideration of dimensions, and a slightly harder argument gives the same conclusion for finitely generated modules over a Noetherian ring. (A standard lemma shows that any surjective endomorphism of such a module is an isomorphism.) Now consider the homotopy category of those simply connected CW complexes $X$ for which all homology groups $H_k(X)$ are finitely generated. Any biretraction in this category will give a biretraction of homology groups, which will necessarily be an isomorphism, so the original biretraction is an isomorphism by Whitehead's theorem. Thus, biretractibility is the same as homotopy equivalence in this context, and many similar contexts. -Of course, Karol's example has no good finiteness properties and so is not covered by this analysis. -For a slightly different example, in the category of (possibly infinite) sets, a biretraction need not be an isomorphism, but nonetheless biretractible objects are isomorphic, by the Schröder–Bernstein theorem.<|endoftext|> -TITLE: Why would the category of sets be intuitionistic? -QUESTION [21 upvotes]: This question is probably really naive. And, I hope the title doesn't come off as too combative. I think that topoi of $\mathbf{Set}$-valued sheaves provide an excellent motivation for higher-order intuitionistic logic, because most such topoi aren't boolean. But, I have an extremely hard time accepting that $\mathbf{Set}$ is intuitionistic. -Let me elaborate on this a bit. -Suppose our intuition for the phrase "subset of $X$" comes from the idea of having an effective total function $X \rightarrow \{0,1\}$ that returns an answer in a finite amount of time. In this case, the subsets of $X$ ought to form a Boolean algebra. -Suppose on the other hand that our intuition for the phrase "subset of $X$" comes from the idea of having an effective partial function $X \rightarrow 1$, such that the program for $f(x)$ either halts in a finite amount of time, or else runs forever. Then, with a bit of thought, we see that subsets of $X$ should be closed under union and intersection. But also, there's really no way to take the complement of a subset in general. So, under this viewpoint, the subsets of $X$ form a bounded distributive lattice. -So, I'm willing to believe that $\mathbf{Set}$ is a Boolean topos (the classical viewpoint), and I'm also willing to believe that it's much, much less than that (the "algorithmic" viewpoint), because I think these are both well-motivated and reasonable positions. -The thing about trying to apply intuitionistic logic to $\mathbf{Set}$ is that it seems to straddle a very strange middle ground between these two extremes. It says: yes, you can take the complement of any subset. But no, it may not be the case that $A^c \cup A = X$ (where $A$ is a subset of $X$, and $A^c := X \setminus A.$). And, it may not be the case that $(A^c)^c = A.$ But nonetheless, you can be assured that $((A^c)^c)^c = A^c$. Huh? Look, I realize this makes perfect sense from a BHK perspective; for instance, the complement of the complement of $(0,1) \cup (1,2)$ is $(0,2)$. So, the double negation law fails. But, that's in a topological space! I mean, why would $\mathbf{Set}$ be subject to this kind of reasoning? - -REPLY [5 votes]: There's a third intuition: our intuition for subsets comes from the notion of predicate — a subset $S \subseteq X$ corresponds to the condition on $x \in S$ (for $x$ of type $X$). -The idea of subsets as characteristic functions $X \to \{ 0, 1 \}$ comes from ideas about semantics: - -Predicates on $X$ correspond to truth-valued functions on $X$ -$\{0, 1\}$ is the set of truth values - -Of course, this only holds water if you already believe that truth is two-valued, which you don't if you disbelieve the mathematical universe is classical. -The topological character of truth turns out to follow naturally from an idea about what implication means: - -$P,Q \vdash R$ if and only if $P \vdash Q \to R$ - -Since we equate having two premises $P,Q$ with having a single premise $P \wedge Q$, it follows that the class of predicates on any object — and thus subsets of that object — is a Heyting algebra. -Since it's the very nature of Set that's under discussion, the Heyting algebra is automatically a complete Heyting algebra, and thus a locale. - -All that said, my position is that foundations (and thus Set) are classical, and even if I were persuaded to someday go all-in with intuitionistic logic, it would be in the form of something like working exclusively in the internal language of a nonboolean topos.<|endoftext|> -TITLE: When are finite maps quotients by finite groups? -QUESTION [5 upvotes]: Let $f: X \to Y$ be a finite map of projective varieties. -I'm trying to understand when and how often should i expect $f$ to be a quotient map by a finite group acting on $X$. Even more strictly let $G=Aut(X/Y)$. - - -When is $f$ isomorphic to the quotient map by an action $X \to X^G$? - - -If $X$ and $Y$ are smooth projective curves then I think this always holds but I haven't been able to formalize this. Are there any useful conditions which ensure that (1) holds? -EDIT: I deleted the irrelevant part of the question. I state it as a different question: When is $Aut(X/Y)=Aut(k(X)/k(Y))$? - -REPLY [9 votes]: If $X$ is normal and $G$ acts on $k(X)/k(Y)$ then the $G$ acts also on $X/Y$ (in a way consistent with its action on $k(X)$). - -Because $X$ is integral, its ring of functions on each affine open embeds into $k(X)$, so the action of $G$ on an affine open is determined by the action on the field of fractions. So the $G$-action is unique. Because of this, the existence is a local question, so we may assume $X$ and $Y$ affine. -In this case $k[X]$ consists of all elements of $k(X)$ that are integral over $k[X]$. Because each element of $k[X]$ is integral over $k[Y]$ (by the finiteness assumption), $k[X]$ consists of all elements of $k(X)$ that are integral over $k[Y]$. -From this definition it is clear that every automorphism of $k(X)$ that fixes $k(Y)$, and so fixes $k[Y]$, necessarily fixes $k[X]$, and thus acts on $X$. - -If $Y$ is also normal, and $k(X)^G= k(Y)$ then $Y= X/G$. - -There is certainly a map $X/G \to Y$. Whether or not this map is an isomorphism is again a question local on $Y$ -As we saw before, the elements of $k[X]$ are exactly the elements of $k(X)$ that are integral over $k[Y]$. Among these, the $G$-invariant elements are those that lie in $k(Y)$. But by assumption on $Y$, all the elements of $k(Y)$ that are integral over $k[Y]$ lie in $k[Y]$. -So $k[X/G] = k[X]^G=k[Y]$.<|endoftext|> -TITLE: Lebesgue outer measure -QUESTION [8 upvotes]: Denote the Lebesgue outer measure by $\mu^{\star}$. Is there a subset $X \subseteq [0, 1]$ such that $\mu^{\star}(X) > 0$ and $\mu^{\star} \upharpoonright \mathcal{P}(X)$ is a measure (countably additive set function)? - -REPLY [3 votes]: This question has a negative answer (given by Gregorz Plebanek), which follows from the following theorem of Gitik and Shelah. -Theorem (Gitik-Shelah, 1989): If a set $X$ admits an atomless probability $\sigma$-additive measure $\mu:\mathcal P(X)\to[0,1]$ defined on the algebra of all subsets of $X$, then the Banach space $L_1(\mu)$ has density $>|X|$. -On the other hand, if for some $X\subset\mathbb R$ with $\nu=\mu^*(X)$ the restriction $\mu^*|\mathcal P(X)$ is a measure, then $L_1(\nu)$ is separable (since the $\sigma$-algebra of Borel subsets of $X$ is countably generated), which contradicts Gitik-Shelah Theorem. -A combinatorial proof of Gitik-Shelah Theorem was given in -[A. Kamburelis, A new proof of the Gitik-Shelah theorem. -Israel J. Math. 72:3 (1990) 373–380].<|endoftext|> -TITLE: Partition into sets of positive outer measure -QUESTION [10 upvotes]: Let $\mu^{\star}$ denote Lebesgue outer measure. Suppose $X \subseteq [0, 1]$ and $\mu^{\star}(X) > 0$. Can we divide $X$ into uncountably many sets $\{X_i : i \in I\}$ such that for every $i \in I$, $\mu^{\star}(X_i) > 0$? - -REPLY [7 votes]: Suppose that the cofinality and the uniformity number of the null ideal are equal to $\kappa$. Fix a cofinal family of null sets $\{N_\alpha: \alpha \in \kappa \}$ and a bijection $\varphi=\langle \varphi_1,\varphi_2 \rangle$ from $\kappa$ onto $\kappa \times \kappa$. We define inductively a sequence $\langle x_\alpha : \alpha \in \kappa \rangle$ of distinct elements of $X$ by choosing $x_\alpha$ from the set $X \setminus \left(N_{\varphi_2}(\alpha) \cup \{x_\beta : \beta \in \alpha\} \right)$, which is non-empty since we are taking away from $X$ a null set. Now just let $X_i=\{x_\alpha : \varphi_1(\alpha)=i\}$ for $i \in \kappa$. The $X_i$ are clearly disjoint and since $x_\alpha \in X_i \setminus N_j$ whenever $\varphi(\alpha)=\langle i,j \rangle$, they are non-null sets. -On the other hand, Shelah proved in this article that if it is consistent that there is a measurable cardinal then it is also consistent that there is a non-null set $X$ which cannot be partitioned into uncountably many non-null sets.<|endoftext|> -TITLE: What areas of algebra could be interesting to probability theorists? -QUESTION [14 upvotes]: I would like to find some topic of algebra (beyond linear algebra; algebraic number theory is fine) that would be interesting both to a student that wants to specialize in probability theory and to me (being an "abstract algebraist") so that the student would make a talk on this subject. Any ideas? I don't want this piece of algebra to be "too combinatorial", and I don't think that the student is interested in "specific" applications of probability theory. So, do any important parts of probability theory depend on "advanced" algebra (or number theory)? - -REPLY [21 votes]: I am firstly an algebraist and later shifted to probability somehow, so I think I can answer your question from my own experience. I am a algebraist from the bottom of my heart though...A natural thought is you can teach them some basic random matrices, yet you may think it as "linear algebra" so: -(1) Boolean algebra -This is the basic building block for axiomatic probability theory and the $\sigma$-algebra is actually a set Boolean algebra. Some advanced in probabilty theory is possible only if you are clear about the structure of Boolean algebra. One example is a post I asked earlier. (MO post) Interestingly, P.Halmos also wrote both measure theory as well as Boolean algebras. One personal favorite is his "Lectures" - -Halmos, Paul R. "Lectures on Boolean algebras." (1966). - -(2)Stochastic algebra -This field is relatively young and was proposed mainly by U.Grenander. This algebraic structure is best justified by his comment "...(Kolmogorov)The classical results indicate that such advance should be possible by defining algebraic relations in the space and studying their probabilistic implications. This leads us automatically to think of notions like groups, topological vector spaces and algebras." in the preface(p.13) in - -Grenander, Ulf. Probabilities on algebraic structures. Courier - Corporation, 2008. - -A very readable introduction on this subject of studying probability measure on a algebraic structure is - -Budzban, Gregory, Philip Joel Feinsilver, and Arunava Mukherjea. - Probability on Algebraic Structures: AMS Special Session on - Probability on Algebraic Structures, March 12-13, 1999, Gainesville, - Florida. Vol. 261. American Mathematical Soc., 2000. - -However this is more or less rather analytic-oriented because the basic object the authors had in mind was Lie groups, so I guess this is not what you want. Also mentioned by @Nate Eldredge, some geometric group theory may be what you want if you want to study the group by its representation over random walk space or other kinds of spaces. See Chap 3 of Diaconis1988. -(3)Probabilistic number theory -From my own knowledge, some number theory problems can be solved using probabilistic method. The most well-known example is the distributions of additive functions defined on algebraic number field in number theory can be described using probabilistic argument. However, as pointed out by Kublius in Chap X of his famous monograph, such a probabilistic statement is hardly extensible beyond Gaussian number field. As for algebraic number theory I do not know much so no comments. -(4)Algebraic statistics -I do not know if you think this is "too combinatorial". But it is a rather popular method to study the Markov random fields(which is a probabilistic subject) on a graph using algebraic approach, see - -Garcia, Luis David, Michael Stillman, and Bernd Sturmfels. "Algebraic geometry of Bayesian networks." Journal of Symbolic Computation 39.3 (2005): 331-355. - -There is a whole branch of statistic called "algebraic statistics" which does make use of algebraic geometric notions to proceed, mainly tropical geometry, but that might seem too "combinatorial" to you sometimes. -(5)Free probability and von Neumann algebra -If we regard the free probability as a inexchangeable(non-commutative) version of measure theory, then the free product of the von Neumann algebras can be represented as dependent random variables. In this sense when exchangeability in a sequence of random variables is lost, then we fell into the category of non-commutative product of probability measures, and it is somehow surprising that this field is intrinsically related to the von Neumann algebra(To be more specific, the non-commutative algebra of random matrices equipped with weak topology). See also the MO post about the motivation of free probability.<|endoftext|> -TITLE: List of Gauss Codes? -QUESTION [5 upvotes]: I'm looking for a list of Gauss codes for knots 12 crossings and higher. I downloaded the list of all the knots up to 12 crossings from Charles Livingston's Knot Info. Are there any other online sources of Gauss codes? - -REPLY [3 votes]: You can generate your own Gauss codes: 1. Every Gauss code can be represented faithfully by a signed chord diagram, known as a Gauss diagram (Polyak and Viro 1994). 2. There is an efficient algorithm for generating essentially unique chord diagrams (J. Sawada 2002). 3. There is an criterion for determining the planarity of signed Gauss codes (J. Scott Carter 1991).<|endoftext|> -TITLE: In what sense is GCD an extension of boolean OR? -QUESTION [8 upvotes]: The J Programming langauge has an operator which acts as both the GCD and boolean Or. The J Primer has this note about it: - -The GCD is a useful extension of the domain of the or function to non-boolean arguments. - -As J is a highly mathematical language, I assume this extension has a basis in mathematics as well. -In what sense, if any, can the the GCD be considered an "extension" of boolean "Or"? - -REPLY [18 votes]: In the ordering $\preceq$ of nonnegative integers by divisibility, 1 is the least element and 0 is the greatest, and we have for instance -$$ -1\preceq 2\preceq 6\preceq 12\preceq\dots\preceq 0.$$ -In this ordering, gcd is the same as meet (greatest lower bound), which is dual to least upper bound, which is what boolean OR is for $\{0,1\}$. -So it makes sense if you think of numbers as "degrees of truth", where multiplicative factors are evidence of falsehood. -See also: What is gcd(0,0)?<|endoftext|> -TITLE: Freeness of tensor product -QUESTION [17 upvotes]: Let $G$ be a finite group. Is $\mathbb{Z}G\otimes_{Z(\mathbb{Z}G)}\mathbb{Z}G$ free as a $\mathbb{Z}$-module, where $Z$ denotes the centre? - -REPLY [8 votes]: There's a distinctly non-zero chance that my calculations are wrong, but I think it has a lot of $2$-torsion for $G=A_5$, the alternating group of degree $5$. I'm afraid the method I've used is a little indirect, and I haven't extracted an explicit torsion element. -Certainly $\mathbb{Z}G\otimes_{Z(\mathbb{Z}G)}\mathbb{Z}G$ is a finitely generated abelian group, and so it will be free if and only if -$\operatorname{Hom}_\mathbb{Z}(\mathbb{Z}G\otimes_{Z(\mathbb{Z}G)}\mathbb{Z}G,\mathbb{F})$ -has the same $\mathbb{F}$-dimension for every field $\mathbb{F}$. -But $$\operatorname{Hom}_\mathbb{Z}(\mathbb{Z}G\otimes_{Z(\mathbb{Z}G)}\mathbb{Z}G,\mathbb{F})\cong -\operatorname{Hom}_{Z(\mathbb{Z}G)}\left(\mathbb{Z}G,\operatorname{Hom}_\mathbb{Z}(\mathbb{Z}G,\mathbb{F})\right),$$ -which is isomorphic to $\operatorname{Hom}_{Z(\mathbb{F}G)}(\mathbb{F}G,\mathbb{F}G),$ -the endomorphism algebra of the group algebra $\mathbb{F}G$ as a module over its centre. -For $\mathbb{F}=\mathbb{C}$ this is easy to calculate from the irreducible character degrees. It's the sum of the fourth powers of the degrees, which for $G=A_5$ is $1^4+3^4+3^4+4^4+5^4=1044$. -For $\mathbb{F}$ algebraically closed of characteristic two, $\mathbb{F}A_5$ has a simple non-principal block isomorphic to $M_4(\mathbb{F})$ which contributes $4^4=256$. The centre of the principal block has a fairly simple structure $Z=\mathbb{F}[x,y,z]/(x,y,z)^2$, where $x,y$ and $z$ are in the socle of the principal block, each annihilated by all but one of the conjugacy classes of primitive idempotents, and according to my calculation, as a module for this centre, the principal block is the direct sum of one copy of $Z/(y,z)$, four copies of each of $Z/(x,y)$ and $Z/(x,z)$, and $26$ copies of $Z/(x,y,z)$, and the endomorphism algebra is $1258$-dimensional, so that (taking into account the non-principal block) the endomorphism algebra for the whole group algebra as a module for its centre is $1514$-dimensional. -This can't be the best way to do it, but maybe it provides some clues for a more illuminating answer. -By the way, I tried some smaller groups ($A_4$ in characteristic $2$, $D_{2p}$ in characteristic $p$), and for those the dimensions were the same.<|endoftext|> -TITLE: Positivity of certain Fourier transform -QUESTION [15 upvotes]: Is the Fourier transform of the function -$$ f(\xi) = e^{-t|\xi|^{2m}}$$ -positive for $t>0$ and $m \in \mathbb{N}_0$? - -REPLY [25 votes]: A simple argument that the Fourier transform cannot be nonnegative for any $m>1$, integer or not, is given in my 1991 paper with Odlyzko and Rush: - -Noam D. Elkies, Andrew M. Odlyzko, and Jason A. Rush: On the packing densities of superballs and other bodies, Invent. Math. 105 (1991), 613-639. - -The Fourier transform $f$ of a nonnegative even function must satisfy -$4f(\xi) \leq 3 f(0) + f(2\xi)$ for all $\xi$. This fails for any $m>1$ once $\xi$ is close enough to (but not equal) zero. -Postscripts: -0) The inequality $4f(\xi) \leq 3 f(0) + f(2\xi)$ holds because -the difference is $\int_{-\infty}^\infty \hat f(\eta) \, w(\eta) \, d\eta$ -with $w(\eta) \geq 0$ for all $\eta$ because -$3 - 4\cos \theta + \cos 2\theta = 8 \sin^4 (\theta/2) \geq 0$. -1) The nonnegativity of $3 - 4\cos \theta + \cos 2\theta$ -is also a key ingredient in the proof of thenonvanishing of the Riemann zeta function $\zeta(s)$ on the edge -$s = 1+it$ of the critical strip, and thus of the Prime Number Theorem. -See the last exercise of -this chapter -my notes on analytic number theory. -2) Sergei noted that it is also known that for $0 < m \leq 1$ -the Fourier transform is positive. As I wrote earlier this year -on Math Stackexchange, -the same paper with Odlyzko and Rush also proves that fact as follows, -crediting the argument to B.F. Logan (see Lemma 5 on page 626). -We first prove this for $m=1$, then reduce $01$, the second derivative at the origin vanishes and thus also the variance of $X$. The positive function would be a delta function which is a contradiction. - -REPLY [8 votes]: This is a well-known problem, solved recently in dissertation of Zastavny based on some inequalities for p.d.f. -More: it is a Schonberg problem, positive for $0 -TITLE: measurable linear functionals are also continuous on separable Banach spaces? -QUESTION [6 upvotes]: It is well known continuous linear functionals are (Borel) measurable. I have read, as a remark, the converse is also true for separable Banach spaces, but I could not find any references. - -REPLY [3 votes]: You do not even need seperability. Proposition 1.2.29 in the book Barrelled Locally Convex Spaces of Bonet and Perez Carreras says that every Borel measurable linear map from a Baire locally convex space to a locally convex space is continuous. This result is due to Laurent Schwartz and it is related to his more famous Borelian graph theorem (linear maps from a separable Banach space into Souslin locally convex spaces with borelian graph are continuous).<|endoftext|> -TITLE: Derivatives of $C^{\infty}$ non analytic function -QUESTION [5 upvotes]: Question: Given $f\in C^{\infty}$ which is not analytic on a bounded domain $\Omega \subseteq \mathbb{R}$. What can we say about the sequence $\lbrace f^{(m)} \rbrace _{m=1}^{\infty} $? Specifically - what can we say about the convergence and/or decay of its $L^2$ and $L^{\infty}$ norms? If nothing, why? Of special importance is to bound globally Taylor like terms of the form $f^{(m)}(\xi)/m! $ for some $\xi \in \Omega$. -Motivation, not Neccessary: My question is motivated by the residue terms of numerical integration formulas. We can usually have nice convergence rates for analytical functions due to the exponential decay in their taylor series coefficients $f^{(m)}/m! $. -I have read some of the math.stackexchange posts about it, and while this one does contain some possible leads, I couldn't find the answers to my questions there yet. -Thanks - -REPLY [11 votes]: For the $L^\infty$ norms, nothing at all apart from Alexander Eremenko's answer, because of Borel's theorem: any sequence of real (resp. complex) numbers can be realized as the sequence of Taylor coefficients of some real (resp. complex) valued smooth function $f$ at any given point. By the Sobolev embedding theorem and the fact that $f$ may be chosen compactly supported in the interior of $\Omega$, one can bound the $L^\infty$ norm of $f$ from above by a positive constant times the sum of the $L^1$ norms of the $n$-th order derivatives of $f$, where $n$ is the dimension of the domain $\Omega$. Therefore, if the sequence of $L^\infty$ norms of the derivatives of $f$ blows up, so does the corresponding sequence of $L^1$ norms. Since you assumed $\Omega$ to be bounded, this implies the same property for the $L^p$ norms for all $1\leq p\leq\infty$ by Hölder's inequality. -(side remark: the case of Sobolev's embedding theorem which is relevant to the present context is an easy consequence of the fundamental theorem of Calculus applied once to each variable of $f$, which also shows that the constant above may be taken equal to 1) - -REPLY [8 votes]: We can say is that $\limsup |a_n|^{1/n}=\infty$, where $a_n=\max_{a\in K}|f^{(n)}(a)|/n!$, for every compact $K\subset\Omega$.<|endoftext|> -TITLE: Which quaternary quadratic form represents $n$ the greatest number of times? -QUESTION [27 upvotes]: Let $Q$ be a four-variable positive-definite quadratic form with integer coefficients and let $r_{Q}(n)$ be the number of representations of $n$ by $Q$. The theory of modular forms explains how $r_{Q}(n) = a_{E}(n) + a_{C}(n)$ is broken into the Eisenstein series piece, and the cusp form piece. Knowing what numbers are represented by $Q$ boils down to getting a formula for $a_{E}(n)$ which is of size ''about'' $n$ (as long as local conditions are satified - the formula of Siegel makes this precise), and bounding $|a_{C}(n)|$ (which is $\leq C_{Q} d(n) \sqrt{n}$, where $d(n)$ is the number of divisors of $n$). The hardest part of this is determining $C_{Q}$, and all the methods for doing this rely on bounds for $r_{Q}(n)$. -My question is - -If $n$ is a fixed positive integer, what is ${\rm max}_{Q}~r_{Q}(n)$? Here the max runs over all 4-variable positive-definite integral quadratic forms. - -My suspicion is that this max always occurs for a form $Q$ with low discriminant, and in particular I would guess the max is always $\leq 24 \sigma(n)$. Equality is achieved for odd $n$ with $Q = x^{2} + y^{2} + z^{2} + w^{2} + xw + yw + zw$, the smallest discriminant quaternary form. (Edit: There are some $n$ that are represented more ways with a form of discrimiant $5$. It seems the right conjecture is $r_{Q}(n) \leq 30 \sigma(n)$.) -I would even be satisfied with a bound of the type $r_{Q}(n) \leq C n \log(n)$ with a value of $C$ that doesn't depend on $Q$. It's possible that one could prove something like this for quadratic polynomials by induction on the number of variables, but I don't see how to make that work. (It seems that $2$ variables is the naturally starting point.) -Another possible approach is something like the circle method, which can recover similar bounds to those which the theory of modular forms gives. In the papers on this subject though, the dependence on the form in the error term seems to make a result of the type I seek difficult. - -REPLY [4 votes]: In a comment under Will Sawin's answer, GH says that it should not be too difficult to find a (relatively small) constant $C$ and a proof for -$$ r_Q(n) \leq \, C \; \sigma(n) \, \det(Q)^{-1/9} \; + \; n^{4/5} $$ -which, if $C$ were found to be small enough, would give teeth to the computations below. -I did want to see the behavior of specific forms of low discriminant from Nipp's tables, as Jeremy briefly indicated in an email. To get $r(n) \geq 15 \sigma(n)$ we seem to need discriminant $d \leq 21.$ To get $r(n) \geq 20 \sigma(n)$ we seem to need discriminant $d = 4,5.$ -I should add that there are infinitely many forms that give $r(1) = 12,$ so that this ratio is at least $12.$ Given any positive integer $T \geq 2,$ -$$ ( x^2 + y^2 + z^2 + yz + zx + xy) + T w^2 $$ -represents $1$ twelve times. -Discriminant $4$ achieves the ratio $24.$ For $d=5,$ with prime $p \equiv \pm 2 \pmod 5,$ we get -$$ r(p) = 30 (p-1) = 30 \; \sigma(p) \cdot \left( \frac{p-1}{p+1} \right) $$ -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -d = 5 - n reps sigma - ratio 20 1 20 1 1 = 1 - ratio 10 2 30 3 2 = 2 - ratio 15 3 60 4 3 = 3 - ratio 20 5 120 6 5 = 5 - ratio 22.5 7 180 8 7 = 7 - ratio 25.7142857142857 13 360 14 13 = 13 - ratio 26.6666666666667 17 480 18 17 = 17 - ratio 27.5 23 660 24 23 = 23 - ratio 28.4210526315789 37 1080 38 37 = 37 - ratio 28.6363636363636 43 1260 44 43 = 43 - ratio 28.75 47 1380 48 47 = 47 - ratio 28.8888888888889 53 1560 54 53 = 53 - ratio 29.1176470588235 67 1980 68 67 = 67 - ratio 29.1891891891892 73 2160 74 73 = 73 - ratio 29.2857142857143 83 2460 84 83 = 83 - ratio 29.3877551020408 97 2880 98 97 = 97 - ratio 29.4230769230769 103 3060 104 103 = 103 - ratio 29.4444444444444 107 3180 108 107 = 107 - ratio 29.4736842105263 113 3360 114 113 = 113 - ratio 29.53125 127 3780 128 127 = 127 - -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= - d 4 record ratio 24 number 1 sigma 1 reps 24 - d 5 record ratio 29.6969696969697 number 197 sigma 198 reps 5880 - d 8 record ratio 17.8181818181818 number 197 sigma 198 reps 3528 - d 9 record ratio 12 number 1 sigma 1 reps 12 - d 12 record ratio 19.4117647058824 number 67 sigma 68 reps 1320 - d 12 record ratio 19.7979797979798 number 197 sigma 198 reps 3920 - d 13 record ratio 13.8585858585859 number 197 sigma 198 reps 2744 - d 16 record ratio 8 number 1 sigma 1 reps 8 - d 16 record ratio 12 number 1 sigma 1 reps 12 - d 17 record ratio 8.91 number 199 sigma 200 reps 1782 - d 20 record ratio 17.8181818181818 number 197 sigma 198 reps 3528 - d 20 record ratio 13.7142857142857 number 194 sigma 294 reps 4032 - d 20 record ratio 11.8787878787879 number 197 sigma 198 reps 2352 - d 21 record ratio 15.8383838383838 number 197 sigma 198 reps 3136 - d 21 record ratio 15.84 number 199 sigma 200 reps 3168 - d 24 record ratio 11.88 number 199 sigma 200 reps 2376 - d 24 record ratio 12 number 1 sigma 1 reps 12 - d 24 record ratio 11.8666666666667 number 179 sigma 180 reps 2136 - d 25 record ratio 6 number 1 sigma 1 reps 6 - d 28 record ratio 9.89010989010989 number 181 sigma 182 reps 1800 - d 28 record ratio 12 number 1 sigma 1 reps 12 - d 28 record ratio 9.89583333333333 number 191 sigma 192 reps 1900 - d 29 record ratio 10 number 5 sigma 6 reps 60 - d 29 record ratio 12 number 1 sigma 1 reps 12 - d 32 record ratio 9.8989898989899 number 197 sigma 198 reps 1960 - d 32 record ratio 11.8787878787879 number 197 sigma 198 reps 2352 - d 32 record ratio 12 number 1 sigma 1 reps 12 - d 32 record ratio 8.46031746031746 number 166 sigma 252 reps 2132 - d 32 record ratio 11.8666666666667 number 179 sigma 180 reps 2136 - d 33 record ratio 8 number 1 sigma 1 reps 8 - d 33 record ratio 7.91752577319588 number 193 sigma 194 reps 1536 - d 33 record ratio 7.91666666666667 number 191 sigma 192 reps 1520 - d 36 record ratio 12 number 1 sigma 1 reps 12 - d 36 record ratio 12 number 5 sigma 6 reps 72 - d 36 record ratio 8 number 1 sigma 1 reps 8 - d 36 record ratio 11.9504132231405 number 81 sigma 121 reps 1446 - d 36 record ratio 7.9843137254902 number 128 sigma 255 reps 2036 - d 37 record ratio 12 number 1 sigma 1 reps 12 - d 37 record ratio 7.54166666666667 number 191 sigma 192 reps 1448 - d 40 record ratio 12 number 1 sigma 1 reps 12 - d 40 record ratio 7.75257731958763 number 193 sigma 194 reps 1504 - d 40 record ratio 7.92156862745098 number 101 sigma 102 reps 808 - d 40 record ratio 7.84 number 149 sigma 150 reps 1176 - d 41 record ratio 8 number 1 sigma 1 reps 8 - d 41 record ratio 5.33333333333333 number 2 sigma 3 reps 16 - d 44 record ratio 8.64 number 149 sigma 150 reps 1296 - d 44 record ratio 8.55172413793103 number 173 sigma 174 reps 1488 - d 44 record ratio 12 number 1 sigma 1 reps 12 - d 44 record ratio 8.53658536585366 number 163 sigma 164 reps 1400 - d 45 record ratio 11.8787878787879 number 197 sigma 198 reps 2352 - d 45 record ratio 12 number 1 sigma 1 reps 12 - d 45 record ratio 11.8762886597938 number 193 sigma 194 reps 2304 - d 45 record ratio 11.5555555555556 number 159 sigma 216 reps 2496 - d 45 record ratio 11.8787878787879 number 197 sigma 198 reps 2352 - d 48 record ratio 11.8787878787879 number 197 sigma 198 reps 2352 - d 48 record ratio 9.8989898989899 number 197 sigma 198 reps 1960 - d 48 record ratio 9.8989898989899 number 197 sigma 198 reps 1960 - d 48 record ratio 12 number 1 sigma 1 reps 12 - d 48 record ratio 9.9 number 199 sigma 200 reps 1980 - d 48 record ratio 9.12592592592593 number 178 sigma 270 reps 2464 - d 48 record ratio 11.88 number 199 sigma 200 reps 2376 - d 48 record ratio 12 number 11 sigma 12 reps 144 - d 48 record ratio 9.1 number 158 sigma 240 reps 2184 - d 49 record ratio 4 number 1 sigma 1 reps 4 - d 52 record ratio 6.53061224489796 number 194 sigma 294 reps 1920 - d 52 record ratio 6.40740740740741 number 142 sigma 216 reps 1384 - d 52 record ratio 12 number 1 sigma 1 reps 12 - d 52 record ratio 8.36842105263158 number 151 sigma 152 reps 1272 - d 52 record ratio 5.57894736842105 number 151 sigma 152 reps 848 - d 52 record ratio 6 number 1 sigma 1 reps 6 - d 53 record ratio 8.12903225806452 number 61 sigma 62 reps 504 - d 53 record ratio 12 number 1 sigma 1 reps 12 - d 53 record ratio 8 number 11 sigma 12 reps 96 - d 56 record ratio 8 number 1 sigma 1 reps 8 - d 56 record ratio 7.31578947368421 number 151 sigma 152 reps 1112 - d 56 record ratio 7.46666666666667 number 89 sigma 90 reps 672 - d 56 record ratio 12 number 1 sigma 1 reps 12 - d 56 record ratio 7.15151515151515 number 131 sigma 132 reps 944 - d 57 record ratio 8 number 1 sigma 1 reps 8 - d 57 record ratio 5.67708333333333 number 191 sigma 192 reps 1090 - d 57 record ratio 5.69072164948454 number 193 sigma 194 reps 1104 - d 57 record ratio 6 number 1 sigma 1 reps 6 - d 60 record ratio 9.89690721649485 number 193 sigma 194 reps 1920 - d 60 record ratio 12 number 1 sigma 1 reps 12 - d 60 record ratio 9.89690721649485 number 193 sigma 194 reps 1920 - d 60 record ratio 9.86666666666667 number 149 sigma 150 reps 1480 - d 60 record ratio 9.86666666666667 number 149 sigma 150 reps 1480 - d 60 record ratio 9.9 number 199 sigma 200 reps 1980 - d 60 record ratio 9.9 number 199 sigma 200 reps 1980 - d 60 record ratio 9.88095238095238 number 167 sigma 168 reps 1660 - d 61 record ratio 12 number 1 sigma 1 reps 12 - d 61 record ratio 6 number 5 sigma 6 reps 36 - d 61 record ratio 6 number 1 sigma 1 reps 6 - d 64 record ratio 6 number 1 sigma 1 reps 6 - d 64 record ratio 12 number 1 sigma 1 reps 12 - d 64 record ratio 6.66666666666667 number 5 sigma 6 reps 40 - d 64 record ratio 4 number 1 sigma 1 reps 4 - d 64 record ratio 6 number 1 sigma 1 reps 6 - d 64 record ratio 6 number 1 sigma 1 reps 6 - d 64 record ratio 6 number 3 sigma 4 reps 24 - d 65 record ratio 8 number 1 sigma 1 reps 8 - d 65 record ratio 5.24137931034483 number 173 sigma 174 reps 912 - d 65 record ratio 5.28 number 149 sigma 150 reps 792 - d 65 record ratio 6 number 1 sigma 1 reps 6 - d 68 record ratio 9.6 number 19 sigma 20 reps 192 - d 68 record ratio 12 number 1 sigma 1 reps 12 - d 68 record ratio 8 number 1 sigma 1 reps 8 - d 68 record ratio 6.09523809523809 number 41 sigma 42 reps 256 - d 68 record ratio 6.12371134020619 number 193 sigma 194 reps 1188 - d 68 record ratio 5.33888888888889 number 184 sigma 360 reps 1922 - d 68 record ratio 5.32549019607843 number 128 sigma 255 reps 1358 - d 69 record ratio 8.57142857142857 number 13 sigma 14 reps 120 - d 69 record ratio 12 number 1 sigma 1 reps 12 - d 69 record ratio 7.92 number 199 sigma 200 reps 1584 - d 69 record ratio 7.92 number 199 sigma 200 reps 1584 - d 69 record ratio 7.91919191919192 number 197 sigma 198 reps 1568 - d 72 record ratio 8 number 1 sigma 1 reps 8 - d 72 record ratio 7.33333333333333 number 83 sigma 84 reps 616 - d 72 record ratio 12 number 1 sigma 1 reps 12 - d 72 record ratio 7.16483516483517 number 181 sigma 182 reps 1304 - d 72 record ratio 7.13924050632911 number 157 sigma 158 reps 1128 - d 72 record ratio 7.33333333333333 number 83 sigma 84 reps 616 - d 72 record ratio 7.15151515151515 number 131 sigma 132 reps 944 - d 72 record ratio 7 number 159 sigma 216 reps 1512 - d 72 record ratio 6.96774193548387 number 183 sigma 248 reps 1728 - d 73 record ratio 4 number 1 sigma 1 reps 4 - d 73 record ratio 6 number 1 sigma 1 reps 6 - d 73 record ratio 4 number 2 sigma 3 reps 12 - d 76 record ratio 12 number 1 sigma 1 reps 12 - d 76 record ratio 5.42857142857143 number 139 sigma 140 reps 760 - d 76 record ratio 5.24390243902439 number 163 sigma 164 reps 860 - d 76 record ratio 6 number 1 sigma 1 reps 6 - d 76 record ratio 5.31868131868132 number 181 sigma 182 reps 968 - d 76 record ratio 6 number 1 sigma 1 reps 6 - d 77 record ratio 12 number 1 sigma 1 reps 12 - d 77 record ratio 8.10989010989011 number 181 sigma 182 reps 1476 - d 77 record ratio 7.96153846153846 number 103 sigma 104 reps 828 - d 77 record ratio 8.18181818181818 number 109 sigma 110 reps 900 - d 77 record ratio 8.08 number 149 sigma 150 reps 1212 - d 80 record ratio 12 number 1 sigma 1 reps 12 - d 80 record ratio 9 number 167 sigma 168 reps 1512 - d 80 record ratio 6 number 1 sigma 1 reps 6 - d 80 record ratio 8 number 1 sigma 1 reps 8 - d 80 record ratio 7.4639175257732 number 193 sigma 194 reps 1448 - d 80 record ratio 8.90909090909091 number 197 sigma 198 reps 1764 - d 80 record ratio 5.93939393939394 number 197 sigma 198 reps 1176 - d 80 record ratio 7.58823529411765 number 67 sigma 68 reps 516 - d 80 record ratio 7.42307692307692 number 103 sigma 104 reps 772 - d 80 record ratio 5.87755102040816 number 194 sigma 294 reps 1728 - d 80 record ratio 5.87755102040816 number 194 sigma 294 reps 1728 - d 80 record ratio 7.39285714285714 number 188 sigma 336 reps 2484 - d 81 record ratio 8 number 1 sigma 1 reps 8 - d 81 record ratio 4.7 number 19 sigma 20 reps 94 - d 81 record ratio 4 number 1 sigma 1 reps 4 - d 81 record ratio 6 number 1 sigma 1 reps 6 - d 81 record ratio 6 number 2 sigma 3 reps 18 - d 84 record ratio 12 number 1 sigma 1 reps 12 - d 84 record ratio 9.52747252747253 number 181 sigma 182 reps 1734 - d 84 record ratio 7.35135135135135 number 146 sigma 222 reps 1632 - d 84 record ratio 7.30612244897959 number 194 sigma 294 reps 2148 - d 84 record ratio 8 number 1 sigma 1 reps 8 - d 84 record ratio 6.4 number 179 sigma 180 reps 1152 - d 84 record ratio 6.33333333333333 number 191 sigma 192 reps 1216 - d 84 record ratio 6.35164835164835 number 181 sigma 182 reps 1156 - d 84 record ratio 6.36 number 199 sigma 200 reps 1272 - d 84 record ratio 7.25925925925926 number 142 sigma 216 reps 1568 - d 84 record ratio 7.25925925925926 number 142 sigma 216 reps 1568 - d 84 record ratio 9.56521739130435 number 137 sigma 138 reps 1320 - d 84 record ratio 9.6 number 179 sigma 180 reps 1728 - d 85 record ratio 12 number 1 sigma 1 reps 12 - d 85 record ratio 6.26086956521739 number 137 sigma 138 reps 864 - d 85 record ratio 6.03846153846154 number 103 sigma 104 reps 628 - d 85 record ratio 6.10909090909091 number 109 sigma 110 reps 672 - d 85 record ratio 6.4 number 29 sigma 30 reps 192 - d 85 record ratio 6.10989010989011 number 181 sigma 182 reps 1112 - d 88 record ratio 12 number 1 sigma 1 reps 12 - d 88 record ratio 4.69230769230769 number 103 sigma 104 reps 488 - d 88 record ratio 6 number 1 sigma 1 reps 6 - d 88 record ratio 4.68 number 199 sigma 200 reps 936 - d 88 record ratio 6 number 1 sigma 1 reps 6 - d 88 record ratio 6 number 1 sigma 1 reps 6 - d 88 record ratio 5 number 3 sigma 4 reps 20 - d 89 record ratio 8 number 1 sigma 1 reps 8 - d 89 record ratio 4.22222222222222 number 17 sigma 18 reps 76 - d 89 record ratio 4 number 1 sigma 1 reps 4 - d 89 record ratio 3.71428571428571 number 4 sigma 7 reps 26 - d 92 record ratio 12 number 1 sigma 1 reps 12 - d 92 record ratio 6 number 1 sigma 1 reps 6 - d 92 record ratio 8 number 1 sigma 1 reps 8 - d 92 record ratio 6.46666666666667 number 29 sigma 30 reps 194 - d 92 record ratio 5.93406593406593 number 181 sigma 182 reps 1080 - d 92 record ratio 6 number 1 sigma 1 reps 6 - d 92 record ratio 5.93333333333333 number 179 sigma 180 reps 1068 - d 93 record ratio 12 number 1 sigma 1 reps 12 - d 93 record ratio 7.15714285714286 number 139 sigma 140 reps 1002 - d 93 record ratio 7.125 number 31 sigma 32 reps 228 - d 93 record ratio 8 number 1 sigma 1 reps 8 - d 93 record ratio 7.5 number 23 sigma 24 reps 180 - d 93 record ratio 7.02777777777778 number 71 sigma 72 reps 506 - d 96 record ratio 12 number 1 sigma 1 reps 12 - d 96 record ratio 8.45454545454546 number 43 sigma 44 reps 372 - d 96 record ratio 6.88888888888889 number 107 sigma 108 reps 744 - d 96 record ratio 6.72463768115942 number 137 sigma 138 reps 928 - d 96 record ratio 8 number 1 sigma 1 reps 8 - d 96 record ratio 7.91208791208791 number 181 sigma 182 reps 1440 - d 96 record ratio 7.91208791208791 number 181 sigma 182 reps 1440 - d 96 record ratio 6.6 number 199 sigma 200 reps 1320 - d 96 record ratio 7.92 number 199 sigma 200 reps 1584 - d 96 record ratio 7.92 number 199 sigma 200 reps 1584 - d 96 record ratio 7.92 number 199 sigma 200 reps 1584 - d 96 record ratio 5.73333333333333 number 118 sigma 180 reps 1032 - d 96 record ratio 5.65079365079365 number 166 sigma 252 reps 1424 - d 96 record ratio 5.63333333333333 number 158 sigma 240 reps 1352 - d 96 record ratio 5.63333333333333 number 158 sigma 240 reps 1352 - d 96 record ratio 7.88405797101449 number 137 sigma 138 reps 1088 - d 96 record ratio 8 number 5 sigma 6 reps 48 - d 96 record ratio 7.91111111111111 number 179 sigma 180 reps 1424 - d 97 record ratio 4 number 1 sigma 1 reps 4 - d 97 record ratio 6 number 1 sigma 1 reps 6 - d 97 record ratio 3.5 number 3 sigma 4 reps 14 - d 97 record ratio 3.33333333333333 number 2 sigma 3 reps 10 - d 100 record ratio 12 number 1 sigma 1 reps 12 - d 100 record ratio 6 number 3 sigma 4 reps 24 - d 100 record ratio 6.28571428571429 number 13 sigma 14 reps 88 - d 100 record ratio 6 number 1 sigma 1 reps 6 - d 100 record ratio 7.97435897435897 number 125 sigma 156 reps 1244 - d 100 record ratio 6 number 1 sigma 1 reps 6 - d 100 record ratio 5 number 3 sigma 4 reps 20 - d 100 record ratio 3.9921568627451 number 128 sigma 255 reps 1018 - d 101 record ratio 12 number 1 sigma 1 reps 12 - d 101 record ratio 8 number 1 sigma 1 reps 8 - d 101 record ratio 6 number 19 sigma 20 reps 120 - d 101 record ratio 6 number 5 sigma 6 reps 36 - d 101 record ratio 6 number 1 sigma 1 reps 6 - d 104 record ratio 12 number 1 sigma 1 reps 12 - d 104 record ratio 5.06122448979592 number 97 sigma 98 reps 496 - d 104 record ratio 5.15463917525773 number 193 sigma 194 reps 1000 - d 104 record ratio 5.22448979591837 number 97 sigma 98 reps 512 - d 104 record ratio 6 number 1 sigma 1 reps 6 - d 104 record ratio 8 number 1 sigma 1 reps 8 - d 104 record ratio 5.33333333333333 number 5 sigma 6 reps 32 - d 104 record ratio 5.11392405063291 number 157 sigma 158 reps 808 - d 105 record ratio 8 number 1 sigma 1 reps 8 - d 105 record ratio 6 number 1 sigma 1 reps 6 - d 105 record ratio 5.27835051546392 number 193 sigma 194 reps 1024 - d 105 record ratio 5.37931034482759 number 173 sigma 174 reps 936 - d 105 record ratio 5.30952380952381 number 167 sigma 168 reps 892 - d 105 record ratio 5.375 number 191 sigma 192 reps 1032 - d 105 record ratio 5.26666666666667 number 179 sigma 180 reps 948 - d 105 record ratio 6 number 1 sigma 1 reps 6 - d 105 record ratio 5.28 number 199 sigma 200 reps 1056 - d 105 record ratio 6 number 1 sigma 1 reps 6 - d 105 record ratio 5.29032258064516 number 61 sigma 62 reps 328 - d 108 record ratio 12 number 1 sigma 1 reps 12 - d 108 record ratio 6.6 number 199 sigma 200 reps 1320 - d 108 record ratio 6.609375 number 127 sigma 128 reps 846 - d 108 record ratio 6.65853658536585 number 163 sigma 164 reps 1092 - d 108 record ratio 8 number 1 sigma 1 reps 8 - d 108 record ratio 7.46666666666667 number 29 sigma 30 reps 224 - d 108 record ratio 7.15909090909091 number 129 sigma 176 reps 1260 - d 108 record ratio 7.15909090909091 number 129 sigma 176 reps 1260 - d 108 record ratio 6.74747474747475 number 197 sigma 198 reps 1336 - d 108 record ratio 6.64367816091954 number 173 sigma 174 reps 1156 - d 108 record ratio 6.63636363636364 number 197 sigma 198 reps 1314 - d 108 record ratio 6.66666666666667 number 11 sigma 12 reps 80 - d 108 record ratio 9.9 number 199 sigma 200 reps 1980 - d 108 record ratio 7.22222222222222 number 159 sigma 216 reps 1560 - d 108 record ratio 9.8989898989899 number 197 sigma 198 reps 1960 - d 109 record ratio 12 number 1 sigma 1 reps 12 - d 109 record ratio 6 number 1 sigma 1 reps 6 - d 109 record ratio 5.33333333333333 number 5 sigma 6 reps 32 - d 109 record ratio 6 number 1 sigma 1 reps 6 - d 109 record ratio 4.5 number 3 sigma 4 reps 18 - d 112 record ratio 12 number 1 sigma 1 reps 12 - d 112 record ratio 6 number 3 sigma 4 reps 24 - d 112 record ratio 6 number 1 sigma 1 reps 6 - d 112 record ratio 5.93406593406593 number 181 sigma 182 reps 1080 - d 112 record ratio 6 number 1 sigma 1 reps 6 - d 112 record ratio 6 number 1 sigma 1 reps 6 - d 112 record ratio 5.02083333333333 number 191 sigma 192 reps 964 - d 112 record ratio 5.05494505494505 number 181 sigma 182 reps 920 - d 112 record ratio 5.05494505494505 number 181 sigma 182 reps 920 - d 112 record ratio 6 number 1 sigma 1 reps 6 - d 112 record ratio 5.06122448979592 number 97 sigma 98 reps 496 - d 112 record ratio 4.57142857142857 number 194 sigma 294 reps 1344 - d 112 record ratio 4.57142857142857 number 194 sigma 294 reps 1344 - d 112 record ratio 4.66666666666667 number 2 sigma 3 reps 14 - d 112 record ratio 5.9375 number 191 sigma 192 reps 1140 - d 112 record ratio 4.55 number 158 sigma 240 reps 1092 - d 113 record ratio 8 number 1 sigma 1 reps 8 - d 113 record ratio 4 number 1 sigma 1 reps 4 - d 113 record ratio 6 number 1 sigma 1 reps 6 - d 113 record ratio 3.71428571428571 number 4 sigma 7 reps 26 - d 113 record ratio 3.41666666666667 number 71 sigma 72 reps 246 - d 116 record ratio 8 number 1 sigma 1 reps 8 - d 116 record ratio 4.09523809523809 number 41 sigma 42 reps 172 - d 116 record ratio 4.66666666666667 number 5 sigma 6 reps 28 - d 116 record ratio 6 number 1 sigma 1 reps 6 - d 116 record ratio 4 number 7 sigma 8 reps 32 - d 116 record ratio 12 number 1 sigma 1 reps 12 - d 116 record ratio 6.66666666666667 number 5 sigma 6 reps 40 - d 116 record ratio 6.14285714285714 number 41 sigma 42 reps 258 - d 116 record ratio 6 number 1 sigma 1 reps 6 - d 116 record ratio 4.60215053763441 number 122 sigma 186 reps 856 - d 116 record ratio 4.66666666666667 number 2 sigma 3 reps 14 - d 117 record ratio 8 number 1 sigma 1 reps 8 - d 117 record ratio 7.33333333333333 number 17 sigma 18 reps 132 - d 117 record ratio 12 number 1 sigma 1 reps 12 - d 117 record ratio 6.92783505154639 number 193 sigma 194 reps 1344 - d 117 record ratio 6.94736842105263 number 151 sigma 152 reps 1056 - d 117 record ratio 6.92783505154639 number 193 sigma 194 reps 1344 - d 117 record ratio 6.04615384615385 number 171 sigma 260 reps 1572 - d 117 record ratio 6.06153846153846 number 171 sigma 260 reps 1576 - d 117 record ratio 6.97619047619048 number 167 sigma 168 reps 1172 - d 117 record ratio 6.98550724637681 number 137 sigma 138 reps 964 - d 120 record ratio 8 number 1 sigma 1 reps 8 - d 120 record ratio 6.25287356321839 number 173 sigma 174 reps 1088 - d 120 record ratio 6.30952380952381 number 167 sigma 168 reps 1060 - d 120 record ratio 12 number 1 sigma 1 reps 12 - d 120 record ratio 6.43298969072165 number 193 sigma 194 reps 1248 - d 120 record ratio 6.30927835051546 number 193 sigma 194 reps 1224 - d 120 record ratio 6.36734693877551 number 97 sigma 98 reps 624 - d 120 record ratio 6.48484848484848 number 131 sigma 132 reps 856 - d 120 record ratio 6.54545454545455 number 131 sigma 132 reps 864 - d 120 record ratio 6.4 number 89 sigma 90 reps 576 - d 120 record ratio 6.31111111111111 number 179 sigma 180 reps 1136 - d 120 record ratio 6.41758241758242 number 181 sigma 182 reps 1168 - d 120 record ratio 6.61818181818182 number 109 sigma 110 reps 728 - d 120 record ratio 6.46153846153846 number 181 sigma 182 reps 1176 - d 121 record ratio 4 number 1 sigma 1 reps 4 - d 121 record ratio 6 number 1 sigma 1 reps 6 - d 121 record ratio 4 number 2 sigma 3 reps 12 - d 124 record ratio 12 number 1 sigma 1 reps 12 - d 124 record ratio 6 number 1 sigma 1 reps 6 - d 124 record ratio 4.61904761904762 number 41 sigma 42 reps 194 - d 124 record ratio 4.19354838709677 number 25 sigma 31 reps 130 - d 124 record ratio 5.33333333333333 number 5 sigma 6 reps 32 - d 124 record ratio 4.30769230769231 number 9 sigma 13 reps 56 - d 124 record ratio 6 number 1 sigma 1 reps 6 - d 124 record ratio 4 number 47 sigma 48 reps 192 - d 124 record ratio 4.02439024390244 number 163 sigma 164 reps 660 - d 125 record ratio 8 number 1 sigma 1 reps 8 - d 125 record ratio 12 number 1 sigma 1 reps 12 - d 125 record ratio 6.15625 number 127 sigma 128 reps 788 - d 125 record ratio 6.32432432432432 number 73 sigma 74 reps 468 - d 125 record ratio 6 number 103 sigma 104 reps 624 - d 125 record ratio 6 number 1 sigma 1 reps 6 - d 125 record ratio 9.8989898989899 number 197 sigma 198 reps 1960 - d 125 record ratio 7.89473684210526 number 185 sigma 228 reps 1800 - -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=<|endoftext|> -TITLE: Absoluteness for the Chang model -QUESTION [6 upvotes]: Often, large cardinals imply that many definable inner models and transitive sets have theories which are absolute under forcing. For example, assuming a proper class of Woodin cardinals, the theory of $L(\mathbb{R})$ is absolute under forcing in the following sense: if $V[G]$ is a set-generic extension of $V$, then $H(\omega_1)^V\preccurlyeq H(\omega_1)^{V[G]}$. See this paper of Bagaria for more facts about forcing absoluteness. -I'm curious about a possible absoluteness principle for a much larger model. The Chang model, $C$, is the smallest model of set theory containing all the ordinals, and closed under countable sequences. My question is: - -What is the large cardinal strength of the statement, "The lightface theory of the Chang model is forcing absolute?" - -That is, for every sentence $\varphi$ in the language of set theory without parameters and every set-generic extension $V[G]$, we have $C^V\models\varphi\iff C^{V[G]}\models\varphi$. -It is not at all obvious to me that this is consistent - in fact, right now I suspect it isn't - but I don't see how to get a contradiction from it. -(We can also ask the same question involving some parameters - say, real parameters - of sufficiently low complexity that the obvious way of building a contradiction doesn't work. However, I don't want to get greedy - for now I'm interested in specifically the purely lightface version.) - -REPLY [7 votes]: Corollary 3.1.7 in Larson's Stationary Tower notes states that if $\delta$ is a Woodin limit of Woodin cardinals, then no forcing in $V_\delta$ can change the theory of the Chang model, even with real parameters. Corollary 3.1.10 states the same thing when $\delta$ is strongly compact.<|endoftext|> -TITLE: Coefficients of polynomials appearing in Newton identities relating symmetric polynomials to power sums -QUESTION [10 upvotes]: Let $$p_k(x_1,\ldots,x_n)=x_1^k+\cdots+x_n^k$$ and let -$$e_k(x_1,\ldots,x_n)=\sum_{1\le i_1 -TITLE: The exceptional isomorphism between PGL(3,2) and PSL(2,7): geometric origin? -QUESTION [12 upvotes]: It is well-known there is an isomorphism between $GL(3,2)=PGL(3,2)$, the automorphism group of the Fano plane (i.e. the projective plane over the finite field with two elements), and $PSL(2,7)$, which is the automorphism group of the oriented projective line over the field with seven elements. (More details are on Wikipedia). -What I'd like to know is if there is a finite geometric reason that these two groups are isomorphic. For instance, some combinatorial mapping of these geometries that induces an isomorphism between their automorphism groups. I was talking to Richard Green today about exceptional stuff in low dimensions and he claimed that there wasn't really a nice way to see it, unlike, for instance, the construction of the exceptional (outer) automorphism of $S_6$ using synthemes and duads. - -REPLY [10 votes]: There is an explanation of sorts in Section 1.4 of Elkies's "The Klein quartic in number theory". There is a three-dimensional lattice $L$ over the cyclotomic field $k=\mathbf Q(\zeta_7)$, and $G$ can be defined as its group of isometries. The resulting three-dimensional representation of $G$ has the unusual property of remaining irreducible when reduced modulo every prime of $\mathcal O_k$. Its reduction modulo a prime over $2$ turns out to be $\mathrm{GL}(3,\mathbf F_2)$ acting on $\mathbf F_2^3$, and its reduction modulo a prime over $7$ is $\mathrm{PSL}(2,\mathbf F_7)$ acting on $\mathbf F_7^3$ as the symmetric square of the two-dimensional representation of $\mathrm{SL}(2,\mathbf F_7)$. (Note that since $-1$ acts trivially on the symmetric square, the symmetric square really is a projective representation.) - -REPLY [7 votes]: V. Dotsenko's construction, on math.stackexchange: -https://math.stackexchange.com/questions/1401/why-psl-3-mathbb-f-2-cong-psl-2-mathbb-f-7/1450#1450 -may fit your requirement "combinatorial mapping of these geometries that induces an isomorphism".<|endoftext|> -TITLE: Reference request: correspondence between central simple algebras and quadratic forms -QUESTION [5 upvotes]: Let $A$ be an algebra over $k$, $\operatorname{tr_A}(x, y):=\operatorname{tr}(m_{xy})$ be a trace form on $A$, and $V_A$ be its restriction on the orthogonal complement to $1$. I wonder why a map $A \mapsto V_A$ gives a bijection -$$\left\{ \begin{array}{cc} \text{central simple algebras over } k \\ \text{ of dimension } 4 \end{array} \right\} \leftrightarrow \left\{ \begin{array}{cc} \text{quadratic forms of rank } 3 \\ \text{ with discriminant } 4^3 \end{array} \right\}.$$ -Any help or reference is welcome! - -REPLY [10 votes]: In the formulation, presumably on the right side what is intended are 3-dimensional non-degenerate quadratic spaces (up to isomorphism), with discriminant 1 (same as $4^3$ mod squares as John Ma notes). But to make this work also in characteristic 2, it is better to proceed with a different point of view: that of conformal isometry of quadratic spaces (i.e., isomorphisms $T:V \simeq V'$ such that $q' \circ T = \lambda q$ for some $\lambda \in k^{\times}$). More specifically, we claim that away from characteristic 2, every 3-dimensional non-degenerate quadratic space is conformal to a unique one with discriminant 1. Thus, by working with conformal isometry classes we will be able to work in a fully characteristic-free manner. -To see what is going on, recall that the set of isomorphism classes of central simple algebras of dimension 4 is ${\rm{H}}^1(k, {\rm{PGL}}_2)$, and the set of conformal isometry classes of dimension 3 is ${\rm{H}}^1(k, {\rm{PGO}}_3)$. But ${\rm{GO}}_{2m+1} = {\rm{GL}}_1 \times {\rm{SO}}_{2m+1}$, so ${\rm{PGO}}_{2m+1} = {\rm{SO}}_{2m+1}$. Hence, ${\rm{PGO}}_3 = {\rm{SO}}_3$. Since ${\rm{SO}}_3 \simeq {\rm{PGL}}_2$ through the representation of ${\rm{PGL}}_2$ via conjugation on the 3-dimensional space of traceless $2 \times 2$ matrices equipped with the determinant as the standard split non-degenerate quadratic form $xy - z^2$ (preserved by that conjugation action!), that answers the entire question at the level of isomorphism classes of objects. (The link to ${\rm{SO}}_3$ encodes the link to discriminant 1.) -But we can do better than keep track of isomorphism classes: we can also keep track of isomorphisms, as explained below. This is a refinement of John Ma's answer, as well as that of Matthias Wendt (which appeared at almost exactly the same time as this answer first appeared, so I didn't see it until this one was done). -The following notation will permit considering finite fields on equal footing with all other fields. For a finite-dimensional central simple algebra $A$ over an arbitrary field $k$, let ${\rm{Trd}}:A \rightarrow k$ be its "reduced trace" and ${\rm{Nrd}}:A \rightarrow k$ be its "reduced norm". These are really most appropriately viewed as "polynomial maps" in the evident sense. That is, if $\underline{A}$ is the "ring scheme" over $k$ representing the functor $R \rightsquigarrow A \otimes_k R$ (i.e., an affine space over $k$ equipped with polynomial maps expressing the $k$-algebra structure relative to a choice of $k$-basis) then we have $k$-morphisms ${\rm{Trd}}:\underline{A} \rightarrow \mathbf{A}^1_k$ and ${\rm{Nrd}}:\underline{A} \rightarrow \mathbf{A}^1_k$. -For $A$ of dimension 4 we set $\underline{V}_A$ to be the kernel of ${\rm{Trd}}:\underline{A} \rightarrow \mathbf{A}^1_k$; speaking in terms of kernel of ${\rm{Trd}}$ is a bit nicer than speaking in terms of orthogonal complements so that one doesn't need to separately consider characteristic 2 (where the relationship between quadratic forms and symmetric bilinear forms breaks down). This $\underline{V}_A$ is an affine space of dimension 3 over $k$ on which ${\rm{Nrd}}$ is a non-degenerate quadratic form $q_A$ (i.e., zero locus is a smooth conic in the projective plane $\mathbf{P}(V_A^{\ast})$, where $V_A := \underline{V}_A(k)$): indeed, these assertions are "geometric" in nature, so it suffices to check them over $k_s$, where $A$ becomes a matrix algebra and we can verify everything by inspection. -We will show that the natural map of affine varieties -$$\underline{{\rm{Isom}}}(\underline{A}, \underline{A}') \simeq \underline{{\rm{CIsom}}}((\underline{V}_A, q_A), (\underline{V}_{A'}, q_{A'}))/\mathbf{G}_m$$ -from the "isomorphism variety" to the "variety of conformal isometries mod unit-scaling" is an isomorphism; once that is shown, by Hilbert 90 we could pass to $k$-points to conclude that isomorphisms among such $A$'s correspond exactly to conformal isometries among such $(V_A, q_A)$'s up to unit scaling. It suffices to check this isomorphism assertion for varieties over $k_s$, where it becomes the assertion that the natural map -$$\underline{{\rm{Aut}}}_{{\rm{Mat}}_2/k} \rightarrow {\rm{CAut}}_{({\rm{Mat}}_2^{{\rm{Tr}}=0}, \det)/k}/\mathbf{G}_m$$ -from the Aut-scheme to the scheme of conformal isometries up to unit-scaling is an isomorphism. But this is precisely the natural map -$${\rm{PGL}}_2 \rightarrow {\rm{PGO}}(-z^2-xy) = {\rm{PGO}}(xy+z^2) = {\rm{SO}}(xy+z^2) = {\rm{SO}}_3$$ -between smooth affine $k$-groups that is classically known to be an isomorphism over any field $k$ (can check bijectivity on geometric points and the isomorphism property on tangent spaces at the identity points). -Finally, we want to show that every 3-dimensional non-degenerate quadratic space $(V, q)$ is conformal to $(V_A, q_A)$ for some $A$. -Note that if such an $A$ exists then it is unique up to unique isomorphism in the sense that if $A$ and $A$ are two such equipped with conformal isometries $(V_A, q_A) \simeq (V, q) \simeq (V_{A'}, q_{A'})$ then this composite conformal isometry arises from a unique isomorphism $A \simeq A'$ of $k$-algebras. Hence, by Galois descent (!) it suffices to check existence over $k_s$! But over a separably closed field the smooth projective conic has a rational point, so $(V, q)_{k_s}$ contains a hyperbolic plane and thus is isometric to $xy + \lambda z^2$ on $k_s^3$ for some $\lambda \in k_s^{\times}$. This is conformal to $(-1/\lambda)q_{k_s}$, but $(1/\lambda)xy - z^2$ and $-x'y' - z^2$ are clearly isometric, and the latter is ${\rm{Mat}}_2^{{\rm{Tr}}=0}$ equipped with the restriction of det. - -REPLY [6 votes]: On a conceptual level, I would like to see this as an instance of a sporadic isomorphism of algebraic groups. Take the conjugation action of $GL_2$ on the space of $2\times 2$-matrices with trace 0, equipped with the trace form. One can show that this induces an isomorphism $PGL_2\cong SO(3)$. This is discussed e.g. in this MO-question. From the isomorphism of algebraic groups, you get a bijection -$$ -H^1(k,PGL_2)\cong H^1(k,SO(3)). -$$ -These are étale/Galois cohomology groups classifying étale locally trivial torsors over $\operatorname{Spec} k$ with the appropriate structure group. Now identify $PGL_2$-torsors with central simple algebras of dimension $4$, this is done e.g. in the book of Gille-Szamuely "Central simple algebras and Galois cohomology". On the other side, identify $SO(3)$-torsors with quadratic form of trivial discriminant. This should give the bijection in the question.<|endoftext|> -TITLE: Is there a polynomial-time algorithm for untangling the unknot? -QUESTION [11 upvotes]: I've found assertions that recognising the unknot is NP (but not explicitly NP hard or NP complete). I've found hints that people are looking for untangling algorithms that run in polynomial time (which implies they may exist). I've found suggestions that recognition and untangling require exponential time. (Untangling is a form of recognising.) -I suppose I'm asking whether there exists -(1) a "diagram" of a knot, -(2) a "cost" measure of the diagram, -(3) a "move" which can be applied to the diagram, -(4) the "move" always reduces the "cost", -(5) the "move" can be selected and applied in polynomial time, -(6) the "cost" can be calculated in polynomial time. -For instance, Reidemeister moves, fail on number (4) if the "cost" is number of crossings. -So what is the current status of the problem? -Thanks -Peter - -REPLY [7 votes]: Recently, Marc Lackenby discussed a new algorithm for unknot recognition in a talk at the Newton Institute (see time around 1:03). He conjectures (but indicates at the time that it is work-in-progress) that his algorithm runs in quasi-polynomial time ($c^{\log c}$, where $c$ is the crossing number). In any case, his talk discusses the state of the art of the subject. More recently, he is announcing this as a conjecture, so presumably it is still a work in progress after 6 months.<|endoftext|> -TITLE: Strongly proper ccc forcing -QUESTION [10 upvotes]: Is there an example of a strongly proper ccc forcing that is not equivalent to Cohen forcing? - -REPLY [6 votes]: The answer to your question is yes, and it follows from the following paper of Koppelberg and Shelah Subalgebras of Cohen algebras need not be Cohen. -In this paper, it is shown, for each $\kappa \geq \aleph_2,$ there exists a non-Cohen complete subalgebra of $Add(\omega, \kappa)$. -This subalgebra, is c.c.c and strongly proper, as a projection of Cohen forcing, but is not isomorphic to Cohen forcing.<|endoftext|> -TITLE: Is a normed space which is homeomorphic to a Banach space complete? -QUESTION [30 upvotes]: I have a normed space $(E,||\cdot||)$ which is homeomorphic (as a topological space) to a Banach space $F$. -Does this imply that $(E,||\cdot||)$ is also a Banach space? -I think I read something like this to be true if $E$ (and therefore also $F$) is separable, but I am not totally sure. So, also this special case would be interesting. - -REPLY [31 votes]: Let $\bar{E}$ be the norm completion of $E$, which is a Banach space. Then we can consider $E$ as a dense linear subspace of $\bar{E}$, where the subspace topology and the norm topology on $E$ coincide. In particular, since this topology is homeomorphic to $F$, it is completely metrizable, so $E$ is a $G_\delta$ in $\bar{E}$ (Kechris, Classical Descriptive Set Theory, Theorem 3.11). As a dense $G_\delta$, in particular $E$ is comeager in $\bar{E}$. If $x \in \bar{E} \setminus E$, then $E, E+x$ are disjoint comeager subsets of $\bar{E}$, which is absurd by the Baire category theorem. So $E = \bar{E}$ and thus the norm on $E$ is complete. -I think I did not use separability anywhere. - -REPLY [26 votes]: It is an old result of Victor Klee (answering a question of Banach) that a metrizable topological vector space (i.e., there is a translation invariant metric giving the topology) is a complete topological vector space (i.e., w.r.t. the uniformity induced by the $0$-neighborhoods) if there is some complete metric (not necessarily translation invariant) giving the same topology. -The reference is: Victor Klee, Invariant metrics in groups (solution of a problem of Banach), Proc. Amer. Math. Soc. 3, 484 - 487 (1952). -The proof (which is quite similar to the arguments in Nate's answer) can also be seen in Koethe's book Topological Vector Spaces I §15.11.<|endoftext|> -TITLE: Why is a Topological Field Theory equivalent to a Frobenius algebra? -QUESTION [15 upvotes]: How can a physicist understand a 2-dimensional topological field theory as a Frobenius algebra? Are there some explicit examples in order to understand this relation? -The definition (e.g. on Wikipedia) of the Frobenius algebra is quite clear, it is a finite dimensional associative algebra equipped with a special non-degenerate bilinear form. If this algebra is represented by $n\times n$ matrices then the bilinear form is the trace. -Now, I fail to see how this is a TFT. -In specific two very famous topological field theories in 2-dimensions (that physicists work with a lot) are the topological A-model and the topological B-model (a la Witten). These models are related via mirror symmetry. How are they defined as a Frobenius algebra and what are their differences as a Frobenius algebra? And natural question to ask is wether mirror symmetry in the TFT side relates somehow the two Frobenius algebras on the algebra side. -So I would like to understand intuitively why a Frobenius algebra and a TFT are the same thing and the detail I ask in specific about the A-model and the B-model which are of my main interest. - -REPLY [13 votes]: Arun Debray's answer is good, with one addendum: 2d TQFT is the same thing as a commutative Frobenius algebra. But to truly understand what's going on, one also needs some pictures. A commutative Frobenius algebra can be defined using pictures that just happen to look exactly like the relations holding between 2-dimensional cobordisms. The best place to learn about this is here: - -Joachim Kock, Frobenius Algebras and 2d Topological Quantum Field Theories, Cambridge U. Press, Cambridge, 2004. - -But briefly, a Frobenius algebra has a multiplication and unit obeying associativity and the left and right unit laws: - -and also a comultiplication and counit obeying coassociativity and the left and right counit laws: - -where the multiplication and comultiplication obey the Frobenius laws: - -A Frobenius algebra is commutative if switching two things and then multiplying them is the same as multiplying them: - -Furthermore, a Frobenius algebra is commutative iff it is cocommutative, meaning this: - -The ultimate statement of this fact is that $2\mathrm{Cob}$, the category with compact oriented 2-dimensional cobordisms as morphisms, is the free symmetric monoidal category on a commutative Frobenius monoid object. The story gets much more interesting in 3 dimensions, and for that I recommend this paper: - -Bruce Bartlett, Christopher L. Douglas, Christopher J. Schommer-Pries and Jamie Vicary, Extended 3-dimensional bordism as the theory of modular objects.<|endoftext|> -TITLE: How can the Kalman filter be adapted to handle binary observations? -QUESTION [7 upvotes]: Imagine a coin with a time-varying probability of coming up heads. (For example, perhaps the probability follows a random walk that is constrained to live in $[0, 1]$. And say we have some information about how quickly we expect the true probability to vary over time (analogous to the standard deviation of the random walk described above). -At discrete time points, we are given data about the outcome of a given coin flip. The goal is to come up with a robust procedure to estimate that probability over time. -This sounds like a task for a Kalman filter. But the literature seems to assume that the observations will be normally distributed around a linear function of the state. That feels quite violated by the assumptions above. -What's a reasonable way to proceed? - -REPLY [4 votes]: In principle, this is what nonlinear filtering does. Check this out, also under the name "hidden Markov model". Particle filters can be adapted to deal with -this setup. -In a nutshell, here is what hidden Markov processes are. -You have a two component Markov chain $(X_n,Y_n)$ (I write it in discrete time, there is an analogue in continuous time). You observe only -$\{Y_i\}_{i=1}^n$ and want to estimate $X_n$, i.e. construct the conditional pdf -of $X_n$ given the observations. -In your case, the law of $Y_n$ given the state history $\{X_i\}_{i=1}^n$ depends only on $X_n$ and is a Bernoulli variable with mean depending on $X_n$. $X_n$ itself is what you call $p$. -PS I just noted that my answer is closely related to that of @passerby51.<|endoftext|> -TITLE: Do these matrices have a name? -QUESTION [8 upvotes]: I am looking for matrices $A\in \mathbb{R}^{m\times n}$ with $m>n$, $A^TA=\frac{m}{n}I$ and $diag(AA^T)=(1\ \dots\ 1)$ where $diag$ denotes the diagonal. Do such matrices have a name? An example for such a matrix would be an $8\times 3$ matrix with the coordinates of a cube centered at the origin as rows. Is the matrix $A$ for every pair $(m,n)$ uniquely determined up to right multiplication with an orthogonal matrix? How are the $m$ points given by the rows of $A$ distributed over the $(n-1)$-dimensional sphere? -My motivation is that I have an unknown $x\in \mathbb{R}^n$. I assume that I can measure the scalar product $a^Tx$ with any unit vector $a\in S^{n-1}$. The measurements are assumed to be i.i.d. Now we assume we have $m$ measurements and write the corresponding vectors as the rows of a matrix $A\in \mathbb{R}^{m\times n}$ and the vectors as a matrix $b\in \mathbb{R}^m$. The best possible reconstruction (see Gauss-Markov) of $x$ is given by the least square solution $(A^TA)^{-1}A^Tb$ where $b\in \mathbb{R}^n$ denotes the measurements. The covariance matrix for the reconstruction is $(A^TA)^{-1}$. This means that if $A$ satisfies the properties above then the components of the reconstructed $x$ are uncorrelated. Also I think that $A$ gives us the best possible way to measure $x$, i.e. minimizing the standard deviation. - -REPLY [5 votes]: Assuming that the Hadamard Conjecture is true, if $m$ is a multiple of $4$, then a thin $m \times n$ matrix that satisfies the given constraints is given by -$$\boxed{\mathrm A := \frac{1}{\sqrt n} \mathrm H_m^{\top} \mathrm S_n}$$ -where - -$\mathrm H_m \in \{\pm 1\}^{m \times m}$ is a Hadamard matrix. Thus, the $m$ rows of $\mathrm H_m$ are orthogonal, i.e., $$\mathrm H_m \mathrm H_m^{\top} = m \mathrm I_m$$ -$\mathrm S_n$ is a thin $m \times n$ matrix whose $n$ columns are chosen from the $m$ columns of the $m \times m$ identity matrix. Thus, the $n$ columns of $\mathrm S_n$ are orthonormal, i.e., - -$$\mathrm S_n^{\top} \mathrm S_n = \mathrm I_n$$ -Hence, -$$\mathrm A^{\top} \mathrm A = \frac{1}{n} \mathrm S_n^{\top} \mathrm H_m \mathrm H_m^{\top} \mathrm S_n = \frac{m}{n} \mathrm S_n^{\top} \mathrm S_n = \frac{m}{n} \mathrm I_n$$ -as desired. Let $\mathrm e_k$ and $\mathrm h_k$ denote the $k$-th columns of $\mathrm I_m$ and $\mathrm H_m$, respectively. Hence, -$$\mathrm e_k^{\top} \mathrm A \mathrm A^{\top} \mathrm e_k = \| \mathrm A^{\top} \mathrm e_k \|_2^2 = \frac 1n \| \mathrm S_n^{\top} \mathrm H_m \mathrm e_k \|_2^2 = \frac 1n \| \mathrm S_n^{\top} \mathrm h_k \|_2^2 = \frac 1n \sum_{k=1}^n (\pm 1)^2 = \frac nn = 1$$ -for all $k \in \{1,2,\dots,m\}$, as desired. Note that we used the fact that the entries of $\mathrm h_k$ are $\pm 1$. -If $m$ is a power of $2$, then $\mathrm H_m$ can be built recursively using the Sylvester construction -$$\mathrm H_{2k} = \begin{bmatrix} \mathrm H_k & \mathrm H_k\\ \mathrm H_k & -\mathrm H_k\end{bmatrix} \qquad \qquad \qquad \mathrm H_1 = 1$$ -which builds (symmetric) Walsh matrices. If $m$ is not a power of $2$, we can use the Paley construction instead. - -Example -Let $m = 8$ and $n = 3$. Since $8$ is a power of $2$, we can use the Sylvester construction to build $\mathrm H_8$. -Using MATLAB, ->> H1 = 1; ->> H2 = [H1,H1;H1,-H1]; ->> H4 = [H2,H2;H2,-H2]; ->> H8 = [H4,H4;H4,-H4] - -H8 = - - 1 1 1 1 1 1 1 1 - 1 -1 1 -1 1 -1 1 -1 - 1 1 -1 -1 1 1 -1 -1 - 1 -1 -1 1 1 -1 -1 1 - 1 1 1 1 -1 -1 -1 -1 - 1 -1 1 -1 -1 1 -1 1 - 1 1 -1 -1 -1 -1 1 1 - 1 -1 -1 1 -1 1 1 -1 - -Let the $3$ columns of $\mathrm S_3$ be the first $3$ columns of $\mathrm I_8$ ->> I8 = eye(8); ->> H8 * I8(:,[1,2,3]) - -ans = - - 1 1 1 - 1 -1 1 - 1 1 -1 - 1 -1 -1 - 1 1 1 - 1 -1 1 - 1 1 -1 - 1 -1 -1 - -Note that the last four rows are copies of the first four rows. Hence, let the $3$ columns of $\mathrm S_3$ be the 2nd, 3rd and 5th columns of $\mathrm I_8$ ->> H8 * I8(:,[2,3,5]) - -ans = - - 1 1 1 - -1 1 1 - 1 -1 1 - -1 -1 1 - 1 1 -1 - -1 1 -1 - 1 -1 -1 - -1 -1 -1 - -Note that the $8$ rows are now the $8$ vertices of the cube $[-1,1]^3$. -We build matrix $\mathrm A$ by normalizing the rows ->> A = inv(sqrt(3)) * H8 * I8(:,[2,3,5]) - -A = - - 0.5774 0.5774 0.5774 - -0.5774 0.5774 0.5774 - 0.5774 -0.5774 0.5774 - -0.5774 -0.5774 0.5774 - 0.5774 0.5774 -0.5774 - -0.5774 0.5774 -0.5774 - 0.5774 -0.5774 -0.5774 - -0.5774 -0.5774 -0.5774 - -Is the constraint $\mathrm A^{\top} \mathrm A = \frac 83 \mathrm I_3$ satisfied? ->> A' * A - -ans = - - 2.6667 0 0 - 0 2.6667 0 - 0 0 2.6667 - -It is. Are the diagonal entries of $\mathrm A \mathrm A^{\top}$ equal to $1$? ->> A * A' - -ans = - - 1.0000 0.3333 0.3333 -0.3333 0.3333 -0.3333 -0.3333 -1.0000 - 0.3333 1.0000 -0.3333 0.3333 -0.3333 0.3333 -1.0000 -0.3333 - 0.3333 -0.3333 1.0000 0.3333 -0.3333 -1.0000 0.3333 -0.3333 - -0.3333 0.3333 0.3333 1.0000 -1.0000 -0.3333 -0.3333 0.3333 - 0.3333 -0.3333 -0.3333 -1.0000 1.0000 0.3333 0.3333 -0.3333 - -0.3333 0.3333 -1.0000 -0.3333 0.3333 1.0000 -0.3333 0.3333 - -0.3333 -1.0000 0.3333 -0.3333 0.3333 -0.3333 1.0000 0.3333 - -1.0000 -0.3333 -0.3333 0.3333 -0.3333 0.3333 0.3333 1.0000 - -They are.<|endoftext|> -TITLE: Is there a notion of hyperbolicity for number rings? -QUESTION [10 upvotes]: For algebraic curves over a nice enough field $k$, we have a notion of what it means to be hyperbolic: If $\overline{C}$ is a smooth projective curve of genus $g$ and $P_1,\dots,P_n$ are closed points, then the open curve $C=\overline{C}\setminus\{P_1,\dots,P_n\}$ is hyperbolic if $2-2g-\sum \deg P_i<0$. -The general function-field/number-field analogy suggests the following question: - -Is there a notion of hyperbolicity for rings - of $S$-integers in number fields (possibly along the lines of some - inequality appearing in arithmetic intersection theory)? - -If there is such a notion of hyperbolicity, are most $S$-integer rings hyperbolic? Is it also true that number rings with trivial or finite étale fundamental group (as in this MO-question) are not hyperbolic? -Here is some of the background which motivated the question: In the function field case, I think I have convinced myself that the slightly stronger inequality $2-2g-n<-1$ implies that there will be cuspidal cohomology for the group $SL_2(k[C])$ (in the sense that the quotient of the product of Bruhat-Tits trees for the points $P_1,\dots,P_n$ is rationally non-contractible). I am trying to understand if similar statements can be made in the number-field situation. Therefore, the ideal answer to the question would be some numerical inequality which relates to Harder's Gauß-Bonnet formula for cohomology of arithmetic groups, implying that hyperbolicity forces non-trivial rational cohomology for the arithmetic group $SL_2(\mathcal{O}_{K,S})$. - -REPLY [6 votes]: My intuition is that there will not be a precise definition of a hyperbolic number field. However, there there may be some number fields you can confidently say are hyperbolic. -Consider the following characterization of a hyperbolic complete curve in the function field case - a degree $d$ cover of $\mathbb P^1$ is hyperbolic if its discriminant $\Delta$ is greater than $q^{2d}$. -This is equivalent to $g>1$ by the Riemann-Hurwitz formula $2-2g = 2d - \log_q (\Delta)$. -So it seems like a number field of degree $d$ should be hyperbolic if the discriminant is greater than $C^{d}$ for some constant $C$. If you want to prove arithmetic analogues of a certain consequence of hyperbolicity, you could try to prove it as a consequence of large discriminant. Is your desired result true for quadratic fields of large discriminant, say? -However for a full definition of hyperbolicity you would need to find a single value of $C$ that is the cutoff for all properties of hyperbolic curves. I don't think a plausible value of $C$ is known - I don't know if one is expected to exist. -One property $C$ should have is that every number field of discriminant $ -TITLE: Category without identities? -QUESTION [12 upvotes]: Just as a monoid is a category with a single object, a semigroup may be seen as a non-unital category, still with associative composition. Then an $S$-set for $S$ a semi-group can be seen as a functor from the category corresponding to $S$ into the category of sets. -One of the nice things about the functor category of $M$-sets for $M$ a fixed monoid is that it's a topos, so in particular extensive i.e admitting a nice notion of connectedness. In this context an $M$-set $X$ is connected iff $\forall x,y\in X \exists s,t\in M$ such that $sx=y$ or $ty=x$. -I think this topos is also locally connected (there's a functorial assignment of connected components, left adjoint to discrete stuff) since we can just "take" these connected components for any monoid. -I was wondering whether not having units gets in the way of repeating the same terms with meaning for semigroups? -More generally, which portions of category theory still hold without identities? The most crucial thing that fails, I think, is Yoneda. - -REPLY [8 votes]: which portions of category theory still hold without identities? - -Almost everything. https://arxiv.org/abs/1311.3524v1. (The link is to v1 of the paper, because the author has removed v2 from the arXiv.) -I'm honestly astonished by the fact that this paper hasn't the impact it deserves.<|endoftext|> -TITLE: A result of Schützenberger on commutators and powers in free groups -QUESTION [17 upvotes]: It is an old result of Schützenberger that in a free group, a basic commutator cannot be a proper power. A look at the original reference -M.-P. Schützenberger, Sur l'équation $a^{2+n} = b^{2+m}c^{2+p}$ dans un groupe libre, C. R. Acad. Sci. Paris 248 (1959), 2435–2436 (French). -quickly reveals that a lot of details are missing and some claims appear to be wrong. - -Question: Let $F$ be a free group and $a,b,c \in F$ with $c \neq 1$ and $n \geq 2$. Why is $[a,b] \neq c^n$? - -In particular, it would be nice to have a somewhat geometric proof of this apparently fundamental fact. There is an algebraic proof in -G. Baumslag, Some aspects of groups with unique roots, Acta Math 104(3) (1960), 217–303. -as Lemma 36.4 but it is relies on various technical computations and is hard to grasp. - -REPLY [4 votes]: There is a visualisation of Schützenberger's -observation based on a variant of the -car-crash lemma that says that - -for any multiple motion -on a map on a closed oriented -surface $S$, the -number of points of complete collision is at least -$\chi(S)+\sum\limits_D(d_D-1)$, -where the summation runs over all -faces $D$ of the map -and $d_D$ is the number of cars moving around a face $D$. - -See -[this self-advertisement] -for exact definitions. -So, if a non-identity commutator in a free group $F(x,y,\dots)$ is, e.g., -a cube of a word $c$, we obtain a one-face map on a torus, where the -label of the face is $c^3$. Let three cars move around this face -counter-clockwise -with a -constant speed one edge per a minute; at a moment -$t$ each car moves along an edge labelled by the $[t]$th letter -of the word $c$ (where the integer part $[t]$ -of $t\in\mathbb R$ is counted modulo the length of $c$). - -The car-crash lemma asserts that -$$ -\small -\pmatrix{ -\hbox{the number of points} -\\\\ -\hbox{of complete collision} -} -\geqslant -\begin{pmatrix} -\hbox{the Euler characteristic} -\\\\ -\hbox{of the torus,} -\\\\ -\hbox{i.e. 0} -\end{pmatrix} -+ -\pmatrix{ -\hbox{$d_D$, i.e. the number of cars} -\\\\ -\hbox{moving around the unique face $D$,} -\\\\ -\hbox{i.e. 3} -} --1=2. -%} -$$ -On the other hand, a collision inside an edge is impossible: - -such a collision would imply that some letter of the word -$c$ is $x$ and $x^{-1}$ simultaneously. A similar argument shows -that a collision at a vertex cannot occur too. This contradiction -proves that a commutator cannot be a cube. -This approach was used in -[this self-advertisement], -[this self-advertisement], -and -[this self-advertisement] -to obtain generalisations of -Schützenberger's observation -and -Duncan--Howie's -theorem (mentioned by Henry Wilton).<|endoftext|> -TITLE: The second Milnor $K$-theory of a field -QUESTION [6 upvotes]: Let $\mathbf{Q}^{\mathrm{ab}}$ be the maximal abelian extension of the field of rational numbers $\mathbf{Q}$. I'm interested in the following question: -Is it true that $K^{M}_{2}(\mathbf{Q}^{\mathrm{ab}})/pK^{M}_{2}(\mathbf{Q}^{\mathrm{ab}})=0$ for any prime $p$ ? -where $K^{M}_{\ast}$ is Milnor $K$-theory. - -REPLY [13 votes]: By the Milnor-Bloch-Kato conjecture, this is equivalent to $\mathrm{Br}(\mathbf{Q}^{\mathrm{ab}})[p] = 0$ (by the Kummer sequence $1 \to \mu_p \to \mathbf{G}_m \to \mathbf{G}_m \to 1$, Hilbert 90 $H^1(K,\mathbf{G}_m) = 0$ and $H^2(K,\mathbf{G}_m) = \mathrm{Br}(K)$). This follows from [Neukirch-Schmidt-Wingberg, Cohomology of Number Fields https://www.mathi.uni-heidelberg.de/~schmidt/NSW2e/NSW2.2.pdf ], Proposition (8.1.14) (ii).<|endoftext|> -TITLE: Coincidence in dimensions of Hilbert modular cusp forms -QUESTION [6 upvotes]: I'm wondering if someone can shed some light onto the following, which most likely is just a silly coincidence, but I would be interested to know if there is more to it. -I found myself needing to compute some dimensions of spaces of Hilbert modular cusp forms over $\mathbb{Q}(\sqrt{5})$ of parallel weight $[2k,2k]$ and level $\Gamma_0(\mathfrak{p}_{11})$ where $\mathfrak{p}_{11} $ is a prime ideal dividing $11$. Now I started by doing the naive thing of simply going to MAGMA and asking for Dimension(HilbertModularCuspForms(...)). Now this is not a good idea since it will become very slow very quickly. But still, if you wait a bit you find that the sequence [Dimension, weight] is $[ 1, 4 ],[ 5, 6 ] -,[ 9, 8 ] -,[ 17, 10 ] -,[ 25, 12 ] -,[ 33, 14 ] -,[ 45, 16 ] -,[ 57, 18 ] -,[ 73, 20 ] -,[ 89, 22 ] -,[ 105, 24 ] -,[ 125, 26 ] -,[ 145, 28 ] -,[ 169, 30 ]$ Now I looked at the sequence of dimensions and wondered if there was a pattern, so I did the usual trick of typing OEIS into google and then putting in the dimensions. It turns out the dimensions match up (as far as I've computed) with the number of (integer) solutions of $w+2x+5y=0$ for $x,y,w \in [-k \dots k]$. -So my question is: Is this nothing more than a coincidence? or is there any connection here? The dimension formulas for Hilbert modular forms seem to usually be a bit messy so I can't really see how the two things are connected (but this likely just because I haven't looked at them carefully yet). My hope was to eventually have a quick way of computing these dimensions and having such a description would make computations much easier. -Full disclosure: The sequence on OEIS actually starts $1,1,5,9,17,$.. and the first "1" would correspond to the dimension in weight $[2,2]$, which is actually zero, so the sequences don't match up in the first term. But if you look at the corresponding space on the quarternionic side (via the Jacquet-Langlands correspondence), then in weight $[2,2]$ one needs to quotient out the space by the stuff factoring through the reduced norm map, which in this case is 1-dimensional, so this might be the where the "1" is. -Thank you. - -REPLY [7 votes]: This answer is very late (and D. Loeffler's formula in the comments is certainly correct), but I want to point out that the dimension formula for parallel-weight modular forms over real-quadratic fields is not all that messy after all. -By Shimizu (in particular, the formulation of Theorem 2.15 in Thomas and Vasquez, Rings of Hilbert modular forms, Compos. Math. 48(2) 139-165), for every $k \ge 2$ and every congruence subgroup $\Gamma \le \Gamma_K := \mathrm{PSL}_2(\mathcal{O}_K)$, -$$\mathrm{dim}\, S_{2k,2k}(\Gamma) = 2k(k-1) \zeta_K(-1) [\Gamma_K : \Gamma] + \chi - a_3 \delta_K s_D - \varepsilon_k a_5 /5,$$ where: -$\zeta_K$ is the Dedekind zeta function; -$\chi$ is the arithmetic genus (which is $1 + \mathrm{dim}\, S_{2,2}(\Gamma)$; in particular, this formula is off by one at $k=1$); -$a_n$ counts the elliptic points with stabilizer of size $n$; -$\delta_k$ and $\varepsilon_k$ are defined by $$\delta_k = \begin{cases} 1: & k \equiv 2\, (\text{mod} \, 3); \\ 0: & \text{otherwise}; \end{cases} \quad \varepsilon_k = \begin{cases} 2 : & k \equiv 2,4 \, (\text{mod}\, 5); \\ 1: & k \equiv 3 \, (\text{mod}\, 5); \\ 0: & \text{otherwise}; \end{cases}$$ -and where $s_D$ depends on the discriminant $D$ of $K$: $$s_D = \begin{cases} 1/6: & D \not \equiv 0 \, (\text{mod}\, 3); \\ 4/15: & D > 12 \; \text{and} \; D \equiv 3 \, (\text{mod} \, 9); \\ 1/3: & D = 12 \; \text{or} \; D \equiv 6 \, (\text{mod}\, 9). \end{cases}$$ -There are at most three ``unknowns" in this formula, $\chi$, $a_3$ and $a_5$, and if all else fails you can simply read these off the dimensions of $S_{2k,2k}(\Gamma)$ for $k=1,2,3$. (And $a_5 = 0$ unless $K = \mathbb{Q}(\sqrt{5})$. Unfortunately you are in that exceptional case.) -In the case you looked at, we find $\zeta_K(-1) = 1/30$; $[\Gamma_K: \Gamma] = 12$; $\chi = 1$; $a_3 = 0$; $a_5 = 4$ and therefore $$\mathrm{dim}\, S_{2k,2k}(\Gamma) = \frac{4k(k-1)}{5} + 1 - \frac{4}{5}\varepsilon_k, \; \; k \ge 2.$$<|endoftext|> -TITLE: When do we genuinely need Noetherian conditions? -QUESTION [26 upvotes]: When are Noetherian conditions on a scheme genuinely essential in algebraic geometry? -I am under the impression that most of the time these conditions are imposed for expositional clarity and to simplify the commutative algebra involved (for example throughout Hartshorne and FGA), and that they can be removed by paying more attention to finiteness conditions on the morphisms involved. -So what are the things that genuinely require Noetherian conditions? Even better, is there a "slogan" that tells you when they will be needed? - -REPLY [26 votes]: Here are some examples illustrating the genuine necessity of noetherian assumptions: -1) Every scheme with just one point is the spectrum of a local artinian ring? -This is true for every noetherian one point scheme and false for every non-noetherian one point scheme. -A non-noetherian (and thus non-Artinian) example : $\operatorname {Spec}(\mathbb Q[T_1,T_2,T_3,\cdots]/\langle T_1^2,T_2^2,T_3^2 \rangle )$ -2) A scheme has only finitely many irreducible components ? -A noetherian scheme has only finitely many irreducible components, but a non-noetherian scheme may have infinitely many irreducible components: this is the case for any disjoint union of infinitely many non-empty schemes. -3) Every scheme has a closed point? -This is true for every noetherian scheme (actually for any quasi-compact scheme), but there exist schemes without any closed point: Qing Liu, Chapter 3, Exercise 3.27, page 114. -4) Injective modules give injective sheaves? -If $I$ is an injective module over the ring $A$, then the associated quasi-coherent sheaf $\tilde I$ on $X=\operatorname {Spec}(A)$ is an injective sheaf of $\mathcal O_X$- Modules if $A$ is noetherian but is not necessarily injective for $A$ non-noetherian: SGA6, Exposé II, Appendice I Un contre-exemple de Verdier, page 195. -5) A finitely presented sheaf is coherent? -Given on a scheme $X$ a sheaf of $\mathcal F$ of $\mathcal O_X$-Modules, does the existence of an open covering $(U_i)$ of $X$ for which one has exact sequences $\mathcal O_{U_i}^{n_i}\to \mathcal O_{U_i}^{m_i}\to \mathcal F\vert _{U_i}\to 0$ imply that $\mathcal F$ is coherent? -The answer is yes if $X$ is noetherian (or even locally noetherian) but no in general: there exist non-noetherian rings $A$ such that the structural sheaf $\mathcal O_X$ on $X=\operatorname {Spec}(A)$ is not coherent! -6) A scheme is affine if all its quasi-coherent sheaves are acyclic ? -Serre's criterion is that a noetherian scheme $X$ is affine if and only if $H^p(X, \mathcal F)=0$ for all quasi-coherent sheaves $\mathcal F$ on $X$ and all $p\gt 0$. -This no longer holds if $X$ is not assumed noetherian: given a field $k$, any infinite disjoint sum $X=\coprod \operatorname {Spec}k$ satisfies the cohomology condition but is not affine since it is not quasi-compact.<|endoftext|> -TITLE: in need of a direct combinatorial/bijective proof -QUESTION [9 upvotes]: The following are very familiar and basic items, individually. -(1) The number $a(n)$ of rectangles (parallel to axes) in an $n\times n$ square grid. -(2) The number $b(n)$ of cubes (parallel to axes) in an $n\times n\times n$ cube. -However, I could not find a reference to a direct bijective proof for $a(n)=b(n)$. Can you provide such an argument of reference? - -REPLY [3 votes]: Following @wojowu's suggestion, we have: -Let $h$ be the side of the inner cube, and let $(i,j,k)$ be its corner nearest the origin. Then we have $0\le i,j,k < n-h+1 \le n$. -Let us describe our rectangle by the corners $(x_1, y_1)$ and $(x_2, y_2)$ with $x_1j$}. -\end{cases} -$$ -Going the other way we have -$$ -(x_1,y_1),(x_2,y_2)\to -\begin{cases} -(x_1,x_2,y_1,n-y_2+1) & \text{if $x_2 < y_2$}\\ -(x_1,x_1,y_1,n-y_2+1) & \text{if $x_2 = y_2$}\\ -(x_2,y_1,x_1,n-y_2+1) & \text{if $x_2 > y_2$} -\end{cases} -$$ -It is easy to see that these are inverse.<|endoftext|> -TITLE: Additivity of Kodaira dimension for a nice fibration -QUESTION [5 upvotes]: Consider a surjective holomorphic map between two complex projective manifolds $\pi :X \rightarrow Y$. Iitaka conjectured the subadditivity of Kodaira dimensions: $\kappa(X)\geqslant\kappa(Y)+\kappa(X_y)$ where $X_y$ is a generic fibre. We know already this holds when $dim(X)=dim(Y)+1$ and when $\pi$ is a fibre bundle. In particular, the conjecture holds for surfaces fibred over curves. -I want to know if there are some conditions to garantee the equality in the conjecture. The equality holds for a product, how about fibre bundles? In dimension two, if we have a fibred surface with neither singular nor multiple fibre, does the equality hold? - -REPLY [9 votes]: There is another inequality which says the following: - -Easy addition - (Using the same notation): - $$ -\kappa(X)\leqslant \kappa(X_y) + \dim Y -$$ - -Consequently, if $Y$ is of general type, i.e., $\kappa(Y)=\dim Y$, then the subadditivity conjecture is equivalent with equality instead of inequality. -Subadditivity is also known in the case $X_y$ is of general type (due to Kollár) and of course the inequality is trivially an equality (without easy addition) if both $X_y$ and $Y$ are of general type. -A problematic case is when $\kappa(Y)<0$. Then subadditivity doesn't say much and in general one cannot expect equality (although various hyperbolicity results can help). For instance, any elliptic K3 surface is fibred over $\mathbb P^1$ and gives an example when the inequality is strict. -One should probably add that there is a more sophisticated version of subadditivity (due to Viehweg) which predicts that if $\kappa(Y)\geq 0$, then -$$ -\kappa(X)\geqslant\kappa(X_y) + \mathrm{max}\{\kappa(Y), \mathrm{Var}(f)\}, -$$ -where $\mathrm{Var}(f)$ is the variation of $f$, i.e., the measure of how much the family varies in moduli (if there is a corresponding moduli space then this is the dimension of the image of the corresponding moduli map). So, if $\kappa(Y)\lneq \mathrm{Var}(f)$, then the inequality in the original format is expected to be strict.<|endoftext|> -TITLE: Is $G$ always the automorphism group of the trivial $G$-torsor? -QUESTION [9 upvotes]: If $G$ is just an ordinary set-theoretic group, then the answer to the question in the title is yes: the automorphisms of $G$ as a (left) $G$-set are all of the form "multiply (on the right) by an element of $G$." -I'm trying to understand the case where $G$ is a group scheme; then as I understand it the stack of $G$-torsors is equivalent to the stack quotient $[\ast/G]$, the stackification of the fibered category whose fiber over a scheme $X$ is a single object with $G(X)$ automorphisms. The image of that single object in the stackification is the trivial $G$-torsor on $X$ (namely the base change $G_X$ of $G$ to $X$) and so the automorphisms of that trivial $G$-torsor should again be $G(X)$. -But consider the case $G=\mu_2$, the group of square roots of unity, which is represented by $\mathrm{Spec}\bigl(\mathbb{Z}[x]/(x^2-1)\bigr)$. Letting $R$ be the ring $\mathbb{F}_2[\varepsilon]/(\varepsilon^2)$, we have $G(R) = \{1,1+\varepsilon\}$. These give us two automorphisms of $G_R = R[x]/(x^2-1)$, namely $x\mapsto x$ and $x\mapsto (1+\varepsilon)x$. But $G_R$ has two more automorphisms that commute with the action by $G(R)$; we can send $x\mapsto x+\varepsilon$ or $x\mapsto (1+\varepsilon)x + \varepsilon$. So how can this be the trivial $G$-torsor over $R$ if it has too many automorphisms as a $G$-object? - -REPLY [12 votes]: The two morphisms are no $G$-morphisms. An $R$-algebra homomorphism $f : R[x]/(x^2-1) \to R[x]/(x^2-1)$ commutes with the $G$-action if and only if the diagram -$$\begin{array}{c} R[x]/(x^2-1) & \xrightarrow{f} & R[x]/(x^2-1) \\ \Delta \downarrow ~~~&& ~~~\downarrow\Delta \\ R[x]/(x^2-1) \otimes_R R[y]/(y^2-1) & \xrightarrow{~f \otimes \mathrm{id}~}& R[x]/(x^2-1) \otimes_R R[y]/(y^2-1) \end{array}$$ -commutes, where $\Delta(x) = x \otimes y$. For $x \mapsto x+\varepsilon$ this would mean $$(x \otimes y)+\varepsilon = (x+\varepsilon) \otimes y,$$ which is not correct. For $x \mapsto (1+\varepsilon)x + \varepsilon$ this would mean -$$(1+\varepsilon) (x \otimes y) + \varepsilon = ((1+\varepsilon)x + \varepsilon) \otimes y,$$ -which is not correct either. Actually, you can see directly from the diagram that $G$-morphisms $f$ satisfy $f(x)= ax$ for some $a \in R$.<|endoftext|> -TITLE: Principal ideal subrings of formal power series rings -QUESTION [10 upvotes]: In the formal power series ring $\mathbb{F}[[x]]$ over a field $\mathbb{F}$ of characteristic $p>0$, consider an element of the form $f=\sum_{i=0}^\infty a_ix^{p^i}$. Let $R$ denote the unitary subalgebra of $\mathbb{F}[[x]]$ generated by $x$ and $f$. -In my recent work I came across the following problem: - -QUESTION. When is $R$ a principal ideal domain? - -Of course, this is trivial whenever $x$ belongs to the subalgebra generated by $f$ (or vice versa). - -REPLY [3 votes]: This happens if and only if the coefficients $a_i$ satisfy a simple Frobenius-linear recurrence $a_i = \sum_{j=1}^n a_{i-j}^{p^j} c_j$ for all $i \geq n$. -First assume that $R$ has dimension $1$. -$R$ is an integral domain generated by $x$ and $f$, so it is $\mathbb F[x,f]/I$ for some prime ideal $I$. Because $x$ and $f$ satisfy some relation, $I$ is nonempty, and because $x$ satisfies no relation all on its own, $I$ is maximal, so $I$ is generated by a single element $g(x,f)$. -Observe that $g(x_1+x_2, f(x_1) + f(x_2))= g(x_1+x_2, f(x_1+x_2))$ vanishes as an element of $\mathbb F[[x]] \otimes \mathbb F[[x]]$. So it vanishes as an element of $R \otimes R = F[x,f]/g(x,f) \otimes F[x,f]/g(x,f)$. -Hence $g(x_1+x_2,y_1+y_2)$ lies in the ideal generated by $g(x_1,y_1)$ and $g(x_2,y_2)$. Because these polynomials all have the same degree, and $g(x_1,y_1)$ and $g(x_2,y_2)$ are in independent variables, we have $g(x_1+x_2,y_1+y_2)=c_1 g(x_1,y_1)+ c_2 g(x_2,y_2)$. We can clearly see that $c_1 = c_2 =1$. -In particular $g(x,y) =g(x,0)+g(0,y)$, so $g$ splits into a polynomial in $x$ and a polynomial in $y$. Furthermore $g(x_1+x_2,0)=g(x_1,0)+g(x_2,0)$, so both these are additive. In other words, the only monomials appearing in $g(x,y)$ are $x^{p^i}$ and $y^{p^i}$. -Let $g(x,y) = \sum_{k=1}^m b_k x^{p^k} + \sum_{j=1}^n c_j y^{p^k} $. -Expanding $g(x,\sum_i a_i x^{p^i})=0$, we get for all $i> \max(n,m)$ that $\sum_j c_j a_{i-j}^{p^j} =0$. Prune the coefficients $c_j$ so that the first one is nonzero, then divide the other coefficients by it. Then add additional $0$s to the end until the condition $i> \max(n,m)$ is no longer necessary. This gives the desired recurrence. - -Conversely if $a_i = \sum_{j=1}^n a_{i-j}^{p^j} c_j$ for all $i \geq n$, then $f(x) - \sum_{j=1}^n f(x)^{p^j} c_j$ is a $p$-power polynomial in $x$ of degree at most $p^{n-1}$, i.e. $$f(x) - \sum_{j=1}^n f(x)^{p^j} c_j = \sum_{k=1}^{n-1} b_k x^{p^k}$$ -The relation $$y- \sum_{j=1}^n y^{p^j} c_j = \sum_{k=1}^{n-1} b_k x^{p^k}$$ defines a finite etale cover of $\mathbb A^1$ - etale because the coefficient of $f(x)$ is nonzero. (We actually may want to work with a slightly smllaer finite etale cover, the connected component of $(0,0)$. To check this is a PID, it is sufficient to check that this smooth curve is isomorphic to $\mathbb A^1$. But there is a clear group structure on this curve: $(x_1,y_1) +(x_2,y_2)=(x_1+y_1,x_2+y_2)$. So the smooth affine curve has a transitive group of automorphisms and hence is isomorphic to $\mathbb P^1$ minus one or two points. It can't be $\mathbb P^1$ minus two points as there is no group homomorphism from that to $\mathbb A^1$.<|endoftext|> -TITLE: Definition of the nonlinear part of the drift in a (stochastic) Navier-Stokes equation -QUESTION [6 upvotes]: Let - -$T>0$ -$d\in\mathbb N$ -$\Lambda\subseteq\mathbb R^d$ be bounded and open -$\mathcal V:=\left\{v\in C_c^\infty(\Lambda)^d:\nabla\cdot v=0\right\}$, $$V:=\overline{\mathcal V}^{\left\|\;\cdot\;\right\|_{H^1(\Lambda,\:\mathbb R^d)}}$$ and $$H:=\overline{\mathcal V}^{\left\|\;\cdot\;\right\|_{L^2(\Lambda,\:\mathbb R^d)}}=\overline V^{\left\|\;\cdot\;\right\|_{L^2(\Lambda,\:\mathbb R^d)}}\tag 1$$ -$\operatorname P_H$ denote the orthogonal projection from $L^2(\Lambda,\mathbb R^d)$ onto $H$ - -Please consider $$\left\{\begin{array}{rll}\displaystyle\frac{\partial u}{\partial t}-\nu\Delta u+\left(u\cdot\nabla\right)u+\frac1\rho\nabla p&=&f&&\text{in }(0,T)\times\Lambda\\\nabla\cdot u&=&0&&\text{in }(0,T)\times\Lambda\\u&=&0&&\text{on }(0,T)\times\partial\Lambda\\ u(0,\;\cdot\;)&=&u_0&&\text{in }\Lambda\end{array}\right.\tag 2$$ with $u:[0,T]\times\Lambda\to\mathbb R^d$, $p:[0,T]\times\overline\Lambda\to\mathbb R$, $f:[0,T]\times\overline\Lambda\to\mathbb R^d$ and $\nu,\rho>0$. - -I want to find a mild solution of a stochastic variant of $(2)$. Therefore, I need to reformulate $(2)$ as an evolution equation in $H$, but how do I need to define the nonlinear part of the drift? - -My problem is the following: Note that $\mathcal D(A):=\mathcal V\cap H^2(\Lambda,\mathbb R^d)$ is dense in $V$ and $$Au:=\text P_H\Delta u\;\;\;\text{for }u\in\mathcal D(A)$$ is a densely-defined linear symmetric operator on $H$ and $-A$ is the generator of a contraction $C^0$-semigroup on $H$. -Now, let $H_r:=\mathcal D(A^r)$ denote the domain of the fractional power of $A$ with exponent $r\in\mathbb R$. In order to apply the usual existence theory for mild solutions of SPDEs (see, for example, Assumption 2 on page 12 of A mild Itō formula for SPDEs), I need that the nonlinear part $F$ of the drift is a Lipschitz continuous mapping from $H_\gamma$ to $H_\alpha$ for some $\alpha,\gamma\in\mathbb R$ with $\gamma-\alpha<1$. -Now, I would like to define $F(u)=\operatorname P_H(u\cdot\nabla)u$, but this would require $(u\cdot\nabla)u$ to belong to $L^2(\Lambda,\mathbb R^d)$ (which is the case, for example, for $u\in H^2(\Lambda,\mathbb R^d)$ with $d\le 4$). - -So, I'm confused how I need to define $F$ (and choose $\gamma$ and $\alpha$). - -Let me note that I'm aware of the fact that, if $d\le 4$, $$b(u,v,w):=\int_\Lambda(u\cdot\nabla)v\cdot w\:{\rm d}x\;\;\;\text{for }u,v,w\in H_0^1(\Lambda,\mathbb R^d)$$ is a well-defined bounded trilinear form on $H_0^1(\Lambda,\mathbb R^d)$. Moreover, under suitable regularity assumptions on $\Lambda$, we can show that $H_{-1/2}=V'$ and $H_{1/2}=V$. Maybe that helps. - -REPLY [2 votes]: For $d=2$ the existence and uniqueness of strong solutions for the stochastic Navier–Stokes equation, including the nonlinear drift term, has been proven by Menaldi and Sritharan, Stochastic 2-D Navier–Stokes Equation (2002).<|endoftext|> -TITLE: lattice with Voronoi cell inside a circle -QUESTION [5 upvotes]: This considers real-valued lattices in two dimensions. -I need to find the densest lattice $\Lambda$, i.e., the one with the smallest determinant of its generator matrix, such that the Voronoi cell of its dual lattice $\Lambda^\ast$ is fully contained in the unit disc. -I strongly believe that it should still be the hexagonal lattice (and numerical tests seem to validate this belief). -Is there a simple proof. - -REPLY [4 votes]: Your conjecture is correct. The condition on $\Lambda^*$ means that -the covering radius of $\Lambda^*$ is at most $1$, and it is known that -the best covering lattice is hexagonal, which makes $\Lambda$ hexagonal too. -One way to see that the hexagonal lattice is the best covering lattice -is as follows. In two dimensions, the Voronoi region is always a -centrally-symmetric hexagon (possibly degenerate to a rectangle) -inscribed in a circle whose radius $-$ besides being the covering radius $-$ -is the circumcenter of a triangle with vertices $0,v,w$ for some -vectors $v,w$ that generate the lattice. So in our setting -the sparsity of $\Lambda^*$ is twice the area of a triangle inscribed -in a unit circle, and this area is maximized when the triangle is -equilateral. This makes $\Lambda^*$, and thus also $\Lambda$, -hexagonal, QED.<|endoftext|> -TITLE: A symmetric-like group and the quaternion group $Q_8$ -QUESTION [21 upvotes]: It is well known that the symmetric group $S_n$ admits presentation with -$\{(ij) \mid i\neq j\}$ as the set of generators and the following list of relations -(in every formula distinct letters denote distinct indices): -\begin{align} -(ij) =&\, (ji), \label{Sym0} \tag{S0} \\ -(ij)^2 = &\, 1, \label{Sym1} \tag{S1} \\ -(jk)(ij)(jk) = &\, (ik), \label{Sym2} \tag{S2} \\ -[(ij), (kl)] = &\,1. \label{Sym3} \tag{S3} -\end{align} -If for $n\geq 3$ we drop relation \ref{Sym0} from this list, we will get an extension of $S_n$, denote it by $\widetilde{S}_n$. My question is: is there a standard name for this group, has it already been studied by anyone in any context? Is there any standard representation for it. -Using GAP I was able to compute $\widetilde{S_3}$ and $\widetilde{S_4}$ explicitly, the answers are [48, 41]; [384, 20069], respectively. So it looks like $\widetilde{S}_n$ is a nontrivial extension of $S_n$ by some finite nonabelian group of order $2^n$. As for the kernels $K_n=\mathrm{Ker}(\widetilde{S}_n\to S_n)$ the answers for $n=3,4,5$ are as follows: [8, 4], [16, 12], [32, 50]. In other words, $K_3 \cong Q_8$, $K_4 \cong Q_8 \times C_2$, and $K_5$ is isomorphic to the central product of $Q_8$ and $D_8$ over $C_2$. - -REPLY [6 votes]: To simplify typing I will call these groups $G_n$ rather than $\tilde{S}_n$. Note that $G_2$ is just the free product of two groups of order $2$, so is the infinite dihedral group, and I conjecture that $|G_n| = 2^n n!$ for $n \ge 3$, and $G_n$ is a central product of an extraspecial or symplectic type $2$-group of order $2^n$ with a double cover $2.S_n$ of $S_n$. -I believe that computations within the group $G_3$ provide enough information to prove that $|G| \le 2^nn!$ and, since $G$ maps onto $S_n$ and onto an elementary abelian group of order $2^n$, we have $|G_n| \ge 2^{n-1}n!$, so I have not quite completed the proof. I have checked the conjecture computionally for $n \ge 7$. The alternative would be that $G_n$ is the direct product of elementary abelian $2^{n-1}$ and $S_n$. I cannot see at the moment how to rule that out. -Let $H_{n-1}$ be the image of $G_{n-1}$ in $G_n$ under the map that maps each generator of $H_{n-1}$ to the generator of $G_n$ with the same name. We do not know a priori that this map is injective, and indeed it is not injective for $n=3$. We prove that $|G_n| \le 2^nn!$ by showing that $|G_n:H_{n-1}| \le 2n$ for all $n$. -We claim that the set $$T = \{ 1,\,(1n)(n1)\} \cup \{(in),\, (ni) : 1 \le i \le n-1\}$$ contains a set of right coset representatives of $H_{n-1}$ in $G_n$ and, since $|T|=2n$, this will prove that $|G_n:H_{n-1}| \le 2n$. -We can check that directly computationally (in GAP or Magma) for $n=3$, and also that $(31)(13)$, $(23)(32)$ and $(32)(23)$ all lie in the same coset as $(13)(31)$. -To prove the claim for general $n$, it is enough to show that, for each $t \in T$ and each generator $x$ of $G_n$, $tx$ lies in $H_{n-1}u$ for some $u \in T$. In fact this follows from the calculations in the $n=3$ case together with the commuting relations of $G$. This is because $t$ and $x$ either involve at most $3$ of $1,2,\ldots,n$, in which case it follows from the $n=3$ case, or they involve $4$, in which case $x \in H_{n-1}$ and $tx=xt$. -For example, when $n=5$ if $t=(25)$ and $x=(35)$ then $tx \in H_{n-1}(25)$ from the $n=3$ case applied with $1,2,3$ replaced by $2,3,5$.<|endoftext|> -TITLE: A "Completion" of $ZFC^-$ -QUESTION [6 upvotes]: Let $T_0$ be the set theory axiomatized by $ZFC^-$ (that is $ZFC$ without powerset) + every set is countable + $\mathbb{V}=\mathbb{L}$. -Question 1: Suppose $\phi$ is a sentence of set theory. Must there be a large cardinal axiom $A$ such that $\phi$ is decided by $T_0$ + "there are transitive set models of $A$ of arbitrarily high ordinal height?" -Update: no. -Question 2: Is there a set $T$ of $\Pi_2$ sentences, such that $ZFC^- \cup T$ is complete? -Update: no, essentially. See my answer below. -Questions 3 and 4 after comments. -Comments. - -Question 1 is related to, but different from several previous questions on Mathoverflow, e.g. Nice Algebraic Statements Independent from ZF + V=L (constructibility) or On statements independent of ZFC + V=L or Natural statements independent from true $\Pi^0_2$ sentences. Regarding the first two---Hamkins gives a long list of examples of things independent of $\mathbb{V}= \mathbb{L}$, but it seems that my schema takes care of all of them. (Of course in those questions $ZFC$ was assumed.) -Moreover, Question 1 was partly motivated by Dorais' answer to the second question above, which references a question of Shelah from The Future of Set Theory. -I'm leaving "large cardinal axiom" undefined here (so Question 1 can't be formalized), but obviously we should exclude inconsistent axioms, or things like $ZFC + $ "there are no transitive set models of ZFC." -If we pick a definition of ``large cardinal axiom", then "every set is countable" + $\mathbb{V}=\mathbb{L}$ + "there are transitive set models of $A$ of arbitrarily high ordinal height" (for each large cardinal axiom $A$) is a set of $\Pi_2$ sentences. So a negative answer to Question 2 implies a negative answer to Question 1. -We can ignore (recursive) large cardinal axiom schemas $\Gamma$ because we can just replace them by $ZFC_0$+"there is a transitive set model of $\Gamma$" where $ZFC_0$ is some large enough finite fragment of $ZFC$. In particular we don't have to write "$ZFC + A$". -$T_0$ + this axiom schema (loosely speaking) is of personal significance to me: in fact I believe it to be true. (Namely, given whatever universe of sets $\mathbb{V}$ in which we are working, it seems reasonable to suppose there is a larger universe of sets $\mathbb{W} \models \mathbb{V}=\mathbb{L}$ in which $\mathbb{V}$ is countable, or such that $\mathbb{W}$ is a model of a given large cardinal axiom $A$. Ergo,...) - -Let $\mathcal{L}_{\mbox{set}}$ be the language of set theory $\{\in\}$ and let $\mathcal{L}_1$ be $\mathcal{L}_{\mbox{set}} \cup \{P\}$, $P$ a new unary relation symbol. Let $T_1$ be $T_0$ + the axioms asserting that $P \subseteq \mbox{ON}$ is stationary (for $\in$-definable classes) and for every $\alpha \in P$, $(\mathbb{V}_\alpha, \in) \preceq (\mathbb{V}, \in)$. We insist that large cardinal axioms $A$ be sentences of $\mathcal{L}_{\mbox{set}}$. -Question 3: Suppose $\phi$ is a sentence of set theory (i.e. of $\mathcal{L}_{\mbox{set}}$). Must there be a large cardinal axiom $A$ such that $\phi$ is decided by $T_1$ + "there are transitive set models of $A$ of arbitrarily high ordinal height?" -Question 4: Is there a set $T$ of $\Pi_2$ sentences of set theory, such that $T_1 \cup T$ decides every sentence of set theory? - -REPLY [4 votes]: Without a definition of “large cardinal axiom”, I’m going to ignore Q1 and Q3. -The answers to Q2 and Q4 are negative by the following general principle. (For Q4, we take $T_0$ to be the set of $\mathcal L_{\mathrm{set}}$-consequences of $T_1$.) - -$\DeclareMathOperator\Tr{Tr}\DeclareMathOperator\Wit{Wit}\let\eq\leftrightarrow\def\gonu#1{\ulcorner#1\urcorner}\let\ob\overline$Proposition: Let $T_0$ be an r.e. theory interpreting Robinson’s arithmetic, and $\Gamma$ a set of sentences for which $T_0$ has a truth predicate $\Tr_\Gamma(x)$, that is, - $$\tag{$*$}T_0\vdash\phi\eq\Tr_\Gamma(\ob{\gonu\phi})$$ - for all $\phi\in\Gamma$. Then no extension of $T_0$ by a set of $\Gamma$-sentences is a consistent complete theory. - -Proof: Let $S\subseteq\Gamma$, and assume for contradiction that $T=T_0+S$ is consistent and complete. The basic idea of the proof is that we can define in $T$ a truth predicate $\Tr(x)$ for all sentences as “$x$ is $T_0$-provable from a set of true $\Gamma$-sentences”, contradicting Tarski’s theorem on the undefinability of truth. -In more detail, we fix an interpretation of, say, $S^1_2$ in $T_0$ so that we have basic coding of sequences of integers, and a (polynomial-time) proof predicate for $T_0$. We define $\Wit(w,x)$ to be the formula -“the sequence $w$ is a $T_0$-proof of a sentence $x$ from extra axioms, each of which is a sentence $a\in\Gamma$ such that $\Tr_\Gamma(a)$,” -and we put -$$\Tr(x)\eq\exists w\,(\Wit(w,x)\land\forall w' -TITLE: Can $\mathcal O_X$ be recognized abstract-nonsensically? -QUESTION [17 upvotes]: This question has been asked by Teimuraz Pirashvili many years ago. I forgot about it after a while and remembered only now by accident. He probably knows the answer by now, but I still don't. -In the category of modules over a ring $R$, the module $R$ is a projective generator. This property does not determine it uniquely, even up to isomorphism, but it does when $R$ is commutative: it can be reconstructed as the ring of endotransformations of the identity functor. (For noncommutative $R$, it is still true that the category is equivalent to the category of modules over the endomorphism ring of any of its projective generators that may exist.) -Now what can be said about the structure sheaf $\mathcal O_X$ of a scheme $X$? Can it be detected inside the category of $\mathcal O_X$-modules without using any additional structure? Of course it is the unit of the monoidal structure, but can it be also characterized as an object in a plain category, without invoking any additional structures? Can in fact the category of $\mathcal O_X$-modules possess other non-isomorphic monoidal structures? Can it be equivalent to the category of $\mathcal O_Y$-modules for some other scheme $Y$? Or, say, some analytic space, or whatever? -Note that $\mathcal O_X$ is not even a generator - the subcategory generated by it is the category of quasicoherent sheaves. (NB As @HeinrichD points out in the first comment, even that is wrong - this is only true locally, in the appropriate sense.) In any case, also in this category $\mathcal O_X$ is typically not projective. So the same question arises - is it some particular kind of generator? -There are of course several reconstruction theorems but I still cannot figure out which (if any) of them provide answers to these questions. -LATER - -as pointed out by Will Sawin, I should be more clear. Rather than modifying the question (I don't quite understand how) let me try to formulate it once more: - -Consider the object $\mathcal O_X$ of the abelian category of sheaves on a scheme $X$. What are its properties formulable without the use of tensor product or any other additional structures, just as an object of this abelian category? - -For example, as Will Sawin explains in his answer, a commutative ring inside its category of modules cannot be distinguished among rank one projectives. But at least it is a projective generator with a commutative endomorphism ring (in fact it is isomorphic to that ring), which is a very restrictive property. Are there some similar properties of $\mathcal O_X$? - -REPLY [5 votes]: As Will points out, the best you can hope for is to recognize being a line bundle. Also, I find you're a bit vague about what category of sheaves you want work in; I'm thinking about coherent sheaves, which is an appropriate analogue for finite dimensional $R$-modules. As we'll see below, I'll also want to assume my scheme is locally of finite type (over a field, or other Jacobson ring, such as $\mathbb{Z}$) EDIT: nfdc23 also points out I should assume reduced. -I have an answer for this is which is probably ultimately equivalent to Heinrich's but which I like better (perhaps the difference is he's thinking of quasi-coherent sheaves?). The only simple objects in the category of coherent sheaves are the skyscraper sheaves at the different closed points. Thus, we can actually reconstruct the closed points of the scheme. You can also define the support of a sheaf by which skyscrapers it has maps to, and reconstruct the Zariski topology by defining a closed set to one that appears as the support of a coherent sheaf. A coherent sheaf is a line bundle if and only if the Hom space to every skyscraper sheaf is 1-dimensional. -EDIT: Why is this so? A coherent sheaf is a line bundle if and only every stalk is free of rank 1 over the local ring at each point. A module over a local ring is free of rank 1 if and only if the rank of its residue at the unique closed point and the generic point is 1. So, it's enough to check that you have residue of rank 1 at every point (including non-closed ones). -So, if you have rank 1 at all closed points, by semi-continuity, you have rank 0 or 1 at every point, and the set where you have rank 1 is closed. Thus, you must have rank 1 everywhere if the closed points are dense (if not, the structure sheaf on their closure will be a counterexample). This occurs whenever you are locally of finite type (as discussed here: https://math.stackexchange.com/questions/615709/is-the-set-of-closed-points-of-a-k-scheme-of-finite-type-dense). Presumably outside finite type, one can modify the construction a bit to include the non-closed points.<|endoftext|> -TITLE: What is the standard notation for reversing the order of vector's components? -QUESTION [8 upvotes]: If we have a vector $x=(x_1,x_2,\ldots,x_n)$, is there any standard way to denote the vector $(x_n,x_{n-1},\ldots,x_1)$?. -I think that $x^{-1}$ could be a good option. - -REPLY [10 votes]: An alternative would be to define and use the exchange matrix (see the Wikipedia entry “Exchange matrix”) -$$ - J = \begin{pmatrix} -0 & 0 &\cdots &0 & 1\\ -0 & 0 & \cdots & 1 & 0\\ -\vdots & \vdots &\ddots & 0 & 0\\ -0 & 1 &\cdots & 0 &0\\ -1 & 0 &\cdots & 0 &0 -\end{pmatrix} -$$ -and to note that $(x_n, \dotsc, x_1)=J (x_1, \dotsc, x_n)$.<|endoftext|> -TITLE: Reaching Hecke eigenvalues from a trace formula -QUESTION [7 upvotes]: I am interested in studying equidistribution of Hecke eigenvalues and proving statistical properties of arithmetical objects. On the road, I face the following problem: how to express sums of the form -$$\sum_{c(\pi) -TITLE: Number of graphs on a given set of vertices with maximum degree of $2$ -QUESTION [5 upvotes]: Given a set of $n$ vertices and the fact that none of them is of degree greater than $2$, how many distinct such graphs are there? - -REPLY [5 votes]: This is OEIS A003292 Number of 4-line partitions of n decreasing across rows. - -a(n) is the number of unlabeled graphs on n nodes whose connected components are a path or a cycle. - Geoffrey Critzer, Nov 28 2011 - -OEIS gives generating function and references which may contain additional information. -G.f.: Product (1 - x^k)^-{c(k)}; c(k) = 1, 1, 2, 2, 2, 2, ....<|endoftext|> -TITLE: Convergence of conditional second moments -QUESTION [5 upvotes]: Let $(\Omega, \mathcal{A},P)$ be a probability space, and let $(\mathcal{F}_k)_{k \geq 1}$ be a filtration which converges to $\mathcal{A}$. I suppose it is true that -$$ -E \left( \big(E \left( X | \mathcal{F}_k \right) \big)^2 \right) \to E \left(X^2 \right). -$$ -How to prove this? I guess one will need Jensen's inequality and a convergence theorem for submartingales, but cannot find the right reference (I am working on number theory, not probability). -(If you know an answer, please also provide a citable reference.) - -REPLY [5 votes]: Let us state Corollary 2.1 of these notes. - -Let $p>1$, $X\in\mathbb L^p$ and let $\left(\mathcal F_n\right)_{n\geqslant 1}$ be a filtration. Denote by $\mathcal F$ the $\sigma$-algebra generated by $\bigcup_{n\geqslant 1}\mathcal F_n$. Then the convergence - $$\lim_{n\to +\infty} \mathbb E\left[X\mid\mathcal F_n\right]=\mathbb E\left[X\mid\mathcal F\right],$$ - where the convergence holds almost everywhere and in $\mathbb L^p$. - -The wanted result follows from this corollary and the elementary fact that if $\left\lVert Y_n-Y\right\rVert_2\to 0$ then $\left\lVert Y_n\right\rVert_2^2\to \left\lVert Y\right\rVert_2$.<|endoftext|> -TITLE: Applications of Alexandrov spaces to Riemannian geometry -QUESTION [12 upvotes]: I am an expert neither in Riemannian geometry nor in Alexandrov spaces. I am wondering what are the applications of Alexandrov spaces to more classical Riemannian geometry. -For example one can show that there exist only finitely many homeomorphisms types of closed smooth Riemannian manifolds of dimension $n$, diameter at most $D$, sectional curvature at least $\kappa$, and volume at least $v>0$ for fixed parameters $n,D,\kappa,v$. The proof uses the Gromov compactness theorem and the Perelman stability theorem. - -REPLY [16 votes]: The two sources of applications come from two sources of examples of Alexandrov spaces: - -Limits of Riemannian manifolds with lower curvature bound. -Quotients of Riemannian manifolds by an isometric group action with closed orbits. - -Your example, the finiteness theorem, is of type 1. Another example is the upper bound on integral of scalar curvature of Riemannian manifold in terms of its lower bound on sectional curvature, diameter and dimension, see my paper. -There are many more, in fact (1) provides the main sourse of applications so far. -The examples of the second type include the classification of 4-dimesional Riemannian manifolds with positive/non-negative curvature; it was done by Grove and Wilking here and based on earlier result of Hsiang and Kleiner. -Yet anther example is the optimal bound for the number of certain type finite subgroups up to conjugacy in a crystallographic group, see this paper by Lebedeva (which is build on an observation of Perelman, which is build on an combigeometrical problem of Erdős, Danzer and Grünbaum).<|endoftext|> -TITLE: Intersecting a convex polytope with the unit sphere -QUESTION [5 upvotes]: I have a list of $m$ affine inequalities in $n$ variables of the following form -$$a_1 x_1 + \cdots + a_n x_n \leq c_n$$ -I would like to know whether there is any point on the unit sphere in $\mathbb R^n$ that satisfies all of them. Is there any easy way to check whether this is the case? -(Based on randomly choosing points, it seems like there are no solutions, but I'd like an algorithm that's up to journal standards of proof.) - -REPLY [5 votes]: It was proved by Freund and Orlin that the problem of checking whether a polytope specified by linear inequalities is not entirely contained in a ball specified by its centre and radius is NP-complete. -And it is obvious that one can assume that the ball is a unit ball, for the corresponding transformation (shift+scaling) can be carried out in polynomial time. -Thus there is little hope for an efficient algorithm for the problem at hand.<|endoftext|> -TITLE: Finite Powers of ordered topological spaces -QUESTION [6 upvotes]: Given a totally ordered set $I$. -It is well known that $I$ becomes a normal Hausdorff space when endowed with the order topology. -What can be said about finite products -$I \times \cdots \times I$ ? When are they normal? -Are there sufficient and necessary conditions on the order of $I$ for all finite powers to be normal? - -REPLY [4 votes]: To answer Will Brian's question, notice that the product space $(\omega_1 + 1) \times \omega_1$ is not normal. (This fact is well-known; the Pressing Down Lemma implies that the diagonal and the right edge cannot be completely separated.) Let $I$ be the ordinal space $\omega_1 + \omega_1$. Since $I$ contains each of the spaces $\omega_1 + 1$ and $\omega_1$ as a closed subspace, the product $I \times I$ is not normal.<|endoftext|> -TITLE: $\operatorname{AD}$ and the measurability of $\omega_1$ -QUESTION [6 upvotes]: Are there proofs of the measurability of $\omega_1$ (under $\operatorname{AD}$) that do not use Turing degrees nor the $\Sigma_1^1$ boundedness lemma? -I've been struggling to find an "elementary" proof of this fact. Note that I consider the proof of "Assume $\operatorname{AD}$. Then every ultrafilter is $\sigma$-complete" to be elementary. -The club filter on $\omega_1$ is thus readily seen to be a $\sigma$-complete filter, but the "ultra" part seems not to be so easy to prove (it's for an introductory course on $\operatorname{AD}$, without too much knowledge of recursivity etc.) - -REPLY [4 votes]: I'm not sure this will work for you, but there's a way to recast the Turing argument so that it avoids recursion theory; if this is the reason you want to avoid the Turing argument, this might be the way to go. -Namely, instead of working with Turing degrees, work with a coarser reducibility, which is easier to explain. For anything coarser than Turing reducibility, the cone property holds by the same argument, and we can define an analogous $f$ mapping reals to ordinals. -Specifically, here are the details for one particularly nice notion, relative projectiveness (or projective reducibility). Say a real $r$ is projective relative to a real $s$, and write $r\le_{p}s$, if there is a second-order sentence $\varphi(x, Y)$ in two variables with no parameters - where $x$ is a natural number variable and $Y$ is a set variable - such that $$r=\{n: \varphi(n, s)\}$$ holds. (There are many equivalent ways to phrase this.) -It's easy to check that this is, in fact, a reducibility (in particular, that it's transitive - this isn't hard, but it's worth doing explicitly), and so yields a degree structure $\mathcal{D}_p$. And by the same proof as in the Turing case, we have that any "$\equiv_p$-invariant" set of reals either contains or is disjoint from a cone in $\mathcal{D}_p$. So the "projective cone" filter on $\mathbb{R}$ is an ultrafilter. Moreover, by taking infinite joins, it's clear that this ultrafilter is countably closed. -Now we want to port it over to $\omega_1$. We'll use the same trick as in the Turing case: for a real $r$, let $f(r)$ be the least ordinal $\alpha$ such that $\alpha$ is not projectively definable in terms of $r$ (formally, there is no well-ordering of $\mathbb{N}$ of ordertype $\alpha$ which is projective relative to $r$); such a real exists, since there are only countably many projectively definable ordinals relative to a given real. -Note that this uses a small bit of choice - namely, that $\omega_1$ is regular. But this is provable in ZF+AD, so that's fine. Note that DC is not known to be provable in ZF+AD (although it does follow from ZF+AD+V=L$(\mathbb{R})$) so we can't use it here. -By the same argument as in the Turing case, the filter gotten by "porting over" the projective cone filter via $f$ is a measure on $\omega_1$. - -Another natural reducibility to consider is relative constructibility. In many ways this is actually more natural than relative projectiveness; however, it makes things a bit trickier, since you have to show that $\omega_1$ is inaccessible from reals assuming AD. -However, this isn't hard - if $\omega_1^{L[r]}=\omega_1$ for some real $r$, then we have $\vert\mathbb{R}\cap L[r]\vert=\omega_1^{L[r]}=\omega_1$ (the first equaltiy since $L[r]$ satisfies condensation appropriately); but then $\mathbb{R}\cap L[r]$ is a counterexample to the perfect set property.<|endoftext|> -TITLE: If a Dirichlet series converges Conditionally, how can I apply Euler product? -QUESTION [5 upvotes]: In 1737, Euler discovered that if $ f(n) $ is multiplicative and $ \sum f(n)/n^{s} $ converges absolutely for ${\rm Re}(s) > \sigma_a$ then we have -\begin{equation} -\sum_{n=1}^{\infty} \frac{f(n)}{n^s} ~=~ \prod_p \Bigg\{ 1+\frac{f(p)}{p^s}+\frac{f(p^2)}{p^{2s}}+ \cdots \Bigg\} -\end{equation} -and, especially, if $f$ is completely multiplicative we have -\begin{equation} -\sum_{n=1}^{\infty} \frac{f(n)}{n^s} ~=~ \prod_p \frac{1}{1-f(p)/p^{s}}~~~~~{\rm if} ~~{\rm Re}(s)>\sigma_a. -\end{equation} -I found an example in Wikipedia (https://en.wikipedia.org/wiki/Euler_product) like -\begin{equation} -\frac{\pi}{4}~=~ \sum_{n=1}^{\infty} \frac{f(n)}{n},~~~~~~~~{\rm where}~~~~f(n)=\begin{cases}(-1)^{(n-1)/2} & {\rm if} \ n \ {\rm odd}, \\ 0 & {\rm if } \ n \ {\rm even}, \end{cases} -\end{equation} -so the theorem gives -\begin{equation} -\frac{\pi}{4}~=~ \prod_{p \not= 2} \frac{1}{1-f(p)/p} ~=~ \prod_{p\not=2} \frac{p}{p-(-1)^{(p-1)/2}}~=~\frac{3}{4}\cdot\frac{5}{4}\cdot\frac{7}{8}\cdot\frac{11}{12}\cdot \frac{13}{12}\cdots. -\end{equation} -However, this example does not converge absolutely but conditionally. -If this example holds, how can I prove it though it converges conditionally? -Is there any other additional condition needed or should I apply a different method? - -REPLY [7 votes]: You are right to question this. The product -$\prod_p \left(1 - \chi(p)/p\right)^{-1}$ -(where $\chi = (-1/\cdot)$ is the Dirichlet character mod $4$) -does converge, and the limit is $L(1,\chi) = \pi/4$ as expected; -But this requires justification $-$ indeed it is equivalent to -the non-vanishing of the Dirichlet function $L(s,\chi)$ -on the edge $s = 1+it$ of the critical strip, which is also what you need -to prove the analogue of the Prime Number Theorem for primes in arithmetic -progressions mod $4$. (Taking logarithms, we see that -$\prod_p \left(1 - \chi(p)/p\right)^{-1}$ converges if and only if -$\sum_p \chi(p)/p$ converges, since this sum differs from -the product's logarithm by an absolutely convergent sum $\sum_p O(1/p^2)$; -getting from $\sum_p \chi(p)/p$ to $L(s,\chi)$, and then showing that -the product $\prod_p \left(1 - \chi(p)/p\right)^{-1}$ actually converges -to $L(1,\chi)$, is a classical chapter of analytic number theory.)<|endoftext|> -TITLE: a modification on an infinite Bernoulli convolution -QUESTION [5 upvotes]: The distribution $\nu_{\lambda}$ of the random series $\sum\pm\lambda^n$ is the infinite convolution product of $\frac12(\delta_{-\lambda^n}+\delta_{\lambda^n})$. This problem has been studied extensively. -I am curious about the following modification: - -QUESTION. Let $\mu_{\lambda}$ be the distribution of $\sum_{n=0}^{\infty}\epsilon_n\lambda^n$ where the $\epsilon_n$ are chosen from $\{-1,0,1\}$ independently with probability $\frac13$. For which $0<\lambda<1$ is $\mu_{\lambda}$ absolutely continuous or singular? - -Has this been investigated? If so, I would be grateful for any reference. - -REPLY [6 votes]: As can be inferred from Nikita's answer, the situation for the family $\mu_\lambda$ is very similar to that of the classical Bernoulli convolutions $\nu_\lambda$, and all the known techniques for the latter situation work almost verbatim for the former. -I don't have anything to add to Nikita's points 1-3, so let me add some details of what is known about absolute continuity: - -For any $k$ there is $\varepsilon_k$ such that $\mu_\lambda$ has a $C^k$ density for almost all $\lambda\in (1-\varepsilon_k,1)$. Erdös proved this for Bernoulli convolutions in 1940, and the same proof works for the measures $\mu_\lambda$ (the argument is based in the fact that the Fourier transform of $\mu_\lambda$ has power decay at infinity). -$\mu_\lambda$ is absolutely continuous with an $L^2$ density for a.e. $\lambda\in (1/3,1)$. This can be proved with the transversality technique used by Solomyak in 1995 to prove that $\nu_\lambda$ is absolutely continuous for a.e. $\lambda\in (1/2,1)$. See: [Simon, Károly; Tóth, Hajnal R. The absolute continuity of the distribution of random sums with digits $\{0,1,\dots,m-1\}$. Real Anal. Exchange 30 (2004/05), no. 1, 397--409]; they also consider the similar problem for any number of equally spaced translations. The modification is not trivial because the interval on which transversality holds depends on the number of translations. -It is a special case of Theorem 1.2 in my paper [Shmerkin, Pablo. On the exceptional set for absolute continuity of Bernoulli convolutions. Geom. Funct. Anal. 24 (2014), no. 3, 946--958. ] that $\mu_\lambda$ is absolutely continuous for all $\lambda\in (1/3,1)$ outside of a set of zero Hausdorff dimension (this is strongly based on Mike Hochman's results for dimensions of self-similar measures). -The last two results are now superseded by my recent preprint [On Furstenberg's intersection conjecture, self-similar measures, and the Lq norms of convolutions, arXiv:1609.07802]. It follows from the results there that $\mu_\lambda$ has an $L^q$ density for all finite $q$, for all $\lambda\in (1/3,1)$, again outside of a (possible) exceptional set of zero Hausdorff dimension. This is not explicitly stated in my paper, but see Theorem 9.1 and the discussion afterwards. -What about absolute continuity for specific values of $\lambda$? A recent breakthrough of P.Varju [Absolute continuity of Bernoulli convolutions for algebraic parameters, arXiv:1602.00261] provides many explicit (in terms of a small constant which is not specified) examples of algebraic numbers for which $\nu_\lambda$ is absolutely continuous. I am quite sure the arguments extend to $\mu_\lambda$ but I am less familiar with this. Breuillard and Varju also have some impressive recent results about full dimension of Bernoulli convolutions for many explicit parameters (not only algebraic) that should also extend to more general self-similar measures. - -All of these results hold in fact (for the appropriate range of $\lambda$) for the distribution of the random sums -$$ -\sum_{n=1}^\infty \varepsilon_n \lambda^n, -$$ -where $\varepsilon_n$ are iid random variables taking finitely many integer values (and for most of the results the values don't even have to be integers).<|endoftext|> -TITLE: The Maximal Ergodic Theorem more than once -QUESTION [7 upvotes]: Suppose we have a dynamical system and a sequence of functions $0=f_0\leq f_1\leq\cdots\leq f_k$. Define $J_{r,\lambda}$ to be the set of points $x$ such that there are $j_0\lambda$. Using the maximal ergodic theorem, we see that $\mu(J_{1,\lambda})\leq ||f_k||_1/\lambda$. Is there a similar bound on $\mu(J_{r,\lambda})$ (maybe even $||f_k||_1/r\lambda$)? -Note that this doesn't seem to trivially reduce to multiple applications of the maximal ergodic theorem, because the sequences of $j$'s might overlap between different points, and one point might see $r$ jumps from $0$ to $j_r$, while another point doesn't see any jumps until some $j'>j_r$. -(We have some ideas that might show this, at least for a slightly more restrictive definition of $J_{r,\lambda}$---that there are $n_0<\cdots\lambda$---but they're messy, and it seems like a clever use of the maximal ergodic theorem should give a slick proof, but we don't see how to do it.) - -REPLY [4 votes]: You can't do substantially better than $\|f_k\|_\lambda$. Here's a simple example: Consider $X=\{0,1,\ldots,2^{N}-1\}$, equipped with normalized counting measure. The transformation is $T(x)=x+1\bmod 2^{N}$. The functions are -$$ -f_i(x)=\begin{cases} -2&\text{if $x -TITLE: Root number of the Rankin-Selberg convolution of two newforms -QUESTION [5 upvotes]: Let $p$ and $q$ be two distinct primes. Let $f\in \mathcal{S}_k^{\ast}(pq,\psi)$ be a holomorphic newform of level $pq$, nebentypus $\psi$, and weight $k$, where $\psi = \chi_p \chi_{0(q)}$, with $\chi_p$ a primitive character modulo $p$ and $\chi_{0(q)}$ the principal character modulo $q$ (i.e. $\psi$ is an imprimitive character induced from the primitive character $\chi_p$). Let $g$ be another holomorphic newform of level $q$, weight $k_g$, and with trivial nebentypus. -Let $L(s,f \otimes g)$ be the Rankin-Selberg convolution $L$-function. Let $\Lambda(s,f\otimes g)=Q(f\otimes g)^{s/2} L_\infty(s, f \otimes g)L(s, f \otimes g)$ be the complete $L$-function, where $Q(f\otimes g)$ is the conductor of $L(s, f\otimes g)$. Then we have the functional equation -\begin{equation} -\Lambda(s, f\otimes g)=\epsilon(f \otimes g) \overline{\Lambda(f\otimes g,1-\bar{s})}. -\end{equation} -My question is: -What are the root number $\epsilon(f\otimes g)$ and the conductor $Q(f\otimes g)$ in this case (the conductor equals $p^2 q^2$, I think)? Can anyone calculate the local $\epsilon_v$-factor at the place $v \mid q$ for me? I know a good reference is http://tan.epfl.ch/files/content/sites/tan/files/PhMICHELfiles/RSfinal.pdf, but most of their statements are for the levels of $f$ and $g$ to be co-prime. - -REPLY [5 votes]: You need to do this via a local argument. A good reference for local components of $\mathrm{GL}_2 \times \mathrm{GL}_2$ automorphic representations is Gelbart and Jacquet, "A relation between automorphic representations of $\mathrm{GL}(2)$ and $\mathrm{GL}(3)$". For just the $\mathrm{GL}_2$ theory, see Schmidt, "Some remarks on local newforms for $\mathrm{GL}(2)$". - -At all primes $v \nmid pq$, the local epsilon factors and conductor exponents are trivial. -The local component of $f$ at $p$ is a principal series representation $\pi_{f,p} = \omega_{f,p,1} \boxplus \omega_{f,p,2}$, where $\omega_{f,p,1}, \omega_{f,p,2}$ are character of $\mathbb{Q}_p^{\times}$ of conductor exponent $c(\omega_{f,p,1}) = 1$ and $c(\omega_{f,p,2}) = 0$, while the local component of $g$ is a spherical principal series representation $\pi_{g,p} = \omega_{g,p,1} \boxplus \omega_{g,p,2}$ with both characters unramified, so that $c(\omega_{g,p,1}) = c(\omega_{g,p,2}) = 0$. Then -\[\pi_{f,p} \otimes \pi_{g,p} = \omega_{f,p,1} \omega_{g,p,1} \boxplus \omega_{f,p,1} \omega_{g,p,2} \boxplus \omega_{f,p,2} \omega_{g,p,1} \boxplus \omega_{f,p,2} \omega_{g,p,2}.\] -The conductor exponent is -\[c(\pi_{f,p} \otimes \pi_{g,p}) = c(\omega_{f,p,1} \omega_{g,p,1}) + c(\omega_{f,p,1} \omega_{g,p,2}) + c(\omega_{f,p,2} \omega_{g,p,1}) + c(\omega_{f,p,2} \omega_{g,p,2}),\] -which is -\[1 + 1 + 0 + 0 = 2.\] -The epsilon factor $\epsilon_p(s,\pi_{f,p} \otimes \pi_{g,p},\psi_p)$ ostensibly depends on an additive character $\psi_p$ of $\mathbb{Q}_p$, which we may choose to be unramified (i.e. $c(\psi_p) = 0$), though the global epsilon factor is independent of this. Anyway, $\epsilon_p(s,\pi_{f,p} \otimes \pi_{g,p},\psi_p)$ is equal to -\[\epsilon_p(s,\omega_{f,p,1} \omega_{g,p,1},\psi_p) \epsilon_p(s,\omega_{f,p,1} \omega_{g,p,2},\psi_p) \epsilon_p(s,\omega_{f,p,2} \omega_{g,p,1},\psi_p) \epsilon_p(s,\omega_{f,p,2} \omega_{g,p,2},\psi_p),\] -which is -\[\left(\omega_{g,p,1}(p) \epsilon_p(s,\omega_{f,p,1},\psi_p)\right) \cdot \left(\omega_{g,p,2}(p) \epsilon_p(s,\omega_{f,p,1},\psi_p)\right) \cdot 1 \cdot 1 = \epsilon_p(s,\pi_{f,p},\psi_p)^2.\] -(See Proposition 1.4 of Gelbart and Jacquet and equations (4) and (6) of Schmidt and use the fact that $g$ has principal nebentypus means that the central character $\omega_{\pi_{g,p}} = \omega_{g,p,1} \omega_{g,p,2}$ of $\pi_{g,p}$ is trivial, so that $\omega_{g,p,1}(p) \omega_{g,p,2}(p) = 1$.) As the conductor exponent is $2$, -\[\epsilon_p(s,\pi_{f,p} \otimes \pi_{g,p},\psi_p) = \epsilon_p\left(\frac{1}{2},\pi_{f,p} \otimes \pi_{g,p},\psi_p\right) p^{-2\left(\frac{1}{2} - s\right)}.\] -The local components of $f$ and $g$ at $q$ are special representations $\pi_{f,q} = \omega_{f,q} \mathrm{St}_q$ $\pi_{g,q} = \omega_{g,q} \mathrm{St}_q$ with $\omega_{f,q}, \omega_{g,q}$ characters of $\mathbb{Q}_q^{\times}$ that are unramified and either trivial or quadratic. The conductor exponent of $c(\pi_{f,q} \otimes \pi_{g,q})$ is $2$, as $\pi_{f,q} \otimes \pi_{g,q}$ is the isobaric sum of an unramified character of $\mathrm{GL}_1(\mathbb{Q}_p)$ and the Steinberg representation of $\mathrm{GL}_3(\mathbb{Q}_p)$, with the conductor exponent of the former being $0$ and the latter being $2$. By Proposition 1.4 of Gelbart and Jacquet and equation (11) of Schmidt, the epsilon factor $\epsilon_q(s,\pi_{f,q} \otimes \pi_{g,q},\psi_q)$ is -\[\epsilon_q\left(s + \frac{1}{2},\omega_{g,q} \omega_{f,q} \mathrm{St}_q, \psi_q\right) \epsilon_q\left(s - \frac{1}{2},\omega_{g,q} \omega_{f,q} \mathrm{St}_q, \psi_q\right) = \epsilon_q(s,\omega_{g,q} \omega_{f,q} \mathrm{St}_q, \psi_q)^2.\] -One can further show that this is equal to -\[\epsilon_q(s,\pi_{g,q}, \psi_q)^2 = \epsilon_q(s,\pi_{f,q}, \psi_q)^2 = p^{-2(\frac{1}{2} - s)}\] -(the reference for this is Section 11.12 of Goldfeld and Hundley's book combined with Schmidt's paper). -Finally, the local component of $f$ at $\infty$ is the discrete series representation $\pi_{f,\infty} = D_{k-1}$ of weight $k$, while $\pi_{g,\infty} = D_{k_g - 1}$. Then $\pi_{f,\infty} \otimes \pi_{g,\infty} = D_{|k - k_g|} \boxplus D_{k + k_g}$, and so -\[\epsilon_{\infty}(s, \pi_{f,\infty} \otimes \pi_{g,\infty},\psi_{\infty}) = \epsilon_{\infty}(s,D_{|k - k_g|},\psi_{\infty}) \epsilon_{\infty}(s,D_{k + k_g},\psi_{\infty}),\] -which is -\[i^{|k - k_g| + 1} i^{k + k_g + 1} = (-1)^{\max\{k,k_g\}+1}.\] -The best reference for this is Knapp, "Local Langlands Correspondence: The Archimedean Case". - -So defining the global epsilon factor -\[\epsilon(s,\pi_f \otimes \pi_g) = \epsilon_{\infty}(s, \pi_{f,\infty} \otimes \pi_{g,\infty},\psi_{\infty}) \prod_{p'} \epsilon_{p'}(s, \pi_{f,p'} \otimes \pi_{g,p'},\psi_{p'}),\] -where $p'$ runs over all primes, the functional equation for the completed Rankin-Selberg $L$-function $\Lambda(s, \pi_f \otimes \pi_g)$ is -\[\Lambda(s, \pi_f \otimes \pi_g) = \epsilon(s,\pi_f \otimes \pi_g) \Lambda(1 - s, \widetilde{\pi}_f \otimes \widetilde{\pi}_g).\] -By the above discussion, -\[\epsilon(s,\pi_f \otimes \pi_g) = (pq)^2 (-1)^{\max\{k,k_g\}+1} \epsilon_p\left(\frac{1}{2},\pi_{f,p},\psi_p\right)^2.\] -Finally, we note that $\epsilon_p\left(\frac{1}{2},\pi_{f,p},\psi_p\right) = \eta_f(p)$, the pseudo-eigenvalue of $f$ corresponding to the Atkin-Lehner operator $W_p$; see equation (7.6) of Templier, "Voronoï Summation for $\mathrm{GL}(2)$", and equation (7.8) shows that -\[\eta_f(p) = \overline{\lambda_f}(p) \tau(\chi_p) p^{-1/2}\] -(note that $|\lambda_f(p)| = 1$).<|endoftext|> -TITLE: Why are affine functions called "affine" functions? -QUESTION [5 upvotes]: I am learning about affine functions and I do not understand why a certain type of function (functions that are in the form of $f(x)=ax+b$) are called affine functions. I read about the word "affine" and I know that it means related, but I do not understand how is it related to this type of function and what was the reason behind this naming. I believe that understanding the reason behind namings can help one better understand a concept. - -REPLY [6 votes]: According to Jeff Miller, the adjective affinis (in Latin: "bordering", "adjacent", "linked by marriage") was introduced in geometry by Euler, in analogy to similis, to denote a weaker equivalence relation than the latter: "Because curves originated this way do keep a certain Affinity between them, we will name these curves affine". - -AFFINE. Affinis and affinitas were first used by Leonhard Euler in - Introductio in analysin infinitorum (1748) Chapter XVIII: De - similitudine et affinitate linearum curvarum. He also wrote (II. - xviii. 239): "Quia Curvae hoc modo ortae inter se quandam Affinitatem - tenent, has Curvas affines vocabimus."<|endoftext|> -TITLE: Large power of an adjacency matrix -QUESTION [7 upvotes]: The adjacency matrix I have at the start is -[0,1,0,0,0] -[0,0,1,0,0] -[1,0,0,1,1] -[0,0,0,0,0] -[0,0,0,0,0] -I don't understand why this matrix^9999 equals -[1,0,0,1,1] -[0,1,0,0,0] -[0,0,1,0,0] -[0,0,0,0,0] -[0,0,0,0,0] -or why this matrix^33334 equals -[0,1,0,0,0] -[0,0,1,0,0] -[1,0,0,1,1] -[0,0,0,0,0] -[0,0,0,0,0] -Can someone please explain what is happening? - -REPLY [6 votes]: Vertex $1$ is connected to $2,$ $2$ to $3,$ $3$ to $1, 4, 5,$ and $4$ and $5$ have no out edges, so your graph is a directed cycle with a couple of hairs pointing out. The number of paths of length $k$ from $1$ to $2$ is $1$ if $k = 1 \mod 3,$ and $0$ otherwise. same from $2$ to $3,$ same from $3$ to $1.$ Similarly for paths from $1$ to $3,$(except $k$ has to be $2 \mod 3$), etc. - -REPLY [5 votes]: Since the adjacency matrix $\mathrm A$ is not symmetric, we have a directed graph. - -Given a positive integer $k$, the $(i,j)$-th entry of $\mathrm A^k$ gives us the number of directed walks of length $k$ from $i$ to $j$. Given the cycle $1 \to 2 \to 3 \to 1$, the entry $(\mathrm A^k)_{11}$ should be $1$ when $k$ is a multiple of $3$ and $0$ when $k$ is not a multiple of $3$. Using SymPy, we can verify this: ->>> A = Matrix([[0,1,0,0,0], - [0,0,1,0,0], - [1,0,0,1,1], - [0,0,0,0,0], - [0,0,0,0,0]]) ->>> A**2 -[0 0 1 0 0] -[ ] -[1 0 0 1 1] -[ ] -[0 1 0 0 0] -[ ] -[0 0 0 0 0] -[ ] -[0 0 0 0 0] ->>> A**3 -[1 0 0 1 1] -[ ] -[0 1 0 0 0] -[ ] -[0 0 1 0 0] -[ ] -[0 0 0 0 0] -[ ] -[0 0 0 0 0] ->>> A**4 -[0 1 0 0 0] -[ ] -[0 0 1 0 0] -[ ] -[1 0 0 1 1] -[ ] -[0 0 0 0 0] -[ ] -[0 0 0 0 0] ->>> A**5 -[0 0 1 0 0] -[ ] -[1 0 0 1 1] -[ ] -[0 1 0 0 0] -[ ] -[0 0 0 0 0] -[ ] -[0 0 0 0 0] ->>> A**6 -[1 0 0 1 1] -[ ] -[0 1 0 0 0] -[ ] -[0 0 1 0 0] -[ ] -[0 0 0 0 0] -[ ] -[0 0 0 0 0] - -Note that $9999$ is a multiple of $3$, whereas $33334$ is not.<|endoftext|> -TITLE: What's the cokernel of a monoid homomorphism? -QUESTION [8 upvotes]: Let $f:A\to B$ be a monoid homomorphism. Where can I find an explicit description of the its cokernel? Are there any books on this topic? -By the cokernel of $f$, I mean the universal arrow which postcomposes with $f$ to give the trivial homomorphism. (Sorry for not including this from the start, I just thought there's no risk of ambiguity.) -If anyone cares, here's my motivation. In the category of groups, the cokernel of the kernel of a group homomorphism $f$ is the quotient of the domain by the kernel, which is comprised of the cosets of the kernel. The first isomorphism theorem says this quotient is isomorphic to the image. This makes sense because the multiplicative kernel action has strongly connected components (because of the existence of inverses), so the cosets of the kernel are the fibers. -For monoids there's no first isomorphism theorem because the kernel is largely uninformative. However, some monoid epimorphisms are known to be the cokernels of their kernels (namely Schreier split monoid epimorphisms), and I would like to see what this means concretely. - -REPLY [5 votes]: First of all, the construction is as for all (pointed) algebraic structures. Let $\sim$ be the congruence relation generated by $f(a) \sim 1$ for $a \in A$. Here, congruence relation means an equivalence relation on the underlying set of $B$ satisfying $b \sim b' \Rightarrow x b \sim x b' \wedge b x \sim b' x$ for all $b,b',x \in B$. Then $\mathrm{coker}(f)$ is the quotient monoid $B/{\sim}$, i.e. the set of equivalene classes of $\sim$ equipped with the monoid structure which is uniquely determined by the property that $B \to B/{\sim}$, $b \mapsto [b]$ is a homomorphism. -It remains to give an explicit description of $\sim$. In the commutative case (more generally, when $f:A \to B$ is central), we have the following: -$b \sim b' \Longleftrightarrow \exists a,a' \in A ( f(a) b = f(a') b')$ -In the non-commutative case, the description is much more complicated. You basically have to work with chains of relations and longer products as in the general description of generated congruences. More explicitly, $\sim$ is the transitive closure of the following relation: -$b \approx b' \Longleftrightarrow \exists n \in \mathbb{N} \, \exists a_i,a'_i,c_i,c'_i \in A \, \exists b_i,b'_i \in B: \\ b = b_1 f(a_1) \dotsc b_n f(a_n), ~ b' = b'_1 f(a'_1) \dotsc b'_n f(a'_n),\\ b_1 f(c_1) \dotsc b_n f(c_n) = b'_1 f(c'_1) \dotsc b'_n f(c'_n).$ -That being said, it should be pretty clear that you are lost when you want (or have) to use elements instead of the universal property. -Let me mention that the Grothendieck group resp. the universal enveloping group (this is what it's sometimes called in the non-commutative case) of a monoid $A$ is just the cokernel of the diagonal $A \to A \times A$.<|endoftext|> -TITLE: existence of an elliptic curves with given number of points over finite field -QUESTION [6 upvotes]: Is there a theorem which guarentees the existance of an elliptic curve with given number of points over $\mathbf{F}_p$ for a given $p$. -Thanks - -REPLY [10 votes]: Deuring proved that for every $a, |a| < 2\sqrt{p}$, there exists an elliptic curve with $p+1-a$ points over $\mathbb{F}_p$. -M Deuring, Die Typen der Multiplikatorenringe elliptischer Funktionenkörper, Abh. Math. Sem. Univ. Hamburg 14 (1941), 197-272. - -REPLY [4 votes]: Perhaps you are looking for Honda-Tate theory, see http://projecteuclid.org/download/pdf_1/euclid.jmsj/1260463295 Honda, Taira (1968), "Isogeny classes of abelian varieties over finite fields", Journal of the Mathematical Society of Japan, 20: 83–95. (The eigenvalues of the Frobenius determine the number of $\mathbf{F}_{q^n}$-rational points by the Lefschetz trace formula.)<|endoftext|> -TITLE: Cohen real without a minimal support -QUESTION [6 upvotes]: We start with a model of $\sf ZFC$, $V$, for simplicity we can imagine that $V$ satisfies $\sf GCH$ or even $V=L$. -Let $\Bbb P$ be $\operatorname{Add}(\omega,\omega_1)$, and let $G$ be a $V$-generic filter for $\Bbb P$. For $E\subseteq\omega_1$, we will write $G\restriction E$ as the restriction of $G$ to the reals with coordinates in $E$. -By the fact that the Cohen forcing is ccc, if $x\in\Bbb R^{V[G]}$, then there is some $E\in[\omega_1]^{\omega}\cap V$ such that $x\in V[G\restriction E]$. However, I was told there will be such $x$ without a minimal $E$. - -What is an example for a real $x$ such that if $x\in V[G\restriction E]$, then there is some $E'\subsetneq E$ for which $x\in V[G\restriction E']$? - -REPLY [6 votes]: Define $x \in 2^{\omega} \cap V[G]$ by $x(n) = $ the first bit of the $n$th Cohen real. Then for every $E \subseteq \omega_1$, $E \in V$, we have $x \in V[G \upharpoonright E]$ iff $\omega \setminus E$ is finite. So there is no such minimal $E$ for $x$.<|endoftext|> -TITLE: Which intrinsic invariants of a projective variety can you deduce from its Hilbert polynomials? -QUESTION [28 upvotes]: Given a projective variety $X$, each of its embeddings $i:X\hookrightarrow \mathbb P^N$ gives rise to an integer valued Hilbert polynomial $P_{X,i}(t)\in \mathbb Q[t]$. -These polynomials depend however on the chosen embedding $i$: for example $\mathbb P^1$ linearly embedded into $\mathbb P^2$ has Hilbert polynomial $H_{\mathbb P^1,\operatorname {lin}}(t)=t+1$ but when embedded as a conic by $i: \mathbb P^1 \hookrightarrow \mathbb P^2: (u:v)\mapsto (u^2:uv:v^2)$ it has Hilbert polynomial $H_{\mathbb P^1,i}(t)=2t+1$. -However, whatever embedding one chooses, all the polynomials $P_{X,i}(t)$ have the same degree, namely $\operatorname {dim} X$, and the same constant term $P_{X,i}(0)$, namely the arithmetic genus $p_a(X)=(-1)^{\operatorname {dim} X}(\chi(X,\mathcal O_X)-1)$. -My rather naïve question is then: which features of the Hilbert polynomial $P_{X,i}(t)$ are intrinsic to $X$, that is independent of the embedding $i:X\hookrightarrow \mathbb P^N$, apart from the degree and the constant term of that polynomial? -Edit -Another way of explaining what I'm after might be the question: -Given a projective variety $X$, which Hilbert polynomials does one obtain by embedding it into various projective spaces $X\hookrightarrow\mathbb P^N$ in all possible ways? - -REPLY [24 votes]: The OP encouraged me to post my comment, along with some examples, as an answer, so here goes. - -Lemma. Let $X$ be a smooth projective variety over an algebraically closed field $k$. Then there exists a numerical polynomial $p \in \mathbb Q[x_1,\ldots,x_r]$ such that for every (ample) divisor $H$ on $X$, there exist $e_1,\ldots,e_r \in \mathbb Z$ such that - $$P_H(t) = p(e_1t,\ldots,e_rt).$$ - -Proof. If $D_1 \equiv D_2$ are numerically equivalent divisors, then Grothendieck–Riemann–Roch gives -$$\chi(X,\mathcal O_X(D_i)) = \int_X \operatorname{ch}(\mathcal O_X(D_i)) \cdot \operatorname{Td}(X).$$ -Since the right hand side only depends on certain intersection numbers of powers of $c_1(D_i)$, it only depends on the numerical class of $D_i$. Hence so does the left hand side, so -$$\chi(X,\mathcal O_X(D_1)) = \chi(X,\mathcal O_X(D_2)).$$ -Replacing $D_1$ and $D_2$ by powers, we conclude that -$$P_{D_1}(t) = P_{D_2}(t).$$ -Now choose a set of generators $D_1,\ldots,D_r$ of $\operatorname{NS}(X)$. By Snapper's lemma (see FGA Explained, Thm B.7), the function -\begin{align*} -\mathbb Z^r &\to \mathbb Z\\ -(m_1,\ldots,m_r) &\mapsto \chi(X,\mathcal O_X(m_1D_1 + \ldots + m_rD_r)) -\end{align*} -is a numerical polynomial $p \in \mathbb Q[x_1,\ldots,x_r]$. The result follows since any divisor is numerically equivalent to $\sum e_i D_i$ for some $e_i$. $\square$ -Remark. Note that we don't need $H$ or the $D_i$ to be ample. This gives some extra flexibility in computations, as we will see in the example below. - - -Remark. It turns out it's actually pretty hard to compute this $p$, for example because it depends on a choice of generators of $\operatorname{NS}(X)$. Let's carry out a nontrivial example. - -Example. Let $X = \mathbb F_n$ be the $n$-th Hirzebruch surface. Then we have $\operatorname{Pic}(X) = \mathbb Z H \oplus \mathbb Z F$, with $H^2 = n$, $F^2 = 0$, and $F\cdot H = 1$ (see e.g. Beauville's Complex Algebraic Surfaces, Prop IV.1). Thus, the intersection product is given by the symmetric bilinear form -$$\begin{pmatrix} n & 1 \\ 1 & 0 \end{pmatrix}.$$ -Moreover, the canonical class is $K = -2H + (n - 2)F$ (see loc. cit., Prop III.18). Finally, we have $\chi(X,\mathcal O_X) = 1$, for example since $X$ is rational and $\chi(X,\mathcal O_X)$ is a birational invariant (or: compute). -This allows us to compute everything: by Riemann–Roch, we have -\begin{align*} -\chi(X,\mathcal O_X(aH+bF)) &= \tfrac{1}{2}(aH+bF)\cdot(aH+bF-K) + \chi(X,\mathcal O_X).\\ -&= \tfrac{1}{2}\begin{pmatrix}a & b\end{pmatrix}\begin{pmatrix} n & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} a+2 \\ b+2-n \end{pmatrix} + 1\\ -&= \tfrac{1}{2}(na(a+2) + a(b+2-n) + b(a+2)) + 1\\ -&= \tfrac{1}{2}(na^2 + 2ab + (n+2)a + 2b) + 1. -\end{align*} -That is, -$$p(x,y) = \tfrac{n}{2}x^2 + xy + (\tfrac{n}{2}+1)x + y + 1.$$ -Substituting $(x,y) = (at,bt)$ gives -$$P_{aH+bF}(t) = (\tfrac{n}{2}a^2+ab)t^2 + (\tfrac{an}{2}+a+b)t + 1.$$ -Thus, the polynomials occurring as Hilbert polynomial of some divisor are parametrised by -$$P(t) = f_2(a,b)t^2 + f_1(a,b)t + 1,$$ -where -\begin{align*} -f_2(a,b) &= \tfrac{n}{2}a^2+ab\\ -f_1(a,b) &= (\tfrac{n}{2}+1)a+b. -\end{align*} -In general, this method gives $\dim X$ homogeneous polynomials $f_1, \ldots, f_d$ in $\rho(X) = \operatorname{rk} \operatorname{NS}(X)$ variables of degrees $1, \ldots, d$ respectively (as noted, the constant term is always $\chi(X,\mathcal O_X)$). -If one then wishes to restrict to ample divisors only, one firstly has to classify which ones they are (one might try to use Nakai–Moishezon, but one has to think about how to find the effective curve classes algorithmically). -Ample divisors on $\mathbb F_n$. We want to use the Nakai–Moishezon criterion to find the ample divisors. Note that the only irreducible curve in $aH + bF$ for which $a$ or $b$ is negative is $B \in H - nF$ (see loc. cit., proof of Prop IV.1). Note that $B \cdot (aH + bF) = b$ for all $a, b \in \mathbb Z$. -Claim. A divisor $D = aH + bF$ is ample if and only if $a > 0$ and $b > 0$. -Proof. If $D$ is ample, then $a = D \cdot F > 0$ and $b = D \cdot B > 0$. -Conversely, let $D = aH + bF$ with $a > 0$ and $b > 0$. Then $D \cdot B > 0$, and $D \cdot (cH + dF) > 0$ whenever $c \geq 0$ and $d \geq 0$ and $(c,d) \neq (0,0)$. This includes $D^2$ and $D \cdot C$ for $C$ any irreducible curve other than $B$. Thus, the Nakai–Moishezon criterion implies that $D$ is ample. $\square$ -(Note in particular that $H$ is not ample.) -Closing remark. While in theory this gives a complete answer to the question which polynomials are Hilbert polynomials of divisors, we already see from this (relatively easy) example that this method is very laborious. -Moreover, it does not work nicely in families, because the Picard rank might jump. Even for nicely behaved varieties like abelian varieties or K3 surfaces, this phenomenon occurs, making it virtually impossible to answer the question in full generality for those varieties. -On the other hand, for surfaces the answer depends only on the Néron–Severi lattice of $X$, together with the canonical class and the effective cone of curves (to find back the ample divisors). In higher dimensions, this picture is complicated a little bit in that it needs the full knowledge of the Todd class (which now contains more data than just the canonical class).<|endoftext|> -TITLE: Maps out of Eilenberg-Mac Lane Spaces -QUESTION [20 upvotes]: Is anything known about the maps out of an Eilenberg Mac-lane Space $K(G,n)$? -Obviously I'm interested in extensions of Miller's resolution of the Sullivan conjecture, that $Map_*(K(G,1),X)\simeq\ast$ for $G$ a discrete, locally finite group and $X$ a connected, finite complex. On the other hand Gray has shown that there are uncountably many phantom maps $K(\mathbb{Z},2)\rightarrow S^3$ so it seems like there is nothing much to say without some specialisation. So allowing for some method of getting rid of phantom maps, say, p-completion, rationalisation, etc... is anything known? -Recently there have started to emerge some novel applications for the nullificiation and cellularisation functors of Dror Farjoun in the context of $BZ_p$-null homotopy theory. I'd like to know what $K(\mathbb{Z},2)$-, $K(\mathbb{Z}_p,2)$- and $K(\mathbb{Z},3)$-null homotopy theory looks like. - -REPLY [20 votes]: This question was totally answered by Alex Zabrodsky, right after Haynes Miller proved the Sullivan conjecture. See the paper: "On phantom maps and a theorem of H. Miller", Israel J. Math. 58 (1987), 129-143. -In summary, all maps are phantom, and all the homotopy groups of the space of maps can be determined from rational information.<|endoftext|> -TITLE: Lachlan on topology for priority arguments -QUESTION [7 upvotes]: There is a set of notes by Lachlan from 1973 on casting priority arguments in topological language; references to these notes are few and far between, but one source refers to them as "Topology for Priority Arguments," which for now I'll assume is the title. Regardless, they seem to have never been published; my understanding is that their line of inquiry never caught on. -I did see them long ago, in a computability theory class; and I've recently become re-interested in them. The instructor, unfortunately, no longer has a copy. I was wondering if anyone knew how I could find these notes, or had a copy. - -REPLY [5 votes]: After about my third priority argument in a graduate class, I too was interested in finding a general approach for handling them. You can find something similar from Lachlan in published form as The priority method for the construction of recursively enumerable sets in Lecture Notes in Mathematics 337. -The most developed general framework for priority arguments is A framework for priority arguments by Lerman. It's more order-theoretic than topological, but I doubt it would be too difficult to recast it in topological language.<|endoftext|> -TITLE: Does the classification of reductive groups follow from that of semisimple groups? -QUESTION [5 upvotes]: I had a question for anyone familiar with the proofs of the classification of reductive groups. I skipped most of the details of classification when I originally learned linear algebraic groups, and now I'm trying to go back and fill the gaps in my knowledge. -Let $G$ be a connected algebraic group over an algebraically closed field. If $G$ is reductive, then $G$ is classified up to isomorphism by its root datum, and conversely every root datum belongs to some connected, reductive group. -If $G$ is semisimple, then $G$ is classified up to isomorphism by its root system and its fundamental group, and conversely every pair of a root system and a subgroup of the weight lattice modulo the root lattice belongs to some semisimple group. -Taking the classification theorem for semisimple groups for granted (say existence and uniqueness), is it possible to deduce the classification theorem for connected, reductive groups? Or does the classification theorem for reductive groups need to be done again from scratch. - -REPLY [6 votes]: As indicated in the comments, there is no need to redo the entire classification argument when passing from "semisimple" to "reductive". But it's useful to recall some of the history. The emphasis in the Chevalley seminar 1956-58 was on achieving a uniform classification of semisimple algebraic groups over an algebraically closed field of arbitrary characteristic. For this a slightly more general treatment of isogenies between such groups is more natural. (Chevalley discovered for example that the simple groups of types $B_\ell$ and $C_\ell$ are isogenous in characteristic 2 while their groups of rational points are isomorphic even though the underlying algebraic groups are not). -A little later, Demazure and Grothendieck translated most of this into the more flexible language of group schemes in SGA3, while Borel and Tits by 1965 expanded the framework by defining reductive algebraic groups over an arbitrary field. (For such groups Tits achieved a classification, modulo some later reformulation of his theorem stated in the proceedings of the 1965 Boulder AMS Institute. The main unsolved problem is to classify the $k$-anisotropic groups, which vary a lot for different fields of definition $k$.) -The Chevalley seminar and the other sources in French are available online from numdam, but note that SGA3 has been -re-edited in recent years while a corrected typeset version of the Chevalley seminar was published in 2005 by Springer. (In my 1975 textbook, I mostly followed the Chevalley seminar; but when the characteristic is not 2 or 3, my 1966 thesis showed that it's also possible to rely more on the Lie algebra as in the classical characteristic 0 case.) -Variants were found by M. Takeuchi and T.A. Springer, using for example ideas of Serre and Steinberg about generators and relations for the groups. But in spite of the differences in the published approaches, all require a lot of detail about the internal structure of simple algebraic groups (those with no proper closed normal subgroups): -Bruhat decomposition, generation by tori along with root subgroups. -A key conclusion is that two such algebraic groups are isomorphic precisely when they have the same root system (or Dynkin diagram) and the same fundamental group, except in type $D_\ell$ with $\ell >2$ even, when you have to distinguish the half-spin and special orthogonal groups. -(Note too that the work of Tits gave a more precise picture of the internal structure: when $G$ is a simple algebraic group, its only proper normal subgroups are those contained in the finite center.) -Using the methods of Chevalley (1955), one further shows the existence of all possible types of simple algebraic groups. The classification of possible semisimple groups is then a routine but slightly messy exercise: start with a product of simply connected simple groups, then factor out a subgroup of the (finite) center. There may be a great many possibilities. -Translating all of this into the language of reductive groups is then a matter of reading between the lines in Springer's textbook: given an isomorphism of root data, one gets an isomorphism of root systems along with a comparison of fundamental groups, etc. Unfortunately, there is no single source in the literature for a truly unified treatment of all these matters, including isogenies. But the core of it all is the study of the Borel/Chevalley structure theory. The transition to reductive groups is needed mainly because these are more natural for induction purposes than the semisimple ones: Levi subgroups of parabolics are reductive but seldom semisimple. However, central tori over an algebraically closed field are fairly innocuous.<|endoftext|> -TITLE: characterization of subalgebras of universal enveloping algebra coming from Lie subalgebras -QUESTION [13 upvotes]: Let $\mathfrak{g}$ be a Lie algebra and $\mathfrak{g}'$ its subalgebra. Then the universal enveloping algebra $U(\mathfrak{g}')$ can be canonically embedded into $U(\mathfrak{g})$, that of $\mathfrak{g}$. -Now I'm interested in the reverse direction. Given a subalgebra $Y$ of $U(\mathfrak{g})$, under what conditions on $Y$, $Y$ is the universal enveloping algebra $U(\mathfrak{g}')$ of some Lie subalgebra $\mathfrak{g}'$ of $\mathfrak{g}$? -By considering the center of $U(\mathfrak{g})$, not all subalgebras of $U(\mathfrak{g})$ arise as universal enveloping algebras of Lie subalgebras. -I searched the internet but couldn't find anything related to this question. Much appreciated for any answer or reference. - -REPLY [15 votes]: In characteristic $0$, if $Y$ is a Hopf subalgebra of $U(\mathfrak{g})$, then $Y = U(\mathfrak{g}')$ for some Lie subalgebra $\mathfrak{g}'$ of $\mathfrak{g}$ (and conversely, every such subalgebra is a Hopf subalgebra). This is a consequence of Proposition 6.13 and Theorem 5.18 of John W. Milnor and John C. Moore's 1965 paper "On the structure of Hopf algebras." -If you are working over a field of positive characteristic, then the result is also true if you replace the universal enveloping algebra by the restricted enveloping algebra (see Theorem 6.11 of loc. cit.).<|endoftext|> -TITLE: Dyson's invitation: Opportunities in juxtaposition of incompatibles -QUESTION [6 upvotes]: "Up to now, my examples of missed opportunities have been mathematical discoveries which actually occurred, although they could have occurred a long time earlier. In such cases one can be sure that an opportunity existed, but it existed only in the past. I now come to the more difficult task of identifying missed opportunities that are still open. Here one can no longer be sure that the opportunity is real, but if it is real then it has the virtue of existing in the present. The past opportunities which I discussed have one important feature in common. In every case there was an empirical finding that two disparate or incompatible mathematical concepts were juxtaposed in the description of a single situation. Taking the four examples in turn, the pairs of disparate concepts were respectively: modular functions and Lie algebras, field equations and particle dynamics, Lorentz invariance and Galilean invariance, quaternion algebra and Grassmann algebra. In each case the opportunity offered to the pure mathematician was to create a wider conceptual framework within which the pair of disparate elements would find a harmonious coexistence. I take this to be my methodological principle in looking for opportunities that are still open. I look for situations in which the juxtaposition of a pair of incompatible concepts is acknowledged but unexplained." - from Missed Opportunities by Freeman Dyson. - -Dyson made a name for himself by showing the compatibility of the Schwinger, Feynman, and Tomonaga approaches to QED. He presents two unresolved juxtapositions of incompatible mathematical elements still under intense investigation: -A) General relativity and quantum mechanics -B) Feynman's sum over histories and existing theories of normed vector spaces. -Can you pose other Dyson pairs of incompatible mathematical structures (perhaps not as grand as Dyson's) that might present some interesting opportunities for integration under a yet-to-be-discovered (or invented) overarching theory? That is, can you extend Dyson's list? - -REPLY [7 votes]: My favourite "juxtaposition of a pair of incompatible concepts that is acknowledged but unexplained": the Riemann hypothesis and spectral theory, as phrased in the Hilbert-Pólya conjecture that the imaginary parts of the zeroes of the Riemann zeta function correspond to eigenvalues of an unbounded self-adjoint operator. -This is actually a topic on which Dyson himself has worked: When confronted with the statistical distribution of the zeroes on the critical line, which exhibited an anti-bunching or repulsion effect, Dyson noted that this seemed to coincide with that of the eigenvalues of a random Hermitian matrix, drawn from the Gaussian unitary ensemble. -The topic is nicely explained in The spectrum of Riemannium: - -In the 1970's the mathematician Hugh Montgomery found that the - statistical distribution of the Riemann zeroes on the critical line has a - certain property that they tend not to cluster too closely together, but to - repel. During a visit to the Princeton Institute for Advanced Study in - 1972, he showed this result to Freeman Dyson, one of the founders of - the theory of random matrices. Dyson realized that the statistical - distribution found by Montgomery appeared to be the same as the pair - correlation distribution for the eigenvalues of a random and large Hermitian matrix. These distributions are of importance in - physics, because for example the energy levels of an atomic nucleus - satisfy such statistics. -In a posterior development that has given substantive force to this - approach to the Riemann hypothesis through functional analysis and - operator theory, the mathematician Alain Connes has formulated a - “trace formula” using his non-commutative geometry framework that is - actually equivalent to a certain generalized Riemann hypothesis. However, the - mysterious operator believed to provide the Riemann zeta zeroes remains - hidden yet. Even worst, we don’t even know on which space the Riemann - operator is acting on. - -This graph illustrates the highly suggestive juxtaposition of Riemann zeroes and spectral statistics.<|endoftext|> -TITLE: How to find Suleimanova's work on the Nonnegative Inverse Eigenvalue Problem? -QUESTION [12 upvotes]: Many papers cite the work of Suleimanova when studying inverse eigenvalue problems - in particular, the nonnegative inverse eigenvalue problem (NIEP). However, I cannot seem to find her work anywhere. She proves an important result for lists of real eigenvalues. There is apparently one paper in Russian... somewhere. -Does anyone know where I can find her work? -How can people cite her work, if there's only one known paper that's in Russian? -Thanks. - -REPLY [5 votes]: A multi-set of real numbers $\Lambda = \{\lambda_1,\dots,\lambda_n\}$ is called a Suleĭmanova spectrum if $\Lambda$ contains one positive element and $\sum_{i=1}^n \lambda_i \geq 0$. -Suleĭmanova [Doklady Akad. Nauk SSSR (N.S.) 66, 343–345 (1949); MR0030496] loosely proved that every such set is realizable, i.e., that it is the spectrum of an $n$-by-$n$ nonnegative matrix. -Appendix B of Joanne Swift's 1972 master's thesis Location of the Characteristic Roots of Stochastic Matrices at McGill University contains an English translation of Suleĭmanova's seminal paper. -Friedland [Israel J. Math. 29, no. 1, 43–60 (1978); MR0492634] and Perfect [Duke Math. J. 20, 395–404 (1953); MR0055969] proved Suleĭmanova's result via companion matrices. However, constructing the companion matrix of a Suleĭmanova spectrum is computationally prohibitive. Recently, I gave a constructive proof of Suleĭmanova's result [Electron. J. Linear Algebra 31, 306–312 (2016); MR3504411] using permutative matrices.<|endoftext|> -TITLE: Unbeatable clique covers -QUESTION [7 upvotes]: Let $G=(V,E)$ be a simple, undirected graph. A clique cover is a set ${\cal C}\subseteq {\cal P}(V)$ such that - -every element of ${\cal C}$ is a clique, and -$\bigcup {\cal C} = V$. - -We call a clique cover ${\cal C}$ beatable if there is a clique cover ${\cal C}_1$ such that $$|{\cal C}_1 - {\cal C}| < |{\cal C} - {\cal C}_1|.$$ -Non-beatable clique covers are called unbeatable. -It is easy to prove that every finite graph has an unbeatable clique cover. But what about infinite graphs? - -REPLY [8 votes]: I think it is a known conjecture. The special case that there always exists a clique cover such that the union of any two of those cliques is not a clique, can be proved from Zorn's lemma (or without it), cf. my book with Totik: Problems and Theorems in Classical Set Theory, Problem 14.6.(m). This special case is actually equivalent to the axiom of choice: F. Galvin, P. Komjáth: Graph colorings and the axiom of choice, Periodica Math. Hung. 22 (1991), 71–75.<|endoftext|> -TITLE: Extending homeomorphisms from closed countable sets to S^2 -QUESTION [10 upvotes]: Let $A, B \subset S^2$ be closed, countable sets and $\phi \colon A \rightarrow B$ be a homeomorphism. Can we extend $\phi$ to a homeomorphism from $S^2$ to itself? -It is well-known that the answer is yes if $A$ and $B$ are finite. When $A$ and $B$ have only one accumulation point, the question is equivalent to the extension of a bijection between closed discrete sets in $\mathbb R^2$ to an homeomorphism of $\mathbb R^2$ to itself. This is also not too difficult to construct. Similary, I think I can show it in the case that $A$ and $B$ have only finitely many accumulation points. - -REPLY [3 votes]: $A\cup B$ is countable and closed, so there is a point $n\in S^2\setminus (A\cup B)$ with an open disk centered at $n$ that does not meet $A\cup B$. Using stereographic projection from the pole $n$, we consider $A$ and $B$ as closed, bounded subsets of the plane. Moreover, each set is totally disconnected. We then apply the following theorem: -Theorem 13.7 [page 93 of Moise's Geometric topology in dimensions 2 and 3]: -Let $M$ and $M'$ be totally disconnected compact sets in $\mathbb{R}^2$, and let $f:M\to M'$ be a homeomorphism. Then $f$ has an extension $F:\mathbb{R}^2\to \mathbb{R}^2$. -The extension $\tilde{F}: S^2\to S^2$ of $F$ to the sphere is still a homeomorphism.<|endoftext|> -TITLE: Integer roots of a symmetric polynomial -QUESTION [6 upvotes]: The question is very simple and I apologize for that, but I am not an expert of this kind of problem. -Given the polynomial -$$ P(x_1,\ldots,x_{2n})=x_1^2+\ldots+x_n^2-x_{n+1}^2-\ldots-x_{2n}^2,$$ -I would like to know if there are non trivial integer roots $(y_1,\ldots, y_{2n})$ such that -$$y_1+\cdots+y_{n}=y_{n+1}+\cdots+ y_{2n}.$$ -With non trivial I mean the ones like -$$y_1=y_{n+1},\ldots,y_{n}=y_{2n},$$ -or their permutations. - -REPLY [17 votes]: Fix large $N$ and consider all $n$-tuples $(x_1,\dots,x_n)\in \{1,\dots,N\}^n$. There are $N^n$ such $n$-tuples, at least $N^n/n!$ tuples modulo permutations, and for them the pairs $(x_1+\dots+x_n,x_1^2+\dots+x_n^2)$ take at most $n\cdot N\cdot n\cdot N^2=n^2N^3$ possible values. Thus by pigeonhole principle some value is obtained at least $N^{n-3}/(n^2\cdot n!)$ times. This is greater than 1 if $n>3$ and $N$ is chosen large enough.<|endoftext|> -TITLE: Are there any solutions to the diophantine equation $x^n-2y^n=1$ with $x>1$ and $n>2$? -QUESTION [12 upvotes]: This problem arose when considering storage of cannonballs in n-dimensional pirate ships, as explained in this MSE post. This MO question can also be reduced to the $n=3$ case. If $x,y$ is a solution then $$0<\frac{x}{y}-2^\frac1n<\frac{2^\frac1n}{2ny^n}$$ then by Roth's theorem this has finitely many solutions for fixed $n$. Let $$2^{1/n}=a_0+\frac{1}{a_1+\dots}$$ -be the canonical continued fraction of $2^{1/n}$, then $a_0=1$ and $a_1\in\{\lfloor\frac{n}{\ln(2)}\rfloor,\lfloor\frac{n}{\ln(2)}\rfloor-1\}$, and since $\frac{x}{y}$ is a convergent of this continued fraction, $y>\frac{n}{\ln(2)}-1$. There are no solutions with $x^{n}<2^{64}$. It is also sufficient to only consider $n=4$ and odd primes, in FLT fashion. - -REPLY [2 votes]: The case $n = 4$ and so $n$ even can be done by hand. Darmon and Merel proved (in 1997) the stronger statement that there aren't even any rational solutions to this equation for $n$ odd besides $(x^n,y^n) = (1,1)$, See their paper "Winding quotients and some variants of Fermat's last theorem," which can be found here: -http://www.math.mcgill.ca/darmon/pub/Articles/Research/18.Merel/paper.pdf<|endoftext|> -TITLE: Does every Riemann surface with boundary immerse in C? -QUESTION [7 upvotes]: Does every connected, compact Riemann surface $\Sigma$ with boundary, $\partial \Sigma\not =\emptyset$, admit a holomorphic function (smooth on the boundary) $f:\Sigma\to\mathbb C$ whose derivative is everywhere non-zero? - -REPLY [12 votes]: A more general result is proven in -Gunning, R. C., Narasimhan, R., Immersion of open Riemann surfaces. Math. Ann. 174, 103–108 (1967). -As for compact surfaces with boundary, it is essentially a part of the definition that they embed holomorphically into open Riemann surfaces. -Incidentally, it is an open problem for $n$-dimensional Stein manifolds (with trivial complex tangent bundles), $n\ge 2$, if they admit locally biholomorphic immersions in ${\mathbb C}^n$. The best (to my knowledge) result in this direction is due F.Forstenič (Acta Math., 2003) who proved that every complex parallelizable $n$-dimensional Stein manifold admits a holomorphic submersion to ${\mathbb C}^{n-1}$.<|endoftext|> -TITLE: Covering measure one sets by closed null sets -QUESTION [5 upvotes]: (The following question arose in a joint research with Adam Przeździecki and Boaz Tsaban.) -For a $\sigma$-ideal $\mathcal{I}$ of subsets of the unit interval -$[0,1]$, define -$$\newcommand{\card}[1]{\left|#1\right|}\newcommand{\cov}{\operatorname{cov}}\newcommand{\sub}{\subset} \cov(\mathcal{I}):=\min\{\card{\mathcal{A}}: \mathcal{A}\sub \mathcal{I}, \bigcup\mathcal{A}=[0,1]\}. -$$ -Let $\mathcal{E}$ be the family of all $F_\sigma$ Lebesgue null subsets of the unit interval, and $\mathcal{N}$ be the family of all Lebesgue null subsets of the unit interval. - -Let $\kappa_\mathcal{E}$ be the minimal cardinal number such that some measure one set is covered by $\kappa_\mathcal{E}$ elements of $\mathcal{E}$. -Similarly, let $\kappa_\mathcal{N}$ be the minimal cardinal number such that some measure one set is covered by $\kappa_\mathcal{N}$ elements of $\mathcal{N}$. - -We have $\kappa_\mathcal{E}\leq \cov(\mathcal{E})$, and $\kappa_\mathcal{N}=\cov(\mathcal{N})$. -Question 1: Is it provable that $\kappa_\mathcal{E}=\cov(\mathcal{E})$? -Question 2: Does the cardinal number $\cov(\mathcal{E})$ change if we work in a closed positive subset of the unit interval instead of the whole interval? -A negative answer for Question 2 implies a positive answer for Question 1. - -REPLY [3 votes]: The answer to the first question is yes: it is always true that $\kappa_{\mathcal E} = \mathrm{cov}(\mathcal E)$. The answer to the second question is no. -As Piotr points out, a negative answer to the second question implies a positive answer to the first. [This is because if $A \subseteq [0,1]$ is a set of measure one, then it contains a closed $K \subseteq [0,1]$ of positive measure (indeed, we can get the measure of $K$ as close to $1$ as we like). A negative answer to the second question means that it takes at least $\mathrm{cov}(\mathcal E)$ sets in $\mathcal E$ to cover $K$; since $K \subseteq A$, it takes at least $\mathrm{cov}(\mathcal E)$ sets in $\mathcal E$ to cover $A$ too.] -So it suffices to show that the answer to the second question is no: the $\mathcal E$-covering number is the same for any positive-measure closed subsets of $[0,1]$. -To see this, fix a closed $K \subseteq [0,1]$ of positive measure $\alpha$. Let $\mathcal F$ be a family of sets in $\mathcal E$ that covers $K$. Our goal is to cover $[0,1]$ with a family of sets in $\mathcal E$ of size $|\mathcal F|$. This shows that the $\mathcal E$-covering number for $[0,1]$ is no bigger than the analogous number for $K$. It is obvious that the $\mathcal E$-covering number for $[0,1]$ is no smaller than the analogous number for $K$, since any cover of $[0,1]$ by sets in $\mathcal E$ is also a cover of $K$. -Define a map $\phi: K \rightarrow [0,1]$ by -$$\phi(x) = \frac{1}{\alpha} \cdot m(K \cap [0,x])$$ -for all $x \in K$, where $m$ denotes the Lebesgue measure. Using the fact that $K$ is closed, one can show that $\phi$ is surjective. In particular, -$$\phi(\mathcal F) = \{\phi[A \cap K] : A \in \mathcal F\}$$ -is a family of sets that covers $[0,1]$. To finish the proof, it is enough to show that the map $A \mapsto \phi[A \cap K]$ sends sets in $\mathcal E$ to sets in $\mathcal E$: then $\phi(\mathcal F)$ is the desired $\mathcal E$-cover of $[0,1]$. -From the definition of $\phi$, we see -$$m(\phi[(0,b)]) = \frac{1}{\alpha} \cdot m(K \cap (0,b))$$ -and taking complements we get -$$m(\phi[(a,1)]) = \frac{1}{\alpha} \cdot m(K \cap (a,1)).$$ -and taking intersections, we see that for any interval $(a,b) \subseteq [0,1]$, -$$m(\phi[(a,b) \cap K]) = \frac{1}{\alpha}(m((a,b) \cap K)).$$ -That is, $\phi$ can increase the measure of an interval by at most a factor of $\frac{1}{\alpha}$. With a little more (routine) work, this shows that the map $A \mapsto \phi[A \cap K]$ sends null sets to null sets. -Observe that $\phi$ is continuous (the inverse image of an open interval in $[0,1]$ is an open interval in $K$). Therefore $\phi$ maps compact sets to compact sets. Therefore the map $A \mapsto \phi[A \cap K]$ sends compact sets to compact sets. Since $[0,1]$ is compact, this means that it also sends $F_\sigma$ sets to $F_\sigma$ sets. So we have showed that the map $A \mapsto \phi[A \cap K]$ sends null $F_\sigma$ sets to null $F_\sigma$ sets, and we are done.<|endoftext|> -TITLE: Uniquely 4-colorable Unit Distance Graphs -QUESTION [5 upvotes]: A graph is "unit distance" if it can be embedded in the plane -in such a way that whenever two vertices are connected by an edge, -their distance is $1$. -A graph is uniquely $4$-colorable if there is a unique way -(up to renaming of colors) of coloring its vertices with $4$ -colors in such a way that no two adjacent vertices have the -same color, and if no such coloring with $3$ colors exists. -Question: Is there finite graph that is both uniquely -$4$-colorable and unit distance? - -REPLY [5 votes]: EDIT the original answer was wrong, thanks to Jan Kyncl for pointing the mistake, hopefully it is fixed now. -I think this is open, since existence will improve the lower bound of chromatic -number for unit distance graphs from $4$ to $5$ almost surely. -Assume uniquely $4$ colorable finite unit distance graph $G$ exists. Color it -with colors $a,b,c,d$. Take two copies of $G$: $G_1$ and $G_2$. For vertices -colored $a$, $a_1 \in G_1, a_2 \in G_2$ merge $a_1$ and $a_2$ to vertex $A$ to -get unit distance graph $G'$ with the property that all vertices in $G_1,G_2$ which are colored $a$ have the same color in all $4$ colorings of $G'$. For vertices $a_1' \in G_1, -a_2' \in G_2$ colored with the $a$ color, rotate $G_2$ with center $A$ trying -to get $a_1',a_2'$ at distance $1$. The triangle inequality is enough. If you -can do this, add edge $(a_1',a_2')$ to get unit distance graph, which is not $4$ -colorable. I think such four vertices exist with high probability and comment by Pat Devlin confirms this.<|endoftext|> -TITLE: Are most Kähler manifolds non-projective? -QUESTION [8 upvotes]: Since now-a-days lots of research activities are happening to prove many results for compact Kähler manifolds which are known for projective varieties, I was wondering are there plenty of non-projective Kähler manifolds? If yes, where can I find some explicit examples? I am aware of the theorem that a generic complex torus $\mathbb{C}^g/\Lambda$ is non-projective. - -REPLY [9 votes]: Expanding a bit on Francesco Polizzi's remark, complex K3 surfaces are Kähler [1], the moduli space $M$ of complex K3 surfaces is an irreducible 20-dimensional complex variety, and within $M$, the set of K3 surfaces that are algebraic varieties is a countable union of disjoint subvarieties $F_g$ for integers $g\ge2$. Here $F_g$ has dimension 19, and the K3 surfaces in $F_g$ are those that have a primitive line bundle $\mathcal L$ satisfying $c_1(\mathcal L)^2=2g-2$. -So this is the analogue for K3 surfaces of the example you know for complex tori, i.e., in the moduli space of complex tori of a given dimension, the projective ones form a countable union of proper subvarieties. -[1] Siu, Y. T. (1983), Every K3 surface is Kähler, Inventiones Mathematicae 73 (1): 139–150<|endoftext|> -TITLE: There is no arcwise isometry from a high dimensional manifold into a low dimensional manifold -QUESTION [9 upvotes]: $\newcommand{\al}{\alpha}$ -$\newcommand{\ga}{\gamma}$ -$\newcommand{\e}{\epsilon}$ -Let $X,Y$ be Riemannian manifolds, such that $\dim(X) > \dim(Y)$. -I am trying to prove the following statement (mentioned by Gromov in his book on metric geometry): -There is no arcwise isometry (i.e length preserving map) from $X$ to $Y$. -However, the naive attempt to prove this hits an obstacle which I do not see how to pass: -Suppose by contradiction $f:X \to Y$ is an arcwise isometry. Then $f$ is $1$-Lipschitz, hence differeniable almost everywhere (by Rademacher's theorem). -Question: Let $p \in X$ be a point where $f$ is differentiable. Is $df_p:T_pX \to T_{f(p)}Y$ an isometry? -(This will imply the claim of course). -Here is what happens when trying to show this naively: -Let $v \in T_pX$, and let $\al:[0,1] \to X$ be a smooth path s.t $\al(0)=p,\dot \al(0)=v$. -Then $|\dot \alpha(s)| = |\dot \alpha(0)|+\Delta(s)=|v|+ \Delta(s)$ where $\lim_{s \to 0} \Delta(s) =\Delta(0)= 0$, thus -$$ (1) \, \, L(\alpha|_{[0,t]})=\int_0^t |\dot \alpha(s)| ds=t|v|+\int_0^t \Delta(s) ds.$$ -$\al$ is Lipschitz and $f$ is $1$-Lipschitz, so $\ga:= f \circ \al$ is Lipschitz. By theorem 2.7.6 in the book ``A course in metric geometry'' ( Burago,Burago,Ivanov) it follows that: -$$ (2) \, \, L(\ga|_{[0,t]})=\int_0^t \nu_{\ga}(s) ds, $$ -where $\nu_{\ga}(s):=\lim_{\e \to 0} \frac{d\left( \ga(s),\ga(s+\e) \right)}{|\e|}$ is the speed of $\ga$. -Note that $\nu_{\ga}(0)= |\dot \ga(0)|=|df_p(v)|$. -Using the assumption $f$ preserves lengths, we would now like to compare $(1),(2)$ and take derivatives at $t=0$, to get $$|v|=\frac{d}{dt}\left. \right|_{t=0} L(\al|_{[0,t]})=\frac{d}{dt}\left. \right|_{t=0} L(\ga|_{[0,t]})=|df_p(v)|.$$ -However, it seems that the last equality is false in general; Even when the speed of a Lipschitz curve and the derivative of its length exist at a point, they do not need to be equal. -It seems that the only thing we can say is that $ \frac{d}{dt}\left. \right|_{t=0} L(\ga|_{[0,t]}) \ge \nu_{\ga}(0) =|df_p(v)|$, so we are left with $|v| \ge |df_p(v)|$ which doesn't help. -Is there a way to "fix" the proof above? - -REPLY [2 votes]: I am completenig some details based on Anton's answer: -We prove the following theorem: -Let $X,Y$ be Riemannian manifolds, and let $f:X \to Y$ be a length preserving map. Then $df$ is an isometry almost everywhere. - -The proof is composed of 3 steps: -Step I: -It is enough to prove the claim where $X=(\mathbb{R}^n,g)$ where $g$ is an arbitrary metric. -Step II: -Assume step I, i.e we consider $f:(\mathbb{R}^n,g) \to Y$. -Let $v \in \mathbb{R}^n$ fixed . Then $|df_x(v)|_{f(x)}=|v|_x$ for almost every $x \in \mathbb{R}^n$. -Step III: -Conclude that $df$ is an isometry almost everywhere. - -We begin with proofs of steps I,III, since these are easy. -Proof of step I: -$f$ is $1$-Lipschitz, hence by Rademacher's Thm it is differentiable a.e. -Since any manifold admits a countable cover of charts, we finished. -Proof of step III: -$f:(\mathbb{R}^n,g) \to Y$. -Let $v_1,...,v_n$ be a basis for $\mathbb{R}^n$. By applying step II for every $v \in Q=\{v_1,...,v_n,v_1+v_2,v_1+v_3,...,v_{n-1}+v_n\}$, we get that -$|df_x(v)|_{f(x)}=|v|_x$ for all $v \in Q$ and for almost every $x \in \mathbb{R}^n$. Now it is a simple linear algebra fact that at all these points $x$, $df_x$ is an isometry. Indeed, -$$ \langle df_x(v_i),df_x(v_j) \rangle = \frac{1}{2}(|df_x(v_i)+df_x(v_j)|^2 - |df_x(v_i)|^2 - |df_x(v_j)|^2) = \frac{1}{2}(|df_x(v_i+v_j)|^2 - |v_i|^2 - |v_j|^2) $$ -$$ = \frac{1}{2}(|v_i+v_j|^2 - |v_i|^2 - |v_j|^2) = \langle v_i,v_j \rangle.$$ -Comment: Here we see why restricting to a chart was so helpful! We can use the "same" basis vectors $v_i$ for different points $x$ (on a general non-parallelizable manifold, this is meaningless). - -Proof of step II: -Fix $v \in \mathbb{R}^n$. Then $|df_x(v)|_{f(x)}=|v|_x$ for almost every $x \in \mathbb{R}^n$. -Proof: -Fix $x \in \mathbb{R}^n$, and define $\alpha_x(t)=x+tv$. -Putting $\gamma_x=f \circ \alpha_x$ (this is a Lipschitz path), we get by theorem 2.7.6 that $$ (1): \, \, \nu_{\alpha_x}(t) =\frac{d}{dt}\left. \right|_{t=0} L(\alpha_x|_{[0,t]})=\frac{d}{dt}\left. \right|_{t=0} L(\ga_x|_{[0,t]})=\nu_{\ga_x}(t) \, \, \text{ for almost every $t$}$$ -For any path $\beta$ in a Riemannian manifold, at every point of differentiability, it holds that $\nu_{\beta}(t)=|\dot \beta(t)|_{\beta(t)}$. -Specializing this for $\beta = \alpha_x$ (and combining $(1)$) we obtain -$$ (1)': \, \, |v|_{\alpha_x(t)}=\nu_{\ga_x}(t) \, \, \text{ for almost every $t$}$$ -Specializing this for $\beta = \gamma_x$ leads to -observation I: -$\nu_{\gamma_x}(t)=|df_{\alpha_x(t)}(v)|_{f(\alpha_x(t))}$ for those $t$ where $f$ is differentiable at $\alpha_x(t)$. - -We define the sets -$ B:=\{(x,t) \in \mathbb{R}^n \times \mathbb{R} | \, \, \nu_{\gamma_x}(t)=|v|_{\alpha_x(t)} \, \, \text{ and } \, f \, \text{ is differentiable at } \, \alpha_x(t) \} \subseteq \mathbb{R}^{n+1}, $ -$B_x=\{ t \in \mathbb{R} | \, \, (x,t) \in B \} \subseteq \mathbb{R}$ and -$B^t= \{x\in\Bbb R^n: (x,t)\in B\} \subseteq \mathbb{R}^n$. -We also define $h:\mathbb{R}^{n+1} \to \mathbb{R}^n$ by $h(x,t)=\alpha_x(t)$. -$h(B)=\{\alpha_x(t) | \, \, \nu_{\gamma_x}(t)=|v|_{\alpha_x(t)} \, \, \text{ and } \, f \, \text{ is differentiable at } \, \alpha_x(t) \}=\{\alpha_x(t) | \, \, |df_{\alpha_x(t)}(v)|_{f(\alpha_x(t))}=|v|_{\alpha_x(t)} \}$ -where the last equality follows from observation I. -Hence, it is enough to show $h(B)^c$ has measure zero in $\mathbb{R}^n$. -This essentially follows from Tonelli's theorem: -Since $h$ is surjective, $h(B)^c \subseteq h(B^c)$. Thus, it is enough to show $\mu(B^c)=0$ in $\mathbb{R}^{n+1}$. -$$ B^c=\{ (x,t) | \, \, \neg ( \nu_{\gamma_x}(t)=|v|_{\alpha_x(t)} ) \} \cup \{ (x,t) | \, \, f \, \text{ is not differentiable at } \, \alpha_x(t) \}:=A \cup D$$ -$(1)'$ implies that $\lambda_1(A_x)=0$ for every $x \in \mathbb{R}^n$, hence (by Tonelli) $\mu(A)=0$. -$D^t = \{ x \in \mathbb{R}^n | f \, \text{ is not differentiable at } \, x+tv \} = W -tv$ (where $W \subseteq \mathbb{R}^n$ is the set of points of non-differentiability of $f$), so $\lambda_n(D^t)=\lambda_n(W)=0$ -Again, Tonelli implies $\mu(D)=0$.<|endoftext|> -TITLE: On Selberg's Proof of the Selberg Integral Formula -QUESTION [11 upvotes]: I'm attempting to read through Mehta's write-up of Selberg's proof of the formula for the Selberg integral formula given below: -$$ -\begin{align} -\operatorname{S}_n(\alpha,\ \beta,\ \gamma) & = \int_{[0,\ 1]^n}\prod_{i=1}^nx_i^{\alpha-1}(1-x_i)^{\beta-1}|\Delta(\vec x)|^{2\gamma}\text{d}\vec x \\ -& = \prod_{j=0}^{n-1}\frac{\Gamma(1+(1+j)\gamma)\Gamma(\alpha+j\gamma)\Gamma(\beta+j\gamma)}{\Gamma(1+\gamma)\Gamma(\alpha + \beta + (n+j-1)\gamma)} -\end{align} -$$ -The proof is from Mehta's book "Random Matrices" (Third Edition) on Page 311. -His proof, to me, uses a lot of hand-wavey arguments to derive bounds on the terms $j_m$ in the ordered partitions of $2\gamma{n\choose 2}$. He arrives at $\gamma(m-1)\le j_m\le \gamma(n+m-2)$. -To start, assume $\gamma$ is an integer. The first problem I have with the proof is that he expands $|\Delta(\vec x)|^{2\gamma}$ as a series: -$$ -\Delta(\vec x)^{2\gamma} = \sum_{\substack{j_1+\cdots+j_n=2\gamma{n\choose 2} \\ j_i\le j_{i+1}}}c(j_1,\cdots,\ j_n)x^{j_1}\cdots x^{j_n} -$$ -where the $c$'s are just integers. My problem with this is that he orders the $j_i$'s. If you do this for even a simple example ($n = 2$,\ $\gamma = 1$) you see that this can't possibly be true, unless I'm missing something brutally obvious: -$$ -\Delta_2(x_1,\ x_2)^2 = x_2^2 - 2x_1x_2 + x_1^2 \\ -\sum_{\substack{j_1+j_2 = 2 \\ j_1\le j_2}}c(j_1,\ j_2)x_1^{j_1}x_2^{j_2} = c(0,\ 2)x_2^2 + c(1,\ 1)x_1x_2 -$$ -and clearly there are no integers $c(0,\ 2)$ and $c(1,\ 1)$ for which $c(0,\ 2)x_2^2 + c(1,\ 1)x_1x_2 = x_2^2 - 2x_1x_2 + x_1^2$. -He goes on to use this ordering of the $j_i$'s to prove other various things, but the whole point of this is to arrive at the bounds on $j_m$ given at the beginning of the question. -Now, nowhere else in the proof is it needed that $j_i\le j_{i+1}$, so I have two questions: -1. How can we justify ordering the $j_i$'s? -2. How can we arrive at the bounds $\gamma(m-1)\le j_m\le \gamma(n+m-2)$ more rigorously? -Edit: I realize now that there's a problem, and that's if we can't justify the ordering of the $j_i$'s then there aren't any bounds on $j_m$ (other than $0$ and $2\gamma{n\choose 2}$), so answering 1 immediately answers 2, and refuting 1 would, as far as I can tell, invalidate Selberg's proof of this formula. I'm confident the formula itself holds, since I've also read through Aomoto's proof and it seems much more solid, but I'm interested in seeing how, if possible, we could justify Selberg's original proof. -Disclaimer: I'm a first year at my university who was fortunate enough to attend pure math courses. I can read and understand generally higher level math, and more often than not understand the logic behind things, but sometimes I miss simple things. If this is a really simple question, please forgive me. - -REPLY [7 votes]: See Greg Anderson's two page proof.<|endoftext|> -TITLE: Equivalent definitions of Cartesian Fibrations between Quasi-Categories -QUESTION [7 upvotes]: In the paper by Verity and Riehl "Fibrations and Yoneda lemma in an ${\infty}$-cosmos" (https://arxiv.org/abs/1506.05500), they prove a Yoneda lemma that holds in any ${\infty}$-cosmos (see Corollary 6.2.13). Therefore, in particular, it holds in $\textbf{qCat}$. -If we unravel their definition of cartesian fibration, we find out that it is almost representably defined, in the sense that an isofibration $p:E \to B$ is a cartesian fibration in an ${\infty}$-cosmos iff $map(A,E) \to map(A,B)$ is (almost) a cartesian fibration of quasicategories for every $A$. More precisely, instead of the usual requirement of a suitable lift for a certain edge which has to be cartesian with respect to the inner fibration $map(A,p)$,they ask for a weaker property: this lift $\chi$ has to be such that $map(A,p)$ has the right lifting property only against the inclusion $\Lambda^2[2] \to \Delta[2]$ when the restriction to $\Delta^{\{1,2\}}$ is $\chi$ itself (so it is Lurie's definition of cartesian edge where we restrict ourselves to n=2). -They also have a conservativity requirement, but I am not exactly sure if and how this fits into my question. -Does this produce an equivalent notion of cartesian fibration (I suspect it does not)? Or at least, is it true that $p:X \to Y$ inner fibration of quasicategories is a cartesian fibration iff $map(A,p)$ is a cartesian fibration in this weaker sense for every quasicategory (or even any simplicial set) A? - -REPLY [3 votes]: Riehl and Verity prove that their definition agrees with Lurie's in Corollary 4.1.24 (cf. Remark 2.4.1.4 in HTT).<|endoftext|> -TITLE: an identity for a sum over partitions -QUESTION [13 upvotes]: Write an integer partition $\lambda\vdash n$ in two different ways: -(1) $\lambda=\lambda_1\geq\lambda_2\geq\lambda_3\cdots\geq\lambda_k\geq1$ -(2) $\lambda=1^{m_1}2^{m_2}3^{m_3}\cdots n^{m_n}$ for some $m_i\geq0$. -Denote the length of a partition, compatible with (1) and (2), by $m_1+m_2+\cdots+m_n=k$. - -QUESTION. According to enough experimental evidence, - $$\sum_{\lambda\vdash n}(-1)^{n-k}\frac{k!}{m_1!\cdots m_n!}\prod_{i=1}^k\binom{n+1}{\lambda_i}=\binom{2n}n.$$ - Is it true? If so, any proof? - -REPLY [14 votes]: Here is a purely combinatorial proof of a more general identity -$$ -\sum_{\lambda\vdash n}(-1)^{n-k}\frac{k!}{m_1!\cdots m_n!}\prod_{i=1}^k\binom{A}{\lambda_i}=\binom{A+n-1}n. -$$ -As Ilya suggests, we rewrite this as $$\sum_{\lambda_i\geq 1, \; \lambda_1+\dots+\lambda_k=n} - (-1)^{n-k}\prod_{i=1}^k{A\choose \lambda_i}.$$ -This is an alternating sum of sequences $(S_1,\dots,S_k)$ of subsets of $\{1,2,\dots,A\}$, $\sum |S_i|=n$. We identify each subset $S_i$ with an increasing sequence $p_{i1}<\dots -TITLE: What is the motivic class of a quotient stack? -QUESTION [7 upvotes]: The Grothendieck ring of complex varieties $K(Var_\mathbb C)$ is the free abelian group generated by isomorphism classes $[X]$ of $\mathbb C$-varieties, modulo the scissor relation $[X]=[Z]+[X\setminus Z]$ for every closed subvariety $Z\subset X$. The product is given by $[X]\cdot [Y]=[X\times_\mathbb CY]$. There is a similar construction for stacks, leading to a ring $$K(St_\mathbb C).$$ One has to restrict to stacks with affine stabilizers. This construction is treated for example in section 3 of this paper by Bridgeland. One can show that $K(St_\mathbb C)$ can be obtained from $K(Var_\mathbb C)$ by inverting the classes of all special groups, or equivalently all classes $[GL_d]$ for $d\geq 1$, or equivalently, the classes $\{\mathbb L,\mathbb L^i-1:i\geq 1\}$, where $\mathbb L=[\mathbb A^1]$. -I am missing some basic point because it seems to me that all classes $[G]$ (for $G$ special) are invertible in -$K(St_\mathbb C)$ so that if I have a quotient stack $X/G$ its class will be $$[X/G]=[X]/[G]\in K(St_\mathbb C)$$ regardless of the action of $G$ on $X$, and this is confusing. Here are a couple of examples I would like to understand: - -What is the class of $GL_2/GL_2$, where $GL_2$ acts on itself by conjugation? I would like this to be different from $[pt]=1$. -What is the class of weighted projective space $\mathbb P(a_0,\dots,a_{n-1})=(\mathbb C^{n}\setminus 0)/\mathbb C^*$? I would like this to be different from the class $[\mathbb P^{n-1}]=1+\mathbb L+\cdots+\mathbb L^{n-1}$. - -Thank you! - -REPLY [7 votes]: Yes: if $G$ is a special group, then $[X/G] = [X]/[G]$ in the Grothendieck group of stacks. This is the analogue of the fact that if $X \to E$ is a $G$-torsor over an algebraic variety (and $G$ is still special) then $[X] = [E]\cdot [G]$ in the Grothendieck group of varieties. In the category of stacks, the map $X \to X/G$ is always a $G$-torsor, regardless of the action of $G$ on $X$. Does this help? -Qiaochu's toy example is very relevant but also somewhat orthogonal to this situation: finite groups are not special, and $[X/G] = [X]/[G]$ is almost never true in the Grothendieck group of stacks (or varieties, for that matter) if $G$ is finite. For example, $\mathbb G_m$ is a finite étale double cover of itself, but its class in the Grothendieck group is nonzero. A more interesting example is that the class of $BG$ is in many cases (but not always!) just the class of a point: see "A geometric invariant of a finite group" by Torsten Ekedahl.<|endoftext|> -TITLE: Which category of sheaves on a manifold remembers the manifold? -QUESTION [14 upvotes]: Given a not too nasty topological space $X$, the category of sheaves of sets on $X$ remembers $X$. -Given a scheme $S$, the category of quasicoherent sheaves on $S$ remembers $S$. -Given a smooth manifold $M$, the category of ____ sheaves on $M$ remembers $M$. - -REPLY [5 votes]: Assuming the manifold is Hausdorff, the category of sheaves of modules over the sheaf of smooth functions do the trick. -This category is equivalent to the category of non-degenerate module over the ring of compactly supported smooth functions on the manifold, as it is a commutative ring, it remembers this ring and hence the manifold. -More simply, the ring of all smooth functions on the manifold can be reconstructed as the automorphism of the identity functor on this category. Moreover, this will also be true for any other subcategory of this that contains the unit module, so for example it also works for the category of vector bundle !<|endoftext|> -TITLE: Given a sequence of reals, we can find a dense sequence avoiding it, but can we find one continuously? -QUESTION [14 upvotes]: Let $S$ be the set of injective sequences in $\mathbb{R}$: -$$S = \{s: \mathbb{N} \rightarrow \mathbb{R}: s(m) \neq s(n) \text{ if }m \neq n\}.$$ -Consider $S$ with the topology of pointwise convergence, and $C(S,S)$ the associated continuous functions on $S$. For any sequence $s$ in $S$, let $\text{ran}(s)$ be the corresponding set of reals. -Is there $f \in C(S,S)$ such that $\text{ran}(f(s))$ is always dense in $\mathbb{R}$ and disjoint from $\text{ran}(s)$? -Without continuity, this would be as simple as listing the intervals with rational endpoints, and choosing one point in each interval minus $\text{ran}(s)$. With continuity, I don't know. -Background: I'd like to show that for any sequence of reals, we can find a dense sequence avoiding it, constructively and without using countable choice. I'd be happy to see an answer on that too. I think the question above, phrased without constructvity, gets at much the same issue. - -REPLY [6 votes]: I think there is not such an $f:S\to S$. Consider the sequence $x^t\in S$ continuously depending on $t\in[0,1]$, such that $x_0^t=-t$ and $x_n^t=1/n $ for all $n\ge1$. Since $f(x^1)$ is dense, for some index, say $17$, we have $-1 -TITLE: Spin bordism group of classifying space $BG$ with a finite Abelian $G$ -QUESTION [5 upvotes]: The spin bordism group for the classifying space $BG$ of group $G$ can be denoted as $\Omega^{Spin}_d(BG)$. -For example, $\Omega^{Spin}_d(pt)$ are computed by Anderson-Brown-Peterson (D. W. Anderson, E. H. Brown, Jr. and F. P. Peterson, The -structure of the Spin cobordism ring, Ann. of Math. 86, 271-298 -(1967).). -Below we denote the finite cyclcic group of order $n$ as $\mathbb{Z}/n\mathbb{Z}=\mathbb{Z}_n$. -If we take $BG=pt$, one get -$\Omega^{Spin}_1(pt)=\mathbb{Z}_2$, $\Omega^{Spin}_2(pt)=\mathbb{Z}_2$, $\Omega^{Spin}_3(pt)=0$, $\Omega^{Spin}_4(pt)=\mathbb{Z}$ -If we take $G=\mathbb{Z}_2$, we get, -$\Omega^{Spin}_1(B\mathbb{Z}_2)=\mathbb{Z}_2^2$, $\Omega^{Spin}_2(B\mathbb{Z}_2)=\mathbb{Z}_2^2$, $\Omega^{Spin}_3(B\mathbb{Z}_2)=\mathbb{Z}_8$, $\Omega^{Spin}_4(B\mathbb{Z}_2)=\mathbb{Z}$ - -Here are my questions, if we take $G=(\mathbb{Z}_2)^2$, or $G=(\mathbb{Z}_2)^3$, $G=(\mathbb{Z}_2)^4$, what are the answers for the following: -$$\Omega^{Spin}_1(B(\mathbb{Z}_2)^2)=?, \Omega^{Spin}_2(B(\mathbb{Z}_2)^2)=?, \Omega^{Spin}_3(B(\mathbb{Z}_2)^2)=?, \Omega^{Spin}_4(B(\mathbb{Z}_2)^2)=?$$ -$$\Omega^{Spin}_1(B(\mathbb{Z}_2)^3)=?, \Omega^{Spin}_2(B(\mathbb{Z}_2)^3)=?, \Omega^{Spin}_3(B(\mathbb{Z}_2)^3)=?, \Omega^{Spin}_4(B(\mathbb{Z}_2)^3)=?$$ -$$\Omega^{Spin}_1(B(\mathbb{Z}_2)^4)=?, \Omega^{Spin}_2(B(\mathbb{Z}_2)^4)=?, \Omega^{Spin}_3(B(\mathbb{Z}_2)^4)=?, \Omega^{Spin}_4(B(\mathbb{Z}_2)^4)=?$$ - -P.S. Part of bordism group data may overlap with the group (co)homology data $H_d[G,\mathbb{Z}]$ or $H_d[G,\mathbb{R}/\mathbb{Z}]$, $H^d[G,\mathbb{Z}]$ or $H^d[G,\mathbb{R}/\mathbb{Z}]$. Fortunately, a Ref here in SUPPLEMENTAL MATERIAL shows the useful data in a table: - -Any partial answers and any Refs are welcome. - -REPLY [8 votes]: The paper of Anderson, Brown and Peterson also shows that the localisation of the spectrum $MSpin$ splits as a wedge of suspensions of the real connective $K$-theory spectrum $kO$ and various closely related spectra. It is also a theorem of Hopkins and Hovey that $KO_*(X)=KO_*\otimes_{MSpin_*}MSpin_*(X)$. This shows that the calculation of the groups $\Omega^{spin}_*(BG)=MSpin_*(BG)$ is very closely related to the calculation of the groups $kO_*(BG)$ and $KO_*(BG)$. Greenlees, Bruner and various collaborators have written extensively about this kind of question. However, the answers are often very complicated.<|endoftext|> -TITLE: Surjectivity of representations in algebraic K-theory -QUESTION [5 upvotes]: Let $G$ be a finite group with finite-dimensional irreducible representations $\rho_i:G\to\mathrm{GL}_{n_i}(k)$ over a field $k$ indexed by $i=1,...,m$. These compose with the canonical map $\mathrm{GL}_{n_i}(k)\to\mathrm{GL}(k)$ to give $\rho_i:G\to\mathrm{GL}(k)$ for $i=1,\cdots,m$. Taking classifying spaces and applying the Quillen plus construction therefore gives maps $\rho_i:BG^+\to B\mathrm{GL}(k)^+$, and hence homomorphisms $\rho_i^n:\pi_n BG^+\to K_n(k)$ for every $1,\cdots,m$, where $K_n(k)$ denotes the $n$th algebraic K-theory group of $k$. Have these maps been studied somewhere? When are these maps surjective? -An easy example is that $\rho_i^{2n}=0$ for $n>0$ if $k=\mathbf{F}_p$. In general, though, I believe studying these maps might be hard, because every path connected space arises as $BG^+$ for some discrete group $G$ (Kan-Thurston). As a special case, what can one say about these maps if $\mathrm{char}(k)=p>0$ and $G$ is a $p$-group? - -REPLY [2 votes]: I don't actually know places in the literature where induced maps from representations have been studied (except from the Brauer lift used in Quillen's computation of K-theory of finite fields). The following are some remarks and observations which partially answer the question. -As a first remark, I assume that the plus-construction applied to $BG$ is the one for the maximal perfect subgroup. If the maximal perfect subgroup is trivial, then $BG\to BG^+$ is a homotopy equivalence and then $\rho^n=0$ for $n>1$. This happens e.g. for abelian groups. It also happens for $p$-groups, since any $p$-group strictly contains its commutator subgroup and therefore the maximal perfect subgroup must be trivial (which answers the last question). -Although the Kan-Thurston theorem allows to write any path-connected space as $BG^+$ for some discrete group, this group will usually be infinite. If you allow infinite groups, then $G=GL(k)$ will give a surjective map trivially. For $G$ finite, there are strong restrictions on the homotopy type of $BG^+$. Some information on the homotopy groups of $BG^+$ for finite groups can be found in - -R. Levi: On finite groups and homotopy theory. Mem. Amer. Math. Soc. 118, 1995. -F.R. Cohen and R. Levi: On the homotopy theory of $p$-completed classifying spaces. In: Group representations: cohomology, group actions and topology. Proc. Sympos. Pure Math. 63, Amer. Math. Soc., 1998. - -The homotopy groups of $BG^+$ for finite perfect groups are torsion, so the maps $\rho^n$ can only be surjective if the K-groups of the field $k$ are finite which will probably be very rare. For $k=\mathbb{F}_q$, one can of course use $GL_n(\mathbb{F}_q)$ to get surjections on K-groups in degrees below $n$ (by the stabilization theorem). -In general, it would seem more appropriate to ask if a particular torsion class lies in the image of a map $\rho^n$ for some finite group. For algebraically closed fields of characteristic $p$, all the torsion classes lie in the image of some representation, this follows from the above observation for $k=\mathbb{F}_q$ and Suslin's rigidity theorem. Maybe the Brauer lift used by Quillen for the computation of K-theory of finite fields allows to show that torsion in the algebraic K-theory of $\mathbb{C}$ can be represented using representations of finite groups.<|endoftext|> -TITLE: How to generate Voronoi diagram with polygons of equal area? -QUESTION [8 upvotes]: I would like to generate some random set of points so that their Voronoi diagram consist of equal-area polygons. Is it possible to impose some constraints on the points in order to have the same areas (or at least almost the same) of polygons? Regular polygons are not interesting. - -REPLY [5 votes]: I think that this paper addresses your question, if you stipulate -a constant density distribution: - -Balzer, Michael, Thomas Schlömer, and Oliver Deussen. Capacity-constrained point distributions: A variant of Lloyd's method. Vol. 28. No. 3. ACM SIGGRAPH, 2009. - (ACM link). - - - - - -"[...] This constraint enforces that each point in a distribution has the same capacity. Intuitively, the capacity can be understood as the area of the point’s corresponding Voronoi region weighted with the given density function. By demanding that each point’s capacity is the same, we ensure that each point obtains equal importance in the resulting distribution." - -I believe this can be viewed as a form of an optimal transport problem.<|endoftext|> -TITLE: Intuition behind results in Mumford's "Lectures on curves on an algebraic surface", I -QUESTION [6 upvotes]: These are some questions concerning Mumford's "Lectures on curves on an algebraic surface". -We concern ourselves with questions of the Picard variety $P$, and its dimension, of a complete nonsingular surface $F$ over an algebraically closed ground field $k$ of arbitrary characteristic. Let $\mathfrak{o}$ be the sheaf of local rings on $F$, $\{C_\alpha: \alpha \in S\}$ a family of curves on $F$ such that $C_0 = C$ is nonsingular and $N$ the sheaf of sections of the normal bundle to $C$ on $F$. Then the equality (A): $\dim P = h^1(\mathfrak{o})$ is equivalent to (B): $\{\text{Tangent space to }S \text{ at }\alpha = 0\} \overset{\rho}{\to} H^0(N)$ is surjective for suitable $\{C_\alpha : \alpha \in S\}$. -I was wondering if someone could give me their explanation/intuitions behind the following, as I am finding the book to be quite terse. - -The algebraic solution of problem (A) for characteristic $0$, following Grothendieck, using a theorem of Cartier on an algebraic group scheme. -Problem (A) in characteristic $p$, i.e. $\dim P$ may not be given in general by $h^1(\mathfrak{o})$ but that the tangent space to $P$ (at a point) corresponds to the subspace of $H^1(\mathfrak{o})$ annihilated by the Bockstein operators. - -Thanks in advance! - -REPLY [7 votes]: I think I can provide some intuition for (A), both in characteristic $0$ and $p > 0$. What follows below is more or less a proof, but with a lot of omissions (and hopefully not too many lies...). -I believe that everything I state works for all (geometrically) integral projective $k$-schemes. For simplicity, let's assume $k = \bar k$. -Reference. A great reference for the Grothendieck-style approach to this theory is FGA Explained, chapter 9, or Néron Models, chapter 8. (For general discussion of deformation theory and Hilbert and Quot schemes, I would recommend FGA Explained, whereas for the actual construction of the Picard scheme maybe Néron Models is a bit more readable.) - -For a $k$-scheme $X$, we can view the tangent space at a closed point $x \to X$ as the extensions -$$x \to \operatorname{Spec} k[\varepsilon]/\varepsilon^2 \to X.$$ -That is, it is the preimage of $x \in X(k)$ under the map $\phi\colon X(k[\varepsilon]/\varepsilon^2) \to X(k)$. If $X$ is a group scheme, then $\phi$ is a group homomorphism, and the tangent space at the identity corresponds to the kernel of $\phi$. -Now apply this to the Picard group scheme $\operatorname{\underline{Pic}}_X$. This is usually defined as the fppf sheafification of the functor $Y \mapsto \operatorname{Pic}(X\times Y)/\operatorname{Pic}(Y)$, but in the projective case it suffices to take the étale sheafification; the big difficult theorem is that this is representable by a scheme. (I believe that the sheafification takes care of dividing out by $\operatorname{Pic}(Y)$, so you could omit that step if you want to.) -Thus, we have to compute line bundles on $X_\varepsilon := X \times \operatorname{Spec} k[\varepsilon]/\varepsilon^2$ that are trivial on the central fibre $X$. (Since $k[\varepsilon]/\varepsilon^2$ has no étale covers, we don't need to worry about the sheafification business.) On $X_\varepsilon$, we have a short exact sequence -$$0 \to \mathcal O_X \to \mathcal O_{X_\varepsilon} \to \mathcal O_X \to 0,$$ -coming from $0 \to k \to k[\varepsilon]/\varepsilon^2 \to k \to 0$. Tensoring with a line bundle $\mathcal L$ trivial on $X$ gives -$$0 \to \mathcal O_X \to \mathcal L \to \mathcal O_X \to 0.$$ -Conversely, any such extension can be given the structure of line bundle on $X_\varepsilon$ by having $\varepsilon$ act by the composition $\mathcal L \to \mathcal O_X \to \mathcal L$. Hence, we are classifying extensions of $\mathcal O_X$ by itself. These are given by -$$\operatorname{Ext}^1_X(\mathcal O_X,\mathcal O_X) = H^1(X,\mathcal O_X).$$ -Thus, we have proven: - -Theorem. The tangent space $T_0 \operatorname{\underline{Pic}}_X$ equals $H^1(X,\mathcal O_X)$. $\square$ - -In characteristic $0$, any finite type group scheme is smooth, so we get the equality of dimensions. In characteristic $p > 0$, there are non-smooth group schemes, the simplest example being -$$\mu_p = \operatorname{Spec} k[x]/(x^p-1) \subseteq \mathbb G_m.$$ -In those cases, you actually get the wrong dimension: -$$\dim H^1(X,\mathcal O_X) \geq \dim \operatorname{\underline{Pic}}_X,$$ -with strict inequality if $\operatorname{\underline{Pic}}_X$ is singular.<|endoftext|> -TITLE: a new representation for Eulerian numbers? -QUESTION [7 upvotes]: The Eulerian numbers enjoy many different presentations among which I write the two-variable recursive definition: $A(n,0)=1$ and $A(n,k)=0$ for $k<0$ so that -$$A(n,k)=(k+1)A(n-1,k)+(n-k)A(n-1,k-1).$$ -However, my curiosity is regarding a certain "vanishing-variables" formulation (summation) over the symmetry (permutation) groups $\mathfrak{S}_n$: - -QUESTION. For $0\leq k\leq n$ and a set of variables $x_1, x_2, \dots, x_{n+1}$, experiments suggest - $$A(n,k)=\sum_{\pi\in\mathfrak{S}_{n+1}}(x_{\pi(1)}+\cdots+x_{\pi(k+1)})^n\prod_{i=1}^n\frac1{x_{\pi(i)}-x_{\pi(i+1)}}.$$ - It seems correct to me. Proof or disproof? I am still hoping and waiting. - -REPLY [4 votes]: Quiet a short algebraic proof may go as follows (I cite my own manuscript). For a polynomial $f(x_1,\dots,x_{n+1})$ of degree at most $n$ we define a linear operator $$\Phi[f]=\text{Sym}\, \frac{f(x_1,\dots,x_{n+1})}{(x_1-x_2)(x_2-x_3)\dots (x_n-x_{n+1})},$$ -where $\text{Sym}\, g(x_1,\dots,x_{n+1})=\sum g(x_{\pi_1},\dots,x_{\pi_{n+1}})$, the summation is over all $(n+1)!$ permutations of the variables. Note that $\Phi[f]$ is actually a polynomial of degree at most 0 (all factors in the denominator are cancelled, see Ilya's comment), in other words, it is a scalar, and the linear operator $\Phi$ is a linear functional. -Several observations about $\Phi$. -1) If $f$ is divisible by $y:=x_1+\dots+x_{n+1}$, then $\Phi[f]=y\Phi[f/y]=0$. (The same holds for any symmetric factor $y$.) -2) Assume that $f=(x_{k}-x_{k+1})h(x_1,\dots,x_k)g(x_{k+1},\dots,x_{n+1})$. Then $$\Phi[f]=\begin{cases}\binom{n+1}k\Phi_k[h]\Phi_{n+1-k}[g],\,\text{if}\deg h=k-1\,\text{and} \deg g=n-k\\ -0,\,\text{otherwise}\end{cases}.$$ -3) If $f$ is divisible by $(x_1+\dots+x_k)(x_{k}-x_{k+1})$, then $\Phi[f]=0$. Indeed, it suffices to consider the polynomials $f=(x_1+\dots+x_k)(x_{k+1}-x_k)h(x_1,\dots,x_k)g(x_{k+1},\dots,x_{n+1})$. For them the claim follows from 1) and 2). -Before proving your Euler numbers claim we prove that $c_k:=\Phi[x_k^n]=(-1)^{k-1}\binom{n}{k-1}$. Induction in $n$. Base is straightforward. Induction step. We have $c_0+\dots+c_n=\Phi(x_1^n+\dots+x_{n+1}^n)=0$, by observation in 1). Next, -$$ -c_k-c_{k+1}=\Phi((x_k-x_{k+1})(x_k^{n-1}+\dots+x_{k+1}^{n-1})= -(-1)^{k-1}\binom{n+1}k -$$ -by 2) and by the induction assumption. We obtain $n+1$ linear relations on $c_1,\dots,c_{n+1}$ which determine -them uniquely and also the numbers $c_i=(-1)^{i-1}\binom{n}{i-1}$ satisfy these relations. This finishes the induction step. -Now we want to calculate $\Phi[(x_1+\dots+x_{k+1})^n]$. By 3) we have -$$ -\Phi[(x_1+\dots+x_{k+1})^n-(x_1+\dots+x_{k}+x_k)^n+x_k^n-x_{k+1}^n]=0, -$$ -thus $$\Phi[(x_1+\dots+x_{k+1})^n-(x_1+\dots+x_{k}+x_k)^n]=(-1)^{k}\binom{n+1}k.$$ -Next, $$\Phi[(x_1+\dots+x_{k-1}+2x_k)^n-(x_1+\dots+x_{k-2}+3x_{k-1})^n]=(-1)^{k-1}2^n\binom{n+1}{k-1}.$$ -Proceeding in this way and summing up we get the standard formula for Eulerian number (1) on this MathWorld page.<|endoftext|> -TITLE: Getting PFA + GCH above $\omega$ -QUESTION [7 upvotes]: The Proper Forcing Axiom kills CH in a particularly specific way: it implies that $2^{\aleph_0}=\aleph_2$. However, its impact on the continuum function above $\aleph_0$ is much less clear. It is known, for example, that it implies the Singular Cardinal Hypothesis, SCH. -In this paper (see page 2), Aspero writes that Magidor showed the following: - -Any model of ZFC+PFA has a forcing extension satisfying ZFC+PFA+"$2^\kappa=\kappa^+$ for all uncountable $\kappa$". - -However, he does not give a reference; the bibliography mentions a set of (unpublished) lectures by Magidor. Googling around, I haven't found anything more specific. -I was wondering whether a proof of this exists in the literature, or - failing that - if someone could sketch an outline of it. - -REPLY [7 votes]: In Koenig and Yoshinobu's paper, they prove (Theorem 6.1) that $\sf PFA$ is preserved under $\omega_2$-closed forcings, and this should give the wanted result, as described by Joel. - -Bernhard König and Yasuo Yoshinobu, Fragments of Martin’s maximum in generic extensions, MLQ Math. Log. Q. 50 (2004), no. 3, 297--302. MR 2050172<|endoftext|> -TITLE: Teaching polarisation formula -QUESTION [13 upvotes]: When teaching about Hilbert spaces, one begins with a polarisation formula, which allows us to reconstruct the scalar product from the norm: -$$\langle u,v\rangle=\frac14(\|u+v\|^2-\|u-v\|^2+\imath\|u+\imath v\|^2-\imath\|u-\imath v\|^2).$$ -Is there a good reason to choose this formula instead of the more symmetric one -$$\langle u,v\rangle=\frac1{2\pi}\int_0^{2\pi}e^{\imath\theta}\|u+e^{\imath\theta}v\|^2d\theta,$$ -or the shortest one (here $\jmath=e^{2\imath\pi/3}$) -$$\langle u,v\rangle=\frac13(\|u+v\|^2+\jmath\|u+jv\|^2+\bar\jmath\|u+\bar\jmath v\|^2) \qquad?$$ - -REPLY [18 votes]: The polarization formula carries over to algebraic settings not directly involving real and complex numbers. Let $L/K$ be a quadratic Galois extension, where the nontrivial element of ${\rm Gal}(L/K)$ is denoted with an overline: $\overline{\alpha}$ for $\alpha \in L$. Let $V$ be a finite-dimensional $L$-vector space [edit: as nfdc23 points out, finite-dimensionality is not needed] and $B \colon V \times V \rightarrow L$ be a Hermitian form relative to the extension $L/K$: it is $K$-bilinear, $B(cv,w) = cB(v,w)$ and $B(v,cw) = \overline{c}B(v,w)$ for $c\in L$, and $B(w,v) = \overline{B(v,w)}$. Set $Q \colon V \rightarrow L$ by $Q(v) = B(v,v)$. Can we reconstruct the two-variable function $B$ from the single-variable function $Q$? -Since -$$ -Q(v+w) = Q(v) +Q(w) + B(v,w) + B(w,v), -$$ -replacing $w$ with $-w$ gives -$$ -Q(v-w) = Q(v) +Q(w) - B(v,w) -B(w,v), -$$ -so subtracting and dividing by $2$ (if we are not in characteristic $2$) gives us -$$ -\frac{Q(v+w)-Q(v-w)}{2} = B(v,w) + B(w,v). -$$ -To get a formula having only $B(v,w)$ on the right, pick $c \in L$ such that $\overline{c} \not= c$. Then -$$ -Q(v+cw) = Q(v) +c\overline{c}Q(w) + \overline{c}B(v,w) + cB(w,v), -$$ -and -$$ -Q(v+\overline{c}w) = Q(v) +\overline{c}cQ(w) + cB(v,w) + \overline{c}B(w,v) -$$ -so -$$ -Q(v+cw) - Q(v+\overline{c}w) = (\overline{c}-c)(B(v,w) - B(w,v)) -$$ -and thus -$$ -\frac{Q(v+cw) - Q(v+\overline{c}w)}{\overline{c}-c} = B(v,w) - B(w,v). -$$ -Since we have formulas for $B(v,w)+B(w,v)$ and $B(v,w)-B(w,v)$, by averaging (if we are not in characteristic $2$), -$$ -\frac{Q(v+w) - Q(v-w)}{4} + -\frac{Q(v+cw) - Q(v+\overline{c}w)}{2(\overline{c}-c)} = B(v,w). -$$ -When $L = \mathbf C$, $K = \mathbf R$, $B(v,w) = \langle v,w\rangle$, $Q(v) = ||v||^2$, and $c = i$ this recovers the classical polarization formula. If in the same setting we take $c = j = e^{2\pi i/3}$, then we get a polarization formula that is different from the one you wrote down with $j$, but it fits into the general pattern described above. Since I showed how the usual polarization formula extends to the general case outside of characteristic $2$ (certainly an integration formula does not), that classical formula is not really specific to the choice $c = i$, but your version using $j$ seems specific to the choice $c=j$ since you are dividing through by $3$. I am not persuaded that a formula using $3$ terms instead of $4$ terms is genuinely simpler: the $4$-term polarization formula is essentially the result of averaging a few times. (Most people, whether by habit or otherwise, would prefer to think of $\mathbf C$ as $\mathbf R + \mathbf R{i}$ rather than as $\mathbf R + \mathbf R{j}$ for both geometric and algebraic reasons, e.g., $\overline{a+bj} = a-b-bj$ for real $a$ and $b$.) -What happens if we are using a Hermitian form in characteristic $2$? Division by $2$ in the classical polarization formula or its generalization above breaks down if we try to reconstruct $B$ from $Q$ in characteristic $2$. We can show $Q$ determines $B$ by the following argument that is valid in all characteristics: for $v$ and $w$ in $V$, and $c \in L$, we have -\begin{eqnarray*} -Q(cv+w) & = & c\overline{c}Q(v) + Q(w) + B(cv,w) + B(w,cv) \\ -& = & c\overline{c}Q(v) + Q(w) + cB(v,w) + \overline{B(cv,w)} \\ -& = & c\overline{c}Q(v) + Q(w) + {\rm Tr}_{L/K}(cB(v,w)). -\end{eqnarray*} -Therefore, when $v$ and $w$ are fixed in $V$, the function $f_{v,w} \colon L \rightarrow K$ given by -$f_{v,w}(c) = Q(cv+w) - c\overline{c}Q(v) - Q(w)$ is completely determined by $Q$ and it is also $K$-linear since it equals ${\rm Tr}_{L/K}(cB(v,w))$. For a quadratic Galois extension $L/K$, each $K$-linear mapping $L \rightarrow K$ looks like $f(x) = {\rm Tr}_{L/K}(xy)$ for a unique $y \in L$ (this could be shown directly, or it is a consequence of non-degeneracy of the trace-pairing $L \times L \rightarrow K$, where $\langle x,y\rangle \mapsto {\rm Tr}_{L/K}(xy)$). Thus $B(v,w)$ is the unique number $y$ in $L$ such that $Q(cv+w) - c\overline{c}Q(v) - Q(w) = {\rm Tr}_{L/K}(cy)$ for all $c \in L$. There is no need for any polarization formula. -[Edit: I realized after writing this up that I wrote something essentially like this a few years ago on math.stackexchange: https://math.stackexchange.com/questions/425173/derivation-of-the-polarization-identities]<|endoftext|> -TITLE: Do topological spaces form a full subcategory of spectra? -QUESTION [6 upvotes]: Let $\Sigma^\infty: Top_* \to Spectra$ be a functor sending a pointed topological space $X$ to its suspension spectrum, that is $(\Sigma^\infty X)_n=\Sigma^nX$ with isomorphisms $\Sigma(\Sigma^\infty X)_n \to (\Sigma^\infty X)_{n+1}$. Is it true that this map make $Top_*$ a full subcategory of $Spectra$, that is -$$Hom_{Top_*}(X, Y)=Hom_{Spectra}(\Sigma^\infty X, \Sigma^\infty Y)?$$ -Maybe, it is true in homotopy categories: $Hom_{hTop_*}(X, Y)=Hom_{hSpectra}(\Sigma^\infty X, \Sigma^\infty Y)$? - -REPLY [13 votes]: As noted in the comments, this is most definitely false in homotopy categories: the set $[S^1, S^0]$ of homotopy classes of based maps is trivial, but the set $[\Sigma^\infty S^1, \Sigma^\infty S^0]$ of maps in the homotopy category of spectra is $\Bbb Z/2$. -The statement about the suspension spectrum functor being fully faithful may or may not be true in the category of spectra and it strongly depends on which model you use for spectra. - -In the category of "sequential" spectra where a map of spectra is a collection of maps $X_n \to Y_n$ commuting with the structure maps (sometimes called Bousfield-Friedlander spectra), then the answer is yes: this is a full embedding. I suspect this is the definition that you are using. -In the category of spectra described in Adams' Stable homotopy and generalized homology, he distinguishes the above (a "function" of spectra) from a "map" of spectra. Maps of CW-complexes do not embed fully faithfully into this definition of maps of spectra. -In the category of symmetric spectra, this is also a full embedding. -In the category of orthogonal spectra, this is also a full embedding. -I remember seeing a (draft?) document at some point in the past proving that the suspension spectrum functor produces a fully faithful embedding into Elmendorf-Kriz-Mandell-May's category of S-modules. However, I can't track this down, and it's a bit unexpected (I believe the result is false for the closely related category of spectra developed by Lewis-May-Steinberger).<|endoftext|> -TITLE: Modal vs First-Order Logic on finite models -QUESTION [6 upvotes]: It is known that Modal Logic can be interpreted in First-Order logic via Standard translation. However, this translation needs a unary predicate for every propositional variable. It is also known that without these propositional variables certain formulas in ML can't be expressed in FOL. For example: - -Löb's formula: $\Box (\Box p \to p) \to \Box p$ -McKinsey's formula: $\Box\Diamond p \to \Diamond \Box p$ -Grzegorczyk's formula: $\Box((p\to \Box p) \to p) \to p$ - -In the proofs of this, one usually constructs an infinite counterexample to some property of FOL (for example to Löwenheim-Skolem property in case of McKinsey's formula) [ML book]. -My question: Is there a formula $\varphi$ in ML which is not expressible in FOL and we can prove this by tools available in finite model theory only? -Further question: Could $\varphi$ have some kind of graph theoretic meaning? -[ML book] Blackburn, de Rijke, and Venema: Modal logic - -REPLY [9 votes]: On p. 30–31 of Van Benthem’s Notes on modal definability (Notre Dame Journal of Formal Logic 30 (1988), #1, pp. 20–35), you can find a (brief!) sketch of a proof that the modal formula -$$(\Diamond\Diamond\top\land\Box(\Diamond\top\to\Diamond p))\to\Diamond(\Diamond\top\land\Box p)$$ -is not FO-definable on finite frames. The reason is, essentially, that it defines parity on a certain family of frames, which is impossible in FO by an Ehrenfeucht–Fraïssé argument. I would consider this to use only tools of finite model theory. -In fact, the same argument applies to the McKinsey formula $\Box\Diamond p\to\Diamond\Box p$ itself, if we take the frames in the proof as non-transitive, and make the leaf nodes reflexive. That is, for any $n\in\mathbb N$, let $F_n$ be the Kripke frame with nodes $\{r,u_0,\dots,u_{n-1},v_0,\dots,v_{n-1}\}$ where $r$ sees each $u_i$; $u_i$ sees $v_i$ and $v_{(i+1)\bmod n}$; and $v_i$ sees itself. Then $F_n$ validates the McKinsey formula iff $n$ is odd, but for any given FO sentence $\phi$, all $F_n$ with sufficiently large $n$ agree on the truth of $\phi$ by an Ehrenfeucht–Fraïssé argument (or, using the fact that $F_n$ is interpretable in the undirected $n$-cycle). - -Finite model theory is closely connected to complexity theory, and one can give an alternative argument that the McKinsey formula is not FO on finite frames by noting that the set of frames it defines is coNP-complete under $\mathrm{AC}^0$ reductions (or even DLOGTIME reductions), whereas all FO properties are decidable in $\mathrm{AC}^0$. (Further assuming $\mathrm{P\ne NP}$, this implies that the McKinsey formula is not expressible even in stronger logics such as IFP.) -This can be conveniently shown by reduction from the NP-complete problem Mon-NAE-SAT (monotone not-all-equal SAT). In a combinatorial formulation, this problem is the following. -Input: A set $S=\{C_1,\dots,C_m\}$ of subsets $C_i\subseteq V=\{1,\dots,n\}$. -Output: Does there exist a set $P\subseteq V$ that splits all sets in $S$, i.e., $C_i\cap P\ne\varnothing$ and $C_i\smallsetminus P\ne\varnothing$ for each $i=1,\dots,m$? -We may assume all $C_i$ to be nonempty. -Now, the reduction goes as follows. We construct a frame $F_S=(W_S,R_S)$, where $W_S=\{r,u_1,\dots,u_m,v_1,\dots,v_n\}$, with the accessibility relation $R_S$ defined so that $r$ is related to each $u_i$; $u_i$ is related to those $v_j$ such that $j\in C_i$; and each $v_j$ is related to itself. -The McKinsey frame condition is: for all $w\in W$ and all $P\subseteq W$, there is a successor $u$ of $w$ such that the successors of $u$ are either all in $P$, or all in $W\smallsetminus P$. It is easy to see that this condition holds in all nodes of $F_S$ except possibly in $r$, and it fails in $r$ iff the original NAE-SAT problem is solvable. -Notice also that if $S=\{\{1,2\},\{2,3\},\dots,\{n-1,n\},\{n,1\}\}$, then $F_S$ is exactly the frame $F_n$ we defined earlier.<|endoftext|> -TITLE: Would you resubmit a research paper after it has been superseded by another as yet unpublished paper? -QUESTION [25 upvotes]: Hope that the following soft question is still appropriate on MathOverflow. I was wondering if there is any communal protocol or etiquette with regard to the resubmission of a research paper after it has been superseded by another as yet unpublished paper. -Here's the situation in detail. Suppose that a paper of yours, call it paper A, regarding the existence of some mathematical object foobar has been rejected by some journal X. Before you received the news of rejection of paper A by journal X, you obtained bounds on the complexity of foobar and submitted these complexity estimates as paper B to journal Y. Given that paper B is still pending the refereeing process, would you resubmit paper A to another journal X'? Is it unethical to do so? -Of course, one question is what is the worth of paper A, given that it has been superseded by paper B. One possible factor to take into consideration is that paper B is substantially longer than paper A (in my case paper B about 40 pp. and paper A is about 20 pp.). There may be readers interested in the existence argument of paper A without bothering about the complexity estimates in paper B. - -REPLY [12 votes]: Yes, you should resubmit. Research is a journey, and having both papers published creates a record which can help others who are tracing the steps of your journey. - -REPLY [6 votes]: There is reasonable chance that these papers will be refereed by the same person. Unless both papers are a pleasure to read, you might be unleashing a justified fury of someone who had to do the same hard (and unpaid) job twice. -On the other hand, I think that the good way to deal with this is to make sure that the overlap of A and B is minimal, and publish A as a preprint (arXiV springs to mind as the right place; I would say, publish B as a preprint, too). In case it's too late, and B contains most of A, I would expect the situation in the 1st paragraph, which would result in both A and B being tainted in editor's views.<|endoftext|> -TITLE: When do 27 lines lie on a cubic surface? -QUESTION [35 upvotes]: Consider $27$ (pairwise distinct!) lines in $\mathbb{P}^3$ whose intersection graph is that expected¹ of the $27$ lines on a smooth cubic surface. Question: Is there a simple necessary and sufficient condition for these $27$ lines to indeed lie on a smooth cubic surface? -For a long time I thought this was always the case, but there is at least one obvious necessary condition²: - -(T) Whenever three lines pairwise intersect (i.e., are pairwise coplanar), all three lie on a common plane. - -(Because the plane through two mutually intersecting lines on a cubic surface cuts the surface as the union of three distinct lines. There are $45$ such tritangent planes on the cubic surface.) -Is this condition (T) sufficient? -Further comment and bonus question: The locally closed subvariety of $\mathrm{Gr}(2,4)^{27}$ (or $\mathrm{Sym}^{27}(\mathrm{Gr}(2,4))$) consisting of configurations of $27$ lines satisfying the incidence conditions (made explicit in note (1) below) is not irreducible (because of the first sentence of note (2) below). What are its irreducible components? -Notes: - -I.e., we can label the lines as $a_1,\ldots,a_6$, $b_1,\ldots,b_6$ and $c_{12},\ldots,c_{56}$ (the latter indexed by the $15$ unordered pairs in $\{1,\ldots,6\}$) such that the $a_i$ mutually don't intersect, the $b_i$ mutually don't intersect, each $a_i$ intersects each $b_j$ except exactly when $i=j$, the $c_{ij}$ intersect exactly when their index pairs are disjoint, and $c_{ij}$ intersects exactly $a_i,a_j,b_i,b_j$ among the $a_k$ and $b_k$. Equivalently, the intersection graph is the complement of the vertex graph of the Gosset $2_{21}$ polytope. Equivalently, the faithful transitive action of $W(E_6)$ on $27$ elements where the incidence relation is given by the orbit on the unordered pairs such that every line is incident to exactly $10$ others. -Three distinct mutually intersecting lines in $\mathbb{P}^3$ either lie on a common plane or, dually, meet at a common point. (Apologies for pointing out something so obvious, but I'm sure I'm not the only one who might have missed this trivial fact.) If we take the $27$ lines on a smooth cubic surface $X$ and dualize (i.e., replace them by their polars w.r.t. some fixed nondegenerate quadric), we get another configuration of $27$ lines (lying on the projective dual $X^\vee$ to the cubic surface) which have $45$ points of intersection of triples of lines, but in general (unless $X$ had some Eckardt points) no planes in which three lines lie; so this configuration does not lie on a cubic surface (and indeed, $X^\vee$ is not a cubic surface). - -REPLY [15 votes]: It turns out that condition (T) is, indeed, sufficient for the $27$ lines (distinct and intersecting as expected) to lie on a cubic surface. -To see this, consider the lines $a_1,a_2,a_3,a_4,a_5$ and $b_6$, where the labeling is as in note (2) of the question: $a_1$ through $a_5$ are pairwise skew, and $b_6$ intersects all of them. Choose $4$ distinct points on $b_6$, and $3$ distinct points on each of $a_1$ through $a_5$ also distinct from the intersection point with $b_6$: this makes $4+5\times3=19$ points in total; since there are $20$ coefficients in a cubic form on $4$ variables, there exists a cubic surface $S$ passing through these $19$ points. Since $S$ contains four distinct points on $b_6$, it contains $b_6$ entirely, and since it contains four points (viz., the intersection with $b_6$ and the three additional chosen points) on each of $a_1$ through $a_5$, it contains these also. -Now $b_1$ intersects the four lines $a_2$ through $a_5$ in four distinct points since $a_2$ through $a_5$ are pairwise skew; and since these four points lie on $S$, it follows that $b_1$ lies entirely on $S$; similarly, $b_2$ through $b_5$, and finally $a_6$ (which intersects $b_1$ through $b_5$ in distinct points), all lie on $S$. So $S$ contains the "double-six" $\{a_1,\ldots,a_6,b_1,\ldots,b_6\}$. -So far, property (T) has not been used, only the incidence relation. Now it remains to see that the $c_{ij}$ lie on $S$. Consider the intersection line $\ell$ of the planes $a_i\vee b_j$ (spanned by $a_i$ and $b_j$) and $a_j\vee b_i$ (spanned by $a_j$ and $b_i$; these planes are well-defined since $a_i$ meets $b_j$ and $a_j$ meets $b_i$, and they are distinct since $a_i$ and $a_j$ are skew): this line $\ell$ must be equal to the $c_{ij}$ of the given configuration, because $c_{ij}$ intersects both $a_i$ and $b_j$ so property (T) implies that it lies in the plane $a_i\vee b_j$, and similarly it lies in the plane $a_j\vee b_i$. But $\ell$ also lies on $S$ for similar reasons¹, in other words, $c_{ij}$ lies on $S$, and all the given lines lie on $S$. -Finally, $S$ must be smooth because it contains the configuration of $27$ lines expected of a smooth cubic surface (it is easy to rule out the case where $S$ is a cubic cone, a reducible surface or a scroll by considering the intersecting and skew lines in the double-six; and every configuration where $S$ has double point singularities has fewer than $27$ lines). - -Let me be very precise here, because at this stage we don't know whether $S$ is smooth (so we can't invoke (T) directly on $S$). We have four distinct lines $a_i,a_j,b_i,b_j$ on $S$ such that $a_i$ meets $b_j$ and $a_j$ meets $b_i$ and all other pairs are skew. Call $\pi := a_i\vee b_j$ and $\pi' := a_j \vee b_i$ the planes generated by the two pairs of concurrent lines, and $\ell := \pi\wedge\pi'$ their intersection. We want to show that $\ell$ lies on the surface $S$. If $\pi$ or $\pi'$ is contained in $S$ (reducible) then the conclusion is trivial, so we can assume this is not the case. So the (schematic) intersection of $S$ with the plane $\pi$ is a cubic curve containing two distinct lines ($a_i$ and $b_j$), so it is the union of three lines: $a_i$, $b_j$ and a third line $m$ (a priori possibly equal to one of the former). Consider the intersection point of $a_j$ and $\pi$ (which is well-defined since $a_j$ is skew with $a_i$ so does not lie in $\pi$): it lies on $\ell$ because it is on both $\pi$ and $\pi'$; and it must also lie on $m$ since it is on $\pi$ but neither on $a_i$ nor on $b_j$ (as the two are skew with $a_j$); similarly, the intersection point $b_i\wedge\pi$ is well-defined and lies on both $\ell$ and $m$; so $\ell=m$ lies on $S$ (and we are finished) unless perhaps the two intersections considered are equal, i.e., $a_j,b_i,\pi$ concur at a point $P$, necessarily on $m$. Assume the latter case: $S$ must be singular at $P$ because the line $m$ through $P$ does not lie on the plane $\pi'$ generated by two lines ($a_j,b_i$) through $P$ (i.e., we have three non-coplanar tangent directions at $P$). Now symmetrically, if we call $m'$ the third line of the intersection of $S$ with $\pi'$ (besides $a_j$ and $b_i$), we are done unless $a_i,b_j,\pi'$ concur at a point $P'$, necessarily on $m'$ and necessarily singular on $S$. The points $P$ and $P'$ are distinct because $a_i$ and $a_j$ are skew; and they are on $\ell$ because they are on $\pi$ and $\pi'$; and the line joining two singular points on a cubic surface lies on the surface, so $\ell = P\vee P'$ lies on $S$ in any case. (Phew!) - -What I still don't know is whether there are configurations of $27$ distinct lines with the expected incidence relations and which satisfy neither condition (T) nor its dual (viz., whenever three lines pairwise meet, all three meet at a common point), and in particular, what are the irreducible components of the space of configurations. (I also don't know if there is a way to substantially simplify the tedious argument given in note (1) above.)<|endoftext|> -TITLE: Smooth circle action, $\chi(M^{S^1}) = \chi(M)$ -QUESTION [6 upvotes]: I need a reference for the following result (which I can prove myself, but my proof is rather ugly and I would prefer to just cite the statement instead or re-proving it): -Let $M$ be a closed smooth manifold and let $S^1$ act smoothly on $M$. Then the set of fixed points of the action, $M^{S^1}$, is again a (not necessarily connected) smooth manifold and $\chi(M^{S^1}) = \chi(M)$. - -REPLY [7 votes]: For the Euler characteristic there is the following. -Consider a tubular neighborhood $T$ of the fixed set (submanifold) $M^{{\mathbb S}^1}$. By Mayer-Vietoris sequence, relatively to the open covering $\{U = M-{\mathbb S}^1, T\}$ of -$M$, we get -$$\chi (M) = \chi (U) + \chi(T) - \chi(U\cap T).$$ -But, $\chi (U) =\chi(U\cap T)=0$ since the circle action is free on $U$. Finally, since $T$ equivariantly retracts on $M^{{\mathbb S}^1}$ then -$$\chi (M) = \chi (T) = \chi(M^{{\mathbb S}^1}). -$$<|endoftext|> -TITLE: What structure do you get if you adjoint a root of $z \bar{z} = -1$ to the complex numbers? -QUESTION [6 upvotes]: A pop-up is informing me that my question is likely to be closed. -Still, recall that the complex numbers $\mathbb{C}$ was conceived by trying to adjoint a root of the equation $x^2 = - 1$ to the field of real numbers $\mathbb{R}$, or so we are told. There are two such roots, now known as $i$ and $-i$, and the conjugation involution $z \mapsto \bar{z}$ is the field automorphism of $\mathbb{C}$ which fixes $\mathbb{R}$. -Suppose we entertain the fantasy to adjointing a root of the analogous equation $z \bar{z} = |z|^2 = -1$ to the complex numbers $\mathbb{C}$. What "structure" will we get? Now, given such a root, call it $f$ for the fantasy unit, then $z= e^{i\theta}\cdot f $ would also be a root of $z \bar{z} = 1$ for all $e^{i\theta}$ in the circle group $U(1)$. - -REPLY [4 votes]: The first thing to realize is that the question must be properly interpreted, which probably can be done in several ways. -It seems reasonable to interpret it as follows: - -Find a ring $R$ with an injective homomorphism $\mathbb{C} \rightarrow R$, such that: - -there is an involution $\sigma: R \rightarrow R$, i.e. an endomorphism of order $2$, such that the restriction of $\sigma$ to $\mathbb{C}$ is the complex conjugation; -there is an element $f \in R$, such that $f \sigma(f) = -1$; -$R$ is generated by $\mathbb{C}$ and $f$. - - -The last condition can possibly be replace by: $R$ is generated by $\mathbb{C}$, $f$ and $\sigma(f)$. -But you probably also want to put other conditions on $R$. -For example, if you require $R$ to be commutative, then $R$ can be the ring $\mathbb{C} \oplus \mathbb{C}$, with $\mathbb{C}$ embedded diagonally, $\sigma(z, w)=(\overline{w}, \overline{z})$, and the element $f=(1, -1)$. -If you don't require $R$ to be commutative, then it can also be the biquaternion $R = \mathbb{C} \oplus \mathbb{C} i \oplus \mathbb{C} j \oplus \mathbb{C} k$. -For the moment I don't have an example where $R$ is still a division algebra and $f$ is algebraic.<|endoftext|> -TITLE: Fiberwise compactification of a LG model -QUESTION [7 upvotes]: It is believed that a mirror of $\mathbb{CP}^2$ is a fiberwise compactification of the family $$W \colon (\mathbb{C}^\times)^2 \rightarrow \mathbb{C}, \quad (x,y)\mapsto x+y+\frac{1}{xy}.$$ Is it obtained by considering a pencil $\mathbb{CP}^2 \rightarrow \mathbb{CP}^1$ -$$ -W_1:[X:Y:Z]\mapsto[X^2Y+XY^2+Z^3:XYZ], -$$ -blowing-up the base locus of $W_1$ and removing the fiber of $0$ and $\infty$? Is it true that a fiberwise compactification is always obtained in this way and it is canonical in a certain sense? - -REPLY [3 votes]: For the mirror of $\mathbb{CP}^2$, a smooth fiber of $W$ is a 3-punctured torus, and you can compactify the fibers by filling in 1, 2 or 3 punctures. This corresponds, in $\mathbb{CP}^2$, to different divisors. -The original LG model you wrote down is a toric mirror of $\mathbb{CP}^2$, which means it's mirror to the toric divisor $D_0$. So after removing it, you get $(\mathbb{C}^\ast)^2$. In this case, mirror symmetry gives you an equivalence -$D^\pi\mathscr{W}(T^\ast T^2)\cong D^b\mathit{Coh}((\mathbb{C}^\ast)^2)$, -which can be obtained from the triangulated equivalence -$D\mathscr{F}(T^\ast T^2,W)\cong D^b\mathit{Coh}(\mathbb{CP}^2)$ -by applying a categorical version of localization, where the left-hand side is the Fukaya-Seidel category. -Filling in one of the punctures compactifies $(\mathbb{C}^\ast)^2$ to the affine algebraic surface $M=\mathit{Spec}(\mathbb{C}[x,y]/(xy-1))$, which is in fact self-mirror, i.e. $M\cong M^\vee$. This is to say, in this case the corresponding divisor $D_1\subset\mathbb{CP}^2$ is the union of a line and a conic. By a theorem of Seidel, we have an equivalence -$D\mathscr{F}(M,W)\cong D\mathscr{F}(T^\ast T^2,W),$ -which in fact follows from the fact that in this case we can choose a trivial bulk term to obtain a trivial deformation of the directed $A_\infty$ category associated to the fiberwise compactification of the Lefschetz fibration. This result holds also in the cases when filling in 2 or 3 points in a smooth fiber. From this we see that these LG models are all mirrors of $\mathbb{CP}^2$. -In the case when 2 points are filled in for each smooth fiber of $W$, the divisor $D_2\subset\mathbb{CP}^2$ is a nodal elliptic curve, while in the case when all the 3 punctures are filled in, $D_3$ is a smooth elliptic curve. -A large class of fiberwise compactifications of Lefschetz fibrations arises in the way you have just described, see the paper of Seidel (https://arxiv.org/pdf/1504.06317v2.pdf) for the general set up. However, it's not true for all fiberwise compactifications, simply because there are Lefschetz fibrations which do not come from Lefschetz pencils.<|endoftext|> -TITLE: $K_0$ of Burnside ring? -QUESTION [6 upvotes]: I am trying to find $K_0(AG)$ for $G$ cyclic of prime order. According to a paper I have, tom Dieck and Petrie's paper "Geometric modules over the Burnside ring" (1978) shows $K_0(AG) = \mathbb{Z} \oplus \mathbb{Z}/((p-1)/2)$ for $p\geq 3$, but I cannot find this anywhere in the paper. Can somebody either explain where this is in the paper or point me to another source? - -REPLY [5 votes]: As $AG$ is noetherian of dimension 1*, finitely-generated projective $AG$-modules are classified by their rank and determinant, and so -$$K_0(AG) = \mathbb{Z} \oplus Pic(AG).$$ -Proposition 10.3.8 of tom Dieck--Petrie gives an exact sequence calculating $Pic(AG)$ in terms of the ghost map, and for $G=C_p$ cyclic of odd prime order this is easy to work out their sequence manually and get -$$Pic(AC_p) \cong (\mathbb{Z}/p)^\times/\{\pm 1\} \cong \mathbb{Z}/((p-1)/2).$$ -* Certainly $AG$ is finitely generated, so noetherian. As the ghost map $\Phi : AG \to CG \cong \mathbb{Z}^{\times N}$ is an injective ring homomorphism, and $\vert G \vert \cdot CG \subset \Phi(AG)$, it follows that $CG$ is a finite $AG$-module. But $CG$ is a product of 1-dimensional rings, so is 1-dimensional, and hence $AG$ is also 1-dimensional. -Addendum -For $G=C_p$ a quite easy direct calculation of $K_0(AC_p)$ is possible. There are only two transtive $C_p$-sets, $x := [C_p]$ and $1 := [C_p/C_p]$, and $C_p \times C_p$ is a free $C_p$-set of cardinality $p^2$, so represents $p[C_p]$. Thus as rings we have -$$AC_p \cong \mathbb{Z}[x]/(x^2-px) =: R.$$ -The homomorphism $f : R \to \mathbb{Z}$ sending $x$ to $p$ sends the ideal $(x)$ of $R$ isomorphically to the ideal $(p)$ of $\mathbb{Z}$, and $R/(x) \cong \mathbb{Z}$ so there is a Mayer--Vietoris sequence (see the K-book, III.2.6) -$$K_1(\mathbb{Z}) \oplus K_1(\mathbb{Z}) \to K_1(\mathbb{Z}/p) \to K_0(R) \to K_0(\mathbb{Z}) \oplus K_0(\mathbb{Z}) \to K_0(\mathbb{Z}/p) \to 0.$$ -Now the two maps $K_1(\mathbb{Z}) \to K_1(\mathbb{Z}/p)$ are equal and are $\mathbb{Z}^\times \to (\mathbb{Z}/p)^\times$ so the claimed calculation follows.<|endoftext|> -TITLE: Countable shifts of closed positive sets -QUESTION [9 upvotes]: Let $\mu$ be the Lebesgue measure, and $+$ be addition modulo $1$ in the interval $[0,1)$. -Question1: Is there a closed set $C\subset [0,1)$ of positive measure such that for any countable set $D\subset [0,1)$, we have $\mu(C+D)<1$? -Question2: Is there a closed set $C\subset [0,1)$ of positive measure such that $\mu(\mathbb{Q}+C)<1$, where $\mathbb{Q}$ is the set of rational numbers? - -REPLY [3 votes]: Easier (well, I mean: requiring less deep knowledge) reason than in the answers by Goldstern and Will Brian. -Your set $C$ may be approximated by a disjoint union of small dyadic intervals (simply disjoint intervals also work) with accuracy 1 percent. Thus by pigeonhole principle one of them consists of points of $C$ by at least 99 percents. Its disjoint (or almost disjoint) rational shifts, which form a partition of the whole segment $[0,1]$, show that $[0,1]$ consists of points of $C+\mathbb{Q}$ by at least 99 percents.<|endoftext|> -TITLE: Equality with binomials -QUESTION [10 upvotes]: I am reading a paper and I am trying to understand an equality which is given without proof: -$$\sum_{k=1}^s\binom{2s-k}{s}\frac{k}{2s-k}v^k(v-1)^{s-k}=v\sum_{k=0}^{s-1}\binom{2s}{k}\frac{s-k}{s}(v-1)^{k} $$ -Here, $s>0$, $k$ and $v$ are positive integers. -The equality in question appears in Lemma 2.1 of -http://web.williams.edu/Mathematics/sjmiller/public_html/ntharm15/handouts/graphs/mckay_EigenvalueLargeRandomGraphs.pdf -Would you be kind and give me some insights on how to derive this equality? -Thank you, -LH - -REPLY [4 votes]: Here is a non-automated proof. We divide by $v$, and then we expand $v^{k-1}$ on the left hand side as -$$v^{k-1}=(1+v-1)^{k-1}=\sum_{j=0}^{k-1}\binom{k-1}{j}(v-1)^j.$$ -Then, looking at the coefficients of $v-1$ on the two sides, we are left with proving the identity -$$ \sum_{\substack{0\leq k\leq s\\0\leq j\leq k-1\\j+s-k=\ell}}\binom{2s-k}{s}\frac{k}{2s-k}\binom{k-1}{j} = \binom{2s}{\ell}\frac{s-\ell}{s},\qquad 0\leq\ell\leq s-1.$$ -With the notation $m:=s-k$ we have $k=s-m$, $j=\ell-m$, and the identity becomes -$$ \sum_{m=0}^\ell\binom{s+m}{s}\frac{s-m}{s+m}\binom{s-m-1}{\ell-m} = \binom{2s}{\ell}\frac{s-\ell}{s},\qquad 0\leq\ell\leq s-1.$$ -Equivalently, -$$ \sum_{m=0}^\ell\binom{s+m-1}{s-1}\binom{s-m}{\ell-m} = \binom{2s}{\ell},\qquad 0\leq\ell\leq s-1, \tag3$$ -i.e., -$$ \sum_{m=0}^\ell\binom{s+m-1}{s-1}\binom{s-m}{s-\ell} = \binom{2s}{2s-\ell},\qquad 0\leq\ell\leq s-1.$$ -We derive this one from the obvious identity of analytic functions -$$ (1-x)^{-s}(1-x)^{\ell-s-1}=(1-x)^{\ell-2s-1},\qquad |x|<1.$$ -Indeed, expanding each factor as a power series around the origin, we find that: - -the coefficient of $x^m$ in $(1-x)^{-s}$ equals $\binom{s+m-1}{s-1}$; -the coefficient of $x^{\ell-m}$ in $(1-x)^{\ell-s-1}$ equals $\binom{s-m}{s-\ell}$; -the coefficient of $x^{\ell}$ in $(1-x)^{\ell-2s-1}$ equals $\binom{2s}{2s-\ell}$.<|endoftext|> -TITLE: Degree of equations of secant varieties of Veronese varieties -QUESTION [7 upvotes]: Let $Sec_r(V)$ be the $r$-secant variety of a Veronse variety $V\subset\mathbb{P}^N$, that is -$$Sec_r(V) = \bigcup_{p_1,...,p_r\in V}\left\langle p_1,...,p_r\right\rangle\subset\mathbb{P}^N$$ -where $V$ is the image of $\mathbb{P}^n$ via the embedding induced by $\mathcal{O}_{\mathbb{P}^n}(d)$. -Is it true that if $Sec_r(V)\neq\mathbb{P}^N$ then any polynomial on $\mathbb{P}^N$ vanishing on $Sec_r(V)$ has degree greater or equal than $r+1$ ? - -REPLY [5 votes]: The answer by JM Landsberg in the link from aginensky's comment precisely says that the $r+1$ lower bound on the degree of generators is true. -See top of page 2 in the article "Prolongations and computational algebra" by Sidman and Sullivant where they point to the article "On the ideal of an embedded join" by Ulrich and Simis. - -Edit: Having thought about the question a bit more, I realized this is trivial (for the Veronese, but not for secants of general varieties as in the above articles) if one knows the symbolic method from 19th century invariant theory. -Consider the secant $\sigma_r(v_d(\mathbb{P}^n))$ for the degree $d$ Veronese embedding. Let $C(F)$ be a homogeneous polynomial of degree $r$ in the coefficients of a generic form $F$ of degree $d$ in $n+1$ variables. -If $C(F)=0$ for all $F$'s which can be written $F=L_1^d+\cdots +L_r^d$ for some linear forms $L_i$, then -the polynomial $C$ vanishes identically. -Indeed, -$$ -C(F)=\left.M(F_1,\ldots,F_r)\right|_{F_1=\cdots=F_r=F} -$$ -where $M$ is the associated symmetric multilinear form given by -$$ -M(F_1,\ldots,F_r)=\frac{1}{r!}\frac{\partial^r}{\partial t_1\cdots \partial t_r} -\ C(t_1 F_1+\cdots+t_r F_r)\ . -$$ -From the hypothesis (and working over a field like $\mathbb{C}$ which contains $d$-th roots of unity) one immediately gets -$$ -M(L_1^d,\ldots,L_r^d)=0 -$$ -for linear all forms $L_1,\ldots,L_r$. -Finally, one has the identity -$$ -C(F)=\frac{1}{d!^r}\ F(\frac{\partial}{\partial L_1})\cdots -F(\frac{\partial}{\partial L_r})\ -M(L_1^d,\ldots,L_r^d) -$$ -where the point coordinates $x_1,\ldots,x_{n+1}$ are replaced by differential operators in the coefficients of the linear forms $L_1,\ldots,L_r$. As a result, $C(F)$ is identically zero. -The last identity is what makes the classical symbolic method work. For more on this, please see this article (or here for the preprint version).<|endoftext|> -TITLE: The set of homotopy classes of maps between compact manifolds is countable -QUESTION [11 upvotes]: I'm reading Ravenel's book Nilpotence and periodicity in stable homotopy theory. In section 1.1, it says the set of homotopy classes of maps of maps between compact manifolds or between algebraic varieties over real or complex numbers is countable. I know that the set of homotopy classes between triangulable spaces is countable. I also know that algebraic varieties over real or complex numbers are triangulable while compact manifolds are not always triangulable. So I want to know how to prove that the set of homotopy classes of maps between compact manifolds is countable. - -REPLY [11 votes]: This may be somewhat heavy-handed but it follows from work of Kirby-Siebenmann that any compact manifold is homotopy equivalent to a finite CW-complex (implying what you want because any finite CW-complex is homotopy equivalent to its regular neighborhood in some Euclidean space, which in turn can be triangulated as a finite simplicital complex). -More precisely, let $M$ be a compact manifold, possibly with boundary. If $M$ has dimension $\ge 6$ and empty boundary, then $M$ is homeomorphic to a CW-complex [p.107 of Kirby-Siebenmann's book "Foundations of Topological manifolds"]. In general, fix $k$ such that $W=M\times D^k$ has dimension $\ge 7$. On the same page 107 it is stated that $W$ is homeomorphic to a mapping cylinder of a map $f:\partial W\to X$, where $X$ is a finite CW-complex. As mentioned above $\partial W$ is also a CW complex. The map $f$ can be homotoped to a cellular map, so the mapping cylinder $W$ is homotopy equivalent to a CW-complex as promised.<|endoftext|> -TITLE: A condition that might force biregularity of a bipartite graph -QUESTION [5 upvotes]: Let $\Gamma$ be a finite connected bipartite graph with colour classes $U$ and $V$ such that: - -every vertex of $U$ has degree $n$, and $n\ge 3$; -every vertex of $V$ has degree at least $4$; -$\Gamma$ has diameter $4$; -for every vertex $u$ of $U$, there are at least $(n+1)/2$ vertices in the neighbourhood $\Gamma(u)$ of $u$ that have degree the minimum degree of $\Gamma(u)$. - -Is $\Gamma$ biregular (i.e., the vertices of $V$ have constant degree)? If not, then under what natural and weak conditions is biregularity forced to hold? - -REPLY [3 votes]: I tried to construct an example that is not biregular. To make it easy, I assumed that n=3 and that the vertices of V all have degree 4 except one of degree 5 (so each vertex of U is automatically adjacent to at least two vertices of degree 4). -One option where the numbers work out is to have 7 vertices in U and 5 vertices in V, so if I set V = {A,B,C,D,E} and then make the 7 neighbourhoods equal to {ABC, ABD, ABE, ACE, ADE, BCD, CDE} then I think that vertex A has degree 5, and all the others degree 4. -I don't know what sort of condition you'd need to rule out examples like this..<|endoftext|> -TITLE: Surjectivity of map between Néron models $\mathcal{E} \to \mathcal{E}'$ -QUESTION [5 upvotes]: My question is related to a previous question on the Mordell-Weil rank of the elliptic curve $E/\mathbf{Q} : y^2 = x^3- 2$ asked here. More precisely, I want to understand the following. Let $E'/\mathbf{Q}$ be the elliptic curve $y^2 = x^3 + 54$. There is a rational $3$-isogeny $\phi : E \to E'$ given by -$$\phi(x,y) = \left( \frac{x^3 - 8}{x^2}, \frac{y(x^3 -+ 16)}{x^3}\right).$$ -Let $\mathcal{E},\mathcal{E}'$ denote the Néron models of $E,E'$ respectively over $\mathbf{Z}_{(3)}$. Let -$$ -\Phi \colon \mathcal{E} \to \mathcal{E}' -$$ -denote the induced map on Néron models. - -My goal: I would like to prove that $\Phi$ is étale surjective, so that the map of representable sheaves $\underline{\mathcal{E}} \to \underline{\mathcal{E}'}$ on $(\mathbf{Z}_{(3)})_{\text{ét}}$ is surjective. - -It is enough to show this is true on special fibers. Let $\widetilde{\Phi} \colon \widetilde{\mathcal{E}} \to\widetilde{\mathcal{E}'}$ denote the map on reductions. Then I have shown by direct computation that on identity components, -$$ -\widetilde{\Phi}|_{\widetilde{\mathcal{E}^0}} \colon \widetilde{\mathcal{E}^0} \to\widetilde{\mathcal{E}'^0} -$$ -is étale surjective. Therefore $\widetilde{\Phi}$ is étale, because we can check it after base changing to $\overline{\mathbf{F}_3}$, in which case every component of $\widetilde{\mathcal{E}}$ is a translate of the identity component by a $\overline{\mathbf{F}_3}$ rational point. -Now for surjectivity, I have the following information (from a table in Silverman's ATEC). We define $k := \mathbf{F}_3$. - - -The component groups of the special fibers of the Néron models are: $$\widetilde{\mathcal{E}}(k)/\widetilde{\mathcal{E}^0}(k) = \widetilde{\mathcal{E}'}(k)/\widetilde{\mathcal{E}'^0}(k) = \mathbf{Z}/2\mathbf{Z}.$$ - - -However, with this information I can't seem to conclude that $\widetilde{\Phi} \colon \widetilde{\mathcal{E}} \to \widetilde{\mathcal{E}'}$ is surjective. The problem seems to be that we don't know what the $\overline{k}$-points of the component groups are. Using things like Lang's Theorem to say that $\widetilde{\mathcal{E}}(k)/\widetilde{E^0}(k) = \widetilde{\mathcal{E}/\mathcal{E}^0}(k)$ don't seem to help too. I would appreciate any insight on how to get surjectivity of $\Phi$. - - -Edit: It seems that my description of component groups is wrong (see Chris's answer below). However, Chris has outlined reasons why $\Phi$ cannot be surjective. In any case, my ultimate goal is to show that as representable sheaves on the site ${\mathbf{Z}_{(3)}}_{\'{e}t}$, the map $\underline{\mathcal{E}} \to \underline{\mathcal{E}'}$ is surjective. Perhaps this can still be shown? - -REPLY [3 votes]: I get that the Kodaira types of $E$ is II and for $E'$ it is IV${}^{*}$. This means that $\mathcal{E}$ is connected (or in simple terms no point over $\mathbb{Q}_3^{\text{unr}}$ reduces to the singular point $(-1,0)$ of the reduced Weierstrass equation) and $\mathcal{E}'$ has component group $\Phi' = \mathbb{Z}/3\mathbb{Z}$ (for instance the point $(3,9)\in E'(\mathbb{Q})$ reduces to the singular point $(0,0)$). It follows that $\hat\phi$ is surjective. The kernel of $\phi$ is étale: the point $(0,1+3+2\cdot 3^2+\cdots)\in E(\mathbb{Q}_3)[3]$ generates the kernel. Instead the kernel of $\hat\phi$ is not even quasi-finite. -One can also see directly that $\phi$ is not surjective: If $(x,y)$ in $E$ is mapped to $(3,9)$ then $x^3-3x^2-8=0$, but that has no solution in $\mathbb{Q}^{\text{unr}}_3$. -Edit: I corrected my statement about the kernel of $\hat\phi$. Yes, subgroups of order $p$ when the elliptic curve has additive reduction at $p$ are nasty.<|endoftext|> -TITLE: Graph with Poisson Clock at each Vertex -QUESTION [19 upvotes]: Let $G$ be a connected, undirected graph, with countably infinite set of vertices and countably infinite set of edges. Assume that the degree of each vertex is finite, and moreover, the degrees of all vertices are uniformly bounded. -Let each vertex carry one of two values: $1$ or $-1$. -Now, equip each vertex with a Poisson clock ($\lambda=1$), all clocks are independent. At each time the clock of a vertex ticks, the vertex updates its value to be the value of most of its neighbors (in case of a draw $-$ the value of the vertex remains unchanged). -Does there exist such a graph as described, with certain initial values at vertices, such that with $\mathbf{positive}$ probability, there will be a vertex where the value is $not$ eventually constant? -Thank you. -$\mathbf{EDIT:}$ If you wish, for a beginning, analyze the example given by domotorp in his comment (which could be a solution): take the $3$-regular tree with initial values as follows: pick one vertex, it will be $1$. The vertices around it will be $-1$. The vertices at distance $2$ from the initial vertex are again $1$. And so on, changing the value layerwise. In this graph and initial values, will there be a vertex, that with positive probability, will not converge? -(Even if the answer for this example is NO, the fact that every vertex in the example almost surely converges is also nontrivial, and a proof of this will also be upvoted.) - -REPLY [3 votes]: Yes, my example is easy to modify after some thought, just take a thick enough layer for each level. -More precisely, let $f$ be a sufficiently fast growing function, and define the initial value on any vertex at distance $f(n)\le d< f(n+1)$ from the root as $(-1)^n$. -Given $f(n)$, one can also pick a large enough $f(n+1)$ such that independently of the later values the root will get $(-1)^n$ in some time that depends only on $f(n+1)$ with at least $50\%$ probability. -This guarantees that with $1$ probability it will switch values infinitely often (just like every other vertex). -In more details, suppose first that $f(n+1)=\infty$, i.e., all but a finite number of vertices have the value $(-1)^n$. -Lemma. If all vertices at distance $>r$ have value $(-1)^n$, then with probability $1$ after a while all vertices will have value $(-1)^n$. -Proof. Vertices further than $r$ can never change value. If a vertex at distance $r$ gets value $(-1)^n$, it will never change again. So eventually all vertices at distance $r$ will also obtain value $(-1)^n$ and we can use induction on $r$. -From this it follows that the root will obtain value $(-1)^n$ in some $T(n)$ time with $90\%$ probability. -If $f(n+1)$ is not $\infty$, but just large enough compared to $T(n)$, then the probability of any vertex at distance $r+1$ from the root changing value is less than $10\%$. -Therefore, this won't affect the probability of the root changing value.<|endoftext|> -TITLE: How to determine whether a polynomial has integer roots? -QUESTION [7 upvotes]: A polynomial -$$f(x) = x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$$ -where $a_{n-1}, a_0$ are known. How to determine whether polynomial $f$ has integer roots? Can we give a bound of $\max\{a_i \mid i=1,\dots,n-2\}$ and $\min\{a_i \mid i=1,\dots,n-2\}$ decide by $n,a_{n-1}$, and $a_0$ such that the polynomial roots are integers. - -REPLY [7 votes]: I'm not sure if I understand just what you're asking, but perhaps it is the following: - -Suppose we know $a_0$ and $a_{n-1}$, and we know that $f(x)$ has only integer roots. Can we use this information to meaningfully bound the coefficients? - -If that's what you're asking, then the answer is no. Consider for example the polynomial -$$ -g(x) = x (x^2 - c^2)^{k}, -$$ -where $c \in \mathbb{Z}$. Then this has $a_0 = a_{n-1} = 0$. But even for fixed $n$, the other coefficients are unbounded if you let $c$ be large. - -This part added in response to Timothy Chow's comment: -Timothy Chow made the suggestion that allowing $a_{0} = a_{n-1} = 0$ feels like cheating, and on reflection, I agree! -If we allow $a_0 = 0$, then we can't get any bound on the other coefficients -Namely, just consider any monic degree $n-2$ polynomial of the form $p(x) = x^{n-2} + b_{n-4} x^{n-4} + b_{n-5} x^{n-5} + \cdots + b_0$. Then look at the polynomial $x (x+a) p(x) = x^n + a x^{n-1} + a_{n-2} x^{n-2} + \cdots + a_1 x + 0$. This will have integer roots, and the coefficient $a_{n-1}$ is free to be any integer you like. Yet the other coefficients can be wild. For concreteness, consider this -$$ -h(x) = x (x+a)(x^2 - c^2)^k, -$$ -which, for fixed $a$ and $k$, will have unbounded coefficients if you pick $c$ large enough [e.g., the sum of these coefficients is $h(1)$ which is unbounded]. -If we disallow $a_0 = 0$, then we do get a bound -However! If we disallow the case $a_0 = 0$, then yes you can get a bound on the remaining coefficients. Namely, if $f(x)$ factors as $f(x) = \prod_{i} (x +r_i)$, then we know $a_0 = \prod_{i} r_i$. If none of the roots is $0$ (i.e., if $a_0 \neq 0$), then each root is at least $1$ in absolute value, and thus $|a_0|$ is an upper bound on every product of the form $\prod_{i \in S} |r_i|$. Therefore, we have -$$ -|a_{n-k}| = \left| \sum_{|S| = k} \prod_{i \in S} r_i \right| \leq \sum_{|S| = k} \prod_{i \in S} |r_i| \leq \sum_{|S| = k} |a_0| = {n \choose k} |a_0|. -$$ -This bound is tight of course by the polynomials $(x+1)^n$ and $(x-1)^n$. (This reminds me of Newton's inequalities, which probably can provide an alternate proof.)<|endoftext|> -TITLE: Uniqueness theorem for conformal mapping -QUESTION [8 upvotes]: Let $f$ and $g$ be analytic functions in the unit disk $D$, continuous in the closed disk and locally univalent, $f'(z)\neq 0,\; g'(z)\neq 0,\; z\in D$. -Assume that each has only finitely many singularities on $\partial D$, so that -the images of $\partial D$ are piecewise analytic. -Suppose that there exists a homeomorphism $\phi:\partial D\to\partial D$ such that -$f(z)=g\circ\phi(z),\; z\in \partial D$. (This means that the images of the boundary under both functions are reparametrizations of the curve). Can we conclude from this that $\phi$ extends to a conformal automorphism of $D$? -Same question about meromorphic locally univalent functions. -Motivation: this is so, if $f$ and $g$ are globally univalent (injective). Indeed then they map the unit disk on plane regions with the same boundaries, so the images coincide. -EDIT. This is related to my other question: -Greatest lower bound for subordination -and using Milnor's counterexample mentioned in Misha's answer, one can construct a counterexample to the other question. - -REPLY [8 votes]: This is a form of the following topological question: Does the boundary value uniquely determine an immersion of the closed disk into the plane, up to precomposition with a diffeomorphism of the disk. The answer is negative -and is due to John Milnor, see Poenaru's paper "Extension des immersions en codimension 1." Séminaire Bourbaki, 10 (1966-1968), Exp. No. 342. -Incidentally, the best results about classification of extensions of immersions $S^1\to R^2$ to immersions $D^2\to R^2$ are (still) due to S.Blank, his 1967 PhD thesis. These results are discussed in detail in Poenaru's paper.<|endoftext|> -TITLE: Proper class of Woodins and $\textsf{AD}_{\mathbb R}$-hypothesis -QUESTION [5 upvotes]: The $\textsf{AD}_{\mathbb R}$-hypothesis is the statement that there is a $\lambda$ which is both a limit of Woodins and a limit of ${<}\lambda$-strongs. Are there any results relating the consistency of this statement to the consistency of a proper class of Woodins? -In this paper by Zeman he mentions that the $\textsf{AD}_{\mathbb R}$-hypothesis is the existence of a proper class of Woodins and strongs. To me that just sounds strictly stronger than the above-stated version - but that's perhaps just a change in terminology over time? - -REPLY [5 votes]: The existence of a proper class of Woodin cardinals and a proper class of strong cardinals is strictly stronger in consistency strength over ZFC than the existence of a cardinal $\lambda$ that is a limit of Woodin cardinals and a limit of $<\lambda$-strong cardinals. The former implies the consistency of the latter and indeed the former implies that there are a proper class of such $\lambda$. -To see this, let $A$ be the class of Woodin cardinals and $B$ be the class of strong cardinals, and assume both of these are unbounded. Thus, both $A$ and $B$ have unboundedly many limit points. Since the limit points of an unbounded class are club and any two club classes intersect in a club, it follows that there are a proper class of $\lambda$ that are limits of Woodin cardinals and limits of strong cardinals. One can construct such a $\lambda$ explicitly as the limit of the following process: start with any Woodin cardinal, then take the first strong above it, the next Woodin above that, the next strong above that, and so on. Let $\lambda$ be the supremum of these choices. So $\lambda$ is a limit of Woodin Cardinals and strong cardinals, as large as desired. -Thus, the assumption of a proper class of Woodin cardinals and a proper class of strong cardinals implies the consistency of the existence of a $\lambda$ that is a limit of Woodin cardinals and $<\lambda$-strong cardinals, simply by chopping off at an inaccessible cardinal. So the consistency strength is strictly stronger. - -REPLY [2 votes]: I think I got a solution. If you find any mistakes or if anything is unclear, please let me know! -Firstly, recall that the $\Omega>0$-hypothesis (or sometimes stated as $\theta_0<\Theta$-hypothesis), is the statement that there is a $\lambda$ which is the limit of Woodins and there is a $\kappa<\lambda$ which is ${<}\lambda$-strong. So this hypothesis is weaker than the $\textsf{AD}_{\mathbb R}$-hypothesis. - -Proposition. The $\Omega>0$-hypothesis is consistency-wise stronger than a proper class of Woodins. - -Proof. Assume $\lambda$ is a limit of Woodins and let $\kappa<\lambda$ be ${<}\lambda$-strong. Let $\delta<\lambda$ be the least Woodin above $\kappa$ and pick some $\gamma\in(\delta,\lambda)$. Then as $\kappa$ in particular is $\gamma$-strong, fix an elementary embedding $j_0\colon V\to\mathcal M_0$ such that $V_\gamma\subseteq\mathcal M_0$, $\text{crit }j=\kappa$ and $j(\kappa)>\gamma$. In $\mathcal M_0$ we still have that $\delta$ is Woodin, as the extenders witnessing Woodinness lie in $V_\delta$ and $V_\delta\subseteq V_\gamma\subseteq\mathcal M$. -We can then continue. At every successor stage $\alpha+1$ we use the $j_{0\alpha}(\gamma)$-strongness of $j_{0\alpha}(\kappa)$ to get a $j_{0\alpha}(\gamma)$-strong embedding $j_{\alpha+1}\colon\mathcal M_\alpha\to\mathcal M_{\alpha+1}$ and again note that $j_{0\xi}(\delta)$ is still Woodin in $\mathcal M_{\alpha+1}$ for every $\xi\leq\alpha$. At limit stages we take direct limits, with the same conclusion. We now have two cases. -Case 1. There is an ordinal $\alpha$ such that for every $\xi\geq\alpha$, $j_{0\xi}(\lambda)=j_{0\alpha}(\lambda)$. -In this case, after we get to stage $\alpha$, we iterate $j_{0\alpha}(\lambda)$ more times, which moves the extender on $j_{0\alpha}(\kappa)$ up to $j_{0\alpha}(\lambda)$. Then $V_{j_\xi(\lambda)}\cap \mathcal M_{\xi+j_\xi(\lambda)}$ satisfies $\textsf{ZFC}$ and it thinks that there is proper class of Woodins (as $j_\xi(\lambda)$ is an inaccessible limit of Woodins inside $\mathcal M_{\xi+j_\xi(\lambda)}$). -Case 2. For every ordinal $\alpha$ there is some $\xi>\alpha$ such that $j_{0\alpha}(\lambda)0)$-hypothesis $\leq$ $\textsf{AD}_{\mathbb R}$-hypothesis < A proper class of Woodins and a proper class of strongs.<|endoftext|> -TITLE: de Rham cohomology of $x^3+y^3 +z^3+c \,xyz= 0$ find representatives -QUESTION [6 upvotes]: The equation $x^3+y^3 +z^3+c\, xyz= 0$ defines a non-singular elliptic curve $X$ in $\mathbb{C}P^2$ projective space. -In fact, how do we prove it has genus 1 both as an algebraic curve and as a Riemann surface? -Since $H^1(X,\mathbb{C})\simeq \mathbb{C}^2$ what are the explicit representatives of the de Rham cohomology? There should be two of them. - -This question is (verbatim) on Math.Stackexchange unanswered even with a bounty. - -despite a change of variables, I am guessing it might be best to maintain the symmetry between $x,y,z$ if possible. -I am asking for explicit sections in a manner that might have been done in the 19th century (but could be done in many situations today). they should correspond to classical functions in some way. -there's an underlying question of whether the "cohomology" of this curve as a topological space matches up with the sheaf cohomology or de Rham cohomology. Especially when the curve becomes singular. -I'm also looking for examples where they disagree - -REPLY [4 votes]: A possible method for getting explicit representatives involves Griffiths's Residues. More in general, let $X \subset \mathbb{P}^{n+1}$ a smooth hypersurface of degree $d$. -Denote by $H^n_{\textrm{prim}}(X)$ the primitive subspace of the n-th cohomology of $X$, that is $$ H^n_{\textrm{prim}}(X)= \textrm{Ker} ( \lambda: H^n(X) \to H^{n+2}(X) ),$$ $$\lambda(\alpha)= \alpha \cup c_1 (\mathcal{O}_X(1)).$$ -The (restriction on) the Hodge filtration $F^p H^n_{\textrm{prim}}(X)$ can be identified with $$H^0(\Omega^{n+1}_{\mathbb{P}}(n-p+1)(X)) / d H^0((\Omega^{n}_{\mathbb{P}}(n-p)(X)).$$ -Now, if $$\omega= \sum_i (-1)^i x_i dx_0 \wedge \ldots d\hat{x_i} \wedge \ldots dx_{n+1}$$ write $$H^0(\Omega^{n+1}_{\mathbb{P}}(k))=\big\{ \frac{g \omega}{f^k}\ | \ g \in \mathbb{C}[x_0, \ldots, x_{n+1}]_{kd-n-2} \big \} $$ -and use this identification in the quotient above. -One has that if $R_f:= \mathbb{C}[x_0, \ldots, x_{n+1}]/J_f$ with $J_f$ being the ideal generated by the partial derivatives of $f$ then $$H^{n-p,p}_{\textrm{prim}}(X):= H^n_{\textrm{prim}}(X) \cap H^{n-p,p}(X) \cong (R_f)_{(p+1)d-n-2},$$ -where the maps sends the class of $g\omega/f^k$ to the class of $g$. -Now this is all a very nice bit of theory, but in this case the situation is very simple. For a curve we have $H^1(C, \mathbb{C}) \cong H^{1,0}(C) \oplus H^{0,1}(C)$ and moreover $H^1_{\textrm{prim}}(C) \cong H^1(C, \mathbb{C})$ (and the same for the subspaces of course). -The degree of the equation is three, therefore one has $H^0(K_C) \cong (R_f)_0$ and $H^1(\mathcal{O}_C) \cong (R_f)_3$. The former is generated by 1 (it is canonically identified with $\mathbb{C}$). For the latter, your equation for $C$ implies that you can pick as generator for $(R_f)_3 \cong \mathbb{C}\cong \langle z^3 \rangle$. Note that $R_3$ is the socle of the Jacobian ring, that is $(R_f)_t=0$ for $t>3$. -Moreover (in the non weighted case) if $\rho$ is the degree corresponding to the socle of this (artinian, Gorenstein) ring one has $R_a \cong (R_{\rho-a})^*$, and this is in turn just another expression of Serre Duality. -Plugging these polynomials in the quotient above, one gets an explicit expression as desired.<|endoftext|> -TITLE: what else is in $\prod_{j=1}^n(1+q^j)$? -QUESTION [25 upvotes]: From time to time, I run into the finite product $\prod_{j=1}^n(1+q^j)$. And, the more it happens, the more fascinated I've become. So, herein, I wish to get help in collecting such results. To give some perspective into what I look for, check out the below examples. First, some nomenclature: $(q)_k=(1-q)(1-q^2)\cdots(1-q^k)$ and $\binom{n}k_q=\frac{(q)_n}{(q)_k(q)_{n-k}}$. -(0) It's almost silly, but the sum of elementary functions of the specialization $\pmb{q}=(q,q^2,\dots,q^n)$: -$$e_0(\pmb{q})+e_1(\pmb{q})+\cdots+e_n(\pmb{q})=\prod_{j=1}^n(1+q^j).$$ -(1) The classical $q$-binomial theorem, which results from counting restricted distinct partitions or number of weighted tilings: -$$\sum_{k=0}^nq^{\binom{k+1}2}\binom{n}k_q=\prod_{j=1}^n(1+q^j).$$ -(2) I can't remember where I saw this (do you?) but -$$\sum_{k=0}^nq^k\binom{n}k_{q^2}=\prod_{j=1}^n(1+q^j).$$ -(3) The $H$-polynomial of a symplectic monoid $MSp_n$ (see this paper, page 13): -$$\sum_{k=0}^n(-1)^kq^{k^2}\binom{n}k_{q^2}^2\prod_{i=1}^k(1-q^{2i})\prod_{j=1}^{n-k}(1+q^j)^2=\prod_{j=1}^{2n}(1+q^j),$$ -although the authors did not seem to be aware of the RHS. - -QUESTION. Can you provide such formulas (in any field) with the same RHS (always a finite product) as in above, together with resources or references? - -Thank you. - -REPLY [2 votes]: Let $exp(z,q)=\sum_{k=0}^{\infty}z^k/[k]_q!$ be the usual $q-$analogue of the exponential function $e^x.$ -The identity -$\sum_{k=0}^n{q^k}{\binom{n}{k}_{q^2}}=\prod_{j=1}^n(1+q^j)$ -can be obtained by comparing coefficients in -$$exp(\frac{z}{1+q},q^2)exp(\frac{qz}{1+q},q^2)=exp(z,q),$$ -which is a natural $q-$analogue of $e^\frac{x}{2} e^\frac{x}{2}=e^x.$ -This identity occurs in Séminaire Lotharingien de Combinatoire, B05a (1981), but is perhaps older.<|endoftext|> -TITLE: Jacobians of genus 2 curves isogenous to a square of a supersingular elliptic curve over $\mathbb{F}_{p^2}$ -QUESTION [6 upvotes]: I would like to construct hyperelliptic curves whose Jacobians are isogenous to the square of a supersingular elliptic curve over $\mathbb{F}_{p^2}$ -My question is motivated by the following example. -Let $H/\mathbb{F}_{5^2}$ be the hyperelliptic curve given by $y^2 = x^6 + 1$ and $E/\mathbb{F}_{5^2}$ an elliptic curve $y^2 = x^3 + 1$, then we have that $J:=Jac(H)$ has characteristic polynomial of Frobenius given by $(t + 5)^4$. -There is a map $\psi:H\to E$ given by $(x,y)\mapsto (x^2,y)$ and $E$ is supersingular. -If $\phi\in End(J)$ is the $5^2$-Frobenius endomorphism then we have that $\phi = [-5]$ -I think this is a direct consequence of $J$ being isogenous to a the square of a supersingular elliptic curve as the Frobenius satifies its characteristic polynomial in the Tate module, but I am not sure if this suffices. -I would like to know if someone can point me out to a characterization of hyperelliptic curves of genus two where the Frobenius in the endomorphism ring of its Jacobian is the $[n]$ map for some $n$. -In fact it will be nice to construct the hyperelliptic curve of genus 2 such that its jacobian is isogenous to the square of a supersingular elliptic curve. -Thanks. -Just for fun, in MAGMA, I have that the curve $H$ is isomorphic to $y^2=\alpha x^5 + \alpha^{13}x$ and the output to check this is using the generic point of $H$ as follows: -Note: I changed the angle brackets by () as I could not find how to show them. - - -> F(a) := FiniteField(5^2); -> P(x) := PolynomialRing(F); -> f1 := x^6 + 1; -> f := a*x^5 + a^13*x; -> H1 := HyperellipticCurve(f1); -> H := HyperellipticCurve(f); -> IsIsomorphic(H1,H); -true Mapping from: CrvHyp: H1 to CrvHyp: H -with equations : -2*$.1 + 4*$.3 -a^5*$.2 -a^8*$.1 + a^22*$.3 -and inverse -a^22*$.1 + $.3 -a^4*$.2 -a^20*$.1 + 2*$.3 -> FH(X,Y) := FunctionField(H); -> Hext := BaseExtend(H,FH); -> Jext := Jacobian(Hext); -> R(z) := PolynomialRing(FH); -> PtJ := Jext![z-X,Y]; -> FrPtJ := Jext![z-X^25,Y^25]; -> FrPtJ; -(x + 4*X^25, (4*X^60 + 2*X^56 + 4*X^52 + 2*X^40 + X^36 + 2*X^32 + 4*X^20 + 2*X^16 + 4*X^12)*Y, 1) -> -5*PtJ; -(x + 4*X^25, (4*X^60 + 2*X^56 + 4*X^52 + 2*X^40 + X^36 + 2*X^32 + 4*X^20 + 2*X^16 + 4*X^12)*Y, 1) -> FrPtJ + 5*PtJ; -(1, 0, 0) -> w(t) := PolynomialRing(Rationals()); -> t^(2*Genus(H))*Evaluate(Numerator(ZetaFunction(H)),1/t); -t^4 + 20*t^3 + 150*t^2 + 500*t + 625 -> Factorization(t^4 + 20*t^3 + 150*t^2 + 500*t + 625); -[ - (t + 5, 4) -] - -REPLY [6 votes]: Such curves are constructed in my paper "Familles de courbes et de variétés abéliennes sur $\mathbb{P}^1$, II", Astérisque vol. 86 (1981).<|endoftext|> -TITLE: Disc bounded by a plane curve -QUESTION [26 upvotes]: Let $\Sigma$ be a sphere topologically embedded into $\mathbb{R}^3$. - -Is it always possible to find a disc $\Delta\subset\Sigma$ which is bounded by a plane curve? - -It is easy to find an open disc which boundary lies in a plane, -but the boundary might be crazy; for example it might be Polish circle shown on the diagram. - -Comments - -A simpler question: Does every surface topologically embedded in the Euclidean space contains a planar arc? - -Curves on potatoes --- another closely related problem. - -REPLY [5 votes]: As requested, here is a related result: -Theorem. There exists a continuous (also also open) real-valued function $R^2\to R$ whose level sets are all homeomorphic to the open pseudo-arc. In particular, level sets contain no nondegenerate arcs. -Similarly, one can construct a continuous monotone map $S^2\to [0,1]$ such that all level sets are homeomorphic to the pseudo-circle, i.e. the unique (up to homeomorphism) hereditarily indecomposable circularly chainable continuum. Again, pseudo-circle contains no nondegenerate arcs. -The story of this theorem is rather interesting. Quoting from -Prajs, Janusz R., A continuous circle of pseudo-arcs filling up the annulus, Trans. Am. Math. Soc. 352, No. 4, 1743-1757 (2000). ZBL0936.54019. -where a proof of this theorem is given: - -Among the results obtained by Knaster during World War II one can find the following announcement, originally presented in Kiev in 1940: - - -There exists a real-valued, monotone mapping from the plane that is not constant on any arc. In the construction Knaster’s hereditarily indecomposable continua were exploited. Actually, Knaster’s result can be reformulated in the following stronger version: -There exists a real valued, monotone mapping from the plane such that all pointinverses are hereditarily indecomposable. - - -Unfortunately, Knaster’s notes concerning this result were burned during the war, Knaster had never written down the result again, and even his closest exstudents do not know his original idea of construction... - - -A result similar to that announced by Knaster (with higher dimensional analogues) was proved by Brown in 1958: - - -M. Brown, Continuous collections of higher dimensional continua, Ph. D. Thesis, University -of Wisconsin, 1958. - -Of course, Brown did not publish his proof either, but Prajs finally did, see above.<|endoftext|> -TITLE: On the geometrization of double branched covers -QUESTION [9 upvotes]: I recently got into Lickorish's paper Prime knots and tangles and a question, which I didn't have the first time I read it, naturally emerged. -The Thurston-Perelman Geometrization Theorem asserts that given a compact, closed, orientable three-manifold $M$, either: -1) $M$ contains an essentrial sphere, or -2) $M$ contains an incompressible torus, or -3) $M$ is Seifert fibered, or -4) $M$ is hyperbolic. -Given a link $L \subset S^3$ we can form a closed three-manifold $M_L$ by considering the double-branched cover of $S^3$ over $L$. - -QUESTION. How do conditions 1-4, regarding the manifold $M_L$, translate to the topology of the link $L$? - -Following Lickorish's paper it is easy to find some sufficient conditions based on the possibility to recognize certain tangle configurations in a diagram of $L$. What about "if and only if" conditions? - -REPLY [7 votes]: As Ian Agol mentioned in his comment, the OP's question can be thought of in terms of the Orbifold Theorem. There are two (contemporaneous and) independent proofs of the Orbifold Theorem: -Daryl Cooper, Craig D. Hodgson, and Steven P. Kerckhoff, MR 1778789 Three-dimensional orbifolds and cone-manifolds, ISBN: 4-931469-05-1. (The relevant background chapter is available here https://projecteuclid.org/download/pdf_1/euclid.msjm/1389985818) -Michel Boileau, Bernhard Leeb, and Joan Porti, MR 2178962 Geometrization of 3-dimensional orbifolds, Ann. of Math. (2) 162 (2005), no. 1, 195--290. -The background section of either paper seems to answer the OP's question. However, here is my attempt to present the necessary details: -Consider the warm up case, if the double branched cover $M_L$ of a link (embedded in $S^3$) $L$ is geometric (in the sense of admitting one the eight Thurston geometries), then $M$ double covers a geometric orbifold $Q$ which has underlying space $S^3$ and the singular locus has cone angle $\pi$ and is isotopic to $L$ (as an embedded link in $S^3$). This covers cases 3) and 4) of the OP's question. Good background for Case 3) is the work of Montesinos, especially his book Classical Tessellations and Three-Manifolds. -Otherwise, there is an obstruction to geometrization. For closed, orientable 3-manifolds, the obstructions to geometrization are embedded $S^2$ which do not bound $B^3$'s and incompressible $T^2$. As a result of the two papers cited above, the obstructions to (closed) orbifold geometrization are more or less analogous. -First, we need to discuss bad 2-orbifolds of $Q$, i.e. 2-orbifolds which have underlying space $S^2$ transversely intersect the singular locus $Q$ in exactly one point. This is equivalent to such a 2-orbifold not having a manifold cover (see for example Chapter 13 of Thurston's notes). -For the rest of the argument we will restrict to the case of an orbifold $Q$ with underlying space $S^3$ and having a singular locus a link $L$ labeled only by cone angle $\pi$. -EDIT: the paragraph below originally forgot to consider split links. Thanks to the OP for pointing this out. -1) In this context $Q$ is reducible if it contains a) an embedded bad sub-2-orbifold, b) an embedded 2-sphere which does not bound a 3-ball, or c) an embedded 2-fold orientable quotient of 2-sphere which does not bound the quotient obtained by rotating a 3-ball along an unknotted arc. This perspective more naturally generalizes to arguments of the above two papers. In the case of the question, we see that $Q$ will contain an embedded 2-sphere which does not bound a 3-ball if and only if $L$ is split and $Q$ will contain an embedded 2-fold orientable quotient of 2-sphere as in case c) if and only if $L$ is not prime. Furthermore, if $L$ is split or non-prime link, then each piece of a prime decomposition of the link $L$ will correspond to a prime decomposition of the orbifold $Q$ and a prime decomposition of $M_L$. (The general case involves consideration of more 2-orbifold quotients of $S^2$.) So the orbifold version of reducible involves both contains an essential 2-sphere quotient and bad 2-orbifolds. -2) The final case to consider is if $Q$ is irreducible but not geometric. In this case, $Q$ is the orbifold equivalent of being toroidal. The second reference uses the term toric for this case, but we will follow the first reference an say that an orbifold $Q$ is orbifold-atoroidal. Before the precise definition, notice that a torus $T^2$ has a unique orientable 2-fold quotient $S^2(2,2,2,2)$ (a 2-sphere with four cone points of order 2, sometimes called a pillow case). A rational tangle in a 3-ball can be obtained by taking two embedded unknotted arcs in 3-ball and then twisting along their boundaries. Assuming $Q$ irreducible, an embedded $T^2$ is incompressible if it does not bound solid torus to one side. In an irreducible 3-orbifold, an embedded $S^2(2,2,2,2)$ is incompressible if it does not bound a rational tangle to one side. If $Q$ contains an incompressible $T^2$ or $S^2(2,2,2,2)$, then we say $Q$ is orbifold-toroidal and otherwise $Q$ is orbifold-atoroidal. (The more general definition of orbifold-atoroidal involves considering more general quotients of $T^2$.) For the specific instance above, $Q$ is orbifold-toroidal if $L$ is a satellite link or if $L$ contains an essential Conway sphere, which is really just a concise restatement of the above paragraph. The decomposition of $L$ along satellite tori and essential Conway spheres leads to a JSJ-decomposition of $M_L$.<|endoftext|> -TITLE: Growth of Class Numbers -QUESTION [6 upvotes]: There is a classical formula stating: -Let $K$ be a number field with ring of integers $\mathcal{O}_K\subseteq K$ and -let $\mathcal{O}\subseteq \mathcal{O}_K$ be any non-maximal order with conductor $\mathfrak{n}$. Then -$$\frac{\#(\mathcal{O}_K/\mathfrak{n}\mathcal{O}_K)^\times}{\#(\mathcal{O}/\mathfrak{n}\mathcal{O})^\times} = [\mathcal{O}_K^\times:\mathcal{O}^\times]\frac{\# \text{Pic}(\mathcal{O})}{\#\text{Pic}(\mathcal{O}_K)}.$$ -My question is: do we know anything about the growth on the RHS, especially in the case K is a CM field? In particular, can we bound from below the LHS by the conductor $\mathfrak{n}$? I am in particular interested when the CM field corresponds to an isotypic CM abelian variety of dimension greater than $1$. - -REPLY [7 votes]: Not a complete answer, but for $K/\mathbb Q$ imaginary quadratic, this is answered by Exercise 4.12 in Shimura's Arithmetic Theory of Automorphic Forms. The exercise reads (in your notation) -$$ -\#\text{Pic}(\mathcal O)=\#\text{Pic}(\mathcal O_K)\cdot n\cdot [\mathcal{O}_K^\times:\mathcal{O}^\times]^{-1}\cdot \prod_{p\mid n} -\left[1-\left(\frac{K}{p}\right)p^{-1}\right], -$$ -where the conductor of the order is $\mathfrak n=n\mathbb Z$ with $n$ a positive integer, and with $\left(\frac{K}{p}\right)$ is $1$, $-1$, $0$ according to whether $p$ splits, is inert, or is ramified in $K$. So the quantity that you have written is -$$ -n\cdot \prod_{p\mid\mathfrak{n}} -\left[1-\left(\frac{K}{p}\right)p^{-1}\right]. -$$ -This gives the desired lower bound of $\gg n$ unless $n$ is composed of a very large number of small primes that split in $K$. In any case, one can use the standard estimate -$$ -\prod_{p\mid n} [1-p^{-1}] \gg \frac{1}{\log\log n} -$$ -to get a lower bound that's just a bit worse than $\gg n$.<|endoftext|> -TITLE: On the definition of "almost-everywhere" for non-complete measure spaces -QUESTION [8 upvotes]: If $(X,\mathcal{B},\mu)$ is a (non-necessarily complete) measure space, we can give two different notions of a property $P(x)$ that is true almost-everywhere : -(D1) There is a measurable set $A$ such that $\mu(X\backslash A)=0$ and such that for all $x\in A$, $P(x)$ is true. -(D2) The subset of all $x\in X$ such that $P(x)$ holds is measurable and its complement has zero measure. -(D1) seems the one commonly used, and is obviously weaker than (D2). -Of course, (D1) and (D2) are equivalent when the measure is complete. -My question is : if we take (D2) as the definition of "almost-everywhere", do we "lose" anything fundamental in the theory of measure and integration ? -It seems to me (but I'm not sure) that everything is still true, for instance, -the integral of a positive measurable function is zero implies that the function is zero almost everywhere in the sense of (D2). -The Fubini theorem seems again to hold with (D2) too. The completeness of $L^p$ spaces (with the equivalence relation associated to almost everywhere equality corresponding to (D2)) seems to hold again (but I'm not completely sure). -I ask this question because even if (D2) is a little more annoying to use when proving that something holds almost-everywhere (we have to check measurability of the set on which the property holds), in the end it gives statements that are more precise (because precisely mesurability is implied). Therefore, I am wondering if I'm not missing anything by using (D2) only. - -REPLY [3 votes]: There are relatively concrete examples of a Borel set $E$ in the plane $\mathbb R^2$ such that the projection -$$ -p_1(E) = \{ x \in \mathbb R | \exists y, (x,y) \in E\} -$$ -is not a Borel set. (But it is a Lebesgue set.) LINK -So, as far as Borel sets in the line are concerned, condition $p_1(E)$ is a non-measurable condition. There is a Borel set $Q \supset p_1(E)$ such that $Q \setminus p_1(E)$ is a null set. And then: -$\qquad$for almost all $x \in Q$ there exists $y$ such that $(x,y) \in E$ -is true in sense (D1) but not in sense (D2).<|endoftext|> -TITLE: Are Diagonally dominant Tridiagonal matrices diagonalizable? -QUESTION [6 upvotes]: I have searched this in the literature but could not find any reference, so I would like to post it here. Hope this is at the research level. -Assume that -$$ - A=\begin{vmatrix} -a_1 & b_1 \\ -c_1 & a_2 & b_2 \\ -& c_2 & \ddots & \ddots \\ -& & \ddots & \ddots & b_{n-1} \\ -& & & c_{n-1} & a_n -\end{vmatrix} -$$ -where $a_i< 0,\quad 0\leq c_i,b_i\quad \forall i \quad $, $a_1+b_1=0$, $c_{n-1}+a_n=0$ and $c_{i-1}+a_i+b_i=0,\quad 2\leq i\leq n-1$. -My test using matlab with random matrices and they all diagonalizable. This makes me believe that $A$ should be diagonalizable, however, I have not found a simple proof for it. -My question here is : is $A$ diagonalizable ? can we find a "simple" proof for it? Thank you for your time! - -REPLY [13 votes]: Counterexample: -$$ -\begin{bmatrix} --1 & 1 & 0 & 0\\ -0 & -1 & 1 & 0\\ -0 & 0 & -2 & 2\\ -0 & 0 & 2 & -2 -\end{bmatrix} -$$ -is defective: its eigenvalues are $-1,-1, 0, -4$ (it is block triangular, so its eigenvalues are those of the $2\times 2$ blocks on the diagonal), but $A+I$ has rank 3. -Strategy to construct it: -(added to the answer on request of @EmilioPisanty) -My first thought was: let's take all the $c_k$ equal to $0$ and make a Jordan block -$$ -\begin{bmatrix} --1 & 1 \\ -& -1 & 1\\ -&& \ddots & \ddots \\ -&&& -1 -\end{bmatrix}. -$$ -This almost works, but the last row has to be adjusted to satisfy the zero-sum condition. My idea was adjusting it by adding rows and columns (but the other natural idea of changing $c_{n-1}$ only can be made to work, too; see below). It is a known fact that if $T_{11}$ and $T_{22}$ have no eigenvalues in common, then -$ -\begin{bmatrix} -T_{11} & T_{12}\\ -0 & T_{22} -\end{bmatrix} -$ -and -$ -\begin{bmatrix} -T_{11} & 0\\ -0 & T_{22} -\end{bmatrix} -$ -are similar (proof: $$ -\begin{bmatrix} -I & X\\ -0 & I -\end{bmatrix} -\begin{bmatrix} -T_{11} & T_{12}\\ -0 & T_{22} -\end{bmatrix} -\begin{bmatrix} -I & X\\ -0 & I -\end{bmatrix}^{-1} -= -\begin{bmatrix} -T_{11} & T_{12}+XT_{22}-T_{11}X\\ -0 & T_{22} -\end{bmatrix}, -$$ -and the Sylvester equation $T_{11}X-XT_{22}=T_{12}$ is solvable whenever $T_{11}$ and $T_{22}$ have disjoint spectra). -So I just took $T_{11}$ a Jordan block, $T_{22}$ any matrix with no eigenvalues in common with it, and I knew that the resulting matrix had to have a Jordan block, too, no matter what $T_{12}$ was. -Alternate idea -Let's adjust the Jordan block by making the minimum possible change: alter $c_{n-1}$: -$$A= -\begin{bmatrix} --1 & 1 \\ -& -1 & 1\\ -&& \ddots & \ddots \\ -&&& -1 & 1\\ -&&&& -1 & 1\\ -&&&&1 & -1 -\end{bmatrix}. -$$ -This matrix is still block triangular, so its eigenvalues are $n-2$ times $-1$, and then whatever the eigenvalues of $\begin{bmatrix}-1 & 1\\ 1 & -1\end{bmatrix}$ are. Does it still have a Jordan block, or is it diagonalizable? I guess it probably has a Jordan block: matrices with a Jordan block have more degrees of freedom, so I expect them to be "generic" among the matrices with multiple eigenvalues. Let's just throw it into Wolfram Alpha to check and hope it works. Bingo! Oh, wait, in retrospect it is obvious that it works, because $A+I$ has clearly rank $n-1$. - -REPLY [9 votes]: All real tridiagonal matrices with $b_kc_k>0$, are diagonalizable, and their spectra are real and simple. -See, for example, Gantmakher and Krein, Oscillation matrices and kernels..., AMS 2002. -Sketch of the proof. Expanding the determinant $|A-\lambda I|$ write a recurrent formula for characteristic polynomials of truncated matrices. It is seen from this formula that the eigenvalues depend only on $a_k$ and the products $c_kb_k$. -Therefore the symmetric matrix with $c_k^\prime=b_k^\prime=\sqrt{c_kb_k}$ has the same spectrum. This shows that the spectrum is real. -To show that all eigenvalues are distinct one applies Sturm's theorem to the -sequence of characteristic polynomials. -On the other hand, if you allow all $c_k=0$, for example, you can have a Jordan cell which is not diagonalizable.<|endoftext|> -TITLE: q-versions of the geometric distribution and their names -QUESTION [9 upvotes]: I'm trying to set straight various $q$-deformations of the standard geometric distribution. -The geometric distribution on $\left\{ 0,1,\ldots \right\}$ is well-known, it has -$$ -\mu_1(X=j)=(1-p)p^j,\qquad j=0,1,2,\ldots. -$$ -Here $0 -TITLE: What is a triangle? -QUESTION [33 upvotes]: So I've been reading about derived categories recently (mostly via Hartshorne's Residues and Duality and some online notes), and while talking with some other people, I've realized that I'm finding it difficult to describe what a "triangle" is (as well as some other confusions, to be described below). -Let $\mathcal{A}$ be an abelian category, and let $K(\mathcal{A})$ and $D(\mathcal{A})$ be its homotopy category and derived categories respectively. -By definition, in either $K(\mathcal{A})$ or $D(\mathcal{A})$, -(1) A triangle is a diagram $X\rightarrow Y\rightarrow Z\rightarrow X[1]$ which is isomorphic to a diagram of the form -$$X\stackrel{f}{\rightarrow}Y\rightarrow Cone(f)\rightarrow X[1]$$ -This is the definition, though I don't really understand the motivation. -Somewhat more helpful for me, is the definition: -(2) A cohomological functor from a triangulated category $\mathcal{C}$ to an abelian category $\mathcal{A}$ is an additive functor which takes triangles to long exact sequences. -Since taking cohomology of a complex (in either $K(\mathcal{A})$ or $D(\mathcal{A})$) is a cohomological functor, this definition tells me that I should think of a triangle as being like "a short exact sequence" (in the sense that classically, taking cohomology of a short exact sequence results in a long exact sequence). This idea is also supported by the fact: -(3) If $0\rightarrow X^\bullet\rightarrow Y^\bullet\rightarrow Z^\bullet\rightarrow 0$ is an exact sequence of chain complexes, then there is a natural map $Z^\bullet\rightarrow X[1]^\bullet$ in the derived category $D(\mathcal{A})$ making $X^\bullet\rightarrow Y^\bullet\rightarrow Z^\bullet\rightarrow X[1]^\bullet$ into a triangle (in $D(\mathcal{A})$). -This leads to my first precise question: Can there exist triangles in $D(\mathcal{A})$ which don't come from exact sequences? If so, is there a characterization of them? Is (3) false in the homotopy category $K(\mathcal{A})$? (certainly the same proof doesn't work). -I sort of expect that the answers to the first and last questions above to both be "Yes", which makes the comparison between triangles and "exact sequences" a bit weird. Of course, $K(\mathcal{A})$ and $D(\mathcal{A})$ are almost never abelian categories, and so it's "weird" to talk about exact sequences there. -I suppose at a fundamental level, I find the "homotopy category" somewhat mysterious. I don't completely understand it's role in the construction of the derived category, since after all homotopy equivalences are quasi-isomorphisms. I also find it difficult to internalize this notion of "homotopic morphisms of chain complexes". To me, it's just a "technical trick" which allows one to do all this magic with mapping cones which allows $K(\mathcal{A})$ to be a triangulated category, whereas the normal abelian category of chain complexes is not. I can prove things with it, but whenever I do, I sort of feel unsettled - as if I'm playing with something 'magical' that could, at any moment, turn on me unexpectedly. -In addition to my specific questions above, I suppose I was hoping that someone would be able to articulate in a nice way how one should think of triangles, why this notion of a triangulated category is so successful, and hopefully alleviate some of my unsettlement regarding homotopy. - -REPLY [6 votes]: This is not an answer to your question, but I think that you'll find this piece of information to be useful: - -A triangle in a stable infinity category is given by finitely many pieces of data: -• Three objects: $A$, $B$, $C$. -• Three morphisms: $f:A\to B$, $g:B\to C$, $h:C\to A[1]$ -• Two 2-morphisms: - $\alpha:0\to gf$ - $\beta:kg \to 0$ -The condition that these objects, morphisms, and 2-morphisms need to satisfy is that the loop $h\alpha\cup\beta f\in \Omega Hom(A,A[1])=Hom(A,A)$ (known as a Toda bracket) - should be homotopic to the element $id\in Hom(A,A)$. - -A very nice consequence of the fact that triangles can be expressed in the above simple way is that any functor between stable infinity categories sends triangles to triangles. -In that sense, triangles in stable infinity categories are rather different from short exact sequences in abelian categories.<|endoftext|> -TITLE: Does the Prime Number Theorem have anything to do with Erdos-Kac law or vice versa? -QUESTION [10 upvotes]: The prime number theorem says on average we can find $\frac n{\log n}$ primes of magnitude $n$. -Erdos-Kac law state a typical number of magnitude $n$ has $\log\log n$ primes. -Somehow the fact $e^{\log\log n}=\log n$ seems to make both facts related. -How related are these two facts in number theory? - -Let me put it in this way. -Suppose we somehow naturally choose a subset of natural numbers where Erdos-Kac law applies but with average factors as $f(\log\log n)$ for some $f(x)=o(x)$ then how many primes of magnitude $n$ can we expect in this subset? Say for instance $f(\log\log n)=(\log\log n)^\alpha$ or $f(\log\log n)=(\log\log\log n)^{1/\alpha}$ where $0<\alpha\ll1$ holds? -I think such sets should typically have $\frac{n}{2^{f(\log\log n)}}\gg\frac{n}{\log n}$ primes. - -REPLY [10 votes]: The right context of your question is probably the area of Beurling primes. An arithmetic semigroup is a semigroup $(G,\cdot)$ together with a norm $\|\cdot\|:G\rightarrow[1,\infty)$, such that $\|gh\|=\|g\|\|h\|$ and for all $x$ we have that $N(x) = \#\{g\in G: \|g\|\leq x\}$ is finite. Define $\pi(x)$ as the number of indecompsable elements of $G$ with norm $\leq x$. Beurling proved that $N(x)=cx+\mathcal{O}(\frac{x}{x^{3/2+\delta}})$ implies the prime number theorem in the form $\pi(x)\sim\frac{x}{\log x}$, and the condition is best possible in the sense that for every $\delta>0$ there are groups satisfying $N(x)=cx+\mathcal{O}(\frac{x}{x^{3/2-\delta}})$ and $\pi(x)\not\sim\frac{x}{\log x}$. -The theory of Beurling primes has two motivations: On one hand it has applications to algebra and logic, see the books "Abstract analytic number theory" by Knopfmacher and "Number theoretic densities and logical limit laws" by Burris. On the other hand it serves as a context in which "equivalence" of statements about primes can be made precise. -In this context your question becomes: Is the set of arithmetic semigroups satisfying the prime number theorem equal to the set of semigroups satisfying Erdos-Kac? Erdos-Kac is essentially equivalent to $\sum_{p\leq x}\frac{1}{p}\sim\log\log x$, which is obviously implied by the prime number theorem, but Pollack ( http://pollack.uga.edu/beurling.pdf Wayback Machine ) showed that $\sum_{p\leq x}\frac{1}{p}\sim\log\log x$ is equivalent to the statement $\zeta_G(s)\sim\frac{A}{s-1}$, where $\zeta_G(s)=\sum_{g\in G}\|g\|^{-s}$. In particular all semigroups with $N(x)\sim Ax$ satisfy Erdos-Kac, while the counterexamples found by Beurling show that Erdos-Kac is strictly weaker then the prime number theory.<|endoftext|> -TITLE: Applications of Schubert calculus -QUESTION [5 upvotes]: Schubert calculus is a venerable field in mathematics where the object of study is the cohomology ring of the Grassmannians. Since it has been around for over a hundred years one might wonder if any real world applications in physics for example have been found. - -QUESTIONS. -(a) Does anybody know of such applications? -(b) Moreover, have applications to other areas of mathematics been found? - -REPLY [3 votes]: This survey paper survey illustrates some particular applications to quantum -information theory. -Another interplay with the six-vertex model in physics can be found in this paper, for instance.<|endoftext|> -TITLE: What is the simplest SFT on $\mathbb{Z}^2$ that has no periodic points? -QUESTION [6 upvotes]: An SFT (shift of finite type) is a set of maps to some finite alphabet that is defined by a finite number of disallowed finite words. -By simple I mean has a small alphabet and a small number of disallowed words. - -QUESTION. What are some of the simplest SFTs on $\mathbb{Z}^2$ that have no periodic points? - -REPLY [10 votes]: Wang tiles are unit squares with edges marked with colors, and the problem of whether a given set of Wang tiles can tile the plane such that edges of adjacent squares match has been studied exhaustively (see https://en.wikipedia.org/wiki/Wang_tile). In particular, there is a set of 11 Wang tiles using four colors which tiles the plane, but there is no periodic tiling using them, and this example is essentially minimal for Wang tilings. -This provides a $\mathbb{Z}^2$ SFT whose alphabet has 11 symbols, and with forbidden words that are pairs of adjacent (vertically or horizontally) tiles whose edges do not match. -I don't know if this qualifies as being "simple", but as these things go it's not too bad: the original construction of a set of Wang tiles with this property had 20,426 tiles in it.<|endoftext|> -TITLE: Phase transitions between Category Theories -QUESTION [8 upvotes]: question: What are the mathematical theories suitable to describe the "continuous Phase transitions between Category Theories"? The phase transitions mean that in terms of the quantum statistical physics, in the sense as the second order or higher order continuous phase transitions between two distinct phases. - -Category Theories can be useful to describe the intrinsic topologically ordered quantum matter. The bulk part phases $A$ and $B$ of the 2+1 spacetime dimensional topological order can be constructed using a unitary tensor category $\mathcal{C}$, denoted $\mathcal{C}_A$ and $\mathcal{C}_B$. The boundary of the topological order described by tensor category $\mathcal{C}$ is associated with a module category over $\mathcal{C}$. See for example, the work of Kong-Kitaev. -One can also consider domain walls (or defect lines) between different bulk topological orders, between $\mathcal{C}_A$ and $\mathcal{C}_B$. If the bulk topological orders $\mathcal{C}_A$ and $\mathcal{C}_B$ are the -unitary modular tensor categories, then the gapped domain walls can be regarded as the bimodule categories between modular tensor categories. One may use the fusion space dimension $\mathcal{W}_{ab}$ that tunneling between two topological orders $\mathcal{C}_A$ and $\mathcal{C}_B$, to label different types of gapped domain walls. Here $a$ and $b$ are the objects (the anyon or quasiparticle contents) of the unitary modular tensor categories theories. For example, the work of Lan-Wang-Wen and Kawahigashi, and the work of Rehren (the coupling matrix), Davydov, Müger, Nikshych and Ostrik. -However, the domain walls (or defect lines) between different bulk topological orders are the spatial junction between different bulk topological orders, or the spatial junction between different Category Theories. Do we have methods to describe the temporal junction between different bulk topological orders, say along the phase transitions along the time direction by tuning some coupling parameters in the quantum Hamiltonian? Namely, the temporal junction between different two (unitary modular tensor) category $\mathcal{C}_A$ and $\mathcal{C}_B$? This is the (temporal) phase transition between the two Category Theories. What are the mathematical theories suitable to describe the "(temporal) Phase transitions between Category Theories"? Such that the (temporal) Phase transitions is continuous and smooth? As the second order or higher order continuous phase transitions in terms of the statistical physics. - - - -Other information: -This table bridges the simple terminology between physical excitations and Category Theories - -This table bridges the terminology between topological orders and Category Theories: - -REPLY [5 votes]: In general, we expect field theories to be described by some higher categorical structures, where bulk models are assigned objects (also called 0-morphisms), domain walls are assigned morphisms (also called 1-morphisms), and higher-codimension defects are assigned higher morphisms. For the example of Levin-Wen models, there is a 3-category of tensor categories, where the morphisms are bimodule categories, and 2-morphisms are functors between bimodule categories. In this context, there does not seem to be a substantial difference between spacial domain walls and temporal domain walls. -If you want a continuously parametrized phase change, then you may want to consider simplicial maps from the singular complex of the interval to the $(\infty,1)$-category of "tensor categories" (or whatever your bulk structures happen to be). For Levin-Wen, each point in the interval is assigned a tensor category, and there are compatible morphisms (i.e., bimodules) between each pair of points. More generally, you may replace the interval with your spacetime manifold to get a fully varying family of tensor categories. Smoothness requires some infinitesimal structure that I do not know how to derive. -I do not know of a reference for your precise question, but I think Gaitsgory-Lurie's work on chiral categories has some information on families or local systems of $(\infty,1)$-categories. There is also a brief treatment of "constructible sheaves" in an appendix of Lurie's "Higher Algebra", but the topos language may be difficult to translate.<|endoftext|> -TITLE: Changing the signs of the coefficients of a polynomial to make all the roots real -QUESTION [24 upvotes]: We are given a polynomial -$$P_n(x):=a_nx^n + a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$ -with real coefficients. - -Questions. -$\boldsymbol{(i)}$ How can we determine if there are $\epsilon_1,\ldots,\epsilon_n\in\{-1,+1\}$ such that - $$P_n(x;\pmb{\epsilon}):=\epsilon_n a_nx^n + \epsilon_{n-1} a_{n-1}x^{n-1}+\cdots+\epsilon_1 a_1x+a_0$$ - has only real roots? Here $\pmb{\epsilon}=(\epsilon_1,\dots,\epsilon_n)$. -$\boldsymbol{(ii)}$ If there is a solution, how can we find it? - -Obviously an exhaustive search is hopelessly slow. This question might just possibly (on a good day) be of relevance to a problem on graph polynomials. - -REPLY [9 votes]: Perhaps it is useful to observe that if $P_n$ has only real roots, -the same is true for its derivative?<|endoftext|> -TITLE: A determinant inequality -QUESTION [17 upvotes]: Notation: Suppose $\mathbf{A}$ and $\mathbf{B}$ are positive definite matrices in $\mathbb{R}^{n\times n}$ such that $\mathbf{A} \succeq \mathbf{B}$ (Loewner order). Let $\mathcal{S}(n,k)$ be the set of all $k$-subsets of $\{1,2,\dots,n\}$. For any $\mathcal{Q} \subset [n] \triangleq \{1,2,...,n\}$, $\mathbf{M}_\mathcal{Q}$ is the matrix obtained by deleting the rows and columns of matrix $\mathbf{M}$ whose indices are in $\mathcal{Q}$. Similarly, $\mathbf{x}_\mathcal{Q}$ is the vector obtained by deleting the rows of vector $\mathbf{x}$ whose indices are in $\mathcal{Q}$. -Conjecture: For any such $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{x} \in \mathbb{R}^n$ and for all $k \in [n-1]$: -$$ -\frac{\sum_{\mathcal{Q} \in \mathcal{S}(n,k)} \det(\mathbf{A}_\mathcal{Q} + \mathbf{x}_\mathcal{Q}\mathbf{x}_\mathcal{Q}^\top)}{\sum_{\mathcal{Q} \in \mathcal{S}(n,k)} \det(\mathbf{A}_\mathcal{Q})} -\leq -\frac{\sum_{\mathcal{Q} \in \mathcal{S}(n,k)} \det(\mathbf{B}_\mathcal{Q} + \mathbf{x}_\mathcal{Q}\mathbf{x}_\mathcal{Q}^\top)}{\sum_{\mathcal{Q} \in \mathcal{S}(n,k)} \det(\mathbf{B}_\mathcal{Q})} -\tag{1} -$$ -Progress so far: - -We have $\det(\mathbf{A}_\mathcal{Q}+\mathbf{x}_\mathcal{Q}\mathbf{x}_\mathcal{Q}^\top) = \det(\mathbf{A}_\mathcal{Q})\,(1+\mathbf{x}_\mathcal{Q}^\top \mathbf{A}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q})$. Therefore, (1) can be rewritten as (2): -$$ -\tag{2} -\sum_{\mathcal{Q} \in \mathcal{S}(n,k)}\frac{ \det(\mathbf{A}_\mathcal{Q})\,}{\sum_{\mathcal{Q} \in \mathcal{S}(n,k)} \det(\mathbf{A}_\mathcal{Q})} \, -\mathbf{x}_\mathcal{Q}^\top \mathbf{A}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q} -\leq -\sum_{\mathcal{Q} \in \mathcal{S}(n,k)}\frac{ \det(\mathbf{B}_\mathcal{Q})\,}{\sum_{\mathcal{Q} \in \mathcal{S}(n,k)} \det(\mathbf{B}_\mathcal{Q})} \, -\mathbf{x}_\mathcal{Q}^\top \mathbf{B}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q} -$$ -Lemmas: It is easy to show that for any $\mathcal{Q} \subset [n]$: - -(2.1) $\quad \mathbf{A} \succeq \mathbf{B} \Rightarrow \mathbf{A}_\mathcal{Q} \succeq \mathbf{B}_\mathcal{Q}$ -(2.2) $\quad \det(\mathbf{A}_\mathcal{Q}) \geq \det(\mathbf{B}_\mathcal{Q})$ -(2.3) $\quad \mathbf{A}^{-1}_\mathcal{Q} \preceq \mathbf{B}_\mathcal{Q}^{-1} \Rightarrow \mathbf{x}_\mathcal{Q}^\top \mathbf{A}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q} \leq \mathbf{x}_\mathcal{Q}^\top \mathbf{B}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q}$ -(2.4) $\quad \det(\mathbf{A}_\mathcal{Q})\,\mathbf{x}_\mathcal{Q}^\top \mathbf{A}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q} \geq \det(\mathbf{B}_\mathcal{Q})\,\mathbf{x}_\mathcal{Q}^\top \mathbf{B}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q} $ - -Through recursion, it suffices to show that (1) or (2) hold for the special case of $\mathbf{A} = \mathbf{B} + \mathbf{p}\mathbf{p}^\top$ for some $\mathbf{p} \in \mathbb{R}^n$. -A very special case is posted here ($\mathbf{B} = \mathbf{I}$, $\mathbf{A} = \mathbf{I} + \mathbf{p}\mathbf{p}^\top$). -I haven't encountered any counterexample after running "many" simulations (I know this is not a strong argument). - -Update: Let me explain the motivation as requested. -Motivation: -Suppose $\{\mathbf{a}_i\}_{i=1}^{m}$ are some vectors in $\mathbb{R}^n$ ($m \geq n$) and $\mathbf{A}_0 \succ \mathbf{0}$ is a given matrix in $\mathbb{R}^{n \times n}$. Now consider, -$$ -\begin{align} -f_k : 2^{[m]} &\to \mathbb{R}, \\ - \mathcal{S} & \mapsto c_k(\mathbf{A}_0 + \sum_{i \in \mathcal{S}}\mathbf{a}_i\mathbf{a}_i^\top) -\end{align} -$$ -where $c_k(\mathbf{A})$ is the coefficient of $x^k$ in the characteristic polynomial of $\mathbf{A}$, i.e., $\det(x\mathbf{I} - \mathbf{A})$. -Conjecture: $f_k$ is monotone log-submodular (multiplicative submodular) for all $k \in \{0,1,\dots,n\}$. -I have proved the monotonicity (maybe for $|f_k|$). -Special Cases: This holds for $k=n-1$ (trace), $k=0$ (determinant is log-submodular) and $k=n$ (constant, $f_n(\mathcal{S}) = 1$). -Now (1) emerges from the proof of log-submodularity (multiplicative submodularity) of $f_k$ after expressing $c_k$ as the sum of determinants of principal minors. -Applications: - -I came across this when working on graph Laplacian matrices. $f_k$ for Laplacian matrices (and $\mathbf{a}_i = \mathbf{e}_s - \mathbf{e}_r$ where $\{\mathbf{e}_s\}_{s=1}^n$ is the standard basis) is related to the weighted number of spanning trees. I recently showed that the weighted number of spanning trees is a monotone log-submodular function of the edge set (see a draft here). Other coefficients can be also be nicely related to the weighted number of spanning trees as shown by Alexander Kelmans (as a generalization of Kirchhoff's matrix tree theorem). For Laplacian matrices, (2) has a beautiful interpretation in terms of the expected value of the effective resistance ("distance") between two vertices after performing some random operations on the graph. -(1) and (2) also arise in $k$-DPPs (determinantal point process). - -REPLY [6 votes]: As suspected, the desired inequality actually holds for all hyperbolic polynomials; the inequality in the OP follows as corollary (Corollary 1) to Theorem 2 (which seems to be new). -We will need the following remarkable theorem to obtain our result. - - -Theorem 1 (Bauschke, Güler, Lewis, Sendov, 2001) Let $p$ be a homogenous hyperbolic polynomial; let $v$ be a vector in the strict interior of the hyperbolicity cone $\Lambda_{++}$ of $p$. Then, - \begin{equation*} - g(x) := \frac{p(x)}{Dp(x)[v]} -\end{equation*} - is concave on $\Lambda_{++}$. - - -This theorem helps prove the more general inequality (also conjectured by Denis Serre above). - - -Theorem 2. Let $p$ be a homogenous hyperbolic polynomial with hyperbolicity cone $\Lambda_{++}$. Let $a, b, c \in \Lambda_{++}$. Then, $p$ satisfies the (conic log-submodularity) inequality: - \begin{equation*} - \tag{0} - p(a)p(a+b+c) \le p(a+b)p(a+c). - \end{equation*} - - -Proof. -Let $c \in \Lambda_{++}$ be arbitrary. Consider the function $f(a) := \frac{p(a+c)}{p(a)}$. Inequality (0) amounts to showing that $f(a)$ is monotonically decreasing on the cone $\Lambda_{++}$. Equivalently, we consider $\log f$ and show that its derivative is negative in the direction $v$. That is, for an arbitrary direction vector $v\in \Lambda_{++}$, we show that - \begin{equation} - \tag{1} - \frac{Dp(a+c)[v]}{p(a+c)} - \frac{Dp(a)[v]}{p(a)} \le 0\quad\Longleftrightarrow\quad - \frac{p(a+c)}{Dp(a+c)[v]} \ge \frac{p(a)}{Dp(a)[v]}. - \end{equation} - But from Theorem 1, we know that $\frac{p(x)}{Dp(x)[v]}$ is concave. Moreover, since $p$ is homogenous, from its concavity we obtain its superadditivity - \begin{equation*} - \frac{p(a+c)}{Dp(a+c)[v]} \ge \frac{p(a)}{Dp(a)[v]} + \frac{p(c)}{Dp(c)[v]}, - \end{equation*} - which is stronger than the desired monotonicity inequality (1) (since all terms are nonnegative). - - -Corollary 1. Let $E_k(A) = e_k \circ \lambda(A)$ denote the $k$-th elementary symmetric polynomial of a positive definite matrix $A$. Then for any positive definite $A, B, C$ we have - \begin{equation*} - E_k(A)E_k(A+B+C) \le E_k(A+B)E_k(A+C). - \end{equation*} - This log-submodularity, immediately implies the log-submodularity of $f_k(S) := E_k(A+\sum\nolimits_{i\in S}v_iv_i^T)$.<|endoftext|> -TITLE: Matrix diagonalization and eigenvector computation constructively -QUESTION [8 upvotes]: Assuming Bishop's constructive mathematics, is it true that any real-valued square matrix with distinct roots of the characteristic polynomial can be diagonalized? By distinct, I mean apart: $x \neq y \triangleq \exists q \in \mathbb{Q}.|x - y| > q$. (There may be similar defnitions) -For the case $n=2$, it seems true since we can deduce $a_{11} - \lambda_j \neq 0 \lor a_{22} - \lambda_j \neq 0$ for any $j = 1, 2$ where $a_{ii}, i=1,2$ are the diagonal elements and $\lambda_j$s are the roots of the characteristic polynomial. It then allows multiplying by a nonzero number and, using the property that $(a_{11} - \lambda_j)(a_{22} - \lambda_j) = a_{12}a_{21}$, solving the respective system of linear equations and consequently finding an eigenvector: $A v = \lambda_j v$. -Already for the case $n=3$, the argument seems not to work and one cannot proceed without some case distinction on reals. This work addresses the problem in Lemma 1.5, but they seem to assume $xy = 0 \implies x =0 \lor y=0$ which is not valid constructively. -Coquand and Lombardi in Theorem 2.3 constructed an effective procedure of finding eigenvectors of a projection matrix. I suspected that something like this could be done for general matrices where it is known beforehand that the roots of the characteristic polynomial are distinct. -Lombardi and Quitte addressed the problem on page 100 (top). Also in Proposition 5.3, they claim that -$$\prod_{i=1}^n ( (A - \lambda_i I)_{1 \dots n-1, 1 \dots n-1} ) \neq 0$$ -by "exhibiting" the companion matrix of $\lambda^n - 1$. Their language is a bit obscure. However, it seems they imply the following lemma: -Let $A = M(\mathbb{R},n)$, and suppose that the roots $\lambda_1, \dots, \lambda_n$ of the charteristic polynomial for $A$ are mutually apart. Then, for every root $\lambda_j$ of the characteristic polynomial, there is a principal submatrix of $A - \lambda_j I$ of size $(n-1) \times (n-1)$ which is invertible. - -REPLY [3 votes]: The following works constructively over an arbitrary local ring $R$ (constructively, $\mathbb{R}$ is a local ring). -Assume that you matrix $M$ is canceled by a polynomial $Q$, of degree $m$ (with leading coefficent $1$), that $Q$ can be factored into -$$Q= \prod^m_{i=1}(X-q_i)$$ -and that for each $i \neq j$, $(q_i - q_j)$ is invertible (in the case of $\mathbb{R}$ it just means that $q_i$ and $q_j$ are appart). -Then one can diagonalize $M$. In particular it can be applied when you have some multiplicites in your eigenvalue as long as the minimal polynomial has simple roots. -Let -$$P_i = \frac{\displaystyle \prod_{j \neq i} (X-q_j)}{ \displaystyle \prod_{j \neq i} (q_i-q_j)} $$ -i.e. $P_i(q_i)=1$ and if $i \neq j$ , $P_i(q_j)=0$ -It is constructive that a polynomial is divisibe by $Q$ if and only if it vanishes at all the $q_i$ (basically because polynomial division by a polynomial with unit leading coefficient works well). Hence you can easily check that: -$(P_i)^2 - P_i$ -$P_i P_j$ -$1-\sum_i P_i$ -$X-\sum_i q_i P_i$ -are all divisible by $Q$ and hence cancel $M$. -So the $P_i(M)$ form a complete family of projection, and $M = \sum q_i P_i(M)$ and you get the spectral decomposition of $M$. -At this point, if you already know how to find eigen vectors for projections then you are done: the range of each $P_i$ are in direct sum so you just find a basis of each of these subspaces and you get your diagonal basis. -I have not read the paper by Coquand & Lombardi you are quoting so I'm not sure what is their method for the last step. I know of a process for this that work for local ring of 'zero residual characteristic' (i.e. in which integer are invertible sot ) I'm not sure what happen in the fully general case, but it is already enough for the case of $\mathbb{R}$.<|endoftext|> -TITLE: Hierarchy of Grothendieck's SGA, EGA, FGA -QUESTION [7 upvotes]: I was thinking about a possible hierarchy for the top three Grothendieck's works: EGA,SGA,FGA. But I haven't read all these works, and so I'm asking if there is actually such a hierarchy. -Here the word hierarchy means a possible "reading path" of all these three works, which would be the first to be read in order to make basis for the second work and so on. -Here the publication periods: -EGA - 1960-1967 -SGA - 1960-1969 -FGA - 1957-1962 - -REPLY [18 votes]: In a few words: EGA is previous to everything, though one can use SGA 1 to complement some aspects of EGA IV. FGA goes "in between". There is a complicated tree for SGA. SGA 3 is independent of the rest while SGA 4, SGA 5 and SGA 7 are a full saga. SGA 6 is more or less independent but you need Verdier's thesis (or its resume at the end of SGA 4 1/2). In fact, SGA 4 1/2 can be considered as an introduction of certain aspects of SGA 4. FGA has three topics: formal schemes, duality and representable functors and should be read "as needed". The new edition of EGA I is a very nice reference. -It is perhaps interesting to point out that it is unrealistic try to master all of this material in a short amount of time. Perhaps one should study one of the manuals and then rely on EGA and SGA for further topics and additional details. -To mention a few good references (without being exhaustive): Hartshorne, Görtz-Wedhorn, Mumford-Oda, Liu and Bosch.<|endoftext|> -TITLE: Semisimple representations of discrete groups -QUESTION [12 upvotes]: Let $G$ be a discrete group. Let $V$ and $W$ be two finite dimensional complex simple $G$ representations. - -QUESTION. Must the tensor product $V\otimes_{\mathbb{C}} W$ with the diagonal action be a semi-simple representation? (that is- can we write it as the direct sum of simple representations?). - -In case the group is abelian every simple representation is one dimensional, and the answer is therefore yes. In case $G$ is finite, all representations of $G$ are semi-simple. What happens in the general case? - -REPLY [14 votes]: This is a classic result of Chevalley, and I think the following purely algebraic proof technique is due to Serre. The beautiful trick, which grew into a powerful method, is to take Zariski closures of the images of $G$ in various ${\rm{GL}}_n$'s and to exploit the structure theory of algebraic groups (including the finiteness of component groups thereof) to separately reduce the problem to the easy cases of finite groups and tori and the case of representation theory of semisimple Lie algebras. It is the possibility to bring in Lie algebra considerations for representations of a given "discrete" group that is the great idea. -I prefer to write $\Gamma$ rather than $G$ to denote the given discrete group, so our setup will be a pair of semisimple finite-dimensional representations $(V, \rho)$ and $(V', \rho')$ of $\Gamma$ over a field $k$ of characteristic 0. We want to show that $V \otimes V'$ equipped with the natural representation $r$ defined by the condition $r(\gamma): v \otimes v' \mapsto (\gamma v) \otimes (\gamma v')$ is also semisimple. -Let $G \subset {\rm{GL}}(V \otimes V')$ be the Zariski closure of $r(\Gamma)$ (this is why I changed the original notation for the discrete group with which we begin). This is a smooth affine $k$-group, possibly disconnected, but we claim that we can say something non-trivial about the structure of $G$: it is reductive (in the sense that the unipotent radical of $G_{\overline{k}}$ is trivial). -Consider the natural map $$f:{\rm{GL}}(V) \times {\rm{GL}}(V') \rightarrow {\rm{GL}}(V \otimes V')$$ that satisfies $r = f \circ (\rho \times \rho')$. Let $H \subset {\rm{GL}}(V)$ and $H' \subset {\rm{GL}}(V')$ denote the respective Zariski closures of $\rho(\Gamma)$ and $\rho'(\Gamma)$, so $H \times H'$ contains the Zariski closure $\mathscr{H}$ of $(\rho \times \rho')(\Gamma)$. The magic begins with: -Lemma. The $k$-group $\mathscr{H}$ is a reductive group. -Proof: Under the projections $H \times H' \rightrightarrows H, H'$, the images of $\mathscr{H}$ are dense (contain $\rho(\Gamma), \rho'(\Gamma)$ respectively) and closed (as any homomorphism between smooth affine groups over a field has closed image), so both maps $\mathscr{H} \rightrightarrows H, H'$ are surjective. Recall a real theorem: under surjective homomorphisms between smooth affine groups over an algebraically closed field, the image of the unipotent radical of the course is the unipotent radical of the target. Thus, if we can show that $H$ and $H'$ are reductive then $\mathscr{R}_u(\mathscr{H}_{\overline{k}})$ would have trivial images in both $H_{\overline{k}}$ and ${H'}_{\overline{k}}$, yet $\mathscr{H}$ is a $k$-subgroup of $H \times H'$, so this would force $\mathscr{R}_u(\mathscr{H}_{\overline{k}}) = 1$ as desired. -So it suffices to show that $H$ and $H'$ are reductive, and this is exactly where we will use the assumptions that $\rho$ and $\rho'$ are semisimple. Let $U = \mathscr{R}_{u,k}(H)$ denote the so-called $k$-unipotent radical of $H$; i.e., the maximal smooth connected unipotent normal $k$-subgroup. Since $k$ is perfect (even char. 0), it follows by Galois descent that $U_{\overline{k}} = \mathscr{R}_u(H_{\overline{k}})$, so our task is equivalent to showing that $U=1$ (as the case of $H'$ goes in exactly the same way). -Let $V = \oplus V_i$ be the decomposition of $V$ into a direct sum of irreducible representations of $H$ over $k$. Since $U$ is unipotent, by the Lie-Kolchin Theorem the subspaces $V_i^U$ are each nonzero. But $U$ is normal in $H$, so each $V_i^U$ is an $H$-subrepresentation of $V_i$. Irreducibility of $V_i$ for $H$ then forces $V_i^U = V_i$, so $V^U=V$. In other words, $U$ acts trivially on $V$. But this action is defined through the inclusion $U \subset H \subset {\rm{GL}}(V)$, so $U=1$. -QED Lemma -Since $r = f \circ (\rho \times \rho')$, we have $G = f(\mathscr{H})$ because the image under any homomorphism between smooth affine groups (such as $f$) of a Zariski-closed subgroup of the source is a Zariski-closed subgroup of the target. But we also know that any quotient of a reductive group is always reductive (no connectedness hypotheses!), so the Lemma implies that $G$ is reductive (which is equivalent to that for its identity component $G^0$, since $\mathscr{R}_u(\mathscr{G}) = \mathscr{R}_u(\mathscr{G}^0)$ for general smooth affine groups $\mathscr{G}$ over any algebraically closed field and we recall that $(G^0)_{\overline{k}} = (G_{\overline{k}})^0$). -By design, the image of $\Gamma$ in $G(k)$ is Zariski-dense in $G$. Thus, a subspace of $V$ is $\Gamma$-stable if and only if it is $G$-stable. Thus, the semisimplicity of $V$ as a $\Gamma$-representation is equivalent to its semisimplicity as a $G$-representation. In view of the established reductivity of $G$ (equivalently, of $G^0$), now we are finally cooking with gas: we have transformed the original problem entirely into a question about the representation theory of (possibly disconnected!) reductive groups. -Up to here, we have only used that $k$ is perfect, not that it has characteristic 0. The special features of characteristic 0 now appear with the following fact that completes the proof: -Proposition Any finite-dimensional linear representation of a $($possibly disconnected$)$ reductive smooth affine group $G$ over a field $k$ of characteristic $0$ is completely reducible. -Proof: We need to show that of $W$ and $W'$ are finite-dimensional linear representations of $G$ over $k$ then any extension of $W$ by $W'$ as representations of $G$ is split. Since ${\rm{Hom}}_G(W,\cdot) = {\rm{Hom}}_k(W,\cdot)^G$, it suffices to show that the formation of $G$-invariants is right-exact on finite-dimensional linear representations. -That is, if $W \twoheadrightarrow \overline{W}$ is a surjective map between finite-dimensional $G$-representations then we want to show -$W^G \rightarrow \overline{W}^G$ is surjective. If $N \subset G$ is a smooth closed normal $k$-subgroup (so $N$ is also reductive, as is $G/N$) and we can settle the general right-exactness for $N$-invariants and $G/N$-invariants then we get the same for $G$. Moreover, since $(W^G)_{\overline{k}} = (W_{\overline{k}})^{G_{\overline{k}}}$, for the purpose of proving this right-exactness property we may and do assume $k$ is algebraically closed. -If $Z$ is the maximal central $k$-torus in $G$ then $G/Z$ has semisimple identity component $G^0/Z$ and $G/G^0$ is finite etale. Thus, we are reduced to separately treating three cases over algebraically closed fields of characteristic 0: $\mathbf{G}_m$, finite groups, and connected semisimple groups. The case of $\mathbf{G}_m$ works in any characteristic (as linear representations are the same as $\mathbf{Z}$-gradings), and the case of finite groups is well-known (in characteristic 0!). -Finally, we come to the real substance of the matter: connected semisimple $G$. The structure theory of such groups over $k$ of characteristic 0 tells us that the Lie algebra $\mathbf{g}$ is semisimple in the sense of Lie algebras, so the representation theory of $\mathfrak{g}$ is completely reducible in characteristic $0$. -It therefore suffices to show that if $W$ is a finite-dimensional linear representation of a connected linear algebraic group $G$ over $k$ of characteristic 0 (semisimplicity or even reducivity is not relevant anymore) and we also view $W$ as a linear representation of $\mathfrak{g}$ in the usual way (by passing to Lie algebras on the given homomorphism $G \rightarrow {\rm{GL}}(W)$) then $W^G = W^{\mathfrak{g}}$. -More generally, we claim that for any subspace $V \subset W$, (i) $V$ is $G$-stable if and only if $V$ is $\mathfrak{g}$-stable, and (ii) when those equivalent properties hold, the representation of $G$ on $V$ is trivial if and only if the $\mathfrak{g}$-representation on $V$ is trivial. -For the implication "$\Leftarrow$" in (i) (the converse being obvious), we want to show that if $H \subset {\rm{GL}}(W)$ is a Zariski-closed subgroup (such as the stabilizer of $V$) then a $k$-homomorphism $f:G \rightarrow {\rm{GL}}(W)$ factors through $H$ if and only if it does so on the level of Lie algebras. The scheme-theoretic preimage $f^{-1}(H)$ has Lie algebra that is the preimage of $\mathfrak{h}$ in $\mathfrak{g}$, so the closed $k$-subgroup scheme $f^{-1}(H) \subset G$ has full Lie algebra. But $f^{-1}(H)$ is smooth by Cartier's theorem (char. 0!), so by connectedness and smoothness of $G$ this forces $f^{-1}(H)=G$ by dimension considerations, which is to say that $f$ factors through $H$ as desired. -Assertion (ii) is a special case of the general assertion that any homomorphism between connected linear algebraic groups in characteristic 0 (such as $G \rightarrow {\rm{GL}}(V)$) is determined uniquely by the induced map between Lie algebras, and via consideration of graphs it reduces to showing that a Zariski-connected Zariski-closed subgroup in characteristic 0 is uniquely determined by the corresponding Lie subalgebra. This in turn in a fact proved in many books on algebraic groups (or one can appeal to the analytic theory over $\mathbf{C}$ provided one shows that Zariski-connectedness implies analytic connectedness, which requires a real argument and cannot be considered to be elementary). -QED<|endoftext|> -TITLE: Finding order of vanishing for Jacobi Theta function -QUESTION [6 upvotes]: From Rademacher's book (Topics in Analytic Number Theory) I'm using the functional equation of $\vartheta_2(0|\tau) = 2\sum_{m=0}^\infty q^{\left(m+\frac12\right)^2} = \vartheta_2(\tau)$ and the fact that it vanishes at the cusps $\tau = \frac{a}{b}$, $a$ odd, $b$ even. -The order of vanishing at infinity is just $\frac14$. I'd like to find the order of vanishing at the cusp $\tau = \frac12$ (no particular reason for this choice). Using say $A = \left(\begin{array}{lr} 1 & 0 \\ 2 & 1 \end{array}\right) \in \Gamma_0(2)$ and plugging into the functional equation I get $$\vartheta_2\left(\frac{\tau}{2\tau+1}\right) = i e^{-\pi i/4} \sqrt{-i(2\tau+1)}\vartheta_2(\tau)$$ -As I understand it I'd want to plug $\tau = i\infty$ into this, which would give me $\vartheta_2\left(\frac12\right)$ in terms of $\vartheta_2(i\infty) = q^\frac{1}{4}+...$ However, I don't know how to handle the $\sqrt{2\tau+1}$ term if I was to put $\tau = i\infty$, so I feel like there's a big hole in my understanding. -The equation from Rademacher is given as -$$\vartheta_2\left(\frac{\upsilon}{c\tau+d}\bigg|\frac{a\tau+b}{c\tau+d}\right) = i^{3-d-(a+1)/2}e^{\pi i (dc+ab-ac-1)/4}\sqrt{\frac{c\tau+d}{i}}e^{\pi i c\upsilon^2/(c\tau+d)}\vartheta_{1-c, 1-d}(\upsilon|\tau)$$ -on page 182, for $c>0$, $a$ odd. - -REPLY [6 votes]: The convention when working with modular forms $f(\tau)$ of weight $k$ is for the order of vanishing at a cusp $a/c$ to mean the order of vanishing of $(c\tau+d)^{-k} f\left(\frac{a\tau+b}{c\tau+d}\right)$ as a function of $q = e^{2 \pi i \tau}$. One also speaks of the value of a modular form at a cusp as the constant term in the $q$-expansion. This is a bit different than the usual notion of plugging in points, or examining Taylor expansions that one usually things about in complex analysis. -For example, for the regular Jacobi theta function $\theta(\tau) = \sum_{n \in \mathbb{Z}} q^{n^{2}}$, one has $\theta(-1/\tau) = (-i \tau)^{1/2} \theta(\tau)$. One says that $\theta(\tau)$ is holomorphic and non-vanshing at the cusp at zero because $\tau^{-1/2} \theta(-1/\tau)$ has an expansion in integral powers of $q$ with a nonzero constant term. However, for $r$ real and positive, we have $\theta(ir) \to \infty$ as $r \to 0$, as is fairly clear from the $q$-expansion.<|endoftext|> -TITLE: Continuity of Alexander-Spanier cohomology -QUESTION [7 upvotes]: Suppose that a paracompact space $X$ is the inverse limit of paracompact -spaces $X_{i}$ (that is $X=\varprojlim X_{i}$) and $H^{\ast }$ is Alexander-Spanier -cohomology with closed supports. Then the equality $H^{\ast }\left( X,k\right) =\varinjlim -H^{\ast }\left( X_{i},k\right) $ is true? (where $k$ is a field of characteristic -zero.) - -REPLY [4 votes]: First of all, I am not sure if the below answers the question because of a terminological issue. I only know "closed support" in the context of Borel–Moore homology. For Alexander–Spanier cohomology, being defined in terms of functions where support is by definition closed, "closed support" seems to me to refer to the full complex where all functions are allowed – and this is what I am talking about in the below. -The Alexander–Spanier cohomology can be identified with Čech cohomology. On paracompact spaces, this is proved in Spanier's "Algebraic topology", Chapter 6, Section 8 "Fine presheaves". This actually holds for general spaces by a theorem of Dowker: - -C. Dowker. Homology groups of relations. Ann. Math. 56 (1952), 84–95. - -So the continuity question for Alexander–Spanier cohomology is equivalent to the continuity question for Čech cohomology. However, in my comment I was a bit too hasty about the continuity for Čech cohomology. The classical references (e.g. Bredon's "Sheaf theory", Spanier's "Algebraic topology", or the original papers of Steenrod and Spanier) prove continuity for Čech cohomology on the category of compact Hausdorff spaces. -However, Čech cohomology satisfies a weak continuity property for paracompact Hausdorff spaces, by a result of Lee and Raymond: - -C.N. Lee and F. Raymond. Čech extensions of contravariant functors. Trans. Amer. Math. Soc. 133 (1968), 415–434. (paper available here) - -Specifically, this is Theorem 5 in their paper. Note, however, that all the spaces in the inverse system are required to be embedded into one big space, with maps given by subspace inclusions. They define nested systems to additionally be those where the inverse limit is given by intersection and where every neighbourhood of the limit in the ambient space contains one of the spaces of the system. Weak continuity gives isomorphisms in cohomology only for inverse limits of nested systems. This is all discussed in Section 4 of the Lee–Raymond paper. -There's also a paper of Watanabe - -T. Watanabe. The continuity axiom and the Čech homology. Geometric topology and shape theory (Dubrovnik, 1986), 221–239, Lecture Notes in Math., 1283, Springer, 1987. - -In the paper he discusses at length continuity properties of extensions of cohomology theories. If I'm reading his Corollary 22 right, it says that Čech cohomology is continuous on the category of pairs $(X,A)$ where X is a paracompact Hausdorff space with a closed subset $A$. Note, however, that there Čech cohomology is defined in terms of normal open coverings, i.e., those which have a partition of unity.<|endoftext|> -TITLE: Commutative and Cocommutative Quantum Groups -QUESTION [6 upvotes]: I am using this definition: - -An algebra of functions on a finite quantum group $\mathbb{G}$ is a finite dimensional $C^\ast$-Hopf algebra $A=:F(\mathbb{G})$. - -I have the following (very well known --- folklore --- result) - -(Classification Theorem) -Let $A$ be the algebra of functions on a finite quantum group - $\mathbb{G}$: - -if $A$ is commutative then $\mathbb{G}\cong \Phi(A)$. -if $A$ is cocommutative then $A=F(\mathbb{G})\cong \mathbb{C} \Phi(A)=:F(\widehat{\Phi(A)})$. - - -Here $\Phi(A)$ is the set of characters on $A$. -I want to reference these results but am struggling somewhat to find good, old, authoritative references. Any help would be appreciated. - -REPLY [2 votes]: I took a quick look into Timmermann's book "An invitation to Quantum Groups". -It refers to: -Saad Baaj; Georges Skandalis -Unitaires multiplicatifs et dualité pour les produits croisés de C*-algèbres -Annales scientifiques de l'École Normale Supérieure (1993) -Volume: 26, Issue: 4, page 425-488 -ISSN: 0012-9593 -They describe finite dimensional C${}^*$-Hopf algebras in terms of multiplicative unitaries. -Theorem 2.2 shows that if you have commutative multiplicative unitaries you get a locally compact group and therefore in the finite case, a finite group. -This gives the first statement. The second statement is equivalent to the first by duality. -*edit* -Such a result is already stated as Theorem 3.3 in -L. I. Vaĭnerman and G. I. Kac, Nonunimodular Ring Groups and Hopf-Von Neumann algebras, -Mathematics of the USSR-Sbornik, Volume 23, Number 2<|endoftext|> -TITLE: Does the following class of functions have a name? -QUESTION [7 upvotes]: Consider the functions $\mathbb{Z}\to\mathbb{C}$ that can be expressed as $\mathbb{C}$-linear combinations of functions of the form $g(n)=n^d\zeta^n$, where $d\geq 0$ is an integer and $\zeta$ is a root of unity. They are the functions which satisfy a homogeneous linear recurrence relation with constant coefficients such that the characteristic roots are roots of unity. -Do such functions have a name? Are they familiar things in any corner of mathematics? - -REPLY [3 votes]: A linear recurrence is often said to be degenerate if the ratio of two of its characteristic roots is a root of unity. Many arithmetic results on values of linear recurrences only hold for non-degenerate sequences, e.g., the Skolem-Mahler-Lech theorem. So although I haven't seen the term used before, it seems as if totally degenerate linear recurrence might be appropriate (although your sequences even have the individual characteristic roots as roots of unity, so maybe "extra-totally..."!).<|endoftext|> -TITLE: Who discovered this definition of Stiefel-Whitney classes? -QUESTION [9 upvotes]: I would define Stiefel-Whitney classes as the pullbacks of generators of $H^*(BO, \mathbb{Z}/2)$ under a classifying map, and I gather this is a pretty common definition. -However, the book "Characteristic classes" by Milnor-Stasheff contains a different definition, as ``eigenvalues'' for the Steenrod squares acting on the fundamental class of a manifold in its normal bundle. I want to attribute this construction to somebody for a paper, but I'm not sure what the correct attribution is. Who is the discoverer of this definition? - -REPLY [5 votes]: The AMS has spent a lot of money and effort over many years to develop a fine tool that easily allows for the investigation of such questions: it is called Math Reviews (MathSciNet). The oldest mention of SW classes in the reviews seems to be a review by Whitney of a 1947 paper by Wu. There is a fantastic review (in German) by Hirzebruch of Milnor and Stasheff's book, giving some history - e.g. the SW classes go back to the mid 1930's. I believe Milnor and Stasheff themselves discuss how to think about SW classes as obstructions, and I am pretty sure that this finding this interpretation was quite close to the discovery of the formula for the SW classes via Steenrod squares: this type of question is how the `reduced squares' were discovered in the first place. -Added later: ... see Thom, René -Espaces fibrés en sphères et carrés de Steenrod. (French) -Ann. Sci. Ecole Norm. Sup. (3) 69, (1952). 109–182. -In particular, section 3 of chapter 2 carefully develops the formula for the SW classes via Steenrod squares, and compares it to earlier constructions. Chapter 1 discusses what folks might now call the Thom isomorphism (though the Gysin sequence is already known). -Remark: the basic theory of fiber bundles and classifying spaces is just being figured out at this same time.<|endoftext|> -TITLE: Theorems demoted back to conjectures -QUESTION [63 upvotes]: Many mathematicians know the Four Color Theorem and its history: there were two alleged proofs in 1879 and 1880 both of which stood unchallenged for 11 years before flaws were discovered. -I am interested in more such examples, especially, in former (famous or interesting) "theorems" where no remedy could be found to the present day. - -REPLY [11 votes]: Carmichael's totient function conjecture: - -For every positive integer $m$, there is at least one integer $n \neq m$ such that $\varphi(m)=\varphi(n)$. - -Carmichael stated this as a theorem in 1907, but he retracted his proof in 1922 and stated this conjecture as an open problem.<|endoftext|> -TITLE: Finite subgroups of GL_n of polynomial rings over finite fields -QUESTION [5 upvotes]: I am surprised that I didn't find a reference for the following question. -Q: Is there any characterization of the finite subgroups of $GL_n( \mathbb{F}_p [T_1, \dots, T_n])$? Can we do so more generally for $GL_n(\mathbb{Z}/ p^n \mathbb{Z} \ [T_1, \dots, T_n] )$. -Edit: I had previously state an incorrect statement. Thank you for the clarifying answers. - -REPLY [11 votes]: Well, polynomial rings are very complicated. I assume the question is more generally about $GL_n(\mathbb{F}_q[T_1,\dots,T_m])$. -The case of one variable can be deduced from the following paper: - -C. Soulé: Chevalley groups over polynomial rings. In: Homological group theory (Proc. Sympos. Durham) London Math. Soc. Lecture Notes Series 36 (1979), pp. 359-367. - -More details of proof can be found in - -B. Margaux. The structure of the group G(k[t]): variations on a theme of Soulé. Algebra Number Theory 3 (2009), 393-409. - -These results compute the fundamental domain for the action of $GL_n(k[T])$ on the associated Bruhat-Tits building. Any finite subgroup will have a fixed point on the building, so it will appear as subgroup of a stabilizer of the $GL_n(\mathbb{F}_q[T])$-action on the building. The stabilizers are explicitly determined, they are semidirect products $U\rtimes L$ where $L$ is a Levi subgroup of a parabolic in $GL_n(\mathbb{F}_q)$ and $U$ is a subgroup of the upper-triangular matrices with $\mathbb{F}_q[T]$-coefficients (in particular the latter is a $p$-group). Probably everything you want to know about finite subgroups and their conjugacy classification can be deduced from that. -In the case $m>1$, several variables, a lot less is known. One can still use Soulé's theorem, applied to $GL_n(\mathbb{F}_q(T_1,\dots,T_{m-1})[T_m])$. Any finite subgroup will have a fixed point on the building and therefore be conjugate to a subgroup of one of the standard stabilizers described above (but of course for the base field $k=\mathbb{F}_q(T_1,\dots,T_{m-1})$). However, the conjugacy involved will be by matrices in $GL_n(\mathbb{F}_q(T_1,\dots,T_{m-1})[T_m])$ which may not be what you are interested in. Also, there may be some finite subgroups which only appear after some denominators are introduced. Anyway, this confirms the description in YCor's comment. -For $m>1$, I don't know of a comprehensible description of a space on which $GL_n(\mathbb{F}_q[T_1,\dots,T_m])$ would act with finite stabilizers and with identifiable fundamental domain. This is already terribly complicated for $GL_2(\mathbb{F}_q[T_1,T_2])$, see - -S. Krstić and J. McCool. Free quotients of $SL_2(R[X])$. Proc. Amer. Math. Soc. 125 (1997), 1585-1588. (paper can be found here) - -A consequence of their result is that $SL_2(\mathbb{F}_q[T_1,T_2])$ surjects onto a free group of infinite rank. This provides lots of elements of infinite order and probably -also tells us that the conjugacy classification of finite subgroups is very complicated. On the other hand, these results also tell us that finite subgroups don't say that much about the structure of $GL_n(\mathbb{F}_q[T_1,\dots,T_m])$ when $m>1$.<|endoftext|> -TITLE: Diagrams in an Elementary Topos -QUESTION [5 upvotes]: Let $T$ be a sheaf topos and $I$ a small category. Then the functor category $[I,T]$ is also a sheaf topos. Now let $E$ be an elementary topos (cartesian closed category with finite limits + subobject classifier). -Are there additional conditions that we can impose on $E$ to guarantee that $[I,E]$ is an elementary topos? -I think we need completeness at least (since this ensures that $[I,E]$ is cartesian closed). The limits are also computed pointwise, so I guess the only issue is how to guarantee that $[I,E]$ has a subobject classifier... - -REPLY [6 votes]: If $E$ is small complete, then $[I, E]$ is an elementary topos, by the following argument. -If $I_0$ is the discrete category of objects of $I$, then $[I_0, E]$ is just an $I_0$-indexed product of elementary toposes, and such is always an elementary topos. -Then $[I, E]$ is the category of coalgebras for a left exact comonad on a topos $[I_0, E]$, and such is always a topos. In detail, the comonad is a composite -$$[I_0, E] \stackrel{[d_1, E]}{\to} [I_1, E] \stackrel{\Pi_{d_0}}{\to} [I_0, E]$$ -that is right adjoint to a monad -$$[I_0, E] \stackrel{[d_0, E]}{\to} [I_1, E] \stackrel{\Sigma_{d_1}}{\to} [I_0, E]$$ -whose algebras are functors $I \to E$. Here $d_0, d_1: I_1 \rightrightarrows I_0$ are the domain and codomain functions on the set $I_1$ of morphisms of $I$, and $\Pi_{d_0}$ takes a $I_1$-tuple of objects $\langle c_f \rangle_{f \in I_1}$ to the $I_0$-tuple $\langle \prod_{f: d_0(f) = i} c_f \rangle_{i \in I_0}$ (which is where we use completeness). The category of coalgebras of the comonad is equivalent to the category of algebras of its left adjoint monad, by an old result which can be found in the famous 1965 paper of Eilenberg and Moore that introduced the category of algebras construction for decomposing a monad into a left adjoint followed by its right adjoint. -The results I'm mentioning are thoroughly discussed in the Mac Lane-Moerdijk book on topos theory. While I'm at it, let me mention that small completeness of a topos $E$ is equivalent to small cocompleteness of $E$: that completeness implies cocompleteness follows by the theorem of Paré that the power object functor $P: E^{op} \to E$ is monadic so that $E^{op}$ is small-complete if $E$ is, and the converse direction is not difficult.<|endoftext|> -TITLE: Natural classes of compact complex manifolds whose universal covers are bounded domains -QUESTION [5 upvotes]: What are some general classes of compact complex manifolds whose universal covers are bounded domains? One class I know are the Kodaira fibered surfaces. - -REPLY [5 votes]: I'm not aware of a sufficient condition, although many classes have been recognized as carrying the "bounded domain" covers. Here is one interesting paper to look into.<|endoftext|> -TITLE: LDPC codes construction -QUESTION [5 upvotes]: According to Google Scholar original Gallager's article Low-density parity-check codes is cited more than 10000 times. It looks scary for non-experts. -I suspect that the number of algorithms for constructing good sparse matrices for LDPC-codes is much less than 10000. Is there any survey which contains description of such algorithms? -Basically I'm interested in explicit constructions based on number theory methods. - -REPLY [3 votes]: Ryan and Lin's Channel Codes has several chapters (computer-based, finite geometries, finite fields, combinatorial designs) devoted to various constructions of LDPC codes, as well as a chapter devoted to nonbinary codes that also details constructions for that case.<|endoftext|> -TITLE: Closed expression for hypergeometric sum -QUESTION [6 upvotes]: I am trying to simplify an expression and find a closed form for -$$\sum_{m=0}^l \binom{s-m}{s-l} \binom{s-1+m}{s-1}x^m$$ -How could I get rid of this summation? - -REPLY [2 votes]: Mathematica says: -$$\binom{s}{s-l} \, _2F_1(-l,s;-s;x).$$ -(without hypergeometrics for special values of $x,$ like $x=1:$ -$$\frac{(-1)^l (l-2 s-1)!}{l! (-2 s-1)!}$$<|endoftext|> -TITLE: Has anyone seen this graph construction that is similar to the line graph? -QUESTION [6 upvotes]: Given a graph $G$, its line graph, denoted $L(G)$, is the graph whose vertices are the edges of $G$ and where two edges of $G$ are adjacent in $L(G)$ if they are incident to each other, i.e., they share some endpoint. I am interested in the graph, let's call it $L'(G)$, whose vertices are the edges of $G$, two being adjacent if they are incident but not part of a triangle in $G$, i.e., the two vertices not shared by the edges are not adjacent in $G$. Obviously, for a triangle free graph $L(G)$ and $L'(G)$ are the same, but in general $L'(G)$ is just some spanning subgraph of $L(G)$. -Does anyone know if this construction has come up in the literature before? - -REPLY [6 votes]: This graph is known as $\Gamma(G)$ (the corresponding construction in which the edges span a triangle is called $\Delta(G)$) or also the Gallai graph of $G$. -See for example: -V.B. Le -Gallai Graphs and Their Iteration Behavior -Dissertation Thesis, TU Berlin 1994 -V.B. Le -Gallai graphs and anti-Gallai graphs -Discrete Math. 159 1996 179--189<|endoftext|> -TITLE: Is $PRA$ + $TI({\epsilon_0})$ mutually interpretable with some theory in the language of set theory? -QUESTION [5 upvotes]: As is well known, the following theory is equiconsistent with $PA$: - -$ZFC$ with the axiom of infinity replaced by its negation. - -Since this theory is equiconsistent with $PA$, it would seem reasonable to infer (wouldn't it?) that the consistency of '$ZFC$ with the axiom of infinity replaced by its negation' could be provable in "$PRA$ + $TI({\epsilon_0})$. -So what 'theory of sets'(?) is mutually interpretable with "$PRA$ + $TI({\epsilon_0})$? Also, can one define a notion of forcing in the aforementioned theory? -(If this seems too silly a question, please let me know and I will delete....) - -REPLY [6 votes]: Yes, the consistency of "ZFC with the axiom of infinity replaced by its negation" is provable in "PRA + TI($\epsilon_0$)". Technically one has to also show that "PRA + TI($\epsilon_0$)" can prove the equiconsistency result (since it already proves the consistency of PA), but these are fairly natural theories so that shouldn't be a problem. -I'm not aware of a natural fragment of ZFC equiconsistent with "PRA + TI($\epsilon_0$)", and can't think of any reason there would be one. It's sort of remarkable that ZFC has a fragment which matches up perfectly with PA (which is mostly a reflection of how PA is a very natural theory).<|endoftext|> -TITLE: Gauss-Newton vs gradient descent vs Levenberg-Marquadt for least squared method -QUESTION [6 upvotes]: I need to clarify some idea I have in my mind about linear and non-linear regressions. Whatever I know about this topic comes from the book of Taylor "Introduction to error analysis": a set of measurements ${x_i}$ and ${y_i}$ for $i= 1, 2, \dots N$ are assumed to have a trend according to a specific function $y = f(x)$, the discrepancies between the measured value $y_i$ and the function $f(x_i)$ are assumed to follow a Gaussian statistics with variance $\sigma_{y}^2$. Applying the principle of maximum likelihood, the best estimation of the parameters that define $f(x)$ are that ones that minimizes the function. -$$ -\chi^2 = \sum_{i = 1}^{N} \frac{(y_i - f(x_i))^2}{\sigma_y^2} -$$ -This is known as least squared method. -Now in the case of a straight line $f(x) = Ax + B$ the estimation of the parameters is a straightforward job: from a couple of derivatives you figure out $A$ and $B$ and you properly identify $f(x)$. In the more general case $f(x)$ is a polynomial of order $M$, the computation will be more elaborated, but the job is easy at least in principle. In both Matlab and Python there is an implemented function ( polyfit(x, y, M) and np.polyfit(x, y, M) ) that seems to be not difficult to theoretically understand and practically apply to experimental data. -However when the function $f(x)$ is not a polynomial then more complicated numerical methods are necessary in order to figure out the parameters that define $f(x)$. From some google researches I realized that the most popular techniques are - -Gauss-Newton algorithm - -Gradient descent algorithm - -Levenberg-Marquadt algorithm - - -Matlab and Python have an implemented function called "curve_fit()", from my understanding it is based on the latter algorithm and a "seed" will be the basis of a numerical loop that will provide the parameter estimation. -I would like to know in which case it is better to use the first algorithm, in which case the second algorithm is better and in which case the third one is better. -I would be happy if you suggest me any book or other types of material that provide me a (not too) short explanation of those techniques so that each time I have to fit a curve I can understand which is the better method for me. -Finally I would like to know what you would do if you need to provide a Gaussian fit on a set of experimental data. Personally I did a polyfit of second order of the logarithm of the experimental data. I saw on many Matlab and Python webpages that people uses that "curve_fit" that is, from my understanding, the Levenberg-Marquadt method. Which is the best way to perform these fit from your point of view? - -REPLY [12 votes]: The Levenberg-Marquardt method is the most effective optimization algorithm, to be preferred over the methods of steepest descent and Gauss-Newton in a wide variety of problems. You might find this explanation by Henri Gavin instructive: - -The Levenberg-Marquardt curve-fitting method is actually a combination -of the two other minimization methods: the gradient descent method and -the Gauss-Newton method. In the gradient descent method, the sum of -the squared errors is reduced by updating the parameters in the -steepest-descent direction. In the Gauss-Newton method, the sum of the -squared errors is reduced by assuming the least squares function is -locally quadratic, and finding the minimum of the quadratic. The -Levenberg-Marquardt method acts more like a gradient-descent method -when the parameters are far from their optimal value, and acts more -like the Gauss-Newton method when the parameters are close to their -optimal value. - -The Levenberg-Marquardt algorithm may fail to converge if it begins far from a minimum. There exist ways to accelerate the convergence, as explained here.<|endoftext|> -TITLE: If $Y$ is closed in $X$, and $X(F) \cap Y$ is dense in $Y$, then $Y$ is defined over $F$ -QUESTION [7 upvotes]: I've been trying to understand why this result in algebraic geometry is true for a long time. -In the language of classical algebraic geometry, this is what I want to prove: -Let $k$ be a universal field, and let $F \subseteq k$ be a subfield. Let $X \subseteq \mathbb{A}^n$ be an affine (not necessarily irreducible) variety which is defined over $F$. Let $Y$ be a closed subset of $X$. If $Y \cap X(F)$ is dense in $Y$, then $Y$ is defined over $F$. Here $X(F) = X \cap \mathbb{A}^n(F)$ is the set of $F$-rational points. -In the language of commutative algebra: Let $A$ be a finitely generated, reduced $k$-algebra, and let $A_0$ be an $F$-subalgebra of $A$ such that $A = k \otimes_F A_0$. Let $I \subseteq A$ be a radical ideal. Suppose that for every $a \in A$, not in $ I$, there exists a $k$-algebra homomorphism $\phi: A \rightarrow k$ such that $\phi(A_0) \subseteq F$, $I \subseteq \textrm{Ker } \phi$, and such that $\phi(a) \neq 0$. Then $I$ is the extension of an ideal from $A_0$. -This is 11.2.4(ii) in Springer, Linear Algebraic Groups. I do not think the explanation given in the textbook is satisfactory, and was hoping someone may know a complete proof or a reference. Also, I was wondering if there is a version of this result in modern algebraic geometry. - -REPLY [9 votes]: In scheme language, the reformulation of your question (which we'll see really does achieve what you want, and unsurprisingly plays just as essential a role in the scheme version of the theory of linear algebraic groups as it does in the older language) is to show that if $X$ is a scheme of finite type over a field $F$ and if $S$ is a subset of $X(F)$ then the Zariski closure $Z_S \subset X$ of $S$ equipped with its reduced structure is geometrically reduced (over $F$) with the further (crucial) property that for any extension field $F'/F$, the reduced closed subscheme $(Z_S)_{F'} \subset X_{F'}$ is the Zariski closure of $S \subset X(F) \subset X(F')$. -To see that this really settles what you want, if $Y \subset X_{F'}$ is a reduced closed subscheme that is the Zariski closure of a subset $S \subset X(F) \cap Y(F')$ (intersection inside $X(F')$) then clearly $(Z_S)_{F'} \subset Y$ as closed subschemes of $X_{F'}$ yet both are reduced and by design a dense subset $S \subset Y(F')$ lies inside $(Z_S)_{F'}$, so $Y = (Z_S)_{F'}$ is "defined over $F$" (relative to the $F$-descent $X$ of the ambient $X_{F'}$ containing $Y$). -The entire theory of linear algebraic groups is developed from scratch using the scheme framework from the beginning in at least two places: Milne's notes at http://www.jmilne.org/math/CourseNotes/iAG200.pdf, and the two sets of lecture notes at the top of http://math.stanford.edu/~conrad/249BW16Page/handouts.html. In the first of the two sets of lecture notes at the latter link, Proposition 3.2.4(ii) is exactly the above reformulated version of your question (used there for purposes exactly akin to the role played by the result in Springer's book). - -I recommend to forget about the Weil-style language of algebraic geometry used in Springer's book and always work with schemes. My experience as a student was that it is better when reading the standard books on linear algebraic groups to translate all proofs into scheme language and not do anything in the older language. Don't worry about trying to decipher whatever is unclear in Springer's proof of his 11.2.4. -One of the (many) great virtues of the scheme approach is that instead of making smooth constructions over $\overline{F}$ and then having to work hard to descend things to be smooth $F$-schemes (especially tricky at times when $F$ is not perfect, so we cannot use Galois descent), one makes all constructions directly over $F$ and instead the burden of work shifts to proving such $F$-schemes are smooth (using infinitesimal criteria, or flatness arguments, etc.). In particular, with the scheme approach nearly all of Chapter 11 of Springer's book becomes moot. -Springer's book is a wonderful resource over algebraically closed fields, especially for its treatment of the root-system aspects of the structure theory of reductive groups over algebraically closed fields, but you have to be extremely careful with anything done there involving geometric arguments over a ground field that is not algebraically closed; Borel's textbook is more reliable on such matters. A prototypical illustration of the subtle errors that creep in over imperfect ground fields due to the avoidance of schemes in Springer's book is seen in the proof of 15.2.3 in Springer's book: near the end it has to be proved that a certain morphism between $F$-varieties (= geometrically integral $F$-schemes) has finite fibers and one easily reduces to the case $F=F_s$. Springer seems to be misled by the density of $F$-points in an $F$-variety because he checks that fibers over $F$-points are finite and then declares the proof done, completely missing the possibility (which one sees vividly from the scheme perspective, though of course it can also be noticed in the Weil framework too) for imperfect $F$ that the reduced Zariski-closed subset of the target over which fibers are positive-dimensional may not be generically smooth and so may be non-empty but have no $F$-points at all! -Springer's method can't handle fibers not over $F$-points there, so that seems to be a genuine error in the proof of his 15.2.3. A correct proof of a stronger result than 15.2.3 (involving Tits' notion of "pseudo-completeness over $F$" that goes beyond Springer's notion called "($P_F$)") is given as Proposition C.1.6 in the book Pseudo-reductive Groups (using a corrected notion of pseudo-parabolicity since the definition in Springer's book is incorrect, as illustrated by the fact that both parts of the crucial 15.1.2 in his book are false for this notion, the error being in the first sentence when he says one can reduce to the case $\mathscr{R}_{u,F}(G)=1$; this step is fine with the definition of pseudo-parabolicity used in Pseudo-reductive Groups).<|endoftext|> -TITLE: Does any surface of constant curvature admit a cocompact group action? -QUESTION [7 upvotes]: Suppose $S$ is a non-compact and complete surface (2 dimensional smooth Riemannian manifold) of constant curvature. I am wondering if there exists a group $G$ which acts by isometries and properly discontinuously on $S$ such that $S/G$ becomes compact?! Are there maybe any reference where I can find results related to the above situation? -Best wishes - -REPLY [14 votes]: As Uri Bader says in the comments, covering-space theory implies that this happens if and only if $\pi_1S$ is a normal subgroup of $\pi_1\Sigma$, where $\Sigma$ is some compact surface. -The cases of positive and zero curvature are easy, so we may as well assume that $S$ and $\Sigma$ are of constant negative curvature. In this case, a theorem of Greenberg places strong restrictions on normal subgroups of surface groups. - -Theorem (Greenberg): If $H$ is a finitely generated, normal subgroup of the fundamental group of a hyperbolic surface $\Sigma$, then $H$ is either trivial or of finite index - -In particular, if $S$ is non-compact but not the hyperbolic plane, and $\pi_1S$ is finitely generated, then $S$ does not admit such a group action.<|endoftext|> -TITLE: Finding a solution to a simple geometric set of equalities -QUESTION [6 upvotes]: Let $p_1,\dots,p_n$ be a collection of points in the plane $\mathbb{R}^2$ and let $a$ be a positive number such that $a<1$. Is there a good numerical algorithm to find points $x_1,\dots,x_n$ in the plane such that -$$\|x_i - x_{i+1}\| = a(\|x_i - p_i\|+\|p_i-x_{i+1}\|)$$ for all $i$, where we use $n+1\equiv 1$ to simplify notation? A picture for $a=1/3$ is attached, in which the black dots indicate the $p_i$'s, the blue segments indicate the segments from $x_i$ to $x_{i+1}$, and the red segments indicate the $\|x_i - p_i\|+\|p_i-x_{i+1}\|$. - -REPLY [7 votes]: It is easy to find solutions with additional symmetries. Precisely: If $a\neq\cos(\pi m/n)$ there exists exactly one solution such that $|x_k-p_k|=|x_{k+1}-p_k|$ for all $k\in \mathbb{Z} $ (here we denote the sequence $(p_k)$ as an $n$-periodic sequence indicized on $ \mathbb{Z} $). -Indeed, let $u:=e^{i\theta}$ with $\theta=2\arccos a$. By the assumption, $u^n\neq 1$. Starting with $x_0:=x\in\mathbb{C}$, define inductively $x_{k+1}=(x_k-p_k)u+p_k$. The periodicity condition $x_n=x_0$ corresponds then to a non-singular linear equation on $x$ (very easy to write down, which pleasure I won't spoil). -The same procedure produces, more generally, for any $a$ and for any model triangle, with vertices $(0,1 ,u)$ such that $|1-u|=a(1+|u|)$ and $u^n\neq1$, - a unique solution such that the triangles with ordered vertices $(p_k,x_k,x_{k+1} )$ are all similar to the fixed triangle with ordered vertices $(0,1,u)$.<|endoftext|> -TITLE: Ring of invariants of $\operatorname{SL}_6$ acting on $\Lambda^3 \mathbb C^6$ -QUESTION [13 upvotes]: Let $G=\operatorname{SL}_6$ act on $V=\Lambda^3 \mathbb C^6$. I would like to find the ring of invariants $\mathbb C[V]^G$. There is an obvious invariant -$$Sq: V \to \mathbb C, \quad \omega \mapsto \omega^2 \in \Lambda^6 \mathbb C^6 \simeq \mathbb C,$$ -with the last isomorphism $\operatorname{SL}_6$-invariant. Update: $Sq=0$, but there is a unique invariant $\alpha$ of degree $4$, not $2$ (see the comment of Robert Bryant below). -My idea how to prove $\mathbb C[V]^G = \mathbb C[\alpha]$ is to consider -$$W=\mathbb Cx_1 \wedge x_2 \wedge x_3 \oplus \mathbb Cx_4 \wedge x_5 \wedge x_6 \subset V.$$ -If a general $G$-orbit intersects $W$, then a map $f: \mathbb C[V]^G \to \mathbb C[W]$ is injective. But the image lies in $\mathbb C[W]^{\mathbb C^* \rtimes \mathbb Z/2\mathbb Z}$, where the action of $\mathbb C^*$ on $W$ is given by -$$t(w_1 \cdot x_1 \wedge x_2 \wedge x_3 + w_2 \cdot x_4 \wedge x_5 \wedge x_6)=tw_1 \cdot x_1 \wedge x_2 \wedge x_3 + t^{-1}w_2 \cdot x_4 \wedge x_5 \wedge x_6,$$ -and the action of $\mathbb Z/2\mathbb Z$ is given by -$$1(w_1 \cdot x_1 \wedge x_2 \wedge x_3 + w_2 \cdot x_4 \wedge x_5 \wedge x_6)=w_2 \cdot x_1 \wedge x_2 \wedge x_3 - w_1 \cdot x_4 \wedge x_5 \wedge x_6.$$ -Now it is easy to check that $\mathbb C[W]^{\mathbb C^* \rtimes \mathbb Z/2\mathbb Z} \simeq \mathbb C[\beta]$, where $\beta=w_1^2 w_2^2$ is an invariant of degree $4$. Therefore the question is whether a general $G$-orbit intersects $W$ (and also how to construct $\alpha$ with $f(\alpha)=\beta$), and here I am stuck. Could you help me? - -REPLY [2 votes]: Another elementary description of the degree four invariant is as -$$ -\epsilon_{i_1i_2i_3i_4i_5i_6}\ \epsilon_{i_7i_8i_9i_{10}i_{11}i_{12}} -\ \omega_{i_1i_2i_3\ }\ \omega_{i_4i_5i_7}\ \omega_{i_6i_8i_9}\ \omega_{i_{10}i_{11}i_{12}} -$$ -where indices are summed from 1 to 6. The epsilon notation is as in -this MO answer.<|endoftext|> -TITLE: Why should the tensor product of $\mathcal{D}_X$-modules over $\mathcal{O}_X$ be a $\mathcal{D}_X$-module? -QUESTION [13 upvotes]: Let $R$ be a regular algebra over a field $k$ of char 0. Let $D$ be its corresponding algebra of differential operators. -As in the general setting of non-commutative algebra we can tensor right $D$-modules with left $D$-modules to get $R$-modules. However in this case we have more operations available to us. -Let $M$ and $N$ be left $D$-modules. One can define using the leibniz rule a structure of a $D$-module on the tensor product $M \otimes_R N$. The same can be done if we replace one of the factors with a right $D$-modules and flip some signs and similar statements exist for internal Homs over $R$ (i'm not so sure about right tensor right - although I assume that at least in the derived setting one can always use duality to define this structure). -Now, my question is rather vague. I'm trying to understand conceptually what properties of $D$ makes it possible to give a $D$-module structure on $M \otimes_R N$ and maybe understand in what way is this construction canonical (since so far all i've seen is a formula in this context). So to summarize: - -Why do $M \otimes_R N$ and $\mathrm{Hom}_R(M,N)$ have a natural structure of $D$-modules? - -Edit: After being confused by some conflicting answers I've posted a more detailed and exaustive question here: What kind of algebraic object is $\mathcal{D}_X$? (algebra of diifferential operators). What's special about modules over it? -EDIT: Some time has passed and i'm still not satisfied with the my current understanding of this. The original question still remains a mystery: What kind of algebraic object is $\mathcal{D}$? A suitable answer would give a definition of an algebraic object $D$ over a ring $R$ for which all the following holds - -1. The opposite $D^{op}$ is canonically morita equivalent to $D$ (Canonically in the sense that the equivalence should be induced from the algebraic structure on $D$). -2. The ability to form tensor products and hom modules over $R$ between left and right $D$-modules except in two cases: - -Tensor product of a right $D$-module with a right $D$-module. -Hom module (over $R$) from a right $D$-module to a left $D$-module - -3. The forgetful from $D$-modules to $R$-modules is monoidal w.r.t. above tensor product. - -REPLY [12 votes]: OK, I'll give it a shot. The bi-algebra structure on $D$ is something that I found very confusing too, so I will try to spell it out as best I understand. These ideas were explained to me by Pavel Safronov, and I found these notes by Gabriella Bohm be helpful https://arxiv.org/abs/0805.3806 (though they deal with a more general case than we need here). See also the original papers by Sweedler and Takeuchi from the `70's. -The $D$-module set-up -Suppose $X$ is a smooth algebraic variety, and $D=D_X$. The situation we have is the following: the category $D-mod$ and the forgetful functor to $\mathcal O-mod$, are equipped with (symmetric) monoidal structures (the duality of $D$-modules will be discussed later). -There are many ways to understand why this should be the case, as some of the other answers indicate. For example, the category $D-mod$ can be understood as quasi-coherent sheaves on the de Rham space $X_{dR}$ (and the ring $D$ expresses the descent data on the pullback to quasi-coherent sheaves on $X$). Alternatively, if one thinks of $D$ as a deformation quantization of $T^\ast X$, then the monoidal structure arises from the fact that the cotangent bundle is a symplectic group(oid) acting (trivially) on $X$. One can think of $T^\ast X$ as being a commutative group object in the category of symplectic varieties and lagrangian correspondences (I find this last persepective helpful in unpacking the notion of bialgebroid). -However, I think what you are after is not why $D$-modules have this structure, but what structure on the ring $D$ endows $D-mod$ with these structures. The answer (as has already been mentioned) is that $D$ is a bialgebroid over $\mathcal O$. Let me try my best to unpack what that means below. -The categorical structure -(You can ignore this bit if you don't like it). -Consider the following situation: we have monoidal categories $\mathcal C$ and $\mathcal D$ and a monoidal functor -$$F:\mathcal D \to \mathcal C$$ -Suppose also that the functor $F$ is monadic, so that $\mathcal D$ can be expressed as modules for a monad $T$ acting on $\mathcal C$. The monoidal structures on $\mathcal D$, $\mathcal C$ and $F$ must then be reflected in the monad $T$. Such a structure on a monad (acting on a monoidal category $\mathcal C$) is called a bimonad. Rather than saying what this all this means in general, let's consider a special case. -Bialgebroids (over a commutative base) -Suppose $R$ is a commutative ring, and let $\mathcal C = R-mod$. Then a (colimit preserving) monad acting on $R-mod$ is nothing more than a $R$-ring, i.e. a ring $B$ with a ring homomorphism $R\to B$ (note that $R$ need not be central in $B$). In the case we are interested in $B=D$ and $R=\mathcal O$. -Before giving an algebraic definition of a bialgebroid, we note that the point of all this is that a (left) bialgebroid structure on $B$ is exactly equivalent to data of a monoidal structure on $B-mod$ and on the forgetful functor to $R-mod$. Note that if $R$ is central in $B$, this is the usual Tannakian theory, and an $R$-bialgebroid is just an $R$-bialgebra. -So what is an $R$-bialgebroid? Well, we already know that $B$ is an $R$-ring, so there is a product: -$$ -B_{\bullet} \otimes_{R} {}_\bullet B \to B -$$ -where the dots indicate on which side $R$ is acting on $B$. As one might expect, there is also a coproduct, which tells you how $B$ should act on the tensor product $M \otimes_R N$ of two left $B$-modules, but one has to be careful about which monoidal category the coalgebra structure on $B$ lives in. If you unwind the definitions, you see that the coproduct is given by a map -$$ -B \to {}_\bullet B \otimes_R {}_\bullet B -$$ -Note that, unlike in the product map, $R$ is acting on the left on both factors. This is a little confusing at first, but perhaps not so surprising if you consider that in the category $B-mod$ we want to understand how to tensor two left $B$-modules. -Of course, there are some axioms. The one that I found hardest to digest involves something called the Takeuchi product. Let me try to motivate that a bit. -Takeuchi Product -In the usual theory of bialgebras, there is an axiom which says that the coproduct is an algebra map. This doesn't make sense for bialgebroids as ${}_\bullet B\otimes_R {}_\bullet B$ is not an algebra under componentwise multiplication. The Takeuchi product is a certain subspace of this object, defined by: -$$ -B {}_R \times B := \left\{ \sum b_i \otimes b_i' \in {}_\bullet B\otimes_R {}_\bullet B \mid \sum b_i r \otimes b_i' = b_i \otimes b_i'r \right\} -$$ -Note that the the $r$'s in the condition are acting on the right, whereas the relative tensor product is using multiplication on the left. Note also that if $R$ is central in $B$, then the condition is vacuous. One can check that $B {}_R \times B$ is ring under compoentwise multiplication. One of the axioms of a bialgebroid is that the coproduct map factors through the Takeuchi product and is a ring homomorphism. (There is another interesting bialgebroid axiom, which is about the counit map, but for brevity, I won't discuss that). -The Takeuchi product (which in the $R$ commutative case appears to be due to Sweedler?) seemed somewhat mysterious to me until I saw that there is a ring isomorphism: -$$ -B {}_R \times B \simeq End_{B^{op}\otimes B^{op}} (_\bullet B \otimes_R {}_\bullet B ) -$$ -Thus, the comultiplication map is nothing more than the structure of a left $B$, right $B\otimes B$ bimodule on ${}_\bullet B\otimes_R {}_\bullet B$. This fits well with the $D$-modules story: the coproduct on $\mathcal D$ is precisely the transfer bimpdule structure on $$ \mathcal D_{X\to X\times X} = {}_\bullet \mathcal D \otimes_{\mathcal O} {}_{\bullet} \mathcal D$$ -(as it should be, as the transfer bimodule represents the tensor product functor). -The D-module structure on Hom -Let me come back to this another time...<|endoftext|> -TITLE: Computation of $\Omega^{Pin-}_{d}(B\mathbb{Z}_2)$ and Smith isomorphism -QUESTION [5 upvotes]: question: I am looking for the literature with the result or the computation of Pin- bordism group: $\Omega^{Pin-}_{d}(B\mathbb{Z}_2)$. Can someone point out some useful ways to do this or any helpful Refs? - -Some helpful background: There is isomorphism between the following Spin and the Pin- bordism group, known as the Smith isomorphism: -$$ -\Omega^{Spin}_{d+1}(B\mathbb{Z}_2)' \to \Omega^{Pin-}_{d}(pt) -$$ -in particular, the $\Omega^{Spin}_{d+1}(B\mathbb{Z}_2)'$ is not exactly the the usual Spin bordism group $\Omega^{Spin}_{d+1}(B\mathbb{Z}_2)'$, but the reduction, based on a relation: -$$ -\Omega^{Spin}_{d+1}(BG)=\Omega^{Spin}_{d+1}(BG)' \oplus \Omega^{Spin}_{d+1}(pt) -$$ -where the reduction of the spin bordism group $\Omega^{Spin}_{d+1}(BG)$ to $\Omega^{Spin}_{d+1}(BG)'$ gets rid of the $\Omega^{Spin}_{d+1}(pt)$. This part has something to do with the kernel of the forgetful map to $\Omega^{Spin}_{d+1}(pt)$. -In principle, to compute $\Omega^{Pin-}_{d}(B\mathbb{Z}_2)$, we may prove and use the following relations (any comments about this approach): - -$$ -\Omega^{Spin}_{d+1}(B(\mathbb{Z}_2)^2)' \to \Omega^{Pin-}_{d}(B\mathbb{Z}_2)? -$$ - -Some useful info: -$\Omega^{Pin-}_1(pt)=\mathbb{Z}_2, \Omega^{Pin-}_2(pt)=\mathbb{Z}_8, \Omega^{Pin-}_3(pt)=0, \Omega^{Pin-}_4(pt)=0$ -$\Omega^{Spin}_1(B\mathbb{Z}_2)=\mathbb{Z}_2^2, \Omega^{Spin}_2(B\mathbb{Z}_2)=\mathbb{Z}_2^2, \Omega^{Spin}_3(B\mathbb{Z}_2)=\mathbb{Z}_8, \Omega^{Spin}_4(B\mathbb{Z}_2)=\mathbb{Z}$ -This is the reference that I have at hand: Kirby-Taylor, Pin structures on low-dimensional manifolds -I am willing to hear some guidance along this line of thinking, or related issue. - -REPLY [6 votes]: Here's an approach that works up to about dimension 7, outlined by Freed-Hopkins, §10, and explained in more detail by Campbell. -There's a weak equivalence $\Sigma^{-1} \mathrm{MPin}^-\simeq \mathrm{MSpin}\wedge \mathrm{MTO}_1$, where $\mathrm{MTO}_1$ is a Madsen-Tillmann spectrum, the Thom spectrum of the virtual vector bundle $(\underline{\mathbb R} - S)\to B\mathrm O_1$, where $\underline{\mathbb R}$ is the trivial line bundle and $S\to B\mathrm O_1$ is the tautological line bundle. Hence, to understand $\Omega_d^{\mathrm{Pin}^-}(B\mathbb Z/2)$, it suffices to understand the homotopy groups of $\mathrm{MSpin}\wedge\mathrm{MTO_1}\wedge B\mathbb Z/2$. -We'll use the Adams spectral sequence, but there's a key trick that makes it simpler. Let $\mathcal A(1)$ denote the subalgebra of the Steenrod algebra generated by $\mathrm{Sq}^1$ and $\mathrm{Sq}^2$. Then, Anderson, Brown, and Peterson proved that, as $\mathcal A$-modules, -$$ H^*(\mathrm{MSpin};\mathbb F_2)\cong \mathcal A\otimes_{\mathcal A(1)} (\mathbb F_2\oplus M),$$ -where $M$ is a graded $\mathcal A(1)$-module which is $0$ in dimension less than 8. -Thus we can invoke a change-of-rings theorem for the $E_2$-page of the Adams spectral sequence for $\mathrm{MSpin}\wedge\mathrm{MTO_1}\wedge B\mathbb Z/2$: using the Adams grading, when $t -s < 8$, -\begin{align*} -E_2^{s,t} &= \mathrm{Ext}_{\mathcal A}^{s,t}(H^*(\mathrm{MSpin}\wedge\mathrm{MTO_1}\wedge B\mathbb Z/2; \mathbb F_2), \mathbb F_2)\\ -&\cong \mathrm{Ext}_{\mathcal A}^{s,t}((A\otimes_{\mathcal A(1)} \mathbb F_2)\otimes H^*(\mathrm{MTO_1}\wedge B\mathbb Z/2; \mathbb F_2), \mathbb F_2)\\ -&\cong \mathrm{Ext}_{\mathcal A(1)}^{s,t}(H^*(\mathrm{MTO_1}\wedge B\mathbb Z/2; \mathbb F_2), \mathbb F_2). -\end{align*} -Explicitly calculating this is tractable, because $\mathcal A(1)$ is small and we're only going up to dimension 7. - -The $\mathcal A(1)$-module structure on $\tilde H^*(B\mathbb Z/2; \mathbb F_2)$ is standard, and Campbell describes it in Example 3.3 of his -paper. -Campbell also calculates the $\mathcal A(1)$-module structure on $H^*(\mathrm{MTO}_1; \mathbb F_2)$, and describes the answer in Example 6.6 and Figure 6.4. - - -A priori, the above method cannot detect torsion away from the prime 2. But $\Omega_*^{\mathrm{Pin^-}}(B\mathbb Z/2)$ cannot have any $p$-torsion for an odd prime $p$: the $E^2$-page for the Atiyah-Hirzebruch spectral sequence computing $\Omega_*^{\mathrm{Pin^-}}(B\mathbb Z/2)_{(p)}$ is -$$E^2_{q_1,q_2} = H_{q_1}(B\mathbb Z/2; \Omega_{q_2}^{\mathrm{Pin}^-})_{(p)},$$ -but the homology of $B\mathbb Z/2$ has no $p$-torsion when $p\ne 2$, so the spectral sequence vanishes. Therefore its $E^\infty$-page, the $p$-localization of $\Omega_*^{\mathrm{Pin^-}}(B\mathbb Z/2)$, is trivial, so the above method suffices (for $*\le 7$).<|endoftext|> -TITLE: Is the map on etale fundamental groups of a quasi-projective variety, upon base change between algebraically closed fields, an isomorphism? -QUESTION [13 upvotes]: Let $k \subset L$ be two algebraically closed fields of characteristic $0$. Let $U \subset \mathbb P^n_k$ be a smooth quasi-projective variety and let $U_L$ denote the base change of $U$ to $Spec (L)$. Does anyone have a reference for why the map on etale fundamental groups $\pi_1(U_L) \rightarrow \pi_1(U)$ is an isomorphism? -Ultimately, I'm interested in the case that $U$ is normal, but I have a fairly easy argument to deduce that from the case that $U$ is smooth. Brian Conrad has also suggested a promising, though involved, avenue of attack using that topological $\pi_1$ is finitely generated. However since this seems like something that should be known, I was curious whether anyone knows of a reference (or perhaps a very short proof). -Here are some further remarks: It's fairly clear that the map is a surjection, and it is known to be an isomorphism if $U$ is projective (even in characteristic $p$). However, if $U$ is quasi-projective, the map need not be an isomorphism in characteristic $p$ (for example, it fails to be an isomorphism for $\mathbb A^1$, due to Artin Schreier covers). I suspect the isomorphism still holds true without the smoothness hypothesis on U, and true for prime to p parts of the etale fundamental group in characteristic p, but I really only want to apply this when U is smooth (or normal) and $k$ is a subfield of $\mathbb C$. - -REPLY [2 votes]: I ended up writing a proof of the isomorphism $\pi_1(U_L)\rightarrow \pi_1(U)$ in my question above, including a characteristic p version. It is essentially a re-organization of Jason Starr's proof above, together with some suggestions by Brian Conrad. -You can find the write up at -https://arxiv.org/abs/2005.09690<|endoftext|> -TITLE: Kostant's $G$-invariant part in the sym power ring of adjoint representation? -QUESTION [5 upvotes]: Let $g$ be a Lie algebra, say $sl_n(\mathbb C)$. It is considered as the adjoint representation of $G=SL_n(\mathbb C)$. -A famous theorem of Kostant from "Lie Group Representations on Polynomial Rings" states that $$S=I\otimes H,$$ -where $S( g)=\oplus_{m=0}^\infty \vee^m {g}$ is a graded algebra of the symmetric powers of the adjoint representation $g$, -$H$ is $G$-harmonic part, and $I$ is $G$-invariant part. -I would like to see a formula for $I$, or the (degree of) generators of $I$. Is it true $I=\prod_{i=2}^n (1-q^{i})^{-1}$? -$q$ is the symbol for recording the grading. - -REPLY [2 votes]: Unless I misunderstood something, in the case of $sl_n$ the Lie algebra is that of $n\times n$ matrices $M$ with zero trace. The invariants are given by traces of powers of $M$. Here, the algebra of invariants is generated by ${\rm tr}(M^k)$ for $k=2,\ldots,n$. These generators are algebraically independent, so your formula for the Cayley-Sylvester-Hilbert series of $I$, is I believe correct.<|endoftext|> -TITLE: Is $\{x_n\}$ a Cauchy sequence? -QUESTION [9 upvotes]: Let $(X,d)$ be a complete metric space and $f$ a mapping of $X$ into itself. Let $\{f^n(x)\}=\{x_n\}$ be the sequence of iterated transforms. -Suppose $f$ satisfies that for each $\varepsilon >0$,there exists $\delta>0$ such that for all $x,y\in X$, -$$\varepsilon\leq d(x,y)<\varepsilon+\delta\implies d(f(x),f(y))<\varepsilon.$$ -Is $\{x_n\}$ a Cauchy sequence ? - -REPLY [5 votes]: $\let\eps\varepsilon$Yes, that's true, and the argument is somewhat similar to that for your previous question. Surely, completeness is not needed, since you wish to obtain a Cauchy sequence, not a convergent one. -Denote $d_i=d(x_i,x_{i+1})$. Applying the condition for $\eps=d_i$, we get that $d_i$ strictly decrease, so there exists $D=\lim_{i\to\infty} d_i$. If $D>0$, we get a contradiction by setting $\eps=D$; so $D=0$. -Now let us show that for every $\eps>0$ there exists $N$ such that for all $k\geq n>N$ we have $d(x_k,x_n)<2\eps$; this is exactly what we need. Choose $\delta$ for our $\eps$ from the condition; we may assume that $\delta<\eps$. There exists $N$ such that $d_n<\delta$ for all $n>N$. We claim that this $N$ fits. -Indeed, let us show that for all $k\geq n>N$ we have $d(x_k,x_n)<\eps+\delta$; this is clearly sufficient. Induction on $k$. If $k=n$ then there is nothing to prove. Assume now that $k>n$. [ADDED] By the induction hypothesis, we have $d(x_{k-1},x_n)<\eps+\delta$. Now consider the following two cases.[/ADDED] -If $d(x_{k-1},x_n)<\eps$ then -$$ - d(x_k,x_n)\leq d_{k-1}+d(x_{k-1},x_n)<\delta+\eps. -$$ -Otherwise, we have $\eps \leq d(x_{k-1},x_n)<\eps+\delta$, so by the condition $d(x_k,x_{n+1})<\eps$, and -$$ - d(x_k,x_n)\leq d(x_k,x_{n+1})+d_n\leq \eps+\delta. -$$ -The claim is proved.<|endoftext|> -TITLE: Variants and Generalizations of Arf (-Brown-Kervaire) invariants -QUESTION [7 upvotes]: (1) I encounter the Arf invariants in Kirby-Taylor, Pin structures on low-dimensional manifolds. The form that I looked at was: -$$ -S(q)=|H^1(M^2,\mathbb{Z}_2)|^{-1/2} \sum_{x\in H^1(M^2,\mathbb{Z}_2)} \exp[\pi \;i\; q(x)] -$$ -The $M^2$ is an oriented 2 dimensional manifold with spin structures that has $\mathbb{Z}_2$ valued quadratic forms on $H_1(M^2,\mathbb{Z}_2)$, which obeys $$q(x + y) = q(x) + -x ∩ y + q(y) \mod 2,$$ here $x ∩ y$ denotes the $\mathbb{Z}_2$ intersection pairing. The bordism invariant is the Arf invariant. -(2) There is a generalization of the Arf invariant, the Arf-Brown-Kervaire -invariant: -$$S(q)=|H^1(M^2,\mathbb{Z}_2)|^{-1/2} \sum_{x\in H^1(M^2,\mathbb{Z}_2)} \exp[2\pi \;i\; q(x)/4].$$ -It takes values in $\mathbb{Z}_2) \in U(1)$. If $q(x)$ is even, $\forall x$, say $q$ -is $\mathbb{Z}_2$-valued, then the manifold $M^2$ is orientable., it reduces to the Arf invariant. -I hope that I did not make wrong statements above, please correct me if I did it wrong. - -Question: The context I know is only for 2 dimensional manifold. Do we have some analogous Arf and Arf-Brown-Kervaire invariants for (1) higher dimensional manifolds? and (2) analogous form of Arf and Arf-Brown-Kervaire invariants by considering higher homology group $H^d(M,\mathbb{Z}_2)$ or - $H^d(M,\mathbb{Z}_n)$? Do we have such variants and generalizations of the above? - - - -Info from Kirby-Taylor, Pin structures on low-dimensional manifolds: - -REPLY [4 votes]: The Arf invariant and its generalizations occur all over the place in high-dimensional topology, starting with Kervaire's use of his version to construct a topological manifold with no smooth structure. As in the versions you cite, one is dealing with a quadratic form whose underlying form is the intersection form. This was systematized by Kervaire and Milnor in their paper "Groups of homotopy spheres". The Arf invariant also plays a fundamental role in Sullivan's analysis of normal maps, which contains much of the information about a manifold beyond its homotopy type. You can find this summarized nicely in Browder's book on simply-connected surgery, and further developments in Wall's book, Surgery Theory. -You can find a historical treatment of the subject in a paper by Ed Brown, and a discussion of its impact in homotopy theory in Snaith's article in the Notices of the AMS.<|endoftext|> -TITLE: What kind of algebraic object is $\mathcal{D}_X$? (algebra of diifferential operators). What's special about modules over it? -QUESTION [11 upvotes]: Let $R$ be a regular ring over a field of char 0. Let $X=Spec R$ and $D=\mathcal{D}_X$ -the algebra of differential operators over it. -The overall vague question is what kind of algebraic object is $D$ and what kind of category is the category of its modules? Here are some points whose answers could together be considered an answer to this. - - -When (meaning for what kind of $D$-modules $M,N$ left or right) is there a $D$-module structure on $Hom_R(M,N)$? On $M\otimes_RN$? Does this make - $D$-mod into an abelian monoidal category with fiber functor to - $R$-mod? If so is it closed monoidal? -When does a $D$-module $M$ admit a dual $M^*$? (in the sense of monoidal categories). -Is the abelian category of $D$-modules isomorphic to $D^{op}$-mod? (before deriving). If not how are they related? Maybe the correct - thing to consider is the opposite co-opposite $D^{op}_{cop}$? Specifically Why does the dualizing sheaf $\omega_X$ pop up in this context? -Is the $R$-linear dual $Hom_R(D,R)$ a bialgebra? Is it related to the opposite of $D$? -How do I derive correctly the category of $D$-modules? Suppose $M$ and $N$ are $D$-modules and suppose $Hom_R(M,N)$ is the "correct" internal Hom in the abelian monoidal category of $D$-modules. If we hope to have a fiber functor between the derived categories $Hom_R(Q,N)$ (with $Q \to M$ a resolution as a $D$-module) should go to (something quasi isomorphic) to $Hom_R(P,N)$ (with $P \to M$ a resolution as an $R$-module). It doesn't seem to follow easily from the rest of the structure. - -REPLY [10 votes]: Proposition 1.2.9 of http://math.columbia.edu/~scautis/dmodules/hottaetal.pdf explains that if $M$ and $N$ are both left $D$-modules and $M'$ and $N'$ are both right $D$-modules then -(a) $M\otimes_{R} N$ is naturally a left $D$-module; -(b) $M'\otimes_{R} N$ is naturally a right $D$-module; -(c) $\mathrm{Hom}_{R}(M,N)$ is naturally a left $D$-module; -(d) $\mathrm{Hom}_{R}(M',N')$ is naturally a left $D$-module; -(e) $\mathrm{Hom}_{R}(M,N')$ is naturally a right $D$-module. -This Proposition also gives explicit formulae that explain why each of these is the case. Remark 1.2.10 explains why $M'\otimes_{R}N'$ is not naturally a left or right $D$-module in general. -Section 2 of https://arxiv.org/abs/dg-ga/9702008 fits these results into a wider framework. The thing about $D$ that makes these things true is that it is the universal enveloping algebra of a $(k,R)$ Lie-Rinehart algebra. -I think that it is clear from what I've written that $D$-mod (=cat of left $D$-modules) is a (symmetric) monoidal category with $\otimes=\otimes_{R}$ and the forgetful functor $D$-mod to $R$-mod preserves the monoidal product. I'm not sure if a fiber functor requires more than this. -As noted by t3suji in the comments below the internal Hom $\mathrm{Hom}_R(-,-)$ on $R$-mod which induces an internal Hom on $D$-mod via part (c) above does make $D$-mod into a closed monoidal category. Once again this observation works for the enveloping algebra of any Lie-Rinehart algebra. -I think it is probably the case that $M$ has a dual precisely if it is a projective $R$-module of finite rank (that is also a left $D$-module). In this case $M^\ast$ should just be $\mathrm{Hom}_R(M,R)$ which is a left $D$-module by 1(c). -Yes. The reason that $\omega_X$ crops up is that it is a right $D$-module that is a rank $1$ projective $R$-module. It follows that tensoring with it defines an auto-equivalence of categories on $R$-mod (the inverse is given by $\mathrm{Hom}_R(\omega_X,-)\cong \mathrm{Hom}_R(\omega_X,R)\otimes_R-$. It follows from 1(b) that this auto-equivalence sends left $D$-modules to right $D$-modules and from 1(d) that its inverse sends right $D$-modules to left $D$-modules. -Since $D$ is not a finitely generated $R$-module, $\mathrm{Hom}_R(D,R)$ is pretty badly behaved. In particular it is unlikely that the algebra structure on $D$ will induce a coalgebra structure on its 'dual'. Incidentally I'm not sure what you mean by saying that $D$ is an $R\otimes R$ bialgebra (in particular I don't know what the coalgebra structure is nor the counit). -I don't really understand what this part is asking. - -Edited in the light of the comments below by t2suji<|endoftext|> -TITLE: Shellable simplicial complex with restriction on shellings -QUESTION [10 upvotes]: Does there exist a (pure) shellable simplicial complex $\Delta$ with the -following property? There is some facet $F$ of $\Delta$ such that no -shelling can begin with $F$. -This condition is easily seen to be equivalent to the existence of a -shellable simplicial complex $\Delta'$ with at least two facets, such -that for some facet $F'$ every shelling of $\Delta'$ ends with $F'$. - -REPLY [6 votes]: The answer to your question is "yes". Such complexes were first exhibited by Hachimori in his PhD thesis. See the clear explanation on his webpage: -http://infoshako.sk.tsukuba.ac.jp/~hachi/math/library/nonextend_eng.html -The $f$-vector of this complex is (1,7,19,13). I suspect that 7 vertices is minimal. UPDATE: I later noticed that Simon earlier exhibited an example of similar type in Appendix F of his 1994 paper: -Simon, Robert Samuel, Combinatorial properties of “cleanness”, J. Algebra 167, No. 2, 361-388 (1994). ZBL0855.13013.Hachimori's webpage also has a fairly nice presentation of Simon's example: -http://infoshako.sk.tsukuba.ac.jp/~hachi/math/library/simon_eng.html -Adiprasito, Benedetti and Lutz have meanwhile (since you first asked the question) extended an idea similar to Hachimori's to arbitrary dimension in their paper: Adiprasito, Karim A.; Benedetti, Bruno; Lutz, Frank H., Extremal examples of collapsible complexes and random discrete Morse theory, Discrete Comput. Geom. 57, No. 4, 824-853 (2017). ZBL1365.05305. -Complexes having the property of Hachimori's have arisen recently as a tool. They were used as one of the main building blocks to show that the SHELLABILITY problem is NP-complete in this paper of Goaoc, Paták, Patáková, Tancer, and Wagner. (The paper appeared in a conference proceeding, though it doesn't seem to yet be on zbMath.) -I'll mention parenthetically that Goaoc, Paták, Patáková, Tancer, and Wagner require also some other building blocks of the same flavor as Hachimori's complex. Their constructions for these are a bit complicated for my taste, and they don't include all of the details in the above-linked paper. My PhD student Andrés Santamariá Galvis has some simpler constructions that are based on Hachimori's example.<|endoftext|> -TITLE: Generalization of area and coarea formula for fractional Hausdorff measures -QUESTION [7 upvotes]: Let $X,Y$ be Polish spaces, $s,t>0$ and $F:X\to Y$ locally Lipschitz continuous such that $X$ is $\sigma$-finite w.r.t. the $(s+t)$-dimensional Hausdorff measure $\mathcal{H}^{s+t}$. -The Eilenberg inequality shows -$$\int_Y \int_{F^{-1}(y)} \chi_A(x) d\mathcal{H}^s(x) d\mathcal{H}^t(y) \leq Lip(F_{|A})^t \mathcal{H}^{s+t}(A)$$ -for all $\mathcal{H}^{s+t}$-measurable $A\subseteq X$. -In particular the left hand side defines a Borel measure on $X$ which is absolutely continuous w.r.t. $\mathcal{H}^{s+t}$ so that there exists a measurable function $J^{s,t} F: X\to[0,\infty]$ with -$$\int_Y \int_{F^{-1}(y)} \phi(x) d\mathcal{H}^s(x) d\mathcal{H}^t(y) = \int_X \phi(x) J^{s,t}F(x) d\mathcal{H}^{s+t}$$ -for all measurable $\phi: X\to[0,\infty]$. -If $n\leq N$, $X\subseteq\mathbb{R}^n, Y=\mathbb{R}^N$ then the area formula shows $J^{0,n}F(x) = JF(x) = \det(DF(x)^T DF(x))^{1/2}$. -If $n\leq N$, $X\subseteq\mathbb{R}^N, Y=\mathbb{R}^n$ and $F$ is a $C^1$-submersion (or something sufficiently similar) then the coarea formula shows $J^{N-n,n}F(x) = JF(x) = \det(DF(x)DF(x)^T)^{1/2}$. - -Question. Are there other cases of interest in which $J^{s,t}F$ is known or somehow "explicitly" definable from $F$? - -REPLY [3 votes]: There are several intersting consequences of this abstract viewpoint about these formulas. I also would have liked to discuss why the coarea inequality is the backbone of the coarea formula, but let me answer the question for now! -Let us begin with the Euclidean fomulas. Federer proves in [Fe] the following area and co-area formulae (they are 3.2.3 and 3.2.11 in the book): -Theorem (area Formula, Euclidean) Suppose $m \leq n$ and $f:\mathbb{R}^m \to \mathbb{R}^n$ is Lipschitz. Then for any (Lebesgue) measurable set $ A \subset \mathbb{R}^m $, -$$ -\int_A J_m f(x) \, d\mathcal{L}^m(x) = \int_{\mathbb{R}^n} \mathcal{H}^{0}(f^{-1}(z) \cap A) \, d\mathcal{H}^m (z) \, , -$$ -where $J_nf(x):= \left( \det (Df(x)^T Df(x))\right)^{1/2}$. (It is well-defined a.e. by Radamacher's differentiability theorem below.) -Remark: The notation $N(f,A,z)$ is probably more common for $ \text{card} \{ x \in A: f(x) = z\} $ than $ \mathcal{H}^{0}(f^{-1}(z) \cap A) $. However, note the analogy to the coarea formula below. In fact when $m=n$ the two theorems coincide. -Theorem (Coarea Formula, Euclidean) Suppose $m \geq n$ and $f:\mathbb{R}^m \to \mathbb{R}^n$ is Lipschitz. Then for any (Lebesgue) measurable set $ A \subset \mathbb{R}^m $, -$$ -\int_A J_nf(x) \, d\mathcal{L}^m(x) = \int_{\mathbb{R}^n} \mathcal{H}^{m-n}(f^{-1}(z) \cap A) \, d\mathcal{L}^n(z) \, , -$$ -where the existence of $J_nf(x):= \left( \det (Df(x) Df(x)^T)\right)^{1/2}$ is guaranteed by Radamacher's differentiability theorem. -Remark: The Jacobians are well-defined thanks to the Rademacher's theorem: -Theorem (Rademacher, 1919): Let $f:\mathbb{R}^m \to \mathbb{R}^n$ be Lipschitz. Then $f$ is differentiable at $\mathcal{L}^m$-a.e. $x \in \mathbb{R}^m$. -Remark: Measurability and integrability of the functions involved are part of the claims of the theorems. -Remark: Via approximations by characteristic functions the formulas above yield, respectively, integral identities, -$$ -\int_A \phi(x) J_m f(x) \, d\mathcal{L}^m(x) = \int_{\mathbb{R}^n} \left(\int_{f^{-1}(z) \cap A} \phi(y) \, d\mathcal{H}^{0}(y) \right) \, d\mathcal{H}^m (z) \, , -$$ -and -$$ -\int_A \phi(x) J_nf(x) \, d\mathcal{L}^m(x) = \int_{\mathbb{R}^n} \left(\int_{f^{-1}(z) \cap A} \phi(y) \, d\mathcal{H}^{m-n}(y) \right) \, d\mathcal{L}^n(z) \, , -$$ -for integrable $[-\infty,+\infty]$-valued functions $\phi$ defined on $\mathbb{R}^m$, i.e. the domain space in the context of area/coarea formula. Note that -$$ -\int_{\mathbb{R}^n} \left(\int_{f^{-1}(z) \cap A} \phi(y) \, d\mathcal{H}^{0}(y) \right) d\mathcal{H}^m (z) = \int_{\mathbb{R}^n} \left(\sum_{y \in f^{-1}(z) \cap A} \phi(y) \right) \, d\mathcal{H}^m (z) \, . -$$ -A detailed proof of these theorems can be found in Evans and Gariepy's book. [E-G] -In [Fe] these theorems were generalized to maps between Riemannian manifolds and maps between rectifiable subsets of Euclidean spaces. -Definition (Rectifiable Sets): A subset of a metric space $(X,d)$ is said to be (countably) $\mathcal{H}^k$-rectifiable if, up to a set of $\mathcal{H}^k$ measure zero, it is a countable union of images of Lipschitz maps from subsets of $\mathbb{R}^k$ to $X$. Informally, they are the "manifolds" of geometric measure theory! -In 1994 Kirchheim [Kir] introduced the notion of "metric differentiability" for maps $\mathbb{R}^n \to (X,d)$ where $X$ is a metric space. The metric derivative of $f$ at $x$, denoted $mdf(x)$ is a seminorm on $\mathbb{R}^n$. Kirchheim also defined a "Jacoian" for such seminorms. He then proved the following area formula: -Theorem (Kirchheim, area formula): Let $ f: \mathbb{R}^n \to (X,d)$ be a Lipschitz map into a metric space. Then for any measurable subset $ A \subset \mathbb{R}^n $, -$$ -\int_A J_n(mdf(x)) \, d\mathcal{L}^n(x) = \int_X \mathcal{H}^0(f^{-1}(z) \cap A) \, d\mathcal{H}^n(z) \, . -$$ -Of course, again, one easily deduces a change of variables formula from this. -Using the same notion of metric derivative, Karmanova [Kar], and later, independently, Reichel [Rei] defined a coarea factor $C_k(mdf(x))$ and with that proved the following coarea formula. -Theorem (coarea formula): Suppose $(X,d)$ is an $\mathcal{H}^k$-$\sigma$-finite metric space, $A \subset \mathbb{R}^{n}$ is measurable, and $ f:A \to X $ is Lipschitz. Then, - $$ -\int_{A} C_k(mdf(x)) \ d\mathcal{L}^{n}(x) = \int_{X} \mathcal{H}^{n-k}(f^{-1}(z) \cap A) \, d\mathcal{H}^k(z) \, . -$$ -And again it generalizes to a formula involving integration of scalar functions. Reichel also generalizes the theorem to $\mathcal{H}^n$-rectifiable metric spaces in place of $\mathbb{R}^n$. -Let $X$ and $Y$ be separable metric spaces, and let $S \subset E$ be (countably) $\mathcal{H}^n$-rectifiable. For a Lipschitz map $f:S \to Y$, Ambrosio and Kirchheim [A-K] define a notion of "Tangential differential", denoted by $d^Sf(x)$, and then its "Jacobain" $J_n(d^Sf(x))$. They prove then -Theorem (Area Formula): Let $f:X \to Y $ be a Lipschitz function, and let $ S \subset X $ be a (countably) $\mathcal{H}^n$-rectifiable subset. then for any Borel function $\phi:S \to [0,\infty]$, -$$ -\int_S \phi(x) J_n (d^S f(x)) \, d\mathcal{H}^n(x) = \int_{Y} \left( \int_{f^{-1}(z) \cap S)} \phi(y) \, d\mathcal{H}^{0}(y) \right) \, d\mathcal{H}^n (z) \, , -$$ -They then turn to the coarea formula. Let $S$ be an $\mathcal{H}^n$-rectifiable subset of a metric space, and $f:S \to \mathbb{R}^k$ be Lipschitz, where $ n \geq k$. In [A-K], they define a notion of for such maps, denoted by $d^Sf(x)$, and then its "Jacobain" $C_k(d^Sf(x))$. (The letter $C$ is for the coarea factor, a term many authors use in place of Jacobian in the context of the coarea formula.) Using this they prove -Theorem (coarea formual): Let $S$ be an $\mathcal{H}^n$-rectifiable subset of a metric space, and $f:S \to \mathbb{R}^k$ be Lipschitz, where $ n \geq k$. Then for any Borel function $\phi:S \to [0,\infty]$, -$$ -\int_S \phi(x) C_k(d^Sf(x)) \, d\mathcal{H}^n(x) = \int_{\mathbb{R}^k} \left(\int_{f^{-1}(z)} \phi(y) \, d\mathcal{H}^{n-k}(y) \right) \, d\mathcal{L}^k(z) \, . -$$ -The proof uses the classical/Euclidean coarea formula. -Magnani [Mag] has investigated the area and coarea formulae in the context of sub-Riemannian manifolds, Carnot groups, and especially Heisenberg groups. We state only one result. The following is corollary 6.5.4 in Magnani's thesis [Mag]. (Be wary of Magnani's notation where he uses $\mathbb{H}^{2n+1}$ to denote the Heisenberg group of topological dimension $2n+1$.) -Theorem (coarea formula on Heisenberg group): Let $ u : \mathbb{H}^{2n+1} \to \mathbb{R}$ be a Lipschitz map. Then for any non-negative measurable function $ h: \mathbb{H}^{2n+1} \to \mathbb{R}$, -$$ -\int_G h \, |\nabla_H u| = \frac{\alpha_{Q-1}}{\omega_{Q-1}} \int_{\mathbb{R}} \left( \int_{u^{-1}(t)} h \, \, dS^{Q-1} \right) dt -$$ -where $Q=2n+2$ is the Hausdorff dimension of $\mathbb{H}^{n}$, $\nabla_H u $ is the horizontal gradient, $S^{Q-1}$ is the spherical Hausdorff measure (with respect to Carnot-Caratheodory distance), and $\alpha_{Q-1}$ and $\omega_{Q-1}$ are (computable) dimensional constants. -References -[Fe]: H. Federe, Geometric Measure Theory -[E-G]: Lawrence C. Evans, Ronald F. Gariepy: Measure theory and fine properties of functions -[A-K]: Luigi Ambrosio, Bernd Kirchheim, Rectifiable sets in metric and Banach spaces, Math. Ann. 318, 527–555 (2000) -[Kir]: Bernd Kirchheim, RECTIFIABLE METRIC SPACES: LOCAL STRUCTURE AND REGULARITY OF THE HAUSDORFF MEASURE, PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 121, Number 1, May 1994 -[Kar]: Karmanova, Rectifiable sets and coarea formula for metric-valued mappings, Journal of Functional Analysis, Volume 254, Issue 5, 1 March 2008, Pages 1410-1447 -[Rei]: (Doctoral Thesis) Reichel, Lorenz Philip, The coarea formula for metric space valued maps, 2009 https://doi.org/10.3929/ethz-a-005905811 -[Mag]: Valentino Magnani, Elements of Geometric Measure Theory on sub-Riemannian groups<|endoftext|> -TITLE: When and where were Jonquières automorphisms defined first? -QUESTION [5 upvotes]: I have listened to lectures that mention Jonquières automorphisms for affine spaces by name. They don't seem to be found in textbooks on algebraic geometry. -I would like to know the exact reference preferably for the original work where it can be found. -I have access to Hanspeter Kraft's Bourbaki seminar talk Challenging Problems in Affine Spaces (1994-95, exp. 802). The bibliography there has more than hundred entries but not this. -I would be grateful to anyone who can point out the references. - -REPLY [7 votes]: E. de Jonquières: De la transformation géométrique des figures planes, et d'un mode de génération de certaines courbes à double courbure de tous les ordres. Nouv. Ann. (2) 3, 97--111 (1864). You can find it in Numdam.<|endoftext|> -TITLE: History of forcing over admissible sets -QUESTION [6 upvotes]: In his paper "Forcing in admissible sets", Ershov writes - -In unpublished lectures given at Novosibirsk State University in 1976-1977 on the theory of admissible sets, the author - showed that it is possible to use forcing to construct admissible sets. The possibility of using forcing in the theory of admissible sets was suggested in a number of publications, but a precise exposition of this method was never published. The present paper, which is based on notes from the lectures mentioned above, attempts to fill this gap. - -I'm curious about the extent of the "suggestions" that Ershov mentions. Specifically: would it be fair to credit Ershov solely with the result that set forcing preserves admissibility, or should that be Ershov/folklore (or something else)? -Context: I'm writing a paper where I use this result, and would like to attribute it appropriately. - -REPLY [7 votes]: Forcing over admissible sets was first carried out by Jon Barwise in his 1967 Stanford dissertation. -This recent paper of Mathias (it appeared in Fundamenta Mathamaticae in 2015) develops forcing over set theories that are weaker than KP. It also has a nice historical section (page 31) on the topic of forcing over weak set theories.<|endoftext|> -TITLE: Majorana modes and the first Stiefel–Whitney class -QUESTION [8 upvotes]: The first Stiefel–Whitney class of a vector bundle is an element in the first cohomology group of the base space. Namely, the first Stiefel–Whitney class for a vector bundle $E$ over a $d$-dimensional manifold $M^d$ is called $w_1(E) \in H^1(M^d;\mathbb{Z}/2\mathbb{Z})$. - -The Majorana modes are certain "fermions" whose creation and annihilation operators are the same $\hat{\gamma}=\hat{\gamma}^\dagger$. - - -For physical reasons, in the context of 2+1 dimensional $p \pm ip$ superconductors, people have a vague impression that Majorana zero modes in the 2+1 dimensional space-time may be related a nontrivial generator in the cohomology group $H^1(\mathbb{Z}/2\mathbb{Z};\mathbb{Z}/2\mathbb{Z})$, where the first $\mathbb{Z}/2\mathbb{Z}$ can be viewed as $\mathbb{Z}/2\mathbb{Z}$-gauge fields, and the second $\mathbb{Z}/2\mathbb{Z}$ has something to do with the orientation of manifold. This may be similar to the context of the first Stiefel–Whitney class, that has something to do with the orientability of the based manifold. -In contrast, the second Stiefel–Whitney class $w_2(E) \in H^2(M^d;\mathbb{Z}/2\mathbb{Z})$ has something to do with the spin structure of the bundle $E$, which is suitable for defining spinors, including the Dirac spinors. Naively, the Dirac (complex) fermion is a pair of Majorana (real) fermions, say in 2+1 dimensional spacetime. -A side remark is that $H^1(\mathbb{Z}/2\mathbb{Z};\mathbb{Z}/2\mathbb{Z})$ also occurs in the calculation of the 3rd spin bordism group $\Omega_3^{Spin}(B \mathbb{Z}/2\mathbb{Z})$ -for the classifying space $B \mathbb{Z}/2\mathbb{Z}$. -So, my question is the following. - -In mathematics (references/literature are welcome), do the following objects appear in a unified context? - -Majorana modes/fermions; - -the first Stiefel—Whitney classes $w_1(E) \in H^1(M^d;\mathbb{Z}/2\mathbb{Z})$; - -the cohomology group $H^1(\mathbb{Z}/2\mathbb{Z};\mathbb{Z}/2\mathbb{Z})$. - - -What are the math principles and structures behind? - -REPLY [7 votes]: An early paper on this connection is Dirac and Majorana Spinors on Non-Orientable Riemann Surfaces (1987). More recently, -The Stiefel--Whitney theory of topological insulators (2016) relates the existence of an unpaired Majorana zero mode, i.e., geometrically a conical singularity created by the intersection of edge states, to the first Stiefel–Whitney class of a Pfaffian line bundle over the momentum space. (See proposition 1 on page 15.) A follow-up article is Noncommutative topological $\mathbb{Z}_2$ invariant. -I might add that a Majorana zero mode is an altogether different, and more interesting/useful, object than the Majorana fermion mentioned in the OP. Any self-adjoint fermion operator $\gamma$ is a Majorana operator, and any fermion operator $a$ can be decomposed into a pair of Majorana operators, $a=\gamma_1+i\gamma_2$, so the self-adjointness is not a particularly significant restriction. A Majorana zero mode satisfies in addition the requirement that it commutes with the Hamiltonian, $H\gamma=\gamma H$, which is a severe restriction that is only realised approximately in any physical system. As a consequence, when there exist $n$ Majorana zero modes the Hamiltonian has a $2^n$-degenerate ground state manifold, which is potentially useful as a storage of quantum information. The storage is referred to as "protected", because any term in the Hamiltonian that couples only to a single Majorana operator cannot lift the degeneracy. In a physical system this protection is realized by keeping the Majorana zero modes spatially separated.<|endoftext|> -TITLE: MSO2-expressible graph properties unexpressible in MSO1 -QUESTION [9 upvotes]: I would like to ask if there is any good list of graph properties that one can express in the MSO2 (i.e., using quantification over sets of edges) but provably not in MSO1 (i.e., using quantification over sets of vertices only). I know that this should be Hamilton path but I think that there has to be plenty such examples. Ideal are those that are suitable as exercises for non-expressibility results using Ehrenfeucht–Fraïssé games. - -REPLY [3 votes]: I have found a book available online which gives some answers to your question. The book is Graph Structure and Monadic Second-Order Logic: a Language Theoretic Approach by Courcelle and Engelfriet. I am not (at all) an expert on this topic, but the book seems like a good reference. It lists the existence of a Hamiltonian cycle as an example. Some additional interesting examples included in the book are: - -If a graph is simple -Existence of a perfect matching -Existence of a spanning tree with maximum degree 3 - -The most relevant parts of the book are Proposition 5.13, Proposition 5.19, and Remark 5.21. Showing the existence of a perfect matching is not expressible in $MSO_1$ can be with the same arguement on complete bipartite graphs $K_{n,m}$ outlined in Tony Huynh's answer. Also, it seems to me the example of a spanning tree with maximum degree 3 in Remark 5.21 can be modified to show that the existence of a spanning tree with maximum degree $k$ for some given $k$ in not expressible in $MSO_1$ (put more "leaves" on the graph in Figure 5.3).<|endoftext|> -TITLE: A surprising conjecture about twin primes -QUESTION [25 upvotes]: Just for fun, I began to play with numbers of two distinct ciphers. I noticed that most of the cases if you consider the numbers $AB$ and $BA$ (written in base $10$), these have few common divisors: for example $13$ and $31$ are coprime, $47$ and $74$ are coprime. Obviously this is not always the case, because one can take non-coprime ciphers, however I realized that, for $0 \le a n-1$, so the gcd is divisible by $c(d+1)$ and hence greater than $n-1$.<|endoftext|> -TITLE: Why is $C_k(\omega_1)$ Lindelöf? -QUESTION [6 upvotes]: In the thesis "Topology of Function Spaces" by Andrew Marsh (Galway, 2000) it is claimed (page 23, before Definition 30) that the space $C(\omega_1)$ is Lindelöf with respect to the compact open topology. For a proof the reader is refered to -S.P. Gul'ko. On properties of $\sigma$-products. Soviet Math. Dokl. 18, 6 1977. -I searched a while on the net but I could not find another reference to this paper - it seems there is a paper by Gul'ko named "On properties of subsets of $\Sigma$-products" but I do not know wether that is the same paper or how to obtain access to the latter paper. -I wanted to ask Andrew Marsh but I could not find a way to address him (plus there seem to be more people of that name). -Furthermore, I could not find any other source for that claim. -So, my question is: - - -Can you give me a reference (or a proof if it fits on one page) for the fact that $C_k(\omega_1)$ is Lindelöf? It does not need to be the Gul'ko paper above but if you manage to find it and it really contains the proof, that is of course also very fine. -Bonus question (not really part of the question but it would be even nicer): Is it known if $C_k(\omega_1) $ is $k$-Lindelöf in the sense of my last question (see Lindelöf Property for Open Covers for Compact Sets), i.e. does every open $k$-cover admit a countable $k$-subcover? (a $k$-cover is a cover such that every compact subset is contained in one of the members of the cover). - -REPLY [4 votes]: Here is (a bit lengthy and technical) proof of the Lindelof property of the function space $C_k(\omega_1)$. At first some notations. -For any function $f\in C_k(\omega_1)$ and a countable ordinal $\alpha$ let $\|f\|_\alpha=\sup_{x\in[0,\alpha]}|f(x)|$. Let also $\|f\|=\sup_{x\in\omega_1}|f(x)|$. -For every $f\in C_k(\omega_1)$, $\alpha\in\omega_1$ and $\varepsilon>0$ consider the open neighborhood $$B_\alpha[f;\varepsilon):=\{g\in C_k(\omega_1):\|g-f\|_\alpha<\varepsilon\}$$ of $f$ in the function space $C_k(\omega_1)$. -Given an open cover ${\mathcal U}$ of $C_k(\omega_1)$, for every $f\in C_k(\omega_1)$ let $$ -\begin{aligned} -\varepsilon_f&:=\sup\{\varepsilon\in (0,1]:\exists \alpha\in\omega_1\;\exists U\in{\mathcal U}\;\;B_\alpha[f;4\varepsilon)\subset U\},\mbox{ and }\\ -\alpha_f&:=\min\{\alpha\in\omega_1:\exists U\in{\mathcal U}\;\;B_\alpha[f,3\varepsilon_f)\subset U\}. -\end{aligned} -$$ -Choose also a set $U_f\in{\mathcal U}$ such that $B_{\alpha_f}[f;3\varepsilon_f)\subset U_f$. -Claim. For any functions $f,g\in C_k(\omega_1)$ we get $4\varepsilon_f\ge 4\varepsilon_g-\|f-g\|$. -Proof. Assuming that $4\varepsilon_f<4\varepsilon_g-\|f-g\|$, we can choose $\delta>0$ such that $\|f-g\|+4\varepsilon_f+\delta\le 4\varepsilon_g-\delta$ and find an ordinal $\alpha\in\omega_1$ such that $B_\alpha[g;4\varepsilon_g-\delta)\subset U$ for some $U\in{\mathcal U}$. Then $$B_\alpha[f;4\varepsilon_f+\delta)\subset B_\alpha[g;\|f-g\|+4\varepsilon_f+\delta)\subset B_\alpha[g;4\varepsilon_g-\delta)\subset U\in{\mathcal U},$$ which contradicts the definition of the number $\varepsilon_f$. $\square$ -For every ordinal $\alpha\in\omega_1$, identify the Banach space $C_k[0,\alpha]$ with the subspace $\{f\in C_k(\omega_1):f|[\alpha,\omega_1)\equiv const\}$ of $C_k(\omega_1)$, consisting of functions, which are constant on the interval $[\alpha,\omega_1)$. -For two ordinals $\alpha,\beta$ by $\alpha{\vee}\beta$ we denote their maximum $\max\{\alpha,\beta\}$. -We shall construct inductively a non-decreasing sequence of countable ordinals $(\alpha_n)_{n\in\omega}$ and a sequence $(F_n)_{n\in\omega}$ of countable subsets $F_n\subset C_k[0,\alpha_n]$ such that for every $n\in\omega$ the following conditions are satisfied: -$(1_n)$ $C_k[0,\alpha_n]\subset \bigcup_{f\in F_n}B_{\alpha_n\!{\vee}\alpha_f}[f;\varepsilon_f)$; -$(2_n)$ $\alpha_{n+1}=\sup\limits_{f\in F_n}(\alpha_n{\vee}\alpha_f)$. -We start the inductive construction letting $\alpha_0=0$. Assume that for some $n\in\omega$ an ordinal $\alpha_n$ has been constructed. For the open cover $\{B_{\alpha_n\!{\vee}\alpha_f}[f;\varepsilon_f):f\in C_k[0,\alpha_n]\}$ of the separable Banach space $C_k[0,\alpha_n]$, there exists a countable subset $F_n\subset C_k[0,\alpha_n]$ such that $C_k[0,\alpha_n]\subset \bigcup_{f\in F_n}B_{\alpha_n{\vee}\alpha_f}[f;\varepsilon_f)$. Letting $\alpha_{n+1}:=\sup_{f\in F_n}(\alpha_n{\vee}\alpha_f)$ we complete the inductive step. -After completing the inductive construction, consider the countable ordinal $\alpha_\omega=\sup_{n\in\omega}\alpha_n$ and the countable set $F:=\bigcup_{n\in\omega}F_n\subset C_k[0,\alpha_\omega]\subset C_k(\omega_1)$. We claim that $\{U_f:f\in F\}\subset {\mathcal U}$ is a countable subcover of $C_k(\omega_1)$. Given any function $g\in C_k(\omega_1)$, for every ordinal $n\le\omega$ consider the (unique) function $g_n\in C_k[0,\alpha_n]$ such that $g_n|[0,\alpha_n]=g|[0,\alpha_n]$. -By the continuity of the function $g$ at $\alpha_\omega$, there exists a number $n\in\omega$ such that $|g(x)-g(\alpha_\omega)|<\varepsilon_{g_\omega}$ for all $x\in[\alpha_n,\alpha_\omega]$. This implies that $\|g_n-g_\omega\|<\varepsilon_{g_\omega}$ and $4\varepsilon_{g_n}\ge 4\varepsilon_{g_\omega}-\|g_n-g_\omega\|>3\varepsilon_{g_\omega}$ according to the Claim. -By the inductive condition $(1_n)$, for the function $g_n\in C[0,\alpha_n]$ there exists $f\in F_n$ such that $g_n\in B_{\alpha_n\!{\vee}\alpha_f}[f;\varepsilon_f)$ and hence $\|f-g_n\|_{\alpha_n\!{\vee}\alpha_f}<\varepsilon_f$. By the inductive condition $(2_n)$, $\alpha_f\le\alpha_{n+1}\le\alpha_\omega$. Since the functions $f$ and $g_n$ are constant on the interval $[\alpha_n,\omega_1)$, the inequality $\|f-g_n\|_{\alpha_n\!{\vee}\alpha_f}<\varepsilon_f$ implies $\|f-g_n\|=\|f-g_n\|_{\alpha_n\!{\vee}\alpha_f}$. Then Claim yields that $4\varepsilon_{f}>4\varepsilon_{g_n}-\|f-g_n\|>3\varepsilon_{g_n}$ and hence $\varepsilon_{g_\omega}<\frac43\varepsilon_{g_n}<\frac43\frac43\varepsilon_{f}$. -We claim that $g\in B_{\alpha_f}[f;3\varepsilon_f)\subset U_f$. Indeed, for every $x\in [0,\alpha_f]$ we get $$|g(x)-f(x)|\le |g(x)-g_n(x)|+|g_n(x)-f(x)|<\varepsilon_{g_\omega}+\varepsilon_f<\tfrac{16}9\varepsilon_{f}+\varepsilon_f<3\varepsilon_f,$$ - and hence $g\in B_{\alpha_f}[f;3\varepsilon_f)\subset U_f$.<|endoftext|> -TITLE: What is a 'power admissible model'? -QUESTION [5 upvotes]: Q: What exactly is a power admissible model? -Background: Admissible models, introduced by Jon Barwise, form the building blocks of inner model theory. They are transitive models $\mathcal M = (M; \in)$ satisfying a suitable fragment of set theory, namely Kripke-Platek set theory. Sifting through a couple of papers by John Steel, the term power admissible model sprung up a few times and since no definition or reference was given, I more or less assumed that they were just admissible models with enough closure properties such that the given argument would work. -Today I decided to take a closer look at them and after some online research I found a definition of power admissible sets in Cook, Rathjen. Classifying the Provably Total set Functions of $\operatorname{KP}$ and $\operatorname{KP}(\mathcal{P})$, namely Definition 8.2. According to them, a transitive model $\mathcal{M} = (M; \in)$ is power admissible iff it satisfies $\operatorname{KP}^{\mathcal{P}}$, the extension of $\operatorname{KP}$ to formulae in $\mathcal L = \{ \in, \mathcal{P}\}$, interpreting, for any $x, y \in \mathcal{M}$, -$$ -\mathcal{P}^{\mathcal{M}}(x,y) \iff y \subseteq x. -$$ -It seems very likely to me that this is in fact the kind of structure Steel has in mind. However, since I am interested in some technical details of his constructions that rely on the precise properties of $\mathcal{M}$, I'd like to verify this educated guess. - -REPLY [5 votes]: Stefan, in the paper "The Strength of Mac Lane Set Theory," Mathias says that the notion is due to Harvey Friedman, and essentially coincides with what you describe, except that in KP$^P$, foundation is restricted to universal formulae. (See page 47 of Mathias' paper -https://www.dpmms.cam.ac.uk/~ardm/maclane.pdf)<|endoftext|> -TITLE: Cohomology of a projective variety with points removed -QUESTION [7 upvotes]: Take the variety $X$ to be $\mathbb{C}_\infty \times\mathbb{C}_\infty $ with the points $(0,0)$ and $(\infty,\infty)$ removed. Use coordinates $(z,w)\in\mathbb{C}\times \mathbb{C}$ for one chart of the product of the two Riemann spheres. Then $\xi\in\Omega^{0,1}X$given by -\begin{eqnarray*} -\xi\,=\,\frac{z\,\mathrm{d}\bar z+w\,\mathrm{d} \bar w}{|z|^2+|w|^2}\ . -\end{eqnarray*} -has $\bar\partial\xi=0$ so we have $[\xi]\in \mathrm{H}^{0,1}(X;\bar\partial)$. My questions are: -1) What is $ \mathrm{H}^{*,*}(X;\bar\partial)$ ? -2) Is $[\xi]\neq 0$ ? -3) Is there a holomorphic sheaf $S$ on $X$ whose $\mathrm{H}^0(X;S)$ gives $ \mathrm{H}^{0,1}(X;\bar\partial)$ (assuming that the latter is nonzero). -4) What is the place to look for methods used for this sort of problem? -Now the background: First I am obviously not an expert in algebraic geometry, so I apologise to those who are for posing this question. The motivation comes from higher dimensional Lorentz invariant soliton theory, about which almost nothing is known. (Richard Ward's construction of a 2+1 dimensional soliton equation by the mini-twistor method being an exception.) The method I am considering is generalising the inverse scattering method to higher dimensional varieties, and the obvious thing to do is to look for geometric objects 'localised' on singularities of codimension more than one. (In the same sense that a meromorphic function is described by poles 'localised' on a point in dimension one.) The variety $X$ above is an obvious place to start for a 2+1 Lorentz invariant system, as it corresponds to the variety $t^2=x^2+y^2+z^2$, and removing the 2 singular points (not obvious but changing variables checks that other points are regular) for the given 0,1 form. -If anyone would be interested in discussing this possible geometric approach to finding higher dimensional soliton systems I would be very happy to do so. (I use 'soliton system' for the reason that such systems might not be classically integrable systems) - -REPLY [3 votes]: If I understand correctly, you are trying to compute the Dolbeault cohomology of your $X$. Also, I am assuming that by $\mathbb C_\infty$ you mean the Riemann sphere, which in algebraic geometry would be the projective line $\mathbb P^1_{\mathbb C}$. Of course, this is just rephrasing what you are saying, mainly for my own sake. -Let me actually answer in sort of a general way. Let's say that $\bar X$ is a smooth projective variety, $Z\subseteq \bar X$ is a closed subvariety and $X=\bar X\setminus Z$. In your case $\bar X=\mathbb P^1_{\mathbb C}\times \mathbb P^1_{\mathbb C}$ and $Z=\{P,Q\}$ with $P\neq Q\in \bar X$. -Anyway, the simple answer to your first question is that -$$ H^{p,q}(X, \bar\partial)\simeq H^q(X,\Omega_X^p),$$ -where $\Omega_X^p$ is the sheaf of holomorphic $p$-forms on $X$. I realize that you may already know this, so here is a little more. -It might turn out that you can easily compute $H^q(\bar X,\Omega_{\bar X}^p)$ (which is true in your case: this is very easy on $\mathbb P^1_{\mathbb C}$, and then use the Künneth formula). If that's the case, then you can use the following long exact sequence: -$$ -\dots\to H^q_Z(\bar X,\Omega_{\bar X}^p)\to H^q(\bar X,\Omega_{\bar X}^p) \to H^q(X,\Omega_X^p) \to H^{q+1}_Z(\bar X,\Omega_{\bar X}^p)\to\dots -$$ -where $H^q_Z(\bar X,\Omega_{\bar X}^p)$ is the local cohomology supported at $Z$. -Of course, now you need to be able to compute $H^q_Z(\bar X,\Omega_{\bar X}^p)$. In general this may or may not be easy. In your case it is: -Since your $Z$ is the disjoint union of two points, you only need to compute the same with $Z$ replaced by a single point. -Next replace $\bar X= \mathbb P^1_{\mathbb C}\times \mathbb P^1_{\mathbb C}$ by $\mathbb A^2_{\mathbb C}$. The local cohomology will be the same, because it only depends on a neighbourhood of $Z$. $\Omega _{\mathbb A^2_{\mathbb C}}^p$ is trivial, so what you get is that you only need to compute the local cohomology of the polynomial ring $k[x,y]$ at the maximal ideal $(x,y)$. You can probably do this yourself or find it in a book as an example. -So, now you can bootstrap back to your original cohomology group. You will probably have to write down some maps explicitly, but $H^q(\bar X,\Omega_{\bar X}^p)$ is small, so most of your cohomology will come from the local cohomology contribution. Once you do all this, I am sure you will be able to decide whether $[\xi]\neq 0$. -As far as your third question is concerned, I am not entirely sure what you consider a "holomorphic sheaf". Do you mean coherent? If so, then the answer is probably "no". This will be a pretty big group. From the long exact cohomology sequence you can see that most of the cohomology is supported at the missing points. You can write down a (double) skyscraper sheaf on $\bar X$ whose sections will give you that part of the cohomology, but I am not sure that that's what you want. -For #4, I'd say that as you see above, probably the best way to deal with this is via local cohomology.<|endoftext|> -TITLE: About the classification of commutative and of cocommutative, fin. dim. Hopf algebras -QUESTION [10 upvotes]: I want to prove that the cocommutative finite dimensional Hopf algebras over an algebraically closed field of characteristic zero are group algebras (for some finite group) and that the commutative f.d. Hopf algebras (over an algebraically closed field of characteristic zero) are dual to group Hopf algebras (for some finite group). More precisely, I want to prove the following proposition: - -Let $k$ an algebraically closed field of characteristic zero and $Η$ - a finite dimensional $k$-Hopf algebra. Then, we have the following Hopf algebra isomorphisms: - -If $Η$ is commutative, then: $Η \cong (kG)^{*}$ (for some finite group $G$). -If $Η$ is cocommutative, then: $Η \cong kG$ (for some finite group $G$). - - -I understand that the above result can be extracted as a consequence of the the Cartier-Konstant-Milnor-Moore classification theorem, according to which, a cocommutative $k$-Hopf algebra is isomorphic to a smash product between the universal enveloping algebra of the Lie algebra of the primitives of $H$ and the group Hopf algebra of the grouplikes: $H\cong U(P(H))\sharp kG(H)$. ($k$ is of course considered alg. closed with zero characterictic). -However, I want to provide an independent proof. After working a bit on it, I have devised a proof (which I am posting below as an answer) which however makes use of a later result, the so-called Larson-Radford theorem. Could there be some different approach (involving or not the Larson-Radford theorem)? - -REPLY [5 votes]: Let $k$ be a field. Recall: - -Definition: 1. A $k$-algebra $A$ is called étale, if - $A\otimes_k \bar{k} \cong \prod \bar{k}$ - where $\bar{k}$ is an alg. closure of $k$. - -A group scheme $G$ over $k$ is called étale, if the associated Hopf algebra is étale. - - -A classical theorem of Cartier states: - -Theorem (Cartier): Finite group schemes are étale in characteristic zero. - -The essential part in your proof of the classification (where Larson-Radford was used) was to show $H \cong \prod k$ for $k$ alg and $H$ commutative. Since a commutative Hopf algebra defines a finite group scheme, this part also follows immediately from Cartier's theorem. -Hence each alternative proof of Cartier also yields a new proof of your classification. A proof that uses differentials (and no Larson-Radford) can be found for instance in Waterhouse: Introduction to affine group schemes, § 11.4.<|endoftext|> -TITLE: Is there a proof that the Law of Large Numbers is the limit (numerical) for estimation of expected values? -QUESTION [7 upvotes]: For the sake of this question I want to focus on an unfair coin. -Assuming we have a number of $i.i.d.$ samples $X_1, ..., X_n$ and a precision level requirement $\tau$. I search for the optimal estimator $\hat{P_n}$ minimizing the expression $P(|\hat{P_n} - {P(X_1 = 1)} |\geq \tau)$. Also I would like to know the convergence rate in this case. -Intiutively the law of large numbers should be the best choice, which gives exponential convergence speed. But that might not be enough for small $\tau$ and little $n$. So I just wanted to check if in this case we could possibly do any better. I guess the answer should not be due to the CLT, but I can't quite prove it. - -REPLY [3 votes]: And if you want the exact minimax risk rate, take a look at the recent preprint by Iosif Pinelis and myself: https://arxiv.org/abs/1606.08920 -Essentially, the above shows that the optimal estimator is the maximum-likelihood one (i.e., the obvious one obtained by dividing the number of heads by sample size) and the error decays as $c/\sqrt n$, where $c$ is a constant we explicitly compute in the paper.<|endoftext|> -TITLE: Approximating a convex disk by an ellipse -QUESTION [41 upvotes]: For every convex compact set $K$ of area $1$ in $\mathbb{R}^2$, among all ellipses of area $1$ there exists an ellipse $E$ such that the area of the symmetric difference between $K$ and $E$ is smallest possible. - -Questions. -(a) Is $E$ unique? -(b) If the answer is "yes", does the same hold for the analogous question in higher dimensions? - -REPLY [5 votes]: Question (a) has a positive answer in the centrally symmetric case. The proof is involved and I will only summarize the strategy here. Full details can be found in this ArXiv paper. Comments, suggestions, corrections etc are welcome. -Let $K \subset \mathbb{R}^2$ be a compact convex set of area $\pi$. To prove uniqueness of best ellipse approximation, it is sufficient to consider ellipses in the family $E_t := \{(x,y) \in \mathbb{R}^2 ; e^t x^2 + e^{-t} y^2 \le 1 \}$ (because any pair of ellipses can be put inside the family by applying a suitable area-preserving linear map). I actually show that the function $A(t) := \mathrm{area} (K \vartriangle E_t)$ has a unique minimum, and is strictly decreasing (resp. increasing) to the left (resp. right) of it. Equivalently, the function is "strictly quasiconvex". -First, I consider where $K$ is regular: this means that that $K$ has a smooth boundary which is transverse to all ellipses $\partial E_t$, except for a finite number of non-degenerate (i.e. quadratic) tangencies, all of which occur outside the envelope hyperbola $xy = \pm 1/2$. In this regular case, I prove the following more precise "uniform quasiconvexity" properties: - -The function $A$ is $C^1$ everywhere and $C^2$ except at the tangency parameters. -If $t$ is not a tangency parameter, then: -$$ -\max\{|A'(t)|, A''(t)\} > \delta(A(t)) > 0, \tag{$\star$} -$$ -where $\delta$ is some positive continuous function on the interval $(0,2\pi)$ that does not depend on $K$. - -It's not difficult to convince oneself that these properties imply the strict quasiconvexity of the function $A$. Actually any uniform limit of functions with the properties above is also strictly quasiconvex. -Finally, an arbitrary $K$ can always be perturbed with respect to the symmetric difference metric so that it becomes regular. So it follows that the function $A$ is always strictly quasi-convex. -How are the quasiconvexity properties proved (in the regular case)? -By applying a suitable linear map, it is sufficient to consider $t=0$. Suppose this is not a tangency parameter (as the case of tangency is actually easier). Similarly to my older "answer", one can deduce formulas for the derivatives $A'(0)$ and $A''(0)$. Actually $A'(0)$ depends only on the points of intersection between $\partial K$ and the unit circle $\partial E_0$, while $A''(0)$ depends also on the angles of intersection. An inspection of the second formula shows that $A''(0)$ has a strong "tendency" for being positive; for example if no two consecutive intersection points are separated by an angle $>\pi/2$ then $A''(0)>0$. What we need to show in order to obtain the main inequality $(\star)$ is that if $A''(0) \leq 0$ (and $A(0)$ is not too close to $0$ or $2\pi$) then $A'(0)$ cannot be too close to zero. This is done by a case-by-case analysis, combining analytic and geometric arguments -- I can only refer to the paper for the details. -UPDATE (May 2018): The assumption that the ellipses have exactly the same area as $K$ is never used; actually if we consider ellipses whose area is a number $\lambda$ between the areas of the John and the Loewner ellipses of $K$ then the analogous uniqueness theorem holds. So I ultimately obtain a positive answer in dimension $2$ for a question of Artstein-Avidan and Katzin about uniqueness of "maximal intersection ellipsoids".<|endoftext|> -TITLE: Roadmap to Hill-Hopkins-Ravenel -QUESTION [18 upvotes]: How does one go from an understanding of basic algebraic topology (on the level of Allen Hatcher's Algebraic Topology and J.P. May's A Concise Course in Algebraic Topology) to understanding the paper of Hill, Hopkins, and Ravenel on the Kervaire invariant problem? -I have read that some understanding of chromatic homotopy theory and equivariant homotopy theory (neither of which I am familiar with, aside from their basic idea) is required, but it seems that there are still quite a few steps from basic algebraic topology to these two subjects and from these two subjects to the Hill-Hopkins-Ravenel paper. -I do have some idea of a few topics beyond what is discussed in the books of Hatcher and May, such as model categories on the level of Dwyer and Spalinski's Homotopy Theories and Model Categories, as well as some idea of spectral sequences (both of which are required reading for the paper, I believe), but it would be nice if we could come up with some sort of roadmap (with an ordered list of subjects, and even better if we had references) from the Hatcher and May books to the Hill-Hopkins-Ravenel paper. For that matter, I don't even know how spectral sequences and model categories fit in, so it would be nice I guess if there's also an explanation as to how each of these prerequisite subjects fit into the paper. -I'm looking for something like the very nice answer to this query regarding the Norm Residue Isomorphism theorem. - -REPLY [20 votes]: There is one major topic that is missing from your list: spectra. Spectra (and $E_\infty$-ring spectra) are the basics of modern stable homotopy theory and are not treated, if not very cursorily, in the references you've looked at. Seriously, spectra are the bread and butter of a homotopy theorist nowadays and their absence from the standard references is as embarrassing as the absence of schemes would be from the standard material for algebraic geometry. -That said, let me try to put together a list of references that should help you getting started. - -The first place to look at is Adams' amazing Stable homotopy and generalized cohomology. This is a collection of lecture notes held at the university of Chicago and, famously, has to be read in reverse order (first part III, then part II and optionally part I) since they arranged the lectures chronologically rather than logically. Part III will give you a basic understanding of spectra, Part II will introduce the very basics of chromatic homotopy theory. Be warned that the treatment is very old (in fact if I'm not mistaken it is the first published construction of the stable homotopy category) and in particular the construction of the smash product of spectra is very lacking from a modern perspective, so you could skip it and just take the basic properties. In addition, the treatment of localizations has a couple of imprecisions, I suggest you complement that section with Bousfield's The localization of spectra with respect to homology. -To get a more modern perspective, you cannot go much wrong with Schwede's Symmetric spectra. This is a modern treatment of a different (and better) model for spectra, which has the advantage of introducing you to $E_\infty$-ring spectra too, one of the basic structures that play a very important role in Hill-Hopkins-Ravenel (and all the rest of modern homotopy theory). -Next you should get some familiarity with equivariant homotopy theory. A nice starting point is Adams' Prerequisites (on equivariant homotopy theory) for Carlsson's lecture. Again it is a bit old fashioned but having read it helps hitting the ground running. -Since we're talking about equivariant homotopy theory, how could I not mention Schwede's Lectures on stable equivariant homotopy theory? The model he uses for $G$-spectra is slightly different from the one in Hill-Hopkins-Ravenel, but again having read it will certainly help. -Hill-Hopkins-Ravenel's paper itself has a nice self-contained introduction to stable homotopy theory, that I strongly encourage you to read. Maybe skip the proof that the norm is homotopically meaningful in appendix B (it is very technical and while it has some very interesting ideas the only thing you really need is the statement of the theorem). -Chromatic homotopy theory is surprisingly less present in the actual solution of the Kervaire invariant one problem (you basically need only to know about the Adams-Novikov spectral sequence and the statement of a few highly nontrivial computations), but if you want to learn it (you should if you're interested in this circle of ideas), I can recommend Ravenel's Nilpotence and periodicity in stable homotopy theory and Lurie's lecture notes on the subject. If you are interested on how to do those computation that are needed for the HHR paper, your best friend is Ravenel's Complex cobordism and stable homotopy groups of spheres (credits to Lennart Meier for prompting me to add this). Also reading Ravenel's previous paper on the odd Kervaire invariant problem is helpful for understanding how the main HHR paper connects with chromatic homotopy theory. -Since you mentioned spectral sequences, I think you will generally pick them up as you work through the above material. However a paper that really helped me with them is Boardman's Conditionally convergent spectral sequences. Highly recommended. - -I am sure my list has some glaring omissions, but this should at least get you started (and keep you busy for a while I suspect). Good luck and feel free to hang around the homotopy theory chatroom if you have questions!<|endoftext|> -TITLE: Is the discriminant of a free (as a module) $R$-algebra always congruent to a square modulo 4? -QUESTION [12 upvotes]: Let $R$ be a commutative ring. Let $A$ be an $R$-algebra (i.e., an $R$-module - equipped with an $R$-bilinear multiplication map that turns $A$ into a unital - ring). We do not require $A$ to be commutative. Assume that $A$ is free as an $R$-module, with a finite basis. Let - $\left( e_{1},e_{2},\ldots,e_{n}\right) $ be a basis of the $A$-module $R$. -We define a trace map $\operatorname{Tr}_{A/R}:A\rightarrow R$ as - follows: For every $b\in A$, we let $\operatorname{Tr}_{A/R}b$ be - the trace of the endomorphism $A\rightarrow A,\ a\mapsto ba$ of the $R$-module - $A$. (We are using the fact that $A$ has a finite basis here.) Clearly, the - map $\operatorname{Tr}_{A/R}$ is $R$-linear. -Let $\Delta=\det\left( \left( \operatorname{Tr}_{A/R}\left( -e_{i}e_{j}\right) \right) _{1\leq i\leq n,\ 1\leq j\leq n}\right) \in R$. -Is it true that $\Delta=u^{2}+4v$ for some elements $u$ and $v$ of $R$ ? - - -The above is an elaborate generalization of Stickelberger's discriminant -theorem. Indeed, if we assume $A$ to be commutative, then $\Delta -=u^{2}+4v$ is true; this is proven in Remark 5.4 of Owen Biesel, Alberto -Gioia, A new discriminant algebra construction, arXiv:1503.05318v3, -Documenta Mathematica 21 (2016), pp. 1051--1088. (More precisely, they -prove it in the case when $n\geq2$; but the remaining case is obvious.) If we -furthermore assume that $R=\mathbb{Z}$ and $A$ is the integer ring of a number -field $K$, then $\Delta$ becomes the discriminant $\Delta_{K}$ of $K$, and the -$\Delta=u^{2}+4v$ claim becomes $\left( \Delta_{K}\equiv0 \mod % -4\ \vee\ \Delta_{K}\equiv1 \mod 4\right) $, which is the -classical claim of Stickelberger's discriminant theorem. -The claim does not significantly depend on the choice of basis $\left( -e_{1},e_{2},\ldots,e_{n}\right) $, since any change of basis causes $\Delta$ -to be multiplied by a square in $R$ (namely, by the square of the determinant -of the matrix responsible for the change of basis... or of its inverse, -depending on how you define that matrix). -I do not know whether to expect the conjecture to be true or not. All examples -I have tried (the complexity of computing $\Delta$ for high $n$ limits my -abilities here) satisfy $\Delta=u^{2}+4v$ for rather stupid reasons (often, -$\Delta$ will either be a square or be divisible by $4$ to begin with); but -this says more about the poverty of my examples than about the correctness of -the conjecture. For what it's worth, here are my examples: - -If $A$ is the group ring of a group $G$, and $\left(e_1, e_2, \ldots, e_n\right)$ is the standard basis $G$ of $A$ (abusing notation as usual), then $\Delta = \left(-1\right)^{n\left(n-1\right)/2} n^n$. This is either of the form $4v$ or of the form $1+4v$; thus, the conjecture holds here. -If $A$ is the matrix ring $R^{m\times m}$ (so that $n=m^2$), and $\left(e_1, e_2, \ldots, e_n\right)$ is the basis consisting of the matrix units in their usual order, then $\Delta = \left(-1\right)^{m\left(m-1\right)/2} m^n$. This is either of the form $4v$ or of the form $1+4v$; thus, the conjecture holds here. -If $A$ is the (standard) quaternion algebra, then $\Delta = -4^4$; thus, the conjecture holds here. Other quaternion algebras may have different $\Delta$, but the factor $4^4$ will still be present. - -Notice that the trace map $\operatorname{Tr}_{A/R}$ defined above -has a "right analogue": the map $\operatorname{Tr}_{A/R}^{\prime -}:A\rightarrow R$ sending each $b\in A$ to the trace of the endomorphism -$A\rightarrow A,\ a\mapsto ab$ of the $R$-module $A$. If we do not require -commutativity of $A$, these maps will in general not be identical (for a -specific example, extend a left-trivial semigroup by a $1$ to obtain a -monoid, then take the monoid algebra of this monoid). -I do not know whether the $\Delta$ computed via the other map will be different; this is another interesting question. - -REPLY [3 votes]: I see this question is still being viewed. To avoid misleading anyone, let me put a placeholder answer here: The question has been answered positively (with a beautiful proof) by John Voight, Asher Auel and Owen Biesel, who have submitted their proof to the Amer. Math. Monthly and will probably put it on the arXiv once the refereeing process is done.<|endoftext|> -TITLE: Pin$^+$ and Pin$^−$ structure for manifolds in any dimensions -QUESTION [7 upvotes]: For an oriented $d$-manifold $M$, we can ask whether the manifold -admits a Spin structure, say, if the transition functions for the -tangent bundle, which take values in $SO(d)$, can be lifted to $\operatorname{Spin}(d)$ while preserving the cocycle condition on triple overlaps of coordinate charts. -For an unoriented $d$-manifold $M$, we can ask whether the manifold -admits Pin$^+$ or Pin$^−$ structures (that is, lifts of transition functions to -either $\operatorname{Pin}^+(d)$ or $\operatorname{Pin}^−(d)$ from $O(d)$. This is analogous to lifting the transition functions to $\operatorname{Spin}(d)$ from $SO(d)$ for the spin manifold). -If the manifold $M$ is orientable, then the conditions for Pin$^+$ or Pin$^−$ structures are the same. So it reduces to the old condition that $M$ admits a Spin structure or not. -Every 2-dimensional manifold admits a Pin$^−$ structure, but not necessarily a Pin$^+$ structure. -Every 3-dimensional manifold admits a Pin$^-$ structure, but not necessarily a Pin$^+$ structure. - -Question: For some any other $d$, say $d = 0, 1, 4,$ etc. are there statements like: every $d$-dimensional manifold admits a Pin$^-$ structure? Or, every $d$-dimensional manifold admits a Pin$^+$ structure? Or are there useful conditions like SW classes $w_2+w_1^2=0$ like the case for $d = 2, 3$? What are these conditions in other dimensions? - -P.S. The original post on MSE received almost no attention for a week. - -REPLY [11 votes]: For $d \geq 4$, let $M_d = (S^1)^{d-4}\times\mathbb{CP}^2$. As tori are parallelisable, $w(M_d) = w(\mathbb{CP}^2)$, in particular $w_1(M_d) = 0$ and $w_2(M_d) \neq 0$, so $M_d$ does not admit a Spin, Pin$^+$, or Pin$^-$ structure. -Therefore, there is no $d \geq 4$ such that every $d$-dimensional manifold admits a Pin$^+$/Pin$^-$ structure.<|endoftext|> -TITLE: Find the maximum of $|a_{p}|$, if $a_0+a_1x+\dots+a_nx^n:[-1,1]\mapsto [-1,1]$ -QUESTION [6 upvotes]: Let $n$ be a given positive integer, and let $f(x)=\displaystyle\sum_{k=0}^{n}a_{k}x^k$, where $a_{i}\in \mathbb{R}$, $0 \le i \le n$. If -$$|f(x)|\le 1,\qquad \text{for } ~|x|\le 1,$$ -what is the maximum of the $|a_{p}|$ for a fixed $p$？ -I conjecture that the answer is $|[x^p]T_{n}(x)|$, where the $T_{n}(x)$ are the Chebyshev polynomials. Can we find the closed form for it? Thanks - -REPLY [15 votes]: Let $T_{n}(x)=\sum_{\nu=0}^{n}t_{n,\nu}x^{\nu}$ denote the Chebyshev polynomial (of the first kind) of degree $n$ and let $x_{n,\nu}=\cos\nu\pi/n$ for $0\leq\nu\leq n$. -The answer follows from the following two results : -1) Let $f$ be a polynomial (possibly complex), of degree at most $n$, such that -$$|f(x_{n,\nu})|\leq1,\qquad 0\leq\nu\leq n.$$ -Then -$$|a_{n-2\mu}|\leq|t_{n,n-2\mu}|,\qquad0\leq\mu\leq n/2,$$ -which becomes an equality for $f=T_{n}$. -2) Let $f$ be a polynomial (possibly complex), of degree at most $n$, such that -$$|f(x_{n-1,\nu})|\leq1,\qquad 0\leq\nu\leq n-1.$$ -Then -$$|a_{n-2\mu-1}|\leq|t_{n-1,n-2\mu-1}|,\qquad0\leq\mu\leq (n-1)/2,$$ -which becomes an equality for $f=T_{n-1}$. -These are Theorems 16.3.1 and 16.3.2 in : - -Rahman, Q. I.; Schmeisser, G. Analytic theory of polynomials. London - Mathematical Society Monographs. New Series, 26. The Clarendon Press, - Oxford University Press, Oxford, 2002. - -The proof of the first assertion consists in considering a one-parameter family of polynomials constructed from $f$ and $T_{n}$ and depending on a parameter $\theta\in(-1,1)$, and then using Descartes' rule of signs. The second assertion is a simple consequence of the first one.<|endoftext|> -TITLE: Understanding the structure of unitary groups -QUESTION [6 upvotes]: I would like to understand precisely the structure of unitary groups. -Let $F$ be a global number field, $E$ a quadratic extension of $F$, and $U$ a unitary group on $E$ (i.e. the group of automorphism preserving an hermitian form on an $E$-vector space), say of rank $n$ (i.e. the hermitian form is on a squared $n$-dimensional vector space). -What can be said for the local groups, that is for the $U_v$ where $v$ goes through the places of $F$? More precisely: - -are the $U_v$ also unitary groups, or can they be isomorphic to $GL_n(F_v)$? -is there only a finite number of possibility for $U_v$? -what are the obstructions for a unitary group $G$ to have prescribed local groups $G_v$? - -Thanks in advance ! - -REPLY [3 votes]: As you already seem to know, unitary groups are classified by separable quadratic field extensions (in fact one should really work with separable quadratic algebras, i.e. also $F\times F$, corresponding to the ''trivial'' unitary group $\mathrm{GL}_n$). -Thus your questions can essentially be reduced to problems about quadratic field extensions. - -are the $U_v$ also unitary groups, or can they be isomorphic to $GL_n(F_v)$? - -You need to be a bit careful, since, as I already point out, you should really view $GL_n$ itself as a unitary group. But you really want to ask whether $U_v$ can be isomorphic to a general linear group. As already explained in the comments, and following the classification, this happens iff $v$ is split in $E/F$. - -is there only a finite number of possibility for $U_v$? - -You again need to be careful with the phrasing of this question, as it is not clear what you mean, since it is not possible to compare unitary groups over different fields. However, over a given local field $F_v$, there are only finitely many isomorphism classes of unitary groups of given dimension. This follows from the fact that there are only finitely many quadratic extensions of a given local field. This is nothing special about unitary groups, since, more generally, over any local field $F_v$ there are only finitely many isomorphism classes of reductive algebraic groups of given dimension (this can be proved using the classification of reductive algebraic groups over algebraically closed fields combined with the fact that $F_v$ admits only finitely many field extensions of given degree). - -what are the obstructions for a unitary group $G$ to have prescribed local groups $G_v$? - -This is the most interested question. The obvious obstruction is of course the dimension of the groups, so we assume the local groups $G_v$ have the same dimension. Also one cannot hope for a positive answer in general when you consider infinitely many places, so we suppose that we are given only finitely many places $v$ and finitely many groups $G_v$. Then there the answer is: there is no obstruction! There is always some unitary group $G$ which realises the given $G_v$. This follows from the Grunwald-Wang theorem, which implies that you can always realise any given finite collection of quadratic algebras over $F_v$ by some quadratic extension (though of course this elementary case does not require the full force of Grunwald-Wang). -Note that this is an important place where one works with quadratic field extensions; this approximation property fails for field extensions in general. You can read more about this in Section IX.2 of the book "Cohomology of Number Fields" by Neukirch, Schmidt, Wingberg.<|endoftext|> -TITLE: difference between the small and big étale/flat/... site -QUESTION [9 upvotes]: What is the difference between the small and the big étale (or flat or syntomic or ...) site? How does the cohomology vary? When should I use which one? Up to now, I have always used the small sites. - -REPLY [7 votes]: You can always compute the cohomology as the limit over hypercoverings of the final object. But the hypercoverings of the final object are the same in both categories, so it seems that the cohomology is the same (i.e. the cohomology of a sheaf on the big etale site maps isomorphically to the cohomology of its restriction to the small etale site). -Big sites are useful e.g. when you want to consider schemes as etale (or fppf or ...) sheaves. If $S$ is a scheme, then an $S$-scheme $X$ - gives and is determined by its associated sheaf ${\rm Hom}_S(−,X)$ - on the big etale site of $S$. So you can define e.g. the Picard scheme by defining the associated sheaf first and then asking whether it is representable. This utterly fails for the small etale site though: take $X\to S$ a closed immersion, then the restriction of ${\rm Hom}_S(−,X)$ - to the small etale site will usually be the empty sheaf.<|endoftext|> -TITLE: Normal form of elliptic curves via symmetric polynomials -QUESTION [5 upvotes]: First, let me formulate the question in the language of elliptic curves. Almost every elliptic curve has a Weierstrass normal form $y^2=x^3+px+1$, but this form is bad from the point of view of invariant theory, and I have heard that there is another normal form written in terms of symmetric polynomials. -Second, from the point of view of invariant theory, the group $\operatorname{SL}_3$ acts standardly on $S^3\mathbb C^3$, there is an open invariant subset $W \subset S^3\mathbb C^3$ and some subset -$$U=\{ y^2z-x^3-pxz^2-qz^3 \} \subset S^3\mathbb C^3,$$ -such that any $\operatorname{SL}_3$-orbit in $W$ intersects $U \cap W$ exactly once. Namely, $W$ consists of polynomials that define smooth irreducible curve in $\mathbb{CP}^3$, the intersection $U \cap W$ is defined by non-vanishing of the discriminant of $x^3+px+q$, and its elements are called Weierstrass normal forms. -Then the invariants $p$ and $q$ can be defined via some complicated contractions of tensors involving many indices, but it gives little insight. Could you help me with an anologous normal form, but involving symmetric polynomials, and its invariants? I am in particular interesting in the case of $\operatorname{SL}_3$, not just $\operatorname{GL}_3$. It is probably very classical subject, but I can not find the answer, so any reference is welcome! - -REPLY [4 votes]: I think what you're looking for are the standard invariants (and covariants) of the space of non-singular cubic polynomials -$$ f(x,y,z) := ax^3+by^3+cz^3+dx^2y+exy^2+fx^2z+gxz^2+hy^2z+iyz^2+jxyz $$ -relative to the natural action of $\text{SL}_3$. If you're looking for formulas, they're in (among other places) Lessons introductory to the modern higher algebra, George Salmon, 1876 (3rd ed). As Will Sawin noted, these invariants are often denoted by $I(f)$ and $J(f)$. There are also covariants which can be used to define a finite map from the curve $f(x,y,z)=0$ to the elliptic curve $$Y^2Z=X^3+I(f)XZ^2+J(f)Z^3.$$ -Final quibble: It is not true that "Every elliptic curve has a Weierstrass normal form $y^2=x^3+px+1$," although that is almost true over algebraically closed fields of characteristic 0. But even in that case, I think you're missing the isomorphism class of the curve $y^2=x^3+x$, i.e., the curve with endomorphism ring $\mathbb Z[i]$.<|endoftext|> -TITLE: Resolution of an isolated cyclic quotient singularity -QUESTION [7 upvotes]: I am looking for a reference to the following fact which seems to be true and which is probably well-known (at least to experts in resolution of singularities): -Consider an isolated cyclic quotient singularity ${\mathbb C}^n/G$, where $G$ is a finite cyclic group acting on ${\mathbb C}^n$ by linear complex transformations. Then there is a sequence of blow-ups of ${\mathbb C}^n$ with $G$-invariant smooth centers (always lying in the set of points with non-trivial stabilizers under the $G$-action) which yields a complex manifold $M$ so that $M/G$ is smooth and $M/G\to {\mathbb C}^n/G$ is a resolution of the original singularity. -Edit: Jason Starr's example below shows that the claim is actually false. - -REPLY [5 votes]: Edit. As Michael Entov points out, the original example (below with strikethrough) is incorrect. (Sorry!) Here is a corrected example. It is not always possible to find an equivariant resolution satisfying all of the hypotheses you list. The simplest example I know of is as follows. Let $G$ be $\mu_5$, the group of fifth roots of unity. Let the action of $\mu_5$ on $\mathbb{C}^2$ be $\zeta\bullet(x,y) = (\zeta x,\zeta^3 y)$. The unique point with nontrivial stabilizer is the origin. One of the standard affine opens for the blowing up of the origin is $\text{Spec}\ \mathbb{C}[u,x]$ where $ux$ equals $y$. Now the action is $\zeta\bullet(u,x) = (\zeta^2u,\zeta x)$. Up to the automorphism $G\to G$ by $\zeta \mapsto \lambda = \zeta^2$, this action is $\lambda \ast(u,x) = (\lambda u,\lambda^3 x)$. Thus, we have the same type of action for the fixed point $(u,x)=(0,0)$ on the blowing up as for the origin on the original variety. -It is not actually possible to find an equivariant resolution satisfying all of the hypotheses you list. The simplest example I know of is when $G$ equals $\mu_3$, the group of cube roots of unity, and where the action of $\mu_3$ on $\mathbb{C}^3$ is by $\zeta\bullet(x,y,z)=(x,\zeta y, \zeta^{-1}z)$. The only point of $\mathbb{C}^3$ with a nontrivial stabilizer is the origin $(0,0,0)$. For the blowing up of the origin, $\nu:\widetilde{\mathbb{C}^3} \to \mathbb{C}^3$, for the point $p$ on the exceptional divisor corresponding to the tangent direction of the line $\text{Zero}(y,z)\subset \mathbb{C}^3$, there is an affine chart of $p$ in $\widetilde{\mathbb{C}^3}$ that is $\mu_3$-equivariantly isomorphic to the origin in $\mathbb{C}^3$. So it is too much to expect that you can resolve the singularities by only blowing up smooth centers inside the locus of points with nontrivial stabilizers. -What people usually do instead is allow the centers to be not-necessarily-smooth. This can still lead to a smooth manifold $M$ such that $M/G$ is a resolution of $\mathbb{C}^n/G$. -The "canonical" reference for this is Section 4 of Miles Reid's article, paticularly the proof of Theorem 4.6. -MR0927963 (89b:14016) Reviewed -Reid, Miles(4-WARW) -Young person's guide to canonical singularities. Algebraic geometry, Bowdoin, 1985 (Brunswick, Maine, 1985), 345–414, -Proc. Sympos. Pure Math., 46, Part 1, Amer. Math. Soc., Providence, RI, 1987. -14E30 (14B05 14E05 14J10) -This is also discussed in Sections 2.2 and 2.4 of Fulton's book. -MR1234037 (94g:14028) Reviewed -Fulton, William(1-CHI) -Introduction to toric varieties. -Annals of Mathematics Studies, 131. The William H. Roever Lectures in Geometry. -Princeton University Press, Princeton, NJ, 1993. xii+157 pp. -14M25 (14-02 14J30) -Roughly the idea is to perform the blowings up in such a way that the locus of points with nontrivial stabilizer becomes divisorial in some $\widetilde{\mathbb{C}^n}$. In the (corrected!) example above, if you first blow up the ideal $\langle x,y^2 \rangle$, this makes the fixed locus divisorial, but introduces one ordinary double point. Now, if you blow up the ordinary double point, then you have a smooth $M$ such that also $M/G$ is smooth.<|endoftext|> -TITLE: sum, integral of certain functions -QUESTION [5 upvotes]: While working on some research, I have encountered an infinite series and its improper integral analogue: -\begin{align}\sum_{m=1}^{\infty}\frac1{\sqrt{m(m+1)(m+2)+\sqrt{m^3(m+2)^3}}}&=\frac12+\frac1{\sqrt{2}}, \\ -\int_0^{\infty}\frac{dx}{\sqrt{x(x+1)(x+2)+\sqrt{x^3(x+2)^3}}}&=2.\end{align} -The evaluations were guessed using numerical evidence. -Can you provide proofs, or any reference (if available)? - -REPLY [17 votes]: For the integral, notice that the expression under the square root is - $$ x(x+1)(x+2)+x(x+2)\sqrt{x(x+2)} = \frac12\,x(x+2)(\sqrt x+\sqrt{x+2})^2. $$ -Consequently, -\begin{align*} - \frac1{\sqrt{x(x+1)(x+2)+x(x+2)\sqrt{x(x+2)}}} - &= \frac{\sqrt 2}{(\sqrt x+\sqrt{x+2}) \sqrt{x(x+2)}} \\ - &= \frac1{\sqrt 2}\,\frac{\sqrt{x+2}-\sqrt{x}}{\sqrt{x(x+2)}} \\ - &= \frac1{\sqrt 2} \left( \frac1{\sqrt x}-\frac1{\sqrt{x+2}}\right); -\end{align*} -thus, the indefinite integral is - $$ \sqrt{2}\, (\sqrt x-\sqrt{x+2})+C $$ -and the result follows easily. -As Antony Quas noticed, this also works for the sum showing that the partial sum over $m\in[1,M]$ is - $$ \frac1{\sqrt 2} \sum_{m=1}^M \frac1{\sqrt m} - \frac1{\sqrt 2} \sum_{m=3}^{M+2} \frac1{\sqrt m} = \frac1{\sqrt 2} \left( 1+\frac1{\sqrt 2}\right) + o(1). $$<|endoftext|> -TITLE: values of $\zeta$ function are linearly independent? -QUESTION [14 upvotes]: Are the elements of the set $\{\zeta(2n+1)| n\in \mathbb{N}\}$ $\mathbb{Q}$-linearly independent? - -REPLY [20 votes]: People have been exerting steady effort to prove/disprove irrationality of the Riemann zeta values in your list. Of course, $\zeta(3)$ is known to be irrational due to Roger Apery. Such investigations, among others, motivated the question of linear independence. As Felipe commented, however, not much is known, apart from the following article: - -Rivoal, Tanguy. -La fonction zêta de Riemann prend une infinité de valeurs irrationnelles aux entiers impairs. [There are infinitely many irrational values of the Riemann zeta function at odd integers] -C. R. Acad. Sci. Paris Sér. I Math. 331 (2000), no. 4, 267–270. - -The article is available online here. -Wadim Zudilin makes some improvement, see: - -Zudilin. -Arithmetic of linear forms involving odd zeta values. -J. Théor. Nombres Bordeaux 16 (2004), no. 1, 251–291. -Zudilin. -On the irrationality of the values of the Riemann zeta function. Izv. Ross. Akad. Nauk Ser. Mat. 66 (2002), no. 3, 49--102; translation in -Izv. Math. 66 (2002), no. 3, 489–542. - -That is the state-of-the-art for the present issue you raised.<|endoftext|> -TITLE: Infering shapes from overlap with a shifting circle -QUESTION [8 upvotes]: A recent episode of Star Talk Radio discussed among other things the unknown object(s) orbiting Tabby's star (aka "Alien mega structure discovered!" in non-scientific media) and an astronomer said with confidence that we know that whatever is orbiting that star isn't a planet because "it isn't round". Given that the mysterious object(s) were discovered by examining the variations in luminosity coming from the star, this got me thinking about light curves and what they can tell us about transiting objects. -Most generally my question is this: - -Given a closed subset $A\subseteq D\subseteq\mathbb{R}^2$ where $D$ is the closed unit disc, what can we infer about the shape of $A$ when we are given the "light curve" $L_A(t)=\lambda^2(D\setminus (A+te_1))$ or equivalently if we are given $t\mapsto\lambda^2(D\cap (A+te_1))$. - -In particular I would like to know: - -Can we distinguish convex sets from non-convex sets? And if yes, are convex $A$ uniquely determined by their $L_A$ (up to obvious non-uniquenesses like reflection across the x-axis)? - -Doing some elementary geometry one finds that circles are indeed detected by their light curves thus validating Dr. Shostak's claim. - -REPLY [3 votes]: A convex set and a non-convex set can have the same light curve. -Let $r$ be the reflection about the $x$-axis. Then $r(S)$ has the same light curve as $S$. So, let $C$ be a convex set above the x-axis, and let $C = C_1 \cup C_2$. Then $C$ has the same light curve as $C_1 \cup r(C_2)$.<|endoftext|> -TITLE: Is there a geometric realization of $\mathbf{C}((t))$-varieties? -QUESTION [28 upvotes]: Let $MV_F$ be the $\mathbf{A}^1$-homotopy category over the field $F$. Let $H$ be the homotopy category of spaces, and let $H_{/S^1}$ be the homotopy category of spaces over the circle. -When $F = \mathbf{C}$, there's a Betti realization functor $MV_{\mathbf{C}} \to H$. Is there a similar realization functor $MV_{\mathbf{C}((t))} \to H_{/S^1}$? -Etale homotopy theory gives a realization in profinite spaces over a profinite circle. I'm looking for a refinement of that. -If a variety over $\mathbf{C}((t))$ is defined using only convergent power series, I think it is clear which bundle over $S^1$ I want. If the variety is defined using finitely many nonconvergent power series, I can imagine some tricks for defining the bundle over S^1 that I want, even up to homeomorphism. But if a map between varieties has some nonconvergent power series in it, I bet there is no way to write down a map between those bundles, not functorially. I hope that it can be done up to homotopy. - -REPLY [7 votes]: I don't know about $\mathbb A^1$-homotopy categories. Let me try to do something that I think will work just for the category of quasiprojective varieties. -The key lemma is this: - -There is a canonical way to associate to a variety $X$ over $\mathbb C$ and a $\mathbb C((t))$-point of $X$ a homotopy class of maps from the circle to $X$ that is functorial for maps of varieties over $\mathbb C$. - -To do this, by a stratification of $X$ reduce to the case when $X$ is smooth, then by resolution of singularities we may assume that $X$ has a smooth normal crossings compactification. Then extend the $\mathbb C((t))$-point to a $\mathbb C[[t]]$-point of the compactification. Locally near that point, $X$ looks like $\mathbb R^n$ times a torus, so a homotopy class of circle maps is a tuple of integers - you just take the $t$-adic valuations of the coordinates in the obvious way. -Could this depend on the compactification? No. It only depends on the coordinates up to some higher power of $t$, i.e. it is a locally constant function in the $t$-adic topology on $X(\mathbb C((t)))$. Because we may approximate any solution by a solution with algebraic (and hence convergent) power series using Hensel's lemma for the ring of algebraic power series, it is sufficient to check this independence for convergent power series, where it is obvious, because our definition recovers the usual definition given by evaluating the power series on a small circle. -An identical reduction argument proves functorially. - -We view a quasiprojective variety $X - Z \subseteq \mathbb P^n$ over $\mathbb C((t))$ as a $\mathbb C((t))$-point of the flag Hilbert scheme parameterizing pairs of closed subscheme $Z \subseteq X \subseteq \mathbb P^n$. This Hilbert scheme admits a stratification into strata where the fibers have constant topological type. For each stratum $M$ in the Hilbert scheme, we have a fibration of manifolds (or more general spaces) over $M(\mathbb C)$, so $\mathbb C(t))$-point of $M$ defines a homotopy class of circle embeddings into $M$ and hence a homeomorphism class of manifolds over the circle. -For functoriality, we use a more intricate moduli space parameterizing ordered triples of two quasiprojective varieties and a map between them. Again there will be a stratification of constant topological types, and a homotopy class of circle embeddings into a stratum, giving a class of pairs of manifolds w/ a map between them, up to, I believe, homotopy. -You should be able to check that this agrees with composition of maps bby passing to the moduli space of triples of varieties and pairs of morphisms. To check that the homotopies defined their satisfy the appropriate coherence condition, you would have pass to the moduli space of quadruples of varieties with triples of morphisms, and so on.<|endoftext|> -TITLE: On the category of categories with factorization system -QUESTION [6 upvotes]: Is there a reasonable definition for a category of categories equipped with a factorization system, say $({\bf C},(E,M))$ and functors $F\colon ({\bf C},(E,M))\to ({\bf D}, (E', M'))$ that preserve them? I see several possible choices, all unreasonable in some sense: - -$F(E)\subseteq E'$, and $F(M)\subseteq M'$; this is unreasonable, as it seems to strong $F=1_{\bf C}$ does not send $(All, Iso)$ to $(Iso, All)$. Do I have to interpret this as an analogue of the identity being non-continuous if the domain topology is coarser? -$F$ preserves the left class. This is unreasonable, as it is not canonical. Why left, when $F$ is not a left adjoint? -$F$ preserves the right class. This is unreasonable, as it is not canonical. Why right, when $F$ is not a right adjoint? -$F(\mathbb F)\subseteq \mathbb{F}'$, if $\mathbb F =(E,M)$, and $\mathbb F' = (E', M')$. This means that $F(M)\subseteq M'$ and $F(E)\supseteq E'$. - -I must admit I am confused. - -REPLY [12 votes]: The main theorem of Korostenski and Tholen's Factorization systems as Eilenberg-Moore algebras proves that categories with (orthogonal) factorization systems are precisely the normal pseudo algebras for a 2-monad on Cat. The 2-monad in this case is the 2-monad that sends a category $C$ to its arrow category $C^2$ with the rest of the monad structure that you would guess (arising from the fact that the walking arrow category 2, like any category, is canonically a comonoid with respect to the cartesian product). -How do we think about this result? To equip a category $C$ with the structure of an algebra is to define a functor $C^2 \to C$ that is a (strict) retraction of the diagonal functor $C \to C^2$. (This is the meaning of "normal" above.) Some clever diagram chasing proves that such a functor equips $C$ with a functorial factorization, with the functor $C^2 \to C$ sending a morphism to the object through which it factors. To say that this data defines a pseudo-algebra for the 2-monad demands an additional associativity condition, up to natural isomorphism (hence the "pseudo"), and this can be seen to imply that the left factor of a right factor is an isomorphism and that the right factor of a left factor is an isomorphism. These conditions, together with functoriality, suffice to show that the functorial factorization defines an orthogonal factorization system. -Note also that for any 2-monad $T$ on Cat, there is a related 2-monad $T'$ so that strict $T'$-algebras are exactly pseudo $T$-algebras. -So in particular, categories with a factorization system are also the strict algebras for a 2-monad, if you would prefer. -2-monad theory now provides a general framework within which to answer Fosco's question. In general, there are four natural kinds of morphisms for algebras for a 2-monad --- strict, pseudo, lax, and oplax --- with the strict ones (preserving everything on the nose) only rarely encountered in the wild. -Suppose $C$ and $D$ are categories with factorization systems whose functorial factorizations are given by a pair of functors $Q : C^2 \to C$ and $E \colon D^2 \to D$ that send a morphism to the object through which it factors. A lax morphism of algebras specifies a functor $F \colon C \to D$ together with a natural transformation $EF \to FQ$ whose components define a commuting "comparison morphism" between the two factorizations of a map in the image of $F$. The presence of such a comparison implies that $F$ preserves the right classes. -Dually, a colax morphism specifies a functor $F \colon C \to D$ together with a natural transformation $FQ \to EF$ the presence of which implies that $F$ preserves the left classes. By taking mates, one can see that if $F \dashv U$, then $F$ defines a colax morphism of algebras if and only if $U$ defines a lax morphism of algebras, which explains why lax morphism structure feels natural for right adjoints and colax morphism structure feels natural for left adjoints. -Some preliminary remarks along these lines in the more general setting of algebraic weak factorization systems can be found in my thesis or (in a much more satisfying account) in Bourke and Garner's Algebraic weak factorization systems I.<|endoftext|> -TITLE: Betti numbers as characteristic numbers? -QUESTION [9 upvotes]: Let $X$ be a compact differentiable manifold of dimension $m$ or, if you prefer, a smooth complex projective manifold of complex dimension $n=m/2$. -The Euler characteristic $\chi(X):=\Sigma_{i=0}^{m}(-1)^ib_i(X)$ -the alternating sum of Betti numbers of $X$- is related to the Euler class $e(T_X)$ of the tangent bundle $T_X$ by -$$\chi(X)=\int_Xe(T_X)$$ -where $\int_X$ denotes contraction against the fundamental class $[X]$. -Is it possible to write each Betti number $b_i(X)$ as a "characteristic number", i.e. as an integral -$$b_i(X)\stackrel{\text{?}}{=}\int_X\gamma_i(T_X)$$ -of some "characteristic class" $\gamma_i$ of vector bundles? If not, why not? -This question is possibly related. - -REPLY [17 votes]: No. The Stiefel-Whitney and Pontryagin numbers of a closed oriented manifold are cobordism invariants, but the Betti numbers are not. -More explicitly, all closed oriented $3$-manifolds are frameable (and null cobordant), so all of their characteristic numbers vanish, but e.g. the $3$-torus and $3$-sphere have different Betti numbers.<|endoftext|> -TITLE: Does the percentage of groups of order at most $n$ of even order aproach $1$? -QUESTION [7 upvotes]: Let $E_n$ be the number of isomorphism classes of groups of even order at most $n$, let $G_n$ be the number of isomorphism classes of groups of order at most $n$ and $T_n$ be the number of isomorphism classes of $2$-groups of order less than $n$. -I think that it is an open problem if $\frac{T_n}{G_n}$ approaches $1$, so I figured that it is probably not so hard to prove that $\frac{E_n}{G_n}$ approaches $1$, although I was not able to do so. - -REPLY [5 votes]: I think the following works: -Looking at the book enumeration of finite groups, -We obtain that the number of groups of order $p^k$ is at least: $$\frac{p^{\frac{2m^3}{27}}}{p^{\frac{2}{3}m^2}}$$ -We also obtain that the number of groups of order $n$ is: -$$n^{\frac{2\mu^2}{27}}n^{\mathcal O(\mu^{3/2})}$$ -where $\mu$ is the largest exponent for a prime power dividing $N$. -So take a natural $N$, and suppose that $2^m$ is the largest power of $2$ not exceeding $N$, we obtain that the number of groups of order $2^m$ is at least: -$$\frac{2^{\frac{2m^3}{27}}}{2^{\frac{2}{3}m^2}}=\frac{(2^m)^{\frac{2}{27}m^2}}{(2^m)^{\frac{2}{3}m}}\geq\frac{N^{\frac{2m^2}{27}}}{N^{\frac{2}{3}m}2^{\frac{2m^2}{27}}}\geq \frac{N^{\frac{2m^2}{27}}}{N^m}$$ -On the other hand, if $n$ is an odd integer less than $N$, clearly its $\mu$ is at most $m\log_3(2)+1$. For large values of $m$ we can just bound this by $\alpha m$, for some pre-selected $\alpha<1$. -So the number of groups of odd order is : -$$N(N^{\frac{2\alpha^2 m^2}{27}}N^{\mathcal O(m^{3/2})})$$. -So the fraction between the number of groups of order $2^m$ and the number of groups of odd order less than $N$ for large $N$ is at least: -$$\frac{N^{\frac{2m^2}{27}}}{N^mN(N^{\frac{2\alpha m^2}{27}}N^{\mathcal O(m^{3/2})})}=N^{\frac{2(1-\alpha^2)m^2}{27}-\mathcal O(m^{3/2})}$$ -Which clearly goes to infinity.<|endoftext|> -TITLE: Multizeta function values -QUESTION [14 upvotes]: Francis Brown Theorem says that $\zeta(a_{1},\dots ,a_{r})$ the multi-zeta value of weight $N=a_{1}+\dots +a_{r}$ is a $\mathbb{Q}$-linear combination of elements of the set $S=\{\zeta(a_{1},\dots, a_{n})| \text{for any i, } a_{i}\in\{2,3\} \text{ and } a_{1}+\dots +a_{n}=N \}$ -What I don't understand in the statement is the following: are the elements of $S$ $\mathbb{Q}$-linearly independent? -I read that there is a surprising $\mathbb{Q}$-linear relation between some values of the multi-zeta function $\zeta$ of weight 12. The relation I'm talking about is the following: -$$28\zeta(3,9)+ 150\zeta(5,7)+168\zeta(7,5)= \frac{5197}{691}\zeta(12) $$ -Does this relation have some arithmetic-geometric interpretation ? -In his talk (around 5:50) F. Brown made a remark by saying that the relation above relation is related to cusp forms for $SL(2,\mathbb{Z})$. What that means exactly and what is the relation between cusps forms and the $\mathbb{Q}$-linear relation above ? - -REPLY [13 votes]: Vesselin has already given an excellent answer to your first question, so let me just elaborate a little bit on the relation -$$ -28\zeta(3,9)+150\zeta(5,7)+168\zeta(7,5)=\frac{5197}{691}\zeta(12)\tag{1} -$$ -and its connection with modular forms for $\operatorname{SL}_2(\mathbb{Z})$. First of all, it is a rather exotic and surprising relation (I will qualify this statement below). In a nutshell, the coefficients $28,150,168$ on the left-hand side of (1) come from period integrals of the cusp form $\Delta$, while the denominator $691$ of the right-hand side comes from the constant term of the Eisenstein series $G_{12}$. -To any cuspidal Hecke eigenform $f$ of weight $k$, one can associate an even (i.e. invariant under $Y\mapsto -Y$) homogeneous polynomial $r_f(X,Y)^{+} \in K_f[X,Y]_{k-2}$ of degree $k-2$, where $K_f$ is the number field generated by the Fourier coefficients of $f$. Explicitly, -$$ -r_f(X,Y)^{+}=\frac{\operatorname{Re}\left( (2\pi i)^{k-1}\int_0^{i\infty}f(z)(X-zY)^{k-2}dz \right)}{\omega_f^{+}} -$$ -where the integral is along the geodesic path from $0$ to $i\infty$ on the upper half-plane, and $\omega_f^{+} \in \mathbb{R}$ is a suitable ''period''. This construction is part of the very well-known Eichler--Shimura--Manin theory, expositions of which can be found in many places; see e.g. Lang's Introduction to Modular Forms for an introduction, or Brown's Multiple Modular Values and the relative completion of the fundamental group of $M_{1,1}$, Section 7 for a more condensed presentation. -Now in weight $12$ there is a unique cuspidal Hecke eigenform, namely Ramanujan's $\Delta$, and one has (up to scaling by an element of $\mathbb{Q}^{\times}$) -$$ -r_{\Delta}(X,Y)^+=\frac{36}{691}(X^{10}-Y^{10})-X^2Y^2(X^2-Y^2)^3. -$$ -The essential part is the second summand $p(X,Y):=X^2Y^2(X^2-Y^2)^3$, the first one is a coboundary in the sense of group cohomology for $\operatorname{SL}_2(\mathbb{Z})$, and hence can be ignored (for the time being, but see below). Now Gangl--Kaneko--Zagier's result (Theorem 3 in their paper) says in essence that if numbers $q_{r,s}$ are defined by $p(X+Y,Y)=\sum_{r,s \geq 1}\binom{r+s-2}{r-1}q_{r,s}X^{r-1}Y^{s-1}$, then -$$ -\sum_{r+s=12, \, r,s \, odd}q_{r,s}\zeta(r,s) \equiv 0 \mod \zeta(12) -$$ -which, after rescaling, yields (1) modulo $\zeta(12)$. The upshot is that the coefficients $28,150,168$ come from integrals involving the cusp form $\Delta$, and therefore the left-hand side of (1) has a rather clean, arithmetic-geometric interpretation. -To the best of my knowledge, no such interpretation is known for the right-hand prefactor $\frac{5197}{691}$ (in Gangl--Kaneko--Zagier's original approach, they need to compute some kind of beta-integral). However, the occurrence of the denominator $691$ seems to be an artifact of the well-known congruence $G_{12} \equiv \Delta \mod 691$ on the level of Fourier expansions, where $G_{12}$ is the Hecke Eisenstein series of weight $12$ (basically because the numerator of the constant term of G_{12} is divisible by 691). The precise relation should be explained by recent work of Hain--Matsumoto on Universal Mixed Elliptic Motives (in particular, the previously neglected term $\frac{36}{691}(X^{10}-Y^{10})$ resurfaces as the dual of the cocycle of $G_{12}$). As far as I can see, this does not give an explanation about the numerator $5197$, though. -Finally, I would like to point out that there is another connection of (1) to cusp forms, which is not (yet?) arithmetic-geometric but in my opinion very nice: namely, via the theory of double Eisenstein series, which are to usual Eisenstein series what double zeta values are to single zeta values. Since my answer is already much too long, I refer to the original paper of Gangl--Kaneko--Zagier, Section 7 for the details. Some of this has also been generalized by Henrik Bachmann in his PhD thesis (see in particular the bottom of p.34 for details on how to obtain (1) via this theory). - -So why is the relation (1) surprising? It is more generally believed that all algebraic relations between multiple zeta values are implied by the (regularized) double shuffle equations, which are a class of quadratic relations between multiple zeta values. Specialized to the depth two case, they read -$$ -\zeta(r)\zeta(s)=\zeta(r,s)+\zeta(s,r)+\zeta(r+s)=\sum_{m+n=r+s}\left( \binom{m-1}{r-1}+\binom{m-1}{s-1}\right)\zeta(m,n). \tag{2} -$$ -One of the main results of Gangl--Kaneko--Zagier is that (1) cannot be deduced from (2), and one needs to resort to the double shuffle equations in higher depth. If one were able to deduce from (2) that $28\zeta(4)\zeta(8)+\frac{95}{3}\zeta(6)^2 \in \mathbb{Q}\zeta(12)$ (which is of course a consequence of Euler's theorem on even zeta values), this ''depth-defect'' would go away, alas one cannot. This is one of the very perplexing features of the depth filtration on multiple zeta values, and that it can be resolved using modular forms points at some deep relation between the two objects, which we do not completely understand yet.<|endoftext|> -TITLE: Generators of the algebra of symmetric polynomials over finite fields -QUESTION [6 upvotes]: Let $F$ be a finite field and $A$ be the algebra of symmetric polynomials on $x_1,\dots,x_n$ over $F$. What is known about the generators of $A$ except the elementary symmetric polynomials? In particular, given some power sum polynomials, how to determine whether they generate $A$? - -REPLY [2 votes]: The algebra $A$ can only be generated by power sums if the characteristic of $F$ is greater than $n$. In this case, the power sums $p_i = \sum_{j=1}^n x_j^i$ with $1 \le i \le n$ generate $A$, and any other generating system consisting of power sums needs to include $p_1,\ldots,p_n$. -These claims can be justified as follows: Since the algebra of invariants is graded, any set of generators has to include generators of the degrees given by a minimal set of generators. In our case, these degrees are $1,\ldots,n$. So a generating set consisting of power sums has to include $p_1,\ldots,p_n$. Now if the characteristic $l$ is $\le n$, then $p_l = p_1^l$, so $p_l$ isn't a new generator and, in fact, no set of power sums can contain a new generator of degree $l$. On the other hand, if $l > n$, then the Newton formulae say that the elementary symmetric polynomials can be expressed in terms of $p_1,\ldots,p_n$, so the $p_i$ are generators. Alternatively, you could argue that the Jacobian determinant of the $p_i$ is nonzero if $l > n$.<|endoftext|> -TITLE: Formal Definition of Finite Conditions -QUESTION [5 upvotes]: Forcing with finite conditions is a common concept used by set theorists. I was thinking about its meaning, but I couldn't find any exact definition of it. At the first glance it seemed to me that maybe by finite conditions, experts mean that the conditions of a forcing notion are finite however one can naturally turn any arbitrary forcing to another one which are isomorphic as partial orders and the conditions of former are finite. On the other hand, the conditions of many forcings with finite conditions are intrinsically infinite. -By the way implicitly, forcing with finite conditions means every condition has finitely many information about generic object(from forcing point of view). My question is: - - -Is there any formal definition of forcing with finite condition? Or can some one give a formal definition of forcing with finite condition including well-known forcings with finite conditions? - -REPLY [4 votes]: My impression is that this term is used loosely, without formal definition, to refer to the common situation where we have a forcing notion consisting of a partial order whose elements are all finite functions (that is, each on a finite domain) and the forcing order is functional extension. The most canonical example is Cohen forcing $\text{Add}(\omega,1)$ to add a Cohen real or actually any number of Cohen reals. Other examples include the forcing to collapse a given cardinal to $\omega$ or the forcing to add a club set using finite conditions (not with countable conditions, which is another common way to do it). In some cases, there are examples where conditions are augmented with some other finite amount of information that still counts as forcing with finite conditions even though the order isn't literally functional extension. -Every forcing notion is equivalent, of course, to a forcing notion whose conditions are all finite, since we could replace any element $p$ in a partial order with $\{p\}$, which is a singleton set and hence finite, and then redefine the corresponding order on these singletons. So every condition in the new partial order is finite. But this literal interpretation is never what is meant by the phrase, "forcing with finite conditions," and I have never seen a formal definition of the concept going significantly beyond the loose understanding given in the previous paragraph.<|endoftext|> -TITLE: Bounding the matrix norm of a commutator $[A,B]$ in terms of the norms of $A$ and $B$ -QUESTION [13 upvotes]: The setup is as in this question: -Given a norm $N$ over ${\bf M}_n(\mathbb C)$, it is a natural question to find the best constant $C_N$ such that -$$N([A,B])\le C_N N(A)N(B),\qquad\forall A,B\in{\bf M}_n(\mathbb C).$$ -Equivalently, $C_N$ is the maximum of $N(AB-BA)$ provided that $N(A)=N(B)=1$. -Given examples of $C_N$ are - -$C_N=\sqrt{2}$ if $N$ is the Frobenius norm -$C_N=2$ if $N$ is the operator norm $\| \cdot\|_2$ -$C_N=4$ if $N$ is the numerical radius $r(A)=\sup\limits_{x\ne0}\dfrac{|x^*Ax|}{\|x\|^2}$ (See this answer to an MO question). - -if $N$ is the induced $p$-norm, defined for $1\le p\le\infty$ by $\|A\| _p = \sup \limits _{x \ne 0} \frac{\| A x\| _p}{\|x\|_p}$, we have $C_N=2$ for $p=\infty$ (with $\|A\|_\infty $ being just the maximum absolute row sum of the matrix). Indeed, the lower bound $2$ for $\|\cdot\|_\infty $ is obtained by taking e.g. $A=\begin{pmatrix} 1&0\\1&0\end{pmatrix}$ and $B=\begin{pmatrix} 0&1\\0&-1\end{pmatrix}$, and it should be easy to prove that $2$ is also the general upper bound for $\|\cdot\|_\infty $. -Similarly, $C_N=2$ for $p=1$ (with $\|A\|_1 $ being the maximum absolute column sum of the matrix). - -Knowing that $C_N\equiv2$ for $p=1,2,\infty$, is it true that the same holds for the induced $p$-norms for all $p\ge1$? -If $N$ runs over all possible matrix norms, what is the range of $C_N$? In particular, is it bounded below and/or above? - -(To avoid trivialities, let's keep it homogeneous by only considering "normalized" norms, i.e. require $N(I_n)=1$. This does not seem to be part of the standard definition of a norm.) - -REPLY [8 votes]: A somewhat more general setting, namely, finding the best constant $C_{p,q,r}$ in -\begin{equation*} - \|AB-BA\|_p \le C_{p,q,r}\|A\|_q\|B\|_r, -\end{equation*} -for Schatten $p$,$q$,$r$-norms, is studied in this paper. -EDIT (1st Mar'17). See also this recent paper (LAA, 521(15), May 2017, Pages 263–282) that provides sharp bounds on the constant $C_{p,q,r}$ above for real matrices.<|endoftext|> -TITLE: Objects whose morphisms are Lipschitz maps -QUESTION [7 upvotes]: I recently wondered what are the spaces whose morphisms are Lipschitz maps (by which I mean: "locally Lipschitz"). -The answer seems pretty clear, and proceeds like the definition of manifolds: -1) If $X$ is a topological space, a Lipschitz chart is a homeomorphism from an open subset of $X$ to a metric space. -2) Two Lipschitz charts are compatible if the corresponding two transition functions are Lipschitz. -3) A Lipschitz atlas on $X$ is a set of compatible Lipschitz charts whose domains cover $X$. -4) A Lipschitz space is a topological space equipped with a maximal Lipschitz atlas. -So my question is: what are these spaces called? (and why not "Lipschitz spaces"?). I'll be grateful for any reference. - -REPLY [2 votes]: I randomly found an answer to my question two years later! The spaces I described in the question are called "locally metric spaces" in -J. Luukkainen, J. Väisälä, Elements of Lipschitz topology, Ann. Acad. Sci. Fennicae 3 (1977), 85--122. -They are introduced, together with "local metrics", in their Section 3.4. As guessed in the comments above, any paracompact Hausdorff locally metric space is indeed a metric space "up to Lipschitz equivalence" (their Theorem 3.5). They cite -J. H. C. Whitehead, Manifolds with transverse fields in Euclidean space, Ann. Math. 73 (1961), 154--212. -which introduces "local metrics" in Section 2. -Note: to reply to YCor's comment: indeed, I should have written "objects of the category whose morphisms" instead of "objects whose morphisms".<|endoftext|> -TITLE: Continuous functions and infinity -QUESTION [7 upvotes]: Suppose $f(x)$ is continuous on $\mathbb{R}$, for $\forall \delta>0, \forall x\in\mathbb{R}, \lim_{n\rightarrow\infty}f(x+n\delta)=+\infty$. Is it correct that $\lim_{x\rightarrow+\infty}f(x)=+\infty$? - -REPLY [7 votes]: Yes, in fact, -$$\inf_{\delta>0}\ \liminf_{n\to\infty}f(n\delta) =\liminf_{x\to+\infty}f(x).$$ -Assuming w.l.o.g. $\liminf_{x\to+\infty}f(x)<\alpha<+\infty$, the open set $A=\{f<\alpha\}$ is unbounded. Therefore, for any non-empty open interval $(a,b)\subset\mathbb{R}_+$ and any $n\in\mathbb{N}$, the set $\cup_{k> n}(ka,kb)$, that contains a right-unbounded interval, meets $A$. Equivalently, for any $n\in\mathbb{N}$, the open set $ B_n:=\cup_{k> n}{1\over k}A$ meets $(a,b)$, so that $B_n$ is dense in $\mathbb{R}_+$. By the Baire category theorem $\cap_{n\ge0}B_n$ is not empty, actually dense, meaning that there exist $\delta>0$ such that $n\delta\in A$ for infinitely many $n$, an this means $\liminf_{n\to\infty}f(n\delta)\le\alpha.$ Being $\alpha$ arbitrary, the claim follows.<|endoftext|> -TITLE: Platonic Truth and 1st Order Predicate Logic -QUESTION [22 upvotes]: Consider the following simple example as motivation for my question. If it were the case that, say, the Riemann hypothesis turned out to be independent of ZFC, I have no doubt it would be accepted by many as a new axiom (or some stronger principle which implied it). This is because we intuitively think that if we cannot find a zero off of the half-line, then there is no zero off the half-line. It just doesn't "really" exist. -Similarly, if Goldbach's conjecture is independent of ZFC, we would accept it as true as well, because we could never find any counter-examples. -However, is there any reason we should suppose that adding these two independent statements as axioms leads to a consistent system? Yes, because we have the "standard model" of the natural numbers (assuming sufficient consistency). -But can this Platonic line of thinking work in a first-order way, for any theory? Or is it specific to the natural numbers, and its second-order model? -In other words, let $T$ be a (countable, effectively enumerable) theory of first-order predicate logic. Define ${\rm Plato}(T)$ to be the theory obtained from $T$ by adjoining statements $p$ such that: $p:=\forall x\ \varphi(x)$ where $\varphi(x)$ is a sentence with $x$ a free-variable (and the only one) and $\forall x\ \varphi(x)$ is independent of $T$. Does ${\rm Plato}(T)$ have a model? Is it consistent? -The motivation for my question is that, as an algebraist, I have a very strong intuition that if you cannot construct a counter-example then there is no counter-example and the corresponding universal statement is "true" (for the given universe of discourse). In particular, I'm thinking of the Whitehead problem, which was shown by Shelah to be independent of ZFC. From an algebraic point of view, this seems to suggest that Whitehead's problem has a positive solution, since you cannot really find a counter-example to the claim. But does adding the axiom "there is no counter-example to the Whitehead problem" disallow similar new axioms for other independent statements? Or can this all be done in a consistent way, as if there really is a Platonic reality out there (even if we cannot completely touch it, or describe it)? - -REPLY [19 votes]: The phenomenon accords more strongly with your philosophical explanation if you ask also that the sentences have complexity $\Pi^0_1$. That is, the universal statement $\forall x\ \varphi(x)$ should have $\varphi(x)$ involving only bounded quantifiers, so that we can check $\varphi(x)$ for any particular $x$ in finite time. If you drop that requirement, there are some easy counterexamples, hinted at or given already in the comments and other answers. -But meanwhile, even in the case you require $\varphi(x)$ to have only bounded quantifiers, there is still a counterexample. -Theorem. If $\newcommand\PA{\text{PA}}\newcommand\Con{\text{Con}}\PA$ is consistent, then there is a consistent theory $T$ extending $\PA$ with two $\Pi^0_1$ sentences $$\forall x\ \varphi(x)$$ $$\forall x\ \psi(x)$$ -both of which are consistent with and independent of $T$, but which are not jointly consistent with $T$. -Proof. Let $T$ be the theory $\PA+\neg\Con(\PA)$, which is consistent if $\PA$ is consistent. Let $\rho$ be the Rosser sentence of this theory, which asserts that the first proof in $T$ of $\rho$ comes only after the first proof of $\neg\rho$ (see also my discussion of the Rosser tree). Our two $\Pi^0_1$ sentences are: - -every proof of $\rho$ from $T$ has a smaller proof of $\neg\rho$. -every proof of $\neg\rho$ from $T$ has a smaller proof in $T$ of $\rho$. - -The first statement is equivalent to $\rho$, and the second is equivalent over $T$ to $\neg\rho$, since $T$ proves that every statement is provable; the only question is which proof comes first. So both statements are consistent with $T$. -But the sentences are not jointly consistent with $T$, since in any model of $T$, both $\rho$ and $\neg\rho$ are provable from $T$, and so one of the proofs has to come first. QED - -REPLY [8 votes]: This post is not an answer to your question, but it explains the reason that if $\bf GC$ (or $\bf RH$) is independent of $\bf ZFC$, $\bf GC$ (or $\bf RH$) is true in the standard model of natural numbers. -The reason is $\bf GC$ and $\bf RH$ are $\Pi_1$ sentences in the language of arithmetic. -Def. - -$x|y := \exists z(z \leq y \land x\cdot z = y)$ -$Pr(x) := \forall y(y0 \land y|x \to y=1)$ - -Therefor $\bf GC$ can be defined by $\forall x\exists y,z(y+z = 2\cdot x+4 \land Pr(y) \land \Pr(z))$ which is a $\Pi_1$ sentence. For $\Pi_1$ definition of $\bf RH$ see here. -Let $\phi(x)$ be a $\Delta_0$ formula and suppose ${\bf PA} \nvdash \exists x \neg \phi(x)$, then $\mathbb{N}\models \forall x \phi(x)$. This is true because of $\Sigma_1$ completeness of ${\bf PA}$, that is if $\psi$ be a $\Sigma_1$ sentence, then $\mathbb{N}\models \psi$ iff ${\bf PA}\vdash \psi$. -The important thing in this argument is $\Pi_1$ definability of problem. For example consistency of $\bf PA$ is a $\Pi_1$ sentence and by second incompleteness theorem, ${\bf PA} \nvdash Con_{\bf PA}$, but $\mathbb{N}\not\models \neg Con_{\bf PA}$, therefore we can not prove similar theorems for formulas in another level of Arithmetical Hierarchy except $\Pi_1$. - -REPLY [6 votes]: It cannot be done in a consistent way. -Consider a closed statement $\psi$ which is independent of a theory $T$, and take $\forall x . \psi$ and $\forall x . \lnot\psi$. (I made the closed statement have a dummy free variable to satisfy your condition.) Both statements are of the kind you are asking for, but when we add both to $T$ we get an inconsistent theory. -It should be clear that one can come up with examples where the two sentences that contradict each other are not so blatantly in opposition with each other. And with a bit of work we can even come up with examples where the free variable $x$ is doing something.<|endoftext|> -TITLE: A counterexample for Sard's theorem in $C^1$ regularity -QUESTION [26 upvotes]: I can't seem to find an example of a function $f \colon \mathbb{R}^2\to \mathbb{R}$ which is $C^1$ and such that the set of its critical values is not of zero measure. - -What examples are there? - $\phantom{aaa}$ - -REPLY [12 votes]: I decided to challenge myself to make pictures of Piotr Hajlasz's example, partly for fun and partly for the next time I teach this. Let $C_3$ be the standard Cantor middle thirds set: -$$C_3 = \left\{ \sum_{k=1}^{\infty} \frac{a_k}{3^k} : a_1, a_2, a_3, \cdots, \in \{ 0,2 \} \right\}.$$ -Let $C_4$ be the variant "middle halves set" -$$C_4 = \left\{ \sum_{k=1}^{\infty} \frac{b_k}{4^k} : b_1, b_2, b_3, \cdots, \in \{ 0,3 \} \right\}.$$ -Note that every $z$ in $[0,1]$ can be written as $(2/3) x + (1/3) y$ for $x$, $y \in C_4$ in either $1$ or $2$ ways. Here is a drawing of $C_4 \times C_4$ and its intersection with the lines $(2/3)x + (1/3) y = k/16$ for $0 \leq k \leq 16$: - -Here is a $C^1$ function $\phi(x)$ which maps $C_3$ to $C_4$ in an order preserving manner, with derivative $0$ at each point of $C_3$. On each interval of $[0,1] \setminus C_3$, it is an appropriately chosen sine curve. - -And here is the final product, a depiction of the function $f(x,y) = (2/3) \phi(x) + (1/3) \phi(y)$. The level curves show $f(x,y) = k/16$ for $0 \leq k \leq 16$. The black dots are the critical points $C_3 \times C_3$; the red dots demonstrate how every level curve contains $1$ or $2$ critical points.<|endoftext|> -TITLE: Relatively primes spirals -QUESTION [16 upvotes]: When exploring the structure of points of the integer lattice -whose two coordinates are relatively prime -(call these r-prime points),1 -I looked at spirals analogous to "Gaussian prime spirals."2 -Start at an r-prime point $(a,b)$, -walk vertically (North) until you hit another r-prime point, -then walk West until another r-prime point is hit, then South, then East, -continuing to turn counterclockwise -$90^\circ$ at r-prime points until you re-encounter an earlier point, -approached from the same direction as last hit, and so fall -into a cycle. -(The start point is -considered approached from its left.) -Unlike the Gaussian prime spirals, these "relatively primes spirals" -are not generally visually interesting. -Many are just $4$-cycles, e.g.: -$$ -(223, 2), (223, 3), (221, 3), (221, 2) -$$ -Let me illustrate one more before asking a question. -Starting at $(495,2)$ leads to a cycle of length $44$: -$$ -(495, 2), (495, 4), (493, 4), (493, 3), (494, 3), (494, 5), (493, 5), -(493, 4), (495, 4), (495, 7), (494, 7), (494, 5), (496, 5), (496, 7), -(495, 7), (495, 4), (497, 4), (497, 5), (496, 5), (496, 3), (497, 3), -(497, 4), (495, 4), (495, 2), (497, 2), (497, 3), (496, 3), (496, 1), -(497, 1), (497, 2), (495, 2), (495, 1), (496, 1), (496, 3), (494, 3), -(494, 1), (495, 1), (495, 2), (493, 2), (493, 1), (494, 1), (494, 3), -(493, 3), (493, 2) -$$ -Here is an illustration of this cycle: - -          - - -A natural question is: - -Does any start point lead to an infinite path that never cycles? - -A candidate infinite path starts at $(5,2)$: -$$ -(5, 2), (5, 3), (4, 3), (4, 1), (5, 1), (5, 2), (3, 2), (3, 1), (4, 1), (4, 3), -(2,3), (2,1), \ldots -$$ -Here is its first $200$ turns: - -          - - -And here is its first $1000$ turns: - -          - - -And here is its first $10000$ turns: - -          - - -I've tracked it out to $10^6$ turns (reaching out to $(87652,87655)$), and still no cycle. -So, in addition to the general question above, a more specific question -is whether $(5,2)$ ever cycles. -Added animation: - -          - - - -1Arbitrarily long composite anti-diagonals? -2Gaussian prime spirals - -REPLY [11 votes]: First note that, in the path, the coordinates are always positive, because whenever any coordinate decreases to $1$ you hit a relatively prime point, turn, hit another relatively prime point, and then the coordinate starts increasing again, -Next consider the following set of posiitions and next directions: -A $(x,y), y>x+1$, any direction -B $(x,x+1)$, North, East, or West -C $(x,x-1)$, North -This set is closed under your operation. The reason is that if you ever leave type A you must be traveling South or East and hit the R-prime point $(x,x+1)$, so you are then traveling East or North, hence be in type B. If you leave type B and travel North or West you return to A, and traveling east you skip the non-R-prime point $(x+1,x+1)$ (note $x$ is at least $1$) and travel straight to $(x+2,x+1)$ and turn North, hence in a point in C. Finally C returns to B in the same way by skipping $(x,x)$ (note $x$ is at least $2$) and turning West. -Because your spiral enters this closed set, it never leaves it and hence never returns to its initial point. Because the flow is reversible, this means it must run forever without repeating.<|endoftext|> -TITLE: Yoga of six functors for group representations? -QUESTION [22 upvotes]: I'm trying to understand how the six functor philosophy applies to representation theory. Consider the category of classifying stacks $BG$ (assume $G$ discrete for simplicity). To every stack we can assign a triangulated category $D(BG)=D(BG,k):= D(Rep(G))$ the (perhaps bounded) derived category of the abelian category of representations of $G$ (in vector spaces over $k$ algebraically closed of characteristic 0). As far as I understand the formalism of six functors should translate (if at all) as follows: -If $\pi : BG \to pt$ then: - -$\pi_{!}(-)$ is a derived version of $(-)^G$ (invariants) = group cohomology. -$\pi_*(-)$ is a derived version of $(-)_G$ (coinvariants) = group homology . - -Generally if $\pi: BH \to BG$ then: - -$\pi_!(-) = \mathbb{L}Ind^G_H(-)$, -$\pi_*(-) = \mathbb{R}Coind^G_H(-)$ -$\pi^*(-) = \mathbb{L}Res^H_G(-)$. - -I have several questions questions: - - -Are there any mistakes/wrong intuitions in the above? -I'm missing the upper shrieks $\pi^!$ and as such also the duality - functor. How do these look in general? What is the "dualizing - representation" of a group? Is it something familiar? -How much of this carries over to the category of algebraic stacks (on the big etale site of schemes) $BG$ for $G$ linear algebraic group? (suppose all this happens over some fixed field for simplicity). - -REPLY [3 votes]: I'm more familiar with sheaf cohomology than group cohomology, but I think this is at least partly an issue of notation, as Peter McNamara suggests. I will also assume that all groups here are discrete. -To expand on his answer a bit, your notation for the adjoint triple induction $\dashv$ restriction $\dashv$ coindunction (or extension $\dashv$ restriction $\dashv$ coextension) appears to follow the common algebraic geometry convention for ring maps, but they do that because affine schemes are the opposites of commutative rings, and left adjoints are swapped for right adjoints when you take the opposite category. -So to make group cohomology look more like sheaf cohomology, the notation should be $\pi_! \dashv \pi^* \dashv \pi_*$. Then group cohomology is also right derived pushforward $\mathbf{R}\pi_*$, just like sheaf cohomology, and group homology is derived lower shriek, just like compactly supported sheaf cohomology. -Hence, Verdier duality should be given by an adjunction $\pi_! \dashv \pi^!$, but for group reps $\pi_!$ aleady has a right adjoint $\pi^*$, so for discrete group representations we are always in the Wirthmüller context where $\pi^!=\pi^*$.<|endoftext|> -TITLE: When does a Lagrangian dynamical system have an equivalent Hamiltonian description? -QUESTION [9 upvotes]: Let a Lagrangian dynamical system with $n$ degrees of freedom and configuration space $\mathbb{R}^n$ -(i.e. phase space $\mathbb{R}^{2n}$), which is described by $L=L(q_{i},\dot{q}_{i},t)$, $i=1,2,...,n$. -It is well known from classical mechanics, that a sufficient condition for a Lagrangian dynamical system to be equivalently described as a Hamiltonian system, is the Hessian of the Lagrangian w.r.t. the generalized velocities to be non-degenerate: -$$ -\det\Big|\frac{\partial^{2}L}{\partial\dot{q}_{i}\partial\dot{q}_{j}}\Big|\neq 0 -$$ -Now the question is the following: It is frequently mentioned in the literature of classical mechanics (but I have not seen a direct proof of this), that the above sufficient condition is independent of the choice of generalized coordinates and depends only on the dynamical system itself (i.e. only on the Lagrangian function and the underlying configuration space). What kind of direct proof could be provided for this ? -P.S.: the motivation for posting this question was a student's (not my student) question on whether the property of a Lagrangian system to be Hamiltonian as well, is a purely geometrical property. I am not an expert in the subject but I think that essentially, it suffices to answer the question for the case that the configuration space is $\mathbb{R}^n$. In the sense that the independence of the property of the particular coordinate system used, means that in the general case, it will be a property of the underlying differentiable structure on the configuration space manifold. - -REPLY [4 votes]: Here's what I have done: -$\bullet$ Let the Lagrangian $L(q_{i},\dot{q}_{i},t)$, which under the point transformations -$$ -\{q_{i}\}\leftrightsquigarrow\{Q_{i}\} -$$ -given by the invertible relations $Q_{i}=Q_{i}\big(q_{j},t\big)\Leftrightarrow q_{j}=q_{j}\big(Q_{i},t\big)$, $\ \ i,j=1,2,...,n$, (i.e. $\det\Big|\frac{\partial Q_{j}}{\partial q_{i}}\Big|\neq 0$), can be written as: -$$ -L(q_{i},\dot{q}_{i},t)\stackrel{}{\leftrightsquigarrow}L(Q_{i},\dot{Q}_{i},t) -$$ -$\bullet$ Differentiating the point transformations $Q_{i}=Q_{i}\big(q_{j},t\big)$, we get that: -$$ -\frac{dQ_i}{dt}\equiv\dot{Q}_i=\frac{\partial Q_i}{\partial q_j}\dot{q}_i+\frac{\partial Q_i}{\partial t}\ \ \ \ \ \Rightarrow \ \ \ \ \ \frac{\partial\dot{Q}_i}{\partial\dot{q_j}}=\frac{\partial Q_i}{\partial q_j} \ \ \ \ \ \ \ \ \ \ \ \ \ (1) -$$ -$\bullet$ On the other hand, differentiating the Lagrangian we get that: -$$ -\frac{\partial L}{\partial\dot{q}_i}=\sum_{k=1}^{n}\frac{\partial L}{\partial\dot{Q}_k}\frac{\partial\dot{Q}_k}{\partial\dot{q}_i} \Rightarrow -$$ -$$ -\Rightarrow \frac{\partial^2 L}{\partial\dot{q}_j\partial\dot{q}_i}=\frac{\partial}{\partial\dot{q}_j}\Big(\sum_{k=1}^{n}\frac{\partial L}{\partial\dot{Q}_k}\frac{\partial\dot{Q}_k}{\partial\dot{q}_i}\Big)= \sum_{m=1}^{n}\frac{\partial}{\partial\dot{Q}_m}\Big(\sum_{k=1}^{n}\frac{\partial L}{\partial\dot{Q}_k}\frac{\partial\dot{Q}_k}{\partial\dot{q}_i}\Big)\frac{\partial\dot{Q}_m}{\partial\dot{q}_j}= -$$ -$$ -=\sum_{k,m=1}^{n}\frac{\partial^2 L}{\partial\dot{Q}_m\partial\dot{Q}_k}\frac{\partial\dot{Q}_k}{\partial\dot{q}_i}\frac{\partial\dot{Q}_m}{\partial\dot{q}_j}\stackrel{(1)}{\Rightarrow} -$$ -$$ -\stackrel{(1)}{\Rightarrow} -\frac{\partial^2 L}{\partial\dot{q}_j\partial\dot{q}_i}=\sum_{k,m=1}^{n}\frac{\partial^2 L}{\partial\dot{Q}_m\partial\dot{Q}_k}\frac{\partial Q_k}{\partial q_i}\frac{\partial Q_m}{\partial q_j} -$$ -Thus: -$$ -\Big[\frac{\partial^{2}L}{\partial\dot{q}_{i}\partial\dot{q}_{j}}\Big]=\Big[\frac{\partial Q_{j}}{\partial q_{i}}\Big]^{t} -\Big[\frac{\partial^{2}L}{\partial\dot{Q}_{i}\partial\dot{Q}_{j}}\Big]\Big[\frac{\partial Q_{j}}{\partial q_{i}}\Big] -$$ -where $[..]$ stands for the respective Jacobian and hessian matrices and $[..]^t$ stands for the transpo-se matrix. The last relation clearly tells us that the non-vanishing of the hessian determinant is invari-ant under the point transformations, i.e. -$$ -\det\Big|\frac{\partial^{2}L}{\partial\dot{q}_{i}\partial\dot{q}_{j}}\Big|\neq 0 \Leftrightarrow \det\Big|\frac{\partial^{2}L}{\partial\dot{Q}_{i}\partial\dot{Q}_{j}}\Big|\neq 0 -$$ -which concludes the proof. -P.S.: We can easily see why the above, is a sufficient condition for a Hamiltonian description to exist: Generalized momenta $p_{i}$, are defined by -\begin{equation} \label{7.55} -p_{i}=\frac{\partial L}{\partial\dot{q}_{i}}=p_{i}\big(q_{j},\dot{q}_{j},t\big), \ \ \ i,j=1,2,...,n -\end{equation} -If the Jacobian of the generalized momenta w.r.t. the generalized velocities, or equivalently the Hessian of the -Lagrangian w.r.t. the generalized velocities -\begin{equation} \label{7.57} -\det\Big|\frac{\partial p_{i}}{\partial\dot{q}_{j}}\Big|=\det\Big|\frac{\partial^{2}L}{\partial\dot{q}_{i} -\partial\dot{q}_{j}}\Big|\neq 0 -\end{equation} -is non-zero, then, due to the inverse function theorem in $\mathbb{R}^{n}$, the above relations can be solved w.r.t. -the generalized velocities: -\begin{equation} \label{7.58} -\dot{q}_{j}=\dot{q}_{j}\big(q_{i},p_{i},t\big), \ \ \ i,j=1,2,...,n -\end{equation} -Then, the Hamiltonian function is defined through a Legendre transform: -\begin{equation} \label{7.59} -H=H\big(q_{i},p_{i},t\big)=\sum_{i=1}^{n}p_{i}\dot{q}_{i}-L\big(q_{i},\dot{q}_{i},t\big) -\end{equation} -with $i=1,2,...,n$. The Hamiltonian function $H:\mathbb{R}^{2n+1}\rightarrow\mathbb{R}$, is a function, -of generalized coordinates, generalized momenta and time.<|endoftext|> -TITLE: Statements about groups proved using semigroups -QUESTION [15 upvotes]: Question. Has a statement about groups ever been proved using the theory of semigroups? - -By "a proof using the theory of semigroups" I do not mean that some steps in the proof are in fact statements about semigroups rather than groups. What I'm looking for are proofs which involve semigroups in a substantial way, say by constructing some semigroup related to the group of interest. - -REPLY [4 votes]: There is a theorem saying that if we take a Zariski-dense subgroup $\Gamma$ in a semisimple real Lie group $G$, for example $G = \operatorname{SL}_d(\mathbb{R})$, the set of so-called "loxodromic" elements in $\Gamma$ is still Zariski-dense in $G$. (In the case of $G = \operatorname{SL}_d(\mathbb{R})$, "loxodromic" means "diagonalizable with eigenvalues of distinct modulus"). -In order to prove it, one actually proves the same statement for semigroups. See Proposition 5.11 and Remark 5.14 in Y. Benoist and J.F. Quint's book "Random walks on reductive groups".<|endoftext|> -TITLE: Relation between well-orderings of $\mathbb{R}$, and bases over $\mathbb{Q}$ -QUESTION [8 upvotes]: The following question arose from a discussion about the definability of bases of $\mathbb{R}$ as a $\mathbb{Q}$-vector space. -(ZF without AC) something we can note is that the existence of a (definable) well-ordering of $\mathbb{R}$ is easily seen to be equivalent to that of a (definable) well-ordered basis of $\mathbb{R}$ as a $\mathbb{Q}$-vector space. -After these remarks, the following question seems natural : -Is it consistent with ZF that there be a basis of $\mathbb{R}$ over $\mathbb{Q}$ that cannot be well-ordered ? - -REPLY [6 votes]: There are actually two papers: "ZF + there is a Hamel basis" doesn't give that every infinite set of reals has a ctble. subset, see https://ivv5hpp.uni-muenster.de/u/rds/hamel_basis.pdf , and "ZF + DC + there is a Hamel basis" doesn't give that there is a w.o. of R, see https://ivv5hpp.uni-muenster.de/u/rds/hamel_basis_2.pdf<|endoftext|> -TITLE: Value of the Chern-Simons functional for flat connections on $S^3/\Gamma$ -QUESTION [7 upvotes]: Let $\Gamma$ be a finite subgroup of SU(2) and consider the quotient of $S^3$ by $\Gamma$ via its left action. Pick a simply connected compact Lie group $G$ and take a flat connection on this quotient. Or equivalently, choose a homomorphism $\Gamma \to G$. The Chern-Simons action functional should assign a value in $\mathbb{R}/\mathbb{Z}$. How do I compute it? -The paper http://gdz.sub.uni-goettingen.de/dms/load/img/?PID=GDZPPN002334585 by Kronheimer and Nakajima has the formula for the eta invariant which is closely related, but I think it gives the CS functional only as a value in $\mathbb{R}/(\mathbb{Z}/h^\vee(G))$. - -REPLY [6 votes]: For $G=SU(N)$, there is a paper SU(n)–Chern–Simons invariants of Seifert fibered 3–manifolds (Int. J. Math., 09, 295-330 (1998))<|endoftext|> -TITLE: Dieudonné theory over non-perfect base fields -QUESTION [7 upvotes]: Is there a Dieudonné theory for $p$-divisible groups (or for finite flat group schemes of $p$-power order) over non-perfect base-fields? - -REPLY [6 votes]: Johan de Jong sent me an e-mail with a reference to Jong, A.J. de. "Finite locally free group schemes in characteristic $p$ and Dieudonné modules" Inventiones mathematicae 114.1 (1993): 89-138. https://eudml.org/doc/144143.<|endoftext|> -TITLE: How to visualize a category (of "combinatorial" maps) -QUESTION [5 upvotes]: This is a practical and very soft question, with the combinatorial database http://www.findstat.org in mind. -I have a few, around 20, families of combinatorial objects, for example Dyck paths, permutations, perfect matchings, graphs, etc., together with a few, around 200, maps between them. The maps need not be injective or surjective or have any special properties. (Except perhaps that they appear in the literature and are therefore "interesting".) -Some examples of such maps might be the reversal of a Dyck path, various classical maps that send a Dyck path to a 321-avoiding permutation, the map that sends a permutation to its shape under the Robinson-Schensted correspondence, etc. -Thus, we have (a very small) category, which I'd like to visualize. -The aspect that makes this interesting and non-trivial is, that these maps satisfies numerous identities: many maps are involutions, some are idempotent, many maps commute with other maps or are conjugate to other maps, etc. I have all these data. -So, more precisely: I'm looking for a way to visualize (graphically!) these identities. - -REPLY [3 votes]: Despite some interesting proposals given in the comments of the question, I finally ended up with a rather straightforward listing, mainly because the amount of data is quite large. An example can be studied at http://findstat.org/Mp00023.<|endoftext|> -TITLE: What "should" be the absolute galois group of a field with one element -QUESTION [16 upvotes]: As far as I know there is many "suggestions" of what should be a "field with one element" $\mathbf{F}_{1}$. -My question is the following: -How we should think or what should be the "absolute Galois group" of the "hypothetical" $\mathbf{F}_{1}$ ? - -REPLY [13 votes]: The Galois group of the maximal abelian extension of $\mathbb Q$ (or any number field) is given (class field theory) as the quotient of the idele class group by the connected component of the identity which is isomorphic to $\mathbb R$. If there is an ${\mathbb F}_1$ then its extensions provide "constant field extensions" of $\mathbb Q$. So, to be compatible with class field theory and keep the analogy with function fields, $\mathbb R$ must be at least the abelianization of the absolute Galois group of ${\mathbb F}_1$. But finite fields are abelian, so this suggests the answer. A slightly different perspective is that Weil constructed canonically and functorially an extension of the absolute Galois group of any number field by ${\mathbb R}$ (the Weil group) and that should be the extension one would get by allowing constant field extensions again.<|endoftext|> -TITLE: Definite integral involving inverse regularized incomplete beta functions -QUESTION [6 upvotes]: In my research I encountered the following integral -$$J = \int_0^1 I_t^{-1}(a_1,b_1) \: I_t^{-1}(a_2,b_2) \: \mathrm{d}t$$ -which I would like to evaluate as a closed form expression, that is, as a function of the parameters $a_{1}, b_{1}, a_{2}, b_{2} >0$. In words, this is the integral of the product of two inverse regularized incomplete beta functions. -This was posted in Math StackExchange about two months back but received no answer. That link shows my naive attempt via integration by parts. If a closed form (or lack of it) is well known, I would appreciate a reference. Any ideas or approximation for the integral are welcome. - -REPLY [2 votes]: for $a_1=a_2$ and $b_1=b_2$ there is a closed form expression in terms of the incomplete Beta function: -abbreviate $Q_t=I^{-1}_t(a,b)$ and then Mathematica tells me that -$$\int_0^1 Q_t^2 \,dt=\frac{1}{bB(a,b)}$$ -$$\times\left[(1-Q_0){}^b Q_0{}^{a+1}-(1-Q_1){}^b Q_1{}^{a+1}+(a+1) B_{Q_1}(a+1,b+1)-(a+1)B_{Q_0}(a+1,b+1)\right]$$ -I would be surprised if this can be generalized to $a_1\neq a_2$, $b_1\neq b_2$.<|endoftext|> -TITLE: Which compacta contain copies of Cantor cubes? -QUESTION [5 upvotes]: It is well-known that each uncountable compact metrizable space $X$ contains a homeomorphic copy of the Cantor cube $\{0,1\}^\omega$. What about copies of Cantor cubes of larger weight? -Problem. Does every uncountable Dugundji compact space $X$ contain a topological copy of the Cantor cube $\{0,1\}^{\kappa}$ of weight $\kappa=w(X)$ ? -A compact Hausdorff space $X$ is Dugundji if and only if $X$ is an AE(0)-space, which means that each continuous map $f:B\to X$ defined on a closed subspace $B$ of a zero-dimensional compact Hausdorff space $A$ admits a continuous extension $\bar f:A\to X$. - -REPLY [6 votes]: It was proved independently by Efimov and Gerlits that the answer is yes if $\kappa$ has uncountable cofinality. In fact they proved this for any dyadic $X$ (it is well known that Dugundji compacta are dyadic). Their theorem (see "Mappings and imbeddings of dyadic spaces" by Efimov or "On subspaces of dyadic compacta" by Gerlits) actually says: -For a dyadic $X$ with $w(X)=\kappa$ the following are equivalent: -1) $X$ contains a copy of $2^\kappa$. -2) $X$ maps continuously onto $[0,1]^\kappa$. -3) $X$ is not a countable union of closed subspaces of weight less than $\kappa$. -I don't know if this can be improved at all using the stronger hypothesis of $X$ being Dugundgi.<|endoftext|> -TITLE: An inverse spectral problem for Jacobi matrices (or orthogonal polynomials) -QUESTION [5 upvotes]: I will formulate this question in the language of Jacobi operators and spectral measures although it could be entirely rewritten in terms of orthogonal polynomials and measures of orthogonality. - Objects of interest: -Let $J$ be a semi-infinite Jacobi matrix of the form -$$J=\begin{pmatrix} -b_{1} & a_{1}\\ -a_{1} & b_{2} & a_{2}\\ -& a_{2} & b_{3} & a_{3}\\ -& & \ddots & \ddots & \ddots -\end{pmatrix},$$ -where $b_{n}\in\mathbb{R}$ and $a_{n}>0$. Let $\mathcal{M}$ denotes the set of Jacobi matrices $J$ such that $b_{n}\to 0$ and $a_{n}\to 1$ as $n\to\infty$ (the so-called Szegö class in OG polynomials terminology). -Let $J_{0}$ be the Jacobi matrix with $b_{n}=0$ and $a_{n}=1$ for all $n\in\mathbb{N}$. It is well-known that the spectrum of $J_{0}$ is -$$\sigma(J_{0})=[-2,2].$$ -A Jacobi matrix $J\in\mathcal{M}$ iff $J-J_{0}$ is compact. -For any self-adjoint operator $J$ (assume $J$ is bounded), there exists unique measure $\mu$ (the spectral measure) determined by the equality -$$\int_{\mathbb{R}}\frac{d\mu(x)}{x-z}=\langle e_{1},(J-z)e_{1}\rangle, \quad \forall z\in\mathbb{C}\setminus\mathbb{R},$$ -where $e_{1}$ is the first vector of the standard basis of $\ell^{2}(\mathbb{N})$. It holds that $\mbox{supp}\mu=\sigma(J)$. -Question: Assume $\mu$ to be a probability measure with $\mbox{supp}\mu=[-2,2]$. Then there exists a unique Jacobi operator $J$ whose spectral measure coincides with $\mu$. I am interested in what additional conditions (to $\mbox{supp}\mu=[-2,2]$) is to be imposed on the measure $\mu$ to guarantee that $J\in\mathcal{M}$. The condition (if any) should be weaker than the Szegö condition which gives us more, see below. -What is known: There is an answer to this kind of question known for a long time but it implies even more than just $J\in\mathcal{M}$. Namely, if $\mbox{supp}\mu=[-2,2]$ and the density $\rho$ of the the Lebesgue -absolutely continuous component of $\mu$ satisfies the so-caled Szegö condition: -$$\int_{-2}^{2}\frac{\log\rho(x)}{\sqrt{4-x^{2}}}dx>-\infty,$$ -then -$$\sum_{n=1}^{\infty}b_{n}^{2}<\infty \quad \mbox{ and } \quad -\sum_{n=1}^{\infty}(a_{n}-1)^{2}<\infty.$$ -In particular, $J\in\mathcal{M}$. Proof and references for these claims (and much more) can be found here. - -REPLY [2 votes]: One such condition (which is weaker, though not dramatically so perhaps) is that $\rho(x)>0$ a.e. on $(-2,2)$. This is usually called the Denisov-Rakhmanov theorem; see here. In fact, it will give the conclusion also if $J$ has spectrum outside $[-2,2]$ as long as this part of the spectrum is discrete. -This condition is clearly not necessary for $J-J_0$ to be compact; a compact perturbation can easily have purely singular spectrum. I very much doubt that a characterization of compact perturbations (as in the case of Hilbert-Schmidt) is possible. -However, if you keep the $a_n$'s equal to $1$, then the obvious necessary condition from Weyl's theorem (namely, $\sigma_{ess}=[-2,2]$) becomes sufficient also. This lovely result is due to Damanik, Hundertmark, Killip, and Simon.<|endoftext|> -TITLE: Does every space curve lie on a rational surface? -QUESTION [12 upvotes]: Let $C\subset \mathbb P^3$ be a smooth projective curve over $\mathbb C$. Is there a (singular) rational surface $S$ such that $C\subset S\subset \mathbb P^3$? - -I'm also interested in the following higher-dimensional analogue: - -Let $X\subset \mathbb P^n$ be a smooth projective variety. Is there a rational variety $Y$ of dimension $\dim X+1$ such that $X\subset Y\subset\mathbb P^n$? - -REPLY [9 votes]: Edit. There is a much simpler argument than the original answer below, and it also answers the second question. I will add that argument in a moment. However, I will leave the "rational simple connectedness" argument as well. The point is that the "rational simple connectedness" argument holds for any embedding of a curve $C$ in a rationally simply connected variety $P$, e.g., embedded in a general hypersurface of degree $m$ in $\mathbb{P}^n$ so long as $m^2\leq n$. -New answer to both questions. The following technique is usually called "retract rationality". The argument can be summarized as "rational varieties, such as the ambient $\mathbb{P}^n$, are retract rational". In fact, if you do not mind that $Y$ is very singular at the generic point of $X$, you can easily adapt the following argument to $X$ a subvariety of a unirational variety (of codimension $>1$). This actually leads to one of the proposed strategies for the open conjecture that there exist rationally connected varieties that are not unirational: find a rationally connected variety $P$ such that for some $X$ of codimension $>1$, there exists no rational $Y$ containing $X$. For $X$ a subvariety of $\mathbb{P}^n$, or any rational variety $P$, the following argument produces a rational subvariety $Y$ that contains $X$ and that is smooth at the generic point of $X$. -Denote by $\ell$ the dimension of $X$. If $\ell+1$ equals $n$, the result is trivially true by choosing $Y=\mathbb{P}^n$. Thus, assume that $n> \ell+1$. By Bertini, for a general linear projection $$\rho:\mathbb{P}^n \dashrightarrow \mathbb{P}^{\ell + 1},$$ the regular locus of $\rho$ intersects $X$ in a dense open subset of $X$, and the induced rational transformation $$\rho_X:X\dashrightarrow \mathbb{P}^{\ell + 1},$$ is birational to its image. Denote by $U\subset \mathbb{P}^{\ell+1}$ a dense, affine open subset such that $\rho_X^{-1}(U)\subset X$ is a dense open subset and such that $$\rho_{X,U}:\rho_X^{-1}(U)\to U,$$ is a closed immersion. Up to shrinking $U$ further, assume that the invertible sheaf $\mathcal{O}_{\mathbb{P}^n}(1)|_X$ is trivialized on $\rho_X^{-1}(U)$. Since $\rho_{X,U}$ is a closed immersion, the pullback $k$-algebra homomorphism is surjective, $$\rho_{X,U}^\#:\mathcal{O}_{\mathbb{P}^{\ell+1}}(U) \twoheadrightarrow \mathcal{O}_X(\rho_X^{-1}(U)).$$ Thus, the restrictions to $\rho_X^{-1}(U)$ of the $n+1$ defining global sections of $\mathcal{O}_{\mathbb{P}^n}(1)$ lift to $n+1$ global sections of $\mathcal{O}_{\mathbb{P}^{\ell+1}}$ on $U$. By a Bertini-type argument, for a general choice of these $n+1$ lifts, after replacing $U$ by a dense open that still intersects $X$, the $n+1$ global sections have empty base locus, and the associated rational transformation $$j:U\to \mathbb{P}^n,$$ is birational to its image, say $Y=\overline{j(U)}\subset \mathbb{P}^n$. Thus $Y$ is a rational subvariety of $\mathbb{P}^n$ of dimension $\ell+1 = \text{dim}(X)+1$. Since the $n+1$ global sections are chosen to extend the tautological sections on $\rho_X^{-1}(U)$, the composition $j\circ \rho$ is a well-defined rational transformation that agrees with the inclusion $i:X\to \mathbb{P}^n$. Thus, $X$ is contained in an $(\ell+1)$-dimensional rational subvariety $Y$ of $\mathbb{P}^n$. -Original answer to first question. -The answer to the first question is positive by "rational simple connectedness". Let $B$ be a copy of $\mathbb{P}^1$. Let $f:C\to B$ be a finite morphism that is étale over a dense open subset $B^o\subset B$. Let $F$ be a zero-dimensional, reduced $k$-scheme of length $d$ whose connected components are an ordered $d$-tuple of $k$-rational points. Denote by $$I^o = \text{Isom}_{B^o}(B^o \times F,C^o)\to B^o$$ the relative Isom scheme. For $e\geq 1$, denote by $\overline{\mathcal{M}}_{0,d}(\mathbb{P}^3,e)$ the moduli stack of $d$-pointed, genus $0$ stable maps to $\mathbb{P}^3$ of degree $e$. Denote the evaluation $1$-morphism by $$\text{ev}:\overline{\mathcal{M}}_{0,d}(\mathbb{P}^3,e)\to (\mathbb{P}^3)^d.$$ There is a natural action of the symmetric group $\mathfrak{S}_d$ on both source and target, and the $1$-morphism is equivariant for these actions. There is also a natural action of $\mathfrak{S}_d$ on $I^o$, and this action makes $I^o$ into a $\mathfrak{S}_d$-torsor over $B^o$. The inclusion $i:C\hookrightarrow \mathbb{P}^3$ determines a morphism $$i_F: \text{Isom}_{B^o}(B^o\times F,C^o) \to (\mathbb{P}^3)^d.$$ This morphism is also $\mathfrak{S}_d$-equivariant. Form the $2$-fibered product, $$\overline{\mathcal{M}}_{0,d}(\mathbb{P}^3,e)_i = \overline{\mathcal{M}}_{0,d}(\mathbb{P}^3,e)\times_{\text{ev}, i_F} I^o,$$ together with its projection morphism, $$\text{pr}_{I^o}:\overline{\mathcal{M}}_{0,d}(\mathbb{P}^3,e)_i \to I^o.$$ This morphism is $\mathfrak{S}_d$-equivariant. Thus, forming the quotients of domain and target gives a $1$-morphism, $$\text{pr}_{B^o}:\overline{\mathcal{M}}_{0,f^o}(\mathbb{P}^3,e) \to B^o.$$ The geometric generic fiber of this morphism equals the geometric generic fiber of $\text{pr}_{I^o}$, namely, the stack parameterizing $d$-pointed, degree $e$ stable maps to $\mathbb{P}^3$ whose $d$ marked points map to the $d$ points of the geometric generic fiber of $f^o$. -By "rational simple connectedness" (or more elementary arguments), for $e\gg 0$, the geometric generic fiber of $\text{pr}_{B^o}$ is rationally connected. By the Rationally Connected Fibration Theorem, together with the Weak Approximation Theorem of Hassett-Tschinkel, there exists a rational section of $\text{pr}_{B^o}$, $$\sigma^o:B^o\to \overline{\mathcal{M}}_{0,f^o}(\mathbb{P}^3,e),$$ mapping the geometric generic point of $B^o$ to a point parameterizing a stable map with smooth domain mapping isomorphically to its image in $\mathbb{P}^3$. In particular, up to shrinking $B^o$, assume that $\sigma^o$ is a regular $1$-morphism with image in the non-stacky locus. The pullback via $\sigma^o$ of the universal stable map is a quasi-projective surface $S^o$ together with a proper, smooth morphism $\pi^o:S^o\to B^o$ with connected, genus $0$ geometric fibers, a morphism $u^o:S^o\to \mathbb{P}^3$, and a factorization $j^o:C^o\to S^o$ of $f^o$ such that $u^o\circ j^o$ equals $i^o$. -Altogether, $S^o$ is a quasi-projective rational surface with a morphism $u^o$ to $\mathbb{P}^3$ such that $i^o:C^o\to \mathbb{P}^3$ factors through $u^o$.<|endoftext|> -TITLE: Is the hom in derived category of a dg-algebra compatible with base field extension? -QUESTION [6 upvotes]: Let $k$ be a field and $A$ be an ordinary $k$-algebra. Let $M$ and $N$ be two left $A$-modules then we could consider the Ext group $\mathrm{Ext}^i_A(M,N)$ which is actually a $k$-vector space. Let $l/k$ be a field extension and we consider $A_l$, $M_l$, and $N_l$ be the base-change algebras and modules, i.e. $A_l=A\otimes_kl$, etc. We can also consider $\mathrm{Ext}^i_{A_l}(M_l,N_l)$ and we have the natural map $$ \mathrm{Ext}^i_A(M,N)\otimes_kl\to \mathrm{Ext}^i_{A_l}(M_l,N_l).$$ We know it is an isomorphism if $M$ is a finite present $A$-module. -Now let $A$ be a differential graded (dg) $k$-algebra and consider the derived category of dg-$A$-modules $D(A)$. Let $M$ and $N$ be two objects in $D(A)$. For a field extension $l/k$ we can also form $A_l$, $M_l$, and $N_l$ in the similar way. -My question is - -Is the natural map - $$ -Hom_{D(A)}(M,N)\otimes_kl\to Hom_{D(A_l)}(M_l,N_l) -$$ - an isomorphism for $M$ a perfect complex of dg-$A$-modules and arbitrary $N$? - -It seems to be similar to the above result but I cannot find any reference. - -REPLY [3 votes]: Yes. -The full subcategory consisting of those $M$ for which this natural map is an isomorphism (for all $N$) is a thick subcategory by a five lemma argument, and the thick subcategory generated by $A$ is precisely the category of perfect complexes, so it is sufficient to prove it for $M=A$. -But $\text{Hom}_{D(A)}(A,N)=H^0(N)$, so for $M=A$ the map is the natural map -$$H^0(N)\otimes_kl\to H^0(N_l),$$ -which is an isomorphism.<|endoftext|> -TITLE: (Co)complete topoi that are not Grothendieck? -QUESTION [13 upvotes]: Recall that an elementary topos is a cartesian closed category with finite limits and a subobject classifier. A Grothendieck topos is a category equivalent to the category of sheaves on a site. -Are there examples of (co)complete elementary topoi that are not Grothendieck? On the "Cocomplete" side of things, such a category cannot be accessible, since a locally presentable elementary topos is automatically Grothendieck. - -REPLY [17 votes]: A classical example is $G$-$Set$ for a large group $G$. That this is a cocomplete elementary topos is not hard to see. Limits and colimits are formed at the underlying set level, and exponentials $Y^X$ are formed as usual as the set of functions $f: X \to Y$ with the $G$-action $(g, f) \mapsto g f$ defined by $g f: x \mapsto g f(g^{-1} x)$. The subobject classifier is the 2-element set with trivial $G$-action (again, as usual). -Thus $E = G$-$Set$ is a cocomplete (elementary) topos. -It remains to see $E$ is not Grothendieck. If it were, then every continuous functor such as the underlying-set functor $U: E \to Set$ would have a left adjoint $F$, by the special adjoint functor theorem (the hypotheses for the SAFT are satisfied: a Grothendieck $E$ is well-powered and complete and has a cogenerator $\Omega^c$ where $c$ is the coproduct of associated sheaves of objects in a small site presentation). In particular, we would have $\hom_E(F(1), -) \cong U$, but the only candidate for the representing object $F(1)$ would be $G$, which is ruled out by largeness. - -Edit: Adam Epstein pointed out in a comment that the intuition in the last sentence, regarding the only candidate for the representing object, is not correct for some $G$ such as a large simple group. The following addendum patches up this oversight. -Let $G$ be a large free group, say the union of the diagram of free groups $F(\alpha)$ obtained by applying the free group functor to von Neumann cardinals $\alpha$ and initial segment inclusions between them. Then the topos $Set^G$ of $G$-sets is not Grothendieck. -Supposing it is, then the underlying-set functor $U: Set^G \to Set$ is representable, by the special adjoint functor theorem. Hence $U \cong \hom(X, -)$ for some $G$-set $X$. We will show $U$ has a proper class of non-isomorphic representable subfunctors, i.e., the object $X$ has a proper class of quotients, which is impossible in a Grothendieck topos (or even in a locally small topos -- quotients of $X$ are in bijection with equivalence relations on $X$, forming a subcollection of a hom-set $[X \times X, \Omega]$). -To begin with, for the class of canonical inclusions $i_\alpha: F(\alpha) \to G$ we may uniformly exhibit retractions $r_\alpha: G \to F(\alpha)$ (retractions in the sense of group homomorphisms). We may then view $F(\alpha)$ as a $G$-set with action $G \times F_\alpha \to F_\alpha$ taking $(g, x)$ to $r_\alpha(g) \cdot x$, and the representable functor $\hom(F(\alpha), -)$ is naturally a subfunctor of $U$. Indeed, a map $f: F(\alpha) \to A$ is uniquely determined by the value $a = f(1)$ at the identity element $1 \in F(\alpha)$, as the intertwiner condition yields $f(r_\alpha(g)) = g a$, whence $f(x) = i_\alpha(x)a$ for general $x \in F(\alpha)$. Denoting the idempotent map $i_\alpha r_\alpha: G \to G$ by $p_\alpha$, the subfunctor inclusion is given componentwise by -$$\hom(F(\alpha), A) \cong \{a \in A: \forall_{g \in G}\; g a = p_\alpha(g) a\} \hookrightarrow U(A)$$ -and all these subfunctors are non-isomorphic, for the simple crude reason that the $F(\alpha)$ have generally different cardinalities as sets (as soon as $\alpha$ is in the uncountable range).<|endoftext|> -TITLE: Do symmetric monoidal groupoids model all connective spectra? -QUESTION [10 upvotes]: Thomason showed that symmetric monoidal categories model all connective spectra. But can it be done with groupoids? In order for this to be the case, the group completion prices process must be very drastic. But after all, the sphere spectrum is the group completion of the symmetric monoidal groupoid of finite sets, so maybe anything is possible. - -REPLY [5 votes]: Theorem 5.3 of - -Daniel Fuentes-Keuthan, Modelling Connective Spectra via Multicategories, arXiv:1909.11148. - -answers this question positively!<|endoftext|> -TITLE: Can all lines in the euclidian plane be ordinary? -QUESTION [5 upvotes]: Is there a set $X \subset \mathbb{R}^2$ such that every straight line in the plane is ordinary in relation to it? i.e. if $r$ is any straight line then $|r \cap X|=2$. - -REPLY [12 votes]: The answer is yes, by an argument using the axiom of choice. -There are exactly continuum many lines in the plane, and so by the -axiom of choice, we may enumerate them in a well-ordered sequence -of length continuum. -Let's build the set $X$ in stages, so that by stage $\alpha$ we've -included two points from all the lines in the enumeration up to -$\alpha$ and furthermore, we've done so in such a way that no three -points of the set $X$ are collinear. -If we've done this up to stage $\alpha$, then consider the next -line $\ell$. If we've already got two points from $\ell$ in $X$, -then we're done; we've already handled this line. Otherwise, -consider the fewer than continuum many lines through any two points -that we've already placed into $X$. These lines intersect $\ell$ in -fewer than continuum many points, leaving plenty of other points on $\ell$ that we are free to add to $X$ without violating the no-three-points-collinear requirement. If we've already got one point -from $\ell$, then we can add another in such a way that it is not -on any of these other lines, thereby maintaining the -no-three-points-collinear property. And if we have no points yet -from $\ell$ in $X$, then we simply add two such points from $\ell$ -not on any of those other lines. In this way, we add points -to $X$ so as to make $X\cap\ell$ of size two, while maintaining the -no-three-points-collinear property. -Thus, by this transfinite recursive process, we build the desired set $X$, -which has exactly two points from every line. -Essentially the same argument works in $\mathbb{R}^3$ or in any finite dimension, or indeed, even in $\mathbb{R}^\omega$. -The construction also works, suitably modified, with circles or other kinds of shapes instead of lines. The important point being that distinct circles intersect in at most two points. So there is a set in the plane containing exactly three points from every circle (or exactly four, whatever is desired).<|endoftext|> -TITLE: Decomposing a matrix into a product of sparse matrices -QUESTION [5 upvotes]: How to study the decomposition of a square matrix into a product of sparse matrices? -There are no restrictions on the number of matrices in the product, but the fewer the better. - -REPLY [10 votes]: Given an invertible $n \times n$ matrix $\mathrm A$, we perform Gaussian elimination until we obtain a (nonsingular) diagonal matrix. In other words, we left-multiply $\mathrm A$ by permutation matrices (whose inverses are their transposes) and by elementary matrices of the form -$$\mathrm E_{ij} := \mathrm I_n + \beta_{ij} \, \mathrm e_i \mathrm e_j^{\top}$$ -whose inverse is simply -$$\mathrm E_{ij}^{-1} = \mathrm I_n - \beta_{ij} \, \mathrm e_i \mathrm e_j^{\top}$$ -until we obtain a (nonsingular) diagonal matrix $\mathrm D$, i.e., -$$\mathrm E_{i_m j_m} \mathrm P_m \cdots \mathrm E_{i_2 j_2} \mathrm P_2 \mathrm E_{i_1 j_1} \mathrm P_1 \mathrm A = \mathrm D$$ -where $m$, the number of $\mathrm E_{i j}$ elementary matrices to the left of $\mathrm A$, is at most $n (n-1)$. Thus, one decomposition of $\mathrm A$ into a product of sparse matrices is -$$\boxed{\mathrm A = \mathrm P_1^{\top} \mathrm E_{i_1 j_1}^{-1} \mathrm P_2^{\top} \mathrm E_{i_2 j_2}^{-1} \cdots \mathrm P_m^{\top} \mathrm E_{i_m j_m}^{-1} \mathrm D}$$ -where the $\mathrm E_{i j}^{-1}$ matrices are sparse and easy to compute. The fraction of nonzero entries in these matrices is $n^{-1} + n^{-2} \approx n^{-1}$ for the $\mathrm E_{i j}^{-1}$ matrices and $n^{-1}$ for matrix $\mathrm D$ and for each permutation matrix. Hence, the larger $n$, the sparser the matrices. -Multiplying the sparse elementary matrices pertaining to the same column (e.g., $\mathrm E_{21}, \mathrm E_{31}, \dots, \mathrm E_{n1}$ for the first column), we obtain less sparse Frobenius matrices. To buy a smaller number of factors in the decomposition, we pay with sparsity. -Though the $\mathrm E_{i j}$ and $\mathrm E_{i j}^{-1}$ matrices have $n+1$ nonzero entries, only one of them, the $(i,j)$-th entry, is in general different from $1$. Therefore, we only need one real number, $\beta_{ij}$, and one pair of positive integers, $(i,j)$, to fully specify the elementary matrix $\mathrm E_{i j}$ (or its inverse). Taking into account also the $n$ nonzero entries on the main diagonal of the diagonal matrix $\mathrm D$, much of the data needed to build the factors in the decomposition can be stored in the following $n \times n$ matrix -$$\begin{bmatrix} d_1 & \beta_{12} & \dots & \beta_{1n}\\ \beta_{21} & d_2 & \dots & \beta_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ \beta_{n1} & \beta_{n2} & \dots & d_n\end{bmatrix}$$ -Each permutation matrix can be specified using $n$ positive integers. Thus, at most $n^2$ reals and $n^2$ integers are needed to specify the decomposition. - -Example -Consider the (invertible) $3 \times 3$ magic square -$$\mathrm A := \begin{bmatrix} 2 & 9 & 4\\7 & 5 & 3\\6 & 1 & 8\end{bmatrix}$$ -Using SymPy, ->>> from sympy import * ->>> A = Matrix([[2,9,4],[7,5,3],[6,1,8]]) - -Eliminate two entries below the $(1,1)$-th entry: ->>> E21 = eye(3) + Matrix([[0,0,0],[-7.0/2.0,0,0],[0,0,0]]) ->>> E31 = eye(3) + Matrix([[0,0,0],[0,0,0],[-3.0,0,0]]) ->>> E31 * E21 * A -Matrix([ -[2, 9, 4], -[0, -26.5, -11.0], -[0, -26.0, -4.0]]) - -Eliminate the entry below the $(2,2)$-th entry: ->>> E32 = eye(3) + Matrix([[0,0,0],[0,0,0],[0,-26.0/26.5,0]]) ->>> E32 * E31 * E21 * A -Matrix([ -[2, 9, 4], -[0, -26.5, -11.0], -[0, 1.33226762955019e-15, 6.79245283018868]]) - -Numerical noise is now appearing. Eliminate two entries above the $(3,3)$-th entry ->>> E23 = eye(3) + Matrix([[0,0,0],[0,0,11.0/6.79245],[0,0,0]]) ->>> E13 = eye(3) + Matrix([[0,0,-4.0/6.79245],[0,0,0],[0,0,0]]) ->>> E13 * E23 * E32 * E31 * E21 * A -Matrix([ -[2.0, 9.0, -1.66666736145515e-6], -[ 0, -26.5, 4.58333524377963e-6], -[ 0, 1.33226762955019e-15, 6.79245283018868]]) - -Eliminate the entry above the $(2,2)$-th entry: ->>> E12 = eye(3) + Matrix([[0,9.0/26.5,0],[0,0,0],[0,0,0]]) ->>> E12 * E13 * E23 * E32 * E31 * E21 * A -Matrix([ -[2.0, 1.99840144432528e-15, -1.10062938096789e-7], -[ 0, -26.5, 4.58333524377963e-6], -[ 0, 1.33226762955019e-15, 6.79245283018868]]) - -We extract the diagonal: ->>> D = diag(2.0,-26.5,6.79245) ->>> D -Matrix([ -[2.0, 0, 0], -[ 0, -26.5, 0], -[ 0, 0, 6.79245]]) - -Finally, we reconstruct the magic square: ->>> (E21**-1) * (E31**-1) * (E32**-1) * (E23**-1) * (E13**-1) * (E12**-1) * D -Matrix([ -[2.0, 9.0, 4.0], -[7.0, 5.0, 3.0], -[6.0, 1.0, 7.99999716981132]]) - -The reconstruction is not exact due to numerical noise, but it is close enough. Thus, magic square $\mathrm A$ can be approximately factored as follows -$$\underbrace{\begin{bmatrix} 1 & 0 & 0\\ 3.5 & 1 & 0\\3 & 0 & 1\end{bmatrix}}_{= \mathrm E_{21}^{-1} \mathrm E_{31}^{-1}} \, \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\0 & 0.98 & 1\end{bmatrix} \, \underbrace{\begin{bmatrix} 1 & 0 & 0.59\\0 & 1 & -1.62\\0 & 0 & 1\end{bmatrix}}_{\approx \mathrm E_{23}^{-1} \mathrm E_{13}^{-1}} \, \begin{bmatrix} 1 & -0.34 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix} \, \underbrace{\begin{bmatrix} 2 & 0 & 0\\ 0 & -26.5 & 0\\0 & 0 & 6.79\end{bmatrix}}_{\approx \mathrm D}$$ -where the 1st and 3rd of the five matrix factors are Frobenius matrices. In this particular example, no permutation matrices (other than $\mathrm I_n$, that is) were needed.<|endoftext|> -TITLE: Is the matrix $\left({2m\choose 2j-i}\right)_{i,j=1}^{2m-1}$ nonsingular? -QUESTION [26 upvotes]: Suppose we have a $(2m-1) \times (2m-1)$ matrix defined as follows: -$$\left({2m\choose 2j-i}\right)_{i,j=1}^{2m-1}.$$ -For example, if $m=3$, the matrix is -$$\begin{pmatrix}6 & 20 & 6& 0 & 0\newline 1 & 15 & 15 & 1 & 0 \newline 0 & 6 & 20 & 6 & 0 \newline 0 & 1 & 15 & 15 & 1 \newline 0 & 0 & 6 & 20 & 6 \end{pmatrix}$$ -Can anyone tell me how to prove it is nonsingular? - -REPLY [21 votes]: Here is a very low-brow answer to the original question. -Consider the lower-triangular matrix -\begin{equation*} -V = [V_{ij}] = \left[\binom{i-1}{j-1}\right]\quad \text{for}\quad i \ge j. -\end{equation*} -Let $A$ be the matrix in the OP. Then, a quick induction shows that $V^{-1}AV$ is upper triangular (use that $V^{-1}$ has entries $(-1)^{i-j}V_{ij}$), with the diagonal: $[2^n,2^{n-1},\ldots,2]$, which not only establishes invertibility of $A$, but also yields its eigenvalues.<|endoftext|> -TITLE: Is a pointwise " simple tensor" valued continuous map a tensor product of two continuous maps? -QUESTION [5 upvotes]: A matrix $A\in M_{4}(\mathbb{C})$ is called a simple tensor if $A=B\otimes C$ for two $2\times 2$ matrices $B,C$. -Assume that $X$ is a Hausdorff topological space.Assume that $f:X\to M_{4}(\mathbb{C})$ is a continuous map such that $f(x)$ is a simple tensor, for every $x\in X$. - -Are there continous maps $g,h:X\to M_{2}(\mathbb{C})$ with $f=g\otimes h$? - -REPLY [6 votes]: The answer is 'no'. For example, let -$$ -X = \{\ B\otimes C\ | \det(B)\det(C) = 1\ \}\subset \mathrm{SL}(4,\mathbb{C}). -$$ -Then $X$ is isomorphic to the group $\mathrm{SO}(4,\mathbb{C})$ and hence $\pi_1(X)\simeq \mathbb{Z}_2$, while the set -$$ -G = \{\ (B, C)\ |\ \det(B)\det(C) = 1\ \}\subset -\mathrm{GL}(2,\mathbb{C})\times\mathrm{GL}(2,\mathbb{C}) -$$ -is a connected Lie group that satisfies $\pi_1(G)\simeq \mathbb{Z}$, with generator given by the subgroup -$$ -\{\ (B, B^{-1})\ |\ B\in T\ \} -$$ -where $T\subset\mathbb{GL}(2,\mathbb{C})$ is a circle subgroup that generates $\pi_1\bigl(\mathbb{GL}(2,\mathbb{C})\bigr)\simeq \mathbb{Z}$. -Now, the map $\tau:\mathrm{GL}(2,\mathbb{C})\times\mathrm{GL}(2,\mathbb{C})\to \mathrm{GL}(4,\mathbb{C})$ given by $\tau(B,C) = B\otimes C$ is a group homomorphism, and the preimage under $\tau$ of the space $X$ is the two-component subgroup -$$ -G^+ = \{\ (B, C)\ |\ \det(B)\det(C) = \pm1\ \}\subset -\mathrm{GL}(2,\mathbb{C})\times\mathrm{GL}(2,\mathbb{C}). -$$ -(This is because of the identity $\det(B\otimes C) = \det(B)^2\det(C)^2$.) -Note that $\pi_1(G^+) = \pi_1(G) = \mathbb{Z}_2$. -The map $\tau:G\to X$ given by $\tau(B,C) = B\otimes C$ is a group homomorphism. If there were a continuous map $\sigma = (g,h):X\to G^+$ that satisfied $\tau\circ\sigma = \mathrm{id}_X$, this would induce a homomorphism $\sigma_*:\pi_1(X)\to\pi_1(G^+)$ that satisfied -$$ -\tau_*\circ\sigma_* = \mathrm{id}:\pi_1(X)\to\pi_1(X)\simeq\mathbb{Z}_2. -$$ -However, then $\sigma_*$ would be a nontrival homomorphism $\mathbb{Z}_2\to\mathbb{Z}$, and this does not exist.<|endoftext|> -TITLE: How "strong" is the existence of a non trivial ultrafilter on $\omega$? -QUESTION [5 upvotes]: Obviously the question in the title alone doesn't make sense so I'll develop on the context and then I'll ask my question : -Studying $AD$ (axiom of determinacy) I had to prove that $AD$ and $AC$ are incompatible (mod $ZF$). So to prove this I show that under $AC$, there are some undetermined games, and in order to prove this I use the fact that the Fréchet filter can be extended to an ultrafilter (necessarily non principal). As a remark, I thought "This shows that $AD$ is incompatible, not only with $AC$, but also with strictly weaker choice principles, such as the $BPI$ or the statement 'There exists a non-principal ultrafilter over $\omega$' ". Thinking this I wondered how strong that last statement (let $\Omega$ denote said last statement) was, and whether it could imply $BPI$. The answer seems to be "Obviously, no", as $\Omega$ is only about a specific set ($\omega$), whereas the $BPI$ is much more general. But then I wondered whether there was any "general choice principle", strong enough to prove $\Omega$, but not strong enough to prove the $BPI$. I then looked for a way to make "general choice principle" precise in order to look for an answer. So here are my questions : - -Is there a satisfactory way to make the notion of "general choice principle" precise, i.e. characterize certain sentences such that $AC$, $BPI$, etc. fall under this scope, but $\Omega$ doesn't ? -If there is, let $\phi$ be such a general choice principle. Can we have $ZF + \phi \vdash \Omega$, but $ZF + \phi$ doesn't prove $BPI$ ? - -I first thought of defining it as "$ZF^{-}+ \phi$ ($ZF^{-}$ being $ZF$ minus the axiom of infinity) does not prove the axiom of infinity", but 1.it was only for this particular example, 2. it didn't work, as "$Inf \implies \Omega$" would fall under this notion, but clearly wouldn't be satisfcatory, so a solution would have to be more clever than that. - -REPLY [9 votes]: It seems difficult to make the notion of "general choice principle" precise, but I would guess that the principle "every infinite set admits a nonprincipal ultrafilter" would qualify. If so, then it answers your second question. It obviously implies the existence of a nonprincipal ultrafilter on $\omega$, and, unless I'm making a stupid mistake, it doesn't imply BPI because it follows from the conjunction of "there is a nonprincipal ultrafilter on $\omega$" and countable choice. (The point is that countable choice implies that there are no infinite, Dedekind-finite sets, so every infinite set includes a copy of $\omega$.)<|endoftext|> -TITLE: When are there infinitely many primes in a sequence generated by a simple recurrence relation? -QUESTION [17 upvotes]: tl;dr summary: am I right in thinking that we expect $2^n-1$ to be prime infinitely often, but $2^n+1$ to be prime only finitely often? What's the general story here? - -An applied mathematician asked me a question about whether there were infinitely many primes in a certain sequence generated by a linear recurrence relation, and I was quick to respond "if there is not a trick to show that there are only finitely many using elementary methods, then there will probably be infinitely many but we probably won't be able to prove it". Too quick, I suspect. -Here is a general, but vague, question that I realise I don't understand properly. If $a,b,c$ are integers, with $a>1$ and $c>0$, and if the sequence $x_n=\frac{a^n+b}{c}$ ($n=1,2,3,\ldots$) is a sequence of positive integers, when do we expect this sequence to contain infinitely many primes? -There are two tricks I know. -There's a trivial trick "there exists a prime $p$ such that $x_n$ is always a multiple of $p$" which tells me e.g. that the sequence $3^n-1$ only contains finitely many primes. -There's a slightly more subtle trick which tells me that $\frac{4^n-1}{3}$ only contains finitely many primes, and this is that $4^n-1=(2^n+1)(2^n-1)$ which will prove that $\frac{4^n-1}{3}$ is composite when $n$ is large enough. And that's all I know. -I've seen the covering congruences trick used by Graham (R. L. Graham, -A Fibonacci-like sequence of composite numbers) to prove that there are general recurrence relations that contain no primes for slightly less obvious reasons, but I'm not sure if I can get this to work in my case. (edit Thanks to user102986, who reminded me of a covering congruence argument which was posted to this site 7 years ago in response to a question of mine which I'd completely forgotten about -- see comments below). -I initially suspected that if none of these tricks worked then perhaps there would be infinitely many primes but that we couldn't prove this; however this article by Kent Boklan and John Conway adds more meat to the suggestion, probably long believed, that there are only finitely many Fermat primes and hence only finitely many primes of the form $2^n+1$. Conversely GIMPS (motto: "$2^p-1$ may be prime"!) seems to be producing a steady stream of Mersenne primes. There are standard proofs that if $2^n+1$ is prime then $n$ is a power of 2, and that if $2^n-1$ is prime then $n$ is prime, and there are more primes than powers of 2, but this is a little simplistic in my mind and I suspect that people have a better understanding of what is, or should be, going on nowadays. My idle question: what is the general picture? -edit: in response to user102986's comments to make my question more precise, I will ask a random specific question. -Q) Is $2^n+2017$ is prime infinitely often? [warning: $2^n+9262111$ is never prime -- see link in the comments to a 2009 question of mine [further warning: $2^{104}+9262111$ is prime; $9262111$ should be replaced by $1517871$] ]. If it is prime infinitely often we probably can't prove it -- but should we be conjecturing it? - -REPLY [16 votes]: Am I right in thinking that we expect $2^n - 1$ to be prime infinitely - often, but $2^n+1$ to be prime only finitely often? What's the general - story here? - -Yes, you are. The (simplest) reasons for those conjectures come from the Borel-Cantelli Lemmas, the Prime Number Theorem, and the basic divisibility properties of $2^n - 1$ and $2^n + 1$. - -The first Borel-Cantelli Lemma says that if $E_1, E_2, \ldots$ is a sequence of events then -$$\sum_{n} \operatorname{Pr}[E_n] < +\infty$$ -implies that -$$\operatorname{Pr}[\text{infinitely many } E_n \text{ occur}] = 0,$$ -while the second Borel-Cantelli Lemma says that is $E_1,E_2, \ldots$ are independent and -$$\sum_{n} \operatorname{Pr}[E_n] = +\infty$$ -then -$$\operatorname{Pr}[\text{infinitely many } E_n \text{ occur}] = 1.$$ -The Prime Number Theorem tell us that, in some very imprecise sense, the "probability" that $N$ is a prime number is $\approx 1 / \log N$. -It is easy to prove that $2^n - 1$ is prime only if $n$ is prime, while $2^n + 1$ is prime only if $n$ is a power of two. - -In conclusion, the heuristic are the following: There should be infinitely many Mersenne primes, since -$$\sum_{p \text{ prime}} \frac1{\log(2^p - 1)} \approx \sum_{p \text{ prime}} \frac1{p \log 2} = + \infty ,$$ -while there should be only finitely many Fermat primes, since -$$\sum_{k = 1}^{+\infty} \frac1{\log(2^{2^k} - 1)} \approx \sum_{k = 1}^{+\infty} \frac1{2^k \log 2} < + \infty .$$<|endoftext|> -TITLE: How does Jacquet's "Generic Representations" classify tempered representations? -QUESTION [8 upvotes]: Let $L$ be a $p$-adic field $G = GL_n(L)$. Let $P$ be a standard parabolic subgroup with Levi decomposition $P = MU$, where $M \cong G_1\times \ldots \times G_r$, for $G_i \cong GL_{n_i}(L)$. -The following are classical: -1). If $\rho_i$ are discrete series representations of $G_i$, $\rho$ is the $M$ representation given by $(\rho_1 \otimes \ldots \otimes \rho_r)$, and $\pi$ is the image of $\rho$ under the normalized induction functor $I_P^G$, then $\pi$ is tempered. -2). All irreducible tempered representations arise in this way. To clarify, if $\pi$ is an irreducible tempered representation there is a parabolic subgroup $P$ with Levi subgroup $M$ and a discrete series representation $\rho$ of $M$ such that $\pi \cong I_P^G \rho$. -I've read this in multiple sources, and whenever they cite a source they point to Jacquet's "Generic Representations"; Rodier cites section 3 ("Tempered Representations") specifically. I must be missing something, as I don't see where in the paper these facts are proved. -What am I missing here? - -REPLY [5 votes]: Do you assume that $\pi$ is irreducible ? You can find a proof in the book of David Renard, available on his webpage, p.343, VII.2.6.<|endoftext|> -TITLE: GOE/GSE duality and Bott periodicity -QUESTION [12 upvotes]: Many papers in random matrix theory make passing references to duality between eigenvalue statistics of the GOE and GSE, for which the most concrete reference I can find is https://arxiv.org/pdf/math-ph/0206011.pdf (page 4 has the relevant discussion). Though I'm somewhat unclear on the actual matrix integral specifics of this paper, they also note that the GUE is self-dual by their definition. -In Bott periodicity, we see something that at least superficially looks very similar, namely the isomorphisms -$$\pi_k(U) \cong \pi_{k+2}(U)$$ -$$\pi_k(O) \cong \pi_{k+4}(Sp)$$ -$$\pi_k(Sp) \cong \pi_{k+4}(O)$$ -of homotopy groups of the infinite classical groups. -Question: is there any known relation between these two things? (I'd also appreciate any independent insight on the GOE/GSE and GUE/GUE duality). - -REPLY [8 votes]: The entire set of correspondences can be read off from this table: - -Listed are the 10 symmetric spaces and for each space in the left column the dual space is shown in the right column, as explained in On duality and negative dimensions in the theory of Lie groups and symmetric spaces, by Ruben L. Mkrtchyan, Alexander P. Veselov (2010). The Bott periodicity modulo 2 in complex K-theory is indicated by the numbers in blue, and the Bott periodicity modulo 8 in real K-theory is indicated by the numbers in red. -The connection between the symmetric spaces $X$ and the Gaussian ensembles of random Hermitian matrices $H$ is that $e^{iH}$ is in $X$ (uniformly distributed in the limit of infinite matrix size). I have indicated the name of the random-matrix ensemble in green. In addition to the three Wigner-Dyson ensembles GOE, GUE, GSE there appear counterparts with a chiral symmetry (meaning that the Hamiltonian is block-offdiagonal). The four ensembles labeled "AZ" are the Altand-Zirnbauer ensembles referred to by John Mangual. (They describe the spectral statistics in superconductors.) -The dualities in the OP for the GUE, GOE and GSE can be read off directly from this table; in addition we find dualities between the seven remaining symmetric spaces and between their corresponding (three chiral and four AZ) random-matrix ensembles. - -shameless plug -If you're interested in the physics behind the AZ ensembles and how this relates to topological superconductivity/Majorana fermions, etc. you might find this review useful.<|endoftext|> -TITLE: Kasteleyn's formula for domino tilings generalized? -QUESTION [22 upvotes]: It seems a marvel when a bunch of irrational numbers "conspire" to become rational, even better an integer. An elementary example is $\prod_{j=1}^n4\cos^2\left(\pi j/(2n+1)\right)=1$. -Kasteleyn's formula reveals $\prod_{j=1}^n\prod_{k=1}^n \left( 4\cos^2(\pi j/(2n+1))+4\cos^2(\pi k/(2n+1))\right)$ is an integer because it enumerates the domino tilings of a $2n$-by-$2n$ square. -These results prompt me to ask for more. First, let's introduce the $r$-product -$$K_r(n):=\prod_{\ell_1=1}^n\cdots\prod_{\ell_r=1}^n\left( -4\cos^2\left(\frac{\pi\ell_1}{2n+1}\right)+\cdots+4\cos^2\left(\frac{\pi\ell_r}{2n+1}\right)\right).$$ - -Questions. This is based on experimental assesment. -(a) Is $K_r(n)$ always an integer? -(b) Is there perhaps a higher-dimensional combinatorial interpretation of Kasteleyn for $K_r(n)$? -(c) Why do $K_r(n)$ feature "small primes" with high-power factorizations? For example, $K_3(2)=3^2(19)^3,\,\, K_3(3)=3^35^6(83)^3(97)^3\,\, K_3(4)=2^63^{34}(17)^6(19)^6(37)^6\,\,$ and - $$K_3(5)=3^5(43)^6(1409)^3(2267)^3(2707)^3(3719)^6(3389)^6.$$ - -I'm not aware of such a generalization, but any reference would be appreciated. -Thanks. -UDPATE The comments have answered (a). Is there a more direct (elementary) proof? Any suggestions for parts (b) and (c)? - -REPLY [2 votes]: The fact that the values $K_2(n)$ in Kastelyn's formula are rational integers is not so amazing (once we see the Galois Theory proof). It is amazing that that they count what they do. However that same integrality proof applies to simpler formulas. -It is worth mentioning that from $$\prod_{j=1}^n4\cos^2\left(\pi j/(2n+1)\right)=1$$ we also have $$\prod_{j=1}^n2\cos\left(\pi j/(2n+1)\right)=1$$ We can't simply use the fact that $$K_r(n):=\prod_{\ell_1=1}^n\cdots\prod_{\ell_r=1}^n\left( -4\cos^2\left(\frac{\pi\ell_1}{2n+1}\right)+\cdots+4\cos^2\left(\frac{\pi\ell_r}{2n+1}\right)\right) \in {\large \mathbb{Z}}$$ to see that also -$$J_r(n):=\prod_{\ell_1=1}^n\cdots\prod_{\ell_r=1}^n\left( -2\cos\left(\frac{\pi\ell_1}{2n+1}\right)+\cdots+2\cos\left(\frac{\pi\ell_r}{2n+1}\right)\right) \in {\large \mathbb{Z}}.$$ However the same reasoning which gives $K_r(n) \in \mathbb{Z}$ also gives $J_r(n) \in \mathbb{Z}.$ And we have yet to see that there is any reason to favor $K_r(n)$ over $J_r(n)$ othet than that we know an amazing fact about $K_2(n).$ -The factors $J_r(n)$ would seem to have as much reason to be small as those of $K_r(n).$. Of course one would be more confident asserting that after computing some of the values, which I have not done.<|endoftext|> -TITLE: Second page of Leray-Serre spectral sequence for fiber bundles -QUESTION [5 upvotes]: Given a fiber bundle $(E,B,p,F)$ with path connected base $B$ and fiber $F$, both closed smooth manifolds of finite dimensions. The second page $E_2^{p,q}$ of the Leray-Serre spectral sequence over $\mathbb{Z}_2$ is give by $H^p(B;\mathcal{H}^q(F;\mathbb{Z}_2))$, where $\mathcal{H}^q(F;\mathbb{Z}_2)$ is the local system (sheaf) of $\mathbb{Z}_2$-vector spaces on $B$ given by $\mathcal{H}^q(F;\mathbb{Z}_2)|_b=H^q(p^{-1}(b);\mathbb{Z}_2)$ for any $b\in B$. In the case this local system is trivial, namely, $\pi_1(B)$ acts trivially on it, we have $$E_2^{p,q}=H^p(B;\mathcal{H}^q(F;\mathbb{Z}_2))=H^p(B;\mathbb{Z}_2)\otimes H^q(F;\mathbb{Z}_2).$$ -But if the local system is nontrivial, is there still any relation between two vector spaces $E_2^{p,q}$ and $H^p(B;\mathbb{Z}_2)\otimes H^q(F;\mathbb{Z}_2)$? More generally, is it true that the bi-graded $\mathbb{Z}_2$-algebra $E_2^{*,*}$ is isomorphic as bi-graded $\mathbb{Z}_2$-algebras to $\left(H^*(B;\mathbb{Z}_2)\otimes H^*(F;\mathbb{Z}_2)\right)/I$ for some ideal $I$? - -REPLY [2 votes]: Let me first explain why your intuition is wrong, then let me provide a counterexample. You claim that "It is hard to imagine that $E_2^{*,*}$ has generators that cannot be expressed as a polynomial in generators of $H^*(B;\mathbb Z_2)$ and $H^∗(F;\mathbb Z_2)$". But it is not so hard to imagine, as the generators of $H^∗(F;\mathbb Z_2)$ in general do not correspond to any elements of $E_2^{*,*}$ at all. If they are not there, obviously elements that are there do not correspond to polynomials in them. -There are maps, but in the other direction, from $H^0(B;\mathcal H^q(F, \mathbb Z_2))$ to $H^q(F,\mathbb Z_2)$, but those don't help you. -A counterexample is provided by any space where $H^0(B;\mathcal H^1(F, \mathbb Z_2))$ vanishes but $H^1(B;\mathcal H^1(F, \mathbb Z_2))$ is nonzero. Then there is no such ideal $I$, because it would have to contain all of $H^1( F;\mathbb Z_2)$ but not all of $H^1(B;\mathbb Z_2) \otimes H^1(F,\mathbb Z_2)$. -We can do this using a suitable $T^2$-bundle on a Riemann surface $S$of genus $g \geq 2$. (If you used $\mathbb Q$-coefficients, an $S^1$-bundle would suffice.) Indeed, we can choose the torus bundle so the monodromy representation of $\pi_1(S)$ on $H^1(T_2,\mathbb Z)$ (and hence $H^1(T_2,\mathbb Z_2)$) factors through a cyclic subgroup of order $3$ acting by an order $3$ element of $GL_2(\mathbb Z)$. So $\mathcal H^1(F,\mathbb Z_2)$, as a sheaf, comes from a nontrivial irreducible two-dimensional representation of $\pi_1$. Then there are no global sctions of $\mathcal H^1(F,\mathbb Z_2)$, so $H^0(S, \mathcal H^1(F,\mathbb Z_2)) =0$, but by the Euler characteristic formula -$$\dim H^0(S, \mathcal H^1(F,\mathbb Z_2)) - \dim H^1(S, \mathcal H^1(F,\mathbb Z_2)) + \dim H^2(S, \mathcal H^1(F,\mathbb Z_2)) =2 (2-2g)$$ -we can see that $H^1(S, \mathcal H^1(F,\mathbb Z_2)) $ is nonvanishing.<|endoftext|> -TITLE: Cardinality of the set of functions commuting with $f:X\to X$ -QUESTION [13 upvotes]: If $X$ is an infinite set and $f:X\to X$, do we have $$\big|\{g:X\to X: g\circ f =f \circ g\}\big|\geq |X| ?$$ - -REPLY [2 votes]: This isn't a proof for the question itself, but a demonstration that you don't need to assume $X$ infinite. From now on, assume $X$ finite. -Assume $X$ connected, that is, $\forall x, y \exists m, n f^m(x) = f^n(y)$. By the pigeonhole principle, we have some $j \neq k$ with $f^j = f^k$, so eventually $\text{Im} f^i$ is constant. Let $X_0$ be that image. -Claim 1: $X_0$ is a loop. In other words, $\forall x, x' \in X_0 \exists n f^n(x) = x'$ -Proof: As $X_0$ is the eventually constant image, we have that $f(X_0) = X_0$, so $f|_{X_0}$ is a bijection. Therefore, there is some $N$ with $f^N|_{X_0} = \text{Id}|_{X_0}$. Combining this with connectedness gives the claim. -Inductively let $X_i = f^{-1}(X_{i - 1})$. By the definition of $X_0$, we have that $X = \cup X_i$. Then there is a smallest $i_0$ with $X_{i_0} = X$. Choose $x_i \in X\backslash X_{i_0 - 1}$. -Claim 2: For any $y \in X$, let $m_y$ be the smallest integer such that there is some $n$ with $f^{m_y}(y) = f^n(x_0)$. Then $n \geq m_y$ for all such $n$. -Proof: Assume $m_y > n$. For some smallest $j$, we have $f^n(x) \in X_j$. Then $x_0 \in X_{j + n}$. By the definition of $x_0$, we must also have that $y \in X_{j + n}$, so $f^n(y) \in X_j$. But as $m_y > n$, we must have that $f^{m_y}(y) \in X_{j - 1}$ or that $j = 0$. By the assumption that $j$ is the smallest, we must have that $j = 0$, so $f^n(y) \in X_0$. But then there is some $n'$ such that $f^{n'}(x) = f^n(y)$, by claim 1, which contradicts the assumption. Therefore, the current claim is proved. -Claim 3: For any $z \in X$, there is a map $g: X \rightarrow X$ such that $g_z \circ f = f \circ g_z, g_z(x_0) = z$. -Proof: For $y \in X$, define $n_y$ as the smallest integer such that $f^{m_y}(y) = f^{n_y}(x_0)$, and define $\iota_{x_0}(y) = n_y - m_y$. By claim 2, $\iota_{x_0}(y) \geq 0$. Then for $z \in X$, define $g_z(y) = f^{\iota_{x_0}}(z)(x_0)$. For $y = x_0$, we have $m_y = n_y = 0$, so $g_z(x_0) = z$. -I claim that $g_z \circ f = f \circ g_z$. -If $m_y \neq 0$, then $m_{f(y)} = m_y - 1, n_{f(y)} = n_y$. We therefore only need to check the case where $m_y = 0$. -If $m_y = 0$, then $y = f^{n_y}(x_0)$, so $f(y) = f^{n_y + 1}(x_0)$. If $n_{f(y)} = n_y + 1$, this is clear: $g_z(f(y)) = f^{n_y + 1}(z) = f(f^{n_y}(z)) = f(g_z(y))$, so we only need to check for $y$ such that $n_{f(y)} \neq n_y + 1$. -It's clear that this can only happen if $f(y) = f^i(x_0)$, recalling that $i$ is the smallest number such that $X_i = X$, and if $y \in X_0$ is the element that precedes it in the "loop". Then $y = f^{p - 1}(f(y))$, where $p$ is the length of the loop. But then $g(y) = f^{i + p - 1}(z)$, so $f(g(y)) = f^p(f^i(z)) = f^i(z) = g(f(y))$ by the definition of $i$ and the loop. So we are done. -Now consider $X$ not connected. We proceed by induction. Let $Y$ be a component of $X$, and $X'$ the rest of $X$. Assume $Y$ consists of a single element $\{y\}, f(y) = y$. Then we know that $|\{g: X' \rightarrow X'|g \circ f = f \circ g\} \geq |X'|$. For each such $g$, we can construct $g': X \rightarrow X$ such that $g'|_{X'} = g, g'(y) = y$; by inspection, this commutes with $y$. We also have the constant map $g_0: g_0(x) = y$. This gives us enough distinct maps that commute with $f$. -Assume neither $Y$ nor $X'$ consists of a single element. Then $\{g: X \rightarrow X|g \circ f = f \circ g\} \supset \{g_1: X' \rightarrow X', g_2: Y \rightarrow Y| g_1 \circ f = f \circ g_1, g_2 \circ f = f \circ g_2\}$, so by induction, $|\{g: X \rightarrow X|g \circ f = f \circ g\}| \geq |\{g: X' \rightarrow X'|g \circ f = f \circ g\}| |\{g: Y \rightarrow Y|g \circ f = f \circ g\}| \geq |X'||Y| \geq |X'| + |Y| = |X|$. -Edit: To picture this, define by a "lasso" a finite set $L = \{f^n(x)\}$. Lassos are characterized by two numbers: the length of the loop, and the length of the "rope" before the loop. Effectively, this proof "collapses" $X$ to the lasso with the longest rope, then shows that for any $i, i', j, j'$ such that $j'|j, i' \leq i$ there is a "lasso map" $L_{ij} \rightarrow L_{i'j'}$.<|endoftext|> -TITLE: What does $K_1(R)$ tell us about $GL_n(R)/E_n(R)$? -QUESTION [7 upvotes]: Let $D$ be a division ring, and $R=D[t_1,\ldots,t_n]$. If $GL_m(R)$ is the usual group of invertible matrices over $R$, then by $E_m(R)$ I mean the subgroup of $GL_m(R)$ generated by the elementary matrices of the form $I_m+r\epsilon(i,j)$, $r\in U(R)$. It is known that the whitehead group $K_1(R)$ of $R$ is isomorphic to the commutator quotient group of $U(D)$. Is there any way we can extract from this, anything about $GL_m(R)/E_m(R)$? - -REPLY [7 votes]: In general what you are asking about is the problem of $K_1$-stabilization, that is the study of maps $GL(n, R)/E(n, R)\to GL(n+1, R)/E(n+1, R)$. For rings of finite dimension all such maps are bijective starting with some $n$ depending on the dimension. -The first relevant link is "On the stabilization of general linear group" by L. Vaserstein. I believe that any paper on the subject cites this one, so you can explore them by following the link. -For the particular problem about polynomials over a division ring, take a look at "The general linear group of polynomial rings over regular rings" by T. Vorst. Here's the abstract: - -In this note we shall prove for two types of regular rings $A$ that every element of $GL_r(A[X_1,\ldots,X_n])$ is a product of an element of $E_r(A[X_1,\ldots,X_n])$ (the group of elementary matrices) and an element of $GL_r(A)$, for $r\geq3$ and $n$ arbitrary. This is a kind of $GL_r$-analogue of results of Lindel and Mohan-Kumar and is an extension of a result of Suslin.<|endoftext|> -TITLE: Sum of the ratios of Schur functions -QUESTION [7 upvotes]: There are simple expressions for the sums of linear and quadratic combinations of Schur functions over all partitions (including the empty one) -$$ -\sum_\lambda s_\lambda(x)=\prod_{i}\frac{1}{1-x_i}\prod_{i -TITLE: Diophantine equation for 2016: triangular $|{\rm GL}_2({\bf F}_q)|$ -QUESTION [38 upvotes]: For a prime power $q$ the group ${\rm GL}_2({\bf F}_q)$ has -$(q^2-1)(q^2-q)$ elements. This happens to be a triangular number for -$q=2$ (being $6 = 1+2+3$), and $-$ more notably, especially this year :-) $-$ -for $q=7$ (being $1+2+3+\cdots+63 = 2016$). Indeed $2016$ is surely -the largest triangular number of the form $(q^2-1)(q^2-q)$ for $q \in \bf Z$, -even without the assumption that $q$ be a prime power. (The others are -$300 = 1+2+3+\cdots+24$, for $q=-4$, and the trivial $0$ for $q=0$ and -$q=\pm1$.) -As far as I know this question has no serious mathematical -significance or application. But it seems surprisingly hard to prove that -$2016$ is the last such number. I know how to do it, but not easily; -this Diophantine problem might be just a bit beyond the available tools -for an elementary proof, and thus a good test case for extending those tools. - -How accessible / elementary a proof can we give for the fact that - $-4,-1,0,1,2,7$ are the only integers $q$ for which - $(q^2-1)(q^2-q)$ is triangular (equivalently: for which - $8(q^2-1)(q^2-q)+1$ is a square)? - -Here's what I know about this. -For starters, there are only finitely many such $q$ by -Siegel's -theorem -on integral points, because the equation $(q^2-1)(q^2-q) = (s^2-s)/2$ -gives a curve $C$ of positive genus $1$. But not only is Siegel's theorem -far from elementary, it's also ineffective: -one needs several large solutions to prove that no further solution exists, -and usually there isn't even one solution that's large enough. -(Or is there a proof of Siegel's theorem for which -the $q = -4$ and $q=7$ solutions of our equation -are enough to prove that the list of solutions is complete?) -At this point we're tempted to throw $C$ into a standard routine for -finding integral points on an elliptic curve via Baker's effective bounds. -But those bounds are even farther from elementary than Siegel's theorem. -Moreover, they do not immediately apply here, because our curve $C$ -is not in Weierstrass form $y^2 = x^3 + ax^2 + bx + c$. -Now Magma does have a function - -IntegralQuarticPoints -for finding all integer points on a curve $y^2 = Q(x) = ax^4+bx^3+cx^2+dx+e$ -given also a rational point. Here we have $x=q$ and $Q(x) = 8(q^2-1)(q^2-q)+1$, -and we already know some rational points such as $(x,y)=(0,\pm 1)$; -and indeed the input -time IntegralQuarticPoints([8,-8,-8,8,1],[0,1]) - -yields output ending with -[ - [ 7, 127 ], - [ 2, -7 ], - [ -4, 49 ], - [ 0, -1 ], - [ -1, 1 ], - [ 1, 1 ] -] -Time: 0.460 - -(the Time is reported in seconds). -But I don't know what Magma did to check that there are no other -$q \in \bf Z$, and thus don't know how accessible the resulting proof might be. -Since $C$ is a genus-$1$ curve with rational points, -we can transform $C$ birationally into Weierstrass form; -for example, starting from the rational point $(q,s)=(0,0)$, where -$s = -2q + O(q^2)$, we may write $s = -2q + (x+1)q^2$ to obtain -$$ -(x^2+2x-1)q^2 - (4x+2) q + 5-x = 0, -$$ -and this quadratic in $q$ has discriminant $4(x^3+x^2-7x+6)$ $-$ -so we have identified $C$ with the curve -$$E : y^2 = x^3 + x^2 - 7x + 6.$$ -But this transformation does not take integral points of $E$ to -integral points of $C$. (Geometrically, integrality depends -on both the curve and the point(s) at infinity, and here we have -the same curve but different distinguished points.) -We can still use this Weierstrass form $E$ -to compute the rational points of $C$. If the group of rational points -were finite we would be done; if it had rank $1$ we could usually, -with more effort, use $p$-adic techniques to find all the integral points -on $C$. (It may be a stretch to call these techniques elementary $-$ -and even computing the rank of $E$ requires at least the arithmetic of -a cubic number field, because the polynomial $x^3 + x^2 - 7x + 6$ -is irreducible $-$ but this still feels like a much smaller hammer than -the Baker bounds.) Unfortunately mwrank finds that $E$ has rank $2$, -with generators $(x,y) = (1,1)$ and $(-3,3)$. -Magma reported this too en route to computing its list of integral points, -which must be why the input had to include some rational point of $C$. -(This curve has conductor $2^3 331 = 2648$, -and is listed in the LMDFB as elliptic curve -2648.a1.) -That leaves me with two backup options: -First, we can apply Baker over the quadratic field $K = {\bf Q}(\sqrt 2)$. -The two points at infinity of $C$ are defined over $K$, so we can choose -one of them as the origin of the group law to identify $C$ with $E$ over $K$, -and then integral points of $C$ do go to integral points in $E(K)$. -[There can be yet more integral points, because we're ignoring the -other point at infinity of $C$; but those spurious can be eliminated at the end, -or ignored using the condition that our point in $E(K)$ -must come from $C({\bf Q})$. -The group $E(K)$ has rank $3$, since it contains with finite index -the direct sum of $E({\bf Q})$ and $E_2({\bf Q})$ where -$E_2$ is the quadratic twist $2y^2 = x^3 + x^2 - 7x + 6$, -isomorphic with the rank-$1$ LMFDB curve -21184.d1 -of conductor $2^6 331$.] -Second $-$ and possibly requiring only a smaller "hammer" $-$ we can -reduce to some quartic Thue equations over $K$. In the course of -computing $E({\bf Q})$ and the relevant part of $E(K)$, we usually -do a $2$-descent, and thus find a finite list of rational functions -$f_i = N_i/D_i \in K(X)$ of degree $4$ such that every $(x,y) \in C({\bf Q})$ -has $x = f_i(x')$ for some $i$ and $x' \in K$. At this point we don't -really need to finish computing $E(K)$ by searching for rational points, -because it's enough to find all integral values of each $f_i$ $-$ -and this amounts to solving a finite list of quartic Thue equations -$\Delta_i(t,u) = d$, where $\Delta_i$ is a homogeneous quartic corresponding -to $D_i$, and the variables $t,u$ and target value $d$ are in -$O_K = {\bf Z}[\sqrt 2]$. With luck all of those equations might be solvable with Skolem's $p$-adic method instead of invoking Baker. -But that is still a long way from elementary, and will take orders of magnitude -more than $0.460$ seconds... - -Is there a simpler proof? - -REPLY [6 votes]: This solution works iff $q$ is a prime power, so it's not a full solution, but I guess it's better than nothing, as it is completely elementary. -If $q$ is a prime power, then $q$ must fully divide either $s$ or $s-1$. We may WLOG assume it divides $s$ (since otherwise we may take $t=1-s$ as $s$). Then we have that, letting $s=mq$, -$$2(q-1)^2(q+1)=m(mq-1).$$ -Taking $\bmod q$, we have that $m\equiv -2\bmod q$, so we may set $m=(x+1)q-2$, giving $s=-2q+(x+1)q^2$ and -$$(x^2+2x-1)q^2 - (4x+2) q + (5-x) = 0,$$ -as noted in the question. From this, we have $x\equiv 5\bmod q$. Let $x=nq+5$. Then -$$s=-2q+(nq+6)q^2=nq^3+6q^2-2q=6q^2+q(nq^2-2).$$ - -We will prove the following lemma: -Lemma: If $q$ is a nonzero integer and -$$(2s-1)^2 = 8(q-1)^2q(q+1)+1,$$ -then $|s|<3q^2.$ -Proof: First, note that -$$3x^4-(x-1)^2x(x+1)=2x^4+x^3+x^2-x=(2x^2-x)(x^2+x+1)$$ -is positive for all $x$ outside of $[0,1/2]$, and specifically is positive if $x$ is a nonzero integer. Thus, -$$(x-1)^2x(x+1)<3x^4.$$ -$$(2s-1)^2=8(q-1)^2q(q+1)+1<24q^4+1\leq 25q^4.$$ -So, -$$|2s-1|<5q^2.$$ -$$|2s-1|+|1|<5q^2+1\leq 6q^2.$$ -$$2|s|< 6q^2,$$ -implying the result. - -If $n=0$, then $s=-2q+6q^2$, and we have -$$(6q^2-2q)(6q^2-2q-1)=2(q-1)^2q(q+1).$$ -This quartic has a triple root at $q=0$ and a single root at $q=11/17$, so we can safely ignore this case. Thus, $|n|\geq 1$. If $|q|\geq 11$, then $|nq|\geq 11$. So, -$$|nq^2|\geq 11|q|.$$ -$$|nq^2|\geq 9|q|+2.$$ -$$|nq^2-2| \geq 9|q|.$$ -But -$$s=6q^2+q(nq^2-2).$$ -So, since $|nq^2-2|\geq 9|q|$, we have that $|s|\geq 3q^2,$ which contradicts our lemma. Thus, we have reduced the problem to a finite case check on integers $-10\leq q\leq 10$, which can easily be done.<|endoftext|> -TITLE: Who first proved that we can prove that we prove things we prove? -QUESTION [16 upvotes]: Sorry about the title, I couldn't resist. -It's a classic fact that, not only does $PA$ prove every true $\Sigma_1$ sentence, but $PA$ proves that $PA$ proves every true $\Sigma_1$ sentence! In particular, restricting attention to $\Sigma_1$ sentences of the form "$PA$ proves ---", in the modal logic of $PA$-provability we have $$\Box(\Box p\implies \Box\Box p).$$ -Indeed, even more is true: in the paper Oracle bites theory, Visser states - -It is well known that, in the context of EA, all theories extending - the very weak arithmetic R prove all true $\Sigma_1$-sentences. - -And various proofs of these facts can be found in various places. -My question is: who first proved (and where) that a strong enough theory of arithmetic proves every true $\Sigma_1$ sentence, and moreover proves that it proves every true $\Sigma_1$ sentence? - -REPLY [9 votes]: The theorem $\Box P \Rightarrow \Box\Box P$ is due to Martin Löb and first appears in his 1955 paper "Solution of a Problem of Leon Henkin", J. Symb. Logic 20 115–118: it appears as condition (V) (page 116) in the paper in question, and whereas conditions (I)–(IV) are referred there to the earlier (1939) book by Hilbert and Bernays, Grundlagen der Mathematik, condition (V) (although easily deduced from the others) is new. -The reasoning "that a strong enough theory of arithmetic proves every true $\Sigma_1$ sentence" is exactly the one which Löb uses in his proof (if we grant that "$\exists x.(f(x)=0)$" for a recursive $f$ qualifies as "every $\Sigma_1$ sentence"). -As evidence that Löb was the first to state this fact, I offer the following quote from G. Boolos in The Logic of Provability (1995), chapter 2: "Hilbert and Bernays had listed three somewhat ungainly conditions […]. The isolation of (the attractive) (i), (ii), and (iii) [essentially as on Wikipedia] is due to Löb." As well as the entry "Provability Logic" from the Stanford Encyclopedia of Philosophy: "In the same paper, Löb formulated three conditions on the provability predicate of Peano Arithmetic, that form a useful modification of the complicated conditions that Hilbert and Bernays introduced in 1939 for their proof of Gödel's second incompleteness theorem [again the same conditions as on Wikipedia]".<|endoftext|> -TITLE: Regular elementary abelian subgroups of primitive permutation groups -QUESTION [9 upvotes]: A finite group $B$ is said to be a B-group if every primitive permutation group having a regular (transitive) subgroup isomorphic to B is $2$-transitive. -Schur proved that a cyclic group of composite order is a B-group. Wielandt showed that no group of the form $B_1 \times \cdots \times B_d$ with $|B_1| = \ldots = |B_d| \ge 3$ and $d \ge 2$ is a B-group. (See Theorems 25.3 and 25.7 in Wielandt, Finite permutation groups.) -Exercise 3.5.6 in Dixon and Mortimer Permutation groups asks for a proof that no elementary abelian $p$-group is a B-group. However this is false: the permutation groups of degree $4$ containing $C_2 \times C_2$ as a regular subgroup are $\langle (12)(34), (13)(24)\rangle$ and $\mathrm{Dih}(4)$ (both imprimitive) and $A_4$, $S_4$ (both $2$-transitive). My question is: - -For which $d \in \mathbb{N}$ is the elementary abelian group $C_2^d$ a B-group? - -Necessary conditions for a primitive permutation groups to have a regular subgroup are given in Corollary 3 of Liebeck, Praeger, Saxl, Transitive Subgroups of Primitive Permutation Groups, J. Alg 234, 291–361 (2000). The main theorem of Li, The finite primitive permutation groups containing an abelian regular subgroup, Proc. London Math. Soc. 87, 725–747 (2003), gives a sharper result. It seems non-trivial to use either of these papers to answer the question, but I'd welcome a correction. -If $H \le \mathrm{GL}(\mathbb{F}_2^d)$ then the subgroup $\mathbb{F}_2^d \rtimes H$ of the affine general linear group $\mathrm{AGL}(\mathbb{F}_2^d)$ is primitive if and only if $H$ is irreducible. So if there exists an irreducible $H$ acting intransitively on $\mathbb{F}_2^d \backslash \{0\}$, then $C_2^d$ is not a B-group. Such groups exist whenever $d$ is composite, but not always when $d$ is prime. For example, the three irreducible subgroups of $\mathrm{GL}(\mathbb{F}_2^3)$ all contain a Singer element of order $7$, so give $2$-transitive subgroups of $\mathrm{AGL}(\mathbb{F}_2^3)$. (The only other primitive permutation groups of degree $8$ containing a subgroup isomorphic to $C_2 \times C_2 \times C_2$ are $A_8$ and $S_8$, so it follows that $C_2 \times C_2 \times C_2$ is a B-group.) This motivates my second question: - -For which primes $p$ is there an irreducible subgroup $H$ of $\mathrm{GL}(\mathbb{F}_2^p)$ such that $H$ acts intransitively on $\mathbb{F}_2^p \backslash \{0\}$? - -REPLY [2 votes]: This post gives a complete proof of the final box in Nick Gill's excellent answer. Let $V = \mathbb{F}_2^n$.$\newcommand{\GL}{\mathrm{GL}}$ - -Theorem. $C_2^n$ is a B-group if and only if $2^n-1$ is a Mersenne prime and the only simple groups having an $n$-dimensional irreducible representation over $\mathbb{F}_2$ are $C_{2^n-1}$ and $\GL(V)$. - -The key results we need are - A transitive permutation group of prime degree is either solvable (and so affine) or $2$-transitive; this is due to Burnside. - - If $H \le \GL(V)$ is $2$-transitive then either $n=4$ and $H \cong A_7$ or $H = \GL(V)$; this was proved by Cameron and Kantor in 1979. - - An almost simple group of degree $2^n$ is $2$-transitive; this follows from Guralnick's Theorem in Nick's answer. - -Proof. Suppose $2^n-1$ is composite. Let $g \in \mathrm{GL}(V)$ be a Singer element of order $2^n-1$. By Zsigmondy's Theorem, there is a power $g^r$ such that $\langle g^r \rangle$ acts irreducibly but intransitively on $V$. (This was noted by Derek Holt in a comment on the question.) Hence $V \rtimes \langle g^r \rangle$ is simply transitive and primitive and so $C_2^n$ is not a B-group. -Conversely, suppose $2^n-1$ is prime and that $G$ is a primitive permutation group of degree $2^n$. By the O'Nan–Scott theorem, $G$ is either almost simple, or of affine type, or of diagonal type, or of twisted wreath or product type. Following Nick's answer, in the final two cases, we have $G \le S_a \wr S_b$ with $a^b = 2^n$. If $a=2^m$ then $n= mb$. But $n$ is prime, so this is impossible. The diagonal case does not arise because $2^n$ is not the order of a simple group. By (3) if $G$ is almost simple then $G$ is $2$-transitive. -We are left with the affine case. Suppose that the simple group $S$ has an irreducible $n$-dimensional representation over $\mathbb{F}_2$. Then $V \rtimes S$ is primitive and simply transitive, so $C_2^n$ is not a B-group. -Finally, suppose that $C_2^n$ is not a B-group. We must find a proper non-cyclic irreducible simple subgroup $S$ of $\GL(V)$. We are in the affine case, so there exists an irreducible $H \le \GL(V)$ such that $H$ is intransitive on $V \backslash \{0\}$. Let $M$ be a maximal subgroup of $\GL(V)$ containing $H$. The 10 irreducible Aschbacher classes all involve groups preserving a decomposition of $V$ that can exist only when $\dim V = n$ is composite. Therefore $M$ is almost simple. -If $M$ is transitive on $V \backslash \{0\}$ then, since $|V \backslash \{0\}| = 2^n-1$ is prime, (1) implies that either $M \cong C_{2^n-1} \rtimes C_s$ for some $s$ dividing $2^n-1$, or $M$ is $2$-transitive. In the first case, since $H$ is irreducible, $H \ge C_{2^n-1}$. In the remaining case $V \rtimes M$ is $3$-transitive, and by (2), $M = \GL(V)$, and so $H = \GL(V)$. In both cases this contradicts the intransitivity of $H$. Hence we did not lose simple transitivity in passing from $H$ to $M$. Let $S$ be the simple subgroup of $M$. By Clifford's Theorem, the restriction of the $\mathbb{F}_2 M$-module $V$ to $S$ is a direct sum of conjugate irreducible submodules. But $\dim V = n$ is prime. Hence $S$ acts irreducibly. Since $M$ is intransitive on $V \backslash \{0\}$, so is $S$. Hence $S$ is as required. $\Box$ -Remark. It seems plausible that the only simple groups having an $n$-dimensional irreducible representation over $\mathbb{F}_2$ for $n$ such that $2^n-1$ is a Mersenne prime are $C_{2^n-1}$ and $\GL(V)$. From the modular Atlas data in GAP, the only sporadic simple groups that could be exceptions are $J_4$, $Ly$, $Th$, $Fi_{24}$, $B$ and $M$. Tables of small dimensional representations of quasi-simple groups in non-defining characteristic and Chevalley groups in defining characteristic, show there are no exceptional representations if $n \le 250$. So if $n \le 250$ then $C_2^n$ is a B-group if and only if -$$ n \in \{1,2,3,5,7,13,17,19,31,61,89,107,127\}. $$ -Edit. I had to edit this answer to remove the entirely false claim in the final remark that all irreducible representations of alternating groups over $\mathbb{F}_2$ are self-dual (and so have an invariant symplectic form and are even dimensional). The isomorphism $A_8 \cong \mathrm{GL}_4(\mathbb{F}_2)$ shows this is false.<|endoftext|> -TITLE: in search of a transformation between determinants -QUESTION [13 upvotes]: Motivated by this MO question. Consider the two matrices $A_n$ and $B_n$ with entries $\binom{2j}i$ and $\binom{n+1}{2j-i}$, respectively; for $1\leq i, \,j\leq n$. -I can show $\det A_n=\det B_n=2^{\binom{n+1}2}$; it is not a problem (see, for instance, my answer here). - -Question. -(1) Is there a transformation converting $A_n$ into $B_n$ (or vice-versa)? -(2) Is there a combinatorial interpretation/bijection revealing the counts by $\det A_n$ and $\det B_n$? - -My asking for (2) is due to the fact that $2^{\binom{n+1}2}$ enumerates domino tilings of an Aztec diamond. - -(3) Let $s(k)=$the number of $1$'s in the binary expansion of $k$. Then, $A_n$ and $B_n$ share the same Smith normal form (showing the diagonal vector) given by -$$[2^{\max(4k-2n+s(n-k)-s(k),0)}:\, 1\leq k\leq n].$$ -This claim is based on data from Noam Elkies' comments seen below. Any proof? - -Remark. Let $\lceil x\rceil=$the smallest integer greater than or equal to $x$ (ceiling function). So, (3) implies -$$\sum_{k=1}^n\max(4k-2n+s(n-k)-s(k),0)=\frac{n(n+1)}2;$$ -Or, equivalently $\sum_{k=\lceil\frac{n}2\rceil}^n(s(k)-s(n-k))=\lceil\frac{n}2\rceil$. - -REPLY [9 votes]: There is such a transformation, of the form predicted in Linear transformation that preserves the determinant. -Denoting $R$ the involution matrix $e_i\mapsto e_{n+1-i}$, it turns out that the matrix $A$ has an $LU$-decomposition in which $U:=\left[{j\choose i}\right]_{{1\le i\le n}\atop{1\le j\le n}}$, and the lower triangular part is $L=RCR$, where $C$ is the upper triangular matrix in Suvrit's decomposition , -$B=VCV^{-1}$ (warning: $B$ of this question is named "$A$" there). So $B= (VR) A(VRU)^{-1}$, with $\det(VR)=\det(VRU)^{-1}=1$. -$$*$$ -[edit] The description becomes a bit more gracious if we include the indices $i=0$ and $j=0$. So, if we define the $n\times n$ matrices with integer entries -$$A_n:=\left[{2j\choose i}\right]_{{0\le i< n}\atop{0\le j< n}}\qquad B_n:=\left[{n\choose 2j-i}\right]_{{0\le i< n}\atop{0\le j< n}}$$ -$$U_n:=\left[{j\choose i}\right]_{{0\le i< n}\atop{0\le j< n}}\qquad L_n:=\left[2^{2j-i}{j\choose2j- i}\right]_{{0\le i< n}\atop{0\le j< n}}$$ -$$N_n:=\Big[ \delta_{i+1,j}\Big]_{{0\le i< n}\atop{0\le j< n}}\qquad R_n:=\Big[ \delta_{n-i,j}\Big]_{{0\le i< n}\atop{0\le j< n}}$$ -Then, (hiding the subscript $n$) -$$A=LU$$ -and -$$B=VCV^{-1}$$ -with -$$V:=U^{T}R\qquad C:=(I+N)AU^{-1}\ .$$<|endoftext|> -TITLE: Automorphisms of the hyperreals over the rationals and nontrivial automorphism groups -QUESTION [19 upvotes]: A classic result says the automorphism group of $\mathbb{R}$ (over $\mathbb{Q}$) is trivial. The proof is simple: every automorphism preserves squares, and hence fixes the positive reals, so it must be order preserving. Since it must fix $\mathbb{Q}$, and $\mathbb{Q}$ is dense in $\mathbb{R}$, if any real number were not fixed, this would yield a contradiction. -In larger real-closed fields where $\mathbb{Q}$ is not dense, automorphisms are still order preserving, but the argument that they are trivial does not work. I haven't found any examples of a non-trivial automorphism of a real-closed field, but I also can't prove they don't exist. So, the first question I'd like to ask is whether all automorphism groups of real-closed fields are trivial. -If they aren't all trivial, then I want to know what, if anything, we can say. To ask a less vague question, can we classify the automorphisms of the hyperreals? How many are there, and what is their groups structure like? - -REPLY [10 votes]: This note complement Joel's, who pointed out that every first order theory with an infinite model has a model with many automorphisms (this was first proved by Ehrenfeucht and Mostowski in their famous 1956 paper on indiscernibles). Joel also outlined why saturated models and countable computably saturated ones have nontrivial automorphisms. -It is well-known that if $\cal{M}$ is computably saturated and countable, then as soon as we know that some element $m$ of $\cal{M}$ is not first order definable in $\cal{M}$, then we can build an automorphism of $\cal{M}$ that moves $m$. Indeed, we can build continuum-many such automorphisms. This makes for a fun exercise. -Furthermore, it was shown by F. Körner (1998) that if $\cal{M}$ is a countable structure in a countable signature, then there is an elementary extension of $\cal{M}$ that carries an automorphism that is "maximal", i.e., it moves every nonalgebraic element (she showed that $\cal{M}$ can be chosen as any arithmetically saturated elementary extension of M). -In the above, an element $m$ of $\cal{M}$ is said to be algebraic if there is a formula $\phi$ with one free variable such that $\phi(m)$ holds in $\cal{M}$ and $\phi$ has only finitely many solutions in $\cal{M}$. -Körner's theorem was generalized to all structures (whose universe and signature can have any cardinality) by G. Duby in 2003, with a different, more powerful proof technique. -Here are the References: -Friederike Körner, Automorphisms moving all non-Algebraic points and an application to NF, The Journal of Symbolic Logic,Vol. 63, No. 3 (Sep., 1998), pp. 815-830. -Grégory Duby, Automorphisms with only infinite orbits on non-algebraic elements, Archive for Mathematical Logic, July 2003, Volume 42, Issue 5, pp 435–447.<|endoftext|> -TITLE: The Sylvester-Gallai theorem over $p$-adic fields -QUESTION [31 upvotes]: The famous Sylvester-Gallai theorem states that for any finite set $X$ of points in the plane $\mathbf{R}^2$, not all on a line, there is a line passing through exactly two points of $X$. - -What happens if we replace $\mathbf{R}$ by $\mathbf{Q}_p$? - -It is well-known that the theorem fails if we replace $\mathbf{R}$ by $\mathbf{C}$: the set $X$ of flexes of a non-singular complex cubic curve has the property that every line passing through two points of $X$ also passes through a third. -For example, the flexes of the Fermat curve $C:X^3+Y^3+Z^3=0$ are given by the equation $XYZ=0$ and are all defined over the field $\mathbf{Q}(\zeta_3)$ generated by a cube root of unity $\zeta_3$. As a consequence, if a field $K$ contains $\mathbf{Q}(\zeta_3)$ then the set of flexes of $C$ gives a counterexample to the Sylvester-Gallai theorem over $K$. For example, for any prime $p \equiv 1 \pmod{3}$, the field $\mathbf{Q}_p$ contains $\mathbf{Q}(\zeta_3)$ so that Sylvester-Gallai fails over $\mathbf{Q}_p$. -I don't know what happens in the case $p=3$ or $p \equiv 2 \pmod{3}$. Note that the set of flexes of $C$ is not defined over $\mathbf{Q}_p$ anymore, but nothing prevents more complicated configurations of points giving counterexamples to the theorem over $\mathbf{Q}_p$. - -More generally, what happens over an arbitrary field $K$? Is it true that the Sylvester-Gallai theorem holds over $K$ if and only if $K$ does not contain the cube roots of unity? - -EDIT. David Speyer's beautiful example shows that the Sylvester-Gallai theorem fails over $\mathbf{Q}_p$ for any prime $p \geq 5$. Furthermore, regarding the problem of deciding whether SG holds over a given field $K$ (which looks like a difficult question, at least to me), this and Gro-Tsen's example show that the condition that $K$ does not contain the cube roots of unity clearly needs to be refined. In order for SG to hold over a characteristic $0$ field $K$, it is necessary that $K$ does not contain any root of unity of order $\geq 3$. I don't know whether this is also a sufficient condition. - -REPLY [31 votes]: If $n \geq 3$ and $K$ is a field of characteristic not dividing $n$, containing a primitive $n$-th root of unity $\zeta$, then the $3n$ points of the form $(1:-\zeta^a:0)$, $(0:1:-\zeta^b)$, $(-\zeta^c:0:1)$ are a Sylvester-Gallai configuration. In particular, taking $n=p-1$, this gives an SG configuration over $\mathbb{Q}_p$ for $p \geq 5$.<|endoftext|> -TITLE: What is the error in this disproof of the $\Omega$-conjecture? -QUESTION [18 upvotes]: I was reading about the $\Omega$-conjecture and have thought of a refutation of it, which seems too simple to not have been noticed since the $\Omega$-conjecture has been around, so i'm skeptical and want to see whether anyone can spot a flaw in it. -I will assume there is a proper class of hyper-huge cardinals. This assumption implies a proper class of Woodin cardinals and by Usuba implies the existence of the bedrock model $W$, and since $V$ is a set-generic extension of $W$ and our assumption (as well as the $\Omega$-conjecture) is invariant throughout the set-generic multiverse, we can work in $W$. Suppose for a contradiction that the $\Omega$-conjecture holds. Since $V = W$ is $\Sigma_2$ definable in any set-forcing $V[G]$, we can uniformly evaluate $\Sigma_2$-truths of $V$ in any set-forcing of $V$ by other recursively given $\Sigma_2$ sentences. Since the $\Sigma_2$-laws of the set-generic multiverse are definable in $H(\delta_0^+)$ where $\delta_0$ is the least Woodin cardinal (by the $\Omega$-conjecture), and they can be used to compute all the $\Sigma_2$ truths of $V$ (including the theory of $H(I_0^{+})$ and beyond), this violates Tarski's undefinability of truth. Thus the $\Omega$-conjecture must fail assuming our large cardinal hypothesis. $\square$ -Is this argument really valid? (I'm worried 2016 will end by refuting $V = \text{Ultimate }L$) - -REPLY [15 votes]: Woodin's theorem says that assuming the $\Omega$ Conjecture and the existence of a proper class of Woodin cardinals, the set $\mathcal V_\Omega$ of $\Pi_2$ sentences that hold in every universe of the generic multiverse is lightface definable over $H_{\delta^+}$ where $\delta$ is the least Woodin cardinal. You claim there is a Turing reduction from the $\Pi_2$ theory of the bedrock to $\mathcal V_\Omega$, obtaining in this way a $\Sigma_2$ definition of $\Pi_2$ truth (assuming the Ground Axiom), a contradiction. The proposed reduction sends a formula $\phi$ to the sentence $f(\phi)$ expressing "The bedrock satisfies $\phi$." Tell me if I'm misunderstanding you. -One problem: it isn't clear that $f(\phi)$ is $\Pi_2$. Note that if $W$ is a $\Sigma_2$ or $\Pi_2$ or even $\Delta_2$ inner model and $\phi$ is a $\Pi_2$ sentence, the sentence $W\vDash \phi$ is not obviously $\Pi_2$. We can write $\phi$ as $$\forall x \ (x\notin W \vee \exists y\ (y\in W\wedge \psi(x,y))$$ for some $\Delta_0$ formula $\psi$. This seems to be no simpler than $\Pi_3$ even if $W$ is $\Delta_2$. -For example, we claim that the statement $$\Psi\equiv \text{CH}\text{ fails in the bedrock}$$ is not equivalent over ZFC + the Bedrock Axiom to a $\Pi_2$ formula, even though $\neg\text{CH}$ is $\Delta_2$. (The Bedrock Axiom just asserts that the generic multiverse has a bedrock.) Suppose towards a contradiction that this statement is equivalent to a $\Pi_2$ formula, which we may assume is of the form $\forall \alpha\ (V_\alpha\vDash \psi)$. Fix a model $$M\vDash\text{the Bedrock Axiom}+\neg\text{CH} + \neg\Psi$$ (so the bedrock of the generic multiverse of $M$ satisfies $\text{CH}$). Since $M\vDash \neg \Psi$, there is some ordinal $\alpha$ of $M$ such that $M_\alpha\vDash \neg \psi$. Now pass to a class forcing extension $M[H]$ satisfying the Ground Axiom and such that $M[H]_{\alpha} = M_{\alpha}$ and $M[H]_{\omega+\omega} = M_{\omega+\omega}$. (See Reitz's thesis, Theorem 12.) We have $M[H]\vDash \neg\text{CH}$ (since $M\vDash \neg \text{CH}$ and $M[H]_{\omega+\omega} = M_{\omega+\omega}$). On the other hand since $M[H]_\alpha = M_\alpha\vDash \neg \psi$, $M[H]\vDash \exists \alpha\ (V_\alpha\vDash \neg \psi)$. Thus $M[H]\vDash \neg \Psi,$ so the bedrock of $M[H]$ satisfies $\text{CH}$, which contradicts the fact that $M[H]$ is its own bedrock.<|endoftext|> -TITLE: Existence of Randomized polynomial time algorithm and some arithmetic analog of $ACC^0$ circuits for Factoring of primitive polynomials before LLL? -QUESTION [8 upvotes]: Before LLL came along in $1982$ there was no deterministic polynomial (in degree and number of bits in coefficients) way to factor square free primitive polynomials in $\Bbb Z[x]$. -However was there a probabilistic polynomial (in degree and number of bits in coefficients) way to factor square free primitive polynomials in $\Bbb Z[x]$? Is there a reference? - -The reason for the query is following. The LLL algorithm is highly iterative and in circuit complexity parlance seems to be polynomial size in $O(\log n)$ depth . So I am wondering if there was a randomized or deterministic algorithm which can factor at least a fraction of primitive polynomials in $\Bbb Z[x]$ and that which would still be in $O(1)$ depth and polynomial size. - -REPLY [3 votes]: The Zassenhaus algorithm (see the Wikipedia article on factoring) Is polynomial, except for the exponential dependence on the number of irreducible factors of the polynomial. However, the expected number of irreducible factors is $O(1)$ (see this nice UChicago REU by Alegra Juarez), so Zassenhaus is expected polynomial time. The worst case is horrible, though, and so the Novocin-van Hoeij algorithm (see Novocin's very nice description) cleverly circumvents the problems.<|endoftext|> -TITLE: Toda's book on homotopy groups of spheres -QUESTION [12 upvotes]: Yesterday one my friend told about recent book of Hiroshi Toda, where the computations of 3-torsion in homotopy groups of spheres are given up to a very high stem (about 75). This book is published with a rare publisher and is overlooked by MathSciNet and ZentralblattMath. Does anybody know the explicit reference: title, publisher, year? -Thanks. -Roma Mikhailov. - -REPLY [13 votes]: If I've got the right reference, this appears to be listed as: - -"Unstable 3-primary homotopy groups of spheres, Study of Econoinformatics (Vol. 29), Himeji: Himeji Dokkyo University" in the book "Gottlieb and Whitehead Center Groups of Spheres, Projective and Moore Spaces" by Marek Golasiński and Juno Mukai, -"Unstable 3-primary homotopy groups of spheres, Econoinformatics 29 (2003)" in the new edition of Ravenel's green book, and -"Unstable 3-Primary Homotopy Groups of Spheres, Faculty of Econoinformatics, Himeji Dokkyo University, 2003" in Guozhen Wang's thesis proposal. - -But I can't find it.<|endoftext|> -TITLE: A 4-manifold with Stiefel-Whitney classes $w_1\neq 0$, $w_3 \neq 0$, and $w_1^2=0$? -QUESTION [18 upvotes]: Is there a 4-manifold whose Stiefel-Whitney classes satisfy $w_1\neq 0$, $w_3 \neq 0$, and $w_1^2=0$? -This question follows on from this one where the condition $w_3 \neq 0$ is replaced by $w_2 \neq 0$. On a closed four-manifold $\operatorname{Sq}^1(w_2) = w_3$ so $w_3 \neq 0$ implies $w_2 \neq 0$. Note however that not every example with $w_2 \neq 0$ will have $w_3 \neq 0$; this is the case for both answers to the previous question. - -REPLY [17 votes]: Let's begin by reformulating the question a bit. Note that any orientable 4-manifold is spin-c, so in particular has $w_3 = 0$. The condition that $w_1 \not= 0 $ is thus redundant. -Wu's theorem gives that on a closed 4-manifold - -$w_1 = v_1$, so $Sq^1(c) = w_1 c$ for any $c \in H^3$ ($\mathbb{Z}/2$ coefficients everywhere) -$w_2 = v_2 + v_1^2$, so $v_2 = w_2 + w_1^2$. Thus $(w_2 + w_1^2)z = z^2$ for any $z \in H^2$ -$w_3 = v_3 + Sq^1(v_2) + Sq^2(v_1)$. Because $Sq^2(v_1) = 0$ and $v_3 = 0$ for degree reasons, this implies $w_3 = Sq^1(w_2)$. - -We also have universally that $w_3 = Sq^1(w_2) + w_1 w_2$, so for closed 4-manifolds we get $w_1 w_2 = 0$. Putting these together we find -$$w_3 x = Sq^1(w_2) x = Sq^1(w_2 x) + w_2 Sq^1(x) = w_1 w_2 x + w_2 x^2 = x^4$$ -on any closed 4-manifold with $w_1^2 = 0$. Hence the question is equivalent to looking for a closed 4-manifold with $w_1^2 = 0$ and some $x \in H^1$ such that $x^4 = 1$. -We can now approach the problem by the following strategy. Look for closed 4-manifolds with the right characteristic numbers. If there are none, then the answer to the question is negative. If there is one, then try to obtain a solution to the problem by surgery. -In this case we need a closed 4-manifold with a distinguished class $x \in H^1$, and the relevant characteristic numbers are $x^4 = 1$, $w_1^4 = w_1^3x = w_1x^2 = 0$; actually one condition is redundant because $w_1^3 x = Sq^1(w_1^2 x) = w_1^2 x^2$. -$RP^2 \times RP^2$ has $w_1^4 = 0$, and if we take $x$ to be the generator from one of the two factors then $w_1^2 x^2 = 1$ while $x^4 = 0$. On the other hand $RP^4$ has $w_4 = 1$, and if we take $x$ to be the generator then $w_1^2 x^2 = x^4 = 1$, while if $x = 0$ then of course $w_1^2 x^2 = x^4 = 0$. Hence if we let $M$ be the connected sum of $RP^2 \times RP^2$ and two copies of $RP^4$ then $M$ is a closed 4-manifold with a class $x \in H^4(M)$ such that $w_1^4 = w_1^2 x^2 = 0$ and $x^4 = 1$. -$H^1(M) \cong H_1(M)$ has rank 4. Any $\alpha \in H_1(M)$ with $w_1(\alpha) = 0$ can be represented by an embedded circle with trivial normal bundle. Doing surgery on this circle cuts down the rank of $H^1$ by 1. If in addition $x(\alpha) = 0$, then the result of the surgery will still have a class $x$ with the desired properties $x^4 = 1$, $w_1^4 = w_1^2 x^2 = 0$. So by doing surgery twice on $M$ we obtain an $X$ with the right characteristic numbers, but $H^1(X)$ of rank 2 generated by the essential classes $w_1$ and $x$. -In particular $w_1^3$ and $w_1^2 x$ both pair to 0 with all elements of $H^1(X)$, so must vanish. Thus, if we let $S \subset N$ be an embedded smooth surface Poincare dual to $w_1^2$ then $w_1$ and $x$ both vanish on $S$. It would be nice if we could deduce that $S$ can be taken to be an embedded sphere so that it can be killed by surgery, but that's not obvious to me. -However, we can cut out a tubular neighbourhood of $S$ to obtain a 4-manifold $Y$ with boundary. Then $x \in H^1(X)$ has a preimage $y \in H^1(Y, \partial Y)$. Now glue $Y$ to itself along the boundary to obtain a closed 4-manifold $Z$, and let $z \in H^1(Z)$ be the image of $y$ from one side. Then $w_1(Z)^2 = 0$ and $z^4 = 1$, so $w_3(Z)z = 1$.<|endoftext|> -TITLE: What does the Zariski topos of $\mathbb{P}^1$ classify? -QUESTION [10 upvotes]: (The word "geometric theory" below is used in the sense of logic / classifying topos.) -We know that the (big) Zariski topos over $\text{Spec}\mathbb{Z}$ classifies the theory of local rings. My question is, what geometric theory does the (big) Zariski topos over $\mathbb{P}^1$ classify? -From algebraic geometry point of view it should classify something like a local rings together with a map to $\mathbb{P}^1$, i.e. something like isomorphism classes of $(A, L, e_0, e_1)$ where $A$ is a local ring, $L$ is an locally free $A$-module of rank 1, and $e_0, e_1\in L$ spans $L$. Is there a good way to describe a geometric / coherent theory of it? -I saw some discussions here but it's mostly about the affine case. I'm curious on what role properness may play here. If possible I'd like to see some description on the geometric theory for a general scheme (or algebraic curve). -(On "locally free $A$-module of rank 1": in this case L must be free, but that uses another result, merely asking for being locally free of rank 1 seems more natural to me.) - -REPLY [9 votes]: First note that a morphism $\operatorname{Spec}(A) \to \mathbb{P}^1$ is just given by an element of the "classical projective space" $\mathbb{P}^1(A) = \{ [a:b] \,|\, \text{$a$ is invertible or $b$ is invertible} \}$, if $A$ is a local ring. The description you gave is valid as well, but since (as you remark) "locally free" is equivalent to "free" over a local ring, it can be simplified further. -The big Zariski topos of $\mathbb{P}^1$ classifies the theory of a "local ring together with a point $[a:b]$". This theory can be explicitly described as follows: - -A sort $R$ together with function symbols, constants, and axioms expressing that $R$ is a local ring. -A sort $P$ (to be thought of as the set of $[a:b]$ with $a,b:R$ where at least one coordinate is invertible) together with a relation $\langle\cdot,\cdot,\cdot\rangle$ on $R \times R \times P$ and the following axioms: - - -$\text{$a$ is invertible} \vee \text{$b$ is invertible} \dashv\vdash_{a,b:R} \exists p:P.\ \langle a,b,p \rangle$ -$\langle a,b,p \rangle \wedge \langle a,b,p' \rangle \vdash_{a,b:R,\, p,p':P} p = p'$ -$\top \vdash_{p:P} \exists a,b:R.\ \langle a,b,p \rangle$ -$\langle a,b,p \rangle \wedge \langle a',b',p \rangle \dashv\vdash_{a,a',b,b':R,\, p:P} \exists s:R.\ \text{$s$ is invertible} \wedge a' = s a \wedge b' = s b$ - -A constant of sort $P$. - -In a topos $\mathcal{E}$, a model of this theory is given by a local ring $A$ in $\mathcal{E}$ together with a point of the "classical projective space of $A$" as defined above. (Equivalently, by a rank-1 quotient of $A^2$ up to isomorphism of quotients.) -A different theory which the big Zariski topos of $\mathbb{P}^1$ classifies is the theory of a "homogeneous filter $F$ of $\mathbb{Z}[X,Y]$ meeting the irrelevant ideal together with a local ring over the degree-zero part $A := \mathbb{Z}[X,Y][F^{-1}]_0$ which is local over $A$". -For an arbitrary $\mathbb{N}$-graded ring $S$, the big Zariski topos of $\operatorname{Proj}(S)$ classifies the theory of a "homogeneous filter $F$ of $S$ meeting the irrelevant ideal together with a local ring over the degree-zero part $A := S[F^{-1}]_0$ which is local over $A$". (This in turn can be rewritten using the universal property of the degree-zero part of graded localization.) This description follows by combining the description of the little Zariski topos of $\operatorname{Proj}(S)$ (which classifies the theory of a "homogeneous filter $F$ of $S$ meeting the irrelevant ideal") and the description of the big Zariski topos as a topos over the little Zariski topos (which classifies the theory of a "local ring over $\mathcal{O}_S$ which is local over $\mathcal{O}_S$"). -Proofs can be found in these note of mine (Section 12.8 and Section 16.1 at the time of writing). -If you care about issues of size, note that all the statements made here are only valid if we define the big Zariski topos of a scheme $X$ using the site consisting of only the locally finitely presented $X$-schemes. (It doesn't matter whether we restrict to affine schemes or not.) If we employ the site consisting of all $X$-schemes, the construction will not be well-defined in ordinary set theory, and if we restrict to those $X$-schemes contained in some universe (either Grothendieck or partial as in the Stacks project), then the resulting topos will have much more points.<|endoftext|> -TITLE: H_3 of SL(n,Z) and SL(n,F_p) -QUESTION [13 upvotes]: Can anyone tell me what $H_3(SL_n(\mathbb{Z});\mathbb{Z})$ and $H_3(SL_n(\mathbb{F}_p);\mathbb{Z})$ are? It is easy to find references for $H_1$ and $H_2$, but it turns out that I need $H_3$ as well. All I care about are the stable values. - -REPLY [14 votes]: Summarizing the comments, the stable ($n\geq 3$) values of $H_3(SL_n;\mathbb Z)$ are - -$H_3(SL_\infty(\mathbb Z);\mathbb Z) = \mathbb Z/24$ -$H_3(SL_\infty(\mathbb F_q);\mathbb Z) = \mathbb Z/(q^2-1)$ - -and can be found in Weibel's The $K$-book. -Namely, one has -$$ -K_2(R)\xrightarrow{[-1]}K_3(R)\to H_3(E(R))\to0 -$$ -for any ring (Corollary IV.1.20 there), with $K_2(R)=H_2(E(R);\mathbb Z)$ and $K_3(R)=H_3(St(R);\mathbb Z)$, where $E(R)\subseteq GL(R)$ is the subgroup generated by elementary matrices and $St(R)\twoheadrightarrow E(R)$ is its universal central extension. -For most decent rings, including integers and fields, $E_n=SL_n$; for $R$ the integers, $K_2(\mathbb Z)=H_2(SL_n(\mathbb Z))=\mathbb Z/2$ (Milnor), $K_3(\mathbb Z)=H_3(St_n(\mathbb Z))=\mathbb Z/48$ (Lee & Szczarba 1976), and the map $[-1]$ is nonzero. For $R=\mathbb F_q$, $K_2$ is zero and $K_3$ is $\mathbb Z/(q^2-1)$ (Quillen).<|endoftext|> -TITLE: A question on Cheeger-Gromov compactness theorem -QUESTION [5 upvotes]: The Cheeger-Gromov compactness theorem says the following. Let us fix $n\in \mathbb{N}$ and positive constants $K,D,v$. Let $\{(M_i^n,g_i)\}$ be a sequence of closed infinitely smooth $n$-dimensional Riemannian manifolds with $|Sec(M_i)|\leq K$, diameter at most $D$ and volume at least $v$. Then, after a choice of a subsequence, there exist a closed smooth $n$-dimensional manifold $M$ with a Riemannian metric $g$ of class $C^{1,\alpha}$ for any $\alpha\in(0,1)$, and diffeomorphisms $\phi_i\colon M\to M_i$ such that $\phi_i^*(g_i)$ converges to $g$ in $C^{1,\alpha}$ for any $\alpha\in (0,1)$. -Question 1. What is a reference for this result in this particular form? -Question 2. I have an impression that I have heard that there are some refinements of the above result saying that the diffeomorphisms $\phi_i$ can be chosen in such a way that the pull-back under them of the Riemann curvature tensor of $g_i$ converges to some tensor in a Sobolev space. Is that correct? What is the precise statement? A reference? - -REPLY [8 votes]: Question 1: -Peters, Stefan -Convergence of Riemannian manifolds. -Compositio Math. 62 (1987), no. 1, 3–16. -and -Greene, R. E.; Wu, H. -Lipschitz convergence of Riemannian manifolds. -Pacific J. Math. 131 (1988), no. 1, 119–141. -Question 2: -This follows by the proofs of Peters and Greene-Wu using standard Sobolev estimates, instead of Schauder estimates, for elliptic PDE's (see, for example, the book of Gilbarg-Trudinger).<|endoftext|> -TITLE: Quotient-free monoidal categories -QUESTION [8 upvotes]: Let $(\mathsf{Rel},\otimes,1)$ denote the monoidal category of sets and relations, where $1$ is the one-element set. I once conjectured (with a little help from Jamie Vicary) that $\mathsf{Rel}$ is "quotient-free" in the sense that if a strong monoidal functor $F\colon\mathsf{Rel}\to S$ identifies any parallel pair of morphisms, then $F$ identifies every parallel pair of morphisms, and hence it factors through the terminal monoidal category (since $\mathsf{Rel}$ has a zero-object). [I'd be happy to hear suggestions for a better name than "quotient-free monoidal category", or for a better way of thinking of such things.] -Definition: We say a monoidal category $M$ is quotient-free if for any monoidal category $S$ and strong monoidal functor $F\colon M\to S$, if $F(f_0)=F(f_1)$ for distinct morphisms $f_0\neq f_1\colon A\to B$ then $F$ factors through a terminal monoidal category. -Explaining the conjecture to Tobias Fritz, he quickly proved it (by contradiction) as follows. -Proposition: The monoidal category $(\mathsf{Rel},\otimes,1)$ is quotient-free. -Proof (Fritz): Suppose that $A$ and $B$ are sets and that $R_0,R_1\colon A\to B$ are relations such that $R_0\neq R_1$. Then there exists $a\in A$ and $b\in B$ such that $(a,b)\notin R_0$ and $(a,b)\in R_1$ (without loss of generality). -Let $e_a\colon 1\to A$ and $e_b\colon 1\to B$ correspond to the relations characterizing the subsets $\{a\}\subseteq A$ and $\{b\}\subseteq B$, respectively, and let $e'_b\colon B\to 1$ be the transpose of $e_b$. Then we have two different relations -$$1\xrightarrow{e_a}A\xrightarrow{R_0\ ,\ R_1}B\xrightarrow{e'_b}1.$$ -These ($e'_bR_0e_a$ and $e'_bR_1e_a$) are the only two relations $1\to 1$, equaling the "null" relation $\emptyset_{1,1}$ and the identity $\mathrm{id}_1$, respectively. -Assuming now that $F(R_0)=F(R_1)$, we have $F(\mathrm{id}_1)=F(\emptyset_{1,1})$. It follows that $F$ identifies any given relation $X\colon C\to D$ with the null relation $\emptyset_{C,D}\colon C\to D$, because -$$ -FX\cong -F(X)\otimes F(\mathrm{id}_1)= -F(X)\otimes F(\emptyset_{1,1})\cong -F(X\otimes\emptyset_{1,1})= -F(\emptyset_{C,D}). -$$ -Thus for any set $A$, we obtain an isomorphism $F(A)\cong F(\emptyset)$, where $\emptyset$ is the zero-object of $\mathsf{Rel}$. $\square$ -Question: What are other examples of quotient-free monoidal categories? -Question: Might we consider quotient-free monoidal categories as acting like fields, which are also somehow quotient-free? That is, maps to quotient-free monoidal categories would be analogous to points? Any thoughts on this would be useful. - -REPLY [4 votes]: I would call these monoidal categories simple. But one should exclude the terminal monoidal category, which are too simple to be simple. -Here is a baby example: A commutative monoid, considered as a strict one-object monoidal category (for this to make sense, we need commutativity!), is simple iff it has exactly two regular quotient monoids, the trivial monoid and the monoid itself. Equivalenty, it has exactly two congruence relations. -The only simple commutative groups are $C_p$ for primes $p$. The only finite simple (commutative) monoid, which is not a group, is the multiplicative monoid $(\{0,1\},\cdot,1)$; see Proposition 2.1 in the paper The kernel of monoid morphisms by Rhodes and Tilson. Unfortunately, I could not find more about simple commutative monoids.<|endoftext|> -TITLE: Do topological commutative monoids model all 0-connective spectra (after group completion)? -QUESTION [6 upvotes]: Of course, before group completion, topological commutative monoids do not model all connected $E_\infty$ spaces -- among the grouplike ones, they model only products of Eilenberg-Mac Lane spaces. But after group completion, perhaps they do. Similarly, I wonder: Do strictly symmetric monoidal (i.e. the symmetry isomorphisms $a \otimes b \cong b \otimes a$ are identities) topologically-enriched categories model all connective spectra? -Maybe something even stronger is true: Do strictly symmetric monoidal groupoids (or equivalently, 1-types with a commutative monoid structure) model all connective spectra after group completion? Do strictly symmetric monoidal groups (equivalently, $K(G,1)$'s with a commutative monoid structure) model all 0-connective spectra after group completion? -If the answer to some version of this question is affirmative, I'd liken it to the fact that every space is the realization of a category -- you can get some extra strictness in a model at the expense of some noninvertibility. -This question obviously has some affinities to another question I asked a few days ago. - -REPLY [5 votes]: Do strictly symmetric monoidal topologically-enriched categories model all connective spectra? -If your modeling functor is some reasonable variant of the ordinary classifying space functor, I think the answer must be No. -What I know for sure about this question is written (with Angélica Osorno) in this paper: Modeling stable one-types. In that paper we study how to model the Postnikov data of spectra with just two nonzero homotopy groups (in dimension 0 and 1). The Postnikov data in such a low-dimensional case consists only of the two (abelian) homotopy groups $\pi_0$ and $\pi_1$, together with a Postnikov invariant. -Therefore we restrict not only to symmetric monoidal categories, but to those symmetric monoidal categories where all morphisms are invertible and all objects are invertible up to isomorphism. (The standard term for these is "Picard categories", but a more descriptive name would be -"grouplike symmetric monoidal groupoids".) -The sloganized summary of the paper is that the symmetry of the Picard category is the data which corresponds to the Postnikov invariant of the stable 1-type. (There's a neat connection with the theory of quadratic maps, and some calculations of Eilenberg and Mac Lane from the 1950's, which partially explain why you might expect this.) I'll skip the precise details here -- you can try the paper, or just ask me later. -For a specific example, we write down a symmetric monoidal category which models the Postnikov truncation of the zeroth space of the sphere spectrum. It has a strictly associative and unital monoidal structure, but the symmetry is nontrivial. - -Now, maybe there's some other way of modeling connective spectra with strictly symmetric monoidal but topologically enriched categories. But I think our work proves it can't be done in a way that generalizes the ordinary classifying space functor when the enrichment is discrete.<|endoftext|> -TITLE: Around Brunn-Minkowski inequality -QUESTION [9 upvotes]: Let me recall the Brunn-Minkowski inequality, which states concavity of ${\rm vol}^{1/d}$ for domains in ${\mathbb R}^d$: -$${\rm vol}(A+B)^{1/d}\ge{\rm vol}(A)^{1/d}+{\rm vol}(B)^{1/d},$$ -with equality only if $A$ and $B$ are homothetic (that is $B=\lambda A+v$). -Suppose now that $A$ is star-shaped about the origin. It can be described from a function $f:S^{d-1}\rightarrow(0,+\infty)$ by -$$A=A^f:=\{r\omega\,|\,\omega\in S^{d-1},\,0\le r>\mu^{\frac{1}{2}}(A)+\mu^{\frac{1}{2}}(B)$. -$\sqrt{\mu^{\frac{1}{2}}(A\oplus B)\cdot \mu^{\frac{1}{2}}(A+B)}\sim O(\delta^{\frac{1}{4}})>>\mu^{\frac{1}{2}}(A)+\mu^{\frac{1}{2}}(B)$<|endoftext|> -TITLE: "Correct" definition of stratified spaces and reference for constructible sheaves? -QUESTION [13 upvotes]: It seems that the theory of constructible sheaves (in particular anything that goes into proving that they form an abelian category) requires some technical statements about existence of certain stratifications and yet at the same time the notion of a "stratified space" seems to have several inequivalent definitions in the literature. - - -Is there (as of today) an agreement among specialists in the field for what is the "correct" definition of a "stratified space"? - -What would be a good reference to read about stratified spaces, constructible sheaves and their six functors? (in particular a complete construction of the abelian and derived categories of constructible sheaves). (Perhaps with a little bit of stratified Morse theory too). - - - -My end goal is to have a concrete understanding of perverse sheaves on algebraic varieties and their stratifications. I don't want to get bogged down in the technicalities of stratified spaces but I would like to know how stratifications look and how to construct them in this context. However if there's a "natural theory" (and in particular a natural category) of stratified spaces I'd love to read about it. - -REPLY [6 votes]: A bit more reader friendly than Kashiwara-Schapira is Borel's Intersection Cohomology. It doesn't treat perverse sheaves, but it has a good overview of stratified spaces in the first few chapter and develops constructible sheaves and the six functors in Chapter V. You might also try Dimca's Sheaves in Topology, which does talk about perverse sheaves. Neither of these present the most general possible development, but they're pretty accessible. -I'll also plug the discussion of stratified spaces in my book-in-progress, though I take a more topological approach that's not particularly oriented toward sheaves. Here's a link: http://faculty.tcu.edu/gfriedman/IHbook.pdf<|endoftext|> -TITLE: Is a complex vector bundle over a punctured closed surface trivial? -QUESTION [5 upvotes]: Let $M$ be a connected closed surface (possibly with non-zero genus) and let $P\subset M$ be a nonempty finite set of points. Set $\dot{M} = M \setminus P$. Let $\pi : E \rightarrow \dot{M}$ be a complex vector bundle of rank $2$. Is this vector bundle trivial? - -REPLY [11 votes]: For any topological group $G$, there is a classifying space $BG$ and a principal $G$-bundle $EG \to BG$ called the universal principal $G$-bundle which is determined up to isomorphism by the fact that $EG$ is weakly contractible. On a paracompact topological space $X$, any principal $G$-bundle $P \to X$ admits a map $f : X \to BG$, called a classifying map, so that $P \to X$ is isomorphic to $f^*EG \to X$. Moreover, two principal $G$-bundles $P_1, P_2 \to X$ are isomorphic if and only if their classifying maps $f_1, f_2 : X \to BG$ are homotopic. In particular, a principal $G$-bundle is trivial if and only if its classifying map is nullhomotopic. -Complex rank $n$ vector bundles can be identified with principal $U(n)$-bundles, so your problem reduces to showing that every map $f : \dot{M} \to BU(2)$ is nullhomotopic. -The long exact sequence in homotopy applied to the universal principal $G$-bundle, together with the weak contractibility of $EG$, shows that $\pi_{k+1}(BG) \cong \pi_k(G)$. In particular, $\pi_1(BU(2)) \cong \pi_0(U(2)) = 0$ as $U(2)$ is path-connected. -The surface $\dot{M}$ deformation retracts onto a bouquet of circles. Restricting $f$ to one of these circles, we get a map $S^1 \to BU(2)$ which is nullhomotopic as $BU(2)$ is simply connected. It follows that $f$ is nullhomotopic and therefore every rank two complex vector bundle on $\dot{M}$ is trivial. -As $U(n)$ is path-connected for every $n$, $BU(n)$ is always simply connected so the argument above would still work if we replace $BU(2)$ by $BU(n)$. Therefore we see that every complex vector bundle on $\dot{M}$ is trivial. In fact, as $SO(n)$ is connected for all $n$, the same argument shows that all real orientable vector bundles on $\dot{M}$ (which correspond to principal $SO(n)$-bundles on $\dot{M}$) are trivial.<|endoftext|> -TITLE: How can I construct D-modules over projective space using explicit differential equations? -QUESTION [7 upvotes]: Over $\mathbb{A}^n$, it is easy to construct D-modules by writing down an explicit linear system of PDE's and then writing a presentation of the associated D-module -$$ -\mathcal{D}^n \xrightarrow{} \mathcal{D}^m \to \mathcal{M} \to 0 -$$ -Is there an analogous process for constructing D-modules over projective space as an explicit system of differential equations? - -REPLY [7 votes]: You can construct $\mathcal{D}$ modules on projective space (or actually any space) by precisely the same process: write down a system of PDE's (using differential operators on the whole of $\mathbb{P}^n$) and consider the corresponding quotient. The key caveat is that on a general space, not every $\mathcal{D}$-module will be of this form: some of them can be defined using systems on different coordinate patches which are consistent, but not globally (I think a degree 0 line bundle on an elliptic curve should give a counterexample). However, $\mathbb{P}^n$ is special, and every $\mathcal{D}$-module can be defined this way (by the Beilinson-Bernstein theorem).<|endoftext|> -TITLE: Choosing subsets of $\mathbb R$ of cardinality $\frak c$, who wins? -QUESTION [9 upvotes]: Consider the following infinite game: two players, I and II, are alternating and choosing a descending sequence of subsets of $\mathbb R$ of cardinality $\frak c$, so I chooses a set $A_1\subseteq\mathbb R$, II chooses a set $A_2\subseteq A_1$, I chooses $A_3\subseteq A_2$ etc., all having the size continuum. Let $A=\cap_{i=1}^\infty A_i$. I wins if $A$ is nonempty, otherwise II wins. - -Who has a winning strategy in this game? - -This game was posed to me as a puzzle from a person I don't have any means of contact with anymore (so I can't ask them about this). However, I don't know whether he himself knew the answer or not (as far as I know it might be independent of ZFC or unknown) which is why I'm posting this on MO and not Math.SE (the same person has posed as a puzzle a combinatorial problem which I happened to know is open). -All I know is that it's consistent with ZF that II has the winning strategy: if $\mathbb R$ had countable cofinality, i.e. was a countable union of sets $S_i,i\in\mathbb N$ of cardinality $<\frak c$, then II could play on his turns $A_{2k}=A_{2k-1}\setminus(S_1\cup\dots\cup S_n)$, which would have size $\frak c$. This doesn't go through in ZFC, since there $\frak c$ has uncountable cofinality. -One can also consider the problem with $\mathbb R$ replaced by some other set, for example an ordinal. For some sufficiently well-behaved sets (not Dedekind-finite, for example) the above reasoning shows that if a cardinal has countable cofinality, I has a winning strategy. I am especially interested in the case of $\omega_1$, which is equivalent to the problem at hand if we assume CH. -I am interested in this problem both in context of ZF and ZFC, and I am willing to assume CH to resolve the original problem if it is of help there. - -REPLY [8 votes]: This problem (with inconsequential differences in the rules of play) was posed by S. Banach as Problem 67 in The Scottish Book, which is available here. The following text is copied from R. Daniel Mauldin's edition: - -(A MODIFICATION OF MAZUR'S game, [see Problem 43]). -We call a half of the set $E$ [in symbols, $(1/2)E$] an arbitrary subset $H\subset E$ such that the sets $E,H,E-H$ are of equal power. -(1) Two players $A$ and $B$ give in turn sets $E_i, i=1,2,\dots$ ad inf. so that $E_i=(1/2)E_{i-1}\ i=1,2,\dots$ where $E_0$ is a given abstract set. Player $A$ wins if the product $E_1E_2\dots E_iE_{i+1}\dots$ is vacuous. -(2) The game, similar to the one above, with the assumption that $E_i=1/2[E_0-E_1-\dots-E_{i-1}]\ i=2,3,\dots$ ad inf., and $E_1=(1/2)E_0.$ Player $A$ wins if $E_1+E_2+\dots=E_0.$ -Is there a method of win for player $A$? If $E_0$ is of power cofinal with $\aleph_0,$ then player $A$ has a method of win. Is it only in this case? In particular, solve the problem if $E_0$ is the set of real numbers. -Addendum. There exists a method of play which will guarantee that the product of the sets is not vacuous. The solution was given by J. Schreier. - -The problem is dated August 1, 1935; the addendum is dated August 24, 1935; Schreier's solution (using the well-ordering theorem, see Joel David Hamkins' answer) was published in the paper - -J. Schreier, Eine Eigenschaft abstrakter Mengen, Studia Math. 7 (1938), 155–156, eudml.<|endoftext|> -TITLE: $f'=e^{f^{-1}}$, again -QUESTION [31 upvotes]: This question is a spin-off of this one, in which the OP asks whether there is a solution $f:\mathbb R\to\mathbb R$ of the functional equation (not exactly an ODE) $f'=e^{f^{-1}}$, where $f^{-1}$ is the compositional inverse of $f$. The posted answer exploits the growth of $f(x)$ when $x\to-\infty$ and obtains a contradiction, which resolves the question nicely, but also invites the following question: what if we restrict to $f:\mathbb R_{\ge0}\to\mathbb R_{\ge0}$ and impose $f(0)=0$? This idea has been explored in the comments, where a formal power series expansion is obtained for $f$ which does not seem to converge for any $x\ne0$. -Taking another approach, we can use an iteration scheme starting from $f_1(x)=x$ and inductively solve the ODE $f_{n+1}'=e^{f_n^{-1}}$ with the initial condition $f_{n+1}(0)=0$ to obtain $f_{n+1}$, much in the spirit of Picard iteration. Explicitly, for example, we have -$f_2'=e^x$ and $f_2=e^x-1$; -$f_3'=e^{\ln(x+1)}=1+x$ and $f_3=x+x^2/2$; -$f_4'=e^{\sqrt{1+2x}-1}$ and $f_4=e^{\sqrt{1+2x}-1}(\sqrt{1+2x}-1)$ -and the next iteration produces non-elementary functions. It is clear that the sequence $(f_{2k-1})_{k\ge1}$ is increasing, $(f_{2k})_{k\ge1}$ is decreasing, and $f_{2k-1}g'(0)=1$ and $g(x)>x$ for all $0x$ on $(0,a)$, this inclusion is only possible if $a=+\infty$, so that $g$ is unbounded. On the other hand, arguing as in Christian Remling's previous answer, since $e^{-g(g(t))}g'(t)=1$ and $g(t)\ge t$, we have for any $x\ge0$ -$$x=\int_0^{x}e^{-g(g(t))}g'(t)dt=\int_0^{g(x)}e^{-g(s)}ds\le \int_0^{+\infty}e^{-s}ds=1 ,$$ -a contradiction. -$$*$$ -Rmk 1. To justify the total monotonicity of $g$, note that, as a general elementary fact, a real analytic function on an interval $I$, whose Taylor series at some point $x_0\in I$ has non-negative coefficients, has Taylor series with non-negative coefficients ay any point $x\in I$, $x\ge x_0$. Indeed, this is clear for $x_1\ge x_0$ within the radius of convergence of $x_0$, and since there is a uniform radius of convergence at any $y\in [x_0,x]$, one reaches $x$ by finitely many steps $x_0 -TITLE: Waldhausen $K$-theory before group completion -QUESTION [10 upvotes]: $K$-theory is often billed as the "universal way to split exact sequences". But it seems we're too anxious to group-complete things to actually take the slogan at face value. -Consider the following $\infty$-categories: - -$\mathcal{W}$ - Waldhausen categories (or Waldhausen $\infty$-categories, if you prefer) -$\mathcal{C}_1$ - symmetric monoidal $\infty$-categories -$\mathcal{C}_2$ - $E_\infty$-spaces -$\mathcal{C}_3$ - infinite loop spaces - -Say that a functor $F: \mathcal{W} \to \mathcal{C}_i$ is additive if $F\mathcal{E} W\to F W \times F W$ is an equivalence for all $W \in \mathcal{W}$, where $\mathcal{E}W$ the Waldhausen category of exact sequences $w' \to w \to w''$ in $W$ and the map projects onto $(w',w'')$. Consider the following functors: - -$K_\oplus^1: \mathcal{W} \to \mathcal{C}_1$ - sending $W$ to its simplicial localization, with $E_\infty$ structure given by coproduct -$K_\oplus^2: \mathcal{W} \to \mathcal{C}_2$: - sending $W$ to the nerve of its category of weak equivalences (the core of $K_\oplus^1 W$) -$K_\oplus^3 : \mathcal{W} \to \mathcal{C}_3$: - sending $W$ to the group completion of $K_\oplus^2 W$ - -The universal property of $K$-theory is that it is a functor $K^3: \mathcal{W} \to \mathcal{C}_3$ which constitutes a reflection of $K^3_\oplus$ into the category of additive functors $\mathcal{W} \to \mathcal{C}_3$ (I think this is Clark Barwick's formuation). Analogously, I ask: -Questions - -Does $K^1_\oplus$ admit a reflection $K^1$ into the category of additive functors $\mathcal{W} \to \mathcal{C}_1$? If so, is it modeled by a variant of the $S_\bullet$ construction? -Does $K^2_\oplus$ admit a reflection $K^2$ into the category of additive functors $\mathcal{W} \to \mathcal{C}_2$? If so, is it modeled by the $S_\bullet$ construction itself? - -REPLY [6 votes]: I'm not sure about Waldhausen categories in general, but if you restrict attention to stable $\infty$-categories (with trivial Waldhausen structure in which all maps are cofibrations) then group completion is essentially forced on you by the splitting of exact sequences. In principle, this happens because for every object $X \in {\cal C}$, splitting the exact sequence -$$ X \to 0 \to \Sigma X $$ -means that $\Sigma X$ becomes the inverse of $X$. In particular, taking the maximal subgroupoid of a stable $\infty$-category ${\cal C}$ (considered as an $\mathbb{E}_\infty$-monoid in spaces with respect to direct sum), and then forcing the relations given by splitting of exact sequences, yields an $\mathbb{E}_\infty$-monoid which is already group-like. -This idea can also be phrased from the perspective of additivity. Observe first that the $\infty$-category ${\rm Cat}^{\rm ex}_\infty$ of small stable $\infty$-categories is semi-additive in the sense that it has a zero-object (an initial object which is also final) and biproducts (coproducts which are also products). Since the $\infty$-category of $\mathbb{E}_\infty$-monoids is also semi-additive it is somewhat natural to restrict attention to functors $F:{\rm Cat}^{\rm ex}_\infty \to {\rm Mon_{\mathbb{E}_\infty}}$ which are themselves semi-additive, in the sense that they preserve zero-objects and biproducts. Note that so far we have not enforced any group-likeness condition. Note also that such a semi-additive $F$ will be additive in the sense you describe if and only if it sends the map $p:{\cal EC} \to {\cal C}\times{\cal C}$ given by $[x' \to x \to x''] \mapsto (x',x'')$ to an equivalence of $\mathbb{E}_\infty$-monoids. We now claim the following: if a semi-additive functor $F:{\rm Cat}^{\rm ex}_\infty \to {\rm Mon_{\mathbb{E}_\infty}}$ is additive then it automatically takes values in group-like $\mathbb{E}_\infty$-spaces. To see this observe that when ${\cal C}$ is stable the $\infty$-category ${\cal EC}$ of fiber sequences has an automorphism $T: {\cal EC} \to {\cal EC}$ sending $[x'\to x \to x'']$ to $[x \to x'' \to {\rm cof}(x\to x'')]$. Furthermore, the functor $p:{\cal EC} \to {\cal C}\times{\cal C}$ above has a one-sided inverse $i: {\cal C} \times {\cal C} \to {\cal EC}$ sending $(x,y)$ to $[x \to x\oplus y \to y]$. In particular, if $F(p)$ is an equivalence then $F(i)$ is an equivalence as well. It then follows that if $F$ is additive then it sends the composed functor -$$ {\cal C} \times {\cal C} \stackrel{i}{\to} {\cal EC} \stackrel{T}{\to} {\cal EC} \stackrel{p}{\to} {\cal C} \times {\cal C} \stackrel{\sigma}{\to} {\cal C} \times {\cal C} $$ -to an equivalence, where $\sigma$ is the equivalence $\sigma(x,y) = (\Omega y,x)$. But this composed functor is just the shear map $(x,y) \mapsto (x,x\oplus y)$ and since $F$ is semi-additive it follows that the shear map of the $\mathbb{E}_\infty$-monoid $F({\cal C})$ is an equivalence, i.e., $F({\cal C})$ is group-like.<|endoftext|> -TITLE: Homology of the product of spaces with integer coefficients and the Massey products -QUESTION [6 upvotes]: Consider -$H_*(X\wedge Y;Z)$, where $X=Y=BZ/2$ for concreteness' sake. If we write $e_i$ the generator of $H_i(BZ/2;Z/2)$., we see that the $E_2=E_{\infty}$ term of the Bockstein spectral sequence -for $X\wedge Y$ is trivial, thus the permanent cycles are the image of $\beta$. As we have -$$\beta e_{2j}=e_{2j-1}$$ one can choose a basis of the set of permanent cycles as -$$\{e_{2i-1}\otimes e_{2j-1}\}\cup \{e_{2i}\otimes e_{2j-1}+e_{2i-1}\otimes e_{2j}\}.$$ -If we still denote by $e_{2i-1}$ the lift of $e_{2i-1}\in H_{2i-1}(BZ/2;Z/2)$ to $H_{2i-1}(BZ/2;Z)$, then we see that the elements of the first set lifts simply to the "products $e_{2i-1}\otimes e_{2j-1}$" of integral homology classes, whereas those in the second lift to "some sort of Massey products $\langle e_{2i-1} , 2, e_{2j-1}\rangle $". -Now, my questions are - -Is there any reference for this kind of facts, that is the description of the homology of the product of spaces using "Massey products"? -Is there a setting in which one can "really" consider the obvious lifts of the elements $e_{2i}\otimes e_{2j-1}+e_{2i-1}\otimes e_{2j}$ as Massey product? - -REPLY [9 votes]: Here's a very general form. Suppose that we have six chain complexes $A_0, A_1, A_2, A_{01}, A_{12}, A_{012}$, with bilinear "multiplication" pairings of chain complexes: -$$ -\begin{align*} -A_0 \otimes A_1 &\to A_{01}\\ -A_1 \otimes A_2 &\to A_{12}\\ -A_{01} \otimes A_2 &\to A_{012}\\ -A_0 \otimes A_{12} &\to A_{012} -\end{align*} -$$ -Let's assume that we also have a chain homotopy $H$ between the two composite maps $A_0 \otimes A_1 \otimes A_2 \rightrightarrows A_{012}$. Given cycles $x \in A_0, y \in A_1, z \in A_2$ of degrees $p$, $q$, and $r$ respectively such that $[x \cdot y] = 0 \in H_{p+q}(A_{01})$ and $[y \cdot z] = 0 \in H_{q+r}(A_{12})$, then we can form a Massey product: choose elements $u$ and $v$ such that $\partial u = x \cdot y$ and $\partial v = y \cdot z$, and form -$$ -\langle x,y,z\rangle = u \cdot z - (-1)^p x \cdot v + H(x\otimes y \otimes z) -$$ -which represents an element in $H_{p+q+r+1}(A_{012})$. This depends in the usual way on the choices of $u$ and $v$, so there is indeterminacy. (If you use cohomological indexing then for elements in cohomological degrees $p$, $q$, and $r$ you get a Massey product in $H^{p+q+r-1}(A_{012})$.) -The "usual" definition of a Massey product is when all six chain complexes are equal to a single complex $A$ equipped with a strictly associative multiplication (allowing us to choose $H = 0$) that makes it into a differential graded algebra. -In the case you describe, we can take - -$A_0 = A_{01} = C_*(X,pt)$ the relative singular chain complex of $X$, -$A_2 = A_{12} = C_*(Y,pt)$ the relative singular chain complex of $Y$, -$A_{012} = C_*(X \wedge Y, pt)$ the relative singular chain complex of $X \wedge Y$, and -$A_1 = \Bbb Z$, the constant chain complex $\Bbb Z$ in degree 0. - -Then two of the multiplications are just the unit isomorphism and the other two multiplications $C_*(X,*) \otimes C_*(Y,*)$ are induced by the Eilenberg-Zilber shuffle map. The unit isomorphism is associative and so we can take $H = 0$ in this case, recovering your formula. -If you like, the entire framework of these six objects, pairings, homotopy, and chosen elements asks for a DG-category (really an $A_\infty$ DG-category) with four objects and chosen maps -$$ -a \stackrel{z}{\to} b \stackrel{y}{\to} c \stackrel{x}{\to} d -$$ -so that the double composites are null; in this description we can also think of this secondary operation $\langle x,y,z\rangle$ as related to Toda's "bracket" construction. The case of a differential graded algebra is the one-object DG-category case.<|endoftext|> -TITLE: Intuition behind the Dehn Invariant -QUESTION [12 upvotes]: EDIT: as pointed-out below, this has been posted on math.stackexchange. I'll leave it up to the community whether or not to delete this question, but I do think there is room for a more technical answer than the one posted on math.stackexchange. - -A famous question related to Hilbert's third problem: - -Given any two polyhedra of equal volume, is it always possible to cut the first into finitely many polyhedral pieces which can be reassembled to yield the second? - -This was answered in the negative by Max Dehn, who showed that it was impossible to cut a cube into a finite number of polyhedral pieces that can be reassembled into a tetrahedron. This was accomplished by associating to each polyhedra $P$, an element $D(P)$ of the group $\mathbb{R}\otimes_{\mathbb{Q}}(\mathbb{R}/\mathbb{Q}\pi)$ that is invariant under cutting and pasteing. To obtain his result, Dehn then proved that the cube had value 0 in the group above, and the tetrahedron had a non-zero value. -My question is this: is there a more conceptual explanation for why this question should be answered in the negative? In particular, the corresponding statement about polygons is true: given two polygons with the same area, there is a decomposition of the first polygon into polygonal pieces that may be reassembled into the second (see the Wallace-Bolya-Gerwien Theorem). Is there a specific property of polygons that we are exploiting to prove WBG, that fails when we go to three dimensions? - -REPLY [13 votes]: The Dehn functional of a polyhedron $P$ with edge lengths $\ell_i$ and exterior dihedral angles $\theta_i$ is $D(P) = \sum_i \ell_i \otimes_{\mathbb Q} (\theta_i\, \mathrm{mod}\, \mathbb{Q}\pi)$. It looks similar to the (twice) the total mean curvature $S(P) = \sum_i \ell_i \theta_i$. Both of them satisfy the "valuation property": -$$ -F(P \cup Q) = F(P) + F(Q) - F(P \cap Q) -$$ -for any two convex polyhedra whose union is convex. In particular, if $P \cap Q$ is a polygon, then $S(P \cap Q)$ is $\sum_i \ell_i \cdot \pi$. Dehn's idea is to take $\theta\, (\mathrm{mod}\, \pi)$ to make this vanish. Since we need linearity and distributivity, the product of $\ell$ with $\theta$ becomes the tensor product over $\mathbb{Q}$. Because of $D(P \cap Q)=0$, Dehn's functional is invariant with respect to cutting a polyhedron along a plane, so on equidecomposable polyhedra the functional has the same value. -Dehn almost surely knew about the total mean curvature (Minkowski studied more general functionals shortly before), so this might be the way he invented his functional. -As for the polygons, a similar argument leads to a functional that is the angle sum modulo $\pi$, so always vanishes.<|endoftext|> -TITLE: does this set of permutations form a group? And more -QUESTION [5 upvotes]: Consider the group of $mn\times mn$ permutation matrices $\mathfrak{S}_{mn}$ and partition each such matrix $P$ into $n^2$ blocks of $m\times m$ matrices $Q_{i,j}$. Now, transpose each $Q_{i,j}$ (independently) to form a new $mn\times mn$ matrix denoted $P^t$ (with an abuse of notation). Let's construct the set $U_{mn}:=\{P^t\in\mathfrak{S}_{mn}:\, P\in\mathfrak{S}_{m,n}\}$. -It's clear that if $m=1$ then $U_{mn}=\mathfrak{S}_{mn}$; the same if $n=1$. - -Question 1. For which $m$ and $n$, does $U_{mn}$ form a group? -UPDATE. Negative answer shown below. - -A cute note: $\frac{\#\mathfrak{S}_{2n}}{\# U_{2n}}=C_n$ is the Catalan number. Is $\# U_{2n}$ a subgroup? -Answer. No. See below. Another counterexample to the converse of Lagrange's theorem: $\# U_{2n}$ divides $\#\mathfrak{S}_{2n}$ but $U_{2n}$ is not a subgroup of $\mathfrak{S}_{2n}$. - -Question 2. View $P\in U_{2n}$ as $P\leftrightarrow\sigma$ as a $1$-line permutation $\sigma=(\sigma_1,\dots,\sigma_{2n})$. In this way, what is an equivalent (to the above "transpose") characterization of $P$ in the language of $\sigma$? -UPDTE. I am Still waiting for an answer. - -If you're interested in enumeration then visit here. - -REPLY [10 votes]: For $m, n > 1$ it's never a group. -Let $\sigma \in \mathfrak{S}_n$, $\sigma$ switches $1, m + 1$, and in fact $\sigma = \sigma^t$. Clearly, $\sigma \in U_{mn}$. Similarly, let $\tau$ switch $1, 2$; similarly, $\tau = \tau^t$. Then $\sigma \tau \sigma$ switches $2, m+1$. This clearly is not in $U_{mn}$.<|endoftext|> -TITLE: Weber's class number problem and real quadratic fields of class number one -QUESTION [9 upvotes]: (A) It is an old, outstanding problem to show that there are infinitely many real quadratic fields with class number one. -(B) On the other hand, Weber's class number problem (for $p=2$) asks to show that the degree $2^n$ cyclotomic extension $F_n=\mathbb Q(\cos(2\pi/2^{n+2}))$ of $\mathbb Q$ has class number one, for all $n$. -According to p.41 of this thesis, showing (B) for infinitely many $n$ implies (A), but no idea of the proof is given. This does not seem to follow immediately for me, but perhaps I am missing something simple here. Certainly $F_n$ contains a real quadratic subfield, but why should its class number also be 1? - -REPLY [8 votes]: This statement is most certainly nonsense; I guess what she meant to write was that in order to prove the existence of infinitely many number fields with class number $1$ it is sufficient to prove $h(F_n) = 1$ for infinitely many (and therefore for all) $n$.<|endoftext|> -TITLE: Sum of square roots of binomial coefficients -QUESTION [7 upvotes]: Question: let -\begin{equation} -c_n=\sum_{i=0}^n \sqrt{\binom{n}{i}}. -\end{equation} -How does $c_n$ grow with $n$? -My conjecture is that $c_n=\Theta(2^{0.5n}n^{0.25})$.This is because -\begin{equation} - 0.5^{0.5n} c_n = \sum_{i=0}^n \sqrt{\binom{n}{i}0.5^n}=\sum_{i=0}^n\sqrt{Pr(X=i)}, -\end{equation} -for $X\sim Bin(n,0.5)$. The binomial distribution $Bin(n,0.5)$ is approximately the normal distribution $N(0.5n, 0.25n)$. Also, if $Y\sim N(0.5n, 0.25n)$, it is not hard to see that -\begin{equation} -\int \sqrt{f(y)} dy =\Theta(n^{0.25}). -\end{equation} -Therefore I believe that it is also true that $0.5^{0.5n} c_n=\Theta(n^{0.25})$. I played with matlab to get some evidence and found that -\begin{equation} -\frac{0.5^{0.5n} c_n}{n^{0.25}} \approx \frac{\pi}{2}, -\end{equation} -for $n=100,1000,10000,100000,1000000$. So the conjecture should be true. So anyone can prove it? -(It turns out that $\frac{0.5^{0.5n} c_n}{n^{0.25}} \approx (2\pi)^{0.25}$ instead of $\frac{\pi}{2}$). - -REPLY [9 votes]: For smallish $k$, we have -$$ \binom{n}{n/2+k} \approx \binom{n}{n/2} \exp(-2k^2/n). $$ -So -$$\sum\sqrt{\binom{n}{n/2+k}} - \approx \sqrt{\binom{n}{n/2}} \int_{-\infty}^\infty e^{-k^2/n}\,dk - \approx 2^{n/2} (2\pi n)^{1/4}.$$ -Of course I did some hand-waving here, but all of this is rigorously justifiable. Use the Euler-Maclaurin theorm to justify replacing the sum by an integral.<|endoftext|> -TITLE: rigorous derivation of isoperimetric inequality from ideal gas equation -QUESTION [5 upvotes]: I'm an undergraduate math student that learned about classical ideal gases and the associated maxwell-boltzmann distribution for particle velocities in a statistical physics course. Now, starting from this knowledge and assuming that there are no collisions between particles, I tried to show that an ideal gas maximises the enclosed volume. -More formally for any compact Euclidean manifold $M$ in $\mathbb{R}^3$ with elastic boundary $\partial M$ containing a macroscopic number of gas particles, a gradual increase in temperature ultimately results in a sphere. -My method of proof is to take for granted that $\partial M$ allows a maximum pressure $P^*$ and that if we gradually increase the temperature $T$ of the particles inside $M$, -$$ P^* \text{ everywhere on $\partial M$} \implies M \text{ is spherical} \tag{*}$$ -I think this method of proof would work but I realised that it's not easy to fill in the details considering that I assume that this balloon $M$ is contained in a gas with a different temperature and pressure. For this reason, I'd like to know whether this result has already been proven rigorously for a macroscopic number of particles in order to compare my work. -I think this might be as hard as proving the double bubble conjecture so it might be that this result has not yet been rigorously demonstrated. -Note 1: I am interested in the special case of Hookean elasticity as I think that Hookean elasticity would be a sufficiently good approximation. I also assume that the boundary $\partial M$ has negligible thickness. -Note 2: So far I have the following handwavy argument that points to a proof. Assuming that $P^*$ has been attained uniformly on $\partial M$, the magnitude of the force on $\partial M$ must be approximately constant all over $\partial M$. In consequence, by the Maxwell-Boltzmann distribution and momentum conservation, the distance between opposite extremities of the balloon must be approximately constant. - -REPLY [4 votes]: For the $n$-dimensional phase space of $n$ particles on a line the isoperimetric inequality of an ideal gas was derived in: Phase space measure concentration for an ideal gas (2009). - -We point out that a special case of an ideal gas exhibits - concentration of the volume of its phase space, which is a sphere, - around its equator in the thermodynamic limit. The rate of approach to - the thermodynamic limit is determined. Our argument relies on the - spherical isoperimetric inequality of Lévy and Gromov.<|endoftext|> -TITLE: Why are spherical representations subquotients of unramified principal series? -QUESTION [6 upvotes]: I'm trying to learn the basics of the representation theory of $p$-adic groups and I'm stuck on a few things: -Let $G$ is a connected split reductive group over a non-archimedean local field $F$, and $K=G(\mathcal{O})$, a hyperspecial maximal compact subgroup of $G(F)$. Then there are two facts that apparently go hand in hand: - -The Iwahori-Hecke algebra $H(G//K)$ is commutative. -If $\pi$ is any irreducible admissible (complex) representation of $G(F)$, then dim$(\pi^K) \leq 1$. - -For the first statement there is a trick: one uses the Cartan decomposition to show that there is an anti-automorphism on the Hecke algebra that is the identity. But I don't see how to prove the second fact -- does it follow quickly from the first fact? -There is also this statement: For any irreducible admissible $\pi$, $\pi^K \neq 0$ if and only if $\pi$ is a Jordan-Holder component of an unramified principal series representation of $T(F)$, where $T$ is a maximal split torus in $G$. -I would appreciate any help on how these statements are proven. Part of the reason I'm confused is that the source I'm looking at seems to me to suggest that these statements follow from the first statement, but after looking on the internet, it seems that the statement about unramified principal series is a deep result of Satake, reproven by Casselman. - -REPLY [9 votes]: Statement 2. comes from the following classical fact whose proof can be found in e.g. Bushnell and Kutzko, "The admissible dual of ${\rm GL}(N)$ via compact open subgroups". This is a particular case of Proposition (4.2.3), page 147 of loc. cit. I state it in your case: -The following sets are in natural bijection: -(i) equivalence classes of irreducible representations $\pi$ such that $\pi^K \not= 0$; -(ii) Isomorphism classes of simple ${\mathcal H}(G//K)$-modules. -Of course, in your case, since ${\mathcal H }(G//K)$ is commutative, its simple left modules are $1$-dimensional. -For your last statement. There are several good references for the unramified principal series. Among them: -Casselman's lecture notes : Introduction to the theory of admissible representations of $p$-adic reductive groups (available on his web page). -Casselman, W. The unramified principal series of -adic groups. I. The spherical function. Compositio Math. 40 (1980), no. 3, 387–406. -Important remark. Your last statement, as it is stated is false, or, to be true, it depends on what you mean by "unramified principal series". If you take as a definition that an unramified principal series is with no restriction a representation parabolically induced from an unramified character of a maximal split torus, then for your statement to be true you have to replace $K$ by an Iwahori subgroup of $G$. In that case the corresponding theorem is due to A. Borel (Invent. Math. 35, 1976, 233--259). For the case of your maximal compact subgroup $K$, only the direct implication holds. A classical counter-example is the steinberg representation. It is an irreducible (sub)quotient of some unramified principal series, but it is not $K$-spherical (but of course it is Iwahori-spherical).<|endoftext|> -TITLE: On a result of Kawamata on ampleness and nonexistence of rational curves -QUESTION [5 upvotes]: Recently I read the following results. -(1) Zheng, F.. Kodaira dimensions and hyperbolicity of -nonpositively curved compact K\"ahler manifolds. Comment. Math. -Helv. 77 (2002), no. 2, 221-234. -(2) Jahnke, P.; Peternell, T.; Radloff, I.. Some recent -developments in the classification theory of higher dimensional -manifolds. Global aspects of complex geometry, 311-357, Springer, -Berlin, 2006. -Theorem [Zheng Thm 2 p232] Let $M^2$ be a compact -K\"ahler surface of nonpositive sectional curvature. If it is of -general type, then it is Kobayashi hyperbolic. -Theorem [Jahnke-Peternell-Radloff Cor 6.10 p334] Let -$X_n$ be a projective manifold whose universal cover is Stein (or -has no positive-dimensional subvariety). Then either $K_X$ is ample -or $\chi(\mathcal{O}_X) = 0$, $K_X$ is nef and $K_X^n=0$. -It seems that both results rely on the following step: -Proposition: Any minimal surface or variety $X_n$ of -general type ($K_X$ big and nef), Assume $X_n$ is projective (no -need when $n=2$), then $K_X$ is ample if $X_n$ admits no rational -curve. -Both two papers refer this as a 1991 or 1992 result of Kawamata, -however, I failed to find a precise statement. Can anyone give a -precise statement of Kawamata's result or point out a reference? - -REPLY [12 votes]: I suppose they might mean the basepoint-free theorem. If you definitely want a reference to Kawamata, I believe it's in his Annals paper, "The cone of curves...", but probably the standard reference is Kollár-Mori (1998). -So, the point is, if $K_X$ is nef and big, then by the basepoint-free theorem some multiple of it is basepoint-free and hence defines a morphism. It is easy to see that the image of that morphism is the canonical model of $X$ and hence has an ample canonical class. -Since it is a canonical model, it has rational Gorenstein singularities which implies that the (non-trivial) fibers of this morphism contain rational curves. In fact, a lot more is true, these fibers are rationally chain connected, so not only contain rational curves, but are covered by them. In its currently known form this was originally conjectured by Shokurov and proved by Hacon-McKernan. Their result is from 2007, so newer than the ones you are reading. Then again, you only need existence of rational curves which can be derived from the (relative) cone theorem for the morphism. Behind all of these is the phenomenal result of Mori which is probably best known as Bend-and-Break. That shows how the failure of the positivity of the canonical class leads to the existence of rational curves. -To finish the argument one notes that if the original $X$ does not contain rational curves, then the above morphism must be an isomorphism and hence $K_X$ is ample.<|endoftext|> -TITLE: Finite surjective morphism of normal varieties and Galois coverings -QUESTION [7 upvotes]: Let $f:X\to Y$ be a finite surjective morphism of normal varieties. We say that $f$ is a Galois covering if the extension $k(X)/k(Y)$ is Galois. -I have the following questions: -1.- Why we need to assume normality of the varieties for the definition of a Galois covering? -2.- Is it true that a finite surjective morphism between normal varieties is always flat? -3.- Is it true that $f$ is a Galois covering if and only if the group of automorphism $Aut(X/Y)$ acts transitively on all fibers of $f$? - -REPLY [9 votes]: I'm just going to make a quick additional comment on #2 (just to add on to Timo Keller's explanations and Sasha's counter example). -Theorem (see Matsumura, page 179) If $R \subseteq S$ is a finite extension of Noetherian rings with $R$ regular and $S$ Cohen-Macaulay, then $S$ is a flat $R$ module. -So you just need $Y$ to be regular and $X$ to be Cohen-Macaulay (Matsumura actually has a more general statement which also works in the non-finite case). -On the other hand, if $Y$ is not regular then $X/Y$ is very rarely flat. Indeed, since the map is finite, flat is the same as $f_* O_Y$ being a locally free $O_X$-module and we can ask how many (local) $O_X$-summands $f_* O_Y$ has as an $O_X$-module. -Fact If $R \subseteq S$ is a finite extension of normal local rings with the same residue field which is etale in codimension 1 (ie, the sort of group quotient that Sasha wrote above), then $S$ has at most one $R$-summand as an $R$-module (in particular, it is nowhere near flat). See the question I asked earlier: Number of free summands of finite local extensions<|endoftext|> -TITLE: minimum-maximum entries matrix -QUESTION [5 upvotes]: Let $M(n)$ be an $n\times n$ matrix in the variables $x_1,\dots,x_n$ with entries -$$M_{i,j}(n)=\frac{x_{\max(i,j)}}{x_{\min(i,j)}}, \qquad 1\leq i,j\leq n.$$ -I'm interested in the following: - -Questions. -(1) Is there a neat or "closed form" evaluation for the determinant $\det M(n)$? -(2) Is there an explicit formula for the inverse of $M(n)$? - -Thank you. - -REPLY [3 votes]: I wish to advertise the Method of Condensation in proving determinantal evaluation. The key is guess the answer, which is thanks to Ehud Meir. Then, generalize it a bit. Let $x_1, x_2,\dots$ be an infinite set of variables and modify the original matrix (by shifting variables) to $M^{a,b}(n)$ so that -$$M_{i,j}^{a,b}(n):=\frac{x_{\max(i+a,j+b)}}{x_{\min(i+a,j+b)}}.$$ -Convention: $M^{0,0}(n)=M(n)$. -Claim. If $a\neq b$ then $\det M^{a,b}(n)=0$, and if $a=b$ then -$$\det M^{a,a}(n)=\prod_{r=2}^n\frac{x_{k-1+a}^2-x_{k+a}^2}{x_{k-1+a}^2}.\tag1$$ -Proof. The case $a\neq b$ is easy - simply factor out a variable from $n^{th}$-column/row and another variable from the $(n-1)^{th}$-column/row. These new columns/rows are identical. -Inductive proofs neatly work with this Dodgson's recursive relation -$$\det Z^{0,0}(n)=\frac{\det Z^{1,1}(n-1)\det Z^{1,1}(n-1)-\det Z^{0,1}(n-1)\det Z^{1,0}(n-1)}{\det Z^{1,1}(n-2)}$$ -satisfied by any matrix (so long as denominators do not vanish). Thus, it holds for $\det M^{a,b}(n)$. -So, it remains to prove that the (explicit) formula on the RHS of (1) does satisfy the same equation. However, this is quite a routine simplification (preferably with symbolic sofwares). The proof follows. $\square$<|endoftext|> -TITLE: Applications of microlocal analysis? -QUESTION [28 upvotes]: What examples are there of striking applications of the ideas of microlocal analysis? -Ideally I'm looking for specific results in any of the relevant fields (PDE, algebraic/differential geometry/topology, representation theory etc.) which follow as natural consequences of microlocal analysis and are not known to be obtained easily without it. - -REPLY [5 votes]: A pretty recent and impressive application of microlocal analysis is to analytic number theory or automorphic forms, via the orbit method as developed by Nelson and Venkatesh. In loc. cit. they set the stage for the theory specifically thought to be applied to problems in the area of automorphic forms (trace formulas, moments, subconvexity, and more widely problems on large families of automorphic forms). -A brilliant outcome of this setting is the proof of subconvexity for all L-functions of general linear groups, by Nelson.<|endoftext|> -TITLE: Trivialization of the vector bundle of centralizers of regular elements -QUESTION [10 upvotes]: Let $\mathfrak{g}$ be a complex semisimple Lie algebra of rank $r$. Let $\mathfrak{g}^{\mathrm{reg}}$ be the set of $X\in\mathfrak{g}$ whose centralizer $Z_{\mathfrak{g}}(X):=\{Y\in\mathfrak{g}:[X,Y]=0\}$ has dimension $r$ (called the set of regular elements here). - -Question. Can we find polynomial maps - $$f_1,\ldots,f_r:\mathfrak{g}\to\mathfrak{g}$$ - such that for all $X\in\mathfrak{g}^{\mathrm{reg}}$, $\{f_1(X),\ldots,f_r(X)\}$ is a basis for $Z_{\mathfrak{g}}(X)$? - -The answer is trivially yes for $\mathfrak{g}=\mathfrak{sl}(2,\mathbb{C})$. Simply take $f_1:\mathfrak{sl}(2,\mathbb{C})\to\mathfrak{sl}(2,\mathbb{C})$ to be the identity map. -Using Mathematica, I also found two such polynomials $f_1,f_2$ for $\mathfrak{g}=\mathfrak{sl}(3,\mathbb{C})$, but they are a bit too messy to write down here. - -Edit. In other words, if we let -$$E=\{(X,Y)\in\mathfrak{g}^{\mathrm{reg}}\times\mathfrak{g}:[X,Y]=0\}$$ -then $E$ is a vector bundle of rank $r$ over $\mathfrak{g}^{\mathrm{reg}}$ and the question is if it is trivial in the algebraic category. - -REPLY [7 votes]: The answer is 'yes', and it follows from Theorem 0.1 in B. Kostant, "Lie group representations on polynomial rings" (American Journal of Mathematics, 85 (1963), 327–404). Here is the argument/construction: -Let $\phi_1,\ldots,\phi_r$ be generators of the ring of ad-invariant polynomial functions on $\frak{g}$. (We can assume that $\phi_1(X)=\tfrac12\langle X,X \rangle$, where $\langle, \rangle$ is the Killing form on $\frak{g}$, and we can assume that $\phi_i$ is homogeneous of degree $m_i+1$ where $1=m_1\le m_2\cdots\le m_r$ are the exponents of $\frak{g}$. By Kostant's Theorem 0.1 quoted above, the differentials -$$ -\mathrm{d}\phi_1,\ldots, \mathrm{d}\phi_r -$$ -are linearly independent at each point $X\in\frak{g}^{\mathrm{reg}}$. -Define a polynomial map $f_i:\frak{g}\to\frak{g}$ homogeneous of degree $m_i$ for $i=1,\ldots,r$ as the 'gradient' of $\phi_i$ with respect to the Killing metric, i.e., -$$ -(\mathrm{d}\phi_i)_X(Y) = \langle f_i(X),Y\rangle. -$$ -Because $\phi_i$ is ad-invariant, it satisfies $(\mathrm{d}\phi_i)_X\bigl([X,Y]\bigr)=0$ for all $Y\in\frak{g}$. Consequently, -$$ -\bigl\langle [X,f_i(X)],Y\bigr\rangle = -\bigl\langle f_i(X),[X,Y]\bigr\rangle = 0 -$$ -for all $Y$, and, since the Killing form is nondegenerate, this implies that $[X,f_i(X)]$ vanishes identically on $\frak{g}$. (Note that $f_1(X) = X$.) -By Kostant's result, the vectors $f_1(X), f_2(X),\ldots, f_r(X)$ are linearly independent and hence are a basis of $Z_{\frak{g}}(X)$ for all $X\in\frak{g}^{\mathrm{reg}}$. -Thus, these polynomial vector fields furnish a solution to the OP's problem.<|endoftext|> -TITLE: When is an analytic function in $L^2(\Bbb R)$? -QUESTION [9 upvotes]: I asked this question on Math Stack Exchange some time ago and a similar question recently appeared regarding $L^1$ instead see here This has prompted me to bring it to this community in the hopes of getting an answer, partial answer, or even perhaps literature that addresses this question. Text from initial post below. Original post. - -Suppose $f:\Bbb R\to\Bbb R$ is (real?) entire. In order for $f$ to be in $L^2(\Bbb R)$, clearly all terms in the power series cannot be positive since $f$ would diverge at $\pm\infty$. Likewise, the distribution of negative terms cannot go to zero so we see that the power series for $f$ must be alternating (in some fashion). However this does not tell us much. -Taking $f(x) = \sum\limits_{n=0}^{\infty} \dfrac{(-1)^n}{n!}x^{2n}$ gives $f(x) = \exp(-x^2)$ and is in $L^2(\Bbb R)$. In this case, the coefficients have factorial decay but it is not immediately obvious what kind of decay the coefficients can have while still giving rise to an $L^2$ function. -Are there sufficient conditions on the power series coefficients that will ensure that the function is in $L^2(\Bbb R)$? For instance, are there asymptotic bounds on the coefficients that will ensure that the function is in $L^2(\Bbb R)$ or is this an impossible task? - -REPLY [6 votes]: Since you ask about "real analytic functions", should it be $f:\mathbb{R}\to \mathbb{R}$ instead of $f:\mathbb{R}\to \mathbb{C}$? -I think analyticity is a "local" concept according to its definition. -Because of 2, $f$ may not have a global power series expansion. For example, $1/(1+x^2)$ is real analytic over $\mathbb{R}$, but the radius of convergence for its power series expansion at $0$ is $1$. -Regardless of 1-3, even assuming you were talking about real analytic functions with a global power series expansion, we have -$$\cos x = \sum_0^\infty (-1)^n\dfrac{x^{2n}}{(2n)!},$$ -where the absolute values of coefficients decay much faster than the example you gave. However, $\cos x \notin L^1(\mathbb{R})\cup L^2(\mathbb{R})$. -Actually, intuitively, the faster it decays, the less possible it will be in $L^2(\mathbb{R})$. Consider $f(x)=x$, where all coefficients vanish for terms $x^n (n > 1)$. Of course, $f$ is real analytic with a globally convergent power series expansion but $f\notin L^p(\mathbb{R})$ for any $0 -TITLE: The difference between KP+Inf+Pow and Z -QUESTION [5 upvotes]: I'd like to clarify some details about the theories Z (= Zermelo set theory) and KP+=KP+Infinity+Powerset (KP is Kripke-Platek set theory). -In this paper (M1), Mathias claims that Z+KP is consistent relative to Z; in particular, this implies that KP+ does not prove Con(Z). Moreover, by explicitly working in Z+KP, Mathias implies that Z does not contain KP. For the former fact he refers to this other paper (M2); however, I can't find that fact in (M2). But I could be missing something obvious - there are a ton of subtheories of ZFC discussed in (M2), and the claim could easily be a trivial consequence of one of the theorems Mathias proves, even if it's not stated explicitly. -I'd like to first double-check these claims: - -Question 1: Is it true that Z+KP is consistent relative to Z, and that Z does not contain KP? (If the answer to the first point is "yes", is this proved in (M2)? If not, where?) - -And then ask a couple related questions, comparing the strengths of Z and KP. -First proof-theoretically: - -Question 2: Does Z prove Con(KP+)? - -And then set-theoretically: - -Question 3: Is the smallest height of a transitive model of KP+ strictly smaller than the smallest height of a transitive model of Z? - -I suspect the answers to all three questions are "yes": that is, Z is not stronger than KP+, but it is "morally" stronger. However, I would like to check this. - -REPLY [4 votes]: Questions 2 and 3 both have a negative answer. -For 3, $V_{\omega + \omega}$ is a model of Z but KP + Inf proves the existence of all computable ordinals, so the height of a transitive model of KP+ must be much taller than $\omega + \omega$. -For 2, we in fact have that KP+ proves the consistency of Z. This is because from Powerset, $\Sigma_1$-Replacement, and the existence of $V_\omega$ we can prove the existence of $V_{\omega + \omega}$.<|endoftext|> -TITLE: Countable partitions of Cantor space mod meager -QUESTION [5 upvotes]: Let $I$ be an index set. Given $A\subseteq I\times 2^\omega$ and $i\in I$, set $(A)_i = \{x\in 2^\omega: (i, x)\in A\}$. -Now let $I\times 2^\omega = \bigcup_{n<\omega} A_n$. How large must $I$ be to ensure that there are indices $i\neq j\in I$ and $n< \omega$ with $(A_n)_i \cap (A_n)_j$ non-meager? In particular, is it consistent that there is a partition of $\omega_1\times 2^w$ into countably many pieces not satisfying the above? - -REPLY [2 votes]: Let $\kappa$ be least such that whenever $\{A^i_n : n < \omega\}$, for $i < \kappa$ are coverings of $2^{\omega}$, there are $i < j < \kappa$ and $n < \omega$ such that $A^i_n \cap A^j_n$ is non meager. -Claim: CH implies $\kappa > \omega_1$. -Proof: Using an Ulam matrix. -Claim: Assume MA plus not CH. Then $\kappa = \omega_1$. -Proof: Let $\{A^i_n: n < \omega\}$ be coverings of $2^{\omega}$ for $i < \omega_1$. Towards a contradiction, assume that for every $i < j < \omega_1$ and $n < \omega$, $A^i_n \cap A^j_n$ is meager. Put $B^i_n = A^i_n \setminus \bigcup \{A^j_n : j < i\}$ and $B^i = \bigcup \{B^i_n : n < \omega\}$. Then each $B^i$ is comeager and so is $B = \bigcap \{B^i : i < \omega_1\}$ (by MA plus not CH). But now if $x \in B$, then $(\forall i < \omega_1)(\exists n < \omega)(x \in B^i_n)$. Choose $i < j < \omega_1$ and $n < \omega$ such that $x \in B^i_n \cap B^j_n$: Contradiction. -So the lower bound $\omega_1 \leq \kappa$ cannot be improved in ZFC. I am not sure about the optimal upper bound for $\kappa$.<|endoftext|> -TITLE: An interesting variant on the maximum independent set problem. -QUESTION [5 upvotes]: Suppose i have a graph $G=(V,E)$ with $|V|=n$. Furthermore suppose i give you a maximum independent set $\mathcal{I}$ in $G$. Now suppose i obtain a new graph $G'$ from $G$ by removing a single vertex $v \in \mathcal{I}$ from $V$. Now of course $\alpha(G') \in \{|\mathcal{I}|,|\mathcal{I}|-1\}$ but can we say more? Is it an NP-hard to find a maximum independent set in $G'$? - -REPLY [3 votes]: Following the suggestion in Ben Barber's comment, any algorithm for solving your problem (of finding a maximum independent set in $G'$) can be used to find a maximum independent set in any finite graph. -Let G be a graph of order $n.$ For $0\le k\le n$ let $G_k$ be a graph of order $n+k$ obtained by adding to $G$ an independent set $\mathcal I_k$ of $k$ new vertices, and adding edges joining all the vertices in $\mathcal I_k$ to all the vertices of $G.$ -Since $\mathcal I_n$ is a maximum independent set in $G_n,$ we can use your algorithm to find a maximum independent set in $G_{n-1}.$ If the maximum independent set that we find in $G_{n-1}$ is $\mathcal I_{n-1},$ we use the algorithm again to find a maximum independent set in $G_{n-2},$ and so on. Eventually we will find a maximum independent set of some $G_{k}$ which is contained in $V(G).$<|endoftext|> -TITLE: Theorems that led to very successful research programs in Geometry and Topology -QUESTION [19 upvotes]: In the recent times I have heard a lot about the following: - -The Atiyah-Singer Index theorem -H-principle of Gromov ( and others ) - -It seems to me that these results led to decades of successful research in terms of improving the result, giving a simpler proof or new applications. -This motivates me to ask: What are the other such results which led to very successful research programs? I would add Bott Periodicity also. -Edit: With my thin background in geometry and topology I cannot make the question more precise or add more examples but I can tell why I would like to know such programs. After I heard these two theorems and their applications I thought every mathematician should know a little about these results even if they don't try to completely understand its proof. For example, Hamilton's 1982 result mentioned by PVAL is something I want to read about. Thanks for many good answers. Again, I emphasize that I have a very thin background in these things. - -REPLY [2 votes]: Work done by many in low-dimensional topology that laid the groundwork for Thurston's work and his conjecture, which was proved by Perelman. I don't want to list names, since I'm sure I'll miss some key people.<|endoftext|> -TITLE: Are there recursive sets $X$ with Property A that contain infinitely many incompressible strings? -QUESTION [6 upvotes]: Let us say a set $X$ satisfies Property A if$$\liminf_{n \to \infty} {{\left|X^{\le n}\right|}\over n} = 0.$$Are there recursive sets $X$ satisfying Property A that contain infinitely many incompressible strings? - -REPLY [3 votes]: I'm assuming you call `incompressible' integers whose binary representation $x$ satisfy $K(x)>|x|-c$ for a fixed $c$? -If that's the case, the answer is no. If $X$ is a computable set of integers and some integer $k$ of representation $x$, with $|x|=N$, belongs to $X$, then $K(x|N) \leq \log |X^{\leq 2^{N+1}}| + c'$ for a fixed $c'$. (Indeed one can describe $x$ by just giving its position as a string of length $N$ inside $X$). But by your assumption on $X$, $\log |X^{\leq 2^{N+1}}| - N$ tends to $-\infty$, so almost all strings in $X$ satisfy, say, $K(x\mid|x|) < |x| - 3c$. You can then remove the the condition $|x|$ and get $K(x) < |x| - 2c$ by a usual padding technique (see the book by Li and Vitanyi, I can elaborate if this is important).<|endoftext|> -TITLE: Alternative proof of a theorem of Riesz -QUESTION [9 upvotes]: My question is not research level, but I have not received any feedback on Mathstack; so I am posting it here. I am aware of the traditional proof of the Riesz Theorem that relates linear functionals on spaces of continuous functions on a locally compact space to measures. I want to do it in a different way, and would appreciate any comments. -Let $X$ be a compact Hausdorff space, $C(X)$ the space of continuous functions on $X$, and $\mathcal B_X$ the Borel sets on $X$. Then, if $\phi:C(X)\to \mathbb C$ is a positive linear functional, there is a measure $\mu:\mathcal B_X \to \mathbb R$. such that $\phi (f)=\int_X fd\mu$ for each $f\in C(X)$. -I want to prove this using the fact that every compact metric space $X$ is the continuous image of a surjection, $p$, from $2^{\mathbb N}$. Of course, $2^{\mathbb N}$ has the product topology which is also induced by the metric $d(x,y)=1/n$ where $n$ is the least integer such that $x_n\neq y_n$. -Here is an outline, and then questions at the end. -1) First, suppose $X=2^{\mathbb N}$. Then $\mathcal A=\left \{ \pi^{-1}_n(E):E\in 2^{n} \right \}$ is a subbase for the topology of $2^{\mathbb N}$, and so is an algebra that generates $\mathcal B_{2^{\mathbb N}}$. We have that $\chi_A$ is continuous for each $A\in \mathcal A$, so we set $\mu (A)=\phi (\chi_A)$. Then $\mu $ is finitely additive, and in fact countably additive vacuously (because each element of $\mathcal A$ is compact). Using Carathéodory and a density argument now gives the result if $X=2^{\mathbb N}$. -2) Let $X$ be any compact metric space. We may assume that $\left \| \phi \right \|=1$. Moreover, it is not hard to show that in general $\phi$ is a positive linear functional if and only if $\left \| \phi \right \|=\phi (1)$. Now the map $f\mapsto f\circ p$ is an isometry and $M=\left \{ f\circ p:f\in C(X) \right \}$ is a subspace of $C(2^{\mathbb N})$. Then $\tilde{\phi}_M(f\circ p):=\phi (f)$ is a functional on $M$ which extends, via Hahn-Banach, to $\tilde {\phi}$ on all of $C(2^{\mathbb N})$. Since $\left \| \tilde { \phi }\right \|=\left \| \tilde { \phi_M }\right \|=\left \| \phi \right \|=\phi (1)=\tilde { \phi }(1\circ p)=\tilde { \phi (1) }$, $\tilde { \phi }$ is a positive linear functional; by 1), there is a measure $\tilde { \mu }$ such that $\tilde {\phi}(g)=\int _{2^{\mathbb N}}gd\tilde {\mu}$. Thus we compute $\phi (f)=\tilde {\phi}(f\circ p)=\int _{2^{\mathbb N}}f\circ pd\tilde {\mu}=\int _{X}fd(\tilde {\mu}\circ p^{-1})$. All that remains is uniqueness of the measure $\nu =\tilde {\mu}\circ p^{-1}$, but this follows exactly as in the traditional proof. -Now I want to extend this to an arbitrary compact Hausdorff space, and from there treat the locally compact case. I only have a very vague idea of how to proceed, but here is my idea: -Look first at the Baire sets of $X$; if $f:X\to Y$ is continuous to a compact metric space, and if $E$ is Baire (in this case it is Borel as well) in $Y$, then $f^{-1}(E)$ is Baire in $X$. Consider the collection $\mathcal C$ of all pairs $(f,Y)$ such that $f:X\to Y$ is a continuous map into a compact metric space. Now consider all sets of the form $f_Y^{-1}(E)$ where $Y$ is a compact metric space, $f:X\to Y$ is continuous, and $E$ is Baire in Y. Next I would show that the union of all these constitutes the Baire sets in $X$. Then, for each compact metric space $Y$ and each $f:X\to Y$, and using the trick in 2), I find a measure $\mu_Y$ such that $\int_Y gd\mu_Y=\tilde {\phi}(g):=\phi (g\circ f)$. Finally, I patch these measures together to get the measure on the Baire sets of $X$ and from there, to the Borel sets. - -REPLY [5 votes]: A proof along the lines that you describe was worked out by V.S. Sunder here. -There are a few different approaches that you can take to reduce the Riesz Representation Theorem to a class of simpler spaces. For compact spaces, you can either reduce to compact metric spaces like Sunder does or directly to projective (i.e. extremally disconnected) compact spaces like Carothers. -Sunder's reduction to the compact metric case relies on the fact that every Baire set in a compact space is a continuous preimage of a Baire set in $[0, 1]^\mathbb{N}$. From here, you have two choices. You can either reduce to the case of $2^\mathbb{N}$ as you describe, or you can work directly with $[0, 1]^\mathbb{N}$. Sunder does the former, but I think you should also be able to do the latter and use an explicit proof for $[0, 1]$, e.g. in terms of Bernstein polynomials. -Carothers takes a more direct approach. Given a compact space $X$, he considers the projective cover of $X$ given by the Stone-Čech compactification of the discretization of $X$, where the clopen sets generate the Baire sets and finite additivity on clopen sets implies countable additivity. I personally feel that this better motivates advanced mathematics than the metric space proof, since the use of projective covers of topological spaces corresponds to the use of injective envelopes of $C^*$-algebras. -To generalize from compact spaces to locally compact spaces, you have a few choices. You can use the fact that locally compact spaces are compactly generated, which is roughly what Sunder does. You can also use compactifications, e.g. the one-point or Stone-Čech compactifications. -The use of the one-point compactification corresponds to the unitization of $C(X)$ as a $C^*$-algebra, where positive linear functionals have unique extensions to unitizations by taking a limit over a contractive approximate identity, i.e. again by approximating from compact subsets. If you worked it out explicitly in terms of topological spaces, it would look pretty similar to Sunder's proof. -The approach using the Stone-Čech compactification is a bit more interesting because it generalizes to arbitrary completely regular spaces. The best treatment of this I have seen is Measures on Topological Spaces by J.D. Knowles. The approach there uses both the Markov-Alexandrov approach of representing linear functionals on $C_b(X)$ in terms of finitely additive measures, as well as the Riesz-Kakutani approach on $C(\beta X)$ in terms of countably additive measures, and then studies the relationship between finite additivity, countable additivity, and inner approximations by compact sets.<|endoftext|> -TITLE: SOLVED: How to retrieve Eigenvectors from QR algorithm that applies shifts and deflation -QUESTION [5 upvotes]: After having googled for several days without locating a definitive answer, I will try my luck here! -I have implemented a version of the QR algorithm to calculate Eigenvalues and hopefully Eigenvectors of a matrix $A$ of dimension $n\times n$. -In order to speed up the convergence rate i have applied a version of the algorithm that utilizes several improvenments, mainly inspired by https://www.math.kth.se/na/SF2524/matber15/qrmethod.pdf. -Firstly, we calculate $H=Hessenberg(A)$, which transforms $A$ to upper Hesenberg form $H_{n\times n}$. -Now, using $H$ we estimate the Eigenvalues using a QR-algorithm applying a Wilkinson shift and deflation. This can loosely be described as the following pseudo procedure, where $\lambda_i$ denotes the i'th Eigenvalue: -$\text{set}\ H_0:=H\\ -\text{for}\ m=n,\ldots,2 \ \text{do} \\ -\quad k=0\\ -\qquad \text{repeat}\\ -\qquad \quad k=k+1\\ -\qquad \quad \sigma_k=Wilkinson(H_{k-1})\\ -\qquad \quad H_{k-1}-\sigma_kI=:Q_kR_k\quad (*)\\ -\qquad \quad H_k=R_kQ_k+\sigma_kI\\ -\qquad \text{until}\ \vert h^{(k)}_{m,m-1}\vert<\epsilon\\ -\qquad \lambda_m=h^{(k)}_{m,m}\\ -\qquad H^{(0)}=H^{(k)}_{1:(m-1),1:(m-1)}\\ -\text{end for}$ -The last step is simply a deflation step that drops the last row and column of $H$ upon satisfactory convergence towards an eigenvalue. -The function $Wilkinson$ calculates the shift, and the QR factorization $(*)$ is done using Givens rotations, which is a standard procedure. -My question is, how do i determine the corresponding eigenvectors? In the standard QR-algorithm this can be computed as $\Pi _iQ_i$, however due to the deflation step I don't know how to proceed? -I have verified that eigenvalues are calculated correctly. -In advance, thank you very much for any help! -EDIT: Implementing Federico's solution below works as intended. Make sure you use your initial similarity transform $H=UAU^*$, where $H$ is upper Hessenberg, i.e. let $\bar{Q}_H$ be the eigenmatrix of $H$ then $\bar{Q}_A=U^*\bar{Q}_H$ yields the eigenmatrix of $A$. - -REPLY [2 votes]: Instead of dropping one row and one column, compute at each step a $(n-1)\times(n-1)$ orthogonal transformation (or $(n-k)\times(n-k)$, after $k$ deflation steps) $Q$ by working to the reduced matrix, and then apply it to the full matrix as -$$ -\begin{bmatrix} -Q^* \\& I -\end{bmatrix} -\begin{bmatrix} -H_{11} & H_{12}\\ -0 & H_{22} -\end{bmatrix} -\begin{bmatrix} -Q \\& I -\end{bmatrix} -= -\begin{bmatrix} -Q^*H_{11}Q & Q^*H_{12}\\ -0 & H_{22} -\end{bmatrix}. -$$ -In practice all you have to do is operating on the leading $(n-k)\times(n-k)$ block as you were doing before, and then multiplying $H_{12}$ by the orthogonal transformation $Q$ that you have generated. -In this way, your algorithm computes explicitly a sequence of $n\times n$ orthogonal transformations $Q_1, Q_2, \dots, Q_m$ that turns $A$ into a triangular matrix (Schur form). You can accumulate the product $Q_1Q_2\dotsm Q_m$ with $O(n^2)$ additional operations per step (so $O(n^3)$ in total during the algorithm, under the usual assumptions that $O(1)$ iterations per eigenvalue are sufficient). After that, all you have to do is recover the eigenvectors from the Schur form. -I hope this is sufficiently clear!<|endoftext|> -TITLE: Is there a useful limit or co-limit of a diagram that has only a single object? -QUESTION [9 upvotes]: I'm starting to study category theory kind of informally and everytime I read about the definitions of limits and co-limits, the first three examples are always the same: - -terminal/initial objects, -products/coproducts, -and pullbacks/pushouts. - -It's always explained how terminal (initial) objects are the limit (colimit) of a an empty diagram and products (coproducts) are the limit (colimit) of a diagram with only two objects with no additional structure. -I always get that feeling of "what about the diagram with a single object?", but since that's never mentioned, I figured it must be something trivial or not useful. In a Poset it feels it would be some kind of lowest upper bound/greatest lower bound of a single object or something like that, but that looks like it would be trivially equal to the object itself. So, first question: is this limit interesting or is it trivial or maybe not used anywhere? -A related question: -Recently I started reading studying the book Mathematical Physics by Robert Geroch (*) and when he defines a free group it feels a lot like what the limit of a diagram with a single object would be: "a free group on the set $S$ is a group $G$ together with a mapping $\alpha$ from $S$ to $G$ such that for any other group $G'$ with a mapping $\alpha'$ from $S$ to $G'$, there exists a unique homomorphism $\mu: G \to G'$ such that the diagram commutes", meaning that $\alpha' = \mu \cdot \alpha$. But of course that's not a limit because G and S belong to different categories and $\alpha$ and $\alpha'$ are not morphisms. The books defines other kinds of free things after that in a similar way -- free vector spaces, free topological spaces, etc. So, the second question is: are those free constructions related to limits? Is there some category where those constructions are actual limits of a diagram with a single object? -(*) I'm a physicist and my mathematical training was always focused on mathematical methods for physics and very little in formal, rigorous mathematics - since this book doesn't assume a lot of education in things like topology, algebraic geometric, etc, it feels more adequate for my background. Would love other suggestions. - -REPLY [2 votes]: Let $|G|$ be set of elements of a group $G$. Similarly, if $f : G \to H$ is a group homomorphism, let $|f|$ be the underlying mapping on elements. A diagrammatic translation of that definition of free group is as follows. -Consider the category whose objects are: - -Pairs $(G',\alpha')$ consisting of a group $G'$ together with a function of sets $\alpha' : S \to |G'|$ -Arrows $(G',\alpha') \to (G'', \alpha'')$ are group homomorphisms $\beta : G' \to G''$ satisfying $|\beta| \circ \alpha' = \alpha''$ - -It turns out this category has an initial object (which can be seen either as the colimit of an empty diagram or the limit of the entire category), and the pair $(G,\alpha)$ is taken to be an initial object. -So, I think your intuition was mashing together two separate facts: - -The free group is indeed the colimit of a fairly small diagram — the empty one! -All of the objects in this category involve maps from the single set $S$ - - -Incidentally, this sort of pattern comes up a lot — constructing a category whose objects are collections of interesting data, and then doing things in that category. -The particular category above is an example of a comma category from the constant functor $* \to \mathbf{Set}$ whose image is $S$ to the forgetful functor $\mathbf{Grp} \to \mathbf{Set}$. In fact, one of the things you can do with comma categories is precisely to make sense of arrows between objects of different categories. -The construction above is also an example of an adjunction; the "free group" functor $\mathbf{Set} \to \mathbf{Grp}$ is the left adjoint to the "underlying set of elements" functor $\mathbf{Grp} \to \mathbf{Set}$.<|endoftext|> -TITLE: Is there a nonelementary hyperbolic group without this transitivity property? -QUESTION [8 upvotes]: Some background (see e.g. the books by Ghys & de la Harpe or Bridson & Haefliger for more information): -Let $\Gamma$ be a group with a finite symmetric generating set $S$. Recall that $\Gamma$ is called a (word) hyperbolic group if the Cayley graph of $(\Gamma, S)$ is hyperbolic (in the sense of Gromov). This notion is independent of the choice of the generating set $S$. -Assuming $\Gamma$ to be hyperbolic, there is a finite directed graph canonically associated to $(\Gamma,S)$ called the geodesic automaton (GA), whose arrows are labelled by elements of $S$. Paths in the GA graph starting from a special basepoint correspond to segments of geodesics in the Cayley graph. -Now consider the recurrent vertices of the GA graph, i.e., the set of vertices that belong to loops. Following this paper by Haissinsky, Mathieu, and Mueller, let us say that the GA graph is strongly connected if given any two recurrent vertices, there is one path from one to the other. -For example, if $\Gamma = \mathbb{Z}$ and $S = \{1,-1\}$ then the GA graph is: -which is obviously not strongly connected. - -Question. If $\Gamma$ is a nonelementary hyperbolic group, is the geodesic automaton graph strongly connected for some (all?) choice of $S$? - -REPLY [5 votes]: If I understand correctly, you are asking if the automaton graph is recurrent, that is there is only one non trivial recurrence class (a recurrence class is trivial if it is reduced to one vertex). -For Fuchsian groups, you can always find a generating set such that the automaton graph is recurrent. This was proved by Caroline Series in the paper The infinite word problem and limit sets in Fuchsian groups (Ergodic theory and dynamical system, 1981). - -However, in general, we do not know how to construct such an $S$. -Applying Cannon's method does not necessarily yields a recurrent graph. -Here is an example. Take the free group with two generators $a,b$. -Taking the generating set $S=\{a,b,a^{-1},b^{-1}\}$ would give you a connected automaton. Indeed, two elements have the same cone-type if and only if their normal forms end with the same letter. So you have 5 cone-types which gives you 5 states in the automaton : $e$ (the neutral element), $a$, $b$, $a^{-1}$ and $b^{-1}$. You can go from $e$ to any other cone-type, except itself, and in fact you can go from any state to any other state except $e$ and its inverse. -Now, take the generating set $S'$ consisting of every element with one or two letters, that is $S'$ is the ball of radius 2 minus $e$ for the word metric coming from $S$. -To compute the cone-types, first notice that if $g$ has odd $S$-length $2n+1$ (that is, it is at distance $2n+1$ from $e$ for the generating set $S$), then any $S$'-geodesic from $e$ to $g$ will have length $n+1$. More precisely, it will consist of $n$ increments of $S$-length 2 and 1 element of $S$-length 1. In particular, you can only extend such a geodesic adding increments of $S$-length 2. What it means is that once you reached a odd $S$-length element, you can only extend geodesics to odd $S$-length elements, while for an even $S$-length element, you can either extend it to another even $S$-length element, or add a $S$-length 1 increment and get stuck in the odd $S$-length component. -Note also that the only thing (besides parity) that counts, to know how you can extend geodesics, is the last letter in the normal form. -Combining all this, you see there are 9 distinct cone-types : that of $e$, those of $a,b,a^{-1},b^{-1}$ and those of $ab,ab^{-1},ba,ba^{-1}$. In particular, note that $ab$ and $a^{-1}b$ have the same cone-type. -You can go from $a$ to any other state with one letter (for example you can go from $a$ to $a^{-1}$ adding the element $ba^{-1}$, which you could not do using the generating set $S$). More generally, the sub-graph consisting of states with one letter is strongly connected. Similarly, the sub-graph consisting of states with two letter is strongly connected. You can also go from $ab$ to any state with one letter, except $b^{-1}$, etc... -The only thing you need to do is to choose an order on elements of $S'$ and to consider only geodesics that are shortest for the lexicographical order. But even if you need to remove some of the edges, this does not change the components. What you get in the end is a graph with exactly two recurrent classes (odd and even $S$-lengths cone-types). You can go from the even-type reccurent class to the odd one, but not the other way around. And of course, there is also one edge from $e$ to any other state. -I hope this was clear enough. You see here that even for the free group, given some generating set, it is very hard to tell whether you will get a recurrent automaton or not. Here it was possible to compute the cone-types, but this might be a very difficult task in general. Asking around, it seems to me that some people believe that it should be possible to always find a good $S$ (I mean with a recurrent associated automaton). Some others think that this is impossible to tell. - -Now, I'll try to explain how you can get around this difficulty. I strongly recommend the survey of Danny Calegari The ergodic theory of hyperbolic groups (available on his webpage). -If you want to do dynamics on the path space of the automaton, say you have a function defined on this path space, you can look at the so-called maximal components. Those are the recurrent components of you graph such that the spectral data of your function (usually the dominant eigenvalue of the associated transfer operator) is maximal. What can go wrong is if you have a path from one maximal component to another one (but not the other way around, otherwise these would not be two different components). -This was explored in the paper of Danny Calegari and Koji Fujiwara Combable functions,quasimorphisms, and the central limit theorem (Ergodic theory and dynamical systems, 2010). See also the paper of Sébastien Gouëzel Local limit theorem for symmetric random walks in Gromov-hyperbolic groups (Journal of the AMS, 2014).<|endoftext|> -TITLE: modulus-related analytic functions -QUESTION [7 upvotes]: Let $D\subset\mathbb{C}$ be the open unit disk. Suppose $f,g,F,G:D\rightarrow\mathbb{C}$ are analytic functions linked by -$$\vert f(z)\vert^2+\vert g(z)\vert^2=\vert F(z)\vert^2+\vert G(z)\vert^2; \qquad \forall z\in D.$$ - -Question 1. If $f\neq \alpha g$ and $g\neq\beta f$ for any $\alpha, \beta\in\mathbb{C}$ then is the same true for $F$ and $G$? - -Caveat. Not true for real analytic functions: -take $f=\sin x, \,g=\cos x, \, F=\frac1{\sqrt{2}}=G$. -EDIT. Encouraged by Christian's positive answer, let's upgrade the problem. - -Question 2. Suppose $f, g, h$ are linearly independent (over $\mathbb{C}$) such that - $$\vert f\vert^2+\vert g\vert^2+\vert h\vert^2=\vert F\vert^2+\vert G\vert^2+\vert H\vert^2.$$ - Should $F, G, H$ be linearly independent, too? - -REPLY [2 votes]: In fact, more is true. If $f_j$ are linearly independent and $F_i$ are -linearly independent, and -$$\sum_{j=1}^n|f_j|^2=\sum_{i=1}^m|F_i|^2,$$ -then $m=n$ and $F_i$ are obtained from $g_i$ by a unitary transformation. -See, for example, https://arxiv.org/pdf/math/0007030.pdf, section 3. This is called the "Calabi rigidity".<|endoftext|> -TITLE: Conditions for a finite group to be isomorphic to its automorphism group -QUESTION [7 upvotes]: So in the interest of gaining a better understanding of a conjecture (due to Scott, 1960) on the automorphism series (first part of the automorphism tower, no direct limits) of a finite group that every finite group has an automorphism series that is eventually constant (periodic w/ period 1). I used GAP to create a complete list up to order $511$ of those finite groups which satisfy $G\simeq Aut(G)$ (henceforth called $Aut$-stable groups). Studying the list briefly made me notice a number of interesting patterns. One of the most striking features is that all of these groups were either centerless, or had $Z(G)\simeq\mathbb{Z}_2$. These are precisely the two groups which have trivial automorphism group, and I suspect that this fact is true of all (finite) $Aut$-stable groups. The condition is not sufficient on its own (e.g. there are plenty of centerless groups that aren't $Aut$-stable, but thanks to the classic theorem of Wielandt, 1939, every centerless group has an automorphism series that stabilizes in finitely many steps (sidenote: if anyone has access to a detailed English version of the proof, I'd love to see it; original is in German)), so this leaves me to the task of attempting to ascertain what conditions are necessary and sufficient for $Aut$-stability. -So my (somewhat broad) question is: What conditions, for a finite group, are necessary and/or sufficient for $Aut$-stability? I am not necessarily expecting a full answer as this question could very well be open at present, but anything that might help shed some light on the why and how is welcome. (Additionally, if this topic has been studied in the literature, references would be appreciated.) -The list of $Aut$-stable groups of order up to $511$, with GAP ids and structure descriptions: -$(1,1)\simeq\mathbb{Z}_{1}$ -$(6,1)\simeq S_{3}$ -$(8,3)\simeq D_{8}$ -$(12,4)\simeq D_{12}$ -$(20,3)\simeq\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4}$ -$(24,12)\simeq S_{4}$ -$(40,12)\simeq\mathbb{Z}_{2}\times(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})$ -$(42,1)\simeq(\mathbb{Z}_{7}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$ -$(48,48)\simeq\mathbb{Z}_{2}\times S_{4}$ -$(54,6)\simeq(\mathbb{Z}_{9}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$ -$(84,7)\simeq\mathbb{Z}_{2}\times((\mathbb{Z}_{7}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2})$ -$(108,26)\simeq\mathbb{Z}_{2}\times((\mathbb{Z}_{9}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2})$ -$(110,1)\simeq(\mathbb{Z}_{11}\rtimes\mathbb{Z}_{5})\rtimes\mathbb{Z}_{2}$ -$(120,34)\simeq S_{5}$ -$(120,36)\simeq S_{3}\times(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})$ -$(144,182)\simeq((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes\mathbb{Z}_{8})\rtimes\mathbb{Z}_{2}$ -$(144,183)\simeq S_{3}\times S_{4}$ -$(156,7)\simeq(\mathbb{Z}_{13}\rtimes\mathbb{Z}_{4})\rtimes\mathbb{Z}_{3}$ -$(168,43)\simeq((\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{7})\rtimes\mathbb{Z}_{3}$ -$(216,90)\simeq(((\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$ -$(220,7)\simeq\mathbb{Z}_{2}\times((\mathbb{Z}_{11}\rtimes\mathbb{Z}_{5})\rtimes\mathbb{Z}_{2})$ -$(240,189)\simeq\mathbb{Z}_{2}\times S_{5}$ -$(252,26)\simeq S_{3}\times(\mathbb{Z}_{7}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$ -$(272,50)\simeq\mathbb{Z}_{17}\rtimes\mathbb{Z}_{16}$ -$(312,45)\simeq\mathbb{Z}_{2}\times(\mathbb{Z}_{13}\rtimes\mathbb{Z}_{4})\rtimes\mathbb{Z}_{3}$ -$(320,1635)\simeq((\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2)\rtimes\mathbb{Z}_5)\rtimes\mathbb{Z}_4$ -$(324,118)\simeq S_{3}\times(\mathbb{Z}_9\rtimes\mathbb{Z}_3)\rtimes\mathbb{Z}_2)$ -$(336,208)\simeq PSL(3,2)\rtimes\mathbb{Z}_2$ -$(342,7)\simeq (\mathbb{Z}_{19}\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_2$ -$(384,5677)\simeq((((\mathbb{Z}_{4}\times\mathbb{Z}_{4})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2})\rtimes\mathbb{Z}_{2})\rtimes\mathbb{Z}_{2}$ -$(384,5678)\simeq((((\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2})\rtimes\mathbb{Z}_{2})\rtimes\mathbb{Z}_{2}$ -$(432,520)\simeq(((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes\mathbb{Z}_{3})\rtimes Q_{8})\rtimes\mathbb{Z}_{2}$ -$(432,523)\simeq(((\mathbb{Z}_{6}\times\mathbb{Z}_{6})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2})\rtimes\mathbb{Z}_{2}$ -$(432,533)\simeq\mathbb{Z}_{2}\times((((\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2})$ -$(432,734)\simeq(((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes Q_{8})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$ -$(480,1189)\simeq(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})\times S_{4}$ -$(486,31)\simeq(\mathbb{Z}_{27}\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_{2}$ -$(500,18)\simeq(\mathbb{Z}_{25}\rtimes\mathbb{Z}_{5})\rtimes\mathbb{Z}_{4}$ -$(506,1)\simeq(\mathbb{Z}_{23}\rtimes\mathbb{Z}_{11})\rtimes\mathbb{Z}_{2}$ - -REPLY [4 votes]: Your "centerless" observation is on mark. -If $G$ is a complete group then $G\cong Aut(G)$. A complete group is in particular centerless. -Incidentally, the converse is not true: for instance take the Dihedral group $D_8$.<|endoftext|> -TITLE: Is $x^{2k+1} - 7x^2 + 1$ irreducible? -QUESTION [28 upvotes]: Question. Is the polynomial $x^{2k+1} - 7x^2 + 1$ irreducible over $\mathbb{Q}$ for every positive integer $k$? - -It is irreducible for all positive integers $k \leq 800$. - -REPLY [43 votes]: Here is a proof, based on a trick that can be used to prove that -$x^n + x + 1$ is irreducible when $n \not\equiv 2 \bmod 3$. -We work with Laurent polynomials in $R = \mathbb Z[x,x^{-1}]$; note that -$R$ has unit group $R^\times = \pm x^{\mathbb Z}$. -We observe that for $f \in R$, the sum of the squares of the coefficients -is given by -$$\|f\| = \int_0^1 |f(e^{2 \pi i t})|^2\,dt - = \int_0^1 f(e^{2 \pi i t}) f(e^{-2 \pi i t})\,dt - = \int_0^1 f(x) f(x^{-1})\big|_{x = e^{2 \pi i t}}\,dt .$$ -Now assume that $f(x) = g(x) h(x)$. Then, since -$f(x) f(x^{-1}) = \bigl(g(x)h(x^{-1})\bigr)\bigl(g(x^{-1})h(x)\bigr)$, -$G(x) := g(x) h(x^{-1})$ satisfies $\|G\| = \|f\|$ -and $G(x) G(x^{-1}) = f(x) f(x^{-1})$. -Now we consider $f(x) = x^n - 7 x^2 + 1$; then $\|f\| = 51$. -If $f = g h$ as above, then write -$G(x) = \pm x^m(1 + a_1 x + a_2 x^2 + \ldots)$ -and $G(x^{-1}) = \pm x^l(1 + b_1 x + b_2 x^2 + \ldots)$. -The relation $G(x) G(x^{-1}) = f(x) f(x^{-1})$ translates into -(equality of signs and) -$$(1 + a_1 x + \ldots)(1 + b_1 x + \ldots) = 1 - 7 x^2 + O(x^{n-2}).$$ -Assuming that $n > 40$ and considering terms up to $x^{20}$, one can -check (see below) that the only solution such that -$a_1^2 + a_2^2 + \ldots + a_{20}^2 + b_1^2 + b_2^2 + \ldots + b_{20}^2\le 49$ -is, up to the substitution $x \leftarrow x^{-1}$, given by -$1 + a_1 x + \ldots = 1 - 7x^2 + O(x^{21})$, -$1 + b_1 x + \ldots = 1 + O(x^{21})$. Since the $-7$ (together with the -leading and trailing 1) exhausts our allowance -for the sum of squares of the coefficients, all other coefficients -must be zero, and we obtain that $G(x) = \pm x^a f(x)$ -or $G(x) = \pm x^a f(x^{-1})$. Modulo interchanging $g$ and $h$, -we can assume that $g(x) h(x^{-1}) = \pm x^a f(x)$, -so $h(x^{-1}) = \pm x^a f(x)/g(x) = \pm x^a h(x)$, -and $x^{\deg h} h(x^{-1})$ divides $f(x)$. -This implies that $h(x)$ divides $x^n f(x^{-1})$, so $h(x)$ must divide -$$f(x) - x^n f(x^{-1}) = 7 x^2 (x^{n-4} - 1).$$ -So $h$ also divides -$$f(x) - x^4 (x^{n-4} - 1) = x^4 - 7 x^2 + 1 = (x^2-3x+1)(x^2+3x+1).$$ -Since $h$ also divides $f$, it must divide the difference $x^n - x^4$, -which for $n \neq 4$ it clearly doesn't, since the quartic has no -roots of absolute value 0 or 1; contradiction. -The argument shows that $x^n - 7 x^2 + 1$ is irreducible for $n > 40$; -for smaller $n$, we can ask the Computer Algebra System we trust. -This gives: -Theorem. $x^n - 7 x^2 + 1$ is irreducible over $\mathbb Q$ -for all positive integers $n$ except $n=4$. -ADDED 2017-01-08: -After re-checking the computations, I realized that there -was a small mistake that ruled out some partial solutions prematurely. -It looks like one needs to consider terms up to $x^{20}$. -Here is a file with MAGMA code that verifies that -there are no other solutions.<|endoftext|> -TITLE: Radon-Nikodym theorem for non-sigma finite measures -QUESTION [11 upvotes]: Let $(X,\mathcal M, \mu)$ be a measured space where $\mu$ is a positive measure. -Let $\lambda$ be a complex measure on $(X,\mathcal M)$. When $\mu$ is sigma-finite, the Radon-Nikodym theorem provides a decomposition of $\lambda$ in a sum of an absolutely continuous measure wrt $\mu$ plus a singular measure wrt $\mu$. -Question. Is that statement still true without the sigma-finiteness of $\mu$? It does seems to be so easy for the case of the Hausdorff measures of dimension $d0$ there exists $\delta>0$ such that for all $A\in\Sigma$, $\mu(A)\leq\delta$ implies $|\nu(A)|\leq\epsilon$. -We say that $\nu$ is truly continuous with respect to $\mu$ if and only for all $\epsilon>0$ there exists $\delta>0$ and $F\in\Sigma$ with $\mu(F)<\infty$ such that for all $A\in\Sigma$, $\mu(A\cap F)\leq\delta$ implies $|\nu(A)|\leq\epsilon$. -If $(X,\Sigma,\nu)$ is $\sigma$-finite, then a real-valued signed measure $\nu$ on $\Sigma$ is truly continuous with respect of $\mu$ if and only if it is absolutely continuous with respect to $\mu$. -Theorem: Let $(X,\Sigma,\nu)$ be a measure space and $\nu$ be a real valued signed measure on $\Sigma$. Then $\nu$ is truly continuous with respect to $\mu$ if and only if there exists a measurable function $f:\Sigma\to \mathbb{R}$ such that $$\nu(A)=\int_A f~\text{d}\mu$$ for all $A\in\Sigma$. -This formulation of the Radon-Nikodym is from the second volume of David Fremlin's treatise on measure theory. Fremlin doesn't assume $\nu$ to be countably additive, which requires additional distinctions. There is also a treatment in Section 5.4 of the book Measure and Integration by Dietmar Salamon (preprint available here), which also contains a useful characterization by Heinz König of being truly continuous in terms of absolute continuity and an inner regularity property<|endoftext|> -TITLE: The minimum of alpha times omega for vertex-transitive graphs -QUESTION [6 upvotes]: Definitions: For a graph $G$ we denote the independence number by $\alpha(G)$ and the clique number by $\omega(G)$. A graph is vertex-transitive if for every pair of vertices $x,y \in V(G)$ there is an automorphism that takes $x$ to $y$. -Question: What is the order of magnitude of the minimum of $\alpha(G)\times \omega(G)$ for vertex-transitive graphs? -Motivation/Additional information: I was thinking about a conjecture of János Körner. Let us say that two permutations have a flip if there are two coordinates that contain the same two elements in both permutations but the order of the two elements is different. Körner conjectures that the maximal size of a set of permutations where each pair have a flip is only exponential in the length of the permutations. One way to look at this problem is as follows. Define a graph $H$ where the vertices are the permutations, and we connect two vertices if the corresponding permutations have a flip. Then we have $|V(H)|=m!$. Körner conjectures that $\omega(H)$ is only exponential in $m$. Josef Cibulka determined the order of magnitude of the independence number of this graph: $\alpha(H) \sim m^ \frac{m}{2} $ (The construction is very nice, he calls a "flip" a "reverse", the paper can be found here https://pdfs.semanticscholar.org/aa5f/7de6b1727bb7618d6674f2aa848bf69335c6.pdf). The following inequality holds for all vertex transitive graphs, hence also for $H$: $$\alpha(H) \times \omega(H) \leq |V(H)|.$$ -If the conjecture of Körner is true then $\alpha(H) \times \omega(H) \sim m^{\frac{m}{2}-o(m)}$ and $|V(H)|\sim m^m$. If we normalize the number of vertices of $H$ to be $n$ (as usual), we have a vertex transitive graph where $\alpha \times \omega$ is as small as $n^{\frac{1}{2}-o(1)}$. But I could not construct such a graph. -For not necessarily vertex transitive graphs $\alpha(G)\times \omega(G)$ can be as small as $\log(n)^2$, an example is an Erdős-Rényi random graph for $p=1/2$. (I am not sure whether this is indeed the minimum for general graphs, but this is not relevant for the conjecture that I am interested in.) -I could construct vertex-transitive graphs where $\alpha \times \omega \sim n^{c-o(1)}$ for $c \in (1/2,1)$ as follows: The pentagon has $\alpha \times \omega=4$ and $n=5$. We can blow up a graph as follows. If $G$ has $n$ vertices, take $n$ vertex disjoint copies of $G$. Fix an arbitrary pairing of these vertex disjoint copies with the vertices of the original graph. And if two vertex disjoint copies have their pairs connected in $G$, then add every edge between these disjoint copies. It is easy to see that if $G'$ is the blowup of $G$, then $\alpha(G')=\alpha(G)^2$ and $\omega(G')=\omega(G)^2$ and $|V(G')|=|V(G)|^2$ and if $G$ was vertex-transitive, $G'$ is also. Thus blowing up the pentagon many times we get a sequence of graphs where (after normalizing the number of vertices) we have $\alpha \times \omega \sim n^{0.8613}$. - -REPLY [5 votes]: One very promising place to look is at Paley Graphs which are self dual (so $\alpha=\omega$.) This answer to a question suggests that, based on prime $n \lt 10000,$ it might be that $\alpha \omega =O(log^4 n).$ Although it been the subject of a fair amount of research, all that is known for sure is that $\log n \lt \alpha=\omega \le \sqrt{n}.$ The upper bound can, evidently, be reduced to $\sqrt{n}-1$ infinitely often. -Any other circulant graphs could be considered and those with degree $\frac{n-1}{2}$ (so edge density $\frac12$) do seem optimal. A pentagon is a circulant graph (in fact a Paley Graph) and the tensor product of circulant graphs is a circulant graph. -Any graph of $R(3,3)=6$ vertices or more has $\max(\alpha,\omega) \ge 3.$ In general, one less than a Ramsey number $R(s,s)-1$ is the largest size for a graph with $\max(\alpha,\omega) \lt s.$ -$R(4,4)-1=17$ and among all graphs on 17 vertices (regular or not) the Paley Graph on $17$ vertices is the unique example with $\alpha \lt 4$ and $\omega \lt 4.$ -No larger Ramsey numbers are known although $42 \lt R(5,5) \le 49.$ No graph on $42$ vertices could be a Paley graph and there are at least $656$ known with $\alpha=\omega=4.$ The Paley Graph on $37$ vertices has $\alpha=\omega=4$ but the Paley Graph on $41$ vertices has $\alpha=\omega=5.$ -So Paley graphs are perhaps not the absolute best but they are easily described explicit (self dual, vertex transitive) graphs that seem quite good. As far as I know, the largest know graph with $\alpha \lt 6$ and $\omega \lt 6$ is the Paley Graph on $101$ vertices. -A final quote: - -Paley graphs are quasi-random (Chung et al. 1989): the number of times - each possible constant-order graph occurs as a subgraph of a Paley - graph is (in the limit for large q) the same as for random graphs, and - large sets of vertices have approximately the same number of edges as - they would in random graphs. - -LATER It occurs to me that a random circulant graph with edge density $\frac12$ might be a better choice in some ways. On one hand the chance of $\{{1,2,3\cdots,k\}}$ being a clique in that model is $\frac1{2^k}$ but, due to the multiplicative nature of quadratic residues, in the Paley graph this just requires the primes up to $k$ to be residues hence it is $\frac1{2^{\ln k}}=\frac1{k^{\ln2}}.$ On the other hand, if we somehow knew that the expected value of $\alpha \omega$ is less than $1000 (\log n)^2,$ that wouldn't give us any explicit graph.<|endoftext|> -TITLE: When is every injective module $\Sigma$-injective? -QUESTION [5 upvotes]: I have been looking for a couple of days for the answer to this question to no avail. Let me define what $\Sigma$-injective is. - -Let $R$ be a unital, not necessarily commutative ring. A left $R$-module $I$ is called $\Sigma$-injective if an arbitrary direct sum of copies of itself is again injective. - -The Bass-Papp theorem states that $R$ is left noetherian if and only if the direct sum of an arbitrary family of injective left $R$-modules is again injective, so noetherian rings have the property I am looking for, but I was wondering if the fact that every injective is $\Sigma$-injective is equivalent to some classical ring-theoretical property. - -REPLY [2 votes]: A result of Faith and Walker (page 205 in C. Faith, E. A. Walker, Direct-sum representations of injective modules, J. Algebra 5 (1967), 203-221) answers your question: - -If each injective left $R$-module is $\Sigma$-injective, then $R$ is left noetherian. - -(In fact, a sufficient condition for $R$ to be left noetherian is that each injective cogenerator of the category of left $R$-modules is $\Sigma$-injective - see Proposition 2.2 in this article)<|endoftext|> -TITLE: Subgroups of residually finite groups -QUESTION [8 upvotes]: Let $\Gamma$ be a finitely generated residually finite group. For a subgroup of finite index $\Lambda<\Gamma$ let us denote by $\pi_\Lambda:\Gamma\rightarrow \Gamma/\Lambda$ the quotient map. Is it possible to find a subgroup $H<\Gamma$ such that the restriction of $\pi_\Lambda$ to $H$ is surjective for every finite index subgroup $\Lambda<\Gamma$? -Does $SL_n(\mathbf Z)$ have such subgroups? -Can we find such $H$ finitely generated? - -REPLY [7 votes]: A general class of such groups are maximal subgroups $H<\Gamma$ of infinite index. Such a subgroup $H$ must surject any finite quotient of $\Gamma$. For finitely generated linear groups like $SL_n(\mathbb{Z})$ which are not virtually solvable, maximal subgroups of infinite index were constructed by Margulis-Soifer. I don't know in what generality it is known which residually finite groups have maximal subgroups of infinite index (and this is a much stronger property than being dense in the profinite topology). Note, however, that if $H<\Gamma$ is dense in the profinite topology, then a maximal proper subgroup $H < K < \Gamma$ containing -$H$ must be of infinite index (since $H$ is not contained in any finite-index proper subgroup, and hence $K$ cannot be finite-index). So the question boils down to which finitely generated residually finite groups contain a maximal subgroup of infinite index (however, note that the existence of maximal subgroups in a finitely generated group requires Zorn's lemma see the comments for how to find a maximal subgroup containing $H$). -If $\Gamma$ is LERF (like a finitely-generated Kleinian group), then $H$ must be infinitely generated to have this property. However, for $n>2$, as Henry indicates in his comment, one has many finitely-generated subgroups of $SL_n(\mathbb{Z})$ which are dense in the profinite topology. -Finitely generated subgroups are contained in a proper finite-index subgroup is called the engulfing property. Your question is about whether $\Gamma$ has the engulfing property for all subgroups $H$. -Remark: Incidentally, in your statement, you don't assume that $\Lambda$ is normal in $\Gamma$. But since $\Gamma$ is finitely generated, there is an normal subgroup $Core(\Lambda)= \cap_{g\in\Gamma} g\Lambda g^{-1}$ which is also of finite index, so that there is a factorization $\Gamma \to \Gamma/Core(\Lambda)\to \Gamma/\Lambda$. Then if $H$ surjects $\Gamma/\Lambda$ for every normal $\Lambda\lhd \Gamma$ of finite index, then $H$ surjects $\Gamma/\Lambda$ for every finite-index subgroup $\Lambda < \Gamma$. I was using this implicitly in the first paragraph.<|endoftext|> -TITLE: Region of convergence of Eisenstein series is a union of Weyl chambers when groups have discrete series? -QUESTION [7 upvotes]: Let $G$ be a reductive algebraic group, and let $M$ be a Levi subgroup of $G$. In Urban's paper Eigenvarieties for Reductive Groups, it seems to be assumed that if $G(\mathbb{R})$ and $M(\mathbb{R})$ both have discrete series, then the region of convergence of Eisenstein series for the pair ($G$,$M$) is (up to translation) a union of Weyl chambers. (Otherwise it would not make sense to define the set $W^M_{Eis}$.) This assumption seems to be true in all cases that I have checked but I am not sure how to prove it except by exhaustive case analysis. Is a proof written down anywhere? -EDIT: I discussed this with Urban and it seems that the assumption is not true in general. For example, it is not true when $G=Sp(6)$ or $SO(4,3)$ and $M=GL(2) \times GL(1) \subset GL(3)$. However, the assumption does seem to be true when $M$ is maximal (and we still require $G$ and $M$ to have discrete series). I would still be interested in seeing a proof in that case. - -REPLY [2 votes]: The assumption in Urban's paper is incorrect: the region of convergence need -not be a union of Weyl chambers. Hence the character distribution of the -Eistenstein series coming from a single cusp generally does not have a -unique $p$-adic interpolation. However, the sum of all of the character -distributions will still have a unique $p$-adic interpolation. This -can be seen as follows. The sum in Lemma 4.6.2 groups together the parabolic -subgroups that are Weyl conjugates. If one instead groups together the -parabolic subgroups that contain the same Levi, then the portions of the regions of convergence that satisfy $w=w_0$ fit together to form exactly one Weyl chamber. -In the case where M is maximal and G and M have discrete series, Corollary 5.6 in On the Eisenstein Series of Cohomology of Arithmetic Groups by Li and Schwermer and its Poincaré dual imply that the region of convergence is the union of translates of the fundamental Weyl chamber by elements of $W^M$ whose length is less than half of the largest length. Here $W^M$ is the set of minimal length coset representatives of $W_G/W_M$ and $W_G$ and $W_M$ are the Weyl groups of $G$ and $M$ respectively.<|endoftext|> -TITLE: Can we use the Rogers-Ramanujan cfrac to parameterize the Fermat quintic $x^5+y^5=1$? -QUESTION [28 upvotes]: Define $\color{blue}{q=e^{2\pi i \tau}}$ and Dedekind eta function $\eta(\tau)$. Note: I found these relations empirically, but their consistent forms suggest they can be rigorously proven. - -I. $p=2$ - -We have, -$$\left(\frac{8}{\alpha^8+8}\right)^2+\left(\frac{\beta^8}{\beta^8+32}\right)^2=1\tag1$$ -$$\alpha=\frac{\eta(\tau/2)}{\eta(2\tau)},\quad\beta=\frac{\eta(\tau)}{\eta(4\tau)}$$ -If $\tau=\sqrt{-n}$, then, -$$\frac{_2F_1\Big(\tfrac14,\tfrac34;1;\,1-\big(\tfrac{8}{\alpha^2+8}\big)^2\Big)}{_2F_1\Big(\tfrac14,\tfrac34;1;\,\big(\tfrac{8}{\alpha^8+8}\big)^2\Big)}=\color{red}{\sqrt{2n}}$$ - -II. $p=3$ - -We have, -$$\left(\frac{3}{\gamma^3+3}\right)^3+\left(\frac{\delta^3}{\delta^3+9}\right)^3=1\tag2$$ -$$\gamma=\frac{\eta(\tau/3)}{\eta(3\tau)},\quad\delta=\frac{\eta(\tau)}{\eta(9\tau)}$$ -If $\tau=\sqrt{-n}$, then, -$$\frac{_2F_1\Big(\tfrac13,\tfrac23;1;\,1-\big(\tfrac{3}{\gamma^3+3}\big)^3\Big)}{_2F_1\Big(\tfrac13,\tfrac23;1;\,\big(\tfrac{3}{\gamma^3+3}\big)^3\Big)}=\color{red}{\sqrt{3n}}$$ -Note: The identity, -$$\left(\frac{x^3y + y}{x y^3 + x}\right)^3+\left(\frac{x^3 - y^3}{x y^3 + x}\right)^3=1,\quad \text{if}\; x^3+y^3=1$$ -should guarantee infinitely many eta parametrizations to $(2)$. Note also that, -$$\gamma^3+3=4C^2(q)+\frac1{C(q)}\\ -\delta^3+3=4C^2(q^3)+\frac1{C(q^3)}$$ -with cubic continued fraction, -$$C(q)=\frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}=\cfrac{q^{1/3}}{1+\cfrac{q+q^2}{1+\cfrac{q^2+q^4}{1+\cfrac{q^3+q^6}{1+\ddots}}}}$$ -$\color{blue}{Update:}$ -I just realized that the cubic eta identity $(2)$ is equivalent to the Borweins' cubic theta identity, -$$\frac{c^3(q)}{a^3(q)}+\frac{b^3(q)}{a^3(q)} = 1$$ -where, -$$a(q) = 1+6\sum_{n=0}^\infty\left(\frac{q^{3n+1}}{1-q^{3n+1}}-\frac{q^{3n+2}}{1-q^{3n+2}}\right)$$ -$$b(q) = \tfrac{1}{2}\big(3a(q^3)-a(q)\big)$$ -$$c(q) = \tfrac{1}{2}\big(a(q^{1/3})-a(q)\big)$$ -from Ramanujan's Notebooks Vol. V, p.93. - -III. $p=4,8$ - -$$\left(\frac{\vartheta_2(0,q)}{\vartheta_3(0,q)}\right)^4+\left(\frac{\vartheta_4(0,q)}{\vartheta_3(0,q)}\right)^4 = 1\tag{3a}$$ -with Jacobi theta function $\vartheta_n(0,q)$. If $q=e^{2\pi i \tau}$, then equivalently, -$$\left(\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\right)^8+\left(\frac{\eta^2(\tau)\,\eta(4\tau)}{\eta^3(2\tau)}\right)^8 = 1\tag{3b}$$ -hence the addends of $(3a)$ and $(3b)$ are equal. Expressed as, $$U^8(\tau)+V^8(\tau) =1$$ -If $\tau=\sqrt{-n}$, then, -$$\frac{_2F_1\big(\tfrac12,\tfrac12;1;\,1-U^8(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12;1;\,U^8(\tau)\big)}=\color{red}{\sqrt{4n}}$$ -This has a beautiful octic continued fraction studied by Ramanujan, -$$U(\tau)=\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}= \cfrac{\sqrt{2}\,q^{1/8}}{1+\cfrac{q}{1+q+\cfrac{q^2}{1+q^2+\cfrac{q^3}{1+q^3+\ddots}}}}$$ -and can solve the general quintic. - - -IV. Question - -Q: Are there analogous eta quotients to parameterize the Fermat quintic $x^5+y^5=1$? And can we also use the Rogers-Ramanujan cfrac to do so? - -REPLY [6 votes]: (Courtesy of a comment by Nemo who suggested Huber's paper.) - -Part I. $x^5+y^5 = 1$ - -In "A Theory of Theta Functions to the Quintic base", -Tim Huber defines four theta functions which can be ultimately expressed in terms of the Rogers-Ramanujan identities. Define $q=e^{2\pi i z}$ and, -$$P(z):=q^{11/60}H(q)=q^{11/60}\prod_{n=1}^\infty \frac1{(1-q^{5n-2})(1-q^{5n-3})}$$ -$$Q(z):=q^{-1/60}G(q)=q^{-1/60}\prod_{n=1}^\infty \frac1{(1-q^{5n-1})(1-q^{5n-4})}$$ -and the Rogers-Ramanujan continued fraction, -$$R(z)=q^{1/5}\frac{H(q)}{G(q)}=\cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}}$$ -then Huber's four functions in simplified form are, -$$\begin{aligned} -a(\tau) &=\eta^{2/5}(\tau)\;P(\tau)\\[2mm] -b(\tau) &=\eta^{2/5}(\tau)\;Q(\tau)\\[2mm] -c(\tau) &= 5^{1/4}\phi^{1/2}\,\eta^{2/5}(5\tau)\;P\big(\tfrac{-1}{5\tau}\big)\\[2mm] -d(\tau) &= \frac{5^{1/4}}{\phi^{1/2}}\,\eta^{2/5}(5\tau)\;Q\big(\tfrac{-1}{5\tau}\big)\end{aligned}$$ -with golden ratio $\phi$. These obey, -$$\Big(\frac{a\,\phi}{b}\Big)^5+\Big(\frac{c}{b}\Big)^5 = 1\tag1$$ -$$-\Big(\frac{a}{b\,\phi}\Big)^5+\Big(\frac{d}{b}\Big)^5 = 1\tag2$$ -It then follows that the ratio of $a,b$ as well as $c,d$ are Rogers-Ramanujan cfracs, -$$\frac{a(\tau)}{b(\tau)} = R(\tau)\tag3$$ -$$\frac{c(\tau)}{d(\tau)} = \phi\, R\big(\tfrac{-1}{5\tau}\big)\tag4$$ -so $R(\tau)$ in a way can parameterize the Fermat quintic. Some manipulation will also show that, -$$\frac{c(\tau)}{d(\tau)} = \phi\,\frac{1-\phi\, R(5\tau)}{\phi+R(5\tau)}\tag5$$ - -Part II. $x^5+y^5+z^5 = 1$ - -It turns out that just like, -$$x^3+y^3=1$$ -can be solved by the cubic continued fraction $C(q)$ and $C(q^3)$, its quintic analogue -$$x^5+y^5+z^5 = 1$$ -has a beautiful solution using the Rogers-Ramanujan cfrac $R(q)$ and $R(q^5)$. This is given by, -$$\alpha^5\phi^5+\beta^5\phi^5+\alpha^5\beta^5 = 1\tag6$$ -where, -$$\alpha = R(q),\quad\beta = \frac{1-\phi R(q^5)}{\phi+R(q^5)}$$ -and which can be derived by combining $(1),(2)$.<|endoftext|> -TITLE: On the definition of Hilbert spaces and real structures on Hilbert spaces -QUESTION [5 upvotes]: Let us consider the space $L^2:=L^2(\mathbb{R}^n,\mathbb{C})$ and the associated scalar product $S(f,g):=\int f \overline g$. In distribution theory, we have a situation where we have to deal with two different identifications (which makes things a little bit tricky) : - -if we identify $L^2$ with its antidual, an element $f\in L^2$ is an antilinear form $g\mapsto S(f,g)$ -if we identify $L^2$ as a distribution space, an element $f\in L^2$ is a linear form $g\mapsto \int f g $. - -I am aware of the following facts : - -we can anti-identify (through a anti-isomorphism) $L^2$ to its dual (as an Hilbert space can always be), but this just shifts the problem, since the first identification becomes the map $f\mapsto (g\mapsto \int \overline f g)$. -We can define distributions as antilinear maps (and not linear), and define the identification of a function as a distribution by : $f\mapsto \int f \overline g$. This way, the two identifications coincide, and there is no need to keep track of the two different identifications made. - -These observations made me think that maybe the discrepancy can be solved in another way, where instead of redefining distributions, we redefine (or more precisely refine) the abstract definition of Hilbert spaces. -More precisely, let us say that a (complex) Hilbert space with real structure is a triple $(H,b,C)$ where $C$ is a antilinear involution on $H$, and $b$ is a bilinear map such that -$\langle \cdot,\cdot\rangle = b(\cdot,C(\cdot))$ is a scalar product on $H$ giving $H$ the structure of a Hilbert space. -With this definition it seems that a Hilbert space with real structure is naturally isomorphic (through a bijective linear isometry, namely $f\mapsto b(f,\cdot)$) to its dual (and not its antidual), which is indeed the natural situation we have in the case of $L^2$ (where $C:f\mapsto \overline f$ and $b(f,g):=\int f g$). -My question is then : why do we study (complex) Hilbert spaces in general and not (complex) Hilbert spaces with real structure (as previously defined) ? Are there Hilbert spaces that appear naturally without an obvious real structure ? -Or again, formulated differently : if we formulated the whole theory of Hilbert space with the definition given before (with a real structure), would we gain anything in applications to abstract it further to the case of Hilbert space without real structure ? - -REPLY [4 votes]: Good question. It does seem like most naturally occurring complex Hilbert spaces come equipped with a natural real structure. Incidentally, I pointed out in this paper that the same is true of Hilbert modules over C*-algebras, and in this case you can use the natural involution to turn a left action into a right action and vice versa. So many naturally occurring Hilbert modules are automatically Hilbert bimodules. -The first example that comes to mind of a Hilbert space without an obviously natural real structure is the Bargmann-Segal space of entire analytic functions on $\mathbb{C}^n$ whose integral against $e^{-|z|^2}$ is finite. I guess there is a natural isomorphism of this space with $L^2(\mathbb{R}^n)$, and you can then import the real structure of the latter, so maybe it's not a completely convincing example. -Edit: better example, the Bergman space of holomorphic functions on a domain in $\mathbb{C}$ which are square integrable against Lebesgue measure. I don't see any obvious real structure for a general domain.<|endoftext|> -TITLE: This is not a dyadic cosine-product -QUESTION [9 upvotes]: The double-angle formula, $\sin2x=2\sin x\cos x$, turns the scary-looking integral -$$\int_0^{\infty}dz\prod_{k=1}^{\infty}\cos\frac{z}{2^k}$$ -into fun once you realize $\prod_k\cos\frac{z}{2^k}=\frac{\sin z}z$, because then it's well-known that $\int_0^{\infty}\frac{\sin z}zdz=\frac{\pi}2$. -I've found the following variant intriguing and curious. - -Question. Is this valid? If not, what is the value of the integral? - $$\int_0^{\infty}dz\prod_{k=1}^{\infty}\cos\frac{z}{k}=\frac{\pi}4.$$ - -In case such is known, what is a reference? - -REPLY [24 votes]: It's not quite $\pi/4$ . . . -Using the same formula -$\prod_{m=1}^\infty \cos(x/2^m) = \frac{\sin x}{x} = \text{sinc}\,x$, -we write the integrand as -$$ -\prod_{n=1}^\infty \text{sinc}\,\frac{2z}{2n-1}, -$$ -and then the integrals -$$ -I_N := \int_0^\infty dz \prod_{n=1}^N \text{sinc}\,\frac{2z}{2n-1} -$$ -of the partial products are $1/2$ of the notorious -Borwein integrals: -$I_N$ is exactly $\pi/4$ for $N \leq 7$, but strictly (albeit minutely) -decreasing once $N \geq 8$ -(ultimately because that's when $\sum_{n=2}^N 1/(2n-1)$ exceeds $1$). -EDIT I see that all this and more appears in the -Mathworld entry for -"Infinite -Cosine Product Integral" (see formulas (4) ff.), -with a reference to pages 101-102 of - -J. Borwein, D. Bailey, and R. Girgensohn: Experimentation in Mathematics: Computational Paths to Discovery, Wellesley, MA: A K Peters, 2004.<|endoftext|> -TITLE: Generalization of winding number to higher dimensions -QUESTION [14 upvotes]: Is there a natural geometric generalization of the winding number to higher dimensions? - -I know it primarily as an important and useful index for closed, plane curves -(e.g., the Jordan Curve Theorem), -and for its role in Cauchy's theorem integrating holomorphic functions. -I would be interested to learn of generalizations that essentially -replace the role of the circle $\mathbb{S}^1$ with $\mathbb{S}^n$. -I've encountered references to the Fredholm index, -the Pontryagin index, -and to Bott periodicity, -but none seem to be straightforward geometric generalizations of winding number. -This is an entirely naive question, and references and high-level descriptions -would be appreciated, and more than suffice. - -REPLY [2 votes]: In your question you mentioned the word "Fredholm index". -So I would like to say that in the circle case there are two different interpretations of Fredholm index of certain linear operators in terms of winding number. So it would be interesting to consider a possible generalization of these $1$ dimensional facts to higher dimensional spheres. -1)If I remember the following theorem correctly, there is a Theorem by Veku in "Topology and analysis, the Atiyah-Singer index formula and gauge-theoretic physics, by B. Booss and D. D. Bleecker" which says: -Theorem: If $X$ is a non vanishing vector field along $S^1\subset \mathbb{R}^2$, not necessarily tangent to $S^1$, then the fredholm index of the pair operator $(\Delta, \partial/\partial X)$ is equal to the "winding number of $X$. In the above pair operator $\Delta$ is the standard Laplace operator on the interior of circle and the derivational operator $\partial/\partial X$ is restricted to the boundary. -So it would be a good idea to consider an appropriate generalization of this fact. For every smooth self map on sphere one can consider an arbitrary extension to whole $\mathbb{R}^{n+1}$ and try to find an appropriate generalization. However an immediate plain generalization is not true but one should consider a modified generalization. I was thinking to this question about 7 years ago and I observed that a plain generalization is not true since the corresponding operator on $S^3$ is not a Fredholm operator. I had intension to discuss these materials in my following talk I presented in Timisoara but I did not had enough time to present all of the materials of this abstract, since my time was 20 minutes: -http://at.yorku.ca/c/b/d/z/37.htm -(Sorry, the talk abstract I wrote is very snafu) -2) Many years ago I learned from a speaker in Non commutative geometry -that the multiplication by a non vanishing complex map $f$ on $S^1$ is a fredholm operator on $L^1(S^1)$ whose index is $- W(f)$, the winding number of $f$. So it would be interesting to consider an $S^3$ analogy or even more a compact Lie group analogy. -The following paper contains an explanation of a similar situation -http://www.ams.org/journals/proc/2005-133-05/S0002-9939-04-07642-7/S0002-9939-04-07642-7.pdf<|endoftext|> -TITLE: On a dual of Kaplansky's $2^{nd}$ conjecture: admissible algebras? -QUESTION [5 upvotes]: Kaplansky's second conjecture (on Hopf algebras) deals with "admissible" coalgebras: He calls a coalgebra admissible, if there is an algebra structure making it a Hopf algebra. The conjecture states that: - -a coalgebra $C$ is admissible if and only if any finite subset of $C$ lies in a finite dimensional admissible subcoalgebra. - -As far as I know there are counterexamples refuting the conjecture. (the first one was stated by Larson if I remember correctly). -Inspired by the above, let us lay the following definition of the notion of admissible algebra: - -Definition: An algebra $A$ will be called admissible if there is a coalgebra structure on $A$ and a suitable linear map $S:A\to A$, such that all these data (the algebra, the coalgebra and the linear map $S$) constitute a Hopf algebra with antipode $S$. - -Now, my question is whether there is some criterion (necessary, sufficient or both) regarding to when an algebra is admissible (or non-admissible) in the sense of the above definition. In other words: - -Given an algebra $A$, under which conditions can a coalgebra structure be found on $A$ such that it will be turned into a Hopf algebra? -When would such a coalgebra structure be unique? -What could be examples of a non-admissible algebras, in the sense of the above definition? - -REPLY [2 votes]: Note that a Hopf algebra is augmented. Let $A$ be an augmented algebra over a commutative ring $k$. A necessary condition for $A$ having a Hopf algebra structure is that the cohomology ring -$$H^\ast(A,k) := Ext_A^\ast(k,k)$$ -is graded commutative under Yoneda composition. -In fact, the coproduct of a Hopf algebra (or, more generally, of a bialgebra) induces a graded comutative cup product on cohomology and one can show that Yoneda composition and cup product agree on $Ext_A^\ast(k,k)$. -When I have time, I'll look for an example there this criterion applies.<|endoftext|> -TITLE: Multivariate quasipolynomials and where to find them -QUESTION [9 upvotes]: This question is inspired from thinking about David Speyer's question about complex variable Ehrhart theory. -In one variable, Ehrhart theory has been vastly generalized. For example it has been established that counting any set that can be defined in Presburger Arithmetic and a single parameter is quasipolynomial! As a corollary, in David's question, the counting function is a quasipolynomial when we specialize the value of one of the variables. -The multiparameter version of Ehrhart theory is bound to get a bit trickier. For example the number of lattice points in the segment between $(s,0)$ and $(0,t)$ is $1+\gcd(s,t)$. This function is a quasipolynomial when you specialize either value, but it is not a quasipolynomial in both variables, meaning that there is no integer $N$ so that when restricting $s,t$ to residue classes $(\text{mod } N)$ we get a polynomial in $s,t$. -My question is motivated from a basic property of polynomials (what Richard Palais calls analogs of Hartog's theorem): If a multivariate function is a polynomial of bounded degree in each variable separately, then it is a multivariable polynomial. -Question: Can we describe bivariate functions that are quasipolynomial of bounded degree in each variable? Are they bivariate quasipolynomials plus perhaps linear combinations of terms like $\gcd(P,Q)$, for bivariate polynomials $P,Q$? - -REPLY [5 votes]: There is a multivariate version of an Ehrhart quasipolynomial (but it is necessarily only piecewise defined). Namely, fix an integer matrix A, the number of integer lattice points in the rational polytope { x : Ax ≤ b } is a piecewise-defined quasipolynomials in the components of b (viewed as an integer vector). Geometrically, what we're doing is fixing the normal vectors of a set of (possible) facets of a polytope and then vary the position of the facets. (Note that in your example, the normals are not preserved.) -It is a short step (using slack variables) to convert ≤ into =, and then the counting function above goes by the name of vector partition function (associated with A). See this seminal paper by Dahmen & Micchelli as a starting point.<|endoftext|> -TITLE: Does existence of midpoints imply intrinsic? -QUESTION [14 upvotes]: It is well-known, that a complete metric space, where any two points have a midpoints ($\forall x,y~ \exists z:~d(x,z)=d(y,z)=\frac{d(x,y)}{2}$) is strictly intrinsic, in the sense that any $x,y$ can be joined by a path of length $d(x,y)$. -Also, completeness and existence of $\varepsilon$-midpoints ($\forall x,y~ \exists z:~|d(x,z)-\frac{d(x,y)}{2}|+|d(y,z)-\frac{d(x,y)}{2}|\le \varepsilon$) for any $\varepsilon$ imply the space being intrinsic, i.e. any $x,y$ can be joined by a path of length $d(x,y)+\varepsilon$ for any $\varepsilon$. -Can we replace completeness by something else? -It seems that $\varepsilon$-midpoints are completely useless without completeness (a disc without one radius is an example). -On the other hand, it seems that the existence of midpoints is a rather strong condition. So far I have the following examples: -$\mathbb{R}\times(0,+\infty)\bigcup \mathbb{Q}$ with the induced metric is a space with midpoints, which is not strictly intrinsic, but still it is intrinsic. -$\{(x,y)\in \mathbb{R}^{2}, x-y\in\mathbb{Q}\}\bigcup\{x=y\}$ with the induced $L_{\infty}$ metric is a connected space with midpoints but not intrinsic. -However these space are bad: neither of them is locally compact, and the latter is not even locally path connected (local compactness implies local completeness and so local strict intrinsicness, and so local path connectedness). -Now let us assume that in connected locally compact (or merely locally path connected) metric space $X$ for any two points there is a midpoint. Is it true that this space is (strictly) intrinsic? -There were numerous edits, because I have added two counterexamples, one of which later proved to be completely incorrect, and another one had a mistake. - -REPLY [7 votes]: The answer is no. -There is a locally compact, connected metric space, where any two points have a midpoint, that is not intrinsic. -The example I have in mind looks a little bit like this: - -[The little circles are meant to indicate holes in my rectangles (not little arcs).] -The metric for this space is the "modified taxicab metric", by which I mean the metric that measures the shortest path between two points, but where I allow paths that jump over holes. (One could imagine a taxicab in New Orleans, which by necessity has to learn to ignore potholes and just run right over them.) -I say that my example looks "a little bit" like this because I have only drawn two iterations of the construction, but I would like to imagine that we have done infinitely many iterations. -A little more precisely, let's call the big rectangle $B_0$. Notice that its bottom left corner has been removed, making a hole. Let's consider this holey rectangle to be stage $0$ of our construction. To get stage $1$, we build an infinite sequence of "bridges" accross the hole, which we'll call $B_1^n$, $n \in \mathbb N$, and remove a point from each bridge, making more holes. The exact manner in which the bridges are constructed is irrelevant, provided we do two things: -(1) Letting $p$ and $q$ denote the points where a bridge $B_1^n$ meets $B_0$, design the bridge so that the distance from $p$ to $q$ is the same whether we take the shortest "path" (path with a hole in it) through $B_0$, or if we choose to take the bridge instead. -(2) With $p$ and $q$ as before, choose the hole in our bridge so that the distance from $p$ (or $q$) to the hole in $B_0$ is different from its distance to the hole in $B_1^n$. (This is true the way I've drawn it, for example, because I used rectangles that are not squares.) -This completes stage $1$ of the construction. Further stages work the same way: for each hole from stage $n$, we construct an infinite sequence of holey bridges at stage $n+1$ (and we do it in such a way that the above two properties are satisfied). Ultimately, the space I have in mind is the union of all these stages. -Let's check that this space has the properties you require. -It is locally compact. -Every point in our space has a neighborhood that is homeomorphic to either an interval or a triod (the letter T). This makes the space locally compact. -It is connected (even better: it is path connected). -A straightforward induction on $n$ shows that if we add a point to our space at stage $n$ of our construction, then there is a path from that point to $B_0$. Since $B_0$ is path connected (because I didn't remove the top right corner!), it follows that the whole space is path connected. -Any two points have a midpoint. -Let $p$ and $q$ be any two points in our space, both occuring in stage $n$ of the construction. If we temporarily ignore the holes in our space, then there is a (not necessarily unique) shortest path connecting $p$ to $q$, and it is contained in stage $n$ of the construction. Because of the metric we've chosen, the midpoint $r$ of this path should be a midpoint between $p$ and $q$ (the only potential problem being that $r$ might not be in our space -- it could be a hole). If $r$ is not a hole then it is a midpoint for $p$ and $q$. If $r$ is a hole, then we can modify our path between $p$ and $q$ using one of the bridges we built around the hole. Because of our first bridge-building condition, this modified path has the same length as the original one. Therefore its midpoint, $s$, should be a midpoint for $p$ and $q$ as well. Our second bridge-building condition guarantees that $s$ is not a hole. -It is not intrinsic. -Let $p$ be a point in (the interior of) the left edge of $B_0$ and let $q$ be a point in its bottom edge. Any path connecting $p$ and $q$ must go through the top right corner of $B_0$ (Why? Because removing this point disconnects the whole space!). It is clear that any such path is significantly longer than the distance from $p$ to $q$. Thus this space is not intrinsic.<|endoftext|> -TITLE: Memorable ordinals -QUESTION [11 upvotes]: Suppose $V=L$ + reasonable hypotheses (e.g. "ZFC has a countable transitive model"). Call a countable ordinal $\alpha$ memorable if for some countable $\beta$, $\alpha$ is definable without parameters in every $L_\gamma$ with $\beta<\gamma<\omega_1$. -My question is: - -Are there uncountably many memorable ordinals? - - -Some comments, right off the bat: - -The answer is trivially "no" for uniformly memorable ordinals, that is, ordinals which are coboundedly definable without parameters by the same formula. However, since the defining formula is allowed to vary with $\gamma$, this doesn't work. -Towards a positive answer, note that there are uncountably many countable $\theta$ such that $L_\theta$ is pointwise definable (call such an ordinal "insightful"); this was proved by Hamkins, Linetsky, and Reitz. However, this doesn't actually resolve the issue: for a fixed $\alpha$, there may be many countable $\eta>\alpha$ with no insightful ordinals $>\alpha$ which are definable without parameters in $L_\eta$ (e.g. if there is no greatest insightful ordinal $<\eta$, this seems a distinct possibility). - - -Interestingly, this question has possibly interesting variants even if $L$ is a very tiny subclass of $V$! Given any hierarchy $(M_\alpha)_{\alpha\in\omega_1}$ with $M_\alpha\cap ON=\alpha$, we can ask whether uncountably many $\alpha$ satisfy "For all sufficiently large $\beta<\omega_1$, $\alpha$ is parameter-free definable in $M_\beta$." Now, we can trivially construct examples where the answer is "yes": namely, let $M_\alpha$ be $L_{\eta_\alpha}$ where $\eta_\alpha$ is the $\alpha$th insightful ordinal; so the general existence question isn't interesting. However, I would be curious if there are any "natural" hierarchies of $M_\alpha$s which do satisfy this property; especially if their union is $H_{\omega_1}$ in some $V$ which is very far from $L$. - -REPLY [3 votes]: REPAIRED ARGUMENT -The memorables are strictly bounded between the first $\Sigma_2$-stable in $\omega_1$ and the first $\Sigma_3$-stable in $\omega_1$. The least non-memorable is thus a $(\Sigma_2\wedge \Pi_2)$-singleton. -Let $\delta_0$ be this first non-memorable. The previous erroneous argument yields a characterisation or perhaps a paraphrase of $\delta_0$. -For $\beta<\omega_1$ let -(A) $H(\beta)$ be the Skolem Hull inside $L_\beta$ of the empty set. -(B) Let $\omega_1(\beta):= (\omega_1)^{L_\beta}$ if the latter is defined, - $= \beta$ otherwise. -Then (i) $H(\beta)$ is the set of pointwise definable objects in $L_\beta$. (ii) $H(\omega_1(\beta))=$ $L_\tau$ for some $\tau\leq \omega_1(\beta)$. (iii) For unboundedly many $\beta$, $\beta = \omega_1(\beta)$. -Claim Let $\delta_1 =$ the least $\delta< \omega_1$ so that for unboundedly many $\beta$ $H(\beta)=L_{\delta_1}$. Then $\delta_1=\delta_0$. -Proof: It is easy to argue that $\delta_1$ is defined. Then, by definition $\delta_1$ is not memorable (as $\delta_1 -\notin H(\beta)$ for arbitrarily large $\beta$). So it suffices to show that $\tau<\delta_1 \rightarrow \tau$ is memorable. -By definition, and countability, of $\delta_1$: -$\exists \beta_0\forall \beta> \beta_0\forall\tau<\delta_1\,\, H(\beta)\neq L_\tau \quad (*).$ -As $\beta \longrightarrow \omega_1$ so does $\omega_1(\beta) \longrightarrow \omega_1$ non-decreasingly. Thus there is $\beta_1>\beta_0$ so that: -$\forall\beta>\beta_1\,\, \omega_1 -(\beta)>\beta_0.$ -Then, using (ii), for any $\beta>\beta_1\,\, H(\omega_1(\beta))=L_\gamma$ for some $\gamma \leq\omega_1(\beta)$ but by $(*)$ $\delta_1\leq -\gamma.$ Thus any $\tau< \delta_1$ is pointwise definable in $H(\omega_1(\beta))$ and so (using a definition of $\omega_1$), it is pointwise definable in $L_\beta$. So all $\tau$ less than $\delta_1$ are memorable as required. $\quad $ QED. -So something similar would work if CH holds, or in CH models. Eg let $A\subseteq\omega_1$ be such that $H_{\omega_1}=L_{\omega_1}[A]$. Consider the least $\delta$ with $L_\delta[A\cap \delta]\prec L_{\omega_1}[A]$.<|endoftext|> -TITLE: A neat evaluation of an infinite matrix? -QUESTION [11 upvotes]: Let $M_n$ be an $n\times n$ matrix defined as -$$M_n -=\left[\frac{2i+1}{2(i+j+1)}\binom{i-1/2}i\binom{j-1/2}jx^{i+j+1}\right]_{i,j=0}^n.$$ -With $I_n$ the identity matrix, consider $A_n:=I_n-M_n^2$. When I computed $\det A_n$, it looks rather "ugly". However, its infinite dimensional counterpart $\det A_{\infty}$ seems to reach a neat evaluation. To avoid issues on what it means by "determinant of infinite matrix", I simply work with the following convention: $\det(A_{\infty})=\lim_{n\rightarrow\infty}\det(A_n)$. So, - -Question. Is this determinantal evaluation true? - $$\det(A_{\infty})=\sqrt[4]{1-x^2}.$$ - -NOTE 1. The fractional values $\binom{i-1/2}i$ are (as usual) computed via Euler's Gamma function, $\Gamma(z)$. -NOTE 2. If it helps, we make two observations: (a) both $\det A_n$ and $\sqrt[4]{1-x^2}$ are functions of $y:=x^2$; (b) as functions of $y$, the taylor series for $\det A_n$ and $\sqrt[4]{1-y}$ agree up to degree $n$. - -REPLY [6 votes]: The claimed expression for the determinant indeed holds. In fact, one may explicitly determine the eigenvalue decomposition of the infinite matrix $M_\infty$ viewed as a self-adjoint operator on a Hilbert space. Its eigenvalues are -$$ \left(\frac{2}{q_x^{p+\frac12}+q_x^{-p-\frac12}}\right)_{p=0}^\infty,\qquad q_k = e^{-\pi\frac{K\left(\sqrt{1-x^2}\right)}{K(x)}}, \tag{1} -$$ where $q_k$ is the elliptic nome of modulus $x$ and $K(x)$ the complete elliptic integral of the first kind. -Once this is known one may compute -$$ -\begin{aligned} -\det(I - M_\infty^2) &= \prod_{p=0}^\infty\left(1-\left(\frac{2}{q_x^{p+\frac12}+q_x^{-p-\frac12}}\right)^2\right) \\ -&=\prod_{p=0}^\infty\left(\frac{1-q_x^{2p+1}}{1+q_x^{2p+1}}\right)^2 = \frac{\theta_4(0,q_x)}{\theta_3(0,q_x)}=\sqrt[4]{1-x^2}, -\end{aligned} -$$ -where the last two equalities follow from standard properties of the Jacobi theta functions $\theta_i$. -To see how the eigenvalue decomposition can be obtained, it is convenient to look at the Dirichlet space $\mathcal{D}$ of complex analytic functions $f$ on the open unit disk that vanish at $0$ and that have finite norm with respect to the Dirichlet inner product -$$\langle f,g\rangle_{\mathcal{D}} = \frac{1}{\pi} \int_{|z|<1} \overline{f'(z)}g'(z) \mathrm{d}^2z = \sum_{n=1}^\infty n\,\overline{[z^n]f(z)}\,[z^n]g(z).$$ -A basis is given by the monomials $(e_n:=z^n)_{n\geq 1}$, which satisfy $\langle e_n,e_m\rangle_{\mathcal{D}}=n\, \delta_{n,m}$. Let us also introduce the bounded linear operator on $\mathcal{D}$ given by -$$\mathbf{R}_x f = f \circ \psi_x, \qquad \psi_x(z) = \frac{1-\sqrt{1-x z^2}}{\sqrt{x}\,z}.$$ -By power series expansion one may check that -$$ \mathbf{R}_x e_p = \sum_{\ell=p}^\infty \left(\frac{x}{4}\right)^{\ell/2} \frac{p}{\ell}\binom{\ell}{(\ell+p)/2}\, 1_{\{\ell+p\text{ even}\}} e_\ell$$ -and the adjoint with respect to $\langle \cdot,\cdot\rangle_{\mathcal{D}}$ is determined by -$$ \mathbf{R}_x^\dagger e_p = \sum_{\ell=1}^p \left(\frac{x}{4}\right)^{p/2} \binom{p}{(\ell+p)/2}\, 1_{\{\ell+p\text{ even}\}} e_\ell.$$ -In particular, $\mathbf{R}_x\mathbf{R}_x^\dagger$ is a self-adjoint operator that preserves the even and odd functions in $\mathcal{D}$. -Up to a factor of $2$ its matrix elements on the odd monomials are precisely $[M_\infty]_{i,j}$: -$$ -\begin{aligned} -2\mathbf{R}_x\mathbf{R}_x^\dagger e_{2i+1} &= \sum_{j=0}^\infty \sum_{\ell=0}^{\min(i,j)} 2\left(\frac{x}{4}\right)^{i+j+1} \binom{2i+1}{i+\ell+1}\frac{2\ell+1}{2j+1} \binom{2j+1}{j+\ell+1} e_{2j+1}\\ -&= \sum_{j=0}^\infty\frac{2i+1}{2(i+j+1)}\binom{i-1/2}i\binom{j-1/2}jx^{i+j+1}e_{2j+1}\\ -&= \sum_{j=0}^\infty [M_\infty]_{i,j} e_{2j+1}. -\end{aligned} -$$ -In T. Budd, Winding of simple walks on the square lattice, arXiv:1709.04042, Section 2.1, I examined a closely related operator $\mathbf{J}_x = \mathbf{R}_x^\dagger\mathbf{R}_x$, whose matrix elements count diagonal walks on $\mathbb{Z}^2$ starting on the positive $x$-axis and hitting the $y$-axis at prescribed height. By a convenient elliptic parametrization of the disc one can determine all eigenvectors of $\mathbf{J}_x$. Since $\mathbf{R}_x$ is injective, $\mathbf{R}_x\mathbf{R}_x^\dagger$ has the same eigenvalue decomposition (after applying $\mathbf{R}_x$ to the eigenvectors of $\mathbf{J}_x$) given by Proposition 9, of which the eigenvectors with odd label $m=1,3,5,\ldots$ -span the odd functions in $\mathcal{D}$. The corresponding eigenvalues are precisely the ones given above in (1). -Added (22 Jan): One may also check the formula for $\det(I\pm M_\infty)$ proposed in the post by Hucht. Using the theta function product formulas we find -$$ -\det(I-M_\infty) = \prod_{p=0}^\infty \frac{(1-q_x^{p+\frac12})^2}{1+q^{2p+1}} = \left(\frac{\theta_2(0,\sqrt{q_x})}{\theta_1'(0,\sqrt{q_x})}\right)^{\frac14} \frac{\theta_4(0,\sqrt{q_x})}{\theta_3(0,q_x)^{\frac12}}. -$$ -With the help of $\theta_1'(0,\sqrt{q_x}) = \theta_2(0,\sqrt{q_x})\theta_3(0,\sqrt{q_x})\theta_4(0,\sqrt{q_x})$ and -$$ -\theta_3(0,\sqrt{q_x})^2 = (1+x)\theta_3(0,q_x)^2,\quad \theta_4(0,\sqrt{q_x})^2 = (1-x)\theta_3(0,q_x)^2 -$$ -this yields -$$ -\det(I- M_\infty) = \left(\frac{(1-x)^3}{1+x}\right)^{1/8}. -$$ -Similarly -$$ -\det(I-M_\infty) = \prod_{p=0}^\infty \frac{(1+q_x^{p+\frac12})^2}{1+q^{2p+1}} = \left(\frac{\theta_2(0,\sqrt{q_x})}{\theta_1'(0,\sqrt{q_x})}\right)^{\frac14} \frac{\theta_3(0,\sqrt{q_x})}{\theta_3(0,q_x)^{\frac12}}= \left(\frac{(1+x)^3}{1-x}\right)^{1/8}. -$$<|endoftext|> -TITLE: Explicit invariant of tensors nonvanishing on the diagonal -QUESTION [18 upvotes]: The group $SL_n \times SL_n \times SL_n$ acts naturally on the vector space $\mathbb C^n \otimes \mathbb C^n \otimes \mathbb C^n$ and has a rather large ring of polynomial invariants. The element $$\sum_{i=1}^n e_i \otimes e_i \otimes e_i \in \mathbb C^n \otimes \mathbb C^n \otimes \mathbb C^n$$ is known to be GIT-semistable with respect to this action. In other words, there is a homogeneous $SL_n \times SL_n \times SL_n$-invariant polynomial, of nonzero degree, which is nonvanishing on this element. The proof I know (Theorem 4.7) uses the Hilbert-Mumford criterion and so does not explicitly construct the polynomial. - -Can you give an explicit homogeneous $SL_n \times SL_n \times SL_n$-invariant polynomial that is nonzero on this element. - -One can check from the definition of the hyperdeterminant that the hyperdeterminant vanishes on this element as long as $n>2$. So that won't work. -I don't know how many other natural ways there are for defining an invariant function on $\mathbb C^n \otimes \mathbb C^n \otimes \mathbb C^n$ for all $n$ simultaneously there are. -My motivation for this problem is to understand semistability of tensor powers of tensors better, and so to understand slice rank of tensor powers of tensors better. I want to give explicit criteria to show that all a tensor's tensor powers are semistable. Tensor powers of the diagonal tensor remain, so the diagonal tensor certainly has this property. Given an explicit invariant, perhaps one could find an explicit neighborhood of the diagonal tensor consisting of tensors that also have this property. - -REPLY [9 votes]: Let me start with some remarks about the classical symbolic method (without which one cannot understand 19th century invariant theory) -and multisymmetric functions. -I will use an example first. Take four series of three variables $a=(a_1,a_2,a_3)$, $b=(b_1,b_2,b_3)$, -$c=(c_1,c_2,c_3)$ and $d=(d_1,d_2,d_3)$. -Now define the polynomial $\mathcal{A}$ in these 12 indeterminates given by -$$ -\mathcal{A}=(bcd)(acd)(abd)(abc) -$$ -where we used a shorthand notation for determinants -$$ -(bcd)=\left| -\begin{array}{ccc} -b_1 & b_2 & b_3 \\ -c_1 & c_2 & c_3 \\ -d_1 & d_2 & d_3 \\ -\end{array} -\right| -$$ -and similarly for the other "bracket factors" $(acd)$, etc. -This is an example of multisymmetric function in the "letters" $a,b,c,d$ which is multihomogeneous of multidegree $[3,3,3,3]$. -Let me write $M_{[3,3,3,3]}^3$ for the space of such multisymmetric functions. The subscript is the multidegree while the superscript refers -to the number of variables in each series. -The symmetric group acting is $\mathfrak{S}_4$ which permutes these series. If each series had one variable in it instead of -three we would be in the familiar realm of the theory of ordinary symmetric functions. The analogous theory of multisymmetric -functions is much more complicated and has been studied classically by Euler, Lagrange, Poisson, Schläfli, Brill, Gordan, Junker -and more recently by Angeniol, Dalbec, Briand, Vaccarino, Domokos and others. -A nice yet old reference on this is the book on elimination theory by Faa di Bruno I mentioned in this MO answer. -Consider now a generic ternary cubic form $F=F(x_1,x_2,x_3)$. One can write it as -$$ -F(x)=\sum_{i_1,i_2,i_3=1}^3 F_{i_1,i_2,i_3} x_{i_1} x_{i_2} x_{i_3} -$$ -where $(F_{i_1,i_2,i_3})$ is a symmetric tensor (by which I mean a "matrix" rather than a Bourbaki tensor). -This form lives in the symmetric power $S^3(\mathbb{C}^3)$ but I will be (intentionally) sloppy and write $S^3$ instead. -The space of degree 4 homogeneous polynomials in the coefficients of $F$ will be denoted by $S^4(S^3)$. -Let me define a linear map $\Phi:S^4(S^3)\rightarrow M_{[3,3,3,3]}^3$ as follows. -Given a degree four homogeneous polynomial $C(F)$ first construct its full polarization -$$ -\widetilde{C}(F_1,F_2,F_3,F_4)=\frac{1}{4!} \frac{\partial^4}{\partial t_1\partial t_2\partial t_3\partial t_4} -C(t_1F_1+t_2 F_2+t_3 F_3+t_4 F_4) -$$ -which satisfies $\widetilde{C}(F,F,F,F)=C(F)$. -Now set $F_1(x)=a_{x}^3$ where $a_x$ is the classical notation for the linear form $a_1x_1+a_2x_2+a_3x_3$ and similarly -specialize to the cubes of linear forms $F_2(x)=b_{x}^3$, $F_3(x)=c_{x}^3$ and $F_4(x)=d_{x}^3$. -Finally, define $\Phi(C)=\widetilde{C}(F_1,F_2,F_3,F_4)$. -I will also define another linear map $\Psi: M_{[3,3,3,3]}^3\rightarrow S^4(S^3)$. -Let $\mathcal{S}=\mathcal{S}(a,b,c,d)$ be a multisymmetric function in $M_{[3,3,3,3]}^3$. -We define the differential operator -$$ -\mathcal{D}=\frac{1}{3!^4} -F\left(\frac{\partial}{\partial a_1},\frac{\partial}{\partial a_2},\frac{\partial}{\partial a_3}\right) -F\left(\frac{\partial}{\partial b_1},\frac{\partial}{\partial b_2},\frac{\partial}{\partial b_3}\right) -$$ -$$ -F\left(\frac{\partial}{\partial c_1},\frac{\partial}{\partial c_2},\frac{\partial}{\partial c_3}\right) -F\left(\frac{\partial}{\partial d_1},\frac{\partial}{\partial d_2},\frac{\partial}{\partial d_3}\right) -$$ -and let -$\Psi(\mathcal{S})=\mathcal{D}\ \mathcal{S}$, i.e., the result of applying the differential operator $\mathcal{D}$ -to the polynomial -$\mathcal{S}$. -It is not hard to see that $\Phi$ and $\Psi$ are inverses of one another. -In the particular case of $\mathcal{A}$ above, the image $\Psi(\mathcal{A})$ is Aronhold's degree four $SL_3$ invariant of plane cubics -and $\mathcal{A}$ itself is called the (German) symbolic form of that invariant. The classics usually omitted "$\Psi(\cdots)$" but they -knew perfectly well what they were doing. Also note that this formalism, in an essentially equivalent form, was invented earlier by Cayley -as I briefly explained in Section 1.11.4 of my article -"A Second-Quantized Kolmogorov-Chentsov Theorem". -One can easily write $\Psi(\mathcal{A})$ applied to $F$, in "physicsy" tensor contraction notation as -$$ -\epsilon_{i_{1} i_{2} i_{3}}\epsilon_{i_{4} i_{5} i_{6}}\epsilon_{i_{7} i_{8} i_{9}}\epsilon_{i_{10} i_{11} i_{12}} -F_{i_{4} i_{7} i_{10}} F_{i_{1} i_{8} i_{11}} F_{i_{2} i_{5} i_{12}} F_{i_{3} i_{6} i_{9}} -$$ -with summation over all twelve indices in the set $\{1,2,3\}$. I used the standard physics notation for the Levi-Civita tensor -$\epsilon_{i_1 i_2 i_3}$ equal to zero if some indices are equal and otherwise to the sign of the permutation $i_1,i_2,i_3$ -of $1,2,3$. This notation is consistent with the one I used in my previous answers to these MO questions: -Q1, -Q2, -and Q3. -At bottom, this is an intrinsically diagrammatic language and if the reader had trouble following what I said so far, it's probably because -I did not draw any pictures or "finite-dimensional Feynman diagrams" (I don't know how to do this on MO). To see these pictures, -please have a look at this article about an invariant of cubic threefolds or Section 2 -of this one mostly on binary forms. -In a nutshell, the above contraction can be encoded in a bipartite graph with 4 $\epsilon$-vertices and 4 $F$-vertices, all of degree three. -Of course the above formalism generalizes with no difficulty (except managing notations) to plethysms of symmetric powers -$S^p(S^q(\mathbb{C}^n))$ and similar $\Phi$, $\Psi$ maps giving the isomorphism with the relevant space of multisymmetric functions, i.e., -$M_{[q,q,\ldots,q]}^{n}$ (with $q$ repeated $p$ times) or in simpler notation $M_{[q^p]}^{n}$. -There is a canonical $GL_n$-equivariant map -$$ -H_{p,q}:S^p(S^q(\mathbb{C}^n))\rightarrow S^q(S^p(\mathbb{C}^n)) -$$ -with "H" standing for Hermite, Hadamard or Howe. I will not define it here but it's the first thing that would come to mind -if asked to produce such a map -(see this review by Landsberg for a detailed discussion in the wider context of geometric complexity theory). -It is not hard to see that a homogeneous multisymmetric function in $M_{[p^q]}^{n}\simeq S^q(S^p(\mathbb{C}^n))$ is in the image -of $H_{p,q}$ iff it is a polynomial in the homogeneous elementary multisymmetric functions which belong to $M_{[1^q]}^{n}$. -In the context of the Aronhold invariant above, these elementary functions are the coefficients of the polynomial -$$ -P(x_1,x_2,x_3)=a_x b_x c_x d_x -$$ -and similarly in general. -Finally, I can start discussing Will's excellent question. Let $T_{i_1,i_2,i_3}$ denote the tensor of interest with the action of -$(g,h,k)\in (SL_n)^3$ given by -$$ -[(g,h,k)\cdot T]_{i_1,i_2,i_3}=\sum_{j_1,j_2,j_3=1}^{n} g_{i_1 j_1} h_{i_2 j_2} k_{i_3 j_3} T_{j_1,j_2,j_3}\ . -$$ -If Will's tensor had an even number $r$ of indices with the action of $(SL_n)^r$ the answer would be trivial. -Take $n$ copies of $T$, contract all first indices of these copies with one $\epsilon$, -then all second indices with another $\epsilon$, etc. For $r=3$ this tentative invariant vanishes identically. -One can however keep the first indices free, and this gives a map $T\mapsto \Gamma(T)=F$ producing an $n$-ary $n$-ic -$F(x_1,\ldots,x_n)$ corresponding to the symmetric tensor -$$ -F_{i_1\ldots i_n}=\sum_{j_1,\ldots,j_n, k_1\ldots,k_n=1}^{n} -\epsilon_{j_1\ldots j_n} \epsilon_{k_1\ldots k_n} T_{i_1,j_1,k_1}\cdots T_{i_n,j_n,k_n}\ . -$$ -For symmetric $T$'s, such as Will's diagonal tensor $D_{ijk}=\delta_{ij}\delta_{ik}$, this is the Hessian covariant. -Jason's idea can thus be rephrased as finding an $SL_n$-invariant $I=I(F)$ of $n$-ary $n$-ics $F$ -and proposing $I(\Gamma(T))$ as an answer to Will's question. The main issue is to show $I(\Gamma(D))\neq 0$. -Now $\Gamma(D)$ is the form $F(x_1,\ldots,x_n)=n!\ x_1 x_2\cdots x_n$. -For $n$ even, the smallest degree for an invariant of $n$-ary $n$-ics $F(x_1,\ldots,x_n)$ is $n$ (unique up to scale). Its symbolic form is -$$ -\mathcal{S_n}=(a^{(1)} a^{(2)}\cdots a^{(n)})^n -$$ -where $a^{(1)}=(a_{1}^{(1)},a_{2}^{(1)}\ldots,a_{n}^{(1)}),\ldots,a^{(n)}=(a_{1}^{(n)},a_{2}^{(n)}\ldots,a_{n}^{(n)})$ -are $n$ series of $n$ variables. Let $I_n=\Psi(\mathcal{S_n})$ denote this invariant. One can also write -$$ -I_n(F)=\epsilon_{i_{1,1}\ldots i_{1,n}}\epsilon_{i_{2,1}\ldots i_{2,n}}\cdots\epsilon_{i_{n,1}\ldots i_{n,n}} -\ F_{i_{1,1}\ldots i_{n,1}}F_{i_{1,2}\ldots i_{n,2}}\cdots -F_{i_{1,n}\ldots i_{n,n}} -$$ -with summation over indices understood. -The evaluation of $I_n(\Gamma(D))$ is a multiple of the number of row-even minus the number of row-odd $n\times n$ -Latin squares. Via work of Huang, Rota and Janssen showing that the result is nonzero is equivalent to the (widely open) Alon-Tarsi conjecture. -The following may be new (if not please let me know): -For $n$ odd, the smallest degree for an invariant of $F(x_1,\ldots,x_n)$ is $n+1$ (unique up to scale). Its symbolic form is -$$ -\mathcal{S}=(a^{(2)} a^{(3)}\cdots a^{(n+1)})(a^{(1)} a^{(3)}\cdots a^{(n+1)})\cdots(a^{(1)} a^{(2)}\cdots a^{(n)})\ . -$$ -Again let me define $I_n=\Psi(\mathcal{S_n})$. This is nonzero because the polynomial in the $a^{(i)}$ is not only multihomogeneous -of multidegree $[n^{n+1}]$ but also multisymmetric. For instance the transposition $a^{(1)}\leftrightarrow a^{(2)}$ exchanges the first -two bracket factors while the other ones pick up a factor $(-1)$. This gives $(-1)^{n-1}=1$ since $n$ is odd. Other elementary -transpositions $a^{(i)}\leftrightarrow a^{(i+1)}$ can be treated in the same way. -Of course $I_3$ is Aronhold's degree 4 invariant. I don't know if $I_5$ could be of use for the study of quintic threefolds. -I boldly (or foolishly?) make the following conjecture which I think is a natural analog of the Alon-Tarsi conjecture in the odd case. -Conjecture: For $n$ odd, $I_n$ does not vanish on $F(x_1,\ldots,x_n)=x_1 x_2\cdots x_n$. -Here is a proof for $n=3$. The Aronhold invariant of degree 4 is the defining equation for the Veronese secant variety -$\sigma_{3}(v_3(\mathbb{P}^2))$. -Hence, it does not vanish on forms with border rank $>3$. -The border rank of the form $x_1 x_2\cdots x_n$ is exactly $2^{n-1}$ as shown by Oeding in this article. The case $n=3$ relevant to the present proof was shown earlier by -Landsberg and Teitler. QED. -A final remark regarding the endgame in Jason's argument with a view to producing an explicit invariant of moderate degree: it follows from what I mentioned above that saying that any homogeneous multisymmetric -function in $M_{[p^q]}^{n}\simeq S^q(S^p(\mathbb{C}^n))$ can be expressed as a polynomial in the elementary ones is equivalent -to the Hermite-Hadamard-Howe map $H_{p,q}$ being surjective and this is known for $p$ large with an explicit bound due to Brion (see -the review by Landsberg I mentioned). The problem I was discussing is closer to the $p=q$ case related to the Foulkes-Howe conjecture. -The latter is known to be false, but not by much! (see this article). -Edit: I learned from C. Ikenmeyer that the conjecture about the nonvanishing of $I_n$ for $n$ odd is not new and is stated in Remark 3.26 in -https://arxiv.org/abs/1511.02927 -However, I noticed recently that this is equivalent to the Alon-Tarsi conjecture for $n$ even. Indeed, the above invariants $I_n$ satisfy the following identity: -$$ -I_n(F(x_1,\ldots,x_{n-1})x_n)=(-1)^{\frac{n}{2}}\frac{n!}{n^n}\ I_{n-1}(F) -$$ -for all $n$ even and all forms $F$ which only depend on the first $n-1$ variables.<|endoftext|> -TITLE: What is the main failure in using Naive Chow group in Artin Stack -QUESTION [8 upvotes]: I'm reading Andrew Kresch's paper, Cycle groups in Artin Stacks. -The author defined Chow groups of Artin stacks by very technical way, instead of ordinary ways which he called 'naive chow group', quotient of free group generated by integral substacks by rational equivalence. -I don't understand why he used technical definition instead of ordinary one. What is the main failure arises when using 'naive chow group's? - -REPLY [15 votes]: Already for a finite group $\Gamma$, the only integral closed substacks of $B\Gamma$ are the empty stack and all of $B\Gamma$. So your naive Chow group would be cyclic (typically zero) in each degree. On the other hand, the "correct" Chow group of divisor classes should be the Pontrjagin dual group $\text{Hom}_{\text{Group}}(\Gamma,k^\times)$ (here $k$ is the ground field, which I guess I am assuming is characteristic $0$ and algebraically closed). Typically this is not cyclic, e.g., for $\Gamma = (\mathbb{Z}/r\mathbb{Z})^n$, the Pontrjagin dual group is $(\mu_r)^n$, and this is not cyclic for $n>1$. -The correct Chow groups of $B\Gamma$, i.e., the $\Gamma$-equivariant Chow groups of $\text{Spec}(k)$, were defined by Totaro and Edidin-Graham. Their theory works for every Deligne-Mumford stack of the form $[X/\Gamma]$ for $X$ a quasi-projective $k$-scheme with a $k$-action of $\Gamma$. The paper of Kresch extends these Chow groups to all Deligne-Mumford stacks that have a stratification whose associated strata are each of the form $[X_i/\Gamma_i]$. -This was essential at that time, since it was not then known that the stacks of stable maps could be written as $[X/\Gamma]$, yet it was known that the stacks have stratifications with strata of the form $[X_i/\Gamma_i].$ The Behrend-Fantechi construction of a virtual fundamental class for (closed) Gromov-Witten invariants in algebraic geometry depends on the existence of Chow groups for stacks satisfying certain axioms (in particular, Vistoli's rational equivalence). Kresch proved that the stacks of stable maps have such a Chow theory.<|endoftext|> -TITLE: Efficient method to write number as a sum of four squares? -QUESTION [8 upvotes]: Wikipedia states that there randomized polynomial-time algorithms for writing $n$ as a sum of four squares -$n=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}$ -in expected running time $\mathrm {O} (\log^{2} n).$ -My question is can someone give the efficient algorithm( $\mathrm {O} (\log^{2} n)$ ) to represent $n$ as sum of four squares. - -REPLY [3 votes]: One of their methods for $n=4k+2$ is as follows: -Randomly select an even number $a$ and an odd number $b$ such that $a^2+b^2 < n$. Then, we hope $p=n-a^2-b^2$ is a prime (you can show there's about a $1/(A \log n \log \log n)$ chance of $p$ being prime ); $p$ is of the form $4r+1$, so if $p$ is prime there's a solution to $c^2+d^2=p$. -To find that, we try to solve $m^2+1 \equiv 0 \pmod{p}$; I'm actually going to describe a slightly different method. Select $x$ at random from $1$ to $p-1$; then, $x^{2r}=\pm 1$ depending on whether $x$ is a quadratic residue so calculate $x^r$ by repeated squaring, and with a $1/2$ chance if $p$ is prime (a smaller one if $p$ is composite) you'll find a valid $m$. -Given such an $m$, $m+i$ is a Gaussian integer with norm divisible by $p$ but smaller than $p^2$; use the Euclidean algorithm on the Gaussian integers with $p$ and get $c+di$ with norm $p$, and $a^2+b^2+c^2+d^2=n$. -For an odd number $n$, solve for $a^2+b^2+c^2+d^2=2n$; note that by mod $4$ considerations exactly two of $a,b,c,d$ must be odd, and without loss of generality assume $a,b$ are odd and $c,d$ are even. Then,$(\frac{1}{2}(a+b))^2+(\frac{1}{2}(a-b))^2+(\frac{1}{2}(c+d))^2+(\frac{1}{2}(c-d))^2=n.$ -For $n$ a multiple of $4$, solve for $n/4$ recursively, and multiply all values by $4$.<|endoftext|> -TITLE: $p$-adic numbers in physics -QUESTION [14 upvotes]: As far as I know, in modern physics we assume that the underlying field of work is the field of real numbers (or complex numbers). Imagine one second that we make a crazy assumption and suggest that the fundamental equations of physics can be expressed with $p$-adic numbers. What could be really rewritten formally? Does it make sense in the physical world to work over $p$-adic numbers. Is there such attempt in the history of mathematical physics ? What is the most convincing justification (in physics) that we need to work over the field of real numbers ? - -REPLY [23 votes]: Since the question on physics.stackexchange that was referred to in the comments above has been closed, let me essentially repeat my answer here together with some updates. - -I think there are mainly two reasons for motivating the introduction of $p$-adic models in physics. - -They could exist in nature. -They provide insightful toy models for physical phenomena. - -There is a vast physical literature on the subject which cite Reason 1 as justification, this is sometimes called the Vladimirov Hypothesis. Namely, we do not know the texture of spacetime at the Planck scale therefore it is possible that it might look more like $\mathbb{Q}_p^d$ than $\mathbb{R}^d$. -It is a seductive idea, but there is no evidence for it. Moreover, this would beg the question of why would Nature choose a particular prime number. A perhaps better hypothesis is the Adelic one advocated by Manin, i.e., all primes should occur democratically. -In any case, this is quite speculative at present. -The way I see it, -$p$-adic numbers are useful in physics mainly because of Reason 2. -By exaggerating only a little, one could argue that $p$-adic toy models is what guided Kenneth Wilson when making his great discoveries in the theory of the renormalization group -which entirely revolutionized physics in the early seventies. -When studying complex multiscale phenomena it is often important to decompose functions into time-frequency atoms which live on a tree, e.g., when using a wavelet decomposition. -Unfortunately for most questions of interest the metric which governs how these atoms interact with each other is not the natural (from the tree point of view) ultrametric distance, but the Euclidean metric of the underlying continuum. Hierarchical models in physics amount to changing the model so it is the ultrametric distance which defines atomic interactions. The same idea also appears in mathematics where such toy models are often called "dyadic models". See this wonderful post by Tao for a nice discussion of this circle of ideas. Given a problem in Euclidean space, there are lots of ways of setting up a simplified hierarchical model for it. -The $p$-adics, in some sense provide the most canonical, structured and principled way of doing this. -To go back to Wilson and the RG, he also used a hierarchical model -and gave it yet another name "the approximate recursion". The importance of this toy model in Wilson's path to discovery is clear from his quote: - -"Then, at Michael's urging, I work out what happens near four dimensions for the approximate recursion formula, and find that d-4 acts as a small parameter. Knowing this it is then trivial, given my field theoretic training, to construct the beginning of the epsilon expansion for critical exponents." - -which can be found here (about one third from the bottom of the page). -For example, the use of $p$-adics gives a toy model for rigorously understanding conformal field theories. This is explained in my article "Towards three-dimensional conformal probability" (these slides are perhaps easier to understand since they have lots of pictures). -More recently, the following articles explored similar issues: - -"$p$-adic AdS/CFT" by Gubser, Knaute, Parikh, Samberg and Witaszczyk, -"Tensor networks, $p$-adic fields, and algebraic curves: arithmetic and the ${\rm AdS}_3$/${\rm CFT}_2$ correspondence" by Heydeman, Marcolli, Saberi and Stoica, -"Edge length dynamics on graphs with applications to $p$-adic AdS/CFT" by Gubser, Heydeman, Jepsen, Marcolli, Parikh, Saberi, Stoica and Trundy. - -As another example, in the article "A second-quantized Kolmogorov-Chentsov theorem via the operator product expansion" I discovered a general result for making sense of products of random Schwartz distributions starting from Wilson's operator product expansion. The statement of the theorem and the method of proof was done over $\mathbb{Q}_p$ -first. -Another remark is that exotic numbers like $p$-adics can appear in two different conceptual roles. Suppose one is interested in a "scalar field" $\phi:X\rightarrow Y$ where $X$ is space-time. The toy models I mentioned make $X$ $p$-adic but keep $Y$ real or complex. As in Dyson's quote from Carlo's answer, things should still be Archimedean-valued. This kind of distinction is well known to researchers in the area called the Langlands Program which splits in two separate arenas depending on whether one's favorite $L$ or zeta functions take values in $\mathbb{C}$ or $\mathbb{Q}_p$. -There are works in $p$-adic physics (e.g. by Khrennikov) where $Y$ is taken $p$-adic, but I am not familiar with the motivation behind this. -Finally, and in relation to Laie's comment above, one possible use of $p$-adics is to try to exploit an Adelic product formula in order to compute a real quantity of interest as the inverse of the product of similar quantitites over $\mathbb{Q}_p$. This was observed for tree level Veneziano amplitudes and was one of the first motivations for developing $p$-adic physics. However, these adelic formulas break down for higher loop amplitudes. See this Physics Report article for more on this story.<|endoftext|> -TITLE: The missing link: an inequality -QUESTION [44 upvotes]: I've been working on a project and proved a few relevant results, but got stuck on one tricky problem: - -Conjecture. If $2\leq n\in\mathbb{N}$ and $00$ amounts to the positivity of the polynomial -$$V:=PQ^2P''+(PQ')^2-P^2QQ''-(P'Q)^2.$$ - -REPLY [3 votes]: This is only a comment to @DimaPasechnik, but I cannot put the picture in a comment. The surface to the right of the $x=y$ is a plot of Dima's function (barring mistakes); clearly not convex.<|endoftext|> -TITLE: What is the density of integers of the form $a^2+nb^2$? -QUESTION [9 upvotes]: Landau proved that the mean density of integers of the form $a^2+b^2$ up to $x$ is $K\frac{x}{\sqrt{\log x}} (1+o(1))$, where $K$ is an explicit constant. One proof is based on the fact that a prime $p$ is of this form iff $p\neq 3\mod 4$, and so we can express the generating function for these integers in terms of the zeta function and the $L$-function of the non-trivial character modulo $4$. From there one proceed using complex analysis. -For integers of the form $a^2 + nb^2$, one should have similar results. We know what primes has this form by class field theory, and then in principle one can continue as before. -I am sure this is classical, but unfortunately I've failed to find it in the literature (including google). Does anyone know a reference for this? - -REPLY [3 votes]: R. W. K. Odoni has his own proof of the theorem, see this paper on the problem, - -"On norms of integers in a full module of an algebraic number field and the distribution of values of binary integral quadratic forms", - -published in Mathematika 22 (1975), no. 2, 108–111. See Narkiewicz's review. Odoni's paper does not appear in the paper mentioned by Daniel. -The concluding remarks in Odoni's paper are very interesting and refer to some earlier partial results.<|endoftext|> -TITLE: generating function for Dyck paths avoiding a pattern -QUESTION [6 upvotes]: This is a follow up to Counting QF-1 algebras inside quotient algebras of upper triangular matrices via Dyck paths. For $k\geq 1$ let $P_k=x+O(x^{k+2})$ be the formal power series that satisfies -$$ -P_k = \frac{x}{1-P_{k+1}} - \frac{x^2(1-x^k)}{1-x}. -$$ -I am interested in $P_1$, is there a nicer formula? - -REPLY [4 votes]: Here is one possible expression. I have not tried to simplify it, maybe there actually is much simpler formula, I don't know. -$$ -P_k(q)=1-(1-q)\frac{H(q^{k+2})}{H(q^{k+3})} -$$ -with -$$ -H(z)={}_1\phi_1\left(\begin{smallmatrix}0\\\left(\frac q{1-q}\right)^2\end{smallmatrix},q,\frac z{(1-q)^2}\right), -$$ -where $_1\phi_1$ is the basic hypergeometric function; thus -\begin{align*} -H(z)&=1-\frac1{(1-q)((1-q)^2-q^2)}z\\ -&+\frac q{(1-q)(1-q^2)((1-q)^2-q^2)((1-q)^2-q^3)}z^2\\ -&-\frac{q^3}{(1-q)(1-q^2)(1-q^3)((1-q)^2-q^2)((1-q)^2-q^3)((1-q)^2-q^4)}z^3\\ -&\cdots\\ -&+(-1)^n\frac{q^{\binom n2}}{(1-q)\cdots(1-q^n)((1-q)^2-q^2)\cdots((1-q)^2-q^{n+1})}z^n\\ -&\pm\cdots -\end{align*} -Here is a plot of the modulus for $P_1(q)$ inside the unit disc (color $=$ phase); I decided to add it because of a suggestive feature: strings of alternating poles and zeros along lines converging to $1$. It hints at a possible infinite product maybe somehow related to theta-functions. - -At any rate the plot shows the pole nearest to the origin at $q=0.46305364318045766...$, so the leading asymptotics for coefficients must be of base reciprocal to this, i.e. $2.159577005228934...$. Indeed empirically the $n$th coefficient seems to be $\sim0.02700488448532407\times2.159577005228934^n$ -What follows is not so much a proof, but rather sort of a heuristic explanation of how did I arrive at this; still I believe it is reasonably reliable. -From -$$ -P_k(x)=-\frac{1-x^k}{1-x}x^2+\frac x{1-P_{k+1}(x)} -$$ -we get -$$ -P_k(x)= --\frac{1-x^k}{1-x}x^2+\cfrac x{1+\frac{1-x^{k+1}}{1-x}x^2-\cfrac x{1+\frac{1-x^{k+2}}{1-x}x^2-\cfrac x{1+\frac{1-x^{k+3}}{1-x}x^2-\cfrac x{\qquad\ddots}}}}. -$$ -Let us introduce -$$ -F_k(z,q)=-\frac{1-q^kz}{1-q}q^2+\cfrac q{1+\frac{1-q^{k+1}z}{1-q}q^2-\cfrac q{1+\frac{1-q^{k+2}z}{1-q}q^2-\cfrac q{1+\frac{1-q^{k+3}z}{1-q}q^2-\cfrac q{\qquad\ddots}}}}, -$$ -so that $P_k(q)=F_k(1,q)$. Then we get the functional equation -$$ -F_k(z,q)=-\frac{1-q^kz}{1-q}q^2+\frac q{1-F_k(qz,q)}. -$$ -To linearize this, let us introduce another function $A_k(z)$ given by -$$ -\frac{1-q}{1-F_k(z,q)}=\frac{A_k(qz)}{A_k(z)}, -$$ -so that $P_k(q)=1-(1-q)\frac{A_k(1)}{A_k(q)}$. -The above functional equation then becomes -$$ -(1-q)\frac{A_k(z)}{A_k(qz)}=1+\frac{1-q^kz}{1-q}q^2-\frac q{1-q}\frac{A_k(q^2z)}{A_k(qz)}, -$$ -so -$$ -(1-q)^2A_k(z)=(1-q+(1-q^kz)q^2)A_k(qz)-qA_k(q^2z). -$$ -If $A_k(z)=1+a_1z+a_2z^2+...$, then for the coefficients $a_n$ (with $a_0=1$) we get -$$ -a_n=-\frac{q^{k+n+1}}{(1-q^n)((1-q)^2-q^{n+1})}a_{n-1}. -$$ -This gives -$$ -A_k(z)={}_1\phi_1\left(\begin{smallmatrix}0\\\left(\frac q{1-q}\right)^2\end{smallmatrix},q,\left(\frac q{1-q}\right)^2q^kz\right), -$$ -or in our notation $A_k(z)=H(q^{k+2}z)$. Then the above equality $P_k(q)=1-(1-q)\frac{A_k(1)}{A_k(q)}$ gives the claimed expression. -PS And here is the Mathematica code that produced the plot, in case somebody would like to play with it -P1[q_]:=With[{c=(q/(1-q))^2}, - 1-(1-q) QHypergeometricPFQ[{0},{c},q,q c]/QHypergeometricPFQ[{0},{c},q,q^2 c]] -ListPlot3D[Flatten[Table[ - With[{q=r E^(I a)},{r Cos[a],r Sin[a],Abs[P1[q]]}], - {r,Table[1-1/n^2,{n,1.01,10,.01}]},{a,-\[Pi],\[Pi],\[Pi]/180} - ],1], - ColorFunction->(Hue[(\[Pi]+Arg[P1[#1+I #2]])/(2\[Pi])]&), - ColorFunctionScaling->False, MeshFunctions->{(#3&)},Mesh->20]<|endoftext|> -TITLE: Category of bicomodules of a cosemisimple Hopf algebra -QUESTION [5 upvotes]: A cosemisimple Hopf algebra $H$ is one which is equal to the direct sum of its subcoalgebras. As is well-known, this is equivalent to its category of $H$-comodules being semisimple. Is this also true for the category of bicomodules of the Hopf algebra? (See here for the definition of a bicomodule.) - -REPLY [5 votes]: The answer is yes, if we are talking about finite dimensional, Hopf algebras over a field: -$\bullet$ $H$ being cosemisimple (as a coalgebra) is equivalent to the dual hopf algebra $H^*$ being semisimple (as an algebra). But a semisimple Hopf algebra is also separable (see Ex.5.2.12). Now, $H^*$ being separable is equivalent (see th.6.1.2) to $H^*\otimes (H^*)^{op}$ being semisimple thus any (right) $H^*\otimes (H^*)^{op}$-module is semisimple. Since an $H^*\otimes (H^*)^{op}$-module is essentially the same thing as an $H^*$-bimodule (in the sense that the corresponding categories are isomorphic), we conclude that any $H^*$-bimodule is semisimple. This means that the category ${}_{H^*}\mathcal{M}_{H^*}$ of $H^*$-bimodules is semisimple. -Now, to answer your question, recall (see th.2.3.3) that the category $Rat({}_{H^*}\mathcal{M}_{H^*})$ of rational $H^*$-bimodules is isomorphic to the category ${}^{H}\mathcal{M}^H$ of $H$-bicomodules and that for a finite dimensional Hopf algebra $H$, all $H^*$-modules are rational, i.e. -$$ -{}^{H}\mathcal{M}^H\cong Rat({}_{H^*}\mathcal{M}_{H^*})={}_{H^*}\mathcal{M}_{H^*} -$$ -thus, the category of $H$-bicomodules ${}^{H}\mathcal{M}^H$ is isomorphic to the (semisimple) category ${}_{H^*}\mathcal{M}_{H^*}$ of $H^*$-bimodules. (In general $Rat({}_{H^*}\mathcal{M}_{H^*})$ is a full subcategory of ${}_{H^*}\mathcal{M}_{H^*}$). -$\bullet$ For the converse, the above argument can easily run all the way back, recalling that any separable algebra is -by definition- semisimple. -Thus, the above shows that, for a finite dimensional hopf algebra $H$: - -$H$ being cosemisimple is equivalent to the category ${}^{H}\mathcal{M}^H$ of $H$-bicomodules being semisimple.<|endoftext|> -TITLE: attaching maps in CW complexes -QUESTION [5 upvotes]: Suppose I have a finite CW complex $X$ with $p$-skeleton $X^{(p)}$. - -Let $\varphi_f \colon S^p \to X^{(p)}$ be part of the attaching map of a $(p+1)$-cell $f$. -Let $\Phi_e \colon D^p \to X^{(p)}$ be part of the attaching map of a $p$-cell $e$ and let $q_e \colon X^{(p)} \to S^p$ be the map that collapses everything outside of $\Phi_e(Int(D^p)) \subset X^{(p)}$ to a point. - -Suppose that the map $\theta = q_e \circ \varphi_f \colon S^p \to X^{(p)} \to S^p$ is surjective. Let $x_0 = \Phi_e(0) \in X^{(p)}$. - -Is it possible to change $\varphi_f$ by a homotopy to a map $\hat{\varphi}_f$, such that $\hat{\varphi}_f^{-1}(x_0)$ is a finite set of points? - -Since $\theta$ is a continuous map between smooth manifolds, it is homotopic to a smooth map. Hence, I can arrange that $\theta^{-1}(x_0)$ is a finite set of points after a homotopy. But the result of the homotopy may not lift to a map $S^p \to X^{(p)}$. -I tried to change $\varphi_f$ only inside the preimage of the open cell in $X^{(p)}$ and keep the rest fixed, but I did not manage to ensure that there are no new points created in the process that are mapped to $x_0$. - -REPLY [3 votes]: This is essentially Lemma 4.10 of Hatcher's book, which is the key step in proving the cellular approximation theorem. It shows that you can homotope $\varphi_f$ so that there is an open set $U$ in your $p$-cell $e$ such that $\varphi_f$ is piecewise linear on the inverse image of $U$. Then a general element of $U$ has only finitely many preimages.<|endoftext|> -TITLE: What is precisely still missing in Connes' approach to RH? -QUESTION [51 upvotes]: I have read Connes' survey article http://www.alainconnes.org/docs/rhfinal.pdf -and I am somewhat familiar with his classic paper on the trace formula: http://www.alainconnes.org/docs/selecta.ps -Very roughly speaking the idea is to describe a dictionary which translates the concepts and techniques used in the proof of the analogue of the Riemann Hypothesis for function fields. -This translation uses various techniques from tropical geometry and topos theory. At first I was hopeful I might understand the key issues with this translation, since I have some experience with the theory of Grothendieck toposes (or topoi). -Nevertheless I have been lost when it comes to understanding precisely what the remaining problems are. As already briefly discussed in this thread: Riemann hypothesis via absolute geometry in the proof of the Riemann hypothesis for a curve $X$ over $F_p$ the surface $X \times X$ plays an important role. According to a new preprint of Connes / Consani there is now a suitable analogy for the surface $X \times X$ which is given by a topos called the "scaling site", cf. https://arxiv.org/abs/1603.03191 -I would like to know what are the issues that are left open to complete the analogy with the proof of RH in the case of a curve over $F_p$? - -REPLY [4 votes]: Terence Tao answered this question as follows: -" He (Connes) also discussed his recent paper with Consani in which they managed to view the (completed) Riemann zeta function as the Hasse-Weil zeta function of a curve over F_1, where F_1 is interpreted as the tropical ring {0,1}. So the Riemann hypothesis for the integers is now expressed in a very similar fashion to the Riemann hypothesis for curves, which has a number of proofs. Unfortunately, they are missing a huge ingredient in the dictionary, namely they have no Riemann-Roch theorem over F_1. Still, it is a very suggestive similarity, and something to keep an eye on in the future... "<|endoftext|> -TITLE: An explicit series representation for the analytic tetration with complex height -QUESTION [27 upvotes]: Tetration is the next hyperoperation after more familiar addition, multiplication and exponentiation. It can be seen as a repeated exponentiation, similar to how exponentiation can be seen as a repeated multiplication, and multiplication — as a repeated addition. Tetration is denoted $^n a$, where $a$ is called the base and $n$ is called the height, and is defined for $n\in\mathbb N\cup\{-1,\,0\}$ by the recurrence -$$ -{^{-1} a} = 0, \quad {^{n+1} a} = a^{\left({^n a}\right)},\tag1$$ -so that -$${^0 a}=1, \quad {^1 a} = a, \quad {^2 a} = a^a, \quad {^3 a} = a^{a^a}, \, \dots \quad {^n a} = \underbrace{a^{a^{{.^{.^{.^a}}}}}}_{n\,\text{levels}},\tag2$$ -where power towers are evaluated from top to bottom. -To simplify matters, we restrict our attention to real bases in the interval $1-2$ to an analytic function that satisfies the functional equation for tetration $t(z+1)=a^{t(z)}$ for all arguments in its domain of analyticity. -Conjecture 2. The function $t(z)$ is periodic with a purely imaginary period $\tau = 2\pi n i/\ln q$, and admits an analytic continuation to the half-plane $\Re(z) \le -2$ by repeated backward application of the functional equation $(6)$. There it has a countable set of isolated singularities at $z = m + \tau\,n, \,m,n\in\mathbb Z, \, m \le -2$, and a countable set of branch cuts that can be made along horizontal rays at $z = x + \tau\,n, \, n\in\mathbb Z, \, x\in\mathbb R, \, x < -2$. -Conjecture 3. For real $x>-2$, the derivative $t'(x)$ is a completely monotone function. -Conjecture 4. There is only one analytic function that vanishes at $z=-1$ and satisfies both the functional equation for the tetration $(6)$, and the complete monotonicity condition from the Conjecture 3. - -Related questions: [1][2][3]. - -REPLY [5 votes]: I have written and submitted for publication The Existence and Uniqueness of the Taylor Series of Iterated Functions. It has cleared an initial peer review with three knowledgeable mathematicians and now is in further peer review for publication. The intent is to have a single authoritative work on extending tetration and all the higher hyperoperators to the complex numbers. The uniqueness and existence theorems in the paper guarantee a unique solution to the problems. So at best, other extensions of tetration must be consistent with the extension provided in the paper. Basing the mathematics on iterated functions provides the generality needed to extend all the higher hyperoperators to complex numbers, not just tetration.<|endoftext|> -TITLE: 12th grade - Ramanujan Partition theory -QUESTION [6 upvotes]: I've been really trying to prove Ramanujan Partition theory, and different sources give me different explanations. -Can someone please explain how Ramanujan (and Euler) found out the following equation for the number of partitions for a given integer? -Any help is appreciated thank you so much! -$$P(n)\sim\frac1{4n\sqrt3}\exp(\pi\sqrt{2n/3})$$ - -REPLY [3 votes]: Another reference is Atle Selberg's paper "Reflections around the Ramanujan centenary". More precisely, the appendix, if I remember. It appears in the book "Ramanujan: essays and surveys". My recommendation would be to read that before Hardy, for reasons that Selberg himself explains.<|endoftext|> -TITLE: Taylor expansion of cumulant generating function -QUESTION [8 upvotes]: For the characteristic function $\mathbf E e^{i t X}$ of a random variable $X$ with $n+1$ finite moments, there is the well known and easy to prove bound on the remainder of the Taylor series -$$\left\lvert\mathbf E e^{i t X}-\sum_{k=0}^n \frac{(it)^k}{k!}\mathbf E X^k\right\rvert\le\min\left\{\frac{\lvert t\rvert^{n+1}\mathbf E \lvert X^{n+1}\rvert}{(n+1)!},\frac{2\lvert t\rvert^n\mathbf E\lvert X^n\rvert}{n!}\right\}.$$ -Can something similar be said for the remainder of the cumulant generating function $\log\mathbf E e^{itX}$ with an error bound in terms of cumulants? I.e., I am hoping for a bound of the form -$$\left\lvert \log\mathbf E e^{itX}-\sum_{k=0}^n\frac{(it)^k}{k!} \kappa_k(X)\right\rvert\leq a(t) \kappa_n(X) + b(t) \kappa_{n+1}(X),$$ where $\kappa_k(X)$ is the $k$-th cumulant of $X$. -I am not even sure whether such an bound even exists -- partly because the cumulants can exhibit certain cancellation effects not present in the moments. - -REPLY [3 votes]: Can something similar be said for the remainder of the cumulant - generating function $\log\mathbf E e^{itX}$ with an error bound in - terms of cumulants? - -Yes and no. -Your question is whether the union bound for Fourier transform can be somehow generalized to log Fourier transform for the class of $L^{n+1}(\mathbb{R})$ (for simplicity I do not discuss $\mathbb{R}^n$ domain case). -"Yes part". -The result you are asked for is basically a Tauberian-type theorem for log Fourier transform that controls the residue in terms of coefficients($\kappa_n$ as in OP). For log Laplace transform $\log\mathbf E e^{tX}$, the result is readily available in Theorem 16.1 of [Korevaar]. -We write the density $p(\xi)=dP(\xi)$ and the Laplace transform -$\mathcal{L}dP(\xi)=\mathcal{L}dP(\frac{1}{x})=\int_{0^{-}}^{\infty}e^{-\frac{v}{x}}dP(x)$ according to Part IV Thm 16.1, we have -$log\mathcal{L}dP(\xi)=log\mathcal{L}dP(\frac{1}{x})\sim(\alpha-1)\frac{x^{\alpha}L(x)}{x}$ as $x\rightarrow\infty$ where $L(x)$ is the Laplace transform. For harmonic $P$, there are some results similar contained near the end (possibly title "random functions" or something like that, do not have the book at hand now) of [Paley-Wiener]. [Safarov] provides some concrete applications under certain assumptions on $P$'s (say 1.20-1.21). In this regard, "complex Tauberian" or "Fourier Tauberian" are the correct keywords to search for. -"No part". -Although there are some results in spirit of Tauberian theorems as mentioned above, I do not know there is a result concerning a class as general as $L^{n+1}(\mathbb{R})$. -As you already noticed there is such a potentiality of singularities that prevents you from bounding the residues, say Schwartz-class measures and their power polynomials. If you translate the bound on moment generating function $\mathbf E e^{i t X}$ onto $\log\mathbf E e^{itX}$ using the relation between moments and cumulants (see, Mobius (inverse) transform [wiki, Rota]), you will still have $\pm$ cancellation in these relations; -However, there is still arising research concerning bounding the tail behavior of a given density using cumulant expansions [Valz et.al], so I guess this is not completely a dead end. -This is also (approximately) a question puzzled me a year ago. This answer is my exploration, and I really hope someone more knowledgeable can provide a more complete answer. Hope this helps! -Reference -[Korevaar]Korevaar, Jacob. Tauberian theory: A century of developments. Vol. 329. Springer Science & Business Media, 2013. -[Paley-Wiener]Paley, Raymond Edward Alan Christopher, Norbert Wiener, and Norbert Wiener. Fourier transforms in the complex domain. Vol. 19. New York: American Mathematical Society, 1934. -[Safarov]Safarov, Yu. "Fourier Tauberian theorems and applications." Journal of Functional Analysis 185.1 (2001): 111-128. -[wiki] https://en.wikipedia.org/wiki/Cumulant -[Rota]Rota, Gian-Carlo, and Jianhong Shen. "On the combinatorics of cumulants." Journal of Combinatorial Theory, Series A 91.1-2 (2000): 283-304. -[Valz et.al]Valz, Paul D., A. Ian McLeod, and Mary E. Thompson. "Cumulant generating function and tail probability approximations for Kendall's score with tied rankings." The Annals of Statistics (1995): 144-160. -[Bazant]Martin Bazant, Lecture 5: Asymptotics with Fat Tails. https://ocw.mit.edu/courses/mathematics/18-366-random-walks-and-diffusion-fall-2006/lecture-notes/lec05.pdf<|endoftext|> -TITLE: Cartier-Kostant-Milnor-Moore theorem -QUESTION [10 upvotes]: If $k$ is an algebraically closed field of characteristic zero and $H$ is a cocommutative Hopf algebra, then -$$ -H \cong U(P(H)) \ltimes kG(H). -$$ -What happens if the field is not algebraically closed? Is the theorem still true or is there any counterexample? What about characteristic different from zero? - -REPLY [10 votes]: When $k$ fails to be algebraically closed the theorem is false but the discrepancy can be understood in terms of Galois descent and so in principle understood in terms of Galois cohomology. -Suppose $H$ is a cocommutative Hopf algebra over $k$, not algebraically closed but characteristic $0$. Then the classification theorem applies to the base change $H \otimes_k \bar{k}$ of $H$ to the algebraic closure, which must therefore take the form of the semidirect product of a universal enveloping algebra and a group algebra. So $H$ is a $k$-form of such a thing, but need not be such a thing itself. -For example, suppose $H$ is finite-dimensional, so $H \otimes_k \bar{k}$ is a group algebra $\bar{k}[\Lambda]$ of a finite group $\Lambda$. The most obvious $k$-form of $\bar{k}[\Lambda]$ is, of course, $k[\Lambda]$, but other $k$-forms are possible. In this setting $H$ necessarily splits over a finite Galois extension $L$ in the sense that we must already have -$$H \otimes_k L \cong L[\Lambda].$$ -Now, the Galois group $G = \text{Gal}(L/k)$ acts on the LHS by Hopf algebra automorphisms (over $k$), so also acts on the RHS by Hopf algebra automorphisms (over $k$), and hence acts on the group of grouplike elements $\Lambda$ on the RHS. If we already had $H \cong k[\Lambda]$ then this action would be trivial, but in fact every possible action of $G$ on $\Lambda$ occurs, and conjugacy classes of such actions classify the possible choices of $H$. Given such an action $\rho : H \to \text{Aut}(\Lambda)$ we can recover $H$ as -$$H \cong L[\Lambda]^G$$ -where the action of $G$ on $L[\Lambda]$ extends the action of $G$ on $\Lambda$. -Very explicitly, let $k = \mathbb{R}, L = \mathbb{C}$, and $\Lambda = C_3$, where the Galois group $G = C_2$ acts via inversion. Then we have -$$H \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C}[C_3] \cong \mathbb{C}[x]/(x^3 - 1)$$ -where $x$ is grouplike but the action of complex conjugation sends $x$ to $x^{-1} = x^2$. The fixed point subalgebra $H$ has a basis given by the identity $1$ and the "real and imaginary parts" -$$c = \frac{x + x^{-1}}{2}$$ -$$s = \frac{x - x^{-1}}{2i}$$ -(it helps to think of $x$ as secretly being $\omega = e^{\frac{2\pi i}{3}}$ here) of $x$, with comultiplication given by the sine and cosine angle addition formulas -$$\Delta(c) = c \otimes c - s \otimes s$$ -$$\Delta(s) = c \otimes s + s \otimes c.$$ -So neither $c$ nor $s$ is grouplike, and in fact $H$ has no nontrivial grouplike elements except the identity. In $H \otimes_{\mathbb{R}} \mathbb{C}$ the nontrivial grouplike elements are $c + is$ and $c - is$.<|endoftext|> -TITLE: Combinatorial interpretation of series reversion coefficients -QUESTION [8 upvotes]: In a recent paper studying some generalizations of Stirling numbers, my coauthors and I needed the following result: -If $f(x)=\sum_{n \geq 1} a_n x^n/n!$ is a power series with $a_1 \neq 0$, and $g(x) = \sum_{n \geq 1} b_n x^n/n!$ is its series reversion (so $f(g(x))=g(f(x))=x$), then -$$ -b_n = \sum_{T} w(T) -$$ -where the sum is over phylogenetic trees on $\{1, \ldots, n\}$. Here a phylogenetic tree on $\{1,\ldots, n\}$ is a rooted tree with $n$ leaves, no vertex with exactly one child, and leaves labelled from the set $\{1, \ldots, n\}$ (but no other vertex receiving a label); see page 3 of the paper linked above for a picture. The weight $w(T)$ is computed as follows: if $T$ has $n$ leaves and $m$ non-leaves then -$$ -w(T) = (−1)^m a_1^{-(m+n)} \prod \{a_{d(v)}: v~\mbox{a non-leaf}\} -$$ -where $d(v)$ denotes the number of children of $v$ (and when $n=1$ the isolated vertex is considered a leaf). -The proof is short, and this feels like something that should have been written down prior to now, but we could find no reference. -Question: Has the above result appeared in the literature? -(There is a paper of Chen from 1990 with a similar result but with a distinctly different and not obviously equivalent formulation. Chen shows that $b_n$ is a weighted sum over Schroeder trees: trees on $n$ labelled vertices (as opposed to leaves), with no restriction on numbers of children, in which each non-leaf is endowed with an ordered partition of its children. The count of Schroeder trees by number of vertices is A053492. The count of phylogenetic trees by number of leaves is A000311.) - -REPLY [6 votes]: This result can also be found as Theorem 3.3.9 in Ginzburg-Kapranov: "Koszul duality for operads" (Duke Math J 1994). Set $r=1$ in their result and rescale so that $a_1=1$. They had discovered it independently, but thank D. Wright for informing them that the result was known before due to J. Towber. They also give a reference to Wright: "The tree formulas for reversion of power series" (JPAA 1989), where this result appears as Theorem 3.10. -The context of this result in the setting of operads is that any dg operad $P = \{P(n)\}$, say with $P(0)=0$ and $P(1)$ a copy of the base field, has a generating series which encodes the Euler characteristics of the chain complexes $P(n)$. The generating series of $P$ is the compositional inverse of the generating series for the "bar construction" on the operad $P$. The bar construction of operads is defined by a sum over trees whose internal vertices are decorated by the chain complexes of the operad $P$; in fact, the trees involved are exactly those that you call phylogenetic trees.<|endoftext|> -TITLE: Reeb flows on $S^3$ versus volume preserving flows -QUESTION [9 upvotes]: Is there an example of a smooth vector field $v$ on $S^3$ such that $v$ preserves a volume form and $v$ is not a Reeb vector field? -Recall that $v$ is a Reeb vector field if there exists a contact $1$-form $\alpha$ such that $\alpha(v)=1$ and $v$ belongs to the kernel of $d\alpha$. - -REPLY [10 votes]: I took the time to give a clean version of the answer and to eliminate the dependence on really hard theorems. -Construction of geodesible, volume-preserving flows on $S^3$ that are not Reeb flows for any contact form. -Ingredients: - -A contact form $\alpha$ whose Reeb vector field $X$ has no nonconstant integrals of motion and which carries a closed characteristic. Such a form can be obtained from a Finsler or Riemannian metric on the two-sphere with ergodic geodesic flow (examples by Katok and Donnay). -A smooth function $f$ that is nowhere zero and such that for some closed characteristic $\gamma$ in $(S^3, \alpha)$ we have that -$$ -{ \int_{S^3} f^{-1} \ \alpha \wedge d\alpha \over \int_{S^3} \alpha \wedge d\alpha} -\neq -{\int_\gamma f^{-1} \ \alpha \over \int_\gamma \alpha}. -$$ - -Theorem. -The vector field $fX$ preserves the volume form $f^{-1} \alpha \wedge d\alpha$, but it is not the Reeb vector field of any contact form on $S^3$. -The proof is by contradiction and we will need the following -Claim. -If $fX$ is the Reeb vector field of a contact form $\beta$, then for some nonzero constant $\lambda$ and some smooth function $g$ we have that $\beta = \lambda \alpha + dg$. -Proof of the claim. Since the kernels of the $2$-forms -$d\alpha$ and $d\beta$ coincide and the space is three-dimensional, the forms are multiples: -$d\beta = \lambda d\alpha$, with $\lambda$ some smooth function. Moreover -$\lambda$ must be an integral of motion of $X$, -$$ -0 = {\cal L}_X d\beta = {\cal L}_X \lambda d\alpha = X(\lambda) \alpha . -$$ -By hypothesis $\lambda$ must be a constant and, therefore, -$d(\beta - \lambda \alpha) = 0$ or $\beta = \lambda \alpha + dg$. -Now we proceed with the proof of the theorem. First we show that $\lambda$ is the -average of the function $f^{-1}$ over $S^3$. Indeed, -$$ -1 = \beta(fX) = \lambda \alpha (fX) + dg(fX) = \lambda f + fX(g), -$$ -and so -$$ -\int_{S^3} f^{-1} \ \alpha \wedge d\alpha = -\int_{S^3} (\lambda f + fX(g))f^{-1} \ \alpha \wedge d\alpha. -$$ -Since $X$ preserves the form $\alpha \wedge d\alpha$, the integral of -$X(g)\alpha \wedge d\alpha$ is zero and we obtain -$$ -\int_{S^3} f^{-1} \ \alpha \wedge d\alpha = \lambda \int_{S^3} \alpha \wedge d\alpha . -$$ -Now we notice that $\lambda$ is also the average of $f^{-1}$ over every closed leaf of the characteritic foliation of $(S^3, \alpha)$. If $\gamma$ is a closed leaf, then -$$ -\int_\gamma f^{-1} \ \alpha = \int_\gamma (\lambda f + fX(g)) f^{-1} \ \alpha. -$$ -As before, using that $X$ preserves $\alpha$, we obtain that the integral of -$X(g)\alpha$ along $\gamma$ is zero, and hence -$$ -\int_\gamma f^{-1} \ \alpha = \lambda \int_\gamma \alpha. -$$ -This contradicts our assumption of the existence of a closed characteristic for which -$$ -{ \int_{S^3} f^{-1} \ \alpha \wedge d\alpha \over \int_{S^3} \alpha \wedge d\alpha} -\neq -{\int_\gamma f^{-1} \ \alpha \over \int_\gamma \alpha}. -$$ -%%%%%%%%%%%%%%%%%% Old, messy answer %%%%%%%%%%%%% -Edit. Long answer and a bit "thinking aloud" or "typing while thinking" in its organization (I'll reorganize it when I'll have the time), but at the end you have the following result: -Theorem. Consider a contact form $\alpha$ on the three-sphere and so that its Reeb vector field $X$ does not admit any smooth integral of motion except constants (e.g., the lift of some ergodic Finslerian or Riemannian geodesic flow on the two-sphere). If $f$ is a nowhere zero smooth function on $S^3$ such that it is not constant, and such there exist two closed Reeb orbits of $\alpha$ over which the averages of $1/f$ are distinct, then the vector field $fX$ preserves the volume form $(1/f) \alpha \wedge d\alpha$, but it is not the Reeb vector field of any contact form on $S^3$. -Consider a contact form $\alpha$ on the three-sphere and so that its Reeb vector field $X$ does not admit any smooth integral of motion except constants (e.g., the lift of some ergodic Finslerian or Riemannian geodesic flow on the two-sphere). Let $f$ be a nowhere zero smooth function on $S^3$ that is not constant. -Claim. The vector field $fX$ preserves the volume form $(1/f) \ \alpha \wedge d\alpha$, but it is not the Reeb vector field of any contact form DEFINING THE SAME CONTACT STRUCTURE AS $\alpha$ -The proof is by contradiction: -Assume $fX$ is the Reeb vector field of the contact form $\beta$. This means that the characteristic distributions of $d\alpha$ and $d\beta$ coincide. Since these are $2$-forms in a three-dimensional space, we conclude that these forms are multiples: $d\alpha = \lambda d\beta$. Moreover $\lambda$ must be an integral of motion of $X$, -$$ -0= \mathcal{L}_X d\alpha = \mathcal{L}_X \lambda d\beta = X(\lambda) d\beta . -$$ -By hypothesis, this means that $\lambda$ is a constant and, since both forms define the same contact structure, that $\beta = \lambda \alpha$. In this case, the Reeb vector field of $\beta$ is $fX = \lambda X$ and $f = \lambda$, contrary to the assumption that $f$ was not constant. $\square$ -Remark. If $\alpha$ and $\beta$ do not define the same contact structure, the above proof still shows that $\alpha$ and $\beta$ differ only in a multiplicative constant and the addition of the differential of some function. I have to finish grading so I'll come back to this later if someone has not given some nice explicit example. -Continuation. If we drop the assumption that both forms define the same contact structure, the preceding proof give us that $\beta = \lambda \alpha + dg$, where $\lambda$ is a nonzero constant and $g$ is some smooth function on the three-sphere. The relation between $\lambda$ and $f$ is easy to establish. Indeed, if we note that -$$ -1 = \beta(fX) = \lambda \alpha (fX) + dg(fX) = \lambda f + fX(g) , -$$ -multiply both sides by the volume form $(1/f) \ \alpha \wedge d\alpha$, and integrate over the three-sphere, we obtain that -$$ -\int (1/f) \ \alpha \wedge d\alpha = \lambda \int \ \alpha \wedge d\alpha -$$. -Note that the "missing" integral is zero because $X$ preserves the form $ \alpha \wedge d\alpha$. So $\lambda$ is the average of $1/f$ with respect to the volume form $ \alpha \wedge d\alpha$. -Going back to the equation -$$ -1 = \beta(fX) = \lambda f + fX(g) = f(\lambda + X(g)) \ \ {\rm or} \ 1/f = \lambda + X(g), -$$ -observe that if we multiply both sides by $\alpha$ and integrate over any closed characteristic $\gamma$, we obtain -$$ -\int_\gamma (1/f) \alpha = \lambda \int_\gamma \alpha -$$ -Therefore $\lambda$ is also the average of $1/f$ over every single closed characteristic of $(S^3,\alpha)$. That is a bit too much to ask of $\lambda$! -Thanks to D. Cristofaro-Gardiner and M. Hutchings we know that every Reeb vector field in the sphere has alt least two closed characteristics. We can be more restrictive with our original choice of $f$ and require that the average of $1/f$ be different for two closed characteristics of $(S^3,\alpha)$. The form $\beta$ then does not exist. -We have then: -Theorem. Consider a contact form $\alpha$ on the three-sphere and so that its Reeb vector field $X$ does not admit any smooth integral of motion except constants (e.g., the lift of some ergodic Finslerian or Riemannian geodesic flow on the two-sphere). If $f$ is a nowhere zero smooth function on $S^3$ such that it is not constant, and such there exist two closed Reeb orbits of $\alpha$ over which the averages of $1/f$ are distinct, then the vector field $fX$ preserves the volume form $(1/f) \alpha \wedge d\alpha$, but it is not the Reeb vector field of any contact form on $S^3$.<|endoftext|> -TITLE: Does the Hasse norm theorem easily imply the global squares theorem? -QUESTION [10 upvotes]: The Hasse norm theorem says a nonzero element $x$ of a number field $K$ is a norm for a cyclic extension $L/K$ iff $x$ is a norm for all completions $L_p/K_p$ at primes $p$ of $L$ (writing $p$ also for the restriction to $K$). -The global squares theorem says $x$ is a square in number field $K$ iff it is a square in every completion $K_p$ at a prime of $K$. -Is there some quick proof of the global squares theorem from the global norm theorem? Hilbert proved the global norm theorem for the case where $L/K$ is a quadratic extension. Is that special case enough to quickly prove the global squares theorem? -I would especially like to know a constructive proof that if $x$ is not a square in number field $K$ then there is a quadratic extension $L/K$ for which $x$ is not a norm. - -REPLY [2 votes]: Well, I would say that this crucially depends on what you define to be "quick". If you admit global class field theory, at least in its idelic formulation, the fact that all primes in $K$ are split in the extensions $K(\sqrt{x})/K$ proves that $x$ is a square quite rapidly (as Timo Keller has observed in the comments, this is actually easier than Chebotarev; and Franz Lemmermeyer's answer says even more). I give a short proof at the end of my answer, although it does not "need" the global norm theorem (which is anyhow hidden beneath the idelic formulation of global class field theory). I am pretty sure that this is the proof you have in mind when you ask for a quicker one. -Without class field theory, I could come up only with an "elementary" proof assuming that -[H1] $K$ admits at least one real embedding, let it be $\sigma\colon K\hookrightarrow \mathbb{R}$, and -[H2] $K/\mathbb{Q}$ is unramified at $2$. -Consider then the following -Claim An element in $\mathcal{O}_K\setminus\{0\}$ is a square if and only if it is a (global) norm from every quadratic extension $L/K$ -Admit the claim: we are going to deduce that if an element $x=r/s$ with $r,s\in\mathcal{O}_K$ is locally a square everywhere, then $x\in(K^\times)^2$. Let $y=xs^2$: then $y\in\mathcal{O}_K\setminus\{0\}$ is locally a norm from every quadratic extension $L/K$, since it is a square in $K_\mathfrak{p}^\times$ for all $\mathfrak{p}$. By the global norm theorem it is a global norm from every quadratic extension, and the claim tells you $y\in (K^\times)^2$: hence the same holds for $x=y/s^2$. -To prove the claim, observe that one direction is obvious: squares are norms from every quadratic extension. On the other hand, let $x\in\mathcal{O}_K\setminus\{0\}$ be a norm from every quadratic extension. I will use the real embedding $\sigma$. Consider the function -$$ -f\colon \mathcal{O}_K\times\mathcal{O}_K\setminus \{0\}\longrightarrow \mathbb{R} -$$ -sending $(a,b)$ to $f(a,b)=\sigma\left(\left(\frac{a}{b}\right)^2-\frac{x}{b^2}\right)$. Then there is a bound $B\in\mathbb{R}$ such that $\mathrm{Im}(f)\subseteq [B,+\infty)$. Indeed, $(a/b)^2\geq 0$, while $b\mapsto x/b^2$ is bounded from below, since $\sigma(\mathcal{O}_K)$ is discrete. Chose now a prime number $\ell\equiv 3\pmod{4}$ which is unramfied in $K/\mathbb{Q}$ and such that $-\ell -TITLE: approximate closed form for infinite series (no power series) -QUESTION [5 upvotes]: The electric potential of a charge between two infinite conducting planes can be expressed as an infinite series (of image charges) [Kellogg1929] as -$$ -V=\sum_{n=-\infty}^{\infty}\left( -\frac{1}{\sqrt{(z-2na-z_c)^2+x^2}} -- -\frac{1}{\sqrt{(z-2na+z_c)^2+x^2}} -\right) -$$ -(for $a > 0$, $0 < z < a$, $0 < z_c < a$, and $x > 0$) -No closed form for this series seems to exist. Yet, when inspecting sums up to $n=100,000,000$ I found the result to closely follow -$$ -V = \frac{P}{a}\sin\left(\frac{z}{a}\pi\right)\sin\left(\frac{z_c}{a}\pi\right)e^{-\frac{Q}{a}x} -$$ -in the range where $x > a$. The parameters I found empirically are $P=2.5$ and $Q=3.25$. The closed form is not equal to the infinite series but it comes close enough (it deviates by 10%-20% in absolute value over 12 orders of magnitude ($a < x < 10a$) that to my intuition there must be a way to show that the infinite series has a predominantly exponential behavior. Is that possible and how could that be shown? I understand that exponential functions are infinite power series, but the first series is not a power series. - -REPLY [6 votes]: This approximate solution can be obtained by transforming the infinite sum to a contour integral and then making an asymptotic expansion. This is worked out in Application of Sommerfeld-Watson Transformation to an Electrostatics Problem (1969). The result is slightly different from your surmise, -$$V\approx\sqrt{\frac{8}{xa}}\sin(\pi z/a)\sin(\pi z_c/a)e^{-\pi x/a}\;\;\text{if}\;\;x\gtrsim a,$$ -so your factor $Q=3.25$ in the exponential decay has the exact value $Q=\pi$, but more significantly there is a pre-exponential decay $\propto 1/\sqrt{x}$. -An alternative way to arrive at this asymptotic expression is to rewrite the sum over image charges identically as -$$V=\frac{4}{a}\sum_{n=1}^\infty \sin(n\pi z/a)\sin(n\pi z_c/a)K_0(n\pi x/a)$$ -and then make an asymptotic expansion of the Bessel function.<|endoftext|> -TITLE: Is a base-change of an integral domain by an extension of its base field without algebraic elements still a domain? -QUESTION [5 upvotes]: Let $K$ be a field, and let $L/K$ be an algebraically closed field extension (i.e. the only elements of $L$ that are algebraic over $K$ are already in $K$). Let $R$ be a $K$-algebra that is an integral domain. Does it follow that $R \otimes_K L$ is an integral domain? I'm particularly interested in the case where $R$ is a finitely generated $K$-algebra. -My question is closely related to this question, where Will Sawin gives a yes answer when $L$ is purely transcendental over $K$. Also (at least when $R$ is finitely generated as a $K$-algebra), the answer seems to be yes when $K$ is algebraically closed, according to a recent preprint. Furthermore, the answer is typically 'no' if $L$ has algebraic elements over $K$, even when $R$ itself is a field. For instance, let $f$ be an irreducible polynomial in $K[t]$ that has a root in $L$, and let $R := K[t]/(f)$. - -REPLY [8 votes]: No. Let $K = \mathbf{F}_p(s,t)$, $A = K[x,y]/(sx^p + t y^p - 1)$. One checks $A$ is a domain, even Dedekind, so we can define $L = {\rm{Frac}}(A)$. By exploring residue fields of $A$ at maximal ideals and using that $A$ is Dedekind one shows with some thought that $K$ is algebraically closed in $L$. But for the field $R = K[s^{1/p}, t^{1/p}]$ of degree $p^2$ over $K$ we have $R \otimes_K L = L[x,y]/(h^p)$ for $h = s^{1/p}x + t^{1/p}y - 1$, so this is non-reduced. In this case $A$ is a domain of finite type over $K$ that is a geometrically irreducible $K$-algebra that is geometrically everywhere non-reduced. This example is due to MacLane, but I don't know a literature reference for it.<|endoftext|> -TITLE: Fundamental groups of hyperbolic $4$-manifolds and $\rm CAT(0)$ cube complexes -QUESTION [6 upvotes]: Suppose $M^4$ is a compact hyperbolic (i.e. curvature $-1$) $4$-manifold and $\Gamma\cong\pi_1(M^4)$. -Is there any expectation whether $\Gamma$ acts properly and co-compactly on a $\rm CAT(0)$ cube complex? -Note that by work of Bergeron Wise based on work of Kahn-Markovic the answer to the question is always positive for $\Gamma=\pi_1(M^3)$, where $M^3$ is any hyperbolic $3$-manifold. However, if instead $\Gamma$ is a co-compact lattice in $U(2,1)$, the answer is always negative (by Delzant Py (I guess)). - -REPLY [8 votes]: This question is certainly open in general. I don't know if anyone has formally expressed an 'expectation' in print, but you might be interested in the following pieces of positive evidence. -I believe that all known examples of such 4-manifolds essentially come either from arithmetic constructions or from Coxeter groups. Haglund--Wise showed that all hyperbolic Coxeter groups are cocompactly cubulated (indeed virtually special) [3], and Bergeron--Haglund--Wise showed that 'standard' cocompact arithmetic lattices in $SO(n,1)$ are virtually special [1]. So most known examples are certainly cocompactly cubulated. -One might worry that closed hyperbolic 3-manifolds are too rigid to be constructed in large numbers, in the way that Kahn--Markovic did with surfaces. But Calegari and I were able to construct very many acylindrical hyperbolic 3-manifolds in random groups [2]. (They don't cubulate for other reasons, but it makes the point that a single group may contain many rigid subgroups.) -[1] Bergeron, Nicolas(F-PARIS6-IMJ); Haglund, Frédéric(F-PARIS11-M); Wise, Daniel T.(3-MGL) -Hyperplane sections in arithmetic hyperbolic manifolds. (English summary) -J. Lond. Math. Soc. (2) 83 (2011), no. 2, 431–448. -[2] Calegari, D; Wilton, H. -3-manifolds everywhere, https://arxiv.org/abs/1404.7043 -[3] Haglund, Frédéric(F-PARIS11-M); Wise, Daniel T.(3-MGL) -Coxeter groups are virtually special. (English summary) -Adv. Math. 224 (2010), no. 5, 1890–1903<|endoftext|> -TITLE: Beauty of some numbers discovered by Ramanujan -QUESTION [6 upvotes]: I am a graduate PhD student and my topic is analytic number theory. I am also a mathematics teacher. I am planning to give a course to pupils in high school that motivates them to study arithmetic and see the beauty of numbers. For instance, I have enough informations about the golden ratio. -I watched Professor Ken Ono on national geographic speaking about the beauty of Pi discovered by Ramanujan. I also heared about the story of the taxi cab number $1729.$ Could you help me by providing me with some references or informations about some discovered facts about some special numbers that Ramanujan did. I wanted my pupils to see the beauty and the genuis in Ramanujan's work. -Thanks in advance. - -REPLY [5 votes]: I'm sure the list you seek would be almost endless. One may suggest that you browse the books by Bruce C. Berndt, Ramanujan's Notebooks, Part I, II, etc, Springer. -Euler's formula $e^{\pi i}+1=0$ is everyone's favorite. In the same spirit, but to show the massive computational power of Ramanujan, here is special case from Entry 17, page 435, Part V, of the above-mentioned series. -$$\sum_{k=1}^{\infty}\frac{(-1)^{k-1}k}{e^{k\pi\sqrt{3}}-(-1)^k}=\frac1{4\pi\sqrt{3}}-\frac1{24}.$$ -Notice the interplay of the two famous constants $\pi$ and $e$. -In view of Robert Israel's reasonable comment, perhaps we could go for the modest expressions: -$$\sqrt{2\left(1-\frac1{3^2}\right)\left(1-\frac1{7^2}\right)\left(1-\frac1{11^2}\right)\left(1-\frac1{19^2}\right)} -=\left(1+\frac17\right)\left(1+\frac1{11}\right)\left(1-\frac1{19}\right)$$ -found in S. Ramanujan, Notebooks of Srinivasa Ramanujan, Volume II, -Tata Institute of Fundamental Research, Bombay, 1957. See pp. 309 and 363. -Regarding the second example, it is feasible to encourage students to find similar results of the same kind because there are many. They would be able experiment.<|endoftext|> -TITLE: Non-Reeb vector fields on the three-sphere -QUESTION [6 upvotes]: Let $X$ be the Hopf vector field on the three-sphere. Is there a smooth nowhere zero function $f$ so that the modified vector field $fX$ is not the Reeb vector field of any contact form on the three-sphere? -This question is a follow-up on aglearner's interesting question on how to construct volume-preserving flows on the three-sphere that are not Reeb vector fields. -I'm also looking for evidence to support the following guess: If $X$ is a vector field on the three-sphere, there exists a nowhere zero smooth function $f$ so that the vector field $fX$ is not a Reeb vector field for any contact form on the three-sphere. - -REPLY [2 votes]: The answer is yes. -Proposition. Let $\alpha$ be the standard contact form on the three-sphere (for which the Reeb vector field is the Hopf vector field $X$). If $f$ is a strictly positive function on $S^3$ such there exist two Hopf circles $\gamma_1$ and $\gamma_2$ for which -$$ -\int_{\gamma_1} f^{-1} \alpha \neq \int_{\gamma_2} f^{-1} \alpha \ , -$$ -then the vector field $fX$ is not the Reeb vector field of any contact form on $S^3$. -The key idea is that all solutions of $\dot{x}(t) = f(x(t))X(x(t))$ are periodic and that if $fX$ were a Reeb vector field, all these solutions would have the same period. -Indeed, assume $fX$ is the Reeb vector field of a contact form $\beta$. If $\gamma_1$ and $\gamma_2$ are integral curves of $fX$ with periods $T_1$ and $T_2$, then -$$ -T_2 - T_1 = \int_{\gamma_2} \beta - \int_{\gamma_1} \beta = \int_\Sigma d\beta \ , -$$ -where $\Sigma$ is a cylinder with boundary $\gamma_2 - \gamma_1$ and foliated by Hopf circles (I'm being a bit sloppy with the word foliated). It follows that this last integral is zero and so $T_1 = T_2$. -Now we just have to notice that the period of $\gamma_i$ ($i = 1, 2$) equals -the integral of $f^{-1}\alpha$ over $\gamma_i$: -$$ -\int_{\gamma_i} f^{-1} \alpha = \int_0^{T_i} f^{-1}(\gamma_i(t))\alpha(\dot{\gamma_i}(t)) \, dt = \int_0^{T_i} f^{-1}(\gamma_i(t))\alpha(f(\gamma_i(t)) X(\gamma_i(t))) \, dt = T_i -$$<|endoftext|> -TITLE: Computation of $\pi_4$ of simple Lie groups -QUESTION [13 upvotes]: Below we assume any simple Lie group $G$ to be simply connected. -$\pi_3(G)=\mathbb{Z}$ for any simple Lie group $G$ and there is a uniform proof for that. -Now the textbooks say $\pi_4(G)$ is trivial except for $G=Sp(n)$, for which it is $\mathbb{Z}/2\mathbb{Z}$. -My question is the following: is there a uniform way to derive this fact, in such a way to show which property of the symplectic group is so special? - -REPLY [3 votes]: There is actually a collection of more general statements that detail the structure of the $\pi_4$. -Firstly it was Browder [The Cohomology of Covering Spaces of H-Spaces] who proved that a simply connected finite H-space is 2-connected. Clark [On $\pi_3$ of Finite Dimensional H-Spaces] showed that the first non-vanishing homotopy group of a finite loop space appeared in degrees 1 or 3 (the result that $\pi_2=0$ for a simply connected finite H-space also appears in Browders [Torsion in H-Spaces], see also Lin's [The First Homotopy Group of a Finite H-Space]). Kane and Hubbuck [On $\pi_3$ of a Finite H-Space] prove that $\pi_3$ is torsion free, building on previous work of Thomas and Lin which required stricter assumptions (the Lie group $E_8$ does not fall under Thomas's requirements, for instance). -Finally Hubbuck [Homotopy Groups of Finite H-Spaces] studied $\pi_4$, showing that it was a direct sum of finite groups of order 2. The proof uses JHC Whitehead's Certain Exact Sequence (see Baues [Homotopy Type and Homology] for a good discussion), identifying the groups that appear within using the work of others. The end result is an exact sequence -$H_5(X;\mathbb{Z})\xrightarrow{\rho_2} H_5(X;\mathbb{Z}_2)\xrightarrow{Sq^2_*}H_3(X;\mathbb{Z}_2)\rightarrow \pi_4X\rightarrow0$ -And we know that $H_3(X;\mathbb{Z}_2)=\oplus\,\mathbb{Z}_2$ from the Hurewicz theorem and the work of Kane and Hubbuck so the result on $\pi_4X$ follows. -Harper claims that if $X$ is simply connected and $H_*(\Omega X;\mathbb{Z})$ is torsion free then the result follows from the previous work of Bott and Samelson [Applications of the Theory of Morse to Symmetric Spaces].<|endoftext|> -TITLE: Has the Jacobson/ Baer radical of a group been studied? -QUESTION [12 upvotes]: On groupprops, the Jacobson or Baer radical of a group $G$ is defined to be the intersection of all maximal normal subgroups of $G$. This is similar to, but distinct from, the Frattini subgroup which is the intersection of all maximal subgroups of $G$. -Obviously the Jacobson radical of rings and the Frattini subgroup have been well-studied and are very useful. -Are there any significant results concerning or uses of the Jacobson radical of groups? -For comparison, significant results for the Frattini subgroup would include the facts about in on the Wikipedia page. So equivalent characterisations, its properties and so on. - -REPLY [4 votes]: In addition to applications to finite/finitely generated groups, it's also useful in the study of locally compact groups (taking a definition that only allows closed normal subgroups). Other names for it include the cosocle, the Mel'nikov subgroup or the normal Frattini subgroup. -In a connected Lie group $G$, $J(G)$ is the largest (not necessarily connected) soluble closed normal subgroup of $G$, and $G/J(G)$ is a direct product of simple Lie groups. Given the Gleason-Yamabe theorem, similar statements apply more generally to connected locally compact groups. -If $G$ is a profinite group, then every proper closed normal subgroup is contained in a maximal open normal subgroup, and $G/J(G)$ is the largest quotient that is a Cartesian product of finite simple groups. Moreover, the series $G > J(G) > J(J(G)) > \dots$ always has trivial intersection, so it can be a useful characteristic series, especially if the terms are known to have finite index (equivalently, if $J(H)$ has finite index for every open subgroup $H$). Zalesskii showed that if $G$ is an infinite profinite group and $J(H)$ has finite index for every open subgroup $H$, then $G$ has a just infinite quotient. -If $G$ is a compactly generated locally compact group with no infinite discrete quotient, then Caprace--Monod showed that there is a smallest cocompact normal subgroup $N$ of $G$, and $N/J(N)$ is generated topologically by a finite set of topologically simple closed normal subgroups, with $N > J(N)$ unless $N$ is connected and soluble.<|endoftext|> -TITLE: Irreducible representations of simple complex groups -QUESTION [8 upvotes]: Let $G$ be a simple complex algebraic group. What are its complex irreducible finite-dimensional representations? -Before you start voting to close the question, I never said "rational". I am asking about abstract representations of $G$ as a group. -I can a make an obvious conjecture but I have no idea how to prove it, even for $SL_2$. Yet, it could be something well-known. Pick a rational irreducible representation $(V,\rho)$ and a field endomorphism $\tau: {\mathbb C} \rightarrow {\mathbb C}$. The field endomorphism extends to an endmorphism of $G$: just apply $\tau$ to each matrix entry of of $g\in G\subseteq GL_n(\mathbb C)$. Thus, we get a twisted rational representation $(V,\tau\circ \rho)$ that clearly remains irreducible. -STUPID WABBIT CONJECTURE: all irreducible representations are twisted rational representations. - -REPLY [5 votes]: Just to complete the answer, every complex irreducible representation of $G(\mathbb{C})$ ($G$ is simply connected) is an $r$-fold tensor product of the form $\rho _1\circ \sigma _1\otimes \cdots \otimes \rho _r \circ \sigma _r$, where each $\rho _i$ is an algebraic irreducible representation of $G$ and $\sigma _i$ are $distinct$ embeddings of the field $\mathbb{C}$ into itself. This readily follows from the results of Borel-Tits (see Theorem (10.3) of the article of Borel-Tits mentioned by nfdc23).<|endoftext|> -TITLE: Have this generalization of Indifference graphs been studied before? -QUESTION [6 upvotes]: Indifference graphs are those graphs $G=(V,E)$ for which there exists a real-valued function $f$ defined on $V(G)$ such that, if $u$ and $v$ are distinct vertices, $|f(u)−f(v)| \lt 1$ if and only if $\{u,v\}\in E$. -A famous result of Roberts (1969) shows that this graphs are equivalent to the unit interval graphs (or the proper interval graphs), and they are further equivalent to the $K_{1,3}$-free interval graphs. -I need know about those graphs $G=(V,E)$ for which there exists two real-valued function $f$ and $g$ defined on $V(G)$ such that, if $u$ and $v$ are distinct vertices, $|f(u)−f(v)| \lt |g(u)−g(v)|$ if and only if $\{u,v\}\in E$. ISGCI didn't know of such a class. - -Have this class of graphs been studied before? What kind of properties do they have? Are they perhaps equivalent to some well-known (or a better known) graph class? - -REPLY [7 votes]: Seeing that indifference graphs are the same as unit interval graphs is very easy: we simply identify each vertex $v$ with a unit interval centred at $f(v)$. -We can try to do something similar here. Identify each vertex $v$ with the point $(f(v), g(v))$ in $\mathbb R^2$. Then two points are adjacent if and only if the line joining them is closer to vertical than horizontal. Rotating the picture 45 degrees anticlockwise, two points are adjacent if and only if the line joining them has negative slope. If we assume that in this rotated picture no $x$ or $y$ coordinate is repeated, then this is precisely the definition of a permutation graph. -If some coordinates are repeated, then we can perturb the points slightly so that the associated vertical or horizontal lines joining pairs of points become lines of positive slope, so we also get a permutation graph in this case.<|endoftext|> -TITLE: Poincare Recurrence by Mean Ergodic Theorem -QUESTION [5 upvotes]: I have a question regarding a confusion from reading the Princeton Companion to Mathematics on the topic of Ergodics Theorems. It is about proving a stronger version of Poincare Recurrence Theorem using Neumann's Mean Ergodic Theorem. I apologize if this question is too easy for this site. -Let $A$ be a subset of positive measure, $T$ a measure preserving transformation and $U$ a unitary operator defined by $Uf(x):=f(Tx)$. Define the ergodic average $A_{N,M}$ to be -$$ -A_{N,M}(f):=\frac 1{N-M}\sum_{n=M}^{N-1}U^nf. -$$ -It is not hard to verify that for $f:=1_A$, $\langle f,U^nf\rangle = \mu(A\cap T^{-n}A)$. Here's the passage that confuses me: - -It follows that - $$ -\langle f,A_{N,M}(f)\rangle = \frac 1{N-M}\sum_{n=M}^{N-1}\mu(A\cap T^{-n}A). -$$ - If we let $N − M$ tend to infinity, then $A_{N,M}f$ tends to a $U$-invariant function $g$. Since $g$ is $U$-invariant, $\langle f,g\rangle = \langle U^nf,g\rangle$ for every $n$ and therefore $\langle f,g\rangle=\langle A_{N,M}(f),g\rangle$ for every $N$ and $M$ and finally $\langle f,g\rangle=\langle g,g\rangle$. By the Cauchy–Schwarz inequality, this is at least - $$ -\left(\int g(x)d\mu\right)^2 = \left(\int f(x)d\mu\right)^2=\mu(A)^2. -$$ - Therefore we deduce - $$ -\lim_{N-M\to\infty}\frac 1{N-M}\sum_{n=M}^{N-1}\mu(A\cap T^{-n}A)\ge \mu(A)^2. -$$ - -How does one arrive at the final result from Cauchy–Schwarz inequality? -I understand everything prior to the statement "By the Cauchy-Schwarz ..." but not after that. Using Cauchy–Schwarz I could deduce $||g||\le ||f||$ and get -$$ -\lim_{N-M\to\infty}\frac 1{N-M}\sum_{n=M}^{N-1}\mu(A\cap T^{-n}A)\le \mu(A)^2. -$$ -which is the complete opposite of what was stated. -Here is the link to the same question posted on MSE with no answer. -Edit: As pointed out by John Griesmer, the inequality I derived should read -$$ -\lim_{N-M\to\infty}\frac 1{N-M}\sum_{n=M}^{N-1}\mu(A\cap T^{-n}A)\le ||f||^2 = \mu(A). -$$ -instead. A silly mistake there by me. - -REPLY [5 votes]: $A_{N,M}(f)$ converges to some $U$-invariant function $g$ that satisfies $\langle g, f\rangle$ = $\langle g, g\rangle$. -We also have $\langle g, 1\rangle = \lim \langle A_{N,M}(f), 1\rangle = \langle f, 1\rangle$ since $\langle A_{N,M}(f), 1\rangle = \langle f, 1\rangle$ for all $N, M$. -So we have $\lim \ \langle A_{N,M}(f) , f \rangle = \langle g, f \rangle \ =\langle g, g \rangle = \langle g, g \rangle \langle 1, 1 \rangle \geq \langle g , 1 \rangle^2$ -the last inequality following from the Cauchy-Schwarz inequality. For $f = {\bf 1}_A$, we have $\langle f, 1 \rangle = \int {\bf 1}_A d\mu = \mu(A).$<|endoftext|> -TITLE: Topological obstruction for the existence of spin$^c$ structure -QUESTION [5 upvotes]: Recently I asked on stack exchange the following question: https://math.stackexchange.com/questions/2088888/vanishing-of-certain-cohomology-class-and-existence-of-spin-structure -I would like to know whether there is a similar construction for the spin$^c$ manifolds: namely whether one can construct some cohomology class indpendent from the choice of transition functions and liftings to $spin^c(n)$ with the property that this class is trivial iff $M$ is spin$^c$ manifold. -Remark: I know $C^*$-algebraic approach which gives the so called Dixmier-Douady class in $H^3(M,\mathbb{Z})$. However I would like to understand whether one can proceed purely geometrically. - -REPLY [13 votes]: The obstruction to spin$^c$ is often denoted by $W_3(M) \in H^3(M,\mathbb{Z})$. It is obtained as follows: Let -$$ -\beta \colon H^2(M,\mathbb{Z}/2\mathbb{Z}) \to H^3(M,\mathbb{Z}) -$$ -be the Bockstein homomorphism obtained from the short exact sequence -$$ -0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0 -$$ -Then we have $W_3(M) = \beta(w_2(M))$, i.e. the class obtained by applying the Bockstein homomorphism to the second Stiefel-Whitney class. This agrees with the Dixmier-Douady class of the (stabilisation of the) complex Clifford bundle $\mathbb{C}l(M)$. -Geometrically you can do the following: There is an exact sequence -$$ -1 \to U(1) \to Spin^c(n) \to SO(n) \to 1 -$$ -Let $M$ be an oriented $n$-manifold and let $\pi \colon P \to M$ be the oriented frame bundle. This is a principal $SO(n)$-bundle. Let -$$ -P^{[2]} = \{ (p_1,p_2) \in P^2 \ | \ \pi(p_1) = \pi(p_2) \} -$$ -Then we have a principal $U(1)$-bundle obtained as follows: -$$ -L = \{ (p_1,p_2,g) \in P^{[2]} \times Spin^c\ | \ p_1\cdot q(g) = p_2 \} -$$ -where $q \colon Spin^c \to SO(n)$ is the canonical homomorphism. If we identify this $U(1)$-bundle with the associated line bundle, we have the following isomorphism -$$ -\mu \colon \pi_{12}^*L \otimes \pi_{23}^*L \to \pi_{13}^*L -$$ -where $P^{[3]}$ is defined analogously to $P^{[2]}$ and $\pi_{ij} \colon P^{[3]} \to P^{[2]}$ is the projection to the $i$th and $j$th factor. This should be thought of as a multiplication and satisfies an associativity constraint over $P^{[4]}$. This structure is called the lifting bundle gerbe. -How do you get the class $W_3(M)$ from this? -Choose a open cover $U_i$ of $M$, such that there are sections $\kappa_i \colon U_i \to P$, let -$$ -L_{ij} = (\kappa_i,\kappa_j)^*L . -$$ -Choose trivialisations $\theta_{ij} \colon U_{ij} \times \mathbb{C} \to L_{ij}$ (we can always choose the cover, such that these exist), where $U_{ij} = U_i \cap U_j$. Over the triple intersections $U_{ijk}$ we now have two trivialisations of $L_{ik}$. One is -$\theta_{ik} $, -the other one is -$$ -\mu_{ijk} \circ (\theta_{ij} \otimes \theta_{jk}) -$$ -where $\mu_{ijk}$ is the isomorphism $L_{ij} \otimes L_{jk} \to L_{ik}$ induced by $\mu$. The trivialisations differ by a continuous map -$$ -\omega_{ijk} \colon U_{ijk} \to U(1) -$$ -which turns out to be a Cech $2$-cocycle due to the associativity of the multiplication $\mu$. Hence, it represents an element -$$ -[\omega] \in \check{H}^2(M,U(1)) \cong H^3(M,\mathbb{Z}) . -$$ -This is the Spin$^c$ obstruction class.<|endoftext|> -TITLE: Expressing properties of graded algebras in terms of the $\mathbb{G}_m$action -QUESTION [5 upvotes]: Let us fix a base ring $k$. The category of $\mathbb{Z}$-graded $k$-algebras is equivalent to the category of $\mathbb{G}_m$ equivariant affine $k$-schemes. The following 2 properties often come up when constructing the $Proj$ of a graded ring: Let $A$ be a graded $k$-algebra. - -The zeroth graded piece is the base ring $A_0 =k$ -The algebra $A_\bullet$ is positively graded. i.e. $A_j=0$ for all $j<0$. -The canonical multiplication homomorphism $Sym^\bullet A_1 \to A_\bullet$ is surjective. - -I'm trying to find equivalent characterization of the above properties which don't use the aforementioned equivalence. In other words I want the above 3 properties expressed in terms of properties of $\mathbb{G}_m$-equivariant schemes (orbits, stabilizers, fixed points etc.). - -REPLY [2 votes]: Expanding on Dylan Wilson's answer, condition (2) is indeed just a continuation of $\mathbb{G}_m$-action to an $\mathbb{A}^1$-action, since $\mathcal{O}(\mathbb{A}^1) \hookrightarrow \mathcal{O}(\mathbb{G}_m): \mathbb{k}[t] \hookrightarrow \mathbb{k}[t, t^{-1}]$. Condition (1) is trickier, since the action of $\mathbb{G}_m$ on $V$ will have inseparable orbits --- see the action on $\mathbb{A}^n$. Thus we can't think of it as transitivity (not of the group action at least, the monoid $\mathbb{A}^1$ action is transitive in a sense that any two points can be mapped to a same point). Instead we should look at it together with condition (2), which implies that the action of $\mathbb{G}_m$ has fixed points: for any $x \in V$ the point $0\cdot x \in V$ will be fixed. Here $0 \in \mathbb{A}^1$ and its action exists by (2). Condition (1) then reduces to the statement that the $\mathbb{G}_m$-action has a unique fixed point. On the level of functions it is given by a graded morphism $\mathcal{O}(V) \to \mathbb k$ as graded algebras. Condition (3) means that the action is free if we remove that single fixed point. We can see it because the action on $A_1^*$ is a free action apart from $0 \in A_1^*$ and the embedding $V \hookrightarrow A_1^*$ is $\mathbb{G}_m$-equivariant, as a morphism of graded algebras. -The simplest example where you can see all those statements is the obvious action on $\mathbb{A}^n$. On the other hand, consider the obvious action of $\mathbb{G}_m$ on itself. It corresponds to the natural grading on $\mathbb{k}[t,t^{-1}]$. The reasons (2) fails here is obvious. Since there is no morphism $\mathbb{k}[t,t^{-1}] \to \mathbb{k}$, the action has no fixed points. In this case (1) indeed reduces to transitivity. Instead of (3) we have that $Sym(A_1 \oplus A_{-1}) \to A_\bullet$ is surjective. This also implies that the action is free, for the same reasons as above.<|endoftext|> -TITLE: Orthogonality of positive (semi-)definite matrices -QUESTION [7 upvotes]: Given positive semidefinite matrices $A,B \succeq 0$, $A, B \in \mathbb{R}^{n \times n}$, if we have $$\langle A, B \rangle = 0,$$ where $\langle \cdot, \cdot \rangle$ denotes the Frobenius inner product, then - -What are tight necessary conditions of ranks of $A, B$? For example, $\mbox{rank} (A) + \mbox{rank}(B) \leq n$? How to prove that? -What can be say about the orthogonality of the eigenvectors of $A$ and $B$? For example, are they pairwise orthogonal? - -REPLY [9 votes]: If $A$ and $B$ are positive semidefinite and $A^{1/2}$ and $B^{1/2}$ their positive semidefinite square roots, -$$\text{Tr}(AB) = \text{Tr}(A^{1/2} B A^{1/2}) = \text{Tr}((B^{1/2} A^{1/2})^T B^{1/2} A^{1/2})$$ -and this is $0$ if and only if $B^{1/2} A^{1/2} = 0$. -That in turn is equivalent to $\text{Ran}(A) \subseteq \text{Ker}(B)$. -So indeed $\text{rank}(A) + \text{rank}(B) \le n$, and this bound is tight. -Moreover, the eigenvectors of $A$ for nonzero eigenvalues are in $\text{Ker}(B)$ and therefore are orthogonal to the eigenvectors of $B$ for nonzero eigenvalues.<|endoftext|> -TITLE: What is the set of all "pseudo-rational" numbers (see details)? -QUESTION [21 upvotes]: Define a “pseudo-rational” number to be a real number $q$ that can be written as -$q=\sum_{n=1}^{\infty} \frac{P(n)}{Q(n)}$ -Where $P(x)$ and $Q(x)$ are fixed integer polynomials (independent of n). All rational numbers are pseudo-rational, as is $\pi^2$ using $P(x)=6,Q(x)=x^2$. There must exist numbers that are not pseudo-rational (defined as "pseudo-irrational") because the set of pseudo-rationals is countable. Is $e$ pseudo-rational? Is $\sqrt{2}$? - -REPLY [7 votes]: Also possibly of interest: -$$ \sum_{n=1}^\infty \left(-\frac{2}{b(n-1/b)(n+1/b)} + \frac{b}{n(n+1)}\right) = \pi \cot(\pi/b)$$ -$$ \sum_{n=1}^\infty {\frac {t \left( {t}^{2}{n}^{2}+2\,{n}^{2}+2\,n+1 \right) }{ \left( n+ -1 \right) n \left( {t}^{2}{n}^{2}+1 \right) }} - = \pi \coth(\pi/t) $$ -EDIT: -And, if I'm not mistaken, -$$ \sum_{n=1}^\infty \left(\frac{1-m}{mn} + \sum_{k=1}^{m-1} \frac{1}{mn-k}\right) = \ln(m) $$ -for positive integers $m$.<|endoftext|> -TITLE: Flatness from constancy of dimension of fibers -QUESTION [9 upvotes]: Let $M$ be a finitely presented module over the ring $R$. Suppose that for all primes $P\subset R$ the $k(P)$ vector space $M\otimes _Rk(P)$ has a dimension $d(P)$ independent of $P$. - Can I conclude that $M$ is flat over $R$? -I am asking because I want to better understand the criterion for a family of projective schemes to be flat over some base in terms of invariance of their Hilbert polynomial. -I checked some examples. -For example let $M$ be the ideal of a point $P$ of an affine curve $C$. -Here $d(P)$ is the dimension of the Zariski tangent space of $C$at $P$ so that the proposed criterion for flatness holds. - -REPLY [14 votes]: Yes if $R$ is reduced (no nilpotent elements). This is (for instance) Lemma 1 p. 51 in Mumford's Abelian Varieties (2nd edition). No in general: just take $R=k[\varepsilon ]/(\varepsilon ^2)$, $M=k$.<|endoftext|> -TITLE: Type of place versus type of unitary group -QUESTION [6 upvotes]: Let $F$ be a totally real number field, $E$ a totally imaginary quadratic extension over $E$, and $V$ an hermitian $n$-dimensional vector space over $F$. I assume $n=2m$ is even. Let $U$ be a unitary group, i.e. the group au automorphism preserving a given hermitian form on $V$. -The structure of local components of $U$ at the finite places of $F$ is among the following ones for instance, see those notes of Michael Harris: - -$U_v \cong GL_n(V)$, this happens iff $v$ splits in $E$ -$U_v \cong U(n)$, the only one non-quasi-split unitary group of rank $n$ over $V_v$, and this happens a finite number of time -$U_v \cong U(m,m)$, the only one quasi-split unitary group of rank $n$ over $V_v$ - - -I wonder if there is a characterization of those last two possibilities in terms of the behaviour of $v$ in $E$. Since there is only a finite number of places where $U_v$ is non-quasi-split, could we wait for it to happen iff the place ramifies? - -REPLY [10 votes]: Things are perhaps a bit messier than you hope. In particular it is not true that the unitary group is non-quasi-split if and only if $v$ ramifies. Disclaimer: I did not know the answer to this question off the top of my head but I did want to know, so I just figured it out below; hopefully there are no errors (hopefully an expert will glance over it and let me know if there are). -From the way you write (talking about "the only one non-quasi-split unitary group...") you seem to be assuming that $v$ is a finite place of $F$ (of course the case of infinite places is very well-known). As you know, if $v$ splits in $E$ then $E\otimes_F F_v$ is isomorphic to $F_v\oplus F_v$ and the unitary group becomes $GL_n$ at $v$. If $v$ does not split (i.e. we are in the inert or ramified case) then there is one prime $w$ above $v$ in $E$, and we have a local extension $E_w/F_v$ of degree 2. So we need to understand unitary groups over local fields in order to answer your question. -So now let $L/K$ be a degree 2 extension of $p$-adic fields and say $V$ is a vector space over $L$ equipped with a Hermitian form (Hermitian for the action of $Gal(L/K)$ of course). If we choose an $L$-basis for $V$ then this form gives rise to a Hermitian matrix $\Phi$ (so $\overline{\Phi}^t=\Phi$). The determinant of $\Phi$ is an element of $L^\times$ which is equal to its Galois conjugate, so it's in $K^\times$. Let $c(\Phi)$ be the image of this determinant in the group $K^\times/N_{L/K}(L^\times)$, a group of order 2. It turns out that this invariant $c$ parametrises isomorphism classes of Hermitian matrices -- so in particular there are two isomorphism classes of Hermitian forms on $V$. For each isomorphism class we get a unitary group. -The next step depends on whether $n$ is odd or even. If $n$ is odd then it turns out that even though the two forms are not isomorphic, the unitary groups are (this can be easily seen -- scaling the form by an element of $K^\times$ can change the isomorphism class of the form but clearly doesn't change the corresponding unitary group). But you are interested in the case $n$ even, and in this case it turns out we get two unitary groups, one quasi-split and one not quasi-split. In particular we see that it is easy to build a non-quasi-split unitary group over $K$ even if $L/K$ is unramified, and it is easy to build a quasi-split unitary group over $K$ even if $L/K$ is ramified -- we just need to get the determinants right. Moreover, we can do all these things over $E/F$ as well, and this is why life is not as simple as you think. -To understand things better and to see what is true, we next need to understand when our local unitary group is quasi-split. You have some global Hermitian form giving rise to a global Hermitian matrix whose determinant is $d\in E$, and what we know so far is that if $v$ is a finite place of $F$ which is not split, and $w$ the unique place of $E$ above $v$, then whether or not $U$ is quasi-split at $v$ depends only on what the image of $d$ is in $F_v^\times/N_{E_w/F_v}(E_w^\times)$. So to see exactly what is going on, I need to tell you which element of $F_v^\times/N_{E_w/F_v}(E_w^\times)$ corresponds to the quasi-split case, and then we also need to check (for our own sanity) that in the global situation $d$ will give us a quasi-split local unitary group for all but finitely many $v$. -Now here's the bad news. It turns out that the story locally (if I worked it out correctly) is the following. We can write $L=K(\sqrt{k})$ for some $k\in K$, and if my calculations are correct, the element of $K^\times/N_{L/K}(L^\times)$ corresponding to the quasi-split unitary group is $k^m$, where $n=2m$. This is because (if I got it right) if we let our Hermitian form be anti-diagonal with entries $+\sqrt{k},-\sqrt{k},+\sqrt{k},\ldots$ (this is Hermitian if I got it right) then this form gives us a quasi-split unitary group because the upper triangular matrices are a Borel. In particular, the naive guess that the quasi-split group corresponds to the identity element of $K^\times/N_{L/K}(L^\times)$ is not always true. The norm of the element $\sqrt{k}\in L$ is $-k$, so it seems to me that the element of $K^\times/N_{L/K}(L^\times)$ corresponding to the quasi-split unitary group is $(-1)^m$, which of course is the identity element iff $(-1)^m$ is the norm of an element of $L$. This happens if $m$ is even or if $L/K$ is unramified (because then any unit is a norm) but not in general. -However, going back to the global situation, we have a global determinant $d\in F^\times$ and at all but finitely many places $d$ will be a local unit, and at all but finitely many places $E_w/F_v$ will be unramified, so it is true that the global unitary group is quasi-split at all but finitely many local places. -In general then, to figure out which places your unitary group is not quasi-split you need to figure out what the determinant $d$ of a corresponding Hermitian form is, and then for each place of $F$ which is either ramified in $E$ or for which this determinant is not a unit (there are only finitely many of these), you need to figure out whether $(-1)^md$ is a local norm or not; the places for which this number is not a local norm are the places where the unitary group is ramified.<|endoftext|> -TITLE: Selberg trace formula, quadratic L-values, and generalization -QUESTION [7 upvotes]: It is known that the geometric side of the Selberg trace formula on GL(2) is related to values of quadratic L-functions (due to Sarnak, Zagier, etc). -Are there any conjectures or results about its generalization to other groups, such as GL(3) and GL(n)? - -REPLY [3 votes]: If I understand correctly what you are looking for, then yes, a fair amount of work has been done. Deitmar and Hoffman use a simple trace formula on SL(3) to get asymptotics of class number of cubic orders, which can also be interpreted as a prime geodesic theorem. This is a higher-dimensional analogue of quadratic class numbers (which can be viewed as special $L$-values) appearing in a trace formula for SL(2). Deitmar also treats higher rank analogues in other papers. -Another way in which zeta/$L$-functions arise in the geometric side of trace formulas (maybe not what you are looking for, but perhaps still interesting) is discussed in general in this paper of Hoffman and for Sp(4) in his joint paper with Wakatsuki. Namely, one gets Shintani zeta functions arising as coefficients in the unipotent geometric terms of the Arthur-Selberg trace formula.<|endoftext|> -TITLE: Derived push-forwards of structure sheaves for morphisms of algebraic stacks -QUESTION [7 upvotes]: Given a morphism of algebraic stacks $f: X \to Y$ (possibly non-representable) what sufficient conditions will guarantee that the derived push-forward of the structure sheaf on $X$ is isomorphic to the structure sheaf on $Y$? I am looking for a stacky version of a result of Buch-Mihalcea (Thm 3.1 of Quantum K-theory of Grassmannians https://arxiv.org/abs/0810.0981, which itself is a variation of a result of Kollar) -Thm 3.1 of BM: Let $f : X \rightarrow Y$ be a surjective map of projective varieties with rational singularities. Assume that the general fiber of $f$ is rational, i.e. $f^{−1}(y)$ is an irreducible rational variety for all closed points in a dense open subset of $Y$. Then $f_∗[\mathcal{O}_X]=[\mathcal{O}_Y]$ in the K-theory of Y. -I am happy to assume, in any proposed generalization, that the stacks are proper DM stacks (even smooth) and that the general fiber is an irreducible rational variety. - -REPLY [3 votes]: I am amplifying the above comments. The theorem of Buch-Mihalcea also holds for stacks. -Let $k$ be a field of characteristic $0$. Let $Y$ be a Deligne-Mumford stack that is finite type over $k$ and that is normal: there exists a surjective, étale morphism $h:\widetilde{Y}\to Y$ with $\widetilde{Y}$ a normal scheme. Let -$f:X\to Y$ be a proper, surjective morphism of Deligne-Mumford stacks. Assume that $X$ is integral, and assume that $X$ has rational singularities: there exists a surjective, étale morphism $g:\widetilde{X}\to X$ such that $\widetilde{X}$ is a $k$-scheme that has rational singularities. Finally, assume that the geometric generic fiber of $f$ is isomorphic (as a stack) to a smooth proper variety that is rationally connected, or even just $\mathcal{O}$-acyclic. -Proposition. The natural map $\mathcal{O}_Y\to f_*\mathcal{O}_X$ is an isomorphism, and $R^q f_*\mathcal{O}_X$ equals $0$ for every $q>0$. -Because $h$ is flat, $h^*R^qf_*\mathcal{O}_X$ equals $R^q \widetilde{f} \mathcal{O}_{X\times_Y \widetilde{Y}}$, where $\widetilde{f}:X\times_Y \widetilde{Y} \to \widetilde{Y}$ is the projection. Thus, without loss of generality, assume that $Y$ is a normal scheme. As in the comment, there exists a coarse moduli space $p:X \to |X|$. The morphism $f$ factors as the composition of $p$ and a surjective, proper $k$-morphism, $$|f|:|X|\to Y.$$ Because the characteristic is $0$, $X$ is a tame stack. Thus, $\mathcal{O}_{|X|} \to p_*\mathcal{O}_X$ is an isomorphism and $R^qp_*\mathcal{O}_X$ equals $0$ for all $q>0$. Therefore, it suffices to prove that $\mathcal{O}_Y\to |f|_*\mathcal{O}_{|X|}$ is an isomorphism and $R^q|f|_*\mathcal{O}_{|X|}$ equals $0$ for all $q>0$. Via Buch-Mihalcea, it suffices to prove that $|X|$ has rational singularities. This follows from Proposition 5.13, p. 157 of "Birational Geometry of Algebraic Varieties" by Kollár and Mori.<|endoftext|> -TITLE: A seemingly simple inequality -QUESTION [18 upvotes]: Let $a_i,b_i\in\mathbb{R}$ and $n>1$, does the inequality -$$ -\left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right)+\left(\sum_{i=1}^na_i b_i\right)^2\ge \sqrt{\left(\sum_{i=1}^n a_i^4\right)\left(\sum_{i=1}^n b_i^4\right)}+\sum_{i=1}^na_i^2b_i^2 -$$ -hold true? -Observe that $a_i$ and $b_i$ are not required to be non-negative. -I ran an extensive number of numerical simulations and no counterexample showed up yet. -Note 1. The inequality holds true for $n=2$, as showed here. -Note 2. This conjecture was formulated by Fedor Petrov in an attempt to provide a solution to a particular case of this question. -Note 3. I have posted this question on math.SE some time ago but it has received no answer, so I cross-posted it here. -Note 4. As Fedor Petrov rightly observed in his answer below, the inequality also follows by an argument used in one of his answers in the above-cited question. However, I decided to accept Markus Sprecher's answer because of its conciseness and clarity. - -REPLY [5 votes]: Just to clarify. As I remember, this is simply equivalent to the (partial case of) the question you cite. Since the cited question was solved, I would not call it a conjecture. But the question to find an independent proof makes sense. -Well, let me elaborate. Here proving the partial case $N=2$ of your inequality I prove the following inequality: $$\frac{\|x\|^2\cdot \|y\|^2+(x,y)^2}{\|Tx\|^2 \|Ty\|^2}\geqslant \frac2{{\rm tr}\, T^4}$$ -for any self-adjoint positive definite operator $T$ and any two vectors $x,y$ in $\mathbb{R}^n$. If we denote $x=(a_1,\dots,a_n)$, $y=(b_1,\dots,b_n)$, $p=(a_1^2,\dots,a_n^2)$, $q=(b_1^2,\dots,b_n^2)$, $T^2=diag(s_1,\dots,s_n)$, $s=(s_1,\dots,s_n)$, we rewrite this as $$\left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right)+\left(\sum_{i=1}^na_i b_i\right)^2\ge 2\frac{(p,s)\cdot (q,s)}{(s,s)},$$ -and maximizing over $s$ gives you $$\sqrt{\left(\sum_{i=1}^n a_i^4\right)\left(\sum_{i=1}^n b_i^4\right)}+\sum_{i=1}^na_i^2b_i^2$$ in RHS, this is the equality case in the lemma in the cited answer.<|endoftext|> -TITLE: When is a compact Lie group endowed with a left-invariant complex structure a Kähler manifold of balanced manifold? -QUESTION [5 upvotes]: Let $G$ be a compact Lie group having a left-invariant complex structure $J$. -Is there a hermitian metric $h$ in $G$, compatible with the complex structure $J$, such that $G$ is a Kähler manifold? -In the case it is not possible to turn $G$ into a Kähler manifold, does it hold that exist a hermitian metric such that $G$ is a balanced manifold? That is, $\Delta_d = 2 \Delta_\overline{\partial}$? - -REPLY [7 votes]: To answer the first question: Only when $G$ is abelian. In fact, if $G$ is not abelian and has dimension $2d$, then there is no element of $H^2(G,\mathbb{R})$ whose $d$-th power is nonzero in $H^{2d}(G,\mathbb{R})$ (just look at the representation by bi-invariant forms), and there would have to be such an element if there were going to be even a symplectic structure on $G$, much less a Kähler structure. Conversely, if $G$ is abelian, then, yes, since $G=\mathbb{R}^{2d}/\mathbb{Z}^{2d}$, and, for any left-invariant complex structure, $G$ is $\mathbb{C}^d/\Lambda$, where $\Lambda\subset\mathbb{C}^d$ is a lattice (i.e., a discrete subgroup of maximal rank $2d$). Thus, there is a left-invariant Kähler structure, for example, (though there are others). -Added on 21 Jan 2017: As for your second question, I believe I have a partial answer, which is 'If the center of $G$ is finite (i.e., $G$ is semisimple), then $G$ does not possess a balanced metric'. Here is why: It is my understanding that, if $G$ is a compact Lie group and $J$ is a left-invariant complex structure, then $J$ is one of the complex structures constructed by Samelson and Wang. (See H.C. Wang, Closed Manifolds with Homogeneous Complex Structure, American Journal of Mathematics 76 (1954),1–32, which seems to indicate this though it doesn't state it explicitly. Also, see the article mentioned by François in the comments below.) -In such cases, there exists a maximal toral subgroup $T\subset G$ that is $J$-complex, and the quotient $G/T$ inherits a complex structure such that the quotient map $\pi:G\to G/T$ is holomorphic. Moreover, $G/T$ is not only Kähler, but an algebraic complex manifold to boot (in fact, it's the complete flag variety of $G$). -In particular, $G/T$ contains an algebraic (complex) hypersurface $Z\subset G/T$, whose preimage $X=\pi^{-1}(Z)\subset G$ is therefore a compact complex hypersurface in $G$. Now if $\omega$ were a positive $(1,1)$-form on $G$ (with respect to the complex structure $J$) such that $\omega^{d-1}$ were closed (where $2d$ is the dimension of $G$), then the integral of $\omega^{d-1}$ over $X$ would be positive. In particular, it would follow that the homology class of $X$ in $H_{2d-2}(G,\mathbb{R})$ were nonzero. -However, when $G$ is semi-simple (i.e., when its center is finite), we know by the Whitehead Lemma that $H_2(G,\mathbb{R}) = 0$, so Poincaré Duality implies that $H_{2d-2}(G,\mathbb{R})=0$ as well, contradicting the fact that the real homology class of $X$ must be nonzero. -I believe a more careful analysis of the cohomology ring of $G$ when the center of $G$ has positive dimension will lead to the result that $G$ can carry a balanced metric only when $G$ is abelian.<|endoftext|> -TITLE: Notations for dual spaces and dual operators -QUESTION [9 upvotes]: I'm asking for opinions about the 'best' notations for: - 1. the algebraic dual of a vector space $X$; - 2. the continuous dual of a TVS; - 3. the algebraic dual (transpose) of an operator $T$ between vector spaces; - 4. the dual (transpose) of a continuous operator between TVS; - 5. the adjoint of a bounded operator $T$ between Hilbert spaces. -My problem is that I would like to use these notions in the same context. The standard notations tend to overlap but I am forced to use different notations for each of these entities. Of course it is easy to come up with notations, but some traditions are well established and it is not trivial to respect them and at the same time keep them apart, with some elegance. -What I'm using now: - 1. $X'_{alg}$ - 2. $X'$ - 3. $^tT$ - 4. $T'$ - 5. $T^*$ -Thank you for your advice. - -REPLY [4 votes]: Given your situation of having to juggle all these notational traditions at the same time, I would recommend for a space $X$ and an operator $A$: - -$X^{\vee}$, 2. $X'$, 3. $A^{\rm T}$, 4. $A^{\rm T}$, 5. $A^*$ - -My rationale is as follows. -For 1: Algebraic geometers especially use the "vee' notation for the algebraic dual so I think the cultural association helps the brain automatically make the association with the algebraic notion of dual. -For 2: One usually does not write the space of tempered distributions as $S^*(\mathbb{R}^d)$ but rather as $S'(\mathbb{R^d})$. -For 3 and 4: I would use the same notation as per the comment by user95282 since the maps are related by restriction. One could use ${}^{\rm t}A$ for both but I prefer the matrix algebra notation if only because it is easier to type. -For 5: As yuggib said, it is standard in the theory of $C^*$-algebras and spectral theory. I don't see any reason to be a contrarian and not follow what everyone else does.<|endoftext|> -TITLE: Mid-Square with all bits set -QUESTION [6 upvotes]: Is there a positive 128-bit integer whose square has all middle bits equal to 1? -(The "middle bits" are naturally the 65th bit through the 192nd bit, defining -the 1st bit as the least significant bit of the full integer.) - -REPLY [7 votes]: Is there a positive 128-bit integer whose square has all middle bits equal to 1? - -YES. One is AAAAAAAAAAAAAAAB555555555555555516, which square is 71C71C71C71C71C7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF8E38E38E38E38E3916. - -It's asked if there are integers $n$ such that $0 -TITLE: Unusual symmetries of the Cayley-Menger determinant for the volume of tetrahedra -QUESTION [54 upvotes]: Suppose you have a tetrahedron $T$ in Euclidean space with edge lengths $\ell_{01}$, $\ell_{02}$, $\ell_{03}$, $\ell_{12}$, $\ell_{13}$, and $\ell_{23}$. Now consider the tetrahedron $T'$ with edge lengths -$$\begin{aligned} -\ell'_{02} &= \ell_{02} & - \ell'_{13} &= \ell_{13}\\ -\ell'_{01} &= s-\ell_{01} & - \ell'_{12} &= s-\ell_{12}\\ -\ell'_{23} &= s-\ell_{23}& - \ell'_{03} &= s-\ell_{03} -\end{aligned} -$$ -where $s = (\ell_{01} + \ell_{12} + \ell_{23} + \ell_{03})/2$. -If the edge lengths of $T'$ are positive and satisfy the triangle inequality, then the volume of $T'$ equals the volume of $T$. In particular, if $T$ is a flat tetrahedron in $\mathbb{R}^2$, then $T'$ is as well. This is easily verified by plugging the values $\ell'_{ij}$ above into the Cayley-Menger determinant. -In fact, it's possible to show that the linear symmetries of $\mathbb{R}^6$ that preserve the Cayley-Menger determinant form the Weyl group $D_6$, of order $2^5 * 6! = 23040$. This is a factor of $15$ times larger than the natural geometric symmetries obtained by permuting the vertices of the tetrahedron and negating the coordinates. -The transformations don't always take Euclidean tetrahedra to Euclidean tetrahedra, but they do sometimes. For instance, if you start with an equilateral tetrahedron $T$ with all side lengths equal to $1$, then $T'$ is also an equilateral tetrahedron. Thus if $T$ is a generic Euclidean tetrahedron close to equilateral, $T'$ will also be one, and $T$ and $T'$ will not be related by a Euclidean symmetry. -I can't be the first person to observe this. (In fact, I vaguely recall hearing about this in the context of quantum groups and the Jones polynomial.) What's the history? How to best understand these transformations (without expanding out the determinant)? Are $T$ and $T'$ scissors congruent? Etc. - -REPLY [3 votes]: The following comment in the question intrigued me: - -In fact, it's possible to show that the linear symmetries of -$\mathbb{R}^6$ that preserve the Cayley-Menger determinant form the -Weyl group $D_6$, of order $2^5 * 6! = 23040$. - -In particular, this suggests that it should be possible to perform a change of basis such that the Cayley–Menger determinant is a symmetric polynomial in the six variables, which is also invariant under even numbers of sign changes. -Some experimentation confirmed that the following variables work in the sense that the geometric and Regge symmetries all preserve the 12-element set $\{ \pm u, \pm v, \pm w, \pm x, \pm y, \pm z \}$: - -$u = \ell_{01} + \ell_{23}$; -$v = \ell_{02} + \ell_{13}$; -$w = \ell_{03} + \ell_{12}$; -$x = \ell_{01} - \ell_{23}$; -$y = \ell_{02} - \ell_{13}$; -$z = \ell_{03} - \ell_{12}$; - -More succinctly, the variables are the three sums of opposite edges and the three differences of opposite edges. The Cayley–Menger determinant can then be shown to be equal to the following 57-term polynomial: -$$ uvwxyz - \dfrac{1}{16} \sum_\text{sym}^{6} u^6 + \dfrac{1}{16} \sum_\text{sym}^{30} u^4 v^2 - \dfrac{1}{8} \sum_\text{sym}^{20} u^2 v^2 w^2 $$ -where the ‘symmetric sums’ are over all distinct monomials obtained by permuting the six variables (and the number above the sum is the number of terms). This is visibly invariant under the Weyl group $D_6$ generated by arbitrary permutations and even numbers of sign changes.<|endoftext|> -TITLE: The conic bundle of a cubic threefold -QUESTION [5 upvotes]: Let $X$ be a smooth cubic threefold over $\mathbb{C}$ and let $L \subset X$ be a line. -The projection from $L$ yields a rational map $X \dashrightarrow \mathbb{P}^2$. Resolving the indeterminacy by blowing up $L$, we obtain a conic bundle morphism $\pi: \tilde{X} \to \mathbb{P}^2$. -Consider the relative anticanonical bundle $\omega_{\pi}^{-1}$. By standard properties of conic bundles we obtain an embedding -$$\tilde{X} \hookrightarrow \mathbb{P}(\mathcal{E})$$ -which respects $\pi$. Here $\mathcal{E} := \pi_*\omega_{\pi}^{-1}$ is a vector bundle of rank $3$ on $\mathbb{P}^2$ and $\mathbb{P}(\mathcal{E}) \to \mathbb{P}^2$ is the associated $\mathbb{P}^2$-bundle over $\mathbb{P}^2$. - -Is there is an explicit description of the vector bundle $\mathcal{E}$? For example, is it independent of $X$ or $L$? Is it determined by the normal bundle of $L$? Is it a direct sum of lines bundles? - -Note that $\mathbb{P}(\mathcal{E}) \cong \mathbb{P}(\mathcal{E} \otimes \mathcal{O}(k))$ for any $k \in \mathbb{Z}$. In particular, I am quite happy with determining $\mathcal{E}$ up to twist by a line bundle. - -REPLY [10 votes]: Choose coordinates $(U,V,X,Y,Z)$ in $\mathbb{P}^4$ so that $L$ is the line $X=Y=Z=0$. The equation of $X$ is of the form -$$ AU^2+2BUV+CV^2+2D U + 2EV +F=0\ ,$$where $A,B,\ldots F$ are homogeneous forms in $X,Y,Z$ of degree $1,1,1,2,2,3$. You can view this equation as a section $s \in H^0(\mathbb{P}^2,\mathrm{Sym}^2\mathcal{E} \otimes \mathcal{O}_{\mathbb{P}^2}(1))$, where $\mathcal{E}$ is the vector bundle $\mathcal{O}_{\mathbb{P}^2}\oplus \mathcal{O}_{\mathbb{P}^2}\oplus \mathcal{O}_{\mathbb{P}^2}(1)$. Put $P:=\mathbb{P}_{\mathbb{P}^2}(\mathcal{E})$, and let $p:P\rightarrow \mathbb{P}^2$ be the structure map. Then $s$ defines a section of the line bundle $\mathcal{O}_{P}(2)\otimes p^*\mathcal{O}_{\mathbb{P}^2}(1)$ on $P$, whose zero locus is $\tilde{X} $. -So up to a twist, your vector bundle is just $\mathcal{O}_{\mathbb{P}^2}\oplus \mathcal{O}_{\mathbb{P}^2}\oplus \mathcal{O}_{\mathbb{P}^2}(1)$. You may then compute $\omega _{\pi }$ by the adjunction formula and see which twist is needed to get $\pi _*\omega _{\pi }$.<|endoftext|> -TITLE: Ideas in the elementary proof of the prime number theorem (Selberg / Erdős) -QUESTION [31 upvotes]: I'm reading the elementary proof of prime number theorem (Selberg / Erdős, around 1949). -One key step is to prove that, with $\vartheta(x) = \sum_{p\leq x} \log p$, -$$(1) \qquad\qquad \vartheta(x) \log x+\sum_{p\leq x}(\log p) \vartheta(x/p) = 2x\log x+O(x)$$ - -What's the idea behind this identity? -Here is the heuristic argument I understand : if $p$ is a prime number, the average gap before the next prime is $\approx \log p$. Thus if we set a weight $0$ for non-prime numbers, and $\log p$ for prime numbers, then when we sum all these weights for $n = 1 ... x$, we should get $\vartheta(x) = \sum_{p \leq x} \log p \sim x$, and that's equivalent to the PNT. Ok. -But where comes the idea from, to add on the left hand side of the identity (1) the part $\sum_{p\leq x}(\log p) \vartheta(x/p)$ ? What's the idea here? -What's the idea to go from (1) to the prime number theorem? - - -Note: the identity (1) can be replaced by an equivalent identity, with $\psi(x) = \sum_{p^k\leq x} \log p = \sum_{n \leq x} \Lambda(n)$ (where $\Lambda(n) = \log p$ if $n=p^k$ and $0$ otherwise): -$$(2) \qquad\qquad \psi(x) \log x+\sum_{n\leq x}\Lambda(n) \psi(x/n) = 2x\log x+O(x)$$ -Another equivalent form is (just in case (2) or (3) might be easier to understand): -$$(3) \qquad\qquad \sum_{p \leq x} (\log p)^2 + \sum_{pq\leq x} \log p \log q= 2x\log x+O(x)$$ - -REPLY [3 votes]: Two very nice answers have already been given, but I would like to add that Michel Balazard has a book in preparation on this topic (written for undergrads), in which he tries to give a deduction of the PNT from Selberg's identity which as simple as possible. Also, a student here has a nice write-up of the proof. It is very short, it is elementary, but it is in french.<|endoftext|> -TITLE: Example of non-holonomic D-module and explicit computation of characteristic variety -QUESTION [6 upvotes]: I'm currently trying to have a better understanding of the concepts of characteristic variety and holonomic $D$-modules (let us assume that they are coherent) on a holomorphic manifold $X$. I know that for a system of differential equations $P$, the holonomicity of the $D$-module $D_X / D_X P$ means that the system $P$ is maximaly overdetermined. In order to see that, we compute the characteristic variety $\text{char}(D_X/D_X P)$ and if it is lagrangian in $T^* X$, the $D$-module is holonomic by definition. -I'd like to compute explicitely the charactistic variety of two "easy" coherent $D$-modules in order to see the holonomicity. So to be precise, my question is - -Find two systems of differential equations $P$ and $Q$ on, let's say $\mathbb{C}^2$, such that $\text{char}(D_X/D_X P)$ and $\text{char}(D_X/D_X Q)$ can be explicitely computed in a short time and such that $\text{char}(D_X/D_X P)$ is lagrangian and $\text{char}(D_X/D_X Q)$ is not. - -For example, I've also studied the notion of regularity and for that concept, the examples are easy : $z\partial_z -z$ is regular and $z\partial_z -1$ is not. But I couldn't find such easy examples for holonomicity. Perhaps it is linked with the complexity of the computation of characteristic varieties. (I know it is linked to Gröbner basis, but I don't know very well this theory) -Any examples or literature recommendations will be highly appreciated. - -REPLY [8 votes]: (These examples are shamelessly pilfered from Gröbner Deformations of Hypergeometric Differential Equations by Saito, Sturmfels and Takayama, which is perhaps the place to learn computational D-module stuff.) -Holonomic: $D\cdot\{z_1\partial_2,z_2\partial_1\}$. This left ideal has characteristic ideal of dimension 2, so it is holonomic. -Non-holonomic: let $f=(z_1^3-z_2^2)$; $D\cdot\{f\partial_1+\partial f/\partial z_1, f\partial_2+\partial f/\partial z_2\}=D\cdot\{\partial_1 f,\partial_2 f\}$. This left ideal has char.ideal of dimension 3, so it isn't holonomic. -(Edit: if you want to actually compute stuff, I recommend the Dmodules package for Macaulay2; it has tools to do most D-module things, including characteristic variety/ideal, gröbner bases etc.)<|endoftext|> -TITLE: When the automorphism group of an object determines the object -QUESTION [74 upvotes]: Let me start with three examples to illustrate my question (probably vague; I apologize in advance). - -$\mathbf{Man}$, the category of closed (compact without boundary) topological manifold. For any $M, N\in \mathbf{Man}$ there is the following theorem (Whittaker) which says that - - -$$\mathrm{Homeo}(M)\cong_{\text{as groups}} \mathrm{Homeo}(N) \, \textit{ if and only if } \, M\cong_{\text{as manifold}} N$$ - - -$\mathbf{NFields}$, the category of Number fields. For any $F, K\in \mathbf{NFields}$ there is a theorem (Neukirch-Uchida) which says that - - -$$\mathrm{Gal}(\overline{K}/K)\cong_{\text{as progroups}} \mathrm{Gal}(\overline{F}/F) \textit{ if and only if } K\cong_{\text{as fields}} F $$ - - -$\mathbf{Vect}$, the category of finite dimensional vector spaces. For any $V, W\in \mathbf{Vect}$ we have that - - -$$\mathrm{GL}(V)\cong \mathrm{GL}(W) \textit{ if and only if } V\cong W $$ - -In the first and the third examples we see that the automorphism group of an object determines the object. The second example seems to be similar in some sense however it does not admit the same naive interpretation. - -$\textbf{Question:}$ - What are other non-trivial examples of interesting categories where the automorphism group of an object determines the object itself? Is there a name for such categories? Is there a way to compare and characterize these kind of categories? - -REPLY [3 votes]: Geometric Complexity Theory: - -Theorem (Mulmuley and Sohoni [MS]) - The permanent (respectively the determinant) polynomial is characterized by its symmetry group. - -That is if $P$ is a homogeneous polynomial of degree $m$ in $m^2$ variables and its symmetry group $G_P$ also fixes the permanent (respectively the determinant), then $P$ must be a scalar multiple of the permanent (respectively the determinant). -Landsberg and Ressayre [LR] made progress on Valiant's version of P vs NP using this result.<|endoftext|> -TITLE: Is there a model structure on (strict?) Monoidal Categories? -QUESTION [8 upvotes]: Basically, what the title says. -Presumably, one could use the fact that monoidal categories (resp. strict monoidal categories) are one-object bicategories (resp. 2-categories) and use the Lack model structure on those, but I am unsure if this would work or not. - -REPLY [4 votes]: There is a model structure on the category of monoidal categories and strict monoidal functors. A strict monoidal functor is a fibration / weak equivalence just when its underlying functor is a fibration / weak equivalence in Cat. -More generally for T a 2-monad with rank on Cat you can lift the model structure on Cat to the category of strict T-algebras and strict T-algebra morphisms: this generalises the above example. -For these results, see the discussion in Section 1.7 and Theorem 4.5 of Steve Lack's paper "Homotopy theoretic aspects of 2-monads": https://arxiv.org/abs/math/0607646<|endoftext|> -TITLE: Technical lemma on root systems, reduced to linear algebra -QUESTION [9 upvotes]: Update: I have posted the case of $G = SL(n)$ as a different question here. -This is a technical lemma I am currently stuck at. Any suggestions about how to proceed are welcome. -Let $G$ be a split semisimple group over a number field. Let $B, T$ be a chosen minimal parabolic, torus respectively, and $\theta$ be an automorphism of $G$ of finite order, preserving $B$ and $T$. By the theory of algebraic groups, we can associate to $(G, B, T)$, a root datum $(X^*, \Delta, X_*, \Delta^\vee )$, where $X^*$ (resp. $X_*$) is the lattice of characters (cocharacters) and $\Delta$ (resp. $\Delta^\vee$) the simple roots (coroots). The Weyl group $W$ acts on the root space $V^* = X^* \otimes \mathbb R$ and co-root space $V_*$ as usual and the action of $\theta$ descends to a permutation of the finite sets $\Delta, \Delta^\vee$. For every $\beta \in \Delta$, let $\varpi_\beta \in \hat\Delta$ be the corresponding weight so that $\hat\Delta$ is a basis of $V^*$ dual to $\Delta$. -Question: Fix $w \in W, w \neq 1$. Does there exist a cone $\Omega$ inside the positive Weyl chamber of $V^*$ such that if $\lambda \in \Omega$ and $\lambda - \theta w \lambda = \displaystyle\sum_{\beta \in \Delta} {d_\beta} \beta$, then $d_\beta > 0$? -(Edit: To clarify, by positive Weyl chamber, I mean a positive span of weights, not roots.) -This is not true per se but I would like the coordinates $d_\beta$ positive only in some directions: Let $Q(w)$ be the smallest standard (i.e., containing $B$) parabolic subgroup containing a representative of $w$. By a correspondence between standard parabolic subgroups and subsets of $\Delta$, we have the subset $\Delta^{Q(w)}$ of $\Delta = \Delta^G$ corresponding to $Q(w)$. I need $d_\beta>0$ whenever $\beta \in \Delta^{Q(w)}$. -Example: Let $G = SL(n)$ with usual Borel and torus and let $\theta(x) = {}^Tx^{-1}$. We can identify $V^*$ with the subspace of $\mathbb R^n$ of vectors with coordinates adding to zero so that $\Delta = \{ \beta_1 = (1, -1, 0, \cdots, 0), \cdots, \beta_{n-1} \}$ and $\{ \varpi_\beta \in \hat\Delta\}$ is the dual basis (of weights). We have $\theta(\beta_i) = \beta_{n-i}$. In the case of $SL(3)$ taking $\lambda = c_1 \varpi_1+ c_2 \varpi_2$ and $w$ to be the permutation $(1,2)$ gives -$$ \lambda - \theta w\lambda = \left(\frac{c_1 - c_2}{3}\right)\beta_1 + \left(\frac{2c_1 + c_2}{3}\right)\beta_2 $$ -so we can define the cone to be $c_1 > c_2 > 0$. -I have written a Python code and verified this till $SL(7)$ but unable to prove so far for general $G$. -EDIT: When $\theta = 1$ we can take $\Omega$ to be the full positive Weyl chamber; see Bourbaki's Lie Groups and Lie Algebras Chapter 6 $\S 1.5$ (Edit: 1.6, not 1.5) Proposition 18 for a proof based on induction on $\ell(w)$. -EDIT': I am elaborating on the case $\theta = 1$. Fix a root $\gamma \in \Delta^{Q(w)}$ and $\lambda$ in the positive Weyl chamber. The coefficient of $\gamma$ in $\lambda - w \lambda$ is given by -$\langle \lambda - w \lambda, \varpi_\gamma^\vee \rangle = \langle \lambda, (1 - w^{-1}) \varpi_\gamma^\vee \rangle$ -and by Bourbaki's aforementioned lemma, the vector $(1 - w^{-1}) \varpi_\gamma^\vee$ is a non-negative combination of co-roots. - -REPLY [4 votes]: Maybe it's helpful to add a longer comment, in community-wiki format. The original question is not well-formulated, I think, as shown in the later convoluted remarks on the case $\theta =1$. It's probably better here to follow Bourbaki (Chapter VI), since the essential problem concerns just an irreducible root system in a vector space $V$ having as basis a fixed choice of simple roots. This choice then determines a partition of the complement in $V$ of reflecting walls for the various roots into Weyl chambers. One of these (call it $C$) is the dominant Weyl chamber (called "positive" in the question) defined by $\langle \lambda, \alpha_i^\vee \rangle >0$ for all simple $\alpha_i$ with $\lambda \in V$. Its closure $\overline{C}$ is then a fundamental domain for the action of the Weyl group $W$. -The fundamental weights $\varpi_i$ lie in the walls of $C$ and are at acute (or right) angles. On the other hand, the simple roots $\alpha_i$ are at obtuse (or right) angles and determine a "positive root cone" (call it $D$) consisting of positive linear combinations of simple roots. Then $D$ contains $C$ because each $\varpi_i$ is a positive $\mathbb{Q}$-linear combination of the $\alpha_i$. But $D$ is usually larger than $C$. -We are given an automorphism $\theta$ of the Dynkin diagram, for example the one of order 2 for type $A_\ell$ when $\ell \geq 2$ (coming from $\mathrm{SL}_{\ell+1}$) which switches $\alpha_i$ and $\alpha_{\ell-i+1}$. The question then concerns $(*) \:\lambda - \theta w \lambda$ for a fixed $w \in W$ and any dominant weight $\lambda$ (say in $\overline{C}$). Bourbaki's Prop. 18 says for $\theta =1$ that $\lambda - w \lambda$ lies in $\overline{D}$, but of course usually not in $\overline{C}$. -First write $\lambda$ as a $\mathbb{Z}^+$-linear combination of the $\varpi_i$, say with coefficients $c_i$. Then $(*)$ is a $\mathbb{Q}$-linear combination of simple roots. Maybe the intended question for arbitrary $\theta$ is whether there is a cone inside the $\mathbb{Q^+}$-span of $D$ (or $\overline{D}$) consisting of those elements $(*)$ defined by conditions on the $c_i$ such as those in the example. (A fixed denominator may occur, coming from the index of the root lattice in the weight lattice.) Of course, the cone need not lie in $\overline{C}$, as shown by the case $\theta =1$. Anyway, the version I've stated seems likely to have a positive answer, but I don't know how to prove it.<|endoftext|> -TITLE: Most important mathematical results in last 30 years -QUESTION [10 upvotes]: Which results from the last 30 years, in any area of mathematics, do you think are the most important ones? -Specifically, which are the ones that will have more impact across all math and/or settle significant questions, in your opinion? It does not matter if they are big, lond-proofed results like a classification theorem, or little, elementary lemmas proved in a paragraph. -To keep this reasonable, I propose that every contributor posts at most 10 results. -Please, write one result per post, so that we can all vote properly and have a real ranking afterwards. Also, please don't just write the name, write a little description (hopefully, self-contained) and put some link if desirable. -(My personal aim with this question is to learn. I hope to be a bit better versed after studying your answers!) - -REPLY [6 votes]: Perelman's proof of the Geometrization conjecture (see here, here and here) was the crowning achievement of decades of work. It was the most important of Thurston's conjectures about the topology of 3-manifolds. -Oh, and I nearly forgot to mention that the Poincare Conjecture is a consequence. - -REPLY [5 votes]: Agol's proof of the Virtual Haken conjecture was a wonderful application of the tools developed by Wise and his coauthors in geometric group theory to 3-manifold topology. The Virtual Haken conjecture, which can be thought of as the topological classification of compact 3-manifolds, is a fundamental result. -Agol's theorem: - -Every cubulable hyperbolic group is virtually special. - -also has other dramatic consequences for hyperbolic groups. It's arguably the most important theorem proved in both algebra and topology in the last five years. - -REPLY [3 votes]: Kahn--Markovic's proofs of the Surface Subgroup conjecture and the Ehrenpreis conjecture.<|endoftext|> -TITLE: Is the fusion argument on trees of uncountable height consistent? -QUESTION [8 upvotes]: In the countable context where we are given a perfect subtree $T$ of $2^{<\omega}$ and a sequence of colorings $f_i: T\to 2, i\in \omega$, it is possible to obtain a perfect subtree $T'\subset T$ and an infinite set of levels $A\in [\omega]^\omega$ such that for each $i\in \omega$ there exists $n_i\in A$ such that for all $t\in T$ with $ht(t)\in A\backslash (n_i+1)$, $f_i(t)=f_i(t\restriction n_i)$. -This is easily obtainable via 1-dimensional Halpern-Läuchli theorem (https://en.wikipedia.org/wiki/Halpern–Läuchli_theorem) along with a typical fusion argument. -My question is: is the generalization to trees on larger cardinals consistent? -More precisely, is the following consistent: -Let $\kappa$ be at least an inaccessible cardinal. -Given a perfect $<\kappa$-closed subtree $T$ of $2^{<\kappa}$ and a sequence of colorings $f_i: T\to 2, i\in \kappa$, it is possible to obtain a perfect $<\kappa$-closed subtree $T'\subset T$ and a set of levels $A\in [\kappa]^\kappa$ such that for each $i\in \kappa$ there exists $n_i\in A$ such that for all $t\in T$ with $ht(t)\in A\backslash (n_i+1)$, $f_i(t)=f_i(t\restriction n_i)$. -Some remark: - -When the number of colorings is $<\kappa$, this is true (in fact stronger consequence is true, namely the exists a perfect subtree that is simultaneously homogeneous for all the colorings) -It seems the naïve generalization of classical fusion argument will just fail at the limit stage. - -Any thoughts will be appreciated. - -REPLY [2 votes]: The answer is yes. In fact, more general statements are true. See https://arxiv.org/abs/1704.06827 for more detail (in particular Theorem 3.1).<|endoftext|> -TITLE: Enriched slice categories -QUESTION [8 upvotes]: $\require{AMScd}$Are there references for a construction of the enriched slice category of $\mathcal A \in \mathcal{V}\text{-Cat}$? A reasonable definition should be - -Fix an object $a\in\mathcal A$ and let the objects of ${\cal A}/a$ be the set ${\cal V}(J, {\cal A}(x,a))$ for all $x\in\cal A$. $J$ is the monoidal unit of $\cal V$. -Let ${\cal A}/a(p,q)$ be the $\cal V$-object resulting from the pullback -$$ -\begin{CD} - P @>>> {\cal A}(x,y) \\ - @VVV {}@VVq_*V\\ -J @>>p> {\cal A}(x, a) -\end{CD} -$$ -for $p : x\to a, q : y \to a$ and $J$ the monoidal unit. - -Does it work for every $\cal V$? - -REPLY [7 votes]: There is at least one sense in which this "works". Namely, it is the comma object in the 2-category $\mathcal{V}$-Cat of $\mathrm{Id} : \mathcal{A} \to \mathcal{A}$ over $[a] : \mathcal{J} \to \mathcal{A}$, where $\mathcal{J}$ is the unit $\mathcal{V}$-category with one object and $J$ as its hom-object. It's true that it doesn't capture as much information in the enriched case as it does in the ordinary case, but given that it has this universal property I don't know what else one might mean by "enriched slice category".<|endoftext|> -TITLE: Is Turaev-Viro-Barrett-Westbury stronger than homotopy? -QUESTION [9 upvotes]: I've heard that Reshetikhin-Turaev (RT) is stronger than homotopy, and it can distinguish certain homotopy-equivalent, but non-homeomorphic Lens spaces (I think $L(7,1)$ and $L(7,2)$). Now the Turaev-Viro-Barrett-Westbury (TVBW) invariant for a spherical fusion category $\mathcal{C}$ is the Reshetikhin-Turaev invariant for $\mathcal{Z}(\mathcal{C})$, which is a restriction, so in principle, RT could be stronger than TVBW. -Are there calculations that show explicitly how TVBW is stronger than homotopy? What category do you have to use to achieve this? -Otherwise, I'm very happy to hear expert opinions stating that this is an open problem. - -REPLY [8 votes]: I think the following example seems to work: consider a special class of TVBW, the Dijkgraaf-Witten invariant whose input are a finite $G$ and a 3-cocycle from $H^3(G, C^*)$ (together they define a spherical fusion category). More specifically, let $G$ be an Abelian group. The invariant evaluated on $L(p,q)$ is given by the following: -$\displaystyle\mathcal{Z}(L(p,q))=\frac{1}{|G|}\sum_{g\in G, g^p=1}\prod_{j=1}^{p-1}\omega(g, g^j, g^n)$, -here $n=q^{-1}$ mod $p$. -Set $G=\mathbb{Z}/14$, and take a representative cocycle: -$\omega(a,b,c)=\exp\Big[\frac{2\pi i}{14^2}[a]([b]+[c]-[b+c]) \Big]$, -where $[a]\equiv a \text{ mod } 14$. -One can check that with this cocycle, $\mathcal{Z}(L(7,1))\neq \mathcal{Z}(L(7,2))$.<|endoftext|> -TITLE: Mayer Vietoris Spectral sequence for topological K theory -QUESTION [7 upvotes]: In Sheaf theory one can obtain the Mayer Vietoris spectral sequence for cohomology. For $\mathcal{U}$ an open cover of $X$ we get the convergence -$E_2^{pq} = \check H^p(\mathcal{U},H^q(-,F)) \Longrightarrow H^{p+q}(X,F)$. -In topological K theory it is a fact (see for exemple Karoubi's book II.4.18) that we have the classical Mayer Vietoris long exact sequence for a cover of $X$ by two open (or closed) sets. This can be seen as the collapsing of the previous spectral sequence adapted to K theory at the stage 1. -My question is : is there a way to build a MV spectral sequence for topological K theory ? - -REPLY [4 votes]: As suggested by Denis Nardin I am moving my comment here. However I don't know details well enough, so I am making this cw in case somebody can fill them in. -So, choose a cohomology theory $h^*$ like e. g. $K$-theory, and, given a cover $(U_i)_{i\in I}$ of $X$, let $p:Y\to X$ be the canonical map $\coprod_{i\in I}U_i\to X$. Then consider the associated simplicial space -$$ -\check C(p):=\left(Y\begin{smallmatrix}\leftarrow\\\rightarrow\\\leftarrow\end{smallmatrix} Y\times_XY\begin{smallmatrix}\leftarrow\\\rightarrow\\\leftarrow\\\rightarrow\\\leftarrow\end{smallmatrix}Y\times_XY\times_XY\begin{smallmatrix}\leftarrow\\\vdots\\\leftarrow\end{smallmatrix}\cdots\right), -$$ -so that $\check C(p)_n=\coprod_{i_0,...,i_n}U_{i_0}\times_X\cdots\times_XU_{i_n}$. -A result of Dugger (found in "Topological Hypercovers") shows that geometric realization of this simplicial space is weakly homotopy equivalent to $X$, and so is the homotopy colimit of it. This induces isomorphisms on $h^*$ between the geometric realization of $\check C(p)$ to $X$, and between the homotopy colimit of $\check C(p)$ to $X$. -If $\check C(p)$ were Reedy cofibrant, Theorem 5.83, on page 163 of "Generalized Cohomology" by Kōno and Tamaki, would say there is a spectral sequence converging (again under some mild conditions) to $h^*$ of the geometric realization, with the second page given by cohomology of the cochain complex corresponding to the cosimplicial abelian group $h^*(\check C(p))$. This would be the Moore normalization of the latter complex (given by restricting to nondegenerate simplices of $\check C(p)$) has $\prod_{\{i_0,...,i_n\}\subseteq I}h^*(U_{i_0}\cap\cdots\cap U_{i_n})$ in the $n$th degree, with understandable differentials. -But $\check C(p)$ is not in general Reedy cofibrant (See "Topological Hypercovers" again; the discussion in the beginning of Secion 3). The idea is that Reedy cofibrancy would imply that the finite intersections are cofibrant, which isn't necessarily the case. -One way to recover the result would be to take a cofibrant replacement of $\check C(p)$ first and -then take the cohomology spectral sequence of that resulting simplicial space. This is precisely the homotopy colimit spectral sequence, but unfortunately -does not have as nice a description at the second page; it involves taking the homology of the Moore normalization of the simplicial replacement. -This generalizes (Dugger's Primer, Prop. 18.17): given a spectrum $\mathscr E$, -and a diagram $F : \mathbf C \to \textbf{Top}$, there is a spectral -sequence describd by taking the left derived colimit -$$ -E_{s, t}^2 = H_s(\mathbf C, \mathscr E_t(F)) = -\operatorname{colim}_s \mathscr E_t(F) \Rightarrow -\mathscr E_{s + t}(\operatorname{hocolim}_{\mathbf C}(F)) \cong \mathscr E_{s + t}(X). -$$ -Here, we would choose the diagram to be the above simplicial space, and the desired K-Theory spectrum.<|endoftext|> -TITLE: Can all crossings in a graph be moved to one point? -QUESTION [5 upvotes]: Consider a graph $G$ with at least two unavoidable crossings, say, the disjoint union of two copies of $K_5$. Can such a graph always be drawn so that there is only one singular point (where all crossings happen)? I guess there is an easy proof that this is not possible. - -REPLY [17 votes]: No, this is not always possible. -Lemma. Let $G$ be an $n$-vertex graph with at least $3n-2$ edges. Then $G$ cannot be drawn in the plane so that all crossings occur at the same point. -Proof. We make the standard assumption that every pair of edges which intersect in a drawing are not 'tangent' at the point of intersection. Suppose $D$ is a drawing of $G$ where all edges cross at the same point. Let $G'$ be the graph obtained from $D$ by introducing a new vertex at the crossing point and then suppressing all parallel edges. We claim that $|E(G')| \geq |E(G)|$. Let $H$ be the subgraph of $G$ induced by the edges which pass through the crossing point. Since no two crossing edges are tangent, $H$ contains at most one cycle (which must be a triangle). Therefore, $H$ has average degree at most $2$. It follows that $|E(G')| \geq |E(G)|$, as claimed. Thus, $G'$ is a planar graph with $n+1$ vertices and at least $3n-2$ edges. But this contradicts the fact that every $n$-vertex planar graph has at most $3n-6$ edges. -In particular, the above lemma implies that $K_7$ cannot be drawn so that all edges cross at the same point. -Acknowledgement. Many thanks to bof for help in making this answer converge to its present form (see the comments below).<|endoftext|> -TITLE: Class number formula for binary hermitian forms -QUESTION [5 upvotes]: The class number of integral binary quadratic forms of discriminant $D$ correspond to the class number of $\mathbb Q(\sqrt{D})$. In this case the class number formula is the classical one of Dirichlet. -For the class number of binary hermitian forms of discriminant $D$, on the other hand, there is no such correspondence I am aware of. Still, there is a similar formula proved by G. Humbert around 1920, but it does not give $h(D)$ itself (see here). In a special case: -$$ -\sum_{j=1}^{h(D)}\frac{D}{4}\sigma(F_j)=\frac{\pi}{4}\prod_{p|D,p\neq2}(1+\Big(\frac{-1}{p}\Big)\frac{1}{p}) -$$ -where $\sigma(F_j)$ is a certain collection of fundamental domains. -I'd like to know if there is a more precise formula available. - -REPLY [2 votes]: May be you can consult Latimer's paper On ideals in generalized quaternion algebras and Hermitian forms., Trans. Amer. Math. Soc. 38 (1935), no. 3, 436–446. -The main theorem (Theorem 3) establishes a bijection between what he calls "regular classes of ideals" in the maximal order of a certain quaternion algebra, and the classes of binary hermitian forms of discriminant $\alpha$ (in his notation) with coefficients in $\mathbb{Z}$. Actually, he works in greater generality, over a number field $K$ and replacing $\mathbb{Z}$ with $\mathcal{O}_K$ (which he calls $G$). Latimer himself says in the Introduction that what he's doing is a generalization of the classical relation between quadratic forms and class groups of quadratic fields.<|endoftext|> -TITLE: The most outrageous (or ridiculous) conjectures in mathematics -QUESTION [191 upvotes]: The purpose of this question is to collect the most outrageous (or ridiculous) conjectures in mathematics. -An outrageous conjecture is qualified ONLY if: -1) It is most likely false -(Being hopeless is NOT enough.) -2) It is not known to be false -3) It was published or made publicly before 2006. -4) It is Important: -(It is based on some appealing heuristic or idea; -refuting it will be important etc.) -5) IT IS NOT just the negation of a famous commonly believed conjecture. -As always with big list problems please make one conjecture per answer. (I am not sure this is really a big list question, since I am not aware of many such outrageous conjectures. I am aware of one wonderful example that I hope to post as an answer in a couple of weeks.) -Very important examples where the conjecture was believed as false when it was made but this is no longer the consensus may also qualify! -Shmuel Weinberger described various types of mathematical conjectures. And the type of conjectures the question proposes to collect is of the kind: - -On other times, I have conjectured to lay down the gauntlet: “See, -you can’t even disprove this ridiculous idea." - -Summary of answers (updated: March, 13, 2017 February 27, 2020): - -Berkeley Cardinals exist - -There are at least as many primes between $2$ to $n+1$ as there are between $k$ to $n+k-1$ - -P=NP - -A super exact (too good to be true) estimate for the number of twin primes below $n$. - -Peano Arithmetic is inconsistent. - -The set of prime differences has intermediate Turing degree. - -Vopěnka's principle. - -Siegel zeros exist. - -All rationally connected varieties are unirational. - -Hall's original conjecture (number theory). - -Siegel's disk exists. - -The telescope conjecture in homotopy theory. - -Tarski's monster do not exist (settled by Olshanski) - -All zeros of the Riemann zeta functions have rational imaginary part. - -The Lusternik-Schnirelmann category of $Sp(n)$ equals $2n-1$. - -The finitistic dimension conjecture for finite dimensional algebras. - -The implicit graph conjecture (graph theory, theory of computing) - -$e+\pi$ is rational. - -Zeeman's collapsing conjecture. - -All groups are sofic. - - -(From comments, incomplete list) 21. The Jacobian conjecture; 22. The Berman–Hartmanis conjecture 23. The Casas-Alvero conjecture 24. An implausible embedding into $L$ (set theory). 25. There is a gap of at most $\log n$ between threshold and expectation threshold (Update: a slightly weaker version of this conjecture was proved by Keith Frankston, Jeff Kahn, Bhargav Narayanan, and Jinyoung Park!; Further update: the conjecture was fully proved by Jinyoung Park and Huy Tuan Pham ). 26. NEXP-complete problems are solvable by logarithmic depth, polynomial-size circuits consisting entirely of mod 6 gates. 27. Fermat had a marvelous proof for Fermat's last theorem. (History of mathematics). - -REPLY [3 votes]: I think the Erdős-Szekeres conjecture (see the Wikipedia article on the Happy Ending Problem), which says that the smallest $N$ for which any $N$ points in general position in the plane contain the vertices of some convex $n$-gon is $N=2^{n-2}+1$, is a little bit ridiculous in that there is not a lot of reason to believe it. Indeed, only the cases $n=3,4,5,6$ are known, and while it is known that $2^{n-2}+1$ is a lower bound for this $N$, I really am not aware of any other substantial evidence in favor of this conjecture other than that it's a very nice pattern.<|endoftext|> -TITLE: Einstein Manifolds in dimension five? -QUESTION [8 upvotes]: My teacher of Riemannian Geometry told the class about the problem of finding Einstein Manifolds in 5 dimension. My question is, what is the difficulty of this problem? Possessing a first course in Riemmanian geometry can I understand this? -Thanks - -REPLY [4 votes]: Simply connected 5-manifolds are classified by simple invariants, second homology and second Stiefel-Whitney class. So one can consider the existence problem, i.e. which 5-manifolds admit an Einstein metric. Many are know to have an Eintein metric. See the paper by János Kollár "Einstein metrics on 5-dimensional Seifert bundles".<|endoftext|> -TITLE: The expectation of two sides of rectangle is equal. Can we deduce that in the expectation the rectangle is not very far from being a square? -QUESTION [5 upvotes]: Let $T$ be a set of $n\ge 3$ points in the plane such that not all of them lie in a common line. Pick two distinct points $\{a=\left( \begin{array}{c} a_{1} \\a_{2} \end{array} \right) ,b=\left( \begin{array}{c} b_{1} \\b_{2} \end{array} \right)\}$ uniformly at random from $T$. Let $R$ be the rectangle whose sides are horizontal and vertical, and whose two diagonally opposite vertices are $\{a,b\}$. Let $X$ be the length of horizontal side of $R$ and $Y$ be the length of its vertical side (i.e. $X=|a_{1}-b_{1}|$ and $Y=|a_{2}-b_{2}|$). - -Assuming $\mathbb{E}(X)=\mathbb{E}(Y)$, how small can be expected value of random variable $Z := \min(X,Y)$? - What lower bounds can one get for $\mathbb{E}(Z)$? - -REPLY [8 votes]: Since you can scale everything linearly, let's suppose $\mathbb E(X) = \mathbb E(Y) = 1$. -I'm not sure this is optimal, but I think it must be close. -Consider a case -where all but two of the points are at the origin, with one point (call it $c$) at - $(n/2, 0)$ and one ($d$) at $(0, n/2)$. (You actually want $n$ distinct points, so make all but two points very close to the origin, and insert the word "approximately" as needed below). -Then $X = 0$ unless one of your two points is $c$ (which has probability $2/n$ and makes $X = n/2)$, so $\mathbb E(X) = 1$, and similarly $Y = 0$ unless -one of your points is $d$. $Z = 0$ unless both $c$ and $d$ are chosen, which has probability $2/(n(n-1))$, so $\mathbb E(Z) = 1/(n-1)$.<|endoftext|> -TITLE: Galois Representations and Rational Points -QUESTION [11 upvotes]: Suppose $X$ and $Y$ are two connected smooth projective varieties over $\mathbb{Q}$ (of the same dimension) that have the same $\ell$-adic Galois representations (up to semisimplification). What is the relation between $X(\mathbb{Q})$ and $Y(\mathbb{Q})$, even conjecturally? - -REPLY [19 votes]: In general one can say very little. There are some positive results (as indicated in the comments) in special cases, but the below example kills any hope that one can say something in general. NB "they have the same $\ell$-adic Galois representations" is vague -- I will interpret as "they have isomorphic $\ell$-adic etale cohomology in all degrees" which is the strongest reasonable interpretation I can think of. -So let $E$ be an elliptic curve over $\mathbb{Q}$ with positive rank and non-trivial Tate-Shaferevich group. Let $C$ be a torsor for $E$ corresponding to a non-trivial element of the group. Then $E(\mathbb{Q})$ is infinite and $C(\mathbb{Q})$ is empty. -However I think that the etale cohomology groups of $E$ and $C$ are all isomorphic. $H^0$ is trivial, $H^2$ is the cyclotomic character, and $H^1(C)$ is isomorphic to the dual of the Tate module of the Jacobian of $C$, and the Jacobian of $C$ is isomorphic to $E$ again, so $H^1(C)$ and $H^1(E)$ are both isomorphic to the dual of the Tate module of $E$. -In this example $E(\mathbb{Q})$ and $C(\mathbb{Q})$ are completely different, but at least $E(\mathbb{Q}_p)$ and $C(\mathbb{Q}_p)$ are the same for all $p$. So as another example, let $X$ be projective 1-space over $\mathbb{Q}$ and let $Y$ be a smooth plane conic over $\mathbb{Q}$ with no local points at some prime $p$. The problem here is that the Galois representations tell you very little about the geometry of the variety. $H^0$ is trivial, $H^2$ is cyclotomic, and $H^1$ is zero; again $X(\mathbb{Q})$ is infinite, $Y(\mathbb{Q})$ is empty, and even worse $X(\mathbb{Q}_p)$ is infinite and $Y(\mathbb{Q}_p)$ is empty. Note also that in this case $\pi_1(X)$ and $\pi_1(Y)$ are also isomorphic as Galois modules; the comments by Will Sawin and Piotr Achinger do not apply because $X$ and $Y$ do not have high enough genus to be anabelian.<|endoftext|> -TITLE: Why presheaves with transfer? -QUESTION [14 upvotes]: Presheaves with transfer are a main technical ingredient in Voevodsky's construction of his category of mixed motives. From the perspective of motivic homotopy theory, the only difference between SH and DM is that objects in the latter have transfers. -I'd like to know some non-artificial examples of presheaves with transfer. I know that all qfh sheaves have transfers, but I'm looking for concrete motivational examples. -Is algebraic K-theory a presheaf with transfer? I've seen people talking about transfer maps in K-theory, but K-theory does not live in DM(k) (while it lives in SH(k)). What's going on there? -Mainly, I'd like to understand how Suslin and Voevodsky realized that motivic cohomology was supposed to have transfers. - -REPLY [17 votes]: I hope someone can provide a better answer with more of an eye towards the motivic world, but for now let me outline that the exact same phenomenon exists in classical stable homotopy theory. Here the analogue of $SH$ is the category $Sp$ of spectra. There are plenty of ways to define it, so let me assume it exists for now. One way to think about them is as space-valued homology theories, that is functors $\mathrm{Top}\to \mathrm{Top}_*$ satisfying some conditions. -This is a symmetric monoidal category, with a unit $\mathbb{S}$ called the sphere spectrum. For any space $X$ we can associate a spectrum, called its pointed suspension spectrum $\Sigma^\infty_+X$. Then we get a homology theory called stable homotopy defined by -$$\pi_n^sX := [\Sigma^n\mathbb{S}, \Sigma^\infty_+X]$$ -This is an interesting cohomology theory, and in fact it does have transfer maps: for any map $X\to Y$ with finite and discrete homotopy fibers we get a map called Becker-Gottlieb transfer [1] $\Sigma^\infty_+Y\to \Sigma^\infty_+X$ which, roughly speaking, sends a point $y\in Y$ to the sum of the points of the fiber. However this transfer has a very complicated functoriality. Intuitively the problem is that the for $X\xrightarrow{f}Y\xrightarrow{g} Z$ the equation $g^*f^*=(gf)^*$ takes place in the space $\mathrm{Map}(\Sigma^\infty_+Z,\Sigma^\infty_+X)$, and so instead of a simple equality you should think of it as the datum of a homotopy between the two maps, and then you need to add coherences to these homotopies when you try to write down associativity and so on and so forth... -But sometimes we would like these equalities to hold in a more strict sense. So let us try to make them do so. Instead of spaces let us work with the category $\mathrm{Cor}$ of correspondences of spaces. Its objects are spaces, but now the maps are closed subspaces $Z\subseteq X\times Y$ where the projection to $X$ is a covering space map. -Then we can try to repeat the construction of spectra using the category of correspondence of spaces, that is studying functors $\mathrm{Cor}\to \mathrm{Top}_*$ satisfying some conditions. It turns out that the category we get doing this is $D(\mathbb{Z})$, the derived category of $\mathbb{Z}$ and that the obvious forgetful functor $D(\mathbb{Z})\to \mathrm{Sp}$ is the functor providing the identification of $D(\mathbb{Z})$ with $H\mathbb{Z}$-modules in $\mathrm{Sp}$. The functor from spaces to $D(\mathbb{Z})$ realizing the "strict stable homotopy type" is just the functor sending $X$ to $C_*(X)$, its complex of chains, and the resulting homology theory is just ordinary homology. Now the coherences conditions on the transfers hold on the nose. -In the motivic world the story is more complicated, but the idea is essentially the same. The reason we are introducing transfers in the definition is not to get transfers (we would have them anyway!) but to force their behaviour with respect to functoriality. The reason why this is a reasonable idea to get motivic homology is that when you do it in the classical setting you get ordinary homology. - -[1] In fact the Becker-Gottlied transfer exists every time the fibers have finite stable homotopy type, but this introduces a lot more issues in the functoriality.<|endoftext|> -TITLE: Looking for Arnol'd quote about Russian students vs western mathematicians -QUESTION [11 upvotes]: I think I once saw a sentence in an article by V.I. Arnol'd saying something like: here is a problem that every Russian schoolchild can solve, but no western mathematician can solve. But I can't find it now. - -Does anyone know the quote and where it was written? - -I ask because I'm preparing a talk on research by high school students, and I might include this quote if the actual wording isn't too inflammatory. - -REPLY [15 votes]: You possibly mean this 'two volumes' problem from the book ``Problems for children from 5 to 15'': -Russian original: - -На книжной полке рядом стоят два тома Пушкина: первый и второй. Страницы каждого тома имеют вместе толщину 2 см, а обложка –– каждая –– 2 мм. Червь прогрыз (перпендикулярно страницам) от первой страницы первого тома до -последней страницы второго тома. Какой путь он прогрыз? -[Эта топологическая задача с невероятным ответом –– -4 мм –– совершенно недоступна академикам, но некоторые -дошкольники легко справляются с ней.] - -English translation: - -Two volumes of Pushkin, the first and the second, are side-by-side on -a bookshelf. The pages of each volume are 2 cm thick, and the cover – front -and back each – is 2 mm. A bookworm has gnawed through (perpendicular to -the pages) from the first page of volume 1 to the last page of volume 2. How -long is the bookworm’s track? -[This topological problem with an incredible answer – 4 mm – is absolutely -impossible for academicians, but some preschoolers handle it with ease.] - -REPLY [5 votes]: Here's Arnold's essay on teaching mathematics. -http://pauli.uni-muenster.de/~munsteg/arnold.html -"For example, these students have never seen a paraboloid and a question on the form of the surface given by the equation $xy = z^2$ puts the mathematicians studying at ENS into a stupor. Drawing a curve given by parametric equations (like $x = t^3 - 3t$, $y = t^4 - 2t^2$) on a plane is a totally impossible problem for students (and, probably, even for most French professors of mathematics)." - -REPLY [4 votes]: Maybe you are referring to his book with problems for schoolchildren? See problem 13 there, and also a remark in the preface.<|endoftext|> -TITLE: Adams Spectral sequence for computing some $B$-bordism groups -QUESTION [10 upvotes]: As the title suggests, I'm trying to apply the Adams Spectral sequence to get some insights of the bordism group -$$ \Omega_4(\xi)= \pi_4(M\xi)$$ -where $\xi \colon BSpin \times K(D_{2n},1) \to BSO$ is a stable vector bundle. I'm trying to use ASS because after an application of the James Spectral Sequence (kind of twisted AHSS) I was able to conclude that -$$ \Omega_4(\xi)= \mathbb{Z} \ \text{ or } \mathbb{Z}\oplus \mathbb{Z}_2$$ -Here you can find a little bit of context and some description of my previous attempts. -My idea is that a computation of $_{(2)}{\pi_4(M\xi)}$ should give me the right choice for $\Omega_4(\xi)$, therefore I try to run an ASS. My lack of expertise in this field lead me to ask a question about how to start the ASS, since I think it's easier to study the ASS with a clear problem in mind, otherwise I wouldn't understand the importance of a lot of technical lemmas done in many of the books covering it. -So the ASS I'm interested in should look like this $$E^{s,t}_2\cong Ext_{\mathcal{A}_2}^{s,t}(H^*(M\xi;\mathbb{Z}_2); \mathbb{Z}_2)$$ - -where the yellow diagonal is the one I'm interested in. A first glance to it lead me to this question: - -(1) How can I conclude something if the diagonal contains infinitely many non-zero stable terms? - -Even computing the $2$-page is troublesome. According to what I know (I've read the chapter about ASS in Fomenko-Fuchs Homotopical Topology), I should find a (minimal)-projective resolution of the the $\mathcal{A}_2$-module $H^*(M\xi;\mathbb{Z}_2)$ which via Thom iso I think can be seen as $H^*(BSpin ;\mathbb{Z}_2)\otimes H^*(D_{2n};\mathbb{Z}_2)$. Problem is that I'm supposed to find an infinitely long (minimal) resolution $B_{\bullet}\to H^*(M\xi)$, since for example $E^{s,t}_2=\hom_{\mathcal{A}_2}(B_s,\mathbb{Z}_2)$ so I really need to compute $B_s$ for all $s$ (and for every internal grade $t$). - -(2) How can I (cleverly) compute at least the second page of this ASS and conclude something out of it? - -The reason I'm asking these questions is that (as you can see in the linked question) I was suggested to use ASS, so I believe something could be said in this case and I'm really interested in learning how to use this powerful tool. The big amount of algebra (at least for a non-algebrist grad student like me) involved in the ASS is making difficult to me getting used to such a tool. -I'm aware that there are plenty of material about the ASS in the books and online, but I prefer learning them by getting my hands dirty with concrete examples I'm really interested in computing, otherwise I fear I will get lost in the ocean of literature about ASS - -REPLY [6 votes]: You should look at Hatcher's notes on spectral sequences, -and if you're fluent enough in German, I can recommend -§§5-6 of: -Stolz, -Hochzusammenhängende Mannigfaltigkeiten und ihre Ränder. -With an English introduction. Lecture Notes in Mathematics, 1116. Springer-Verlag. -The upshot is that since you are interested only in low dimensions, you need to construct the resolution only in low dimensions.<|endoftext|> -TITLE: On the consistency of the definition of the conductor for automorphic forms -QUESTION [6 upvotes]: Let $\pi$ be an irreducible admissible representation of $\mathrm{GL}_2(F)$, where $F$ is local non-archimedean. The local conductor associated to $\pi$ can be defined in two usual manners: -By its associated L-function -Godement and Jacquet associate to it the automorphic L-function $L(s, \pi)$. This L-function satisfies a functional equation of the form -$$L(1-s, \tilde{\pi}) = \varepsilon(s, \pi) L(s, \pi)$$ -This $\varepsilon$-factor is essentially of the form $c^{s - \frac{1}{2}}$, and we define the appearing real number $c(\pi)=c$ to be the conductor of $\pi$. -By some "depth" property -The other way to define the conductor is to follow the work of Casselman for $\mathrm{GL}_2$, and more generaly JPSS for $\mathrm{GL}_n$. We consider the decreasing sequence of compact open congruence subgroups, for $r \geqslant 0$: -$$ -K_{0, \mathfrak{p}}\left(\mathfrak{p}^r\right) = -\left\{ -g \in \mathrm{GL}_2\left(\mathcal{O}_{\mathfrak{p}}\right) \ : \ -g \equiv -\left( -\begin{array}{cc} -\star & \star \\ -0 & \star -\end{array} -\right) -\mod \mathfrak{p}^r -\right\} \subseteq \mathrm{GL}_2(F). -$$ -The conductor of an irreducible admissible infinite-dimensional representation $\pi_\mathfrak{p}$ of $\tilde{G_\mathfrak{p}}$ with trivial central character (to ease notations) is then defined by: -$$ -c(\pi) = N\mathfrak{p}^{f(\pi_{\mathfrak{p}})} , -$$ -where: -$$ - f\left(\pi_{\mathfrak{p}}\right) = \min \left\{r \geqslant 0 \ : \ \pi_{\mathfrak{p}}^{K_{0, \mathfrak{p}}\left(\mathfrak{p}^r\right)} \neq 0\right\} -$$ -Are they consistent? -Here is my question: is it obvious that those two definitions are the same? - -REPLY [11 votes]: These definitions are consistent, though it's not immediate. -The conductor quantifies the extent to which $\pi$ is ramified. As an aside, I prefer to write $c(\pi)$ for the conductor exponent of $\pi$, which is a nonnegative integer, so that $\mathfrak{p}^{c(\pi)}$ is the conductor of $\pi$, and $q^{c(\pi)}$ is the absolute conductor of $\pi$, where $q = N(\mathfrak{p}) = \# \mathcal{O}_F / \mathfrak{p}$. This isn't standard terminology though, as these are all unfortunately called the same thing by different people. -The conductor exponent is tied to a distinguished vector in $\pi$, called the newform of $\pi$. (It is also called the Whittaker newform, the essential Whittaker function, the newvector, or the essential vector; the terminology here hasn't reached a consensus, but newform is best in my opinion because for $\mathrm{GL}_2$, it is the local component of a classical newform.) There are two definitions of the newform of $\pi$, which are equivalent. I'll state them for $\mathrm{GL}_n$ instead of $\mathrm{GL}_2$. Throughout, $n \geq 2$ and $F$ is a nonarchimedean local field. - -Recall that for a spherical (that is, unramified principal series) representation $\pi'$ of $\mathrm{GL}_{n-1}(F)$, there exists a spherical vector $W'^{\circ}$ in the Whittaker model $\mathcal{W}(\pi',\overline{\psi})$ of $\pi'$, so that $W'^{\circ}(1_{n-1}) = 1$ and $W'^{\circ}(gk) = W(g)$ for all $g \in \mathrm{GL}_{n-1}(F)$ and $k \in K_{n-1} = \mathrm{GL}_{n-1}(\mathcal{O}_F)$; that such a vector exists and is unique is classical. Now let $\pi$ be a (possibly ramified) generic irreducible admissible representation of $\mathrm{GL}_n(F)$. Then there exists a unique Whittaker function $W^{\circ}$ in the Whittaker model $\mathcal{W}(\pi,\psi)$ of $\pi$ satisfying -$$W^{\circ} \left(g \begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}\right) = W^{\circ}(g), \qquad W^{\circ}(1_n) = 1$$ -for every $g \in \mathrm{GL}_n(F)$ and $k \in K_{n-1}$ such that for every spherical representation $\pi'$ of $\mathrm{GL}_{n-1}(F)$ with associated spherical vector $W'^{\circ}$, the Eulerian integral -$$\Psi(s,W^{\circ},W'^{\circ}) = \int\limits_{N_{n-1}(F) \backslash \mathrm{GL}_{n-1}(F)} W^{\circ} \begin{pmatrix} g & 0 \\ 0 & 1 \end{pmatrix} W'^{\circ}(g) |\det g|^{s - 1/2} \, dg$$ -is equal to the Rankin-Selberg $L$-function $L(s, \pi \times \pi')$. $W^{\circ}$ is called the newform of $\pi$. (In general, $\Psi(s,W,W')/L(s,\pi \times \pi')$ is a polynomial in $q^{-s}$; the point is that there exists a distinguished choice of $W \in \mathcal{W}(\pi,\psi)$, $W' \in \mathcal{W}(\pi',\overline{\psi})$ for which this polynomial is equal to $1$.) -For a nonnegative integer $m$, let $K_1(\mathfrak{p}^m)$ denote the congruence subgroup of $K_n = \mathrm{GL}_n(\mathcal{O}_F)$ given by -$$\left\{\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in K_n : c \in \mathrm{Mat}_{1 \times (n-1)}(\mathfrak{p}^m), \\ d - 1 \in \mathfrak{p}^m \right\}.$$ -(Note that necessarily $a \in K_{n-1}$ and $b \in \mathrm{Mat}_{n-1 \times 1}(\mathcal{O}_F)$.) Let $\pi$ be a (possibly ramified) generic irreducible admissible representation of $\mathrm{GL}_n(F)$. Then there exists a minimal $m$ for which the vector subspace -$$\pi^{K_1(\mathfrak{p}^m)} = \left\{v \in \pi : \pi(k) \cdot v = v \quad \text{for all $k \in K_1(\mathfrak{p}^m)$}\right\}$$ -of $\pi$ is not equal to $\{0\}$. We denote by $c(\pi)$ this minimal $m$. Then $\pi^{K_1(\mathfrak{p}^{c(\pi)})}$ is one dimensional, so that there exists a unique Whittaker function $W^{\circ}$ in the Whittaker model $\mathcal{W}(\pi,\psi)$ of $\pi$ satisfying -$$W^{\circ} (gk) = W^{\circ}(g), \qquad W^{\circ}(1_n) = 1$$ -for every $g \in \mathrm{GL}_n(F)$ and $k \in K_1(\mathfrak{p}^{c(\pi)})$. - -The original proof of the existence and uniqueness of the newform $W^{\circ}$ of $\pi$ is via (1), in the paper Conducteur des répresentations du group linéaire by Jacquet, Piatetski-Shapiro, and Shalika. However, it was noticed by Matringe several years ago that the proof was in fact incomplete. He gave a correct proof, as did Jacquet. -In the last section of Jacquet, Piatetski-Shapiro, and Shalika's paper, they show that (1) implies (2) via the functional equation for $L(s,\pi \times \pi')$. They make use of the fact that the epsilon factor $\epsilon(s,\pi \times \pi',\psi)$ is equal to $\epsilon(1/2, \pi \times \pi',\psi) q^{-(n - 1) c(\pi) (s - 1/2)}$ for some nonnegative integer $c(\pi)$, and it is precisely this nonnegative integer $c(\pi)$ that satisfies $\pi^{K_1(\mathfrak{p}^{c(\pi)})} \ni W^{\circ}$ and $\pi^{K_1(\mathfrak{p}^m)} = \{0\}$ for all $m < c(\pi)$. -I don't know if the fact that (2) implies (1) has appeared anywhere in print, but I do know how to prove it. There is a paper by Miyauchi that shows that the (Whittaker) newform $W^{\circ}$ given by (2) is such that -$$\Psi(s,W^{\circ}) = \int_{F^{\times}} W^{\circ} \begin{pmatrix} x & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & & \cdots & 1 \end{pmatrix} |x|^{s - \frac{n - 1}{2}} \, d^{\times} x$$ -is equal to the $L$-function $L(s,\pi)$, and the same method of proof (using Hecke operators) shows that $\Psi(s,W^{\circ},W'^{\circ}) = L(s, \pi \times \pi')$ for all spherical representations $\pi'$ of $\mathrm{GL}_{n-1}(F)$. Now you can work backwards to show that $\epsilon(s,\pi \times \pi',\psi)$ is equal to $\epsilon(1/2, \pi \times \pi',\psi) q^{-(n - 1) c(\pi) (s - 1/2)}$, as in Martin Dickson's answer.<|endoftext|> -TITLE: What is a finite Haken cover of the Seifert–Weber space? -QUESTION [11 upvotes]: It's known that the Seifert–Weber space (obtained from a dodecahedron by gluing opposite faces with a 3/10 turn) is an example of a non-Haken 3-manifold. Since every closed 3-manifold is virtually Haken, I was wondering: is there was a known finite cover of the Seifert-Weber space that is Haken? - -REPLY [9 votes]: I just checked using SnapPea that there is a cover of the Seifert-Weber dodecahedral space of index 25 (a 5-fold cyclic cover of a 5-fold cyclic cover) which has positive first betti number hence is Haken. -The Seifert-Weber space is a 5-fold cyclic branched cover over the Whitehead link complement. There are two such 5-fold covers (up to homeomorphism), which one may compute using SnapPea (perform $(5,0)$ surgery on each cusp of the Whitehead link, then compute all 5-fold cyclic covers of this orbifold, giving four manifold covers, with two isometry types). One may then compute the 5-fold cyclic covers of these two manifolds. One of them (not Seifert-Weber) has a 5-fold cyclic cover with positive betti number, whereas the 5-fold cyclic covers of the Seifert-Weber space have trivial betti number. However, one of them will be a 5-fold cover of its sibling, and hence will have a 5-fold cyclic cover which has positive betti number. -There are many other ways that we know the Seifert-Weber space to have a finite cover with positive first betti number (Hempel's paper pointed out by Igor shows that there is a 5-fold irregular cover), but it is an interesting question whether given a manifold $M$ with $b_1(M,\mathbb{F}_p)\geq 4$, is there a $p$-cover which has positive first betti number (I asked this as question 5 in a survey paper). Since all 5-fold covers of the Seifert-Weber space have $b_1(*; \mathbb{F}_5)\geq 4$, then this computation shows that a single 5-fold cyclic cover works in some cases. - -REPLY [8 votes]: This is constructed (reasonably explicitly) in John Hempel's 1982 paper. -John Hempel, MR 664329 Orientation reversing involutions and the first Betti number for finite coverings of $3$-manifolds, Invent. Math. 67 (1982), no. 1, 133--142.<|endoftext|> -TITLE: Relationship between the eigenvalues of a matrix and its symmetric or antisymmetric part -QUESTION [6 upvotes]: First, we have a real matrix $N$ $(m \times m)$, and we don't have much constrains on N. -For simplicity, we can normalize $N$ as -$$ -\frac N{\sqrt{Tr(N^T N)}} -$$ -Later on we are always using the normalized $N$. Normalized $N$ has a good property: -$$ -\sum_{i,j} N_{i,j}^2 = 1 -$$ -or in the form of trace: -$$ -Tr(N^T N) = 1 -$$ -We can always separate the matrix $N$ by its symmetric part and its antisymmetric part: -$$ -N = \frac 1 2 (N+N^T) + \frac 1 2 (N - N^T) = N^s + N^a -$$ -Then -$$ -Tr(N^T N) = Tr((N^s)^2 + [N^s,N^a] - (N^a)^2) = 1 -$$ -$[N^s,N^a]$ is traceless, so that -$$ -Tr((N^s)^2) - Tr((N^a)^2) =1 -$$ -We can set $Tr((N^a)^2) $ as $- \omega$, and it is obvious that $0\le \omega \le 1$, so that $Tr((N^s)^2) = 1- \omega$ -Then we can set the eigenvalues of $\sqrt{N^T N}$ as $\lambda_0^1, \dots \lambda_0^m$ (the singular values of $N$); the eigenvalues of $N^s$ as $\lambda_s^1, \dots, \lambda_s^m$. -The question is what is the relationship between $\lambda_0^1, \dots, \lambda_0^m$ and $\lambda_s^1, \dots, \lambda_s^m$ and $\omega$? -for example, when $\omega = 0$, $\lambda_0^1, \dots \lambda_0^m$ and $\lambda_s^1 \dots \lambda_s^m$ must be the same, because when $\omega =0$, $N^a$ is a zero matrix. -Thanks! - -REPLY [2 votes]: Assume that $N$ is a real valued matrix. Let $x$ be an eigenvector corresponding to $\lambda_s$, i.e. $N_sx = \lambda_sx$. Note that $N_ax$ is always orthogonal to $x$. Therefore $||Nx||^2 = {\lambda_s}^2 + ||N_ax||^2$. This means that ${\lambda_0^i}^2 \ge {\lambda_s^i}^2 + ||N_ax_i||^2$, where $x_i$ is the corresponding eigenvector. -I don't think interlacing can be established since we don't really have control over $N_a$ beyond the fact that $||N_a||_f = \sqrt{1 - ||N_s||_F^2}$. If the norm of $N_s$ is small then $N_a$ can have significant effect. For example if $||N_ax_2||^2 \ge {\lambda_s^1}^2 + ||N_ax_1||^2 - {\lambda_s^2}^2$, then no interlacing can happen.<|endoftext|> -TITLE: More on multiply transitive permutation groups -QUESTION [5 upvotes]: This question is a sort of a follow-up to these two: reference on classfication of multiply transitive permutation groups and Multiply transitive groups, continued -The question is simply: is it true that the set of $n,$ for which any doubly transitive subgroup of $S_n$ is one $S_n$ or $A_n$ is of density one (since any doubly transitive group is either affine or almost simple, this seems clear, but I might be missing something silly). - -REPLY [11 votes]: Yes. In the paper -P.J. Cameron, P.M. Neumann, and D.N. Teague, On the degrees of primitive permutation groups, Math. Z. 180 (1982), 141-149, -the stronger statement is proved that, for almost all $n$, the only primitive permutation groups of degree $n$ are $A_n$ and $S_n$. (By a fortunate coincidence I saw a reference to this in a paper I was reading a day or two ago.)<|endoftext|> -TITLE: Smoothness of a projective variety via the derived category -QUESTION [7 upvotes]: Let $X$ be a smooth projective integral variety over an algebraically closed field $k$. Let $Y$ be a (not necessarily smooth) projective integral variety over $k$. -Assume that $D^b(X) \cong D^b(Y)$. -Does it follow that $Y$ is smooth? -Edit: Here $D^b(X)$ (resp. $D^b(Y)$) is $D^b(Coh(X))$ (resp. $D^b(Coh(Y))$). - -REPLY [5 votes]: I shall elaborate on the comments and provide a sketch of the proof (for which I unfortunately don't know a reference). As $k$ is a perfect field, $X$ is smooth if and only if $X$ is regular. By a well-known theorem of Serre $X$ is regular -if and only if for every closed point $x\in X$ the local ring $\mathscr{O}_{X,x}$ is of finite homological dimension, -Theorem. The following are equivalent: -(i) $X$ is regular; -(ii) $Coh(X)$ has finite homological dimension; -(iii) $D^b(X)$ is Ext-finite. -Proof. (i)$ \Rightarrow $(ii): Serre duality. -(ii)$ \Rightarrow $(iii): By assumption the condition required by Ext-finiteness holds for all complexes concentrated in one degree, and it follows for all bounded complexes for instance by use of standard spectral sequences (see remark 3.7 in Huybrecht's `Fourier-Mukai transforms'). -(iii)$ \Rightarrow $(i): Suppose that $X$ is not regular, let $x\in X$ be a closed singular point. Let $\mathscr{O}_x$ be the skyscraper sheaf with stalk $\kappa(x)$ at $x$. Then the local-to-global spectral sequence yields isomorphisms $\mathrm{Ext}^n_X(\mathscr{O}_x,\mathscr{O}_x)\simeq\mathrm{Ext}^n_{\mathscr{O}_{X,x}}(\kappa(x),\kappa(x))$. The latter is nonzero for infinitely many $n$ because $\mathscr{O}_{X,x}$ has infinite homological dimension. Hence $D^b(X)$ is not Ext-finite.<|endoftext|> -TITLE: Transcendence of $e^{\frac{\pi^2}{12 \log 2}}$ -QUESTION [6 upvotes]: Is it known whether $e^{\frac{\pi^2}{12 \log 2}}$ is transcendental or algebraic? -This number showed up in this other question. - -REPLY [11 votes]: This is most likely open, since alredy $e^{\pi^2}$ is not known to be transcendental. -As an added difficulty, I don't think that $\frac{\pi^2}{12 \log 2}$ is known to be transcendental either. -There are very few, very limited, tricks to prove this kind of result: things like taking $(-1)^{-i}$ and $i^i$ and applying Gelfond–Schneider, or building the Weierstrass $\wp$-function of $\mathbb{Q}(\sqrt{-d})$ to get the transcendence of $e^{\pi\sqrt{d}}$ from its invariants.<|endoftext|> -TITLE: Smooth structures on $M\times S^1$ -QUESTION [12 upvotes]: Let $M$ be a smooth $n$-manifold. Here are three constructions which produce manifolds which are homeomorphic to $M\times S^1$, but might not be diffeomorphic to it: - -Take $M'\times S^1$, where $M'$ is homeomorphic to, but not diffeomorphic to $M$. -Take a connected sum of $M\times S^1$ with an exotic $n+1$-sphere. -Take the mapping torus of a self-diffeomorphism of $M$ which is homotopic, but not (differentiably) isotopic to the identity. - -I would like to know two things: -Question A: when do these constructions actually produce new smooth structures? -Question B: which manifolds have the property that every smooth structure on them can be obtained by operations 1-3 above? -I would be interested both in theoretical answers with references to the literature, and explicit answers for specific manifolds, such as $S^6$, or other large-but-not-too-large dimensional spheres. -Remark 1: question A for construction 2 (connected sums) was essentially already asked here. This problem is apparently known among experts as "computing inertia groups". So I think part of my question is: are the inertia groups of $S^n\times S^1$ known? -Remark 2: question A for construction 3 is closely related to this question The difference is that I am asking about the diffeomorphism type of the total space, not fibered equivalence. - -REPLY [7 votes]: I assume that you are interested in this question in high dimensions, which should be 6 (or possibly 5) for $M \times S^1$. Construction 2 is actually a special case of construction 3, for if you take $(M \times S^1) \# \Sigma$ (the last being a homotopy sphere) then you can split along a copy of $M$ to get an s-cobordism from $M$ to itself. Trivializing this s-cobordism gives a diffeomorphism $f$ of M to itself such that $(M \times S^1) \# \Sigma$ is the mapping torus of $f$. If you exhibit (as Smale tells us you can) $\Sigma$ as the union of two disks glued by a diffeomorphism $g$ of the boundary, then you get that $f$ is essentially the identity map of $M$ connected sum with $g$. -I'm not sure it exactly answers your question, but Farrell's fibering theorem (Farrell, F. T. The obstruction to fibering a manifold over a circle. -Indiana Univ. Math. J. 21 1971/1972 315–346) seems relevant. It implies that a smooth manifold homeomorphic to $M \times S^1$ is diffeomorphic to a fiber bundle over $S^1$, with fiber that is homotopy equivalent to (in fact topologically h-cobordant to) $M$.<|endoftext|> -TITLE: Expanding into monomials -QUESTION [6 upvotes]: Given a multi-variable function $F$, denote the number of monomials by $N(F)$. For example, $N(x(x+y))=N(x^2+xy)=2$ and -$$ -N(x(x+y)(x+y+z))=N(x^3+2x^2y+x^2z+xy^2+xyz)=5. -$$ -Define the functions $f_n=\prod_{i=1}^n(y_1+\cdots+y_i)$ and $g_n=\prod_{i=1}^n(y_1+\cdots+y_i+x_{i+1})$ and -$$ -h_n=\prod_{i=1}^n(y_1+\cdots+y_i+x_{i+1}+x_{i+2}). -$$ -This work - -Tewodros Amdeberhan and Richard P. Stanley, Polynomial Coefficient Enumeration, preprint (2008) arXiv:0811.3652 (pdf) - -shows (and it is easy) that $N(f_n)=C_n$ (the Catalan numbers) and with a little bit of effort $N(g_n)=\frac1n\sum_{k=0}^n2^k\binom{n}k\binom{n}{k-1}$ (the large Schröder numbers). - -Question. What about $N(h_n)$? - -REPLY [3 votes]: Let $\mathrm{CC}_n$ denotes the set of Catalan codes of length $n$, i.e. -$$\mathrm{CC}_n = \left\{ (c_0, \dots, c_{n-1})\in \mathbb{Z}_{\geq 0}^n\ :\ c_0+\dots+c_i\leq i\ \text{for all}\ i=0,1,\dots,n-1 \right\}.$$ -For $c\in \mathrm{CC}_n$, denote $s(c) = n-c_0-\dots-c_{n-1}$. -Then -$$N(h_n) = \sum_{k=0}^n [x^k]\ \sum_{c\in \mathrm{CC}_{n-k}} (1-2x)^{-s(c)-1} \cdot\prod_{i=0}^{n-k-1} \frac{1+(1-2x)^{-c_i-1}}{2},$$ -where $[x^k]$ is the operator taking the coefficient of $x^k$. (I did not think much about simplifying this expression.) -Notice that there is a similar expression for $N(g_n)$, which indeed simplifies to the large Schröder number $S_n$: -$$ -\begin{split} -N(g_n) &= \sum_{k=0}^n [x^k]\ \sum_{c\in \mathrm{CC}_{n-k}} (1-x)^{-s(c)-1} \cdot\prod_{i=0}^{n-k-1} (1-x)^{-c_i-1} \\ -& = \sum_{k=0}^n [x^k]\ C_{n-k}\cdot (1-x)^{-2(n-k)-1} \\ -& = \sum_{k=0}^n C_{n-k}\cdot \binom{2n-k}{k} \\ -& = S_n. -\end{split}$$ -This is my SAGE code implementing the aforementioned formula for $N(h_n)$: -def myNh(n): - R. = PowerSeriesRing(ZZ) - return sum( \ - sum( \ - ( prod( \ - 1 + (1-2*x+O(x^(k+1)))^(-c[i]-1) \ - for i in range(n-k) ) \ - ) * (1-2*x+O(x^(k+1)))^(-(n-k-sum(c))-1) / 2^(n-k) \ - for c in map(lambda x: x.to_Catalan_code(), DyckWords(n-k)) )[k] \ - for k in range(n+1) ) - -For $n=0,1,\dots,14$, it gives -1, 3, 11, 46, 210, 1018, 5150, 26889, 143829, 784167, 4341843, 24348352, 138007784, 789375504, 4550522248 -which match the values computed by others in the comments. I've added it as the sequence A281548 to the OEIS.<|endoftext|> -TITLE: Chern-Weil homomorphism and classifying space -QUESTION [8 upvotes]: Let $G$ be a real or complex Lie group with Lie algebra $\mathfrak g$, and let $\mathbb C[\mathfrak g]$ be the algebra of $\mathbb C$-valued polynomials on $\mathfrak g$. Denote by $$\mathbb C[\mathfrak g]^G=\{f\in \mathbb C[\mathfrak g] | ~f(Ad_g x)=f(x),~ \forall g \in G,~ \forall x\in \mathfrak g\}$$ -the subalgebra of fixed points in $\mathbb C[\mathfrak g]$ under the adjoint action of $G$. -Now, given principal $G$-bundle $P$ on a manifold $M$, the Chern-Weil homomorphism is an homomorphism of $\mathbb C$-algebras $\mathbb C[\mathfrak g]^G \to H^*_{dR}(M, \mathbb C)$, which is given by $f\mapsto f(F_\nabla)$ for some connection $\nabla$ on the given bundle. -Suppose $G$ is compact, then according to Wikipedia, we have an isomorphism of $\mathbb C$-algebras $$\mathbb C[\mathfrak g]^G \cong H^*_{dR}(BG, \mathbb C) $$ -Question: Is this isomorphism given by the Chern-Weil homomorphism? -If so, a problem is that $BG$ may fail to be a smooth manifold, so additional conditions should be needed. And, what $G$-bundles on $BG$ should be taken into consideration? And, can we construct explicitly the inverse homomorphism? -If not, how to understand this isomorphism? - -REPLY [2 votes]: Is this isomorphism given by the Chern-Weil homomorphism? - -Yes, see Theorem 7.20 in the paper of Freed and Hopkins, -which computes the de Rham complex of B∇G -as C[g]G equipped with the zero differential. -Some important remarks: -1) B∇G is the stack of principal G-bundles with connection. -Connections are necessary to define the Chern-Weil homomorphism. -2) The statement already holds on the level of chain complexes, -not just individual cohomology groups. -3) B∇G is not a smooth manifold, but a stack -(namely, a simplicial presheaf) on the site of smooth manifolds.<|endoftext|> -TITLE: Conjugate points on cut locus -QUESTION [14 upvotes]: Let $M$ be a Riemannian with nonempty boundary $\partial M$. -Define multiplicity of $x\in M$ as the number of minimizing geodesics from $x$ to $\partial M$. -The following fact seems to be standard: - -The set of points with multiplicity $\ge 2$ is dense in the cut locus of $M$ with respect to $\partial M$. - -Is it proved somewhere? -Comments - -I know how to prove it, but if you have a one-line proof I am very interested. -Any point of multiplicity 1 in the cut locus is conjugate, but not the other way around. Say for a hemisphere, the cut locus contains a single point — it is a conjugate point of multiplicity $\infty$. - -REPLY [5 votes]: I've got a letter from Stephanie Alexander with a complete answer. Let me summarize it here. -The first proof is given in 4.8 of "Schnittort und konvexe Mengen..." by Hermann Karcher (1968). (The formulation is slightly weaker, but from the proof proves our statement follows; the idea is the same as in the answer of Ian Agol.) -An other proof was given in "Decomposition of cut loci" by Richard Bishop (1977). -Latter it appears as Lemma 2 in "Distance function and cut loci..." by Franz-Erich Wolter (1979). The proof is basically -the same as Karcher's.<|endoftext|> -TITLE: Functions with vanishing Kahler differentials along a subscheme -QUESTION [10 upvotes]: Question: Let $k$ be an algebraically closed field of characteristic zero. -Let $X=\mathrm{Spec}\!~R$ be a scheme of finite type over $k$. Suppose $I\subset R$ is -an ideal and $f\in I$. -Assume that $df = 0$ in $\Omega_{R/k}\otimes_R R/I$. -Does it follow that there is some positive integer $k$ such that $f^k \in I^{k+1}?$ -Motivation: Suppose that $X$ and $V(I)$ are both smooth, then we have the short exact sequence -$$ - 0 \to I/I^2 \to \Omega_{R/k}\otimes_R R/I \to \Omega_{(R/I)/k} \to 0. -$$ -Thus if $f\in I$ and $df \equiv 0$ mod $I$, then $f\in I^2$. -In general this fails. For example, we can take $R=k[x]$, $I=(x^m)$ and $f = x^{m+1}$. Then $f\not\in I^2$ if $m>1$. However -$$ - f^{m}\in I^{m+1}. -$$ -Geometrically, the statement says that if a function vanishing on the subscheme also has vanishing differential, then -the function is nilpotent on the normal cone. -A Quick Reduction and Some Partial Results: If the statement is true for some $R$, then it's also true for any quotient of $R$. Hence we may assume that $R$ is the polynomial ring $k[x_1,\dots,x_n]$. -Under this hypothesis, the statement becomes - -If $f\in I$ and $\partial f/\partial x_i\in I$ for $i=1,\dots,n$. Does it imply that $f^k\in I^{k+1}$ for sufficiently large $k$? - - -if $I=(g)$ is principal, then we can prove the statement by factorizing $g$ intro irreducible polynomials -$$ -g=g_1^{a_1}\cdots g_r^{a_r}. -$$ -If we write $f=gh$. Then it's easy to show that $g_1\cdots g_r$ divides $h$. -Hence we can take $k=\max \{a_i\}$. -if $I$ is a graded ideal (generated by monomials), then we may assume that $f$ is a monomial, say $f=x_1^{a_1}\cdots x_n^{a_n}\in I$ and set $k=a_1+\cdots +a_n$. Then we have -$$ -f^{k} = \sum_{a_i\neq 0} \left(\frac{\partial f}{\partial x_i} -\cdot -\frac{x_i}{a_i} -\right)^{a_i} -\in I^k\cdot f\subset I^{k+1}. -$$ -a comment from S.Li below: Without loss of generality, we can simply take $I$ to be the ideal generated by $f$ and all the $\partial f/\partial x_i$. - -REPLY [2 votes]: I am writing up the comments above as an answer, partly because they involve a couple of fun lemmas about blowing up. I tried to find a proof that does not use resolution of singularities, but I could not find one. -Let $X$ be a scheme, and let $\mathcal{J}$ be a quasi-coherent sheaf of ideals. Then the blowing up, $\nu:Z\to X$, is final among morphisms to $X$ such that the inverse image ideal sheaf $\nu^{-1}\mathcal{J}\cdot \mathcal{O}_Z$ is everywhere locally principal and generated by a nonzerodivisor (this condition is automatically true for the unique morphism from the empty scheme). There is also a construction of $Z$ as $$Z=\underline{\text{Proj}}_X\ \bigoplus_{n \geq 0} \mathcal{J}^n .$$ In particular, if $\mathcal{J}$ is everywhere locally finitely generated, then $\nu$ is proper. In that case, for every $n\geq 0$, there is a natural homomorphism of locally finitely generated, quasi-coherent $\mathcal{O}_X$-modules, $$\alpha_n:\mathcal{J}^n \to \nu_*(\nu^{-1}(\mathcal{J}^n)\cdot \mathcal{O}_Z).$$ It need not be the case that every $\alpha_n$ is an isomorphism, e.g., for $X=\text{Spec}\ k[s,t]$ and $\mathcal{J} = (\langle s^d, s^{d-1}t,st^{d-1},t^d \rangle)^\widetilde{\ \ \ }$, then $\alpha_n$ is an isomorphism precisely for $n\geq d-2$. -Lemma 1. If $X$ is Noetherian and $\mathcal{J}$ is coherent, then there exists $n_0\geq 0$ such that for all $n\geq n_0$, $\alpha_n:\mathcal{J}^n\to \nu_*(\nu^{-1}\mathcal{J}^n\cdot \mathcal{O}_Z)$ is an isomorphism. -The proof is basically Exercise II.5.9 from Hartshorne's "Algebraic Geometry". -Now let $X$ be an integral, Noetherian scheme, let $\mathcal{I}$ and $\mathcal{J}$ be nonzero coherent ideal sheaves, and let $\mu:Y\to X$, resp. $\nu:Z\to X$, be the blowing up of $X$ along the ideal sheaf $\mathcal{I}$, resp. $\mathcal{Z}$. Each blowing up a dominant morphism of integral schemes such that the pullback of the ideal sheaf is locally principal, and universal among all such morphisms. Then both the blowing up of $Y$ along $\mu^{-1}\mathcal{J}\cdot \mathcal{O}_Y$ and the blowing up of $Z$ along $\nu^{-1}\mathcal{I}\cdot \mathcal{O}_Z$ are universal among dominant morphisms of integral schemes to $X$ such that the pullback ideal sheaves of both $\mathcal{I}$ and $\mathcal{J}$ are locally principal. Thus, there is a canonical isomorphism of these two blowings up; call the common scheme $W$. So there is a commutative diagram $$ \begin{array}{lcr} W & \xrightarrow{\widetilde{\nu}} & Y \\ \widetilde{\mu}\downarrow & & \downarrow \mu \\ Z & \xrightarrow{\nu} & X \end{array} $$ where every morphism is a blowing up. In fact, this is the blowing up of $\mathcal{I}\cdot \mathcal{J}$, cf. http://stacks.math.columbia.edu/tag/085Y Denote the composite morphism from $W$ to $X$ by $\lambda:W\to X$. Denote by $\widetilde{\mathcal{I}}$, resp. $\widetilde{\mathcal{J}}$, the inverse image ideal sheaf $\nu^{-1}\mathcal{I}\cdot \mathcal{O}_Z$, resp. $\mu^{-1}\mathcal{J}\cdot \mathcal{O}_Y$. Since $\mu^{-1}\mathcal{I}\cdot \mathcal{O}_Y$ is already locally principal generated by a nonzerodivisor, the natural homomorphism, $$\beta_n:\mu^{-1}\mathcal{I}^n\cdot \mathcal{O}_Y \to \widetilde{\nu}_*(\widetilde{\nu}^{-1}(\mu^{-1}\mathcal{I}_n\cdot \mathcal{O}_Y)\cdot \mathcal{O}_W),$$ is an isomorphism for every $n\geq 0$. Then $\lambda^{-1}\mathcal{I}\cdot \mathcal{O}_W$ equals $\nu^{-1}(\mu^{-1}\mathcal{I}\cdot \mathcal{O}_Y)\cdot \mathcal{O}_W$. Thus, by Lemma 1 above, for every $n\geq n_0$, the natural map, $$\mathcal{I}^n \to \lambda_*(\lambda^{-1}\mathcal{I}^n\cdot \mathcal{O}_W),$$ is an isomorphism. Again by Lemma 1 above, for every $n\geq n_1$, $$\widetilde{\mathcal{I}}^n \to \widetilde{\mu}_*(\widetilde{\mu}^{-1}\widetilde{\mathcal{I}}^n\cdot \mathcal{O}_W),$$ is an isomorphism. Altogether, this proves the following. -Lemma 2. For an integral, Noetherian scheme $X$, for a nonzero coherent sheaf $\mathcal{J}$ with blowing up $\nu:Z\to X$, for every coherent sheaf $\mathcal{I}$, there exists an integer $n_0\geq 0$ such that for every $n\geq n_0$, the natural homomorphism $\mathcal{I}^n\to \nu_*(\nu^{-1}\mathcal{I}^n\cdot \mathcal{O}_Z)$ is an isomorphism. -Thus, in your setting, for every $m\geq n_0-1$, to prove that $f^m$ is a global section of $\mathcal{I}^{m+1}$, it suffices to prove that $\nu^*f^m$ is a global section of $\nu^{-1}(\mathcal{I}^{m+1})$. Here I am assuming that $X$ is a smooth, finite type $k$-scheme, where $k$ is an algebraically closed field of characteristic $0$. By strong resolution of singularities, there exists a blowing up $\nu:Z\to X$ such that $\nu^{-1}\mathcal{I}\cdot \mathcal{O}_Z$ is everywhere locally principal generated by an element of the form $g_1^{a_1}\cdots g_r^{a_r}$, where $(g_1,\dots,g_r)$ is part of a regular system of parameters locally. By your proof, locally, $\nu^*f^m$ is a section of $\nu^{-1}\mathcal{I}^{m+1}\cdot \mathcal{O}_Z$ for every $m\geq \max(a_1,\dots,a_r)$. -Since the quasi-compact scheme $Z$ is covered by finitely many open affines on which $\nu^{-1}\mathcal{I}\cdot \mathcal{O}_Z$ is locally generated by $g_1^{a_1}\cdots g_r^{a_r}$, it follows that there exists $m_1$ such that for every $m\geq m_1$, $\nu^* f^m$ is a global section of $\nu^{-1} \mathcal{I}^{m+1}\cdot \mathcal{O}_Z$. Thus, by the previous paragraph, there exists $n_1=\max(n_0-1,m_1)$ such that for every $m\geq n_1$, $f^m$ is a section of $I^{m+1}$.<|endoftext|> -TITLE: Axiom of choice, Banach-Tarski and reality -QUESTION [61 upvotes]: The following is not a proper mathematical question but more of a metamathematical one. I hope it is nonetheless appropriate for this site. -One of the non-obvious consequences of the axiom of choice is the Banach-Tarski paradox and thus the existence of non-measurable sets. -On the other hand, there seem to be models of Zermelo-Fraenkel set theory without axiom of choice where every set would be measurable. -What does this say about the "plausibility" of the axiom of choice? Are there reasons why it is plausible (for physicists, philosophers, mathematicians) to believe that not all sets should be measurable? Is the Banach-Tarski paradox one more reason why one should "believe" in the axiom of choice, or is it on the opposite shedding doubt on it? - -REPLY [4 votes]: I think it is interesting to re-phrase this question relative to other axioms and/or theorems of ZFC. What Andrej Bauer's answer suggests is that it may not be the axiom of choice per se that is the culprit, but rather the underlying structure. -For example, it is provable that the existence of non-Lebesgue measurable sets and the Banach–Tarski paradox are both an implication over ZF (without Choice) of the Hahn–Banach theorem (HB) in functional analysis. This means that we can prove those "pathologies" as a theorems from ZF+HB. -Another way to put it is to analyze the structure within the context of different orders of logic. In ordinary first-order logic Choice is provably equivalent to the well-ordering theorem (WO) over ZF. What is different in second-order logic is that WO is strictly stronger than choice: WO $\vdash_{ZF}$ Choice, but Choice $\nvdash_{ZF}$ WO. -In other words, one may get the feeling that perhaps there has been too much emphasis put on Choice. Other axioms and theorems can prove just as problematic. The approach described above is the essence of Reverse Mathematics (RM). While RM has been traditionally carried out at a much lower proof-theoretic strength level (subsystems of second-order arithmetic), in my opinion this provides a very useful framework for analyzing the foundations of other parts of mathematics.<|endoftext|> -TITLE: Research directions in persistent homology -QUESTION [24 upvotes]: I am interested in what are the possible directions for new research in persistent homology (more of the mathematical theoretical aspects rather than the computer algorithm aspects). -So far from googling, proving stability (not affected by small changes) seems to be one direction. -Another direction I have seen is the inverse problem of persistent homology: to what extent can the original space be reconstructed from the persistent homology. - -Are there any other research or generalizations of persistent homology? - -REPLY [14 votes]: Persistent homology is related to spectral sequences. There is already a discussion Persistence barcodes and spectral sequences about if this leads to new theoretical insides and there are stated further references (for example, there is stated the book of Edelsbrunner and Harer, which is a standard book for Persistent homology and can find further discussions about some research directions in Edelsbrunner and Harer's works as well). -Furthermore, Persistent homology leads to discrete Morse theory, what can be used for computing cellular sheaf cohomology (if you want to know more about it, i.e. Justin Curry's work could be interesting for you).<|endoftext|> -TITLE: Homotopy type of a complex affine variety -QUESTION [8 upvotes]: Let $X$ be an affine variety of dimension $n$ over $\mathbb{C}$. - -Does the analytic space associated with $X$ have the homotopy type of a $n$-dimensional CW complex? - -REPLY [11 votes]: If $X$ is smooth, and if you ask to have the same homotopy type of a CW complex of real dimension at most $n$, this is precisely the statement of the Andreotti-Frankel theorem. -It is true, more generally, for a Stein manifold of complex dimension $n$. -If $X$ is arbitrarily singular, the same theorem holds, provided $X$ is irreducible. This was shown by Karčjauskas for algebraic varieties and by Hamm for Stein spaces.<|endoftext|> -TITLE: "Joyal type" model structure for (n,1)-categories? -QUESTION [6 upvotes]: The Joyal model structure on the category of simplicial sets, has monomorphisms as cofibrations and quasi-categories as fibrant objects (these model $(\infty,1)$-categories). In HTT (section 2.3.4) Lurie defines the notion of an $n$-category (these model $(n,1)$-categories) and proves some theorems about them. For example, $1$-categories are just simplicial sets which are isomorphic to a nerve of a category (prop. 2.3.4.5). Moreover, every quasi-category $\mathcal{C}$ has an $n$-truncation $h_n\mathcal{C}$ which is an $n$-category with a map $\mathcal{C} \to h_n\mathcal{C}$ which is universal from $\mathcal{C}$ to an $n$-category (see prop. 2.3.4.12 for a precise statement). - -Qusetion: Is there a model structure on the category of simplicial sets with monomorphisms as cofibrations, $n$-categories as fibrant objects and such that two quasi-categories are equivalent in this model structure if and only if their $n$-truncations are equivalent in the Joyal model structure? - -I ask this mainly out of pure curiosity and haven't put too much effort in thinking about it myself, in the hope that someone already knows the answer. -Hence, perhaps there are simple reasons that this can't happen, in this case (and any other case), I'm open to considering some variations. -Remark: In fact, as I learned from this question, the cofibrations and fibrant objects determine the model strucute, so the question can be broken in two. First, whether there is a model structure with the specified cofibrations and fibrant objects, and if so, how can we interpret the weak equivalences. - -REPLY [2 votes]: You may be interested in having a look at this: https://arxiv.org/abs/1810.11188.<|endoftext|> -TITLE: Can a convex polytope with $f$ facets have more than $f$ facets when projected into $\mathbb{R}^2$? -QUESTION [14 upvotes]: Let $P$ be a convex polytope in $\mathbb{R}^d$ with $n$ vertices and $f$ facets. -Let $\text{Proj}(P)$ denote the projection of $P$ into $\mathbb{R}^2$. -Can $\text{Proj}(P)$ have more than $f$ facets? -In the general case, each successive projection can increase the number of facets from $f$ to $\left\lfloor \frac{f^2}{2} \right\rfloor$, but I'm wondering if $\mathbb{R}^2$ is a special case. - -REPLY [23 votes]: Your question is essentially about extension complexity. In general, the extension complexity of a polytope $P$ is the minimum number of facets over all polytopes $Q$ which project to $P$. You are interested in the extension complexity of polygons. Fiorini, Rothvoß, and Tiwary proved that regular $n$-gons have extension complexity $O(\log n)$. For lower bounds, they give examples of $n$-gons which have extension complexity $\sqrt{2n}$. -It is an open question if there exists for infinitely many $n$, an $n$-gon with extension complexity $\Omega(n)$. -Edit. Shitov has recently shown that every convex $n$-gon has extension complexity at most $147 n^{2/3}$, which resolves the open question in the negative. -As a bonus, here is a picture of the polytope in Nate's answer (courtesy of Samuel Fiorini).<|endoftext|> -TITLE: Neighborhoods of the hyperdiagonal in a product space -QUESTION [5 upvotes]: We recently need the following result about product space in an applied analysis paper: -Let $(M,d)$ be a metric space, and $\bigotimes^{k+1}M = \overbrace{M \times M \times \dots \times M}^{(k+1)-{\rm times}}$ be equipped with the product topology. -Let $U \subset \bigotimes^{k+1}M$ be a neighborhood of the hyperdiagonal -$ \{ (x,x,\dots,x) : x\in M\}$. When $k\geq 2$, there is a smaller open neighborhood -$V' \subset U$ of the hyperdiagonal -of the form -$ -V' = \{(x_0,x_1,\dots,x_k) \,:\, (x_i,x_{i+1}) \in V \,, 0 \leq i < k\} -$ -for $V \subset M \times M$ an open neighborhood of the diagonal of $M\times M$. -The proof (see Appendix B of http://www.math.drexel.edu/~tyu/Papers/SE7.pdf) relies on a metric; we wondered if any form of this result is written down somewhere, and if the metrizability of the space $M$ is required for the result to hold true. Thanks. - -REPLY [2 votes]: The result does not hold for general topological spaces. -For a counterexample, equip $M = \{a, b, c\}$ with the topology whose open sets are $\emptyset$, $M$, $\{a, b\}$, $\{b\}$ and $\{b, c\}$. -The set $U= \{a,b\}^3\cup \{b,c\}^3\subseteq M^3$ is an open neighborhood of the ternary diagonal. -Now let $V\subseteq M^2$ be any open neighborhood of the binary diagonal. Since (i) $V$ contains $(a,a)$, (ii) $V$ is open in the product topology, and (iii) $\{a,b\}$ is the smallest open set of $M$ containing $a$, it follows that $\{a,b\}^2\subseteq V$. Similarly $\{b,c\}^2\subseteq V$. But now $V$ is already too big for our purposes. If $V'$ is defined to be -$\{(x_0,x_1,x_2) \,:\, (x_i,x_{i+1}) \in V \,, 0 \leq i \lt 2\}, -$ -then $(a,b,c)\in V'-U$. That is, there is no neighborhood $V$ of the binary diagonal that induces a neighborhood $V'$ of the ternary diagonal such that $V'\subseteq U$.<|endoftext|> -TITLE: Why is every Hamiltonian system locally integrable? -QUESTION [5 upvotes]: It is common knowledge that every Hamiltonian system is locally integrable (away from singular points of the Hamiltonian), meaning that, in a neighborhood of each point of the $2n$-dimensional symplectic manifold on which the Hamiltonian vector field is defined, it is possible to find $n$ integrals of motion in involution. -In general these do not globally extend to give the compact Lagrangian fibration/foliation appearing in the Arnold-Liouville theorem, but here I want to focus on this purely local situation. -What is the simplest rigorous argument you can give me of this fact? - -REPLY [9 votes]: For non-singular Hamiltonian systems, you can see it as a consequence of a slight generalization of the Darboux Theorem which is known as the Carathéodory--Jacobi--Lie Theorem (see, e.g. Libermann, Marle, Symplectic Geometry and Analytical Mechanics). - -Carathéodory--Jacobi--Lie Theorem Let $(M,\omega)$ be a $2n$-dimensional symplectic manifold, with associated Poisson bracket $\{-,-\}$. Assume to have, for some $0\leq - k\leq n$, smooth functions $f_1,\ldots,f_k$, defined on a neighborhood - $U$ of a point $x$ in $M$, which are independent and - Poisson-commuting, i.e. -$$df_1\wedge\ldots\wedge df_k\neq 0,$$ -$$\{f_i,f_j\}=0,\text{ for all }0\leq i -TITLE: Modular forms from counting points on algebraic varieties over a finite field -QUESTION [15 upvotes]: Suppose we are given some polynomial with integer coefficients, which we regard as carving out an affine variety $E$, for example: -$$ 3x^2y - 12 x^3y^5 + 27y^9 - 2 = 0 \tag{$*$} $$ -(We might consider a bunch of equations, we might work over projective space, but let's keep it simple for now). -We are interested in the number of points on $E$ when we reduce modulo $p$, i.e. over the finite field $\mathbb{F}_p$ as the prime $p$ varies. For our single equation in two variables, as a rough approximation, we would expect $p$ points in general. So for each prime $p$, we define numbers $a_p$ which measure the deviation from this, -$$a_p := p - \text{number of solutions to $(*)$ over $\mathbb{F}_p$}$$ -Question: When is it true that the numbers $\{a_p\}$ are "modular", in the sense that there exists a modular form -$$f = \sum_{n=1}^\infty b_n q^n$$ -such that $b_p = a_p$ for almost all primes $p$? (The "almost all" is to avoid problems with bad primes. Note that $f$ is still uniquely determined by the above requirement.) -The Modularity Theorem of Breuil, Conrad, Taylor and Diamond says that this is true when $E$ is an elliptic curve, i.e. takes the form $y^2 = 4x^3 - g_2 x - g_3$ for some integers $g_2, g_3$. In that case, $f$ is a weight 2 modular form of level $N$ where $N$ is the "conductor" of $E$. -But is it true for more general varieties? -(Note: I am aware of a generalized "Modularity Theorem" for certain Abelian varieties, but it's not clear to me that what people mean by "Modular" in that context is the same as the simple-minded notion I'm using --- that an adjusted count of points mod $p$ gives the Fourier coefficients of a modular form.) - -REPLY [21 votes]: The correct setting for this construction turns out to be projective varieties, so let me suppose we have a smooth variety $X$ inside $\mathbf{P}^N$, for some $N \ge 1$, defined by the vanishing of some homogenous polynomials $F_1, \dots, F_r$ in variables $x_0, \dots, x_N$, with the $F_i$ having coefficients in $\mathbf{Q}$. Actually, let me assume the $F_i$ have coefficients in $\mathbf{Z}$, which is no loss since we can just multiply up. Then we can make sense of the reduction of $X$ modulo $p$; and we want to study the point counts $\#X(\mathbf{F}_p)$ as a function of $p$, possibly neglecting some finite set $\Sigma$ containing all primes such that the reduction of $X$ mod $p$ is singular. -Thanks to Grothendieck, Deligne, and others, we have a very powerful bunch of tools for analysing this problem. The setup is as follows. Choose your favourite prime $\ell$. Then the theory of etale cohomology attaches to $X$ a bunch of finite-dimensional $\mathbf{Q}_\ell$-vector spaces $H^i_{\mathrm{et}}(X_{\overline{\mathbf{Q}}}, \mathbf{Q}_\ell)$ (let me abbreviate this by $H^i_\ell(X)$ to save typing). The dimension of $H^i_\ell$ is the same as the $i$-th Betti number of the manifold $X(\mathbb{C})$; but they encode much more data, because each $H^i_\ell(X)$ is a representation of the Galois group $\operatorname{Gal}(\overline{\mathbf{Q}} / \mathbf{Q})$, unramified outside $\Sigma \cup \{\ell\}$; so for every prime not in this set, and every $i$, we have a number $t_i(p)$, the trace of Frobenius at $p$ on $H^i_\ell$, which turns out to be independent of $\ell$. -Theorem: $\#X(\mathbf{F}_p)$ is the alternating sum $\sum_{i=0}^{2 dim(X)} (-1)^i t_i(p)$. -Now let's analyse $H^i_\ell$ as a Galois representation. Representations of Galois groups needn't be direct sums of irreducibles, but we can replace $H^i_\ell$ by its semisimplification, which does have this property and has the same trace as the original $H^i_\ell$. This semisimplification will look like $V_{i, 1} + \dots + V_{i, r_i}$ where $V_{i, j}$ are irreducible; and the $V_{i, j}$ all have motivic weight $i$, so the same $V$ can't appear for two different $i$'s. So we get a slightly finer decomposition -$\#X(\mathbf{F}_p) = \sum_{i=0}^{2 \mathrm{dim} X} (-1)^i \sum_{j=1}^{k_i} t_{i, j}(p)$ -where $t_{i,j}(p)$ is the trace of $Frob_p$ on $V_{i,j}$. -Let me distinguish now between several different types of irreducible pieces: - -$V_{i, j}$ is a one-dimensional representation. Then $i$ must be even, and the trace of Frobenius on $V_{i, j}$ is $p^{i/2} \chi(p)$ where $\chi$ is a Dirichlet character. -$V_{i, j}$ is two-dimensional and comes from a modular form. Then $t_{i,j}(p) = a_p(f)$, and $f$ must have weight $i+1$. -$V_{i,j}$ is two-dimensional and doesn't come from a modular form. This can happen, but it's rare, and it's expected that all examples come from another kind of analytic object called a Maass wave form; in particular this forces $i$ to be even. -$V_{i, j}$ has dimension $> 2$. Then $V_{i, j}$ cannot be the Galois representation coming from a modular form, because these always have dimension 2. - -You seem to want your varieties to have $X(\mathbf{F}_p)$ = (polynomial in $p$) + (coefficient of a modular form). From the above formulae, it's clear that this can only happen if all the $V_{i,j}$ have dimension 1 or 2; there is exactly one with dimension 2 and it comes from a modular form; and the one-dimensional pieces all come from the trivial Dirichlet character. This always happens for genus 1 curves, because the $H^0$ and $H^2$ are always 1-dimensional for a curve, and the genus condition forces the $H^1$ to be two-dimensional. -However, once you step away from genus 1 curves, this is totally not the generic behaviour, and it will only occur for unusual and highly symmetric examples, such as the rigid Calabi-Yaus and extremal $K_3$ surfaces in the links you've posted.<|endoftext|> -TITLE: Totally convex, convex and locally convex sets -QUESTION [5 upvotes]: Consider the following definitions : -$C\subset M$ is convex if any $p,\ q\in C$ all minimizing - geodesic between $p$ and $q$ are in $C$ -$C$ is totally convex if for $p,\ q\in C$, every geodesic between $p$ and $q$ are in -$C$ -$C$ is locally convex if for $p\in C$ there is open set $U$ around $p$ s.t. $U\cap C$ is convex -So on $S^2(1)$ closed hemisphere $C$ is locally convex but not convex And $C$ is not totally geodesic -And on a suitable sphere which has no constant curvature there is a convex set which is not totally geodesic -Hence these are distinct -But are there connected locally convex set $C$ s.t. in $C\subset C_1\subset C_2$ where $C_1$ is smallest convex set containing $C$ and $C_2$ is smallest totally geodesic set containing $C_1$, inclusions are strict? -If $C$ is not connected, then we have an example -Thank you in advance - -REPLY [3 votes]: Let $M$ be the cylinder $\{x^2+y^2=1,\ |z| < 2\}$ plus hemispherical caps on both ends. -Let $C$ be the set $\{x < \frac{1}{2},\ x^2+y^2=1,\ |z| < 1\}$. $C$ is locally convex. -Then $C_1$ is the cylinder $\{x^2+y^2=1,\ |z| < 1\}$, with all shortest geodesics between points in $C$. -And $C_2$, with all geodesics between points in $C$, is all of $M$.<|endoftext|> -TITLE: Submodules of $({\mathbb Z}/6{\mathbb Z})^n$ intersecting $\{0,1\}^n$ trivially -QUESTION [14 upvotes]: $\newcommand{\F}{{\mathbb F}}$ -$\newcommand{\Z}{{\mathbb Z}}$ -Suppose that $\F$ is a finite field of prime order $p:=|\F|$, and let $n$ be a positive integer. I consider the regime where $\F$ is fixed and $n$ grows. Assuming for simplicity that $n$ is divisible by $p-1$, the linear subspace $V<\F^n$ determined by the equations - $$ x_1+\dotsb+x_{p-1}=x_p+\dotsb+x_{2p-2}=\dotsb=x_{n-p+2}+\dotsb+x_n=0 $$ -has dimension $\dim V=\big(1-\frac1{p-1}\big)n$ and does not contain any zero-one vector with the obvious exception of the vector $(0,\dotsc,0)$. On the other hand, any subspace $V\le\F^n$ of dimension $d>\big(1-\frac1{p-1}\big)n$ must contain such a vector. To see this, one can choose $n-d$ homogenous linear forms $L_1,\dotsc,L_{n-d}$ such that $V$ is determined by the equations - $$ L_j(x_1,\dotsc,x_n)=0,\ j=1,\dotsc,n-d $$ -and apply the Chevalley-Warning theorem to the polynomials $L_j(x_1^{p-1},\dotsc,x_n^{p-1})$. The sum of the degrees of these polynomials is $(n-d)(p-1)$, which is strictly smaller than $n$ (the number of variables) if $d>\big(1-\frac1{p-1}\big)n$; in this case the polynomials are guaranteed to have a non-zero common root $(x_1,\dotsc,x_n)\in\F^n$, and then the zero-one vector $(x_1^{p-1},\dotsc,x_n^{p-1})$ lies in $V$. -How different is the situation where the field $\F$ gets replaced with a ring, like $\Z/6\Z$? - -What is the largest possible size of a submodule $V<(\Z/6\Z)^n$ that does not contains any zero-one vector, save for the vector $(0,\dotsc,0)$? - -The construction above shows that one can have $|V|=6^{(4/5)n}$ while $V\cap\{0,1\}^n=\{0\}$; is this best possible? -A very basic upper bound can be obtained by observing that if $V\cap\{0,1\}^n=\{0\}$ and $e_1,\dotsc,e_n$ form the "standard basis'' of $(\Z/6\Z)^n$, then $V,e_1+V,e_1+e_2+V,\dotsc,e_1+\dotsb+e_n+V$ are pairwise disjoint; hence $|V|\le 6^n/(n+1)$. -A version of the problem with a non-orthodox notion of size: - -How large must an integer $d$ be to ensure that if $V_2\le\F_2^n$ and $V_3\le\F_3^n$ are linear subspaces with $\dim V_2+\dim V_3\ge d$, then there is a non-zero integer vector $x\in\{0,1\}^n$ satisfying $x\!\pmod 2\in V_2$ and $x\!\pmod 3\in V_3$? - -$\ $ - -Curiously, it is easy to determine precisely the largest possible size of a submodule $V<(\Z/6\Z)^n$ such that $V$ has a trivial intersection with the set $\{0,\pm 1\}^n$. Namely, for any such submodule $V$, every element of $(\Z/6\Z)^n$ has at most one representation as a sum of an element from $V$ and an element from $\{0,1\}^n$. Consequently, $|V|\cdot|\{0,1\}^n|\le|(\Z/6\Z)^n|$; that is, $|V|\le 3^n$. On the other hand, $V:=\{0,2,4\}^n$ intersects $\{0,\pm1\}^n$ trivially, showing that the bound $|V|\le 3^n$ is sharp. -Also, it is easily seen that if $V<(\Z/6\Z)^n$ has a trivial intersection with $\{0,3\}^n$, then $|V|\le 3^n$. As a slightly more difficult exercise, if $V<(\Z/6\Z)^n$ has a trivial intersection with $\{0,2\}^n$ (equivalently, has a trivial intersection with $\{0,4\}^n$), then $|V|\le(2\sqrt 3)^n$. Thus, only the problem where $V\cap\{0,1\}^n=\{0\}$ remains open. - -REPLY [11 votes]: A better upper bound of $5^n$ can be obtained via Haemer's upper bound for the Shannon capacity, applied to the Sperner capacity of a suitable directed graph, along the lines of Croot-Lev-Pach. As $6^{4/5} = 4.193$, this closes a significant chunk of the gap between the upper bound and the lower bound, though by no means all of it. -The method is to observe that, for such a $V$, $$\left\{x,y \in V \mid x-y \in \{0,1\}^n \right\} = \left\{x,y \in V \mid x=y\right\}$$ so the matrix $M$ whose entry $M_{x,y}$ is $0$ if $(x-y) \not\in \{0,1\}^n$ and is $(-1)^{ \sum_i (x_i - y_i)}$ otherwise is equal to the identity matrix when restricted to $V \times V$. Thus the cardinality of $V$ is at most the rank of $M$. Because $M$ is the $n$-fold tensor product of a rank $5$ matrix, its rank is $5^n$, so $|V| \leq 5^n$. -All the exponential loss in this argument comes from the first step where we drop the condition that $V$ is a subspace and remember only that $\{x,y \in V \mid x-y \in \{0,1\}^n\} = \{x,y \in V \mid x=y\}$. The upper bound of $5^n$ on $V$ satisfying this weaker condition is sharp up to a subexponential factor, as can be demonstrated by the set of vectors in $\{0,1,2,3,4\}^n$ that sum to $2n$, which satisfies that condition and has size $\sim 5^n / \sqrt{n}$.<|endoftext|> -TITLE: Lower bound for $\frac{\sum_{i,j}\min((f_i-f_j)^2,(g_i-g_j)^2)}{\sum_{i,j}\max((f_i-f_j)^2,(g_i-g_j)^2)}$ -QUESTION [8 upvotes]: Let $f\in\mathbb{R}^n$ and $g\in\mathbb{R}^n$ be two orthogonal unit vectors such that $\sum_{i}{f_i}=\sum_{i}{g_i}=0$. - -Question. Can we prove this? - $$\frac{\sum_{\{i,j\}}\min((f_i-f_j)^2,(g_i-g_j)^2)}{\sum_{\{i,j\}}\max((f_i-f_j)^2,(g_i-g_j)^2)} \ge \frac{1}{2n-1}$$ - -REPLY [2 votes]: The example below shows that the bounds are tight. -Take the vectors $v=[1,2,\dots,(n-2),-\binom{n-1}2,0]$ and $w=[-1,-1,\dots,-1,(n-1)]$, then normalize them, with $\Vert v\Vert^2=\frac{3n-5}2\binom{n}3$ and $\Vert w\Vert^2=n(n-1)$, to get -$$f=\frac{v}{\Vert v\Vert}\qquad \text{and} \qquad -g=\frac{w}{\Vert w\Vert}.$$ -Then, $\sum_{i,j}\max_{i,j}\{(f_i-f_j)^2,(g_i-g_j)^2\}$ equals -\begin{align} 2\sum_{1\leq i -TITLE: group actions in dimension 2 and 3 -QUESTION [9 upvotes]: I am looking for a reference to the following claims: - -Any compact group (connected or not) acting on $S^2$ is differentiably conjugate to a linear action. This must be classical. -A circle $S^1$ acting on $RP^3$ (and supposedly any spherical space form) is differentiably conjugate to a linear action. -This is probably true for every compact group acting on a $3$-dimensional spherical space form? - -Wolfgang Ziller - -REPLY [7 votes]: For (1), the usual trick is to average a Riemannian metric (using -Haar measure on the group) to get a fixed metric. Hence the group -will be a subgroup of $SO(3)$ (e.g., by uniformization, the metric -on $S^2$ is conformal to the round $S^2$, so the group must act -by conformal transformations, i.e. $PSL_2(\mathbb{C})$. The -maximal compact subgroup of $PSL_2(\mathbb{C})$ is $SO(3)$). -For (2), the quotient space is a 2-dimensional orbifold (maybe -with boundary corresponding to the fixed locus). Away from the -boundary, it will be a Seifert fibration, and one may use -the classification of Seifert fibrations to see that the action -is conjugate to a linear action.<|endoftext|> -TITLE: Why are Fuchsian groups interesting? -QUESTION [28 upvotes]: I keep hearing that fuchsian groups are interesting for other reasons than the Fuchsian model for hyperbolic Riemann surfaces. -What are those reasons? -Are the Fuchsian groups with fixed points interesting from a geometric perspective? -Where do Fuchsian groups appear besides hyperbolic geometry? -I also read somewhere about a relation between fuchsian groups and fractals. Does someone know more about that and/or has a good reference? - -REPLY [7 votes]: Are the Fuchsian groups with fixed points interesting from a geometric perspective? - -Yes, notice $PSL(2,\mathbb{Z})$ fixes points. The fundamental domain in this case is isometric to a hyperbolic structure on the disk with two cone points. -More generally, if a Fuchsian group $G$ has finite co-volume in $H^2$, then $H^2/G$ will be a hyperbolic 2-orbifold. For example, all of the orientation preserving subgroups of the hyperbolic triangle groups are Fuchsian groups of this type. -In particular, hyperbolic triangle groups have a unique hyperbolic structure, so if 3-manifold has quotient which contains a (orientable) hyperbolic triangle orbifold then the original 3-manifold contains a totally geodesic surface.<|endoftext|> -TITLE: Could the Riemann zeta function be a solution for a known differential equation? -QUESTION [43 upvotes]: Riemann zeta function is a function of complex variable $s$ that analytically continous the sum of Dirichlet series .defined as :$$\zeta(s)=\sum_{n=1}^{\infty}\displaystyle \frac{1}{n^s} $$ for when the real part is greater than $1$. - - -My question here is: - Could the Riemann zeta function be a solution for a known differential equation? - - -Note: I would like if there is a paper or ref show that zeta function presented a solution for known Differential equation. - -REPLY [21 votes]: The Riemann zeta function is "hypertranscendental" in the sense shown HERE -It is not the solution $y(x)$ of a differential equation of the form -$$ -F(x,y,y',y'',\dots,y^{(n)})=0 -$$ -where $F$ is a polynomial (with constant coefficients). - -REPLY [20 votes]: The fact that $\zeta$ satisfy no algebraic differential equation is due to its famous relation with the Gamma function which was proved by Hölder not to satisfy such an equation. -Detailed answer can be found here with five (commented and linked) references. -On the other hand, this function is linked with many other transcendental special functions like polylogarithms which satisfy Fuschian non commutative differential equations.<|endoftext|> -TITLE: Jacobi's elliptic functions and plane sections of a torus -QUESTION [10 upvotes]: In $\mathbb R^3$ with Cartesian coordinates $(x,y,z),$ revolve the circle $(x-\sqrt 2)^2+z^2 =1,\ y=0$ about the $z$-axis. -This yields a torus embedded in $3$-space that is conformally equivalent to the flat torus $(\mathbb R/(2\pi))^2,$ via a mapping that takes $(x,y,z)=(1+\sqrt 2,0,0)$ to $(0,0)$ and $(x,z) = (-1-\sqrt 2,0, 0)$ to $(0,\pi),$ and $(-1+\sqrt 2,0,0)$ to $(\pi,0)$ and $(1-\sqrt 2,0,0)$ to $(\pi,\pi).$ -One of Jacobi's elliptic functions gives us a $2$-to-$1$ mapping from the flat torus into $\mathbb C\cup\{\infty\}$ that takes $(0,0)$ to $0$ and $(0,\pi)$ to $\infty.$ -Then the inverse of the stereographic projection takes $\mathbb C\cup\{\infty\}$ to the sphere $x^2+y^2+z^2=1$ while taking $0$ to $(1,0,0)$ and $\infty$ to $(-1,0,0).$ -The composition of mappings takes the embedded torus to the sphere. -Fact 1: This composition of mappings takes the largest and the smallest of the four $x$-intercepts of the torus respectively to the largest and smallest of the two $x$-intercepts of the sphere. -Fact 2: This composition of mappings is conformal except at four points on the torus (corresponding to points at which the derivative of Jacobi's function is $0$). -(Unless I got some of the nitpicking details wrong about which points map to which) I believe I have reason to suspect the following really startling proposition: - -The intersection of this sphere with each plane containing the $x$-axis is the image under this composition of mappings of the intersection of the torus with the same plane. - -So I'm wondering: - -Is this true? -Is it known? -Can it be proved by using some known fact about Jacobi's function and maybe some elbow grease? - -APPENDIX: The inverse of the conformal mapping from the embedded torus to the flat torus is this: -\begin{align} -\text{For }\alpha,\beta\in\mathbb R/(2\pi), \\[4pt] -x & = \cos\beta/(\sqrt2-\cos\alpha) \\[4pt] -y & = \sin\beta/(\sqrt2-\cos\alpha) \\[4pt] -z & = \sin\alpha/(\sqrt2-\cos\alpha) \\[4pt] -& {} \qquad \uparrow \\ -& {} \qquad \text{same denominator in all three coordinates} -\end{align} -POSTSCRIPT: If I'm not mistaken, I've reduced this to the following. -You have a doubly periodic holomorphic function with periods $2\pi$ and $2\pi i$ that takes $0$ and $\pi + i\pi$ to $0$ and takes $\pi$ and $i\pi$ to $\infty.$ -Now look at the inverse-images of straight lines through $0$ corresponding to arguments (i.e. angles from the real axis) between $0$ and $45^\circ$. Those inverse-images should be graphs of functions of the form $y=\arcsin(c\sin(x))$ where $x$ and $y$ are respectively the real and imaginary parts. -Is that a known proposition about Jacobi's functions? -Postscript: Greg Egan created a visual illustration of what I described above. What is seen below is a compressed version. His original looks better. But it still moves as fast as this one. A version that moves much more slowly would be better. - -REPLY [12 votes]: This is more comment and question than answer, but I'm writing it as an answer because it's too long to fit as a comment. -Michael Hardy kindly pointed me to a reference on the original mapping between the plane and the sphere: -“A Quincuncial Projection of the Sphere”, C. S. Peirce, American Journal of Mathematics, Vol. 2, No. 4 (Dec., 1879), pp. 394-396 -Here, Peirce defines the mapping between a point $(x,y)$ in the plane and a point on the sphere with co-latitude $p$ and longitude $\theta$ as: -$$\operatorname{cn}\left(x + i y \mid \frac{1}{2}\right) = \tan\left(\frac{p}{2} \right)\left( \cos(\theta) + i \sin(\theta) \right)$$ -This $\operatorname{cn}$ function has periods $4 K$ and $2(1+i)K$, where $K=F\left(\frac{1}{2}\right)$. It has zeroes at $(2q+1) K + 2r K i$ and poles at $2q K + (2r+1) K i$, for integers $q, r$. To be clear, the parameters for the Jacobi $\text{cn}$ function and the complete elliptic integral $F$ are what is usually called $m$, as opposed to the modulus, usually called $k$, where $m=k^2$. -The tables in Peirce are rescaled to make the factor of $K$ go away, but they also seem to be flipped and translated relative to his initial definition. I can reproduce Peirce’s tables with Mathematica if I take them to be describing $x',y'$ where, in relation to his original $x, y$ we have: -$$x'+ i y' = 1 - \frac{(x + i y)}{K}$$ -Now, Michael Hardy requires a function on $(\mathbb{R}/(2\pi))^2$, construed as a square in the complex plane, that has periods $2\pi$ and $2 \pi i$, zeroes at $0$ and $\pi+i \pi$ and poles at $\pi$ and $i \pi$. We can write this as: -$$H(z) = \operatorname{cn}\left(K \left(\left(\frac{1+i}{\pi}\right) z + 1 \right) \mid \frac{1}{2}\right)$$ -He defines a conformal map between a torus in $\mathbb{R}^3$ and this square in the complex plane via: -$$\phi_0^{-1}(\alpha + \beta i) = \frac{(\cos(\beta),\sin(\beta),\sin(\alpha))}{\sqrt{2}-\cos(\alpha)}$$ -We also have the stereographic map, projecting from the south pole of the unit sphere onto a plane through the equator, whose inverse can be written: -$$\psi_0^{-1}\left(\tan\left(\frac{p}{2}\right)\left( \cos(\theta) + i \sin(\theta) \right)\right) = (\cos(\theta) \sin(p), \sin(\theta) \sin(p), \cos(p))$$ -[Edited to add, 21 Feb 2017: I’ve just belatedly noticed that in the question, the stereographic projection is defined with the poles of the unit sphere on the $x$-axis, rather than the usual convention of putting the poles on the $z$-axis. This explains why I was unable to verify the original claims, and ended up reorienting the torus to make things work.] -Now, the conjecture is that there is a family of planes $\scr{P}$ such that for any plane $P \in \scr{P}$ we have: -$$\psi^{-1}(H(\phi(\scr{T} \cap P))) = \scr{S} \cap P$$ -where $\scr{T}$ is the torus, $\scr{S}$ is the sphere, and I've dropped the subscripts $0$ from the maps because I want to allow the original versions to be adjusted slightly. -To see how to adjust these maps, note that we want $\scr{P}$ to include both the $x y$-plane and the $x z$-plane. The stereographic projection of the equator of the sphere is just the unit circle in the complex plane, and the preimage of the unit circle under $H$ consists of the four lines $\alpha=\frac{\pi}{2}$, $\alpha=\frac{3\pi}{2}$, $\beta=\frac{\pi}{2}$ and $\beta=\frac{3\pi}{2}$, where we’re writing points in the fundamental domain as $z=\alpha + \beta i$. -The map $\phi_0^{-1}$ from the fundamental domain to the torus maps the lines $\alpha=0$ and $\alpha=\pi$ to the two circles on the torus that lie in the $x y$-plane. So we need to define: -$$\phi^{-1}(\alpha + \beta i) = \phi_0^{-1}\left(\left(\alpha + \frac{\pi}{2}\right) + \beta i\right) = \frac{(\cos(\beta),\sin(\beta),\cos(\alpha))}{\sqrt{2}+\sin(\alpha)}$$ -To accommodate the $x z$-plane, note that the intersection of the torus with this plane corresponds to the lines $\beta=0$ and $\beta=\pi$, the first of which is mapped by the initial linear transformation in $H$ to a line at 45 degrees that passes through the point $z=K$ in the domain of the original Jacobi function, and contains alternating zeroes and poles of the function. If we take a midpoint between a zero and a pole, such as $\frac{3}{2}K+\frac{1}{2}K i$, the Jacobi function evaluates to $-\sqrt{\frac{1}{2}}(1+i)$, which corresponds to a longitude of $\frac{5\pi}{4}$ on the sphere. We can’t remedy this by offsetting $\beta$ in the torus map, because the only lines of constant $\beta$ in the fundamental domain of $H$ that ultimately map to meridians on the sphere are $\beta=0$ and $\beta=\pi$. So we need to redefine the map from the complex plane to the sphere by adding a rotation of $-\frac{\pi}{4}$: -$$\psi^{-1}(z) = \psi_0^{-1}(\exp(-i \pi/4) z)$$ -Having made these changes, let’s see exactly what happens for the two planes. -The lines $\alpha=\frac{\pi}{2}$ and $\alpha=\frac{3\pi}{2}$ clearly map under $\phi^{-1}$ to the two circles in the $x y$-plane where it intersects the torus $\scr{T}$. Under $H$, they map 2-to-1 to two opposite quadrants of the unit circle (those with arguments between $-\frac{\pi}{2}$ and $-\pi$, and between $0$ and $\frac{\pi}{2}$), and then under $\psi^{-1}$ to two opposite quadrants of the equator. So they don't map to the complete equator; the preimage of the full equator in the fundamental domain of $H$ includes $\beta=\frac{\pi}{2}$ and $\beta=\frac{3\pi}{2}$. But certainly they map into the same plane as the image on the torus. -For the lines $\beta=0$ and $\beta=\pi$, it’s easy to see that for the torus $\phi^{-1}$ gives us two circles in the $x z$-plane. Under $H$ and $\psi^{-1}$, the two lines each separately cover the entire circle where the sphere intersects the $x z$-plane. So in this case, the conjecture is exactly true. -However, given the maps that seem to have been forced on us by the need to make things work for these two planes, I can’t see how to make the case $x=\pm z$ work, or $y =\pm z$ (I'm not sure whether the claim that the conjecture is true for the former is a typo and the latter was intended, because the initial conjecture was for planes containing the $x$-axis, and $x=\pm z$ do not contain the $x$-axis). -For the case $y =\pm z$, consider the curve (in fact, a straight line) in the fundamental domain of $H$ given by: -$$\alpha = \cos^{-1}(\gamma)$$ -$$\beta = \sin^{-1}(\gamma)$$ -Since $\sin(\beta)$ and $\cos(\alpha)$ will be equal, the image of this line under $\phi^{-1}$ lies in the plane $y=z$. -Suppose we put $\gamma=\frac{1}{\sqrt{2}}$, giving us the point: -$$p_0 = \frac{(1+i)\pi}{4}$$ -We have: -$$\phi^{-1}(p_0) = \left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$$ -However: -$$H(p_0) = -\sqrt{\sqrt{2}-1} i$$ -$$\psi^{-1}(H(p_0)) = \left(-\sqrt{\sqrt{2}-1},-\sqrt{\sqrt{2}-1},\sqrt{2}-1\right)$$ -which is not in the plane $y=z$. -Edited to add: -These plots might be helpful. They show the $(\alpha, \beta)$ domain, coloured according to which plane the image of each point lies on, when sent either to the sphere or to the torus. Here, I am using the original maps, $\psi_0^{-1} H$ and $\phi_0^{-1}$ to take the $(\alpha, \beta)$ domain to the surfaces of the sphere and the torus. The hue of each point depends on the angle measured around the specified axis in $\mathbb{R}^3$, modulo $\pi$, with points coloured black when the angle lies close to a multiple of $\pi/4$. - - - - - - -Edited to add: Noting the similarity between the plot for the torus with planes around the $x$-axis and the plot for the sphere with planes around the $z$-axis, I've found that the two can be brought into very close alignment by a change of basis for the torus that preserves the axis, but measures the angles in the opposite direction and translates them all by $\pi/4$. To be precise, if we use the standard basis for $\mathbb{R}^3$ for the sphere, and the basis: -$$\left\{\left(0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right),\left(0,-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right),\left(-1,0,0\right)\right\}$$ -for the torus, and then compare the azimuthal angles in polar coordinates adapted to the respective bases, then these angles are very similar, differing from each other by an average of about 0.025. -And the angles agree exactly when they are multiples of $\pi/4$. -I guess this must correspond to Michael Hardy's original observation. -Edited to add: This is what the final map from the torus to the sphere looks like: - -Meridians pulled back from the sphere are in black, lines of latitude in red. The pullbacks of the meridians come very close to lying in the same planes as the original meridians on the sphere, if we define: -$$\phi_1^{-1}(\alpha + \beta i) = \frac{(\frac{1}{\sqrt{2}}(\sin(\alpha)-\sin(\beta)),\frac{1}{\sqrt{2}}(\sin(\alpha)+\sin(\beta)),\cos(\beta))}{\cos(\alpha)-\sqrt{2}}$$ -The 2-to-1 map from the torus to the sphere is then $\psi_0^{-1} H \phi_1$, where $\psi_0^{-1}$ and $H$ are as previously defined.<|endoftext|> -TITLE: Hausdorff approximating measures and Borel sets -QUESTION [5 upvotes]: Suppose $ 1 \leq m \leq n $ are integers and for each $ 0 < \delta < \infty $ let $\mathscr{H}^{m}_{\delta} $ be the size $ \delta $ approximating measure of the $ m $ dimensional Hausdorff measure $ \mathscr{H}^{m} $ of $ \mathbf{R}^{n} $. Recall $ \mathscr{H}^{m}(A) = \sup_{\delta > 0} \mathscr{H}^{m}_{\delta}(A) $ whenever $ A \subseteq \mathbf{R}^{n} $. -Is it true that for every Borel (compact) subset $ B \subset \mathbf{R}^{n} $ there exists $ 0 < \delta < \infty $ such that $ B $ is $ \mathscr{H}^{m}_{\delta} $ measurable? -Of course it is well known (and easy to prove) that if $ m < n $ (the case $ m = n $ is excluded because $ \mathscr{H}^{n}_{\delta} $ is equal the $ n $ dimensional Lebesgue measure for every $ 0 < \delta < \infty $) then for every $ 0 < \delta < \infty $ there exists a Borel set that is not $ \mathscr{H}^{m}_{\delta} $ measurable. For example if $ n = 2 $ and $ m= 1 $ it is not difficult to see that the boundary of an open ball with radius $ \delta/2 $ is not $ \mathscr{H}^{1}_{\delta} $ measurable. - -REPLY [3 votes]: There is a difference between $m = n$ and $m < n$. In case $m=n$, all the outher measures $\mathscr H^n_\delta$ are equal to the Lebesgue measure. So each Borel set in $\mathbb R^n$ is $\mathscr H^n_\delta$-measurable for all $\delta > 0$. -For $m < n$, the following is given as Exercise 2.4.5 in the book "Topics on analysis in metric spaces" by Ambrosio and Tilli. It is written there that this is due to Kirchheim: -If $\delta > 0$ and $A$ is $\mathscr H^m_\delta$-measurable, then $\mathscr H^m(A) = 0$ or $\mathscr H^m(\mathbb R^n \setminus A) = 0$. -As a particular consequence, the closed unit ball in $\mathbb R^n$ is not $\mathscr H^m_\delta$-measurable if $\delta > 0$ and $0 \leq m < n$.<|endoftext|> -TITLE: Advantage of discrepancy -QUESTION [6 upvotes]: In the definition of Minimal model of projective variety, some authors use of discrepancy, and some others omit this condition. I am wondering to know the advantage of discrepancy In the definition of Minimal model. - -REPLY [9 votes]: Take $(X,B)$ and $(Y,B_Y)$, be the log canonical pairs and assume we have birational map $φ:X⟶Y$, we need to take $d(E,X,B)≤d(E,Y,B_Y)$ for any prime divisor $E$ on $X$. In fact discrepancy is measure of singularities, and this condition $d(E,X,B)≤d(E,Y,B_Y)$ saying that the singularities of $Y$ is at least as good as singularities of $X$ -This is equivalent to say that there exists a common resolution $g:W→X$, and $h:W→Y$ s.t, $g^∗(K_X+B)≥h^∗(K_Y+B_Y)$ to preserve sections . -Heuristically we want always there exists a natural map from $$H^0(X,m(K_X+B))→H^0(Y,m(K_Y+B_Y))$$ and if that condition of discrepancy holds then we have such natural injective map<|endoftext|> -TITLE: Does there exist a curve defined over a number field $K$ such that every map to $\mathbb{P}^1_K$ is ramified over at least 4 points? -QUESTION [5 upvotes]: Belyi's theorem says that if $X$ is a curve defined over a number field $K$ then $X_{\overline{K}}$ admits a map to $\mathbb{P}^1_{\overline{K}}$ which at most 3 branch points. Must there also exist a map to $\mathbb{P}^1_K$ defined over $K$? -This may be a slightly lazy question, but I figure it's probably got a quick answer. - -REPLY [2 votes]: To record nfdc23's answer: -The following is true: -If $X$ is a curve over a number field $K$, then it admits a map to $\mathbb{P}^1_K$ defined over $K$ branched over at most 3 points. -This follows from Theorem 2.5 in Mochizuki's 2004 paper here.<|endoftext|> -TITLE: Is the exponential function the sole solution to these equations? -QUESTION [13 upvotes]: Let us take the exponential function $\lambda^z$ where $0 < \lambda < 1$. There are many great uniqueness conditions this holomorphic function satisfies. For example, it is the only function holomorphic in the right half plane, bounded in the right half plane, that interpolates $\lambda^{z} \Big{|}_{\mathbb{N}}$. It is also the only function to satisfy the differential equation $\frac{d}{dz}f(z) = \log(\lambda)f(z)$ when $f(0) =1$. Also it is the only function such that $f(z)f(w) = f(z+w)$ and $f(1) = \lambda$ and $f:\mathbb{R}^+\to\mathbb{R}$. It is also the only function holomorphic such that -$f(z)f(w) = f(z+w)$ and $f(1) = \lambda$ and $f$ has a period of $2 \pi i /\log(\lambda)$. -I'm asking this question because of this similar question on tetration An explicit series representation for the analytic tetration with complex height However when approaching the uniqueness of this tetration function on the real positive line, what is barring our path is a lemma (if you will) about the exponential function. This lemma is rather simple to state, and leads me to believe there must be a solution of this question. -I am not conjecturing this lemma to be true by calling it a lemma (I have a feeling its true); it'd just be nice to see a proof in the positive. If someone has references or suggestions or even a straight proof about why this must be true/not true, that'd be outstanding. The lemma has to do with completely monotonic functions, and is rather tricky. It can be stated plainly as -If a function $F$ is analytic on $\mathbb{R}^+$ and satisfies -$$F(1) =\lambda$$ -$$\lambda F(x) = F(x+1)$$ -$$(-1)^n\frac{d^n}{dx^n}F(x) >0$$ -must $F(x) = \lambda^x$? -This can also be stated, if $F(x) = \lambda^{x}\phi(x)$ for some $1$-periodic function $\phi(x)$, and $(-1)^n\frac{d^n}{dx^n}F(x) > 0$, must $\phi(x)$ be constant? -It can also be stated in the manner I think is most likely to have an answer ready at hand. I also believe this to be the most valuable way of stating the question. -Letting $F$ be analytic on $\mathbb{R}^+$, if $F(1) = e$ and $$\frac{d^n}{dx^n}F(x) > 0$$ $$e\cdot F(x)= F(x+1)$$ must $F(x) = e^{x}$? - -REPLY [9 votes]: Here's a proof that also uses Bernstein's representation theorem for totally monotone functions (I wondered if that result was needed, and Alexander Eremenko's answer now makes me think, that it is). -If $F:\mathbb{R}_+\to\mathbb{R}_+$ satisfies $F(x+1)=eF(x)$, it is of the form $F(x)=\exp(x+h(x))$, with $h$ a $1$-periodic function. But a totally monotonic function is logarithmically convex (see below) and $x+h(x)$ can only be convex if $h$ is a constant (actually zero, if also $F(0)=1$ or $F(1)=e$). -$$*$$ -Totally monotonic functions are logarithmically convex: -From $F(x)=\int_{\mathbb{R}_+}e^{\lambda x}d\mu(\lambda)$, differentiating under the integral sign and applying the Cauchy-Schwarz inequality -$$F'(x)^2=\bigg(\int_{\mathbb{R}_+}\lambda e^{\lambda x}d\mu(\lambda)\bigg)^2=\bigg(\int_{\mathbb{R}_+}\lambda e^{\lambda x/2}\cdot e^{\lambda x/2}d\mu(\lambda)\bigg)^2\le$$$$ -\int_{\mathbb{R}_+}\lambda^2 e^{\lambda x}d\mu(\lambda)\int_{\mathbb{R}_+}e^{\lambda x}d\mu(\lambda)=F''(x)F(x)\ . $$ -Therefore $(F'/F)'\ge0,$ that is $\log F(x)$ is convex.<|endoftext|> -TITLE: Proper coverings as a quotient of the coverings -QUESTION [5 upvotes]: Let $X\neq \emptyset$ be a set. We say that $U\subseteq {\cal P}(X)\setminus \{\emptyset\}$ is a covering of $X$ if $\bigcup U = X$. We call a covering proper if for $a\neq b\in U$ we have $a\not\subseteq b$. -Let $\text{Cov}(X)$ denote the collection of all coverings of $X$. For $A, B\in \text{Cov}(X)$ we set $A\leq B$ if $A$ refines $B$, that is for all $a\in A$ there is $b\in B$ such that $a\subseteq b$. -This relation defines a pre-order on $\text{Cov}(X)$; it is easily seen that it is not anti-symmetric. However, the refinement relation is anti-symmetric on $\text{PropCov}(X)$, the set of proper coverings. -On $\text{Cov}(X)$ we set $A \simeq B$ if $A\leq B$ and $B\leq A$. It is easy to verify that the relation $\leq_q $ on $\text{Cov}(X)/\simeq$ defined by $$[A]_\simeq \leq_q [B]_\simeq \text{ if and only if } A \leq B \text{ in Cov}(X)$$ is a partially ordered set. -Are $(\text{Cov}(X)/\simeq, \leq_q)$ and $(\text{PropCov}(X),\leq)$ isomorphic as posets? - -REPLY [3 votes]: Are $(\text{Cov}(X)/\simeq, \leq_q)$ and -$(\text{PropCov}(X),\leq)$ isomorphic as posets? - -If $X$ is finite, then $(\text{Cov}(X)/\simeq, \leq_q)$ -and $(\text{PropCov}(X),\leq)$ -are isomorphic, but if $X$ is infinite they are not. -To see why they are isomorphic when $X$ is finite it is enough to -note that, in the case of finite $X$, each $\simeq$-class -of covers has the property that all members share -the same maximal elements. Moreover, this common set -of maximal elements constitutes a proper cover, -and it is the unique proper cover in this $\simeq$-class. -This bijection between $\simeq$-classes of covers -and proper covers preserves and reflects order, hence establishes -an isomorphism between $(\text{Cov}(X)/\simeq, \leq_q)$ and -$(\text{PropCov}(X),\leq)$. -To see why $(\text{Cov}(X)/\simeq, \leq_q)$ -and $(\text{PropCov}(X),\leq)$ -are not isomorphic whene $X$ is infinite, -it will suffice to point out that -$(\text{Cov}(X)/\simeq, \leq_q)$ is a lattice -while $(\text{PropCov}(X),\leq)$ is not. - -I first argue the easy part: -$(\text{Cov}(X)/\simeq, \leq_q)$ -is a lattice. -Consider the following closure operator -on the set of covers: -given $B\in \text{Cov}(X)$, let $B^* = \bigcup_{A\leq B} A$. -It is easy to see that $B^*$ is a cover of $X$, and that -an alternative definition for it is -$$ -B^* = (\bigcup_{b\in B} {\mathcal P}(b))\setminus \{\emptyset\}. -$$ -Moreover, $A\leq B$ holds iff $A\subseteq B^*$ iff $A^*\subseteq B^*$, -so $A\simeq B$ iff $A^*=B^*$. -Now, a subset $Z\subseteq {\mathcal P}(X)$ is of the form $Z=B^*$ -iff (i) $\emptyset\notin Z$ and (ii) $Z\cup\{\emptyset\}$ -is an order ideal of ${\mathcal P}(X)$ which contains -all the singleton subsets of $X$. This is enough to show that -$(\text{Cov}(X)/\simeq, \leq_q)$ is isomorphic to the -interval of the lattice of order ideals of ${\mathcal P}(X)$ -above the order ideal consisting of singleton sets. -This is a distributive, algebraic lattice. - -Now the less-easy part. I argue that -$(\text{PropCov}(X),\leq)$ is not a lattice. -In fact, I will write out only the case -$X=\omega$, but afterwards I will explain how to extend the argument to -any superset of $\omega$. - -Let's name some subsets of $\omega$: -$E$: the set $\{0,2,4,\ldots\}$ -of all even natural numbers. -$E_n$: the set $\{0,2,4,\ldots,2n\}$ -of even numbers up to $2n$. -$E_n^*$: the set $\{0,2,4,\ldots,2n,2n+1\}$ -of even numbers up to $2n$ -along with one odd number $2n+1$. -Now let's name some proper covers of $\omega$: -$C_E$: the proper cover consisting of $E$ and all odd singletons. -$C_{E^*}$: The proper cover consisting of all $E_n^*$'s. -$C_n$: The proper cover including $E_n$ and all singletons -of elements from $\omega \setminus E_n$. -It is not hard to see that -(1) $C_E$ and $C_{E^*}$ are incomparable under $\leq$. - -(2) $C_n\leq C_E$ and $C_n\leq C_{E^*}$ for all $n$. -My goal is to show that $C_E$ and $C_{E^*}$ do not have a -meet (= greatest lower bound) in $(\text{PropCov}(X),\leq)$. -Suppose that $D$ is a proper cover of $\omega$ -that is a candidate for the meet -of $C_E$ and $C_{E^*}$ in $(\text{PropCov}(X),\leq)$. -Then $D\leq C_E$, $D\leq C_{E^*}$, and $D$ is above all other -common lower bounds of -$C_E$ and $C_{E^*}$, for example $C_n\leq D$ for all $n$. -Since $E_n=\{0,2,4,\ldots,2n\}\in C_n$, -there must be a set $d_n\in D$ such that $E_n\subseteq d_n$. -Since $D\leq C_E$, it follows that $d_n$ consists of even -integers. -Since $D\leq C_{E^*}$, it follows that $d_n$ is a finite set. -Thus $d_n$ is properly contained in some -$E_m=\{0,2,\ldots,2m\}$ for sufficiently -large $m$. But now repeat this argument to find -some $d_m\in D$ containing $E_m$. We now have -$d_n\subsetneq E_m \subseteq d_m$, contradicting the -properness of $D$. -\\\\\ -Finally, to extend the -proof that $(\text{PropCov}(\omega),\leq)$ is not a lattice -to a proof that $(\text{PropCov}(X),\leq)$ is not a lattice, -for any superset $X\supseteq \omega$, repeat the argument -above, but extend all the covers of $\omega$ which were used -in the argument to covers -of $X$ by adding to those covers -all singletons $\{x\}$, $x\in X\setminus\omega$.<|endoftext|> -TITLE: Mathematical expectation of minimum of k random variables with fixed sum n -QUESTION [6 upvotes]: We have $n$ independent identically distributed random variables $X_1$, $X_2$, ..., $X_N$, $X_i=j$ with probability $1/k$ for $j=1, 2, ... k$. Let $Y_j$ be a number of random variables $X_i$, which are equal to $j$, i.e. $Y_j=|i:X_i=j|$. -I'm interested in mathematical expectation of $\min(Y_1, Y_2, ..., Y_k)$ as $n\to\infty$. -My hypothesis is as follows -$$ -{\mathbf E} \min(Y_1, Y_2, ..., Y_k)=\frac{n}{k}-c_k\sqrt{\frac{n}{k}}+o\left(\sqrt{\frac{n}{k}}\right),\quad n\to\infty. -$$ -I proved that formula for $k=2$ and found $c_2=\frac{1}{\sqrt{\pi}}$. Computer calculation shows that $c_3\approx 0.84$, $c_4\approx 1.02$. - -REPLY [5 votes]: Your formula is correct. -Let $Y^{(n)}_j:=\#\{i\leq n: X_i=j\}$. Then, $Y^{(n)}=(Y_1^{(n)},\dots,Y_k^{(n)})$ can be viewed as a sum of $k$-dimensional i. i. d. random variables, with terms uniform among $(1,0,\dots,0)$, $(0,1,\dots,0)$, .... Note that $$\mathbb{E}Y^{(n)}=\left(\frac{n}{k},\dots,\frac{n}{k}\right),$$ and by Central Limit theorem, $\frac{1}{\sqrt{n}}(Y^{(n)}-\mathbb{E}Y^{(n)})$ converges in distribution to the $k$-dimensional Gaussian $\mathcal{N}(0,\Sigma)$ with covariance matrix given by $\Sigma_{ii}=\frac{1}{k}-\frac{1}{k^2}$, $\Sigma_{ij}=-\frac{1}{k^2}$ for $i\neq j$. -The minimum of the coordinates is a continuous function on $\mathbb{R}^k$. If it were bounded, we could conclude straight away that -$$ -\frac{1}{\sqrt{n}}\left(\min_{1\leq i\leq k}\mathbb{E}Y^{(n)}_i-\frac{n}{k}\right)=\mathbb{E}\min_{1\leq i\leq k}\left[\frac{1}{\sqrt{n}}(Y^{(n)}-\mathbb{E}Y^{(n)})\right]_i\to \mathbb{E}\min_{1\leq i\leq k}[\mathcal{N}(0,\Sigma)]_i, -$$ -which is equivalent to your formula. To complete the proof, you can, for example, truncate the minimum of the coordinates to make it bounded, and then apply Chernoff's bounds to show that the probability to end up in a truncated part is exponentially small as the cut-off goes to infinity, uniformly in $n$.<|endoftext|> -TITLE: Why is generalised complex structure defined to be a reduction of structure group to $O(n,n) \cap Gl(n,\mathbb{C})$? -QUESTION [5 upvotes]: It is a basic and "intuition request" question. I have asked it on StackExchange yet it is probably to specialized for it since there were no answears. -Generalised complex structure is defined to be a field of endomorphisms $\mathcal{J}$ of the big tangent bundle $T^{big}M=TM \oplus T^*M$ such that $\mathcal{J}^2=-I$ and being orthogonal with respect to "natural paring" - neutral metric - $=\xi(Y)+\eta(X)$. The first condition gives reduction of structure group of $T^{big}M$ to complex group, my question is why do we require the second condition? - -REPLY [2 votes]: This condition implies that $J $ preserves the natural pairing. This is quite typical; if one puts a metric on a complex manifold one usually requires the analogous condition to hold, making it a Kahler manifold.<|endoftext|> -TITLE: Closed formula for sine powers -QUESTION [7 upvotes]: I am looking for a closed formula for the expressions -$$ \sum_{k=1}^{n-1} \sin\left(\pi \frac{k}{n}\right)^m,$$ -with $n \in \mathbb{N}$ and $m \in \mathbb{N}$ odd. -Playing with these sums a bit, I got the impression that the result is always a fairly simple algebraic expression, the dependence on $m$, $n$ however looking pretty arbitrary. - -REPLY [5 votes]: Letting $\xi=e^{\frac{2\pi i}{2n}}$, a primitive root of unity $z^{2n}=1$. Then, $\sin\left(\frac{\pi k}n\right)=\frac{\xi^{k}-\xi^{-k}}{2i}$ and hence -\begin{align} -\sum_{k=0}^{n-1}\sin^m\left(\frac{\pi k}n\right) -&=\frac1{(2i)^m}\sum_{k=0}^{n-1}\sum_{j=0}^m\binom{m}j(-1)^j\xi^{-jk}\xi^{(m-j)k} \\ -&=\frac1{(2i)^m}\sum_{j=0}^m(-1)^j\binom{m}j\sum_{k=0}^{n-1}\xi^{(m-2j)k} \\ -&=\frac1{(2i)^m}\sum_{j=0}^m(-1)^j\binom{m}j\frac{\xi^{(m-2j)n}-1}{\xi^{m-2j}-1} \\ -&=\frac1{(2i)^m}\sum_{j=0}^m(-1)^j\binom{m}j\frac{\xi^{mn}-1}{\xi^{m-2j}-1}. -\end{align} -The outcome is parity-dependent. For example, if $m\rightarrow 2m$ is even then the average value is -$$\frac1n\sum_{k=0}^{n-1}\sin^{2m}\left(\frac{\pi k}n\right) -=\frac1{2^{2m}}\binom{2m}m,$$ -as long as $n>m$. This is because $\xi^{2mn}-1=0$ and the only time $\xi^{2m-2j}-1=0$ is when $j=m$ (since $n>m$). Therefore, -$$\sum_{k=0}^{n-1}\sin^{2m}\left(\frac{\pi k}n\right) -=\frac1{(2i)^{2m}}\cdot (-1)^m\binom{2m}m=\frac1{2^{2m}}\binom{2m}m.$$ -By the way, -$$\frac1{\pi}\int_0^{\pi}\sin^{2m}x\,dx=\frac1{2^{2m}}\binom{2m}m.$$ -Contrary to what I thought initially, things are a bit tricky (or unlikely) that there is a closed formula in the case $m\rightarrow 2m+1$ is odd. What we get with the above approach is just an identity: -$$\sum_{k=0}^{n-1}\sin^{2m+1}\left(\frac{\pi k}n\right) -=\frac1{2^{2m}}\sum_{j=0}^m(-1)^j\binom{2m+1}{m-j}\cot\left(\frac{2j+1}{2n}\pi\right).$$<|endoftext|> -TITLE: Does the class of Hausdorff spaces have a shared "Coordinate space"? -QUESTION [15 upvotes]: Given topological spaces $X$ and $C$ we call $C$ a coordinate space for $X$ to mean that every open set $U \subset X$ is of the form $f^{-1}(V)$ for some open $V \subset C$ and continuous $f \colon X \to C$. -More generally if $\mathscr X$ is a class of spaces we call $C$ a coordinate space for $\mathscr X$ to mean it is a coordinate space for every member of $\mathscr X$. -Two familiar examples of coordinate spaces: - -The Seirpinski space $\{0,1\}$ with exactly one closed point and one open point is a coordinate space for the class of all topological spaces. -The closed unit interval is a coordinate space for the class of all completely-regular spaces. Moreover the closed unit interval is itself completely-regular. - -But what about the class of all Hausdorff spaces? Obviously the Seirpinski space is still a coordinate space here. But what if we demand the coordinate space itself be Hausdorff -- as with example 2? -I would reckon such a space does not exist but have no idea how to prove that? -Does this notion already have a name and has the question been addressed before? - -REPLY [3 votes]: A classical notion was considered and solved in an elegant paper by Stanisław Mrówka: (1) instead of arbitrary open subspaces of $\ X\ $ one simply considers just X itself; and (2) instead of a single space $\ C,\ $ one considered the class of all powers $\ C^S.\ $ Then Stanisław Mrówka answered that this kind of embeddability problem is characterized by the property of separating points from closed subsets of $\ X\ $ by functions into $\ C^F,\ $ where this time the exponents $\ f\ $ are just finite sets--a very elegant conceptual theorem. -Acknowledgment--thank you, Will Brian, for pointing to my earlier error.<|endoftext|> -TITLE: A roadmap for understanding perfectoid spaces -QUESTION [23 upvotes]: Perfectoid spaces are this year's subject for the Arizona Winter School (link) and, as preparation, I am currently trying to understand the subject better. There are wonderful explanatory accounts (What is a perfectoid space, What are "perfectoid spaces"?) but I would like to learn the technical details. An additional problem would be that I don't know much (anything) about rigid analytic geometry. I have found many lecture notes on the internet treating non-archimedean geometry, adic-spaces, Berkovich spaces etc but I don't know where to start learning. Moreover, I don't know which things are important and which things can be skipped at first reading. I am familiar with schemes and local fields at the least. -Is there a natural way to progress through the material? - -REPLY [10 votes]: I can tell from personal experience that it is possible to learn perfectoid spaces without knowing rigid geometry, just like it is possible to learn schemes or even stacks without knowing much about varieties over complex numbers. In fact, it's even possible to successfully transition to research with this approach. Of course, for the approach to be meaningful/successful you need some "mathematical maturity" (in the sense of being able to clearly distinguish easy/formal parts of the theory from the real meat); for instance, you need a good command of commutative algebra/algebraic geometry (to say the least). -If you want to pursue this, I suggest reading (more or less line by line) Wedhorn's "Adic spaces" for the basics and then (or in parallel) Scholze's "Perfectoid spaces." -Before wise elders start reprimanding me for giving such "irresponsible" advice, let me issue a couple of caveats: - -This approach is not for everyone. Specifically, if you feel you need "motivating examples" at every step and find it difficult to swallow dry theories for the mere sake of their own intrinsic beauty, then I think you're better off studying in a more linear fashion (i.e., learning rigid geometry first). -You'll need to take some parts of section 2 of "Perfectoid spaces" on faith because you will not know much about Berkovich spaces (but the point is, they are not really needed!). For instance, you will need to ignore the description of the types of points on $(\mathbb{P}^1)^{ad}$ (an overrated example anyway, from my somewhat limited experience). -If you take this approach and want to at some later point transition to research, you have to be acutely aware of the fact that for a long time your intuition will derive from a good command of dry aspects of the theory (or from analogies with related theories that you know well) and not from examples or of concrete computations. In particular, you need to be aware of the danger of making big blunders when forming intuitions in this way about what should be true or how to prove something. Try to continuously bridge the gap as you move on. - -I do agree with the others though that it may be a little bit too late for the AWS to be entirely meaningful. Even if you do not completely understand the lectures, try to isolate the contact points with the material that you've been studying in order to get something out of them.<|endoftext|> -TITLE: Finite groups with all irreducible representations one dimensional -QUESTION [6 upvotes]: Let $k$ be a field of characteristic $p$ and $G$ a finite group. This question might be a dulicate of this question: -Which finite groups have no irreducible representations other than characters? -but I think it is still something that remained to explain. -If $G$ has all irreducible representations one dimensional then $G$ is an extension of a $p$-group (namely $G'$) by an abelian group as Ehud proved in his answer. -It still remains to show that this extension splits in order to deduce that $G$ is a semidirect product. Could you please explain why this extension should split? -I couldn't understand Pablo's the argument with the triangularizable matrices. - -REPLY [2 votes]: I decided to answer myself the question after Geoff Robinson's comments. -From Ehud's answer of the previous question it follows that -$$ -1 \rightarrow G'\rightarrow G \rightarrow G/G'\rightarrow 1 -$$ -is an exact sequence where $G'$ is a p-group and $G/G'$ is abelian. On the other hand, if $G/K$ is abelian and $K$ is a p-group then it can been easily verified that a p-Sylow subgroup of $G$ containing $K$ is normal. -Thus as Kevin Buzzard suggests choosing a p-Sylow subgroup of $G$ containig $G'$ it follows that -$$ -1 \rightarrow P \rightarrow G \rightarrow G/P \rightarrow 1 -$$ -is an exact sequence where $P$ is a $p$-group and $G/P$ is abelian since $P$ contains the commutator subgroup $G'$. -Thus one can conclude that over any field of characteristic $p$ a finite group has all its irreducible representations one dimensional if and only if $G$ has a normal $p$-Sylow subgroup with abelian quotient.<|endoftext|> -TITLE: reference request: rational points on the unit sphere -QUESTION [12 upvotes]: I wonder what would be a good/early reference for the fact: - -rational points on the unit sphere (centered at the origin) are dense. - -Stereographic projection (from a rational point in the sphere) provides a bijection between rational points on the sphere and rational points in euclidean space, where the rationals are dense. (This is a special case of a rational line intersecting a quadric in two rational points) -In many places in the literature the above statement is made, but no reference is given. I am looking for (early) references that provide this fact, perhaps only for the circle or the 2-sphere first. -While related, this question does not quite answer my question; I am looking for early references. - -REPLY [16 votes]: The earliest reference is surely Diophantus' Arithmetica. His "method of adequality" can be used to construct rational points on quadrics that approximate real points arbitrarily well (that is, starting from the existence of a rational point). -This is not of course how Diophantus phrases it, but that is what it comes down to. For example, in Book V, Problem 10, he treats the problem of finding rational $x$ and $y$ satisfying $x^2+y^2=9$ and additionally $x^2>2$ and $y^2>6$. Problem 11 asks for rational $x,y,z$ with $x^2+y^2+z^2=10$, with each of $x^2$, $y^2$, and $z^2$ greater than $3$. Similar problems occur a couple of times more in the same book, and it is easy to satisfy oneself that the method works in the generality described above. Quite an accomplishment for a mathematician working in the Hellenistic era! -For more, see pp. 95-98 of the excellent monograph -Thomas L. Heath. Diophantus of Alexandria; a study in the history of Greek algebra. -as well as the paper -Mikhail G. Katz, David M. Schaps, and Steven Shnider. Almost equal: the method of adequality from Diophantus to Fermat and beyond. Perspectives on Science, Vol. 21, No. 3, pp. 283-324.<|endoftext|> -TITLE: Uniform Faltings -QUESTION [10 upvotes]: Suppose I give you positive integers $g\geq 2$ and $N.$ Is it always possible to find an absolutely irreducible curve of genus $g$ over $\mathbb{Q}$ which has at least $N$ rational points? For that matter, what if $g=2?$ I assume that the answer is YES, but what do I know? - -REPLY [16 votes]: On the contrary, some conjectures suggest that the answer is NO! It follows from the Bombieri-Lang conjecture (sometimes known as Lang's conjectures) that a uniform bound should exist. -More precisely, given a number field $F$ and a genus $g\geq2$ BL implies that there's some bound $N(F,g)$ such that every smooth projective genus $g$ curve over $F$ has at most $N(F,g)$ points. Even better -- the bound depends only on the degree of $F$ over $\mathbb{Q}$, not $F$ itself. -The fact that Bombieri-Lang implies this was proved by Caporaso, (Joe) Harris and Mazur in 1997 and my recollection at the time was that some people regarded this as evidence against Bombieri-Lang rather than for the uniform bound. However, at this stage, the question is open, even for curves of genus 2 over $\mathbb{Q}$. -I once saw a talk of Elkies where he exhibited a curve of genus 2 over $\mathbb{Q}$ with something like 588 rational points, but this may not be the record any longer.<|endoftext|> -TITLE: A curious series related to the asymptotic behavior of the tetration -QUESTION [18 upvotes]: The tetration is denoted $^n a$, where $a$ is called the base and $n$ is called the height, and is defined for $n\in\mathbb N\cup\{-1,\,0\}$ by the recurrence -$$ -{^{-1} a} = 0, \quad {^{n+1} a} = a^{\left({^n a}\right)},\tag1$$ -so that -$${^0 a}=1, \quad {^1 a} = a, \quad {^2 a} = a^a, \quad {^3 a} = a^{a^a}, \, \dots \quad {^n a} = \underbrace{a^{a^{{.^{.^{.^a}}}}}}_{n\,\text{levels}},\tag2$$ -where power towers are evaluated from top to bottom. -Let $a$ be a real number in the interval $1 -TITLE: How do i show that $\displaystyle\frac{\prod_{k=1}^np_k}{\sum_{k=1}^{n}p_k}$ is an integer for finitely many $n$? -QUESTION [8 upvotes]: I have run some computations for some finitely primes to know the nature of the ratio below (the product of the first few primes over the sum of them), specifically if it is an integer for finitely many positive integers $n$. - -My question is: How do i show thiat $\displaystyle\frac{\prod_{k=1}^np_k}{\sum_{k=1}^{n}p_k}$ is an integer for finitely many $n$'s? - -Note 1: $p_1 -TITLE: Connected graphs isomorphic to themselves plus any non-edge? -QUESTION [8 upvotes]: Does there exist an infinite, connected, locally finite graph $G=(V,E)$ such that whenever $u,v$ are distinct vertices of $G$, the graph $G'=(V,E\cup \{\{u,v\}\})$ is isomorphic to $G$? -I conjecture that the answer is no, but I have not been able to find an obvious proof. -If the answer is no, all the assumptions will be needed: infinite since we could otherwise take a complete graph, locally finite since we could otherwise take an infinite complete graph (or more interestingly the Rado graph), and connected since otherwise we could take a disjoint union of infinitely many copies of every finite graph. -Note that it's easy to construct (connected, infinite, locally finite) graphs that are isomorphic to themselves plus a specific non-edge, e.g. a bi-infinite ladder missing all of its positive rungs. - -REPLY [6 votes]: The answer is yes, such a graph does exist, and I'll sketch how to construct such a graph. The ideas for the construction come from the paper: -Bowler, N., Erde, J., Heinig, P., Lehner, F. and Pitz, M., 2016. A counterexample to the reconstruction conjecture for locally finite trees. arXiv preprint arXiv:1606.02926. -Let's suppose we have already built a locally finite connected graph $H_i$, and we have a list of the non-edges (where we've left countably many gaps for later steps) such that for the first $i$ non-edges $e_1,e_2, \ldots e_i$ there is an isomorphism from $H_i \cup e_j$ to $H_i$ for $j \leq i$. -Let us further suppose that there is an infinite set of leaves in $H_i$, coloured red and blue, such that each of the isomorphisms send red leaves to red leaves and blue leaves to blue leaves. The idea will be that, if we extend $H_i$ to a graph $H' \supset H_i$ by only gluing bits on at coloured leaves, and gluing the same bit at each red leaf, and the same bit at each blue leaf, then these isomorphisms will extend to ones from $H' \cup e_j$ to $H'$. -Take the next edge in the ordering $e_{i+1}$ (which we may assume is not adjacent to any red or blue leaf). We will build a locally finite connected graph $H_{i+1} \supset H_i$ such that - -For each $j \leq i+1$ there is an isomorphism from $H_{i+1} \cup e_j$ to $H_{i+1}$, and for $j \leq i$ this extends the isomorphism from $H_i \cup e_j$ to $H_i$. -There is an infinite set of leaves in $H_{i+1}$, coloured green and yellow, such that all the above isomorphisms preserve the colours of the leaves. - -To do so, let us write $\overline{H_i}$ for $H_i \cup e_{i+1}$. We build a graph as follows, we take a copy of $H_i$ and a copy of $\overline{H_i}$ and we join them from a red leaf in $H_i$ to a blue leaf in $\overline{H_i}$, adding a green and yellow leaf in the middle. - -We want to extend this to a graph in such a way that, 'behind' every leaf coloured red in the picture, the same graph appears, and similarly for blue leaves. We can achieve this by recursively sticking a copy of the part of the figure behind the red leaf (so, the path in the middle and $\overline{H_i}$) onto every red leaf in the picture and a copy of the part behind the blue leaf (the path in the middle and $H_i$) onto every blue leaf, where in each copy we keep the colour of the leaves (so that later in the recursion we glue further copies onto those leaves) -After a countable number of steps we obtain a graph $H_{i+1} \supset H_i$. By construction we have glued the same graph behind each red leaf of $H_i$ and similarly behind each blue leaf of $H_i$, and so the isomorphisms from $H_i \cup e_j$ to $H_i$ will extend to isomorphisms from $H_{i+1}$ to $H_{i+1} \cup e_j$. Furthermore, these will preserve the colour of the yellow and green leaves. -However there is also an isomorphism from $H_{i+1} \cup e_{i+1}$ to $H_{i+1}$. Indeed, in the figure if we map $H_i \cup e_{i+1}$ to $\overline{H_i}$ (via the 'identity'), this will preserve the colour of the leaves, and the graph $H_{i+1}$ has the property that the same graph appears `behind' each red leaf, and similarly for the blue leaves, and so this map extends to an isomorphism of $H_{i+1}$. Again, it's easy to see that this isomorphism preserves the colour of yellow and green leaves. -Since at each stage our graphs are locally finite, there are only countably many non-edges, and so we can add the new non-edges to our list, still leaving countably many gaps, in such a way that the next non-edge we want to deal with isn't adjacent to any yellow or green leaf. Then we repeat the process with the next edge $e_{i+2}$ in the list, with the red and blue edges now yellow and green. -In this way we build a sequence of graphs $H_0 \subset H_1 \subset H_2 \ldots$, and if we take the direct limit of this sequence then we claim the resulting graph $H$ will have the desired property. -Indeed, any non-edge $e$ in $H$ was a non-edge in some $H_i$, and so was on our list of non-edges at stage $i$, and so in a later stage, say $H_k$ we ensured that $H_k \cup e$ was isomorphic to $H_k$. However, by our construction there is a `compatible' sequence of isomorphisms from $H_j \cup e$ and $H_j$ for each $j \geq k$ (in that, they restrict down to each other), and so the direct limit of these maps is an isomorphism from $H \cup e$ to $H$.<|endoftext|> -TITLE: does this sum have a limit? -QUESTION [8 upvotes]: Define the sequence given by the finite sum -$$a_n:=\sum_{k=2}^{n+1}\binom{2k}k\binom{n+1}k\frac{k-1}{2^k\binom{4n}k}.$$ - -Questions. -(1) Is $0 -TITLE: Morphisms of principal bundles with different structure groups and associated bundles -QUESTION [7 upvotes]: Consider a pair of principal bundles $P \to M$ and $P' \to M'$ with groups $G$ and $G'$, respectively. A morphism from $P$ to $P'$ is a pair $(\Phi, \phi)$ where $\phi: G \to G'$ is a Lie group homomorphism and $\Phi: P \to P'$ is a fibre bundle map which is equivariant with respect to $\phi$ in the sense that $\Phi(u.g)= \Phi(u).\phi(g)$. This gives us a category $\mathsf{Prin}$ of principal bundles and their homomorphisms. If we require that $G = G'$ and $\phi= \mathrm{id}_G$ then we get the subcategory $\mathrm{Prin}G$ of $G$-principal bundles. -Now, we know that the construction of associated bundles is a bifunctor -$\mathsf{Prin} G \times G\mathsf{Man} \to \mathsf{Bund}$, where $G\mathsf{Man}$ is the category of manifolds with left $G$-action and $\mathsf{Bund}$ is the category of fibre bundles. -Now, suppose that we pick a morphism of principal bundles $(\Phi, \phi)$ as above, where $G$ and $G'$ might differ. Then what is the relationship between the corresponding associated bundle functors? Do we get a natural transformation between the functors, for example $P' \times_{G'} (-) \Rightarrow P \times_{G}\phi^*(-)$, where $\phi^*: G'\mathsf{Man} \to G\mathsf{Man}$ is the restriction functor? If so, under what conditions would this be a natural isomorphism? -For a concrete example, suppose $(\Phi, \phi)$ is a morphism from a $\mathrm{Spin} (n)$ principal bundle $P$ to an $\mathrm{SO}(n)$-principal bundle $P'$, where $\phi: \mathrm{Spin}(n) \twoheadrightarrow \mathrm{SO}(n)$ is the universal covering space. Typically, $P'$ will often be the orthonormal frame bundle of an oriented semi-Riemannian manifold. -My go-to reference for these sort of things is Natural Operations in Differential Geometry by Kolar, Michor, and Slovak, but I couldn't find an answer there. I have this question tagged as spin geometry because I figure people working in this field have thought about this sort of question before. - -REPLY [5 votes]: New answer: Here is a more abstract solution: tensor product of modules and balanced products are both (enriched) coends, defined in exactly the same way. So if you pretend all of the topological groups here are noncommutative rings, and the spaces with group action are (bi)modules, then almost everything here is obvious from basic abstract algebra and the extension-restriction of scalars adjunction. -Why are tensor products and balanced products coends? Well a coend can be used to define as a tensor product of functors. Given a covariant functor $F: \mathsf{C} \to \mathsf{D}$ and a contravariant functor $G: \mathsf{C}^{\mathrm{op}} \to \mathsf{D}$ we can take their coend $F \otimes_\mathsf{C} G$ provided $\mathsf{D}$ is monoidal and sufficiently nice. -A ring $R$ can be thought of as a one-object ringoid $R$, and a left $R$-module $V$ is then precisely an additive functor $V: R \to \mathsf{Ab}$. A right $R$-module $W$ is a contravariant additive functor $W: R^{\mathrm{op}} \to \mathsf{Ab}$. Then the coend $V \otimes_R W$ is precisely the usual tensor product of modules. -Likewise a group is a one-object groupoid, and a left $G$-space $X$ is a functor $X: G \to \mathsf{Top}$. Thinking of them as functors, given a left $G$-space $X$ and a right $G$-space $Y$ their coend is the usual balanced product $X \times_G Y$. -Original answer: -Thanks to Stephen Mitchell's notes on principal bundles, I think I have figured out the relationship. Since any morphism of principal bundles decomposes as a reduction of structure group followed by a pullback, it suffices to consider the case where $\Phi: P \to P'$ is a morphism of principal bundles covering the identity $\mathrm{id}_M$ on the base. As above, $\Phi(u.g)=\Phi(u).\phi(g)$ where $\phi: G \to G'$ is a Lie group homomorphism. -In fact, such a morphism $\Phi: P \to P'$ exists if and only if $P \times_G G' \cong P'$ as $G'$-principal bundles. Indeed, the composite $P \times G \xrightarrow{\Phi \times \mathrm{id}_G} P' \times G \xrightarrow{\rho} P'$ is balanced with respect to $G$, where $\rho$ is the right action on $P'$, so it descends to a map $P \times_G G' \to P'$ out of the balanced product, which is in fact $G'$-equivariant. And since this is a morphism of principal $G'$-bundles covering the identity, it is an isomorphism. The converse is easy to obtain by sticking the $G$-equivariant inclusion $P \xrightarrow{\langle \mathrm{id}_P, e \rangle} P \times G'$ in the obvious spot in the above diagram. -Moreover, it looks like (though I have not had a chance to check yet) that in fact this construction is part of an adjunction. This is analogous to the extension and restriction of scalars adjunction for modules. -Just like the tensor product for bimodules over noncommutative rings, the balanced product is associative up to natural isomorphism. And by analogy with the classical situation in algebra, the group $G$ is the tensor unit for $(-) \times_G (-)$. -It follows that for any $G'$-rep $V$ (or more generally any $G'$-space $X$) that $$P' \times_{G'} V \cong (P \times_G G') \times_{G'} V \cong P \times_G (G' \times_{G'} V) \cong P \times_G \phi^* V.$$ -Since all the isomorphisms here are natural, we have an isomorphism of the associated bundle functors: $P' \times_{G'} (-) \cong P \times_G \phi^* (-)$.<|endoftext|> -TITLE: Inner product of columns of a matrix -QUESTION [6 upvotes]: What information about a (square) matrix we earn if the inner product of its columns are known? - -REPLY [8 votes]: A more geometric supplement to Federico's answer: if you think of columns as vectors, then knowing the inner products determines the set of vectors up to rigid motion, so any geometric information is determined by the inner products (including the volume of the simplex they span [determinant], singular values [the semi-axes of the ellipsoid which is the image of the unit ball by your matrix], the areas of the faces of the simplex determined by them, etc, etc.<|endoftext|> -TITLE: Does minimum of an analytic map restricted to analytic curves implies minimum? -QUESTION [5 upvotes]: Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be an analytic function such that its restriction to any arbitrary analytic curve $\gamma$ passing through the origin $0\in \mathbb{R}^n$ attains a local minimum in $0$ and let say that $f(0)=0$. -Is $0$ a point of local minimum for $f$? If the answer is positive, does the same statement generalize to the case in which $f:\mathbb{H} \rightarrow \mathbb{R}$, where $\mathbb{H}$ is an arbitrary Hilbert space? -To me the problem seems that, a priori, there could be a sequence of points $x_n$ approaching to $0$, laying on a continuous but non analytic curve $\overline{\gamma}$ and such that $f(x_n)>f(0)$. The answer to the first question should be yes and one can see it using some quite strong results on the local structure of the zero set of an analytic map, like Lojasiewicz's theorem. But the problem is that this argument does not generalize to infinite dimensional Hilbert spaces. - -REPLY [5 votes]: As to the generalization to infinite dimensional Hilbert spaces, the lack of compactness makes the property fail for not even too delicate reasons. Consider the cubic functional $$f(x):=\sum_{n\ge1} {1\over n}x_n^2 - \sum_{n\ge1} {x_n^3}\ ,$$ on the Hilbert space $\ell_2$. For any $C^2$ curve $\gamma:(-1,1)\to \ell_2$ with $\gamma(0)=0$ and $\dot\gamma(0)\neq0$ we have -$${d\over dt}f(\gamma(t))=\sum_{n\ge1} \Big({2\over n} \gamma_n\dot\gamma_n -3 \gamma_n^2\dot\gamma_n\Big) ,$$ -$${d^2\over dt^2}f(\gamma(t))=\sum_{n\ge1} \Big({2\over n} \dot\gamma_n^2+{2\over n} \gamma_n\ddot\gamma_n- 3\gamma_n^2\ddot\gamma_n-6\gamma_n \dot\gamma_n^2\Big) \ .$$ -Therefore ${d\over dt}f(\gamma(t)){\big|_{t=0}}=0$ and ${d^2\over dt^2}f(\gamma(t)){\big|_{t=0}}>0$, so $f$ has a strict local minimum in $0$ along $\gamma$. On the other hand, for any $n\in\mathbb{N}$ we have $f\big({2\over n} e_n\big)=-{4\over n^3},$ so that $0$ is not a local minimum on $\ell_2$.<|endoftext|> -TITLE: VC dimension of axis-parallel boxes on the torus -QUESTION [5 upvotes]: First the basic definitions: Let $H$ be a family of sets, and let $P$ be a set of points. Then $H$ is said to shatter $P$ if $\{ h \cap P:~h \in H\}=2^P$, that is, if every subset of $P$ can be obtained by intersecting $P$ with an element of $H$. The Vapnik-Chervonenkis dimension of $H$ is the maximal cardinality of a point set $P$ that is shattered by $H$. See also https://en.wikipedia.org/wiki/VC_dimension. -Let $A$ be the family of axis-parallel boxes in the $d$-dimensional unit cube $[0,1]^d$ having one vertex at the origin. It is known that the VC dimension of $A$ is $d$. Let $B$ be the family of all axis-parallel boxes in $[0,1]^d$ (not necessarily anchored at the origin). The VC dimension of $B$ is known, it is $2d$. -Now the question: Let $C$ be the class of all axis-parallel boxes on the $d$-dimensional unit torus. You could also thing of $C$ as the class of all sets in $[0,1]^d$ which are the $d$-dimensional Cartesian product of elements of $D$, where $D$ is the collection of all subintervals and all complements of subintervals of $[0,1]^d$. Now, what is the VC dimension of $C$? - -REPLY [2 votes]: Update (May 5, 2020) -Gillibert, Lachmann, and Müllner have just posted an arXiv preprint in which they show that the VC-dimension of $C$ is on the order of $d\log_2 d$. So they prove that the VC-dimension of $C$ is not linear in $d$, which is remarkable given the behavior of similarly defined geometric set systems in $\mathbb{R}^d$. -Their proof techniques appear quite involved, especially the lower bound, which is arguably the more interesting part. But I wanted to point to Lemma 3.1 of their paper, which highlights an easy (and obvious in retrospect) way to both improve the upper bound I gave below, and also eliminate the appeal to Sauer-Shelah in this particular case. -For $1\leq i\leq d$, let $C_i$ be the set of periodic boxes of the form $I_1\times\ldots\times I_d$ such that $I_j=[0,1]$ for all $j\neq i$. So $C=C_1\wedge\ldots\wedge C_d$ in my notation below. The easy part of the Lemma below is that, if $\pi_C$ denotes the shatter function for $C$, then $\pi_C(m)\leq \pi_{C_1}(m)\ldots\pi_{C_d}(m)$ for all $m$. It's also easy to see that $C_i$ has VC-dimension $3$, and thus my original argument used Sauer-Shelah to bound $\pi_{C_i}(m)=(em/3)^3$ (for $m\geq 3$). So we get $\pi_C(m)\leq (em/3)^{3d}$ and one optimizes $(em/3)^{3d}<2^m$ to get my upper bound of $(3+o(1))d\log_2d$ on the VC-dimension of $C$. -But it's easier (and much more sensible) to see directly that $\pi_{C_i}(m)$ grows on the order of $m^2$. In particular, $\pi_{C_i}$ coincides the shatter function of intervals on the circle, and so $\pi_{C_i}(m)$ is the number of subsets of $m$ points on $S^1$, which themselves are cyclic intervals. This is easily bounded above by $m^2$ (I think the precise figure is $m^2-m+2$). -So altogether we have $\pi_C(m)\leq m^{2d}$ and so if $m^{2d}<2^m$ then the VC-dimension of $C$ is less than $m$. If particular, for any $c>2$, the VC-dimension of $C$ is less than $cd\log_2 d$ if $d$ is sufficiently large (depending on $c$). -Another way of putting this is that we get a better bound using the fact that, while each $C_i$ has VC-dimension $3$, it has VC-density $2$. -Still, this argument (like the one below) only uses geometry in dimension $1$. In their preprint, the authors above use more sophisticated techniques to bring down the upper bound to the order of $d\log_2d$. And, of course, their argument for the lower bound is another story. - -Original post (December 10, 2019) -This answer doesn't provide the exact VC-dimension of $C$, but does provide an almost linear upper bound. -The observation that a periodic box is a union of at most $2^d$ boxes could be used get a bound of $(2d^2+o(d^2))2^d$ for the VC-dimension of non-periodic boxes (using the fact that boxes have VC-dimension $2d$). I will use a similar idea to obtain a much better bound (but still not linear in $d$, unfortunately). - -Lemma. Suppose $H_1,\ldots,H_n$ are set systems on a set $X$, each of VC-dimension at most $k$. Suppose $H$ is one of the following two set systems: - -$H_1\vee\ldots\vee H_n:=\{S_1\cup\ldots\cup S_n:S_i\in H_i\}$, or -$H_1\wedge\ldots\wedge H_n:=\{S_1\cap\ldots\cap S_n:S_i\in H_i\}$. - -Then $H$ has VC-dimension at most $k(1+o(1))n\log_2(n)$ (where $o(1)$ depends only on $n$). -Proof. Let $\pi_i$ be the shatter function for $H_i$, i.e., -$$ -\pi_i(m)=\max\{|\{Y\cap S:S\in H_i\}|:Y\subseteq X,~|Y|=m\}. -$$ -Let $\pi$ be the shatter function for $H$. One can show $\pi(m)\leq\pi_1(m)\cdot\ldots\cdot\pi_n(m)$ for any $m$. By the Sauer-Shelah lemma, we have $\pi_i(m)\leq (em/k)^k$ for all $i$ and $m\geq k$. So $\pi(m)\leq (em/k)^{kn}$ for any $m\geq k$. In particular, if $m\geq k$ and $(em/k)^{kn}<2^m$, then $\pi(m)<2^m$, and so the VC-dimension of $H$ is less than $m$. So we just need to optimize $m$ satisfying these inequalities. The following works and is of the form stated in the lemma: -$$ -m:=kn\log_2(cn\log_2(cn)) -$$ -where $c=e+\log_2(e)$. - -So if we use case $(1)$ and write a periodic box as a union of at most $2^d$ boxes, then we get $(2d^2+o(d^2))2^d$. -Instead, we can use case $(2)$ to get a better bound. In particular, for $i\leq d$, let $C_i$ be the set of periodic boxes of the form $I_1\times\ldots\times I_d$ such that $I_j=[0,1]$ for all $j\neq i$. Then the collection $C$ of all periodic boxes is precisely $C_1\wedge\ldots\wedge C_d$. Moreover, each $C_i$ has the same VC-dimension as the set of $1$-dimensional periodic boxes, which is $3$. So altogether, this yields: - -Corollary. The VC-dimension of $C$ is at most $(3+o(1))d\log_2(d)$. - -The precise VC-dimension of $C$ appears to still be an open problem. I only found one paper discussing it, which only conjectures that the VC-dimension is linear in $d$ (but gives no known bounds). One major defect in my argument as that geometry is only being used in dimension $1$, and then the rest is just abstract combinatorics. One would expect intersections of elements from the $C_i$'s to be much better behaved than intersections of arbitrary sets. Probably the bound in the lemma can be improved a little bit, but not to something linear.<|endoftext|> -TITLE: Counterexample to Riesz representation for Hilbert modules -QUESTION [8 upvotes]: For a Hilbert space $H$, the Riesz representation theorem states that $H$ is isomorphic to its dual $H^*$ via $x \mapsto \langle x, -\rangle$. -It is often stated in the literature that this does not work in full generality for a Hilbert module over a $C^*$ algebra. For example, attempts to define the adjoint of a morphism between Hilbert modules run into the lack of such a representation theorem. To avoid this, we insist that a morphism between Hilbert modules admit an adjoint. -Is there a simple counterexample to Riesz representability for Hilbert modules? -To be more precise, take $E$ an Hilbert module over $A$ and $\phi : E \to A$ being $A$ linear. The question is on the existence of an $x$ in $E$ such that $\phi = \langle x, - \rangle$. - -REPLY [5 votes]: Take $A= \mathcal{C}([0,1])$ and $H$ the ideal of $A$ of functions that vanish at $0$. -$H$ is a Hilbert $A$ module (as any ideal, with the natural multiplication of $A$ and the scalar product $(x,y)=x^*y$ of $xy^*$ depending on if you are talking of right or left modules) the inclusion of $H$ into $A$ is a continuous $A$ linear map and it has no adjoint. -Indeed an adjoint would be a map $p$ from $A$ to $H$ such that for all $y$ in $H$ and $x$ in $A$, $y^* p(x) = y^*x$ hence $p(1)$ would be an element of $H$ which is a unit for the ideal $H$, that does not exists.<|endoftext|> -TITLE: Where does one go to learn about DG-algebras? -QUESTION [19 upvotes]: The theory of differential graded algebras (in char 0) and their modules has numerous applications in rational homotopy theory as well as algebraic geometry. -I'm looking for a reasonably complete reference book/collection of articles that treat the basic algebraic and homotopical aspects of dg-algebras akin to Atiyah-MacDonald's book on commutative algebra. -Hopefully containing the following topics - -Model structures on dg algebras and simplicial algebras (presenting the "correct" infinity category) and relation between them (concrete description of fibrations and cofibrations). -Resolutions and minimal models -Localizations and Completions -Analogs of the classical types of morphisms (flat, smooth, unramified, etale, open immersion, closed embedding, finite, finite type, etc.). -Derived categories of modules (Triangulated / Stable $\infty$) and functors on them (push-pull-shriek functors, perfect complexes vs. bounded vs. unbounded derived categories, dualizing complex, cotangent complex, localization and completion, fourier mukai transforms). -Derived descent theorems - descent for dg-modules in the derived categories. (hopefully making it clear what kind of information goes into glueing a dg-module). -Hochshild and Cyclic (co-)homology -DG lie algebras and Koszul duality - -I realize that one can always go to Lurie's books for a comprehensive and much more general treatment but I hope that a more elementary reference exists. I do believe that some aspects of this theory are a lot less technical than they appear. - -REPLY [7 votes]: There is a study of DG algebra (DG rings, DG modules, DG categories, DG functors) in the book below. With the related derived categories, etc. - -Derived Categories<|endoftext|> -TITLE: If normal with respect to prime base then normal for all bases -QUESTION [9 upvotes]: I tried to find it on internet but couldn't so m asking this here. I want to ask if a number is normal with respect to all prime number base then do we know that it is normal with respect to any base. Obviously we know that if a number is normal with respect to base b then for any $ b^k$ it is normal. -Thanks - -REPLY [9 votes]: Gerry Myerson already answered in the comments, but let me be a little more explicit. Take any number $m\in\mathbb{N}$ with at least two prime factors (for example $6$). Then construct a Cantor set $C=C_m$ similar to the ternary Cantor set, except that instead of keeping two intervals of relative length $1/3$ we keep two intervals of relative length $1/m$ at each step. Alternatively, $C_m$ is the set of points in $[0,1]$ whose base $m$ expansion has only digits $0, m-1$. Clearly, no point in $C$ is remotely normal to base $m$. -On the other hand, $C$ carries a natural measure $\mu$, which can be defined by the property that is assigns the same mass $2^{-k}$ to all $2^k$ intervals in the $k$-th stage of the construction ($\mu$ is also a multiple of Hausdorff measure of the appropriate dimension on $C$). -It is then known that $\mu$-almost all points are normal to any base $p$ such that $\log p/\log m$ is irrational. In particular, this is true for all prime numbers $p$. So $\mu$-almost all points are counterexamples to the question. -The proof of this essentially goes back to Cassels [Cassels, J. W. S. On a problem of Steinhaus about normal numbers. Colloq. Math. 7 1959 95--101], who proved the special case $m=3$, by looking at the Fourier transform of $\mu$ along sequences $p^n \xi$. I am quite sure the proof works for any $m$. See also our paper [Hochman, Michael; Shmerkin, Pablo. Equidistribution from fractal measures. Invent. Math. 202 (2015), no. 1, 427--479.] for more discussion and generalizations. - -REPLY [7 votes]: Wolfgang Schmidt, Über die Normalität von Zahlen zu verschiedenen Basen, Acta Arith. 7 1961/1962 299–309, MR0140482 (25 #3902), according to the review by N. G. de Bruijn, proved this: Let the set of all integers $>1$ be split into two disjoint classes $R$ and $S$, such that for every $n>1$ all powers of $n$ belong to the same class as $n$ itself. Then there exists a continuum of numbers $\xi$ which all have the property that $\xi$ is normal with respect to every $r\in R$ and non-normal with respect to every $s\in S$. -In particular, let $R$ be the set of all primes and prime powers, $S$ the set of all integers with two or more distinct prime factors. Then there is a continuum (that is, an uncountable infinity) of numbers normal to every prime base and not normal to any base that isn't prime.<|endoftext|> -TITLE: Bishop quote stating that axiom of choice is constructively valid -QUESTION [14 upvotes]: This is about constructive mathematics, but it is not a research question. But since it may also be of interest for research mathematicians, I hope this question is appropriate for this forum. As Andrej Bauer writes in his recent paper Five stages of accepting constructive mathematics: - -As exciting as a multiverse of mathematics may be, the working mathematician - has no time to spend their career wandering from one world to another. Nevertheless, - they surely are curious about the newly discovered richness of mathematics, - they welcome ideas coming from unfamiliar worlds, and they strive to make their - own work widely applicable. They will find it easier to accomplish these goals if they - speak the lingua franca of the mathematical multiverse—constructive mathematics. - - -So, I found a quote by mathematician Errett Bishop I don't quite understand. It says: - -A choice function exists in constructive mathematics, because a choice is implied by the very meaning of existence. - -(Wikipedia refers to the book Constructive analysis by Errett Bishop and Douglas S. Bridges, published in 1985.) -Isn't constructive mathematics all about not asserting that something exists until we can construct a witness? But what the axiom of choice is doing is that it states that there is always a choice function, although we can't always construct one. In view of that fact, I think that the axiom of choice is not constructively justified. Could you clarify what Bishop is trying to say in this quote? -Surprisingly, in 1967 (that's well before he said the quote from above), in one of his books, Bishop gives the exercise to prove that the axiom of choice implies the law of excluded middle. So it seems Bishop knew that the axiom of choice is not constructively valid. How does that match historically with the quote from above? - -REPLY [26 votes]: According to the BHK interpretation of intuitionistic logic we have that: - -A proof of $\exists x \in A . \phi(x)$ consists of a pair $(a, p)$ where $a \in A$ and $p$ is a proof of $\phi(a)$. -A proof of $\forall x \in A . \phi(x)$ is a method which takes as input $a \in A$ and outputs a proof of $\phi(a)$. -A proof of $\psi \Rightarrow \xi$ is a method which takes proofs of $\psi$ to proofs of $\xi$. - -Here "method" should be understood as an unspecified, pre-mathematical notion. It could be algorithm, or continuous map, or mental process, or Turing machine, etc. -The axiom of choice can be stated, for any sets $A$, $B$ and relation $\rho$ on $A \times B$, as: -\begin{equation} -(\forall x \in A . \exists y \in B . \rho(x,y)) - \Rightarrow - (\exists f \in B^A . \forall x \in A . \rho(x, f(x)). -\tag{AC} -\end{equation} -This is equivalent to the usual formulation (exercise, or ask if you do not see why). Let us unravel what it means to have a proof of the above principle. First, a proof of -$$\forall x \in A . \exists y \in B . \rho(x,y) \tag{1}$$ -is a method $C$ which takes as input $a \in A$ and outputs a pair $$C(a) = (C_1(a), C_2(a))$$ such that $C_1(a) \in B$ and $C_2(a)$ is a proof of $\rho(a, C_1(a)).$ Second, a proof of -$$\exists f \in B^A . \forall x \in A . \rho(x, f(x)) \tag{2}$$ -is a pair $(g, D)$ such that $g$ is a function from $A$ to $B$ and $D$ is a proof of $\forall x \in A . \rho(x, g(x))$. -Therefore a proof of (AC) above is a method $M$ which takes the method $C$ which proves (1) and outputs a pair $(f, D)$ which proves (2). Is there such an $M$? It looks like we can take $f = C_1$ and $D = C_2$, and viola the axiom of choice is proved constructively! Well, not quite. We were asked to provide a function $f : A \to B$ but we provided a method $C_1$. Is there a difference? That depends on the exact meaning of "method" and "function". There are several possibilities, see below. -The important thing is that now we can understand what Bishop meant by "a choice is implied by the very meaning of existence". If we ignore the difference between "method" and "function" then under the BHK interpretation choice holds because of the constructive meaning of $\exists$: to exist is to construct, and to construct a $y \in B$ depending on $x \in A$ is to give a method/function that constructs, and therefore chooses, for each $x \in A$ a particular $y \in B$. -It remains to consider whether a "method" $C_1$ taking inputs in $A$ and giving outputs in $B$ is the same thing as a function $f : A \to B$. The answer depends on the exact formal setup that we use to express the BHK interpretation: -Martin-Löf type theory -In Martin-Löf type theory there is no difference between "method" and "function", and therefore choice is valid there (but the exact argument outlined above). -Bishop constructive mathematics -In Bishop constructive mathematics a set is given by an explanation of how its elements are constructed, and when two such elements are equal. For instance, a real number is constructed as sequence of rational numbers satisfying the Cauchy condition, and two such sequences are considered equal when they coincide in the usual sense. This means, in particular, that two different constructions may represent the same element (both $n \mapsto 1/n$ and $n \mapsto 2^{-n}$ represent the real number "zero"). -Now, importantly, we distinguish between operations and functions. The former is a mapping from a set $A$ to a set $B$, and the latter a mapping which respects equality (we say that it is extensional). To see the difference, consider the operation from $\mathbb{R}$ to $\mathbb{Q}$ which computes from a given $x \in \mathbb{R}$ a rational $q \in \mathbb{Q}$ such that $x < q$: since $x$ is a Cauchy sequence, we may take $q = x_i + 42$ for a large enough $i$ (which can be determined explicitly once we make our definition of reals a bit more specific). The operation $x \mapsto q$ does not respect equality: by taking a different Cauchy sequence $x'$ which represents the same real, we get a rational upper bound $q'$ which is not equal to $q$. In fact, in Bishop constructive mathematics it is impossible to construct an extensional operation that computes rational upper bounds of reals. -In Bishop constructive mathematics method is understood as operation, and function as extensional operation. Choice is then valid only in some instances, but not in general. In particular, if $A$ has the property that every element is canonically represented by a single construction, then every operation from $A$ to $B$ is automatically extensional, and choice from $A$ to $B$ is valid. An example is $A = \mathbb{N}$ because each natural number is represented by precisely one construction: $0$, $S(0)$, $S(S(0))$, ... - -The moral of the story is that the devil hides important details in the passage from informal, pre-mathematical notions to their mathematically precise formulation.<|endoftext|> -TITLE: an algebra generated by some known series -QUESTION [5 upvotes]: Denote the e.g.f. for the number of (unordered) rooted labeled trees on $n$ nodes by -$$\Phi(x)=\sum_{n\geq1}\frac{n^{n-1}}{n!}x^n.$$ -And, the related series $\Psi(x)=\sum_{n\geq1}\frac{n^n}{n!}x^n$. Designate the operator $D:=x\frac{d}{dx}$. Some properties: -(1) $\Psi=D\Phi$; -(2) both $\Phi$ and $\Psi$ have radius of convergence $\frac1e$; -(3) (from Lagrange) $\Phi(x)$ is the inverse of $x(v)=ve^{-v}$; -(4) $(1-\Phi)(1+\Psi)=1$; -(5) $\Phi^k(x)=k\sum_{n\geq j}\frac{n^{n-k-1}}{(n-k)!}x^n$. - -Questions. -(a) Is there anything interesting that can be said about the algebra $\mathcal{A}=\mathbb{Q}[\Phi,\Psi]$? -(b) In view of (4) above, what about the quotient of $\mathcal{A}$ by the relation $\Phi\Psi-\Psi+\Phi$? - -REPLY [5 votes]: This has been considered by Dimitri Zvonkine, see his article "An algebra of power series arising in the -intersection theory of moduli spaces of curves -and in the enumeration of ramified coverings -of the sphere" (https://arxiv.org/pdf/math/0403092.pdf).<|endoftext|> -TITLE: Geodesics on Homogeneous Spaces of $SU(n)$ -QUESTION [5 upvotes]: Consider the homogeneous space $SU(n)/K$, where $K$ is a sub-group of $SU(n)$ and the bi-invariant metric on $SU(n)$. -What is the appropriate quotient metric on the homogeneous space and what are the geodesics? Clearly, for example, $K=U(n-1)$ gives $\mathbb{C}P^{n-1}$ and the Fubini-Study geodesics which can be expressed in terms of the original geodesics on $SU(n)$. Is any similar formula available for the more general case and is the resulting metric unique (upto constant positive multiple) as in the FS case? -I'm not seeking homogeneous geodesics specifically. - -REPLY [5 votes]: I believe the answers to your and related questions are covered in Ch.X of -[1] Foundations of Differential Geometry, vol. II, Kobayashi, Nomizu -More specifically, let $\mathfrak g:=Lie(SU(n))$ and $\mathfrak k:=Lie(K)$ be Lie algebras of the corresponding groups. Since $SU(n)$ is a compact simple Lie group there is unique (up to multiple) invariant metric $\kappa$ on $\mathfrak g$ coming from the Killing form. -Let us fix a point $p=[eK]\in M=SU(n)/K$; any invariant tensor on $M$ is completely determined by its value at $p$. In particular, invariant metric on $M$ is determined by a metric on $T_p M\simeq \mathfrak g/\mathfrak k$, where isomorphism is given by the infinitesimal action along $X\in\mathfrak g$. One of the ways to equip $\mathfrak g/\mathfrak k$ with a metric is to identify it with a $\kappa$-orthogonal complement of $\mathfrak k$. This way you get so-called naturally reductive homogeneous space. For $K=SU(n-1)$ this gives Fubini-Study metric on $\mathbb P(n-1)$; for $K=S(U(k)\times U(n-k))$ this metric makes $Gr(n,k)$ a symmetric space. -There are two important connections on $M$ with a naturally reductive metric. One is the standard Levi-Civita connection $\nabla^{LC}$ corresponding to the metric defined above, another one is canonical connection $\nabla$ (see Ch X.2 of [1]). Important facts are: - -$\nabla^{LC}$ and $\nabla$ have the same geodesics (Theorem 2.10 and Theorem 3.3 in [1]) -Geodesic of $\nabla$ in the direction $X\in \mathfrak k^\perp\simeq T_p M$ is given by the curve $\exp (tX)p$ (Corollary 2.5 in [1]). -Any invariant tensor field on $M$ is $\nabla$-parallel. - -Finally, I would like to mention that naturally reductive metric is not the only invariant metric on $M$ and other invariant metrics might have different geodesics.<|endoftext|> -TITLE: From a coend over a pair to a coend over the tensor product -QUESTION [5 upvotes]: I'm trying to find a simple proof that a monoid under Day convolution is equivalent to a lax monoidal functor (see nCatLab). For simplicity, consider functors from $C$ to $Set$. For $F$ to be a monoid there has to be a natural transformation: -$ \mu : F \otimes F \to F $ -where the tensor product is the Day convolution: -$F\otimes F := \int^{x y} C(x \otimes y, -) \times F(x) \times F(y)$ -A lax monoidal functor has a natural transformation: -$F(x) \times F(y) \to F(x \otimes y)$ -which we can plug in to get: -$\int^{x y} C(x \otimes y, -) \times F(x \otimes y)$ -If I could only make a "change of variables" from $(x, y)$ to $x\otimes y$, like this: -$\int^{x \otimes y} C(x \otimes y, -) \times F(x \otimes y)$ -I could use the ninja Yoneda lemma to perform the integral and get $F$. How do I convert this handwaving argument to a proof? - -REPLY [8 votes]: This can be seen via a sequence of isomorphisms involving ends and co-ends, using the presentation of natural transformations via ends: -$$ -\begin{array}{cl} - & \int_{xy} Fx \times Fy \to F (x \otimes y) \\ -\cong& \int_{xy} Fx \times F y \to (\int_z C(x \otimes y,z) \to Fz) & (1) \\ -\cong& \int_{xyz} Fx \times Fy \to C(x \otimes y,z) \to Fz & (2) \\ -\cong& \int_{xyz} Fx \times Fy \times C(x \otimes y,z) \to Fz & (3) \\ -\cong& \int_z (\int^{xy} Fx \times Fy \times C(x \otimes y,z)) \to Fz & (4) \\ -\cong& \int_z (F \otimes F)z \to Fz & (5) \\ -\cong& F \otimes F \Rightarrow F & (6) -\end{array} -$$ -where $(1)$ is by Yoneda; $(2)$ is commuting of ends with exponentials; $(3)$ is currying; $(4)$ is currying with ends/co-ends; $(5)$ is the definition of the Day product; and $(6)$ is natural transformations as ends.<|endoftext|> -TITLE: Determined, finite games -QUESTION [7 upvotes]: What is the simplest way to prove that each finite game is also determined? I know that a game is said to be determined if one of the players has a winning strategy. I was hoping to prove by contradiction that assuming in a finite game Player 1 does not have a winning strategy it needs to be the case that Player 2 has the winning strategy, but maybe there is an easier way. - -REPLY [14 votes]: There are numerous proofs of what I call the fundamental theorem of finite -games. -Theorem. (Fundamental theorem of finite games) - In any finite two-player game of perfect information, one of the players has a winning strategy. -Proof 1. Back-propagation through the game tree. Label the -nodes with the player who has a winning strategy from that -position. Every node will get a label, by recursively labeling from the terminal nodes of the tree (which amounts to working backwards from the won-game positions). What is the label on the root node? The winning strategy -is to stay on the nodes with that label.QED -Proof 2. Let $W$ be the nodes in the game tree for which player -I has a winning strategy in the game proceeding from that position. -If this includes the top node, then I wins. Otherwise, it is not difficult to see that player -II can avoid the nodes in W, and therefore win. QED -Proof 3. The theorem amounts to the de Morgan law. The -assertion "player II has a winning strategy" is simply $$\forall -x_1\exists x_2\dots \vec x\text{ is a win for II}.$$ So if player II -does not have a winning strategy, we negate that assertion, and -push the negation through all the quantifiers, by de Morgan's law, -arriving at: $$\exists x_1\forall x_2\dots\vec x\text{ is a win for -I}.$$ And this is what it means for player I to have a winning -strategy. QED -Proof 4. First prove the Gale-Stewart theorem that every open -game is determined, which itself has several classic proofs. For example, -you can see some of them explained in the introduction of my recent -article, - -V. Gitman, J. D. Hamkins, Open determinacy for class -games, -to appear. - -One of these proofs involves transfinite game values, but in the case of bounded-length finite games, all the game values will be finite. Finally, note that any finite game is an open game. QED -Most of these arguments (all except the de Morgan argument) generalize to prove the determinacy of clopen games, rather than merely uniformly bounded-length finite games. A clopen game is a game whose game tree is well-founded, or in other words, every play terminates in a win for one of the players.<|endoftext|> -TITLE: Where is Seminaire Bourbaki on-line? -QUESTION [11 upvotes]: Where is Seminaire Bourbaki on-line? -The reason I ask is that for years it was complicated-enough already to find it in traditional libraries, as it would be catalogued according to its venue or ephemeral names of various aspects of it. -With the advent of the internet, because most of the Sem. Bourb. archive was so old, there was no obvious, easy way to convert it to digital, and, obviously, scanning-in cannot be done overnight. -EDIT: and, to make clear(er) what has been so obvious to me for decades that I'd failed to explain it in the question... for many decades, Sem. Bourb. consisted of not-too-long superb expositions by superb mathematicians of other superb mathematicians' important work. (Although at some point many of these transcripts were in English, not French), as I say to my students, the existence of decades of Sem. Bourb. transcripts is reason enough to learn how to read mathematical French. - -REPLY [5 votes]: Here you have a complete and updated set of resources -https://www.bourbaki.fr/index-en.html - - Elements of mathematics -and, in particular, the two new volumes (Topologie Algébrique and Théories Spectrales) - Séminaire Betty B. -The junior Bourbaki Seminar - N. Bourbaki Seminar -The very one -Previous sessions - Recorded talks of the séminaire N. Bourbaki -(on youtube) - Earlier Texts -Volumes of the Seminar since its creation. -Texts of the Bourbaki Seminar lectures until 2010 (on Numdam) - Poincaré Seminar -If you like Physics.<|endoftext|> -TITLE: Affine spaces as algebras for an operad? -QUESTION [10 upvotes]: (at.algebraic-topology because I don't know who else thinks about operads) -Let $A$ be an affine space, i.e. a torsor over (the abelian group underlying) a vector space $V$ over a field $K$. Then for any $a_1,\dotsc,a_n \in A$ and any $\lambda_1,\dotsc,\lambda_n\in K$ with $\sum \lambda_i = 1$, we can form the weighted average -$$ \lambda_1 a_1 + \dotsb + \lambda_n a_n \in A. $$ -I'm trying to work out exactly what structure this exhibits on $A$. -Let's let $Pr_K(n) \subset K^n$ denote the subset of probability vectors, i.e. $n$-tuples which sum to 1. I believe this forms an operad: the map -$$ Pr_K(n)\times Pr_K(i_1) \times \dotsb \times Pr_K(i_n) \to Pr_K(i_1 + \dotsb + i_n)$$ -sends -$$\langle \{\lambda_j\}_{j=1}^n; v_1,\dotsc, v_n\rangle \mapsto \lambda_1 v_1 \mathbin| \dotsb \mathbin| \lambda_n v_n$$ -where by $|$ I mean "concatenate these tuples". Then I think $A$ should be an algebra over the operad $Pr_K$. -Is this correct? Is it written down somewhere? What's the standard name for this operad? Can you do anything with this perspective? - -REPLY [5 votes]: For $K=\mathbb{R}$, the positive part of your operad (mentioned in Gabriel's comment), and its algebras have been discussed by Tom Leinster and others in connection to entropy. See, for example, -https://golem.ph.utexas.edu/category/2011/05/an_operadic_introduction_to_en.html -http://www.maths.ed.ac.uk/~tl/b.pdf<|endoftext|> -TITLE: Conjecture about harmonic numbers -QUESTION [11 upvotes]: I was inspired by this topic on Math.SE. -Suppose that $H_n = \sum\limits_{k=1}^n \frac{1}{k}$ - $n$th harmonic number. Then -Conjecture - -Let $M$ be a set of all $n$ such that - $$H_n - \lfloor{H_n\rfloor} < \frac{1}{n^{1+\epsilon}}.$$ - Then - $$\forall\epsilon>0 : |M| = \bar\eta(\epsilon) < \infty.$$ - -Picture below illustrates my conjecture, where I have checked this conjecture for $n < 10^6$ for each $\epsilon\in(0,1.1)$ with step 0.01 - -The following picture based on data provided by @GottfriedHelms for $n \approx 10^{100}$ (see answer below). - -REPLY [3 votes]: This is a comment at the comments of Gerhard "Still Computing Oh So Slowly" Paseman's answer, giving just indexes n for more record-holders. -Let $\small h_n$ denote the n'th harmonic number, $\small A_n=\{h_n\}$ its fractional part. The sequence of $A_n$ has a remarkable shape like a sawtooth-curve with increasing wavelength when n is increased and with sharp local minima - a shape which can be exploited when we seek for possible n to be included in M. -I use $\small f(n) = w_n = \{h_n\} \cdot n$ and $\small f(n,\varepsilon)=w_n \cdot n^\varepsilon $ with the OP's condition rewritten as $\small f(n,\varepsilon)<1 $ as criterion for the inclusion of n into the set M . Of course the cardinality of M is limited by an upper bound $ \small n \le N$ with some N that can numerically be handled, so actually we should write explicitely $\small |M(\varepsilon,N)| $ instead of M only. I could manage to use $\small N \approx e^{2000}$ with the help of Pari/GP. -The local minima of $\small A_n$ occur near $\small x_k=e^{k-\gamma}, k \in \mathbb N^+$ and the n to be tested is one of the next integers enclosing the $\small x_k$ - so this exact n must be empirically be determined. (Note that my n are the n+1 used in the comments at the OP denoting the previous high value of A ) -The harmonic numbers $\small h_n$ can be computed in Pari/GP using h(n) = psi(1+n) + Euler ; however, this seems to be limited to something like $ \small n \lt e^{600}$ and so I had to introduce the Euler-McLaurin-formula for the larger n and implemented the switch from one method to the other at n = 1e50 - -The following table shows the first 40 entries of my 2000-row table containing $\small n \approx 10^{840}$ which was possible to compute using Pari/GP in just a couple of seconds. -The basic option according to the focus in the OP's question is to test $\small f(n,\varepsilon) = w_n \cdot n^\varepsilon \lt 1$. For an example with $\small \varepsilon=0.1$ see the 6'th column and the 7'th column allowing to sum to the cardinality of M where we find $\small | M(0.1,2000) | = 7$ with $\small n \le N \approx e^{2000}$ -A second, much nicer, option is to define the function $\small e(n) = -\log_n(w_n) $ and test $\small e(n) \gt \varepsilon$ whether to include this n or not. This is simply possible when looking into the 5'th column and simply compare. -This latter method allows to compute the cardinalities of $\small M(\varepsilon)$ for arbitrary $\varepsilon$ really fast, for instance to create informative scatter- or lineplots, when first a list for $e(n)$ of length $\small N$ is created and then the successful comparisions with the intended $\varepsilon$ are summed to determine the cardinality. - n | h_n | A=frac(h_n) | w=A*n | e(n) | w*n^0.1|in M? - -----------------+--------+-------------+----------+-----------+--------+--- - 1 1.00000 0.00000 0.00000 1.000000 0.00000 1 - 4 2.08333 0.0833333 0.333333 0.792481 0.382899 1 - 11 3.01988 0.0198773 0.218651 0.634006 0.277901 1 - 31 4.02725 0.0272452 0.844601 0.0491822 1.19066 . - 83 5.00207 0.00206827 0.171667 0.398793 0.267051 1 - 227 6.00437 0.00436671 0.991243 0.00162136 1.70523 . - 616 7.00127 0.00127410 0.784844 0.0377178 1.49191 . - 1674 8.00049 0.000485572 0.812848 0.0279149 1.70759 . - 4550 9.00021 0.000208063 0.946686 0.00650460 2.19790 . - 12367 10.0000 0.0000430083 0.531883 0.0670005 1.36472 . - 33617 11.0000 0.0000177086 0.595311 0.0497632 1.68811 . - 91380 12.0000 0.00000305167 0.278861 0.111798 0.873923 1 - 248397 13.0000 0.00000122948 0.305399 0.0954806 1.05775 . - 675214 14.0000 1.36205E-7 0.919678 0.00623806 3.52030 . - 1835421 15.0000 3.78268E-7 0.694281 0.0252988 2.93703 . - 4989191 16.0000 9.54538E-7 0.476237 0.0481002 2.22652 . - 13562027 17.0000 1.48499E-8 0.201395 0.0975770 1.04059 . - 36865412 18.0000 3.71993E-9 0.137137 0.114033 0.783098 1 - 100210581 19.0000 9.73330E-9 0.975380 0.00135314 6.15552 . - 272400600 20.0000 1.61744E-9 0.440592 0.0421997 3.07297 . - 740461601 21.0000 4.01333E-9 0.297172 0.0594162 2.29065 . - 2012783315 22.0000 1.38447E-10 0.278664 0.0596444 2.37389 . - 5471312310 23.0000 1.97920E-11 0.108288 0.0991384 1.01951 . - 14872568831 24.0000 2.27220E-11 0.337935 0.0463183 3.51618 . - 40427833596 25.0000 6.07937E-12 0.245776 0.0574601 2.82623 . - 109894245429 26.0000 7.60776E-12 0.836049 0.00704359 10.6250 . - 298723530401 27.0000 1.82203E-12 0.544283 0.0230213 7.64454 . - 812014744422 28.0000 5.52830E-13 0.448906 0.0292071 6.96806 . - 2207284924203 29.0000 1.00870E-13 0.222650 0.0528504 3.81951 . - 6000022499693 30.0000 2.16954E-14 0.130173 0.0692963 2.46795 . - 16309752131262 31.0000 3.65111E-14 0.595487 0.0170391 12.4772 . - 44334502845080 32.0000 1.81005E-15 0.080247 0.0802804 1.85827 . - 120513673457548 33.0000 4.59281E-15 0.553496 0.0182434 14.1651 . - 327590128640500 34.0000 2.31992E-15 0.759983 0.00821173 21.4950 . - 890482293866031 35.0000 2.71425E-17 0.024169 0.108145 0.755506 1 - 2420581837980561 36.0000 2.55560E-16 0.618603 0.0135588 21.3700 . - 6579823624480555 37.0000 1.20561E-17 0.079326 0.0695767 3.02860 . - 17885814992891026 38.0000 4.26642E-18 0.076308 0.0687541 3.21976 . - 48618685882356024 39.0000 7.23781E-19 0.035189 0.0871101 1.64093 . -132159290357566703 40.0000 2.02186E-18 0.267208 0.0334763 13.7708 . - -The set $\small M(0.1,e^{40})$ (represented by 40 rows) is here the set of all $n$ where $\small f(n,0.1)=w_n \cdot n^{0.1} \lt 1$ . The number of such entries for $\small n=132159290357566703 \approx 10^{17}$ is here 7 . -All numerical tests which I've done indicate that the cardinality of $\small M(\varepsilon \gt 0,\infty)$ is finite and roughly reciprocal to $\small \varepsilon$ and only if $\varepsilon=0$ is surely infinite. - -Pari/GP tools -This is the Pari/GP-program which I used. -Using the functions - h(n) = Euler+if(n<1e50, psi(1+n),log(n)+1/2/n-1/12/n^2+1/120/n^4-1/252/n^6+1/240/n^8) - A(n) = if(n==1,return(0)); frac(h(n)) - w(n) = A(n)*n - e(n) = if(n==1,return(1)); -log(w(n))/log(n) - -The following needs only two steps to find the n with the next local minimum when called with index $\small k \in \mathbb N$ : - {find_n(k)=local(n1,n2,a1,a2); - n1=floor(exp(k-Euler)); n2=n1+1; a1 = A(n1);a2 = A(n2); - if(a1eps) - - \\ apply, note: for long lists we need high internal precision - list = makeList(40) \\ the max N is here about e^40 - print(cardM(0.1,list) ) - - -[table 2]: This are sample data for the OP's plot of the cardinality of $\small M(\varepsilon)$ by the argument $\varepsilon$ where $\small N \approx \exp(250) \approx 10^{108}$ : - eps | c=#M | r=c*eps | c= card(M) for n<= N approx 10^108 - ------+------------------ - 0.01 104 1.040 - 0.02 55 1.100 - 0.03 36 1.080 - 0.04 29 1.160 - 0.05 21 1.050 - 0.06 16 0.960 - 0.07 12 0.840 - 0.08 12 0.960 - 0.09 10 0.900 - 0.10 7 0.700 - 0.11 6 0.660 - 0.12 4 0.480 - 0.13 4 0.520 - 0.14 4 0.560 - 0.15 4 0.600 - 0.16 4 0.640 - 0.17 4 0.680 - 0.18 4 0.720 - 0.19 4 0.760 - 0.20 4 0.800 - - -Pictures -[Picture 1] : I've extended the search-space for $n$ to the range $\small 1 \ldots e^{1000} \approx 10^{434} $ and to check sanity to the range $\small 1 \ldots e^{2000} \approx 10^{868} $. For epsilons $\varepsilon$ from $0.001$ in $200$ steps up to $0.2$ I made the following graph: - -Remarks: for the very small epsilon the increase of the search-space gives slightly higher results, which also illustrates, that for "larger" epsilon the cardinality of $\small M(\varepsilon)$ is finite - -[Picture 2]: Indicates uniformity of the $\small f(n) =w_n = frac(h_n) \cdot n^1$ at the n, where $\small w_n $ has a local minimum (on request of @GerhardPaseman): - -[Picture 3]:It is also convenient to show a rescaling of the $\small f(n)$ so that we can immediately determine the cardinalities $\small |M(\varepsilon)|$ just by counting the number of dots $\small e(n)$ above $\small \varepsilon$. The derivation of the formula is -$$ \small{ \begin{array}{lll} - A(n) &\lt & {1\over n^1\cdot n^\varepsilon} & \text{ from OP}\\ - f(n) \cdot n^\varepsilon &\lt & 1\\ - \ln(f(n)) + \varepsilon \cdot \ln(n) &\lt & 0 \\ - {\ln(f(n)) \over \ln(n)} + \varepsilon &\lt & 0 \\ - \varepsilon & \lt& -\ln_n(f(n)) -\end{array} }$$ -and we simply count, how many dots in the picture show $\small e(n) = - \ln_n(f(n)) \gt \varepsilon$ - - -Data for cardinalities-plot -epsilon |M(eps,N)||M(eps,N)| - N~e^1000 N~e^2000 -------------------------------- -0.0000 1000 2000 -0.0010 633 862 -0.0020 430 482 -0.0030 315 330 -0.0040 258 264 -0.0050 204 205 -0.0060 171 171 -0.0070 151 151 -0.0080 135 135 -0.0090 120 120 -0.0100 109 109 -0.0110 100 100 -0.0120 98 98 -0.0130 89 89 -0.0140 84 84 -0.0150 79 79 -0.0160 73 73 -0.0170 69 69 -0.0180 62 62 -0.0190 57 57 -0.0200 55 55 -0.0210 50 50 -0.0220 47 47 -0.0230 44 44 -0.0240 42 42 -0.0250 41 41 -0.0260 39 39 -0.0270 39 39 -0.0280 37 37 -0.0290 37 37 -0.0300 36 36 -0.0310 36 36 -0.0320 36 36 -0.0330 36 36 -0.0340 34 34 -0.0350 34 34 -0.0360 34 34 -0.0370 33 33 -0.0380 32 32 -0.0390 31 31 -0.0400 29 29 -0.0410 29 29 -0.0420 29 29 -0.0430 26 26 -0.0440 26 26 -0.0450 25 25 -0.0460 25 25 -0.0470 24 24 -0.0480 24 24 -0.0490 23 23 -0.0500 21 21 -0.0510 21 21 -0.0520 21 21 -0.0530 20 20 -0.0540 20 20 -0.0550 20 20 -0.0560 20 20 -0.0570 19 19 -0.0580 18 18 -0.0590 18 18 -0.0600 16 16 -0.0610 16 16 -0.0620 16 16 -0.0630 16 16 -0.0640 16 16 -0.0650 16 16 -0.0660 16 16 -0.0670 16 16 -0.0680 15 15 -0.0690 14 14 -0.0700 12 12 -0.0710 12 12 -0.0720 12 12 -0.0730 12 12 -0.0740 12 12 -0.0750 12 12 -0.0760 12 12 -0.0770 12 12 -0.0780 12 12 -0.0790 12 12 -0.0800 12 12 -0.0810 11 11 -0.0820 11 11 -0.0830 11 11 -0.0840 11 11 -0.0850 11 11 -0.0860 11 11 -0.0870 11 11 -0.0880 10 10 -0.0890 10 10 -0.0900 10 10 -0.0910 10 10 -0.0920 10 10 -0.0930 10 10 -0.0940 10 10 -0.0950 10 10 -0.0960 9 9 -0.0970 9 9 -0.0980 8 8 -0.0990 8 8 -0.1000 7 7 -0.1010 7 7 -0.1020 7 7 -0.1030 7 7 -0.1040 7 7 -0.1050 7 7 -0.1060 7 7 -0.1070 7 7 -0.1080 7 7 -0.1090 6 6 -0.1100 6 6 -0.1110 6 6 -0.1120 5 5 -0.1130 5 5 -0.1140 5 5 -0.1150 4 4 -0.1160 4 4 -0.1170 4 4 -0.1180 4 4 -0.1190 4 4 -0.1200 4 4 -0.1210 4 4 -0.1220 4 4 -0.1230 4 4 -0.1240 4 4 -0.1250 4 4 -0.1260 4 4 -0.1270 4 4 -0.1280 4 4 -0.1290 4 4 -0.1300 4 4 -0.1310 4 4 -0.1320 4 4 -0.1330 4 4 -0.1340 4 4 -0.1350 4 4 -0.1360 4 4 -0.1370 4 4 -0.1380 4 4 -0.1390 4 4 -0.1400 4 4 -0.1410 4 4 -0.1420 4 4 -0.1430 4 4 -0.1440 4 4 -0.1450 4 4 -0.1460 4 4 -0.1470 4 4 -0.1480 4 4 -0.1490 4 4 -0.1500 4 4 -0.1510 4 4 -0.1520 4 4 -0.1530 4 4 -0.1540 4 4 -0.1550 4 4 -0.1560 4 4 -0.1570 4 4 -0.1580 4 4 -0.1590 4 4 -0.1600 4 4 -0.1610 4 4 -0.1620 4 4 -0.1630 4 4 -0.1640 4 4 -0.1650 4 4 -0.1660 4 4 -0.1670 4 4 -0.1680 4 4 -0.1690 4 4 -0.1700 4 4 -0.1710 4 4 -0.1720 4 4 -0.1730 4 4 -0.1740 4 4 -0.1750 4 4 -0.1760 4 4 -0.1770 4 4 -0.1780 4 4 -0.1790 4 4 -0.1800 4 4 -0.1810 4 4 -0.1820 4 4 -0.1830 4 4 -0.1840 4 4 -0.1850 4 4 -0.1860 4 4 -0.1870 4 4 -0.1880 4 4 -0.1890 4 4 -0.1900 4 4 -0.1910 4 4 -0.1920 4 4 -0.1930 4 4 -0.1940 4 4 -0.1950 4 4 -0.1960 4 4 -0.1970 4 4 -0.1980 4 4 -0.1990 4 4<|endoftext|> -TITLE: Exact bin packing the harmonic series: references? -QUESTION [5 upvotes]: Given $n$ and $B_n= \lceil H_n \rceil$, where the latter is the $n$th harmonic number $\sum^n_{i=1} 1/i$, for most $n$ it is easy to pack the first $n$ terms of the harmonic series into $B_n$ many unit bins. An interesting question is for which $n$ one cannot perform such a packing. My guess is there are no such $n$, and I would like to know of references to this problem. -However, even more I would like to know about exact packing. There is no exact packing for $n=4$ or $5$, but (because $1= 1/2 + 1/3 + 1/6$) there are exact packings for other $n \leq 10$. Here a packing is exact if all but (at most) one bin is exactly filled to capacity. -What is the next $n \gt 10$ for which there is an exact packing? -Edit 2017.03.09 GRP: http://oeis.org/A101877 has more information. Call a subset $D$ of positive integers at most $n$ good for $n$ if the harmonic sum formed from $D$ Is floor of $H_n$. Hugo van der Sanden computed some of the good subsets below for $n=24,65,184$ and higher numbers, and Paul Hanna asked the stronger question if every subset good for $n$ had a subset good for (some number close to ) $n/e$. I am asking if there are some subsets good for enough $n$ that contain a chain of successively smaller subsets good for smaller integers so that an exact packing results. Ernie Croot III has shown (as told by Greg Martin in his ArXiv post on Denser Egyptian Fractions) a stronger result which implies that for all but finitely many integers $k$ there is an $n$ and a subset good for $n$ for which $k$ is floor of $H_n$. So far none of these references address the exact question asked above. (Double entendre intended.) -End Edit 2017.03.09 GRP. -Are there any references to this specific problem? -Gerhard "Not Going On A Trip" Paseman, 2017.01.25. - -REPLY [3 votes]: Thanks to Jeremy Rouse, I have developed some heuristics and a nomogram-like approach to computing (bounds on a candidate for) the next smallest $n$ which admits an exact packing. -I call them heuristics, but they are modest extensions of known facts which should be easy to prove. The primary observation is that a prime power cannot occur just once in an integral packing, and if it occurs more than once, it must occur in a cancellative combination. Thus $1/25$ or $1/27$ can't be the only denominators in an integral packing with that high a power of $5$ or $3$, nor can we just have both of $1/25$ and $1/50$, but we can have $1/27$ and $1/54$ together, and we can have $1/50$ and $1/75$ together, or we can have any of the above with enough high powers of the same prime. This follows from looking at the $p$-adic value of $a/b + c/(dp^n)$ for $c$ and $d$ coprime to $p$ and $p^n$ not dividing $b$. -To help in determining which $n$ are feasible for an exact packing, I arrange the positive integers in several rows, with the $i$th row containing those numbers that are both multiples of the $i$th prime and have that prime as its largest prime factor. Part of the table looks like this with prime powers marked: -2 4* 8* 16* - 3 6 9* 12 18 24 27* - 5 10 15 20 25* 30 - 7 14 21 28 - -I then use it to find a good subset of denominators, namely a set of integers at most $n$ whose - reciprocal sum is $B_n - 1$. The bonus is that if this subset contains another which helps form an exact packing into $B_n - 2$ bins, then this good subset helps form an exact packing for $n$. -I can use this to find Jeremy Rouse's solution pretty quickly. As $H_{11}$ is near enough to $3 +1/11$, I first determine unfeasible $n\gt 11$ by noting that if $n\lt 18$, then $9$ cannot be part of a good subset,if $n \lt 24$ this excludes $8$, and if $n\lt 28$, then $7$ and its small multiples are excluded. Similarly, multiples of primes greater than 10 and prime powers greater than $10$ are excluded. -This leads to the conclusion that for $n\lt 18$ the sum of reciprocals of allowed numbers is less than $3$, and exceeds $3$ by $1/24$ only when $n=20$. When $n=24$ we get a potential good subset which excludes $12$ (because $12$ is needed to handle the excess) and Jeremy's solution pops out. -I can use the table also to determine quickly that after $30$, the next feasible number to admit an exact packing must be greater than $51$, primarily by balancing potential members greater than $31$ of a good subset against members less than $31$ which can't be part of a good subset for $n$: if the smaller denominators overpower the larger ones, then that $n$ is not feasible. -I then group the potential members by line, seeing if I can find subsets of each line which may be part of a good subset. Leaving out those found by Jeremy, for $n=66$ I find $16$ (contributing $1/16$), $12,27,36,48,54 (3/16)$, $30,60 (1/20),$ $7$ through $42 (7/20)$, $11,22,33,55,66 (1/5)$, and $13,26,52,65 (3/20)$, which gives an exact packing for $n=66$. -I have not determined if $65,63,$ or $60$ are feasible, but I suspect not as multiples of $17$ and higher primes are excluded already by $n \lt 68$, and the pair $55,66$ seems hard to replace. Note that these examples were found by hand, and that analysis of nice subsets of multiples of primes $p_i$ can be done programmatically. I disagree with Jeremy Rouse regarding the level of intractability of this problem, and thank him again for his inspiring example. -Edit 2017.03.06 GRP: After finding a packing for $n=65$ (leaving $ 21,39,44,$ and $55$ to handle the excess), I find that leaving out $65$ makes it impossible to produce an exact packing for $n=63$, as $13$ needs to be left out of the good subset, and so do a multiple of $7$ and a multiple of $11$, which is too much for the excess over $4$ of the smooth denominators that are candidates for a good packing. So after $30$, the next $n$ to admit an exact packing are $65$ through $82$. -Based on gut feel and some trial nomogramming, I suspect the next lowest admissible $n$ will be around $170$ or greater. So far the solutions above are found by hand. The exploration continues. End Edit 2017.03.06 GRP. -Gerhard "Who Wants To Go Further?" Paseman, 2017.03.03.<|endoftext|> -TITLE: Colimits of cofibrations and homotopy colimits -QUESTION [7 upvotes]: Say C is a left proper model structure. I have a diagram where all maps are cofibrations. Is its colimit a homotopy colimit? -I know this is true for pushouts. Is it true for sequential colimits? Filtered colimits? Sifted colimits? I am most interested in the sequential case. -This question is obviously related to, and is a slight generalization of, this previous question. - -REPLY [8 votes]: In general, this is certainly not true. Take for example a space $X$ with an action by a group $G$. As a group acts by isomorphisms, it acts in particular by cofibrations. But the map $X/G \to X_{hG}\simeq EG \times_G X$ from the orbits to the homotopy orbits is usually not an equivalence if the action is not free. -A sequential colimit of cofibrations is always a homotopy colimit if the first object is cofibrant. This follows for example by the general theory of Reedy model structures: In the Reedy model structure a sequential diagram as above is cofibrant. If your model category is left proper, you can indeed drop the assumption that the first object is cofibrant. -In general, if you have a direct diagram category $\mathcal{D}$ such that for every $X \in \mathcal{D}$ the category of $Y\neq X$ mapping to $X$ has a terminal object or is empty, then the colimit of every $\mathcal{D}$-shaped diagram of cofibrant objects and cofibrations is a homotopy colimit. This follows again by the Reedy model structure on $\mathcal{D}$-shaped diagrams (as the Latching objects are very easy to determine).<|endoftext|> -TITLE: $\sum_{d\leq x} (\mu(d)/d) \log(x/d)$: is (the analogue of) Mertens' conjecture still false? -QUESTION [8 upvotes]: It is known to be false that $\sum_{m\leq x} \mu(m) \leq \sqrt{x}$ for all $x$ (Mertens' conjecture), and it is generally believed that $\lim \sup_{x\to \infty} |M(x)|/\sqrt{x} = \infty$. From the latter, it would follow that $$\lim \sup_{x\to\infty} \sqrt{x} \left|\sum_{m\leq x} \frac{\mu(m)}{m}\right| = \infty,$$ -by partial summation. However, what about a smoothed sum, such as -$$\sum_{m\leq x} \frac{\mu(m)}{m} \log \frac{x}{m}?$$ -Is it clear that -$$\lim \sup_{x\to\infty} \sqrt{x} \left|\sum_{m\leq x} \frac{\mu(m)}{m} \log \frac{x}{m} - 1\right| = \infty?$$ -If one can't deduce the truth of this easily from $\lim \sup_{x\to \infty} |M(x)|/\sqrt{x} = \infty$, I'd be interested in what standard random models imply on the matter. - -REPLY [9 votes]: We have that -\[\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} = \frac{1}{2\pi i} \int_{\sigma_0 - i\infty}^{\sigma_0 + i\infty} \frac{1}{\zeta(s + 1)} \frac{x^s}{s^2} \, ds\] -for $\sigma_0$ sufficiently large; see the bit about Riesz typical means in Section 5.1 of Montgomery and Vaughan. -Now move the contour to the left (I'm ignoring the issue of the horizontal contours - these can be dealt with, though it takes a little effort). We pick up a pole at $s = 0$ with residue $1$. From here, the standard methods (Section 15.1 in Montgomery and Vaughan) show that -\[\limsup_{x \to \infty} \sqrt{x} \left|\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} - 1\right| > 0.\] -EDIT: Lucia is right that the Riemann hypothesis together with Linear Independence hypothesis do not imply that -\[\limsup_{x \to \infty} \sqrt{x} \left|\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} - 1\right| = \infty.\] -Rather, they imply that -\[\limsup_{x \to \infty} \sqrt{x} \left|\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} - 1\right| = \sum_{\rho} \frac{1}{|\zeta'(\rho)| |\rho - 1|^2}.\] -To see this, assume the Riemann hypothesis and the simplicity of the zeroes of $\zeta(s)$, and mimic the proof of Lemma 4 of Ng's paper on the summatory function of the Möbius function to get an explicit expression more or less of the form -\[\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} - 1 = \sum_{\rho} \frac{x^{\rho - 1}}{\zeta'(\rho) (\rho - 1)^2} + o\left(\frac{1}{\sqrt{x}}\right),\] -where the sum is over the nontrivial zeroes of $\zeta(s)$. This gives the bound -\[\limsup_{x \to \infty} \sqrt{x} \left|\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} - 1\right| \leq \sum_{\rho} \frac{1}{|\zeta'(\rho)| |\rho - 1|^2}.\] -If one additionally assumes the Linear Independence hypothesis, then the Kronecker-Weyl equidistribution theorem implies that this is in fact an equality; this method goes back to the work of Ingham. -A conjecture due to Gonek and Hejhal states that -\[J_{-1/2}(T) := \sum_{0 < \gamma < T} \frac{1}{|\zeta'(\rho)|} \asymp T (\log T)^{1/4}.\] -Using methods from random matrix theory, Hughes, Keating, and O'Connell conjecture more generally that $J_k(T) := \sum_{0 < \gamma < T} |\zeta'(\rho)|^{2k} \sim c_k T (\log T)^{(k + 1)^2}$ whenever $\Re(k) > -3/2$, where -\[c_k = \frac{1}{2\pi} \frac{G(k + 2)^2}{G(2k + 3)} \prod_p \left(1 - \frac{1}{p}\right)^{k^2} \sum_{m = 0}^{\infty} \left(\frac{\Gamma(m + k)}{m! \Gamma(k)}\right)^2 \frac{1}{p^m},\] -with $G(z)$ being the Barnes $G$-function. So by partial summation, the Gonek-Hejhal conjecture implies that -\[\sum_{\rho} \frac{1}{|\zeta'(\rho)| |\rho - 1|^2}\] -converges. -I'm not surprised that your numerical calculations show that this is very small; Kotnik and van de Lune do numerical calculations for a similar sum over zeroes (see Table 5 of their paper) and obtain a sequence of partial sums that seem to converge to something very small. -On the other hand, if the Riemann hypothesis is false or if $\zeta(s)$ has a zero of order greater than $1$, then -\[\limsup_{x \to \infty} \sqrt{x} \left|\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} - 1\right| = \infty.\] -(These can both be proved using the methods in Section 15.1 of Montgomery and Vaughan.)<|endoftext|> -TITLE: Existence of solution to these inequalities -QUESTION [10 upvotes]: They say that all mathematics problems eventually reduce to linear algebra or combinatorics. I have reduced mine to proving a solution exists for the following set of inequalities but have no idea how to proceed. -Question: For $n \geq 2$, fix once and for all, a permutation $\tau \in S_n, \tau \neq (1,n)(2,n-1)\cdots$, i.e., $\tau$ isn't the long element. Let -$$ \Delta(\tau) = \{ i \in \{1, 2, \cdots, n-1 \} : \tau(\{ 1, 2, \cdots, i \}) \neq \{ n, n-1, \cdots, n-i+1 \} \}$$ -Do there exist real numbers $b_1 > b_2 > \cdots > b_{n-1} > b_n = 0$ such that the inequalities -$$ \frac{b_1 + b_2 + \cdots + b_i + b_{\tau(1)} + \cdots + b_{\tau(i)}}{2i} > \frac{b_1 + b_2 + \cdots + b_n}{n} \quad \dots \text{Eq. }(i)$$ -hold simultaneously for every $i \in \Delta(\tau)$? -Motiation: My original question is posted here and the above question is the case when $G = SL(n)$, after some simplifications and explicit computation with the Cartan matrix. -Remark: I have explicitly verified this holds for $n \leq 8$. These calculations suggest that the $b_i$'s cannot be independent of $\tau$. - - -MAJOR EDIT: I made a mistake while posing the question. The $\tau$ in Equation $(i)$ should actually be $\tau^{-1}$. The mistake came -because $\sum b_i \varpi_{\tau(i)}$ was wrongly written as -$\sum b_{\tau(i)} \varpi_i$ instead of $\sum b_{\tau^{-1}(i)} \varpi_i$. I will reward the bounty to the -answer by @Pietro Majer but the question I really want answered is: - -Do there exist real numbers $b_1 > b_2 > \cdots > b_{n-1} > b_n = 0$ such that the inequalities -$$ \frac{b_1 + b_2 + \cdots + b_i + b_{\tau^{-1}(1)} + \cdots + b_{\tau^{-1}(i)}}{2i} > \frac{b_1 + b_2 + \cdots + b_n}{n} \quad \dots \text{New Eq. }(i)$$hold simultaneously for every $i \in \Delta(\tau)$? - -I can award some more bounty points for proving the new inequality. - -REPLY [8 votes]: For a given $\tau\in S_n$ and for any $j\in[n]$ define the numbers -$$a_j:=\chi_{\Delta}(j)-\chi_{\Delta}(j-1)-\chi_{\Delta }(n-j)+\chi_{\Delta}(n-j+1),$$ -where $\chi_\Delta:\mathbb{Z}\to\{0,1\}$ denotes the characteristic function of the set $\Delta:=\Delta(\tau)\subset\mathbb{Z}$. Also, with $c:=5n-a_n$, define -$$b_j:=a_j-5j+c.$$ -We may note right away that since $|a_j|\le2$, the $b_j$ are strictly decreasing, and that $b_n=0$, by the choice of the constant $c$. -For any $E\subset[n]$, for simplicity of notation we put - $$ \alpha(E):=\sum_{j\in E} a_j,\qquad \beta(E):=\sum_{j\in E} b_j$$ -(so we may think $ \alpha$ and $\beta$ as discrete signed measures supported in $[n]$). -For $i\in[n]$, summing over $j=1,\dots i$ we have -$$ \alpha([i])=\chi_\Delta(i)-\chi_\Delta(n-i).$$ -Incidentally, for any $i\in[n]$ we have $i\in\Delta$ if and only if, by definition, $\tau([i])\neq[i]+n-i$ thus also, since $\tau$ is bijective, if and only if $\tau([i]^c)\neq([i]+n-i)^c$, that is $\tau([n-i]+i)\neq [n-i]$ or $\tau^{-1}([n-i])\neq [n-i]+i$, which means $n-i \in \Delta^{-1}:=\Delta(\tau^{-1})$. Hence the last formula also writes -$$ \alpha([i])=\chi_{\Delta}(i)-\chi_{\Delta^{-1}}(i).$$ -Also note that, since $n\not\in\Delta$ -$$ \alpha([n])=0,$$ -and $$ \alpha([i]+n-i)=- \alpha([n-i]) =-\chi_\Delta(n-i)+\chi_\Delta(i)= \alpha([i]).$$ -We proceed showing the inequalities on the arithmetic means. -Case I. Assume $i\in\Delta\setminus\Delta^{-1}$. Then by definition of $\Delta^{-1}$, $\tau^{-1}([i])=[i]+n-i$, so that -$$ {\alpha([i])+ \alpha(\tau^{-1}[i])\over 2i}= {\alpha([i])+ \alpha([i]+n-i)\over2i}={\chi_{\Delta}(i)-\chi_{\Delta^{-1}}(i)\over i}={1\over i}>0, $$ -and summing the arithmetic means of $-5j+c$ on the same sets we have plainly -$${\beta([i])+ \beta(\tau^{-1}[i])\over 2i}>{\beta([n])\over n}.$$ -Case II. Assume $i\in\Delta\cap\Delta^{-1}$. Thus $\tau^{-1}([i])\neq[i]+n-i$ and, just because $b_j$ are strictly decreasing -$${\beta([i])+ \beta(\tau^{-1}[i])\over 2i}>{\beta([i])+ \beta([i]+n-i)\over2i}$$ -and since we have $\alpha([i])=\alpha([i]+n-i)=\alpha([n])=0$ because $\chi_{\Delta}(i)=\chi_{\Delta^{-1}}(i)=1$, summing as before the arithmetic means of the affine part of $b_j$, $${\beta([i])+ \beta([i]+n-i)\over2i}={\beta([n])\over n},$$ -concluding the proof.<|endoftext|> -TITLE: implementing Propp-Wilson's Coupling From the Past on Lozenge Tilings of a Hexagon -QUESTION [5 upvotes]: I'm trying to write a program (with javascript or python) that samples a random lozenge tiling of a hexagon with Propp - Wilson's coupling from the past algorithm. I'm quite clear of the framework of the algorithm, but I don't know which is the most efficient way to encode a tiling, or whether different encoding methods would result in non-uniform samplers. -I have at least 3 possible way to encode a lozenge tiling of a hexagon: - -View it as a plane partition, i.e. piling cubes in a 3D room. This is quite intuitive, but not easy to implement the add - remove step. -View it as a non-intersecting path system. Each path is further represented by a 0-1 array. -encode it as a interlacing array (which I have not understand it yet). But I have seen at least 3 people that mentioned this approach. For example here - -So my question is: which is the best way to encode a tiling, or can anyone explain the 3rd approach? - -REPLY [3 votes]: You should represent it as a plane partition. But you shouldn't represent a plane partition as a 3D array of 0/1 for absent/present. Most people think of a partition as a set of natural numbers, not a Ferrers diagram. The Ferrers diagram has more symmetry, but usually the set (or decreasing sequence) of numbers is the better choice. Similarly, Wikipedia defines a plane partition as a 2D array of natural numbers, subject to constraints that correspond to the Ferrers diagram not falling over. This is well suited to a computer: it is a compact representation; the compatibilities are inequalities between adjacent entries; adding a box corresponds to incrementing a number. -CFTP requires that you have a Gibbs sampling Markov chain. The way Gibbs sampling works is that you erase part of the diagram, compute all possible ways of filling it in and uniformly choose from this small set. You could probably create a Gibbs sampling Markov chain from any of your representations whose stationary distribution would be uniform. -But CFTP uses more structure, specifically a lattice structure, a partial order such that sup and inf exist. Plane partitions have this structure by product partial order — one partition is bigger than another if all of its components are bigger. If you want to use a different representation, you must check that it has this structure. -Actually, there is a slightly different representation that might be better and has been popular in the past. In the case of regular partitions, it turns the partition 45 degrees, so that the gravity pulling boxes into the corner is now the $y$ axis. The $x$-axis parameterizes diagonal stacks of boxes, touching corner to corner. We record an array index by $x$ of the heights of these stacks. This restores the symmetry of the axes. Similarly, we can represent a (boxed) plane partition as a hexagonal grid of heights, where height is $x+y+z$.<|endoftext|> -TITLE: State of the art knowledge about homology of $SL_2(k[t,t^{-1}])$ -QUESTION [8 upvotes]: What is the current state of knowledge of the group homology of $SL_2(k[t,t^{-1}])$? - -I am mostly interested in the case $k$ is algebraically closed of characteristic zero. The most recent work I am aware of is these two papers of Knudson from 1996-7: -http://www.ams.org/mathscinet-getitem?mr=1375567 -http://www.ams.org/mathscinet-getitem?mr=1443493 -Has there been any more work in the time since? - -REPLY [7 votes]: There are two papers that could be interesting to you. First, there is a paper of Kevin Hutchinson: - -K. Hutchinson. On the low-dimensional homology of ${\rm SL}_2(k[t,t^{-1}])$. -J. Algebra 425 (2015), 324–366. - -He uses the amalgamation sequence from Knudson's paper and computes the boundary maps to show that second and third homology split (in a way similar to the fundamental theorem of K-theory) - one summand is homology of ${\rm SL}_2(k)$ and the other summand is something else: $K^{\rm MW}_1(k)$ for degree $2$ and an oriented scissors congruence group for degree $3$. -This splitting (with $\mathbb{Z}[1/2]$-coefficients) is generalized to all degrees in - -M. Wendt. Homology of ${\rm SL}_2$ over function fields I: parabolic subcomplexes. to appear in J. Reine Angew. Math. arXiv:1404.5825 - -Proposition 6.11 shows that the homology of ${\rm SL}_2(k[T,T^{-1}])$ splits as a copy of homology of ${\rm SL}_2(k)$ and the remainder in degree $i$ is given by groups of oriented configurations of $i+1$ points on the projective line. The computations in the latter paper are not done using Knudson's amalgamation sequence, but rather by computing the quotient of a product of two trees (for the two valuations at $T=0$ and $T=\infty$) by the action of ${\rm SL}_2(k[T,T^{-1}])$. While this is now a 2-dimensional space (and so more complicated than the tree you have to deal with when using the amalgamation sequence), the stabilizers come out nicer (making the analysis and homology computations for the stabilizers easier than in Knudson's paper). -However, as far as I know, there is not much that can be said about these groups of configurations of points on the projective line in general. The degree 3 groups (4 points on $\mathbb{P}^1$) can be "computed" over $\overline{\mathbb{Q}}$ via K-theory and Borel regulators but even over $\mathbb{C}$ they would be unknown; and the situation is worse for $\geq 5$ points.<|endoftext|> -TITLE: Is the dual of a compact generator also a compact generator of the derived category of a variety? -QUESTION [5 upvotes]: Let $X$ be a variety (or more generally a quasi-compact, separated scheme) and $D(X)$ be the derived category of complexes of $\mathcal{O}_X$-modules with quasi-coherent cohomologies. Let $\mathcal{E}$ be a compact generator of $D(X)$. It is well-known that $\mathcal{E}$ is a perfect complex (Generators and representability Theorm 3.1.1). Moreover it is also well-known that $\mathcal{E}$ is quasi-isomorphic to a strictly perfect complex $\mathcal{S}$, i.e. each $\mathcal{S}^n$ is a locally free $\mathcal{O}_X$-module with finite rank and $\mathcal{S}^n=0$ for $|n|\gg 0$. So without loss of generality we can assume $\mathcal{E}$ itself is strictly perfect. -Now we consider the dual of $\mathcal{E}$, $\mathcal{E}^{\vee}$, i.e. $\mathcal{E}^{\vee,n}=\mathcal{H}om(\mathcal{E}^{-n},\mathcal{O}_X)$ and the differential is defined naturally. In fact $\mathcal{E}^{\vee}$ could be considered as $\mathcal{RH}om(\mathcal{E},\mathcal{O}_X)$, see Stack Project Lemma 21.35.9. - -My question is: is $\mathcal{E}^{\vee}$ also a compact generator of the derived category $D(X)$? Why? - -REPLY [2 votes]: Let me try to sketch an argument (though I am not quite sure in details). -A theorem of Neeman implies that a compact object $M$ in a compactly generated triangulated category $T$ is a generator if and only if the smallest full dense subcategegory $T$ containing $M$ equals with the subcategory of compact objects of $T$. Now, the latter condition remains true when you apply duality (since compactness is preserved by duality; isn't it in your case?).<|endoftext|> -TITLE: Does the category of Lawvere theories have products? -QUESTION [5 upvotes]: I know Law has a tensor product, is closed with respect to that tensor product, and it has coproducts. Does it have products? -My best guess at the cartesian product of Lawvere theories is the "intersection" of the theories: -say $Th_1$ has a sort $X,$ function symbols $f_i\colon X^{n_i} \to X$ and a set of equations $R$; say $Th_2$ has a sort $Y,$ function symbols $g_j\colon Y^{m_j} \to Y$ and a set of equations $S$. -Then the product will have a sort $X\times Y$, function symbols given by all pairs $(f_i, g_j)\colon (X\times Y)^{n_i = m_j} \to (X\times Y)$ with the same signature, and the "intersection" $R \cap S$ of the equations. -For instance, suppose $Th_1$ is the theory of abelian groups with sort $A$ and $Th_2$ is the theory of monoids with sort $M$; then the result would have one sort $A\times M$, function symbols $(+, \cdot)\colon (A\times M) \times (A\times M) \to (A\times M)$ and $(0, e)\colon 1 \to (A\times M)$ subject to associativity and unit laws, but not commutativity. That is, the product is just the theory of monoids. - -REPLY [4 votes]: According to -Fajtlowicz, S. -Birkhoff's theorem in the category of non-indexed algebras. -Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 17 1969 273-275. -a product of algebras and varieties was introduced by W. Narkiewicz. The terminology ``nonindexed product'' was used. It is stated in the above paper that the nonindexed product is the category-theoretic product in the -category of nonindexed algebras. -This is basically the thing that you are describing: the product of the clones of the varieties within the category of clones, or the product of their algebraic theories within the category of algebraic theories. Given varieties $\mathcal U$ and $\mathcal V$, the models of ${\mathcal U}\times {\mathcal V}$ are those algebras isomorphic to a set-theoretical product of some algebra $A\in \mathcal U$ with some algebra $B\in \mathcal V$, whose $n$-ary operations are pairs $(f,g)$ where $f$ is an $n$-ary operation of $A$ and $g$ is an $n$-ary operation of $B$. It is clear how these operations should act on $A\times B$, namely -$$ -(f,g)((a_1,b_1),\ldots,(a_n,b_n)) = (f(a_1,\ldots,a_n),g(b_1,\ldots,b_n)). -$$ -The conjecture that the product of the variety $\mathcal A$ of abelian groups with the variety $\mathcal M$ of monoids is the variety of monoids is not correct. There are two isotypes of monoids of size $2$, but there are $3$ isotypes of algebras in $\mathcal A\times \mathcal M$ that have size $2$. - -Let me edit this to respond to comments: "isotype" = "isomorphism type". To expand on the preceding paragraph, let $A_2$ be the $2$-element group. Let $MA_2$ be the $2$-element monoid that is a group, and let $MS_2$ be the $2$-element monoid that is a semilattice. Let $*$ denote a $1$-element algebra of any type. Then the $3$ isotypes of $2$-element algebras in ${\mathcal A}\times {\mathcal M}$ are $A_2\times *$, $*\times MA_2$ and $*\times MS_2$. Observe that $A_2\times *$ and $*\times MA_2$ are not isomorphic, since there is a binary operation of ${\mathcal A}\times {\mathcal M}$ of the form $(f,g)(x,y)=(x+y,x)$. This agrees with the group operation on $A_2\times *$ but not on $*\times MA_2$. - -Let me edit again to respond (slightly) to a question about identities satisfied by a product ${\mathcal U}\times {\mathcal V}$. Given varieties $\mathcal U$ and $\mathcal V$ and their identities, the identities of ${\mathcal U}\times {\mathcal V}$ have been worked out. I learned about a number of sources from Walter Taylor: -Taylor, Walter The fine spectrum of a variety. Algebra Universalis 5 (1975), no. 2, 263-303. [See Proposition 0.9] -McKenzie, Ralph -On spectra, and the negative solution of the decision problem for identities having a finite nontrivial model. -J. Symbolic Logic 40 (1975), 186–196. -García, O. C.; Taylor, W. The lattice of interpretability types of varieties. Mem. Amer. Math. Soc. 50 (1984), no. 305, v+125 pp. [See Definition preceding Proposition 3.] -I won't list the details here, but will note that this issue (identities defining the product) has a complicated history. Taylor mentioned the names: Newman, -Foster, Pixley, Knoebel, Gratzer, Lakser, Plonka, Hu, Kelenson, Bernardi, Draskovicova, Chang, Jonsson, Tarski, Fajtlowicz, Lawvere, McKenzie, Garcia, Taylor.<|endoftext|> -TITLE: A game: building a bounded degree graph -QUESTION [11 upvotes]: Given positive integers $n$ and $d\leqslant n-1$. Two players build a graph, starting with $n$ vertices and no edges. On each turn, a player joins two yet not joined vertices by an edge. It is forbidden to get a vertex of degree greater than $d$. The player who has no legal move loses. Who wins? -If I am not mistaken or missing other source, this was proposed for $d=2$ and $n=2001$ by Dmitry Maximov to the math circle high school students. It appeared to be harder than expected, but I hope that I may prove that, for $d=2$, for $n=4m+1$ vertices the second player wins and for $n=4m+2$ vertices the first player wins. - -REPLY [2 votes]: When $d=2$, for the purposes of determining the legal moves we only care about whether each component is an isolated vertex, a single edge, or a path of length at least two. This structure is simple enough that we can solve the game efficiently by dynamic programming. -Here's an example in Haskell. -{-- - (vertices, edges, paths) - - We only guard explicitly against there being no path components, - as moves involving edge and vertex components use up those components, - leading to illegal game states that will be detected in the next - round. ---} - -win' (0,0,0) = False -- nothing at all -win' (1,0,0) = False -- isolated vertex -win' (0,1,0) = False -- single edge -win' (v,e,0) = not $ all win [(v-2,e+1,0), -- join two vertices - (v-1,e-1,1), -- extend edge by a vertex - (v,e-2,1)] -- join two edges -win' (v,e,p) = not $ all win [(v,e,p-1), -- close a cycle or join two paths - (v-1,e,p), -- extend path by a vertex - (v,e-1,p), -- extend path by an edge - (v-2,e+1,p), -- join two vertices - (v-1,e-1,p+1), -- extend edge by a vertex - (v,e-2,p+1)] -- join two edges - -cache = [ [ [win' (v,e,p) | p <- [0..] ] | e <- [0..] ] | v <- [0..] ] - -win (v,e,p) | v < 0 || e < 0 || p < 0 = True -- opponent made an illegal move -win (v,e,p) = cache !! v !! e !! p - -If we believe that there are no lingering bugs, the game appears to be decided by parity for $n$ sufficiently large (but not the parity that you would expect from the final state being a union of cycles). -*Main> [win (n,0,0) | n <- [1..100]] -[False,True,True,False,False,True,True,True,False,True, - False,True,False,True,False,True,False,True,False,True, - False,True,False,True,False,True,False,True,False,True, - False,True,False,True,False,True,False,True,False,True, - False,True,False,True,False,True,False,True,False,True, - False,True,False,True,False,True,False,True,False,True, - False,True,False,True,False,True,False,True,False,True, - False,True,False,True,False,True,False,True,False,True, - False,True,False,True,False,True,False,True,False,True, - False,True,False,True,False,True,False,True,False,True] - -This presumably leads to a formal proof by induction.<|endoftext|> -TITLE: (Homotopy)-Fibre of H-spaces -QUESTION [8 upvotes]: By H-space I mean a unital magma in $Ho(\mathsf{Top}_*)$, i.e. the homotopy category of pointed spaces (feel free to use any of your favourite model :) ). -This means that we have a pointed space $X$ and a map $\mu_X:X \times X \to X$ such that $\mu_X \circ j \simeq \nabla$, where $\nabla : X \vee X \to X$ is the fold (or co-diagonal) map, and $j:X \vee X \to X \times X$ is the pushout-product of the map $* \to X$ with itself. -An H-map between H-spaces is a map $f:X \to Y$ of the underlying spaces such that the obvious square involving $f$ and the multiplication maps is homotopy commutative. -Here comes the question: how to prove that the homotopy fibre of $f$ is again an H-space (in a sensible way, of course)? -Homotopy pullbacks of spaces are weak pullbacks in the homotopy category, and that's how you get a multiplication on the fibre compatible with that of $X$. However, I can't seem to find how to check the desired (homotopy) commutativity, as that would seem to require that weak pullback to be a strict one, which is known to be false. -Thanks in advance for any answer/comment! - -REPLY [3 votes]: This is done explicitly by Zabrodsky in his book Hopf Spaces, although I cannot find the exact theorem. -First note that we need a more strict definition of an H-map. An H-map $(f,F):(X,\mu_X)\rightarrow (Y,\mu_Y)$ is a map $f:X\rightarrow Y$ and a homotopy $F:X\times X\rightarrow Y^I$ relative $X\vee X$, satisfying $e_0\circ F=f\circ\mu_x$ and $e_1\circ F=\mu_Y\circ (f\times f)$, where for $a=0,1$, the map $e_a:Y^I\rightarrow Y$ is the evaluation at $a$. -Now the homotopy fibre $F_f$ of $f$ is the (categorical) pullback of $(X\xrightarrow{f} Y\xleftarrow{e_0} PY)$ where $PY=\{l:[0,1]\rightarrow Y\,|\,l(1)=\ast \}$ is the path space over $Y$. That is, $F_f=\{(x,l)\in X\times PY\,|\,f(x)=l(0)\}$. Define $\mu_f:F_f\times F_f\rightarrow F_f$ by -$\mu_f\left((x,l),(y,m)\right)=\left(\mu_X(x,y),F(x,y)+P\mu_Y(l,m)\right)$ -It is straightforward to check that this gives a well defined multiplication on $F_f$. Moreover the canonical projection $F_f\rightarrow X$ and the fibre inclusion $\Omega Y\rightarrow F_f$ are H-maps. The multiplication $\mu_f$ depends on the choice of homotopy $F$. If $\mu_X$, $\mu_Y$ have other properties (homotopy associativity, homotopy commutativity, etc) and $f$ preserves these, then often $\mu_f$ too can be shown to inherit these properties too.<|endoftext|> -TITLE: Is the Hurewicz theorem ever used to compute abelianizations? -QUESTION [32 upvotes]: The Hurewicz theorem tells us that if $X$ is a path-connected space then $H_1(X, \, \mathbb{Z})$ is isomorphic to the abelianisation of $\pi_1(X)$. This gives a potential method for computing the abelianisation of a (sufficiently nice) group $G$: realise it as the fundamental group of a space $X$ and then compute $H_1(X, \, \mathbb{Z})$ by your favourite means. - -Is this method ever used in practice? Are there nice examples of abelianisations which are easily (best?) computed in this way? - -REPLY [13 votes]: The mapping class group of a smooth manifold $M$ is the group of all its self diffeomorphisms up to isotopy, i.e. $\pi_0(\operatorname{Diff}(M))\cong \pi_1(B\operatorname{Diff}(M))$. -A large portion of geometric topology is concerned with gaining a better understanding of the homotopy type of $B\operatorname{Diff}(M)$, in particular of its cohomology as the latter is the ring of characteristic class of smooth $M$ bundles. -In the last 15 years, initiated by Madsen and Weiss' solution of the Mumford conjecture, the understanding of the homology of $B\operatorname{Diff}(M)$ has fundamentally improved which led, among others, to calculations of $H_1(B\operatorname{Diff}(M))\cong \pi_0(\operatorname{Diff}(M))^{ab}$ for some $M$, e.g. by Galatius and Randal-Williams in the following paper. -Abelian quotients of mapping class groups of highly connected manifolds, Mathematische Annalen -June 2016, Volume 365, Issue 1, pp 857–879<|endoftext|> -TITLE: A question on Riemannian manifolds without conjugate points -QUESTION [13 upvotes]: Consider the plane $\mathbb{R}^2$ with a complete metric $g$ without conjugate points. We will denote by $K$ the Gaussian curvature. -Question: Is there a constant $C >0$ such that $K(p) \leq C$, for every $p \in \mathbb{R}^2$? - -REPLY [9 votes]: It boils down to the following statement. - -There a rotationally symmetric $g$ on $\mathbb{R}^2$ without conjugate points, with huge positive curvature at $0$ and curvature $\equiv -1$ outside of the unit ball. - -If such metrics are constructed, then we can cut pieces from it and glue them together along domains of curvature $\equiv -1$. -This way we construct a Riemannian manifold without conjugate points and no upper curvature bounds. -(We need to spread the unit balls far apart, so globally the glued space looks like Lobachevsky plane, and yet make sure no geodesic pass thru 3 balls — this is easy to achieve.) -To construct such examples, -prescribe the curvature depending on the distance to the origin, so that at $0$ it is huge positive, but quickly becomes $-1$ and stay constant. -The no conjugate points condition follows from the Jacobi equation.<|endoftext|> -TITLE: Expected number of lines meeting four given lines or "what is 1.72..." -QUESTION [14 upvotes]: Given four random lines in $\mathbb{R}P^3$, how many lines intersect all of those lines? -In the recent paper Probabilistic Schubert Calculus, Peter Bürgisser and Antonio Lerario -discuss this question and much more general versions of it. -In Proposition 6.7, they determine the expected number of lines meeting four random lines in $\mathbb{R}P^3$ to be: -$$\operatorname{edeg}G(2,4) =\\ 2^{-13}\int_{[0,2\pi]^6}\left| - \det\begin{pmatrix} - \sin{t_1}\sin{s_1}&\sin{t_2}\sin{s_2}&\sin{t_3}\sin{s_3}\\ - \cos{t_1}\sin{s_1}&\cos{t_2}\sin{s_2}&\cos{t_3}\sin{s_3}\\ - \sin{t_1}\cos{s_1}&\sin{t_2}\cos{s_2}&\sin{t_3}\cos{s_3} - \end{pmatrix}\right|dt_1dt_2dt_3ds_1ds_2ds_3$$ -The integrand can be expanded to -$$\left|\cos(s_2)\sin(s_1)\sin(s_3)\sin(t_2)\sin(t_1 - t_3)- \sin(s_2)\big(\cos(s_1)\sin(s_3)\sin(t_1)\sin(t_2 - t_3) + \cos(s_3) -\sin(s_1)\sin(t_3)\sin(t_1 - t_2)\big)\right|$$ -The absolute value in the integrand and the integration over six variables seem to make it difficult to evaluate this integral to high precision. -I am interested in the exact value of this number. Bürgisser and Lerario computed it to be $1.72$ rounded to two digits; -I ran $10^{11}$ Monte Carlo evaluations to obtain $1.726225\dots$ -with an estimated error of $7.3\cdot 10^{-6}$ -My questions are: - -Can the exact value of $\operatorname{edeg}G(2,4)$ be determined; is there perhaps a relation with other constants; is it, for example, algebraic over $\mathbb{Q}[\pi]$? -If this is too difficult, what are ways of evaluating the above integral to higher accuracy numerically? - -Update: In AdamP.Goucher's great answer, he provides more digits ($1.726230876$) and also uses a reformulation described by Matt F. in a comment to get an integrand without any absolute values: -$$G(2,4) = 2^{-6} \int \dfrac{h\left((y-u)^2+(v-x)^2\right)^2}{\left((1+(u + (y-u)c - (v-x)h)^2)(1+(x + (v-x)c + (y-u)h)^2)\prod_{\alpha\in\{u,y,x,v\}}(1+\alpha^2)\right)^{3/2}}$$ -where we integrate over $\mathbb{R}^5\times \mathbb{R}^+$ (and $h$ is the positive variable). -Perhaps this form could come in handy for evaluating this integral to higher accuracy?! - -Update 2 The new formula from L.Mathis and Antonio Lerario is very useful for calculating digits! The following mpmath code can returns in less than $3$ minutes -$1.7262312489219034885256331685361697650475579915479447$ -(the last few digits might not be accurate) -I expect to make that even much faster when solving the two integrals $F$ and $G$ symbolically first. -import functools -from mpmath import mp -@functools.lru_cache(maxsize=1000) -def F(u): - return mp.quad(lambda phi: (u*mp.sin(phi)**2)/(mp.sqrt(mp.cos(phi)**2 + u**2*mp.sin(phi)**2)), [0, mp.pi/2]) -@functools.lru_cache(maxsize=1000) -def G(u): - return mp.quad(lambda phi: (mp.sin(phi)**2)/(mp.sqrt(mp.sin(phi)**2 + u**2*mp.cos(phi)**2)), [0, mp.pi/2]) -def H(u): - return F(u)/G(u) -def L(u): - return F(u)*G(u) -def integrand3(u): - return L(u)**2*(1/H(u) - H(u))*mp.diff(H, u)/H(u) - -dps = 50 -mp.dps = dps - -%time z = 3*mp.quad(integrand, [0,1]); z - -REPLY [4 votes]: If anyone is still interested, Antonio Lerario and I recently published a paper: Probabilistic Schubert Calculus: Asymptotics, -in which we give a more convenient formula to compute this number. -What was called before $\operatorname{edeg}G(2,4)$ is now denoted by $\delta_{1,3}$ and we give a line integral formula in Proposition 24: -\begin{aligned} -\delta _{1,3}=-6 \pi ^{4}\int _{0}^1 L(u)^{2}\mathrm {sinh}(w(u))w'(u)\mathrm{d}u \end{aligned} -where $L=F⋅G$ and $w=log(F/G)$ with -\begin{aligned} -F(u)&:=\int _{0}^{\pi /2} \frac{u \ \sin ^2(\theta )}{\sqrt{\cos ^2\theta +u^2 \sin ^2\theta }}\mathrm {d}\theta \\ G (u)&:=\int _{0}^{\pi /2} \frac{\sin ^2(\theta )}{\sqrt{\sin ^2\theta +u^2 \cos ^2\theta }}\mathrm {d}\theta . -\end{aligned} -We didn't run any advanced numerical evaluation but this looks much nicer than the previous formula. I hope this will help. It is still unknown to us if this number can be expressed as a closed formula using special functions.<|endoftext|> -TITLE: Monoid under Day convolution and lax monoidal functor: Is strength necessary? -QUESTION [13 upvotes]: nCatLab states that a monoid in the Day-convolution monoidal category is equivalent to a lax monoidal functor. In the Haskell community this result is used to explain the construction of the applicative functor. Rivas and Jaskelioff show that a monoid in the endofunctor category $[Set, Set]$ with Day convolution is equivalent to a strong lax monoidal functor. They need strength to get the natural transformation $Id \to F$ from the (lax) preservation of the unit $1 \to F(1)$. I'm trying to understand why they need strength even though the nCatLab result doesn't. Granted, nCatLab deals with enriched functors, and self-enrichment is related to strength, but functors in $Set$ are both strong and self-enriched. -Instead of starting from $[Set Set]$, I'd like to start from a monoidal category $C$ and work with the functor category $[C, Set]$ with Day convolution. A monoid is a functor $F$ in this category, together with two natural transformations. I'm interested in one of them. -The unit under Day convolution is $C(I, -)$ where $I$ is the unit object in $C$. Consider the set of natural transformations from $C(I, -)$ to $F$: -$[C, Set](C(I), -), F) = \int_a C(I, a) \to F(a)$ -By Yoneda, this is equal to $F(I)$. -A function from $1$, the singleton set, will pick up a natural transformation in $[C, Set](C(I, -), F)$ and an element of $F(I)$: -$1 \to F(I)$ -This is exactly one of the morphisms defining the lax monoidal functor. It seems like, in this formulation, strength is not needed. (It's not needed in the rest of the proof either.) What am I missing? - -REPLY [2 votes]: You are correct. -The most straightforward structure that you have on $[C, Set]$ is that of a multicategory (https://ncatlab.org/nlab/show/multicategory). Given functors $F_1, F_2, ..., F_n$ (n can be 0) and $F$ the set of multimorphisms $(F_1, F_2, ..., F_n) \rightarrow F$ is defined as the set -$$Nat(F_1(X_1)\otimes ... \otimes F_n(X_n), F(X_1\otimes ... \otimes X_n)).$$ -The multicategory structure is obvious. -The notion of monoid makes sense in any multicategory. Particularly, a monoid in the multicategory $[C, Set]$ is exactly a lax monoidal functor (not necessarily strong). -Furthermore, via Day convolution $[C, Set]$ happens to be a representable multicategory. More concretely, pretty much by (one of) the definitions of the Day convolution we have universal natural transformations -$$F_1(X_1)\otimes ... \otimes F_n(X_n) \rightarrow (F_1\otimes_{Day}...\otimes_{Day}F_n)(X_1\otimes ... \otimes X_n)$$ -which is the property guaranteeing representability (See C. Hermida, Representable multicategories). Specifically, the multicategory structure on $[C, Set]$ is represented by the Day monoidal structure on $[C, Set]$, which of course just means that -$$[C, Set]_{Mult}((F_1, F_2, ..., F_n), F) \cong [C, Set](F_1\otimes_{Day}...\otimes_{Day}F_n, F).$$ -This implies that a monoid in the multicategory [C, Set], i.e. a lax monoidal functor as noted before, is the same as a monoid in the Day-convolution monoidal category [C, Set]. There is no need of further conditions.<|endoftext|> -TITLE: Riemann Theta Function On Hyperbolic Riemann Surfaces -QUESTION [6 upvotes]: The Riemann theta function for a genus $g$ closed Riemann surface with period matrix $\tau=[\tau_{ij}]$ is defined by -$$\theta(\{z_1,\cdots,z_g\}|\tau)=\Sigma_{n\in\mathbb{Z}^g}e^{\pi i(n\cdot\tau\cdot n + 2n\cdot z)}$$ -The zeroth of this function is given by theta divisor, a divisor class $\Theta=W_{g-1}+\mathcal{K}$ in which $W_{g-1}$ is a degree $g-1$ divisor and $\mathcal{K}$ is the Riemann constant vector. I have two questions: - -How can Riemann theta function be defined for a hyperbolic Riemann surface? -How can the theta divisor i.e. the locus of the zeroes of Riemann theta function be characterized in hyperbolic geometry? More specifically, Is there a way to characterize the theta divisor in terms of Fenchel-Nielsen coordinates on the Teichmuller space of hyperbolic Riemann surfaces? - -REPLY [2 votes]: To amplify on ElucisusFTW's answer, the relationship between the hyperbolic metric and the complex structure is the so-called "accessory parameters" problem, and very little is known about it, except in the very special setting of hyperbolic punctured spheres (and the related punctured tori), where there is a large number of papers by Zograf and Takhtajan (warning, the latter name is spelled in many different ways). This is closely related also to string theory (which I know next to nothing about), and the determinants of Laplacians, which CAN be computed in terms of hyperbolic invariants. See, for example, Pollicott-Rocha and Strohmaier-Uski (note that the latter paper says that the former paper is wrong...) -Pollicott, Mark; Rocha, Andr\'e C., A remarkable formula for the determinant of the Laplacian, Invent. Math. 130, No.2, 399-414 (1997). ZBL0896.58067. -Strohmaier, Alexander; Uski, Ville, An algorithm for the computation of eigenvalues, spectral zeta functions and zeta-determinants on hyperbolic surfaces, Commun. Math. Phys. 317, No. 3, 827-869 (2013). ZBL1261.65113.<|endoftext|> -TITLE: speeding up Gosper and WZ algorithms -QUESTION [5 upvotes]: In our ongoing work to speed up symbolic summation and other similar algorithms in Sagemath, we notice that naive implementations of Gosper and Wilf-Zeilberger (a.k.a. WZ) algorithms are usually quite slow. -What is the state of the art here? E.g. is there software available that can derive Clausen $_4 F_3$ identity (see (V) on p.43 of the A=B book) quickly (few seconds on a laptop, say?) using the the WZ algorithm? - -REPLY [4 votes]: The group at RISC has been working aggressively towards developing improved algorithms for many computer algebra problems, including the Zeilberger's. So, it is a good place to ask. -Meantime, I just wanted to say this regarding Clausen $_4 F_3$ identity that the OP mentioned. If you read on in the book A=B, on page 127 there is a rewrite of the identity and a procedure outlined as well. I just checked it myself using a much earlier version from Zeilberger, called EKHAD, and executed in Maple. The verification of Clausen is instant!<|endoftext|> -TITLE: Stable homotopy category of complexes of sheaves -QUESTION [9 upvotes]: Let $X$ be an Hausdorff space, which is locally compact and locally connected. Let $R$ be a commutative and unitary ring and let $D(X)$ be the derived category of unbounded complexes of sheaves of $R$-modules. -Question: is $D(X)$ a stable homotopy category, and is it unital algebraic in the sense of Hovey, Palmieri and Strickland (def 1.1.4 in their AMS memoir "Axiomatic Stable Homotopy Theory")? -Edit: A. Neeman proved that if $M$ is a non-compact manifold of $dim\geq 1$, and $R=\mathbb{Z}$, the only compact object of the category $D(M)$ is the zero object (see his paper "On the derived category of sheaves on a manifold"). Thus in that case $D(M)$ cannot be an algebraic stable homotopy category. - -REPLY [10 votes]: There are results about quasicoherent sheaves on schemes in the paper "The derived category of quasi-coherent sheaves and axiomatic stable homotopy" by Alonso, Jeremías, Pérez and Vale. This refers to another paper "Localization -in categories of complexes and unbounded resolutions" by Alonso, Jeremías and Souto, which works with more general abelian categories satisfying Grothendieck's AB5 condition (filtered colimits are exact). You should be able to recover what you need from these.<|endoftext|> -TITLE: Is the Hodge Conjecture an $\mathbb{A}^1$-homotopy invariant? -QUESTION [18 upvotes]: Let $X$ and $Y$ be two nonsingular projective varieties defined over the complex numbers. - -If $X$ and $Y$ are $\mathbb{A}^1$-homotopy equivalent, then does $X$ satisfying the Hodge conjecture imply that $Y$ satisfies the Hodge conjecture? - -REPLY [15 votes]: Isn't this very easy? If the varieties are $\mathbb{A}^1$-homotopy equivalent, then their Voevodsky motives are isomorphic also (since there is a connecting functor making the obvious diagram commutative). So it remains to note that all the ingredients of your question are "motivic". For this purpose one may recall that Chow motives embed into Voevodsky one and note that both the cycle classes and the Hodge classes are determined by Chow motives of the corresponding varieties.<|endoftext|> -TITLE: Simple characterization of Postnikov & Whitehead towers? -QUESTION [7 upvotes]: I'm asking this question in the most model-ambiguous way I can since this is the kind of answer i'm looking for. -There are various explicit constructions of the Whitehead and Postnikov towers. I'm trying to understand what exactly characterizes these construction. - -Postnikov tower: A Postnikov tower of (a possibly non-pointed) space is a factorization of the terminal morphism $X \to *$ to a directed limit diagram: -$$X \cong\underset{\rightarrow_n}{\lim}X_n \to \dots \to X_2 \to X_1 \to X_0 \cong *$$ -Such that $X_k$ is $(k-1)$-truncated and each morphism $X_{k} \to X_{k-1}$ is a $k$-equivalence (isomorphism on homotopy groups in degree smaller than $k$ and surjection on $k$). -Question 1: Does this property determine the Postnikov tower up to weak equivalence of diagrams? - -A similar question about the Whitehead tower follows naturally - -Whitehead tower: A Whitehead tower of a space is a factorization of the initial morphism $* \to X$ to a directed limit diagram: -$$* \cong\underset{\rightarrow_n}{\lim}X_n \to \dots \to X_2 \to X_1 \to X_0 \cong X$$ -Such that $X_k$ is $(k-1)$-connected and each morphism $X_{k} \to X_{k-1}$ is an isomorphism on homotopy groups in degree larger than $k-1$). -Question 2: Does this property determine the Whitehead tower up to weak equivalence of diagrams? - -REPLY [5 votes]: Indeed, the properties you stated characterize Postnikov and Whitehead towers. A nice conceptual way of justifying this is by using $k$-connected / $k-$truncated factorization systems. -To fix terminology, a $k$-connected map is one whose all homotopy fibers are $k$-connected and a $k$-trunceted map is one whose all homotopy fibers are $k$-truncated. (Note that in the case of connectedness this disagrees with the traditional definition according to which a map is $k$-connected if and only if all its homotopy fibers are $(k-1)$-connected.) -Then every map factors as a composite of a $k$-connected map followed by a $k$-truncated map. Moreover, certain lifting property holds. On the point-set level you could say that $k$-connected cofibrations have the LLP with respect to $k$-truncated fibrations. However, more properly, a homotopy coherent / higher categorical lifting property is satisfied, i.e. for any $k$-connected map $A \to B$ and a $k$-truncated map $X \to Y$ the square of (derived) mapping spaces -$\require{AMScd}$ -\begin{CD} -\mathrm{Map}(B, X) @>>> \mathrm{Map}(A, X) \\ -@VVV @VVV \\ -\mathrm{Map}(B, Y) @>>> \mathrm{Map}(A, Y) \\ -\end{CD} -is a homotopy pullback. (In other words, any lifting problem with $A \to B$ on the left and $X \to Y$ on the right has a contractible space of solutions.) -It follows directly that any map factors uniquely as a composite of a $k$-connected map followed by a $k$-truncated map, i.e. the space of such factorizations is contractible. Your statements are recovered as special cases of that. The $k$th Postnikov section of a space $X$ is obtained by factoring $X \to *$ as a composite of a $k$-connected map followed by a $k$-truncated map. In the based case, the $k$th connected cover of $X$ is obtained by factoring $* \to X$ as a composite of a $(k-1)$-connected map followed by a $(k-1)$-truncated map. (There is an offset here since $X \to *$ is $k$-connected if and only if $* \to X$ is $(k-1)$-connected.) -It should also be noted that a map $A \to B$ is $k$-connected if and only if it induces isomorphisms on homotopy groups up to dimension $k$ and an epimorphism in dimension $k + 1$. Similarly, a map $X \to Y$ is $k$-truncated if and only if it induces isomorphisms on homotopy groups above dimension $k + 1$ and a monomorphism in dimension $k + 1$. I believe your description of the Whitehead tower should be adjusted accordingly (although in this particular case, injectivity in dimension $k + 1$ may be vacuous).<|endoftext|> -TITLE: Classification of $p$-groups, what after it? -QUESTION [9 upvotes]: In finite group theory, $p$-groups or simple groups can be considered as building blocks of all the groups. What is known about these families of groups is that - -The classification of simple groups is done. -The classification of $p$-groups is much difficult problem. - -Due to classification of simple groups, many problems from different areas of mathematics have been solved; there is a book Atlas of Finite Simple Groups: Ten years on, which perhaps contains some work with this thought. -On the other hand, although classification of $p$-groups is difficult, many researchers are focused on classifying $p$-groups having a specific property. For example, such work is collected in five big volumes (each $\geq$ 600 page, multiple of thousands of problems in total), only devoted to finite $p$-groups by Berkovich and Janko, in which, the authors state in preface that their aim is classification w.r.t. some properties. -I wondered many times, but didn't get any definite answer to following question while discussing with some people working on $p$-groups. I thought it may not be good to post this question on group-pub-forum to de-motivate some group-theoriests. The question is - -As long as classification of some types of $p$-groups is concerned, has its application appeared in some other branch of mathematics (or at least, in algebra, or at least in group theory itself!)? - -REPLY [3 votes]: Let me speak to part of your question: What after it? We can not reasonably expect to classify the $p^{2n^3/27 +O(n^{8/3})}$ groups of order $p^n$ for large $n$. We can classify $p$-groups with particular properties focusing on those with applications to other problems (as Geoff and Alireza mention). Alternatively, we can focus on -understanding small and highly symmetric examples. Certainly in graph theory small and highly symmetric examples have played an important role e.g. $s$-arc transitive graphs (with $2\leqslant s\leqslant 7$) have applications to incidence geometry and abstract polytopes. This philosophy was behind our paper -Maximal linear groups induced on the Frattini quotient of a p-group. -Indeed, one could argue that most large $p$-groups with small automorphism groups are uninteresting in the same way that most large graphs with few symmetries are uninteresting.<|endoftext|> -TITLE: Intuition for the construction of the space $M_G=EG\times _G M$ -QUESTION [5 upvotes]: Reference: Atiyah & Bott, The moment map and Equivariant cohomology -Question: What could be the motivation and the intuition behind the construction of the space $M_G=EG\times _G M$? When I am studying the equivariant cohomology, I have no idea about why we would like to define the equivariant cohomology as the ordinary cohomology of $M_G$. -A simple case is, if $G$ is a compact Lie group and acts freely on $M$, then $M_G$ is simply $M/G$, the space of all $G$-orbits. -But, in general cases, how to understand the picture of $M_G$? And, how does $M_G$ relate to $M/G$? -For example, take $G=S^1\times S^1$ and $M=\mathbb P^1$, and the action is given by $(t_0,t_1): [x,y]\mapsto [t_0x, t_1y]$. Then this action is not free and it has two fixed point $[1,0]$ and $[0,1]$. In this case, what is the difference between $M/G$ and $M_G$? - -REPLY [6 votes]: In homotopy theory and in homological algebra, there a general idea that for any given operation that you might want to perform, there's a class of "nice" objects for that operation. It is a good idea to replace your object by another one which is "nice" and "equivalent" in some appropriate sense before performing the operation. -If the operation is $-/G$ for some group $G$, then the class of "nice" objects is $G$-spaces with a free action. It's a good idea to replace $M$ by $M\times EG$ before performing $-/G$. -If the operation is $\times_BX$ for some map $X\to B$, the the class of "nice" objects are the spaces equipped with a map to $B$ which is a fibration. -If the operation is $-\otimes_RM$ for some ring $R$ and some $R$-module $M$, then the class of "nice" objects are the complexes of projective modules. -If the operation is $Hom_R(M,-)$ for some ring $R$ and some $R$-module $M$, then the class of "nice" objects are the complexes of injective modules.<|endoftext|> -TITLE: A generalized log inequality for positive definite trace-one matrices -QUESTION [8 upvotes]: Let $\{V_i\}_{i=1}^N$ be a set of $n\times m$, $n\geq m$, real matrices of full column rank and let $X=X^\top\in\mathbb{R}^{n\times n}$ be a positive definite trace-one matrix. Moreover, let $A^{1/2}=(A^{1/2})^\top$ denote the (unique) symmetric square root of a positive semi-definite matrix $A$, $\mathrm{tr}\, A$ the trace of $A$, and $\log A$ the matrix logarithm of $A$. - -Question: Does the inequality - $$\tag{$1$} \label{1} -\sum_{i=1}^N \mathrm{tr}\log \left(\frac{1}{mN}\sum_{j=1}^N V_i^\top X^{1/2}V_j(V_j^\top X V_j)^{-1} V_j^\top X^{1/2}V_i\right) - \sum_{i=1}^N \mathrm{tr} \log \left(V_i^\top X V_i\right) \ge 0 -$$ - hold true? - -My (very little) progress so far. -Using the fact that $\mathrm{tr}\log Y+\mathrm{tr}\log Z=\mathrm{tr}\log YZ$ and $-\log\, Y=\log Y^{-1}$ for positive definite $Y$, $Z$, we can rewrite \eqref{1} as -$$\tag{$2$} -\sum_{i=1}^N \mathrm{tr}\log \left(\frac{1}{mN}\sum_{j=1}^N V_i^\top X^{1/2}V_j(V_j^\top X V_j)^{-1} V_j^\top X^{1/2}V_i\left(V_i^\top X V_i\right)^{-1}\right) \ge 0. -$$ -Also, exploiting the formula $\mathrm{tr}\log Y =\log \det Y$, we get -$$\tag{$3$} -\sum_{i=1}^N \log\det \left(\frac{1}{mN}\sum_{j=1}^N V_i^\top X^{1/2}V_j(V_j^\top X V_j)^{-1} V_j^\top X^{1/2}V_i\left(V_i^\top X V_i\right)^{-1}\right) \ge 0, -$$ -or, equivalently, -$$\tag{$3$} -\prod_{i=1}^N \det \left(\frac{1}{mN}\sum_{j=1}^N V_i^\top X^{1/2}V_j(V_j^\top X V_j)^{-1} V_j^\top X^{1/2}V_i\left(V_i^\top X V_i\right)^{-1}\right) \ge 1. -$$ -I'm stuck at this point. I believe that the problem can be further simplified to some well-known determinant inequality. However, at the moment, I don't know how to proceed. - -Note 1. This question can be seen as a generalization of this problem. Indeed, for $m=1$, \eqref{1} reduces exactly to the case treated in that question, which has been proved to be true. -Note 2. Numerical simulations suggest that \eqref{1} is very likely to be true. -Edit. I edited the inequality by replacing the factor $1/N$ in the first logarithm by $1/(mN)$. This is a somewhat "stronger" version which I believe is the "right" generalization of the aforementioned problem. - -REPLY [3 votes]: Fedor's proof can be generalized : -Let $S = X^{1/2}$, $U_i = V_i S^{1/2}$ and $W_i = U_i (U_i^T S U_i)^{-1/2}$ . -Then $W_i^T S W_i = I_m$ where $I_m$ is the $m\times m$ identity . -Since $log(x) \ge 1 - 1/x$ it is enough to show that -$$tr \sum_{i=1}^N \mu_i \le 1$$ -where -$$\mu_i = (\sum_{j=1}^N W_i^T W_j W_j^T W_i)^{-1}$$ . -Then -$$tr \sum_{i=1}^N \mu_i = tr \sum_{i=1}^N \mu_i^2 \mu_i^{-1}$$ -$$= tr [\sum_{i=1}^N \mu_i^2 \sum_{j=1}^N W_i^T W_j W_j^T W_i]$$ -$$\ge tr \sum_{i=1}^N \sum_{j=1}^N W_i \mu_i W_i^T W_j \mu_j W_j^T$$ -$$= tr [(\sum_{i=1}^N W_i \mu_i W_i^T)^2]$$ -$$\ge [tr (S \sum_{i=1}^N W_i \mu_i W_i^T)]^2$$ -$$= (tr \sum_{i=1}^N \mu_i)^2$$ , -where I have used that -$$tr (\mu_i^2 W_i^T W_j W_j^T W_i) + tr (\mu_j^2 W_j^T W_i W_i^T W_j) \ge 2 tr (W_i \mu_i W_i^T W_j \mu_j W_j^T)$$ . -This follows from -$$0 \le tr[(W_i^T W_j \mu_j - \mu_i W_i^T W_j) (\mu_j W_j^T W_i - W_j^T W_i \mu_i)]$$ .<|endoftext|> -TITLE: Confusion in some notations in Lie sub-algebras of exceptional Lie algebra -QUESTION [5 upvotes]: I was following Humphrey's Lie algebra for study, and came to study of Weyl groups of root systems. The book has stated orders of Weyl groups of exceptional Lie algebras, and there were no comments or exercises on their structures. I searched other books and came to Knapp's Lie algebra some exercises, and finally came to a confusion. -$E_8$ contains roots of $\mathbb{R}^8$ of the form $\pm (e_i \pm e_j)$ and $\frac{1}{2} (c_1e_1 + \cdots + c_8e_8)$ with $c_i\in \{1,-1\}$ for $i=1,2,\cdots,8$ and $c_i=1$ for even number of $i$'s. -$E_6$ contains those roots of $E_8$ which are orthogonal to $e_8+e_7$ and $e_8+e_6$. These are -$$\pm \frac{1}{2} (e_8-e_7-e_6 + c_1e_1 +\cdots + c_5e_5), c_i\in \{1,-1\}, c_i=1 \mbox{ for odd no. times}.$$ -and $\pm (e_i\pm e_j)$ for $i,j$ distinct from $1,2,...,5$. -Then Knapp asks for following: - -Consider roots of $E_6$ orthogonal to $\frac{1}{2}(e_8-e_7-e_6+e_5+e_4+e_3+e_2+e_1).$ Show that they form a root system of type $A_5$. - -Q.1 My simple question is whether it should be $A_5$ or $A_4$? Because, the answer (I think) to question gives roots -$$\pm (e_i-e_j) , \hskip5mm i,j=1,2,3,4,5, \mbox{ and } i\neq j.$$ -I confused, whether answer is incorrect or it is notational difference of $A_n$ in Humphrey's and Kanpps books? (According to Humphrey's $A_n$ contains roots of $\mathbb{R}^{n+1}$ of form $e_i-e_j$.) -Q.2 Where can I see structure description of Weyl groups of exceptional root systems? I was following Bourbaki's Algebra 4-6, in that some order formula for Weyl group is given. But I want to study structures also. Can you suggest some reference for it? - -REPLY [6 votes]: [Answer restated for clarity:] -Though it's a little awkward to combine two separate questions in one posting, the answer to Q2 is somewhat scattered in the literature due to the fact that exceptional Weyl groups tend to show up in various different places. A short survey, with references, is given in Section 2.12 of my 1990 Cambridge book Reflection Groups and Coxeter Groups. For example, the Atlas of Finite Groups has some relevant entries cited. -To answer Q1, Knapp's conclusion is correct, though I don't have his book at hand: the subsystem has type $A_5$. The confusion comes from the fact that the standard description of the root system $A_n$ starts with a euclidean space of dimension $n+1$ and then constructs the desired root system on a hyperplane. So the usual notation for roots is not directly comparable to that used for $E_8$ or its subsystem $E_6$. -Note that the given root for $E_6$ in the shaded box is just the highest root relative to the usual system of simple roots; this is called $\tilde{\alpha}$ by Bourbaki. Then it's clear from a look at the extended Dynkin diagram how to characterize the roots orthogonal to this one (or its negative) as a subsystem of type $A_5$. -Concerning Victor's answer to Q1, it would have prevented some confusion if he had tried to give a reference for his "general principles" remark. -[Small linguistic comment: When an English name like Jones or Humphreys ends in the letter 's', it's always tricky to follow the usual rule about forming a possessive. Probably the best solution is just to add a single mark such as Jones' (but sometimes people write Jones's instead). It's all the fault of our ancestors, who were mostly illiterate anyway.]<|endoftext|> -TITLE: simple questions on topological rings arising in the context of Perfectoid Spaces -QUESTION [5 upvotes]: (I apologize in advance for these simple questions, I am a beginner trying to go through Scholze's paper Perfectoid Spaces). -Let $(R, R^+)$ be an affinoid $k$-algebra as defined in Scholze's paper Perfectoid Spaces, Definition 2.6. In particular, there exists a subring $R_0 \subset R$ such that $aR_0$, $a \in k^\times$ forms a basis of open neighborhoods of $0$. -I'm struggling to understand the role $R_0$ plays, and I suppose my real question is Question 3 below, so if you just want to answer that one, that's fine. -Question 1: What are some good examples (by good I perhaps mean those that arise in applications of perfectoid spaces) of $R$ and especially $R_0$ to keep in mind? I know examples of $R$ are quotients $Q$ of the Tate algebra $k\langle T_1, \ldots, T_n\rangle$. The topology on $Q$ I believe is induced from the Gauss/sup norm on the Tate algebra, I think in this case we can take $R_0=\{r \in Q \mid |r|\leq 1 \}$ -Question 2. For any $x \in Spa(R, R^+)$, does $x$ induce the topology on $R$? I don't think so, because for example take the trivial valuation, which I believe is an element of $Spa(\mathbb{Q}_p, \mathbb{Z}_p)$, but doesn't induce the p-adic topology on $\mathbb{Q}_p$. Correct me if I'm wrong. -Question 3: Let $w \in k$ be topologically nilpotent,i.e. $|w|<1$. and let $h \in R$. In Remark 2.8, Scholze says that for sufficiently large $N$, $w^Nh \in R^+$, as $R^+ \subseteq R$ is open. Why? I tried reasoning as follows: let $R_0$ be as above. There exists $a \in k^\times$ such that $aR_0 \subseteq R^+$. We can find $N$ sufficiently large such that $|w^N| < |a|$. Got stuck. - -REPLY [15 votes]: Firstly, I don't think you should start to learn about adic spaces by thinking about perfectoid spaces. The sensible examples of adic spaces for a beginner to think about are sane Noetherian things like the closed polydisc etc. It's quite hard/messy to do any explicit basic calculations (of the sort one would like to do when getting the hang of thingS) with a general perfectoid ring because when you complete you lose a certain amount of control. I'll try to answer your questions. -The role of $R_0$ is that it defines a topology on $R$. Have you thought about examples like closed polydiscs or more general affinoids in the original sense of Tate? If $R$ is a reduced affinoid then one good choice for $R_0$ would be the functions with supremum norm at most 1; conversely, if you know this ring then (using powers of a fixed element $\pi\in k$ with norm less than 1) you know what it means for a function to be small (it's in $\pi^NR_0$) so now you know what it means for two functions to be close (their difference is small; the topology on a reduced affinoid is defined this way, so two functions are close iff they're close everywhere) so now you know the topology on $R$. More abstractly you can define the topology on $R$ to have basis $r+\pi^NR_0$ as $r,N$ vary, but the geometric intuition is what I said above. So to give $R_0$ is to give the topology. -Q1) Good examples: forget perfectoids. Let $R$ be $\mathbb{Q}_p\langle T_1,T_2,\ldots,T_n\rangle$ and let $R_0$ be $\mathbb{Z}_p\langle T_1,\ldots,T_n\rangle$. If you're not happy with the topology on $R$ then convince yourself that $r+p^nR_0$ give a basis for the topology. Next try a reduced affinoid. The spectral seminorm is a norm on that; try examples like an annulus $\mathbb{Q}_p\langle X,Y\rangle/(XY-p)$. Figure out what a general element of that ring looks like. Figure out which elements have supremum seminorm at most 1. That's a good example for $R_0$. Try $XY-p^n$ or $XY-1$ as the relationship instead. Check that $R_0$ is what you think it is. Check $r+p^nR_0$ give a basis for the topology -- do you see why this is obvious? -Now try a non-reduced affinoid, because here there is an extra phenomenon which is less geometric -- nilpotents in the structure sheaf. The simplest example is $\mathbb{Q}_p[X]/X^2$ with its topology induced from the obvious isomorphism with $\mathbb{Q}_p^2$, and already this exhibits phenomena that you need to understand. Check that the supremum seminorm is not a norm. Check that if $S$ is the elements of seminorm at most 1 then $r+p^nS$ do not form a basis for the topology. We fix this by letting $R_0$ be $\mathbb{Z}_p[X]/X^2$. Check that $R_0$ defines the topology in the usual way. In particular your "Q" example in your question is not right in general -- only when the affinoid is reduced. -Now try perfectifying one of these (not the non-reduced one, but any or all of the other ones). If you want. -Q2) No. This sort of question just shows that you're currently blind to standard behaviour of adic spaces so shouldn't be thinking about crazy non-Noetherian perfectoid spaces yet. Think about the closed unit disc. Adic spaces have lots of strange points, points corresponding to higher rank valuations, weird points if the underlying field isn't spherically complete etc. Think about "standard" points first; if you can't understand standard points like the $\mathbb{Q}_p$-points of the closed unit discs you'll never get anywhere. Think about the point $T=0$ on the closed unit disc. Figure out what the corresponding valuation is, understand the induced topology on $R$, understand it geometrically. Two functions are close for the valuation topology corresponding to the point $T=0$ if the functions take values at $T=0$ which are close to each other. That's what the valuation is telling you about -- behaviour at 0. The topology on $R$ is telling you something much more global -- it's telling you if two functions are close everywhere. We want the valuations to be continuous -- and that has a geometric meaning, and that meaning is something you should understand and it is not saying that the topology on $R$ (a global thing, telling us about when a function is small everywhere) is determined by behaviour near one point (indeed, behaviour at a point is only telling us about the functions at this point). The valuations are continuous, which means geometrically that the collection of functions $g$ close to $f$ for the valuation topology (i.e. such that $g(0)$ is close to $f(0)$) is open (i.e. contains the functions which are close to $f$ everywhere -- because "close" means "uniformly close" -- that's the definition of the topology on a reduced affinoid). Explained this way, continuity is clear, and it's also clear that the valuation topology is very different to the norm topology. You have to understand the closed unit disc very well before you can understand a perfectoid space. Indeed, you should probably understand the closed disc rigid space, then the closed disc Berkovich space, then the closed disc adic space, before you understand the perfectoid "closed disc" obtained by throwing in all the $p$-power roots of $T$. -Q3) $R^+$ is open and contains 0. So $R^+$ contains all functions which are globally small. Now $h$ is a function on an affinoid so it's globally bounded, so clearly $w^Nh$ for some large $N$ is globally small. This is the geometric idea. If you did Q1 then translate this into algebra -- it's not too hard. If you need a hint, think about the basis $r+w^NR_0$ of the topology. -Good luck! I had read all of Bosch-Guentzer-Remmert and some notes on Berkovich spaces, so I had a good feeling for both classical rigid spaces and Berkovich spaces, before I learnt adic spaces (from Wedhorn's notes), and then after that was when I dared to take on perfectoid spaces. You need the geometric intuition to guide you through the algebra.<|endoftext|> -TITLE: Is the "Moebius Stairway" Graph Already Known? -QUESTION [14 upvotes]: It is a wellknown fact, that Moebius Ladder Graphs have $2n$ vertices, but nowhere could I find any hint of how to generalize them to Graphs with $2n+1$ vertices. -Last week I had the idea of giving up the restriction to cubic graphs and arrived at glueing together the two ends of triangle strips with $2n+1$ triangles in the "Moebius manner", i.e. with a twist. -The result is - - -a $4$-regular graph -with exactly two edge-disjoint Hamiltonian cycles if $2n+1\ge 7$ , which contrasts the situation of Moebius Ladders, where the number of Hamiltonian cycles is different for each size and given by A124356 - OEIS and none contains a pair of edge-disjoint Hamiltionian cycles. -the chromatic number is $5$ for $5$ vertices and $4$ for all other cases of $2k+1$ vertices, again contrasting the situation of Moebius Ladders, where it is $2$ for $4k+2$ vertices and $3$ for $4k$ vertices (for 4 vertices it would also be $4$, but $K_4$ is normally not considered to be a Moebius Ladder) - - -Question: -Have those Moebius Stairway graphs been described or studied already, i.e. are further special properties known? - -As a remark let me explain the name "Moebius Stairway" graph: if the triangles are chosen to be isosceles right triangles and the strip is then drawn in an ascending $45^{\circ}$ angle, it looks somewhat similar to a stairway and, besides that, I liked the idea of providing an alternative to ladders. - -REPLY [2 votes]: This is the square of an odd cycle. If you found it as a spanning subgraph of another graph you might call it the square of a Hamilton cycle. There are lots of results about powers of cycles from this opposite perspective; for example, they appear at minimum degree $2n/3$ or in random graphs at $p = 1/\sqrt n$.<|endoftext|> -TITLE: What is the status on this conjecture on arithmetic progressions of primes? -QUESTION [25 upvotes]: The Green-Tao theorem states that for every $n$, there is an arithmetic sequence of length $n$ consisting of primes. -For primes, $p$, let $P(p)$ be the maximum length of an arithmetic progression of primes whose least element is $p$. - -Is it known whether $P(p)=p$ for every prime? - -(This clearly generalizes the Green-Tao theorem, asserting that long progressions show up "as soon as possible." Note that $P(p) \leq p$ by viewing the progression mod $p$.) - -REPLY [27 votes]: Yes, this is unknown; it is even unknown (as GH from MO suspected in a comment) whether $P(p) \ge 3$ always. An equivalent statement to $P(p) \ge 3$ is that there exists an integer $x>0$ such $p+x$ and $p+2x$ are both prime. This is a twin-prime-like problem: nobody has ever proved a statement saying that two fixed linear polynomials $ax+b$ and $cx+d$ are infinitely often simultaneously prime, or even that they must generally be simultaneously prime once. (The Green-Tao theorem converts into a statement about linear polynomials $x,x+d,x+2d,...$ in two variables $x$ and $d$; when we fix $p$ here, we have only one variable.) -On the other hand, the prime $k$-tuples conjecture does imply that $P(p)=p$ for every prime $p$: the corresponding polynomials are $p+x,\dots,p+(p-1)x$, and these polynomials form an admissible set (their product is not identically zero modulo any prime).<|endoftext|> -TITLE: Special values of the modular J invariant -QUESTION [9 upvotes]: A special value: -$$ -J\big(i\sqrt{6}\;\big) = \frac{(14+9\sqrt{2}\;)^3\;(2-\sqrt{2}\;)}{4} -\tag{1}$$ -I wrote $J(\tau) = j(\tau)/1728$. -How up-to-date is the Wikipedia listing of known special values for the modular -j invariant ? -Value (1) is not on it. -Alternatively, is there a compilation of special values elsewhere? -The value (1) is related to this hypergeometric value, too: -$$ -{}_2F_1\left(\frac{1}{6},\frac{1}{3};1;\frac{1}{2}\right) = -\eta(i\sqrt{6}\;)^2 \,2^{1/2} \,3^{3/4} \,(1+\sqrt{2}\;)^{1/6} -\tag{2}$$ -Here, $\eta(\tau)$ is the Dedekind eta function -(I came to this while working on my unanswered question -at math.SE) - -REPLY [3 votes]: You are at $z=i\sqrt{6}$. Hence you can use singular modulus $k_r$. It holds in general (when $z=i\sqrt{r}$) -$$ -j_r=\frac{256(k_r^2+(k'_r)^4)^3}{(k_rk'_r)^4}\textrm{, }k'_r=\sqrt{1-k_r^2}\textrm{, }\forall r>0. -$$ -But $k_6=\lambda^*(6)=(2-\sqrt{3})(\sqrt{3}-\sqrt{2})$ (see here). Hence you get your value.<|endoftext|> -TITLE: Group cohomology version of Deligne-Beilinson cohomology -QUESTION [10 upvotes]: I appreciate Deligne-Beilinson cohomology as a topological cohomology generalization of de Rham cohomology, which concerns the topological structure of manifolds. -On the other hand, we know that there is Group Cohomology theory, suitable for describing the classifying spaces $BG$ of a group $G$, for example, see -ncatlab: group+cohomology and ncatlab: Dijkgraaf-Witten gauge theory. -Is there a group cohomology version of Deligne-Beilinson cohomology, concerning classifying spaces $B^n G$ of a group $G$? What are some key intro References? Thank you in advance. - -REPLY [6 votes]: Sorry, I just saw this question. -One can consider Deligne cohomology for simplicial manifolds in the rather obvious way, namely by adding an additional "simplicial manifold" direction to the ususal double complex, and then take the total cohomology. As resolutions one can use certain hypercovers. -For the simplicial manifold $BG$, I have described and done this is Section 2.2. of my paper -Waldorf, Konrad, Multiplicative bundle gerbes with connection, Differ. Geom. Appl. 28, No. 3, 313-340 (2010). ZBL1191.53022. -As explained there, the underlying version of "smooth group cohomology" is Brylinski's "Differentiable cohomology of gauge groups", which remained unpublished. It is equivalent to the Segal-Mitchison smooth group cohomology. -References on general simplicial Deligne cohomology are: -Brylinski, J.-L.; McLaughlin, D. A., The geometry of degree-four characteristic classes and of line bundles on loop spaces. I, Duke Math. J. 75, No. 3, 603-638 (1994). ZBL0844.57025. -Gomi, Kiyonori, Equivariant smooth Deligne cohomology, Osaka J. Math. 42, No. 2, 309-337 (2005). ZBL1081.14030.<|endoftext|> -TITLE: What is the geometric significance of the definition of supermanifold? -QUESTION [8 upvotes]: We know that a supermanifold $M$ is a locally ringed space $(M,O_M)$ which is locally isomorphic to $(U,C^\infty(U) \otimes \wedge W^\ast)$, where $U$ is an open subset of $\mathbb{R}^n$, $W$ is a finite dimensional real vector space and the above isomorphism defined in the category of $\mathbb{Z}_2$ graded algebra i.e. the parity $\bigoplus_{k \geq 0}C^\infty(U) \otimes \wedge^k W^\ast \rightarrow \mathbb{Z}_2$ defined by $f \otimes x \rightarrow |f \otimes x|:=|x|=k \mod 2$. I would like to know how can we geometrically think of this. For example we know that the algebra of differential forms $\Omega(M)$ on a manifold $M$ which is locally isomorphic to $C^\infty(U) \otimes T^\ast_x M$ for some $x \in U$, therefore the sheaf of differential forms on a manifold corresponds to a supermanifold. How can we geometrically visualize this? Moreover what is the significance of defining a supermanifold structure for the sheaf of differential forms for a manifold $M$. - -REPLY [3 votes]: The Wikipedia page on supermanifolds does a decent job of presenting different definitions and their relations. See in particular the statement of Batchelor's theorem. -Forgetting about the "super" part, already an ordinary manifold can be described as a ringed space $(M,O_M)$, locally isomorphic to $(U,C^\infty(U))$. When $U = \mathbb{R}^n$, with standard coordinates $x^i$, the coordinates $x^i$ play the role of commuting generators (up to taking limits of polynomial expressions) of $C^\infty(\mathbb{R}^n)$. The point of this description is that in the pair $(M,O_M)$, the sheaf of algebras $O_M$ should be interpreted as the sheaf of algebras of functions on open subsets of $M$. But since you are looking into ringed spaces at all, I suspect that you have already seen this interpretation. -Supergeometry allows the sheaf $O_M$ to be a supercommutative algebra (not just a commutative one). The heuristic is that the supermanifold $M$ should be covered by charts of the form $\mathbb{R}^n_{even} \times \mathbb{R}^m_{odd}$ with corresponding algebras of functions $C^\infty(\mathbb{R}^n_{even} \times \mathbb{R}^m_{odd}) := C^\infty(\mathbb{R}^n_{even}) \otimes \bigwedge^\bullet (\mathbb{R}^m_{odd})^*$, where the second tensor factor is the algebra of functions generated by the odd (hence supercommuting) "coordinates" $\theta^j$ on $\mathbb{R}^m_{odd}$. Now, since $\mathbb{R}^n_{even} \times \mathbb{R}^m_{odd}$ does not exist as a topological space, we simply consider the algebra $C^\infty(\mathbb{R}^n_{even} \times \mathbb{R}^m_{odd})$ as assigned to the $\mathbb{R}^n_{even}$ factor, which does exist as a topological space. -This is exactly analogous to how the sheaf of functions on a fibered manifold $N \to M$, locally modeled on charts of of the form $V\times U \to U$, gives rise to the ringed space $(M,O_M)$ where which locally looks like $(U, C^\infty(V\times U))$. Batchelor's theorem makes this precise, by identifying any supermanifold $(M,O_M)$ with the total space of a fibered supermanifold $M \to M_0$, where $M_0$ is an ordinary manifold and the fibers of $M \to M_0$ are purely odd. -In this sense, the example that you gave, $(M,\Omega^\bullet)$, corresponds to the total space of the fibered supermanifold $(\Pi T)M \to M$, where $\Pi$ denotes the parity shift (from even to odd) of the cotangent fibers. There might be a common name for this supermanifold, but I can't recall it at the moment.<|endoftext|> -TITLE: Weak*-norm continuous operators on von Neumann algebras -QUESTION [8 upvotes]: Let $M$ be a von Neumann algebra with predual $M_*$, and let $T\colon M\to M$ be a bounded, linear map. Let us say that $T$ is (sequentially) weak*-norm continuous if for every net (sequence) $(a_j)_j$ in $M$ that weak* converges to $a$ in $M$ we have $\lim_j \| T(a_j) - T(a) \| = 0$. -In particular, given $h\in M$, we may ask when the multiplication operator $L_h\colon M\to M$, $a\mapsto ha$, is (sequentially) weak*-norm continuous. -I suppose such operators on $M$ and elements of $M$ have been considered before. Is there an established terminology or some papers where this has been used? -Is it known when $M$ has an approximate unit $(h_j)_j$ such that multiplication by each $h_j$ is weak*-norm continuous? - -REPLY [3 votes]: Your example of a multiplication operator has the property of being weak$^*$-continuous, and in that setting, we can say quite a lot, with an elementary proof. -Claim: Let $E$ be a Banach space and $T:E^*\rightarrow E^*$ be a weak$^*$-continuous operator (that is, the adjoint of some $S:E\rightarrow E$). Then the following are equivalent: - -For any bounded net $(a_j)$ in $E^*$ which is weak$^*$-convergent to 0, we have that $\|T(a_j)\| \rightarrow 0$. -$S$ is a compact operator. -$T$ is a compact operator. - -Proof: That $S$ is compact if and only if $T$ is compact is Schauder's theorem. Set $X:=\{S(\omega):\omega\in E, \|\omega\|\leq 1\}$ and notice that (1) is equivalent to: - -For any bounded net $(a_j)$ in $E^*$ which is weak$^*$-convergent to 0, we have that $\langle a_j,\omega\rangle\rightarrow 0$ uniformly on $X$. - -Then $S$ is compact if and only if $X$ is compact. If $X$ is compact, then a simple $\epsilon$-net argument shows that (4) holds (using that the net $(a_j)$ is assumed bounded). Conversely, suppose that $X$ is not compact, so there is $\epsilon>0$ such that $X$ admits no $2\epsilon$-net. For any finite-dimensional subspace $N\subseteq E$, any closed and bounded subset of $N$ is compact, and so the distance from $N$ to $X$ must be at least $\epsilon$, say. By Hahn-Banach we can find $a_N\in E^*$ and $\omega_N\in X$ with $\langle a_N,\omega \rangle=0$ for all $\omega\in N$, with $|\langle a_N, \omega_N \rangle| \geq \epsilon$ and with $\|a_N\|\leq 1$. The (bounded) net $(a_N)$ hence shows that (4) does not hold. -If $E$ is separable then you can consider (bounded) sequences in place of nets. -Thus, I believe, your original question is equivalent to asking when the (left) multiplication operator given by $h\in M$ is compact. -[ I stressed the word "bounded". If you allow unbounded nets then the condition becomes that $X$ has finite-dimensional span, i.e. that $S$ (equivalently $T$) is a finite-rank operator. In the sequence case, the Principle of Uniform Boundedness says you have to be bounded anyway. ]<|endoftext|> -TITLE: Is there a name for this property Weil saw for integer polynomials? -QUESTION [16 upvotes]: Andre Weil noticed as a student in 1925 that the polynomial ring $\mathbb{Z}[x]$ comes close to being a PID, and he describes this as `` the embryo of my future thesis.'' -He observed that, given $f(x),g(x)\in\mathbb{Z}[x]$, the Euclidean algorithm computes a sequence of polynomials where each is a linear combination of the preceding ones and which either: -1) Ends in a term which divides the preceding term and thus is something like a GCD for $f$ and $g$; or -2) Ends in an integer $d$ which does not divide the preceding term but bounds the common divisors of values pointwise in this way: For any integer $n$, an integer common divisor of $f(n),g(n)$ must divide $d$. -I have not seen this property discussed anywhere. Does it have a name? -There is related discussion in the question The resultant and the ideal generated by two polynomials in $\mathbb{Z}[x]$ -The number $d$ is called the reduced resultant of $f,g$. - -REPLY [18 votes]: I do not know of a name. But Weil's observation follows from properties of the resultant, and generalizes to other rings $A[x]$ ($A$ a GCD domain). -The resultant $R(f,g)$ of two polynomials is defined either as a certain determinant or as a certain product over pairs of roots of $f,g$, see this Wikipedia page. -From the two definitions, we see that: - -The resultant vanishes iff $f,g$ have common root (this is basically the definition of the resultant). -The resultant has the property that there exist $p,q \in \mathbb{Z}[x]$ ($\deg p < \deg g$, $\deg q < \deg f$) such that $$(*) p(x)f(x)+q(x)g(x)=R(f,g)$$ identically. In particular, $R(f,g) \in (f(x),g(x))$ - this ideal contains a non-zero integer! - -So, - -If the resultant is 0, there is a genuine GCD. -Otherwise, we end up with a non-zero integer $R(f,g)$. Plugging $x=n$ in $(*)$ we see that indeed the GCD of $f(n),g(n)$ divides $R(f,g)$. - -The above observations generalize to $A[x]$ when $A$ has the GCD property. -As explained in the wikipedia page, the Euclidean algorithm differs from the resultant calculation by a simple factor. - -It is interesting to note that the resultant $R(f,g)$ is not the necessarily the generator of $(f(x),g(x)) \cap \mathbb{Z}$, which we will denote $c(f,g)$. Myerson has shown that $R(f,g)$ divides an (effective) power of $c(f,g)$. -When $R(f,g)$ is squarefree and $f,g$ are monics it may be shown that $c(f,g)=R(f,g)$. This may be seen from the results of this recent paper of Frenkel and Pelikán, who study the possibly values of $\gcd(f(n),g(n))$ as $n$ ranges over $\mathbb{Z}$.<|endoftext|> -TITLE: A roadmap to Hairer's theory for taming infinities -QUESTION [59 upvotes]: Background -Martin Hairer gave recently some beautiful lectures in Israel on "taming infinities," namely on finding a mathematical theory that supports the highly successful computations from quantum field theory in physics. -(Here are slides of a similar talk at Heidelberg. and a video of a related talk at UC Santa Cruz.) -I think that a relevant paper where Hairer's theory is developed is : A theory of regularity structures along with later papers with several coauthors. -Taming infinities -Quantum field theory computations represent one of the few most important scientific successes of the 20th century (or all times, if you wish) and allow extremely good experimental predictions. They have the feature that computations are based on computing the first terms in a divergent series, and a rigorous mathematical framework for them is still lacking. This issue is sometimes referred to as the problem of infinities. -Here is one relevant slide from Hairer's lecture about the problem. - -And here is a slide about Hairer's theory. - -The Question -My question is for further introduction/explanation of Hairer's theory. -1) What are these tailor-made space-time functions? -2) What is the role of noise? -3) Can the amazing fact be described/explained in little more details? -4) In what ways, does the theory provides a rigorous mathematical framework for renaormalization and to physics' computations in quantum field theory. -5) How is Hairer's theory compared/related to other mathematical approaches for this issue. (Renormalization group, computations in quantum field theory, etc.) - -REPLY [29 votes]: Let me comment on points 4) and 5). The problem with infinities in QFT or traditional equilibrium statistical field theory -is related to the one addressed by Martin's theory but there are some differences. For concreteness let me talk about the $\phi^4$ model -only. Mathematically, the problem it poses is to make sense of the probability measure -$$ -\frac{1}{\mathcal{Z}}\exp\left( --\int_{\mathbb{R}^d}\{ -\frac{1}{2} (\nabla\phi)^2(x)+\mu \phi(x)^2+g\phi(x)^{4} -\} d^dx -\right)\ D\phi -$$ -on the "space of all functions" $\phi:\mathbb{R}^d\rightarrow\mathbb{R}$. This is the kind of heuristic formulas -one finds in physics QFT textbooks. The symbol $D\phi$ stands for Lebesgue measure on this space of functions -and $\mathcal{Z}$ is a normalization constant so the full space has measure one as befits a probability measure. -Now let's turn this into a well posed mathematical question. -First remove the $\phi^2$ and $\phi^4$ terms, i.e., consider the case $\mu=g=0$. Then this measure $\mu_{C_{-\infty}}$ -makes perfect sense. It is the centered Gaussian measure on the space of temperate distributions $S'(\mathbb{R}^d)$ -and with covariance $C_{-\infty}$ given by -$$ -C_{-\infty}(f,g)=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^d}\frac{\overline{\widehat{f}(\xi)} -\widehat{g}(\xi)}{|\xi|^{2}} d^d\xi -$$ -for all test functions $f$ and $g$ in $S(\mathbb{R}^d)$. Using this first rigorous step, one can reformulate -the problem as that of making sense of -$$ -\frac{1}{\mathcal{Z}}\exp\left( --\int_{\mathbb{R}^d}\{ -\alpha (\nabla\phi)^2(x)+\mu \phi(x)^2+g\phi(x)^{4} -\} d^dx -\right)\ d\mu_{C_{-\infty}}(\phi) -$$ -with a new normalization constant $\mathcal{Z}$ that I will still keep calling $\mathcal{Z}$. -I also introduced the "wave function renormalization coupling constant" $\alpha$ for more generality. -We made a bit of progress (we avoided the problematic Lebesgue measure $D\phi$), -but this still does not make mathematical sense because $\mu_{C_{-\infty}}$ is supported on nasty Schwartz distributions -and pointwise powers like $\phi^2$ and $\phi^4$ are ill-defined, just like $\Phi^3$ in Martin's answer. This is the source -of the UV (ultraviolet) infinities. There are also IR (infrared) problems due to the integration inside the exponential -being over $\mathbb{R}^d$ instead of a compact set. To address these issues, we need what the French call troncature et -régularisation. Let $\rho_{\rm UV}$ be a mollifier, i.e., a compactly supported $C^{\infty}$ function -$\mathbb{R}^d\rightarrow\mathbb{R}$ with $\int \rho_{\rm UV}=1$. -Let $\rho_{\rm IR}$ be a cut-off function, i.e., a nonnegative compactly supported $C^{\infty}$ function -$\mathbb{R}^d\rightarrow\mathbb{R}$ which is equal to 1 in a neighborhood of the origin. To slice Fourier momenta -into (Littlewood-Paley) shells we introduce an integer $L>1$, not necessarily equal to 2 as is customary in -harmonic analysis. For $r,s\in\mathbb{Z}$, define the rescaled functions $\rho_{{\rm UV},r}(x)=L^{-dr}\rho_{\rm UV}(L^{-r}x)$ -and $\rho_{{\rm IR},s}(x)=\rho_{\rm IR}(L^{-s}x)$, and consider the probability measure $\nu_{r,s}$ given by -$$ -\frac{1}{\mathcal{Z}}\exp\left( --\int_{\mathbb{R}^d}\rho_{{\rm IR},s}(x)\{ -\alpha (\nabla\phi)^2(x)+\mu \phi(x)^2+g\phi(x)^{4} -\} d^dx -\right)\ d\mu_{C_{r}}(\phi) -$$ -where $\mu_{C_r}$, or regularized Gaussian measure, is the direct image of $\mu_{C_{-\infty}}$ -by the convolution map $\phi\mapsto \rho_{{\rm UV},r}\ast\phi$. -In other words, $\mu_{C_r}$ is the centered Gaussian measure with covariance -$$ -C_{r}(f,g)=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^d}\frac{|\widehat{\rho_{{\rm UV},r}}(\xi)|^2\ \overline{\widehat{f}(\xi)} -\widehat{g}(\xi)}{|\xi|^{2}} d^d\xi\ . -$$ -A good metaphor would be to say that your orginal flat-screen TV was too smart. The linear size of the -screen was $L^s=\infty$ and that of a pixel was $L^r=0$. Instead one should make $r$ and $s$ finite so that $\nu_{r,s}$ -is mathematically well defined, and then study the limit where $r\rightarrow-\infty$ and $s\rightarrow\infty$ -in the sense of -weak convergence of probability measures on the topological space $S'(\mathbb{R}^d)$. -Now renormalization theory in physics tells us that unless we allow the couplings $(\alpha,\mu,g)$ to depend on the -UV cut-off scale $r$, the following is more likely to happen: 1) we don't converge (e.g., loss of tightness), 2) -we converge to something utterly uninteresting like the atomic measure on the singleton $\{\phi=0\}$, 3) we -converge to something less trivial but still uninteresting, namely, a Gaussian measure like the GFF $\mu_{C_{-\infty}}$ -or white noise, or massive free fields interpolating between the two. -Therefore the weak limit we need to study depends on the choice of bare ansatz $(\alpha_r,\mu_r,g_r)_{r\in\mathbb{Z}}$ -(or rather the germ of this sequence at $r=-\infty$). Finally, the well posed mathematical question I promised, -regarding trying to make sense of the original $\phi^4$ functional integral is the following. -Problem: Find an explicit parametrization of all weak limits (of probability measures on $S'(\mathbb{R}^d)$) -given by $\lim_{r\rightarrow-\infty}\lim_{s\rightarrow\infty}\nu_{r,s}$ for all possible choices of bare ansatz -$(\alpha_r,\mu_r,g_r)_{r\in\mathbb{Z}}$. -What renormalization theory in physics also tells us is that although it seems one has a hugely infinite-dimensional -amount of freedom in choosing the sequence $(\alpha_r,\mu_r,g_r)_{r\in\mathbb{Z}}$, the set $\mathscr{T}$ -of weak limit points -is a finite-dimensional variety. For $d=3$, one expects three parameters or "renormalized coupling constants" -$(\alpha_{\rm R},\mu_{\rm R},g_{\rm R})$ suffice. -One can even get rid of $\alpha_{\rm R}$ if one quotients by taking constant -multiples of the random field $\phi$. -There are rigorous renormalization group techniques for constructing portions of $\mathscr{T}$. Kupiainen's work mentioned -by Ofer is an adaptation of these techniques to the time-dependent rough SPDE setting. -The above is what I call the non-anchored Gibbsian way -of trying to construct elements $\nu\in\mathscr{T}$. There is a completely different approach which I call the -anchored stochastic quantization approach. There one need to make sense of the SPDE in Martin's answer, which he -did locally in time. Then one needs to understand this SPDE globally in time and construct an invariant -measure which gives $\nu\in\mathscr{T}$. This is also fraught with difficulties but there has been quite a bit of progress in this direction -(see, e.g., this article related to invariant measures and Martin's comment below). -The key difference between the two approches is that in the non-anchored setting one does not have a fixed probability space -to work with. In the second anchored situation one does since all fields are functionals of the driving noise. In the anchored setting, -$L^2$ estimates involving second moments only are enough to prove convergence in probability and thus in -law for the random fields of interest. In the non-anchored situation one needs to control all moments -(correlation functions) with uniform $n!$ bounds on these moments. -It is hard to say more as an MO answer, but you can see these papers for further explanations: - -"A second-quantized Kolmogorov-Chentsov theorem via the operator product expansion", published here. -"Towards three-dimensional conformal probability", published here. - "QFT, RG, and all that, for mathematicians, in eleven pages" - -Before reading these articles it may helpful to consult the slides of my recent Colloquium talk "A Toy Model for Three-Dimensional Conformal Probability". They should be much easier to follow since they contain lots of pictures. -The article in bullet point 1) provides an alternate way of defining pointwise products of random distributions -(like $\phi^2$ and $\phi^4$ above) using Wilson's operator product expansion (OPE). Regarding 5) in the OP's question, -I believe it would be of great interest to prove that the moments of the solution $\Phi$ constructed by Martin -satisfy the (dynamical version) of Wilson's OPE, and then compare $\Phi^3$ obtained by the theory of regularity structures -with the one constructed (from the OPE) in my 2nd quantized KC paper. - -Update (Jan 27, 2018): The article by Mourrat and Weber mentioned by Martin in his comment below has now appeared in CMP, see here. It provides a new construction of the scalar $\phi^4$ model in three dimensions, in finite volume. - -Update (Jan 21, 2019): Progress in the area has been quite fast and, since my last update, the limitation to finite volume has been overcome as mentioned in Martin's comment below. For the infinite volume treatment of $\phi_3^4$ via the stochastic quantization method see this article by Gubinelli and Hofmanová (with the paracontrolled approach) and the one by Moinat and Weber mentioned below (with the regularity structures approach).<|endoftext|> -TITLE: Reference: Stochastic Analysis on Hilbert Manifolds -QUESTION [5 upvotes]: I'm looking for a reference to a book which develops an It\^{o} lemma for semi-martingales with values in infinite dimensional Hilbert-Manifolds. I expect the techniques to be the same but still I appreciate a reference as there are hints in some of the literature that these things are known. -So far I've only found finite dimensional stuff. Can any one point me in the right direction? - -REPLY [2 votes]: As the question is general and observes the wide of the subject; I have chosen some reference. -Path Integrals on a Compact Manifold with Non-negative Curvature - -Foundations of the Theory of Semilinear Stochastic Partial Differential Equations - -Stochastic differential equations on manifolds<|endoftext|> -TITLE: Spaces that are finitely covered by manifolds -QUESTION [10 upvotes]: Suppose $X \to Y$ is a finite-sheeted covering of CW-complexes. Moreover, assume that the total space is homotopy equivalent to a (closed, connected, smooth) manifold $M$. I am interested in conditions to impose on the map $X \to Y$ so that we can find a manifold $N \simeq Y$ and a covering $M \to N$ making the diagram $$ \begin{array}{ccc} - X & \simeq & M \\ - \downarrow && \downarrow \\ - Y & \simeq & N - \end{array} -$$ -commute. In general, this is not possible, for example we could take any finite group $G$ and $$X = M \times EG \to M \times BG = Y.$$ Are there counterexamples for $Y$ a finite CW-complex? On the other side, are there conditions to impose so that we actually can deduce the existence of $N$? - -REPLY [6 votes]: Conditions where $N$ exists are for example when $M$ is a non-positively curved locally symmetric manifold, by a theorem of Mostow (for hyperbolic manifolds) and Margulis in general. Mostow's theorem is generalized to manifolds with hyperbolic fundamental group whose Gromov boundary is a sphere of dimension at least 5. So if one started with such a manifold, any group containing the fundamental group with finite index and torsion-free would have to be a manifold group. -More generally, it is conjectured that finitely presented groups which are $PD(n)$ groups are the fundamental groups of closed $n$-manifolds (Conjecture 3.4). If true, this would show that $N$ exists when $X$ is aspherical. This conjecture would follow for $n\geq 5$ from the Novikov conjecture. Many cases of this conjecture are known (for groups satisfying certain restrictions). See also Sections 5 and 8 of this paper.<|endoftext|> -TITLE: Is the dualizing sheaf on a Cohen-Macaulay scheme reflexive? -QUESTION [10 upvotes]: I am reading the paper Frobenius splitting of Hilbert schemes -of points on surfaces by Kumar and Thomsen. At the end of Lemma 11, they seem to imply that the dualizing sheaf on a Cohen-Macaulay scheme is reflexive. I am not familiar with reflexive sheaves, so I would like to ask whether this statement is true or not. - -REPLY [5 votes]: Just for the record, this does not need CM. The following is true: - -Let $Z$ be an excellent scheme that admits a dualizing complex. Then $ω_Z$ is torsion-free and $S_2$ on $Z$. If in addition Z is normal, then $ω_Z$ is a reflexive $\mathscr O_Z$ -module. - -Also note that torsion-free and $S_2$ seems to be the "right" notion to replace reflexivity on non-normal schemes. -If there is interest in this, I will add a proof. - -As requested, here is a proof: -(The following is Lemma 3.7.5 in this paper.) - -Lemma Let $Z$ be an excellent scheme that admits a dualizing complex. Then $\omega_Z$ is torsion-free and $S_2$ on $Z$. If in addition $Z$ is normal, then $\omega_Z$ is a reflexive $\mathscr O_Z$-module. - - -Proof. The statement is local, so we may assume that $Z$ is a noetherian affine local scheme. -Then, since it admits a dualizing complex, it can be embedded into a finite dimensional Gorenstein affine local scheme W as a closed subscheme by [Kaw02, Cor. 1.4]. -Being Gorenstein and local, $W$ must be pure dimensional. Let $r = \mathrm{codim}(Z, W)$. Then -by [Mat80, (16.B) Theorem 31(i)] there exists a length $r$ regular sequence in the ideal of $Z$ -in $W$ and let $W'$ be the common zero locus. Then $W'$ is also Gorenstein, -$Z\subseteq W$ is a closed subscheme and $\dim Z = \dim W$ . - - -It follows that $\omega_ {W'}^\bullet \simeq - \omega_{W'}[m]$, where $\omega_{W'}$ is a line bundle on ${W'}$. By Grothendieck -duality $\omega^\bullet_ Z\simeq RHom_{W'}(\mathscr O_Z, \omega^\bullet_{W'})$, and since $\dim Z=\dim W'$, -$\omega_Z\simeq Hom_{W'}(\mathscr O_Z, \omega_{W'})$ hence it is indeed torsion-free on -$Z$. - - -By [Stacks Project, Tag 0AWE] -$\omega_Z$ is $S_2$. In particular, if $Z$ is normal, then it is reflexive by [Stacks Project,Tag 0AVB]. - - -As far as non-trivial examples of schemes admitting dualizing complexes are concerned, try anything that can be embedded (locally) into a Gorenstein scheme. In fact, those are exactly the ones you are looking for.<|endoftext|> -TITLE: A special binomial identity in need of a proof -QUESTION [7 upvotes]: I've encountered a curious identity as a codicil in some work. Is there a proof or reference? -$$\sum_{k=-n}^n\frac{2k+1}{n+k+1}\binom{2n}{n-k}\frac{x^k}{1+x^{2k+1}}=\frac{x^n}{1+x^{2n+1}}.$$ - -REPLY [23 votes]: The $k=j-1$ and $k=-j$ terms cancel, so all that's left is the $k=n$ term.<|endoftext|> -TITLE: A simple requirement for a degree sequence to be graphical -QUESTION [5 upvotes]: The following theorem about the degree sequences of finite simple graphs is quite easy to prove from the Erdos-Gallai theorem. -Let $0 \lt \alpha \le \beta \lt n$ be integers. Call $(\alpha,\beta,n)$ paragraphical if every integer sequence $\alpha \le d_1,\ldots,d_n \le \beta$ with even sum is the degree sequence of a simple graph. -Then $(\alpha,\beta,n)$ is paragraphical iff -$$ n \ge \biggl\lfloor\frac{ (\alpha+\beta+1)^2 }{4 \alpha} \biggr\rfloor.$$ -I find it really hard to believe that nobody published this anywhere, but can I find it? The question is: where is it? -Thanks! - -REPLY [5 votes]: Answering my own question: Almost exactly this result was proved in: -Igor E Zverovich and Vadim E Zverovich. Contributions to the theory of graphic sequences. Discrete Mathematics, 105(1):293–303, 1992. -Having now found this, I recall that someone gave me the tip "Zverovich" at a conference last year, but I lost the tip and also forgot who it was that gave it to me (sorry). -An improvement, allegedly the best possible improvement, appeared in G. Cairns, S. Mendan, Y. Nikolayevsky, A sharp refinement of a result of Zverovich–Zverovich, Discrete Mathematics, Volume 338, Issue 7, 6 July 2015, Pages 1085-1089. -The result of Cairns et al. is similar to what is in my question, but not exactly the same. After looking very closely I can see that my version is wrong. The simplest proof is that it is not closed under complementation. My apologies to anyone who tried to prove it.<|endoftext|> -TITLE: Understanding (statement of) a theorem of Jack McLaughlin -QUESTION [6 upvotes]: In the book Twelve sporadic groups, Griess states - -If $A$ is an abelian group, $G$ acts on $A$, $z\in Z(G)$ satisfies $z-1\in$ Aut$(G)$, then $H^n(G,A)=0$ for $n\geq 0$. This is an observation of Jack McLaughlin. - -My first simple question is what do we mean here by $z-1\in$ Aut$(G)$? -Second question, I didn't get reference of it in google search; can one state a reference for a modern, elementary proof of it? (I mean, if a proof of this fact appeared in some book or recent papers, please state it.) - -REPLY [7 votes]: The statement is not identical, but it is the "center kills" argument, as in 149 page 42 of these lecture notes. -For $g \in Z(G)$, the map $m \mapsto gm$ is a ${\mathbb Z}G$-automorphism of $A$. The map it induces on the cohomology groups $H^n(G,A)$ is the so-called conjugation map $c_g$, and since $g \in G$, this is the identity map on $H^n(G,A)$. Of course, the same map with $g=1$, $m \mapsto m$ also induces the idenity on $H^n(G,A)$, so $g-1$ induces the zero map. -But if $g-1$ is an automorphism of $A$, then it is a ${\mathbb Z}G$-automorphism, and induces an automorphism of $H^n(G,A)$. So $H^n(G,A)=0$.<|endoftext|> -TITLE: Are all partial consecutive harmonic subsums distinct? -QUESTION [14 upvotes]: Let $b \gt a \geq 0$ be integers, and as elsewhere let $H_n$ be $\sum^n_{i=1} 1/i$. A partial consecutive harmonic subsum is a number $H(a,b)$ of the form $H_b - H_a$ (with $ H_0=0$). If $c=a$ and $d=b$ are two other integers we have of course -$$H(a,b)=H(c,d).$$ -Question: Are there any other cases where we have equality , especially where $d \gt c \gt b \gt a$ ? Note that $d=c$ and $b=a$ Is ruled out. -The following shows a connection with prime numbers. If $p$ is a prime number with $2p \gt b \geq p \gt a$, then $p$ is a factor of the denominator of the (reduced) fraction that is $H(a,b)$. There are more elaborate conditions that imply that a prime belongs to the denominator. If there is a nontrivial solution, then the (products of the numbers in the) intervals $(a,b]$ and $(c,d]$ have most if not all of their prime factors in common. In particular $(c,d]$ should have all composite numbers. -I am hoping for a distinctness result to help with a question about (non-exact) packing of the harmonic series. Again, references on this question to the literature are appreciated. -Edit 2017.02.05 GRP: -I appreciate the comments and links offered so far. Inspired by a suggestion of -Włodzimierz Holsztyński, I offer a conjecture toward his comparison of $H(ka,kb)$ and -$H(na,nb)$. The idea is to break both sums into $nk$ many pieces and compare the partial -sums. Toward showing $H(na,nb) \gt H(ka,kb)$ for integers $n \gt k \gt 0$, I conjecture -the following possibly stronger condition. Given integers $b \gt 0$ and $n \gt k \gt 0$, for all -$0 \lt j \leq nk$ the following inequality holds: -$$ \sum^j_{i=1} \frac1{k(bn + \lceil i/k \rceil)} \gt \sum^j_{i=1} \frac1{n(bk + \lceil i/n \rceil)} .$$ -I would be pleased to see this resolved. -End Edit 2017.02.05 GRP -Gerhard "Maybe Tie-in With Grimm Later" Paseman, 2017.01.30. - -REPLY [5 votes]: I thank Gerry Myerson. His post put me on watch for papers coauthored by Silberger, of which a recent one (http://arxiv.org/abs/1702.01316) refers to a paper of Erdos and Niven (Some properties of partial sums of the harmonic series) from 1946 which answers the title question affirmatively. Thanks also to the Silbergers and Erdos and Niven. -I thank Michael Stoll for his interest in the problem and for clarifying some statements I made in the post. -I also thank Włodzimierz Holsztyński for inspiring the following proof sketch, which solves the version in the edit. (I begin the sketch now, and will finish it and provide another in a subsequent edit.) The motivation is to show for integers $0 \lt k \lt n$ and $0 \lt a \lt b$ that $H(ka,kb) \neq H(na,nb)$. Using $a=0$ and $b=1$ for guidance, I show $H(ka,kb) \lt H(na,nb)$ for $b=a+1$, which will imply the motivated result. -I start by splitting each of the $k$ summands of $H(ka,ka+k)$ into $n$ equal parts, and similarly the $n$ summands of $H(na,na+n)$ each into $k$ equal parts. That suggests comparing two sums each of $nk$ terms. Except that I am using $a$ instead of $b$, and that I want $j=nk$, the sum comparison looks exactly like the comparison of the question edit above. Of course, if I prove the conjecture in the edit, I get the result I want. -I change notation now replacing $a$ and $a+1$ with $b$ and $b+1$, and subtract the two sums term by term. It is now enough for me to show that the following holds for $0 \lt j \leq nk$: -$$\sum^j_{i=1} \frac{n \lceil i/n \rceil - k \lceil i/k \rceil}{k(nb + \lceil i/k \rceil)n(kb + \lceil i/n \rceil)} \gt 0$$ -This is a straightforward induction on $j$ when $t_j = n \lceil j/n \rceil - k \lceil j/k \rceil$ is positive. -Since $t_1$ is positive, the partial sum can be shown positive also when $t_j$ is zero and the induction holds. It remains to show that the sum is still positive for those $j$ for which $t_j$ is negative. This will be provided in the edit to come. -Edit 2017.02.07 GRP: -To handle the case when $t_j$ is negative, I prove something even stronger. I restrict the indices $i$ and $j$ to range between $mn+1$ and $mn+n$ for an integer $m$ with $0 \leq m \lt k$, and I show that this portion (the sum for $i=mn+1$ to $j$) is also positive. I look for the largest $i$ in this range with $t_i \geq 0$ and note that it is not far from $mn+n$. If $r =(mn+n) \bmod k$ with $0 \leq r \lt k$ then this largest $i$ is $mn+n-r$. We now have $(n-r)$ positive terms of weight $r$ or more against $j-(n-r) \leq r$ negative terms of weight at most $(k-r)$ (I am suppressing mention of the denominators and the exact numerators of the summands), and since $n \gt k$, this means the positive terms outweigh the negative terms. Thus the subsubsums are also positive, so the subsums are positive, giving the desired conclusion and the end of this sketch. -I told you that story to tell you this one. The just completed sketch shows promise in giving an elementary proof of the title problem in a similar way. The hitch is that an additional term shows in the numerator. I will start this sketch and hope to finish it in a subsequent edit. -Recall that it suffices to show that subsums sharing no terms are disjoint. Now choose positive integers $a$ and $x$, and then choose $b \geq a+x$. There will be a unique integer $y \gt x$ such that $H(b,b+y-1) \lt H(a,a+x) \leq H(b,b+y)$. I want to show $H(a,a+x) \lt H(b,b+y)$ again by splitting the $1/(a+i)$ summands into $y$ equal pieces, and similarly each $1/(b+i)$ in $x$ equal pieces. I then line up the sums of the $xy$ many pieces and hope to prove positivity of the difference by proving positivity of the subsums as in the previous sketch. -When I do this, I get the following as a summand, where I use $d=ya-xb$: -$$ \frac{d+ y\lceil i/y \rceil -x\lceil i/x \rceil}{y(a+ \lceil i/y \rceil)x(b + \lceil i/x \rceil)} .$$ -Now if $d \geq 0$, the sketch above goes through and we can celebrate. If $d$ is negative and small, I think a variant of the above sketch will work, right down to the subsubsum portion. However I do not have that to put in this edit. -If $y/b$ is small (and indeed I can use Bertrand to make it less than 1 and other results in prime gaps to make it less than 1/5) then I can show $d$ has small size (perhaps less than $(y-x)/2$), and I may be able to tweak the sketch to provide a new and weaker proof of the Erdos Niven result. I would like to make it even more elementary, so with this edit I invite others to play with it. -End Edit 2017.02.07 GRP -Gerhard "Serializing Answers Seems Somewhat Easier" Paseman, 2017.02.07.<|endoftext|> -TITLE: Linear programming is continuous -QUESTION [6 upvotes]: Consider an arbitrary linear program: -$$\max \vec c \cdot \vec x$$ -subject to: -$$\textbf{A}\cdot \vec x = 0, \quad \vec a \le \vec x \le \vec b$$ -Assume that this program is feasible and bounded. Now suppose I perturb the bounds by small amounts: -$$\vec a' = \vec a + \delta \vec p, \quad \vec b' = \vec b + \delta \vec q$$ -where $\vec p,\vec q$ are unit vectors and $\delta > 0$. Assume that the modified problem remains feasible and bounded. -Prove or disprove: For arbitrary $\epsilon > 0$, there exists a $\delta > 0$ such that there exist optimal solutions $\vec x, \vec x'$ to the original and perturbed linear programs (respectively) satisfying $|\vec x - \vec x'| < \epsilon$. -Extra: Is there a way to define uniform continuity in this context, such that Linear Programming is, or isn't, uniformly continuous? - -REPLY [5 votes]: You can get this kind of continuity if the optimal solution is nondegenerate in the following sense. Let the coefficient matrix be $m \times n$, and suppose there is a subset $B$ of $[1,\ldots,n]$ with cardinality $m$ (the "basic variables") such that the submatrix $A_B$ for columns in $B$ is invertible, -and in your optimal solution the $x_j$ for $j \notin B$ are each equal to either $a_j$ or $b_j$, and -the $x_j$ for $j \in B$ are strictly between $a_j$ and $b_j$.<|endoftext|> -TITLE: Silver's approach to the inconsistency of $\mathrm{ZFC}$ -QUESTION [39 upvotes]: As all probably know, Jack Silver passed away about one month ago. The announcement released, with delay, by European Set Theory Society includes a quote by Solovay about his belief on inconsistency of measurable cardinals and $\mathrm{ZFC}$: - - -As Prof. Robert Solovay recently put it: "For at least the last 20 years, Jack was convinced that measurable cardinals (and indeed $\mathrm{ZFC}$) was inconsistent. He strove mightily to prove this. If he had succeeded it would have been the theorem of the century (at least) in set theory." - - -I was curious to find out what convinced him to not believe consistency of $\mathrm{ZFC}$ and what I kind of attempts he tried, of course I found nothing. - -Is there any published or unpublished note about his belief and approach, or possibly his philosophy toward it? - -REPLY [19 votes]: There is rumor that Silver's efforts started as an attempt to come up with a flawless argument showing the main "theorem" in Jensen's "A modest remark." That main "theorem" says that in ZF, there is no measurable cardinal. Magidor found the mistake a few days after Jensen released his manuscript. The manuscript still exists, but I'm pretty sure Jensen is happy if it's no longer circulated, even though it's historically also interesting as it has the first account of what's now called the Dodd-Jensen core model.<|endoftext|> -TITLE: Is it true that $X\times I\sim Y\times I\implies X\sim Y$? -QUESTION [12 upvotes]: So I asked this question a few weeks ago on MSE and I was suggested to repost it here. - -Let $I$ be the unit interval. Suppose that $X$ and $Y$ are topological spaces such that $X\times I$ is homeomorphic to $Y\times I$. Does it follow that $X$ is homeomorphic to $Y$? - -As pointed out in the comments in the other thread, there are counterexamples to analogous questions with $I$ replaced by the circle or the real line. Therefore I expect the answer to my question to be negative too. I would be also interested in what one can assume about $X$ and $Y$ to make the implication true. - -REPLY [7 votes]: @WlodekKuperberg (perhaps) and I (for sure) were exposed to this kind of examples by Karol Borsuk, or possibly Karol Borsuk simply had an example like the one I will present below: - -\begin{equation} -D\ :=\ \{z\in\mathbb C: |z|\le 1\}\ \subseteq\ \mathbb C -\ -\end{equation} -\begin{equation} -A\,\ :=\,\ D\times\{0\}\ \cup\ \{1\ \ \ i\ \ -\!1\ \ -\!i\}\times [-1;0]\,\ \subseteq\,\ \mathbb C\times\mathbb R -\end{equation} -\begin{equation} -X\,\ :=\,\ A\,\ \cup\,\ \{-1\ \ 1\}\times [0;1]\,\ \subseteq\,\ \mathbb C\times\mathbb R -\end{equation} -\begin{equation} -Y\,\ :=\,\ A\,\ \cup\,\ \{i\ \ \ 1\}\times [0;1]\,\ \subseteq\,\ \mathbb C\times\mathbb R -\end{equation} -Then $\ X\ $ and $\ Y\ $ are not homeomorphic while $\ X\times I\ $ -and $\ Y\times I\ $ are. - -REMARK One may check Karol Borsuk's series of publications about the uniqueness of topological decomposition into Cartesian products, and a paper by Hanna Patkowska about the uniqueness of the decomposition of ANRs into 1-dimensional ANRs. - -A kind request (I'd greatly appreciate): Wlodek Kuperberg, please add a picture to my analytic description; let the pictures of $\ X\ $ and $\ Y\ $ be embedded into $\ \mathbb C;\ $ I mean homeomorphic copies of $\ X\ $ and $\ Y$. - -ACKNOWLEDGEMENT I am grateful to Wlodek Kuperberg for providing such a very nice graphics (so cute and psychologically loaded; it's the first graphics illustration in my MO posts)). *** Włodek, congratulation on your another NICE answer (Gauss said, a few but ripe).<|endoftext|> -TITLE: Zariski density of conjugates of $SL_2(\mathbb{Z})$ in $Sp_{2g}$ -QUESTION [6 upvotes]: Let $Sp_2g$ be the symplectic group defined over $\mathbb{Q}$. Consider $SL_2(\mathbb{Z})$ as a subgroup of $Sp_{2g}(\mathbb{Z})$ (the embedding that I have in mind is $A\to \begin{pmatrix} -A& 0\\ -0& I_{2g-2} -\end{pmatrix}$, but you can take any algebraic group embedding which yields a positive answer to the following question). -Is it true that there eixst finitely many $g_i\in Sp_{2g}(\mathbb{Q})$ such that the $\mathbb{Q}$-Zariski closure of $\cup_i g_i SL_2(\mathbb{Z})g_i^{-1}$ is equal to $Sp_{2g}$? i.e. that this union is Zariski-dense in $Sp_{2g}$? If yes, what conditions and restrictions should these elements satisfy? How should the $g_i$ be chosen? I would appreciate any result, comments or reference in this direction. - -REPLY [5 votes]: Even for any algebraic group embedding of $SL_2$ into $Sp_{2g}$, with an arbitrary amount of conjugation by arbitrary elements, the answer is no. Consider the matrix $\begin{pmatrix} \lambda & 0 \\ 0 & \lambda^{-1} \end{pmatrix}$ in $SL_2$. Its image under the algebraic group embedding is some $2g \times 2g$ matrix whose eigenvalues are powers of $\lambda$. The characteristic polynomial of this matrix is some $2g$-tuple of polynomials in $\lambda$, all defined over $\mathbb Z$, whose image is a $1$-dimensional subvariety, defined over $\mathbb Q$, of the space of polynomials. -Because elements conjugate to this one are dense in $SL_2(\mathbb C)$, the image of every element of $SL_2(\mathbb C)$ under the characteristic polynomial map is contained in this $1$-dimensional $\mathbb Q$-subvariety. The same is true for any number (even an infinite number) of conjugate copies of $SL_2$. Because the characteristic polynomials of elements of $Sp_{2g}$ form a $g$-dimensional variety, there is no way this union can be $\mathbb Q$-Zariski dense.<|endoftext|> -TITLE: Does Spin cobordism vanish in dimension $4k-1$? -QUESTION [21 upvotes]: For the purposes of a remark in a paper in preparation, I would like to know if anyone can confirm that $\Omega^{spin}_{4k-1} = 0$. -In the Atiyah-Patodi-Singer paper, Spectral asymmetry and Riemannian geometry: II (Math. Proc. Camb. Phil. Soc., 78, (1975), 405–432) they give an analytical interpretation of the Adams $e$-invariant; see page 422. This starts with the statement that the spin cobordism group $\Omega^{spin}_{4k-1} = 0$, with a reference to Stong's book on cobordism. I didn't find this statement there, and so went back to the original Anderson-Brown-Peterson papers ([1] The Structure of the Spin Cobordism Ring, Annals 86, No. 2 (Sep., 1967), pp. 271-298 and research announcement [2] Spin Cobordism, Bull. AMS (1966), 256-260) on which Stong's treatment is based. Extracting the answer from those papers seems fairly strenuous, so I hope someone more expert than I can help. -One possibly confusing item (pointed out to me by Vitaly Lorman) is in Theorem 1.7 of [2], which reports some computer calculations of the image of $\Omega^{spin}_n$ in unoriented cobordism. This image is reported to be non-trivial for $n = 39, 43$, which seems to contradict the APS statement. Perhaps those calculations are wrong, although the authors say that they confirm the main results of the paper. -By the way, the same vanishing statement also appears in the paper of Atiyah-Smith (Compact lie groups and the stable homotopy of spheres, Topology 13, pp. 135-142, 1974). - -REPLY [9 votes]: The spin cobordism groups $\Omega^{spin}_n$ have been computed for $n \leq 127$; see section 10 of Secondary Invariants for String Bordism and tmf by Bunke and Naumann. They use MAPLE together with the decomposition of the 2-completion of $MSpin$ found by Anderson, Brown, and Peterson in their paper Spin Cobordism. -In particular, one sees that $\Omega^{spin}_{4k-1}$ is zero for $1 \leq k \leq 9$, but $\Omega^{spin}_{39} \cong \mathbb{Z}_2\oplus\mathbb{Z}_2 \neq 0$. In fact, $\Omega^{spin}_{4k-1} \neq 0$ for $10 \leq k \leq 32$.<|endoftext|> -TITLE: Properness of schemes with $\mathbb C^*$-action via properness of orbits and fixed points locus -QUESTION [5 upvotes]: Question: Let $X$ be a scheme separated and of finite type over $\mathbb C$ with a $\mathbb C^*$-action. Suppose that - -The fixed point locus is proper. -Each orbit has proper closure. - -Does this imply that $X$ is proper? Proper means proper over $\mathbb C$, of course. -Some comments: It would be handy if we had this criterion for properness. I have been trying to find a counterexample but did not succeed. I tried to start with a proper scheme and then throw away some non-constant orbits. But it seems that this will always give us something like $\{(x,y)\in \mathbb A^2, y= 0 \text{ when }x=0\}$, which is not a scheme. I am not sure if this picture is correct in general, especially when $X$ is highly singular. Also there could (a priori) be examples which can not be compactified respecting the $\mathbb C^*$-action. - -REPLY [8 votes]: Edit. I am making the answer shorter by eliminating the repeated argument. -Let $k$ be a perfect field. Let $X$ be a finite type, separated $k$-scheme. Let $$m:\mathbb{G}_m \times_{\text{Spec}\ k} X \to X$$ be a $k$-action of $\mathbb{G}_m$ on $X$. Since $k$ is perfect, $m$ induces a $k$-action of $\mathbb{G}_m$ on $X_{\text{red}}$, the reduced scheme of $X$. By the universal property of the normalization, this in turn induces an action of $\mathbb{G}_m$ on the normalization, $X^{\text{nor}}_{\text{red}}$. The $k$-scheme $X$ is proper if and only if $X^{\text{nor}}_{\text{red}}$ is proper. Thus, without loss of generality, assume that $X$ is integral and normal. -For every $k$-scheme, $q:T\to \text{Spec}\ k$, and for every $k$-morphism to $X$, $f:T\to X$, as in Geometric Invariant Theory, denote by $$\psi_f:\mathbb{G}_m\times_{\text{Spec}\ k} T \to X\times_{\text{Spec}\ k} T$$ the $T$-morphism whose projection to $X$, $\text{pr}_X\circ f_m$, equals the composition, $$\mathbb{G}_m\times_{\text{Spec}\ k} T \xrightarrow{(\text{Id}_{\mathbb{G}_m},f)} \mathbb{G}_m\times_{\text{Spec}\ k} X \xrightarrow{m} X.$$ Let $\mathbb{G}_m\times_{\text{Spec}\ k} \mathbb{P}^1 \to \mathbb{P}^1$ be the standard extension to $\mathbb{P}^1$ of the regular action of $\mathbb{G}_m$ on the dense open $\mathbb{G}_m = \mathbb{P}^1\setminus\{0,\infty\}$. Then the morphism $\psi_f$ is a $T$-rational transformation, $$\psi_f:\mathbb{P}^1\times_{\text{Spec}\ k} T \dashrightarrow X\times_{\text{Spec}\ k} T.$$ -Closed Orbit Hypothesis. For every morphism $f$ with $T = \text{Spec}\ K$ for $K$ a field extension of $k$, the morphism $\psi_f$ extends to a regular morphism $\mathbb{P}^1\times_{\text{Spec}\ k} \text{Spec}\ K \to X\times_{\text{Spec}\ k} \text{Spec}\ K$. -Assume the closed orbit hypothesis. Then when $T$ is Spec of a DVR, say $R$ with residue field $\text{Spec}\ \kappa$ and with fraction field $\text{Spec}\ K$, then the rational transformation $\psi_f$ is regular on the generic fiber $\mathbb{P}^1\times_{\text{Spec}\ k}\text{Spec}\ K$. In particular, restricting this regular morphism to $\{0\}\times_{\text{Spec}\ k} \text{Spec}\ K$ gives a $k$-morphism, $$\psi_{f,0}: \text{Spec}\ K \to X\times_{\text{Spec}\ k} \text{Spec}\ R.$$ Since $0\in \mathbb{P}^1$ is a $\mathbb{G}_m$-invariant point, also $\psi_{f,0}$ maps into the fixed locus $X^{\mathbb{G}_m}$. -Fixed Locus Hypothesis. For every morphism $f$ with $R$ a DVR containing $k$, the associated regular morphism $\psi_{f,0}:\text{Spec}\ K \to X\times_{\text{Spec}\ k} \text{Spec}\ R$ extends to a regular morphism on $\text{Spec}\ R$. -Via the valuative criterion of properness, this is equivalent to properness of the fixed locus $X^{\mathbb{G}_m}$. It is convenient to modify $\psi_f$ a bit. Denote by $\phi_f$ the unique morphism, $$(\text{pr}_{\mathbb{P}^1},\psi_f):\mathbb{P}^1 \times_{\text{Spec}\ k} \text{Spec}\ K \to \mathbb{P}^1\times_{\text{Spec}\ k} X\times_{\text{Spec}\ k} \text{Spec}\ K.$$ -This is a locally closed immersion. Denote by $W_f$ the Zariski closure of the image of $\phi_f$. This is a finite type, separated, integral $R$-scheme, but it may be that $W_f$ is not normal. Thus, denote by $W_f^{\text{nor}}\to W_f$ the normalization. By the fixed locus hypothesis, $W_f$ contains the image of the regular morphism, '$$\phi_{f,0}:\text{Spec}\ R \to X\times_{\text{Spec}\ k} \text{Spec}\ R =\{0\} \times X\times_{\text{Spec}\ k} \text{Spec}\ R.$$ Thus the closed fiber of $W_f^{\text{nor}}$ is not empty. -Claim. After a finite sequence of blowings up of $\mathbb{P}^1\times_{\textbf{Spec}\ k}\text{Spec}\ R$ at $\mathbb{G}_m$-fixed closed points, the rational transformation $\phi_f$ extends to a regular $\mathbb{G}_m$-equivariant morphism to $W_f^{\text{nor}}$. In particular, the strict transform of $\{1\}\times \text{Spec}\ R$ gives an extension of $f$ to a regular morphism $\text{Spec}\ R \to X$. -Altogether, the projection, $$\text{pr}_{\mathbb{P}^1,R}:W_f^{\text{nor}} \to \mathbb{P}^1\times_{\text{Spec}\ k} \text{Spec}\ R,$$ is a finite type, separated, $\mathbb{G}_m$-equivariant $R$-morphism that is an isomorphism on $\text{Spec}\ K$-fibers. The remainder of the argument is an analysis of this morphism. However, to prove boundedness, it is convenient to use an ad hoc, auxiliary morphism. Since $W_f^{\text{nor}}\to \text{Spec}\ R$ is a flat, separated, finite type morphism of relative dimension $1$, it is a quasi-projective relative curve. Thus, there exists a dense open $R$-immersion, $\iota_f:W_f^{\text{nor}} \to \overline{W}_f$ where $\overline{W}_f\to \text{Spec}\ R$ is a projective curve. Denote by $\mathcal{L}$ a very ample invertible sheaf on $\overline{W}_f$. Moreover, let $s:R^{\oplus 3} \to H^0(\overline{W}_f,\mathcal{L})$ be global sections that generate $\mathcal{L}$. Denote by $d$ the degree of $\mathcal{L}$ on the fibers of $\overline{W}_f\to \text{Spec}\ R$. -By the valuative criterion of properness, the $R$-rational transformation, $$\phi_f:\mathbb{P}^1\times_{\text{Spec}\ k}\text{Spec}\ R \dashrightarrow \overline{W}_f,$$ extends to a regular morphism away from finitely many points. -The pullback of $\mathcal{L}$ by this regular morphism extends to an invertible sheaf on all of $\mathbb{P}^1\times_{\text{Spec}\ k} \text{Spec}\ R$, and the global sections $s$ generate the extension away from the finitely many indeterminacy points. The degree of the extension of $\mathcal{L}$ equals $d$ since $\phi_f$ is an isomorphism of $\text{Spec}\ K$-fibers. -The restriction of $\phi_f$ to the $\text{Spec} \ \kappa$-fiber is an $R$-rational transformation $\mathbb{P}^1\times_{\text{Spec}\ k} \text{Spec}\ \kappa \to \overline{W}_f$. By the valuative criterion, this extends to a regular morphism. The pullback of $\mathcal{L}$ is a globally generated invertible sheaf. Thus the degree $d_0$ of this invertible sheaf satisfies $0\leq d_0\leq d$. The number of blowings up necessary to regularize $\phi_f$ is $\leq d - d_0$. -We will iteratively blow up the $\mathbb{G}_m$-invariant closed points in the indeterminacy locus of $\phi_f$ over $0$. Thus, let -$$P_m \xrightarrow{\nu_m} P_{m-1} \xrightarrow{\nu_{m-1}} \dots \xrightarrow{\nu_1} \mathbb{P}^1 \times_{\text{Spec}\ k} \text{Spec}\ R$$ be a sequence of blowings up of a single $\mathbb{G}_m$-fixed point that maps to $\{0\}\times \text{Spec}\ \kappa$ and that is in the indeterminacy locus of $\phi_f$. By Zariski's Main Theorem, each exceptional divisor $E_i$ of $\nu_i$ is mapped quasi-finitely to $\overline{W}_f$. By the valuative criterion of properness, the restriction to $E_i$ of $\phi_f$ extends to a regular morphism, and this is finite to its image. Thus, the pullback of $\mathcal{L}$ on $E_i$ by this regular, finite morphism has degree $\geq 1$. Since the sum of $d_0$ and the degrees of the exceptional divisors is $\leq d$, the number $m$ of such blowings up is $\leq d-d_0$. So for some $m\leq d-d_0$, the associated rational transformation $$\phi_f:P_m \dashrightarrow \overline{W}_f$$ is regular at every $\mathbb{G}_m$-equivariant point lying over $0$. -Denote by $P_m^o\subset P_m$ the maximal open subscheme on which $\phi_f$ is regular and maps $P_m^o$ into the open subset $W_f^{\text{nor}} \subset \overline{W}_f$. This is a $\mathbb{G}_m$-invariant open subset of $P_m$. -In particular, $\phi_f$ is regular on the strict transform of $\{0\}\times \text{Spec}\ R$. Since $\phi_{f,0}$ maps this into $W_f^{\text{nor}}$, $P_m^o$ contains the strict transform of $\{0\}\times \text{Spec}\ R$. Thus, $P_m^o$ also intersects the unique $\nu$-exceptional component $E_0$ of $P_m\times_{\text{Spec}\ R} \text{Spec}\ \kappa$ -that intersects this strict transform. The complement of $P_m^o$ in this component consists of at most $1$ closed point, and this is $\mathbb{G}_m$-equivariant. By hypothesis, $\phi_f$ is regular at this closed point. And then, by the closed orbit hypothesis, this closed points is contained in $W_f^{\text{nor}}$. Now we can repeat this argument with every $\nu$-exceptional irreducible component $E_{i+1}$ of $P_m\times \text{Spec}\ \kappa$ that intersects an irreducible component $E_i$ on which it has already been proved that $\phi_f$ is regular and maps $E_i$ into $W_f^{\text{nor}}$. In particular, since the strict transform of the original closed fiber $\mathbb{P}^1 \times \text{Spec}\ \kappa$ intersects one of these exceptional components, $\phi_f$ is regular at the unique closed point of the strict transform that maps to $\{0\} \times \text{Spec}\ \kappa$, and this closed point maps into $W_f^{\text{nor}}$. So, by the same argument, the complement of $P_m^o$ in this strict transform consists of at most one point, the $\mathbb{G}_m$-invariant closed point $\infty$, and then $\phi_f$ is not regular at $\infty$. -Now we can repeat the argument above with $\infty$ in place of $0$. Instead of using the strict transform of $\{0\}\times \text{Spec}\ R$, use the strict transform of $\mathbb{P}^1\times \text{Spec}\ \kappa$. -Thus, let -$$P_n \xrightarrow{\nu_n} P_{n-1} \xrightarrow{\nu_{n-1}} \dots \xrightarrow{\nu_{m+1}} P_m$$ be a sequence of blowings up at $\mathbb{G}_m$-fixed points that map to $\{\infty\}\times \text{Spec}\ \kappa$ and each of which is in the indeterminacy locus of $\phi_f$. By Zariski's Main Theorem, each exceptional divisor $E_i$ of $\nu_i$ is mapped quasi-finitely to $\overline{W}_f$. Thus, the pullback of $\mathcal{L}$ on $E_i$ has degree $\geq 1$. Since the sum of $d_0$ and the degrees of the exceptional divisors is $\leq d$, the number $n$ of such blowings up is $\leq d-d_0$. So for $n\leq d-d_0$, the associated rational transformation $$\phi_f:P_n \dashrightarrow \overline{W}_f$$ is regular at every $\mathbb{G}_m$-equivariant point lying over $\infty$. -By the same argument as above, the open subset $P_n^o\subset P_n$ contains the entire closed fiber $P_n\times_{\text{Spec}\ R} \text{Spec}\ \kappa$. Since $P_n$ is proper over $\text{Spec}\ R$, $P_n^o$ equals all of $P_n$. This proves the claim, and this implies that $X$ is proper. -Comment. For a product of multiplicative groups, e.g., $G\times H \cong \mathbb{G}_m\times \mathbb{G}_m$, the same result holds. The point is that the closed orbit hypothesis for $G\times H$ implies the closed orbit hypothesis for both $G$ and $H$. Thus, for an action of $G\times H$ on $X$, the $G\times H$-closed orbit hypothesis for $X$ implies the $H$-closed orbit hypothesis for the $H$-invariant closed subset $X^G\subset X$. Of course the $H$-fixed locus inside $X^G$ equals the $G\times H$-fixed locus. Thus the $G\times H$-fixed locus hypothesis for $X$ implies the $H$-fixed locus hypothesis for $X^G$. By the argument above, $X^G$ is proper. Therefore $X$ satisfies both the $G$-closed orbit hypothesis and the $G$-fixed locus hypothesis. So by the argument above, $X$ is proper. -Please note, however, that the result does not hold for more general reductive groups. For $G$ a nontrivial semisimple group, for a parabolic subgroup $P$, for the natural action of $G$ on $G/P$, for any separated, finite type $k$-scheme $Y$ that is not proper, the induced action of $G$ on $X=(G/P)\times_{\text{Spec}\ k} Y$ satisfies the $G$-closed orbit hypothesis. Moreover, it vacuously satisfies the $G$-fixed locus hypothesis since the $G$-fixed locus is empty. Yet $X$ is not proper.<|endoftext|> -TITLE: Albanese variety over non-perfect fields -QUESTION [17 upvotes]: It is a result of Serre (Morphismes universels et varietes d'albanese) that the Albanese (abelian) variety, i.e. an initial object for morphisms to (torsors over) abelian varieties, exists for any reduced scheme over a perfect field. -I have a counterexample for non-reduced schemes, but how about the perfectness of the base field? So my questions are: - -(1) Where does Serre's proof break down for non-perfect base fields? -(2) Does anyone know a counterexample over non-perfect fields? - -NB: Serre does not state the hypothesis (maybe because at that time varieties were always integral over an algebraically closed field), but every author who quotes him states those hypothesis. - -REPLY [13 votes]: The answer is affirmative (in Serre's formulation via principal homogeneous spaces) for proper, geometrically reduced, and geometrically connected schemes $X$ over any field $k$, giving a strong mapping property relative to working over all $k$-schemes. That is: -Theorem There exists a map $f:X \rightarrow E$ to a torsor for an abelian variety $A$ over $k$ such that for any $k$-scheme $S$ and any $S$-map $F:X_S \rightarrow T$ to a torsor for an abelian scheme $B$ over $S$, there exists a unique $S$-map $g:E_S \rightarrow T$ such that $g \circ f_S = F$. -Remark. The necessity of assuming geometric connectedness is illustrated in the Example at the end below. It is a useful weakening of geometric integrality (e.g., reducible semistable curves!), and Grothendieck's formulation in his Bourbaki expose appears to accidentally miss this necessary hypothesis. -Remark. In the above setting, there is a unique map of $S$-groups $A_S \rightarrow B$ making $g$ equivariant for their respective actions. Indeed, we claim the same for any $S$-map between torsors over $S$ for any two abelian schemes (no need for one of them to have arisen by base change from $k$), and by descent it suffices to prove such existence and uniqueness over an fppf cover of $S$. That reduces the task to the case of trivial torsors, so the claim is that if $B$ and $B'$ are abelian schemes over $S$ then any map of $S$-schemes $F:B \rightarrow B'$ (not necessarily respecting identity sections) is equivariant for a unique map between the abelian schemes. But $F = t_{b'} \circ f$ for $b' = F(e)$ and $f$ an $S$-homomorphism (with $t_{b'}:x \mapsto b'+x$), so all is clear. -The abelian variety $A$ turns out to be the dual of the maximal abelian subvariety of the (possibly non-reduced and non-proper) Picard scheme of $X$. The only literature reference I'm aware of is a Bourbaki expose by Grothendieck that treats "just" the case of geometrically normal proper $k$-schemes. In lieu of that, I give the proof of the more general result below, after reviewing what is in Grothendieck's paper (also identifying the step within which he missed the need for a geometric connectedness hypothesis). -To build up to this in stages, let's first go back to the ur-reference on this matter: the case of schemes that are proper and geometrically normal (and by necessity also geometrically connected, to be explained below) over any field is treated in Grothendieck's TDTE VI, Seminaire Bourbaki Expose 236, Cor. 3.2 and Theorem 3.3 (esp. part (iii), where $A$ refers to the unnamed abelian scheme in part (i), as in the discussion just before Theorem 3.3, and has nothing to do with $A$ in part (ii) that is not defined there but is the coordinate ring of a local scheme $S$). -In these results Grothendieck is assuming the existence of certain Picard schemes and dual abelian schemes (which of course he knew in many cases), and he is aiming to prove a stronger mapping property for the Albanese (over general $k$-schemes, not just over $k$ or field extensions thereof). The existence hypotheses that Grothendieck makes on Pic and on the dual abelian scheme in his statements of 3.2 and 3.3 were of course proved by him in the projective and geometrically integral case. Below we'll comment on the verification of such existence hypotheses more widely. -Note that in Corollary 3.2 when Grothendieck writes "normal et proper sur $k$" he means for "sur $k$" to qualify normality too (not just to qualify properness), with normality "over $k$" meaning by definition what is now called "geometrically normal" over $k$. One can see this intent from the use of 2.1(ii) in his proof (though it is also consistent with Cor. 6.7.8(vi) and Def. 6.8.1(vi) in EGA IV$_2$, where "normal over $k$" terminology is introduced with the same meaning). -The proofs in TDTE VI (as in other TDTE exposes) are sometimes just sketches, such as for 3.3(iii). The reason for his geometric normality hypothesis in 3.2 is to ensure via the valuative criterion for properness that ${\rm{Pic}}^0_{X/k}$ is proper, in which case he can use its finite etale $n$-torsion for all $n$ not divisible by ${\rm{char}}(k)$ to show that the underlying reduced scheme of this commutative finite type $k$-group scheme is a smooth $k$-subgroup scheme (hence an abelian variety). -The existence of Pic$_{X/k}$ as a locally finite type $k$-scheme for $X$ proper (not just projective) and geometrically reduced over a field $k$ was proved independently by Murre and Oort (if I remember correctly), though also recovered later by Artin as a special case of his existence results over general base schemes for Pic as a algebraic space quasi-separated and locally of finite presentation over the base (together with the fact that an algebraic space group locally of finite type over a field is a scheme, which is a consequence of Knutson's result that any locally noetherian quasi-separated algebraic space contains a dense open subspace that is a scheme, combined with careful translation arguments over finite extensions of the field). -In Theorem 3.3(iii) Grothendieck has to assume the existence of the dual abelian scheme. Over a field (all you need) this was of course known at that time in general since abelian schemes over fields are projective. In general it is due independently to Raynaud and Delinge, and is proved via algebraic-space techniques early in the book of Faltings and Chai. In general, the latter argument shows if $B \rightarrow S$ is an abelian scheme then the locally finitely presented algebraic space group ${\rm{Pic}}_{B/S}$ is $S$-separated (using valuative criterion) and the union of its fibral identity components is an open and closed $S$-subgroup ${\rm{Pic}}^0_{B/S}$ that is the dual abelian scheme, denoted $B^{\vee}$. -The upshot is that if a $k$-proper $X$ is geometrically reduced over $k$ then Grothendieck's existence hypotheses on Pic over $k$ and dual abelian schemes over any $k$-scheme are verified (and such objects over $k$ of course yield analogous representing objects after base change to any $k$-scheme). But what about the geometric normality hypothesis imposed in Cor. 3.2? -Grothendieck wants to build an abelian subvariety $C \subset P := {\rm{Pic}}_{X/k}$ such that any $S$-homomorphism to $P_S$ from an abelian scheme factors through $C_S$. The dual of such a $C$ will turn out to satisfy the desired Albanese property (for reasons we will explain below, using the further condition of geometric connectedness for $X$ over $k$). He follows the natural strategy of seeking conditions under which $P^0_{\rm{red}}$ is a smooth $k$-subgroup scheme of $P^0$ that is proper and hence is an abelian variety, in which case that would do the job (as one sees by considerations with relative schematic density of finite etale torsion levels in abelian schemes Zariski-locally on the base, or other reasons via $P^0/P^0_{\rm{red}}$). It is this strategy that leads him to impose the geometric normality condition on $X$. -But we can do better, by assuming only that $X$ is proper and geometrically reduced and geometrically connected. This involves applying the following result to ${\rm{Pic}}_{X/k}$. -Proposition. Let $G$ be a commutative group scheme locally of finite type over a field $k$. Let $C \subset G$ be the maximal abelian subvariety. For any $k$-scheme $S$ and abelian scheme $B$ over $S$, any $S$-homomorphism $B \rightarrow G_S$ factors through $C_S$. -Before we give the proof, we warn that $G_{\rm{red}}$ is generally not a $k$-subgroup of $G$ (when $k$ is not perfect and $G$ is not smooth), even if $G$ is geometrically connected; see Example 1.3.2(2) in Exp. VI$_{\rm{A}}$ of SGA3 for Raynaud's commutative affine example of such over any imperfect field. -Also, since any quotient of an abelian variety by a closed subgroup scheme is an abelian variety, we see that such a maximal $C$ exists for dimension reasons (and any abelian subvariety is contained in the identity component $G^0$ that is of finite type). Of course, one has no idea about the dimension of $C$ or even its non-triviality in general. -Proof. The main work is to show that the formation of such $C$ commutes with any extension of the ground field. Equivalently, after passing to $G/C$, we claim that if $C=0$ then $G_K$ contains no nonzero abelian subvariety for any extension field $K/k$. By Galois descent it is clear that the maximal abelian subvariety of $G_{k_s}$ vanishes. For any $K/k$, let $B \subset G_K$ be an abelian subvariety. We want to prove $B=0$. We can ramp up $K$ to be separably closed and hence contains $k_s$, so for the purpose of proving $B=0$ we can assume $k=k_s$. -For any $n > 0$ not divisible by ${\rm{char}}(k)$, $B[n]$ is a subgroup of $G_K[n]=G[n]_K$, and $G[n]$ is a finite etale $k$-group since it is killed by $n$. Thus, $G[n]$ is constant since $k=k_s$, so $B[n]$ descends uniquely to a $k$-subgroup $H_n \subset G[n]$. Let $H \subset G$ be the Zariski closure of the collection of $H_n$'s. The formation of $H$ commutes with extension of the ground field, so clearly $H_K = B$. Hence, $H$ is a $k$-subgroup of $G$ that is an abelian variety, so $H=0$ and hence $B=0$. -Now let $S$ be a $k$-scheme and $f:B \rightarrow G_S$ be an $S$-homomorphism from an abelian scheme $B$ over $S$. We want this to factor through $C_S$. By composing with the natural map to $(G/C)_S$ that represents the fppf quotient of $G_S$ modulo $A_S$, we can replace $G$ with $G/C$ so that $C=0$. Thus, for all $s \in S$ the $k(s)$-group scheme $G_s$ contains no nonzero abelian subvariety over $k(s)$ due to the compatibility with ground field extension discussed above. -In particular, the map $f_s: B_s \rightarrow G_s$ vanishes for all $s \in S$ since $B_s/(\ker f_s)$ is an abelian variety that is a $k(s)$-subgroup of $G_s$. But then by Corollary 6.2 in GIT (after passing to the case when $S$ is noetherian, as we may certainly do) it follows that $f=0$! -QED Proposition -Now we are ready to prove the Theorem stated at the start. The first thing we recall is that any torsor $E$ for an abelian variety $A$ over $k$ has ${\rm{Pic}}^0_{E/k}$ canonically isomorphic to the dual abelian variety $A^{\vee}$. Indeed, we can pick a finite Galois extension $k'/k$ so that $E(k')$ is non-empty, so $E_{k'} \simeq A_{k'}$ as torsors, well-defined up to $A_{k'}$-translation on $A_{k'}$. But the effect of $A_{k'}$-translation on $A_{k'}$ makes $A_{k'}$ act on the $k'$-group scheme ${\rm{Pic}}_{A_{k'}/k'}$, and that action is trivial since $A_{k'}$ is smooth and connected and ${\rm{Pic}}^0_{A_{k'}/k'}$ is an abelian variety (so its automorphism scheme as a $k'$-group is etale and thus receives no nontrivial homomorphism from an abelian variety). It follows that the pullback isomorphism $A^{\vee}_{k'} = {\rm{Pic}}^0_{A_{k'}/k'} \simeq {\rm{Pic}}^0_{E_{k'}/k'} = ({\rm{Pic}}^0_{E/k})_{k'}$ is independent of the choice of point in $E(k')$ that underlies the torsor isomorphism with $A_{k'}$. Thus, the composite pullback isomorphism is equivariant for the Galois descent data on both sides and hence it descends to a $k$-isomorphism $A^{\vee} \simeq {\rm{Pic}}^0_{E/k}$ that is independent of all choices. -The upshot of having the canonical isomorphism $A \simeq {\rm{Pic}}^0_{E/k}$ is that $A$ is not "extra data" in the statement of the Theorem, but rather is canonically determined by $E$. By using the etale topology in place of Galois theory, the exact same argument shows that for any torsor $T$ for an abelian scheme $B$ over a scheme $S$, the algebraic space ${\rm{Pic}}_{T/S}$ contains a unique open and closed $S$-subgroup ${\rm{Pic}}^0_{T/S}$ with (geometrically) connected fibers and that this $S$-subgroup is canonically isomorphic to the dual abelian scheme $B^{\vee}$, so in particular it is a scheme. (Any action of an abelian scheme on another abelian scheme over any base is trivial, due to triviality locally on the base for the action on the finite etale torsion levels that are collectively relatively schematically dense in the sense of EGA IV$_3$, 11.10.) -By the effectivity of Galois descent for (quasi-)projective $k$-schemes (such as $E$, whatever it may be) it suffices to prove the Theorem after making a preliminary finite Galois extension on $k$. Here we are using the earlier observation that any $g$ as in the Theorem determines a unique $S$-homomorphism $A_S \rightarrow B$ over which $g$ is equivariant. We are going to build the desired $A$ as dual to the maximal abelian subvariety $C$ of ${\rm{Pic}}_{X/k}$. -It is at this point that the need for geometric connectedness arises, and to clarify that we avoid using that condition for a moment. So -upon making a suitable finite Galois extension on $k$ we can arrange that the geometrically reduced (and hence generically smooth!) $k$-scheme $X$ has the form $\coprod X_i$ with each $X_i$ not only geometrically reduced but also geometrically connected and having a $k$-point $x_i \in X_i$. -Fix such choices of $x_i$ (an auxiliary device in the proof, not part of the statement of the Theorem). -Any $S$-map $F:\coprod (X_i)_S = X_S \rightarrow T$ to a torsor for an abelian scheme $B$ thereby has $T$ canonically identified with $B$ using the section $F((x_1)_S)$, and likewise for the sought-after universal $k$-map $f:\coprod X_i = X \rightarrow E$. Provided that there is only a single $X_i$ (i.e., the original $X$ is geometrically connected over $k$), our task is reduced to the analogue of the Theorem using abelian schemes and pointed maps (with $x_1 \in X(k)$) rather than torsors for abelian schemes. It is precisely at this step that geometric connectedness is essential, because otherwise making $(X_1)_S \rightarrow T$ a pointed map by identifying $T$ with $B$ via the image of $(x_1)_S$ would not have any impact on making the resulting $S$-maps $(X_i)_S \rightarrow B$ be pointed (and it is precisely here that counterexamples to the Theorem in the absence of geometric connectedness over $k$ creep in; see the Example at the end). -So now $X = X_1$ and we rename $x_1$ as $x$. There is a universal $x$-rigified line bundle $L$ on $X \times P$, so for any abelian scheme $B$ the pointed $S$-maps $F:X_S \rightarrow B = {\rm{Pic}}^0_{B^{\vee}/S}$ correspond exactly to isomorphism classes of $x$-rigidified line bundles $M$ on $X_S \times B^{\vee}$ that are trivialized along $X_S \times \{0\}$. The latter trivialization is unique upon requiring (as we may, via an $O(X_S)^{\times}$-scaling) it agrees with the effect of the $x$-rigidification along the $S$-point $(x_S,0)$. Thus, such data corresponds exactly to a pointed $S$-map $B^{\vee} \rightarrow {\rm{Pic}}_{X_S/S} = P_S$ ($P := {\rm{Pic}}_{X/k}$). Such a map is an $S$-homomorphism by Corollary 6.4 in GIT, so if $C$ denotes the maximal abelian subvariety of $P$ (via the Proposition above) then such maps correspond exactly to $S$-homomorphisms $B^{\vee} \rightarrow C_S$. -But by double-duality for abelian schemes, those in turn correspond exactly to $S$-homomorphisms $C^{\vee}_S \rightarrow B$. Unraveling, this procedure gives the result with $A := C^{\vee}$. -QED Theorem -Example. Consider $X = {\rm{Spec}}(k')$ for a finite etale $k$-algebra $k'$ having degree $> 1$, so this is geometrically reduced (and even geometrically normal) over $k$ but not geometrically connected. We claim that the conclusion of the Theorem cannot hold for such $X$ over $k$. Indeed, if there were a universal $f:X \rightarrow E$ over $k$ -with the property as in the Theorem (for all $k$-schemes) then it would retain this property after scalar extension to a finite separable extension of $k$ that splits $k'$. -That is, it suffices to show the Theorem cannot hold when $X$ is a disjoint union of $d>1$ copies of ${\rm{Spec}}(k)$. Labeling the points as $x_1, \dots, x_d$, in the context of the mapping properties we can use the image of $x_1$ to trivialize the torsors and thereby drop it from consideration by using pointed maps to abelian schemes. That is, it suffices to show that if $Y$ is a disjoint union of $d-1>0$ copies of ${\rm{Spec}}(k)$ there is no $k$-map $f:Y \rightarrow A$ to an abelian variety over $k$ such that for any $k$-scheme $S$ any $S$-map $F:Y_S \rightarrow B$ to an abelian scheme $B$ over $S$ factors uniquely through an $S$-homomorphism $A_S \rightarrow B$. But $F$ is nothing other than the data of an $S$-point of $B^{d-1}$, and we aim to show that there is no abelian variety $A$ over $k$ equipped with a $k$-point $a \in A^{d-1}(k)$ with the magical property that every $S$-point of $B^{d-1}$ has the form $g^{d-1}(a_S)$ for a unique $S$-homomorphism $g:A_S \rightarrow B$. Even limiting ourselves to $S = {\rm{Spec}}(\overline{k})$ and varying $B$, it is clear no such amazing pair $(A, a)$ exists.<|endoftext|> -TITLE: Which kind of foundation are mathematicians using when proving metatheorems? -QUESTION [19 upvotes]: Zermelo-Fraenkel set theory (with choice) is commonly accepted as the standard foundation of mathematics. It is a material set theory. For every two objects/sets $a,b$ one can ask whether $a=b$ or not. Also, one can always ask whether $a\in b$ is true or not. So $\in$ is a global element relation. -As an alternative foundation for set theory, Shulman proposed SEAR. It is a structural set theory. That is, elements have no internal structure, i.e. are just "abstract dots". One has a type declaration $a\colon A$ for saying that $a$ is an element of $A$. But this can't be negated, so $\colon$ is no relation. But if $A$ is a set (should I say "abstract set" for emphasizing the point that I mean "set in a structural set theory"?), then one has a local element relation $\in_A$: for each element $a$ in $A$ and each subset $B$ of $A$ (= function $A\to 2:=\{0,1\}$), the statement $a\in_A B:\iff B(a)=1$ is either true or false. -On the SEAR-page I linked to above, there is a proof (due to Shulman I guess) showing that SEAR and ZF are basically equivalent: from a model of ZF one can construct a model of SEAR and vice versa. This is a meta theorem. But in which foundation does the proof of such a meta theorem take place? Is this meta foundation a structural or a material set theory? - -REPLY [19 votes]: Your question is much more specific than your title suggests. As to the question itself, my answer is that it doesn't matter. The proof is given in mathematics, not in any formal system. A foundation for mathematics, in order to count as a foundation of mathematics, must be able to formalize most ordinary mathematical arguments. Thus, any foundation for mathematics could be used to formalize such a metatheorem. -With that said, some people care about whether metatheorems of this sort can be formalized in very weak theories. There's nothing wrong with that, but it's not something I spend my time thinking about, and in particular I didn't think about it when writing the proof you refer to. The closest I came was noting that instead of a "construction of a model" the proof can equivalently be regarded as giving a translation of first-order formulas. If you're interested in this sort of question, then Nik's answer seems reasonable to me, but I'm not familiar with the details of how this sort of thing goes.<|endoftext|> -TITLE: On Nash equilibrium in $2 \times 2 \times ... \times 2$ games -QUESTION [6 upvotes]: Game theory post-grad reporting. -In my current research I have stumbled upon a problem that seems simple, but managed to block me for weeks already. -Let's say we have $n$-player game with $2$ pure strategies available to each player, and we know for sure that this game has at least one Nash equilibrium in totally mixed strategies. What would be good sufficient conditions for such game to have exactly one equilibrium in totally mixed strategies? Please, take note, I'm talking about totally mixed strategies for all players, and do not care of situations where one or more players use pure strategies. -On a hunch it feels like a class of $2 \times 2 \times ... \times 2$ games with exactly one Nash equilibrium in totally mixed strategies should be very large. But to my surprise I can't outline even single meaningful subclass of it. -EDIT 1: My current best guess is that at least following should be provable: - -If in any two situations that are distinct by action of only one player no player have same payoff, then game can't have more then one equilibrium in totally mixed strategies. - -But unfortunately, even for such weak assumption I don't see how. - -REPLY [4 votes]: I hope the following will provide a nontrivial game with continuum of completely mixed equilibria. -Let us consider the payoff function of a single player, and later we will extend the analysis to a multiplayer game. -Consider the 2x2 game in which the payoff is 1 in the two diagonal entries, and 0 in the two off-diagonal entries. Once the other player plays each pure strategy with probability 1/2, you are indifferent among your pure strategies. -We can have a similar property in a 2x2x2 game: denoting the two pure strategies of each player by $\{A,B\}$, the payoff to the player is 1 for the following strategy profiles: $(A,B,A), (A,A,B), (B,A,A), (B,B,B)$; and 0 otherwise. In this game, once at least one of the other players plays each pure strategy with probability 1/2, our player is indifferent among his two pure strategies. -We can extend this definition to any number of players: the payoff of our player is 1 if the number of players who choose A is even, and 0 otherwise. Suppose that this is the payoff function of ALL players. Then any strategy profile in which at least two players play $(1/2,1/2)$ is an equilibrium. This hints of a plethora of other nontrivial games that have multiple strictly mixed equilibria.<|endoftext|> -TITLE: Homeomorphisms of $S^n\times S^1$ -QUESTION [22 upvotes]: Is every homeomorphism $h$ of $S^n\times S^1$ with identity action in $\pi_1$ pseudo isotopic to a homeomorphism $g$ of $S^n\times S^1$ such that $g(S^n\times x)=S^n\times x$ for each $x\in S^1$? I would be satisfied with an answer for $n>3$. - -REPLY [3 votes]: I attempt to answer the question: - -What is the group of homeomorphisms of $S^n\times S^1$, up to pseudo-isotopy? - -which seems of greater general interest. I will not completely answer it, but just do the easy part, producing an upper bound. I am not able to rule out the existence of a single pseudo-isotopy class of homeomorphisms homotopic to the identity but not pseudo-isotopic to the identity, which would be a counterexample to Olga’s question. -After writing this answer, I noticed that the paper of Browder cited by Danny Ruberman actually fully answers this question. And, in fact, it uses pretty much this method. But I think my answer is a little cleaner because it isolates surgery as an existing machine, while Browder was still working on the foundations of the theory. In particular, I have no restriction on the fundamental group of the manifold. He answers the whole question, ruling out the $\mathbb Z/2$ I was unable to address, but I don’t see how he handles it. - -Absolute problems are difficult. Relative problems are easier. To attack homeomorphisms, put them in a long exact sequence involving homotopy equivalences and some kind of relative problem. Even after one computes the groups on either side in the long exact sequence, one is left with determining the image and quotient that contribute to the group of interest and the extension problem between them. But it gives pretty good upper and lower bounds. -Specifically, the group of homeomorphisms up to isotopy $\pi_0 Top(M)$ maps to the group of homotopy equivalences up to homotopy $\pi_0 hAut(M)$. The latter group is difficult to compute, let alone the image, though in your case it is easy. The kernel is the subgroup of homeomorphisms that are homotopic to the identity, up to isotopy. This receives a surjection from $\pi_1(hAut(M)/Top(M))$. I can’t compute that, but I mention it because of the exact sequence -$$\pi_1hAut(M)\to \pi_1(hAut(M)/Top(M)) -\to \pi_0 Top(M) \to \pi_0 hAut(M)$$ -which suggests that $\pi_1hAut(M)$ will matter; indeed, it will reappear in an analogous role. Coarsening the equivalence relation from isotopy to pseudo-isotopy changes the group and makes it accessible. I will prove that the relative surgery structure set $S(M\times I;\partial)$ (which is a group under concatenation) surjects to the group of homeomorphisms homotopic to the identity, modulo pseudo-isotopy and the kernel receives a surjection from $\pi_1(hAut(M))$. And surgery structure sets are computable. -In your case it is easy to see that -$$\pi_0hAut(S^n\times S^1) -=(\mathbb Z/2)^3 -=\pi_0O(2)\times\pi_0O(n+1)\times\pi_1O(n+1)$$ -and thus that every homotopy equivalence lifts to a homeomorphism. I will compute $S(S^n\times S^1\times I;\partial)=\mathbb Z/2$. So that shows that the answer is either $(\mathbb Z/2)^3$ or an extension of that group by $\mathbb Z/2$. -Homotopy equivalences -First, compute the group of homotopy equivalences. Reduce from automorphisms to based automorphisms: in any geometric category the choice of a basepoint gives a map from the automorphism group to the space, with fiber are basepoint-preserving automorphisms: $Aut_*(M)\to Aut(M)\to M$. Based mapping spaces are easier to work with. Also, the coset space $hAut(M)/Top(M)$ doesn’t depend on whether there is a basepoint, so if we are interested in the image of $\pi_1hAut(M)$ we can replace it with the based version; $\pi_1hAut(M,*)=\mathbb Z/2$. Based mapping spaces are computed cell by cell. It is easy to compute $Map(S^n\vee S^1,S^n\times S^1)=\Omega S^1\times \Omega S^n\times \Omega^nS^n\times \Omega^nS^1=\mathbb Z\times\Omega S^n\times\Omega^n S^n$ because it is a map from a coproduct to a product. Adding a cell gives a fibration sequence: $Map((D^{n+1},\partial),S^n\times S^1)\to Map_*(S^n\times S^1,S^n\times S^1)\to Map_*(S^n\vee S^1,S^n\times S^1)$. And $Map((D^{n+1},\partial),X)$ is basically a torsor over $Map_*(S^{n+1},X)$. What makes this example easy, other than the small number of cells, is that two of the factors of the base split, so the homotopy groups in low degree are the sum of those of the base and the fiber. -Why Surgery? -Surgery computes the set of manifold structures on a given space. That’s something like the components of the homotopy fiber of the map from the category of manifolds and homeomorphisms to the category of spaces and homotopy equivalences. The components of the homotopy fiber have the homotopy type of the space $hAut(M)/Top(M)$ that I mentioned above. That isn’t the homotopy type of the space that surgery computes, but it gives a heuristic why surgery is relevant. A more precise, but more complicated reason is mapping tori. If $\varphi$ is a homeomorphism of $M$ then its mapping torus is the quotient $M\times I/(x,1)\sim(\varphi(x),0)$. If a homemorphism is homotopic to the identity, then its mapping torus is homotopy equivalent to the trivial bundle $M\times S^1$, and thus gives an element of the surgery structure set. If a homeomorphism is pseudo-isotopic to the identity, its mapping torus is homeomorphic to the trivial bundle. Thus homeomorphisms that are homotopic to the identity, up to pseudo-isotopy are relevant to $S(M\times S^1$). In fact, Shaneson’s splitting theorem says that $S(M\times S^1)=S(M\times I;\partial)\times S^h(M)$. The key ingredient is the relative surgery problem $S(M\times I;\partial)$ and this is also the fundamental group of the surgery space, connecting this to the prior heuristic. -Relative surgery -Leaving aside the motivation, I will show that $S(M\times I;\partial)$ is exactly what we want. It is a group under concatenation. It surjects to the group of homeomorphisms that are homotopic to the identity, up to pseudoisotopy; and the kernel of this map receives a surjection from $\pi_1hAut(M)$. Then I will compute $S(S^n\times S^1\times I;\partial)=\mathbb Z/2$. -For $W$ a manifold with boundary $S(W;\partial)$ is the set of maps $N\to W$ from a manifold to $W$ that are simple homotopy equivalences and such that they are homeomorphisms on the boundary, up to the equivalence relation of homeomorphisms $N\to N’$ that make triangle commute, up to homotopies which are constant on the boundary. Let us specialize to $W=M\times I$. By the $s$-cobordism theorem, any such $N$ is an $s$-cobordism, thus is homeomorphic to a cylinder $M\times I$. So this surgery set is really about the homotopy equivalence from $M\times I$ to itself. There is a homomorphism from $S(M\times I;\partial)$ to the group of homeomorphisms modulo pseudo-isotopy: given $N\in S(W;\partial)$ choose a cylinder structure $N=M\times I$ and compose with a constant homeomorphism of $M$ to make the structure map $N\to M\times I$ be the identity at time 0; and read off the homeomorphism at time 1. This is well-defined up to homeomorphisms of $M\times I$ that are the identity at time 0, ie, up to pseudo-isotopy. The image must be homotopic to the identity, because the structure map gives a homotopy. Conversely, every homeomorphism that is homotopic to the identity is in the image because a choice of homotopy gives a structure map. Finally, if we have $N\in S(M\times I)$ that yields the identity homeomorphism, its structure map gives a loop in -$\pi_1(hAut(M))$. So we have an exact sequence -$$\pi_1(hAut(M))\to S(M\times I;\partial)\to G \to \pi_0(hAut(M))$$ -where $G$ is the sought group, the group of homeomorphisms up to pseudo-isotopy. -(This argument feels like I am identifying $S(M\times I;\partial)$ with $\pi_1(hAut(M)/H)$ for some group $H$, but I can’t see what it should be.) -Computation -Surgery theory computes $S(W;\partial)$ for a $w$-manifold $W$ by putting it in a long exact sequence -$$[\Sigma W/\partial,G/Top]\to L_{w+1}(\pi_1W)\to S(W;\partial)\to [W/\partial,G/Top]\to L_w(\pi_1W)$$ -Borel conjectured that a homotopy equivalence between aspherical closed manifolds is homotopic to the identity, ie, $S(B\pi)=*$. From the point of view of surgery theory, the natural generalization is that $L_*(\pi)=H_*(B\pi;L(e))$ for torsion-free groups $\pi$; and that $G/Top$ is the connected component of $L=L(e)$. This is known for our fundamental group $\mathbb Z$ and even though our space is not the circle, it means that there will be a lot of cancelation. -Specializing to $W=S^n\times S^1\times I$, we have $L_{n+2}(\mathbb Z)=L_{n+2}\oplus L_{n+1}$, while $S^n\times S^1$ is stably a wedge of spheres $S^{n+1}\vee S^n\vee S^1$. Thus -$$[S^n\times S^1\times I/\partial,G/Top] -=\pi_{n+2}G/Top\oplus \pi_{n+1}G/Top -\oplus \pi_2G/Top\oplus \pi_1G/Top\\ -=L_{n+2}\oplus L_{n+1}\oplus L_2\oplus L_1$$ -which contains $L_{n+2}(\mathbb Z)$ as a summand. The Borel conjecture does not just say $L_{n+2}(\mathbb Z)$ is a summand, but that the map is compatible; and similarly for the terms a dimension up. Thus the structure set is the other summand: $S(S^n\times S^1\times I;\partial)=L_2\oplus L_1=\mathbb Z/2\oplus e=\mathbb Z/2$. -So $\mathbb Z/2$ is an upper bound to the size of the group of homeomorphisms homotopic to the identity, up to pseudo-isotopy. But my guess is that $\pi_1(hAut(S^n\times S^1,*))=\mathbb Z/2$ hits it and the actual group is trivial. -Addendum: Diffeomorphisms -There is a version of surgery theory for smooth manifolds. It replaces $G/Top$ with $G/O$. The space $G/O$ differs from $G/Top\sim L$ by the groups of exotic spheres. Thus there is less cancelation and it is generally more difficult to compute the smooth structure set. But in this case, with the stable splitting, it is fairly easy to compute, at least in terms of exotic spheres: -$$S^{Diff}(S^n\times S^1\times I;\partial) -=S^{Diff}(S^{n+2})\oplus S^{Diff}(S^{n+1})\oplus \mathbb Z/2$$ -The $\mathbb Z/2$ is as before and I don’t know whether it gives exotic diffeomorphisms or comes from $\pi_1hAut(S^n\times S^1)$, but the others must yield diffeomorphisms not pseudo-isotopic to the identity. In fact, we can identify exotic spheres with exotic diffeomorphisms of lower dimensional spheres or disks: $S^{Diff}(S^{n+1})=\pi_0Diff^+(S^{n})=\pi_0Diff(D^{n};\partial)$. Thus $S^{Diff}(S^{n+1})$ gives diffeomorphisms of $S^n$ and thus $S^n\times S^1$; and $S^{Diff}(S^{n+2})$ gives diffeomorphisms of $D^{n+1}$ and thus any $n+1$-dimensional manifold, such as $S^{n}\times S^1$.<|endoftext|> -TITLE: Monomorphisms, epimorphisms, (co-)images and factorizations in $\infty$-categories -QUESTION [6 upvotes]: Several of the many notions that don't work the same way when passing to $\infty$-categories are the ones mentioned in the title. I'm trying to understand the conceptual picture around these notions in the world of higher category theory (conceptual answers will be more useful to me than model specific answers). -My question are conceptually very simple: - - -What are ($n$-?)monomorphisms and ($n$-?)epimorphisms in an $\infty$-catgory? What are the fundamental differences with $1$-categories? -What are ($n$-?)coimages and ($n$-?)-images in an $\infty$-category? (it seems that on n-lab the definitions for both of them [1] [2] are exactly the same which is very confusing) What are the fundamental differences with $1$-categories? Which of these coincides with the Postnikov decomposition in $\infty$-groupoids? -What is an $n$-ary factorization system for an $\infty$-category? Is this a direct analog of epi-mono factorization? If so In what sense? - - -I could elaborate but I tried and concluded that it will probably only serves to confuse me and the readers. I'm still confused by many of these notions. - -REPLY [12 votes]: An $n$-monomorphism is a map $A\to B$ for which $Map(X,A)\to Map(X,B)$ has all homotopy fibers $n$-truncated, for all $X$; a space (=$\infty$-groupoid) is $n$-truncated if its homotopy groups all vanish in dimensions $>n$. -An $n$-epimorphism is an $n$-monomorphism in the opposite $\infty$-category. -When $n=-1$ you recover Denis's definition of monomorphism and epimorphism: note that a space is $(-1)$-truncated if and only if it is empty or contractible. -The phenomenon of "$n$-epi/mono factorization" in an $\infty$-category does not actually involve factorization as ($n$-mono)($n$-epi), but rather as ($n$-mono)($n$-connected map). The class of $n$-connected maps is usually defined as a kind of left complement to $n$-monomorphisms, so the existence of such factorizations is basically the assertion that your $\infty$-category admits a factorization system where the right-class is the class of $n$-monos. (These exist in all "reasonable" examples, e.g., presentable $\infty$-categories.) -This same issue shows up in 1-category theory: epi/mono factorization tends to be rare, and more commonly one considers factorizations of the form: (regular epi)(mono). -The factorization ($n$-mono)($n$-connected) can be thought of as factorization through an "$n$-image". It is clearly not a self-dual notion, hence there is a dual but distinct notion of factorization through an "$n$-coimage".<|endoftext|> -TITLE: How can a Roomba turn as little as possible? -QUESTION [10 upvotes]: Suppose I have a convex polygon $C$ and a radius $r>0$, and I seek a path $P$ that "covers" $C$, in the sense that any point $C$ is within distance $r$ of $P$: $$d(x,P)\leq r~\forall x\in C~,$$ -where $d(x,P) := \min_{x'\in P} \|x-x'\|$. My question: are there any analytical results (lower/upper bounds, etc.) that describe the minimum amount of turning needed to cover $P$? The figure below shows three covering paths of a rectangle, and the upper path has a total of $9\pi$ radians worth of rotation whereas the lower two have $3\pi$. - -REPLY [4 votes]: The following paper studies this "milling" problem (generalized) from a complexity viewpoint: - -Arkin, E. M., Bender, M. A., Demaine, E. D., Fekete, S. P., Mitchell, J. S., & Sethia, S. (2005). Optimal covering tours with turn costs. SIAM Journal on Computing, 35(3), 531-566. (Preliminary arXiv version.) - -Among many results, they prove - -that the covering tour problem with turn costs is NP-complete, - even if the objective is purely to minimize the number of turns, the pocket is - orthogonal (rectilinear), and the cutter must move axis-parallel. - -The provide several approximations algorithms for variants of the -problem. For the Roomba variant in which the orthogonal polygon -may have holes and the tour is axis-parallel, they achieve -a $3.75$-approximation ratio. - -                - - -        -Fig.5.2.b: An optimal tour: square with square hole. - -There is literature on NC milling of convex shapes, but I cannot access -any the papers, so I am not certain of their relevance: - - -Wang, Hsu-Pin, Heng Chang, and Richard A. Wysk. "An analytical approach to optimize NC tool path planning for face milling flat convex polygonal surfaces." IIE transactions 20.3 (1988): 325-332. -Prabhu, Prasad V., Anand K. Gramopadhye, and Hsu-pin Wang. "A general mathematical model for optimizing NC tool path for face milling of flat convex polygonal surfaces." The International Journal of Production Research 28.1 (1990): 101-130. -Deshmukh, Abhijit V., M. M. Barash, and Hsin-Pang Wang. On selection of tool path orientations for generating prismatic features. School of Industrial Engineering, Purdue University, 1993.<|endoftext|> -TITLE: What is a "scholium"? -QUESTION [22 upvotes]: In a paper I'm working on, I'm tempted to write something like: - -Note that the argument above also proves the following result: -Scholium. bla bla - -Is this ok? Is it correct to say that a "scholium" is a "corollary of a proof"? - -REPLY [14 votes]: Bourbaki defines a "scholie" in the preface of the Éléments de mathématiques as follows: - -Sous le nom de « scholie », on trouvera quelquefois un commentaire d'un théorème particulièrement important. - -I.e., a scholium (for Bourbaki) is a commentary to a particularly important theorem. -There aren't many in the treaty, but those that are seem to be non-mathematical, or meta-mathematical: essentially, they are a guidance on how to use the theorem or when to apply it, or an indication of a general proof technique. Something like "this theorem is useful for deriving results about foobars from the general theory of bazquxes by applying the frobnification functor and using the theorem to transfer the property". Not something formalizable as a mathematical statement. -Examples of scholia in Bourbaki are in A IV §2 nº3 or FVR VI §1 nº1 or INT IX §1 nº8. There is also one in AC VIII §3 nº3, but the latter seems to be just a corollary (of a corollary), so apparently younger Bourbaki authors didn't get the memo on what a scholium was supposed to be.<|endoftext|> -TITLE: What can be said about this double sum? -QUESTION [16 upvotes]: Question. Can this number be expressed in terms of classical values? - $$\sum_{n,m=1}^{\infty}\frac1{(n^2+m^2)^{\frac32}}=1.056348517615643291\dots$$ - -UPDATE. I'm encouraged by Noam, Kevin and Igor's directional replies. To spice things up, I ask: is this true? -$$\sum_{n,m=1}^{\infty}\frac1{(n^2+m^2)^s}=\zeta(s)\beta(s)-\zeta(2s),$$ -wherever convergence occurs. Here, $\zeta(s)$ and $\beta(s)$ are the Riemann zeta and Dirichlet beta functions, respectively. - -REPLY [6 votes]: Exhaustive analysis of lattice sums can be found in the book Lattice sums then and now. This book can be found for free on the web. -OP's sum is given by formula 1.3.14 on page 33: -L. Lorenz. Bidrag tiltalienes theori. Tidsskrift Math., 1:97–114, 1871.<|endoftext|> -TITLE: Finitely generated group every 2-generated subgroup of which is finite -QUESTION [9 upvotes]: I know of Tarski monsters and the Burnside Problem. I would like to know if there is an infinite finitely generated group $G$ such that for any $g$ and $h$ in $G$, the subgroup generated by $\{g,h\}$ is finite. -I am also interested in related questions: - -For which $(m,n)$ does there exist an infinite $m$-generated group every $n$-generated subgroup of which is finite? -For which $(m,n,k)$ does there exist an infinite $m$-generated group every $n$-generated subgroup of which has cardinality at most $k$? - -Thank you. - -REPLY [9 votes]: Questions 1 and 2 are answered by Golod's theorem, see Theorem 3.3 in -the survey: -M. Ershov, Golod-Shafarevich groups: a survey. Internat. J. Algebra and Computation, vol. 22 (2012). -Namely, for every $d\ge 2$ there exists an infinite $d$-generated group such that every $d-1$-generated subgroup is finite. -(He even gets a finite $p$-group for the given prime number $p$ but this is irrelevant.) -As for the last question, I am not sure, most likely it is unknown. I suggest to ask Ershov directly.<|endoftext|> -TITLE: what is the status of this problem? an equivalent formulation? -QUESTION [10 upvotes]: R. Guy, Unsolved problems in number theory, 3rd edition, Springer, 2004. -In this book, on page 167-168, Problem C5, Sums determining members of a set, discusses a question Leo Moser asked: suppose $X\subset\mathbb{Z}$ is an $n$-element set and $A$ be set of all $k$-element sums of subsets of $X$. Is there any other $Y\neq X$ such that $Y\subset\mathbb{Z}$ such that its $k$-element sums are exactly $A$? -Only small cases were known: $k=2,3,4,5$ and any $n$. - -Question. What is the general status of this problem for $k>5$? Max Alekseyev provided some references. -Update. I'm still interested in the question: are there (non-cosmetic) equivalent formulations of the above problem? - -REPLY [17 votes]: In a recent paper A set of 12 numbers is not determined by its set of 4-sums (in Russian), Isomurodov and Kokhas identify an error in the argument of Ewell (1968) and present two distinct sets of 12 integers with the same multiset of 4-sums. -They also give a nice overview of the area and provide useful references. For the general case $(n,k)$, they mention the following results: - -Selfridge and Straus (1958) describe a union of Diophantine equations such that for $(n,k)$ not satisfying any of the equations (a typical situation), the set $X$ is unique. For $(n,k)$ satisfying at least one of the equations, the uniqueness of $X$ remains an open question (the case $(12,4)$ was such). -Fraenkel, Gordon, and Straus (1962) prove that for each $k$, there exists only a finite number of values of $n$, for which $X$ may be not unique.<|endoftext|> -TITLE: Picard group of derived category of sheaves -QUESTION [6 upvotes]: Let $X$ be a topological space and $R$ be a commutative ring with unit, $D(X,R)$ is the derived category of unbounded complexes of sheaves of $R$-modules. Moreover we suppose that $X$ is a stratified space and $D_c(X,R)\subset D(X,R)$ is the derived category of complexes of sheaves with constructible cohomology sheaves. The categories $D(X,R)$ and $D_c(X,R)$ are both closed symmetric monoidal categories, thus we can speak of their Picard groups as the group of isomorphism classes of invertible objects for the derived tensor product. -I was wondering if we know some computations of the Picard group of $D(X,R)$ or of $D_c(X,R)$? - -REPLY [6 votes]: Thanks to Drew Heard's comment I was able to find answers to my questions. In his paper "Picard groups of derived categories" H. Fausk proves the following theorem (see Theorem 4.2). -Theorem: Let $(\mathcal{E},\mathcal{O})$ be a commutative unital ringed Grothendieck topos with enough points such that for all points $p$ -of $\mathcal{E}$ the ring $\mathcal{O}_p$ has a connected prime -ideal spectrum. Then there is a natural split short exact sequence: -$$0\rightarrow Pic(\mathcal{O}-Mod)\rightarrow Pic(D(\mathcal{O}-Mod))\rightarrow C(pt(\mathcal{E}))\rightarrow 0$$ -Let us apply this theorem to the case of a connected and locally connected topological space $X$, together with the constant sheaf $\underline{R}$ (i.e. $\mathcal{O}=\underline{R}$) and let us suppose that $R$ is an integral domain (its prime ideal spectrum is connected), then we have the corollaries. -Corollary A: We have a split short exact sequence -$$0\rightarrow Pic(\underline{R}-Mod)\rightarrow Pic(D(X,R))\rightarrow \mathbb{Z}\rightarrow 0$$where the section is the morphism that sends the integer $n\in\mathbb{Z}$ to the constant sheaf $\underline{R}[n]$. -Corollary B: Moreover if $X$ is stratified then we have a split short exact sequence -$$0\rightarrow Pic(\underline{R}-Mod_c)\rightarrow Pic(D_c(X,R))\rightarrow \mathbb{Z}\rightarrow 0$$ where $\underline{R}-Mod_c$ is the category of constructible sheaves of $\underline{R}$-modules. -Let us look to the case of a point then we have a split short exact sequence (with the hypothesis that $Spec(R)$ connected): -$$0\rightarrow Pic(R)\rightarrow Pic(D(R)) \rightarrow \mathbb{Z}\rightarrow 0.$$<|endoftext|> -TITLE: Can we control where the critical point of an elementary embedding is mapped? -QUESTION [5 upvotes]: At the risk of this question having an easy answer, I am asking the following: -Suppose we have a measurable cardinal $\kappa$ and a (possibly regular) cardinal $\lambda>\kappa$. Is there any condition, like the existence of an appropriate ultrafilter, that implies the existence of an elementary embedding $j:V\to M$ with critical point $\kappa$ so that $j(\kappa)=\lambda$? -An example is when $\kappa$ is huge: if there is a normal $\kappa$-complete ultrafilter $U$ on $P(\lambda)$ with the property that $\{X\subseteq P(\lambda)\mid ot(X)=\kappa\}\in U$, then the ultrapower embedding $j:V\to M\simeq Ult(V,U)$ satisfies $j(\kappa)=\lambda$ (and $~{}^{j(\kappa)}M\subseteq M$).. -However, I am interested in the case where $\kappa$ is only measurable, maybe not even strong. -I am assuming Choice and whenever I write $j:V\to M$ I mean that $M\subseteq V$ is a transitive model of ZFC. - -REPLY [10 votes]: Here is a necessary and sufficient criterion: $\lambda>2^\kappa$. This is easily seen to be necessary, since if $j:V\to M$ has critical point $\kappa$, then the power set $P(\kappa)$ is contained in $M$, and from this it follows that $j(\kappa)>2^\kappa$ since $j(\kappa)$ is inaccessible in $M$. Conversely, we can hit every cardinal above $2^\kappa$, by the following. -Theorem. If $\kappa$ is a measurable cardinal, then every cardinal $\lambda>2^\kappa$ is the image $\lambda=j(\kappa)$ of some elementary embedding $j:V\to M$ with critical point $\kappa$. -Proof. Let $\mu$ be a normal measure on $\kappa$ and consider the class of images $j_\alpha(\kappa)$, where $j_\alpha$ is the $\alpha$-iterated ultrapower by $\mu$. These are the ordinals that form the critical sequence. This class of ordinals is closed and unbounded. It is clearly unbounded, since by iterating further, we can push $j(\kappa)$ as high as desired. It is closed, since the iterations are defined to take the direct limit at limit stages, and this makes the critical sequence continuous. -Finally, I claim that every cardinal above $2^\kappa$ is on the critical sequence. To see this, it suffices to argue that at each stage, we don't jump over the next cardinal. If $j_\alpha:V\to M_\alpha$ is the $\alpha^{th}$ iterate, then the extender representation shows that every element of $M$ has the form $j(f)(s)$, where $f:\kappa\to V$ and $s$ is a finite sequence from the critical sequence below $\kappa_\alpha$, which are ordinals below $j_\alpha(\kappa)$. If we go one more step, to $j_{\alpha+1}:V\to M_{\alpha+1}$, then we only need to add one more generator or seed, namely $\kappa_\alpha$ itself, and so $|j_{\alpha+1}(\kappa)|^V\leq |\kappa^\kappa|\cdot|(\alpha+1)^{<\omega}|$, which has the same size as $j_\alpha(\kappa)$. So at successor stages, we don't get to the next cardinal, and so we reach all the cardinals at limit stages. So we'll get every cardinal above $2^\kappa$. -QED -Lastly, note that although you had asked about cardinals $\lambda$ that are the images of $\kappa$ under an embedding, nevertheless such images are not always cardinals. For example, if $\mu$ is a measure on $\kappa$ with ultrapower $j:V\to M$, then $j(\kappa)$ is never a cardinal, since it is strictly between $2^\kappa$ and $(2^\kappa)^+$. So one might want a criterion for recognizing when an ordinal $\lambda$ is the image of $\kappa$ under an embedding, and I think that is a much subtler question.<|endoftext|> -TITLE: B-invariant subvarieties -QUESTION [5 upvotes]: Let $B$ be a connected, solvable algebraic group (of dimension 2) acting on a projective variety $Y$ (of dimension $3$): for instance, let $B$ be a Borel subgroup of a reductive algebraic group $G$. -Question1: Is it always true that there is a $B$-invariant complete flag (i.e. codimension $1$, irreducible and $B$-invariant subvarieties), $Y \supset Y_1\supset Y_2 \supset Y_3$? For my purposes I can assume that there is a point in $Y$ with finite stabliser group and therefore can choose the orbit-closure of that point to be $Y_2$. How about $Y_1$? Is there always a point with $1$-dimensional stabiliser group of the $B$-action on the just described $Y_2$? Obviously there will always exist a $B$-fixed point $Y_3$, by Borel's fixed point theorem, if $Y_1$ is a projective variety on which $B$ acts. -Question2: Does this generalize to arbitrary dimension $n$ of $Y$ and any reductive algebraic group $G$ with a Borel subgroup $B$? - -REPLY [7 votes]: Here is an argument which avoids the use of Hilbert schemes and works for any complete (irreducible) variety $Y$. -As in Jason's answer it suffices to construct a $B$-stable subset $Y_1$ of codimension $1$, provided that $\dim Y\ge1$. For this we consider two cases: - -$Y$ does not contain an open $B$-orbit. By Rosenlicht's theorem, there exists an open $B$-stable subset $Y_0\subseteq Y$ such that the orbit space $\pi:Y_0\to Z:=Y_0/B$ exists. Our assumption implies $\dim Z>0$. Therefore, $Z$ contains a subvariety $Z_1$ of codimension $1$. Now take for $Y_1$ the closure of $\pi^{-1}(Z_1)$ in $Y$. -$Y$ does contain an open $B$-orbit $Y_0$. Then the above argument is not applicable since $Z$ is a point. Instead we use the well-known fact that homogeneous spaces for solvable groups are affine. Thus $Y_0$ is affine. But it not complete since $\dim Y_0\ge1$. Thus the complement $C$ of $Y_0$ in $Y$ is non-empty. We conclude with another well-known fact, namely that the complement of an affine open set is always pure of codimension one. So, take for $Y_1$ any component of $C$.<|endoftext|> -TITLE: Are all the mappings which satisfy this equation scaled isometries? -QUESTION [5 upvotes]: Let $M,N$ be smooth oriented $d$-dimensional Riemannian manifolds, $\, f:M \to N$ a smooth map. Let $\Omega^1(M,f^*TN)=\Gamma(T^*M \otimes f^*TN)$ be the space of $f^*TN$-valued one-forms. -Let $d$ be the covariant exterior derivative associated with the pullback connection of the Levi-Civita connection on $N$ (via $f$), and let $\delta$ be its adjoint. -Finally, define the section $Q(df) \in \Gamma(T^*M \otimes f^*TN)$ as the closest orientation-preserving isometry to $df$. (here "closest" is w.r.t the natural norm induced on $T^*M \otimes f^*TN$ by the metrics on $M,N$). -I am interested in mappings $f$ which satisfy: -$(1) \, \, f$ is an orientation-preserving immersion. -$(2) \, \, \delta (Q(df))=\delta(df)=0$. -(The motivation comes from studying the critical points of a distortion functional. The equations in $(2)$ are related to the associated Euler-Lagrange equations). -Recall that $\delta(df)=0$ is harmonicity, and that every isometry is harmonic. Thus, if $f$ is an orientation-preserving isometry it satisfies $(2)$. -Moreover, if $f$ is harmonic and there exist a constant $\lambda > 0$ such that $df$ is always $\lambda $ times an (orientation-preserving) isometry, then $Q(df)=\frac{1}{\lambda}df$, so again $(2)$ is satisfied. (In fact the harmonicity of $f$ follows from the other condition). -Question: Suppose $f:M \to N$ satisfies $(1),(2)$. Is it true that it is a "scaled isometry"? (i.e is it conformal with a "constant scaling factor")? Or at least conformal? - -REPLY [6 votes]: Here's a simple counterexample: Let $M=N=T^2$ (the standard torus, thought of as $\mathbb{R}^2/\mathbb{Z}^2$). Let $f:M\to N$ be the identity, and let the metrics on $M$ and $N$ be any two translation-invariant metrics on $T =\mathbb{R}^2/\mathbb{Z}^2$ (so, in particular, they don't need to be conformal). Then one easily sees that $f$ satisfies your equations, but if the two metrics aren't constant multiples of one another, $f$ won't be a 'scaled isometry'.<|endoftext|> -TITLE: Did Hilbert discuss his 23 problems with Felix Klein? -QUESTION [16 upvotes]: Hilbert's lecture at the ICM in Paris in 1900 presented 10 of the famous 23 open problems. It is well known that the idea of the lecture came from Hermann Minkowski. Hilbert was at Gottingen at the time where he was hired through untiring efforts of Felix Klein. As detailed by historian David Rowe and others, both Hilbert and Klein were involved in a battle against the Berliners at the time. The Berlin school dominated by followers of Kummer, Weierstrass, and Kronecker was known for its focus on arithmetized analysis. Hilbert's 23 open problems sought to broaden the scope of mathematics beyond such narrow focus. It seems as though it would have been natural for Hilbert to have discussed the 23 problems with Klein. Is there any evidence of such discussions in published work or private correspondence? -Here is what Minkowski wrote in a letter to Hilbert: -"Most alluring would be the attempt to look into the future, in other words, a characterization of the problems to which the mathematicians should turn in the future. With this, you might conceivably have people talking about your speech even decades from now. Of course, prophecy is indeed a difficult thing" (Minkowski 1973, 5 January 1900; see German original). -The reference is -Hermann Minkowski, Briefe an David Hilbert, Hg. L. Ru¨denberg und H. Zassenhaus, New York: Springer-Verlag, 1973. -This information comes from page 16 of Rowe's article -Rowe, D. "Mathematics made in Germany: on the background to Hilbert's Paris lecture." Math. Intelligencer 35 (2013), no. 3, 9--20. -Beyond the issue of possible correspondence concerning Hilbert's Paris lecture, Frei's book on the Klein-Hilbert correspondence may contain further evidence that Klein and Hilbert were, first of all, allied against the Berliners, and second of all both moderns contrary to the thrust of the Mehrtens hypothesis on Klein being allegedly countermodern: -Der Briefwechsel David Hilbert-Felix Klein (1886-1918). [The correspondence between David Hilbert and Felix Klein (1886-1918)] Edited, with comments, by Guether Frei. Arbeiten aus der Niedersachsischen Staats- und Universitatsbibliothek Gottingen [Publications of the Lower Saxony State and University Library in Gottingen], 19. Vandenhoeck & Ruprecht, Gottingen, 1985. -Note 1. As Jan Peter Schäfermeyer pointed out here, Klein not only published six papers by Cantor in Mathematische Annalen but also used Cantor as a referee for the journal. From the modern perspective this would indicate a progressive attitude on Klein's part. Any further details would be appreciated. - -REPLY [2 votes]: As Constance Reid points out, Klein had lost his interest in pure mathematics around 1900 and had devoted himself to projects in applied mathematics and teaching, which Hilbert had scarcely any interest for. -To Runge, whom he would make the first full professor in applied mathematics in Germany in 1904, he had already written in 1894 that he thought that mathematicians were too often occupied with artificial problems that were bred in university rooms, a view that was shared by Runge, who had already in the 1880s defected from pure mathematics.<|endoftext|> -TITLE: Finding largest triangular submatrix in binary matrices -QUESTION [6 upvotes]: The input of my problem is a 0/1 matrix. -The problem consists in finding the largest triangular submatrix. -In my problem, a square matrix is called triangular if all the entries below (or above) the main diagonal (or the off diagonal) are zero and all the entries on the diagonal are equal to one. -For example, in the following matrix -$$M=\begin{pmatrix} 1&1&1&1&1&1\\1&0&1&1&1&1\\1&1&1&0&0&1\\0&0&1&0&1&1\\1&1&1&1&1&1\\1&0&1&0&0&1 \end{pmatrix}, $$ -there is an obvious 3x3 triangular submatrix $M([3,4,5],[4,5,6])$ but there also is a less obvious 4x4 triangular submatrix $M([2,3,6,4],[4,2,1,5])$. -For a given matrix, is there a known algorithm to find the largest triangular submatrix? -Thank you! - -REPLY [2 votes]: Actually, it seems my problem has been studied in the following paper: - -M. C. Golumbic,T. Hirst, M. Lewenstein, Uniquely Restricted Matchings, Algorithmica, Volume 31, Issue 2, October 2001. - -The authors proved the problem of finding the largest triangular submatrix is NP-complete. However, they do not provide any algorithm.<|endoftext|> -TITLE: Applications of algebra to analysis -QUESTION [27 upvotes]: EDIT: I would like to make a list of modern applications of algebra in analysis. By "modern" I will mean developments since the beginning of the 20th century. It is well known that classical linear algebra of the 19th century (or earlier), such as vector spaces, determinants, diagonalization and Jordan form of matrices, have many applications in other areas of mathematics and beyond that, in particular in analysis. Below I will give an example of a modern application. -The choice of areas is motivated by my personal taste. -Now let me describe one of my favorite examples. Let $P$ be real a polynomial in $n$ variables. For any smooth compactly supported test function $\phi$ consider the integral $\int_{\mathbb{R}^n}|P(x)|^\lambda\phi(x)dx$. It converges absolutely for $Re(\lambda)>0$ and is holomorphic in $\lambda$ there. The question posed by I.M. Gelfand was whether it has meromoprhic continuation in $\lambda$ to the whole complex plane. -This problem has positive answer. In full generality it was solved by J. Bernstein and S. Gelfand, and independently by M. Atiyah, in 1969, see http://www.math.tau.ac.il/~bernstei/Publication_list/publication_texts/Bern-Gel-P-lam-FAN.pdf They used the Hironaka's theorem on resolution of singularities. That is a deep important result from algebraic geometry for which Hironaka was awarded Fields medal. -A little later, in 1972, J. Bernstein invented a different proof of the meromorphic continuation theorem which did not use Hironaka's result. The main step was the following statement. There exists a differential operator $D_\lambda$ on $\mathbb{R}^n$ whose coefficients depend polynomially on $x$ and are rational functions of $\lambda$ such that -$$|P|^\lambda=D_\lambda (|P|^{\lambda +1}) \mbox{ for } Re(\lambda) >0.$$ -Using this functional equation one can recursively extend the above integral to the whole complex plane. -In order to prove the above functional equation J. Bernstein developed a purely algebraic method which later on became fundamental in the theory of algebraic D-modules. - -REPLY [3 votes]: The second theorem below was first proved by Wiener (1933). Later Gelfand (1941) found a wonderful algebraic proof based on Banach algebras, which I included since it is so simple and elegant. -Theorem (Gelfand-Mazur) If $A$ is a complex Banach algebra with unity $e$ whose nonzero elements are invertible, then $A\cong\mathbb{C}$. -Proof (sketch). For any $a\in A$ there exists $\lambda(a)\in\mathbb{C}$ such that $a=\lambda(a)e$: otherwise $a-\lambda e$ would be invertible for any $\lambda\in\mathbb{C}$ and, choosing any $\phi\in A^*$ (the dual space) such that $\phi(a^{-1})\neq 0$, the holomorphic map $\lambda\mapsto\phi((a-\lambda e)^{-1})$ would contradict Liouville's theorem. -Clearly $a\mapsto\lambda(a)$ defines an isomorphism with $\mathbb{C}$. -Theorem (Wiener). If $f:S^1\to\mathbb{C}\setminus\{0\}$ has the form $f(e^{i\theta})=\sum_{n\in\mathbb{Z}}c_n e^{in\theta}$ with $\sum|c_n|<\infty$, -then $\frac{1}{f}$ has the same form (i.e. $\frac{1}{f(e^{i\theta})}=\sum_{n\in\mathbb{Z}}c_n'e^{in\theta}$ with $\sum|c_n'|<\infty$). -Proof (Gelfand). The set of functions $g(e^{i\theta})=\sum_{n\in\mathbb{Z}}a_n e^{in\theta}$ with $\sum|a_n|<\infty$ forms a commutative Banach algebra $B$ with the norm $\|g\|:=\sum|a_n|$ (and multiplicative identity $1$). -The thesis amounts to show that $f$ is invertible in $B$. -If this does not happen, then $f$ is contained in some maximal ideal $M$, which has to be closed by maximality (since invertible elements form an open set). By Gelfand-Mazur theorem $B/M\cong\mathbb{C}$, so there exists a homomorphism -$\phi:B\to\mathbb{C}$ satisfying $\phi(f)=0$. Notice that, for any $b\in B$, $|\phi(b)|\le\|b\|$: indeed, whenever $|\lambda|>\|b\|$ the element $b-\lambda\cdot 1=-\lambda(1-\lambda^{-1}b)$ is invertible in $B$ and thus $\phi(b)-\lambda=\phi(b-\lambda\cdot 1)\neq 0$. In particular, $\phi$ is continuous. -Let $h(e^{i\theta}):=e^{i\theta}$. Observe that $h\in B$ is invertible and that, for any $n\in\mathbb{Z}$, -$\|h^n\|=1$. So $|\phi(h)|^n=|\phi(h^n)|\le 1$ for all $n$, i.e. $\phi(h)\in S^1$. Thus we arrive at the contradiction -$$ 0=\phi(f)=\sum c_n\phi(h^n)=\sum c_n\phi(h)^n=f(\phi(h)). $$<|endoftext|> -TITLE: Poles and residues of degenerate Eisenstein series on GL(n) -QUESTION [6 upvotes]: Suppose $P \subseteq GL(n)$ is a parabolic subgroup. How do I find (or what is a reference for) the poles and residues of degenerate Eisenstein series associated to $P$? -More precisely, suppose $P_1, \ldots, P_r$ are parabolic subgroups of $GL(n)$ containing $P$, $\delta_{P_i}$ denotes the modulus character of $P_i$, $s_1, \ldots, s_r$ are complex parameters, and $$f(g, s_1, \ldots, s_r) \in Ind_{P}^{GL(n)}(\delta_{P_1}^{s_1} \cdots \delta_{P_r}^{s_r}).$$ The degenerate Eisenstein series associated to this data is -$$E(g,f,s_1, \ldots, s_r) = \sum_{\gamma \in P(F)\backslash GL_n(F)}{f(\gamma g, s_1, \ldots, s_r)}.$$ - -Question. Where are the poles of this Eisenstein series located, and what are the residues at these poles? - -If $P$ is the maximal parabolic of $GL(n)$ that stabilizes a line in the $n$-dimensional representation of $GL(n)$, the above question may be answered using Poisson summation. Outside of this special case, any comments or references would be helpful. - -REPLY [2 votes]: First, because the kind of induction that creates Eisenstein series can be done in stages, and commutes (after meromorphic continuation) with taking residues (even in the generalized sense required in this story), all these degenerate Eisenstein series are residues of the minimal-parabolic Eisenstein series. -Second, the poles (or polar divisors) of Eisenstein series (of all sorts) only occur at poles of their constant terms. The minimal-parabolic Eisenstein series has a very symmetrical minimal-parabolic constant term (and all the other constant terms are related by another sort of induction). -(Thus,) examination of that minimal-parabolic constant term of the minimal-parabolic Eisenstein series shows the hyperplanes along which that Eisenstein series, and any of its residues, can have poles. -An appendix in Langlands 544 treats that minimal parabolic Eis via Poisson summation and Bochner's analytic continuation lemma. Essentially the same argument is given in http://www.math.umn.edu/~garrett/m/v/gln_cont_spec.pdf -Then, for subtler reasons (basically looking at Plancherel), I think none of the residues of these Eis except the constant function are square-integrable. Going farther out "on a limb", it may be that the only residues in the cone analogous to a right half-plane are even-more-degenerate Eisenstein series. Poles outside that cone are more complicated, due to the functional equations.<|endoftext|> -TITLE: Differential calculus of functions of self-adjoint operators -QUESTION [6 upvotes]: Let $H$ be a Hilbert space over $\mathbb{C}$. Fix a self-adjoint operator $A:D(A)\rightarrow H$ and a Borel function $f:\mathbb{R}\rightarrow\mathbb{C}$. The operator $f(A)$ is defined by the spectral theorem. -I wonder whether under certain assumptions on $f$ and $A$, something of the following form $$f(A+B）=f(A)+L(B)+o(\|B\|)$$ can be concluded for $B$ bounded and symmetric. Here $L$ denotes a bounded ($\mathbb{R}$-)linear map from the space of (symmetric) bounded operators to the space of bounded operators on $H$. (In general, self-adjointness is lost under scaling by a complex number, so I can only hope to deal with real scalars.) -In my specific problem, $H$ is finite-dimensional, $f$ is smooth but not necessarily analytic. I am also interested in what happens for infinite-dimensional $H$ (and thus possibly stronger assumptions on $f$.) - -REPLY [3 votes]: Suppose that $A$ has discrete spectrum consisting of eigenvalues with finite multiplicities -$$ 0<\lambda_1 < \lambda_2<\cdots $$ -with $\lambda_n\to\infty$ as $n\to \infty$. Denote by $(\psi_n)$ an orthonormal basis consisting of e-vectors of $A$. In this basis $A$ is an (infinite) diagonal matrix $\DeclareMathOperator{\diag}{diag}$ -$$ A=\diag(\lambda_1,\lambda_2,\dotsc, ...).$$ -Then $\newcommand{\ve}{\varepsilon}$ -$$f(A+\ve I)=\diag( f(\lambda_1+\ve), f(\lambda_2+\ve),\dotsc),..). $$ -Assume that $\newcommand{\bR}{\mathbb{R}}$ $f:\bR\to\bR$ is $C^2$. Then -$$ f(\lambda_n+\ve I)=f(\lambda_n)+\ve f'(\lambda_n)\ve+ \frac{1}{2}f''(\xi_n)\ve^2,\;\;\xi_n\in (\lambda_n,\lambda_n+\ve). $$ -Suppose more concretely that $f(x)=x^3$. The above shows that $f(A+\ve I)-f(A)$ is not bounded. -This may not be as surprising since $A^3$ is not a bounded operators. Here is a more interesting examples. -Suppose for simplicity that $\lambda_n=n$ and $f$ is a $C^2$ function such that for $|x-n|<0.1$ we have $f(x)=n^2\cos (x-n)-n^2+1$. Note that -$$f(\lambda_n)=1,\;\;\forall n $$ -so $f(A)=I$. Consider the bounded operator -$$ B=\diag( 1^{-1/2}, 2^{-1/2},\dotsc, n^{-1/2},\dotsc, ). $$ -Then -$$f(A+\ve B)-f(A)=\diag(\dotsc, f(n+\ve n^{-1/2})-f(n),\dotsc), $$ -and we observe that, if $\ve<0.1$, then -$$ f(n+\ve n^{-1/2})-f(n) =n^2\cos\ve n^{-1/2}. $$ -We deduce that $f(A+\ve B)-f(A)$ is not bounded. -Remark (a) Here is a possible reformulations. There are several natural topologies on the space of closed selfadjoint operators; see this paper. One of them, called gap topology in the above paper is defined by a certain metric $\gamma$ (distance between the graphs) and has the property -$$\gamma(A_n,A)\to 0 \,\Longleftrightarrow\; \Vert f(A_n)- f(A)\Vert\to 0,\;\;\forall f\in C_0(\bR), $$ -where $C_0(\bR)$ denotes the space of continuous functions $f:\bR\to\bR$ such that -$$\lim_{t\to\pm\infty} f(t)=0. $$ -It may be the case that if $f\in C^2(\bR)$ is such that $f,f', f''\in C_0(\bR)$ then a conclusion of the type you've formulated could be true. -(b) Let me mention a closely related question. The set of closed selfadjoint Fredholm operators on $H$, equipped with the above gap topology can be organized as a Banach manifold. More precisely it is an open dense subset $\newcommand{\eO}{\mathscr{O}}$ $\eO$ of $\DeclareMathOperator{\Lag}{Lag}$ $\Lag(H\oplus L)$ Grassmannian of Lagrangian subspaces of $H\oplus H$; see this paper or this paper for more details. The above remark shows that any $f\in C_0(\bR)$ defines a continuous function $\hat{f}:\eO\to\bR$, $A\mapsto f(A)$ and I am $52$% sure that it extends to $\Lag(H\oplus H)$. Is it true that if $f\in C_0(\bR)$ is such that $f'\in C_0(\bR)$ that $\hat{f}$ is a $C^1$-function on $\eO$?<|endoftext|> -TITLE: How many multiplications are needed to generate a matrix algebra? -QUESTION [9 upvotes]: This is similar to an earlier question but I hope that it will be seen as being sufficiently distinct to merit separate consideration. -Let $M_d(\mathbb{C})$ denote the set of all $d \times d$ complex matrices and let $X \subset M_d(\mathbb{C})$ be nonempty. Then the set -$$\mathcal{A}(X):=\mathrm{span}\bigcup_{n=1}^\infty \{A_1\cdots A_n \colon A_j \in X\}$$ -is a subalgebra of $M_d(\mathbb{C})$, usually called the algebra generated by $X$. My question is: - - -What is the smallest number $N(d)$ such that - $$\mathcal{A}(X)=\mathrm{span}\bigcup_{n=1}^{N(d)} \{A_1\cdots A_n \colon A_j \in X\}$$ - for every nonempty set $X\subseteq M_d(\mathbb{C})$? - - -Let me remark that $N(d)=d^2$ is such a number, although I do not know if it is the smallest such number. To see this let us define -$$\mathcal{A}_N(X):=\mathrm{span}\bigcup_{n=1}^{N} \{A_1\cdots A_n \colon A_j \in X\}$$ -for each $N \geq 1$. We note that if $\mathcal{A}_N(X)=\mathcal{A}_{N+1}(X)$ for some integer $N$ then clearly $\mathcal{A}_N(X)=\mathcal{A}_{N+m}(X)$ for all $m \geq 1$ and therefore $\mathcal{A}(X)=\mathcal{A}_N(X)$. Since -$$\mathcal{A}_1(X) \subseteq \mathcal{A}_2(X) \subseteq \cdots \subseteq \mathcal{A}_{d^2}(X) \subseteq \mathcal{A}_{d^2+1}(X)$$ -and all of these sets are nonzero vector subspaces of the $d^2$-dimensional space $M_d(\mathbb{C})$, by the pigeonhole principle there is $N \leq d^2$ such that $\dim \mathcal{A}_N(X)=\dim \mathcal{A}_{N+1}(X)$ and therefore $\mathcal{A}_N(X)=\mathcal{A}_{N+1}(X)=\mathcal{A}(X)$. In particular we may take $N(d)= d^2$. -If we take $X$ to be a singleton set containing a permutation matrix of order $d$ then it is clear that $\mathcal{A}_N(X)=\mathcal{A}(X)$ for $N=d$ but not for any smaller $N$, so we cannot in general take $N(d) -TITLE: A question on the Hecke L-function -QUESTION [6 upvotes]: For a Hecke L-function, if all of the local eigenvalues are roots of unity, is it an Artin L-function? - -REPLY [12 votes]: Whatever your answer to my question in comment, the answer to the title question is yes. Take the case of the $L$-function attached to a modular form $f$, which we assume an eigenform for almost all Hecke operators (since otherwise there is no decomposition of the L-function into Euler product and the phrase "local eigenvalues" has no meaning.) -By Deligne, we know then that the local eigenvalues at $p$ for almost every $p$ are Weil numbers, more precisely they are algebraic numbers whose every embedding in $\mathbb C$ has absolute values $p^{(k-1)/2}$ where $k$ is the weight of your modular form (cf. the last corollary of this link for instance). If just one of those eigenvalue is a root of unity, this already forces $k=1$, in which case we know by Deligne-Serre that the Galois representation attached to $f$ has finite image, that is that $L(f,s)$ is an Artin $L$-function.<|endoftext|> -TITLE: Double series problems -QUESTION [6 upvotes]: How to calculate$$\sum_{n=-\infty}^{\infty}{\sum_{m=-\infty}^{\infty}{\frac{\left(-1\right)^n}{\left(6m\right)^2+\left(6n+1\right)^2}}}.$$Follow this,we first get $$\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{{{\left( {6k + 1} \right)}^2} + {l^2}}}} = \frac{{{l^2}}}{{{l^4} + 1}} - \frac{1}{{{l^2} + 1}} - \frac{\pi }{{12l}}\left( {\frac{{\sinh \frac{{\pi l}}{6}}}{{\cosh \frac{{\pi l}}{6} + \frac{{\sqrt 3 }}{2}}} + \frac{{\sinh \frac{{\pi l}}{6}}}{{ - \cosh \frac{{\pi l}}{6} + \frac{{\sqrt 3 }}{2}}}} \right).$$But it seems difficult to continue! - -REPLY [6 votes]: The paper Two-dimensional series evaluations via the elliptic -functions of Ramanujan and Jacobi deals exactly with double sums of this kind and shows how to evaluate them in terms of elliptic functions. -The book Lattice sums then and now gives alternative methods to calculate lattice sums.<|endoftext|> -TITLE: If a finite poset supports a Cohen-Macaulay ASL, how far can it be from Cohen-Macaulay? -QUESTION [5 upvotes]: By the fundamental work of De Concini, Eisenbud, and Procesi, an algebra with straightening law (ASL) must be Cohen-Macaulay if it is built on a Cohen-Macaulay poset. I would like to understand the state of the art regarding the converse question: If $A$ is a Cohen-Macaulay ASL over a field $k$ generated by the poset $P$, how far from Cohen-Macaulay can $P$ be (over $k$)? More precisely: - -(a) Is there a known lower bound on the depth of the Stanley-Reisner ring $k[P]$? -(b) What is the "worst" known example? -(c) What are the known conditions on $P$ or $A$ that force $P$ to be Cohen-Macaulay? - -I have been able to extract very little information about this from the literature I can find online. Here is what I know: -1) If $P$ is Buchsbaum and $A$ is Cohen-Macaulay, $P$ is Cohen-Macaulay (Miyazaki 2010). This is a partial answer to (c), but the Buchsbaum assumption is very strong. -2) Terai 1994 claims that $\operatorname{depth} k[P] \geq \operatorname{depth} A - 1$; however, Miyazaki 2010 claims Terai's proof is in error. This would have been a good answer to (a) if Terai had been correct, or to (b) if Miyazaki had given a counterexample. -3) In the more general setting of Hodge algebras, Hibi 1986 gives an example of a Cohen-Macaulay hodge algebra whose discretization is not Cohen-Macaulay. However, this example is not an ASL (or this would have been a partial answer to (b)). -This is all I have been able to find so far. I don't even know an example of a Cohen-Macaulay ASL whose discrete counterpart is not Cohen-Macaulay. I would appreciate any guidance you can offer about what is known about this. -Update 2/13/17: I reached Eisenbud and Terai by email. Neither is aware of an example of a C-M ASL whose discretization is not C-M, though Eisenbud says he would expect such a thing to exist. Terai says the problem is very difficult. I am willing to assume at this point that more or less nothing is currently known beyond the result of Miyazaki linked above. But if you do know something, I'm all ears. - -REPLY [4 votes]: Aldo Conca and Matteo Varbaro have posted a preprint that purports to answer this question, at least for graded ASLs: -Squarefree Gröbner degenerations -It appears that if $A$ is Cohen-Macaulay, then $k[P]$ must also be Cohen-Macaulay! See Corollary 3.9. -Addendum June 25, 2020: Conca and Varbaro's paper has now been published in Inventiones Mathematicae.<|endoftext|> -TITLE: Algebraic Surfaces -QUESTION [7 upvotes]: Is there any example of a smooth, projective surface $S$ over $\mathbb{C}$, with Picard group $\mathbb{Z}$ and such that $H^1(S, L)$ is not zero for some ample line bundle $L$ ? - -REPLY [6 votes]: Edit. My original exact sequences were wrong. I straightened them out now. -Let $X$ be a smooth, projective hyper-Kähler fourfold with $\text{Pic}(X)$ isomorphic to $\mathbb{Z}$, e.g., a sufficiently general deformation of the pair $(\text{Hilb}^2_{M/\mathbb{C}},\mathcal{O}(1))$ for $M$ a polarized K3 surface and $\mathcal{O}(1)$ the natural Plücker invertible sheaf. Let $Y\subset X$ be a smooth ample hypersurface such that $h^r(X,\mathcal{O}_X(\underline{Y}))$ vanishes for $r=1,2$. Let $Z\subset X$ be a very general hypersurface that is "sufficiently" ample. In particular, assume that $h^r(X,\mathcal{O}_X(\underline{Z}))$ vanishes for $r=1,2$ and $h^2(X,\mathcal{O}_X(\underline{Z}-\underline{Y}))$ vanishes. Let $i:S\hookrightarrow X$ be $Y\cap Z$, and denote by $\mathcal{I}$ the ideal sheaf of $S$ in $X$. Let $L$ be $\mathcal{O}_X(\underline{Y})|_S$. -There are short exact sequences, -$$ 0 \to \mathcal{I}(\underline{Y}) \to \mathcal{O}_X(\underline{Y}) \to i_* L \to 0,$$ -$$ -0 \to \mathcal{O}_X(-\underline{Z}) \to \mathcal{O}_X\oplus \mathcal{O}_X(\underline{Y}-\underline{Z}) \to \mathcal{I}(\underline{Y}) \to 0.$$ -By hypothesis, $h^r(X,\mathcal{O}_X(\underline{Y})) = 0$ for $r= 1,2$. Thus, via the long exact sequence of cohomology, the following connecting map is an isomorphism, $$\delta_\Sigma^1:H^1(S,L)\xrightarrow{\cong} H^2(X,\mathcal{I}(\underline{Y})).$$ By Serre duality, $h^{2+r}(X,\mathcal{O}_X(-\underline{Z}))$ equals $h^{2-r}(X,\mathcal{O}_X(\underline{Z}))$. Thus, for $r=0,1$, both of these vanish. Finally, by Serre duality, $h^2(X,\mathcal{O}_X(\underline{Y}-\underline{Z})))$ equals $h^2(X,\mathcal{O}_X(\underline{Z}-\underline{Y}))$, so both of these vanish. Thus, the long exact sequence of the second short exact sequence gives an isomorphism, -$$ -H^2(X,\mathcal{O}_X)\to H^2(X,\mathcal{I}(\underline{Y})), -$$ -so also $H^1(S,L)$ is isomorphic to $H^2(X,\mathcal{O}_X)$. -Of course for the hyper-Kähler fourfold $X$, $H^2(X,\mathcal{O}_X)$ is a $1$-dimensional vector space. -By the Lefschetz hyperplane theorem, the restriction map $\text{Pic}(X)\to \text{Pic}(Y)$ is an isomorphism (and the same holds for $Z$). By the generalized Noether-Lefschetz theorem (as in SGA7), for $Z$ sufficiently ample and very general, the restriction map $\text{Pic}(Y)\to \text{Pic}(S)$ is also an isomorphism. Thus $\text{Pic}(S)$ is isomorphic to $\mathbb{Z}$. -Edit. I was misremembering what is proved in SGA 7_2. The generalized Noether-Lefschetz theorem was not proved there. The generalized Noether-Lefschetz was proved by Kirti Joshi. -MR1299006 (96f:14005) -Joshi, Kirti(6-TIFR-SM) -A Noether-Lefschetz theorem and applications. -J. Algebraic Geom. 4 (1995), no. 1, 105–135. -https://arxiv.org/pdf/alg-geom/9305001v2.pdf<|endoftext|> -TITLE: Differential forms as a sheaf on the site of all manifolds vs. sheaf on an individual manifold -QUESTION [6 upvotes]: The de Rham complex can be viewed as a sheaf $\Omega$ on the entire site $\mathsf{Man}$ of smooth manifolds via the usual pullback of differential forms $(f: M \to N) \mapsto (f^*: \Omega(N) \to \Omega(M))$. So, the de Rham complex is a smooth space. Diffeological spaces, for example, are concrete sheaves on the site of smooth manifolds, but I don't know enough about sheaves to say whether or not the de Rham complex is a diffeological space. -We can also consider the category of open sets of $M$ as a site $O(M)$, where the arrows are inclusions. Then we have the sheaf $\Omega_M \in \mathsf{Sh}(M)$ of differential forms on $M$. Likewise, we have the sheaf $\Omega_N \in \mathsf{Sh}(N)$ of differential forms on $N$. -Now, as I understand it, there are morphisms $i_M: \mathsf{Man} \to M$ and $i_N: \mathsf{Man} \to N$ of sites, given by the respective inclusion functors $i_M^t: O(M) \to \mathsf{Man}$ and $i_N^t: O(N) \to \mathsf{Man}$. -Can we describe the sheaves of differential forms $\Omega_M$ and $\Omega_N$ on $M$ and $N$ in terms of these morphisms of sites? I figure they should somehow be induced by the sheaf $\Omega$ on all of $\mathsf{Man}$, but I am not very experienced with sheaves on sites yet. Moreover, given a smooth map $f: M \to N$, there should a relationship between the inverse image functor and the pullback of forms. - -REPLY [4 votes]: The sheaf $ \Omega^k $ ($k>0$) on the site $ \mathsf{Man} $ is not a diffeological space, because it is not a concrete sheaf: Obviousely, -$ \Omega^k(\mathbb{R}^k)\ni\alpha\mapsto \underline{\alpha}:\mathrm{hom}(\mathbb{R}^0,\mathbb{R}^k)\rightarrow\Omega^k(\mathbb{R}^0)=0 $ -is not one-to-one. -Sheaves $ \Omega_M $ and $ \Omega_N $ are the restriction of the sheaf $ \Omega $ to the sites $ O(M) $ and $ O(N) $, respectively. -For a smooth map $f:M \rightarrow N $, the induced inverse image functor commutates with the restrictions. -The pullback induced by $ f $ is just the global section map.<|endoftext|> -TITLE: A Combinatorial Game with Integer Sequences -QUESTION [9 upvotes]: Two players, Alice and Bob, take turns constructing a sequence $a_1,a_2,a_3,\dots$, of distinct positive integers, none greater than a given parameter $K$. Alice plays first and makes $a_1=1$. Thereafter, on the $n$-th move, the player whose turn it is plays $a_n=a_{n-1}+n,a_{n-1}\times n,a_{n-1}\div n$, or $a_{n-1}-n$, whichever he or she chooses provided $a_n$ is a positive integer not already in the sequence, and $\le K$. -The player who makes the last legal move wins. -For which values of $K$ does Alice have a winning strategy? - -REPLY [9 votes]: Not an answer, but some quick data that may help. -On Alice's first move, she only has one option, so her Grundy value is either 0 (no winning strategy) or 1 (a winning strategy). More interesting is the first Grundy value that Bob sees. Some quick Mathematica coding gets the Grundy values for $K\leq 100$. -Mex[A_] := Module[{m = 0}, While[MemberQ[A, m], m++]; m]; -opts[K_, chosen_] := - Module[{move = Length[chosen] + 1, last = Last[chosen]}, - next = - Select[{last + move, last - move, last*move, last/move}, - IntegerQ[#] && 1 < # <= K && Not[MemberQ[chosen, #]] &]; - Map[Append[chosen, #] &, next]]; -gv[K_, {}] = 0; -gv[K_, chosen_] := Mex[gv[K, #] & /@ opts[K, chosen]]; -ParallelTable[gv[K, {1}]], {K, 1, 100}] - -The Grundy values that Bob sees are -0, 1, 1, 1, 2, 2, 0, 0, 0, 0, 0, 0, 1, 1, 2, 1, 1, 0, 2, 2, 2, 2, 2, -1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 0, 0, 2, 2, 2, 2, -2, 1, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 1, 0, 2, 2, 2, 2, 1, 1, 1, -1, 1, 1, 0, 1, 0, 0, 0, 2, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 0, -0, 0, 0, 0, 2, 0, 0, 0,.... In other words, Bob cannot win (unless Alice allows it) for $K \in \{1,7,8,9,10,11, 12, 18, 41, 42, 50, 51, 52, 53, 55, 56, 57, 58, 59, 62, 73, 75, -76, 77, 92, 93, 94, 95, 96, 98, 99, 100,\dots\}$. Is there ever a Grundy value of 3 or 4? No, as Bob only has two options $a_2=2$, $a_2=3$ for the second move of the game. But deep inside the game, perhaps 3's and 4's happen. -Neither the sequence of Grundy values nor the good-for-Alice $K$ are in the OEIS.<|endoftext|> -TITLE: What is the universal property of the Weyl group? -QUESTION [13 upvotes]: If $G$ is a group and $H\le G$ a subgroup, let $NH$ denote the normalizer of $H$ in $G$, and let $WH = NH/H$; following May's Concise Course, §3.4 this I call the Weyl group. I have also seen the definition $WH=NH/C(H)$, where $C(H)$ is the centralizer of $H$ in $G$, from https://groupprops.subwiki.org/wiki/Weyl_group. -What is the universal property of $WH$? It would also be good to have a notion of $Wf$ for any morphism $H \mathop{\longrightarrow}\limits^f G$, rather than just inclusions. -I'm asking this in pursuit of an abstract-nonsensical explanation of: - -If $H\le G$ is a subgroup, and $X$ is a $G$-set, then $X^H$ is naturally a $WH$-set. - -For example, since much of the formalism of $G$-sets extends to $\mathbf{Set}^C$ for any small category $C$, I might eventually try to generalize this to any functor between small categories. - -REPLY [10 votes]: If $F : C \to D$ is any functor whatsoever, the objects $F(c), c \in C$ always have a natural action of the automorphism group $\text{Aut}(F)$ of $F$ as a functor, and $\text{Aut}(F)$ is universal with respect to this property. If $F : C \to \text{Set}$ is a representable functor, this automorphism group coincides with the auotmorphism group of the representing object, by the Yoneda lemma. The functor of taking $H$-fixed points is representable by $G/H$, whose automorphism group is the Weyl group. -In general, it's an interesting question to ask what extra structure some objects $d \in D$ acquire by virtue of having been spit out by a functor $F : C \to D$. A very general answer to this question is that under mild hypotheses $F$ has what's called a codensity monad, and the objects $F(c), c \in C$ naturally acquire the structure of an algebra over this monad. If $F$ is the right adjoint of an adjunction this is the usual monad arising from that adjunction. -If $F : C \to \text{Set}$ is a representable functor represented by an object $c$, it has a left adjoint given by sending a set $X$ to the coproduct $\coprod_X c$ of $X$ copies of $c$ (assuming that these coproducts exist), and the codensity monad sends a set $X$ to the set $\text{Hom}(c, \coprod_X c)$. If $c$ is "connected" in the sense that a morphism $c \to \coprod_X c$ factors through one of the inclusions of a copy of $c$ (which is the case here, where $c = G/H$), then this is the monad whose algebras are $\text{End}(c)$-sets; otherwise it's more complicated.<|endoftext|> -TITLE: Is the Cartan matrix of a finite-dimensional (Hopf) algebra invertible over the rationals? -QUESTION [5 upvotes]: This is probably well-known to representation theorists, but this doesn't imply being well-known to me. -Let $k$ be a field, and let $A$ be a $k$-algebra that is finite-dimensional as a $k$-vector space. Let $S_1, S_2, \ldots, S_m$ be a complete list of pairwise non-isomorphic simple $A$-modules (up to isomorphism). Let $P_1, P_2, \ldots, P_m$ be the projective covers of $S_1, S_2, \ldots, S_m$, respectively. (Thus, $P_1, P_2, \ldots, P_m$ is a complete list of pairwise non-isomorphic indecomposable projective $A$-modules.) The Cartan matrix of $A$ is defined to be the $m\times m$-matrix in $\mathbb{Z}^{m\times m}$ whose $\left(i,j\right)$-th entry (for all $i$ and $j$) is the number of composition factors of $P_j$ isomorphic to $S_i$. (In other words, its $\left(i,j\right)$-th entry is $\dfrac{\dim \operatorname{Hom}_A \left(P_i, P_j\right)}{\dim \operatorname{End}_A \left(S_i\right)}$.) See §7.4 of Peter Webb's A Course in Finite Group Representation Theory for proofs and some background. -1. Is it true that the Cartan matrix of $A$ is invertible as a matrix in $\mathbb{Q}^{m\times m}$ ? (I think the answer is "no", but I don't know of a counterexample.) -2. Does this change if we require $k$ to be algebraically closed (or at least to be a splitting field for $A$) ? -3. Does this change if we furthermore require $A$ to be a Hopf algebra? -What I know is that the Cartan matrix of $A$ is invertible if $A$ is the group algebra of a finite group. This is proven in Corollary 10.2.4 of Peter Webb's above-mentioned book, but the proof uses the theory of Brauer characters, which as far as I know is particular to group algebras. (Or not? Is there a notion of Brauer characters for Hopf algebras? I certainly wouldn't find it strange, at least compared with the strangeness of classical Brauer character theory, but I have never seen such a notion.) - -REPLY [3 votes]: Perhaps it's worthwhile to mention one very large and natural class of finite dimensional Hopf algebras (over fields of prime characteristic $p$) for which the Cartan matrix is certainly not invertible: the restricted universal enveloping algebras of the Lie algebras of semisimple algebraic groups. Recall that if $G$ and hence its Lie algebra $\mathfrak{g}$ has dimension $d$, the restricted enveloping algebra has dimension $p^d$; it has for each ordered basis of $\mathfrak{g}$ an ordered basis of PBW type obtained by truncating at $p$th powers according to the $[p]$-operation on $\mathfrak{g}$. See for example the remarks at the end of section 4 in my 1971 paper here. -Take the simplest case, with $\mathfrak{g} = \mathfrak{sl}_2$. Here the $p$ simple modules have dimensions $1,2, \dots, p$. One block (indecomposable 2-sided ideal) involves just the simple/projective Steinberg module of dimension $p$, with Cartan invariant 1. Each other block involves just two simple modules of "linked" highest weights. If weights are identified with non-negative integers $0,1, \dots, p-1$ (with $p-1$ the highest weight of the Steinberg module), such blocks correspond to pairs of weights adding up to $p-2$. Then the Cartan invariants of the block are all 2. -Larger ranks lead to much more complicated (and not easily computed) Cartan invariants, but the theory developed in older work such as my paper yields the same result on the vanishing determinant of the matrix of these invariants.<|endoftext|> -TITLE: Decorated permutations and subset permutations -QUESTION [6 upvotes]: Decorated permutations are defined as permutations where the fix-points come in two colors (say $\overline{\cdot}$ and $\underline{\cdot}$). For example, the 16 decorated permutations of length 3 are -$$\overline{1}\overline{2}\overline{3},\underline{1}\overline{2}\overline{3},\overline{1}\underline{2}\overline{3},\overline{1}\overline{2}\underline{3},\underline{1}\underline{2}\overline{3},\underline{1}\overline{2}\underline{3},\overline{1}\underline{2}\underline{3},\underline{1}\underline{2}\underline{3},$$ -$$\overline{1}32,\underline{1}32,\hspace{10pt} -21\overline{3},21\underline{3},\hspace{10pt} -231,\hspace{10pt} -312,\hspace{10pt} -3\overline{2}1,3\underline{2}1\ .$$ -They play an important role in the context of the positive Grassmannian and are in bijection with many other combinatorial objects such as bounded affine permutations, Grassmann necklaces, Le-diagrams, positroids and equivalence classes of plabic graphcs, see, for example, here or here. These combinatorial objects have certainly been studied in detail in recent years. -Their total number of clearly $\sum_{k=0}^n F(n,k)2^k$, where $F(n,k)$ denotes the number of permutations of length $n$ with $k$ fixed-points. -When looking for these on the OEIS, I saw that neither of the above names was found there, whilst subset permutations are found there as A000522. These are, given a set with $n$ elements, obtained by choosing $k$ elements among them, and then permute these $k$ elements, giving in total $\sum_{i=0}^{n} {n \choose k}k! = \sum_{k=0}^{n} n!/k!\ $. -For example, for $n=3$, we obtain the following 16 elements: -$$-,1,2,3,12,21,13,31,23,32,123,132,213,231,312,321.$$ -I claim (and quickly prove below) that subset permutations also belong to the list of combinatorial objects in bijection with the above. -My questions are now - -Have the later already appeared in this context? - -and - -Is the equality $\sum_{k=0}^n F(n,k)2^k = \sum_{k=0}^{n} n!/k!$ otherwise directly implied? - -Construction of the bijection: -Take a subset permutation. This is, take a set of $k$ element in $\{1,\ldots,n\}$ together with a permutation of these. -Consider this as a permutation of length $n$ with the remaining $n-k$ elements being fixed-points. -Now label these $n-k$ fixed-points by $\underline{\cdot}$ and the additional fixed-points from the permutation of the $k$ elements by $\overline{\cdot}$. -Than this is a bijection from subset permutations to decorated permutations. -For example, $143768$ as a subset permutation of $\{1,\ldots,8\}$ is mapped to $\overline{1}\underline{2}43\underline{5}76\overline{8}$. -Also, the preimages of the 16 decorated permutations at the beginning are -$$123,23,13,12,3,2,1,-,132,32,213,21,231,312,321,31.$$ - -REPLY [3 votes]: Max's answer is neat in using $e^{2x}\frac{e^{-x}}{1-x}=\frac{e^x}{1-x}$. We offer one approach without generating functions. To this end, let $f_n:=\sum_{k=0}^n\frac1{k!}$ and $g_n:=\sum_{k=0}^n\frac{2^k}{k!}\sum_{j=0}^{n-k}\frac{(-1)^j}{j!}$. Assume empty sums to be $0$. Clearly, $f_{n+1}-f_n=\frac1{(n+1)!}$. On the other hand, we have -\begin{align} g_{n+1}-g_n -&=\sum_{k=0}^{n+1}\frac{2^k}{k!}\sum_{j=0}^{n+1-k}\frac{(-1)^j}{j!} --\sum_{k=0}^{n+1}\frac{2^k}{k!}\sum_{j=0}^{n-k}\frac{(-1)^j}{j!} \\ -&=\sum_{k=0}^{n+1}\frac{2^k}{k!}\frac{(-1)^{n+1-k}}{(n+1-k)!} -=\frac{(-1)^{n+1}}{(n+1)!}\sum_{k=0}^{n+1}\binom{n+1}k(-2)^k \\ -&=\frac{(-1)^{n+1}}{(n+1)!}(-1)^{n+1}=\frac1{(n+1)!}. -\end{align} -Therefore, $f_{n+1}-f_n=g_{n+1}-g_n$. Since $f_0=g_0=1$, it follows $f_n=g_n$ and $n!f_n=n!g_n$. Obviously, $n!f_n=\sum_{k=0}^n\frac{n!}{k!}$ and -$$n!g_n=\sum_{k=0}^n2^k\frac{n!}{k!}\sum_{j=0}^{n-k}\frac{(-1)^j}{j!}=\sum_{k=0}^n2^k\binom{n}k\cdot !(n-k)=\sum_{k=0}^nF(n,k)2^k.$$<|endoftext|> -TITLE: Homotopical algebra is not concrete -QUESTION [29 upvotes]: There is this old result by Freyd that "homotopy is not concrete": - -Freyd, Peter. "Homotopy is not concrete." The Steenrod Algebra and Its Applications: A Conference to Celebrate NE Steenrod's Sixtieth Birthday. Springer Berlin Heidelberg, 1970. - -In 2017 we say that - - -there is a homotopical category $(\bf Top, \cal W)$, such that the localization ${\bf Top}\!\!\left[{\cal W}^{-1}\right]$ is not ($\bf Set$-)concrete. - - -The key result here is the following lemma, called "Isbell condition": -For $f\in {\cal A}/A$ an object of the slice category, let $C(f,B)$ be the class of pairs $u,v : A \to B$ such that $uf=vf$ as morphisms $\text{src}(f) \to A \to B$. -Define an equivalence relation $\asymp$ on $({\cal A}/A)_0$ that says $f\asymp g$ iff $C(f,B)=C(g,B)$ for every $B\in\cal A$, and let $\textsf{S}({\cal A}/A)$ the quotient $\left({\cal A}/A\right)_{0,/\asymp}$. - -Isbell condition: $\cal A$ is concrete if and only if $\textsf{S}({\cal A}/A)$ is a set for every $A\in\cal A$. - -If you follow the development of this, you find another, slightly older paper - -Freyd, Peter. "On the concreteness of certain categories." Symposia Mathematica. Vol. 4. 1969. - -that develops quite a bit this technology, and contains (Thm 4.1) a result that in 2017 we write as - - -the localization ${\bf Cat}\!\!\left[\text{Eqv}^{-1}\right]$ of the category of categories to equivalences (i.e. the homotopy category $\textsf{Ho}({\bf Cat}_\text{folk})$ is not concrete. - - -Now, following Christie's meta-theorem it's easy to wonder if there is a pattern, and maybe a proof, here. -Freyd's theorem is as old as Quillen's definition of a model category, so I doubt that Freyd ignored that you can ask the following question: - - -Let $\mho$ be a universe. If $\cal M \in {\bf Cat}$ is locally $\mho$-small and has a model structure, how often is the localization $\textsf{Ho}(\cal M)$ a ($\mho\text{-}\bf Set$-)concrete category? - - -(One could argue that this result really belongs to the world of homotopical categories and should be stated therein: it should, but a model structure is highly tamer to handle). -So: - -Has anybody attacked this problem with modern technology? -Do you think that the above is a valuable question? -I believe it is, because - - -Every category that breaks Isbell condition (let's call it a "non-Isbell category" for the sake of brevity) seems quite nasty. And yet its homotopy theory can be well-understood. Isbell condition itself is stated in terms of the set theory of $\cal A$ and it is (unsurprisingly) linked to $\cal A$ having "nice" factorization systems ("nice" here means proper+something; did somebody explicitly prove this, maybe even Freyd?). So one can "foresee" if $\cal M$ will have a non-concrete localization proving that there is no homotopy-nice factorization system on $\cal M$ (a factorization system on $\cal M$ is homotopy-nice if it is an homotopy FS in the sense of Bousfield, and the FS induced by it on $\textsf{Ho}(\cal M)$ is nice). -All these categories seem to exist (but I will be happy to see you disproving me, especially in the nontrivial cases): - - -a concrete category whose localization is concrete -a non-concrete category whose localization is concrete -a non-concrete category whose localization is non-concrete -a concrete category $\cal M$ whose allocalization ${\cal M}[\text{All}^{-1}]$ is non-concrete -a non-concrete category $\cal M$ whose allocalization ${\cal M}[\text{All}^{-1}]$ is concrete -a non-concrete category $\cal M$ whose allocalization ${\cal M}[\text{All}^{-1}]$ is non-concrete -a concrete category $\cal M$ whose allocalization ${\cal M}[\text{All}^{-1}]$ is non-concrete - -There's an interesting problem: every category is a quotient of a concrete category (Kučera, JPAA 1971, link). Is every category a localization of a concrete one? - -If $\textsf{Ho}(\cal M)$ is not concrete, there should be a property $P$ of $\cal M$ preventing $\textsf{Ho}(\cal M)$ to be Isbell. It seems obvious that every category $\cal N$ which is Quillen equivalent to $\cal M$ has $P$, and yet Freyd's technology seems to be rather context-specific, to the point that it's hard to believe that $P$ can be transported along the adjunction of a Quillen equivalence. Or maybe it is, under suitable assumptions? - -REPLY [9 votes]: This stream of thought led to the following preprint that has been on the arXiv for a few days. https://arxiv.org/abs/1704.00303 -This is not an act of self-promotion, but maybe some of the interested readers of this thread can find some development of these ideas. -Comments are welcome, and in fact encouraged.<|endoftext|> -TITLE: Does Keisler-Shelah isomorphism theorem hold for infinitary logics? -QUESTION [7 upvotes]: In model theory, the Keisler-Shelah isomorphism theorem asserts that two models of a theory are elementary equivalent if and only if they have isomorphic ultrapowers. On the other hand, assuming that $\kappa$ is strongly compact, we can ensure the existence of a $\kappa$-complete ultrafilter on every set (that is, an ultrafilter closed under intersections of less than $\kappa$ elements) and prove a version of Łoś theorem for the infinitary logic $\mathcal{L}_{\kappa, \kappa}$. -My question is whether a suitable version of the Keisler- Shelah theorem also holds for infinitary languages. Shelah's original proof (the first without using GCH), or the similar version in Chang-Keisler book "Model theory", seem to be readily generalizable, but I would like if possible to find a reference for it. The statement I have in mind is as follows: -Let $\kappa$ be a strongly compact cardinal and assume $\mathcal{A}$ and $\mathcal{B}$ are models of a theory in $\mathcal{L}_{\kappa, \kappa}$. Then $\mathcal{A}$ and $\mathcal{B}$ are elementary equivalent (with respect to the language $\mathcal{L}_{\kappa, \kappa}$) if and only there is a set $I$ and a $\kappa$-complete ultrafilter $U$ over $I$ such that the ultrapowers $\Pi_{i \in I}\mathcal{A}/U$ and $\Pi_{i \in I}\mathcal{B}/U$ are isomorphic. -Does this hold or has it been considered in the literature? - -REPLY [9 votes]: This is a very nice question. -The answer is no, it doesn't necessarily hold. Let me describe a counterexample. Consider the language of linear orders. By the pigeon-hole principle, since there are only a set of possible $L_{\kappa,\kappa}$-theories in this language, there must be two ordinals $\alpha<\beta$ such that as linear orders, $\langle \alpha,<\rangle$ and $\langle\beta,<\rangle$ are $L_{\kappa,\kappa}$-elementarily equivalent. -Now consider any $\kappa$-complete ultrafilter $U$ on a set $I$. Let $j:V\to M$ be the ultrapower by $U$. The target model $M$ is well-founded, since $U$ is $\kappa$-complete. So we may take $M$ to be a transitive class. -The ultrapower $\langle\alpha,<\rangle^I/U$ is really just the same as $\langle j(\alpha),<\rangle$, and similarly $\langle\beta,<\rangle^I/U$ is isomorphic to $\langle j(\beta),<\rangle$. Since $\alpha\neq\beta$, it follows that $j(\alpha)\neq j(\beta)$, and since these are ordinals, they are not isomorphic. -So we've found two linear orders that are $L_{\kappa,\kappa}$-elementarily equivalent, but cannot have isomorphic ultrapowers by any $\kappa$-complete ultrafilter on any set.<|endoftext|> -TITLE: Free groups and free restricted Lie algebras -QUESTION [13 upvotes]: If $G$ is any group and $\gamma_k(G)$ denotes the $k$th term in the lower central series of $G$, then the commutator bracket on $G$ endows -$$\mathcal{L}(G) = \bigoplus_{k=1}^{\infty} \gamma_k(G) / \gamma_{k+1}(G)$$ -with the structure of a Lie ring. A famous theorem that should probably be attributed to Magnus says that if $G$ is the free group on $n$ generators, then $\mathcal{L}(G)$ is the free Lie ring on $n$ generators. This result is documented in many places, e.g. Magnus--Karass--Solitar's book on combinatorial group theory and Serre's book Lie Group and Lie Algebras. -Now fix a prime $p$ and let $\delta_k(G)$ be the fastest descending central series of $G$ satisfying the following three properties: - -$\delta_1(G) = G$, and -$[\delta_k(G),\delta_{\ell}(G)] \subset \delta_{k+\ell}(G)$ for all $k,\ell \geq 1$, and -$(\delta_k(G))^p \subset \delta_{pk}(G)$ for all $k \geq 1$. - -This series was first defined by Zassenhaus. The commutator bracket and the $p$th power operation on $G$ endow -$$\Lambda(G) = \bigoplus_{k=1}^{\infty} \delta_k(G) / \delta_{k+1}(G)$$ -with the structure of a restricted Lie algebra over the field $\mathbb{F}_p$. -I'm pretty certain that I can prove that if $G$ is a free group on $n$ generators, then $\Lambda(G)$ is the free restricted Lie algebra on $n$ generators over the field $\mathbb{F}_p$. I need this result for a paper I am writing (on an unrelated topic), but I'm certain that this result is known and would greatly prefer to just cite a reference for it. -Question: Can anyone give me a reference for this? - -REPLY [6 votes]: This is Theorem 6.5 p.130 from Lazard, Michel. -Sur les groupes nilpotent et les anneaux de Lie. -Ann. Sci. Ecole Norm. Sup (3) 71 (1954) 101-190.<|endoftext|> -TITLE: Are the definable hyper-reals, using quantifiers only over the standard reals and natural numbers, the same as the algebraic numbers? -QUESTION [8 upvotes]: This question arose today at Yevgeny Gordon's talk, "Will nonstandard analysis be -the analysis of the future?" at the CUNY Logic -Workshop. Here is my way of asking it. -Consider the ordered real field $\newcommand\R{\mathbb{R}}\R$ with a predicate for the natural numbers, and a -nonstandard version of it $\R^*$, the hyper-reals, which an ordered -field extending $\R$ and having the transfer property, a map -$a\mapsto a^*$, which preserves the truth of any statement in the -language of ordered fields, allowing also the predicate for the -natural numbers. The hyper-real numbers of the form $a^*$ are -referred to as the standard elements of $\R^*$. -Question. Suppose that a hyper-real number $a\in\R^*$ is -definable in $\R^*$ by a formula $\varphi$ in the language of -ordered fields with a predicate for natural numbers, but where the -scope of the quantifiers is only over the standard elements. Must -$a$ be algebraic? -It is easy to see that every algebraic hyper-real number is -definable in this way. For example, the hyper-real $\sqrt{2}$ is -definable in $\R^*$ as the unique $x$ for which $x^2=2$ and $00\ \exists N\ \forall n\geq N \ -|(1+\frac 1n)^n-x|<\epsilon.$$ This definition can be undertaken in $\R$, using the predicate for $\mathbb{N}$, which allows one to define exponentiation and so on. But in $\R^*$, these quantifiers are not only -over the standard numbers; we have to quantify also over the -nonstandard numbers. If you restrict to standard $\epsilon$ and -standard $N$ and $n$ only, then there will be an entire interval of -hyper-reals that are that close to those numbers---anything -infinitesimally close to $e$ will do. So the restricted-scope version of the definition will not succeed as a definition. -Similarly, it is not clear how to define $\pi$ or indeed any other -transcendental hyper-real number while quantifying only over -standard numbers. -I believe that Tarski's theorem on real-closed -fields will prove -the positive result for the special case of the question, where -$\varphi$ does not use the predicate for the natural numbers (but -still has the scope of all quantifiers restricted to the standard -hyper-reals). My reason for this expectation is that I believe we -can apply Tarski's elimination of quantifiers procedure to such a -$\varphi$ and thereby prove that $\varphi(x)$ is equivalent to a -quantifier-free assertion in the language of ordered fields. Then, -using the fact the algebraic numbers form an elementary -substructure of the reals, as ordered fields, it follows that the -existence of a solution in $\R$ is equivalent to the existence of a -solution in the algebraic numbers. And so the given number must be -algebraic. -But I am a little fuzzy on the details of how the scope-restriction -affects this argument, and so if you can affirm or refute it, I -would be grateful. - -REPLY [8 votes]: $\newcommand{\st}{\textrm{st}}\newcommand{\bR}{{\bf R}}\newcommand{\bN}{{\bf N}}$ -I think the limit of any definable convergent sequence $(a_n)$ (including $e$) is definable by the formula -$$\varphi(x)=(\exists x'\in {\bR}^\st\, x'=x)\land \forall N\in\bN^\st\exists n_0\in \bN^\st\forall n\in \bN^\st_{>n_0}(a_n-x)^2 -TITLE: Complexity of computing the number of visible points -QUESTION [5 upvotes]: A visible point is a point $(a, b)\in \mathbb{Z}^2,$ with $gcd(a,b)=1$. It is well-known that the number $V(N)$ of visible points with $00$, since we can compute the totient function in sub-polynomial time. - -REPLY [9 votes]: There is an algorithm for computing $F(N) = \# \{ (a,b) : 1 \leq a, b \leq N, \gcd(a,b) = 1 \}$ in time $O(N^{5/6 + \epsilon})$. This relies on the algorithm of Deleglise and Rivat (see their paper here) that computes $M(x) = \sum_{n \leq x} \mu(n)$ in time $O(x^{2/3} (\log \log x)^{1/3})$. -The idea is to use -$$ - F(N) = \sum_{a=1}^{N} \sum_{b=1}^{N} \sum_{d | \gcd(a,b)} \mu(d) = \sum_{d=1}^{N} \mu(d) \left\lfloor \frac{N}{d}\right\rfloor^{2}. -$$ -Now, break this sum up into the terms with $1 \leq d \leq N^{\alpha}$ and the terms with $N^{\alpha} < d \leq N$. The first term -$$ - \sum_{d=1}^{N^{\alpha}} \mu(d) \left\lfloor \frac{N}{d} \right\rfloor^{2} -$$ -can be computed easily in time $O(N^{\alpha + \epsilon})$, since all $\mu(d)$ for $d \leq N^{\alpha}$ can be computed in time $O(N^{\alpha} \log \log(N))$ (see this preprint). For the range $N^{\alpha} < d \leq N$, let $k = \lfloor N/d \rfloor$. We get the sum -$$ - \sum_{k=1}^{\lfloor N^{1-\alpha} \rfloor} k^{2} \sum_{\substack{e \\ \lfloor N/e \rfloor = k}} \mu(e) = \sum_{k=1}^{\lfloor N^{1-\alpha} \rfloor} k^{2} (M(b_{k}) - M(a_{k})). -$$ -Here $a_{k}$ and $b_{k}$ are the smallest and largest positive integers $N$ for which $\lfloor N/e \rfloor = k$, respectively. There are $O(N^{1-\alpha})$ terms in the sum above, and each can be computed in time $O(N^{2/3+\epsilon})$. The total running time $O(N^{\alpha + \epsilon} + N^{5/3 - \alpha + \epsilon})$ is minimized when $\alpha = 5/6$ and gives $O(N^{5/6 + \epsilon})$. - -REPLY [4 votes]: This is not an answer to your (interesting) question, but instead mentions a possibly related question. - -Cardinal, Jean, and Udo Hoffmann. "Recognition and complexity of point visibility graphs." Discrete & Computational Geometry 57.1 (2017): 164-178. - (earlier arXiv version.) - -Perhaps their ~30 references, or the ~10 subsequent citations, might lead somewhere...? - -"We study the recognition problem for point visibility graphs: given a simple undirected graph, decide whether it is the visibility graph of some point set in the plane. We show that the problem is complete for the existential theory of the reals."<|endoftext|> -TITLE: How to identify localization of categories? -QUESTION [10 upvotes]: Let $C, D$ be categories with finite limits, $F:C\to D$ be a essentially surjective functor that commutes with finite limits, and let $S$ be the set of morphisms of $C$ that become isomorphisms in $D$ under $F$. It can be proved that $S$ is a right multiplicative system (as defined here). Then we have a functor $S^{-1}F: S^{-1}C \to D$. Are there criterion for $S^{-1}F$ to be an equivalence? -(To avoid set-theoretic issues, let's assume $C, D$ are essentially small.) - -REPLY [5 votes]: The criterion for $S^{-1}F$ to be an equivalence is that $F$ satisfies the definition of the localisation, as in Definition 7.1.1 in Categories and Sheaves. -In the above settings, since $F$ is essentially surjective, so is $S^{-1}F$. Then, assuming the axiom of choice, the question degenerates to when is $S^{-1}F$ is fully faithful. The question's assumptions do not seem to simplify the general criterion, given in the definition. One needs hom-sets in $D$ to be isomorphic to hom-sets in $S^{-1}C$, and there are two ways (or a combination of both) in which that may fail: - -If hom-sets in $D$ are quotients of those of $S^{-1} C$: Let $C$ be an essentially small category with finite limits (e.g. small finite sets), let $D$ be the terminal category (with one object and one morphism), and let $F:C\to D$ be the terminal functor. $D$ has finite limits, and $F$ is essentially surjective and commutes with finite limits. $S$ consists of all morphisms in $C$, and hence $S^{-1}C$ is the groupoidification of $C$. -If hom-sets in $D$ are 'bigger' than those of $S^{-1}C$: -Let $F:C\to D$ be the functor $$\mathbb{R}\otimes_{\mathbb{Z}}-:\mathbb{Z}\text{-}\mathbf{Mod}\to\mathbb{R}\text{-}\mathbf{Mod}.$$ -Both $\mathbb{Z}\text{-}\mathbf{Mod}$ and $\mathbb{R}\text{-}\mathbf{Mod}$ have finite limits, $F$ commutes with finite limits that $\mathbb{R}$ is a flat $\mathbb{Z}$ module, and $F$ is essentially surjective as each $\mathbb{R}$-module (i.e. $\mathbb{R}$-vector space) with a base $B$ is isomorphic to the image of $\mathbb{Z}B=\bigoplus_B \mathbb{Z}$ along $F$. The functor, -$$S^{-1}F:S^{-1}C\to D,$$ -that is, -$$\mathbb{R}\otimes_{\mathbb{Q}}-:\mathbb{Q}\text{-}\mathbf{Mod}\to \mathbb{R}\text{-}\mathbf{Mod}$$ -is not fully faithful, as $\mathbb{Q}\text{-}\mathbf{Mod}(\mathbb{Q},\mathbb{Q})\cong \mathbb{Q}\ncong\mathbb{R}\cong -\mathbb{R}\text{-}\mathbf{Mod}(\mathbb{R},\mathbb{R})$. - -The above two examples are the trivial cases where $S^{-1}F$ fails to be an equivalence of categories. More generally, one may have a combination of such deficiencies.<|endoftext|> -TITLE: on a property of minuscules in weight lattice -QUESTION [5 upvotes]: Let $\Phi$ be a root system in inner product space $E=\mathbb{R}^l$ (inner product $(\cdot,\cdot)$ and $\Delta=\{\alpha,_1,\cdots,\alpha_l\}$ be a fundamental root system. -Consider the root lattice $\Lambda_r=\mathbb{Z}\alpha_1 + \cdots + \mathbb{Z}\alpha_l$ in $E$ and weight lattice $\Lambda$ -$$\Lambda =\{\lambda\in E : \frac{2(\lambda,\alpha)}{(\alpha,\alpha)} \in\mathbb{Z} \forall \alpha\in \Delta\}. $$ -Let $\Lambda^+ = \{ \lambda\in\Lambda : \frac{2(\lambda,\alpha)}{(\alpha,\alpha)} \geq 0 \forall \alpha\in\Delta\}$. -The miniscules are specific elements of $\Lambda^+$; they are minimal elements of $\Lambda^+$ w.r.t ordering defined as follows: $\lambda_1\leq \lambda_2$ if $\lambda_2-\lambda_1=\sum_{i=1}^l m_i\alpha_i$ with $m_i\geq 0$. -I wanted to see the proof of following fact (seen here this, page 2, top line) - -For different minuscules $\lambda,\nu$, the cosets $\lambda+\Lambda_r$ and $\nu+\Lambda_r$ are distinct. - -Is this easy to prove? How should we proceed? -I tried a simple case: suppose $\lambda,\nu$ are minuscules with $\lambda=\nu+\alpha-\beta$ where $\alpha,\beta$ are distinct fundamental (i.e. simple) roots. I couldn't get any direction to get contradiction from this. -The actual general problem is that if $\lambda+\Lambda_r=\nu+\Lambda_r$, then $\lambda-\nu$ is integral combination of fundamental roots, so y $\lambda=\nu+m_1\alpha_1 + \cdots + m_l\alpha_l$, so some coefficients could be positive and some negative. The simplest non-trivial case is then exactly one $m_i$ is $+1$, exactly one $m_j$ is $-1$ and rest are zero; I don't get any idea to proceed for contradiction. - -REPLY [4 votes]: To comment on the question here (in community-wiki mode), I should point out first that $\S13$ of my now-ancient book was meant to develop some properties of weights just in the framework of abstract root systems without having developed the representation theory where the ideas first arose. In this experimental spirit, my exercises were a bit speculative (and in some cases not worked out by me in detail until much later, since I knew the conclusions were correct, e.g., Exercise 13.10). -That being said, my Exercise 13.13 is intended to be done using just the tools of this chapter. However, the exercise itself comes from one in Bourbaki, VI, $\S2$, which I've now added (reluctantly) to the informal notes linked in the question. The Bourbaki exercise relies on their development in Chapter V of ideas about reflection groups, applied to affine Weyl groups. This gets complicated, since it uses the dual root system in an essential way. I didn't go into affine Weyl groups in my 1972 book but did include an account in Chapter 4 of my 1990 book on reflection groups. An advantage of that setting is to visualize things better: the affine Weyl group leads to a euclidean simplex (fundamental alcove) whose nonzero vertices are of the form $\varpi/c$ for the fundamental weights $\varpi$ and corresponding coefficients $c$ in the highest root of the (dual) root system. Only for $c=1$ do you get minuscule weights (none in some types of irreducible root system). -Anyway, one is led to the list in my Exercise 13.13 (or in Bourbaki's Chapter VIII). A subtle point is the dualization, which assigns (in Bourbaki's numbering which I adopted) $\lambda_\ell$ to type $B_\ell$ and $\lambda_1$ to type $C_\ell$. Bourbaki originally reversed these, but that seems to be corrected in the English translation. (Note too that they used E. Cartan's symbol $\varpi$, a version of the handwritten $\pi$ for poids, whereas I used $\lambda$. Their notation is more commonly used but is easy to confuse with $\omega$.) -Finally, a comment on the spelling of minuscule. This is the original version, emphasizing MINUS (I guess in contrast to majuscule), -but often rendered as miniscule with emphasis on MINI. The question here mixes both spellings, which is undesirable.<|endoftext|> -TITLE: Is it obvious that analytic torsion is a topological invariant? -QUESTION [10 upvotes]: Ray and Singer proved in their paper that analytic torsion is independent of metric (some details may still need to be checked), and together with Cheeger-Muller theorem this implies analytic torsion is well defined, coincide with Reidemeister torsion. -While it is not difficult to see the formal similarities between analytic torsion and Reidemeister-Franz torsion from the view point of determinants, it is not entirely clear to me why intuitively it is an invariant up to homeomorphism (instead of differeomorphism), as the definition itself clearly depends on the Laplacian. I am sure this is a well-studied topic among experts, but I could not find anything in the literature either. -Except results related to index theorem, I also do not know if there are other analytic invariants defined based on $\Psi DO$s but are topological in nature. So here I ask. - -REPLY [5 votes]: It is kind of awkward to answer my own question: It seems the answer is related to partition function of certain quadratic functional under gauge invariance. The relevant literature is: -The Partition Function of a Degenerate Functional -and -The partition function of degenerate quadratic functional and Ray-Singer invariants -by A. S. Schwarz. I found pointer to this reference via reading John Lott's lecture notes and personal communication with Pavel Mnev. The paper is not hard to read but skipped a lot detail in the proofs. So if someone can give a sketch of the main argument I will be grateful. -There has also been some recent important work by Phillip Andreae, who pointed out in his thesis that it is impossible to have a "Mckean-Singer" type of formula for analytic torsion. So in particular this rules out possible construction of arithemetic cohomology groups in Arakelov theory via integration of local terms arising from heat kernel asymptotics.<|endoftext|> -TITLE: A "prequestion" about meromorphic representations of algebraic groups -QUESTION [5 upvotes]: In a comment exchange around an answer to Is a group scheme determined by its category of representations? there arose the issue of Tannakian reconstruction for non-affine algebraic groups (e. g. abelian varieties, in particular, elliptic curves). Some very naïve approach came to my mind, and then @AntonFetisov mentioned one proper way to do it, via representations of the category of quasicoherent sheaves, in view of Lurie's Tannakian reconstruction theorem. He then suggested to post a separate question, which I promised to do, but formulating it in a comprehensive way takes much more time and effort than I thought. So I decided in the meanwhile to make a preliminary version of the most primitive part of the question. Here it is. -Let $G$ be an algebraic group. Call a meromorphic representation of $G$ a collection of rational functions $(\varrho_{ij})_{1\leqslant i,j\leqslant n}$ on $G$ such that -$$ -\varrho_{ij}(xy)=\sum_{1\leqslant k\leqslant n}\varrho_{ik}(x)\varrho_{kj}(y) -$$ -for all $x,y\in G$ (and all $1\leqslant i,j\leqslant n$). -Have these been studied anywhere? Are there any nontrivial ones (that is, not equivalent, in one way or another, to "ordinary" representations)? Is it possible to classify such representations for $G$, say, an elliptic curve? - -REPLY [3 votes]: Edit. Thanks to nfdc23 for pointing out a couple of corrections. As nfdc23 points out, these kinds of things are important in passing from a birational group law to a (regular) group law, so they may go back to Weil. Also, I vaguely remember something about some of this in SGA 3, so that might also be a reference. -As requested, I am posting my comments as an answer. I cannot quite remember who first proved these things, but it might have been Rosenlicht (I think that is where I learned them, anyway). -Let $G$ be a group scheme over $k$ (edit. and assume that $G$ is of finite type, and also assume that $k$ is perfect to avoid working over Artinian rings). Denote by $m:G\times_{\text{Spec}\ k} G \to G$ the multiplication morphism. Denote by $i:G\to G$ the group inverse morphism. Denote by $m':G\times_{\text{Spec}\ k}G \to G$ the composition $m\circ (i\circ\text{pr}_1,\text{pr}_2)$. For every dense open subscheme $U$ of $G$ (edit. and assume that $U$ is schematically dense), the morphism $$m'|_{U\times U}: U\times U \to G,$$ is surjective. This can be checked on geometric points. Thus, let $y\in G(\text{Spec}\ k)$ be an element. Since $U$ and $Uy^{-1}$ are both dense open subschemes of $G$, also $U\cap (Uy^{-1})$ is a dense open. For every geometric point $x$ in $U\cap (Uy^{-1})$, $x$ is in $U$ and $z=xy$ is also in $U$. Thus $(x,z)$ is in $U\times U$, and $m'(x,z)$ equals $y$. Therefore $m'|_{U\times U}$ is surjective. -Now let $H$ be a separated group scheme over $k$ with multiplication morphism $n:H\times_{\text{Spec}\ k}H\to H$. Let $U\subset G$ be a dense open subscheme. Let $\rho_U:U\to H$ be a $k$-morphism. On the dense open subscheme of $G\times_{\text{Spec}\ k} G$, $V=(U\times_{\text{Spec}\ k} U)\cap m^{-1}(U)$, assume that the composition $\rho_U\circ m|_V$ equals the composition $n\circ (\rho_U\times \rho_U)|_V$. Then there exists a unique morphism $\rho:G\to H$ that extends $\rho_U$, and this is a morphism of $k$-group schemes. -Since $H$ is separated and since $\rho_U$ is defined on a dense open, all extensions are unique if they exist. For the same reason, for the unique extension $\rho$, since $\rho_U\circ m|_V$ equals $n\circ (\rho_U\times \rho_U)|_V$, also $\rho\circ m$ equals $n\circ (\rho\times \rho)$. Thus, the unique extension is a morphism of $k$-group schemes. Thus, it suffices to prove that local extensions exist. Again, using uniqueness to confirm the cocycle condition for fpqc descent, it suffices to construct the extension after an fpqc base change. -Probably it is best to use directly the faithfully flat morphism $m'|_{U\times U}$. However, conceptually it is simpler to use the base change from $k$ to its algebraic closure. Then $U(k)$ is dense. Thus, there exist elements $x\in U(k)$ such that the translates $x^{-1}U$ cover $G$. On each open $x^{-1}U$, the unique extension of $\rho_U|_{U\cap (x^{-1}U)}$ is defined by $$\rho(y) = \rho_U(x)^{-1}\cdot \rho_U(xy)$$ (I am supressing $m$ and $n$ in this formula).<|endoftext|> -TITLE: Some nice functional equations for $q$-continued fractions -QUESTION [9 upvotes]: Given $\large q=e^{2\pi i \tau}$. Define, -$$\alpha(\tau) = \sqrt2\,q^{1/8}\prod_{n=1}^\infty\frac{ (1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})}$$ -$$\beta(\tau) = q^{1/5}\prod_{n=1}^\infty\frac{ (1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}$$ -$$\gamma(\tau) = q^{1/4}\prod_{n=1}^\infty\frac{ (1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-2})(1-q^{6n-4})}$$ -$$\delta(\tau) = q^{1/3}\prod_{n=1}^\infty\frac{ (1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})(1-q^{6n-3})}$$ -$$\epsilon(\tau) = q^{1/2}\prod_{n=1}^\infty\frac{ (1-q^{8n-1})(1-q^{8n-7})}{(1-q^{8n-3})(1-q^{8n-5})}$$ -$$\lambda(\tau) = q^{1/1}\prod_{n=1}^\infty\frac{ (1-q^{12n-1})(1-q^{12n-11})}{(1-q^{12n-5})(1-q^{12n-7})}$$ -each of which has a beautiful $q$-continued fraction. Then we propose the functional equations, -$$\begin{aligned} -\alpha^8\Big(\frac{-1}{4\tau}\Big)&=1-\alpha^8(\tau)\\[2mm] -\beta^5\Big(\frac{-1}{5\tau}\Big)&=\frac{\phi^5\beta^5(\tau)-1}{-\beta^5(\tau)-\phi^5},\quad\phi=\tfrac{1+\sqrt5}2\\[2mm] -\gamma^4\Big(\frac{-1}{6\tau}\Big)&=\frac{\gamma^4(\tau)-\tfrac19}{\gamma^4(\tau)-1}\\[2mm] -\delta^3\Big(\frac{-1}{6\tau}\Big)&=\frac{\delta^3(\tau)-\tfrac18}{-\delta^3(\tau)-1}\\[2mm] -\epsilon^2\Big(\frac{-1}{8\tau}\Big)&=\frac{u^2\,\epsilon^2(\tau)-1}{\epsilon^2(\tau)-u^2},\quad u =1+\sqrt2\\[2mm] -\lambda\Big(\frac{-1}{12\tau}\Big)&=\frac{v\,\lambda(\tau)-1}{\lambda(\tau)-v},\quad v = 2+\sqrt3\end{aligned}$$ -Note 1: Excepting $\gamma^4\big(\tfrac{-1}{6\tau}\big)$ which is in this post and served as the model, I just found the rest empirically, but their consistent forms suggest these are correct. -Note 2: Also, the two order $6$ obey the simple $\displaystyle \frac1{\gamma^4(\tau)}-\frac1{\delta^3(\tau)}=1$. -Questions: - -How do we derive the easiest one from first principles and generalize it for the rest? -Is the list complete? Without requiring representation as a q-continued fraction, is there a similar functional equation for, say, a $p=7$ order? - -REPLY [4 votes]: Throughout this answer, I'll be referencing Kubert-Lang, "Modular Units" Chapter 2, sections 1 and 2, and Chapter 3 section 4. -Each of your functions is a ratio of the Siegel functions of the form -$$A_N(\tau)=\left(\frac{g_{\frac1N,0}}{g_{\frac aN,0}}(N\tau)\right)^m.$$ -If $r=(r_1,r_2)\in \mathbb Q^2$, $B_2(x)=x^2-x+1/6$ is the second Bernoulli polynomial, $q=e^{2\pi i \tau}$ as usual, and $\zeta=e^{2\pi i r_2}$, then the Siegel function $g_a$ is defined by -$$ -g_a(\tau)= -q^{\frac12 B_2(r_1)}e^{2\pi i \frac{r_2(r_1-1)}2}\prod_{n=0}^\infty(1-q^{n+r_1}\zeta)(1-q^{n+1-r_1}\zeta^{-1}). -$$ -The Siegel functions $g_a$ are modular units, meaning they're modular functions whose zeros and poles are supported at cusps. -If $\gamma\in \text{SL}_2(\mathbb Z)$, then -$$ -g_a|_0\gamma=g_{a\gamma}. -$$ -Moreover, we can reduce $r_1$ and $r_2$ modulo $\mathbb Z$, or change signs of both $r_1$ and $r_2$, if we introduce a root of unity. However, the $m$'s in your ratios satisfy $m(1-a^2)\equiv 0 \pmod N$ if $N$ is odd and $\pmod{2N}$ if $N$, and so using Theorem 3.4.1 of Kubert-Lang we find we can ignore this extra factor so the transformations rules becomes this simple. Note, the theorem is written in terms of Klein functions $t_a=g_a\Delta^{-\frac1{12}}$. Using these transformation rules, it's easy to see that all of your functions are invariant under $\Gamma_1(N)/\{\pm1\}$. -The specific groups you have considered are ones where this group is genus $0$, and your functions $A_N$ are Hauptmoduln with a unique zero at the cusp $\infty$, a constant $C_0$ at the cusp $0$, and a unique pole at one other cusp. We can calculate $C_0$ using the formulas above: If $\zeta_{2N}=e^{2\pi i \frac{1}{2N}}$, then the constant is given by $\left(\frac{\zeta_{2N}-\zeta_{2N}^{-1}}{\zeta_{2N}^{a}-\zeta_{2N}^{-a}}\right)^m$. -Since $A_N$ is a hauptmodl, its image under the Fricke involution is a rational function in $A_N$. Since it now has a unique $0$ at the cusp $0$ (Where $A_N$ has the constant $C_0$) and the constant $C_0$ at the cusp infinity (Where $A_N$ has a zero), we must have -$$ -A_N|_0\begin{pmatrix}0&-1\\N&0\end{pmatrix}=\frac{-A_N+C_0}{X\cdot A_N+1}, -$$ -for some $X$, determined by the movement of the pole. If the pole does not move, then $X=0$. Otherwise, $-1/X$ is the constant of $A_N$ at the cusp where $A_N|_0\begin{pmatrix}0&-1\\N&0\end{pmatrix}$ has its pole. -All in all, this suggest you might have hope for $N=7,9$ and $10$, which all give genus $0$ groups. Unfortunately The ratios of the type you've described above aren't hauptmoduln. You might be able to find similar results if you allow for larger order products.<|endoftext|> -TITLE: Is a unitary Hamiltonian TQFT the same as a unitary axiomatic TQFT? -QUESTION [12 upvotes]: Introduction -Axiomatic TQFTs -An axiomatic $n$-dimensional TQFT is a symmetric monoidal functor $\mathcal{Z}\colon \operatorname{Bord}_n \to \operatorname{Hilb}$ from $n$-dimensional oriented bordisms to Hilbert spaces (other targets are certainly possible, but for the purposes of this question, we will stay with Hilbert spaces). -It is unitary if $\mathcal{Z}$ is a unitary, or $\dagger$-functor. Written out, this means: Let $\Sigma\colon M_1 \to M_2$ be an (oriented) bordism. The orientation reversed bordism goes in the other direction, i.e. $\overline{\Sigma}\colon M_2 \to M_1$. The TQFT must then satisfy $\mathcal{Z}(\overline{\Sigma}) = \mathcal{Z}(\Sigma)^\dagger$, where $\dagger$ is the adjoint of maps of Hilbert spaces. -Hamiltonian TQFTs -An $(n+1)$-dimensional Hamiltonian TQFT is a local Hamiltonian lattice quantum system on an $n$-dimensional manifold $M$, that has a gapped ground state which is invariant under certain local perturbations. It is unitary in the sense that the Hamiltonian $H$ is self-adjoint, and thus the time evolution is unitary. -The correspondence to axiomatic TQFTs lies in the lowest eigenspace. Any $n$-manifold has a (possibly degenerate) ground state, and it corresponds to the Hilbert space assigned to that manifold by an $(n+1)$-dimensional axiomatic TQFT. -Question -My question is now, how these two notions of unitarity fit together. Are they equivalent, or is one stronger? -Observations -Observation 1: Cylinders -As a first observation, it's clear that the two notions of unitarity agree when we just consider cylinders, i.e. bordisms of the form $I \times M$. This is the typical situation for Hamiltonian TQFTs. -(I'm not sure how the mapping class group is typically represented on Hamiltonian TQFTs, though.) -Observation 2: Topology changes -Most of the interesting information about axiomatic TQFTs is found in the way they behave on non-cylinder bordisms. If the boundaries to such a bordism are assigned Hilbert spaces of different dimensions, the bordism can never give rise to a unitary map of Hilbert spaces, though! -From the Hamiltonian perspective, it's obvious that it can't. When topology changes, the Hamiltonian changes (even if the lattice stays the same), and thus the lowest eigenspace. Take the orthogonal decomposition of the physical Hilbert space $\mathcal{H}$ into eigenspaces: -$$\mathcal{H} = \mathcal{H}_0 \oplus \mathcal{H}_1 \oplus \cdots$$ -$$1_{\mathcal{H}} = \pi_0 + \pi_1 + \cdots$$ -$$\pi_k\pi_{k'} = 1_{\mathcal{H}_k} \delta_{kk'}$$ -The projection $\pi_0$ onto the lowest eigenspace of the new Hamiltonian should correspond to $\mathcal{Z}(\Sigma)$, where $\Sigma$ is the non-cylinder bordism. But why should this map be unitary in the axiomatic-TQFT-sense? -Question -Assume we are given an axiomatic TQFT that can be modelled by a local Hamiltonian lattice system. (The Hamiltonian is of course assumed to be self-adjoint.) Is the axiomatic TQFT then automatically unitary in the sense of a unitary functor? What are the topology changing maps? - -REPLY [6 votes]: The answer is no: -There are unitary Hamiltonian TQFTs (ie there are gapped lattice Hamiltonian systems in physics) which are not a unitary axiomatic TQFTs. -An example can be given in 3+1-dimension space-time. -A unitary Hamiltonian TQFT can be a stacking of layers of $\nu=1/2$ bosonic fractional quantum Hall states. -Such a unitary Hamiltonian TQFT is not a unitary axiomatic TQFT. -This was discussed by us in arXiv:1406.5090. -Only topological orders (ie gapped quantum liquids) may correspond to a unitary axiomatic TQFT. For a more detailed statement, see Bosonic topological orders and unitary fully dualizable fully extended TQFT (In some sense, we require the lattice Hamiltonian can be defined on any random lattice, and still give rise to ground states in the same phase)<|endoftext|> -TITLE: Compact manifolds with big mapping class group -QUESTION [14 upvotes]: I was wondering if compact surfaces were the only compact manifolds with a "big" or "complicated" mapping class group. -Are there higher dimensional manifolds (which are not in some way reducible to a surface) that have an infinite mapping class group? What kind of group can arise as the mapping class group of a compact manifold? -Thanks for your attention :) - -REPLY [12 votes]: In - Infinitesimal computations in topology Sullivan shows in Theorem 13.3 that if $M$ is a simply-connected manifold of dimension $>5$, then $\pi_0(\mathrm{Diff\,} M)$ is commensurable to an arithmetic group (see p.295 for definitions of an arithmetic group). -The group of homotopy classes of homotopy self-equivalences is also commensurable to an arithmetic one (Theorem 10.3) -In Corollary 13.3 Sullivan explains how the forgetful map between the two is largely controlled by the rational Pontryagin class of the manifold (which is preserved by diffeomorphisms by not by homotopy equivalences).<|endoftext|> -TITLE: Approximating logarithm by a series for large values -QUESTION [5 upvotes]: It is not difficult to show that $\ln(x)$ for real positive $x$ can be approximated by the series -$$ --\gamma+\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k!k}x^k -$$ -up to an absolute error bounded by $e^{-x}/x,$ where $\gamma$ is the Euler-Mascheroni constant. I assume this must be well known and would appreciate any references. - -REPLY [4 votes]: The statement is equivalent to an expansion for the Exponential Integral function $\operatorname{Ei}(x)$, so any text on special functions should do (check e.g. the references in the linked wiki article).<|endoftext|> -TITLE: powered partition function generator: 1/2 of them are zeros? -QUESTION [14 upvotes]: Ramanujan delivered his famous congruences -$$p(5n+4)\equiv_50, \qquad p(7n+5)\equiv_70, \qquad p(11n+6)\equiv_{11}0$$ -for the integer partitions with generating function $F(x)=\prod_{k=0}^{\infty}\frac1{1-x^k}=\sum_{n\geq0}p(n)x^n$. -Let $p\geq5$ be a prime number, and consider the series -$$F(x)^{p-1}=\sum_{n\geq0}a_p(n)x^n.$$ -Experimental evidence suggests (curiously) that, modulo $p$, exactly half of the $p-1$ rows -$$\begin{cases} -a_p(pn+1): n\geq0 \\ -a_p(pn+2): n\geq0 \\ -a_p(pn+3): n\geq0 \\ -\qquad\dots\dots\dots \\ -a_p(pn+p-1): n\geq0 -\end{cases}$$ -are identically zero. - -Question. Is this true? Why? Or, is it known? -Question. Why does this fail to be true for the prime $p=3$? - -For example, $a_5(5n+3)\equiv_50$ and $a_5(5n+4)\equiv_50$ for any $n\geq0$. -Notation. $\equiv_p$ means congruent modulo $p$. - -REPLY [22 votes]: This is true. First, note that by the Pentagonal number theorem due to Euler, -$$\frac{1}{F(x)} = \sum_{k \in \mathbb{Z}} (-1)^k x^{\frac{k}{2}(3k-1)}.$$ -For a given prime $p \ge 5$, the function $f(k)=\frac{k}{2}(3k-1),f:\mathbb{Z}\to \mathbb{F}_p$ has only $\frac{p+1}{2}$ values in its image, one of which is $0$. This follows by completing the square: -$$ \frac{k}{2}(3k-1) = \frac{3}{2}\left( (k-\frac{1}{6})^2 - \frac{1}{6^2} \right).$$ -We deduce that there are $\frac{p-1}{2}$ non-zero distinct values $a_1,\cdots,a_{\frac{p-1}{2}}$ modulo $p$ such that the $n$'th coefficient of $\frac{1}{F(x)}$ is zero if $n \equiv a_i \bmod p$ for some $i$. We conclude by noting that modulo $p$, -$$F(x)^{p-1} \equiv_p \frac{F(x^p)}{F(x)}.$$ -In particular, $a_p(n) = 0$ if $n \equiv a_i \bmod p$. -The result does not hold for $p=3$ because $f$ behaves differently then (it is surjective in that case).<|endoftext|> -TITLE: If $X$ is a degree 3 smooth integral surface in $P^N$, $N > 3$, is it still true that it contains 27 lines? -QUESTION [5 upvotes]: It is a well known fact that a smooth cubic surface in $P^3$ contains 27 lines. One proof proceeds by moving through the parameter space $U$ of smooth cubics until one reaches an cubic that can be understood by simple algebra, such as the Fermat cubic $V(x_0^3 + x_1^3 + x_2^3 + x_3^3)$. This is formalized by introducing an incidence correspondence $X$ of lines on cubics in $U \times G(1,3)$; X turns out to be a $P^{15}$ bundle over $G(1,3)$, and the projection of $X$ onto $U$ can be shown via the Jacobian criterion to be etale and finite. From there one is essentially done. -Fix some $N > 3$. -If one considers the Hilbert scheme of degree 3 surfaces in $P^N$, it is plausible that one can repeat the flavor of this argument. However, it's not clear that the Hilbert polynomials of the smooth integral degree 3 surfaces are the same (it's also not clear to me that they will be different), so it is possible that the Hilbert scheme will have multiple components. However, fixing or giving geometric meaning to the extra coefficient in the Hilbert polynomial will solve this problem. -Moreover, this Hilbert scheme may stop being smooth, and the incidence correspondence may not be so easily analyzable. I don't know how to compute the dimensions of these Hilbert schemes - I recall one can compute $H^0$ of the normal bundle of a given cubic surface to find the tangent space on the Hilbert scheme at that cubic. In the case of a complete intersection, it's possible to describe the normal bundle using adjunction, but computing global sections doesn't seem straightforward. -I am wondering what is known in this direction. Let me some specific questions: -1) How many components does the Hilbert scheme of degree 3 surfaces have? What if we pass to the (open?) subscheme of smooth, irreducible surfaces? What are their dimensions? Are they smooth? -2) How many lines are on a smooth irreducible degree 3, cubic surface in $P^N$? What are the possibilities? (Irreducibility imposed to rule out the disjoint union of a plane and a degree 2 hypersurface in $P^5$, which has infinitely many lines.) -3) Can someone give an example of a smooth, irreducible degree 3 surface in $P^N$ ($N > 3$) that is not the blow up of $P^2$ at 6 points? How many lines does it have? Is there an example with no lines? -[4) Is this suitable for MO?] - -REPLY [12 votes]: If $X\subset \mathbb P^n$ is a smooth and non-degenerated variety then -$$ - \deg X \ge 1 + \mathrm{codim} X -$$ -as you can learn from Varieties of Minimal Degree by Eisenbud and Harris. -Thus to understand smooth cubic surfaces, it suffices to consider -cubic surfaces in $\mathbb P^3$ and $\mathbb P^4$. Smooth degree $3$ surfaces -in $\mathbb P^4$ are rational normal scrolls (loc. cit. Theorem 1), and as such contains infinitely many lines.<|endoftext|> -TITLE: Algorithm to generate random commuting permutations -QUESTION [7 upvotes]: I am seeking to understand the properties of a typical pair of permutations $(\sigma,\tau) \in \mathrm{Sym}(n)^2$ chosen uniformly at random from all pairs such that $\sigma$ and $\tau$ commute. In particular, I would like to prove that the following holds. -$(\ast)$ With high probability, the graph on $\{1,\ldots,n\}$ induced by $\sigma$ and $\tau$ has few 'unnecessary' short cycles (i.e. short cycles which do not correspond to commutators). -It seems that the best way to do this would be to have a straightforward algorithm to generate such a random pair. I have been unable to find such an algorithm in the literature and I would appreciate if someone could tell me what is known about this subject. -I would be equally happy to have a way of establishing $(\ast)$ for a random pair of approximately commuting permutations. To make the last statement precise, for each $\epsilon > 0$ I would like to know that if $n$ is large enough then $(\ast)$ holds for a pair chosen uniformly at random from of the set of all pairs $(\sigma,\tau) \in \mathrm{Sym}(n)^2$ such that $\sigma \tau \sigma^{-1} \tau^{-1}$ has at least $(1-\epsilon)n$ fixed points. - -REPLY [4 votes]: Here's a method for sampling two commuting elements of a group $G$, given that you can sample a random element of $G$ and that you can also sample a random conjugacy class of $G$ (and assuming as well that for any two elements $a,x \in G$ which are in the same conjugacy class, you can efficiently compute some $z \in G$ with $zxz^{-1} = a$). It's based on Theorem IV of the paper "On some problems of a statistical group-theory. IV" by Erdős and Turán. -Step 1: Let $C$ be a uniformly random conjugacy class of $G$, and let $x$ be an arbitrary element of $C$. -Step 2: Let $y$ be a uniformly random element of $G$. -Step 3: Let $a = yxy^{-1}$. -Step 4: Compute an element $z \in G$ such that $zxz^{-1} = a$, such that $z$ depends only on $a$ and $x$. -Step 5: Let $b = yz^{-1}$. -Then $ab = ba$ and $(a,b)$ is uniformly distributed among ordered pairs of commuting elements of $G$.<|endoftext|> -TITLE: Several conjectured identities for polylogarithms -QUESTION [15 upvotes]: I asked a question at M.SE a couple of years ago about polylogarithms $\!^{[1]}$$\!^{[2]}$$\!^{[3]}$$\!^{[4]}$ where I conjectured -$$720\,\operatorname{Li}_4\!\left(\tfrac12\right)-2160\,\operatorname{Li}_4\!\left(\tfrac13\right)+2160\,\operatorname{Li}_4\!\left(\tfrac23\right)+270\,\operatorname{Li}_4\!\left(\tfrac14\right)+540\,\operatorname{Li}_4\!\left(\tfrac34\right)+135\,\operatorname{Li}_4\!\left(\tfrac19\right)\\ -=19\pi^4+30\left(\pi^2-\beta^2\right)\left(10\alpha^2-12\alpha\beta+3\beta^2\right)-30\,\alpha^2\left(19\alpha^2-24\alpha\beta+8\beta^2\right)\!,\\ -\text{where $\alpha=\ln2,\,\,\beta=\ln3$.}$$ -I put a bounty on it recently, but it still remains unanswered. So, I decided to re-post it here. Also, I recently discovered several similar conjectures for $\operatorname{Li}_5(z)$ involving powers of the golden ratio $\phi=\frac{1+\sqrt5}2$: -$$4455\,\operatorname{Li}_5\left(\phi^{-2}\right)-1215\,\operatorname{Li}_5\left(\phi^{-4}\right)-360\,\operatorname{Li}_5\left(\phi^{-6}\right)+15\,\operatorname{Li}_5\left(\phi^{-12}\right)\\ -=3015\,\zeta(5)-702\ln^5\phi+180\,\pi^2\ln^3\phi-38\,\pi^4\ln\phi$$ - -$$45000\,\operatorname{Li}_5\left(\phi^{-2}\right)-16875\,\operatorname{Li}_5\left(\phi^{-4}\right)-144\,\operatorname{Li}_5\left(\phi^{-10}\right)+9\,\operatorname{Li}_5\left(\phi^{-20}\right)\\ -=28944\,\zeta(5)-6000\ln^5\phi+1600\,\pi^2\ln^3\phi-356\,\pi^4\ln\phi$$ - -$$15660\,\operatorname{Li}_5\left(\phi^{-2}\right)-19440\,\operatorname{Li}_5\left(\phi^{-4}\right)+7680\,\operatorname{Li}_5\left(\phi^{-6}\right)+2430\,\operatorname{Li}_5\left(\phi^{-8}\right)-15\,\operatorname{Li}_5\left(\phi^{-24}\right)=5025\,\zeta(5)+2088\ln^5\phi-240\,\pi^2\ln^3\phi-28\,\pi^4\ln\phi$$ -How can we prove these conjectures? Are there other similar identities of this kind? - -REPLY [10 votes]: (Updated answer): -Upon further research, it turns out your three equations involving $\phi$ are special cases of three polylogarithm ladders of index $12,\,20,\,24$ that can be found in "The Polylogarithm in Algebraic Number Fields" (1985) by M. Abouzahra and L. Levin. Let $\color{red}{\rho = \frac1{\phi}}$. -I. Ladder of index 12 -$$A = \frac{\text{Li}_n(\rho^{12})}{12^{n-1}}- \frac32\frac{\text{Li}_n(\rho^6)}{6^{n-1}}- \frac{\text{Li}_n(\rho^4)}{4^{n-1}}+ \frac{11}{48}\frac{\text{Li}_n(\rho^2)}{2^{n-1}}$$ -$$B = -\frac{13}{48}\frac{\ln^n\rho}{n!}+\frac{\zeta(2)}{48}\frac{\ln^{n-2}\rho}{(n-2)!}-\frac{19\,\zeta(4)}{1728}\frac{\ln^{n-4}\rho}{(n-4)!} $$ -then for $n=1,2,3,4,5$, -$$A+B= \frac{67\,\zeta(5)}{6912}\frac{\ln^{n-5}\rho}{(n-5)!} $$ -with $A+B = 0$ for $n<5$. If $n=5$, then we recover your experimentally found equation. Incidentally, a consequence (or is it the other way round?) of this ladder is the high-degree cyclotomic equation satisfied by the golden ratio, -$$(1 - \phi^{-12}) = (1 - \phi^{-6})^{3/2}(1 - \phi^{-4})(1 - \phi^{-2})^{-11/48}\,\phi^{13/48}$$ -Notice that the powers of $(1 - \phi^{k})$ are coefficients in $A,B$. -II. Ladder of index 20 -Similar to the ladder above. For $n=5$ yields the OP's second formula but, with modifications, can be extended to $n=7$. -III. Ladder of index 24 -Likewise. If modified, can be extended even to $n=9$. This also explains the index $24$ formulas below for $n=6,7$ that I found empirically. - - -(Old answer): As to your question if there are similar identities, it seems there is a whole family, -$$5\,\text{Li}_2(\phi^{-1}) -=-5\ln^2\phi+\,3\,\color{blue}{\zeta(2)}$$ - -$$15\,\text{Li}_3(\phi^{-2})=10\ln^3\phi\,-2\pi^2\ln\phi+12\,\color{blue}{\zeta(3)}$$ - -$$144\text{Li}_4(\phi^{-1})-9\text{Li}_4(\phi^{-2})-64\text{Li}_4(\phi^{-3})+4\text{Li}_4(\phi^{-6})=27\ln^4\phi-6\pi^2\ln^2\phi+80\,\color{blue}{\zeta(4)}$$ - -$$4455\,\text{Li}_5\left(\phi^{-2}\right)-1215\,\text{Li}_5\left(\phi^{-4}\right)-360\,\text{Li}_5\left(\phi^{-6}\right)+15\,\text{Li}_5\left(\phi^{-12}\right)\\ -=-702\ln^5\phi+180\,\pi^2\ln^3\phi-38\,\pi^4\ln\phi+3015\,\color{blue}{\zeta(5)}$$ - -$$m_1\text{Li}_6(\phi^{-2})+m_2\text{Li}_6(\phi^{-4})+m_3\text{Li}_6(\phi^{-6})+m_4\text{Li}_6(\phi^{-8})+m_5\text{Li}_6(\phi^{-12})+m_6\text{Li}_6(\phi^{-24})\\ -=m_7\ln^6\phi+m_8\pi^2\ln^4\phi+m_9\pi^4\ln^2\phi +32155\,\color{blue}{\zeta(6)}$$ - -$$n_1\text{Li}_7(\phi^{-2})+n_2\text{Li}_7(\phi^{-4})+n_3\text{Li}_7(\phi^{-6})+n_4\text{Li}_7(\phi^{-8})+n_5\text{Li}_7(\phi^{-12})+n_6\text{Li}_7(\phi^{-24})\\ -=n_7\ln^7\phi+n_8\pi^2\ln^5\phi+n_9\pi^4\ln^3\phi+n_{10}\pi^6\ln\phi +44689995\,\color{blue}{\zeta(7)}$$ - -and where the integers $m_i$ and $n_i$ were found using Integer Relations but are too tedious to write here. -Does this go on forever? Note the exponents of $\phi$ in $\text{Li}_k(\phi^{-n})$ match up for $k=4$ as $c=1\times6=2\times3$ then for $k=5$ as $c=2\times12=6\times4$ then for $k=6,7$ as $c=2\times24=4\times12 = 6\times8$. -P.S. Based on Abouzahra and Levin's paper cited in the update, I guess this can't be brought to arbitrarily high $k$.<|endoftext|> -TITLE: generalizing Wilf's conjecture: Uppuluri-Carpenter numbers -QUESTION [8 upvotes]: The complementary Bell numbers have the exponential generating function -$$\sum_{n\geq0}\tilde{B}_nx^n=e^{1-e^x}.$$ -Herb Wilf conjectured that $\tilde{B}_n=0$ only for $n=2$. By now, there are a few papers with a proof. -I wish to generalize this in the same spirit. Let's introduce super exponentials recursively by $e(x)=e^{(0)}(x):=1-e^x$ and $e^{(k+1)}(x):=e(e^{(k)}(x))$. Introduce the generating functions -$$\sum_{n\geq0}\tilde{B}_n^{(k)}\frac{x^n}{n!}=1-e^{(k)}(x)$$ -and call $\tilde{B}_n^{(k)}$ super complementary Bell numbers. Experimental evidence prompts: - -Question. For each odd integer $k$, the numbers $\tilde{B}_n^{(k)}=0$ iff $n=2$. Is this true? - -Note. $e^{(1)}(x)=1-e^{1-e^x}$ and hence $\tilde{B}_n^{(1)}=\tilde{B}_n$. -Examples. $e^{(2)}(x)=1-e^{1-e^{1-e^x}}$ and $e^{(3)}(x)=1-e^{1-e^{1-e^{1-e^x}}}$. The first few values of $\tilde{B}_n^{(3)}$: -$1, -1, 0, 2, 2, -14, -53, 129, 1668, 1607, \dots$. - -REPLY [2 votes]: Just a comment, not an answer. -It might be helpful to get an idea when one can see, how the coefficients of the odd iterates can be expressed as polynomials on the iteration-index. Here is the table of coefficients of the exponential generating functions of the first 8 odd iterates (row 1: Uppuluri-Carpenter-numbers, row 2: numbers as given in the example in the OP for the 3'rd iterate) with the iteration index $h$: - h | j | x^0 x x^2 x^3 x^4 x^5 x^6 x^7 x^8 ... - ---+---+----------------------------------------------------- - 1 | 1 | 1 -1 0 1 1 -2 -9 -9 50 ... - 3 | 2 | 1 -1 0 2 2 -14 -53 129 1668 - 5 | 3 | 1 -1 0 3 3 -36 -132 764 7514 - 7 | 4 | 1 -1 0 4 4 -68 -246 2246 20248 - 9 | 5 | 1 -1 0 5 5 -110 -395 4925 42530 - 11 | 6 | 1 -1 0 6 6 -162 -579 9151 77020 - 13 | 7 | 1 -1 0 7 7 -224 -798 15274 126378 - 15 | 8 | 1 -1 0 8 8 -296 -1052 23644 193264 - ...|... ... - -The entries along the columns can be calculated by polynomials depending on $h$ or even better on the consecutive rowindex $j$. The generating polynomials for the columns, in terms of $h$, are - coef - at : polynomial in h - ----------------------- - x^0 : 1 - x^1 : -1 - x^2 : 0 - x^3 : 1/2 + 1/2*h - x^4 : 1/2 + 1/2*h - x^5 : 1/4 - 1 *h - 5/4*h^2 - x^6 : -1/8 - 9/2*h - 35/8*h^2 - x^7 : -1 -283/24*h - 7/2*h^2 + 175/24*h^3 - x^8 :-19/4 -365/12*h +119/4*h^2 + 665/12*h^3 - ... : .... - -and in terms of the rowindex $j$ are - coeff - at : polynomial in j - ----------------------- - x^0 : 1 - x^1 : -1 - x^2 : 0 - x^3 : 0 + 1 *j - x^4 : 0 + 1 *j - x^5 : 0 + 3 *j - 5 *j^2 - x^6 : 0 + 17/2*j - 35/2*j^2 - x^7 : 0 + 205/6*j - 203/2*j^2 + 175/3*j^3 - x^8 : 0 + 458/3*j - 546 *j^2 + 1330/3*j^3 - ... : ... - - -A piece of code in Pari/GP to reproduce the first table is -list=matrix(12,16) -tmp = exp(x); -tmp = exp(1-tmp); \\ first iterate -list[1,] = polcoeffs(serlaplace(tmp),16); \\ "polcoeffs" - user function - \\ gives the first few coefficients as a vector (here of length 16) -{ for(k=2,12, - tmp = exp(1-tmp);tmp = exp(1-tmp); \\ use the odd iterations only - list[k,]=polcoeffs(serlaplace(tmp),16) - ); } - printp (list ) - - -[Update] -It might also be of interest, that the iterates of the Uppuluri-Carpenter-numbers can be created using the matrix-exponential of the (lower triangular) Pascalmatrix where the diagonal is removed. That removal of the diagonal makes the exponential-series for the matrix-exponential a finite sum for any truncation of size (which might in turn be of help for the finding of the proof of the conjecture) . -Let P be the lower triangular Pascalmatrix truncated to some size $n \times n$, and $I$ the identity matrix of same size. -Then the iterates can be computed by: -$$ U_1 = \operatorname{Exp}( I - P) \\ -U_2 = \operatorname{Exp}( I - U_1) \\ -U_3 = \operatorname{Exp}( I - U_2) \\ -... -$$ -The numbers of interest occur in the first column of the odd-indexed iterates $U_1,U_3,U_5,...$. -Here are top-left snippets of $U_1,U_3,U_5$: - U_1 U_3 U_5 - -----------------------------+------------------------+------------------- - 1 . . . . | 1 . . | 1 . . | - -1 1 . . . | -1 1 . | -1 1 . | - 0 -2 1 . . | 0 -2 1 | 0 -2 1 | - 1 0 -3 1 . | 2 0 -3 | 3 0 -3 | - 1 4 0 -4 1 | 2 8 0 | 3 12 0 | - -2 5 10 0 -5 | -14 10 20 | -36 15 30 | - -9 -12 15 20 0 | -53 -84 30 | -132 -216 45 | - -9 -63 -42 35 35 | 129 -371 -294 | 764 -924 -756 | - 50 -72 -252 -112 70 | 1668 1032 -1484 | 7514 6112 -3696 | - 267 450 -324 -756 -252 | 1607 15012 4644 | -14838 67626 27504 | - 413 2670 2250 -1080 -1890 | -58515 16070 75060 | -547854 -148380 338130 | - - - - - - + - - - + - - - +<|endoftext|> -TITLE: What is needed to prove the consistency of Tarski's Euclidean geometry? -QUESTION [10 upvotes]: This question might be too elementary for MO, in which case I would gladly move it to math.stackexchange.com -Consider Tarski's axiomatization of Euclidean Geometry. It is stated in the wikipedia page linked and many other places that Tarski proved this first-order theory to be complete and consistent. My question concerns consistency: the proofs I have seen in model-theoretic literature reduce to the triviality that Euclidean geometry has a model in set theory (namely $\mathbb R^2$), so these proofs are actually proofs of the relative consistency of Euclidean Geometry with respect to set theory. But surely there must be a proof of the consistency of Tarski's axioms using only a considerably less powerful theory than set theory (whose consistency may itself be in doubt). - -What (hopefully weak) meta-theory is needed to prove that Tarski's axioms are consistent? - -For example, what about Tarski's original proof (that I can't track down on the internet)? Or perhaps more recent proofs? More philosophically, is there a proof of the consistency of these axioms that Hilbert would call finitist? - -REPLY [15 votes]: In 1999, Harvey Friedman showed how to prove the consistency of Tarski's axioms for geometry in EFA. This is Elementary Function Arithmetic, otherwise known as $I\Delta_0(exp)$, a subtheory of PRA with functions bounded by towers of exponentials.<|endoftext|> -TITLE: is this a familiar gen. fn. for partitions? -QUESTION [6 upvotes]: The $2$-adic valuation of $n\in\mathbb{N}$, denoted $\nu(n)$, is the largest power $t$ such that $2^t$ divides $n$. The number of integer partitions of $n$, denoted by $p(n)$, has generating function -$$\sum_{n\geq0}p(n)x^n=\prod_{k\geq1}\frac1{1-x^k}.$$ -However, I am finding an alternative: -$$\sum_{n\geq0}p(n)x^n=\prod_{k\geq1}(1+x^k)^{\nu(2k)}.$$ - -Question. Is this known? If so, any reference? If not, then any proof? - -Caveat. I would be surprised if this is new. - -REPLY [4 votes]: This can be proved from the famous: - -Distinct parts <-> Odd parts - -which can be found in Hardy & Wright : An Introduction to the Theory of Numbers. -This states: -$$(1+x)(1+x^2)(1+x^3)\dots=\frac1{(1-x)(1-x^3)(1-x^5)\dots}$$ -If you substitute $x\to x^{2^k}$ for $k=1,\dots$, and multiply together, the RHS becomes the usual partition generating function, and the LHS takes your alternative form.<|endoftext|> -TITLE: Two combinatorial identities -QUESTION [8 upvotes]: In my research, I found two combinatorial identities. Mathematica can give the answers immediately, but I don't know how to prove them. Could someone help me? Thank you! -Here are they: -Let $\alpha$, $\beta$ be two arbitrary complex numbers, and $k$, $l$ be two positive integers, then: -\begin{align*} -\sum_{m=1}^k m \binom{\frac{\alpha+\beta}{\beta}k}{k-m}\binom{\frac{\alpha+\beta}{\beta}l}{l+m} -=&\frac{\alpha\,k\,l}{(k+l)(\alpha+\beta)}\binom{\frac{\alpha+\beta}{\beta}k}{k}\binom{\frac{\alpha+\beta}{\beta}l}{l}\\ -\sum_{m=1}^k m \binom{\frac{\alpha+\beta}{\beta}k}{k-m}\binom{\frac{\alpha+\beta}{\alpha}l}{l-m} -=&\frac{\alpha\,\beta\,k\,l}{(k\,\alpha+l\,\beta)(\alpha+\beta)}\binom{\frac{\alpha+\beta}{\beta}k}{k}\binom{\frac{\alpha+\beta}{\alpha}l}{l} -\end{align*} -Actually, the positive integer $l$ can be also a complex number. Then, if one replaces $\frac{\alpha}{\beta}l$ by $\tilde{l}$ in the first identity, it becomes equivalent to the second identity. Besides this observation, I can only prove some trivial cases. - -REPLY [5 votes]: This is inspired by Fedor's answer: -Consider - $$ -f(m):=-\frac{\binom{ab}{m}\binom{cd}{m}(m-bc)(m-ad)}{\binom{ad-1}{m}\binom{bc-1}{m}(a-c)(b-d)}. - $$ -Then - $$ -f(m+1)-f(m)=-\frac{\binom{ab}{m+1}\binom{cd}{m+1}(m+1-bc)(m+1-ad)}{\binom{ad-1}{m+1}\binom{bc-1}{m+1}(a-c)(b-d)}+\frac{\binom{ab}{m}\binom{cd}{m}(m-bc)(m-ad)}{\binom{ad-1}{m}\binom{bc-1}{m}(a-c)(b-d)}=- -\frac{\binom{ab}{m}\binom{cd}{m}}{\binom{ad-1}{m}\binom{bc-1}{m}(a-c)(b-d)} -\left(\frac{\frac{ab-m}{m+1}\frac{cd-m}{m+1}}{\frac{ad-1-m}{m+1}\frac{bc-1-m}{m+1}}(m+1-bc)(m+1-ad)-(m-bc)(m-ad)\right)=- -\frac{\binom{ab}{m}\binom{cd}{m}}{\binom{ad-1}{m}\binom{bc-1}{m}(a-c)(b-d)}(m^2-m(ab+cd)+abcd-m^2+m(ad+bc)-abcd)=m\frac{\binom{ab}{m}\binom{cd}{m}}{\binom{ad-1}{m}\binom{bc-1}{m}}. - $$ -Thus, the sum - $$ -\sum_{m=0}^n m\frac{\binom{ab}{m}\binom{cd}{m}}{\binom{ad-1}{m}\binom{bc-1}{m}} - $$ -is equal $f(n+1)-f(0)$, which implies the identity you want.<|endoftext|> -TITLE: Torsion-free abelian group $A$ such that $A \not \simeq A \oplus \Bbb Z \simeq A \oplus \Bbb Z^2$ -QUESTION [15 upvotes]: Is there a torsion-free abelian group $A$ such that $A \not \simeq A \oplus \Bbb Z \simeq A \oplus \Bbb Z \oplus \Bbb Z$ (as groups)? - -Notice that $\Bbb Z$ is not cancellable, so -$A \oplus \Bbb Z \simeq (A \oplus \Bbb Z) \oplus \Bbb Z$ doesn't imply that $A \simeq A \oplus \Bbb Z$. Combined with this question, such a group $A$ would possibly provide an answer to that question. -An example of torsion free abelian group $A$ such that $A$ is isomorphic to $A \oplus \mathbb{Z}^2$, but not to $A \oplus \mathbb{Z}$ was given there. -The example was the additive group of bounded sequences of elements of $\mathbb{Z}[\sqrt{2}]$, i.e. -$$A = -\left\{ -(x_n)_{n \geq 1} \subset \Bbb Z[\sqrt 2] \;\;:\;\; -\exists C>0,\; \forall n \geq 1,\; |x_n| \leq C -\right\}$$ -I wasn't able to adapt this example in order to answer my question. - -REPLY [17 votes]: $\mathbb{Z}$ is cancellable for abelian groups. This was proved in the 1950s by Walker and Cohn (independently) and is often called "Walker's cancellation theorem". The proof is only a few lines. -So if $A$ is an abelian group with $A\oplus\mathbb{Z}\cong A\oplus\mathbb{Z}^2$, then $A\cong A\oplus\mathbb{Z}$.<|endoftext|> -TITLE: Approximation of sum of the first binomial coefficients for fixed N -QUESTION [10 upvotes]: I'd like to compute $\sum_{i=0}^k {{N}\choose{i}}$. Is there a computable approximation for that? - -REPLY [16 votes]: One of the more convenient and popular approximations of the sum is -$$\frac{2^{nH(\frac{k}{n})}}{\sqrt{8k(1-\frac{k}{n})}} \leq \sum_{i=0}^k\binom{n}{i} \leq 2^{nH(\frac{k}{n})}$$ -for $0< k < \frac{n}{2}$, where $H$ is the binary entropy function. (The upper bound is exactly what Aryeh Kontorovich mentions.) You can find its proof in many textbooks, but probably I first learned it from Chapter 10 of The Theory of Error-Correcting Codes by MacWilliams and Sloane. -Also, this post on MO asks a similar question and is a good resource for the sum in my opinion. You can find several other useful bounds there.<|endoftext|> -TITLE: Non-isomorphic projective planes on $\omega$ -QUESTION [5 upvotes]: Let $X\neq \emptyset$. A set $L\subseteq {\cal P}(X)$ is said to be a projective plane on $X$ if the following conditions are met: - -if $x\neq y\in X$ there is a unique $l\in L$ such that $x, y \in l$, -if $l\neq m \in L$, then we have $|l\cap m | = 1$, -there are four distinct elements of $X$ such that no member of $L$ contains more than $2$ of the four. - -What is the maximal cardinality of a collection ${\cal C}$ of projective geometries on $\omega$ such that no two distinct members of ${\cal C}$ are isomorphic? (The notion of ismorphism of projective planes is defined below.) - -Note. If $L, M$ are projective planes on $X$, we say they are isomophic if there is a bijection $\varphi: X\to X$ such that $l\in L$ if and only if $\varphi(l) \in M$. - -REPLY [10 votes]: You ask for the number of isomorphism classes of projective planes on $\omega$. I claim that it is exactly $2^{\aleph_0}$. -It is at most $2^{\aleph_0}$. -Indeed, a projective plane on $\omega$ can be encoded by the set of $\{x,y,z\}\subset\omega$ of cardinal $3$ which are aligned (i.e., such that $z$ lies on the unique line containing $x,y$), and the set of $3$-element subsets of $\omega$ is countable, so the set of its subsets has cardinal $2^{\aleph_0}$. -It is at least $2^{\aleph_0}$. -For this, it is enough to show that there are at least $2^{\aleph_0}$ pappian projective plane, i.e., planes of the form $\mathbb{P}^2(K)$ for some countable field $K$ : note that $K$ is determined, up to isomorphism by the incidence structure (choose $0,1,\infty$ on some line and construct addition and multiplication by the usual projective constructions). So it is enough to show that there are at least $2^{\aleph_0}$ isomorphism classes of countable fields. But this is easy: take any subset $S$ of the set of prime numbers and consider the field $K_S := \mathbb{Q}(\sqrt{p} : p\in S)$ (the set $S$ can be recovered as the set of primes $p$ having a square root in $K_S$).<|endoftext|> -TITLE: Calculations of cup products in Bredon cohomology -QUESTION [12 upvotes]: Let $G$ be a finite group. In [1], Bredon defines an equivariant cohomology theory for $G$-CW complexes $H^*_G(X;M)$. The coefficients are taken in modules over the orbit category of $G$, that is, contravariant functors $M:\mathcal{O}_G\to \mathcal{Ab}$ from the category of finite $G$-sets and $G$-maps to the category of abelian groups. This Bredon cohomology has now been generalized and extended in many ways by many authors. Although this question is manily about the "classical" setting described in Bredon's book, I would be interested also in answers which pertain to more sophisticated versions. -I've recently started trying to do some computations of Bredon cohomology, and have been able to compute the groups $H_G^i(X;M)$ in particular cases. Now I need to compute some cup products -$$ -H^i_G(X;M)\otimes H^{j}_G(X;N)\to H^{i+j}_G(X;M\otimes N), -$$ -and find I am lacking in tools, or guiding examples, which are available to me in the non-equivariant case. Therefore I ask: - -What are some examples in the literature of calculations of cup products in the Bredon cohomology of specific $G$-spaces? - -Perhaps one needs to resort to finding a $G$-equivariant approximation of the diagonal, or extending to an $RO(G)$-graded cohomology theory. Then I would appreciate references where this is done to produce calculations of products in the $\mathbb{Z}$-graded theory. -[1] Glen E. Bredon, MR 214062 Equivariant cohomology theories, Lecture Notes in Mathematics, No. 34 . - -REPLY [5 votes]: Frankly, there aren't many calculations out there. Most of the work I know of is on the calculation of the $RO(G)$-graded cohomology of a point, of a projective space, or of $B_GO(n)$. Here are some references and notes on them: -[1] L. G. Lewis, Jr., The $RO(G)$-graded equivariant ordinary cohomology of complex projective spaces with linear $\mathbb{Z}/p$ actions, in Algebraic topology and transformation groups, Lecture Notes in Math. v. 1361, 1988. (MR 979507) -In an appendix, this has the first published account of Stong's calculation of the $RO(G)$-graded cohomology of a point for $G= \mathbb{Z}/p$, where $p$ is prime. Stong calculated the multiplicative structure for $p = 2$ and $3$, and Lewis extended that to all primes. The coefficient system used is the Burnside ring system, which evaluates to the Burnside ring $A(H)$ at the orbit $G/H$. -The main purpose of the paper, though is the calculation of the cohomology of complex projective spaces, that is, the spaces $\mathbb{C}P(V)$ where $V$ is a finite- or countably infinite-dimensional complex representation of $G$. The calculation includes the multiplicative structure. -[2] W. C. Kronholm, The $RO(G)$-graded Serre spectral sequence, Homology, Homotopy Appl. 12 (2010), pp. 75-92. (MR 2607411) -Kronholm gives a similar calculation for real projective spaces, for $G=\mathbb{Z}/2$ and with constant $\mathbb{Z}/2$ coefficients. -[3] D. Dugger, Bigraded cohomology of $\mathbb{Z}/2$-equivariant Grassmannians, Geom. Topol. 19 (2015), pp.113-170. (MR 2240234) -Dugger uses Kronholm's result to calculate the $RO(G)$-graded cohomology of $B_GO(n)$ for $G = \mathbb{Z}/2$ and constant $\mathbb{Z}/2$ coefficients. -There are a couple of other papers out there with partial calculations of the cohomology of a point, but for solid calculations of the cohomology of interesting spaces, this is all I've got.<|endoftext|> -TITLE: A dynamical system defined by the Riemann zeta function -QUESTION [6 upvotes]: Let $\zeta$ be the classical Riemann zeta function. -We define a differential equation on $\mathbb{R}^{2} \setminus \{1\}$ by $\dot Z= \zeta(Z)$. From a foliation point of view this vector field can be counted as a smooth vector field on whole $\mathbb{R}^{2}$ with the following equivalent formulation(They have the same trajectories). -$$\dot Z= \lVert z-1\rVert^2 \zeta(Z)$$ -Then the field has a saddle point at $1$. -Are there some researches about this dynamical system?Are there closed orbits for this equation?The latter is equivalent to ask: "Are there zeroes of the Riemann Zeta function whose Taylor expansion (after translation to the origin and real rescalling ) is in the form $"iz+...."$. Every zero of a holomorphic map with this linear part is necessarily a center, a singularity surrounded by a band of closed orbits. - -REPLY [2 votes]: A colleague gives me this article: -Does the Riemann zeta function satisfy a differential equation? -where the next formula appear -$$\zeta'(z)=\zeta(z)\sum\limits_{n=1}^\infty{\ln p_n\over1-p_n^z}.$$ -So, your original problem symplifies to the system -$$\dot{y}=F(z)y^2$$ -$$\dot{z}=y$$ -(of course $F(z)=\sum\limits_{n=1}^\infty{\ln p_n\over1-p_n^z}$). -This system have been studied for $F$s in many spaces. But I can't tell what kind of function is it. -I hope it helps. About closed orbits, I really don't know (but I don't think they are).<|endoftext|> -TITLE: A cochain-level Adem relation? -QUESTION [18 upvotes]: The original paper on Steenrod squares, Steenrod's "Products of cocycles and extensions of mappings", 1947, uses an explicit combinatorial formula for the squares in terms of simplicial cochains: given a simplicial cochain $\alpha$ on some simplicial set, Steenrod defines a cochain $\mathrm{Sq}^i\alpha$; if $\alpha$ is a cocycle, so is $\mathrm{Sq}^i\alpha$, and the cohomology class of $\mathrm{Sq}^i\alpha$ only depends on that of $\alpha$, and so $\mathrm{Sq}^i$ descends to a (linear!) map on cohomology. (In general, $\mathrm{Sq}^i$ is not linear on cochains and does not commute with $\mathrm d$, so this is a nontrivial statement. But it is not too hard to find an operation that agrees with $\mathrm{Sq}^i$ on cocycles and that does commute with $\mathrm d$, and its failure to be linear, it is not hard to show, is a total derivative.) (Steenrod makes his definition with $\mathbb Z$-coefficients, but I will only care about $\mathbb Z/2$ coefficients, and so assume that throughout.) -The well-known Adem relations, due originally to Adem, "The iteration of the Steenrod squares in algebraic topology", 1952 and beautifully explained by Bullett and Macdonald, "On the Adem relations", 1982, are relations of the form $0 = \sum_{\text{certain pairs }(i,j)} \mathrm{Sq}^i \mathrm{Sq}^j$ that hold in cohomology. The first few are - $\mathrm{Sq}^1 \mathrm{Sq}^1 = 0$, $\mathrm{Sq}^1\mathrm{Sq}^2 = \mathrm{Sq}^3 \mathrm{Sq}^0$, $\mathrm{Sq}^1 \mathrm{Sq}^3 = 0$, $\mathrm{Sq}^3 \mathrm{Sq}^1 = \mathrm{Sq}^2 \mathrm{Sq}^2$, ... -At the cochain level, these relations do not hold on the nose. The fact that they hold on cohomology just says that, if $\alpha$ is a cocycle, then $\sum_{\text{certain pairs }(i,j)} \mathrm{Sq}^i (\mathrm{Sq}^j(\alpha))$ is a coboundary, which of course depends on $\alpha$ (and on the particular relation in question). -Is there a combinatorial formula (like the one Steenrod gives for $\mathrm{Sq}^i$) that takes in $\alpha$ a cocycle and produces a primitive for $\sum_{\text{certain pairs }(i,j)} \mathrm{Sq}^i (\mathrm{Sq}^j(\alpha))$? Where can I find it? I care only about the early Adem relations up to $\mathrm{Sq}^3\mathrm{Sq}^1 = \mathrm{Sq}^2\mathrm{Sq}^2$. -A related question was asked by Kapustin. Note that I am not after more categorical descriptions of the Adem relations and the Steenrod squares like those available in the many excellent answers to this old MO question — I really do want a combinatorial description that fits into the framework of Steenrod's original paper. - -REPLY [4 votes]: Explicit homotopies for a cochain-level Adem relation were first worked out in: -Greg Brumfiel, Anibal M. Medina-Mardones, John Morgan. A Cochain Level Proof of Adem Relations in the Mod 2 Steenrod Algebra. 2020 -The odd-prime case is still open.<|endoftext|> -TITLE: Does the existence of a derived functor imply existence of model structure? -QUESTION [9 upvotes]: this is my first thread on mathoverflow, and apologies if this is a trivial question. -In Dwyer and Spalinski's Homotopy theories and model categories, they gave the definition of derived functors, and based on their definition, we require the source C of a functor F : C $\to$ D to be a model category, and if a derived functor (no matter left or right) exists, denoted as DF, then -DF : Ho(C) $\to$ D -The notes also gives a sufficient condition of the existence of left derived functor, but because of duality, we can rephrase the same for right derived functor: if the functor F is as above, and F(f) is an isomorphism whenever f is a weak equivalence between fibrant objects in C, then the right derived functor (RF, t) of F exists. -I think I accept the concepts above fairly well, but then in Section 3.1 of Hartshorne's Algebraic Geometry, he introduced the concept of derived functors of covariant left exact functors F : A $\to$ B, where A and B are abelian categories, with A having enough injectives. Though Hartshorne didn't mention the condition of existence, he wrote if F: A $\to$ B is a covariant left exact functor between abelian categories, and A has enough injectives, then we construct the right derived functor... -So, does the concepts in the two scenarios above correspond? For example (this might be very wrong, just my guess), does "an abelian category with enough injectives" automatically have a model structure? -If the answer is "yes" and we have a model structure compatible with the cohomology objects, does "a covariant left exact functor" correspond to "a functor sending fibrant objects to isomorphisms", and do other properties of derived functors, such as the universal property, also hold? -If the answer is "no", then does it mean we can construct derived functors without model structure and the associated homotopy category? If so, how? Also, are there any relationships between model categories and abelian categories? -Thank you very much for reading my long and tedious thread. I appreciate any insights. - -REPLY [3 votes]: This is not quite the answer to your question as you pose it. I hope it will be useful anyway. By and large I am just expanding user337830 comments. Everything will use homological grading (what can I say, I am a homotopy theorist :)). -The relationship between derived functors in homological algebra and derived functors in homotopy theory is very close. In fact the notion of derived functor in homotopy theory is a (very successful) attempt to generalize the homological algebra notion. -Let me talk about right derived functors for one second. Let $A$ be a Grothendieck abelian category. Then it is known it has enough injectives. This allows us to define a model structure on the category $Ch(A)$ of chain complexes in $A$ such that - -weak equivalences are quasi-isomorphisms; -cofibrations are chain maps $C_*\to D_*$ that are levelwise injective; -fibrations are chain maps $C_*\to D_*$ that are levelwise surjective and such that the kernel $K_*$ is a dg-injective complex (that is all $K_n$ are injective objects of $A$ and for all acyclic complexes $E_*$ every chain map $E_*\to K_*$ is nullhomotopic, bounded above complexes of injectives are an example of such). - -This model structure is sometimes called the injective model structure on $Ch(A)$. -In particular if $M\in A$, an injective resolution of $M$ gives a fibrant replacement for the chain complex consisting of $M$ in degree 0. So it is easy to see that if $A,A'$ are two Grothendieck abelian categories and $F:A\to A'$ is a left-exact functor, the derived functor of $F$ is precisely the right-derived functor of $F_*(-):Ch(A)\to Ch(A')$. -Similarly if $A$ has enough projectives you can form a projective model structure that will allow you to describe left derived functors of right-exact functors. -Now, these two model structures in fact present the same homotopy theory (e.g. they have the same weak equivalences! More to the point, they are Quillen equivalent), so you can pass from one to the other with a minimum of fuss. Their common homotopy category is called the (unbounded) derived category of $A$. -There are also other model structures on $Ch(A)$ with the same weak equivalences that are more convenient for other purposes, e.g. the flat model structure (where the fibrant objects are complexes of flat modules $C_*$ such that $C_*\otimes-$ preserves quasi-isomorphisms). In general there is a correspondence between algebraic model structures on $Ch(A)$ and cotorsion pairs on $A$. See this paper by Hovey for more details.<|endoftext|> -TITLE: Difference between the completed group algebra and the profinite completion of a group ring -QUESTION [13 upvotes]: Let $G$ be a reasonably nice group, say residually finite if need be. -We may consider the group algebra $\mathbb{Z}[G]$. -Let $\widehat{\mathbb{Z}[G]} := \varprojlim_I\mathbb{Z}[G]/I$ be the profinite completion, where $I$ runs over all ideals of finite index. -Let $\widehat{\mathbb{Z}}[[\widehat{G}]] := \varprojlim_{n,U}(\mathbb{Z}/n)[G/U]$ be the "completed group algebra", where $n$ runs over all positive integers, and $U$ all finite index normal subgroups of $G$. -When are these two objects the same? (Are they always the same? Perhaps they're the same when $G$ is abelian?) -This seems like a natural question, but I cannot find it addressed anywhere, and I can't seem to deduce anything nontrivial in either direction. - -REPLY [8 votes]: It seems to me that these are indeed isomorphic. Namely, if $n$ is an integer and $U$ is a finite index normal subgroup of $G$, then the kernel $I_{n,U}$ of the natural map $\mathbb ZG\to \mathbb Z/n[G/U]$ is an ideal of $\mathbb ZG$ of finite index and hence part of the inverse system defining $\widehat{\mathbb ZG}$. On the other hand, if $I$ is a finite index ideal of $\mathbb ZG$, then $R=\mathbb ZG/I$ is a finite ring of some characteristic $n$ and the image of $G$ under the natural map is finite hence of the form $G/U$ with $U$ a finite index normal subgroup of $G$. Thus $I_{n,U}$ is contained in $I$. Therefore, the collection of ideals $I_{n,U}$ form a cofinal system of finite index ideals and hence the two completions are isomorphic by standard properties of inverse limits.<|endoftext|> -TITLE: Is $\widehat{\mathbb{Z}}[[t]]\cong\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]$? -QUESTION [22 upvotes]: Let $\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]] := \varprojlim_{n,m}(\mathbb{Z}/n)[x]/(x^m-1)$ be the complete group algebra of the profinite free group of rank 1. In Corollary 5.9.2 of Ribes-Zalesski's Profinite Groups, they state that $\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]\cong \widehat{\mathbb{Z}}[[t]]$, citing a paper of Lim which doesn't seem to prove exactly what they claim (though I'm sure it must follow, if one is sufficiently familiar with the theory). -Here, let's try giving $\widehat{\mathbb{Z}}[[t]]$ the topology corresponding to the product topology on $\prod_{n\ge 0}\widehat{\mathbb{Z}}$, indexed by the coefficients of $t^n, n\ge 0$. -I would like to show directly that $\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]\cong \widehat{\mathbb{Z}}[[t]]$. If this is true, then the map should be given by identifying a generator "$x$" of the group algebra with $1+t$. -Relative to the (product) topology described above on $\widehat{\mathbb{Z}}[[t]]$, a neighborhood basis of 0 is given by the ideals $(n,t^m)$ for $n,m\ge 1$. For every such ideal, one can find some $N,M\ge 1$ such that the map "$x\mapsto 1+t$" defines a quotient map -$$\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]\rightarrow(\mathbb{Z}/N)[x]/(x^M-1) \rightarrow \widehat{\mathbb{Z}}[t]/(n,t^m)$$ -(this follows from divisibility properties of binomial coefficients). However when I try to go in the other direction, it seems what I need is - for every $N,M\ge 1$, to find an $n,m$ such that $t\mapsto x-1$ induces a map -$$\widehat{\mathbb{Z}}[t]/(n,t^m)\rightarrow(\mathbb{Z}/N)[x]/(x^M-1) $$ -However, this is clearly impossible, since $t$ is always nilpotent on the left, and yet $x-1$ is rarely nilpotent on the right. -Thus, either the "product" topology on $\widehat{\mathbb{Z}}[[t]]$ is not sufficient, or the map is more tricky than just sending "$x\mapsto 1+t$". -On the other hand, $\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]$ is a commutative profinite ring, and hence a product of commutative profinite local rings. It certainly admits $\mathbb{Z}_p[[\mathbb{Z}_p]]$ as quotients for all primes $p$, which is known to be isomorphic to the local ring $\mathbb{Z}_p[[t]]$, so it seems hard to imagine any other possibility for $\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]$. -If it turns out the titular question has a negative answer, then naturally question is: -$$\text{What are the local direct factors of the profinite ring $\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]$?}$$ - -REPLY [5 votes]: There may be some things to check here, but I think the following is correct and should answer your last question. -$\newcommand{\ZZ}{\mathbb{Z}}$ -$\newcommand{\Zhat}{\widehat{\mathbb{Z}}}$ -I believe the final result is -$$\Zhat[[\Zhat]] \cong \prod_q\ZZ_q[[t]]$$ -as $q$ ranges over all prime powers $p^r$ with $r$ coprime to $p$, each appearing $N_{p,r}$ times, where $N_{p,r}$ is the number of Frobenius orbits of generators of $\mathbb{F}_q^\times$ -Firstly, $\Zhat[[\Zhat]]\cong\prod_p\ZZ_p[[\Zhat]]$. To see this, note that in every $(\ZZ/n)[x]/(x^m-1)$, the distinct prime powers dividing $n$ generate comaximal ideals, and hence if $n = \prod_i p_i^{r_i}$ then -$$(\ZZ/n)[x]/(x^m-1)\cong\prod_i(\ZZ/p_i^{r_i})[x]/(x^m-1)$$ -This decomposition at every finite stage should extend to the limit, and so it suffices to analyze $\ZZ_p[[\Zhat]]$. Let -$$\ZZ' :=\prod_{p'\text{ prime}\\p'\ne p}\ZZ_{p'}$$ -then since $\Zhat = \ZZ_p\times\ZZ'$, and since $\ZZ_p[A\times B] = \ZZ_p[A]\hat{\otimes}\ZZ_p[B]$ (completed tensor product over $\ZZ_p$), we have: -$$\ZZ_p[[\Zhat]] = \ZZ_p[[\ZZ_p]]\hat{\otimes}\ZZ_p[[\ZZ']] = \ZZ_p[[t]]\hat{\otimes}\ZZ_p[[\ZZ']]$$ -By definition, -$$\ZZ_p[[\ZZ']] = \varprojlim_m \ZZ_p[x]/(x^m-1)$$ -where $m$ is coprime to $p$. Since such $x^m-1$ factors into distinct irreducibles mod $p$, by Hensels lemma we find that $x^m-1 = \prod_i f_{m,i}$, where each $f_{m,i}\mid x^m-1$ and is irreducible in both $\ZZ_p[x]$ and $\mathbb{F}_p[x]$. These $f_{m,i}$ are actually pairwise comaximal (see this answer), and hence we have a decomposition -$$\ZZ_p[x]/(x^m-1) = \prod_i \ZZ_p[x]/(f_{m,i})$$ -where each term in the product is now isomorphic to $\ZZ_q$, where $q = p^{\deg f_{m,i}}$ (ie, the unique domain unramified over $\ZZ_p$ of degree $\deg f_{m,i}$) Taking the limit over all $m$ coprime to $p$, we get: -$$\ZZ_p[[\ZZ']] = \prod_f\ZZ_p[x]/(f)$$ -where $f$ ranges over the set: -$$\{f\in\ZZ_p[x] \text{ irreducible} : \exists m\text{ coprime to $p$ such that } f\mid x^m-1\}$$ -Thus, we get -$$\ZZ_p[[\Zhat]] = \ZZ_p[[t]]\hat{\otimes}\prod_f\ZZ_p[x]/(f)$$ -By Proposition 7.7.5 in Wilson's book Profinite Groups, the completed tensor product commutes with arbitrary direct products, so we get -$$\ZZ_p[[\Zhat]] = \prod_f(\ZZ_p[[t]]\hat{\otimes}\ZZ_p[x]/(f))$$ -Since each $\ZZ_p[x]/(f)$ is a finite $\ZZ_p$-algebra, the completed tensor product coincides with the usual tensor product (c.f. Ribes-Zalesski prop 5.5.3(d)), so we get -$$\ZZ_p[[\Zhat]] = \prod_f(\ZZ_p[[t]]\otimes\ZZ_p[x]/(f)) = \prod_f (\ZZ_p[x]/(f))[[t]] = \prod\ZZ_q[[t]]$$ -as $q$ ranges over all prime-to-$p$ powers of $p$, each appearing multiple times in the product. -Thus, I believe we have an isomorphism -$$\Zhat[[\Zhat]] \stackrel{\varphi}{\longrightarrow}\prod_q\ZZ_q[[t]] = \left(\prod_q\ZZ_q\right)[[t]]$$ -For every $r$ coprime to $p$, the number of times each $\ZZ_q = \ZZ_{p^r}$ appears in the product is precisely the number $N_r$ of irreducible degree $r$ factors of $x^{p^r-1}-1$ over $\mathbb{F}_p$, or equivalently the number of $p$-power Frobenius orbits of generators of $\mathbb{F}_q^\times$. Let $a\in\Zhat$ come from within the $[[\cdots]]$ of $\Zhat[[\Zhat]]$. Then, $\varphi(a) = a'(t+1)^{a_p}$, where $a_p$ is the image of $a$ in $\ZZ_p$, and $a'\in\prod_{p'\ne p}\ZZ_{p'}$ is defined as follows. For every $q = p^r$ with $(r,p) = 1$, let $\overline{a}$ be the residue of $a$ mod $p^r-1$. Then, the images of $a'$ in the $N_r$ copies of $\ZZ_q$ are in bijection with a complete set of representatives of the Frobenius orbits of primitive $(q-1)$th roots of unity in $\ZZ_q$, each raised to the $\overline{a}$-th power. -In particular, when $r = 1$, we find that the number of copies of $\ZZ_p[[t]]$ in the product is precisely the number of linear factors of $x^{p-1}-1$ over $\mathbb{F}_p$, which is precisely $p-1$. Thus, the projections onto each of these factors, followed by quotienting by $(t)$, yields the $p-1$ homomorphisms onto $\ZZ/p\ZZ$ referred to by Tom Goodwillie, and it seems like there can't be any others.<|endoftext|> -TITLE: Are $\partial$-exact and $\bar\partial$-exact forms also $\partial\bar\partial$-exact? -QUESTION [14 upvotes]: $M$ is non-Kähler complex manifold. Assume that $\omega$ is $\partial$-exact and $\bar\partial$-exact $(p,q)$-form. - -Question. Is $\omega$ also $\partial\bar\partial$-exact? - -Based on fabulous David Speyer's answer to this MO question I suspect that the answer is elementary or very hard. - -REPLY [10 votes]: I think that the answer is in general no and that a counterexample can be constructed as follows. I will refer to the paper by D. Angella, G. Dloussky, A. Tomassini -On Bott-Chern cohomology of compact complex surfaces, Annali di Matematica Pura ed Applicata 195 (2016), pp 199–217. -Let us consider a non-Kähler surface $S$ of Inoue type. Then we have (see Table 1 p. 210 of the aforementioned paper) -$$h^{1, \, 1}_{\bar{\partial}}(S)=0, \quad \textrm{hence} \quad h^{1, \, 1}_{\partial}(S)=0 \quad (\spadesuit) $$ -where the second equality is obtained by conjugation and duality induced by the Hodge $\ast$-operator associated to a given Hermitian metric. -Moreover, we have $$h^{1, \,1}_{\textrm{BC}}(S)=1 \quad (\clubsuit)$$ -where $h^{\bullet, \, \bullet}_{\textrm{BC}}$ denotes the dimension of Bott-Chern cohomology group $H^{\bullet, \, \bullet}_{\textrm{BC}}$, namely $$H^{\bullet, \, \bullet}_{\textrm{BC}}(S) = \frac{\ker \partial \cap \ker \bar{\partial}}{\mathrm{im}\, \partial \bar{\partial}} .$$ -Now, $(\clubsuit)$ means that there exists a $(1, \, 1)$-form $\omega$ on $S$ which is both $\partial$-closed and $\bar{\partial}$-closed, but not $\partial \bar{\partial}$-exact. -On the other hand, $(\spadesuit)$ implies that any $\partial$-closed (resp. $\bar{\partial}$-closed) $(1,\, 1)$-form on $S$ is actually $\partial$-exact (respectively, $\bar{\partial}$-exact). -Summing up, $\omega$ is a $(1, \, 1)$-form on $S$ which is both $\partial$-exact and $\bar{\partial}$-exact, but not $\partial \bar{\partial}$-exact.<|endoftext|> -TITLE: No Tonelli or Fubini -QUESTION [15 upvotes]: Whenever we can interchange summation (perhaps due to Tonelli-Fubini), good things happen. Otherwise, one has to struggle evaluating double sums in just one way, because the alternative results in a divergent series. Having said that, I'm currently interested in the following: - -Have you encountered in your own research or do you recall from someone's research paper "interesting" double sums that are convergent where reversing order of summation does not work? Please provide references. - -REPLY [6 votes]: At L. Spice's suggestion, I hereby reify (!) my comment "Fourier inversion on the real line? Too mundane?" :) -But while we're here, I'd want to remark that to my mind much modern analysis (e.g., post-Schwartz, post-Grothendieck, et al) amounts to reconsideration of seemingly illegitimate symbol manipulation that mysteriously gives reliable answers (e.g., in Heaviside's work in the 1890's and thereafter, in Dirac's physics beginning c. 1928, H. Bethe's and others' work in the early 1930s). -In that context, one might argue that the ground-breaking work of the Italian, German, and French analysts in the late 19th and early 20th century was wonderful, but did not go far enough. In particular, although (as I've ranted elsewhere or other occasions) the precisification of the notion of "function" in set-theoretic terms was an excellent thing, it was/is arguably too restrictive (witnessed already by the "almost everywhere" kludge/revelation in Lebesgue-et-al theory). -In particular, although Fubini-Tonelli's theorem can fail due to "pointwise" problems, if sufficiently recast about not-literal integrals but continuous functionals, it can be designed-to-succeed in many situations where pointwise failure is irrelevant. Thus, in many practical situations, a "literal failure"'s incorrect conclusions can sometimes be salvaged by not being sucked down into the miasma of (irrelevant) pointwise issues. -To my mind, an archetype for this is Clairault's theorem on the interchangeability of partial derivatives ... under some conditions. Naturally, at the time, and still nowadays for many, this is a pointwise issue. However, taking Fourier transforms, we find that distributionally the derivatives are always interchangeable. -Back to the immediate question: another number-theoretic example (which I included as a prank-question in my old book on Hilbert modular forms) is about the "inner product" (which it cannot literally be...) of a holomorphic Poincare series and a holomorphic Eisenstein series. The seeming paradox is that if we unwind the Eisenstein series, we (correctly) compute the zeroth Fourier coef of the Poincare series, which is (in the holomorphic case only!) $0$. But, if we unwind the Poincare series (which is not ok... ) we seem to compute a higher Fourier coefficient of the Eisenstein series, thus, seeming to show that Eisenstein series are constants. The fallacy is that the Poincare series cannot be unwound here because there was cancellation in the summing of it: when unwound, Fubini-Tonelli does not apply, and, indeed, the seeming conclusion is incorrect. -The non-frivolousness of that example resides in a standard procedure in analytic number theory (and in automorphic extensions of the classical versions thereof) where both Eisenstein series and Poincare series are extended by a sort of Hecke-summation device to depend on a complex parameter $s$, and wherein a fairly arbitrary thing is "wound up" to make an automorphic form, and then spectrally decomposed, etc. It is common practice to simply ignore convergence issues in such spectral expansions, with a few exceptions. And, indeed, even in the tiniest cases, $L^2$ convergence is some distance from pointwise, but the discussion is all too often pointwise. -Another archetype is the spectral decomposition of pseudo-Eisenstein series (in various contexts) in terms of genuine Eisenstein series. This decomposition in the simplest cases is derived from ordinary Fourier inversion applied to test-function data for a pseudo-Eisenstein series, with the idea to wind up this expression to an integral of Eisenstein series. The obstacle is convergence. Thus, the Fourier-Mellin inversion path is moved sufficiently far to the right to legitimize the winding up (by Fubini-Tonelli, for example), and then the path is moved back to the critical line. The small surprise, which may seem innocuous in the simplest cases, is that residues of the Eis appear. For $SL(2,\mathbb Z)$, the residues are essentially constants, which may be misleading. Namely, for higher-rank groups (e.g., $SL(4)$ or $Sp(4)$, the residues are highly-nontrivial automorphic forms (Speh forms). -(The previous are arguably even-simpler issues than difficulties in evaluating traces in the obvious fashion.)<|endoftext|> -TITLE: a Hankel matrix of involution numbers -QUESTION [11 upvotes]: Let $I_k$ denote the enumeration of involutions among permutations in $\mathfrak{S}_k$. I always enjoy these numbers. Of course, here is yet another cute experimental finding for which I ask validity. - -Question. Let $n!!=1!2!\cdots n!$. Is the following true? - $$\det\left[I_{i+j}\right]_{i,j=0}^n=n!!$$ - -REPLY [7 votes]: As in arXiv:0902.1650 it suffices to show that $a(n,0)=I_n$ if $a(n,j)$ satisfies $a(n,j)=a(n-1,j-1)+a(n-1,j)+(j+1)a(n-1,j+1)$ with $a(n,-1)=0$ and $a(0,j)=[j=0]$. -But it is easily verified that $a(n,j)=\binom{n}{j}I_{n-j},$ because then the above recursion reduces to $I_n=I_{n-1}+(n-1)I_{n-2}.$ -Edit: Another proof along the same lines follows immediately from formula (2.7) in this paper entitled Involutions and their progenies: -$\sum_k\binom{n}{k}I_{n-k} \binom{m}{k}I_{m-k} k!=I_{m+n}.$<|endoftext|> -TITLE: On the relation between two definitions of torsion functors -QUESTION [5 upvotes]: Let $R$ be a commutative ring, and let $\mathfrak{a}\subseteq R$ be an ideal. For an $R$-module we consider the sub-$R$-modules $$\Gamma_{\mathfrak{a}}(M)=\{x\in M\mid\exists n\in\mathbb{N}:\mathfrak{a}^n\subseteq(0:_Rx)\}$$ and $$\widetilde{\Gamma}_{\mathfrak{a}}(M)=\{x\in M\mid\mathfrak{a}\subseteq\sqrt{(0:_Rx)}\}$$ of $M$. These definitions give rise to left exact subfunctors $\Gamma_{\mathfrak{a}}$ and $\widetilde{\Gamma}_{\mathfrak{a}}$ of the identity functor on the category of $R$-modules, and $\Gamma_{\mathfrak{a}}$ is a subfunctor of $\widetilde{\Gamma}_{\mathfrak{a}}$. -Recall that a subfunctor $F$ of the identity functor is called a radical if $F(M/F(M))=0$ for every $R$-module $M$. Moreover, for every subfunctor $F$ of the identity functor there exists the smallest radical containing $F$. -Now, one can show that $\widetilde{\Gamma}_{\mathfrak{a}}$ is a radical, while $\Gamma_{\mathfrak{a}}$ need not be so. I conjecture but am unable to prove the following: - -Conjecture: $\widetilde{\Gamma}_{\mathfrak{a}}$ is the smallest radical containing $\Gamma_{\mathfrak{a}}$. - -Is anything known about this problem? -(Motivation: In the literature about torsion functors and their right derived functors (i.e., local cohomology) both definitions are used. Since most authors work over noetherian rings this does not matter: For an ideal of finite type, the two functors coincide. But if we wish to study non-noetherian situations it might be helpful to understand the precise relation between the two definitions.) - -REPLY [2 votes]: The conjecture is not true. -By Quý's comment, the conjecture implies that $\Gamma_{\mathfrak{m}}$ is not a radical if $R$ is a $0$-dimensional local ring whose maximal ideal $\mathfrak{m}$ is idempotent but not nilpotent. This contradicts the following result. -Lemma If $R$ is a ring and $\mathfrak{a}\subseteq R$ is an idempotent ideal, then $\Gamma_{\mathfrak{a}}$ is a radical. -Proof: Let $M$ be an $R$-module, and let $x\in M$ be such that $x+\Gamma_{\mathfrak{a}}(M)\in\Gamma_{\mathfrak{a}}(M/\Gamma_{\mathfrak{a}}(M))$. Then, $\mathfrak{a}x\in\Gamma_{\mathfrak{a}}(M)$, and therefore $\mathfrak{a}x=\mathfrak{a}^2x=0$, implying $x+\Gamma_{\mathfrak{a}}(M)=0$. $\square$<|endoftext|> -TITLE: How to solve the system of PDEs defining Killing vectors -QUESTION [7 upvotes]: Recently I came across the following problem. Here's the setting: -Let $(M^n,g)$ be a Riemannian manifold, $\nabla$ the Levi-Civita connection, and $U$ a coordinate neighbourhood with coordinates $\{x^i\}$. $X = X^i \frac{\partial}{\partial x^i}$ is a vector field. We denote by $\nabla_i$ the derivative $\nabla_\frac{\partial}{\partial x^i}$. Denote also $X^i_j := \nabla_j X^i$. Consider $\{ X^i, X^j_k\}$ to be unknown functions and show that the system $\begin{align} \nabla_i X^j &= X_i^j \\ \nabla_k X_i^j &= -R_{pki}^j X^p \end{align}$ with conditions $X_{ij} = -X_{ji}$, where $X_{ij} = X_i^kg_{kj}$, is completely integrable iff $M^n$ is of constant curvature. In this case the solution depends on $\frac{n(n+1)}{2}$ arbitrary constants. -The vectors in the solution space would be the Killing vector fields. The part with the number of constants is clear, but I couldn't acquire the curvature condition, no matter how I tried to attack it. Applying Frobenius' theorem didn't do the job for me, so I suppose I am making a mistake somewhere, be it logical, or computational. The problem is taken from Kentaro Yano's book "Integral Formulas in Riemannian Geometry", Marcel Dekker Inc., NY, 1970, p.35, Problem 19. Similar problems (21, 23, 25) follow, concerning different infinitesimal transformations. According to the convention of the book, the Riemannian curvature tensor is defined as $R_{ijk}^l \frac{\partial}{\partial x^l}= R(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j})\frac{\partial}{\partial x^k}$ - -REPLY [8 votes]: The system of PDEs in the original post actually defines a connection $D$ on the bundle $\mathcal{E} = TM \oplus \mathfrak{so}(TM)$, by -$$ D_X \begin{pmatrix} Y \\ A \end{pmatrix} = \begin{pmatrix} \nabla_X Y + A(X)\\ \nabla_X A - R(X,Y) \end{pmatrix} $$ -for all vector fields $X,Y$ and field $A$ of skewsymmetric endomorphisms of $TM$. Killing vectors are in bijective correspondence with $D$-parallel sections of $\mathcal{E}$, a result due to Kostant and rediscovered by Geroch, who coined the term Killing transport. -Complete integrability is simply the flatness of $D$. The curvature of $D$ is not difficult to calculate and one finds that $D$ is flat if and only if $R(X,Y)Z = \lambda \left( g(Y,Z)X - g(X,Z)Y \right)$, for some $\lambda \in \mathbb{R}$. This is precisely the statement that the metric has constant sectional curvature.<|endoftext|> -TITLE: Gcd of Fibonacci and Catalan -QUESTION [5 upvotes]: Let $F_n=1,1,2,3,5,\ldots$ (starting with $n=1$) be the Fibonacci sequence and let $C_n=\frac{1}{n+1}\binom{2n}{n}$ be the Catalan sequence. -Define $B_z$ to be the cardinality of $$B_z := \#\bigl\{ n \leq z | \gcd(F_n,C_n)=1 \bigr\}.$$ -It seems $$\lim_{z\to\infty} \frac{B_z}{z}=\frac14.$$ -Posted a related question here: -https://math.stackexchange.com/questions/2131648/gcd-of-catalan-and-fibonacci-numbers, but it got no answer and, thinking a bit about it, the question might be more research related than elementary (but I have no real training in such questions). -So question is: Does $B_z/z$ converge, and if so, to what? -Here some values of $g_z := B_z/z$, -$$g_{1000}=\frac{71}{250}=0.284, -\quad g_{2000}=\frac{13}{50}=0.260, -\quad g_{3000}= \frac{187}{750}\approx0.249, -\quad g_{4000}=\frac{979}{4000}\approx0.245.$$ -The sequence of integers such that the gcd is one starts with $$1,2,3,4,5,8,10,11,13,14,17,22,23,25,\ldots$$ and is probably not known in the sequence database. - -REPLY [4 votes]: This is a bit overlong for a comment, and not really an answer, but includes some information about a related problem that might offer approaches to this one. In particular, the fact that the Fibonacci sequence is a divisibility sequence means that gcd properties involving $F_n$ are tied into properties for $F_m$ for $m\mid n$. So you might try looking at the proportion of primes in your set. -The Fibonacci sequence is a linear recursion and a divisibility sequence. For a similar question in which one takes two Fibonacci-type sequences, there are conjectures, but not even an inkling of a proof. For example, Ailon and Rudnick conjectured that $$\{n\ge1 : \gcd(2^n-1,3^n-1)=1\}$$ is infinite, and I published a somewhat dubious heuristic argument that $$\text{Density}\Bigl(\{p~\text{prime}:\gcd(2^p-1,3^p-1)=1\}\Bigr)=1.$$ (The density over all $n\in\mathbb{N}$ seems harder to guess.) More generally, due to the divisibility property of these sequences, for $a,b\ge2$ multiplicatively independent, it is natural to look at $$\{n\ge1 : \gcd(a^n-1,b^n-1)=\gcd(a-1,b-1)\}.$$ I'll also mention that Ailon and Rudnick proved a stronger version of their conjecture when one replaces the integers $a$ and $b$ with polynomials $a(T),b(T)\in\mathbb{C}[T]$.<|endoftext|> -TITLE: What are some of the surprising results of finite sample statistical estimation? -QUESTION [11 upvotes]: I'm trying to familiarize myself with the latest results in finite sample statistics. It seems to me that these results can be classified into two categories: - -Unsurprising results confirm that the asymptotic behavior of a statistic behaves similarly to the finite sample behavior of a statistic. For example: - -It is well known that the maximum likelihood estimator is asymptotically normally distributed with rate $O(1/\sqrt n)$. Many recent papers confirm similar behavior for the MLE in the finite sample regime (e.g. Spokoiny's Parametric estimation. Finite sample theory). These results currently require stronger assumptions than the classical results, but it seems possible that these results will eventually be strengthened to match their classical counterparts. -Many results on random matrices have very similar results in the finite sample and asymptotic regimes. For example, bounds on the minimum and maximum eigenvalues of random matrices are comparable in both regimes. - -Surprising Results show that a statistic has different behavior in the finite sample regime than in the asymptotic regime. For example: - -The empirical mean is an asymptotically normal estimator of the true mean, but the empirical mean is not subgaussian for finite samples. There's been considerable work on finding new mean estimators that are subgaussian for finite samples (e.g. Devroye et. al.'s Sub-Gaussian mean estimators). - - -What are some other surprising results of finite sample statistics? - -REPLY [9 votes]: First of all I have to express my opinion that the gap between large sample behavior and the finite sample behavior should be considered "unsurprising". -Basically speaking, the theory of asymptotics using the framework of decision theory is much tougher than most mathematicians think nowadays if they ever read into [Le Cam]'s exposition. This kind of difficulty mostly arise from two sources, from my perspective. The first one is the typical difficulty in determination of loss function, convexity is convenient yet concavity is a more faithful description of boundary behavior (when sequence of rules tends to $\infty$, in most cases we consider the rules indexed by sample sizes $n$.), this flaws are very carefully(probably over-carefully discussed in pp.16-23 of [Le Cam]). The second one is closer to what you think in OP, that is the inconsistency between local and global asymptotic behaviors, which is also discussed in Chap.10-11 of [Le Cam]. This problem is becoming more and more central concern in recent research, you may also want to read my post Is there a result that provides the bootstrap is valid if and only if the statistic is smooth?, which explained why sometimes seemingly unnecessary "linearity conditions" will be imposed on some asymptotic results. -Although I do not quite agree with your opinion, this question is actually quite interesting, therefore I want to focus on a more specific version of the problem you raise, and then return to your question that "What are some other surprising results of finite sample statistics?" We firstly considered the small sample approximation, and then a correction and discussion about your observation: the different between small/finite sample behavior and large/asymptotic behavior. -Thank you for sharing your thoughts! -$\blacksquare$1.The introduction of small sample asymptotics. -If you are concerned with the asymptotic behavior of finite sample, a relevant field of research that is called "small sample asymptotics" [Field&Ronchetti]. This field has at least two motivations, one is mentioned in [Field&Ronchetti] Preface: - -"...The central question is that of finding good approximations to the - density of statistics in situations where the computation of the exact - density is intractable.... The small sample approximations are simpler - than Edgeworth expansion and can be thought of as a series of low - order Edgeworth expansions. We obtain much better numerical accuracy - than the Edgeworth and our density approximations are always positive - unlike the Edgeworth. " - -The other is actually the demanding need of studying the robustness of certain estimators(M-estimators). When an accurate approximation is available, we do not have to worry about multiple Hadamard differentiation when studying the influence of certain observations. Huber and Ronchetti even updated this contribution into Huber's epic work [Huber] when revised. -Both motivations are kind of vanishing a bit nowadays, the numerical concern is not so central as computation resource is growing; the study of robustness is actually not so popular, mostly due to the rise of MCMC about the same time after Huber's Princeton Robust seminar. -$\blacksquare$2.About random matrix asymptotics. -You said in OP that: - -Many results on random matrices have very similar results in the - finite sample and asymptotic regimes. For example, bounds on the - minimum and maximum eigenvalues of random matrices are comparable in - both regimes. - -This is completely not true. If you view random matrices as sort of representation of the transition kernel topological group, or equivalently dependent stochastic processes, then there are so many discrepancies between asymptotic bound and concentration bound mentioned in [Talagrand]. Things like Berry–Esseen bound(since you talked about empirical mean) is more like a coincidence rather than a universal fact. And the concentration bound is usually looser than asymptotic counterparts, and come with more restrictions. -What [Devroye] showed is a concentration bound (a bound that holds with high probability). This is not uncommon in mathematics, say subharmonic functions sequence do not always converge to subharmonic functions, how can we expect that a subharmonic estimator(empirical mean) sequence converges to a harmonic limit(Gaussian when taking $n\rightarrow \infty$)? -More obviously for finite sample you always have union bound, how about that asymptotically? -$\blacksquare$3.What are some other surprising results of finite sample statistics? -There are quite a few examples that concentration bound disagrees with asymptotic bound as I mentioned above shown in [Talagrand] and [Le Cam]Chap 10-11. But the most famous example that fits your description should be the order statistics from extreme value theory, its asymptotic distribution is exponential (Limiting distribution of the first order statistic of a general distribution) while the distribution of finite sample order statistics $(X_{(1)},\cdots,X_{(n)})$ always depends on the underlying distribution itself. Not surprisingly the Efron-Stein concentration bound is much looser in this case. So what I said is that such disagreement between concentration bound and asymptotic bound is rather common. -Reference -[Le Cam]Le Cam, Lucien. Asymptotic methods in statistical decision theory. Springer Science & Business Media, 2012. -[Field&Ronchetti]Field, Christopher A., and Elvezio Ronchetti. "Small sample asymptotics." Ims, 1990. -[Huber&Ronchetti]Huber, Peter J. Robust statistics. Springer Berlin Heidelberg, 2011. -[Devroye]Devroye, Luc, et al. "Sub-Gaussian mean estimators." The Annals of Statistics 44.6 (2016): 2695-2725. -[Talagrand]Talagrand, Michel. The generic chaining: upper and lower bounds of stochastic processes. Springer Science & Business Media, 2006. -[Hayman2]Hayman, Walter Kurt. Subharmonic functions. Vol. 2. Elsevier, 2014.<|endoftext|> -TITLE: A conjecture about certain values of the Fabius function -QUESTION [10 upvotes]: The Fabius function is a smooth monotone function $F:[0,1]\to[0,1]$, satisfying functional equations -$$F(0)=0, \quad F(1-x)=1-F(x)\tag1$$ -and -$$F'(x) = 2 \,F(2 x) \quad \text{for} \,\, 0 -TITLE: Take a matrix with normalized rows; no column of the inverse has too large a norm -QUESTION [8 upvotes]: For $n>2$, let $X$ be an $n$-by-$n$ invertible matrix where every row has Euclidean norm $1$. Let $Y=X^{-1}$. Let $\Vert y_i \Vert$ be the Euclidean norm of column $i$ of $Y$. -The following conjecture seems to be confirmed by numerical evidence: for every $i$ $$ \frac{\Vert y_i \Vert}{\sum_{j=1}^n \Vert y_j \Vert} < \frac{1}{2}$$ -How can we prove this? -Letting $z_i$ denote row $i$ of $X$, we know that $z_i \cdot y_i = 1$, so by Cauchy-Schwarz, $\Vert z_i \Vert \cdot \Vert y_i \Vert \geq 1$, implying that each column of $Y$ has norm at least $1$. - -REPLY [4 votes]: Indeed, if you think of it, it can be restated in geometric terms. What we need to prove is that the inverse altitudes of a parallelepiped spanned by $n+1$ unit vectors in $\mathbb R^{n+1}$ satisfy the "triangle inequality" (each inverse altitude does not exceed the sum of the rest). Let $v$ be one of the given unit vectors and $v_j$ be the rest. If $h$ is the altitude vector for $v$, then $v-r=h$ where $r=\sum_j a_j v_j$ and $r$ is orthogonal to $h$. Notice that we can use the same linear combination $v-r$ divided by $a_j$ to estimate the altitude $H_j$ for the vector $v_j$ from above by $\|h\|/|a_j|$ and that we can improve that estimate to $\|h-\langle h,v\rangle v\|/|a_j|$ if we modify the coefficient at $v$. Thus it would suffice to show that -$$ -\frac{\|h-\langle h,v\rangle v\|}{\|h\|}\le \sum_j|a_j|\,. -$$ -However the left hand side is the sine of the angle between $v$ and $h$ in the right triangle with sides $v,h,r$, so it equals $\|r\|$. It remains to combine the definition of $r$ with the triangle inequality. -Certainly this all should be very well known since 18.., so if someone can carry out the corresponding literature search and point out the reference, it would be great :-).<|endoftext|> -TITLE: What's the point of a Whittaker model? -QUESTION [15 upvotes]: Let $G$ be a quasi-split connected reductive group over a $p$-adic field $F$. Let $B$ be a Borel subgroup which is defined over $F$, with $B = TU$, $T$ defined over $F$. The choice of $T$ and $B$ gives a set of nonrestricted roots $\tilde{\Delta}$, which together with an $F$-splitting and a nontrivial unitary character $F \rightarrow S^1$ yields a unitary character $\chi$ of $U(F)$. -Let $V$ be a smooth, irreducible, admissible representation of $G(F)$. A linear functional $\lambda: V \rightarrow \mathbb{C}$ is called a Whittaker functional for $\chi$ if for all $u \in U(F)$ and $v \in V$, we have -$$\lambda(u \cdot v) = \chi(u) \lambda(v)$$ -It is a theorem that the space of Whittaker functionals for $\chi$ is at most one dimensional. If this dimension is one, then $V$ is called $\chi$-generic. -Fix a nonzero Whittaker functional $\lambda$, and for $v \in V$, define a function $W_v :G(F) \rightarrow \mathbb{C}$ by $W_v(g) = \lambda(g \cdot v)$. The set $W = W_{\lambda}$ of such functions is closed under addition and scalar multiplication, and becomes a representation of $G(F)$ if we set $g \cdot W_v = W_{g\cdot v}$. Then the representation $W$ is called a Whittaker model, and up to $G(F)$-isomorphism it does not depend on the choice of $\lambda$, since these things are all scalar multiples of each other. -From what I can see, $V \rightarrow W_{\lambda}, v \mapsto W_v$, is an isomorphism of $G(F)$-modules. It is clearly surjective, and for injectivity, if we suppose that $0 \neq v \in V$, and $W_v(g) = 0$ for all $g \in G(F)$, then $\lambda(g \cdot v) = 0$ for all $g \in G(F)$. This is impossible, because $g \cdot v : g \in G(F)$ spans $V$, because $V$ is irreducible. -So my question is, what is the point of defining Whittaker models if the thing we defined is just isomorphic to our original representation? Why would it be important to regard $V$ as a set of functions $G(F) \rightarrow \mathbb{C}$? - -REPLY [10 votes]: In addition to the other good answer, in the context of automorphic forms the uniqueness of local Whittaker models is specifically useful in computations of global integrals: when/if the integral can be expressed in terms of the global Fourier-Whittaker expansion of an automorphic form, then the local uniqueness expresses the thing as a product of local integrals, giving an Euler product, hence, most likely, an L-function of some kind. This is what happens in the adele-group rewrite of Mellin transforms of cuspforms, for example, as well as the adele-group rewrite of the classical Rankin-Selberg integral representation of $L(s,f\times g)$ involving Eisenstein series. These generalize to the Hecke-type integral for L-functions attached to cuspforms on $GL_n\times GL_{n-1}$, as well as the Rankin-Selberg-type integral for $GL_n\times GL_n$.<|endoftext|> -TITLE: Is there literature on finite geometries with ordered lines? -QUESTION [5 upvotes]: A difference between finite geometries and (e.g.) Euclidean space is that "lines" in finite geometries are unordered subsets of the universe, while "lines" in Euclidean space are ordered subsets of the universe, at least insofar as they have an implicit betweenness relation attached. For example, in Euclidean space, we can meaningfully say that point $a$ is between points $b$ and $c$ on line $L$, but this statement makes no sense in a finite projective plane. -Have finite geometries with lines ordered in this way been studied? Are any interesting results gained from this additional structure? And if so, are there any good resources for learning about this? - -REPLY [5 votes]: Yes, this has been studied and is indeed known as ordered geometry or the study of betweenness spaces: -https://en.m.wikipedia.org/wiki/Ordered_geometry<|endoftext|> -TITLE: What do the differential k-forms on a product manifold look like? -QUESTION [8 upvotes]: I am interested in how I could express $\Omega^k( M \times N)$ in terms of $\Omega^i(M)$ and $\Omega^j(N)$ for $i,j = 0,1, \ldots k$. Is there a nice relation? -This question arose in the context of the Freund-Rubin solution to the bosonic equations of motion for 11-dimensional supergravity, where one posits a space-time geometry $AdS_4 \times S^7$ and a 4-form field strength proportional to the volume form on $AdS_4$, considered as an element of $\Omega^4(AdS_4 \times S^7)$. In verifying that such a geometry/field strenth do indeed satisfy the equations of motion, it is necessary to consider its Hodge dual in $\Omega^7(AdS_4 \times S^7)$. I suspect that this is simply the volume form on $S^7$, but can't show this. -I also suspect that the answer to my more general question concerning $\Omega^k( M \times N)$ is not necessary to determine the special case in the second paragraph, but I am quite curious. - -REPLY [13 votes]: Denote by $p_M: M \times N \longrightarrow M$ and $p_N: M \times N \longrightarrow N$ the canonical projections. Then you get an induced bilinear map from $\Omega^i(M) \times \Omega^j(N) \longrightarrow \Omega^{i+j}(M \times N)$ by taking the $\wedge$-product of the pullbacks with $p_M$ and $p_N$. -Hence you have a linear map -\begin{equation} -\Omega^i(M) \otimes \Omega^j(M) \longrightarrow \Omega^{i+j}(M \times N) -\end{equation} -which turns out to be injective. Taking now all $i$ and $j$ with $k = i+j$ gives you already a big part of the differential forms but not all of them (take a look at $k = 0$ where this is not yet surjective). Nevertheless, the image of this map is dense in the natural Frechét topology of differential forms. This gives a bijection of $\Omega^k (M \times N)$ with the sum of the completed projective tensor product of the individual differential forms of the correct degrees.<|endoftext|> -TITLE: Algebraic Geometry in Number Theory -QUESTION [15 upvotes]: It appears to me that there are two main ways by which algebraic geometry is applied to number theory. The first is by studying polynomials over fields of number-theoretic interest (which does not seem to be limited to number fields). Diophantine geometry is part of this circle of ideas, as well as the use of elliptic curves to obtain Galois representations, complex multiplication, the Langlands program, and so on. -The second is to study a number ring as a scheme. Some of the work in this area include the work of Voevodsky proving the Milnor conjecture and the Bloch-Kato conjecture, as well as the duality theorems of Artin and Verdier. -My questions are: - -Are the two connected in some way? (Aside from the fact that they are both applications of algebraic geometry of course - perhaps on a related note, I am also thinking of Grothendieck's Dessins d'Enfants and how it would fit into the big picture.) -What are some modern results on the study of number rings as schemes, aside from the ones mentioned above? What are some good recommendations for literature on this subject? (I have read Algebraic Number Theory by Jurgen Neukirch, which introduces this subject very nicely and develops Riemann-Roch theory for number rings, but I am looking for more, especially more modern work.) -The etale cohomology of number fields is the same as Galois cohomology, which is the subject of the Milnor conjecture and the Bloch-Kato conjecture. There seems to be more literature on the algebraic geometry (particularly the cohomology) of number fields as compared to number rings (Artin-Verdier duality is one example of a result on the cohomology of number rings). Why is this so? - -(I apologize if the title is vague. I find it hard to phrase it in such a way that reflects the nature of the question.) - -REPLY [2 votes]: You could read Milne, Arithmetic Duality Theorems http://jmilne.org/math/Books/ADTnot.pdf and Neukirch-Schmidt-Wingberg, Cohomology of Number Fields http://vg02.met.vgwort.de/na/a877a4fbfcec4aad9721c766bc577bb3?l=http://www.mathi.uni-heidelberg.de/%7Eschmidt/NSW2e/NSW2.2.pdf<|endoftext|> -TITLE: Rigorous introductions to actuarial mathematics -QUESTION [5 upvotes]: Maybe this is not the kind of question for this website, but nevertheless I believe it could be interesting for a large audience. -I am interested to know if there is some book where the subject of actuarial mathematics is treated in a rigorous "mathematical" way, somehow suitable for graduate math students or experienced mathematicians who are interested in this subject for research reasons. -Since I am new in this field, I am interested in a kind of reference where the subject is treated from the basics to advanced topics (of course with a backgroud in probability theory or stochastic process given for granted). -Thank you - -REPLY [2 votes]: Here are some references, where it is made precise that the rigorous mathematical framework leading to the theory is explained. -1)Term-Structure Models , a graduate course , - by Damitir Filipovic (Springer 2009). -2)Risk Analysis in Finance and Insurance (2nd edition), - by Alexander Melkinov (Chapman & Hall/CRC, 2011). -3)Market-Consistent Actuarial Valuation, - by M.V. Wüthrich, H. Bülmann & H. Furrer (Springer 2008). -4) (in French !) : - Mathématique et Assurance (first elements) - by D. Pierre-Loti-Viaud & P. Boulongne (Ellipses 2014).<|endoftext|> -TITLE: Is the bordism from disjoint union to connected sum universal for connected manifolds? -QUESTION [7 upvotes]: Let $M_1$ and $M_2$ be two oriented, connected, closed $n$-manifolds. It is known that the disjoint union $M_1 \sqcup M_2$ and the connected sum $M_1 \# M_2$ are cobordant, via a bordism $\Sigma_{M_1, M_2}$. -I'm wondering whether this bordism universal, in the following sense: - -Conjecture. Let $N$ be any oriented, connected, closed $n$-manifold, and $\Sigma$ a bordism from $M_1 \sqcup M_2$ to $N$. Then there exists an essentially unique bordism $\Sigma'$ from $M_1 \# M_2$ to $N$ such that $\Sigma$ is the gluing of $\Sigma_{M_1, M_2}$ an $\Sigma'$. - -Is this true? -I have a vague feeling that there must be an argument along the lines of shifting around certain critical points of a Morse function, and $\Sigma_{M_1, M_2}$ corresponding to an extremal critical point, but I couldn't find a precise solution. -Edit: $\Sigma$ is required to be connected as well. - -REPLY [3 votes]: TLDR: existence yes, uniqueness no. -Arun has already discussed existence, so let me discuss the uniqueness part of the question. It is a special case of whether cancellation holds in bordism categories, which it usually doesn't except under rather strong assumptions. -The uniqueness part of the question is equivalent to whether all embeddings $\Sigma_{M_1,M_2} \hookrightarrow \Sigma$ relative to $M_1 \amalg M_2$ are in the same orbit under the action of the group of diffeomorphisms of $\Sigma$ restricting to the identity on $\partial \Sigma$. (One direction: given two such embeddings, their complements would be two candidates for $\Sigma'$ and any diffeomorphism rel boundary between these two complements may be glued to the identity map of $\Sigma_{M_1,M_2}$ to get a diffeomorphism of $\Sigma$ which acts one embedding into the other. The converse is similar.) In fact we may construct an example of two embeddings $e, e': \Sigma_{M_1,M_2} \to \Sigma$ where it is not even possible to find a continuous map $f: \Sigma \to \Sigma$ which is the identity on $\partial \Sigma$ and such that $f \circ e \simeq e'$ relative to $\partial \Sigma$. -To build a counterexample we use that for any group $G$, the set of $G \times G^\mathrm{op}$-equivariant maps $G \to G$ (action by multiplying on both sides) is precisely the maps given by multiplication by elements of the center of $G$. Hence if $Z(G) = \{1\}$ it consists of only the identity. This implies the following more complicated looking (but "isomorphic") statement: if $C$ is a groupoid and $m_1$ and $m_2$ are two isomorphic objects in $C$ such that the group $\mathrm{End}_C(m_1)$ has trivial center, then any functor $F: C \to C$ with $F(m_1) = m_1$ and $f(m_2) = m_2$, and $F(f) = f$ for any $f \in \mathrm{End}_C(m_1)$ and any $f \in \mathrm{End}_C(m_2)$, also has $F(f) = f$ for any $f \in \mathrm{Hom}_C(m_1,m_2)$. -Now pick a finitely presented non-trivial group $G$ with trivial center and build a connected oriented cobordism $W$ from $M_1 \amalg M_2$ to any connected $N$, such that there exists an isomorphism $\pi_1(W,w) \cong G$ for some, hence any, $w \in W$, and such that both inclusions $M_1 \to W$ and $M_2 \to W$ induce isomorphisms in fundamental groups. For example, we could first pick a closed connected oriented manifold $M$ with the right fundamental group (this is possible in any dimension $d \geq 4$), then set $M_2 = M_1 = M$ (with opposite orientations), and let $\Sigma$ be the complement of a ball in the interior of $[0,1] \times M$, regarded as a bordism from $M_1 \amalg M_2$ to $N = S^{d-1}$. -Then pick points $m_1 \in M_1$ and $m_2 \in M_2$, which we can think of as objects in the fundamental groupoid $\pi_1(\Sigma)$. Connectedness implies that these objects are isomorphic, and we have arranged that the group of endomorphisms of $m_1$ has trivial center. By what we saw above, any functor $F: \pi_1(\Sigma) \to \pi_1(\Sigma)$ which sends both objects $m_1$ and $m_2$ to themselves and act as the identity on their endomorphism groups, must also act as the identity on the set of isomorphisms from $m_1$ to $m_2$. -The bordism $\Sigma_{M_1,M_2}$ may be constructed by attaching a 1-handle to $M_1 \amalg M_2$ at the points $m_1$ and $m_2$. Then any embedding $\Sigma_{M_1,M_2} \to \Sigma$ gives rise to an isomorphism (namely the path through the attached 1-handle) from $m_1$ to $m_2$ in the fundamental groupoid of $\Sigma$, and any element of this Hom-set arises in this way (since any path may be perturbed to a smooth embedding, which by the orientability assumptions may be thickened up to an embedding of the handle). Then pick two embeddings $e, e': \Sigma_{M_1,M_2} \to \Sigma$ representing distinct elements in the Hom-set. Any diffeomorphism of $\Sigma$ relative to $M_1 \amalg M_2$ which acts $e$ into $e'$ would induce a functor $\pi_1(\Sigma) \to \pi_1(\Sigma)$ acting as the identity on the endomorphisms of $m_1$ and $m_2$ (since these endomorphism groups may be calculated in $M_1$ and $M_2$) but not on the set of morphisms between them. But that's precisely what we ruled out! Hence such a diffeomorphism cannot exist.<|endoftext|> -TITLE: Shape that balances isoperimetric ratio with invisibility -QUESTION [5 upvotes]: I seek a two-dimensional shapes $S$, bounded by a Jordan curve, -that optimally balances its isoperimetric ratio $r(S)$ -against what I call its invisibility index $iv(S)$. -Define the isoperimetric ratio $r(S)$ of $S$ to be -$4 \pi A / L^2$, where $A$ is the area of $S$ and $L$ its -perimeter. This ratio is in $(0,1]$ -and achieves $1$ for $S$ a disk. -See, e.g., the -Wikipedia article on the isoperimetric inequality. -Define invisibility index $iv(S)$ -to be the probability that that a pair $(x,y)$ of random points -in $S$ -(chosen uniformly and independently) -are invisible to one another in the sense that -the segment $xy$ includes a point strictly exterior to $S$. -($iv(S)$ is $1$ minus the -Beer convexity of $S$.) - -Q. What shape (or shapes) $S$ maximize the product - $P(S) = r(S) \cdot iv(S)$? - -If $S$ is a disk, $r(S)=1$ and $iv(S)=0$ so $P(S)=0$. -If $S$ is a thin spiral, then $r(S)$ approaches $0$ -and $iv(S)$ approaches $1$ so $P(S)$ approaches $0$. -In between, $P(S) > 0$. -I've computed $P(S)$ for the very narrow class of -symmetric Ls, unit squares with a square removed -from one corner, as illustrated below. - -          - - -          - -Two symmetric Ls with different parameters $a$. Origin at lowerleft corner. - - -These shapes are determined by one parameter $a$ as illustrated. -Among this class of shapes, it appears that the maximum -product $P(S)$ is achieved when $a \approx \frac{1}{4}$, -the left shape above. Plots $r(\,)$, $iv(\,)$, and $P(\,)$ -are shown below. -The isoperimetric ratio for a square is $r(1) = \pi/4 \approx 0.79$. - -          - - -          - -Red: $r(a)$. Blue: $iv(a)$. Green: Product $P(a)$. - - -Update. Seems like Gerhard Paseman's figure-8 -, with $r=\frac{1}{2}$, $iv=\frac{1}{2}$, $P=\frac{1}{4}$, -is the extreme shape. (In comments I mistakenly said $iv=\frac{1}{4}$.) - -REPLY [2 votes]: A Pacman is a good candidate. -Take a unit circle and cut out a sector with angular width $\alpha$. Then: -\begin{align} -A &= \pi - \frac{\alpha}{2}\\ -L &= 2\pi - \alpha+2\\ -iv &= \frac{(\pi-\alpha)^2}{(2\pi-\alpha)^2}\\ -P &= \frac{2\pi(\pi-\alpha)^2}{(2\pi-\alpha+2)^2(2\pi-\alpha)} -\end{align} -So $P$ approaches $\pi^2/(2\pi+2)^2$ or $0.144$ near $\alpha=0$. This is a Pacman whose mouth is nearly closed.<|endoftext|> -TITLE: Coherent subsheaf of co-admissible modules of Schneider and Teitelbaum -QUESTION [5 upvotes]: Let $M$ be a co-admissible module over a Frechet Stein Algebra $A=\varprojlim A_{q_n}$ as in this paper. Let $N$ be a closed submodule of $M$. I have some difficulty in understanding lemma $3.6$ of the above paper. That is I want to show that $N$ is co-admissible. -For that, consider $M=\varprojlim M_n$. The authors considers $N_n \subset M_n$, the $A_{q_n}$-submodule generated by the images of $N$ under the natural map $M \rightarrow M_n$. Then the authors say that $(N_n)_n$ is a coherent subsheaf of $(M_n)_n$, which is what I have some problem in understanding. -I am having difficulty in showing that $A_{q_{n}} \otimes_{A_{q_{n+1}}} N_{n+1}=N_n$. (I know that it should come from the equality $A_{q_{n}} \otimes_{A_{q_{n+1}}} M_{n+1}=M_n$ which is true by definition.) But somehow I cannot deduce it. Thank you for your help. - -REPLY [5 votes]: The map $A_{q_n} \otimes_{A_{q_{n+1}}} N_{n+1} \rightarrow N_n$ is surjective, by definition of $N_n$. To show that it is injective, it suffices to show that the composition $A_{q_n} \otimes_{A_{q_{n+1}}} N_{n+1} \rightarrow N_n \rightarrow M_n$ is injective. But this composition factors as -$$ -A_{q_n} \otimes_{A_{q_{n+1}}} N_{n+1} \rightarrow A_{q_n} \otimes_{A_{q_{n+1}}} M_{n+1} \rightarrow M_n. -$$ -The first arrow is injective because $A_{q_{n}}$ is a flat $A_{q_{n+1}}$-module, and the second arrow is an isomorphism.<|endoftext|> -TITLE: Further aspects of a Hankel matrix of involution numbers -QUESTION [11 upvotes]: We have two conjectured generalizations of the question asked at -a Hankel matrix of involution numbers -by Tewodros Amdeberhan. Let $n!!=1!\,2!\cdots n!$. -Conjecture 1. Let $I_k$ denote the number of involutions in the - symmetric group $\mathfrak{S}_k$. Then the Smith normal form of the - matrix $[I_{i+j}]_{i,j=0}^n$ has diagonal entries $0!, 1!, \dots, - n!$, -Conjecture 2. Let $J_k=k!\sum_{\lambda\vdash k}s_\lambda$, where - $s_\lambda$ is a Schur function. Set $J_0=1$. When the symmetric - function $\det[J_{i+j}]_{i,j=0}^n$ is expanded in terms of power - sums, then every coefficient is an integer divisible by - $n!!$. (Tewodros' question is equivalent to the coefficient of - $p_1^{n(n+1)}$ being equal to $n!!$.) -Is there a nice formula or combinatorial interpretation of the -coefficients in Conjecture 2? - -REPLY [3 votes]: On Conjecture 1: As remarked by Johann Cigler in the linked question (and shown in the references linked there) the matrix $M:=\left[I_{i+j}\right]_{i\ge0\atop j\ge0}$ diagonalises as $M=U^TDU$ with $D:=\operatorname{diag(k!)}$, and $U$ an upper triangular integer coefficients matrix with unit diagonal elements, hence with integer coefficient inverse: so this is also the SNF of $M$. -For example: -$$ \left[ \begin {array}{ccccc} 1&1&2&4&10\\ 1&2&4&10& -26\\ 2&4&10&26&76\\ 4&10&26&76&232 -\\ 10&26&76&232&764\end {array} \right] =$$$$= \left[ \begin {array}{ccccc} 1&&&&\\ 1&1&&& -\\ 2&2&1&&\\ 4&6&3&1& -\\ 10&16&12&4&1\end {array} \right] \left[ \begin {array}{ccccc} 1& & & & \\ &1& & & -\\ & &2& & \\ & & &6& -\\ & & & &24\end {array} \right] \left[ \begin {array}{ccccc} 1&1&2&4&10\\ &1&2&6& -16\\ & &1&3&12\\ & & &1&4 -\\ & & & &1\end {array} \right] -$$<|endoftext|> -TITLE: Does $q$-Catalan number count subspaces? -QUESTION [11 upvotes]: Consider the $n$-element subsets $\{a_1 -TITLE: Sheaf associated to presheaf Aut -QUESTION [11 upvotes]: Let $S$ be a scheme and let $C$ be the category of schemes flat and locally of finite presentation over $S$. Endow $C$ with the fppf topology (or perhaps any subcanonical topology). Let $\mathcal P$ be a presheaf of sets on $C$ (i.e., an object of $\widehat{C}$) and let $a$ be the associated sheaf functor. Now consider the presheaf $\underline{\rm Aut}(\mathcal P)$ given by $\underline{\rm Aut}(\mathcal P)(T)={\rm Aut}(\mathcal P\times h_{T})$ for all $T\to S$ in $C$, where $h\,\colon C\to \widehat{C}$ is the usual functor (note that $h_{T}$ is, in fact, a sheaf). - -Question. Is it true that there exists a canonical isomorphism of sheaves - $$ -a(\underline{\rm Aut}(\mathcal P))=\underline{\rm Aut}(a(\mathcal P))? -$$ - -This seems plausible, but I haven't been able to find a proof, so either I'm getting old :-) or my guess is false. -Note that if $\mathcal P$ is a sheaf, then my guess is true as it is well-known [SGA 3, IV, Corollary 4.5.13] that, in this case, the presheaf $\underline{\rm Aut}(\mathcal P)$ is, in fact, a sheaf. - -REPLY [3 votes]: A small disclaimer: My knowledge of algebraic geometry is relatively basic so I will not discuss anything related to scheme directly. But it seems that most of what you are asking has very little to do with algebraic geometry so I will answer your question from a purely topos theoretic perspective. -Moreover, when I say geometric morphism, I talking about the inverse image functor (something is preserved by geometric morphism if it is preserved by the inverse image functor of the geometric morphism) -First your definition of $Aut(P)$ does not make sense, because "aut" is not a functor, if you have a morphism from $P'$ and $P$ and an automorphism of $P$ or $P'$ you have no way to transport it into an automorphism of the other. -Here is how one can define the sheaf or presheaf of automorphism of an object: -Definition of Aut(X), and (non) compatibility to geometric morphism : -First any topos $T$ (the category of sheaves over some site) is cartesian closed, i.e. if $X$ and $Y$ are object of $T$ there is an object $Hom(X,Y)$ characterized by the fact for all object $V$ the morphism from $V$ to $Hom(X,Y)$ are exactly the morphism from $V \times X$ to $Y$. -Note that if you are working within a topos of sheaves, this gives an explicit description of $Hom(X,Y)$ as a sheaf: for any object $c$ of the site $Hom(X,Y)(c)$ is the set of morphism from $X \times \tilde{c}$ to $Y$ where $\tilde{c}$ is the representable sheaf associated to $c$ (the sheafification of the presheaf represented by $c$). -In particular, one gets an object $End(X)$ with the universal property that morphism from $V$ to $End(X)$ is the same as a morphism from $V \times X$ to $X$. One can then define the object $Aut(X)$ which will be the subobject of $End(X)$ of invertible endomorphism, this is done using internal logic and after a small computation on obtain that $Aut(X)$ can be described as follow: a morphism from $V$ to $Aut(X)$ is the same as an invertible morphism from $V \times X$ to $V \times X$ which is compatible to the projection to $V$ (i.e. $p \circ f = p)$, here the functoriality is given as follow, if $W \rightarrow V$ is a morphism then $X \times W$ is the pullback of $X \times V$ along the map $W \rightarrow V$, hence any automorphism of $X \times V$ over $V$ can be pulledback into an automorphism of $X \times W$ over $W$. -Now it is a general fact that this construction of $Aut(X)$ is in general not compatible with pullback along geometric morphism or sheafification functor. Hence in general it will not be the case $Aut(a(P)) = a(Aut(P))$ it might be the case for some very special site, but I see no reason for it to be true for the site you are using. -Very informally, the reason this is not the case is because sheafification and geometric morphism only preserve finite product and not infinite product. And $End(X)$ is morally $X^X$ so you unless you have some finiteness condition on $X$ (typically cardinal finite), $End(X)$ will not be preserved by geometric morphism. In your second question you are concerned with $G$ automorphism of an object which is a "finitely generated" $G$ object and this is why it is going to work. -The second case discussed in the comment: -Sill in a general topos, if $G$ is a group object and $H$ is a subgroup, you can always define a quotient $G/H$ which is a $G$-object with the expected universal property: if $X$ is any other $G$-object a morphism of $G$-object from $G/H$ to $X$ is the same as a morphism from $G$ to $X$ which is constant on $H \subset G$. -The existence of $G/H$ can be deduced either from internal logic or just from the co-completness of every topos, but this kind of universal property does not produce a description of $G/H$ as a sheaf anymore (because we don't know what are morphisms from an object $V$ to $G/H$, only morphism out of $G/H$). But one can prove that the construction of $G/H$ is compatible to sheafification and more general geometric morphism, meaning that $f^*(G/H)=f^*(G)/f^*(H)$ which implies that: --the sheaf $G/H$ is the sheafification of the presehaf $G/H$. --the presheaf $G/H$ is the object wise quotient $G(c)/H(c)$ because $X \mapsto X(c)$ is the inverse image functor of a geometric morphism $Set \rightarrow prsh(C)$. -Still in the same situtation $H \subset G$ one can also construct a "normalizer" $N_G(H)$ (also using internal logic) which is a subobject of $G$ satisfying the universal property: -A morphism from $V$ to $N_G(H)$ is the same as a morphism $f : V \rightarrow G$ such that the map from $H \times V$ to $G$ defined by $(h,v) \mapsto f(v) h f(v)^{-1}$ factor into $H$. -i.e. it is just the internal interpretation of $\{ g \in G | \forall h \in H, g h g^{-1} \in H \}$. -This mean that, as for $Aut(X)$ and $End(X)$ above you have a description of $N_G(H)$ as a sheaf (and if $G$ and $H$ are sheave then $N_G(H)$ is indeed a sheaf). -one easily check that $N_G(H)$ is a group object (a subgroup of $G$ ) containing $H$. -Now in any topos, you have the following results: -$$ Aut_G(G/H) \simeq N_G(H)/H $$ -where $Aut_G(X)$ denotes the sub-object of $Aut(X)$ of morphism compatible to the $G$ action, i.e. a morphism from $V$ to $Aut_G(X)$ is an automorphism of $X \times V$ compatible to the projection on the second component and to the $G$ action on the first component. -The isomorphism is implemented by the natural action of $N_G(H)$ on $G/H$ induced by the action of $G$ and $G$. The proof of the isomorphism is very easy when one knows internal logic: the usual set theoretic proof of this isomorphism is fully constructive hence it is true in any topos. -This result is true both in the presheaf topos and in the sheaf topos, but the right hand side is known to be compatible with sheafification and geometric morphism (in the sense that $N_G(H)/H$ in the sheaf topos is the sheafification of $N_G(H)/H$ computed in the presheaf topos). -This answer your question in the comment. -The $aut_G(X)$ you are talking about is $Aut_G(G/H)$ in the sheaf topos and $a(N_G(H)/H )$ is the sheafification of $N_G(H)/H$ in the presheaf topos and hence it is $N_G(H)/H$ in the sheaf topos, and the isomorphism you are aksing is exactly the one I mentioned above.<|endoftext|> -TITLE: Existence of $L^\infty$ function on $\mathbb{T}$ whose Fourier series is $\ell^2$ but no better? -QUESTION [5 upvotes]: I'm sure that this is classical--but can anyone provide a reasonable example of an $L^\infty(\mathbb{T})$ function whose Fourier series is $\ell^2$ but no better? Not even $L^2\log L$? Presumably one exists but nothing has come to my mind. I'm trying to understand just how far one can stretch a (version of a) particular conjecture on Boolean functions, the Fourier Entropy-Influence conjecture. - -REPLY [4 votes]: A theorem due to Kahane, Katznelson and de Leeuw says that for any sequence $(a_n)_{n\in \mathbb{Z}}$ belonging to $\ell_2$ there is a continuous function $f \in C(\mathbb{T})$ such that $|\widehat{f}(n)| \geqslant |a_n|$, so in general you don't get anything better than $\ell_2$. The proof can be found in Appendix B of Katznelson's book -- the construction is somewhat explicit.<|endoftext|> -TITLE: How are material set theory and structural set theory related from the point of view of category theory? -QUESTION [9 upvotes]: In his comments to both cody and Nik Weaver regarding his answer to user7280899's mathoverflow question "What kind of foundation are mathematicians using when proving metatheorems?", Mike Shulman writes: - -I care more about not relative consistency, but "interpretability": that you can directly construct a model of one theory from the other theory in a relatively straightforward way. This tells us how to actually relate (at least in principle) mathematics formalized in the two theories. If someone proves a theorem in $ZFC$, then I understand directly what it means in $SEAR$ [Sets, Elements, And Relations] as a theorem about hereditarily well-founded relations, and vice versa: a theorem in $SEAR$ is a theorem in $ZF$ about the category of sets...Put differently, consider the category of models of $ZF$ and the category of models of $SEAR$. To say $Con(ZF)$ $\Leftrightarrow$ $Con(SEAR)$ is to say that if one of these categories has an object, so does the other. I would much rather know that these categories are related by an adjunction or an equivalence, even if I need a stronger metatheory. - -Which brings me to my question: - -How are material set theory and structural set theory related from the point of view of category theory? More specifically, how is the category of models of $ZF$ related to the category of models of $SEAR$ from the point of view of category theory? And what is the significance of this relation to the foundations of mathematics (since Set Theory and Category Theory have been considered (per se) to be alternate foundations for mathematics)? - -REPLY [8 votes]: I think this question has not been satisfactorily studied, but we can say something. Firstly, note that since a model of SEAR (or other structural set theory) is really a category, such models form not a 1-category but a 2-category. -It's easy to see that the two model constructions act functorially on isomorphisms of models of ZF and on equivalences of models of SEAR. Moreover, any model of ZF can be reconstructed (up to isomorphism) from its category of sets (since by the axiom of foundation and Mostowski's collapsing lemma, transitive sets are the same as well-founded extensional relations), and any model of SEAR+choice can be reconstructed (up to equivalence) from its ZF-model (since by AC, any set can be well-ordered, which is a well-founded extensional structure). So the sets of isomorphism classes of models of ZFC and equivalence classes of models of SEARC are isomorphic. -I don't know to what extent these isomorphisms are natural or 2-natural even on equivalences. (Since well-founded extensional structures admit no nonidentity automorphisms, if they were 2-natural then the 2-groupoid of models of SEARC would in fact be a 1-groupoid; I don't even know whether that's likely to be true.) Moreover, it's not as clear what to do about non-invertible morphisms. The only reference I'm aware of that really discusses this question is Mitchell's Categories of Boolean topoi from 1973, in which he comes close to constructing some kind of adjunction, although he didn't use 2-categorical language.<|endoftext|> -TITLE: Sums of sets in ${[n] \choose k}$ -QUESTION [9 upvotes]: Suppose we look at all sets in ${[n] \choose k}$, for some $k \leq n/2$, and place them in layers according to the sum of their elements. Then, we say that two sets $A$ and $B$ from consecutive layers are connected by an edge, if there is some $i \in [n-1]$ for which $(i,i+1)A = B$. That is, if $A \Delta B = \{i,i+1\}$. Is it true that any two consecutive layers have a perfect matching, in the sense that all the elements of the layer with less sets are matched? For instance, let's take $n=6$, $k=3$. And let's match the layer with sum $12$ to the layer with sum $11$: -$(\{6,5,1\}, \{6,4,1\}), (\{6,4,2\},\{6,3,2\}), (\{5,4,3\}, \{5,4,2\})$. - -REPLY [15 votes]: You are looking at two consecutive ranks of the poset denoted -$L(k,n-k)$. I proved the existence of a matching in -http://math.mit.edu/~rstan/pubs/pubfiles/42.pdf. A more elementary -proof based on linear algebra was later given by R. A. Proctor, -Amer. Math. Monthly 89 (1982), 721-734. Another elementary proof -based on linear algebra appears in my book Algebraic Combinatorics -(built into the proof of Corollary 6.10). It is an open problem to -find a combinatorial proof.<|endoftext|> -TITLE: Given a Kirby diagram of a 4-manifold, what's its homotopy 2-type? -QUESTION [13 upvotes]: It's easy to derive a presentation of the fundamental group of a 4-manifold if you have a Kirby diagram: The 1-handles are generators and the 2-handles are the relations. The 3- and 4-handles, which are invisible in the Kirby diagrams, don't contribute. -Is there anything like this for the homotopy 2-type? -I'm imagining an algorithm that takes a Kirby diagram and spits out a groupoid, or a crossed module. Probably, the 1-handles generate the 1-morphisms, the 2-handles the 2-morphisms, and the higher handles are relations? How can the relations be visualised, though? - -REPLY [8 votes]: Yes: given a CW-complex $X$ the fundamental crossed module $\Pi_2(X,X^1)=(\partial \colon \pi_2(X,X^1) \to \pi_1(X^1))$, where $X^1$ is the 1-skeleton, represents the homotopy 2-type of $X$ (this can be stated in several ways). Moreover $\Pi_2(X,X^1)$ can be calculated combinatorially: $(\partial \colon \pi_2(X^2,X^1) \to \pi_1(X^1))$ is a totally free crossed module, and when you attach three handles you solely need to impose relations on $\pi_2(X^2,X^1)$ in order to get to $\pi_2(X,X^1)$ -Now if you have a Kirby diagram (i.e. a handlebody decomposition), then squashing the handles along their core yields a CW-complex. So in theory the fundamental crossed module of the associated CW-complex can be calculated combinatorially. Except that it might be a non-trivial exercise to determine the attaching maps of the 3-handles in such that way that the relations on $\pi_2(X^2,X^1)$ become transparent. (For complements of knotted surfaces this is quite a doable thing.) -Some discussion is in J. Faria Martins, The fundamental crossed module of the complement of a knotted surface, Trans. Amer. Math. Soc. 361 (2009), no. 9, 4593–4630 https://arxiv.org/abs/0801.3921 -and also in: Higher lattices, discrete two-dimensional holonomy and topological phases in (3+1) D with higher gauge symmetry (https://arxiv.org/abs/1702.00868) section 3.4 -and On 2-Dimensional Homotopy Invariants of Complements of Knotted Surfaces (https://arxiv.org/abs/math/0507239)<|endoftext|> -TITLE: Vectors that are almost orthogonal on average: lower bounds on dimension? -QUESTION [13 upvotes]: Let $v_1,\dotsc,v_k \in \mathbb{R}^d$ be unit-length vectors such that -$$\sum_{1\leq i,j\leq k} |\langle v_i,v_j\rangle|^2 \leq \epsilon k^2.$$ -What sort of lower bound can we give on $d$ in terms of $k$ and $\epsilon$? -Must it be the case that $d\gg \min(\log k,\epsilon^{-1})$, say, or anything of the sort? Is that tight? - -REPLY [7 votes]: The Johnson Lindenstrauss Lemma states that there are $k$ vectors achieving $\epsilon$ provided $d\geq C \epsilon^{-2} \log k.$ -If you actually want to bound the maximum absolute value of the inner product for distinct vectors, instead of the average as you stated which is of course stronger, then tighter bounds apply. -Relevant results for this case are due to Welch, Kabatianski, Levenshtein, Sidelnikov. Welch's applies to arbitrary vectors, real or complex. The others apply to vectors constructed from complex roots of unity of some finite order. -Welch's bound states -Let $e\geq 1$ be an integer and let $a_1,\ldots,a_k$ be distinct vectors in $\mathbb{C}^d.$ Then the following inequalities hold -$$ -\sum_{i=1}^k \sum_{j=1}^k \left| \langle a_i, a_j \rangle \right|^{2e} \geq \frac{\left(\sum_{i=1}^k \lVert a_i \rVert^{2e}\right)^2}{\binom{d+e-1}{e}}, -$$ -If the set of vectors you are interested in is of size roughly $d^u,$ the tightest lower bound is obtained by choosing $e=\lfloor u\rfloor.$ -Edit: -Since you required $\langle a_i,a_i\rangle=1,1\leq i\leq k,$ if I subtract the diagonal inner products, I obtain -$$ -\sum_{1\leq i\neq j\leq k} \left| \langle a_i, a_j \rangle \right|^{2} \geq \frac{k^2-kd}{d}, -$$ -recovering the dependence on $k.$ -Edit 2: -The Johnson Lindenstrauss Lemma is tight up to a constant factor. The Welch bound is tight for some cases, when so-called Welch Bound with Equality sets of vectors exist, which correspond to all unequal innner products being the same in absolute value. -References: -V.M. Sidelnikov, On mutual correlation of sequences, Soviet Math Dokl. 12:197-201, 1971. -V.M. Sidelnikov, Cross correlation of sequences, Problemy Kybernitiki, 24:15-42, 1971 (in Russian) -Welch, L.R. Lower Bounds on the Maximum Cross Correlation of Signals. IEEE Transactions on Information Theory. 20 (3): 397–399, 1974. -Kabatianskii, G. A.; Levenshtein, V. I. Bounds for packings on the sphere and in space. (Russian) Problemy Peredači Informacii 14 (1978), no. 1, 3–25. (A version of this might be available in English translation, in Problems of Information Transmission)<|endoftext|> -TITLE: Can completely multiplicative functions be extended to $\overline{\mathbb{Q}}$ or further? -QUESTION [5 upvotes]: I'm looking for a subject of study that handles the following question. I'm not the most familiar with algebra; I have a strong working knowledge and that's about it, but I've been considering researching more into it. I'm curious if the following exists somewhere in the literature. -Consider a completely multiplicative function $q :\mathbb{Q} \to \mathbb{Q}$ which is also an automorphism of the multiplicative group $\{\mathbb{Q}^{\times},\cdot\}$ (a bijective completely multiplicative function), and $q\big{|}_{\mathbb{N}} = \mathbb{N}$ (so for instance $q(x) \neq \frac{1}{x}$). These functions are essentially "prime permutations" on the unique factorization of elements of $\mathbb{Q}$. We can use a permutation notation, because of this. For instance, $(1\, 2)$ swaps the first prime number with the second; i.e $2$ for $3$. Therein if $x$ is factored, only finitely many $e_i$ are non-zero -$$x = \prod_{i=1}^\infty p_i^{e_i}$$ -then if $q = (1\, 2)$ -$$q(x) = \prod_{i=1}^\infty p_{q(i)}^{e_i}$$ -For example $q(2) = 3$, $q(3) = 2$, $q(2^4 3^2 5^3 7^{-6}) = 3^4 2^2 5^3 7^{-6}$, so on and so forth across all elements of $\mathbb{Q}$. These can be infinite permutations as well, so perhaps the $2n-1$'th prime is switched with the $2n$'th prime for all $n\ge 1$, or something of the sort. We also make the convention $q(0) = 0$, $q(1) = 1$ and that $q$ is bijective so that $q(-1) = -1$. -Now there also exists a natural extension to $\mathbb{Q}$ when equipped with $n$'th roots. That in $q(x^{1/n}) = q(x)^{1/n}$ where we make the convention that positive values are sent to positive values. This quite frankly gives the natural result that $q(\sqrt{2}) = \sqrt{3}$. - -Does there exist a natural extension $\tilde{q}$ to - $\{\overline{\mathbb{Q}}, \cdot\}$--where $\tilde{q}$ is still completely - multiplicative, bijective and $\tilde{q}\Big{|}_{\mathbb{Q}} = q\Big{|}_{\mathbb{Q}}$? I.e: can this be made to work on all algebraic - numbers? - -The reasoning I have for thinking of this is because of field extensions. By defining the field $F = \{\mathbb{Q},\cdot,\oplus\}$ where $x \oplus y = q(q^{-1}(x) + q^{-1}(y))$. We can take the polynomial ring $F[X]$ and talk about $F[X]/p$ for some irreducible polynomial $p$--which is isomorphic to some field extension $F(\alpha)$. -Now let's make the convention $p(\alpha) = 0$, then this can be thought of as -$$0 = p(\alpha) = a_n\alpha^n \oplus a_{n-1}\alpha^{n-1} \oplus...\oplus a_0$$ -$$p(\alpha) = q(q^{-1}(a_1)q^{-1}(\alpha)^n + q^{-1}(a_{n-1})q^{-1}(\alpha)^{n-1} +...+q^{-1}(a_0))$$ -so that -$$0 = q^{-1}(a_1)q^{-1}(\alpha)^n + q^{-1}(a_{n-1})q^{-1}(\alpha)^{n-1} +...+q^{-1}(a_0)$$ -Therefore if $q^{-1}(\alpha) = \beta$ and $b_n = q^{-1}(a_n)$, then $\alpha = q(\beta)$ where $\beta$ satisfies -$$b_n\beta^n + b_{n-1}\beta^{n-1} + ... + b_0 = 0$$ -This definition is consistent with the taking roots version. If $\beta$ satisfies $\beta^2 - 2 = 0$, then $q(\beta) = \alpha$ where $\alpha^2 \ominus 3 = 0$, aka $q(\sqrt{2}) = \sqrt{3}$. (Where the meaning of $\ominus$ should be self evident). -However, as you may have noticed, this does not "choose" which root of the polynomial corresponds to $\beta$. As in $q(\sqrt{2}) = \pm\sqrt{3}$. Nor does it explicitly give a manner of producing said $\alpha$. $\alpha$ just exists in the "isomorphic algebraic numbers" generated by the algebraic closure of $\{\mathbb{Q},\cdot,\oplus\}$ which coincides with roots. And for higher-order polynomials this becomes a serious problem. This led me to two different questions. - -Can we distinguish elements belonging to the algebraic closure of - $\{\mathbb{Q},\cdot,\oplus\}$ and $\{\mathbb{Q}, \cdot, +\}$? I.e, - can we make sense of the correspondence, so that $q(1 +\sqrt{2}) = -> \alpha \in \overline{\mathbb{Q}}$ and not just in some isomorphic copy - of $\overline{\mathbb{Q}}$? - -Secondly, if we consider $q:\mathbb{Q}\to\mathbb{Q}$ and we wish to extend it to $\mathbb{Q}(\alpha)$ then if $q:\mathbb{Q}(\alpha) \to F(\beta)$ there should exist multiple extensions (i.e: a $q_{-}$ and $q_{+}$ where $q_{+}(\sqrt{2}) = \sqrt{3}$ and $q_{-}(\sqrt{2}) = -\sqrt{3}$). If we were to add multiple field extensions, the intuition would say we would compound the amount of extensions of $q$. -For instance, if $q = (1\,2)(3\,4)$ (swap 2 and 3, and swap 5 and 7) then the field extension $Q(\sqrt{2},\sqrt{7})$ has four possible $q$'s extending its definition $q_{ij}(\sqrt{2}) = (-1)^{i}\sqrt{3}$ and $q_{ij}(\sqrt{7}) = (-1)^j\sqrt{5}$ for $i,j = 0\,\text{or}\,1$. But are these definitions compatible when adding multiple field extensions? I was only able to really construct it with a Galois extension (I think I mean a Galois extension). -Would this still continue to work if say, we took $\overline{\mathbb{Q}}$, we added all algebraic numbers to $\mathbb{Q}$. Therein, my gut says to ask: - -Could there be an uncountable number of - $\tilde{q}:\overline{\mathbb{Q}}\to\overline{\mathbb{Q}}$'s which - would extend $q:\mathbb{Q}\to\mathbb{Q}$? (i.e: - $\tilde{q}\Big{|}_{\mathbb{Q}} = q\Big{|}_{\mathbb{Q}}$) - -But I'm doubting this because I think some extensions may be equivalent when adding another element. So, my gut also says to ask: - -Is there only a finite number of $\tilde{q}$, perhaps just one? Does - the number of solutions depend on our choice of $q$? - -And the question I hope is answered no - -Is there no such $\tilde{q}$? - -Any help or comments or suggestions is greatly appreciated. This one has had me stumped for a long time and I'm stuck not knowing what tools to use to approach the problem. If something like this has already been considered in the literature, I'd be very happy to read up on it. Does it already have a name, and a well founded approach? Is it meaningless? Am I garbling out nonsense? Anything is welcome. -Thanks! -EDIT: I should mention this came up as I was thinking about the absolute Galois group. Clearly elements of the absolute Galois group are automorphisms of $\{\overline{\mathbb{Q}},\cdot\}$--however these fix the base field $\mathbb{Q}$. What if we, instead, allow it to permute the basefield $\mathbb{Q}$; does this still make sense. - -REPLY [4 votes]: Such an extension exists. More generally, an automorphism of a subgroup of a divisible (abelian) group extends to the whole group. One can prove this by mimicking the proof that divisible groups are injective (Stacks project): -Let $H \subset G$ with $G$ divisible, and $q \in \operatorname{Aut}(H)$. Then one shows that a pair $(K, \tilde q)$ with $H \subset K$ and $\tilde q \in \operatorname{Aut}(K)$ extending $q$, which is maximal for the obvious ordering, necessarily satisfies $K = G$. -So for example, there exists an extension of $q \in \operatorname{Aut}(\mathbb Q^\times)$ which swaps $\alpha = 1 + \sqrt 2$ with $\beta = 4 + \sqrt[5]{13}$, because we may first extend $q$ to the group $\mathbb Q^\times \oplus \alpha^{\mathbb Z} \oplus \beta^{\mathbb Z}$ in any way we like, and then to $\overline{\mathbb Q}^\times$. -And of course, one can always compose $\tilde q$ with a $\sigma \in \operatorname{Gal}(\overline{\mathbb Q} / \mathbb Q)$. - -A theorem of Mishina gives a characterization of groups with the property that automorphisms of subgroups extend to the whole group: -A. P. Mishina (1962). 'On automorphisms and endomorphisms of Abelian groups', Vestnik Moskov. Univ. Ser. I Mat. Mekh., No. 4, 39–43.<|endoftext|> -TITLE: Correspondence between persistence module and graded module over $R[t]$ -QUESTION [6 upvotes]: In the paper "Computing Persistent Homology" by Zomorodian and Carlsson, it is stated as Theorem 3.1 that: - -The correspondence $\alpha$ defines an equivalence of categories between the category of persistence modules of finite type over $R$ and the category of finitely generated non-negatively graded modules over $R[t]$. - -The proof is just one line: "The proof is the Artin-Rees theory in commutative algebra". -I am curious exactly which part of Artin-Rees theory did they use to conclude that? I know there is a Artin-Rees lemma. -Thanks for any help or references. - -REPLY [8 votes]: Recently, René Corbet and Michael Kerber posted the preprint "The Representation Theorem of Persistent Homology Revisited and Generalized" which gives a proof of this statement with elementary methods. Among other things, the authors also give a short discussion on your question.<|endoftext|> -TITLE: The set of complements equal to the complement of set -QUESTION [8 upvotes]: Consider $A \subset \{0,1\}^n$ -I want $A$ to have two properties. -$1.$ $A$ is increasing, i.e., If $x \in A$ and $x \subseteq y$ then $y \in A$ too. - -[$x \subseteq y$ means that every coordinate of $y$ is greater that or equal to corresponding coordinate of $x$] - -$2.$ $A^c$ is equal to set $B=\{x \mid x^c \in A\}$ -Is there any characterization for such a set? I have to example for it. But I want to find an IFF condition for such sets... -$e1)$ $A=\{x|$ first coordinate of $x$ is $1\}$ -$e2)$ Fix an odd number of coordinates. $A= \{x\mid x$ contains at least half of coordinates equal to $1\}$ -[For even number there is a similar example] - -P.S. Asked It before Here: https://math.stackexchange.com/questions/2135708/when-set-of-complements-is-equal-to-complement-of-set - -REPLY [2 votes]: Although Emil Jeřábek is certainly right in that these, as Bjørn Kjos-Hanssen called them, ultraupsets, seem to be quite complicated, I would like to propose a combinatoric-topological reformulation which I think makes them more "touchable". -If we switch to complements, we are looking at abstract simplicial complexes with a very special property - they contain exactly one from each pair of complementary simplices. -It seems that some of the consequences can be more easily understood with the aid of this geometric intuition. In particular one can visualize such complexes in low dimensions. -Among subcomplexes of an 1-simplex, only single points are possible. -For subcomplexes of a triangle, there are two possibilities: an edge of the triangle, and the discrete 3-element set of its vertices. -For a tetrahedron, one has three (up to isomorphism) possibilities: a 2-face, the disjoint union $($boundary of a triangle$)\cup($point$)$, and three edges meeting at a vertex. -For a 4-simplex one gets: - -a 3-face; -disjoint union $($boundary of a tetrahedron$)\cup($point$)$; -two kinds of complexes with 7 edges: - - -and - - -a complex with 8 edges - - - -one with 9 edges - - - -and the whole 1-skeleton. - -All in all this seems to be a very interesting combinatorial object.<|endoftext|> -TITLE: Conjugacy of Borel subgroups over arbitrary fields -QUESTION [8 upvotes]: Let $k$ be a field and $G$ a connected semisimple algebraic group over $k$. -If $k$ is algebraically closed, then it is well known that all Borel subgroups of $G$ are conjugate by the action of $G(k)$. I would like to know whether this is also true over non-closed fields. - - -Are all Borel subgroups of $G$ over $k$ conjugate by the action of $G(k)$? - - -Recall that a Borel subgroup $B$ of $G$ over $k$ is a closed subgroup of $G$ over $k$ such that the base change of $B$ to the algebraic closure $\bar{k}$ is a Borel subgroup of $G \times_k \bar{k}$. -Of course Borel subgroups need not exist in general; when they do one says that $G$ is quasi-split. If $G$ is not quasi-split then the statement is vacuously true. - -REPLY [11 votes]: Yes. This follows directly from Theorem 20.9 (i) in "Armand Borel, Linear Algebraic Groups, Second enlarged edition, 1991" which goes like this: -Theorem: Let $ G $ be a connected reductive group over a field $ k $. The minimal parabolic $k$-subgroups of $G$ are conjugate under $G(k)$.<|endoftext|> -TITLE: Precise definition/construction of $\text{Sym}(\text{Sym}^2)$? -QUESTION [11 upvotes]: The question title basically has my question. What is the precise definition/construction of$$\text{Sym}(\text{Sym}^2),$$as mentioned in the following? -http://aimpl.org/repnstability/1/ -If possible, I would like this to be self-contained and not defer to other stuff about twisted commutative algebras, as much as they are interesting. - -REPLY [4 votes]: As Dan Petersen points out, it's impossible to answer this question without talking about twisted commutative algebras. Moreover, to talk about TCAs over $\mathbb{Z}$, you need a more general definition of TCAs (which becomes equivalent to the earlier Sam-Snowden definitions when you're over $\mathbb{C}$). I'm not sure this is in the literature, so I'll try to explain. -Definition: A TCA is a commutative algebra inside the tensor category of FB-modules. -Let's expand this definition. FB denotes the category of finite sets and bijections, and an FB-module is a functor $W: \mathrm{FB}\to \mathbb{Z}\mathrm{-Mod}$. In other words, $W$ is essentially a collection of $S_n$-representations $W_n$ (one for each $n\in \mathbb{N}$), with no relation between them. -The disjoint union of finite sets $\sqcup:\mathrm{FB}\times \mathrm{FB}\to \mathrm{FB}$ induces a tensor product $\otimes:\mathrm{FB}\mathrm{-Mod}\times \mathrm{FB}\mathrm{-Mod}\to \mathrm{FB}\mathrm{-Mod}$ (often called the "convolution tensor product". It has the universal property that a map of FB-modules $A\otimes B\to C$ is equivalent to a collection of maps $A_S\otimes_{\mathbb{Z}}B_T\to C_{S\sqcup T}$ for all finite sets $S$ and $T$, natural in $S$ and $T$. (Equivalently, you can say it's a collection of $S_k\times S_\ell$-equivariant maps $A_k\otimes_{\mathbb{Z}}B_\ell\to C_{k+\ell}$, but it's easier to keep track of things like the symmetry $A\otimes B\cong B\otimes A$ with the former definition.) -A TCA is an FB-module $A$ together with a multiplication $A\otimes A\to A$ (which has to be commutative, but this is a property not structure so we don't need to stress it). - -Now we can define $\mathrm{Sym}(\mathrm{Sym}^2)$. Recall that a perfect matching on a set $T$ is an equivalence relation on $T$ where each equivalence class has exactly 2 elements. -Definition: $\mathrm{Sym}(\mathrm{Sym}^2)$ is the FB-module $A$ which assigns to a finite set $T$ the free abelian group $A_T=\mathbb{Z}[\{\text{perfect matchings on }T\}]$. -To make this a TCA, we need to define the multiplication map $A\otimes A\to A$. According to the universal property above, this means we have to specify natural maps $A_S\otimes_{\mathbb{Z}} A_T\to A_{S\sqcup T}$. But this is very easy: if you have a perfect matching of $S$, and a perfect matching of $T$, their union is a perfect matching of $S\sqcup T$. - -From this perspective the name may not be so clear. Here are two explanations. First (and perhaps less satisfying for you?): Define an FB-module $X$ by $X_T=\begin{cases}\mathbb{Z}&\text{if }\lvert T\rvert=1\\0&\text{else}\end{cases}$. Because we have a tensor product, we can make sense of $\textrm{Sym}^k$ in FB-Mod, and for example $\textrm{Sym}^2 X$ turns out to be $(\textrm{Sym}^2 X)_T=\begin{cases}\mathbb{Z}&\text{if }\lvert T\rvert=2\\0&\text{else}\end{cases}$. Moreover, our algebra $A$ is precisely $\textrm{Sym}^*(\textrm{Sym}^2 X)=\bigoplus_{k\in \mathbb{N}}\textrm{Sym}^k(\textrm{Sym}^2 X)$. -Alternately, if you were to work over $\mathbb{C}$ throughout, then there is an alternate description of TCAs over $\mathbb{C}$ in terms of polynomial representations of $\textrm{GL}_\infty(\mathbb{C})$. (See e.g. Section 2.1 of arXiv:1501.06925.) In this perspective, you are looking at nothing other than the commutative ring $\textrm{Sym}^*(\textrm{Sym}^2 \mathbb{C}^\infty)$, with its natural multiplication and natural $\textrm{GL}_\infty(\mathbb{C})$-action.<|endoftext|> -TITLE: Higher dimensional residues in complex analysis -QUESTION [7 upvotes]: Consider a function $f:\mathbb{CP}^1\times\mathbb{CP}^1\to \mathbb{CP}^1 $ defined by $f([x_1,x_2],[y_1,y_2])=[x_1y_1,x_2y_2]$. This function is well defined except at $([0,1],[1,0])$ or vice versa (in therms of the Riemann sphere $\mathbb{C}_\infty$ we do not have a well defined zero times infinity). Now consider a function $g:U\to\mathbb{C}$ analytic in in a neighbourhood $U$ of $([0,1],[1,0])$ in $\mathbb{CP}^1\times\mathbb{CP}^1$. I am looking for a way to define some sort of numerical value of residue to the product $fg$ at the point -$([0,1],[1,0])$. -I am looking for any possible definition of `residue' which make sense and is invariant to change of coordinates. This may be in terms of integral on a polydisk (feel free to introduce a Kahler form if it makes things easier). It may be in terms of cohomology (as long as it is calculable and not completely abstract). Anything will do. -The motivation lies in the theory of integrable systems, and there there is a rather unsatisfactory answer - define a standard example as you wish, and then use the huge number of symmetries of the integrable system to define the other cases which can arise. Of course this may do a particular job, but in general is not a good way to proceed, so I am looking for a more sensible theoretical definition. (I am obviously not an expert in algebraic geometry, all I can say is that from a physical point of view, such a construction seems possible.) - -REPLY [5 votes]: While it's hard to guess exactly what you are after, I believe such residues were first studied by Poincaré (Sur les résidus des intégrales doubles, Acta Math. 9 (1887) 331–380), with your sought invariance encoded in the statement that "residue" takes a closed p-form on the complement of a hypersurface to a closed (p–1)-form on the hypersurface, so that in cohomology it induces a map $\smash{\mathrm{Res}:H^p(X\setminus S)\to H^{p-1}(S)}$. Modern references are, for instance: - -P. A. Griffiths, Poincaré and algebraic geometry, Bull. Amer. Math. Soc. 6 (1982) 147–159. -E. Cattani and A. Dickenstein, Introduction to residues and resultants, pp. 1–61 in Solving polynomial equations, Springer-Verlag, Berlin (2005) (PDF). -A. Yger, The concept of "residue” after Poincaré: cutting across all of mathematics, pp. 225–241 in The scientific legacy of Poincaré, Amer. Math. Soc. (2010) (French PDF).<|endoftext|> -TITLE: If monodromy representation is unipotent then special fiber is snc? -QUESTION [6 upvotes]: We know, if $π : X → S$ is a generically smooth family of complex projective -varieties, such that $X_0 := π^{−1}(0)$ is an snc divisor in $X$, then the -monodromy representation is unipotent. Now assume the monodromy representation is unipotent, then is the central fibre $X_0 := π^{−1}(0)$ simple normal crossing? - -REPLY [8 votes]: Just to be clear, the local monodromy is unipotent if $X_0$ is reduced with simple normal crossings (that's probably what you meant). As Piotr pointed out, your question, as originally formulated, has an easy negative answer. However, as he suggested, the question can be modified to a something more reasonable: - -If the local monodromy is unipotent, does there exist a birational model with $X_0$ reduced snc? - -I think this is also no. Take a Lefschetz pencil of sextics $Y\to S=\mathbb{P}^1$ in $\mathbb{P}^4$, so that $Y_0$ has a node $p$ and no other singularities. By the Picard-Lefschetz formula, the local monodromy is unipotent. Let $X\to S$ be a birational model with $X_0$ snc. We have a morphism $X\to Y$ given by the relative canonical map. This will dominate the blow up $Bl_pY$. The exceptional divisor of $Bl_pY\to Y$ will have multiplicity $2$. So that the components of $X_0$ lying over this also have multiplicity at least $2$.<|endoftext|> -TITLE: Where to publish a new proof of an old theorem? -QUESTION [26 upvotes]: A few months ago I came up with a proof for an old theorem. After being excited for a moment, I then tried to find my proof in the literature. Since I did not find it, then I started to wonder if it was worth publishing it. -I asked a few people about journals that could publish something like this, and they gave me two recommendations: -(1) The Mathematical Gazette, http://www.m-a.org.uk/the-mathematical-gazette -(2)The Plus Magazine, https://plus.maths.org/content/about-plus -First I submitted to the Mathematical Gazette, and my article was rejected because according to the reviewer I was trying to prove something very simple using something much more complex (although I just used undergraduate level math). -Then I submitted to Plus, and it was also rejected by the editors (it probably doesn't fit well with their magazine). -Do you have any suggestions? Thanks. - -REPLY [5 votes]: A new proof of an old theorem, especially using modern machinery, could well be within the scope of the Graduate Journal of Mathematics, which seeks to publish work either by graduate students or that would be of interest to graduate students. From their website: - -The Graduate Journal of Mathematics is an electronic journal that publishes original work as well as expository work of general mathematical interest which add to the literature, have pedagogical value and help make more widely accessible significant mathematical ideas, constructions or theorems...One main aim of our journal is to help researchers in mathematics in general, and graduate students in particular, gain access to important ideas and communicate interesting mathematics. - -Full disclosure: I am on the editorial board. And I'd be happy to receive a submission with a new proof of an old theorem.<|endoftext|> -TITLE: Etale cohomology with coefficients in $\mathbb{Q}$ -QUESTION [9 upvotes]: Let $X$ be a smooth variety of a field $k$. Then is - $$H_{et}^i(X, \mathbb{Q}) = 0$$ - for all $i > 0$? - - -The result is true for $i=1$. This follows from the same argument given for $\mathbb{Z}$-coefficients given here: Etale cohomology with coefficients in the integers. Note that one needs to at the very least assume that $X$ is normal; otherwise $H_{et}^1(X, \mathbb{Q})$ can be non-trivial. -The result is true if $X = \mathrm{Spec}(k)$. This follows from standard properties of Galois cohomology. Note that the analogous result is not true with $\mathbb{Z}$-coefficients; $H^2(k, \mathbb{Z})$ is often non-trivial. -If it helps, I mostly care about the case $i=2$. - -REPLY [2 votes]: You can find this for normal schemes in Deninger, A proper base change theorem for non-torsion sheaves in étale cohomology, Journal of Pure and Applied Algebra 50 (1988), 231–235. http://www.sciencedirect.com/science/article/pii/0022404988901028 (2.1) (proof by reduction to Galois cohomology using the Leray spectral sequence)<|endoftext|> -TITLE: Combinatorics of the Cohomology Ring of the Lagrangian Grassmannians -QUESTION [6 upvotes]: The Lagrangian Grassmannian is an important example in symplectic geometry, see here or here for details. It shares many similarities with the ordinary Grassmannians (as one would expect from the name). As is well-known, the cohomology ring of the Grassmannians has a very nice combinatorial description in terms of partitions, see this very nice M.O. answer for example. Does there exist an analogous description for the cohomology ring of the Lagrangian Grassmannians? -More precisely: -(i) How many generators does the ring have? -(ii) What are its dimensions? -(iii) What is its multiplicative structure? - -REPLY [4 votes]: I'm assuming you mean in $\mathbb{C}^{2n}$ (the answer for real Lagrangian Grassmannians is trickier). In this case, the answer is easy: - -The cohomology is isomorphic to the symmetric polynomials in $n$ variables modulo the relation killing all positive degree symmetric polynomials in the squares of the variables. - -I'm reasonably certain that you need $n$ generators, since the ideal killed lies in the square of the unique graded maximal ideal. -Like the usual Grassmannian, the cohomology of the Lagrangian Grassmannian is generated by the Chern classes of the tautological bundle which correspond to the elementary symmetric functions, but the relations are a bit different. They come from the observation that there is a trivial vector bundle filtered by the tautological bundle $T$ and its dual, so we get the vanishing relations from Whitney sum formula. For isotropic Grassmannians, you don't kill all the symmetric polynomials in the squares, just the complete symmetric polynomials of high enough degree.<|endoftext|> -TITLE: Tachikawa conjecture for commutative algebras proven? -QUESTION [5 upvotes]: The Tachikawa conjecture states that $Ext^i(M,M) \neq 0$ for some $i \geq 1$ for every non-projective module $M$ over a selfinjective finite dimensional algebra. -In theorem 4.6. of http://maths.nju.edu.cn/~huangzy/When%20are%20torsionless%20modules%20projective.pdf , the authors prove something (much stronger!) which implies the Tachikawa conjecture in the commutative case, which would be a sensational result in my opinion. -Question: -Is the proof really true/without gaps? I couldnt understand everything and the authors never replied. Of course it might be extremely hard to give counterexamples to theorem 4.6. there but maybe gaps could be pointed out? - -REPLY [7 votes]: On page 2163, the second line, the authors say "therefore ... is exact". I, and some others, believe this is a gap. The authors have been contacted (in 2010) and have not clarified.<|endoftext|> -TITLE: Can a $W^{1,2}$ map from the disk to the circle restrict to a degree one map on the boundary? -QUESTION [9 upvotes]: The restriction of a continuous map $D^2\to S^1$ to $\partial D^2\to S^1$ must have degree zero. Is that statement true or false if the map is only $W^{1,2}(D^2;S^1)$ and continuous on $\partial D^2$? -In two dimensions, the Sobolev space $W^{1,2}$ is at the borderline regularity and does not embed into the space of continuous functions. However, we can ask that a map's values be on the circle almost everywhere, giving us the space $W^{1,2}(D^2;S^1)$. Such a map has a well-defined restriction to the boundary; the restriction is a $W^{1/2,2}(\partial D^2;S^1)$ map. If it so happens that this restriction is continuous, we can ask about its degree. -A standard example of a discontinuous $W^{1,2}$ function is $f=\log(\log(4/r))$, and we can use it to construct a discontinuous $W^{1,2}$ circle-valued map $g=e^{if}$. However, this map has degree zero on the boundary, and it's not clear to me if one can do a different construction to get a degree one map. -Another good example to consider is $e^{i\theta}$, which is a $W^{1,p}(D^2;S^1)$ map for any $p<2$, and has degree one on the boundary. - -REPLY [4 votes]: The restriction of $f$ to the boundary has degree zero. It is true also in higher dimensions. The proof presented below is based on the proof of density of $C^\infty(M,N)$ in $W^{1,p}(M,N)$, $p\geq \operatorname{dim}M$, due to Schoen and Uhlenbeck [3] (see also Theorem 2.1 in [1]). - -Theorem. If $f\in W^{1,n}(B^n,S^{n-1})$ and $f|_{\partial B^n}\in C^0$, then $f|_{\partial B^n}:S^{n-1}\to S^{n-1}$ has degree zero. - -Here the restriction to the boundary $f|_{\partial B^n}$ is defined as a trace of a $W^{1,n}$ function. -Proof. -Let $B^n=B^n(0,1)$ and let $B^n_{1+\delta}=B^n(0,1+\delta)$ for some small $\delta>0$. First of all, we can extend the mapping $f$ to $\tilde{f}\in W^{1,n}(B^n_{1+\delta},S^{n-1})$ if $\delta>0$ is small enough. Indeed, the Sobolev extension operator $E$ is defined through averages, see [2], so extending $f$ to $B^n_{1+\delta}\setminus B^n$ gives a function $Ef\in W^{1,n}(B^n_{1+\delta},\mathbb{R}^n)$ that is continuous in the annulus $B^n_{1+\delta}\setminus B^n$. If $\delta>0$ is small enough, -$|Ef|>1/2$ on $B^n_{1+\delta}\setminus B^n$ (by continuity and the fact that $|f|=1$ on $\partial B^n$) and hence -$\tilde{f}=Ef/|Ef|$ in $B^n_{1+\delta}\setminus B^n$ and -$\tilde{f}=f$ in $B^n$ belongs to $\tilde{f}\in W^{1,n}(B^n_{1+\delta},S^{n-1})$. If we prove that the degree of $\tilde{f}$ on the boundary of $B^n_{1+\delta}$ is zero, then also degree of $f$ on the boundary of $B^n$ is zero (by homotopy invariace of degree and continuity of $\tilde{f}$ in -$B^n_{1+\delta}\setminus B^n$). -The above construction shows that we can assume that $f$ is continuous in a neighborhood of $\partial B^n$ (because $\tilde{f}$ is continuous near the boundary of the ball $B^n_{1+\delta}$ and the argument given below can be applied to $\tilde{f}$ showing that the degree of $\tilde{f}$ is zero on the boundary of the ball $B^n_{1+\delta}$). -The mapping $f$ takes values into $\mathbb{R}^{n}$ since $f:B^n\to S^{n-1}\subset\mathbb{R}^n$. Given $\epsilon>0$ define -$r_{\epsilon,x}=\epsilon(1-|x|)$ and -$$ -f_\epsilon(x)=\frac{1}{|B(x,r_{\epsilon,x})|}\int_{B(x,r_{\epsilon,x})} f(y)\, dy. -$$ -That is we average $f$ over a ball of radius $\epsilon$ times the distance of $x$ to the boundary of the unit ball $B^n$. -The function $f_\epsilon$ is continuous up to the boundary: as $x$ approaches $\partial B^n$, the radius of the ball over which we average tends to zero and hence $f$ is continuous up to the boundary, because $f$ is continuous in an annulus near the boundary, -$f_\epsilon|_{\partial B^n}=f|_{\partial B^n}$. -According to the Poincare inequality -$$ -\left(\frac{1}{|B(x,r_{\epsilon,x})|}\int_{B(x,r_{\epsilon,x})}|f(y)-f_\epsilon(x)|^n\, dy\right)^{1/n} -\leq C\left(\int_{B(x,r_{\epsilon,x})}|\nabla f|^n\right)^{1/n}. -$$ -Since -$$ -\operatorname{dist}(f_\epsilon(x),S^{n-1})\leq |f(y)-f_\epsilon(x)| -\quad -\text{for all $y$} -$$ -we have -$$ -\operatorname{dist}(f_\epsilon(x),S^{n-1})\leq -C\left(\int_{B(x,r_{\epsilon,x})}|\nabla f|^n\right)^{1/n}. -$$ -The right hand side converges uniformly to $0$ in $x$. That is if $\epsilon$ is small enough, $f_\epsilon(x)\neq 0$ and hence -$$ -g(x)=\frac{f_\epsilon(x)}{|f_\epsilon(x)|} -$$ -is a continuous map $g:B\to S^{n-1}$, -$g|_{\partial B^n}=f_\epsilon|_{\partial B^n}=f|_{\partial B^n}$. This shows that $\operatorname{deg}(f|_{\partial B^n})=0$. -[1] P. Hajłasz, -Sobolev mappings between manifolds and metric spaces. In: Sobolev spaces in mathematics. I, 185–222, Int. Math. Ser. (N. Y.), 8, Springer, New York, 2009. -[2] G. Leoni, A first course in Sobolev spaces. Second edition. Graduate Studies in Mathematics, 181. American Mathematical Society, Providence, RI, 2017. -[3] R. Schoen, K. Uhlenbeck, -Boundary regularity and the Dirichlet problem for harmonic maps. -J. Differential Geom. 18 (1983), 253–268.<|endoftext|> -TITLE: What is the smallest cardinality of a set A whose difference A-A contains $n$ consequtive integer numbers? -QUESTION [11 upvotes]: Problem. What is the smallest cardinality $d(n)$ of a set $A$ of integer numbers such that the difference set $A-A=\{a-b:a,b\in A\}$ contains $n$ consequtive integer numbers? -It can be shown that $(1+\sqrt{4n-3})/2\le d(n)\le \frac32p(\sqrt{n})=\frac32\sqrt{n}+O(n^{21/80})$ where $p(x)$ is the smallest prime number greater or equal to $x$. -These bounds suggest the following more precise questions: -Question 1. Is $d(n)\le (\sqrt{2}+o(1))\sqrt{n}$? -Question 2. Is $d(n)=(1+o(1))\sqrt{n}$? -Comment. Looking at the literature, I discovered that this question has been studied by classics: Erdos, Gal (1948), Redey, Renyi (1949), Leech (1956), Whichmann (1963), Golay (1972). More information (in the context of perfect rulers) can be found here. Wichmann proved that for every $r,s\ge 0$ there exists a set $A\subset \mathbb N\cup\{0\}$ of cardinality $n=4r+s+3$ such that $A-A=[-L,L]$ where $L=4r(r+s+2)+3(s+1)$. -This gives an affirmative answer $d(n)\le \sqrt{2n}$ to Question 1. -On the other hand, much earlier Redei and Renyi (1949) proved the lower and upper bounds $1+\frac2{3\pi}< \lim_{n\to\infty}\frac{d(2n+1)^2}{2n}=\inf_{n\in\mathbb N}\frac{d(2n+1)^2}{2n}<\frac{4}3$. These lower and upper bound were improved a bit by Leech (1956) and Golay (1972). This negatively answers my Question 2 (and completes the answer given by Lucia). - -REPLY [8 votes]: Since you say that only Question 2 is open, I'll only address that. The answer is no, and $d(n)$ must be at least $(1+\delta)\sqrt{n}$ for some positive $\delta$. I won't compute this, but it shouldn't be too hard to find some bound. -Suppose for contradiction that $|A|\le (1+\delta) \sqrt{n}$. Since the difference set is symmetric about $0$, we might as well assume that the consecutive numbers hit in the difference set are from $[-n/2,n/2]$. For any number $k$ let $r(k)$ denote the number of ways of writing $k$ as $a-b$ with $a$, $b$ in $A$. Now -$$ -\sum_{k} r(k) = |A|^2, -$$ -and by hypothesis $r(k) \ge 1$ for $k \in [-n/2,n/2]$. Therefore -$$ -\sum_{k\notin [-n/2,n/2]} r(n) + \sum_{k \in [-n/2,n/2]} (r(n)-1) -\ll \delta n. -$$ -Note also that there must be some interval of length $n/2$ such that -almost the full density (say $1-10\sqrt{\delta}$) of $A$ is contained in that interval. Else there -would be many differences of size larger than $n/2$, contradicting the estimate above. -Set -$$ -{\hat A}(\ell) = \sum_{a\in A} e(a\ell/n). -$$ -Then for $\ell \neq 0 \mod n$, using that $\sum_{k \in [-n/2,n/2]} e(k\ell/n) = O(1)$, -$$ -|{\hat A}(\ell)|^2 = \sum_{k} r(k) e(k\ell/n) \le \sum_{k \notin [-n/2,n/2]} r(k) + \sum_{k \in [-n/2,n/2]} (r(k)-1) +O(1) \ll \delta n. -$$ -If $\delta$ is small enough, then what we have essentially shown is that the elements of $a$ are essentially uniformly distributed $\mod n$ (the smaller $\delta$ is, the better the equidistribution). But that is not possible, because we observed earlier that most of $A$ must fit into an interval of size $n/2$.<|endoftext|> -TITLE: How to calculate the cup product of etale cohomology to prove that in Poincare duality agrees with Weil-pairing? -QUESTION [8 upvotes]: I learn from SGA4$\frac{1}{2}$ Dualité Proposition 3.4 that Tr($\phi$(u)$\cup x$)=$\phi (x)$,where $\phi$ is a homomorphism $\phi$: $H^1_c(X,\mu_n) \to \mathbb{Z}/n$ and u is the (1,1) part of diagonal class $H^2_{\Delta}(X\times \bar{X},j_!\mu_n)$ which can be proved agrees with "de son image réciproque par $f_0$:$X\to Pic_D^0(\bar{X})$"in $H^1_c(X,Pic_D^0(\bar{X})）$ . Maybe we can assume that $\phi$ is $\frac{g(X+S)}{g(X)}$ where g is the function associate with the point $T\in A[n]$ used to define Weil-pairing.My question: How to prove $\phi(u)$ is the divisor corresponding to T and the agreement of the pairing？ - -REPLY [3 votes]: Let $X$ be a smooth projective curve over field $k$. Let $a\in A[n], \mathcal{L}\in A^\vee[n]$. -Recall the definition of Weil pairing (Mumford's Abelian varieties, IV.20): since $\mathcal{L}\in\mathrm{Pic}^0(A)$, we know $n^*\mathcal{L}=\mathcal{L}^{\otimes n}=\mathcal{O}_A$, so after pulling back by $n\colon A\to A$,the line bundle $\mathcal{L}$ becomes trivial, thus corresponds to a factor of automorphy $\chi\colon A[n]\to\mu_n$, the Weil pairing is defined by $e_n(a,\mathcal{L})=\chi(a)$. -To show the Weil pairing coincides with cup product, using SGA41/2 (6.2.2.3)(Duality, Proposition 3.4), it suffices to show the composition $$A^\vee[n]\to H^1(A[n],\mu_n)\to A^\vee[n]$$ is identity. The first map is taking factor of automorphy of the $n$-torsion line bundle (or equivalently, the $\mu_n$-torsor), the second map is collapsing the $A[n]$-torsor $$0\to A[n]\to A\overset{n}{\to} A\to 0$$ to a $\mu_n$-torsor. When we collapse such a torsor, the $n^{2g}$ fibers correspond to $A[n]$ are identified, the scaling is given by a homormophism $A[n]\to\mu_n$. Think this through one see the composition is identity.<|endoftext|> -TITLE: Elementary extensions of direct product -QUESTION [8 upvotes]: I apologize if this question is elementary: Let $A$, $B$, and $A^{\prime}$ be groups such that $A^{\prime}$ is an elementary extension of $A$. Is it true that $A^{\prime}\times B$ is an elementary extension of $A\times B$? Clearly this is true for ultrapowers. - -REPLY [8 votes]: Yes. Let $T$ be the theory of two disjoint groups, in the language $(\cdot_1, \cdot_2, U_1, U_2)$. Note that if $(G_1, G_2) \models T$ then the group operation on $G_1 \times G_2$ is definable without parameters. Thus we can recover the theory of $G_1 \times G_2$ from the theory of $(G_1, G_2)$, which is clearly determined by $(\mbox{Th}(G_1), \mbox{Th}(G_2))$. Thus if $G_1, G_2, G_1', G_2'$ are any groups with each $G_i \equiv G_i'$, then $G_1 \times G_2 \equiv G_1' \times G_2'$. -EDIT: this just shows elementary equivalence. In order to get elementary extensions: let $G_1, G_2$ be given groups, and consider the language $(\cdot_1, \cdot_2, U_1, U_2,c_g: g\in G_1, d_h: h \in G_2)$ where we add constant symbols for elements of $G_1$ and $G_2$. Let $T_*$ assert that $T$ holds, and the elementary diagram of $G_1$ holds in $U_1$, and the elementary diagram of $G_2$ holds in $U_2$. Note that $T_*$ is actually complete. Now let $(G_1', G_2', g, h)_{g \in G_1, h \in G_2} \models T_*$. Then the group operation on $G_1' \times G_2'$ is definable without parameters, and further every element of $G_1 \times G_2$ is definable without parameters. Thus $(G_1 \times G_2, (g, h))_{g \in G_1, h \in G_2} \equiv (G_1' \times G_2', (g, h))_{g \in G_1, h \in G_2}$, since each sentence must be decided by $T_*$; this is the same as saying $G_1 \times G_2 \preceq G_1' \times G_2'$.<|endoftext|> -TITLE: Is simply typed lambda calculus with fixed-point combinator Turing-complete? -QUESTION [8 upvotes]: There are many sources cite that simply typed lambda calculus extended with fixed-point combinator is Turing complete. For example, Does there exist a Turing complete typed lambda calculus? or the following quote from Boltzmann Samplers for Closed Simply-Typed Lambda Terms (2017). - -Extended with - a fix-point operator, simply-typed lambda terms can be used as the intermediate - language for compiling Turing-complete functional languages. - -However, I haven't found proof of this statement. -Could you please suggest where I could find the proof or the guideline of how to prove this statement? Thank you. - -REPLY [5 votes]: Damiano is right, the answer is no: it is not Turing complete. -Supposing to define natural numbers as some type (o->o)->o->o, -even with fixpoints you cannot define the predecessor (that would be -enough to have equality, and then, by Goedel characterization via -mu-recursion, all computable functions). -In fact, by just adding fixpoints you only get partial polynomials -instead of the total polynomials of the traditional calculus. -This can also be seen by generalizing Schwichtenberg approach -(see e.g. Functions Definable in the Simply-Typed Lambda Calculus) -Instead of normal forms you must consider bohm-trees, now, that have -very limited shapes due to -typing constraints. Head variables in terms are the same as in the -traditional case, and the proof proceeds in a very similar way.<|endoftext|> -TITLE: Where exactly is the flaw in this proof of the Kronecker–Weber theorem? -QUESTION [9 upvotes]: Marvin J. Greenberg provided an elementary proof of the Kronecker–Weber theorem here (Amer. Math. Monthly, 81 (1974), no. 6, 601-607). -An argument in the lemma 4 was found to be wrong as noticed in Correction to "An Elementary Proof of the Kronecker-Weber Theorem" (Amer. Math. Monthly, 82 (1975), no. 8, 803): - -Joe L. Mott informed me that the argument for the case $m>1$ in Lemma 4, Volume 81, (1974) 606, is incorrect. What is correct is that $V_i$ is the unique subgroup of $G$ of index $\lambda$, where $i$ is the smallest index such that $V_i \neq G$. - -The lemma 4 is: - -Let $K$ be an abelian extension of $\Bbb Q$ of degree $\lambda^m$, $\lambda$ an odd prime, in which $\lambda$ is the only ramified prime. Then $K/\Bbb Q$ is cyclic. - -The case $m=1$ is dealt correctly. However, I wasn't able to see where exactly Greenberg's proof fails for the case $m>1$ (the numbering and square brackets are mine) : - -Returning to the case $m > 1$, we will show that $K/\Bbb Q$ is cyclic by showing $V_2$ [i.e. the second ramification group of $\lambda$ in $K$] is the unique subgroup of the Galois group $G = V_1$ of index $\lambda$. - Let $H$ be any subgroup of index $\lambda$ in $G = \mathrm{Gal}(K/\Bbb Q)$, $K'$ its fixed field, $G' \cong G /H$ the Galois group of $K'$ over $\Bbb Q$, $V_j'$ the $j$-th ramification group of $K'$. -(1) By restriction to $K'$, $V_j$ maps into $V_j'$. According to the sublemma [i.e. the case $m=1$], $V_2'$ is trivial. Hence $V_2 \leq H$. -(2) Applying this, in particular to the case where $H = V_j$ is the first ramification group which is not all of $G$, we see that $j = 2$ and $V_2$ has index $\lambda$. Hence $V_2$ is the unique subgroup of index $\lambda$. - -I see no problem with (1) : let $f : G \to G' \cong G/H$ be the restriction to $K'$. Then $f(V_j) \subset V_j'$. Since $V_2' = 1$, we get $V_2 \subset \ker(f) = H$. -I see no problem with (2), since $V_{j-1} / V_j$ is non trivial, and embeds in $O_K/(\lambda)$ (fact 3 in Greenberg's paper), which has cardinality $\lambda$ because $\lambda$ is totally ramified in $K$ (see the very beginning of the proof of Lemma 4). So $V_j$ has indeed index $\lambda$ in $G$. - -I tried to find a counter-example of an abelian extension $K/\Bbb Q$ whose degree and discriminant are both powers of an odd prime $\lambda$, such that the second ramification group $V_2$ of $\lambda$ (in $K$) is the whole Galois group, but it was without success. Maybe the following MAGMA code could help: -p := 3; r := 3; -a := RootOfUnity(p^r); -M := MinimalPolynomial(a + 1/a); #here K=NumberField(M) will be the subfield of Q(a) fixed by a subgroup of order p-1 in Gal(Q(a)) -R := PolynomialRing(RationalField()); -K := NumberField(M); -OK := RingOfIntegers(K); OK; -print " ";print " "; - -Gal, _, Map := AutomorphismGroup(K); Gal; -P3 := Decomposition(OK, 3)[1][1]; P3; -print " ";print " "; - -V2 := RamificationGroup(P3, 2); V2; -print "Cardinality of V2 is ", #V2; - -Thank you very much for your help. - -REPLY [6 votes]: As promised, here are my calculations of the higher ramification groups in certain cyclotomic extensions. -Let $G$ denote the Galois group of the field $L$ of $p^n$-th roots of unity, and consider the Hilbert subgroups for the completely ramified prime $p$. The decomposition group and the inertia both are equal to $G$. For computing the higher ramification groups, observe that if -$\sigma(\zeta) - \zeta \in {\mathfrak p}^j$ for some $j \ge 1$, then -$\sigma(\alpha) - \alpha \in {\mathfrak p}^j$ for all $\alpha \in {\mathbb Z}[\zeta]$. Thus we only need to look at the action of $\sigma$ on roots of unity. -Let $\sigma_a: \zeta \to \zeta^a$. Then clearly -$$ (\zeta^{a-1}-1) = \begin{cases} - {\mathfrak p} & \text{ if } a-1 \in {\mathbb Z} \setminus p{\mathbb Z}, \\ - {\mathfrak p}^p & \text{ if } a-1 \in p{\mathbb Z} \setminus p^2{\mathbb Z}, \\ - {\mathfrak p}^{p^2} & \text{ if } a-1 \in p^2{\mathbb Z} - \setminus p^3{\mathbb Z}, \\ - \cdots & \quad \cdots \\ - {\mathfrak p}^{p^{n-1}} & \text{ if } a-1 \in p^{n-1}{\mathbb Z} - \setminus p^n{\mathbb Z}. - \end{cases} $$ -This gives us the higher ramification groups: -$$ \begin{array}{lllll} - V_1 & = \ V_2 & = & \ldots = V_{p-1} & \simeq {\mathbb Z}/p^{n-1}{\mathbb Z}, \\ - V_p & = \ V_{p+1} & = & \ldots = V_{p^2-1} & \simeq {\mathbb Z}/p^{n-2}{\mathbb Z}, \\ - V_{p^2} & = \ V_{p^2+1} & = & \ldots = V_{p^3-1} & \simeq {\mathbb Z}/p^{n-3}{\mathbb Z}, \\ - & \ \ldots & & & \\ - V_{p^{n-2}} & = \ V_{p^{n-2}+1} & = &\ldots = V_{p^{n-1}-1} & \simeq {\mathbb Z}/p{\mathbb Z}, \\ - V_{p^{n-1}} & = \ 1 & & & - \end{array} $$ -Next we use Herbrand's theorem on the determination of ramification groups for subfields. Let $H$ be the subgroup of $G$ fixing the subextension $K$ of degree $p^{n-1}$; then $H$ has order $p-1$, and it only has the neutral element in common with the higher ramification groups. This implies that the higher ramification groups in $L/K$ are all trivial, and that $n_2 = n_3 = \ldots = 1$ in the notation of his article over here (modern representations of his result invoke the "upper numbering" which I haven't looked at for some time), and this in turn implies that the ramification groups "collapse" by a factor of $n_1 = p-1$. By this I mean the following: instead of $p-1$ ramification groups $\simeq {\mathbb Z}/p^{n-1}{\mathbb Z}$ there is only -one such group for $K/{\mathbb Q}$, namely $V_1$. Instead of $p^2-p$ ramification groups $\simeq {\mathbb Z}/p^{n-2}{\mathbb Z}$ there are now $p$ such groups, namely $V_2 = \ldots = V_{p+1} \simeq {\mathbb Z}/p^{n-2}{\mathbb Z}$, and so on. In particular, $V_2 \ne G$ whenever $K$ is cyclotomic of prime power degree. In particular there cannot exist examples with $V_2 = G$ by Kronecker-Weber. -As a sanity check I computed the discriminant of $K$ (in the case $n = 3$) using Hilbert's formula for the different and obtained disc$(K) = p^k$ with $k = 3p^2 - p - 2$, which agrees with calculations done by pari for $p =3, 5, 7$.<|endoftext|> -TITLE: A question about relation of the character degrees of $ G/N $ and $ G $ -QUESTION [6 upvotes]: Let $ G $ be a finite group and $ N $ be a normal subgroup of $ G $. Let $ G/N $ have two irreducible characters of degrees $ p_1$ and $ p_2$, where $ p_1$ and $ p_2$ are different primes. Let $ G/N $ have no irreducible character such that $ p_1p_2\mid \chi (1) $. If $ (p_1p_2, |N|)=1$, can we say that $G $ has no irreducible character $\chi $ such that $ p_1p_2\mid \chi (1) $? -I guess that this is impossible but I could not find any counterexample for it. - -REPLY [6 votes]: It is possible for $G$ to have such a character. The smallest example has order $120$. Let $N = \langle k \rangle \cong C_5$. Let $G = N \rtimes S_4$ where the action of the symmetric group $S_4$ on $N$ is non-trivial but factors through the sign representation. Thus $k^{(12)} = k^{-1}$, $k^{(234)} = k$ and $\langle k, (12)\rangle$ is dihedral of order $10$. The character degrees of $G$ are $\{1,2,3,6\}$ and the character degrees of $G/N \cong S_4$ are $\{1,2,3\}$.<|endoftext|> -TITLE: Is there a good computer program for searching for endomorphisms between finite algebras which make diagrams commute? Is this problem NP-complete? -QUESTION [8 upvotes]: Let $(X,*),(Y,*),(Z,*)$ be finite algebras. The binary operations $*$ are not required to satisfy any identities though I am interested in the special case where $*$ is associative. Suppose that $f:X\rightarrow Z,g:Y\rightarrow Z$ are homomorphisms. Then is there already an optimized computer program that searches for homomorphisms $\phi:X\rightarrow Y$ where $g\circ\phi=f$? Is the problem of finding such a homomorphism $\phi$ an NP-complete problem? Is this problem still NP-complete when all operations are associative? - -REPLY [6 votes]: I mentioned this problem to one of my students, Kevin Berg, who proved that this homomorphism factorization problem is NP-complete when considered for the class of all finite algebras in a language containing at least 2 unary operations or containing at least one operation of arity at least 2. He also proved that the problem remains NP-complete when restricted to algebras with one associative binary operation (i.e. for semigroups). His arxiv preprint is here: -The Complexity of Homomorphism Factorization<|endoftext|> -TITLE: Terminology for a monoid $H$ s.t. $xy \in H^\times$ only if $x, y \in H^\times$ -QUESTION [5 upvotes]: The title has it all. Is there any consolidated terminology for referring to a (multiplicative) monoid $H$ such that $xy \in H^\times$ (if and) only if $x, y \in H^\times$? Here is a short list of monoids with this property: - -Commutative monoids. -Unit-cancellative monoids (so, in particular, cancellative monoids). -Monoids with at least one atom/irreducible element (as proved by Benjamin Steinberg). - -Incidentally, the three classes of monoids from this list have, of course, non-empty intersection, but none of them is contained in the union of the others. -Edit. In a previous version of this post, I was writing that the property in the title holds if $H^\times = \{1_H\}$ (i.e., for reduced monoids). That's not true, as can be seen by looking at $\langle x, y \mid xy = 1 \rangle$. - -REPLY [10 votes]: The correct term is Dedekind finite. A monoid is Dedekind finite of $xy=1$ implies $yx=1$. This is clearly equivalent to your condition.<|endoftext|> -TITLE: Nodal curve in a smooth variety with injective map on fundamental groups -QUESTION [9 upvotes]: Let $C$ be the nodal curve obtained by gluing together the points $0$ and $1$ of $\mathbb{A}^1_{\mathbb{C}}$. The topological fundamental group of $C$ is isomorphic to $\mathbb{Z}$. -One can find an immersion of $C$ in a smooth connected variety such that the map between the fundamental groups is non-trivial. -For example, take $\mathbb{A}^1_{\mathbb{C}}\times \mathbb{Z}/n\mathbb{Z}$ and glue $(1,k)$ with $(0,k+1)$ cyclically for every $k$. This produces a curve $C_n$ with a natural $\mathbb{Z}/n\mathbb{Z}$-action. The quotient of $C_n$ by the action of $\mathbb{Z}/n\mathbb{Z}$-action is isomorphic to $C$. Take an immersion of $C_n$ in a smooth variety $X_n$ with a compatible free $\mathbb{Z}/n\mathbb{Z}$-action. The quotient of $C_n\hookrightarrow X_n$ by the action of $\mathbb{Z}/n\mathbb{Z}$ is an immersion $C\hookrightarrow X$, where $X$ is a smooth variety. The image of $\pi_1(C)$ in $\pi_1(X)$, by construction, has $\mathbb{Z}/n\mathbb{Z}$ as a quotient. In particular it is non-trivial. -Does there exist a smooth connected variety $X$ and an immersion $C\hookrightarrow X$ such that the induced map $\pi_1(C)\rightarrow \pi_1(X)$ is injective? - -REPLY [5 votes]: Let me expand my comment slightly. By Deligne [Théorie de Hodge II, III], the homologies of complex algebraic varieties carry functorial mixed Hodge structures dual to the ones cohomology. Among other things, this means that the (co)homologies carry weight filtrations $W$ which are strictly preserved by induced maps. -If $X$ is smooth then by construction, the possible weights of $H_1(X)$ are $-1,-2$, i.e. $Gr^W_iH_1(X)=0$ unless $i=-1,-2$. On the other hand, $H_1$ of your nodal curve $C$ has pure weight $0$. Therefore, for any map $C\to X$, the induced map on rational homology $H_1(C)\to H_1(X)$ must vanish. Using Morgan [The algebraic topology of smooth algebraic varieties], Malcev completions of $\pi_1$ carry mixed Hodge structures. (This is very roughly the inverse limit of the set of nilpotent quotients of $\pi_1$ upto torsion.) Using this, you can upgrade the argument to show that the map on Malcev completions $\hat{\pi}_1(C)\to \hat{\pi}_1(X)$ must vanish. This would settle your question in good cases.<|endoftext|> -TITLE: Questions about Mitchell forcing and tree property -QUESTION [6 upvotes]: For $\alpha < \beta$, where $\alpha$ is regular and $\beta$ is measurable, let $M(\alpha, \beta)$ be the Mitchell forcing for making $2^\alpha=\beta=\alpha^{++}$ and forcing the tree property at $\beta.$ -Now suppose $\alpha<\beta<\gamma$, with $\alpha$ regular and $\beta, \gamma$ measurable. The product forcing $M(\alpha, \beta) \times M(\beta, \gamma)$ -is the natural forcing for making $\alpha^{++}=\beta$ and $\alpha^{+4}=\gamma$ and forcing the tree property at both $\beta$ and $\gamma.$ -Question 1. Is it known if the same conclusion works with iteration $M(\alpha, \beta) \ast \dot{M}(\beta, \gamma)$? -If not, then I have the following question ($\alpha < \beta<\gamma$ are supposed to be as above): -Question 2. Is there a forcing notion $P(\alpha, \beta)$ which is $\alpha$-directed closed, $\beta$-c.c. of size $\beta$ such that it forces $2^\alpha=\beta=\alpha^{++}$ also it forces tree property at $\beta$ and such that the iteration $P(\alpha, \beta) \ast \dot{P}(\beta, \gamma)$ makes $2^\alpha=\beta=\alpha^{++}$ and forces the tree property at $\beta.$ -(clearly after the iteration tree property holds at $\gamma$, which is supposed to become $\alpha^{+4}$). -Giving references is appreciated. - -REPLY [4 votes]: At least in a variant of Mitchell's forcing, the iteration forces the tree property at both cardinals. -Let $M(\alpha, \beta)$ be the iteration $Add(\alpha, \beta) \ast \mathbb{C}(\alpha, \beta)$ where $p \in \mathbb{C}(\alpha, \beta)$ iff $p$ is a sequence of length $\beta$, with at most $\alpha$ non-trivial coordinates, and $p(\zeta)$ belongs to the collection of names for elements in $Add(\alpha^{+}, 1)^{V^{Add(\alpha, \zeta)}}$. Let $G\subseteq Add(\alpha, \beta)$ be a generic filter. $V[G] \models p \leq q$ iff for every $\zeta < \beta$, $p(\zeta)^{V[G\restriction \zeta]} \leq q^{V[G\restriction \zeta]}$. -The only difference between $M(\alpha, \beta)$ and Mitchell's forcing is that in this model we force at each step "first" with $Add(\alpha^{+}, 1)$ and then with $Add(\alpha, 1)$ and in Mitchell's forcing we use the opposite order. -Let $\alpha < \beta < \gamma$ be regular cardinals, with $\beta$, $\gamma$ measurable (weakly compact is enough, but it is simpler). Let us show that the tree property holds at $\beta$ in the generic extension (the proof for $\gamma$ is easier). -Let $\dot{T}$ be a $M(\alpha, \beta) \ast \dot M(\beta, \gamma)$-name for a $\beta$-tree. By the $\beta^{+}$-distributivity of the $\mathbb{C}$-part of $M(\beta, \gamma)$, $\dot{T}$ is added by $M(\alpha, \beta) \ast \dot{Add}(\beta, \gamma)$. By the chain condition and the homogeneity of $\dot{Add}(\beta, \gamma)$, $\dot T$ is equivalent to an $M(\alpha, \beta) \ast Add(\beta, 1)$-name. Let $\mathbb{Q} = M(\alpha, \beta) \ast Add(\beta, 1)$. -Let $j\colon V\to M$ be an elementary embedding with critical point $\beta$. Let $H\subseteq \mathbb{Q}$ be a generic filter. - $$j(\mathbb{Q}) = M(\alpha, \beta) \ast Add(\alpha^{+}, 1) \ast \mathbb{R} \ast Add(j(\beta), 1)$$ -for some $\mathbb{R}$. -Since $Add(\alpha^{+},1)$ is equivalent to $Col(\alpha^{+}, 2^\alpha)$ and after forcing with $M(\alpha, \beta)$, $2^\alpha = \beta$, it adds a generic filter for $Add(\beta, 1)$. Thus, we can represent $j(\mathbb{Q})$ in the following way: -$$j(\mathbb{Q}) \cong M(\alpha, \beta) \ast Add(\beta, 1) \ast Col(\alpha^{+}, \beta) \ast \mathbb{R} \ast Add(j(\beta), 1)$$ -Using the closure of $Add(j(\beta), 1)$, one can obtain a master condition and force a generic filter $K$ over $M[H]$ such that $H \ast K$ is a generic filter for $j(\mathbb{Q})$ and $j \text{''} H \subseteq H \ast K$. Therefore, in $V[H][K]$ there is an elementary embedding $\tilde{j} \colon V[H] \to M[H][K]$ extending $j$. In particular, $\tilde{j}(\dot{T}^H)$ is $j(\beta)$-tree and thus by taking any element from the $\beta$-th level of $\tilde{j}(\dot{T}^H)$ one can obtain a branch in $\dot{T}$. -Let us show that the forcing $$Col(\alpha^{+}, \beta) \ast \mathbb{R} \ast Add(j(\beta), 1)$$ -(that introduced $K$) cannot add a branch to a $\beta$-tree in $V[H]$. Indeed, this forcing is a projection of the product $Add(\alpha, j(\beta)) \times Col^{V^\mathbb{Q}}(\alpha^{+}, j(\beta))$. By standard arguments, this forcing cannot add a branch to a $\beta$-tree in the model $V[H]$.<|endoftext|> -TITLE: Inequality for Stirling numbers of the second kind -QUESTION [7 upvotes]: I stumbled upon the following inequality which, I believe, is true. I was able to prove it for small k, but I have no proof for the general case. Any help is welcome. - -Let $n\geq k\geq 1$ then - $$\left(1+\frac{1}{n}\right)^k S(n+1,k+1)\geq -\left(1+\frac{1}{k}\right)^n S(n,k)$$ - where $S(n,k)$ is a Stirling number of second kind. - -REPLY [2 votes]: Let me prove a weaker inequality, the proof of which allows further improvements. Note that $S(n,k)=h_{n-k}(1,2,\dots,k)$, where $$h_t(x_1,\dots,x_k)=\sum_{a_1+\dots+a_k=t,a_i\geqslant 0} x_1^{a_1}\dots x_k^{a_k}$$ is a complete symmetric polynomial. Thus $$\frac{S(n+1,k+1)}{S(n,k)}=\frac{h_{n-k}(1,2,\dots,k+1)}{h_{n-k}(1,2,\dots,k)}=\frac{\sum(p_1+1)\dots (p_{n-k}+1)}{\sum p_1\dots p_{n-k}},$$ -where summation is taken over $0\leqslant p_1\leqslant p_2\leqslant \dots \leqslant p_{n-k}\leqslant k$. For any such $(n-k)$-tuple we have -$$(p_1+1)\dots (p_{n-k}+1)\geqslant \left(1+\frac1k\right)^{n-k}p_1\dots p_{n-k},$$ -summing up we get $$S(n+1,k+1)\geqslant \left(1+\frac1k\right)^{n-k} S(n,k),$$ -this is worse than you need, but the inequality was not sharp enough. Roughly speaking, you need to prove that "in average" the ratio $(1+1/p_1)\dots (1+1/p_{n-k})$ grows as $(1+2/k)^{n-k}$, that corresponds to "average" $p_i\sim k/2$.<|endoftext|> -TITLE: An attempt to generalize the previous inequality -QUESTION [10 upvotes]: In my previous MO question, the inequality was about a specific series and nicely answered by Cherng-tiao Perng. After testing with a few more numerical infinite sums, I came to realize that perhaps more is true. - -Does the following hold true for any convergent series? - $$\left(\sum_{k\geq1}t_k\right)^4 -<\pi^2\left(\sum_{k\geq1}t_k^2\right)\left(\sum_{k\geq1}k^2t_k^2\right).$$ - Is the constant $\pi^2$ optimal? - -REPLY [8 votes]: Following Cherng-tiao Perng's proof for the other inequality, here is what I find. I don't have immediate access to Folkmar Bornemann's references, so I'm not sure the ideas here might be similar. No originality is claimed. -Update. A short visit to the library confirms that Hardy's paper contains the below proof, only details are added for the reader's convenience. -Begin by writing $t_k=\frac1{\sqrt{\alpha+\beta k^2}}t_k\sqrt{\alpha+\beta k^2}$, for $\alpha, \beta>0$ to be specified later. Then, we apply the Cauchy-Schwarz inequality as follows -\begin{align}\left(\sum_kt_k\right)^2&=\left(\sum_k\frac1{\sqrt{\alpha+\beta k^2}}t_k\sqrt{\alpha+\beta k^2}\right)^2 -\leq\left(\sum_{k=1}^{\infty}\frac1{\alpha+\beta k^2}\right)\left(\sum_{k=1}^{\infty}(\alpha+\beta k^2)t_k^2\right)\\ -&<\left(\int_0^{\infty}\frac{dx}{\alpha+\beta x^2}\right)\left(\alpha\sum_{k=1}^{\infty}t_k^2+\beta\sum_{k=1}^{\infty}k^2t_k^2\right) -=\frac{\pi}{2\sqrt{\alpha\beta}}\left(\alpha A+\beta B\right); -\end{align} -where we denoted $A:=\sum_kt_k^2$ and $B:=\sum_kk^2t_k^2$. Of course, $\int_0^{\infty}\frac{dx}{\alpha+\beta x^2}=\frac{\pi}{2\sqrt{\alpha\beta}}$ is from Calculus. At this stage, we make a convenient choice of $\alpha=\sqrt{\frac{B}A}$ and $\beta=\sqrt{\frac{A}B}$. Clearly, $\alpha\beta=1$. So, -$$\left(\sum_kt_k\right)^2<\frac{\pi}{2\sqrt{\alpha\beta}}\left(\alpha A+\beta B\right)=\frac{\pi}2\left(\sqrt{\frac{B}A}A+\sqrt{\frac{A}B}B\right)= -\frac{\pi}2\left(\sqrt{AB}+\sqrt{AB}\right)=\pi\sqrt{AB}.$$ -Squaring both sides and replacing $A$ and $B$, we obtain the desired inequality -$$\left(\sum_kt_k\right)^4<\pi^2AB=\pi^2\sum_kt_k^2\sum_kk^2t_k^2.\qquad \square$$ -Remark. $\sum_{k=1}^{\infty}\frac1{\alpha+\beta k^2}<\int_0^{\infty}\frac{dx}{\alpha+\beta x^2}$ is due to Lower Riemann Sums with partition $\{0,1,2,3,4,\dots\}$. -Remark. I've to find Hardy's paper to see why $\pi^2$ is optimal.<|endoftext|> -TITLE: "Good" edge-colorings -QUESTION [16 upvotes]: Let $n >1$ be an integer, and suppose $G = (V,E)$ is a simple undirected graph with $V = \{1,\ldots,n\}$. For $v\in V$ set $N(v) = \{w\in V: \{v,w\} \in E\}$. -It is known by Vizing's theorem that the edges of $G$ can be colored with $\Delta(G)+1$ colors (where $\Delta(\cdot)$ denotes the maximum degree), and of course we have $\Delta(G)+1 \leq n$. -We call an edge-coloring $c:E\to \{1,\ldots,n\}$ good if for all $x\neq y\in V$ with $\{x,y\} \in E$ we have $c(\{x,y\}) \in N(x)\cup N(y).$ - -Does every graph $G=(\{1,\ldots,n\},E)$ have a good edge-coloring? - -REPLY [2 votes]: I don't have a solution yet, but I propose to call this a local edge coloring instead of a good one. -Basically, the nodes are colored with the colors 1 to n here, and the edges should be colored with the same colors. In addition to that, the set of colors that are allowed for an edge $\{u,v\}$ is restricted to $N(u) \cup N(v)$, i.e. to the colors that are locally available in the neighborhood of $u$ and $v$. -We could call an edge coloring strongly local, if the palette for edge $\{u,v\}$ is restricted to $u$ and $v$ themselves, and local, if it is only restricted to $N(u) \cup N(v)$. -Examples: - -Any circular graph $C_n$ has a strongly local edge coloring. -The complete graph $K_4$ does not have a strongly local edge coloring. -Any $K_n$ has a local edge coloring, because $N(u) \cup N(v) = V$, so the locality condition does not impose a real restriction in complete graphs.<|endoftext|> -TITLE: History of optional sampling/stopping theorem -QUESTION [6 upvotes]: Does anyone have a good explanation of the name, and why Doob chose it? It states the following: if $T$ is a stopping time such that $\mathbb{P}(T < \infty)$, and $M_n$ is a uniformly integrable martingale, then -$$ -E[M_T] = E[M_0]. -$$ - -REPLY [4 votes]: "Optional stopping" and "optional sampling" refer to a strategy of peeking while you are sampling and then, based on what you find, exercise the option to quit sampling. -Doob's theorem states that in a fair casino, where your return is a martingale, you cannot increase your expected return if you are given the option to stop betting (= sampling) when you think "it's time to quit". -The names originate from J. L. Doob, Stochastic Processes (Wiley, 1953): page 300, 366. Doob uses "optional sampling" for the stochastic process itself, and "optional stopping" for the random variable that represents the stopping time.<|endoftext|> -TITLE: What is Mumford's example of a normal complex algebraic surface $X$ with non-torsion elements in $H^2_{et}(X,\mathbb{G}_m)$? -QUESTION [6 upvotes]: I have heard that Mumford has constructed an example of a normal complex algebraic surface $X$ such that $H^2_{et}(X,\mathbb{G}_m)$ contains a non-torsion element. But I cannot find the reference. -Then what is the surface $X$ and how to construct a non-torsion element in $H^2_{et}(X,\mathbb{G}_m)$? - -REPLY [4 votes]: In [Milne, Étale Cohomology], p. 146, Remark IV.2.8, there is a reference to [Grothendieck, Le groupe de Brauer. In: Dix Exposés sur la Cohomologie des Schémas], II.1.11b. (see https://webusers.imj-prg.fr/~leila.schneps/grothendieckcircle/DixExp.pdf)<|endoftext|> -TITLE: A differentiable isometry is smooth? -QUESTION [21 upvotes]: I posted this question in MSE but got no response (even after giving a bounty), so I am trying here. -Let $M,N$ be smooth $d$-dimensional Riemannian manifolds. -Suppose $f:M \to N$ is a differentiable isometry ($df_p$ is an isometry at every $p \in M$). I do not assume $f$ is $C^1$. -Is it true that $f$ must be smooth? -This is false when assuming $f$ is only differentiable almost everywhere. -One counter-example is $f(x)=c(x)+x$ where $c$ is the Cantor function. In this case $M=[0,1],N=[0,2]$, $f'=1$ a.e; $M,N$ are not isometric, but both are flat. -Gromov showed similar examples exist for "many" non-flat $M$ and $N=\mathbb{R}^d$. - -Some partial results: -By the inverse function theorem for everywhere differentiable maps, $f$ is a local homeomorphism. -We would like to prove it's a local isometry w.r.t the intrinsic distances, then use the Myers-Steenrod theorem (or use the fact geodesics locally minimize length to deduce $f$ maps geodesics to geodesics, hence it factors through its differential via the exponential maps. this works if $f \in C^1$, see here for details). -The devil is in the details: We need to choose suitable length structures on $M,N$ such that $f$ will become an arcwise isometry. -We cannot use the class of $C^1$ paths since $f$ does not necessarily map $C^1$ maps to $C^1$ maps. -Also, nothing promises us that for a differentiable path $\gamma$ such that $\|\dot \gamma(t)\|$ is integrable, $\|\dot {f \circ \gamma}(t)\|$ will be integrable. - -REPLY [6 votes]: By the nontrivial fact that $f$ is a local homeomorphism, we can assume without loss of generality that $f$ is a bijective homeomorphism. The usual textbook proof of the formula for the differential of the inverse map works here, so that $f^{-1}$ is again a differentiable isometry in the sense of the OP. -So it suffices to show that $f$ is 1-Lipschitz (as a map from $M$ to $N$, which are thought as metric spaces), since then $f^{-1}$ will be 1-Lipschitz as well, i.e. $f:M\to N$ will be a metric space isometry and we will be done thanks to the Myers-Steenrod theorem. -Fix $p,q\in M$ and take any $C^1$ curve $\gamma:[0,1]\to M$ with $\gamma(0)=p$ and $\gamma(1)=q$. Call $L$ the length of $\gamma$. It suffices to prove that $d(f(p),f(q))\le L$ (here $d(\cdot,\cdot)$ denotes the distance), since then, taking the infimum of $L$ among all curves connecting $p$ to $q$, we get $d(f(p),f(q))\le d(p,q)$. -To this aim, fix any $\epsilon>0$. - -Claim: For any $t\in [0,1]$ we can find some $\delta(t)>0$ such that it holds - $$d(f\circ\gamma(s),f\circ\gamma(u))\le(1+\epsilon)d(\gamma(s),\gamma(t))+(1+\epsilon)d(\gamma(t),\gamma(u))$$ - whenever $t\in[s,u]$ and $u-s\le \delta$. - -Indeed, using normal coordinates about $\gamma(t)$ and $f\circ\gamma(t)$, we see that for some $\eta(t)>0$ we have -$$d(x,\gamma(t))\le\eta(t)\Rightarrow d(f(x),f\circ\gamma(t))\le(1+\epsilon)d(x,\gamma(t)),$$ -so we are done by taking $\delta(t)$ such that $\gamma([0,1]\cap \overline B_{\delta(t)}(t))\subseteq \overline B_{\eta(t)}(\gamma(t))$. -Now, by Cousin's theorem, we can find a partition $0=t_0<\cdots -TITLE: What is the term for this type of matrix? -QUESTION [6 upvotes]: Is there an established term for the following type of square matrices? -$\begin{pmatrix} -c & c & c & c & \cdots & c & c \\ -c & a & b & b & \cdots & b & b \\ -c & b & a & b & \cdots & b & b \\ -c & b & b & a & & b & b \\ -\vdots & \vdots & \vdots & & \ddots & & \vdots \\ -c & b & b & b & & a & b \\ -c & b & b & b & \cdots & b & a \\ -\end{pmatrix}$ -The matrix contains just 3 different items $a, b, c$: - -The first row is $c$. -The first column is $c$. -The diagonal is $a$, except for the upper left corner. -The remaining items are $b$. - -Background: $a, b, c$ can be chosen such that the matrix is orthogonal, but has a constant first row. If the dimension is a square (i.e. the matrix is a $r^2 \times r^2$ matrix) then it is possible to choose all entries to be integers - up to a common (usually irrational) factor in front of the matrix for normalization. - -REPLY [7 votes]: When $b = 0$, we have an $n \times n$ symmetric arrowhead matrix. When $b \neq 0$, we have -$$\begin{bmatrix} -c & c & c & \cdots & c & c \\ -c & a & b & \cdots & b & b \\ -c & b & a & \cdots & b & b \\ -c & b & b & \cdots & b & b \\ -\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ -c & b & b & \cdots & a & b \\ -c & b & b & \cdots & b & a \\ -\end{bmatrix} = \begin{bmatrix} -c-b & c-b & c-b & \cdots & c-b & c-b \\ -c-b & a-b & 0 & \cdots & 0 & 0 \\ -c-b & 0 & a-b & \cdots & 0 & 0 \\ -c-b & 0 & 0 & \cdots & 0 & 0 \\ -\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ -c-b & 0 & 0 & \cdots & a-b & 0 \\ -c-b & 0 & 0 & \cdots & 0 & a-b \\ -\end{bmatrix} + b \, 1_n 1_n^{\top}$$ -which is the sum of a symmetric arrowhead matrix and a (nonzero) multiple of the all-ones matrix.<|endoftext|> -TITLE: How are the ratios of successive values of the divisor function distributed? -QUESTION [6 upvotes]: One motivation for this question is a paper by Erdos and Hall "Values of the divisor function on short intervals", in which the authors obtain the leading asymptotics -$$x(\log x)^{2(\sqrt 2-1)-\epsilon}\leq\sum_{n\leq x}\min{(d(n),d(n+1))}\leq x(\log x)^{2(\sqrt 2-1)}$$ and -$$\sum_{n\leq x}\max{(d(n),d(n+1),...,d(n+k-1)})\sim kx\log x.$$ In particular, it is possible that the answer to this question will lead to refinements of these bounds. It is also interesting simply as a curiosity. -Let $r(n)=d(n+1)/d(n)$. Heath-Brown proved that $r(n)=1$ for infinity many $n$, but I don't know how often this is expected to happen. Recent advances in knowledge of the gaps between primes suggests that small and large values are common. - - -What is known about the distribution of values of $r(n)$? - -In particular, is it known that $r(n)$ is neither bounded above or below? - - - -EDIT: $r(n)$ is neither bounded above or below - the proof is easy so I suppose that's not news though. - -REPLY [4 votes]: Under GRH, Titchmarsh showed 1931 that $\sum_{p\leq x}\tau(p+a)\sim C(a)x$, where summation runs over primes only. 1963 Linnik proved the same unconditionally, hence there are many $n$ such that $d(n)=2$, $d(n+1)\gg\log n$ and similarly for $n-1$. Hence $r(n)$ is quite often as big as $\log n$ and as small as $\frac{1}{\log n}$. -In general $r(n)$ is much closer to 1. Using some standard techniques in probabilistic number theory you can show that the joint distribution of $(\omega(n), \omega(n+1))$ is normal with mean and variance $\log\log x$, hence the distribution of $\log r(n)$ is normal with variance $2\log\log n$. -You can construct extreme values of $r(n)$ by taking an integer $n$ with many divisors, and look for a prime $p\equiv \pm 1\pmod{n}$. By Heath-Browns version of Linnik's theorem, for $n>n_0$ such a prime $p$ exists satisfying $p\leq n^{5.5}$, hence $\max_{p\leq x} d(p+1)\geq \max_{n\leq x^{2/11}}d(n)$, thus for some small $c>0$ we have that $r(n)>e^{c\log n/\log\log n}$ has infinitely many solutions.<|endoftext|> -TITLE: Classify all the fields with abelian absolute Galois group -QUESTION [11 upvotes]: I'm wondering if anyone has classified all the fields $K$ such that $Gal(\bar{K}/K)$ is abelian? -The only examples I'm aware of are: finite fields, the real numbers $\mathbb{R}$ and $k((T))$ where $k$ is any algebraically closed field of characteristic 0. (One might also add any algebraically closed field as an example not listed above...) -Is there any other example of such fields? -An possibly easier question would be: can one apriori prove that $Gal(\bar{K}/K)$ is procyclic as suggested by the examples above? - -REPLY [16 votes]: Geyer in Unendliche algebraische Zahlkörper, über denen jede Gleichung auflösbar von beschränkter Stufe ist, Satz 1.13 and the paragraph after that, gives a full characterization of which abelian profinite groups occur as absolute Galois groups: They are either $\mathbb{Z}/2\mathbb{Z}$ or $\prod_p\mathbb{Z}_p^{c(p)}$ for some cardinal numbers $c(p)$. -Side remark: For algebraic extensions of $\mathbb{Q}$, abelian absolute Galois groups are in fact cyclic, see Satz 2.3 in the same paper.<|endoftext|> -TITLE: Product dimension of graphs -QUESTION [8 upvotes]: Given a simple undirected graph $G$, its product dimension is defined as the least positive integer $k$ such that $G$ embeds into a direct product of $k$ complete graphs. In a paper published in 1980, Lovasz et al state the following theorem. -Let $D_k$ denote the graph on $k$ vertices and no edges and $n\ge 2$. If $k>(n-1)!$ then the product dimension of $K_n+D_k=n+1$. -They provide a proof for the case $k> n!$ and for the other case they say that it follows from an unpublished result by Deza and Frankl. I am trying to find a proof for the other case. Hence this question. - -REPLY [4 votes]: Frankl and Deza initiated the study of families of intersecting permutations. There is a lot of literature on this subject and some very hard results but let's focus on an easy result which was one of their early observations. Call two permutations $\pi,\pi'\in S_n$ intersecting if they agree in some position, or in other words $\pi(i)=\pi'(i)$ for some $i$. An intersecting family of permutations is a subset of $S_n$ where any two permutations are intersecting. -Claim: An intersecting family in $S_n$ has size at most $(n-1)!$. -Proof: Notice that if two permutations lie in the same coset of the subgroup generated by an n-cycle, then they can't intersect. Since there are (n-1)! such cosets we get the desired bound. - -Let's explain how this relates to the result about product dimensions. The direct product of $n$ complete graphs $K_{r_1},\dots, K_{r_n}$ can be described by taking vertices to be $n$-tuples $u=(u_1,\dots,u_n)$ and an edge between any pair unless they agree on some coordinate (in other words, it is the complement of the rook graph on the $r_1\times\cdots\times r_n$ grid). -Suppose that $K_n+D_k$ can be embedded in the product of $n$ complete graphs. Let's label the vertices of $K_n$ as $\{w_1,w_2,\dots,w_n\}$ and let's show that to any vertex $v$ of $D_k$ we can associate a permutation in $S_n$. By the interpretation above, $v$ and $w_i$ agree on at least one coordinate, while the $w_i$'s do not agree on any coordinate. Denote by $\pi(i)$ the index of the coordinate where $v$ and $w_i$ agree. This is a permutation that uniquely determines $v$. -Now, the family $\pi_1,\dots,\pi_k$ is intersecting, since there are no edges between any two vertices of $D_k$. Therefore we must have $k\le (n-1)!$ from our earlier claim.<|endoftext|> -TITLE: Isomorphism from $\mathbb{Z}$ to third homotopy group of compact simple Lie group -QUESTION [7 upvotes]: Let $G$ be a compact connected simple Lie group. It is known that its third homotopy group $\pi_3(G)$ is isomorphic to $\mathbb{Z}$. More precisely, there is a Lie group homomorphism -$$\rho:SU(2)\longrightarrow G$$ -which induces an isomorphism -$$\rho_*:\pi_3(SU(2))\overset{\cong}{\longrightarrow}\pi_3(G).$$ -(Recall that $SU(2)\cong S^3$ so $\pi_3(SU(2))=\Bbb Z$.) -Clearly, all conjugates of $\rho$ have the same property. Is there a Lie group homomorphism $\varphi:SU(2)\to G$ such that $\varphi_*=\rho_*$ but $\varphi$ is not conjugate to $\rho$? - -REPLY [11 votes]: No, you cannot have $\varphi_*=\rho_*$ unless $\varphi$ is conjugate to $\rho$. -In fact, as observed by Atiyah-Hitchin-Singer (1978, p. 455), $\varphi_*$ regarded as a map $\mathbf Z\to\mathbf Z$ is just multiplication by the so-called Dynkin index of $\varphi_{\mathbf C}:\text{SL}(2,\mathbf C)\to G_{\mathbf C}$, i.e. "the ratio of the invariant inner products on both Lie algebras, each normalized to make the length of the highest root 2". -And as observed by Dynkin himself (1952, Thm 2.4), all $\varphi$ of index 1 are conjugate (corresponding, as Allen Knutson says, to the minimal nilpotent orbit in $\mathfrak g_\mathbf{C}$): see the translation in (2000, p. 197), or for another proof, Distler-Garibaldi (2010, Lemma 4.5).<|endoftext|> -TITLE: Does the group G(K) have a cocompact solvable closed subgroup? -QUESTION [5 upvotes]: Let $K$ be a (locally compact) local field and $G$ be a linear algebraic $K$-group. -Does the topological group $G(K)$ have a cocompact solvable closed subgroup? -If $\mathrm{char}(K)=0$, it is true that $G(K)$ has a cocompact solvable closed subgroup, see Proposition 9.3 in "A. Borel and J. Tits: Groupes reductifs" (Publ. IHES, 1965: link). So the only case left is when the characteristic is positive. -It is also well-known that for any local field $K$, $\mathrm{GL}(n,K)$ has a cocompact solvable closed subgroup. -What about $G=H_{2n+1}\rtimes \mathrm{Sp}_{2n}$, which is the semidirect product of the Heisenberg group $H_{2n+1}$ with the symplectic group $\mathrm{Sp}_{2n}$? - -REPLY [3 votes]: The answer below (is a detailed edit of a previous answer which) provides a (more-or-less) self contained proof of the fact that $\mathbf{G}(K)$ contains a cocompact solvable closed subgroup. -Let $K$ be a local field and $\mathbf{G}$ a linear algebraic group defined over $K$. I denote $G=\mathbf{G}(K)$ and assume as I may that $G$ is Zariski dense in $\mathbf{G}$. Note that by [Borel's "Linear Algebraic Group, AG14.4] the Zariski closure of $G$ in $\mathbf{G}$ is itself a $K$-algebraic group whose $K$-points set is again $G$, thus replacing $\mathbf{G}$ with the Zariski closure of $G$ justifies this assumption. -Below topological notions, unless otherwise said, are taken with respect to the Hausdorff (ie $K$-analytic) topology. -Lemma 1: -If $G$ is not solvable then $\mathbf{G}$ has an irreducible $K$-representation of dimension $>1$. -Proof: -Assuming every $K$-representation is 1-dimensional, fixing an injective $K$-representation $\mathbf{G}\to \text{GL}_n$ the image of $G$ is easily seen to be in a $K$-Borel subgroup of $\text{GL}_n$ (here we use the fact that $G$ is Zariski-dense in $\mathbf{G}$). -Lemma 2: -If $G$ is not solvable then there exists a $K$-algebraic variety $\mathbf{X}$ endowed with a $K$-action of $\mathbf{G}$ and a point $x\in X=\mathbf{X}(K)$ which is not $G$-fixed such that $Gx$ is compact. -Proof: -Using lemma 1, we fix an irreducible $K$-representation $\mathbf{G}\to \text{GL}_n$ with $n>1$ -and consider the associated $K$-action of $\mathbf{G}$ on $\mathbf{X}=\mathbb{P}^{n-1}$. -Observe that $X=\mathbf{X}(K)$ is compact. -By Zorn Lemma, using compactness, -we can find a minimal closed, non-empty, $G$-invariant subset $Y\subset X$. -We fix such $Y$ and a point $x\in Y$. -By the fact that $n>1$ and the representation is $K$-irreducible we get that $x$ is not $G$-fixed. -Note that $\overline{Gx}$ is a non-empty $G$-invariant closed subspace of $Y$, thus by minimality, $Y=\overline{Gx}$. -We recall that the action of $G$ on $X$ has locally closed orbits (this is proved in the appendix of http://www.math1.tau.ac.il/~bernstei/Publication_list/publication_texts/B-Zel-RepsGL-Usp.pdf). -It follows that $\partial(Gx)$ is a proper closed invariant subset of $Y$, hence by minimality $\partial(Gx)=\emptyset$. -We conclude that $Y=Gx$ and indeed $Gx$ is compact. -Corollary: -If $G$ is not solvable then there exists a proper $K$-algebraic group $\mathbf{H}<\mathbf{G}$ such that $H=\mathbf{H}(K)$ is cocompact in $G$. -Proof: -Fix $\mathbf{X}$ and $x\in X$ as in Lemma 2 and let $H$ be the stabilizer of $x$ in $G$ and $\mathbf{H}$ be the Zariski-closure of $H$ in $\mathbf{G}$. -By [Borel's "Linear Algebraic Group, AG14.4], $\mathbf{H}$ is a $K$-group and clearly $H=\mathbf{H}(K)$ (note however that $\mathbf{H}$ might not be the stabilizer of $x$ in $\mathbf{G}$). -We get a continuous bijective $G$-map $G/H\to Gx$. By a standard argument in the theory of locally compact groups (using Baire category) this map is open, thus $H -TITLE: rate of equidistribution of the horocycle flow for $SL(2, \mathbb{Z})$ -QUESTION [6 upvotes]: I know that for any Fuchsian group $\Gamma$, there is a spectral gap, which leads to -$$ \left| \int_0^1 F(x + iy) \, dx - \int_{\Gamma \backslash \mathbb{H}} F \, \frac{dx \, dy}{y^2} \right| < C_F y^\delta $$ -this is related to the equidistribution of the horocycle flow in the hyperblic plane. -Possibly I need to say that $F$ is smooth ($C^\infty$) or $C^1$ or holomorphic or a cusp form. If I knew more about the theory of modular forms or dynamical systems I could say which one. In the literature the theorems often say, "for reasonable $F$..." -For a particular $\Gamma$ do we know which that $\delta$ should be? Or is it even true that we can choose the same $\delta$ for a wide range of $\Gamma$? The two cases I have in mind are: - -$\Gamma = \mathrm{SL}(2, \mathbb{Z})$ -$\Gamma = \Gamma_0(4)$ -- a congruence subgroup - -As you can see, my knowledge of fuchsian groups, modular forms, dynamical systems are not up to date. -I am looking for $\delta$ that work specifically in these two cases. I think I read somewhere that $\delta \geq \frac{1}{2}$ is always possible. Is this merely the same as a spectral gap for $\Gamma$ ? - -REPLY [7 votes]: While Peter Humphries' answer is entirely correct for the question asked by the OP, the technique indicated there is far from addressing the most general situation. -The most basic technique towards this problem, given by Margulis' in his celebrated thesis for general Anosov flows, is what's known as the flow-box argument, which essentially means (in the case we have at hand) to thicken a period, and then push it along the geodesic flow (split-cartan) direction, as the unipotent direction is horospheric wrt the cartan action, it grows, while the cartan direction stays natural and the opposite unipotent direction gets shrunk (this is the property of the Anosov splitting), using this simple geometrical observation, together with quantitative mixing estimates for the Cartan action (which are merely quantification of the Howe-Moore theorem) gives you this result. The most basic mixing estimate is due to Harish-Chandra (say for $K$-finite vectors, which are the analogues of trigonometrical polynomials in this case), later it has been generalized to practically any smooth vectors in any Sobolov space (HC, Howe, Moore,Ratner), and the most general form (say for any Holder-cont. function) is due to Kleinbock-Margulis and Katok-Spatzier. The exact rate of mixing is obviously reflected in the spectral decomposition of your space (in the non arithmetic case you will need to use Selberg's (or maybe Langlands') results, which are more complicated), as Peter mentioned in his answer. -This approach is described nicely in Manfred's book. -Moreover, as this approach is holistic, it works for any lattice, and your question then transformed into the question of estimating the spectral gap (more correctly, property $(\tau)$) for a set of lattices uniformly, namely if this set of lattices is a family of expanders, and nowadays we know quite a few of those (for example, Selberg's theorem tells you that for principal congruence subgroups). -For $SL_2 (\mathbb{Z})$ and in theory for any principal congruence subgroup, you can do better, by specific analysis of the constant term of the Eisenstein series, and this was first done by Peter Sarnak in one of his first articles, if I recall correctly he shows there that for $SL_{2}(\mathbb{Z})$ $\delta=3/4-\epsilon$ is equivalent to RH (probably you get worse epsilon and need GRH to handle the principal congruence case, but it should work, this relies on explicit description of Eisenstein series in the end). -This approach takes you further than Margulis' argument I've mentioned before, but it has limitations (can handle only principal congruence, very computational heavy, Margulis' method can be bootstrapped to quantify the Furstenberg/Dani-Smillie equidistribution theorem where this spectral approach is quite limited towards this problem). -Burger (and later Strombergsson, Flaminio-Forni) found a different approach, which relies much less on the artihmetics and gives you great results for practically any lattice (this deal with the equidistribution result of Furstenberg (Burger) and Dani-Smillie (Strombergsson), but it is not hard to see that the period theorem follows from this theorem). -Peter Humphries mentioned thin subgroups (infinite index) in his answer, so to make things clear, those are not lattices, and their spectral theory is different (and more complicated) due to Patterson-Sullivan and Lax-Phillips, and we do know many of expander type results for families there due to recent developments in arithmetic combinatorics (say the Bourgain-Gamburd technique). The problem there is to define the equidistirbution problem correctly, namely to which measure (as it is of infinite volume, the constants are not integrable and cannot serve as a test function, and one needs to consider different ways to average, for example consider Hopf's ergodic theorem instead of the regular one, or change to the Burger-Roblin measure, etc.) I will refer you to the magnificent articles of Oh-Mohammadi and Oh-Winter to see the results in those cases.<|endoftext|> -TITLE: Some questions about the map $K_0(\text{Var})\to K_0(\text{Mot})$ -QUESTION [8 upvotes]: Let $k$ be a field. The naive Grothendieck ring of varieties $K_0(\text{Var})$ is generated by isomorphism classes of varieties over $k$ with the scissors relation $[X]=[X-Y]+[Y]$ for $Y$ a closed subvariety of $X$. We might naturally want to compare it with $K_0(\text{Mot})$, where here I want to remain flexible with what I mean by $\text{Mot}$ - for the generic situation, maybe pure effective motives over $k$ with either $\mathbb{Z}$ or $\mathbb{Q}$-coefficients - just as long as we have a symmetric monoidal category to apply $K_0$ to. If the situation is more understood after localizing or with a different equivalence relation, I'd be happy to consider that instead. -In characteristic zero, I can see how to define a map $K_0(\text{Var})\to K_0(\text{Mot})$; we just need "excision for motives," i.e. an agreement of the motives corresponding to $[X]-[Y]$ and $[W]-[Z]$ if $X-Y\cong W-Z$ as affine varieties. This follows from the fairly recent and difficult result of weak factorization in characteristic zero; we get a rational map from $X$ to $W$ which can be factored through series of blow-ups and blow-downs, and then Manin's computation of the motive of a blow-up does the rest. Singular varieties can be handled just fine for the same reason by Hironaka's algorithm. This map then, conjecturally, should be the semisimplification of the "mixed motive" associated to a general variety over $k$. -In characteristic $p$, we don't have resolution of singularities, nor weak factorization - so, is the very existence of this map out of reach? -(1) (How) can we define this map in positive characteristic? -Poonen proved that $K_0$ is not a domain for $k$ of characteristic zero; the case over $\mathbb{C}$ is easiest to illustrate: take two elliptic curves $E_1$ and $E_2$ with CM by $\mathcal{O}_K$ and a nonprincipal ideal $\mathcal{I}$ respectively, in a quadratic imaginary field $K$ of class number two; then $\mathcal{O}_K^2\cong \mathcal{I}^2$, so $([E_1]-[E_2])([E_1]+[E_2])=[E_1]^2-[E_2]^2=0$ exhibits zero divisors, as $[E_1]\ne [E_2]$ since the Albanese variety is a birational invariant, and elliptic curves are their own Albanese varieties. -I think this same argument shows that $K_0(\text{Mot})$ is not a domain for $\mathbb{Z}$-coefficients, and it shows that our map is not injective for $\mathbb{Q}$-coefficients. -(2) If we restrict to the classes of smooth varieties, and take motives with integral coefficients, is this map injective? -(3) If we take motives with rational coefficients, is the resulting Grothendieck group a domain? - -REPLY [4 votes]: (1) As pointed out by Marc Hoyois, there is a natural homomorphism $K_0(Var)\to K_0(DM^{gm})$, where $DM^{gm}$ is the category of geometric Voevodsky motives with coefficients in any (commutative unital) ring $R$; you should only assume (at our current level of knowledge on the resolution of singularities) that the base field characteristic $p$ is invertible in $R$ (to put Borel-Moore motives into $DM^{gm}$). Next, the Voevodsky embedding $Chow\to DM^{gm}$ induces an isomorphism on (the corresponding versions of) their $K_0$-groups; see Proposition 2.3.3 of my "$\mathbb{Z}[1/p]$-motivic resolution of singularities". -(3) I don't know.:) -On the one hand, one can easily prove that the "(more or less) standard" motivic conjectures predict that $K_0(Chow)$ is the free abelian group generated by isomorphism classes of indecomposable numerical motives. Next one can try to proceed using the Tannkian formalism. -On the other hand, one can possibly find zero divisors using an argument similar to Lemma 3 in Poonen's https://arxiv.org/pdf/math/0204306.pdf.<|endoftext|> -TITLE: Which cyclic groups admit a difference set? -QUESTION [5 upvotes]: Problem 1. For which $n$ does the cyclic group $C_n$ admit a difference set $D\subset C_n$, i.e., a set such that each non-unit element $x\in C_n$ can be uniquely written as the difference $x=ab^{-1}$ for some $a,b\in D$? -A necessary condition is that $n=1+d+d^2$ for some $d$. -If $p$ is prime, then for $n=1+p+p^2$ the cyclic group $C_n$ admits a difference set according to the classical result of Singer. What about other numbers $d$? -Problem 2. For which $d$ does the cyclic group $C_n$ of order $n=1+d+d^2$ admit a difference set? -Probably this is known, then give me please a proper reference. - -REPLY [9 votes]: What you called difference sets in cyclic groups are usually called PLANAR cyclic difference sets, namely those with \lambda equal to 1. The question you asked here has been studied for many years. Singer's construction from 1938 shows that for any prime power n, there is a planar cyclic difference set of order n. In the other direction, probably the earliest reference goes back to ``Cyclic projective planes" by Marshall Hall, Jr., published in Duke J. Math. 14 (1947), 1079-1090. The conjecture is that if a planar cyclic difference set of order n exists, then n is a prime power. The conjecture is still open. But many partial results are known. You can find a survey of the currently known results on the conjecture in the following paper: -http://www.combinatorics.org/ojs/index.php/eljc/article/view/v1i1r6/pdf<|endoftext|> -TITLE: Space-tiling convex prisms -QUESTION [10 upvotes]: A convex prism is a subset of $\mathbb{R}^3$ congruent to the Cartesian product of a convex polygon (the prism's base) with the interval $[0,1]$. - -Question. If a family of congruent convex prisms tiles space (not necessarily in a face-to-face manner), must there exist a tiling of - the plane with polygons congruent to the prism's base? - -REPLY [10 votes]: [The following is not quite an answer, but it refutes a natural -generalization suggested in the Comments, and is too long to be -a comment itself.] -Counterexample in ${\bf R}^N \times {\bf R}$ for some $N>2$: -any lattice hexagon $H$ with angles -$90^\circ$, $90^\circ$, $135^\circ$, $135^\circ$, $135^\circ$, $135^\circ$ -in that order. (That is, $H$ is obtained from a lattice rectangle by -truncating two adjacent vertices by two isosceles right lattice triangles, -not necessarily congruent. Alternatively, remove two congruent -lattice triangles, not necessarily isoceles, related by a 90-degree rotation.) -Then $H$ does not tile the plane, but -four copies do tile a (non-simply-connected) polyomino. But it was -recently shown that any polyomino in some ${\bf Z}^n$ -(which need not be simply connected, or even connected at all!) -tiles ${\bf Z}^d$ for some $d$: - -Vytautas Gruslys, Imre Leader, and Ta Sheng Tan: Tiling with arbitrary tiles. - Proc. London Math. Soc. (2016) 112 (6): 1019-1039. - https://doi.org/10.1112/plms/pdw017 $\cong$ http://arxiv.org/abs/1505.03697 - -(I learned about this from Francisco Santos's accepted answer to -Timothy Chow's - Mathoverflow question 49915, which the MO algorithm helpfully put at -the top of its list of questions "Related" to this one.)<|endoftext|> -TITLE: The GCD-matrix: generalizing a result of Smith? -QUESTION [16 upvotes]: Let $M$ be the $n\times n$ matrix, known as the GCD matrix, of entries $M_{ij}=\gcd(i,j)$. In the paper -H J S Smith, On the value of a certain arithmetical determinant, Proc. London Math. Soc. 7:208-212 (1875-76) -it is shown that $\det M=\prod_{k=1}^n\varphi(k)$; where $\varphi(k)$ is the Euler totient function. -Lucas introduced the family of sequences defined recursively by $L_0(s,t)=0, L_1(s,t)=1$ and -$$L_n(s,t)=s\cdot L_{n-1}(s,t)+t\cdot L_{n-2}(s,t).$$ -Remark. My interest in these numbers lies in the fact that they are gcd-compatibile: -$$\gcd(L_i,L_j)=L_{\gcd(i,j)}.$$ - -Question 1. Can Smith's result be extended to the determinantal evaluation - $$\det\left[\gcd(L_i(s,t),L_j(s,t))\right]_{i,j=1}^n \,\,?$$ - -Update. On the basis of Max Alekseyev's calculations depicted below, I state the following claim. - -Conjecture. Define a modified Euler's totient function $\varphi_L(k):=\sum_{d\vert n}L_d(s,t)\cdot\mu\left(\frac{k}d\right)$. Then - $$\det\left[\gcd(L_i(s,t),L_j(s,t))\right]_{i,j=1}^n=\prod_{k=1}^n\varphi_L(k).$$ - -Remark. Observe that $L_n(2,-1)=n$, therefore Question 1 is already answered by Smith. In this case, $\varphi_L=\varphi$. - -Question 2. We may start modest. If $s=2, t=1$ then $L_n(2,1)=P_n$ is the Pell sequence. What is the value of - $$\det\left[\gcd(P_i,P_j)\right]_{i,j=1}^n\,\,?$$ - -REPLY [9 votes]: Suppose $A_n$ is the set of natural numbers that divide $L_n(s,t)$ but don't divide any $L_m(s,t)$ for $m -TITLE: Totally ramified subextension in a finite extension of $\mathbf{Q}_p$ -QUESTION [22 upvotes]: Let $K$ be a finite extension of $\mathbf{Q}_p$. Let $F_d$ be the unramified extension of $\mathbf{Q}_p$ of degree $d$. I would like to know whether there exists some $d \geq 1$ and some $L \subset K \cdot F_d$ such that $L/\mathbf{Q}_p$ is totally ramified and $K \cdot F_d / L$ is unramified. -If $K/\mathbf{Q}_p$ is Galois, this can be done as follows. Take $d=[K:\mathbf{Q}_p]$. Write $Gal( F_d / \mathbf{Q}_p) = \langle \sigma \rangle$ and take any $\tilde{\sigma} \in Gal(K \cdot F_d / \mathbf{Q}_p)$ that lifts $\sigma$. Since $\tilde{\sigma}^d$ is the identity on both $K$ and $F_d$, it is the identity in $Gal(K \cdot F_d / \mathbf{Q}_p)$. We can then take $L = (K \cdot F_d)^{\langle \tilde{\sigma} \rangle}$. -Is the result still true if $K/\mathbf{Q}_p$ is not Galois? - -REPLY [3 votes]: I'm a bit late but here are a few remarks on Kevin's example : - -There is a unique $\mathfrak{A}_4$-extension of $\mathbf{Q}_2$ because there is a unique cyclic cubic extension $C$ (namely the unramified one), the group $G=\mathrm{Gal}(C|\mathbf{Q}_2)=\mathbf{Z}/3\mathbf{Z}$ has a unique irreducible degree-$2$ $\mathbf{F}_2$-representation $\rho$, and $\rho$ occurs with multiplicity $1$ in $C^\times/C^{\times2}$. If $G$ acts on $D\subset C^\times/C^{\times2}$ through $\rho$, then Kevin's $M$ is $M=C(\sqrt D)$. -Kevin's $K$ is not galoisian over $\mathbf{Q}_2$ (its galoisian closure is $M$) and nor is any of its unramified extensions, so no such $L$ exists (because an unramified extension of a quadratic extension is always galoisian). -You can find $\mathfrak{A}_4$-extensions of ramification index $4$ and residual degree $3$ over every local field $F$ with finite residue field of characteristic $2$ (although they might not be unique, and in fact there are infinitely many of them if $F$ has characteristic $2$), so the argument can be made to work over every such $F$. -$\mathfrak{A}_4$-extensions are galoisian closures of primitive quartic extensions (as are $\mathfrak{S}_4$-extensions, and these are the only two possibilities). I'm confident that the same trick can be applied over local fields with finite residue field of characteristic $p$ (arbitrary prime) by working with primitive extensions of degree $p^2$ or perhaps $p^n$ for some $n$. How does one find such extensions ? See https://arxiv.org/abs/1608.04183.<|endoftext|> -TITLE: Tweetable way to see Riemannian isometries are harmonic? -QUESTION [14 upvotes]: $\newcommand{\al}{\alpha}$ -$\newcommand{\euc}{\mathcal{e}}$ -$\newcommand{\Cof}{\operatorname{Cof}}$ -$\newcommand{\Det}{\operatorname{Det}}$ -Smooth Riemannian isometries are harmonic. Can one conclude this just from "staring" at the Dirichlet energy functional? (Ideally without even computing the Euler-Lagrange equations). -(As is the case of constant mappings which are global minimizers) -I am interested in finding a very simple explanation, which involves a minimal amount of computation. -To be honest, I suspect such a simple argument cannot exist (I describe below why), but maybe I am overly pessimistic. -The difficulties: -(1) It's false for isometric embeddings, i.e the result only holds for mappings between equidimensional manifolds. So any argument must be involved enough to address this distinction. -(2) It does not hold in a weak sense. ("Weak isometries are not weakly harmonic"). If we assume $f$ is only weakly differentiable and $df \in O(n)$ almost everywhere, then $f$ need not be weakly harmonic. -Indeed, Gromov constructed arcwise isometries between equidimensional non-flat manifolds and flat ones. Such maps are $1$-Lipschitz, hence weakly differentiable and classically differentiable a.e, and their differentials are isometries a.e. However, they cannot be weakly harmonic, since continuous weakly harmonic maps are smooth, which contradicts the mismatch in curvatures. -Gromov's pathological maps cannot be orientation-preserving or orientation-reversing on any open subset of the domain. My colleagues and I showed that any weakly differentiable map $f \in W^{1,\infty}(M,N)$ satisfying $df \in SO(n)$ a.e is weakly harmonic. -The two points discussed above hint that it's unresonable to expect a "pointwise identity" that will show "isometry $\Rightarrow$ harmonicity". - -I know essentally of 3 different ways to prove isometries are harmonic (all of them requires too much computation, in my opinion): -(I) Prove every totally geodesic map is harmonic ($f:M \to N$ is totally geodesic if it maps geodesics to geodesics). -The equidimensionality enters the game when showing our isometry preserve geodesics. However, the claim "geodesic $\Rightarrow$ harmonic" requires a computation in order to pass from "preservation of geodesics" to $\nabla df =0$ which implies harmonicity since $\tau(f)=\operatorname{trace}_g(\nabla df)$). -(II) Prove every minimal isometric immersion is harmonic. Again, this requires a computation (and the equidimensionality also plays its role). -(III) Showing that for every smooth map $f:M \to N$, $\delta (\Cof df)=0$ where -$\Cof df:TM \to f^*(TN)$ is the cofactor map of $df$ defined by $$ \Cof df= (-1)^{d-1} \star_{f^*TN}^{d-1} (\wedge^{d-1} df) \star_{TM}^1. $$ -($\Cof df$ represents the action of $df$ on $d-1$ dimensional parallelepipeds). -My colleagues and I showed that $\delta (\Cof df)=0$ for arbitrary (smooth) maps $M \to N$ (for details see Proposition 3.4 here). Again, this requires computations (If someone has a more heuristic argument, I would be glad to know). -Since $df$ is an isometry, $df=\Det df \cdot \Cof df$ (See here for heuristics). The smoothnes of $f$ implies $\Det df$ is constant, so -$$ \delta ( df)=\Det df \cdot \delta (\Cof df)=0 $$ -This point of view enlightens the importance of not switching orientation in order to be harmonic. - -"Comment:" -In the case where the target space is Euclidean there is a relatively simple proof: -Suppose $f:(M,g) \to (\mathbb{R}^d,e)$ is a smooth isometry. -Harmonicity of $f$ amounts to the component-wise harmonicity $\Delta f^\al= 0$. -This follows from the pullback property, -$$ -\star_{f^*\euc} df^\al = \star_{f^*\euc}(f^*\,dx^\al) = \pm f^*(\star_\euc dx^\al), \tag{1} -$$ -Where the sign is determined according to whether $f$ preserves or reverses orientation. Since $f$ is smooth this sign is constant (if $M$ is connected). -This implies that -$$ -d\star_{f^*\euc} df^\al = d\big(\pm f^*(\star_\euc dx^\al)\big) \stackrel{(a)}{=} \pm d\big( f^*(\star_\euc dx^\al)\big) = \pm f^*(d\star_\euc dx^\al) \stackrel{(b)}{=} 0, \tag{2} -$$ -Where in equality $(a)$ we used the fact the sign is constant, and equality $(b)$ follows since $$d\star_\euc dx^\al=d \big((-1)^{\al+1}dx^1 \wedge \dots \wedge \widehat{dx^\al} \wedge \dots \wedge dx^d\big)=0.$$ -Since $f^*e=g$, equation $(2)$ becomes $ d\star_{g} df^\al = 0 $ which implies $ \delta (df^\al)=-\star_{g} d\star_{g} df^\al = 0$, so $f^\al$ is harmonic. -Perhaps there is a way to generalize this proof to the general case, using some adapted coordinates. - -REPLY [5 votes]: Let $u: M\rightarrow N$ be a diffeomorphism. By staring at the Dirichlet energy formula and knowing that integration by parts works just as well here as for the classical case for functions, you know that a map is harmonic if and only if $\Delta u = 0$. -However, if $u$ is isometric, then you can see in your head that -$$ -0 = \nabla_k(\partial_iu\cdot\partial_ju) = \nabla^2u_{ik}\cdot\partial_ju + \partial_iu\cdot\nabla^2_{kj}u. -$$ -But you also know, given the symmetries of the indices that the standard argument proving the uniqueness of the Levi-Civita connection (also known as Cartan's lemma) implies that -$$ -\partial_ku\cdot\nabla^2_{ij}u = 0. -$$ -Specifically, -\begin{align*} - \partial_ku\cdot\nabla^2_{ij}u -& = \frac{1}{2}(\nabla_i(\partial_ku\cdot\partial_ju) - + \nabla_j(\partial_ku\cdot\partial_iu) - - \nabla_k(\partial_iu\cdot\partial_ju))\\ -& = \frac{1}{2}(\nabla_ig_{kj} + \nabla_jg_{ki} - \nabla_kg_{ij})\\ -&= 0. -\end{align*} -Since $u$ is a diffeomophism, $\partial_1 u, \cdots, \partial_n u$ span $T_*N$ and therefore, $\nabla^2u = 0$ and $u$ is harmonic. -If $\dim M < \dim N$, then the argument above shows that $\nabla^2u$ is normal to $T_*M$. In fact, it is the second fundamental form (viewed as a normal-vector-valued symmetric tensor).<|endoftext|> -TITLE: What is the normal closure of $\mathbb{Q}_p \cap \bar{\mathbb{Q}}$ over $\mathbb{Q}$? -QUESTION [13 upvotes]: What is the normal closure of $\mathbb{Q}_p \cap \bar{\mathbb{Q}}$ over $\mathbb{Q}$? Is it $\bar{\mathbb{Q}}$? - -REPLY [6 votes]: (Copied from the @nfdc23's comment, because it's an answer.) -For any global field $K$, place $v$ of $K$, and embedding of $K_s$ into a separable closure of $K_v$, the normal closure of $K_v \cap K_s$ over $K$ is $K_s$. Indeed, such an intersection corresponds to the decomposition group of $\operatorname{Gal}(K_s/K)$ at a place of $K_s$, and the intersection of decomposition groups for even just two distinct places of $K_s$ (such as two over a place of $K$) is trivial. See 12.1.3 (and 12.1.9 and 12.1.11) in the book Cohomology of Number Fields by Neukirch, Schmidt, and Wingberg.<|endoftext|> -TITLE: Hankel determinants of harmonic numbers -QUESTION [10 upvotes]: Let $H_n=\sum_{k=1}^n\frac 1 k$ be the $n$-th harmonic number with $H_0=0.$ -Question: Is the following true? -$$\det\left(H_{i+j}\right)_{i,j=0}^n=(-1)^n \frac{2H_{n}}{n! \prod_{j=1}^n \binom{2j}{j} \binom{2j-1}{j}}.$$ -Edit: -Comparing with the orthogonal polynomials whose moments are the numbers $\frac{1}{n+1}$ it suffices to show the following identity: -$$\sum_{j=0}^n (-1)^j\frac{\binom n j \binom{n+j} j}{\binom{2n} n} H_j \prod_{j=0}^{n-1}\frac{(j!)^3}{(n+j)!} = (-1)^n \frac{2H_n}{n! \prod_{j=1}^n \binom{2j}{j} \binom{2j-1}{j}}.$$ - -REPLY [2 votes]: Identities involving harmonic numbers that are of interest for physicists, Utilitas Mathematica 83 (2010), 291-299, H. Prodinger. -This paper contains the identity (1) as well. -Now starts Johann Cigler's big birthday (in 5 minutes). Hereby, I will send my best regards on the occasion.<|endoftext|> -TITLE: Derivative formula -QUESTION [12 upvotes]: While trying to understand a paper of Cayley, he left something unexplained, I managed to show that it is equivalent to the following formula, which I got stuck at: -$k \cdot (f^k)^{(k-1)} = \sum_{j=0}^{k-1} {{k} \choose {j}} (f^{j})^{(j)}(f^{k-j})^{(k-j-1)} $ -Can somebody help? - -REPLY [9 votes]: In addition to Ira Gessel's answer, let me explain how such Pfaff-Cauchy-Hurwitz type identities are connected to polynomial interpolation techniques and to Raney lemma. -Denote ${\bf k}=\{1,\dots,k\}$. -We prove a more general (not polynomial in $f$, but multilinear in $f_1,\dots,f_k$) identity -$$ -k(f_1\dots f_k)^{(k-1)}=\sum_{\emptyset\ne A\subset{\bf k}}\left(\prod_{i\in A} f_i\right)^{(|A|-1)}\left(\prod_{i\notin A} f_i\right)^{(n-|A|)}. -$$ -It suffices to prove this identity when all $f_i$'s are power functions $f_i(t)=t^{x_i}$ (by some standard abstract nonsense). In this case we get a polynomial identity in $x_i's$ (where we denote $(x)_n=x(x-1)\dots(x-n+1)$, and also denote $s(A)=\sum_{i\in A}x_i$): -$$ -k(s({\bf k}))_{k-1}=\sum_{A\sqcup B={\bf k},A\ne \emptyset}(s(A))_{|A|-1}\,(s(B))_{|B|}. -$$ -Both parts are polynomials of degree at most $k-1$, and it suffices to check that they coincide on the following combinatorial simplex $\Delta$: $x_i$ take non-negative integral values such that $\sum x_i\leqslant k-1$. -Note that for any $(x_1,\dots,x_k)\in \Delta$ and any partition ${\bf k}=A\sqcup B$, $A\ne \emptyset$, we have $s(A)+s(B)=s({\bf k})\leqslant k-1$, thus either $s(A)<|A|-1$ or $s(B)<|B|$ or $s(A)=|A|-1,s(B)=|B|$, $s({\bf k})=k-1$. Thus if $s({\bf k}) -TITLE: Which Kahler Manifolds Are Spin? -QUESTION [5 upvotes]: As is well-known (see here for a M.O. question) all Kahler manifolds are $spin^c$. I would like to ask which are in fact $spin$. -Taking my motivation from the case of complex projective space, I make the following (naive) conjecture: - -Conjecture: A compact $2n$-dimensional Kahler manifold $M$ is spin, if there exists a line bundle $L$ over $M$ such that - $$ -L \otimes L \simeq \Omega^{(0,n)}, -$$ - where $\Omega^{(a,b)}$ denotes the space of $(a,b)$-forms. - -Can someone tell me if this is true or nor, and if not, what is an easy counter-example. - -REPLY [5 votes]: By a classical paper of Atiyah (http://www.maths.ed.ac.uk/~aar/papers/atiyahspin.pdf) -the spin structures on a compact complex manifold $(M^{2n},J)$ are in bijective correspondence with isomorphism classes of holomorphic line bundles $\cal{L}$ such that $\cal{L}\otimes\cal{L}=K$ where ${\cal{K}}=\Lambda^{n}(T^{*}M^{1, 0})$ is the canonical line bundle. In particular, an almost complex manifold admits a spin structure if and only if $\cal{K}$ admits a square root, i.e. there exists a complex line bundle $L$ such that $L^{\otimes 2}=\cal{K}$. -Of course not any Kahler manifold is spin, for example there are several flag manifolds which are not spin, since their first Chern class is not even. -added Example: For the coset $M=G/K=SU(n)/S(U(p)\times U(n−p))$ one can show that admits a unique $SU(n)$-invariant spin structure, if and only if $n$ is even.<|endoftext|> -TITLE: Why are there finitely many deformation types of Calabi-Yau threefolds for a given diffeomorhpic type if $b_2 =1$? -QUESTION [9 upvotes]: In an article of Robert Friedman, I came up with a comment: -There are finitely many deformation types of Calabi-Yau threefolds for a given diffeomorhpic type if $b_2 =1$. -And it is said that this is a special case of a result due to Kollar. -(p. 113, Friedman, Robert -{On threefolds with trivial canonical bundle}. Complex geometry and Lie theory) -Can anyone explain why or give some references for it? - -REPLY [7 votes]: I am just posting my comment as an answer. The original form of Matsusaka's Big Theorem is that for a pair $(X,H)$ of a smooth, projective $r$-fold $X$ and an ample divisor $H$, there exists an integer $n_0$ that depends only on $r$ and the Hilbert polynomial $P(n) = \chi(X,\mathcal{O}_X(nH))$ such that for all $n\geq n_0$, the invertible sheaf $\mathcal{O}_X(nH)$ is very ample. The Kollár-Matsusaka Theorem proves that to find a uniform $n_0$, you do not need the entire Hilbert polynomial $P(n)$. You only need the leading coefficient, i.e., the intersection number $(H^r)_X$, and the next coefficient, or equivalently, the intersection numbers $(H^r)_X$ and $(H^r.K_X)_X$. -For a Calabi-Yau, $(H^r.K_X)_X$ equals $0$, so you only need $(H^r)_X$. Once you know that $\mathcal{O}_X(nH)$ is very ample, there is an old inequality (probably due to Castlenuovo) proved by taking hyperplane sections that gives a bound on the integer $m=h^0(X,\mathcal{O}_X(nH))$, namely $m \leq r + d$, where $d=n^r(H^r)_X$ is the degree of the projective embedding of $X$ in $\mathbb{P}^{m-1}_\mathbb{C}$ via the very ample invertible sheaf $\mathcal{O}_X(nH)$. Via the construction of the Hilbert scheme in Grothendieck's original Seminaire Bourbaki articles, or even just via the Chow variety, there is a quasi-compact, locally finite type scheme over $\mathbb{C}$ that parameterizes all smooth, degree $d$ closed subschemes $X\subset \mathbb{P}^{m-1}_{\mathbb{C}}$ that are purely $r$-dimensional. In summary, given an integer $v$, the family of smooth, polarized Calabi-Yau $r$-folds $(X,H)$ with $(H^r)_X\leq v$ is a bounded family. -For a smooth Kähler $r$-fold $X$ with $b_2(X)=1$, by the Hodge identities, $h^{2,0}(X)=h^{0,2}(X)=0$, or else $b_2(X) \geq h^{2,0}+h^{0,2} = 2h^{2,0}$ would be at least as large as $2$. Since $H^2(X,\mathcal{O}_X)$ equals $\{0\}$, the exponential exact sequence gives a surjection $$\text{Pic}(X)\twoheadrightarrow H^2(X,\mathbb{Z}(1)).$$ In particular, every integral cohomology class is the first Chern class of an invertible sheaf. Since $b_2(X)$ equals $1$, the torsion-free quotient of $H^2(X,\mathbb{Z})$ is isomorphic to $\mathbb{Z}$. Denote one generator by $g$. -By surjectivity of the Chern class map, there exists an ample divisor class $H$ on $X$ whose first Chern class maps to $\pm g$. Moreover, $H$ is unique up to adding divisor classes whose image in $H^2(X,\mathbb{Z})$ is a torsion class. The degree of the top self-cup product $\int_X g^r$ equals $\pm (H^r)_X$. This is insensitive to adding torsion-classes to $H$, since $H^{2r}(X,\mathbb{Z})$ is torsion-free. Thus, for one and every such ample divisor class $H$, the top self-intersection number $v=(H^r)_X$ is independent of the choice of ample divisor class, and equals the absolute value of the topological number $\int_X g^r$. -In this argument, it was not actually necessary that $X$ is simply connected. By the Universal Coefficients Theorem, the torsion in $H_1(X,\mathbb{Z})$ does give torsion in $H^2(X,\mathbb{Z})$, but the argument above uses only the torsion-free quotient of $H^2(X,\mathbb{Z})$.<|endoftext|> -TITLE: Complement of a divisor in an affine scheme -QUESTION [7 upvotes]: Let $A$ be a domain, not necessarily noetherian or normal, let $X = {\rm Spec}(A)$ and let $U\subseteq X$ be the complement of a prime divisor of $X$. Is it possible that $\mathcal O_U(U) = \mathcal O_X(X)$? - -REPLY [11 votes]: (Edited to add example) Yes, it may happen that $A=\mathcal{O}_X(U)$: I will give an example at the end. On the other hand, this cannot happen if $A$ is noetherian. More generally: - -Proposition. If $U$ is quasicompact, then $A\subsetneq\mathcal{O}_X(U)$. - -Put $Y:= X\smallsetminus U$, and let $\mathfrak{p}$ be the corresponding prime ideal. The assumption on $U$ means, equivalently, that there is a finitely generated ideal $I$ of $A$ such that $\mathfrak{p}=\sqrt{I}$. -To prove the proposition, observe first that it is clear if there is a principal $I=(t)$ as above. Indeed, in this case $t^{-1}\in \mathcal{O}_X(U)\smallsetminus A$. In particular, it is true if $A$ is local with maximal ideal $\mathfrak{p}$: since $Y$ is a divisor and $A$ is a domain, we have $\mathfrak{p}=\sqrt{tA}$ for any nonzero $t\in \mathfrak{p}$. -For the general case, let $j:U\to X$ be the inclusion. Then $j$ is quasicompact and separated, which implies that $j_*$ preserves quasicoherence (EGA1, (6.7.1)) and commutes with flat base change for quasicoherent modules (EGA1, (9.3.2)). In particular $j_*\mathcal{O}_U$ is quasicoherent, hence determined by its $A$-algebra of global sections $\mathcal{O}_X(U)$. So it suffices to prove that $\mathcal{O}_X\subsetneq j_*\mathcal{O}_U$. By the flat base change to $X':=\mathrm{Spec\,}(A_\mathfrak{p})$, the inclusion $\mathcal{O}_X\subset j_*\mathcal{O}_U$ is transformed into the similar inclusion $\mathcal{O}_{X'}\subset j'_*\mathcal{O}_{U'}$, which is not an isomorphism by the local case. QED -An example where $\mathcal{O}_X(U)=A$. -Take for instance $A=\overline{\mathbb{Z}}$, the ring of algebraic integers. This is a 1-dimensional domain. Every closed point $x\in\mathrm{Max}(A)$ corresponds to a (rank one) valuation $v_x$ on $\overline{\mathbb{Q}}$, with valuation ring $\mathcal{O}_{X,x}$. Fix one such point $y$ and take $U=X\smallsetminus\{y\}$. I claim that $\mathcal{O}_X(U)=A$. -Indeed, take $z\in \mathcal{O}_X(U)$. Then $v_x(z)\geq0$ for each closed point $x\in\mathrm{Max}(A)\smallsetminus \{y\}$. We need to prove that $v_y(z)\geq0$ as well. For this it suffices to prove that for each $t\in\overline{\mathbb{Q}}$ the "pole set" $\{x\in \mathrm{Max}(A)\mid v_x(t)<0\}$ is either empty or infinite. Now, $t$ belongs to some number field $L\subset \overline{\mathbb{Q}}$, and the value of $v_x(t)$ (suitably normalized) depends only on the restriction of $v_x$ to $L$ (i.e. the image of $x$ in $\mathrm{Spec}(A\cap L)$). Since all the closed fibers of $\mathrm{Spec}(A)\to\mathrm{Spec}(A\cap L)$ are infinite sets, we are done.<|endoftext|> -TITLE: Grothendieck group of derived category -QUESTION [9 upvotes]: Let $A$ be an essentially small abelian category, and $D(A)$ it derived category. -Does $K_{0}(D(A)) = 0$? -Thank you! - -REPLY [14 votes]: Yes, it's always zero, assuming $D(A)$ means the unbounded derived category. -My complexes will be cochain complexes, and $X[1]$ will be $X$ shifted down in degree. -First suppose $X$ is bounded below, and let $X'=\bigoplus_{n\geq0}X[-2n]$. Even if $A$ doesn't have infinite direct sums, this is still an object of $D(A)$, since in each degree it only involves a finite direct sum. Then there is a triangle -$$X'[-2]\to X'\to X\stackrel{0}{\to} X'[-1]$$ -and so the class $[X]$ of $X$ in the Grothendieck group is zero. -A similar argument, using shifts in the opposite direction, works if $X$ is bounded above. -Finally, any $X$ fits in a triangle with a bounded below and a bounded above complex, so $[X]=0$ for all $X$.<|endoftext|> -TITLE: Resolvents of Schrodinger operators -QUESTION [6 upvotes]: In the free case one can compute the resolvents of the Laplacian $-\Delta$ in many cases explicitly, in the sense that they are given by an integral operator. Often, one uses the Hille-Yosida theorem or Fourier transform to do so. -Also, the Schwartz-kernel theorem tells us that in a distributional sense there is always an "integral kernel" associated to the resolvent of a Schrodinger operator. -However, examples in the free case suggest that the resolvent of $H:=-\frac{d^2}{dx^2} +V$ ( I am now in dimension $1$ as I would just like to get some conceptual insight) where $V$ is a $C^{\infty}$ and bounded potential with suitable boundary conditions such that this operator is self-adjoint, gives rise to a resolvent -$$(H-z)^{-1}\psi(x) = \int K_z(x,y)\psi(y) dy$$ where -$K_z$ has, for fixed $z$ in the resolvent set, nice properties (almost always smooth) in $x$ and $y$. -See for example this calculation: of the one-dimensional resolvent -or this one of the three-dimensional on math.stackexchange -I would like to ask why this is the case? Which theorem guarantees us that such nice $K_z$ kernels exist? - -REPLY [6 votes]: There is a general answer for any dimension $n\geq1$. Let $\Omega\subseteq \mathbb R^n$ denote the underlying domain. - -For all $z$ in the resolvent set of $H$, $(H-z)^{-1}$ is a (classical) pseudodifferential operator of order $-2$, as $-\Delta+V$ is elliptic. Hence, its kernel $K_z$ is a distribution on $\Omega\times\Omega$ which is conormal with respect to the diagonal $\Delta_\Omega = \{(x,x)\mid x\in\Omega\}$. This implies that the wave front set of $K_z$ is contained in the conormal bundle $N^\ast \Delta_\Omega\setminus0 = \{(\xi,-\xi)\in T_{(x,x)}^*(\Omega\times\Omega)\mid x\in\Omega,\,$ $\xi\in T_x^*\Omega\setminus0\}$. In particular, its singular support is contained in (actually, it is equal to) $\Delta_\Omega$. - -Further properties of the kernel $K_z$ depend on the situation under consideration. For instance, if $\Omega$ is bounded and smooth and $V$ is smooth up to the boundary, then $(H-z)^{-1}$ belongs to Boutet de Monvel's calculus, see e.g. Gerd Grubb Functional calculus of pseudodifferential boundary problems. For $\Omega=\mathbb R^n$, there are (refined) pseudodifferential calculi adapted to the behavior of the potential $V$ as $|x|\to\infty$. - -A general reference for the first part of this answer is Chapter XVIII (especially, Sections 18.1 and 18.2) of Lars Hörmander The analysis of linear partial differential operators III.<|endoftext|> -TITLE: Norms on $\mathbb{R}^d$ whose linear isometries are the hypercube group -QUESTION [6 upvotes]: It is a known fact that for any $2\neq p\in[1,\infty]$, the linear isometries for the corresponding norm $\|\cdot\|_p$ on $\mathbb{R}^d$ is the set of all square-matrices with entries in $\{-1,1,0\}$, with at exactly one nonzero entry for each line and each column, that is : the hypercube group. - -Question : what are the norms that precisely have this set of matrices as linear isometry group ? - -When trying to exhibit such norms $N$, the most general construction that I could guess is the following : -\begin{equation} -N(x) = \int_1^{+\infty} \|x\|_p \,\mathrm{d}\mu, -\end{equation} -where $\mu$ is a finite measure over $[1,+\infty)$. I expect that this should cover all possible such norms, but was not able to prove it. -One track would be to use some Choquet theory for which the previous formula would describe $N$ as a generalized linear combination of extreme points (the norms $\|\cdot\|_p$), but I am not much acquainted with such point of view. -I welcome any reference or suggestion on this question ! -Thanks, -Ayman - -REPLY [3 votes]: This is not answer, rather a summarize of some remarks. -Let $G$ be the hypercube group. A first step towards classifying all norms with isometry group $G$ is to classify all norms with isometry group which contains $G$. This is what we discuss below. -Let $D$ be the set of all seminorms on $\mathbb{R}^d$ for which the average value on the standard basis vector is 1 (this is a natural normalization). Observe that $D$ is a convex set, compact wrt the pointwise convergence topology. Observe that $G$ acts naturally on $D$ and let $P:D\to D$ be the associated averaging operator $N(\cdot)\mapsto \frac{1}{|G|}\Sigma_g N(g^{-1}\cdot)$. -Denoting by $C$ the set of $G$ invariant seminorms (in fact: norms, as $G$ acts irreducibly on $\mathbb{R}^d$) in $D$, $P$ is a projection on $C$. -The $P$-preimage of an extreme point in $C$ is a face in $D$, hence contains an extreme point of $D$. -By this observation, it appears that our task is reduced to classifying the set $E$ consisting of all extreme points in $D$. -Given a non-zero linear functional $\theta\in (\mathbb{R}^d)^*$, up to a suitable normalization we have $|\theta(\cdot)|\in D$ and it is easy to see that this is an extreme point. Let us denote by $F\subset E$ the space of extreme points coming from functionals as above. -Guillaume Aubrun made in his answer the following observation: -Observation: A norm in $N\in D$ is in the closed convex hull of $F$ iff $(\mathbb{R}^d,N)$ is embedable in $L^1$. -Proof: -There exists a probability measure $\mu\in \text{Prob}(F)$ s.t for all $x\in \mathbb{R}^d$, $N(x)=\int |\theta(x)|d\mu(\theta)$ -$\Leftrightarrow$ There exists a linear map $\mathbb{R}^d\to L^1(\mu)$, $x\mapsto \theta(x)$. -It then follows that for $d=2$ indeed $E=F$, as every norm on $\mathbb{R}^2$ is embedable in $L^1$ (right? reference?). -However, for $d>2$ there are norms which are not embedable in $L^1$, e.g $\|\cdot\|_\infty$ (also $\|\cdot\|_p$, $p>2$ - correct?), thus $F\subsetneq E$. -What are the other extreme points of $D$?<|endoftext|> -TITLE: How small (in modulus) can a polynomial get? -QUESTION [8 upvotes]: Question. If $f(z)\in\mathbb{C}[z]$ is a monic polynomial of degree $n$, is it true that - $$\max\{\,\vert f(x)\vert: \, -1\leq x\leq 1\}\geq 2^{1-n} \,\, ?$$ - -Context. This came up while working on some convexity problems. - -REPLY [12 votes]: Yes. Proved by Chebyshev. The extreme case ($\max = 2^{1-n}$) is given by $2^{n-1}\cdot f(x)$ being a Chebyshev polynomial of first kind. -https://en.wikipedia.org/wiki/Chebyshev_polynomials - -REPLY [5 votes]: You are essentially asking to approximate $x^n$ on $[-1,1]$ by a polynomial of degree less than $n$. The Remez algorithm will find the best possible aproximation. In Maple you can enter -numapprox[minimax](x^n,x=-1..1,[n-1,0]); -If you take $n=10$ or so and work to 100 digits you will quickly convince yourself that the maximum error is precisely $2^{1-n}$. It also looks as though the optimal approximation $f(x)$ has $2^{1-n}f(x)\in\mathbb{Z}[x]$. One could probably find a formula for the coefficients experimentally with the help of OEIS and then work back to a proper proof. But sadly I need to go and do less interesting things now.<|endoftext|> -TITLE: Chromatic number of a graph defined by $n$ lines on the plane -QUESTION [10 upvotes]: Given $n$ lines on the plane, consider all their intersection points. Find the minimal number $d=d(n)$ such that they may be always colored in $d$ colors so that on each line any two consecutive points have different color. Of course, $d(n)\leqslant 4$ as this graph is planar. But is it true that $d(n)=3$ for all $n\geqslant 3$? -There is an old olympiad problem that 3 colors are enough if no three lines are concurrent. I gave it to children forgetting this condition, and so now I wonder whether is this still true. - -REPLY [11 votes]: No, here is an example that needs four colours (if I have understood the question correctly): - -(there are other intersection points not shown, of course, but these are irrelevant)<|endoftext|> -TITLE: Almost complex structures on $\mathbb CP^2$ that are not tamed -QUESTION [10 upvotes]: Recall that an almost complex structure $J$ on a manifold $M^{2n}$ is called tamed if there exists a symplectic form $\omega$ on $M^{2n}$ such that $\omega(v,Jv)>0$ for any non-zero tangent vector $v$. -Question. Is there an example of an almost complex structure on $\mathbb CP^2$ such that any $C^{\infty}$ small perturbation of $J$ is not tamed? -Added. It turns out that there exists as well a purely local obstruction for any small perturbations of $J$ to be tamed. The precise statement and the answer is here: Almost complex structures on a 4-ball that are not tamed - -REPLY [5 votes]: Summarising the discussion above and Daniel Ruberman's helpful clarifications below. -Any symplectic structure on $\mathbb{C}P^2$ is standard by a result due to Gromov and Taubes. By Siebert-Tian every symplectic surface in $\mathbb{C}P^2$ of degree at most 17 is smoothly isotopic to an algebraic surface. In particular, there is a unique smooth isotopy class of such surfaces. -Now, take $S \subset \mathbb{C}P^2$ a surface of low degree which is not ambient diffeomorphic to an algebraic surface, but which satisfies the adjunction formula ($g=(d-1)(d-2)/2$). Any almost complex structure for which $S$ is pseudoholomorphic can then not be tamed. (To see that such an almost complex structure exists in the right homotopy class, we can find a smooth homtopy of $\mathbb{C}P^2$ from the identity to a smooth map (not a diffeomorphisms obviously!) which sends $S$ to an algebraic surface.) -Finally, by the automatic transversality of Hofer-Lizan-Sikorav, for a small perturbation of the almost complex structure we can find a pseudoholomorphic surface isotopic to $S$; these almost complex structures hence do not admit taming forms either. (In order to apply the automatic transversality, we must use the assumptions that $c_1([S]) \ge 1$ and that $S$ is immersed.)<|endoftext|> -TITLE: Elementary question about Langlands decomposition -QUESTION [7 upvotes]: Let $G$ be a a complex reductive algebraic group, together with an $\mathbb{R}$-form. Is it true that any continuous homomorphism $G(\mathbb{R}) \to \mathbb{R}^{\times}$ comes from an algebraic homomorphism $G \to \mathbb{C}^{\times}$? -I ask this because I want to see whether in the Langlands decomposition $P = MAN$, if we start from an algebraic group $G$ the group $M$ is still algebraic ($M$ is defined as the intersection of preimages of $\{ 1,-1\}$ under all such homomorphisms $MA \to \mathbb{R}^{\times}$, where $MA = Z_G (\mathfrak{a})$ is a reductive algebraic group). -Edit: As was pointed out in the comments, of course the answer is negative. But then I modify the question, asking instead whether the intersection of kernels of $|\chi|$ for "algebraic" $\chi$ the same as for all continuous $\chi$. -Thanks - -REPLY [9 votes]: Now that the motivation for the question has emerged (algebraicity of $M$ inside $G$), here is how to handle it. Let $G$ be a connected reductive $\mathbf{R}$-group, $P$ a parabolic $\mathbf{R}$-subgroup of $G$, and $S$ a maximal split $\mathbf{R}$-torus in $P$ (so $S$ is also maximal as such in $G$). We may and do choose a minimal parabolic $\mathbf{R}$-subgroup $P_0$ of $G$ contained in $P$ and containing $S$. -The set $\Phi(G,S)$ of nontrivial $S$-weights on ${\rm{Lie}}(G)$ is the (possibly non-reduced) relative root system spanning a finite-index subgroup of the character lattice ${\rm{X}}(S')$ for $S' := (S \cap \mathscr{D}(G))^0$ a maximal split $\mathbf{R}$-torus in $\mathscr{D}(G)$. -The choice of $P_0 \supset S$ corresponds to a positive system of roots in the relative root system, or equivalently a basis $\Delta$ of the relative root system, and there is a natural inclusion-preserving bijective correspondence between the set of parabolic $\mathbf{R}$-subgroups $Q$ of $G$ containing $P_0$ and the set of subsets of $\Delta$. In this way $P$ corresponds to a subset $I \subset \Delta$. Explicitly, $U := \mathscr{R}_u(P)$ we have $P = L_I \ltimes U$ for $L_I := Z_G(S_I)$ with $S_I := (\cap_{a \in I} \ker a)^0$ a subtorus of $S$. Thus, $P(\mathbf{R}) = L_I(\mathbf{R}) \ltimes N$ for the group $N:= U(\mathbf{R})$ that is nilpotent (since $U$ is unipotent). -Let $A = S_I(\mathbf{R})^0$. If I remember correctly, Langlands' definition/construction of $M$ given the above choices is as the unique closed subgroup of $L_I(\mathbf{R})$ with compact center such that it is complementary to the central closed subgroup $A$ of $L_I(\mathbf{R})$. Your question is to show that $M = H(\mathbf{R})$ for a unique closed $\mathbf{R}$-subgroup $H \subset L_I$ with reductive identity component such that $M$ meets every connected component of $H$. Of course, such an $H$ is unique if it exists since its Lie algebra must be that of $M$ (so $H^0$ is uniquely determined in $G$) and it is generated by $H^0$ and any finite subset of $M$ meeting each of its finitely many connected components. The real task is existence of such an $H$. -Define $H = \mathscr{D}(L_I) \cdot T_I \cdot S[2]$ where $T_I$ is the maximal anisotropic central subtorus of $L_I$. Note that $H^0 = \mathscr{D}(L_I) \cdot T_I$, so $H^0$ is reductive, and $H(\mathbf{R})$ meets every connected component of $L_I(\mathbf{R})$ since $S[2](\mathbf{R})$ does (as $S$ is maximal split in the connected reductive $L_I$). Since the connected reductive group $H^0$ is unirational, so $H^0(\mathbf{R})$ is Zariski-dense in $H^0$, the equality $H = H^0 \cdot S[2]$ implies that the center of $H(\mathbf{R})$ meets $H^0(\mathbf{R})$ in the subgroup of $Z_{H^0}(\mathbf{R})$ centralizing $S[2]$. Thus, to show $H(\mathbf{R})$ has compact center it suffices to show that $Z_{H^0}(\mathbf{R})$ is compact. But $Z_{H^0} = T_I$ by design of $H$, so $Z_{H^0}(\mathbf{R})$ is compact since $T_I$ is $\mathbf{R}$-anisotropic. -By design $H \cap S_I$ is finite, so $H(\mathbf{R}) \cap A=1$ since $A$ has no nontrivial finite subgroup. Thus, the multiplication map $H(\mathbf{R}) \times A \rightarrow L_I(\mathbf{R})$ is a closed embedding meeting every connected component, and hence this is an isomorphism if and only if the dimensions agree, or equivalently $\dim H + \dim S_I = \dim L_I$. Once this is shown, it follows that $H(\mathbf{R})$ satisfies all of the properties that uniquely characterize Langlands' construction $M$, so this $H$ would do the job. -(Not only is the Lie group $M$ often disconnected, but $H$ is generally not connected as an $\mathbf{R}$-group, which is to say $S[2]$ is not contained $\mathscr{D}(L_I) \cdot T_I$; one sees this already when $L_I$ is a direct product of ${\rm{GL}}_{n_j}$'s.) -Since $L_I$ is the isogenous central quotient of the direct product of $\mathscr{D}(L_I)$ and the maximal central torus in $L_I$, in view of how $H$ was built it is equivalent to show that the split central torus $S_I$ in $L_I$ and the maximal anisotropic central torus $T_I$ in $L_I$ together generate the maximal central torus of $L_I$. In other words, is $S_I$ actually maximal as a central split torus in $L_I$? -Our task has now been reduced to something in the Borel-Tits structure theory for connected reductive groups over arbitrary fields $k$ as follows. Let $G$ be a connected reductive $k$-group, $S \subset G$ a maximal split $k$-torus, and $\Delta$ a basis of the relative root system $\Phi(G,S)$ of nontrivial $S$-weights on ${\rm{Lie}}(G)$. (This set of weights spans a finite-index subgroup of the character lattice of the maximal split $k$-torus $S' := (S \cap \mathscr{D}(G))^0_{\rm{red}}$ in $\mathscr{D}(G)$.) For a subset $I \subset \Delta$, define the $k$-subtorus $S_I = (\cap_{a \in I} \ker a)^0_{\rm{red}} \subset S$ and let $L_I := Z_G(S_I)$. The task is show that $S_I$ maximal as a central split $k$-torus in $L_I$. -By the centrality of $S_I$ in $L_I$ and the $L_I(k)$-conjugacy of all maximal split $k$-tori in $L_I$ (of which $S$ is one such), it suffices to show that no larger $k$-subtorus of $S$ is central in $L_I$. But $\Delta$ is linearly independent in ${\rm{X}}(S)$, so $S_I$ has codimension $\#I$ in $S$ and hence it suffices to show that the adjoint action on $S/S_I$ on ${\rm{Lie}}(L_I)$ supports $\#I$ linearly independent weights. But $I$ itself is such a set of weights in the subset ${\rm{X}}(S/S_I) \subset {\rm{X}}(S)$.<|endoftext|> -TITLE: What is the smallest cardinality of a self-linked set in a finite cyclic group? -QUESTION [15 upvotes]: A subset $A$ of a group $G$ is defined to be self-linked if $A\cap gA\ne\emptyset$ for all $g\in G$. This happens if and only if $AA^{-1}=G$. -For a finite group $G$ denote by $sl(G)$ the smallest cardinality of a self-linked set in $G$. It is clear that $sl(G)\ge \sqrt{|G|}$. A more accurate lower bound is $sl(G)\ge \frac{1+\sqrt{4|G|-3}}2$. By a classical result of Singer (1938), for any power $q=p^k$ of a prime number $p$, the cyclic group $C_n$ of cardinality $n=1+q+q^2$ contains a self-linked subset of cardinality $1+q$, which implies that $sl(C_n)=1+q=\frac{1+\sqrt{4n-3}}2$. So, for such numbers $n$ the lower bound $\frac{1+\sqrt{4n-3}}2$ is exact. -In this paper we prove the upper bound $sl(C_n)\le \sqrt{2n}$ holding for all $n\ne 4$. -Problem 1. Is $sl(C_n)=(1+o(1))\sqrt{n}$? -This problem is equivalent to -Problem 2. Does the limit $\lim_{n\to\infty}{sl(C_n)}/{\sqrt{n}}$ exist? -If the answer to Problems 1,2 are negative, then we can also ask -Problem 3. Evaluate the constant $\lambda:=\limsup_{n\to\infty}{sl(C_n)}/{\sqrt{n}}$. -At the moment it is known that $1\le\lambda\le\sqrt{2}$. - -REPLY [10 votes]: The difference cover problem has been better studied in the context of $\mathbf{Z}$. Redei, Renyi, and others in the 40s asked for the size of the smallest set $A$ such that $A-A$ covers $\{1,2,\dots,N\}$. They proved an upper bound of roughly $\sqrt{8/3} \sqrt{N}$. To prove this they combined Singer's construction of a perfect difference set with the "perfect ruler" $\{0,1,4,6\}$ (which has difference set $\{-6,\dots,6\}$ each with multiplicity one). This was later improved by Leech and Golay to $\sqrt{8/3 - \epsilon}\sqrt{N}$ (for explicit but not very large $\epsilon$). More interestingly, Redei and Renyi proved a nontrivial lower bound of the form $\sqrt{2 + \frac{4}{3\pi}}\sqrt{N}$. -The upper bound can easily be ported to the cyclic problem by taking $N\approx n/2$ and reducing the set $A$ modulo $n$. This proves an upper bound of roughly $\sqrt{4/3}\sqrt{n}$. However, because of the nontrivial lower bound, this proof technique cannot prove $(1+o(1))\sqrt{n}$. Indeed I think it suggests caution.<|endoftext|> -TITLE: Lagrangian Grassmannian as a Spin Manifold -QUESTION [5 upvotes]: I am trying to better understand this nice answer to a question of mine, which states - -Spin structures on a compact complex manifold $(M^{2n},J)$ are in bijective correspondence with isomorphism classes of holomorphic line bundles ${\cal L}$ such that ${\cal L}\otimes {\cal L} = {\cal K}$ where ${\cal K}$ is the canonical line bundle of $(M,2n)$. - -Now in an answer to another question on the canonical bundle of the Lagrangian Grassmannians, it is stated that - -(paraphrased) The Picard group of each Lagrangian Grassmannian $Sp(2n)/P$ (where $P$ is the maximal parabolic subgroup) is cyclic with generator ${\cal O}(-1)$. Moreover, its canonical bundle is $\cal{O}(-n-1)$. - -Now since the Lagrangian Grassmannian is a flag manifold, and hence a complex manifold, these two facts seem to me to imply that - -The Lagrangian Grassmannian is a spin manifold if and only if $n$ is odd. - -Is my reasoning correct? - -REPLY [6 votes]: The complex Lagrangian Grassmannian $M=G/K=Sp(n)/U(n)$ is an isotropy irreducible Hermitian symmetric space, hence it admits a unique invariant complex structure. It occurs by painting black in the Dynkin diagram of $Sp(n)$ the last simple root $\alpha_{n}$, hence the second Betti number of $M$ equals to 1. Now, $M$ admits a spin structure if and only if $n$ is odd, since for example its first Chern class is given by $c_{1}(M)=(n+1)\Lambda_{n}$, where $\Lambda_{n}$ is the fundamental weight corresponding to the painted black simple root $\alpha_{n}$ and it can be thought of as the generator of $H^{2}(M; \mathbb{Z})$. -When such a spin structure exists, then it will be invariant and unique. -Note: Recall that a complex manifold is spin, if and only if its first Chern class is divisible by 2, i.e. even.<|endoftext|> -TITLE: Concentration Bound of $0/1$ permanent -QUESTION [7 upvotes]: If I pick a random $0/1$ $n\times n$ matrix with $0$ occuring with probability $p$ then what does the distribution of the permanent look like? - -REPLY [5 votes]: Let $X_n$ denote the permanent of your random 0/1 $n\times n$ matrix. Then after normalizing, $X_n$ converges to a log-normal distribution [as does the determinant of the matrix]. That is, $\log(|X_n|)$ has a central limit theorem. -For references, see: -https://www.cambridge.org/core/journals/combinatorics-probability-and-computing/article/div-classtitlethe-numbers-of-spanning-trees-hamilton-cycles-and-perfect-matchings-in-a-random-graphdiv/09FDBF0DCE75F834AEBD831B156753D5 -or -https://mini.pw.edu.pl/~wesolo/publikacje/2002RemWesJTP.pdf -Or see the following recent paper that obtains a more nuanced result (as well as concentration results and bounds on moments). -https://arxiv.org/abs/1605.07570<|endoftext|> -TITLE: What exactly goes wrong with Schwarzschild coordinates at the event horizon? -QUESTION [5 upvotes]: It is well known that if one uses the Schwarzschild coordinates (t, r, $\theta$, $\phi$) to solve Einstein's equations, the components of the metric tensor blow up at the "event horizon", r = 2M (in units where c = G = 1). -It is also well known that this is an "artifact of using bad coordinates", a so-called coordinate singularity. -My questions is: What exactly goes wrong with the coordinate system at r = 2M ? -The standard examples of coordinate singularities that I have seen are something like polar coordinates at the origin of $R^2$ or longitude and latitude at the poles of a sphere and the usual problem is that multiple coordinate lines (of the form $\theta$ = constant) intersect at the same point. -Is that what happens in this case ? Or something else altogether ? -Intuitively, if a manifold is smooth, the only way I can imagine an inner product misbehaving when expressed as a metric tensor, is if the tangent vectors to some coordinate lines are not well-defined. But I can't quite see why exactly this is happening with the Schwarzschild coordinates. - -REPLY [2 votes]: Recall that a coordinate system is for $M$ is given by an open set $\Omega\in \mathbb{R}^n$ and a map $\psi: \Omega\to M$ that is a diffeomorphism between $\Omega$ and its image. Then the inverse $\psi^{-1}$ gives a local coordinate system $\psi^{-1}: \psi(\Omega) \to \mathbb{R}^4$. -In the Schwarzschild case it is useful to think in terms of the corresponding map $\psi$ and not the actual coordinate functions $\psi^{-1} = (r,t,\theta,\psi)$. In other words, what you want to do is look at the map that sends points in $\mathbb{R}^4$ with the coordinate value $(r,t,\theta,\psi)$ to points on the Schwarzschild manifold. -Looking at it this way, the first problem with the Schwarzschild coordinate is that when $r = 2M$, - $$ \psi(2M,t_1,\theta,\psi) = \psi(2M,t_2,\theta,\psi) $$ -for any pair of $t_1, t_2$. This automatically implies that $\partial_t\psi =0 $ and hence $\psi$ is not a diffeomorphism. -Another way to say this is that when $r = 2M$, the mapping $\psi$ corresponding to the inverse of the coordinate functions is independent of $t$; for every $t$, the image of $\psi(2M,t,\cdot,\cdot)$ is a fixed 2-sphere in the Schwarzschild manifold (the bifurcate sphere for the past and future event horizons). - -This is analogous to the situation of the function -$$ u: \mathbb{R}^2 \to \mathbb{S}^2\subset \mathbb{R}^3 $$ -where -$$ (\theta,\phi) \mapsto (\cos\theta, \sin\theta \cos\phi, \sin\theta\sin\phi)$$ -This is a smooth map and is (up to some notational changes) the common way of parametrizing the sphere (with $\theta\in [0,\pi]$ and $\phi\in [0,2\pi)$). When $\theta = 0$ or $\pi$, you get the north/south pole independent of $\phi$. - -Wait, but you ask: the metric for the sphere in the $(\theta,\phi)$ coordinates degenerates, but it doesn't blow up! Why is it that the metric in the Schwarzschild case blows up? -The answer is that the problem with the $t$ coordinate is not the only problem with the coordinate system. The $r$ function can be geometrically defined as the area-radius function on the spherically symmetric manifold that is the Schwarzschild solution. It turns out that the function $r$ has a critical point exactly on the bifurcate sphere. The critical point is a "saddle point". Away from the bifurcate sphere $r$ is non-critical, and so we can define the smooth function $\psi$. But the criticality of $r$ (which is part of the inverse of $\psi$) implies that $\psi$ cannot be extended to a differentiable map near where the image is the bifurcate sphere. - -Putting everything together: there are two things that go wrong with the Schwarzschild coordinates at $r = 2M$. - -From the point of view of the mapping $\psi:\mathbb{R}^4\supset\Omega \to M_S$: when $r = 2M$ the function becomes constant in $t$, and hence is not a diffeomorphism. -Furthermore, since the "inverse" function $r:M_S\to\mathbb{R}$ has a critical point on the bifurcate sphere [which as we recall corresponds to exactly when $r = 2M$ in the Schwarschild coordinates], this means $\psi$ is also not differentiable at $r = 2M$, and so is not a diffeomorphism. - -The first gives rise to the fact that one of the metric coefficients in the Schwarzschild metric being 0, the second gives rise to the fact that one of the metric coefficients become infinite.<|endoftext|> -TITLE: Sum over permutations is 1 -QUESTION [28 upvotes]: This might be easy, but let's see. - -Question 1. If $\mathfrak{S}_n$ is the group of permutations on $[n]$, then is the following true? - $$\sum_{\pi\in\mathfrak{S}_n}\prod_{j=1}^n\frac{j}{\pi(1)+\pi(2)+\cdots+\pi(j)}=1.$$ - -Update. After Lucia's answer, I got motivated to ask: - -Question 2. Let $z_1,\dots,z_n$ be indeterminates. Is this identity true too? - $$\sum_{\pi\in\mathfrak{S}_n}\prod_{j=1}^n\frac{z_j}{z_{\pi(1)}+z_{\pi(2)}+\cdots+z_{\pi(j)}}=1.$$ - -Update. Now that we've an analytic and an algebraic proof, is there a combinatorial argument too? At least for Question 1. - -REPLY [28 votes]: Here is a probabilistic, or, if you wish, combinatorial proof. Assume that we have $n$ baskets containing $z_1,\dots,z_n$ balls respectively (well, let them be positive integers). Choose a random ball (all balls have equal probability) and forbid all the balls from its basket. Repeat with $n-1$ remaining baskets and so on. What is the probability that we consecutively get balls from baskets $\pi_n,\pi_{n-1},\dots,\pi_1$? Yes, it equals $$\prod_{j=1}^n\frac{z_{\pi_j}}{z_{\pi(1)}+z_{\pi(2)}+\cdots+z_{\pi(j)}}$$ -(we start multiplying from $j=n$ downto $j=1$). The sum of all these probabilities equals 1.<|endoftext|> -TITLE: Swappability relation on topological spaces -QUESTION [5 upvotes]: Let $(X,\tau)$ be a topological space. Let us call $x,y\in X$ swappable if there is $f:X\to X$ continuous such that $f(x)=y$ and $f(y)=x$. This relation is obviously reflexive and symmetric, but not necessarily transitive. -What is an example of a set $X$ and a reflexive and symmetric relation $R\subseteq X\times X$ such that there is no topology $\tau$ on $X$ such that $R$ equals the swappability relation of $(X,\tau)$? - -REPLY [5 votes]: If ${\mathbf X}=(X,\tau)$ is a topological space with swappability relation $R$, call $(X,R)$ the swappability graph of ${\mathbf X}$. This graph - -is not the swappability graph of any topological space. -Reasoning: -Assume otherwise that this graph, which I will call ${\mathbf G}$, is the swappability graph of the space $(X,\tau)$ where $X=\{A, B, C, D, E, F\}$. - -This is a finite graph, so $\tau$ is determined by the specialization preorder on $X$, say $x\leq y$ if $x$ is in the closure of $\{y\}$. -If $x\equiv y$ denotes $x\leq y$ and $y\leq x$, then $x\equiv y$ implies that $x$ is swappable with $y$ and also with exactly the same other elements as $y$. This implies that $x$ and $y$ are equal or adjacent in ${\mathbf G}$ and that they have the same neighborhoods in the swappability graph. By examining ${\mathbf G}$ one sees that $x\equiv y$ iff $x=y$. Thus, the specialization preorder is a partial order. - -Now we examine the relationship between the swappability graph and the incomparability graph of the specialization order of $(X,\tau)$. - -For $x\neq y$, if $x$ and $y$ are swappable, then they are incomparable in the specialization order. (Maybe the contrapositive statement is more obvious.) -Call a vertex of ${\mathbf G}$ of degree $2$ a corner ($A, D$ or $F$), and call a vertex of degree $4$ an interior vertex ($B, C$ or $E$). Any interior vertex is adjacent or equal in ${\mathbf G}$ to all but one of the other vertices, hence is swappable with all but one of the other vertices, hence is incomparable with all but POSSIBLY $1$ of the other vertices. If an interior vertex is incomparable with ALL other vertices, then it is isolated in the order, so is swappable with all, which is not the case here. Conclusion: each interior vertex is comparable with exactly one other vertex, namely the opposite corner vertex. -We are at the point where we know that the swappability graph ${\mathbf G}$ looks exactly like the incomparability graph of the specialization order, except for the possibility that the incomparability graph has additional edges between the corners. In particular, the $3$ interior vertices (a clique under swappability/incomparability) must be an antichain in the specialization order. Call this the interior clique. Each vertex in this clique is comparable with exactly one corner, while each corner is comparable with exactly one vertex in the interior clique. -There are now a few cases to check: It may be that, under the specialization order, each corner is above its corresponding vertex in the interior clique. (Call it an above-corner.) In this case it is easy to see that the swappability graph is $K_6$ minus a $1$-factor (which is not ${\bf G}$, contradiction). The same happens if each corner is below its corresponding vertex in the interior clique (a below-corner). Now check what happens if, say, $2$ corners are above-corners while the third corner is a below-corner. Even here there are cases, depending on how the corners compare to each other in the specialization order. The two above-corners cannot be comparable to each other, but may be comparable to the below-corner. There are three cases to check: no above-corner is comparable to a below-corner (swappability graph is $K_6$ minus a $1$-factor); exactly one above-corner is above the below-corner (swappability graph has vertices of degree $3$, so is not ${\mathbf G}$); or both above-corners are above the below-corner (the above-corners are swappable, so the swappability graph cannot be ${\mathbf G}$). This covers all cases up to symmetry.<|endoftext|> -TITLE: Subcomplexes with homotopy type of a sphere in complexes with homotopy type of a wedge of spheres -QUESTION [6 upvotes]: Suppose $X$ is a finite $d$-dimensional simplicial complex which is homotopy equivalent to a wedge of at least two $d$-spheres. Does $X$ contain a subcomplex which is homotopy equivalent to a single $d$-sphere? - -REPLY [12 votes]: EDIT: Wow, I worked a lot harder than necessary. I've added a simplified version of this proof. The original is at the bottom. -Let $A = S^1$, with fundamental group the free group $F$ on a generator $x$. Consider the elements -$$ -\begin{align*} -a &= x^6\\ -b &= x^{10}\\ -c &= x^{15} -\end{align*} -$$ -in $F$, and use them as attaching maps: attach three 2-dimensional cells to $A$ to construct a space $B$. -Here are some properties of $B$. - -It is simply-connected. The fundamental group of $B$ is -$$ -\langle x \mid x^6, x^{10}, x^{15}\rangle -$$ -and this group is trivial. -It has homology $\Bbb Z^2$ in degree $2$, $\Bbb Z$ in degree 0, and 0 elsewhere. You can see this from the cellular chain complex $0 \to \Bbb Z^3 \to \Bbb Z \to \Bbb Z \to 0$ for computing $H_* B$, using the fact that $H_1 B = [\pi_1 B]_{ab} = 0$. -As a result, $\pi_2(B) = \Bbb Z^2$ by the Hurewicz theorem and there is a map $S^2 \vee S^2 \to B$ inducing an isomorphism on $H_*$. By the Whitehead theorem, this means that $B$ is homotopy equivalent to $S^2 \vee S^2$. - -I claim that no proper subcomplex of $B$ is homotopy equivalent to a $2$-sphere because any proper subcomplex that has some 2-dimensional cells is not simply-connected. -The only possibility is for such a subcomplex to take the $1$-cell and some assortment of the $2$-cells. However, taking any two of the 2-cells doesn't give you something simply-connected, because you only get these possible fundamental groups: - -$\langle x \mid x^{10}, x^{15}\rangle \cong \Bbb Z/5$ -$\langle x \mid x^6, x^{15}\rangle \cong \Bbb Z/3$ -$\langle x \mid x^6, x^{10}\rangle \cong \Bbb Z/2$ - -(Using fewer than two of the $2$-cells gives you an even larger fundamental group.) - -The original example had $A = S^1 \vee S^1$, with fundamental group the free group $F$ on generators $x$ and $y$, and used -$$ -\begin{align*} -a &= x^2\\ -b &= y^3\\ -c &= (xy)^5\\ -d &= xyx^{-1}y^{-1} -\end{align*} -$$ -as attaching maps for four 2-dimensional cells.<|endoftext|> -TITLE: When is the local representation associated to an elliptic curve a Steinberg? -QUESTION [7 upvotes]: If $E$ is an elliptic curve over $\mathbb{Q}$, and $\pi$ is the automorphic representation of $\mathrm{GL}_2$ associated to $E$, then one can write $\pi = \otimes_v \pi_v$ with each $\pi_v$ an irreducible representation of $\mathrm{GL}_2(\mathbb{Q}_v)$. -For a non-archimedean place $v$, how do we know the isomorphism class of $\pi_v$, in terms of informations about the curve $E$? -In particular, when is $\pi_v$ a Steinberg or a twisted Steinberg? Some explanation or reference will be welcome. Thank you! - -REPLY [6 votes]: The local representation $\pi_{v}$ attached to an elliptic curve is a Steinburg or a twisted Steinburg if and only if $E$ has potentially multiplicative reduction. This follows from the discussion in Section 15 of Rohrlich's paper "Elliptic curves and the Weil-Deligne group", where it is shown that the corresponding representation of the Weil-Deligne group is a twist of the 2-dimensional special representation if and only if $E$ has potentially multiplicative reduction. (Note: Since we're over $\mathbb{Q}$, we know that $E$ is modular, the local Langlands correspondence will map -a twist of the Steinburg to a twist of the special representation and vice versa.) -Another source is the paper "Euler Factors and Local Root Number of Symmetric Powers of elliptic curves" by Dummigan, Martin, and Watkins, which can be found here. Both the paper of Rohrlich and the paper of Dummigan, Martin and Watkins also discuss the case of potentially good reduction (which can yield a supercuspidal or a principal series representation).<|endoftext|> -TITLE: $\prod_k(x\pm k)$ in binomial basis? -QUESTION [18 upvotes]: Let $x$ be an indeterminate and $n$ a non-negative integer. - -Question. The following seems to be true. Is it? - $$x\prod_{k=1}^n(k^2-x^2)=\frac1{4^n}\sum_{m=0}^n\binom{n-x}m\binom{n+x}{n-m}(x+2m-n)^{2n+1}.\tag1$$ - -The problem came out of simplifying some work which reduced to the RHS of (1), but with a GOOD fortune this appeared to factor neatly into the LHS. - -REPLY [4 votes]: Here is an argument for the leading coefficient (and more). -We use (formal) generating functions. -Let $f_{n,x}(t):= \sum_{m=0}^n {n-x \choose m } {n+x \choose n-m} e^{(x+2m-n)t}$, we are interested -in the polynomial $q_n(x)=\frac{(2n+1)!}{4^n}\;[t^{2n+1}]\,f_{n,x}(t)$. -Clearly -\begin{align*} -f_{n,x}(t)&=[z^n] e^{tx}\,\left(1+ze^t\right)^{n-x}\left(1+ze^{-t}\right)^{n+x}\\ - &=[z^n]\left(\frac{e^{t}\,(1+ze^{-t})}{1+ze^t}\right)^x \left(1+ 2z\cosh(t) +z^2\right)^{n}\\ - &=[z^n] (g_{t}(z))^x\,\left(1+ 2z\cosh(t) +z^2\right)^{n} -\end{align*} -where $g_{t}(z):=\frac{e^{t}\,(1+ze^{-t})}{1+ze^t}$. - Thus (by the Lagrange-Bürmann theorem) -$$G_x(u):=\sum_{n=0}^\infty f_{n,x} u^n = \frac {(g_t(w))^x}{1-u(2\cosh(t)+2w)}$$ -where $w=w(u)$ solves $w=u\left(1+w\cosh(t)+w^2\right)$ with $w(0)=0$. -In the sequel let $c:=\cosh(t), s:=\sinh(t)$. -We find -\begin{align*} - w(u)&=\frac{1}{2u}-c - \sqrt{(c-\frac{1}{2u})^2-1)}\\[0.1cm] - 1-2u(c+w)&=\sqrt{(1-2cu)^2 -4u^2}\\[0.1cm] - %=\sqrt{1-4cu +4s^2 u^2}\\ - g_t(w)&=c-2s^2u +s \sqrt{1-4cu +4s^2 u^2} -\end{align*} -so that -$$G_x(u)=\frac{\left(c-2s^2u +s \sqrt{1-4cu +4s^2 u^2}\right)^x}{\sqrt{1-4cu +4s^2 u^2}}$$ -Observe that the numerator of $G_x$ is of the form $(z+\sqrt{1+z^2})^x$ where $z=s \sqrt{1-4cu +4s^2 u^2}$. -Now consider -$$f(z)=\left(z+\sqrt{1+z^2}\right)^x=\sum_{n\geq 0}c_nz^n$$ -Since $f$ satisfies the linear differential equation -$$(1+z^2)f^{\prime\prime}-x^2f=-zf^\prime$$ -with $f(0)=1,\,f^\prime(0)=x$, the coefficients satisfy the recursion $c_0=1, c_1=x$ and -$$(n+2)(n+1) c_{n+2}=(x^2-n^2) c_n\;.$$ -Thus the coefficients $c_n$ are of the form $c_n=\frac {p_n(x)}{n!}$, where -each $p_n(x)$ is a monic polynomial of degree $n$ in $x$, and we may write -$$G_x(u)=\sum_{k\geq 0}\frac{s^{k}p_k(x)}{k!} \left(\sqrt{1-4cu +4s^2 u^2}\right)^{k-1}\;.$$ -Now extract $[u^nt^{2n+1}]G_x(u)=\frac{4^n}{(2n+1)!} q_n(x)$. -Since $c$ and $s^2$ are even functions of $t$ only the summands with odd powers of $s$ can have -non-vanishing coefficient of $t^{2n+1}$, and since the series of $s$ starts with $t$ only summands -with $k\leq 2n+1$ contribute. Thus the highest power of $x$ in $q_n(x)$ is at most $2n+1$ and can only appear in $p_{2n+1}$, -and since $p_{2n+1}$ is monic the coefficient of $x^{2n+1}$ is -$$\frac{(2n+1)!}{4^n}[t^{2n+1}u^{n}]\frac{s^{2n+1}}{(2n+1)!} \left(1-4cu +4s^2 u^2\right)^{n}=(-1)^n\;,$$ -as desired. -Now Brendan McKay's argument above shows the desired equality.<|endoftext|> -TITLE: Easiest proof for showing finite etale (analytic) quotients of algebraic varieties are algebraic -QUESTION [8 upvotes]: Let $X$ be an algebraic variety over $\mathbb C$. Let $X^{an}\to Y$ be a finite etale morphism with $Y$ a complex analytic space. -I read somewhere that $Y$ algebraizes, ie, $Y=V^{an}$ for some algebraic space $V$ over $\mathbb C$. (Edit: I previously wrote $Y$ algebraic variety. For my purposes, I just want $Y$ to be an algebraic space.) -Why is this, and what is an "easy" proof for this? (Consequently, the morphism $X^{an}\to Y = V^{an}$ also algebraizes by Riemann's existence theorem. Actually, this is not right. It algebraizes in the sense that it comes from an algebraic morphism $W\to V$, but $W$ might not be isomorphic to $X$. ) - -REPLY [6 votes]: Up to passing to the Galois closure $\bar{X} \to X$ and using Riemann Extension Theorem (ensuring that $\bar{X}$ is algebraic) we may assume that $X \to Y$ is a Galois cover, induced by the action of a finite group $G$. Then, quoting M. Roth answer to this MathOverflow question - -In the case of a quotient of a scheme $X$ by a finite group $G$, a necessary and sufficient condition for the quotient scheme $X/G$ to exist is that the orbit of every point of $X$ be contained in an affine open subset of $X$. This is proved in SGA I, Exposé V, Proposition 1.8. - -The condition is in particular satisfied when $X$ is a projective scheme, because the orbit is formed by a finite number of points and we can take the affine subset of $X$ given by the complement of a hyperplane section avoiding all of them (note that the property of being projective is preserved in the Galois closure, since the pullback of an ample divisor by a finite map is again ample).<|endoftext|> -TITLE: Going beyond the Sylvester and Schur theorem with regard to $x,x+1,\dots,x+n-1$ -QUESTION [5 upvotes]: I was recent reading through Paul Erdos's classic elementary proof of Sylvester-Schur. It occurred me that there is a simple argument that when $x$ is sufficiently large and if $p_i$ represents the $i$th prime such that $p_i \le n < p_{i+1}$, then there are least $n - i$ numbers in the sequence $x, x+1, \dots, x+n-1$ with prime divisors greater than $n$. -Have there been any well known result that goes beyond Sylvester-Schur in terms of the count of numbers with a prime divisor greater than $n$ in the integer sequence $x, x+1, \dots, x+n-1$? For example, trivially, $x > (n-1)!$ will have this property of at least $n - i$ numbers in the sequence $x,x+1,\dots,x+n-1$ with a prime divisor greater than $n$. - -Edit: Here is the argument that I spoke of. Please let me know if anything is not clear. - -Let $R(p,n)$ be the power of $p$ such that $p^{R(p,n)} \le n < p^{R(p,n)+1}$ -Let $x > \prod\limits_{p < n}p^{R(p,n)}$ where $p$ is each prime less than $n$. -Let $p_i$ be the $i$th prime such that $p_i \le n < p_{i+1}$ -Let $t_k$ be any integer $x \le x+t_k < x+n$ and $gpf(x+t_k) \le p_i$ where gpf = greatest prime factor. - -Claim 1: There exists a prime $q \le p_i$ such that $q^v \ge n$ and $q^v | x + t_k$ - -This follows directly from $x > \prod\limits_{p < n}p^{R(p,n)}$. If - this were not true, then $x$ must necessarily be less than or equal to - $\prod\limits_{p < n}p^{R(p,n)}$ - -Claim 2: There are at most $i$ such instances of $t_k$ - -Since there are only $i$ distinct primes less than or equal to $n$, it - follows that if there are more than $i$ instances, then at least two must - involve the same prime. -Assume that there exists $k_1 < k_2$ such that: -$0 < k_2 - k_1 < n$ -$q^{v_1} | (x + k_1)$ and $q^{v_1} \ge n$ and $gpf(x+k_1) \le p_i$. -$q^{v_2} | (x + k_2)$ and $q^{v_2} \ge n$ and $gpf(x+k_2) \le p_i$ -There exists integers $a_1 >0, a_2 > 0$ where: -$a_1(q^{v_1}) = x+k_1$ and $a_2(q^{v_2}) = x+k_2$ -if $v_1 < v_2$, then: -$x+k_2 - x + k_1 = q^{v_1}[a_2(q^{v_2 - v_1}) - a_1]$ -Now, $q^{v_1} \ge n$ and $[a_2(q^{v_2 - v_1}) - a 1] \ge 1$ which is - impossible since $x + k_2 - x + k_1 < n$ -if $v_1 > v_2$, then: -$x+k_2 - x + k_1 = q^{v_2}[a_2 - a_1(q^{v1-v2})]$ and the same - argument applies. - -REPLY [6 votes]: I like Larry Freeman's proof so much, I am going to rephrase it, first by weakening it. -Claim (Freeman): Given integer $n$ greater than 1, and given $x$ greater than $n!$, there are at most $\pi(n)$ many $n$-smooth integers in the interval $[x+1,x+n]$. -Indeed, his proof is an injection from the subset of $[x+1,x+n]$ which have largest prime factor at most $n$ into the set of primes in $[1,n]$. -Proof sketch: for each smooth number, pick the largest prime power factor of that number, and map the number to that prime. If two numbers map to the same prime, their difference is divisible by a prime power less than $n$. But then (as Larry points out), one of these smooth numbers must be less than $n!$, as it is then a divisor of $n!$. End of proof sketch. -The sketch above also works with Larry's version where $x$ is allowed to be as small as the least common multiple of the first $n$ numbers. I imagine this is folklore, and appears somewhere in the literature on smooth numbers. I would like to see a reference to this result, as well as improvements. -Fixing $n$, the $n$-smooth numbers get increasingly sparse, and the expected count of these numbers in a small interval decays, probably exponentially. Besides a result of Stormer on consecutive smooth numbers, I am not aware of a result like Larry's that is phrased in a combinatorial fashion. I hope others can show us such references. Even data showing the decay for small values of n would be welcome. -Edit 2017.02.17 GRP -'Prime factors of arithmetic progressions and binomial coefficients' is the title of a paper by Shorey and Tijdeman which has an argument similar to the above. This paper has some answers to the question as well as an extensive bibliography. -End Edit 2017.02.17 GRP -EDIT 2018.01.27 GRP -Many thanks to Jose Brox who in helping with figuring out how primes jump found the following paper: -MICHEL LANGEVIN -Plus grand facteur premier d’entiers en progression arithmétique -Séminaire Delange-Pisot-Poitou. Théorie des nombres, tome 18, no 1 (1976-1977), -exp. no 3, p. 1-7 -in which a nice extension is proved. The interval of n numbers can be an arithmetic progression of n terms with common difference coprime to all terms, with the restriction that all terms are positive and none divide lcm[1..n], and the same map will be injective. The proof is almost the same as above. -Being annoyed at having to leave out divisors of lcm[1..n], I found an argument which allows one to include one divisor, at the possible cost of altering the map at that divisor. It is a lot of work for small gain, but it uses a weak combinatorial estimate on the number of certain perfect powers in an interval, so I am pleased with it. I can for large n bring the forbidden divisors down by a factor of square root of n, and hope to improve this further, using just combinatorial arguments. -END EDIT 2018.02.27 GRP -Gerhard "Too Long For A Comment" Paseman, 2017.02.16.<|endoftext|> -TITLE: How big can a commutative algebra of $n \times n$ matrices be? -QUESTION [6 upvotes]: What's the maximum possible dimension of a commutative subalgebra of the algebra of $n \times n$ complex matrices? -There's a theorem of Burnside saying that any commutative subalgebra of a matrix algebra can be upper triangularized. My friend Bruce Smith pointed out that for $n$ even we can get a commutative subalgebra of dimension $(\frac{n}{2})^2 + 1$. For $n = 4$ its elements look like this: -$$\left( \begin{array}{cccc} -a & 0 & b & c \\ -0 & a & d & e \\ -0 & 0 & a & 0 \\ -0 & 0 & 0 & a -\end{array}\right)$$ -and the same trick works in any even dimension. For $n$ odd we can get dimension $\frac{(n-1)(n+1)}{2} + 1$ using a similar idea, with a rectangle rather than a square of nonzero entries in the upper right corner. -Can one do better? Someone must have figured this out. - -REPLY [2 votes]: Thanks, Christian Remling! There's a much harder question on MathOverflow to which Robin Chapman gave an answer to my question here. -In short: Bruce's guess is indeed the best we can do. This fact was proved by Schur, and there's also a proof here: - -Maryam Mirzakhani, A simple proof of a theorem of Schur, Amer. Math. Monthly 105 (1998), 260–262.<|endoftext|> -TITLE: Closest point in $SU(n) \otimes SU(n)$ to $SU(n^2)$ -QUESTION [7 upvotes]: What is the closest $V_1 \otimes V_2 \in SU(n)\otimes SU(n)$ in the squared trace inner product to a given $U \in SU(n^2)$? I.e. minimize over $V_1, V_2$: -$\min_{V_1, V_2} | V_1 \otimes V_2 - U|$ in terms of a given $U$. - -REPLY [4 votes]: Maybe an example will clarify things a bit: If you think of $\mathrm{SU}(2)$ as the group of complex $2$-by-$2$ matrices of the form -$$ -q = \begin{pmatrix}a&-\bar b\\b&\bar a\end{pmatrix} -$$ -such that $a\bar a + b \bar b = 1$, and you think of $\mathbb{C}^2\otimes\mathbb{C}^2=\mathbb{C}^4$ as the space of $2$-by-$2$ complex matrices, then the representation of $\mathrm{SU}(2)\times \mathrm{SU}(2)$ into $\mathrm{SU}(4)$ can be thought of as the action -$$ -(q_1,q_2)\cdot m = q_1\,m\,q_2^\dagger = q_1\,m\,{q_2}^{-1}. -$$ -This action preserves the $4$-dimensional real subspace $\mathbb{H}\subset \mathbb{C}^4$ consisting of matrices of the form -$$ -p = \begin{pmatrix}a&-\bar b\\b&\bar a\end{pmatrix}, -$$ -and, in fact, as is well-known, the above action of $\mathrm{SU}(2)\times \mathrm{SU}(2)$ on $\mathbb{H}$ is identical with the action of $\mathrm{SO}(4)$ acting on $\mathrm{H}=\mathbb{R}^4$. -Thus, in the case $n=2$ of the OP's question, the subgroup being denoted by $\mathrm{SU}(2)\otimes\mathrm{SU}(2)\subset\mathrm{SU}(4)$ is just $\mathrm{SO}(4)\subset\mathrm{SU}(4)$. The problem then is how to find 'the' (or rather, 'a') closest point in $\mathrm{SO}(4)$ to a given element of $\mathrm{SU}(4)$. -Now, as is known, any element $g\in\mathrm{SU}(4)$ can be factored as -$$ -g = h_1\,\mathrm{e}^{i\delta}\,h_2\tag 1 -$$ -with $h_1, h_2\in \mathrm{SO}(4)$ and $\delta$ a real diagonal matrix with trace zero. If $h_\delta\in\mathrm{SO}(4)$ is a closest element of $\mathrm{SO}(4)$ to $\mathrm{e}^{i\delta}\in\mathrm{SU}(4)$ (i.e., $|\mathrm{e}^{i\delta}-h_\delta|\le |\mathrm{e}^{i\delta}-h|$ for all $h\in\mathrm{SO}(4)$), then $h_1\,h_\delta\,h_2\in\mathrm{SO}(4)$ will be a closest point in $\mathrm{SO}(4)$ to $g = h_1\,\mathrm{e}^{i\delta}\,h_2$. -Unfortunately, $h_\delta$ cannot be chosen to be continuous with respect to $\delta$. For example, if $\delta = \mathrm{diag}(t,-t,0,0)$ then, for $|t|<\pi/2$, one can show that $h_\delta = I_4$ is the closest point in $\mathrm{SO}(4)$ to $\mathrm{e}^{i\delta}$. When $|t|=\pi/2$, there is a whole circle of points in $\mathrm{SO}(4)$ that are at minimum distance from $\mathrm{e}^{i\delta}$. When $\pi/2<|t|\le \pi$, though, the closest point to $\mathrm{e}^{i\delta}$ in $\mathrm{SO}(4)$ is $h_\delta= \mathrm{diag}(-1,-1,1,1)$. -Meanwhile, for all $\delta$ sufficiently small (in the sense that $\mathrm{tr}(\delta^2)$ is sufficiently small), one has $h_\delta = I_4$ is the unique closest element in $\mathrm{SO}(4)$ to $\mathrm{e}^{i\delta}$, so, in that case, the mapping -$$ -g = h_1\,\mathrm{e}^{i\delta}\,h_2 \mapsto h_1h_2 = h(g) -$$ -gives the (unique) closest point in $\mathrm{SO}(4)$ to $g$. This takes care of an open set in $\mathrm{SU}(4)$ for which your problem has a stable solution, provided you know how to perform the factorization (1).<|endoftext|> -TITLE: If many orthogonal vectors are respected (somewhat), are there many eigenvectors with large eigenvalues? -QUESTION [6 upvotes]: Let $V$ be an inner-product vector space (real, of very large but finite dimension, if you wish). Let $S:V\to V$ be a symmetric linear operator. Let $v_1,\dotsc,v_k\in V$ be a collection of vectors of norm $1$, orthogonal to each other. (Here $k$ may be thought of as larger than a constant, but possibly much smaller than $\dim V$.) Assume that -$(S v_i,v_i)>\delta$ for every $1\leq i\leq k$ -and $|S v|_2\leq |v|_2$ for every $v\in V$. -Question: Can we conclude that the space W spanned by all eigenspaces with eigenvalue $\geq \delta/2$ (say; $\delta^2$ or $\delta^{100}/100$ would also do) has large dimension? -By "large" I mean something in the scale of $k$, $\sqrt{k}$, $\delta k$ or thereabouts. It is easy to give a very weak bound ($\gg \log(O(\delta^2 k))$; see below). -Question, version 2: The same question, but with the added condition $(S v_i, v_j)=0$ for all distinct $1\leq i,j\leq k$. - -Sketch of proof of the weak bound: Suppose $W$ had dimension $d=o(\log k)$. By pigeonhole, there is narrow cone in $W$ such that the projections $w_i$ to $W$ of $\geq k/6^d > \sqrt{k}$ vectors $v_i$ lie in that cap. Any two vectors in that cap have inner product $\geq 1/2$ times the product of their norms. Hence $\langle w_i,w_j\rangle \geq |w_1|_2 |w_2|_2/2 \geq |\delta|^2/8$. -Then the projections $u_i$ of those same vectors $v_i$ to the orthogonal complement $U$ of $W$ satisfy $\langle u_i,u_j\rangle \leq -|\delta|^2/8$ for all $i$, $j$ distinct. By a standard argument, this condition can be satisfied by at most $O(|\delta|^{-2})$ vectors. - -REPLY [6 votes]: Write $P$ for the projection onto the $v_j$, and let $N$ denote the number of eigenvalues $\ge \delta/2$ of the matrix $PSP$. Then -$$ -k\delta \le\textrm{tr}\: PSP \le N + \frac{(k-N)\delta}{2} , -$$ -so $N\ge \delta k/3$, say, for small $\delta$. By the min-max principle, the same inequality holds for the original matrix. - -REPLY [3 votes]: Actually, isn't question 2 brutally trivial? The following argument seems to show that the space $W$ spanned by the eigenspaces with eigenvalue $\geq \delta$ has dimension $\geq k$. -Suppose this weren't the case. Then there would be a non-trivial linear combination $\sum_i a_i v_i$ orthogonal to $W$. Thus, on the one hand, -$$\left\langle S\left(\sum_i a_i v_i\right), \sum_i a_i v_i\right\rangle < \delta \left|\sum_i a_i v_i\right|^2 = \delta \sum_i |a_i|^2,$$ -but, on the other hand, -$$\begin{aligned}\left\langle S\left(\sum_i a_i v_i\right), \sum_i a_i v_i\right\rangle &= \sum_{i,j} \overline{a_i} a_{j} \langle S v_i, v_{j}\rangle\\ &= \sum_i |a_i|^2 \langle S v_i, v_i\rangle \geq \delta \sum |a_i|^2.\end{aligned}$$ -Contradiction.<|endoftext|> -TITLE: Co/fibrant replacements via coend calculus -QUESTION [8 upvotes]: In the paper - -Cordier, Jean-Marc, and Timothy Porter. "Homotopy coherent category theory." Transactions of the American Mathematical Society 349.1 (1997): 1-54. - -the authors define a notion of coherent co/end $\oint T(a,a)$ as -$$ -\int_{a',a''} T(a',a'')^{\delta A(a',a'')} -$$ -where $\delta A(a',a'')$ is a suitable "resolution" of the hom functor (linked to the bar construction; follow §1, and in particular the paragraphs above Definition 1.1). -The main claim of this paper is that category theory can be rewritten in a homotopy coherent fashion, building on this definition of homotopy coherent co/end. It is obvious why I like this point of view :-) -One of the first interesting claims is the following: fix a simplicial functor $F : A \to B$ between simplicial categories, and define -$$ -\begin{gather} -\overline F = \{\delta A(a,-),F\} = \oint_{x:A} Fx^{A(a,x)}\\ -\underline F = \delta A(-,a)\otimes F = \oint^{x:A} Fx \otimes A(x,a) -\end{gather} -$$ -This is the counterpart of the "ninja Yoneda lemma" in covariant and contravariant form. -Coend-fu now gives that "underlines and overlines absorb coherence" in that -$$ -\text{CohNat}(F,G) \cong \text{Nat}(F, \overline G)\cong \text{Nat}(\underline F,G) -$$ -This gives natural transformations -$$ \eta_F : F \Rightarrow \overline F \qquad\qquad \eta^F : \underline F \Rightarrow F$$ -as the images of the identity coherent natural transformation of $F$. The authors then prove that these maps are levelwise homotopy equivalences (Prop. 3.4). - -Are $\overline F, \underline F$ replacements for $F$ in suitable model structures on categories of functors? Is $\overline F$ fibrant if $F \in [A, {\bf sSet}]_\text{inj}$? Is $\underline F$ cofibrant if $F \in [A, {\bf sSet}]_\text{proj}$? - -REPLY [7 votes]: Addressing this sort of question was one of the main goals of math/0610194. The best answer I was able to give is that if $B$ has a suitable model structure with respect to which $F$ is objectwise fibrant (resp. cofibrant), then $\overline{F}$ (resp. $\underline{F}$) belongs to a "right (resp. left) deformation retract" (an abstraction of the notion of fibrant (resp. cofibrant) replacement) that is suitable for constructing derived functors of functors such as homotopy limits (resp. colimits) and also the homotopy category of the functor category (which therefore involves homotopy coherent transformations). -A closely related perspective that you may also be interested in can be found in this paper by Gambino: the colimit of $\underline{F}$ is equivalently the colimit of $F$ weighted by a projective-cofibrant replacement of the terminal weight. - -REPLY [5 votes]: This will not answer your question but is to mention that we only defined (and studied in depth) the functors $\underline{F}$ and $\overline{F}$ where the codomain $\mathbf{B}$ of the functor was a 'locally Kan' simplicially enriched category and thus fibrant in the usual model category structure on $\mathcal{S}$-cat. (This eliminates the awkwardness of the example that Dimitri mentions.) This means that the unit and counit maps $\eta_F$ etc, are levelwise homotopy equivalences not just weak equivalences. (N.B. Our philosophy was to show such things as directly as possible so there is no use of spectral sequences (at least explicitly in our contribution), for instance, by showing some map is a weak equivalence, followed by invocation of the Kan-ness of the objects involved so get it is a 'real' homotopy equivalence. This makes some of the arguments more constructive' but at the cost of being heavier on explicit, almost combinatorial or geometric, computation with induction up skeleta etc. as a tool.) -I suspect that some of the ideas explored in the article: -P. J. Ehlers and T. Porter, Ordinal subdivision and special pasting in quasicategories, Advances in Mathematics, 217 (2007), No 2. 489 - 518 -may be useful, but do not, at the moment, see how to prove what you suggest to be the case. It is worth noting that Dan Dugger's construction, mentioned by Dimitri, is very similar to that given by `$F$ goes to $\overline{F}$', so adapting his proof may give the result.<|endoftext|> -TITLE: An identity related to Hankel determinants of $\sum_{k=1}^n \frac{2^k}{k}$ -QUESTION [5 upvotes]: This question is related to Hankel determinants of harmonic numbers. -Let $f(n)=\sum_{k=1}^n \frac{2^k}{k}$ and $r(n)=\sum_{j=0}^n (-2)^{n-j}\binom{n}{j}\binom{n+j}{j}f(j).$ -In order to compute the Hankel determinants $\det\left(f(i+j)\right)_{i,j=0}^n$ I need the following identities: -1) $r(2n)=0$, -2) $r(2n+1)=(-1)^n \frac{(2n+1)!}{((2n+1)!!)^2}2^{2n+2}.$ -Are these identities known? - -REPLY [7 votes]: Define the sequence $a_n(x):=\sum_{j=0}^n(-2)^{n-j}\binom{n}j\binom{n+j}jx^j$ so that -$r(n)=\int_0^2\frac{a_n(x)-a_n(1)}{x-1}dx$. -We need the following fact which follows from the Vandermonde-Chu identity: for $n\geq1$, -\begin{align}\sum_{j=0}^n(-1)^j\binom{n}j\binom{n+j}j\frac1{j+1} -&=\sum_{j=0}^n\binom{n}{n-j}\binom{-n-1}j\frac1{j+1} \\ -&=\frac1{n+1}\sum_{j=0}^n\binom{n+1}{n-j}\binom{-n-1}j=0. \tag1 -\end{align} -At present, the implication of (1) is that -$$\int_0^2a_n(x)dx=2(-2)^n\sum_{j=0}^n(-1)^j\binom{n}j\binom{n+j}j\frac1{j+1}=0. \tag2$$ -Next, we apply Zeilberger's algorithm to generate the two recursive relations -\begin{align} -(n+2)a_{n+2}(x)-2(x-1)(2n+3)a_{n+1}(x)+4(n+1)a_n(x)&=0, \\ -(n+2)a_{n+2}(1)\qquad \qquad \qquad \qquad \qquad \qquad+4(n+1)a_n(1)&=0. -\end{align} -Subtract the $2^{nd}$ equation from the $1^{st}$, divide through by $x-1$ and integrate $\int_0^2(\cdot)dx$. So, -$$(n+2)r(n+2)+4(n+1)r(n)=0\qquad \implies \qquad r(n+2)=-\frac{4(n+1)}{n+2}\,r(n);$$ -where (2) has been utilized effectively. Initial conditions are $r(0)=0$ and $r(1)=4$. The case $n$ even is transparent. The case $n$ odd also follows from induction on $n$ and the fact that if we write the RHS of your claim for $r(n)$ as -$$t(n):=\frac{(-1)^{\lfloor n/2\rfloor}4^n}{n\binom{n-1}{\lfloor n/2\rfloor}}\chi_{odd}(n)$$ -then it is easy to check that $\frac{t(n+2)}{t(n)}=-\frac{4(n+1)}{2n+3}$ when $n$ is odd. The proof is now complete. -Note. Here $\chi_{odd}$ is understood as $\chi_{odd}(odd)=1$ and $\chi_{odd}(even)=0$.<|endoftext|> -TITLE: Are all Grothendieck topologies on Set equivalent? -QUESTION [9 upvotes]: The category $\textbf{Set}$ can be given a Grothendieck topology where the covering families are jointly surjective families of set inclusions $\{X_i\stackrel{\phi_i}{\hookrightarrow} X\}\in\mathrm{Cov}(X)$, $X\in\mathrm{ob}(\textbf{Set})$. - -Are there any other Grothendieck topologies on $\textbf{Set}$, not equivalent to the above one? - -Two Grothendieck topologies on $\mathcal{C}$ are equivalent when they give rise to sheaf categories which are equivalent as reflective subcategories of the category of presheaves on $\mathcal{C}$. - -REPLY [12 votes]: (Much of this has basically been said by someone in the comments already.) -Here is a way of making examples of topologies on Set. Let $\mathcal C$ be a class of sets. Define a topology on Set by saying that a sieve $S$ on an object $X$ is a cover if and only if for every $Y\in\mathcal C$ every morphism $Y\to X$ belongs to $S$. An object is a point (in the sense that the only cover of that object is its maximal sieve) if it belongs to $\mathcal C$, and more generally if it is a retract of such an object. If we enlarge $\mathcal C$ to make it closed under retraction the topology is unchanged. Any presheaf that is a right Kan extension from (the full subcategory) $\mathcal C$ is a sheaf, and conversely any sheaf coincides with the Kan extension of its restriction. -(All of that is valid for any category, not just for Set.) -In the case when $\mathcal C$ consists of just a singleton, this topology on Set is the canonical one, where the sheaves are the representable funtors. If $\mathcal C$ has at least one nonempty set in it then the sheaf is subcanonical. -In general a class of sets closed under retraction must be either: all sets with cardinality less than a fixed cardinal, or all non-empty sets with cardinality less than a fixed cardinal. For example, $\mathcal C$ might be all nonempty sets having at most $n$ elements, or all nonempty finite sets, or all nonempty countable sets, or all nonempty sets (or any of these together with the empty set). -I suppose there are examples of topologies on Set not of this kind. That is, I suppose that there is a topology $T$ on Set such that $T$ does not coincide with the largest one that has the same points as $T$. Does somebody know? -EDIT: I realize now that of course there are perfectly everyday examples not of that kind. For example, the topology generated by finite covers in the ordinary sense. ($X$ is covered by ${Y_i\to X}$ if $X$ is the union of $Y_1,\dots ,Y_n$.)<|endoftext|> -TITLE: What is the relation between the holonomy groupoid of a foliation and the corresponding Haefliger groupoid? -QUESTION [6 upvotes]: Given a foliation, there is a holonomy groupoid and a classifying map -to the Haefliger classifying space via the Haefliger groupid. What is the relation between these groupids? - -REPLY [4 votes]: Write $F$ your foliation, $M$ its ambiant manifold, $q=dim(M)-dim(F)$ its codimension. The holonomy groupoid $H(F)$, if I'm correct, is the set of classes of triples $(x,\gamma,y)$ where $\gamma$ is a tangential path; and $(x,\gamma,y)~(x,\gamma',y)$ iff $\gamma$ and $\gamma'$ have the same holonomy. -I'm not expert enough in groupoids to use the proper vocabulary, but the relation is as follows between the holonomy groupoid $H(F)$ of your foliation and the universal groupoid $H(B\Gamma_q)$. Let $c:M\to B\Gamma_q$ be the Haefliger classifying map of $F$. Then, $F$ is the pullback of the universal foliation on $B\Gamma_q$ (whatever this means) through $c$. In particular, there is an induced groupoid morphism $C: H(F)\to H(B\Gamma_q)$. Moreover, one has a partial injectivity property: for $(x,\alpha,y)$ and $(x,\beta,y)$ in $H(F)$ with the same endpoints, one has $C(x,\alpha,y)=C(x,\beta,y)$ iff $\alpha=\beta$. -Does this help? -On the other hand, if you rather mean to compare the holonomy groupoid $\Gamma=H(F)$ of the given foliation with the holonomy groupoid $H(B\Gamma)$ of the Haefliger classifying space of $\Gamma$, then the continuous classifying map $c:M\to B\Gamma$ induces an equivalence. Precisely, $H(B\Gamma)$ is simple (at most 1 arrow between two units) and $c$ induces a bijection between the set of orbits of $\Gamma$ and the set of orbits of $H(B\Gamma)$.<|endoftext|> -TITLE: Diamond lemma for edge-colored directed graphs -QUESTION [5 upvotes]: Suppose we have a directed graph with colored edges such that (1) every vertex has at most one outgoing edge of each color, (2) for any path of length two, say from $u$ to $v$ to $w$, there exists a vertex $v'$ and a directed path from $u$ to $v'$ to $w$ such that edge $(u,v')$ has the same color as edge $(v,w)$ and edge $(v',w)$ has the same color as edge $(u,v)$, and (3) for all $v$, if we have distinct edges $(v,w)$ and $(v,w')$, there exists $x$ and edges $(w,x)$ and $(w',x)$ with the same color as $(v,w')$ and $(v,w)$ respectively.. Then a standard argument (the "Jordan-Holder argument") shows that, for any starting vertex $s$, EITHER there is no terminating path from $s$ (where a terminating path is one whose final vertex $t$ has no outgoing edges) OR every maximal path from $s$ terminates at a path-independent vertex $t$ and moreover the number of times each color gets used is path-independent. -Can anyone provide a citation for this lemma or something like it? (It occurs in a more specialized form in the theory of abelian sandpiles, and again in the theory of rewrite systems, and again in group theory, but it's really just a general-purpose graph-theory lemma.) - -REPLY [5 votes]: Here's a proof of a more general result that Alex Postnikov presented recently in class. (Actually he did not use colors as you do but that can be incorporated into the proof.) He called the result the "Roman Lemma" because a) it may have been known to the ancient Romans and b) it is based on the principle that all roads lead to Rome. :) (There was also some discussion about its relation to the Roman Catholic Church which I will omit here.) -Recall that a sink of a directed graph is a vertex of outdegree zero. -Lemma: Let $G$ be an edge-colored directed graph, not necessarily finite, without loops but possibly with multiple edges, whose underlying undirected graph is connected. Suppose that whenever $(v,w)$ and $(v,w')$ are edges of $G$ (even with $w = w'$) there is some $k\geq 0$ and two paths $(v,w = w_0), (w_0,w_1),\ldots, (w_{k-1},w_{k} = u)$ and $(v,w' = w'_0), (w'_0,w'_1),\ldots, (w'_{k-1},w'_{k} = u)$ that bring $w$ and $w'$ back together in the same number of steps and whose multisets of edge colors are the same. Then either: -(1) $G$ has no sinks or (2) $G$ has a unique sink $q$, and starting from any fixed $v \in G$, every path from $v$ eventually leads to $q$, in the same number of steps, and moreover with the same multiset of edge colors. -(Note that (2) means that, ignoring edge colors, $G$ is the Hasse diagram of a graded poset with a unique maximal element.) -Pf: Suppose (1) does not hold. So $G$ has some sink; choose a sink $q$. For $i \geq 0$, let $L_i$ be the set of all vertices in $G$ whose shortest path to $q$ is of length $i$. -Subclaim 1: The vertices in $L_i$ have edges only to vertices in $L_{i-1}$. Thus the restriction to $L_0,L_1,\ldots,L_i$ is a Hasse diagram of a graded poset with unique maximal element $q$. -Pf: By induction on $i$. Clear for $i=0$. Suppose to the contrary that $v \in L_i$ has an edge to $w$ with $w \notin L_{i-1}$. By construction there is $w' \in L_{i-1}$ such that $(v,w')$ is also an edge. Then we can join up $w$ and $w'$ in the same number of steps to some $u$. But since by induction all paths from $w'$ to $q$ have length $i-1$, we get a path from $w$ to $q$ of length $i-1$, which contradicts $w \notin L_{i-1}$ (if $w$ is not in any $L_j$ for any $j \leq i-1$ this is clearly a contradiction; if $w$ is in $L_j$ for some $j < i-1$ then this yields a path from $v$ to $q$ via $w$ having length $j+1 -TITLE: Groups determined by their group ring and direct products -QUESTION [9 upvotes]: In the paper [W. Kimmerle - R. Lyons - R. Sandling - D.N. Teague: Composition factors from the group ring and Artin's theorem on orders of simple groups, Proc. London Math. Soc. (3) 60 (1990), no. 1, 89-122] I found the following statement: -" It is well known that if the groups $G_1$ and $G_2$ are determined by their integral group rings, then $G_1 \times G_2$ is determined by $\mathbb{Z}(G_1 \times G_2)$ where $\mathbb{Z}(G_1 \times G_2)$ is the group ring. -I can not prove the above statement and I can not find where it has been proved. I would be grateful if you could give me a proof. -This is the link to the aforementioned paper. - -REPLY [2 votes]: You are asking a reasonable question. I am not an expert on these things, but I think I can help with understanding the missing lemma and with the paper as a whole. -We are considering the integral group ring $\mathbb Z[G]$ of the finite group $G$. Then one can fix a map $\mathbb Z[G] → \mathbb Z$ which will play the role of the augmentation. There is an old result of Glauberman (and possibly someone else) that asserts that one can find the set $S$ of class sums in $Z[G]$. Then one takes sums of elements of $S$. It is easy to see which sums correspond to $\hat N$, the sum of all elements in $N$, for a normal subgroup $N$ of $G$. Since it is easy to see which normal subgroups contain others (they involve a bigger subset of $S$), we wind up describing the lattice of all normal subgroups of $G$ in this manner. An easy proof can be found in Passman’s book “The algebraic structure of group rings” (Lemma 2.3(v)(vi), pages 664–665). -If we look at a particular $\hat N$, then its square is itself times $\lvert N\rvert$, so $\lvert N\rvert$ is determined. Also its annihilator, say $A$, in $\mathbb Z[G]$ is easily seen to be what is called $\Delta(N,G)$, the augmentation ideal of $\mathbb Z[N]$ times $\mathbb Z[G]$. But this is the kernel of the natural epimorphism $\mathbb Z[G] \to \mathbb Z[G/N]$, so $\mathbb Z[G/N] \cong Z[G]/A$. We see that $\hat N$ determines $\lvert N\rvert$ and also $\mathbb Z[G/N]$. -Now for the well known fact. Since we have described the lattice of normal subgroups of $G$ we can look at all pairs $\hat M$, $\hat N$ such that $M \cap N = 1$ and $M N = G$. For each such pair, $G = N \times M \cong G/M × G/N$ and we know $\mathbb Z[G/M]$ and $\mathbb Z[G/N]$. By assumption, there will be at least one “good” pair where both $G/M$ and $G/N$ are determined. Hence G is determined. I guess this is the only way I can interpret the missing result. -Now for the paper as a whole. We are looking for the chief factors of $G$. So start with $\hat N$, where $N$ is a minimal normal subgroup. Since we know $\mathbb Z[G/N]$, induction will give us the chief factors of $G/N$. So we need only determine the isomorphism class of $N$. Of course, we know $\lvert N\rvert$. If $\lvert N\rvert = p^n$, then $N$ is an elementary abelian $p$-group of that order. So assume not. Then $N$, being characteristically simple, must be a finite direct product of isomorphic nonabelian simple groups, say isomorphic to $H$. We know that any finite simple group has some Sylow $p$-subgroup of order $p$. Thus $\lvert N\rvert$ will tell us first the number of factors of $H$ and then $\lvert H\rvert$. The number of factors is the smallest exponent of a prime in $\lvert N\rvert$. If we are lucky, $\lvert H\rvert$ determines $H$, since there is just one family of counterexamples (I mean two infinite families with the same orders). If not, then more work is needed and I am not sure where to go. -One thought is to let $C$ be the centralizer of $N$ in $G$. Since $N$ has trivial center, we have $N \cap C = 1$. Indeed, $C$ is the unique largest normal subgroup with this property. If $C$ is not $1$, then $N$ lives in $G/C$. By induction on $\lvert G\rvert$, we can determine the chief factors of $G/C$ and hence find $N$. On the other hand, if $C = 1$, then $G$ embeds in the automorphism group of N and somehow the paper uses this information. Presumably there exists an old paper which studies the isomorphism question for simple groups and where one could start. I hope this helps.<|endoftext|> -TITLE: multiple zeta values and knots invariants -QUESTION [7 upvotes]: I have heard several times that MZV appear in the context of knot invariants and deformation quantisation. Could anyone explain how and give some references? - -REPLY [2 votes]: If you told me that knot invariants, modular forms, zeta values and quantum field theory were all related, I would not argue or be the slightest bit surprised. And yet, these connections were only disovered in the early 1990s - -Values of Zeta Functions and Applications Don Zagier - -At the very end, mentions a then-recent discover of Kontsevich on Vassiliev invariants and knot theory. And the Knizhnik-Zamolodichikov eq. -Stating these objects are related and pointing to a precise relationship are two different things. It is a topic currently under development.<|endoftext|> -TITLE: Is there an upper bound on dimension of kernel of elliptic operator for a fixed closed manifold M -QUESTION [7 upvotes]: Assume that $M$ is a smooth closed manifold and $E,F$ are fixed smooth vector bundles over $M.$ - -Is there a number $C,$ such that for any elliptic operator $\mathcal{D}:\Gamma(E)\to\Gamma(F)$ - $$\dim\ker\mathcal{D}\leqslant C.$$ - -For simplicity, we may assume that $\Gamma(E)=\Gamma(F)=C^\infty(M).$ - -REPLY [2 votes]: Here is another simple counterexample: every holomorphic vector bundle over a Riemann surface gives rise to an elliptic operator (the $\bar\partial$-operator) whose kernel is the space of holomorphic sections. Two vector bundles over the same compact surface are isomorphic as smooth complex vector bundles if and only if the have the same rank and the same degree. Now take the Riemann sphere $P^1$ and a line bundle $L\to P^1$ of degree $d\geq 1$ and the rank $2$ vector bundle $$E=L\oplus L^*\to P^1.$$ $E$ has degree $0$ for every $d$ but the dimension of holomorphic sections, i.e. the dimension of the kernel is $d+1$. -On the other hand, there is a class of elliptic operators for which such an estimate exists: - every linear elliptic differential operator -$$D\colon\Gamma(M,L)\to\Gamma(M,\tilde L)$$ -of order 1 between complex line bundles $L,\tilde L$ over a surface $M$ is given as the $\bar\partial$-operator of some holomorphic line bundle (with respect to an appropriate Riemann surface structure). The dimension of the kernel is then restricted by $$\dim\ker D\leq \mathrm{deg}(L)+1$$ -because we can always produce a $(k-1)$th order zero at a point for some section in a $k$-dimensional space of holomorphic sections of a line bundle.<|endoftext|> -TITLE: Is there an axiomatic characterization of the entropy of a continuous random variable? -QUESTION [15 upvotes]: Let $X$ be a random variable taking values in $\{1,\ldots,n\}$, and let $p_i$ denote the probability of the event $\{X = i\}$. Shannon defined the entropy of $X$ to be the quantity -$$H(X) = -\sum_i p_i \log p_i$$ -with the convention $x \log x = 0$. This definition is the starting point for all of information theory, and consequently it has been provided with numerous axiomatic characterizations. Shannon gave a characterization in his original paper; Myron Tribus gave another involving how "surprising" a random variable is on average; yet another approach constructs entropy as the objective function for certain optimization problems (turning the principle of maximum entropy into a definition); and Baez, Fritz, and Leinster gave still another characterization involving convexity and functorial properties. I'm sure this list is not exhaustive. -But I have only ever seen axiomatic characterizations for discrete random variables. Shannon himself defined the entropy of a continuous random variable by: -$$H(X) = -\int p(x) \log p(x)\, dx$$ -where $p(x)$ is the density function of $X$. Jaynes argued here that this is the wrong definition because it has the wrong units and it transforms incorrectly under a change of coordinates; he was able to modify the definition accordingly by taking a limit of discrete entropies. Regardless of which definition is correct my question is this: - -Is there an axiomatic characterization of the entropy of a continuous random variable which generalizes a corresponding characterization in the discrete case? If so, what is the "right" class of measure spaces to which this characterization applies? - -REPLY [5 votes]: Let $X$ be a discrete random variable (r.v.) taking distinct values $x_1,x_2,\dots$, and let $p_i:=P(X = x_i)$. Then the entropy of $X$ is defined by the formula -\begin{equation} - H(X): = \sum_i p_i \log \frac1{p_i}. \tag{1} -\end{equation} -Note that it does not matter whatsoever in what set/space the values $x_1,x_2,\dots$ are assumed to be; it does not matter if some of these values are close to, or far away from, one another -- in any sense. -What matters is that the $p_i$'s are the probabilities of distinct values of the r.v. $X$. Clearly, if we aggregate some of these values, then the entropy will go down; and if we split some of these values, then the entropy will go up. -Therefore, the formula $H(X) = \int dx\, p(x) \log \frac1{p(x)}$ will hardly make sense if, say, the integral is understood in the Riemann sense, implying the rather arbitrary grouping of the values $x$ according to the standard metric on $\mathbb R$. Moreover and much more importantly, the Riemann sums $\sum_i p(x_i)\Delta x_i \log \frac1{p(x_i)}\,$ for this integral are quite different from the sums $\sum_i p(x_i)\Delta x_i \log\frac1{p(x_i)\Delta x_i}$ that would genuinely correspond to the reality-based definition (1). Also, these latter sums will usually be very large if the $\Delta x_i$'s are very small, and the values of these sums may fluctuate wildly depending on the choice of the $\Delta x_i$'s. -The more general formula $H(X) = \int p(x) \log \frac1{p(x)}\, \mu(dx)$, where $\mu$ is a measure and the grouping of the values $x$ occurs according to the closeness of the corresponding values of $p(x)$ (!!), will hardly make more sense than the Riemann integral. -The only exception here would be when $\mu$ is the counting measure, with which no actual grouping (or splitting) of any values occurs. Then for the density (say $p$) of the distribution of the r.v. $X$ with respect to the counting measure $\mu$, the condition $\int p\,d\mu=1$ can be rewritten as $\sum_x p(x)=1$, which will imply that $p(x)\ne0$ only for (at most) countably many values of $x$, so that the r.v. $X$ is necessarily discrete -- and then we can write -\begin{equation*} - H(X) = \int p(x) \log \frac1{p(x)}\,\mu(dx) - =\sum_x p(x) \log \frac1{p(x)}, -\end{equation*} -which is the same as (1), up to the change in notation. -So, if the r.v. $X$ is not discrete, then the only reasonable value to assign to the entropy of $X$ appears to be $\infty$, at least from the viewpoint of information theory. -As for the integral $\tilde H(X):=\int p(x) \log \frac1{p(x)}\, \mu(dx)$ in the case when $\mu$ is the Lebesgue measure, the main interest to it seems to be the easily seen fact (see e.g. Barron) that the maximum of $\tilde H(X)$ over all absolutely continuous r.v.'s $X$ with a fixed variance is attained when the distribution of $X$ is normal; moreover, the absolute value of difference of $\tilde H(X)$ from its maximum equals the relative entropy $\int p(x)\log\frac{p(x)}{\varphi(x)}\, dx$, where $\varphi$ is the normal density with the same mean and variance as $p$. However, what is actually used in the proofs is the relative entropy $\int \log\frac{dP}{dQ}\,dP$ (also known as the Kullback--Leibler divergence), which is well defined for any probability measures $P$ and $Q$ such that $P$ is absolutely continuous with respect to $Q$.<|endoftext|> -TITLE: The average of reciprocal binomials -QUESTION [8 upvotes]: This question is motivated by the MO problem here. Perhaps it is not that difficult. - -Question. Here is an cute formula. - $$\frac1n\sum_{k=0}^{n-1}\frac1{\binom{n-1}k}=\sum_{k=1}^n\frac1{k2^{n-k}}.$$ - I've one justification along the lines of Wilf-Zeilberger (see below). Can you provide an alternative proof? Or, any reference? - -The claim amounts to $a_n=b_n$ where -$$a_n:=\sum_{k=0}^{n-1}\frac{2^n}{n\binom{n-1}k} \qquad \text{and} \qquad -b_n:=\sum_{k=1}^n\frac{2^k}k.$$ -Define $F(n,k):=\frac{2^n}{n\binom{n-1}k}$ and $\,G(n,k)=-\frac{2^n}{(n+1)\binom{n}k}$. Then, it is routinely checked that -$$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k),\tag1$$ -for instance by dividing through with $F(n,k)$ and simplifying. Summing (1) over $0\leq k\leq n-1$: -\begin{align} -\sum_{k=0}^{n-1}F(n+1,k)-\sum_{k=0}^{n-1}F(n,k) -&=a_{n+1}-\frac{2^{n+1}}{n+1}-a_n, \\ -\sum_{k=0}^{n-1}G(n,k+1)-\sum_{k=0}^{n-1}G(n,k) -&=-\sum_{k=1}^n\frac{2^n}{(n+1)\binom{n}k}+\sum_{k=0}^{n-1}\frac{2^n}{(n+1)\binom{n}k}=0. \end{align} -Therefore, $a_{n+1}-a_n=\frac{2^{n+1}}{n+1}$. But, it is evident that $b_{n+1}-b_n=\frac{2^{n+1}}{n+1}$. Since $a_1=b_1$, it follows $a_n=b_n$ for all $n\in\mathbb{N}$. -===================================== -Here is an alternative approach to Fedor's answer below in his elaboration of Fry's comment. -With $\frac1{n+1}\binom{n}k=\int_0^1x^{n-k}(1-x)^kdx$, we get $a_{n+1}=2^{n+1}\int_0^1\sum_{k=0}^nx^{n-k}(1-x)^kdx$. So, -\begin{align} -2^{n+1}\int_0^1 dx\sum_{k=0}^nx^{n-k}(1-x)^k -&=2^{n+1}\int_0^1x^ndx\sum_{k=0}^n\left(\frac{1-x}x\right)^k \\ -&=2^{n+1}\int_0^1x^n\frac{\left(\frac{1-x}x\right)^{n+1}-1}{\frac{1-x}x-1}dx \\ -&=\int_0^1\frac{(2-2x)^{n+1}-(2x)^{n+1}}{1-2x}\,dx:=c_{n+1}. -\end{align} -Let's take successive difference of the newly-minted sequence $c_{n+1}$: -\begin{align} -c_{n+1}-c_n -&=\int_0^1\frac{(2-2x)^{n+1}-(2-2x)^n+(2x)^n-(2x)^{n+1}}{1-2x}\,dx \\ -&=\int_0^1\left[(2-2x)^n+(2x)^n\right]dx=2^{n+1}\int_0^1x^ndx=\frac{2^{n+1}}{n+1}. \end{align} -But, $b_{n+1}-b_n=\frac{2^{n+1}}{n+1}$ and hence $a_n=b_n$. - -REPLY [7 votes]: As in the question -$$a_n:=\sum_{k=0}^{n-1}\frac{2^n}{n\binom{n-1}k} \qquad \text{and} \qquad -b_n:=\sum_{k=1}^n\frac{2^k}k.$$ -It is clear that $a_1=b_1$ and $b_{n+1}-b_n=\frac{2^{n+1}}{n+1}$. But we have the same recursive relation for $a_n$ because -\begin{align} -a_n&=2^n\sum_{k=0}^{n-1}\frac{k!(n-1-k)!(k+1+n-k)}{n!(n+1)} \\ -&=2^n\sum_{k=0}^{n-1}\left(\frac{(k+1)!(n-1-k)!}{(n+1)!}+\frac{k!(n-k)!}{(n+1)!}\right) \\ -&=2^n\left(2\sum_{k=0}^{n}\frac{k!(n-k)!}{(n+1)!}-\frac{2}{n+1}\right)=a_{n+1}-\frac{2^{n+1}}{n+1}. -\end{align}<|endoftext|> -TITLE: Coefficient bounds on cusp forms, half-integer weight -QUESTION [10 upvotes]: Let $f(\tau) = \sum_{n=1}^{\infty} a(n) q^n$ be a cusp form on $\Gamma_0(4N)$ of half-integer weight $k \ge 5/2.$ The Ramanujan-Petersson conjecture in this case is that $$a(n) \ll n^{(k-1)/2 + \varepsilon}$$ for any $\varepsilon > 0$. Unlike the integer-weight case, this is unresolved as far as I am aware. -I am interested to know what is currently the best bound on the growth of $a(n)$. The bound I am most familiar with is due to Iwaniec (Theorem 1 of this article) giving $a(n) \ll n^{k/2 - 2/7 +\varepsilon}$. But it seems that Conrey and Iwaniec proved a better bound of the form $a(n) \ll n^{k/2 - 1/3 + \varepsilon}$ here as Corollary 1.3 (it is unclear to me whether this holds generally) -Both of the articles I linked to are fairly old. I would be grateful if an expert could inform me about what has happened since then. - -REPLY [13 votes]: The Conrey-Iwaniec bound is still the best known one. Originally it was restricted to $N=1$, but Petrow and Young extended the result to all square-free $N$'s (this is highly nontrivial). Also, in these results (including Iwaniec's) $n$ itself is assumed to be square-free. Obtaining a bound for all $n$'s is somewhat tricky: see Corollary 2 and the references in Blomer and Harcos. Note that this paper features a weaker bound but in a more general situation (for example, there is no restriction on $N$).<|endoftext|> -TITLE: Distribution that vanishes against approximated delta is zero -QUESTION [7 upvotes]: Suppose we have a Schwartz distribution $\phi$ on $\mathbb{R}^d$ such that $$ \forall x, \ \lim_{\lambda \to 0}| \langle\phi, \psi^{\lambda}_x \rangle| =0$$ -where $\psi^{\lambda}_{x}=\lambda^{-d}{\psi\left(\frac{\cdot - x}{\lambda}\right)}$ approximates a delta in $x$. Here we assume that $\psi$ is in $C_c^{\infty}$ supported in the unit ball. -Can we conclude that $\phi$ is zero? -This statement is obvious if $\phi$ is a function, but I can't prove it if $\phi$ is a generalized function. -EDIT: -I forgot a detail of great importance as shown below: the limit is zero not just for one $\psi,$ but for all $\psi \in C_c^{\infty}$ with compact support in the unit ball. -EDIT #2: -The proof of this fact becomes actually easy in which the limit is zero locally uniformly in $x$, i.e. if for any compact $K \subset \mathbb{R}^d$ we get: $$\lim_{\lambda \to 0} \sup_{x \in K}| \langle\phi, \psi^{\lambda}_x \rangle| =0$$ -In this case we just observe that we actually are looking at a convolution: $\langle\phi, \psi^{\lambda}_x \rangle = \phi * \psi^{\lambda}(- \cdot) \ (x)$. Now observing that $ h_{\lambda} = \psi^{\lambda}(- \cdot)$ still approximates a delta, we can use a famous result stating that $\phi * h_{\lambda} \to \phi$ in the sense of distributions. -Since our uniform limit estimate actually tells us that we are converging to zero uniformly on all compacts, we get that $\phi = 0.$ - -REPLY [2 votes]: I think that the answer is yes. Assume by contradiction that the support $S$ of $\phi$ is nonempty. Since $S$ is closed, it is in particular a complete metric space. -Let $X:=\{\psi\in C^\infty(\mathbb{R}^d):\psi\equiv 0\text{ on }\mathbb{R}^d\setminus B_1\}$, which is a Fréchet space. -Apply now, for any fixed $x\in S$, the uniform boundedness principle to the family of functionals -$$X\to\mathbb{R},\qquad\{\psi\mapsto\langle\phi, \psi^{\lambda}_x \rangle\mid 0<\lambda<1\},$$ obtaining that for some minimal $k(x)\in\{1,2,\dots\}$ it holds -$$ |\langle\phi,\psi_x^\lambda\rangle|\le k(x)\|\psi\|_{C^{k(x)}} $$ -(for any $0<\lambda<1$). The sets $S_N:=\{x\in S:k(x)\le N\}$ are closed and cover $S$, so one of them has nonempty interior: say that $B_r(x)\cap S\subseteq S_N$ for some $x\in S$. -Now, for any positive integer $M$ large enough, we can easily find a collection $(\rho_{M,j})_j\subseteq C^\infty_c(B_r(x))$, with cardinality $O(2^{Md})$, of functions whose supports have diameter less than $2^{-M}$ and satisfying $\|\rho_{M,j}(2^{-M}\cdot)\|_{C^N}=O(1)$, as well as $\sum_j\rho_{M,j}\equiv 1$ on $B_{r/2}(x)$ (say). -Thus, for any $\psi\in C^{\infty}_c(B_{r/2}(x))$, -$$ \langle\phi,\psi\rangle=\sum_j\langle\phi,\rho_{M,j}\psi\rangle $$ -and we can discard all the terms where $\rho_{M,j}$ has support disjoint from $S$. As for the other terms, -let $y_j\in S\cap\text{supp }\rho_{M,j}$; notice that -$$\langle\phi,\rho_{M,j}\psi\rangle=2^{-Md}\langle\phi,\eta_{y_j}^{2^{-M}}\rangle$$ -where $\eta:=\rho_{M,j}(y_j+2^{-M}\cdot)\psi(y_j+2^{-M}\cdot)$. Thus, -$$|\langle\phi,\rho_{M,j}\psi\rangle|\le 2^{-Md}N\|\rho_{M,j}(y_j+2^{-M}\cdot)\psi(y_j+2^{-M}\cdot)\|_{C^N}$$ -$$\le CN 2^{-Md}\|\psi(y_j+2^{-M}\cdot)\|_{C^N}. $$ -Summing in $j$ and letting $M\to\infty$, this easily leads to the bound -$$ |\langle\phi,\psi\rangle|=O(\|\psi\|_\infty), $$ -showing that $\phi$ restricts to a measure on $B_{r/2}(x)$. The hypothesis, plus well-known differentiation theorems, imply that $\phi\equiv 0$ here (see below), contradicting the fact that $x\in S$. -Addendum. Proof of the fact that $\phi$ vanishes on $B_r(x)$: -Let us restrict our attention to $B_r(x)$. We split the measure $\phi=\phi^+-\phi^-$ into positive and negative part and we set $\mu:=\phi^-+\mathcal{L}^d$, $\nu:=\phi^+$. Assume that some $y\in B_r(x)$ satisfies -$$\lim_{\rho\to 0}\frac{\nu(B_\rho(y))}{\mu(B_\rho(y))}=+\infty.$$ -Pick any radial $\psi\in C^\infty_c$ -with $\psi\ge 0$ and $\int_{\mathbb{R}^d}\psi\,d\mathcal{L}^n=1$. Notice that -$$\frac{\int\psi_y^\lambda\,d\nu}{\int\psi_y^\lambda\,d\mu}=\frac{\int_0^\infty\nu(\{\psi_y^\lambda>t\})\,dt}{\int_0^\infty\mu(\{\psi_y^\lambda>t\})\,dt}\to+\infty$$ as $\lambda\to 0$ (since the superlevels are smaller and smaller balls). Thus, -$$\frac{\int\psi_y^\lambda\,d\phi}{\int\psi_y^\lambda\,d\mu}\ge\frac{\int\psi_y^\lambda\,d\nu}{\int\psi_y^\lambda\,d\mu}-1\to+\infty,$$ -but $\int\psi_y^\lambda\,d\mu\ge\int\psi_y^\lambda\,d\mathcal{L}^d=1$, contradicting the hypothesis. So such $y$ does not exist and by Theorem 2.22 (Besicovitch derivation theorem) in Ambrosio, Fusco, Pallara, Functions of Bounded Variation and Free Discontinuity Problems we have $\nu\ll\mu$. -Since $\phi^+$ and $\phi^-$ are mutually singular, we deduce $\phi^+\ll\mathcal{L}^d$. -Repeating all the arguments with $-\phi$ instead of $\phi$, we get $\phi^-\ll\mathcal{L}^d$ as well. -So $\phi$ has a density with respect to $\mathcal{L}^d$. For all $y$ such that $\lim_{\rho\to 0}\frac{\phi(B_\rho(y))}{\mathcal{L}^d(B_\rho(y))}$ exists finite, the ratio $\frac{\int\psi_y^\lambda\,d\phi}{\int\psi_y^\lambda\,d\mathcal{L}^d}$ converges to the same limit. Applying the aforementioned theorem and the hypothesis, we obtain that the density is zero.<|endoftext|> -TITLE: Can this optimization problem be transformed into or approximated by a SOCP? -QUESTION [5 upvotes]: We would like to know if the following optimization problem can be transformed into an SOCP problem or maybe approximated by a SOCP problem. The objective function is defined as -$$ -\mathrm{Obj}(x) = \big({\alpha^{T}x -x^{T}\Lambda x -{x^{\beta}}^{T}\Omega x^{\beta} }\big) -$$ -and we would like to solve -$$ -\max_{x} \mathrm{Obj}(x) -$$ -subject to Linear and Quadratic constraints for the $n$-dimensional vector $x=(x_1,x_2,\ldots,x_n)$. The vector $x^{\beta}$ is defined such that its $i$-th entry is equal to $x_{i}^{\beta}$ and -$$ -\frac{1}{2}\leq\beta<1. -$$ -Both matrices $\Lambda$ and $\Omega$ are positive definite. -Especially important is the case $\beta=\frac{1}{2}$. - -REPLY [5 votes]: First, you should restrict $x$ to be positive or use $|x|^\beta$ instead. -Then I think that the answer is no: -For $\beta\neq 1/2$ you can argue as follows: -The special case of diagonal $\Omega = \mathrm{diag}(w_1,\dots,w_n)$ ($w_i>0$) is simpler as in this case you problem is -$$ -\min_x x^T\Lambda x - \alpha^T x + \sum_i w_i |x_i|^{2\beta}\quad\text{s.t. convex constraints} -$$ -i.e. is is a convex quadratic problem with a weighted $\ell^{p}$ regularizer with $1\leq p=2\beta <2$. -As such it is a fairly simple convex problem. -I am fairly sure that even this special case can not be cast as SOCP (if found this claim in "Mixed norm FIR filter optimization using second-order cone programming" by Dan P. Scholnik (ICASSP 2002) and the report "Second-Order Cone Formulations of Mixed-Norm Error Constraints for FIR Filter Optimization" by Dan P. Scholnik and Jeffrey O. Coleman but no reference is given). -For $\beta=1/2$, i.e. $p=1$ one has to argue differently as the above case can be cast as an SOCP. The case including the spd matrices is equivalent to (neglecting the constraints) -$$ -\min_x \|Ax-b\|_2^2 + \|Lx^{1/2}\|_2^2 -$$ -with some $A$, $b$ and $L$. -Here the penalty looks like -$$ -\|Lx^{1/2}\|_2^2 = \sum_i (\sum_j l_{i,j}x_j^{1/2})^2. -$$ -With $n=2$, and $L=\begin{bmatrix}1 & 0\\1 & 1\end{bmatrix}$ lead to -$$ -\|Lx^{1/2}\|_2^2 = |x_1| + (\sqrt{x_1}+\sqrt{x_2})^2 -$$ -which is not a convex function (simply check that the level sets are not convex), so also here, the answer is no. -There is still a possibility that there may be a clever reformulation/substitution, but I doubt that. A prove that there is no such a reformulation seems very hard…<|endoftext|> -TITLE: Lifting line bundles -QUESTION [6 upvotes]: Let $X$ be a smooth proper geometrically integral scheme over $\overline{\mathbb F_p}$. Assume $X$ is the specialization of a smooth proper scheme over $\mathbb Z_p^{nr}$. Let $L$ be an ample line bundle on $X$. -I would like to show that $L^{\otimes p}$ lifts to characteristic zero. However, the obstruction to lifting $L$ lives in $\mathrm{H}^2(X,\mathcal O_X)$ which might be non-zero. -On the other hand, the obstruction space is an $\overline{\mathbb{F}_p}$-vector space. So the obstruction vanishes after multiplying with $p$. -Does this imply that $L^{\otimes p}$ lifts? - -REPLY [5 votes]: This cannot always be done. If $X$ is a supersingular K3 surface then there are $22$ ample line bundles $L$ -that give independent classes; if every $L^{\otimes p}$ lifted to char. zero, then you'd have a K3 surface in char. zero with $22$ independent line bundles, which is impossible. (Or $X$ could be any liftable unirational surface with $p_g(X)>0$.)<|endoftext|> -TITLE: Clarifying Quillen's comment on why Adams operations commute with transfer -QUESTION [8 upvotes]: Let $f : X \rightarrow Y$ be a finite covering space. In Quillen's paper on the Adams conjecture in Topology, 1970, he gives the following argument for why -$$f_* \psi^p = \psi^p f_* \in KO[p^{-1}]$$ -Quillen's argument is that the map $f_*$ can be interpreted as a Gysin homomorphism, and "since the normal bundle of $f$ has a canonical trivialization, this Gysin homomorphism is essentially the composite of one suspension isomorphism and the inverse of another". -I'd like to flesh out this last part. How does one make it rigorous? - -REPLY [10 votes]: Before I answer the question as stated, let me point out that you can also prove the claim as follows: the transfer is induced by a stable map $Y_+\rightarrow X_+$, and the pth Adams operation is a stable cohomology operation after inverting p, so the result is just naturality of cohomology operations. -Now for Quillen's argument. Embed $X$ into $S^n$ and so factor the map f as $X\rightarrow Y\times S^n \rightarrow Y$. If the normal bundle of the first map is, say, Spin oriented, then the Gysin map is defined as the composite: $KO(X) \rightarrow KO(Th(\nu)) \rightarrow KO(S^n \times Y) \rightarrow KO(Y)$. Here we use the Thom isomorphism, then the collapse map, then suspension isomorphism- I'm writing on a phone so I didn't put in the proper indices on the KO's or that some of those should be reduced K-theory. -Anyway, in this case the normal bundle in question is always trivial, and the Thom isomorphism for a trivial bundle is just the suspension isomorphism. This is what Quillen means by doing a suspension isomorphism and then undoing it. -The two arguments aren't unrelated of course, especially if you recall how the stable map yielding the transfer is constructed.<|endoftext|> -TITLE: A mystery sequence -QUESTION [13 upvotes]: This question arose from the recent one, roots of a polynomial linked to mock theta function?. Let -$$ -g(x):=\sum_{k=0}^\infty x^k\prod_{j=1}^{k-1}(1 + x^j)^2\\=1+x+x^2+3 x^3+4 x^4+6 x^5+10 x^6+15 x^7+21 x^8+30 x^9+43 x^{10}+59 x^{11}+...; -$$ -the sequence $1,1,1,3,4,6,10,15,21,30,43,59,...$ with the generating function $g(x)$ is A059618 on OEIS, it is the sequence of numbers of strongly unimodal partitions. -Now let -$$ -f(q):=g(q)\prod_{n=1}^\infty(1-q^n), -$$ -and let $a_k$ be the $k$th coefficient in the Maclaurin series for $f$, -$$ -f(x)=\sum_{k=0}^\infty a_kx^k\\=1-x^2+x^3+x^6+x^7-x^9+x^{10}-x^{14}+x^{18}-x^{20}+x^{21}+x^{25}+x^{26}-x^{27}\\+x^{28}-x^{30}+x^{33}-x^{35}+x^{36}-x^{39}-x^{40}+x^{42}-x^{44}+2x^{45}-x^{49}+x^{52}-x^{54}\\+x^{55}+x^{56}+x^{57}-x^{60}-x^{65}+... -$$ -The sequence of $a_k$, starting with - -1,0,-1,1,0,0,1,1,0,-1,1,0,0,0,-1,0,0,0,1,0,-1,1,0,0,0,1,1,-1,1,0,-1,0,0,1,0,-1,1,0,0,-1,-1,0,1,0,-1,2,0,0,... - -is not on OEIS. Among the first 1000 terms of the sequence, there are 609 zeroes, 182 ones, 161 -1s, 19 of them are 2 ($a_{45},a_{150},a_{210},a_{221},a_{273},a_{300},...$), 22 are -2 ($a_{77},a_{90},a_{165},a_{225},...$), and two of them ($a_{525}$ and $a_{825}$) are 3; seems like $a_k$ are zero for $k=2^j$ ($j>0$), for $k=p$ or $k=2p$, with $p$ prime $>7$, $k=3p$ and $k=4p$ with $p$ prime $\geqslant23$, $k=5p$ with $p$ prime $>31$, $6p$ for $p>37$, $7p$ and $8p$ for $p>43$, $9p$ for $p>47$, $10p$ for $p>61$, $11p$ for $p>67$,... -What may (or may not) be relevant is another sequence obtained from introducing new variable in the way I learned from a paper by Rhoades linked to from the above OEIS page for $g$. -Let -$$ -g_t(q):=\sum_{k=0}^\infty q^k\prod_{j=1}^{k-1}(1 + q^jt)(1+q^j/t), -$$ -and let -$$ -f_t(q)=g_t(q)\prod_{n=1}^\infty(1-q^n), -$$ -so that $g_1(q)=g(q)$ and $f_1(q)=f(q)$. Then -$$ -f_t(q)=1-q^2+\frac{1+t^3}{(1+t)t}q^3+\frac{1+t^5}{(1+t)t^2}q^6+q^7-\frac{1+t^3}{(1+t)t}q^9+\frac{1+t^7}{(1+t)t^3}q^{10}+...; -$$ -most coefficients have form $\pm\frac{1+t^{2j+1}}{(1+t)t^j}$, except that I cannot figure out how $j$ depends on the number of the coefficient. Exceptions here start from the $15$th coefficient, which is $\frac{1+t^9}{(1+t)t^4}-1$ and the $45$th one which is $\frac{1+t^{17}}{(1+t)t^8}+\frac{1+t^3}{(1+t)t}$. -Despite all these clues, to my shame I've given up searching for an explicit formula for $a_k$. Is there one? I am pretty sure there is, but what is it? - -REPLY [18 votes]: The conjectured identity -$$ -f(q)=(q;q)_\infty\left(1+\sum_{k=1}^\infty q^k(-q;q)^2_{k-1}\right)=\sum_{\substack{m,n\geqslant0\\n\ne1}}(-1)^mq^{\frac{(m+n)(3m+n+1)}2},\tag{1} -$$ -using Euler's pentagonal number theorem $(q;q)_\infty=\sum _{m=-\infty}^\infty (-1)^m q^{\frac{1}{2} m (3 m+1)}$ can be brought to an equivalent form -$$ -(q;q)_\infty\sum_{k=1}^\infty q^k(-q;q)^2_{k-1}=\sum_{\substack{m\ge0}}\sum_{\substack{n\ge 1}}(-1)^mq^{\frac{(m+n)(3m+n+1)}2}.\tag{1a} -$$ -By identity $(4.1)$ in Rhoades' paper the sum on the LHS of $(1a)$ is -$$ -\sum_{k=1}^\infty q^k(-q;q)^2_{k-1}=\frac{1}{2(q;q)_{\infty }} \sum _{n=-\infty}^\infty \frac{q^{\frac{1}{2} n (n+1)}}{q^n+1}-\frac{1}{4}\sum _{k=0}^\infty \frac{q^{k^2}}{(-q;q)_k^2},\tag{2} -$$ -while the double sum in $(1a)$ -\begin{align} -\sum_{\substack{m\ge0}}\sum_{\substack{n\ge 1}}(-1)^mq^{\frac{(m+n)(3m+n+1)}2}&=\sum _{m=0}^\infty \sum _{k=m+1}^\infty (-1)^mq^{\frac{k(k+1+2m)}2}\\ -&=\sum _{k=1}^\infty \sum _{m=0}^{k-1} (-1)^m q^{m k+\frac{1}{2} (k+1) k}\\ -&=\sum _{k=1}^\infty q^{\frac{k^2}{2}+\frac{k}{2}} \frac{1-(-1)^k q^{k^2}}{1+q^k}.\tag{3} -\end{align} -Since $\sum _{n=-\infty}^\infty \frac{q^{\frac{1}{2} n (n+1)}}{q^n+1}=\frac{1}{2}+2 \sum _{n=1}^\infty \frac{q^{\frac{1}{2} n (n+1)}}{q^n+1}$ $(2)$ and $(3)$ contain the same sum $\sum _{n=1}^\infty \frac{q^{\frac{1}{2} n (n+1)}}{q^n+1}$. This sum cancels out after $(2)$ and $(3)$ are substituted in $(1a)$ resulting in -$$ -\frac14-\frac{(q;q)_{\infty }}{4}\sum _{k=0}^\infty \frac{q^{k^2}}{(-q;q)_k^2}=-\sum _{n=1}^\infty \frac{(-1)^n q^{\frac{3 n^2}{2}+\frac{n}{2}}}{1+q^n},\tag{1c} -$$ -and equivalently -$$ -\sum _{k=0}^\infty \frac{q^{k^2}}{(-q;q)_k^2}=\frac{2}{(q;q)_{\infty }}\sum _{n=-\infty}^\infty \frac{(-1)^n q^{\frac{3 n^2}{2}+\frac{n}{2}}}{1+q^n}.\tag{1d} -$$ -$(1d)$ corresponds to the special case $x=-1$ of the identity -$$ -\sum _{k=0}^\infty \frac{q^{k^2}}{(x q;q)_k(q/x;q)_k}=\frac{1-x}{(q;q)_{\infty }}\sum _{n=-\infty}^\infty \frac{(-1)^n q^{\frac{3 n^2}{2}+\frac{n}{2}}}{1-xq^n},\tag{4} -$$ -which can be obtained from Watson-Whipple transformation formula (see the paper "Modular transformations of Ramanujan’s fifth and seventh order mock theta functions", Ramanujan J. 7 (2003), 193–222. by Gordon and McIntosh). $(4)$ also can be proved directly by partial fractions expansion and the following limiting case of q-Gauss summation -$$ -\sum _{k=n}^\infty\frac{q^{k^2}}{(q;q)_{k-n}(q^{n+1};q)_{k}}=\frac{q^{n^2}}{(q^{n+1};q)_n}\sum _{k=0}^\infty\frac{q^{k^2+2kn}}{(q;q)_{k}(q^{2n+1};q)_{k}}=\frac{q^{n^2}}{(q^{n+1};q)_\infty}. -$$<|endoftext|> -TITLE: Existence of a path in a set of subsets of $\omega$ -QUESTION [6 upvotes]: Suppose that $L\subseteq {\cal P}(\omega)$ has the following properties: - -$\omega \notin L$, and for $e\in L$ we have $|e|\geq 2$; -if $e_1\neq e_2 \in L$ then $|e_1\cap e_2|\leq 1$; -if $m,n\in \omega$ there is $e\in L$ such that $\{m,n\}\subseteq e$. - -It is not hard to see that $L$ is countable. Is there a bijection $p:\omega\to L$ such that for all $n\in\omega$ we have $p(n)\cap p(n+1) \neq \emptyset$? - -REPLY [2 votes]: Clearly, $L$ is countably infinite. Thus it will suffice to prove the following lemma, which shows that any finite path $\langle e_1,\dots,e_k,a\rangle$ in $L$ can be extended to include any new element $b\in L.$ -Lemma. Given a finite set $E=\{e_1,\dots,e_k\}\subseteq L$ and $a,b\in L\setminus E,$ we can find $c,d\in L\setminus E$ (not necessarily distinct) such that $a\cap c,\ c\cap d,$ and $d\cap b$ are nonempty. -Proof. We consider two cases. -Case 1. All elements of $E$ are finite. -Choose $m\in a,\ n\in b,$ and $p\in\omega\setminus\bigcup E.$ Find $c,d\in L$ such that $\{m,p\}\subseteq c$ and $\{n,p\}\subseteq d.$ -Case 2. There is an infinite element $u\in E.$ -Choose $m\in a\setminus u,\ n\in b\setminus u,$ and $p\in u\setminus\bigcup(E\setminus\{u\}).$ Find $c,d\in L$ such that $\{m,p\}\subseteq c$ and $\{n,p\}\subseteq d.$<|endoftext|> -TITLE: What are Sylvester-Gallai configurations in the complex projective plane? -QUESTION [15 upvotes]: A Sylvester-Gallai configuration in the the complex projective plane is a finite number of $n\ge 2$ points in the complex projective plane such that there is no line through exactly two of them. Trivial examples are obtained by taking $n\ge 3$ points on the same line. There is also the classical Hessian configuration, obtained by taking the nine inflexion points of a smooth cubic. - -Question. Is there any other example? Is there a way to classify these examples? I would be particularly interested in finding a non-collinear example with an even number of points. Does it exist? - -Remark: In higher dimension, it is known that the configuration of points has to be coplanar. If the points have coordinates defined over $\mathbb{R}$, the Sylvester-Gallai theorem shows that any configuration as above is in fact collinear. Over finite fields one can of course find plenty of configurations by taking all points. - -REPLY [10 votes]: I asked around about this question a while ago and the best answer I got was from Konrad Swanepoel. There are the well-known "Fermat" examples -$$(x^n - y^n)(y^n - z^n)(z^n - x^n) = 0, \qquad n \ge 3.$$ -The $3n$ lines here, together with the $n^2$ points of intersection and the 3 coordinate vertices, form a dual Sylvester–Gallai configuration. So these give rise to a Sylvester–Gallai configuration with $3n$ points and $n^2+3$ lines. If you want an even number of points, just take $n$ to be even. -There are only two other ("sporadic") examples known, one due to Klein and one due to Wiman, which are described for example in this paper. They have 21 and -45 points respectively so they don't give you what you want. -According to Swanepoel the complete classification is still an open problem, but he believes that there are at most finitely many more sporadic examples. -In 1973, Kelly and Nwankpa classified all Sylvester–Gallai designs (a more general concept than configurations) with at most 14 points. Swanepoel warned me that there are some errors in this paper, but if we take its results at face value then it shows that there are no other complex Sylvester–Gallai configurations this small. It should be possible to extend the computational search beyond 14 points but nobody seems have done so (or at least has not published the results). By considering complex reflection groups, a colleague of mine has performed an unpublished computation that, if correct, shows that there are no other examples with reflective symmetry in every line.<|endoftext|> -TITLE: Category-theoretic characterization of $L$ -QUESTION [14 upvotes]: Does there exist a characterization of Goedel's constructible universe $L$ in purely category-theoretic terms, or is constructibility an 'artifact' of material set theory? If, in fact, constructibility is an 'artifact' of material set theory, is there a category-theoretic 'analogue' for $L$? - -REPLY [2 votes]: Consider the following statement from Lawvere's paper, "Cohesive Toposes and Cantor's 'lauter Einsen' (Philosophia Mathematicae (3) Vol. 2 (1994), pp. 5-15): - -It might seem that G$\ddot o$del's $L$ would rest, if anything does, on von Neumann's a priori $\epsilon$-chains, but that was refuted twenty years ago by William Mitchell who showed how to construct it from a suitable given category, structured only by composition of maps; however, to my knowledge this has not been pursued since. - -Though it might not be 'kosher' to answer a question with a question, why does Mitchell's construction not provide the necessary category-theoretic characterization of $L$ I seek (Mitchell's construction is contained in his papers, "Boolean Topoi and the Theory of Sets" (Journal of Pure and Applied Algebra 2(1972), 261-274) and "Categories of Boolean Topoi" (Journal of Pure and Applied Algebra 3 (1973) 193-201))?<|endoftext|> -TITLE: Almost free actions on simply-connected spaces -QUESTION [6 upvotes]: Let $G$ be a simply-connected compact topological group (you can think of $SU(n)$ if you like it more concrete), and let $X$ be a finite-dimensional simply-connected $G$-CW-complex. If we know that all the isotropy groups are finite, does this imply they are trivial? And if not, are they at least bounded in size (in some sense)? - -REPLY [2 votes]: Since it was requested, here is the answer (from the comment) again. -The statement is wrong. For each $k$, one can start with a unique $0$-cell $G/(\mathbb Z/k)$ and attach a $2$-cell $G \times D^2$ by a map $G \times S^1 \to G/(\mathbb Z/k)$ that equivariantly extends a generator $S^1 \to G/(\mathbb Z/k)$ of the fundamental group $\pi_1(G/(\mathbb Z/k)) = \mathbb Z/k$. The resulting $G$-CW complex will be simply-connected, with finite, but non-trivial isotropy subgroups.<|endoftext|> -TITLE: Total space of canonical bundle as resolution of singularity -QUESTION [6 upvotes]: We know for $Y=\mathbb{P}^n$, the total space of the canonical sheaf $Tot(\omega_Y)$ is the resolution of $\mathbb{C}^{n+1}/\mathbb{Z}_{n+1}$ where the generator acts as scalar matrix of multiplying a primitive $n$-th root. I was told this result can be extended to certain types of del-pezzo surfaces but had a hard time to find the reference. -My question: When is the total space of the canonical bundle a resolution of singularity? -I would very much appreciate it if anyone can give some reference to this matter. - -REPLY [6 votes]: I don't think this is special about the canonical bundle. All you need is a line bundle with a section which is contractible inside the total space of the line bundle. -Another way to generate such examples is the following: Take a smooth projective variety $Y$ (embedded in a projective space) and let $X$ be the cone over Y. Blowing up the vertex of $X$ is a resolution of the singularity of $Y$ and the blow up, say $B$, is naturally the total space of a line bundle on $Y$. -You might ask what line bundle you get this way. -Well, we know that if we blow up a point in a smooth variety, then the -exceptional divisor is a projective space and its normal bundle is $\mathscr O(-1)$. This means that when you blow up a cone, the exceptional divisor is isomorphic the original smooth variety which is naturally embedded in the exceptional divisor of the blow up of the affine space which is the cone over the original projective space. So the normal bundle of $Y$ in $B$ is the $\mathscr O(-1)$ corresponding to the original embedding of $Y$ you started with. -The last question one needs to answer is: What is the normal bundle of a section in the total space of a line bundle? The "obvious" answer is that it is that line bundle. Why? Well, the tangent bundle of the section is clearly the restriction of the pull-back of the tangent bundle from $Y$ and so the normal bundle has to be the restriction of the relative tangent bundle. Given that it is a line bundle the relative tangent bundle will be just that line bundle. -So where are we now. Let's say that you have a line bundle $\mathscr L$ on your favorite smooth projective variety whose inverse (dual) is very ample. Using the global sections of this inverse, embed $Y$ into a projective space, take the cone over it and let that be $X$. Blowing up $X$ at its vertex gives you the total space of the line bundle $\mathscr L$ over $Y$. -In case you wonder how the total space of an anti-ample line bundle has a section, then recall that the sheaf of sections of the total space of a line bundle is the dual of the line bundle. -Finally, this shows that this works for the canonical bundle of any Fano variety. Your example appears as the resolution of the cone over the $n+1$-uple embedding of $\mathbb P^n$ which happens to be given by the global sections of the dual of the canonical bundle, but the fact that it is the canonical bundle has not much to do with this. - -The rest is a response to the additional question in the comments below. -Remark. -The following statement is somewhat related to the above problem: - -Lemma - Let $Z\to \mathbb P^n$ be a $\mathbb P^1$-bundle with a section $E\subseteq Z$ and assume that there exists a proper birational morphism $\alpha:Z\to X$ such that $\alpha$ contract $E$, i.e., $\alpha(E)$ is a point and $\alpha|_{Z\setminus E}$ is an isomorphism. If $X$ is normal and factorial (i.e., every Weil divisor is Cartier), then $X\simeq \mathbb P^n$. - -This is an intermediate statement in Mori's proof of Hartshorne's conjecture, see Kollár's book on rational curves. -So, this implies that if $Y=\mathbb P^n$, and $X$ is not the affine space $\mathbb A^{n+1}$ (which happens if $Y$ is embedded as a linear subspace), then $X$ cannot be factorial. The easiest examples of non-factorial singularities are quotients by finite groups. Furthermore, if $Y=\mathbb P^n$ or more generally such that $\mathrm{Pic} Y\simeq \mathbb Z$, then the local class group of the singularity of $X$ is a finite cyclic group (the point is to realize that it is torsion and generated by a single element). This implies that there is a finite (non-flat!) cover of $X$ which is factorial. One could argue that then it is not surprising that we would find finite quotient singularities among these.<|endoftext|> -TITLE: What can be said of a $6$-core Young diagram whose $2$-and $3$-cores are empty -QUESTION [7 upvotes]: Let $\lambda$ be a partition. Suppose that $\lambda$ is both $2$- and $3$-decomposable, in the sense that $\lambda$ admits a total decomposition by both $2$-rim hooks (aka dominos) and $3$-rim hooks. Equivalently, assume that the $2$-core and $3$-core of $\lambda$ is zero. Then what can be said about the $6$-core of $\lambda$, if anything? Has this problem been studied (i.e. some sort of fundamental theorem of arithmetic for tableaux)? Must the $6$-core either be a $2\times 3$ or a $3\times 2$ rectangle? The problem is easily reduced to the classification of $6$-cores whose $2$-and $3$-cores are empty; I had difficulty writing down such a core other than the above two examples. - -REPLY [3 votes]: Let me comment a bit further on the fact that there are other 6-cores with trivial 2 and 3-core, in fact infinitely many of them. The fundamental paper of Garvan, Stanton, Kim, "Cranks and T-cores", gives a bijection between t-cores and integer tuples $(n_0,n_1,\dots,n_{t-1})$ which satisfy $\sum_{i=0}^{t-1}n_i=0$. Under this bijection the corresponding partition has size -$$\frac{t}{2}(n_0^2+\cdots +n_{t-1}^2)+n_1+2n_2+\cdots+(t-1)n_{t-1}.$$ -Now, this means that our set of 6-cores can be identified with 6-tuples $(n_0,n_1,\cdots,n_5)$ which satisfy $\sum_{i=0}^5 n_i=0$. In order to have an empty 2-core they must further satisfy $n_0+n_2+n_4=n_1+n_3+n_5=0$, and in order to have empty 3-core they must satisfy $n_0+n_3=n_1+n_4=n_2+n_5=0$. Therefore we are left with a sublattice of $\mathbb Z^6$ of dimension 2 generated by $(1,1,0,-1,-1,0)$ and $(0,1,1,0,-1,-1)$. These correspond to your two partitions, $2^3, 3^2$. Thus, we can form the generating function for 6-cores with empty 2 and 3 core: -$$\sum_{m,n\in \mathbb Z^2}q^{12(m^2+mn+n^2)-6(m+n)}$$ -This ends up being expressible as an infinite product, or a ratio of eta functions, if you will, as follows -$$\frac{\prod_{k\geq 1}(1-q^{12k})^3(1-q^{18k})^2}{\prod_{k\geq 1}(1-q^{6k})^2(1-q^{36k})}.$$ -Presumably generating functions of partitions that are $n$-cores but have empty $d$-core for all $d|n$ (I would be tempted to call these "pure n-cores") also have infinite product expansions, although I haven't seen this written anywhere.<|endoftext|> -TITLE: Virtually free, torsion-free, and locally free groups -QUESTION [11 upvotes]: A well-known theorem of Stallings says that -any finitely generated virtually free torsion-free group is free. - -Is this true without `finitely generated' condition? - -In other words, - -is every locally free virtually free group free? - -REPLY [10 votes]: This is Theorem B of Swan's famous paper Groups of cohomological dimension one, in which he removes the 'finitely generated' hypothesis from Stallings' theorem. Theorem B states: - -Let $G$ be a torsion-free group. If $G$ has a free subgroup of finite index, then $G$ is free. - -Here's the paper (apparently not behind a paywall).<|endoftext|> -TITLE: The set of largest numbers definable by formulas in different lengths -QUESTION [6 upvotes]: Let $n=\phi(l)$ to be the largest number definable by a first order arithmetic formula $f(x)$ having length at most $l$. By "$n$ is definable by formula $f(x)$" I mean $\mathcal{N}\vDash f(a)$ iff $a=n$, where $\mathcal{N}$ is a structure of natural numbers with the standard interpretation (if it's necessary to be specific, think of it as $\mathcal{N}=\{\mathbb{N}|+,*,1\}$). This function is well-defined. My question is, how fast does it grow? Can we put it in some appropriate framework and say something more about it? -One observation I can make is that no first order arithmetic formula defines the function $n=\phi(l)$, i.e., there exists no formula $\psi(x,y)$ such that $\mathcal{N}\vDash \psi(n,l)$ iff $n=\phi(l)$. This means the set $\{n,l|n=\phi(l)\}$ is beyond arithmetic hierarchy, where I have only very limited knowledge. - -Edit: Issues mentioned in the comments are addressed. - -REPLY [12 votes]: Let me interpret the question asking about what is true in the -standard model $\langle\newcommand\N{\mathbb{N}}\N,+,\cdot,0,1,<\rangle$, which -avoids the non-absoluteness issues mentioned in the comments. In -this case, the definition is well-defined and sensible. We define that $\Phi(n)$ -is the largest number $m$ definable in the structure by a formula -of length at most $n$. -I view the topic here as an analogue of the busy-beaver problem for arithmetic truth. -Theorem. The function $\Phi(n)$ eventually dominates every -arithmetically definable function. -Proof. Suppose that $f:\N\to\N$ is an arithmetically definable -function, so that the relation $f(x)=y$ is definable in the -structure by some formula $\varphi(x,y)$. -Notice that with the powers of two, we can easily define large -numbers with comparatively small formulas. For example, $2^n$ is -definable by a formula of size $n+c$ for some constant $c$. Put -differently, and by iterating this, for any sufficiently large $k$ -we can define a number $k^+$ larger than $k$ with a formula smaller than -$\log(\log(k))$. -Therefore, if $k$ is very large with respect to these constants and -the size of the definition of $f$, then we can define $\max_{x\leq -k^+}f(x)$ using a formula of size less than $k$. Thus, -$f(k)\leq\Phi(k)$, as desired. QED -Meanwhile, the function is Turing computable from $0^{(\omega)}$, -the $\omega$-jump of the halting problem, since that oracle is able -to compute first-order arithmetic truth. So you have a function -that is not computable from any finite jump $0^{(n)}$, but it is -computable from the $\omega$-jump. -I claim that the converse is also true. Your function is -essentially equivalent to arithmetic truth. -Theorem. The function $\Phi$ you have defined is Turing -equivalent to an oracle for arithmetic truth. -Proof. I've already pointed out that $\Phi$ is computable from -arithmetic truth. -Conversely, suppose that we have an oracle for your function -$\Phi$. I claim that we can compute arithmetic truth. The reason is -that we shall be able to compute Skolem witnesses. For example, to -compute the halting problem for a program $p$ on input $n$, we need -only express the formula "$y$ is the length of the compution of $p$ -on $n$" and then apply the function $\Phi$ to get an upper bound -for the length of the computation. If the program doesn't halt by -then, then it won't ever halt, since the length of the computation -is definable. -We can systematically iterate this idea up the arithmetic -hierarchy. Namely, working from the atomic formulas up, every -assertion $\exists x\ -\varphi(x,\dots)$ is replaced with the assertion $\exists -x<\Phi(k)\ -\varphi(x,\dots)$, where $k$ is the size of the expression -appearing to the right. By using a term for $\Phi(k)$, and iteratively applying this procedure to higher quantifiers, in the case that they were nested, we thereby get a computable expression that is equivalent to the original -formula. In this way, arithmetic truth reduces to a computable -process with the function $\Phi$. QED<|endoftext|> -TITLE: About some property of automorphism of octonions -QUESTION [5 upvotes]: Let $f$ be an automorphism of the algebra of octonions. Is it true that $f$ preserves some quaternionic subalgebra? Has the statement an elementary proof? - -REPLY [2 votes]: I would start by finding imaginary octonion $i$ fixed by $f$. This we can conclude from knowing that $f$ is orthogonal map of $R^7$ perpendicular to $1$. Now we apply argument used in Gro-Tsen answer. Each element in $SO_7$ has fixed vector. -Next I would observe that multiplication by $i$ defines complex structure on perpendicular $R^6$. This complex structure is preserved by $f$, because -$f(\color {red}ix)=f(\color{red}i)f(x)=\color{red}if(x)$. -We have proved that $f$ belongs to $U_3$. Now, there exists basis $\color{blue}u,\color{green}v,\color{brown}w$ in $\mathbb C^3$ such that $f$ is diagonal in it. This means $f(\color{blue}u)=e^{\color{red}i\alpha}\color{blue}u$. -The subalgebra $\langle1,\color{red}i,\color{blue}u,\color{red}i\color{blue}u\rangle$ is quaternion subalgebra and it is fixed by $f$. -For example: -1) $\color{red}i\color{blue}u=-\color{blue}u\color{red}i$ because $\color{red}i,\color{blue}u$ are perpendicular and imaginary; -2) $(\color{red}i\color{blue}u)\color{blue}u=\color{red}i\color{blue}u^2=-\color{red}i$ (Moufang identity). -These are geometrical intuitions which I use in octonions. To have strict proof we need to have definition of octonion multiplication. -Side comment - interestingly for two perpendicular imaginary octonions $x,y$ element $(L_xR_y)^2$ is octonion automorphism. It is identity on quaternion subalgebra generated by $x, y$ and minus identity on perpendicular 4-space. Such elements form submanifold in $G_2$ being 8-dimensional exceptional symmetric space $G$ of all subalgebras $\{\mathbb H \subset \mathbb O\}$. -Second comment is that this formula $(L_xR_y)^2$ gives automorphism for octonions over finite fields. We need carefully define what "imaginary octonion" means in this case.<|endoftext|> -TITLE: Topology of the set of Nash equilibria of a normal form game -QUESTION [6 upvotes]: Consider a normal form game with $n$ players (and finitely many options per player) defined by finite option sets $A_1,\ldots,A_n$ and payoff matrices $u_1,\ldots,u_n: \prod_{j=1}^n A_j \to \mathbb{R}$. Let $N$ be the set of Nash equilibria, which is a subset of the set $S := \prod_{j=1}^n S_j$ of mixed strategy profiles where $S_j$ is the linear simplex with vertex set $A_j$ (=set of mixed strategies on $A_j$). -General question: What interesting things can be said about this set $N$ of Nash equilibria, from a topological point of view? -Obviously it need not be connected (it can be a finite set of $>1$ points), and it's pretty clear that it's compact. But, to ask a specific question, is there an example where it has non-trivial $H_1$ (first homology group): can it be homeomorphic to a circle or an annulus? -[I asked this some time ago on MSE but received no response.] - -REPLY [4 votes]: Nash-equilibria satisfy a universality theorem. This is due to Ruchira Datta's paper "Universality Of Nash Equilibria", https://arxiv.org/pdf/math/0301001.pdf. From her abstract: - -Every real algebraic variety is isomorphic to the set of totally mixed Nash equilibria of some three-person game, and also to the set of totally mixed Nash equilibria of an N-person game in which each player has two pure strategies.<|endoftext|> -TITLE: Are there graphs whose matching polynomials are Legendre? -QUESTION [5 upvotes]: It is well-known (at least well-known enough to be on Wikipedia) that there are quite simple graphs whose matching polynomials -$$M(G;x) = \sum_{m\geq 0} (-1)^m \#\{\text{matchings with $m$ edges}\}\, x^{\#V(G)-2m}$$ -are essentially equivalent to the Chebyshev polynomials of both kinds ($C_n$ and $P_n$ respectively), the Laguerre polynomials with integer parameter ($K_{m,n}$), and the Hermite polynomials ($K_n$). -On the other hand, my literature searches don't seem to get any results for graphs closely related to the Legendre polynomials in this way. I was wondering if there is a known construction, or a proof of impossibility, for a statement like: "There exists a family of graphs $G_n$ for which $M(G_n;x)=aP_n(bx)$, where $a$ and $b$ may depend on $n$ but not on $x$." -[Optimally $b$ would not depend on $n$, but at this point I would be happy either way.] - -REPLY [6 votes]: Any family of orthogonal polynomials can be realized as the characteristic polynomials of a sequence of weighted paths, possibly with loops. If the implicit weight function is symmetric about the origin, loops are not needed. In this case the characteristic polynomial coincides with the matching polynomial. It follows that we can realize the Legendre polynomials as matching polynomials of weighted paths - with non-integer weights. -This is not a satisfying solution to your -roblem, at all, but it seems difficult to get further. The matching polynomial of a graph is monic and in many cases we can make our orthogonal polynomials monic by a rescaling. However the cofficients of the leading terms of the Legendre polynomials have complicated factorizations, and so no simple rescaling will help. -Finally, I am not aware of any literature that relates the Legendre polynomial to the matching polynomial of a graph. (For whatever this assertion is worth.)<|endoftext|> -TITLE: Problem with definability in the constructible hierarchy -QUESTION [6 upvotes]: This is a rather technical question. I cannot find my mistake in a proof of the (obviously wrong) following sentence: Every countable ordinal is $\Sigma_2$-definable in $J_{\omega_1 + 1}$ by a formula with no parameter. -I first recall the definition of $J_\alpha$, closely following the notations and ideas of "Fine Structure and Class Forcing" by Sy Friedman. Suppose that $J_\alpha$ is defined (with $J_0 = \emptyset$). Let $\{\phi_e(x)\}_{e \in J_\alpha}$ be an enumeration of first-order formulas with one free variable and parameters in $J_{\alpha}$. We define $J_{\alpha+1}$ by first defining sets $X(n, e)$ for $n \in \omega$ and $e \in J_\alpha$, where each $X(n+1, e)$ is meant to be the set of level $n+1$ and index $e$, having as elements the sets of level $n$ whose indices satisfies the formula $\Phi_e$ in $J_\alpha$. -$$ -\begin{array}{rcl} -X(0, e)&=&e\\ -X(n+1, e)&=&\{X(n, x)\ :\ J_\alpha \vDash \Phi_e(x)\}\\ -\end{array} -$$ -We then define $J_{\alpha+1} = \bigcup \{X(n, e)\ |\ n \in \omega,\ e \in J_{\alpha}\}$. For $\gamma$ limit we define $J_{\gamma} = \bigcup_{\beta < \gamma} J_\beta$. -In the following, everytime we mention $\Sigma_2$-definable, we mean $\Sigma_2$-definable by a formula with no parameter. Fix some ordinal $\alpha$. Suppose some ordinal of $J_{\alpha+1}$ is not $\Sigma_2$-definable in $J_{\alpha+1}$, and let $a$ be the smallest of them. Then $a$ has the following definition: -$$a \text{ is not } \Sigma_2\text{-definable and } \forall b < a\ b \text{ is } \Sigma_2\text{-definable}$$ -Let us try to formalize this properly in $J_{\alpha+1}$. To do so, we use what Friedman calls a universal $\Sigma_1$ predicate, say $W(n, x, y)$. The predicate $W(n, x, y)$ is a $\Sigma_1$ formula which is true in $J_{\alpha+1}$ iff $\phi_n(x, y)$ if true in $J_{\alpha+1}$, where $\phi_n(x, y)$ is the $n$-th $\Sigma_1$ formula of two free variables. Friedman shows that such predicates exist at every level of the hierarchy. -We can now write "$a \text{ is not } \Sigma_2\text{-definable}$" as a $\Delta_3$ formula: -$$\forall n \in \omega\ \ \ \forall x\ W(n, x, a) \text{ or } \exists b \neq a\ \exists x\ \neg W(n, x, b)$$ -It follows that "$a \text{ is } \Sigma_2\text{-definable}$" is also a $\Delta_3$ formula: -$$\exists n \in \omega\ \ \ \exists x\ \neg W(n, x, a) \text{ and } \forall b\ b \neq a \rightarrow \forall x\ W(n, x, b)$$ -Suppose now that $\alpha$ is $\Sigma_2$-definable in $J_{\alpha+1}$. Then the following argument shows that we can lower the complexity of the above formulas to $\Sigma_2$: Now the universal quantifications can be replaced by existential quantifications in $J_{\alpha+1}$, given that we can use $J_{\alpha}$. Indeed, "$\forall x \dots$" (meaning $\forall x \in J_{\alpha+1} \dots$) becomes equivalent to: -$$\forall e \in J_\alpha\ \forall n \in \omega\ \exists x\ x=X(n, e)\text{ and } \dots$$ -It seems that both $J_\alpha$ and $X(n,e)$ have a $\Sigma_1$ definition (using $\alpha$ as a parameter) in $J_{\alpha+1}$. So it follows that if $\alpha$ is $\Sigma_2$ definable in $J_{\alpha+1}$, then the smallest ordinal not $\Sigma_2$ definable in $J_{\alpha+1}$ is $\Sigma_2$ definable in $J_{\alpha+1}$. Then there cannot be such ordinals. -Now it seems that $\omega_1$ is $\Sigma_2$ definable in $J_{\omega_1+1}$ (which contains all the ordinals strictly smaller than $\omega_1 + \omega$), with the following formula of $\alpha$: -$$\forall \beta\ \exists n<\omega\ \beta < \alpha + n\ \text{ and } \forall \alpha' < \alpha\ \exists \beta\ \forall n<\omega\ \beta > \alpha' + n$$ -It follows that every countable ordinal is $\Sigma_2$ definable in $J_{\omega_1}$, by formula with no parameter. This cannot be correct by a cardinality argument. What is wrong ? - -REPLY [3 votes]: I may have found where is the trick hidden: I mean two different things by "being $\Sigma_2$-definable". When I write "the smallest ordinal which is not $\Sigma_2$-definable" I mean being definable by a formula of the form -$\exists \forall \psi$ where $\psi$ only have bounded quantifiers. -But when I define this smallest ordinal with a $\Sigma_2$ definition, I define it with a formula where the bounded quantifiers are spread everywhere in the formula, including (for example) in between two existential quantifiers. -Of course in a model of sufficiently many ZF axioms, these two notions of being $\Sigma_2$-definable are equivalent. You probably need your model to be $\Sigma_1$ or maybe $\Sigma_2$ admissible. But $J_{\omega_1+1}$ is certainly not $\Sigma_1$-admissible... -I think this is what is wrong with this proof.<|endoftext|> -TITLE: Is "conditioning to a sub-$\sigma$-algebra" a measurable operation? -QUESTION [15 upvotes]: Let $\mathcal{B}$ be the Borel $\sigma$-algebra of $[0,1]$, and let $\mathcal{M}$ be the set of probability measures on $([0,1],\mathcal{B})$, equipped with the evaluation $\sigma$-algebra $\ \sigma(\rho \mapsto \rho(A):A \in \mathcal{B})$. -Let $\mathcal{M}_2$ be the set of probability measures on $([0,1] \times [0,1], \mathcal{B} \otimes \mathcal{B})$, again equipped with the associated evaluation $\sigma$-algebra $\ \sigma(\mu \mapsto \mu(A):A \in \mathcal{B} \otimes \mathcal{B})$. -Given a sub-$\sigma$-algebra $\mathcal{G}$ of $\mathcal{B}$, for any $\mu \in \mathcal{M}_2$, define $\mathbb{E}_\mathcal{G}(\mu) \in \mathcal{M}_2$ to be the unique measure such that - -$\mathbb{E}_\mathcal{G}(\mu)$ agrees with $\mu$ on $\mathcal{G} \otimes \mathcal{B}$; -there exists a $\mathcal{G}$-measurable function $\nu \colon [0,1] \to \mathcal{M}$ such that for all $A_1,A_2 \in \mathcal{B}$, -$$ \mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2) \ = \ \int_{A_1 \times [0,1]} \nu(x)(A_2) \, \mu(d(x,y)). $$ - - - -Is the map $\mathbb{E}_\mathcal{G} \colon \mathcal{M}_2 \to \mathcal{M}_2$ a measurable function? Or at the least, is $\mathbb{E}_\mathcal{G}^{-1}(\mathcal{A})$ a universally measurable subset of $\mathcal{M}_2$ for every measurable set $\mathcal{A} \subset \mathcal{M}_2$? - - -Remark: The existence and uniqueness of the measure $\mathbb{E}_\mathcal{G}(\mu)$ follows from the disintegration theorem applied to the measure $\mu|_{\mathcal{G} \otimes \mathcal{B}}$. -UPDATE: A simple "counting" argument as in the comments below yields that there must exist $\mathcal{G}$ such that $\mathbb{E}_\mathcal{G}$ is not measurable with respect to the evaluation $\sigma$-algebra. Nonetheless, the key question remains as to whether $\mathbb{E}_\mathcal{G}$ is necessarily universally measurable (in the sense that the pre-image of every member of the evaluation $\sigma$-algebra belongs to the universal completion of the evaluation $\sigma$-algebra). -[For an explicit example showing that $\mathbb{E}_\mathcal{G}$ is not necessarily continuous: Let $\mathcal{G}$ be the countable-cocountable algebra. Let $\rho_n$ be a sequence of atomless probability measures on $[0,1]$ converging weakly to a non-trivial purely atomic measure $\rho$. Writing $D \colon x \mapsto (x,x)$, we have that $\mathbb{E}_\mathcal{G}(D_\ast\rho_n)=\rho_n \otimes \rho_n$ but $\mathbb{E}_\mathcal{G}(D_\ast\rho)=D_\ast\rho$.] - -A possible approach: Fix $A_1$ and $A_2$. Given a finite or countable partition $\mathcal{H}$ of $[0,1]$ by members of $\mathcal{G}$, we can "approximate" $\mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2)$ by -$$ \mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2) \ \approx \ \mathbb{E}_{\sigma(\mathcal{H})}(\mu)(A_1 \times A_2) \ = \ \sum_{G \in \mathcal{H}} \mu(G \times A_2)\mu((G \cap A_1) \times [0,1]). $$ -Given an increasing sequence of such partitions $\mathcal{H}_n$ such that $\mathcal{G} \subset \sigma(\mathcal{N}_{\pi^1_\ast\mu} \cup \bigcup_{n=1}^\infty \mathcal{H}_n)$, Levy's upward theorem yields that $\mathbb{E}_{\sigma(\mathcal{H}_n)}(\mu)(A_1 \times A_2) \to \mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2)$. -[Here, $\mathcal{N}_{\pi^1_\ast\mu}$ is the set of $\pi^1_\ast\mu$-null sets, where $\pi^1_\ast\mu(A):=\mu(A \times [0,1])$.] -Now such a sequence $\mathcal{H}_n$ always exists---based on the proof of the equivalence between "separable" and "countably generated mod 0" given here, I think the following construction works: Letting $\mathcal{U}$ be a countable algebra generating $\mathcal{B}$, let $\{G_i\}_{i=1}^\infty \subset \mathcal{G}$ be a countable set such that for every $U \in \mathcal{U}$ and $r \in \mathbb{N}$, if $\mathcal{G}_{U,r}:=\{G \in \mathcal{G} : \mu(G \triangle U) < \frac{1}{r} \}$ is non-empty then $\exists \, i$ s.t. $G_i \in \mathcal{G}_{U,r}$. Then take $\mathcal{H}_n$ to be the partition generated by $\{G_1,\ldots,G_n\}$. -So then, to prove the desired result, it is sufficient to show that there is a sequence of "partition-valued" functions -$$ \mathcal{H}_n \colon \mathcal{M}_2 \to \ \{\textrm{finite or countable partitions contained in } \mathcal{G}\} $$ -such that $\mu \mapsto \sum_{G \in \mathcal{H}_n(\mu)} \mu(G \times A_2)\mu((G \cap A_1) \times [0,1])$ is universally measurable for each $n$, and $\mathcal{G} \subset \sigma(\mathcal{N}_{\pi^1_\ast\mu} \cup \bigcup_{n=1}^\infty \mathcal{H}_n(\mu))$ for every $\mu$. -[Remark: One might hope that it could even be possible to take $\mathcal{H}_n$ to be independent of $\mu$. However, this is not possible in general. This follows from the answer to https://math.stackexchange.com/questions/2156998/is-every-sub-sigma-algebra-of-mathcalb-mathbbr-universally-countabl; but also, if it were possible to take $\mathcal{H}_n$ to be independent of $\mu$, then $\mathbb{E}_\mathcal{G}$ would be measurable with respect to the evaluation $\sigma$-algebra, which we have established not always to be the case.] - -Another characterisation of $\mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2)$: -In the above discussion "A possible approach", I gave a sort-of-explicit construction of $\mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2)$, but the difficulty is that it relies on "making a choice" - which could then lead to measurability issues when we allow $\mu$ to vary. -Now one of the comments suggested looking directly at the proof of the disintegration theorem and hoping that the desired measurability might become more clear from there. Disintegration relies fundamentally on the Radon-Nikodym theorem; on the basis of the proof of the Radon-Nikodym theorem, $\mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2)$ can be characterised as follows: -Let $\mathcal{C}$ be the set of $(\mathcal{G},\mathcal{B})$-measurable functions $g \colon [0,1] \to [0,1]$ such that for every $G \in \mathcal{G}$, -$$ \int_{G \times [0,1]} g(x) \, \mu(d(x,y)) \ \leq \ \mu(G \times A_2) \, ; $$ -then -$$ \mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2) \ = \ \sup_{g \in \mathcal{C}} \int_{A_1 \times [0,1]} g(x) \, \mu(d(x,y)). $$ -Again, it is difficult to see any measurability from this. -Increasingly I am starting to suspect that (assuming the axiom of choice) there exists a sub-$\sigma$-algebra $\mathcal{G}$ of $\mathcal{B}$ such that $\mathbb{E}_\mathcal{G}$ is not universally measurable! - -REPLY [4 votes]: Okay, I think I've worked out that the answer is no, i.e. there exists a sub-$\sigma$-algebra $\mathcal{G}$ of $\mathcal{B}$ such that $\mathbb{E}_\mathcal{G}$ is not universally measurable. -(We will write $\lambda$ for the Lebesgue measure on $([0,1],\mathcal{B})$.) -It is well-known that the evaluation $\sigma$-algebra is precisely the Borel $\sigma$-algebra of the topology of weak convergence, which is a Polish topology. So then, to show that (for some given $\mathcal{G}$) $\mathbb{E}_\mathcal{G}$ is not universally measurable, it is sufficient (by Lusin's theorem) to find a probability measure $Q$ on $\mathcal{M}_2$ such that for every measurable set $E \subset \mathcal{M}_2$ with $Q(E)>0$, the restriction of $\mathbb{E}_\mathcal{G}$ to $E$ is not a continuous function (with respect to the topology of weak convergence). -Let $A \subset [0,1]$ be a set such that $A$ and $[0,1] \setminus A$ intersect every Lebesgue-positive measure set. (Assuming the axiom of choice, such a set $A$ exists, as shown here.) Let $\mathcal{G}$ be the $\sigma$-algebra consisting of all countable subsets of $A$ and their complements. -Let $Q$ be the image measure of $\lambda \otimes \lambda$ under the map $D : (x,y) \mapsto \frac{1}{2}(\delta_{x,x}+\delta_{y,y})$. Let $E \subset \mathcal{M}_2$ be any measurable set with $Q(E)>0$, and let $F=D^{-1}(E)$. (So $\lambda \otimes \lambda(F)>0$.) Let $F'=\{x \in [0,1] : \lambda(y:(x,y)\in F)>0\}$. Obviously $\lambda(F')>0$, so fix a point $x \in F' \setminus A$. For any $y \in [0,1]$, we have that -$$ \mathbb{E}_\mathcal{G}(D(x,y)) \ = \ \left\{ \begin{array}{c l} \frac{1}{2}(\delta_{x,x}+\delta_{y,y}) & y \in A \\ \frac{1}{4}(\delta_{x,x} + \delta_{x,y} + \delta_{y,x} + \delta_{y,y}) & y \not\in A. \end{array} \right. $$ -So since $Y:=\{y \in [0,1] : (x,y) \in F\}$ has positive Lebesgue measure, it is clear that $\mathbb{E}_\mathcal{G}(D(x,\cdot))$ is not continuous on $Y$, so $\mathbb{E}_\mathcal{G} \circ D$ is not continuous on $F$, so (since $D$ is obviously continuous) $\mathbb{E}_\mathcal{G}$ is not continuous on $E$.<|endoftext|> -TITLE: Is this sequence of Lie algebra cohomology a part of spectral sequence? -QUESTION [9 upvotes]: There is an exact sequence -$$0 \to H^2(\mathfrak{g}, k) \to H^1(\mathfrak{g}, \mathfrak{g}^*) \to H^0(\mathfrak{g}, S^2\mathfrak{g}) \xrightarrow{d} H^3(\mathfrak{g}, k) \to H^2(\mathfrak{g}, \mathfrak{g}^*) \to H^1(\mathfrak{g}, S^2\mathfrak{g}),$$ -where $\mathfrak{g}$ is a Lie algebra over a field $k$ and $H^i(\mathfrak{g}, M)$ are Lie algebra cohomology. It is interesting because of a map $d$, a Koszul homomorphism, which sends an invariant symmetric bilinear form $\left< \cdot, \cdot \right>$ to a 3-cocycle $\left<[\cdot, \cdot], \cdot\right>$ and is an isomorphism for a semi-simple Lie algebra. -The question is whether it is a part (namely, lower-degree terms) of some spectral sequence? If not, what is a natural way to obtain it? An unclear proof may be found in Neeb, Wagemann The second cohomology of current algebras of general Lie algebras, proposition 7.2. - -REPLY [5 votes]: It's part of the Pirashvili exact sequence, relating Lie algebra and Leibniz cohomologies of Lie algebras. This is discussed (in homology terms) in the end of p2 of this paper of mine on Koszul's homomorphism (arxiv link). Pirashvili's paper is freely accessible here on Numdam; it's also written in terms of homology but the cohomological statement follows.<|endoftext|> -TITLE: Why are $\Gamma_0$ functions called this -QUESTION [6 upvotes]: It is very common to indicate with $\Gamma_0(A)$ the set of lower semicontinuous convex functions from $A$ to $(-\infty,+\infty]$ with nonempty domain. An example of usage of this notation can be found in "Proximal Splitting Methods in Signal Processing" by Combettes and Pesquet. -What is the role of the subscript $0$ in the short notation? - -REPLY [11 votes]: I think Carlo Beenakker digged up the right reference for the notation of the set, but I think more can be said. -First, there is some meaning for the subscript $0$ which can be found in the same paper of Moreau a little later: - -So $\Gamma(H)$ are all functions that are pointwise suprema of continuous affine functions. This is a neat space since it's elements are build from very simple objects, but is also quite general as it includes all convex and lower semi-continuous functions. However, it includes two not-so-nice functions, namely $f(x)\equiv \infty$ and $f(x)\equiv -\infty$. Both are not really needed and complicate the theory a bit, hence $\Gamma_0(H)$ is introduced which is just $\Gamma(H)$ without these two functions (and adding a subscript $0$ to specialize a set seems pretty common). -Also, one can say a bit more about the letter $\Gamma$. I don't think its origin has something to do with projections. I checked Fenchel's lecture notes on convexity from 1951. There he did not use convex functions that were allowed to assume the value $+\infty$, but always includes the domain of definition into the description of the convex functions, i.e. he wrote $[C,f]$ to denote a convex function $f:C\to{}]{-\infty},+\infty[$. -In this lecture notes he describes dualities between convex sets (Chapter II, Sec. 8) and also between functions (Chapter III, Sec. 5). For the duality of sets he used $C$ for sets of points and $\Gamma$ for sets of hyperplanes and then describes the duality of the two descriptions of convexity by either 1) the points in the set or 2) by the hyperplanes that separate the set from outer points. In III.8 where he describes Fenchel duality he used $[C,f]$ for "convex $f$ defined on $C$" and denotes the dual function as $[\Gamma,\varphi] = [C,f]^*$. Here $\Gamma$ somehow describes (a part of) the affine functions minorizing $f$ which may be the motivation for Moreau to use $\Gamma$ for the set.<|endoftext|> -TITLE: Find the inverse of a matrix that is very similar to the Hilbert matrix -QUESTION [10 upvotes]: The standard Hilbert matrix $H$ is given by -$$H_{ij}=\frac{1}{i+j-1},$$ -and it has an inverse given for example in this MO question. -Now I have encountered a matrix $M$ of similar form, namely, -$$M_{ij}=\frac{1+(-1)^{i+j}}{i+j-1}.$$ -Does anyone know the explicit formula for the inverse of M? Please provide some relevant reference if available. -A more general form for this question can be found here. - -REPLY [3 votes]: These matrices have been considered here.<|endoftext|> -TITLE: Total chromatic number and bipartite graphs -QUESTION [7 upvotes]: I'm looking for a proof to the following statement: - -Let G be a simple connected graph -If $\chi''(G)=\chi'(G)+\chi(G)$ holds then the graph should be bipartite, - -where $\chi''(G)$ is the total chromatic number $\chi'(G)$ the chromatic index and $\chi(G)$ the chromatic number of a graph. -I was thinking that it should be easy so i first asked it at mathstackexchange -https://math.stackexchange.com/questions/2141570/graph-with-total-chromatic-number-chig-chig-chig - -REPLY [2 votes]: Let $a=\chi(G)\geq 3$ and $b=\chi'(G)$. Paint the vertices in $a$ colors and edges in $b$ colors properly. Now choose one color class of edges. Repaint each of them into one of the first $a$ colors, distinct from the two colors of the edge's endpoints. Since the repainted edges were pairwise non-adjacent,we get a proper total coloring in $a+b-1$ colors.<|endoftext|> -TITLE: Mazur's article, Notes on etale cohomology of number fields -QUESTION [6 upvotes]: Let $\mathcal{A}$ and $\mathcal{B}$ be two abelian categories. Let $\tau : \mathcal{A} \rightarrow \mathcal{B}$ be a left-exact functor. Let $\mathcal{C}$ be the mapping cylinder category of $\tau$. Then, there are various functors $j_*, i_*$ ... defined on page 524 of Mazur's expository article in 1973, in Ann. Scient. ENS. He insisted that -the derived functor of $j_*$ are given by -$$ R^q j_* M=(R^q \tau M, 0, 0) $$ -(at l.8, page 525). Can someone explain this? - -REPLY [8 votes]: Form an injective resolution -$$ -0 \to M \to I^0 \to I^1 \to I^2 \to \dots -$$ -and apply $j_*$. Then $j_* I^k = (\tau I^k, I^k, id_{\tau I^k})$ in the mapping cylinder category for all $k$, and so to calculate the derived functors you calculate the cohomology of the sequence -$$ -0 \to (\tau I^0, I^0, id) \to (\tau I^1, I^1, id) \to \dots -$$ -Taking kernels or cokernels of a map $(N, M, \varphi) \to (N', M', \varphi')$ in the mapping cylinder category is done by applying it to both $N \to N'$ and $M \to M'$ individually, so to calculate the cohomology of the sequence at hand we take the cohomology of -$$ -0 \to \tau I^0 \to \tau I^1 \to \tau I^2 \to \dots -$$ -(which are the derived functors $R^q \tau M$, by definition) and the cohomology of -$$ -0 \to I^0 \to I^1 \to I^2 \to \dots -$$ -(which is $M$ in degree 0 and 0 in other degrees because this was a resolution of $M$). Therefore, the cohomology of the sequence in question is -$$ -(\tau M, M, id), (R^1 \tau M, 0, 0), (R^2 \tau M, 0, 0), \dots -$$ -which recovers Mazur's statement.<|endoftext|> -TITLE: "Local" Gauss-Lucas theorem? -QUESTION [18 upvotes]: The Gauss-Lucas theorem relates the location of zeros of a polynomial to the location of zeros of its derivative: - -Suppose $f(z)\in \mathbb{C}[z]$ is a non-constant polynomial with roots $\alpha_1,\ldots,\alpha_n$, and let - $$K = K(\alpha_1,\ldots,\alpha_n) \subset \mathbb{C}$$ - denote the convex hull of these points. - Then all roots of $f'(z)$ lie inside $K$. - -If we instead consider a polynomial with real coefficients, Rolle's theorem gives a stronger condition on the location of zeros of the derivative: - -Suppose $f(x)\in\mathbb{R}[x]$ is a non-constant polynomial with real roots $\alpha_1\leq \cdots \leq \alpha_n$, counted with multiplicity. - Then for any $i < j$, the closed interval - $$ I = [\alpha_i, \alpha_j] \subset \mathbb{R} $$ - contains some root of $f'(x)$. - -The second statement in "local" in the sense that knowing only two roots of $f$ gives us some information about where the roots of $f'$ lie. -In the first statement, knowing a subset of the roots will not in general determine the convex hull. - -Question: For $f(z)\in\mathbb{C}[z]$, is there any information we get about the location of roots of $f'(z)$ if we know only the locations of (say) 3 non-collinear* roots of $f(z)$? - -Some guesses which are false: -Given three roots $\alpha_1, \alpha_2, \alpha_3$ of $f(z)$, - -$f'(z)$ must contain a root inside the triangle spanned by $\alpha_1, \alpha_2, \alpha_3$. -$f'(z)$ must contain a root inside the circle passing through $\alpha_1, \alpha_2, \alpha_3$. - -*as suggested by Gerry Myerson, the case when $\alpha_i$ are collinear may require separate analysis. But even this case is unresolved up to my understanding. - -REPLY [5 votes]: Following is information from Marden's Geometry of the Zeros, Sections 25 and 26. Note: Some further information can be found on my paper "Approximate Gauss--Lucas Theorems" (also linked in the comments). - -Theorem: (Kakeya) For $1\leq p\leq n$, there is a function $\varphi(n,p)$ such that if a degree $n$ polynomial has $p$ zeros lying in a disk with radius $R$, then that polynomial has at least $p-1$ critical points lying in the concentric disk with radius $R\cdot\varphi(n,p)$. - -Thus for your example, let $R$ denote one half the diameter of the set of three zeros, and let $C$ denote the disk containing the three zeros, with radius $R$. Then the concentric disk with radius $R\cdot\varphi(n,p)$ contains at least two critical points of the polynomial. Note that in general, for the known bounds, the degree $n$ of the polynomial must be known. -Various estimates on this function $\varphi$ have been given: -Kakeya: $\varphi(n,2)=\csc(\pi/n)$. -Biernacki: $\varphi(n,n-1)\leq(1+1/n)^{1/2}$. -Marden: $\varphi(n,p)\leq\csc(\pi/(n-p+1))$ and $\varphi(n,p)\leq\displaystyle\prod_{k=1}^{n-p}[(n+k)/(n-k)]$. -Richards: $\varphi(n,p)\leq1+\dfrac{8(n-p)^2}{p-8(n-p)^2}$ (subject to the assumption that $p>8(n-p)^2$). (This result is in the linked paper, definitely not in Marden's awesome book!) -Complete citations may be found in my linked paper. Sorry I did not think of this earlier as the answer, I was working on trying to find the smallest radius which would catch at least one critical point, not all $n-p$. And in any case I have no excuse for my (now deleted) earlier answer!<|endoftext|> -TITLE: Irrationality of generalized continued fractions -QUESTION [5 upvotes]: An infinite simple continued fraction -$$\frac{1}{b_1 + \frac{1}{b_2 + \frac{1}{b_3+\dots}}} (b_i\in\mathbb Z)$$ -is irrational. Now for a generalized continued fraction: -$$\frac{a_1}{b_1 + \frac{a_2}{b_2 + \frac{a_3}{b_3+\dots}}} (a_i,b_i\in\mathbb Z),$$ -the same conclusion is apparently not valid. Legendre gave a sufficient condition for irrationality: -$$|a_i|<|b_i|$$ -for any $i$ large enough. -Can this result be strengthened in any way? Especially, might it hold for $a_i, b_i\in\mathbb Q$? -Also, can anyone give an example of an infinite generalized continued fraction that converges to a rational, showing that some condition on the $a_i, b_i$ is needed for irrationality? - -REPLY [4 votes]: There is a family of generalized continued fractions which give irrational or rational values depending on a very simple but nontrivial parametrization. I'd like to use the notation -$$c = a_0 + {b_0\over a_1 + {b_1\over a_2 + ... }} = a + {b\over (a+1) + {(b+1)\over (a+2) + {(b+2)\over (a+3) + ... }... }}$$ -in accordance with the Angell-article (see below). -The following family of generalized continued fraction gives rational or irrational numbers depending on parametrization: -$$\begin{array} {r|r} - \begin{matrix}&a_k=\\b_k=&\end{matrix} - &&0+k & 1+k & 2+k & 3+k & \cdots \\ - \hline - 0+k && 0 & 1 & 2 &3 & \cdots \\ \hline - 1+k && {1\over e-1} & {1\over e-2} & {1\over 2e-5} & {1\over 6e-16} & \cdots \\ \hline - 2+k && {1\over 1} & -{1e-1\over 0e-1} & -{1e-2\over 1e-3}& -{2e-5\over 4e-11} & \cdots \\ \hline - 3+k && {4\over 3} & {2 \over 1} & -{0e-2\over 1e-2}& -{2e-6\over 3e-8} & \cdots \\ \hline - 4+k && {21\over 13} & {9 \over 4} & {3 \over 1} & -{3e-6\over 2e-6} & \cdots \\ \hline - \cdots & & \cdots - \end{array}$$ -I've found that heuristically, using wolframalpha for support in the evaluation. -David Angell describes that family and gives that heuristic the analytic background (at least for the rational results if I got this correctly) see: David Angell - A family of continued fractions (2010) Journal of Number Theory 130 , pg. 904-911 (Elsevier), online: "paywall" -A somewhat larger table is at my mathpages - GenContFrac<|endoftext|> -TITLE: How to make a sandwich from just one piece of bread? -QUESTION [38 upvotes]: I don't know how to go about such questions. It's not exactly my area, so maybe it is stupid, but curiosity is winning. -So I have a piece of bread $P$ of a really non-regular shape (let's make it convex though), and I want to make a sandwich from it, i.e. two pieces of bread with some stuff between them. Obviously I have to cut my piece into two pieces. My knife is straight and I can only make one cut, otherwise the cooking will be too messy. In other words the two pieces that I obtain are the intersections $P^l_1$ and $P^l_2$ of $P$ with half-planes with respect to a certain line $l$. -Then, in order to make a sandwich I have to place $P_1$ above $P_2$ and put stuff in between. The most effective use of bread happens when these pieces are approximately aligned, i.e. only most $P_2$ is covered with $P_1$ and another way around. How to make it that way? -Let's state the questions in the mathematical language. -Let $P$ be a convex body in $\mathbb{R}^{n}$, let $l$ be a hyperplane and let $P^l_1$ and $P^l_2$ be the pieces in which $l$ cuts $P$. Let $\lambda_n$ be the usual volume in $\mathbb{R}^{n}$. -The quantity we are interested in is $$\sup\{ \frac{\lambda_n(P^l_1\bigcap f(P^l_2))}{\lambda_n(P)},f\mbox{ - isometry of }\mathbb{R}^{n},~l\mbox{ - hyperplane}\}.$$ -Q1: Is it bounded from below by some absolute constant? EDIT: As suggested by UriBader John's inscribed ellipsoid provides with an estimate from below with an absolute constant that depends on $n$. Thus, I am leaving the stronger version of the question, i.e. if there is an absolute constant that does not depend on the dimension. -Q2: Is there an algorithm to construct an optimal, or almost optimal $l$ and $f$? - -REPLY [10 votes]: Restricting to the case where $P$ is triangular, if the angles of the triangle are $A$, $B$, $C$ and you fold the triangle along the bisector to the angle $C$, then assuming wlog that $A -TITLE: Manifolds covered by a single disc -QUESTION [6 upvotes]: I feel like this is should be well-known, but I cannot seem to find a straight answer anywhere. Under what conditions can a (closed) manifold be obtained by identifying points on the boundary of the unit disc? Said another way, for which M is is the case that $M$ is diffeomorphic (NOT just homotopy equivalent) to $D^n/x\sim f(x)~$ for some $f: S^{n-1}\to S^{n-1}$? (I don't require f to be smooth, but it shouldn't be too pathological either. Let's say it should be piecewise smooth). -Is there a smooth or topological invariant that would obstruct this? -Any references would also be appreciated. - -REPLY [7 votes]: The answer (to the first question) is yes in the smooth case: If $M^m$ is closed and compact, then there is a Morse-Smale function on $M$ with a single critical point of index $m$. Work by Lizhen Qin shows how to give a CW structure on $M$ such that the interiors of the cells correspond to the unstable manifolds of the gradient flow. The conclusion follows from that. The papers where this is done are: -L. Qin, On moduli spaces and CW structures arising from Morse theory on Hilbert manifolds, J. Topol. Anal., 2 (2010), no. 4, 469–526. (preprint here) -As well as: -L. Qin, On the associativity of gluing, . Topol. Anal., 0, 1 (2017). (preprint here) -L. Qin, An application of topological equivalence to Morse theory<|endoftext|> -TITLE: Extension of $\mathbb Q$ which splits only at primes in $S$ -QUESTION [6 upvotes]: Suppose that I'm given a set of rational primes $S$ with positive Dirichlet density, and a finite set of primes $R$, disjoint from $S$. -Does there exist a number field $K$ that is; - -unramified outside $R$; -splits only at primes in some subset $S'\subseteq S$? - -In my situation, I have a semisimple representation $\rho$ of $G_\mathbb Q =\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)$, unramified outside $R$, and I have information about the action of Frobenius elements $\mathrm{Frob}_p$ for $p\in S$ (let's say I know their characteristic polynomials). I'd like to know if this completely determines my representation on some open subgroup of $G_\mathbb Q$. - -REPLY [12 votes]: For many choices of $R$ and $S$ the answer is obviously no. For example, if $R$ is empty, then the answer is no, because there are no unramified extensions of $\mathbb{Q}$. -For a more interesting example, let $S$ be the set of primes $p \equiv 3 \pmod{4}$ and $R$ be any finite subset of $\{ 2, 5, 13, 17, 29, \ldots \}$. Then there is no number field that splits only at primes $S' \subseteq S$. This is because if $K/\mathbb{Q}$ is such a number field, any prime that splits in $K[i]$ must be $\equiv 1 \pmod{4}$, and $S$ doesn't contain any such primes.<|endoftext|> -TITLE: "Elliptic" proof that Compact Ricci Solitons are Gradient Ricci Solitons -QUESTION [6 upvotes]: I'm concerned with the following -Proposition: If a compact manifold $M$ satisfies $$Rc + \textstyle\frac{1}{2}\mathcal{L}_Xg = \lambda g $$ -where $\lambda$ is a constant (i.e. $M$ is a compact Ricci soliton), then in fact $$Rc + \nabla^2f = \lambda g $$ -so $X = \nabla f + K$ where $K$ is a Killing vector. -(Quick note: I use $\nabla f$ to indicate grad $f$. I know, I know; technically $\nabla f=df$, but $\nabla f$ is too good as notation to waste on something that already has a name, and grad $f$ is unforgivably grotesque.) -One can see this fact quoted in, for example, the second paragraph of this paper: http://www.mathem.pub.ro/bjga/v19n1/B19-1-cd-924.pdf -The reference made there is to Perelman's first paper on the Ricci flow: https://arxiv.org/pdf/math/0211159.pdf -Perelman never comes out and states this theorem explicitly as far as I know, but the outline of his reasoning for it seems to occur in Remark 3.2 on page 9. Apparently it hinges on the monotonicity of his entropy functional, but I'm not much of an expert on this stuff. -I had the following idea for an alternate approach to this proposition: -First, solve the elliptic equation $$div(X)=\Delta f$$ for the function f. (In coordinates this is $\nabla_i X^i=g^{ij}\nabla_i\nabla_j f$, just so my conventions are clear.) Now write $$X=\nabla f+(X-\nabla f)=\nabla f+K$$ -It is clear that $div(K)=0$. Given that we know that the above proposition is true, it must be the case that $K$ is a Killing vector. (This is because a Killing vector is divergence-free, and the decomposition of a vector field $X$ into a gradient and a divergence-free vector field is unique, which is easy to see.) -Unfortunately, I haven't been able to prove that this divergence-free vector field $K$ is a Killing field. One attempt I made begins by integrating: -$$\int |\textstyle\frac{1}{2}\mathcal{L}_K g|^2 dV = \int dV$$ -$$=\intdV$$ -$$= \int dV$$ -$$= -\textstyle\frac{1}{2}\int \mathcal{L}_K g(K, \nabla f) dV $$ -(I've omitted details, but I used a bunch of integration by parts, $div(K)=0$, the second Bianchi identity, and the main equation. The details are for the reader if they are interested.). -So, for example, to prove $\mathcal{L}_K g=0$ it is sufficient to prove $\mathcal{L}_K g(K, \nabla f)=0$, though I can't see any reason why this might be true either. -My question is: Does anyone have any suggestion as to how we might prove $K$ is Killing? I would prefer that the proof be "elliptic" in the sense of relying on the equations as they are and not introducing the parabolic methods of the Ricci flow as Perelman did. Any thoughts at all would be welcome. - -REPLY [3 votes]: There is an "elliptic" proof in this paper by Eminenti, La Nave, Mantegazza, see Theorem 3.1. It does still use Perelman's $\mathcal{W}$ entropy (but it does not use the Ricci flow, it just uses a minimization argument). -The authors there ask in Problem 3.2 whether there is a proof that does not use any "arguments related to Ricci flow".<|endoftext|> -TITLE: Primality test for $2p+1$ -QUESTION [5 upvotes]: In 1750 Euler stated following theorem : - -Let $p \equiv 3 \pmod 4$ be prime then $2p+1$ is prime iff $2p+1 \mid 2^p-1$ . - -In 1775 Lagrange gave a proof of the theorem . -Recently I have formulated following claim : - -Let $p \equiv 5 \pmod 6$ be prime then $2p+1$ is prime iff $2p+1 \mid 3^p-1$ . - -Is this statement known ? -P.S. -You can find my proof attempt here . - -REPLY [8 votes]: The if-part is a special case of Pocklingtons theorem (see https://en.wikipedia.org/wiki/Pocklington_primality_test ). The only if part requires some computation, as you need a quadratic non-reside mod 2p+1. As far as I know in general there is no reasonable algorithm for finding a non-residue is known, which is why you need some congruential restriction. However, in practice you would pick integers at random and check them, which takes on average two tests. Moreover, if you just check integers $1,2,3\ldots$, then after $\mathcal{O}(\log^2)$ tests you either obtain a non-residue, or a disproof of GRH.<|endoftext|> -TITLE: $\zeta$-function regularized determinants -QUESTION [9 upvotes]: In (mathematical) physics in order to compute path integrals one often makes an infinite dimensional change of variables and uses infinite Jacobian as a purely formal expression. This step is done in formal analogy with the finite dimensional case (e.g. in the Faddeev-Popov construction). -Next one often needs to compute this Jacobian. In some situations it is a "determinant" of a self-adjoint elliptic differential operator on a compact manifold. A conventional procedure to compute it is to use the $\zeta$-function regularization. This procedure seems to me to be quite arbitrary. While I am not an expert, I have never seen in literature any motivation of it except that it gives the right answer for finite dimensional transformations. To compare with, in classical analysis there are many different methods to sum divergent series, but they may lead to different answers. -QUESTIONS. 1) Are there other conventional methods which are used in explicit computations of infinite Jacobians? -2) Are there any nice pleausible general properties of the $\zeta$-function regularized determinants which distingush them from other methods? -ADDED: This method leads to the identity mentioned in the answer below by Zurab Silagadze -$$1+2+3+4+\dots=-\frac{1}{12}$$ -which seems to me completely counter-intuitive. - -REPLY [2 votes]: Of course that $1+2+3+4+\ldots =-1/12$ is a "correct" result is somewhat mysterious, as the following excerpt from Ramanujan's second letter to Hardy lively demonstrates: - -"I was expecting a reply from you similar to the one which a - Mathematics Professor at London wrote asking me to study carefully - Bromwich's Infinite Series and not fall into the pitfall of divergent - series. I have found a friend in you who views my labors - sympathetically. This is already some encouragement to me to proceed - with an onward course. I find in many a place in your letter rigourous - proofs are required and so on and you ask me to communicate the method - of proof. If I had given you my methods of proof I am sure you will - follow the London Professor. But as a fact I did not give him any - proof but made some assertions as the following under my new theory. I - told him that the sum of an infinite number of terms in the series - $1+2+3+4+\ldots =-1/12$ under my theory. If I tell you this you will - at once point out to me the lunatic asylum as my goal." - -In fact, as John Baez mentions in http://math.ucr.edu/home/baez/week126.html, Hardy later in his book Divergent Series gave almost a "physicist's proof" of this result. -A good account why physicists consider the zeta-function regularization meaningful and unique is given in https://arxiv.org/abs/hep-th/9308028 (Zeta-Function Regularization is Uniquely Defined and Well, by E. Elizalde). See also https://arxiv.org/abs/0909.0795 (On Twisted Virasoro Operators and Number Theory, by A. Huang).<|endoftext|> -TITLE: Surface dissection of regular tetrahedron to cube -QUESTION [5 upvotes]: Does anyone know what is the fewest-piece - dissection - of the surface of - a regular tetrahedron to the surface of a cube (of the same area)? - -It is well-known that the volume of a regular tetrahedron cannot -be dissected to the volume of a cube, because their Dehn invariants differ. -But their surfaces have a dissection, by applying the -Boyai-Gerwien theorem. -Applying that theorem in one way (among several options) leads to an -$31$-piece dissection (if I calculated correctly), illustrated below. But this is surely very far from optimal (and hardly aesthetically pleasing). -Likely the question has been explored, but I have not found any literature. -It could make a pleasing contrast to the impossibility of a volume dissection. - -Let the cube edge length be $1$, so its surface area is $6$. -A tetrahedron edge length of -$L= 2^{\frac{1}{2}} 3^{\frac{1}{4}} \approx 1.86$ -leads to a surface area of $\sqrt{3} L^2 = 6$. -          - - -          - -Surface dissection of regular tetrahedron to cube. - - -Related: "Covering a Cube with a Square." - -REPLY [5 votes]: I found a solution to this many years ago. Amazingly, this can be solved in as few as 2 pieces. See this webpage for the solution. You will also find a description in Greg Frederickson’s book “Dissections: Plane & Fancy”. (This book is a very good introduction to dissections in general and I highly recommend it.) - - - -In any dissection the two shapes must first be dissected to form strips. These strips are overlaid to give the dissection. The cube is easily converted to a strip element in a number of ways. One is in the form of a zigzag of the six squares. The tetrahedron forms a strip element as a row of four triangles. Overlaying these two strips gives a four piece solution. But the trick with the tetrahedron is in realising that the triangles form a tube, i.e. an infinitely long strip that repeats every four squares. (Try it: cut a paper tetrahedron along one edge and also along the opposite edge. You now have a tube.) This allows the number of pieces to be reduced to just two pieces. -It is hard to believe that my solution can be further improved! -On the above webpage you will also find many other surface dissections including a three piece surface dissection of cube to icosahedron and a four piece surface dissection of tetrahedron to cube to octahedron.<|endoftext|> -TITLE: Maximum cardinal of a set of linearly independent vectors in a module -QUESTION [12 upvotes]: A student asked me this, and I can't believe I never knew the answer to this. -Let $R$ be a commutative ring, and $M$ be an $R$-module. - -If $M$ has a set of $n$ linearly independent vector for each $n\in\mathbb{N}$, does that necessarily imply that $M$ has an infinite set of linearly independent vectors? -More generally, if $\kappa$ is the minimum cardinal such that there is no set of cardinality $\kappa$ of linearly independent vectors. Must $\kappa$ be always a successor cardinal? - -REPLY [8 votes]: Question 1. has a negative answer. -Denote $A := \{(i,j) \in \mathbb{N}^2, j\leq i\}$ and $S := \{$finite subsets of A with at least 2 different first coordinates$\}$. Define the ring $$R := \mathbb{F}_2[t_s: s \in S]\,/\,\big(t_{s_1}t_{s_2}: s_1,s_2 \in S\big),$$ and the $R$-module -$$M := \bigoplus_{(i,j) \in A}Rm_{ij}\,/\,\left(t_s\sum_{(i,j)\in s}m_{ij}: s \in S\right).$$ -For example, $s_0=\{(1,1),(2,1))\} \in S$ and $t_{s_0}(m_{11}+m_{21})=0.$ -Then for each $m$, the subset $\{m_{m1},..m_{mm}\} \subset M$ is a linearly independent subset of size $m$. -Suppose that $(x_k)_{k\in K} \subset M$ is a linearly independent family and let us show that $K$ is finite. -Denote $T := (t_s: s \in S)R$; this is a non-zero ideal of $R$. - -Write $x_k = \Sigma r_{ij}^{(k)}m_{ij}$. Then $\{r_{ij}^{(k)}(0)\} \neq \{0\}$: otherwise $Tx_k=0$. -$\{x_k\}$ is linearly independent mod $T: \Sigma r_kx_k \in TM \Rightarrow \Sigma t_{s_0}r_kx_k =0 \Rightarrow t_{s_0}r_k=0$ all $k \Rightarrow r_k \in T$ all $k$. -$\exists m$ such that $\{x_k\} \subset \sum_nRm_{mn}$ mod $T$: otherwise $\bigcup_k\{(i,j):r_{ij}(0) \neq 0\}$ has at least 2 different first coordinates and some $t_s$ kills some $x_k$ or some $x_{k_1}+x_{k_2}$. -$\#(K) \leq m: \dim_{\mathbb{F}_2}\sum_nRm_{mn}$ mod $T = m$.<|endoftext|> -TITLE: Does equality between sets contradict the philosophy behind structural set theory? -QUESTION [7 upvotes]: Zermelo-Fraenkel set theory (with choice) is commonly accepted as the standard foundation of mathematics. It is a material set theory. This means that for every two objects/sets $a,b$ one can ask whether $a=b$ or not. Also, one can always ask whether $a∈b$ is true or not. So $\in$ is a global element relation. -As an alternative foundation for set theory, Lawvere proposed ETCS (= elementary theory of the category of sets). It is the standard example of a structural set theory. The idea behind structural set theory is that elements—in contrast to material set theory—have no internal structure, i.e. are just "abstract dots". Thus there there is no global element relation, and "objects" are not indepedent things, they always lie in a particular structure (for example, we cannot speak about 2 as an object that exists on its own, but we instead talk about the "2 in the structure IN" or the "2 in the structure IR", and strictly speaking, these are not the "same" objects, but we have a natural injection IN -> IR which maps the "natural number 2" to the "real number 2"). -There have been controversial discussion whether ETCS is more appropriate as a foundation for mathematics than ZFC. I don't want to discuss this here, but want to point you to this paper: -https://arxiv.org/pdf/1212.6543.pdf -in which Tom Leinster introduces ETCS and argues that this foundational system is much nearer to the practice of mathematicians than ZFC. -Quite at the end of the paper, Leinster states that Sets for Mathematics by Lawvere and Rosebrugh is the quite the only book that teaches set theory in the flavour of structural set theory (ETCS)—which makes sense, since Lawvere is the "founder" of structural set theory. -Now, let me come to my question: Does equality between sets contradict the philosophy behind structural set theory? Obviously, when doing set theory in the spirit of structural set theory, we don't need equality between sets. Instead, we talk about isomorphisms, which makes more sense structurally. Also, the typical definition of equality between sets (extensionality) can't be formulated in ETCS. Thus it seems to me that the notion of "equality between sets" doesn't make much sense in structural set theory. Of course, we could say that two sets are equal if there is a bijection between them, or state that equality exists between sets without further specifying what it does (in particular, if we identify all isomorphic sets, whether there are infinitely many isomorphic sets that are not equal, ...). But this wouldn't yield to additional value, and is thus superfluous. Having this thoughts in mind, I was surprised when I read the following in Sets for Mathematics—the standard text book for structural set theory (on page 2!): - -Notation 1.1: The arrow notation f : A -> B just means the domain of f is A and - the codomain of f is B, and we write dom(f) = A and cod(f) = B. - -Here, the authors talk about the equality of two sets (dom(f) = A). They also use the "big bag of morphisms"-definition of category and not the "by pairs (A, B) of objects indexed family of hom-sets". But in this definition, one must talk about operations dom and cod which specify for each morphism a unique domain and codomain. But the word "unique" here presupposes that we have a notion of equality between objects. -Could someone from the foundations of mathematics clarify my confusion? On the nLab (see https://ncatlab.org/nlab/show/category ) there are two definitions of the term "category" ("With one collection of morphisms" and "With a family of collections of morphisms"), and to me it seems the second ("With a family of collections of morphisms") is more appropriate for structural set theory. But then, why does Lawvere—the founder of structural set theory—uses the first one ("With one collection of morphisms") in the only book about structural set theory? - -REPLY [9 votes]: I can't give you an answer that fully addresses what Lawvere and Rosebrugh were thinking, since I haven't asked them. (If you want to ask them, Rosebrugh runs a category-theory mailing list at https://www.mta.ca/~cat-dist/; he is an active participant, and Lawvere has been known to comment from time to time as well.) But I can say something about what can be done with equality between objects in ETCS. -For the most part, I think that ETCS has equality between objects by default. First-order logic, as usually treated, is both untyped and equipped with a global equality predicate, and since ZFC is usually written in such a first-order logic, Lawvere wrote ETCS in that logic too. Using an untyped logic requires him to put all morphisms into a single class (and to put objects in that class as well, identifying an object with its identity morphism), which means that he needs that global equality predicate. -On the other hand, it's quite possible to write ETCS in a typed first-order logic, with a type of sets, a dependent type of morphisms (dependent on two terms of the type of sets), and a dependent equality predicate for morphisms, but no equality predicate for sets. This is how Leinster does it; although he doesn't say things like ‘dependent type’ and ‘equality predicate’, such types and predicates are what appear in his paper. (And while he never denies the existence of an equality predicate between sets, no such predicate appears in his paper either.) -I think that Leinster's way of putting it is more in the structural spirit than Lawvere's, for the reasons in your question. But I don't think that Lawvere made a conscious choice to reject that argument either.<|endoftext|> -TITLE: Non-reduced, non-crystallographic root systems -QUESTION [8 upvotes]: Let $V$ be a $n$-dimensional real vector space with standard inner product $(\cdot,\cdot)$. For any $\alpha \neq 0 \in V$, set $\alpha^\vee := \frac{2}{(\alpha,\alpha)}\alpha$. For $\alpha \neq 0,\beta \in V$ set $n_{\alpha}(\beta) := (\beta,\alpha^\vee)$ and $s_\alpha(\beta) := \beta - n_{\alpha}(\beta)\cdot\alpha$. -A root system in $V$ is a finite set $\Phi$ of non-zero vectors in $V$ satisfying: - -$\Phi$ spans $V$; -$s_{\alpha}(\Phi) = \Phi$ for any $\alpha \in \Phi$; -$\mathrm{Span}\{\alpha\} \cap \Phi = \{\alpha,-\alpha\}$ for any $\alpha \in \Phi$; -$n_{\alpha}(\beta) \in \mathbb{Z}$ for any $\alpha,\beta \in \Phi$. - -This is by now a standard definition and there is a very satisfying classification theory for root systems based on Dynkin diagrams (the so-called Cartan-Killing classification). -Nevertheless, sometimes small modifications to this definition are considered. For instance, sometimes condition 4 above is omitted and root systems satisfying condition 4 are called crystallographic. However, considering non-crystallographic root systems doesn't change much: there are only a few more families of irreducible non-crystallographic root systems. -Similarly, sometimes condition 3 is omitted and root systems satisfying condition 3 are called reduced. Once again this does not change the structure theory so much: from 4 it follows that for any $\alpha \in \Phi$ we have $\mathrm{Span}\{\alpha\} \cap \Phi \subseteq \{2\alpha,\alpha,-\alpha\,-2\alpha\}$, and I think then it is not hard to show that any irreducible non-reduced root system is of the form $A \cup B \cup 2A$ where $A\cup B$ and $2A \cup B$ are irreducible reduced root systems (see Proposition 13, Section 1.4, Chapter VI of Bourbaki's "Lie Groups and Lie Algebras"). -I wonder if anyone has ever considered what happens when we eliminate both 3 and 4 from the above. Now things get a bit worse: even in rank one (i.e. $n=1$) there are infinitely many different root systems- any symmetric set of finite vectors in $\mathbb{R}^1$ is a root system by this definition. From what I can gather from some quick searches on the internet, nobody ever tries to remove both 3 and 4 from the definition of root systems, and maybe that's because the resulting theory is horrible. However, I wonder if this is really the case: is there a nice structural classification of non-reduced, non-crystallographic root systems or not? - -REPLY [8 votes]: Let's stick to the OP's definition of a root system. -Let $\Phi_0$ be the set of normalized roots $\frac{\alpha}{||\alpha||}$, $\alpha\in\Phi$. This is a root system satisfying 1, 2 and 3. Thus it is in the list of not necessarily crystallographic root systems. To reconstruct $\Phi$ we need for every $\alpha_0\in\Phi_0$ the length spectrum $L(\alpha_0)=\{||\alpha||\mid\alpha\in\mathbb{R}_{>0}\alpha_0\}\subseteq\mathbb{R}_{>0}$. These spectra can be chosen arbitrarily, subject to the condition that -$$ -L(w\alpha_0)=L(\alpha_0)\text{ for all $w$ in the Weyl group $W$ of $\Phi_0$} -$$ -Thus root systems with 1 and 2 are classified by a non-crystallographic root system and a finite set of positive numbers for each $W$-conjugacy class of roots. -This can be made more concrete. For this, let $S$ be a set of simple roots of $\Phi_0$. Attached to it is a labeled graph where the label $n_{\alpha\beta}$ above the edge between $\alpha$ and $\beta$ indicates the order of $s_\alpha s_\beta$. Each root is conjugate to a simple root. So it suffices to know $L(\alpha)$ for $\alpha\in S$. If the $n_{\alpha\beta}$ is odd then $\alpha$ is conjugate to $\beta$ (easy exercise). Thus we need -$$ -\alpha,\beta\in S\text{ with }n_{\alpha\beta}\text{ odd}\Rightarrow L(\alpha)=L(\beta). -$$ -A not so easy theorem on Coxeter groups (somewhere in Bourbaki) implies that this condition above is also sufficient.<|endoftext|> -TITLE: Simple homotopy equivalent, non-homeomorphic manifolds -QUESTION [17 upvotes]: What are known examples of two smooth, closed, oriented Manifolds $M,N$ of the same dimension that are simple homotopy equivalent, but not homeomorphic ? -It is well-known that the homotopy type of a given such $2$-manifold is the same as its homeomorphism type, so there won't be any such easy examples. Moreover (and this is where for me, the question becomes really interesting), I've heard at a recent conference that the simple homotopy type of a closed, oriented $3$-manifold is the same as its homeomorphism type, so any known examples must be in even higher dimension. Also, as the Borel conjecture is still open, any known such example must consist of non-aspherical manifolds. - -REPLY [5 votes]: To expand a bit the answers given by others, the machinery of surgery theory outputs many examples of simple homotopy equivalent manifolds that are not homeomorphic, PL homeomorphic, or diffeomorphic (let's just say equivalent to avoid repeating all of these categories.) The only caveat is that what it produces are simple homotopy equivalences that are not homotopic to equivalences. The bible for this is Wall's book, Surgery on Compact Manifolds, although there are other treatments that are perhaps more approachable. Ben Wieland's answer has to do with one aspect of the theory, which has to do with the tangent bundle; the example of Cappell cited by Ian Agol has to do with the Wall surgery groups. I'll suggest some alternative examples along these same lines, although using easier to understand invariants. -If you want the manifolds to be non-equivalent, then a good collection of manifolds to look at are manifolds homotopy equivalent to higher dimensional lens spaces (quotients of $S^{2n-1}$ by a cyclic group $\mathbb{Z}_d$ acting linearly on $\mathbb{R}^{2n}$). In this setting, there are two fundamental sorts of invariants: Reidemeister torsion, and the multisignature (or Atiyah-Singer invariants). If you fix the simple homotopy type, that fixes the Reidemeister torsion, and the multisignature invariants can be varied more or less arbitrarily. This is detailed in Chapter 14E in Wall's book (which is not particularly easy reading). The executive summary of what is going on here is that you are using the action of the Wall surgery group $L^s_{2n}(\mathbb{Z}[\mathbb{Z}_d])$ on the structure set of a lens space $S^{2n-1}/\mathbb{Z}_d$, and the multisignature invariants are really invariants from that surgery group. They have the nice property that they are invariants of the manifold itself (together with an identification of the fundamental group with $\mathbb{Z}_d$) and so don't depend very strongly on a choice of homotopy equivalence.<|endoftext|> -TITLE: Why are anisotropic tori compact? -QUESTION [13 upvotes]: I'll do my best to formulate everything in the modern language. Let $F$ be a local field. A torus $S$ is an $F$-group scheme of finite type such that $S \times_F \overline{F}$ is isomorphic to a finite product of copies of $\mathbb{G}_{m,\overline{F}}$ as $\overline{F}$-group schemes, where $\mathbb{G}_{m,\overline{F}} = \textrm{Spec } \overline{F}[T,T^{-1}]$. A character $\chi$ of $S \times_F \overline{F}$ is a morphism of $\overline{F}$-group schemes $S \times_F \overline{F} \rightarrow \mathbb{G}_{m,\overline{F}}$. -We say that $\chi$ is defined over $F$ if there exists a morphism of $F$-group schemes $\alpha: S \rightarrow \mathbb{G}_{m,F}$ such that $\chi = \alpha \times 1_{\overline{F}}$. If it exists, $\alpha$ is unique. -The trivial character is the morphism $S \times_F \overline{F} \rightarrow \mathbb{G}_{m,\overline{F}}$ coming from the $\overline{F}$-algebra homomorphism $\overline{F}[T,T^{-1}] \rightarrow \overline{F}[T_1^{\pm 1}, ... T_n^{\pm 1}]$, $T \mapsto 1$. It is always defined over $F$. -We say that $S$ is anisotropic if the trivial character is the only character which is defined over $F$. -There is a natural way of topologizing the $F$-rational points $S(F)$ of $S$, such that it is a locally compact topological group. See for example here (http://math.stanford.edu/~conrad/papers/adelictop.pdf). -I have read that $S(F)$ is compact if and only if $S$ is anisotropic, but I have never seen an explanation of this. What is the most natural way to realize this result? - -REPLY [9 votes]: Sorry for adding a second proof, but I think I have now a much clearer one. -Let $k$ be a local field. -Let $\mathbf{T}$ be a $k$-anisotropic torus, that is $\mathbf{T}$ contains no $k$-subgroup $k$-isomorphic to $\mathbf{G}_m$. -It is standard that $T$ also doesn't have a non-trivial $k$-homorphism to $\mathbf{G}_m$. -We argue to show that $T=\mathbf{T}(k)$ is compact. -We assume as we may that $T$ is Zariski dense in $\mathbf{T}$. -We fix a non-trivial irreducible $k$-representation $\mathbf{T}\to\text{GL}_n$. -As mentioned above, $n>1$. We use the fact that the $T$-orbits on $\mathbb{P}^{n-1}(k)$ are locally closed to find a point $x\in \mathbb{P}^{n-1}(k)$ with a $T$-closed orbit. We let $S$ be the stabilizer of $x$ in $T$ and $\mathbf{S}$ be the Zariski closure of $S$ in $\mathbf{T}$. $\mathbf{S}$ is a proper $k$-subtorus. By an induction argument we may assume that $S=\mathbf{S}(k)$ is compact. -We deduce that $T$ is compact. - -For completeness, let me add the remark that in charactersitic 0 the same proof yields the fact that $\mathbf{G}(k)$ is compact when $\mathbf{G}$ is a $k$-anisotropic reductive group. -The only thing to observe is that every $k$-subgroup of $\mathbf{G}$ is again reductive, what enables one to make an inductive argument as above. This follows from Jacobson-Morozov theorem: $\mathbf{G}$ has no unipotent $k$-subgroup.<|endoftext|> -TITLE: On the definition of locally compact for non-Hausdorff spaces -QUESTION [9 upvotes]: It seems that there are different conventions in the literature as to what is a locally compact space (when the space is not supposed Hausdorff). -The two main non equivalent definitions I've seen are : - -(LC1) every point has a compact neighborhood -(LC2) every neighborhood of any point contains a compact neighborhood of the point. - -I am wondering if there is a reason as to why one might prefer one definition to another. More precisely, in practice, what definition is really useful (yields interesting results), or are they are both important in their own right ? In that case why not give a different name to those definitions ? -Naively, when one looks at the definition of locally connected space, one does not use the version (LCn1)"every point has a connected neighborhood", but always (LCn2) "every neighborhood of any point contains a connected neighborhood of the point". Is there a deep reason as to why LCn1 is never considered, but LC1 is ? -I am aware that for Hausdorff space, LC1 and LC2 are equivalent, and since LC1 is easier to check, one might prefer that as a definition, but this argument is unconvincing if LC2 is actually more useful for non-Hausdorff space. - -REPLY [2 votes]: The notion LCn1 just boils down to "the connectedness components of the space are clopen". If this property does indeed show up somewhere, I would expect that the latter is a more convenient way of expressing it. -LC1 on the other hand does indeed seem to capture some intuition about "this space has some properties of compact spaces, but might be too large to be compact". Naming it locally compact is probably a mistake coming from the equivalence to actual local compactness in Hausdorff spaces.<|endoftext|> -TITLE: Is it fine to inquire about a paper that's been under review for around 9 months? -QUESTION [10 upvotes]: I have submitted a paper on applied probability in one of SIAM journals. The paper is under review for 9 months. I asked the editor 1 month ago about it, I was told that one review report has come and they are waiting for the other. Will it be ok if I politely inquire about it now (I have inquired 3 times till now after submission). How frequently I should inquire from now on so that it is not rude. Another question is that, is it the usual time taken for review in SIAM journals? - -REPLY [3 votes]: My experience is in Pure Mathematics, not in Applied Probability. In my area, at least, it is unusual to have a full report recommending acceptance (pending revisions) in under a year, although it can happen much faster. It has happened to me within a few weeks, but that is the exception rather than the rule. My first paper was three pages long and took about 10 months from submission to acceptance. At the time, this seemed quite long to me, but by now I realise it was actually rather reasonable. -It is certainly "fine to inquire about a paper that's been under review for around 9 months". However, if you have already inquired 3 times, and most recently one month ago, then I suggest that you should wait. Editors are busy people, and it can be hard to find willing referees, and even the ones you find are busy people too, so it may take a while to get a report. You don't want to make the editor cranky by inquiring too often. My rule of thumb (again, in Pure Mathematics) is that it is OK to inquire after about 3 months or so, just to make sure the paper has not fallen through the cracks (it happens), and perhaps give a gentle reminder. But usually it would be about 6 months before I inquire, and not more than every three months after that. -It may be that there is an important reason, such as upcoming job applications / interviews etc, that some (even informal/partial) feedback on the status of the paper would be useful. In that case, it is OK to mention this. As long as it is done respectfully, and without appearing impatient, the editor is likely to be sympathetic. However, they are still unlikely to be able to do very much about it, unless you just happen to write at a time where reports have arrived but the editor hasn't yet been able to deal with them. -TL;DR: Be patient and wait another couple of months. It sounds as though your paper is progressing normally.<|endoftext|> -TITLE: On an Inequality of Lars Hörmander -QUESTION [12 upvotes]: Let $P(z)$ be a non-null complex polynomial in $\nu$ variables $z=(z_1,\dots,z_n)$ of degree $\mu$: -\begin{equation} -P(z)=\sum_{|\alpha| \leq \mu} c_{\alpha} z^{\alpha}, -\end{equation} -where as usual for every $\alpha=(\alpha_1,\dots,\alpha_\nu) \in \mathbb{N}^{\nu}$ (here and in the following $\mathbb{N}$ denotes the set of all non-negative integers) we set $|\alpha|=\alpha_1+\dots+\alpha_\nu$, and $z^{\alpha}=z_1^{\alpha_1}\dots z_{\nu}^{\alpha_\nu}$. Consider $P$ as a polynomial function from $\mathbb{R}^\nu$ into $\mathbb{C}$: -\begin{equation} -P(x)=\sum_{|\alpha| \leq \mu} c_{\alpha} x^{\alpha} \quad (x \in \mathbb{R}^{\nu}). -\end{equation} -For any $m \in \mathbb{N}$, any $S \subseteq \mathbb{R}^\nu$, and any $\phi \in \mathcal{D}(\mathbb{R}^\nu)$ set: -\begin{equation} -||\phi||_{m,S} = \sup_{\substack{x \in S \\ |\alpha| \leq m}} |(D^{\alpha} \phi)(x)|. -\end{equation} -Let $M > L > 0$ and put $Q=[-M,M]^\nu$ and $E=Q \backslash (-L,L)^\nu$. I am trying to prove that for any $m \in \mathbb{N}$, there exist $K > 0$ and $m' \in \mathbb{N}$ such that we have -\begin{equation} -||\phi||_{m,E} \leq K ||P\phi||_{m',E} \quad \forall \phi \in \mathcal{D}_{Q} \tag{I}, -\end{equation} -where as usual $\mathcal{D}_{Q}$ is the set of all complex-valued functions $\phi \in C^{\infty}(\mathbb{R}^\nu)$ with support contained in $Q$. See the notes below for an explanation of the origin and relevance of this question. -Thank you very much in advance for your attention. -NOTE (1). If we take $L=0$, so that $E=Q$, then (I) is an immediate corollary of a remarkable result proved by Lars Hörmander in his wonderful work On the Division of Distributions by Polynomials. Indeed, inequality (4.3) of this work (taken with $k=0$) implies that for any $n, m \in \mathbb{N}$, there exist $K > 0$ and $n', m' \in \mathbb{N}$ such that -\begin{equation} -\sup_{\substack{x \in \mathbb{R}^\nu \\ |\alpha| \leq m}} (1+|x|)^n |(D^{\alpha} \phi) (x)| \leq K \sup_{\substack{x \in \mathbb{R}^\nu \\ |\alpha| \leq m'}} (1+|x|)^{n'} |(D^{\alpha} (P\phi)) (x)| \quad \forall \phi \in \mathcal{S}(\mathbb{R}^\nu) \tag{II}. -\end{equation} -We can state (II) in another way. Define the linear subspace $\mathcal{M}_{P}$ of $\mathcal{S}(\mathbb{R}^\nu)$: -\begin{equation} -\mathcal{M}_{P}=\{\psi \in \mathcal{S}(\mathbb{R}^\nu): \psi=P \phi, \phi \in \mathcal{S}(\mathbb{R}^\nu) \}, -\end{equation} -and consider the multiplication map $M_{P}:\mathcal{S}(\mathbb{R}^\nu) \rightarrow \mathcal{M}_{P}$ defined by -\begin{equation} -M_{P}(\phi)=P\phi \quad (\phi \in \mathcal{S}(\mathbb{R}^\nu)), -\end{equation} -Then (II) is equivalent to say that $M_{P}$ has a continuous inverse (this statement is Theorem (1) in Hörmander's work). -NOTE (2). Inequality (I) was stated without proof by ifw in his answer to the post Division of Distributions by Polynomials (see also my own answer for a comment). If true, (I) would allow to give a direct proof of Theorem (4) in Hörmander's paper, which states that every distribution can be divided by a non-null polynomial. In one of his comments, ifw said that (I) could be proved by localizing (II) or by modifying properly Hörmander's original proof. Even though I studied very carefully Hörmander's original proof (which can also be found in Trèves, Linear Partial Differential Equations with Constant Coefficients, $\S$ 5.5), I could not modify it in order to obtain (I) nor I could get (I) by localizing (II). - -REPLY [3 votes]: Finally, I realized how to modify Hörmander's proof in order to prove (I). -I will describe here the necessary changes we have to do to Hörmander's proof. Clearly, all the notation is that of Hörmander's paper. -Let $C$ be a closed, convex set of $\mathbb{R}^\nu$. Then make the following changes in Section (4) of Hörmander's paper On the Division of Distributions by Polynomials: -$\bullet$ take the suprema comparing in all the relations over $C$; -$\bullet$ after relation (4.2), replace everywhere $N^k$ with $N^k \cap C$ and $N^{k+1}$ with $N^{k+1} \cap C$; -$\bullet$ replace $\mathbb{R}^\nu$ with $C$ in (4.13); -$\bullet$ replace in every relation $S(\xi)$ with $S(\xi) \cap C$ and $S(\eta)$ with $S(\eta) \cap C$; -$\bullet$ replace in every relation $S(\xi,\eta)$ with $S(\xi,\eta) \cap C$ and $S_{1}(\xi,\eta)$ with $S_{1}(\xi,\eta) \cap C$. -With these changes, Hörmander's proof shows that to all $n, m \in \mathbb{N}$ and $k \leq \mu$ there are $n', m' \in \mathbb{N}$ and a constant $K > 0$ such that -\begin{equation} -\sup_{\xi \in C} (1+|\xi|)^n |f|_{m, (N^k \cap C)_{\xi}} \leq K -\sup_{\xi \in C} (1+|\xi|)^{n'} |Pf|_{m',\xi}, \quad (f \in C^{m'}(\mathbb{R}^\nu)). -\end{equation} -So, if $C$ is a compact convex set, the previous inequality, taken with $k=0$, implies that to all $m \in \mathbb{N}$ there are $ m' \in \mathbb{N}$ and a constant $K > 0$ such that -\begin{equation} -||f||_{m,C} \leq K ||Pf||_{m',C} \quad (f \in C^{m'}(\mathbb{R}^\nu)) \tag{III}. -\end{equation} -Now, assume that $E$ is a finite union of compact convex subsets $C_1,\dots, C_p$ of $\mathbb{R}^\nu$. By (III), for each $i=1,\dots,p$, there exist $m'_{i} \in \mathbb{N}$ and $K_i > 0$ such that -\begin{equation} -||f||_{m,C_i} \leq K_i ||Pf||_{m'_{i},C_i} \quad (f \in C^{m'_{i}}(\mathbb{R}^\nu)) \tag{III}. -\end{equation} -By taking $K=\max\{K_1,\dots,K_p\}$ and $m'=\max\{m'_{1},\dots,m'_{p}\}$, we then get -\begin{equation} -||f||_{m,E} \leq K ||Pf||_{m',E} \quad (f \in C^{m'}(\mathbb{R}^\nu)) \tag{IV}. -\end{equation} -Clearly (I) is a particular case of (IV), so we are done. -QED -REMARK. One could think that (IV) still holds if we take $E$ to be any compact set, but this is not the case, as the following counterexample shows. Let $\nu=1$, and set -\begin{equation} -E=\{ 0 \} \cup \left\{ \frac{1}{n}: n=1,2,3,\dots \right\}. -\end{equation} -Take $P(x)=x$ and let $K > 0$. Choose $N > K$. By the celebrated Whitney's Extension Theorem (which is Theorem (I) in Analytic Extensions of Differentiable Functions Defined in Closed Sets), there exists $f \in C^{\infty}(\mathbb{R})$, with support contained in $\left[ \frac{1}{N+1},2 \right]$, such that for $n=1,2,3,\dots,N$ -\begin{equation} -f^{(m)}\left( \frac{1}{n} \right) = (-1)^m m! n^{m+1} \quad (m \in \mathbb{N}). -\end{equation} -We have $D^{m}(Pf)(x)=0$ for all $x \in E$ and all $m=1,2,\dots$, so that $||Pf||_{m',E}=1$ for all $m' \in \mathbb{N}$. But $||f||_{0,E}=N > K$, and we conclude that (IV) cannot hold.<|endoftext|> -TITLE: Is there a combinatorial/topological treatment of statistical independence? -QUESTION [9 upvotes]: Is there any reference which studies sets of random variables as independence systems, a type of combinatorial object (see below)? -Motivation: -In particular, since independence systems are abstract simplicial complexes, this would allow one to apply homology theory (I think) to study families of random variables. Moreover, one could cross-apply intuition from other types of independence systems to better understand statistical independence. -For example, two things that I have had difficulty understanding intuitively, that statistical independence is not transitive, and that $(n-1)$-wise independence doesn't necessary imply $n$-wise independence, are facts which have geometrically easy to understand analogs in linear algebra. -Note: The closest thing I can think of having seen are treatments of random variables (with finite support) as standard simplices, since each point of the simplex corresponds to a probability distribution such that $p_1 + \dots + p_n = 1$. Also, this. However, these are not what I am looking for: in what I am describing, only the vertices represent random variables -- the higher-order faces do not represent random variables, they represent relationships of statistical independence between random variables (e.g. a $2$-face connecting three vertices corresponds to the statement that the three random variables symbolized by the vertices are mutually statistically independent). -I also don't think I am looking for a treatise on the theory of random graphs or random matroids or random simplicial complexes -- whether or not two given random variables $X$ and $Y$ are statistically independent is supposed to be a deterministic relationship (at least in simple models). -This question on Math.SE comes the closest to discussing the type of phenomenon I am talking about (the statistical independence of a family of random variables being modeled by an independence system, in this case the independence system of linearly independent vectors). -These two questions on Math.SE are also related, perhaps the first more so than than second: (1) (2). These two questions on MathOverflow also seem possibly related, although to be honest I don't feel I understand them well enough to discern that accurately: (1)(2). -Background: -In chapter 2 of Ghrist's Elementary Applied Topology, the author notes that: - -Example 2.1 (Statistical Independence) Independence occurs in multiple contexts, including linear independence of a collection of vectors or linear independence of solutions to linear differential equations. More subtle examples include statistical independence of random variables: recall that the random variables $\mathcal{X}=\{X_i\}_1^n$ are statistically independent if their probability densities $f_{X_i}$ are jointly multiplicative, i.e., the probability density $f_{\mathcal{X}}$ of the combined random variable $(X_1, \dots, X_n)$ satisfies $f_{\mathcal{X}} = \prod_i f(X_i)$. The independence complex of a collection of random variables compactly encodes statistical dependencies. - -According to Wikipedia, independence systems are equivalent to abstract simplicial complexes. They are a special case of hypergraphs, and more general than matroids. -In section 7.8, p.151 of the same book this idea is referenced once again: - -One clever application of this result [use of discrete Morse theory to study evasiveness in decision tree algorithms as in e.g. this paper, Morse Theory and Evasiveness, by Forman] is to independence tests for random variables. Let $\mathcal{X} = \{X_i\}$ be a collection of random variables and recall from Example 2.1 the independence complex $\mathcal{J}_{\mathcal{X}} \subset \Delta^n$ of $\mathcal{X}$. Given an unknown subcollection $\sigma \subset \mathcal{X}$ of the random variables, how many trials of the form "Is $X_i$ a member of $\sigma$" are required to determine if the collection is statistically independent? According to the results cited above, statistical independence is evasive if $\mathcal{J}_{\mathcal{X}}$ is not acyclic: any nontrivial homology class in $\tilde{H}_{\bullet}(\mathcal{J}_{\mathcal{X}})$ is an obstruction to evasiveness of statistical independence. How many such evasive collections of random variables are there? It is at least twice the total dimension of $\tilde{H}_{\bullet}(\mathcal{J}_{\mathcal{X}})$. - -Unfortunately, despite the fact that most of the book has ample citations and references to further reading, these examples are not accompanied by any references, so I am not sure how to explore this (to me) interesting idea further. Perhaps the reason why is that there are no references discussing this topic, and it is an original idea of the author. -Pages 4-5 here discussing log-linear models seem like they might be related, although it is hard for me to tell based on what is written and it is not expounded upon in much depth. In any case, like an aforementioned related question on Math.SE, it also suggests a relationship between this topic and Bayesian networks and/or algebraic statistics. However, it is worth noting that the edges in a Bayesian network denote conditional dependence, not independence, so they cannot be directly extended to the type of independence/abstract simplicial complex I am referring to. On the other hand, if one forms a new graph by connecting each node with the complement of its Markov blanket, then perhaps this might work. I.e. in other words Bayesian networks may encode the necessary information about conditional independence relationships for this to work -- on the other hand, it might still not work since "conditional independence" refers to a different (conditional) probability measure for each pair of nodes, while the standard example assumes a single choice of probability measure on the probability space on which all of the random variables are defined (I think). The answer is probably obvious but I need to think about it more. -This other document by Robert Ghrist also seems like it might be relevant, but if it is I can't tell for certain (honestly I don't think so but it's better than nothing so I'm including the link anyway). -I was thinking that perhaps one could use log-likelihoods to construct chain complexes and then homologies on these abstract simplicial complexes, although each time I try to work out the details I hit a roadblock/realize I was misunderstanding something. In any case, using the log-likelihood instead of the likelihood seems analogous to the use of the exponential to turn a (geometric/not abstract) simplex into a vector space as outlined in this blog post. Probably a more viable method for calculating the homologies of these complexes would be to use discrete Morse theory, as implied by Professor Ghrist on p.151 of his aforementioned book. -Also I wonder if there is a relation between all of this and information theory: the phenomenon of information gain in decision tree algorithms is well known and is even used to motivate the entire concept of entropy/information in some books, for example. In particular, Ghrist's remarks on p.151 of his book about Forman's paper seem like they could possibly be interpreted in terms of information theory, and being particularly bold, one could imagine that the homologies of these statistical independence complexes have an information-theoretic interpretation. -Disclaimer: This is a revised version of my now-deleted week-old unanswered question on Math.SE -- perhaps this is a current topic of research, which perhaps is why it was unanswered on Math.SE and why it might be on-topic here. If you disagree, please let me know. - -REPLY [2 votes]: This is an old question which already has an accepted answer, but I believe the theory of so-called 'gaussoids' is precisely a combinatorial abstraction of the idea of independence in probability (see e.g. https://arxiv.org/abs/1710.07175).<|endoftext|> -TITLE: The best bound on the growth of $\sum_{n\le X}a(n)$ -QUESTION [11 upvotes]: Let $f(z) = \sum_{n=1}^{\infty} a(n)n^{(k-1)/2} q^n$ be a cusp form, I am interested to know what is currently the best bound on the growth of $\sum_{n\le X}a(n)$ in the following two case : - -When $f$ is a cusp form of integral weight $k$ on $\Gamma_0(N)$ with non-trival character $\chi.$ -When $f$ is a cusp form of half-integer weight $k \ge 5/2$ on $\Gamma_0(4N)$ with non-trival character $\chi.$ - -REPLY [12 votes]: For integral weight $k$ and level $N=1$, I believe that the best result is due to Rankin (1990): -$$ \sum_{n\le X}a(n)\ll_\epsilon x^{1/3}(\log x)^{-\delta+\epsilon}, -\qquad \delta:=\frac{8-3\sqrt{6}}{10}\approx 0.065.$$ -On the other hand, Jutila (1987) proved that in square mean the sum is of size $\asymp x^{1/4}$, so the exponent $1/3$ cannot be lowered to $1/4$. -These results should generalize to arbitrary level and nebentypus. (The quoted book of Jutila (1987) proves the bound $\ll_\epsilon x^{1/3+\epsilon}$ with Voronoi summation, and this technique certainly generalizes. This bound is really due to Walfisz (1933), while the technique goes back to Wilton (1928), upon inserting the famous bound of Deligne (1974).) -For half-integral weight $k$, I don't know the best result from the top of my head, so I skip that part for the time being. -Added. Jutila's book is available online here. The quoted results are (1.5.21) and (1.5.23) on Page 28 (Page 30 in the original printed version).<|endoftext|> -TITLE: Explicit formulas Belyi maps for a rational dessin d'enfants -QUESTION [6 upvotes]: What are the best references for finding explicit formulas for Belyi maps for rational dessin d'enfants? -I am most interested in a formula for the Belyi map that corresponds to a specific rational dessin d'enfant: it is clean, with four black vertices, each of valency 3, and with ramification indices above the infinity of 6,3,2,and 1. (I believe that there is only one such dessin d'enfant). Ideally, it would be great to have tables or software for other similar examples. - -REPLY [2 votes]: You are right that there is only one such dessin. To see that, -type the following in Maple 2019: - -with(GroupTheory): - FindDessins([3,3,3,3],[2,2,2,2,2,2],[6,3,2,1]); - -You can find more information on: -https://www.math.fsu.edu/~hoeij/Heun -In the main file there you can find: - S3: ModuliField: Q. - Obstruction: none - Branch type: 0:[3^4] 1:[2^6] infty:[1,2,3,6] - Decomposition: S47(deg 3) 1/S34(deg 4) - Galois group: A4 x S3. GAP transitive group ID = 12T43. Order = 72 - -The name "S3: ..." refers to rational functions in the file: -https://www.math.fsu.edu/~hoeij/Heun/BelyiMaps -which gives this rational function: -$S_3 := {\frac {4\, \left( {x}^{3}-6\,x-6 \right) ^{3}{x}^{3}}{27\, \left( x+1 - \right) ^{3} \left( 2\,x+3 \right) ^{2} \left( x-3 \right) }}$<|endoftext|> -TITLE: Question about polynomials over finite fields -QUESTION [8 upvotes]: This is a special case of this question. -Let $\mathbb{F}$ be a finite field and $\mathbb{F}_{\leq d}[x,y]$ the set of bivariate polynomials over $\mathbb{F}$ of degree at most $d\ll|\mathbb{F}|$. Do there exist non-empty, disjoint subsets $A,B\subset\mathbb{F}_{\leq d}[x,y]$ such that for all pairs $(\ell,P)$, where $\ell\subset\mathbb{F}^2$ is a line, and $P$ is a univariate polynomial of degree at most $d$ defined on $\ell$ we have: $$\Big|\{Q\in A:Q|_\ell=P\}\Big|=\Big|\{Q\in B:Q|_\ell=P\}\Big|.$$ -Follow-up Question: -Do there exist $A,B\subset\mathbb{F}_{\leq d}[x,y]$ satisfying the above and also $|A|=|B|=|\mathbb{F}|^{\mathcal{O}(1)}$? - -REPLY [2 votes]: The condition that $A$ and $B$ are nonempty disjoint can be replaced with the condition that they are distinct, as we may just replace $A$ and $B$ with $A \cap (A - B)$ and $B \cap (A- B)$ respectively. -Let $q = |\mathbb F|$. -Let $A$ and $B$ be two random linear subspaces of the space of polynomials of degree $\leq d$ each of codimension $k$. As long as $0< k < { d+2 \choose 2}$, with high probability $A$ and $B$ are distinct. As long as $A$ and $B$ intersect transversely the space of polynomials vanishing on $\ell$, the number of elements of $A$ and $B$ taking a given fixed value on $\ell$ is $ q^{ {d+2 \choose 2} - k - (d+1)}$. -Equivalently, this happens when the perpendicular spaces of $A$ and $B$ intersect only at $0$ the space of linear forms on degree $d$ polynomaials that factor through reduction to $\ell$. Each nonzero linear form has a probability of $$\frac{(q^k-1) }{ q^{ {d+2 \choose 2}} -1}$$ of being in the perpendicular subspace, and there are at most $(q^{d+1}-1)$ nontrivial linear forms that factor through restriction to each of $q (q+1)$ lines, so as long as -$$ q(q+1) (q^{d+1}-1) \frac{(q^k-1) }{ q^{ {d+2 \choose 2}} -1}< 1$$ -with high probability $A$ and $B$ are transverse to all these linear subspaces. This happens when $k<{d+2 \choose 2} - d-3$. So it looks to me like even random $d+4$-dimensional subspaces will do the job.<|endoftext|> -TITLE: generating $q$-Catalan numbers -QUESTION [5 upvotes]: An $n$-Dyck path (or a Catalan path) is a lattice path $P$, unit East and North steps, in an $n\times n$ square grid which stays (weakly) above the main diagonal. Let $\square_n$ denote all such paths. Define the area of a path $P$ to be the number of full squares below the path and above the diagonal. Then one variant of a $q$-Catalan number is in the form -$$C_n(q)=\sum_{P\in\square_n}q^{area(P)}.$$ -A few examples: $C_1(q)=1,\, C_2(q)=1+q,\, C_3(q)=1+2q+q^2+q^3$. -There is this notion of $q,t$-Catalan numbers so that $C_n(q)=C_n(q,1)$. -I'm finding (after some fooling around) the following generating function which I've not encountered in the literature. - -Question. Is this known or can you provide a proof? - $$\sum_{n=0}^{\infty}C_n(q)\,z^n -=\frac{\sum_{k=0}^{\infty}q^{k^2}\prod_{j=1}^k\frac{z}{q^j-1}} -{{\sum_{k=0}^{\infty}q^{k(k-1)}\prod_{j=1}^k\frac{z}{q^j-1}}}$$ - -REPLY [10 votes]: The functions -$$ -C_n(q)=\sum_{P\in\square_n}q^{area(P)} -$$ -satisfy the following recurrence relation -$$ -C_n(q)=\sum_{k=1}^nq^{k-1}C_{k-1}(q)C_{n-k}(q).\tag{1} -$$ -Proof. -(taken from the book "The q, t-Catalan Numbers and the Space of Diagonal Harmonics" by J. Haglund, page 14, typo corrected) -We break up our path $P$ according to the “point of first return” to the line y = x. If this occurs at (k, k), then the area of the part of $P$ from (0, 1) to (k − 1, k), when viewed as an element of $\square_{k-1}$, is -$k − 1$ less than the area of this portion of $P$ when viewed as a path in $\square_{n}$.$\qquad\qquad\qquad~~~~~\Box$ -The generating function -$$ -G(z,q)=\sum_{n=0}^\infty C_n(q)z^n -$$ -satisfies the functional equation, which is a direct consequence of $(1)$: -$$ -G(z,q)-1=zG(z,q)G(qz,q). -$$ -We rewrite it for convenience as -$$ -\frac{1}{G(z,q)}=1-zG(qz,q).\tag{2} -$$ -It is known that the generalized Rogers-Ramanijan continued fraction, formula (39), -$$ -F(a,q)=\frac{\sum_{k=0}^\infty\frac{(-a)^kq^{k^2}}{(q;q)_k}}{\sum_{k=0}^\infty\frac{(-a)^kq^{k^2+k}}{(q;q)_k}}=1-\cfrac{aq}{1-\cfrac{aq^2}{1-\cfrac{aq^3}{1-...}}}, -$$ -satisfies the functional equation -$$ -F(a,q)=1-\frac{aq}{F(aq,q)}.\tag{3} -$$ -Since $G(0,q)=F(0,q)=1$ it is clear from comparing $(2)$ and $(3)$ that -$$ -\frac{1}{G(z,q)}=F(z/q,q)=\frac{\sum_{k=0}^\infty\frac{(-z)^kq^{k^2-k}}{(q;q)_k}}{\sum_{k=0}^\infty\frac{(-z)^kq^{k^2}}{(q;q)_k}}, -$$ -as required.<|endoftext|> -TITLE: Does every enriched functor preserve tensors? -QUESTION [5 upvotes]: Let $\cal{P}$ be a $k$-linear semisimple abelian rigid monoidal category with finite dimensional (over $k$) Hom-spaces (for a field $k$). -By a tensored $\cal{P}$-category we mean a $\cal{P}$-category which admits tensors with objects in $\cal{P}$, i.e. for every two objects $X,Y$ in the category and $P$ in $\cal{P}$, there is an object $P\otimes X$ with an isomorphism: -$$[P\otimes X,Y] \cong [P,[X,Y]].$$ -Does every enriched functor $F$ between tensored $\cal{P}$-categories preserve tensors? (i.e. the natural induced map $P\otimes F(X)\to F(P\otimes X)$ is isomorphism) What happens if the categories are also abelian and $F$ is exact? - -REPLY [6 votes]: No. The functor "take a vector space to its double dual" is linear but does not preserve tensors with infinite-dimensional vector spaces. -Enriched functors are automatically "lax tensored". An enriched functor $F$ provides a natural map $[X,Y] \overset F\to [FX,FY]$ for any $X,Y$. Consider setting $Y=P\otimes X$, and study: -$$ \mathrm{id} \in [P\otimes X,P\otimes X] \cong [P,[X,P\otimes X]] \overset {F\circ} \to [P,[FX,F(P\otimes X)]] \cong [P\otimes FX,F(P\otimes X)].$$ -This gives a canonical map $P\otimes FX \to F(P\otimes X)$ which is natural in $P,X$ and compatible with associativity an unit data. It just isn't automatically an isomorphism. - -On first reading, I missed the requirement that $\mathcal P$ be rigid. In that case the answer is Yes, enriched functors are automatically tensored. In general, a-priori-lax constructions become strong when there is enough dualizability. Consider the composition -$$ F(P\otimes X) \to P \otimes P^* \otimes F(P\otimes X) \to P \otimes F(P^* \otimes P \otimes X) \to P \otimes F(X)$$ -where the first map is the unit between $P,P^*$, the second is the lax monoidality constructed above, and the third is $F$ of the counit. This should give an inverse to the lax monoidality.<|endoftext|> -TITLE: Obtaining the Hilbert symbol from cup product on motivic cohomology -QUESTION [6 upvotes]: Let F be a number field. Does the Hilbert symbols at the various places of F arise from the cup product on the motivic cohomology groups of Spec(F)? And if so, is it possible to interpret Moore's reciprocity sequence -$$ -K_2(F)\to \bigoplus_{v}\mu(F_v)\to\mu(F)\to0 -$$ -in this setting? - -REPLY [2 votes]: The Hilbert symbol is a Steinberg symbol, i.e., it can be interpreted as a morphism from $K_2(F)\to\mu_n(F)$ where $F$ is a local field with $n$-th roots of unity $\mu_n(F)$. It can be viewed as composition of three operations, two of which are related to cup products. The first step is to use the cup product for motivic cohomology (albeit in a rather trivial way): for two units $a,b\in F^\times\cong {\rm H}^1(F^\times,\mathbb{Z}(1))$ their cup product is the symbol $\{a,b\}\in {\rm H}^2(F^\times,\mathbb{Z}(2))\cong K^{\rm M}_2(F)$. The second step applies the norm residue isomorphism/Galois symbol/Chern class map (which arises from changing the topology from Nisnevich to étale): $K^{\rm M}_2(F)/\ell\cong {\rm H}^2_{\rm ét}(F,\mu_\ell^{\otimes 2})$. Finally, we need to get from the etale cohomology group to $\mu_\ell(F)$; and this is based on having an inverse of cup-product with the local invariant in the Brauer group. -This description of the Hilbert symbol is the one that gives rise to generalized/higher Hilbert symbols in the following papers of Banaszak and Kahn: - -G. Banaszak: "Generalization of the Moore exact sequence and the wild kernel for higher K-groups" Compositio Math 86 (1993), 281-305. (link to Numdam) -B. Kahn: "Algebraic K-theory and twisted reciprocity laws" K-Theory 30 (2003), 211--240. - -There are also generalizations of the Moore reciprocity sequence to higher K-groups in these papers. Looking at the constructions in these papers may help to better understand in what ways the classical Hilbert symbol and Moore reciprocity sequence fit into the motivic cohomology framework. For now, it seems to me that changing the topology from Nisnevich to etale (and using properties in etale cohomology like the Poitou-Tate sequence) is more fundamental than the use of the motivic cohomology cup product.<|endoftext|> -TITLE: Check irreducibility of an explicit polynomial, without computer -QUESTION [6 upvotes]: I have a polynomial of degree 8 in 6 variables given explicitly by -$$ (\sqrt{1+(x_1+x_2+x_3)^2+(y_1+y_2+y_3)^2}+\sqrt{1+x_1^2+y_1^2}+\sqrt{1+x_2^2+y_2^2}+\sqrt{1+x_3^2+y_3^2})\times\text{the other seven of its Galois conjugates}. $$ -I fed it into Maple and it shows it's irreducible. But being unsure of what is going on behind the scene, I am asking here for a less computation-intensive and more conceptual proof of this fact, in the sense that any computation involved in the proof can be checked by hand in a reasonable amount of time. - -REPLY [8 votes]: For new variables $x$ and $y$ set $x_1=x_2=x_3=x$ and $y_1=y$, $y_2=2y$, $y_3=3y$. Then -$$ -\prod\left(\sqrt{1+9x^2+36y^2}\pm\sqrt{1+x^2+y^2}\pm\sqrt{1+x^2+4y^2}\pm\sqrt{1+x^2+9y^2}\right)=-1024\left((3y^2 + 4)x^6 + (23y^4 + 47y^2)x^4 + (45y^6 + 180y^4)x^2 + 225y^6\right). -$$ -This polynomial has degree $8$, and if it is irreducible, then the original polynomial is irreducible even more. Replacing $x$ with $xy$ and dividing by $y^6$ afterwards reduces to show that -$$ -(3x^6 + 23x^4 + 45x^2)y^2 + 4x^6 + 47x^4 + 180x^2 + 225 -$$ -is irreducible. That holds, because the coefficients of $y^0$ and $y^2$ are relatively prime and non-squares.<|endoftext|> -TITLE: Bimodule categories realized as internal bimodules -QUESTION [6 upvotes]: Let $\mathcal C$ be a finite tensor category, and $\mathcal M$ a finite left $\mathcal C$-module category. By a result of P. Etingof, S. Gelaki, D. Nikshych, and V. Ostrik (http://www-math.mit.edu/~etingof/tenscat1.pdf , Thm. 2.11.6), there is an algebra object $A \in \mathcal C$ and an equivalence of left $\mathcal C$-module categories $\mathcal M \simeq \mathsf{Mod}\mbox{-}A(\mathcal C)$, where $\mathsf{Mod}\mbox{-}A(\mathcal C)$ is the category of right $A$-modules internal in $\mathcal C$. There is a similar result for right $\mathcal C$-module categories. -Now assume that $\mathcal M$ is a $\mathcal C$-$\mathcal C$-bimodule category, e.g. $\mathcal M$ is braided (then there is a canonical right/left $\mathcal{C}$-module structure if $\mathcal M$ was a left/right $\mathcal{C}$-module). (Of course, one can also consider the more general case of a $\mathcal C$-$\mathcal D$-bimodule category.) What is the corresponding result to the one above? Something straight-forward would be the following: If $\mathcal M \simeq \mathsf{Mod}\mbox{-}A(\mathcal C)$ as left module categories and $\mathcal M \simeq B\mbox{-}\mathsf{Mod}(\mathcal C)$ as right module categories, then $\mathcal M \simeq B\mbox{-}\mathsf{Mod}\mbox{-}A(\mathcal C)$ as bimodule categories, where $B\mbox{-}\mathsf{Mod}\mbox{-}A(\mathcal C)$ is the category of $B$-$A$-bimodules in $\mathcal C$. If this is correct, how can it be proven? Is there already a paper covering this? -(By results of Douglas, Schommer-Pries and Snyder, it at least follows that $\mathcal M \boxtimes_{\mathcal C} \mathcal M \simeq B\mbox{-}\mathsf{Mod}\mbox{-}A(\mathcal C)$ as $\mathcal C$-bimodule categories. Does this already show that my claim above does not hold?) -Edit: Rephrasing my question, how can the equivalence $\mathcal M \simeq \mathsf{Mod}\mbox{-}A(\mathcal C)$ be made into an equivalence of bimodule categories, if $\mathcal M$ is a bimodule category? - -REPLY [2 votes]: The equivalence is unfortunately not correct: take for example the category $\mathcal{C}=Vec_G$ for some finite group $G$. Take $\mathcal{M}=Vec$ be the category of vector spaces. Then the forgetful functor from $\mathcal{C}$ to $Vec$ enriches $\mathcal{M}$ with the structure of a $\mathcal{C-C}$ bimodule category. -If we write $A=KG$, the group algebra, then $\mathcal{M}$ is equivalent to the category of right (left) $A$-modules as a left (right) $\mathcal{C}$-module category. However, the category of $A-A$-bimodules in $\mathcal{C}$ is equivalent to the category $Rep-G$, which is not equivalent to $\mathcal{M}$. One possible approach to the problem is to consider $\mathcal{M}$ as a left module category over the category $\mathcal{D}:=\mathcal{C}\boxtimes\mathcal{C}^{op}$. The category $\mathcal{M}$ is then equivalent to the module category of some algebra in $\mathcal{D}$, which will contain $A\boxtimes B$ as a subalgebra.<|endoftext|> -TITLE: $QS^0$ isn't a product of Eilenberg-Mac Lane -QUESTION [12 upvotes]: I am reading Lewis' paper "Is there a convenient category of spectra?". To prove the main result on the non-existence of such a nice category, he shows that otherwise the unit component of $QS^0= \varinjlim \Omega^n S^n$ would have to be weakly equivalent a product of Eilenberg-Mac Lane spaces. So far so good, but it isn't immediately clear to me: -Q: Why can't the unit component of $QS^0$ be a product of Eilenberg-Mac Lane spaces? -This should probably be very easy, given how the paper feels no need to include any justification for it. - -REPLY [22 votes]: If $X$ is a product of Eilenberg-MacLane spaces then the map -$$ \eta^*\colon \pi_2(X) = [S^2,X]\to[S^3,X]=\pi_3(X) $$ is easily seen to be zero (where $\eta\colon S^3\to S^2$ is the Hopf map). However, standard calculations give: -\begin{align*} - \pi_2(QS^0) & =\pi_2^S(S^0)=\mathbb{Z}/2.\eta^2\\ - \pi_3(QS^0) &= \pi_3^S(S^0)=\mathbb{Z}/24.\nu -\end{align*} -with $\eta^3=12\nu$. Thus, $\eta^*$ is nonzero in this context.<|endoftext|> -TITLE: Books on music theory intended for mathematicians -QUESTION [85 upvotes]: Some time ago I attended a colloquium given by Princeton music theorist Dmitri Tymoczko, where he gave a fascinating talk on the connection between music composition and certain geometric objects (as I recall, the work of Chopin can naturally be viewed as walks among lattice points lying in some hyperbolic surface, which gives some sense of canonicity to his compositions). I am asking whether there exist books on music theory that is intended for an audience with a reasonably sophisticated mathematical background. This question is in the same spirit as this one regarding physics. -Any suggestions would be great! - -REPLY [2 votes]: Amiot, Emmanuel. 2016. Music Through Fourier Space. Springer. -From the Springer website: - -This book explains the state of the art in the use of the discrete Fourier transform (DFT) of musical structures such as rhythms or scales. In particular the author explains the DFT of pitch-class distributions, homometry and the phase retrieval problem, nil Fourier coefficients and tilings, saliency, extrapolation to the continuous Fourier transform and continuous spaces, and the meaning of the phases of Fourier coefficients. - -Perspectives of New Music 49/2 (Summer 2011) is devoted almost entirely to "Tiling Rhythmic Canons", including articles by many of the authors mentioned in other posts.1 - -1 The origin of the mathematics in this book can be found in Dan Vuza's four-part article -"Supplementary Sets and Regular Complementary Unending Canons" in Perspectives of New Music (vols. 29/2–31/1). It is the founding article of an area of research ("Tiling Rhythmic Canons") for many of the authors mentioned here, including Tom Johnson, Guerino Mazzola, Emmanuel Amiot, Carlos Agon, and Moreno Andreatta.<|endoftext|> -TITLE: Vanishing of Aronhold S-invariant on the cubic forms on $H^2(X, \mathbb Q)$ -QUESTION [8 upvotes]: I am considering several examples of compact complex threefolds $X$ such that $rk H^2(X)=3$. -Note that we have a cubic form on $H^2(X, \mathbb Q)$ which comes from the cup product. -I calculated the Aronhold S-invariants for those cubic forms and found that they are all zero, -which is very curious to me. -So let me ask some questions: -If S-invariants for a ternary cubic form is zero, what does it mean to the cubic form? -Is it a general phenomenon that S-invariants for cubic forms on such $X$'s are zero? -If it not, what does zero S-invariant possibly mean to $X$? - -REPLY [3 votes]: Here are some references that may be relevant. In the case $b_2=3$ the cubic from coming from the cup product gives rise to a plane cubic curve -- take the vanishing locus of the ternary cubic form on the projective plane $\mathbb{P}({\rm H}^2(X,\mathbb{Q}))$. Some statements concerning (non-)realizability of different types of cubics can be found in Sections 4.3 and 5.3 of - -C. Okonek and A. van de Ven. Cubic forms and complex 3-folds. L'Enseignement Math. 41 (1995), 297-333. - -To connect this with the question, Proposition 5.13.2 (p. 251) of the following paper gives a characterization of the types of plane cubic curves coming from ternary cubics with Aronhold invariant $S=0$: - -I. Dolgachev and V. Kanev. Polar covariants of plane cubics and quartics. Adv. Math. 98 (1993), 216-301. - -(In particular, the types 2,4,6,7 in the Okonek-van de Ven paper have nontrivial Aronhold invariants.)<|endoftext|> -TITLE: Arithmetic Morse theory? -QUESTION [24 upvotes]: Is there any analogue of Morse theory in Number theory? Naive idea arising in my head is that defining a Morse function on scheme and find etale cohomology using that function. Since I'm not an expert about algebraic geometry and Morse theory, I can't advance my thoughts. - -REPLY [45 votes]: One usually considers the analogue of Morse theory in algebraic geometry to be the theory of vanishing cycles and Lefschetz pencils. -Because of the nature of algebraic functions, Morse theory must be a little more complicated. A Morse function on a compact manifold lets us build the manifold up step by step, starting with a local minimum from which the manifold "springs from nothing". Such maps do not exist in complex geometry or algebraic geometry. -Instead, Lefschetz considers a map from a smooth projective variety to $\mathbb P^1$. One demands a Morse-like condition that the critical points of this map are as simple as possible. Applying vanishing cycles theory one can relate the cohomology of the total space to the critical points but one also needs to know as a starting point the cohomology of a smooth fiber. -These were adapted to etale cohomology in SGA 7-2 by Deligne and Katz and were crucial in Deligne's proof of Weil's Riemann hypothesis, so this certainly gives a connection to number theory.<|endoftext|> -TITLE: The smallest number of vertices for a graph with the same endomorphism monoid -QUESTION [5 upvotes]: Let $G$ be a directed graph without loop and suppose that $M$ is its endomorphism monoid. -First how we can create a simple graph $\Gamma$ (using $G$), with the same endomorphism monoid? and then how we can create it with the smallest possible number of vertices? -Thanks for your help - -REPLY [3 votes]: Here's a not terribly efficient method that at least shows that this is possible. Start by replacing each directed edge by a path of length $4$ with a triangle over the second edge of the path. - -These special marker triangles will be the only triangles in the new graph, so this construction almost works. The problem is that in some cases you can't tell from the new graph where the original vertices were, which can be fixed by attaching a path of the same suitably large length to each of the original vertices. Probably length $1$ (i.e. a pendant edge) is sufficient. -By hanging things off the third vertex of the triangle, this construction can be extended to encode directed graphs with labelled edges, which can be used to show that every group is the automorphism group of some graph. -EDIT: I now think you probably have to work a bit harder to get something that works for endomorphisms rather than automorphisms. For example, the long paths I suggested using as vertex tags can be folded any old how onto the rest of the graph, and conceivably you can wrap edges around marker triangles. I'm sure there's something you can do with your bare hands, but you'll need to be a little more careful than I was. - -REPLY [3 votes]: These kinds of results are considered in the paper http://onlinelibrary.wiley.com/doi/10.1002/jgt.20396/epdf<|endoftext|> -TITLE: Recursion theory from the standoint of category theory -QUESTION [6 upvotes]: It is (I believe) a very easy exercise to prove that the general recursive functions over the natural number object $N$ form a category. But what sort of category is it? From the fact that one can prove the Recursion Theorem one might surmise that the category of general recusive functions over $N$ is cartesian closed, but what other category-theoretic properties does it have? - -Question 1: Is it possible (for example) to characterize the category of general recursive functions over $N$ without having to deem specific functions as 'recursive' (as Kleene does in his papers)? - -Furthermore, we now have types of 'recursiveness' over various structures, i.e. $\alpha$-recursive functions, $\beta$-recursive functions ($\alpha$-,$\beta$-recurision over the ordinals); Koepke and Koerwien's $\ast$-recursive functions over the ordinals (which characterize the constructible universe $L$), and Infinite Time Turing Machines; Sacks' $E$-recursion; Recursive functionals and quantifiers of finite type (Kleene); primitive recursive set functions and rudimentary set functions (Jensen, and others for studying the fine structure of $L$); abstract first-order computability (Moschovakis), to name but a few. - -Question 2: Can (or has) category theory (been able to) elucidate the structural interrelations between the various types of recursion? - -Since I am certain that there is a subfield of category theory devoted to answering these types of questions (and probably already have answered some of these questions), I would be interested in obtaining a list of survey articles devoted to this subdiscipline, and (of course) the answers to the two questions I asked (the second can be answered from the survey articles by giving a few concrete examples). -Thanks in advance for any and all help given. - -REPLY [7 votes]: I am not sure this is what you are looking for but I think that the following paper may provide an answer to your question 1: -J. Robin B. Cockett, Pieter J. W. Hofstra: -Introduction to Turing categories. Ann. Pure Appl. Logic 156(2-3): 183-209 (2008) -It gives a categorical axiomatization of computability in arbitrary categories (not just sets and partial maps). The basic notion is that of Turing category which, in a nutshell, is: - -a restriction category, that is, a "category of partial maps": for every $f:A\to B$, there is a monic $\overline f:A\to A$ which morally represents the domain of $f$ as a "partial identity map" (subject to some axioms); total maps are arrows $f:A\to B$ such that $\overline f=\mathrm{id}_A$; -it has a notion of product that interacts well with the partiality structure; -it has a Turing object, which is morally an "object of programs"; technically, it is a sort of "universal internal hom": it is an object $T$ together with a map $\mathrm{eval}_{A,B}:T\times A\to B$ for every objects $A,B$, such that, for all $f:\Gamma\times A\to B$, there is a total map $f^\bullet:\Gamma\to T$ such that -$$\mathrm{eval}_{A,B}\circ (f^\bullet\times \mathrm{id}_A)=f.$$ -The map $f^\bullet$ is called a code of $f$ and is not required to be unique (morally, it is a program parametric in $\Gamma$). - -The category whose objects are powers of $\mathbb N$ and whose arrows are partial recursive functions is the prototypical example of Turing category ($\mathbb N$ is a Turing object, seen as the set of codes of recursive functions). -Apart from the above paper, there is also this survey by Robin Cockett in which he shows how some of the basic theorems of computability theory may be proved in the framework of Turing categories. -By the way, Turing categories are not the only proposal to axiomatize computability categorically; Cockett and Hofstra mention previous work in the introduction to their paper, which you may also find of interest. -As far as your question 2 is concerned, I don't think Cockett and coauthors studied the extensions of recursive functions that you mention. However, they did turn their attention to sub-recursive settings (i.e., complexity theory), and found out that there are Turing categories whose total maps correspond to well-known complexity classes (polynomial time, logspace). If I am not mistaken, this is done in the following two papers: -Robin Cockett, Joaquín Díaz-Boïls, Jonathan Gallagher, Pavel Hrubes: -Timed Sets, Functional Complexity, and Computability. Electr. Notes Theor. Comput. Sci. 286: 117-137 (2012) -J. Robin B. Cockett, Pieter J. W. Hofstra, Pavel Hrubes: -Total Maps of Turing Categories. Electr. Notes Theor. Comput. Sci. 308: 129-146 (2014)<|endoftext|> -TITLE: Conductor as volume of the integers ring -QUESTION [9 upvotes]: I am working on Tate's thesis, and I have some problems with computations, yet the result seems to be a good natural motivation for introducing the arithmetic conductor of a character. -Let $F$ be a non-archimedean local field, and $\psi$ a non-trivial additive character. Define the $\psi$-Fourier transform for a locally constant and compactly supported function on $F$ to be -$$\widehat{\Phi}(x) = \int_F \Phi(y)\psi(xy)dy$$ -We normalize the involved Haar measure so that the Fourier inversion formula holds, that is to say -$$\widehat{\widehat{\Phi}}(x) = \Phi(-x)$$ - -I would like to understand why this implies that the ring of integers - $\mathcal{O}$ have volume $q^{c(\psi)/2}$? - -Recall that $c(\psi)$ is defined by $\psi$ trivial on $\mathfrak{p}^{c(\psi)}$ and not on $\mathfrak{p}^{c(\psi)-1}$, where $\mathfrak{p}$ is the maximal ideal of $\mathcal{O}$. I tried manipulating the integrals splitting into classes modulo $\mathfrak{p}^{c(\psi)}$, but I cannot make it work properly. - -REPLY [15 votes]: Apply the Fourier Inversion Formula to the characteristic function $\Phi(x) = \chi_\mathcal{O}(x)$ of the ring $\mathcal{O}$ of integers in $F$. The Fourier transform is the integral $\widehat{\Phi }(x)=\int_\mathcal{O} \psi (xy)dy$. This integral is zero if and only if $y\mapsto \psi (xy)$ is a non-trivial character on the additive group $\mathcal{O}$ (orthogonality relation for characters). If $\psi (xy)$ is the trivial character, then this means that $x\in {\mathfrak p}^{c(\psi)}=\pi ^{-m}\mathcal{O}$ ($m=-c(\psi)$). We then have -$$ \widehat{\Phi}(x)= \mathrm{vol}(\mathcal{O}) \chi_{\pi^{-m}\mathcal{O}}(x).$$ -Taking Fourier transforms on both sides of this equation, and bearing in mind that the measure is so chosen that Fourier inversion holds, we get -$$\chi_{\mathcal{O}}(x)=\Phi (x)= \Phi (-x)= \widehat{\widehat{\Phi}}(x) = \mathrm{vol}(\mathcal{O})\mathrm{vol}(\pi^{-m}\mathcal{O})\chi_\mathcal{O}(x).$$ We then get the equality -$$\mathrm{vol}(\mathcal{O})\mathrm{vol}(\pi^{-m}\mathcal{O})=1, \quad\text{i.e.}, \quad \mathrm{vol}(\mathcal{O})=q^{-m/2}.$$ This is what was to be proved.<|endoftext|> -TITLE: Is there nontrivial structure to forcing axioms? -QUESTION [12 upvotes]: Suppose that $\cal P$ is a class of forcings, we denote by $\operatorname{FA}_\kappa(\cal P)$ the statement that whenever $\Bbb P\in\cal P$, $\gamma<\kappa$, and $\{D_\alpha\mid\alpha<\gamma\}$ are dense open sets in $\Bbb P$, then there is a filter $G\subseteq\Bbb P$ such that $G\cap D_\alpha\neq\varnothing$ for all $\alpha<\gamma$. -The usual examples are Martin's Axiom where $\kappa=2^{\aleph_0}$, and $\cal P$ is the class of ccc forcings; or the Proper Forcing Axiom where $\cal P$ is the class of proper forcings and $\kappa=2^{\aleph_0}$ (which then implies $\kappa=\aleph_2$). -Let's focus on subclasses of ccc forcings. -We can force Martin's Axiom for only $\sigma$-centered forcings, or other subclasses, and there is a clear hierarchy here. For example, -$$\operatorname{FA}_{2^{\aleph_0}}(\textrm{ccc})\implies\operatorname{FA}_{2^{\aleph_0}}(\sigma\textrm{-centered})\implies\operatorname{FA}_{2^{\aleph_0}}(\sigma\textrm{-linked}).$$ -Of course, there are many, many other subclasses that one can consider. - -Question. Is there any work on separating the various forcing axioms for subclasses of ccc forcings? For example, is it consistent that $\operatorname{FA}_{\aleph_2}(\sigma\textrm{-linked})$ holds, but only $\operatorname{FA}_{\aleph_1}(\sigma\textrm{-centered})$ holds, and $\operatorname{FA}_{\aleph_1}(\textrm{ccc})$ fails altogether? - -(And I cannot stress this enough, the choice of subclasses was entirely arbitrary, and any other subclasses of ccc forcings are of interest here.) - -REPLY [3 votes]: Another reference is the paper "On partial orderings having precalibre-$\aleph_1$ and fragments of Martin's axiom" by Bagaria-Shelah.<|endoftext|> -TITLE: Dodgy Turing degrees -QUESTION [9 upvotes]: This question was originally asked and bountied at MSE, but received no answer there, so I'm asking it again here. -Below, I'm specifically interested in weak truth table (wtt) reducibility, but other reducibilities between truth table and Turing are interesting to me, too, if the question happens to be easier to answer for them. -Let $d$ be a Turing degree, and fix a representative $X\in d$. Say $d$ is (wtt-)dodgy if there is some $d$-computable functional $F=\Phi_e^{X\oplus -}$ such that for all $Y$ with $deg(Y)=d$, we have - -$\Phi_e^{X\oplus Y}=F(Y)$ is total, -$F(Y)\equiv_TY$, but -$F(Y)\not\le_{wtt}Y$. - -(Note that I demand nothing about $F(Z)$ for $Z\not\in d$; in particular, $F$ only needs to output reals when fed elements of $d$, it may fail to be total elsewhere.) -Dodginess is most interesting for "sufficiently large" degrees - for example, above $0'$ every Turing degree splits into infinitely many $wtt$-degrees, so the question is nontrivial. Dodginess is a reasonably definable property, so by Martin's Cone Theorem, either every sufficiently large degree is dodgy or every sufficiently large degree is not dodgy. My question is, Which of these two holds? -My feeling is that a fairly simple trick should show that every sufficiently large (indeed, $\ge_T0'$) degree will not be dodgy; however, I don't see how to do this. In particular, the Recursion Theorem doesn't seem to immediately kill it: suppose $d$ is a sufficiently large degree, and fix $X\in d$. Then $F$ can be identified with a total computable function $g$: $F(\Phi_e^X)\sim\Phi_{g(e)}^X$. Now $g$ is total computable, so it has a fixed point $c$: $\Phi_c^X\sim \Phi_{g(c)}^X$. However, there's no reason to believe that $\Phi_c^X$ is total, let alone an element of $d$, so I don't see how to get any leverage here. - -REPLY [6 votes]: Every sufficiently large degree is dodgy. Assuming $X \geq_T \emptyset'$, we can define $F(Y)$ as -follows. First compute a set $Z$ from $X$ and $Y$: Given $n -= \langle i,j \rangle$, ask $X$ whether $\Phi_i(n)$ converges. If not -then $Z(n)=0$. If yes then ask $X$ whether $\Phi_j^{Y \upharpoonright -\Phi_i(n)}(n)$ converges. If not then $Z(n)=0$. If yes then -$Z(n)=1-\Phi_j^{Y \upharpoonright \Phi_i(n)}(n)$. This ensures that $Z -\nleq_{wtt} Y$. Now let $F(Y)=X \oplus Z$. Then $F$ is total and -$F(Y) \nleq_{wtt} Y$ for all $Y$, and if $Y \equiv_T X$ then also -$F(Y) \equiv_T Y$.<|endoftext|> -TITLE: Correspondence between coverings and field extensions -QUESTION [5 upvotes]: I am self reading from Groups as Galois Group by Helmut Volklein -There is a result on page 94(section 5.4) -Let $G$ be a finite group. Let $P\subset P^{1}$ finite and $q\in P^{1}\P$. There is a natural $1-1$ correspondence between - -The $\mathbb{C}(x)$-isomorphism classes of Galois extensions $L/\mathbb{C}(x)$ with Galois group isomorphic to $G$ and with branch points contained in $P$. -The equivalence class of Galois coverings $f:R\rightarrow P^{1}\P$ with deck transformation group isomorphic to $G$. -The normal subgroups of the fundamental group $\pi_{1}(P^{1}\p,q)$ with quotient isomorphic to $G$. - -I am interested in Inverse Galois problem over $\mathbb{Q}(T)$ as then by Hilbert's Irreducibility theorem I can study IGP over $\mathbb{Q}$ -Is there similar theorem(result) that states the correspondence between field extensions of $\mathbb{Q}(T)$ and coverings of $P^{1}_{Q}$. -Edits -Why I think there is such result -See the attached image [ Topics in Galois Theory, by Jean-Pierre Serre] -He also mentions the same idea, He has written the statement but I want to convince myself (mathematically by a proof) that it is actually true. - -REPLY [5 votes]: Chapter 2 Remark 2.5 of Silvermans' The Arithmetic of Elliptic Curves book gives you a bijection between covers of $\mathbb{P}^1_{\mathbb{Q}}$ by curves over $\mathbb{Q}$ and regular field extensions $K$ over $\mathbb{Q}(T)$ - that is, field extensions such that $\mathbb{Q}\cap K = \mathbb{Q}$. -The problem is that producing such covers is generally a difficult problem. Over $\mathbb{C}$ (or more generally $\overline{\mathbb{Q}}$), one can easily produce such covers topologically using quotients $\pi_1(\mathbb{P}^1_{\mathbb{C}}-S)$ for some varying finite sets of points $S$, then use Riemann's existence theorem to make it algebraic (but still over $\mathbb{C}$). However, to do this over $\mathbb{Q}$, one would need to analyze the finite quotients of $\pi_1(\mathbb{P}^1_{\mathbb{Q}}-D)$ where $D$ is now a $\mathbb{Q}$-rational effective divisor on $\mathbb{P}^1_{\mathbb{Q}}$, and this fundamental group is much more difficult to understand, at least when $D$ contains at least three geometric points. To see this, it's well known that there is a 'homotopy exact sequence' -$$1\rightarrow\pi_1(\mathbb{P}^1_{\overline{\mathbb{Q}}} - D)\rightarrow\pi_1(\mathbb{P}^1_{\mathbb{Q}}-D)\rightarrow\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\rightarrow 1$$ -If $D$ contains $n$ geometric points, then the first term is just the profinite completion of the free group $F_{n-1}$ of rank $n-1$, but to produce covers of $\mathbb{P}^1_{\mathbb{Q}}$, one needs to understand the finite quotients of the middle term, which as you can see contains all the complexity of the absolute Galois group of $\mathbb{Q}$. This isn't so bad when $n = 1,2$, but once $n\ge 3$, producing (nonabelian) covers of $\mathbb{P}^1_{\mathbb{Q}}$ through this method becomes very difficult in general. Most of the progress on the IGP uses more ingenious but indirect methods.<|endoftext|> -TITLE: Are the off-diagonal elements of exp(At) log-concave in t, for nonnegative matrices? -QUESTION [6 upvotes]: Let $A$ be a Metzler matrix, i.e. a real matrix (not necessarily symmetric) whose off-diagonal elements are all non-negative. Then, for $t\ge 0$, the matrix exponential $\exp(At)$ will have all non-negative elements. -Numerically, it seems that given a single off-diagonal element of $\exp(At)$, it is always a log-concave (i.e. log-convex downward) function of $t$, for $t\ge 0$, and that the diagonal elements are always log-convex upward functions. -That is, it looks like the function -$$ -f(t) = \log\Big(\big(\exp(At)\big)_{ij}\Big) -$$ -is a concave function of $t$ for $t\ge 0$ if $i\ne j$, and a convex function of $t$ if $i=j$. My question is whether this indeed is the case. Note that in this expression $\exp$ is a matrix exponential, but $\log$ is just an ordinary logarithm of a positive number. -Here is a typical result. The plot shows each element of $\exp(At)$ as a function of $t$, where $A$ is a matrix whose off-diagonal elements are independently uniformly sampled from $[0,1]$ and whose diagonal elements are independently uniformly sampled from $[-1,0]$. The diagonal elements of $\exp(At)$ are shown in red. - -Showing this should be a simple case of finding the second derivative and showing that it can't be positive, but I haven't seen a way to do that. I haven't been able to find a counterexample either. - -REPLY [8 votes]: The following Matlab code (unless I coded something wrong, which is well possible) finds a few random counterexamples each time I run it, even when restricted to off-diagonal entries: -n = 3; -for trie = 1:100 - A = rand(n); - B1 = log(expm(A)); - B2 = log(expm(2*A)); - B3 = log(expm(3*A)); - - C = B2 - (B3+B1)/2; %its off-diagonal should be >= 0 if the claim holds - C = C - diag(diag(C)); - if not(all(all(C >= 0))) - A - end -end - -For instance: -A = - 0.804449583613070 0.535664190667238 0.989144909700340 - 0.986104241895970 0.087077219900892 0.066946258397750 - 0.029991950269390 0.802091440555804 0.939398361884535 -A = - 0.018177533636696 0.534137567882728 0.625937626080496 - 0.683838613746355 0.885359450931142 0.137868992412558 - 0.783736480083219 0.899004898906140 0.217801593712125 -A = - 0.133503859661312 0.300819018069489 0.286620388894259 - 0.021555887203497 0.939409713873458 0.800820286951535 - 0.559840705872510 0.980903636046859 0.896111351432604 -A = - 0.108016694136759 0.559370572403004 0.848709226458282 - 0.516996758096945 0.004579623947323 0.916821270253738 - 0.143156022083576 0.766681998621487 0.986968274783658 -A = - 0.068357220470829 0.026107108154905 0.961558573103663 - 0.436327077480103 0.954678274080449 0.762414484002993 - 0.173853037365001 0.430596519859417 0.007348661102847 - -Quick remarks with some tips for numerical experimentation: - -For continuous functions, midpoint convexity is equivalent to convexity, so I only tested for that. The $t=1,2,3$ interval is chosen because it looked like the simplest thing to try. -Out of habit, I first coded this with 1000 tries with $5\times 5$ matrices. $100$ and $3\times 3$ are here just for quicker display. It's so fast that it doesn't matter anyway, so it's better to err on the side of more and larger examples. -One should be careful with instructions such as C - diag(diag(C)) (which subtracts its diagonal from a matrix), which could hide numerical mistakes (what if a -1e-16 pops up on the diagonal?). In this case though the subtractions are of the form a - a, which is guaranteed to return 0 even in double-precision arithmetic. Matlab does not have a simpler way to set the diagonal of a matrix to zero or ignore it, unfortunately. (I actually first wrote C(1:n+1:n^2) = 0, but then I replaced it because it is hackish and difficult to read). -There are lots of factoids about matrices (and especially about monotonicity of eigenvalues of nonsymmetric matrices and the matrix exponential) that look true at first sight but have counterexamples. I suggest you to always try some random experiments like these ones. Once one gets in the habit, it's faster to write the code than to think about it. :) - -REPLY [6 votes]: For $2\times2$ matrices, the log-concavity is true. One has $e^{tA}=f(t)I_2+g(t)A$ by Cayley-Hamilton. Writing that the eigenvalues of $e^{tA}$ are the exponentials of those of $tA$, we find -$$g(t)=\frac{e^{t\mu}-e^{t\lambda}}{\mu-\lambda}\,,$$ -where $\mu,\lambda$ are the eigenvalues of $A$. Thus we only have to prove that $g$ is log-concave, that is -$$gg''-g'^2=-e^{t(\mu+\lambda)}\le0.$$ -Notice that the assumption is implicitely used in that it implies $g>0$. -Edit. The formula above seems to be a particular case of a more general one. Suppose $A$ is $n\times n$. With Cayley-Hamilton, we have -$$e^{tA}=f(t)I_n+g(t)A+\cdots+h(t)A^{n-1}.$$ -Let us form the Hankel matrix $M_h(t)=\left(h^{(i+j-2)}(t)\right)_{1\le i,j\le n}$. Then $\det M_h(t)=(-1)^{n+1}e^{t{\rm Tr}\,A}$. -Remark that a smooth function $h$ satisfies a linear ODE of order $n-1$ with some constant coefficients if, and only if, $\det M_h\equiv0$.<|endoftext|> -TITLE: Any "natural" topology on fractional field of a topological ring? -QUESTION [12 upvotes]: Let $R$ be a topological integral domain. Let $K=\mathrm{Frac} R$. Is there any "natural" topology on $K$? Actually, since $K$ can be regarded as a quotient of $R\times R$ quotient some equivalence relation, so maybe we can equip $R\times R$ with the product topology, and that induces topology on $K$? -In particular, suppose $R$ has the complete $(p)$-adic topology, where $(p)$ is the ideal $pR$, then what does the induced topology on $K$ look like? -I did came across an old question, -Is $k(\!(x,y)\!)$ a topological field? -and so, it seems that it is quite a confusing question to topologize a fractional field? - -REPLY [13 votes]: As said Fred Rohrer, exercise 27 exists in the French edition and seems to answer the question of the MO. Let -$$ -s_{frac}\ :\ R\times R'\to Frac(R) -$$ -($R'=R\setminus \{0\}$) be the canonical surjection. The "quotient field equivalence" $\equiv_{frac}$ on $R\times R'$ is that given by $s_{frac}$ i.e. -$$ -(a,p)\equiv_{frac} (b,q) \Longleftrightarrow s_{frac}(a,p)=s_{frac}(b,q) \Longleftrightarrow aq=bp\ . -$$ -As a matter of fact, if the product topology on $R\times R'$ passes to quotient as a topology on $Frac(R)$ compatible with its field structure and inducing the given one on $R$, it is the finest of all topologies (a) compatible with the field structure (b) inducing the given topology on $R$. Bourbaki's exercise gives a sufficient condition ($\equiv_{frac}$ is an open equivalence relation on $R$) so that this occured (i.e. the quotient topology on $Frac(R)$ induces the given topology on $R$). -Update In the general case $S^{-1}R$ can also be constructed as the direct limit $\lim_\rightarrow R_t$ (thanks to David Handelman) where -$$ -R_t:=(t^{\mathbb{N}})^{-1}R\ ;\ t^{\mathbb{N}}:=\{t^n\}_{n\in \mathbb{N}} -$$ -the direct system being preordered by divisibility -$$ -s\prec t\Longleftrightarrow (\exists u\in S)(t=us) -$$ -one can check that the two constructions (by usual quotient or by direct limit) solve the same universal problem (algebraic and topological i.e. in the category of topological rings), their topology is therefore the same. The topology of $(t^{\mathbb{N}})^{-1}R$ could be easier to describe. -Hope this helps.<|endoftext|> -TITLE: Affine vs Yokonuma -QUESTION [8 upvotes]: Let $G=GL_n$. Let us start with the Hecke algebra $H_n$. It acts on K(constructible sheaves on $G/B$) by Hecke correpondences and on K(coherent sheaves on $G/B$) by Lusztig's construction [1]. Now we can extend $H_n$ by adding a commutative algebra in two ways. -In the first way we take constructible sheaves, but replace $B$ by $U$. The finite field analogue of the result is the so-called Yokonuma-Hecke algebra, the algebra of double cosets $U(F_q)\G(F_q)/U(F_q)$. It contains the group algebra of the torus $T(F_q)$ and has a surjective map to $H_n$. -In the second way we take coherent sheaves, so we can tensor by line bundles. This way we recover Lusztig's construction [1] of action of the affine Hecke algebra, which is generated by $H_n$ and a commutative subalgebra, which is the algebra of Laurent polynomials in $n$ variables, in other words the group algebra of the character group of the torus $T$. -So, we have two ways to add a commutative algebra to the Hecke algebra. Is there some kind of well-known duality between them and how to see it on the algebra level? -[1]: Lusztig, George. “Equivariant K-Theory and Representations of Hecke Algebras.” Proceedings of the American Mathematical Society, vol. 94, no. 2, 1985, pp. 337–342. - -REPLY [3 votes]: I am not entirely sure if this is the kind of answer you are looking for, but Chlouveraki in her thesis as well as here explains the differences in the two deformations. Generally speaking, in the finite convolution algebra you describe above we have that the additional generators coming from $T(\mathbb{F}_q)$ have finite order $d=|\mathbb{F}_q^\times|$. In particular, the Yokonuma-Hecke algebra deforms the wreath product algebra $S_n\ltimes (\mathbb{Z}/d\mathbb{Z})^n$ in a way that preserves the wreath product structure but deforms the quadratic Hecke relations. -In some sense an orthogonal deformation of these wreath products is given by the cyclotomic Hecke/Ariki-Koike algebras, which are quotients of the affine Hecke algebra. In these the Hecke relations are preserved but the wreath product structure is deformed instead. The finite Hecke algebra for $GL_n$ embeds naturally inside the Ariki-Koike algebras, but is naturally a quotient of the Yokonuma-Hecke algebras. -Additionally, you could do both deformations at the same time to arrive at cyclotomic and affine Yokonuma-Hecke algebras as in work of Chlouveraki and Poulain d'Andecy. -As to the actual question about a duality between Yokonuma-Hecke and the affine Hecke algebra, I don't know. Geometrically, one might suspect something like it based on Bezrukavnikov-Yun. Also, Weideng Cao has some purely algebraic work proving that the Yokonuma-Hecke algebra is isomorphic to a tensor product of (matrix algebras over) cyclotomic Hecke algebras.<|endoftext|> -TITLE: Search for $A_4$-extension of $\mathbb{Q}$ with particular ramification properties -QUESTION [15 upvotes]: I am looking for an extension $F/\mathbb{Q}$ with the following properties: - -$F/\mathbb{Q}$ is Galois with $\mathrm{Gal}(F/\mathbb{Q}) \simeq A_4$. -$F$ is totally real. -The prime $2$ has full decomposition group. -For every odd prime $p$ that ramifies in $F/\mathbb{Q}$, the primes above $p$ all have decomposition group of odd order. (In other words, every such prime is totally split in the biquadratic sub-extension cut out by the subgroup $V_4$ of $A_4$). - -Using John Jones' number fields database http://hobbes.la.asu.edu/NFDB/ I found 7670 totally real quartic extensions of $\mathbb{Q}$, each of which with normal closure a totally real $A_4$-extension of $\mathbb{Q}$. Using Magma, I found that 792 of these also have full decomposition group at 2. But then none of these also satisfy the last condition. -Is there some reason why no such extension exists, or is it just the case that I haven't looked hard enough? -Note that if no such extensions exist, the reason cannot be purely local in nature. For example, let $F$ be the Galois closure of $x^4 - 108x^2 - 136x - 8$. Then $F/\mathbb{Q}$ is a totally real $A_4$ extension with full decomposition group at $2$. The other primes dividing the discriminant are 17 and 241. The primes above 241 behave in the way that I want, but those above 17 do not. It appears that something similar always happens, i.e., there is always at least one odd prime dividing the discriminant that doesn't behave in the way that I want. -Also note that is there is only one $A_4$-extension of a $p$-adic field (for any $p$), and this is an $A_4$-extension of $\mathbb{Q}_{2}$. That means that no odd prime can have full decomposition group and that there is no local obstacle to the condition on $2$. Moreover, the unique $A_4$ extension of $\mathbb{Q}_2$ has a cubic unramified subextension, so any $A_4$ extension of $\mathbb{Q}$ with full decomposition group at 2 must also be ramified at (at least) one odd prime. - -REPLY [14 votes]: There is no such field. In fact, let $F$ be an $A_4$-extension of the rationals, -and let $K$ denote the cyclic cubic subfield of $F$. Then $F = K(\sqrt{\alpha},\sqrt{\alpha'}\,)$ for some $\alpha \in K$ such that $\alpha\alpha'\alpha''$ is a square in ${\mathbb Q}$. Here $\alpha'$ is the conjugate of $\alpha$ etc. -Every odd prime $p$ that ramifies is assumed to split in $F/K$, hence it must -ramify in $K/{\mathbb Q}$. Thus $F/K$ is unramified outside $(2)$. -The prime $2$ must be inert in $K/{\mathbb Q}$ by assumption 3. But then we can choose $\alpha$ to have odd norm, since $\alpha\alpha'\alpha''$ cannot be a rational square if $2$ divides $\alpha$ exactly. Since no odd prime ramifies in $F/K$, the ideal $(\alpha)$ must be a square of an ideal, and $\alpha$ must be totally positive by assumption 2. -This means that $\alpha \in Sel^+(K)$ (I am using the notation and a few results from https://arxiv.org/pdf/1108.5674.pdf from now on; the Selmer group $Sel^+(K)$ consists of all $\alpha K^{\times 2}$ for which $\alpha$ is totally positive and $(\alpha)$ is the square of an ideal). The $2$-rank of this group is $\rho^+ + s = \rho^+$ in our case (Thm. 2.2) since $K$ is totally real and thus $s = 0$. Here $\rho^+$ is the rank of the $2$-class group in the strict sense. By Thm. 7.2., we have $\rho^+ = \rho$ in our case $(n = 3)$, hence -the $2$-rank of $Sel^+(K)$ is equal to the $2$-rank of the class group. -Since every quadratic unramified extension is generated by the square root of an element in $Sel^+(K)$, and since these groups have the same $2$-rank, every extension $K(\sqrt{\alpha})$ for $\alpha \in Sel^+(K)$ must be a subfield of the Hilbert class field of $K$. But since $2$ is inert in $K$, it is principal, and thus splits in every subfield of the Hilbert class field. This contradicts assumption 3.<|endoftext|> -TITLE: Extra special p-groups -QUESTION [6 upvotes]: Let $P$ be an infinite extra special $p$-group for some prime $p$, namely, $Z(P)=P'=\Phi(P)$ and $P/Z(P)$ is infinite elementary abelian. -Let $C$ be a Prufer $q$-group for some prime $q\neq p$. -Question -Is it possible to find an action of $C$ on $P$ such that $C$ acts irreducibly on $P/Z(P)$? (i.e. $P/Z(P)$ does not contains any $C$-invariant subgroup) -Unfortunately I cannot find any information about the automorphism group of an extraspecial group in the infinite case. Probably, it is just some kind of infinite symplectic group but I cannot figure out if such an action is possible or not. - -REPLY [2 votes]: My construction does not work for all $p$ and $q$. I need to assume that $p$ and $q$ are both odd, that $P$ is of exponent $p$, and that the multiplicative order of $p$ modulo $q$ is even. I don't know whether such an action exists for other $p,q$ - I would guess not. -Let us assume that $P$ is the central product of extraspecial groups $\langle x_i,y_i \rangle$ of order $p^3$ and exponent $p$ for $i \ge 1$. Note that the automorphism group of a central product of $n$ such groups is ${\rm Sp}(2n,p)$. -Now the group ${\rm Sp}(2n,p)$ has elements of order $p^n+1$. (To see that you observe that ${\rm Sp}(2,p^n) = {\rm SL}(2,p^n)$ has such elements and ${\rm Sp}(2,p^n)$ is naturally a subgroup of ${\rm Sp}(2n,p)$.) -Suppose that $2k$ is the order of $p$ modulo $q$, so $q|p^k+1$. Let $t \ge 1$ be maximal such that $q^t|p^k+1$. Then it is an easy exercise to show that the order of $p$ modulo $q^{t+i}$ for any $i \ge 0$ is $2kq^i$. -So, for a given $i > 0$, $q^{t+i}|p^{kq^i}+1$, and so ${\rm Sp}(2kq^i,p)$ has elements $g$ of order $q^{t+i}$. Such elements must act irreducibly on their natural module $V$. (All faithful irreducible representations of a cyclic group $C_m$ over a finite field of order $q$ have dimension the order of $q$ modulo $m$.) Then $g^q$ has order $q^{t+i-1}$ and, under the action of $g^q$, $V$ must decompose into a direct sum of $q$ (isomorphic) irreducible modules $V_i$. -Now the restriction of the symplectic form to each $V_i$ must be non-degenerate. The only other possibility would be that this restriction was totally singular, but that cannot happen with $q$ odd. (I could try and give more details about that if you wanted.) -Now we can put all of these ingredients together to construct an irreducible action of the Prufer $q$-group $C$ on $P$. -Let $g_0 \in C$ have order $q^t$. Then we can define an action of $g_0$ on $P$ by letting it act faithfully and irreducibly on the subgroups $\langle x_i,y_i : 1 \le i \le k \rangle$, $\langle x_i,y_i : k+1 \le i \le 2k \rangle$, etc. -Then, if $g_1$ is an element of $C$ of order $q^{t+1}$ with $g_1^q = g_0$, we can define the action of $g_1$ so that it acts irreducibly on the subgroups $\langle x_i,y_i : 1 \le i \le kq \rangle$, $\langle x_i,y_i: kq+1 \le i \le 2kq \rangle$, etc, and such that the action of $g_1^q$ is the same as that of $g_0$. -Now you can continue this construction by defining actions of $g_i$ consistent with those of $g_{i-1}$ for all $i > 0$. It is easy to see that the resulting action of $C$ on $P$ is irreducible. -Later addition: For the non-degeneracy of the $V_i$, if we could find some non-degenerate invaraint subspace $V_1$ for $g^q$, then their images under $g$ would give the required decomposition $V = V_1 \oplus \cdots \oplus V_q$ with each $V_i$ non-degenerate. -So let us prove by induction on $n$ that any element $x$ of order coprime to $p$ in ${\rm Sp}(2n,p)$ for which irreducible modules have dimension $m$ with $2n/m$ odd must stabilize and act irreducibly on some non-degenerate subspace. If not, then $x$ stabilizes and acts irreducibly on a totally singular subspace $W$ of dimension $m$. Then the subspace $W^\perp$ of $V$ that annihilates $W$ under the symplectic form contains $W$ and has dimension $2n-m$, and the form induced on $W^\perp/W$ is non-degenerate. -Now $x$ fixes the subspace $W^\perp$, so it fixes a complement $X$ of $W$ in $W^\perp$ of dimension $2n-2m$, and the symplectic form induced on $X$ is non-degenerate, so our claim follows by applying induction to the action on $X$, which is an element of ${\rm Sp}(2(n-m)p)$. -That argument seems a bit awkward and their may be a better way of seeing this, but I think it works.<|endoftext|> -TITLE: Smallest prime that is a quadratic residue modulo a fixed prime -QUESTION [5 upvotes]: Let $P$ denote the set of positive primes and let $p$ be a fixed prime. Then define $q_{p}:=\min{q\in P:pp$ are primes such that $(q/p)=1$}\}=1.$$<|endoftext|> -TITLE: Birthday problem with unequal probability: expected number of draws before the $m$-th collision? -QUESTION [5 upvotes]: Let $p$ be an arbitrary distribution over $\mathbb{N}$, and $m\geq 1$ be an integer. Given an infinite sequence of i.i.d. draws $(X_i)_{i\geq 1}$ from $p$, define a collision as a pair $(i,j)$ with $ix)=\mathbb{P}(\mathrm{Poiss}(\tfrac{1}{2}x^2)\leq m-1)$ -and integrating this gives -\begin{align*} - C_m&={m-\tfrac{1}{2} \choose m-1}\sqrt{{\pi \over 2}} &\mbox{ for } \theta_1=0 \\ -%\mbox{ and } C_m&=m+1 &\mbox{ for } \theta_1=1 -\end{align*} -In particular, for the uniform distribution on $\{1,\ldots,n\}$ -$$E(M_2)\sim \tfrac{3}{2}\sqrt{\tfrac{1}{2}\pi n},\; \mathbb{E}(M_3)\sim \tfrac{15}{8} \sqrt{\tfrac{1}{2}\pi n},\ldots\;\;.$$ -(This certainly looks "classical", but I haven't found anything in the literature.)<|endoftext|> -TITLE: Are all Dehn invariants achievable? -QUESTION [17 upvotes]: The Dehn invariant of a polyhedron is a vector in $\mathbb{R}\otimes_{\mathbb{Z}}\mathbb{R}/2\pi\mathbb{Z}$ defined as the sum over the edges of the polyhedron of the terms $\sum\ell_i\otimes\theta_i$ where $\ell_i$ is the length of edge $i$ and $\theta_i$ is its dihedral angle. -Are all vectors in this space realizable as the Dehn invariants of polyhedra? -(To be very specific let's define a polyhedron to be a bounded closed manifold embedded into Euclidean or hyperbolic space as a subset of the union of finitely many planes.) -You can add two representable Dehn invariants by taking the disjoint unions of their polyhedra (or gluing them together on any pair of faces, even with mismatched face shapes, if you prefer a single connected polyhedron). For Euclidean polyhedra, you can multiply the Dehn invariant by any scalar by scaling the polyhedron by the same factor. So the left sides of the products in the sum are controllable but the right sides are not, and getting a nonzero Dehn invariant that represents only a single angle $\theta$ and its rational multiples seems difficult except for some very special cases. -The Dehn invariant also makes sense in hyperbolic space but there you have the opposite problem. You can get a polyhedron whose dihedrals are all rational multiples of the same angle $\theta$, for your favorite $\theta$, by scaling one of the Platonic solids, but scaling the lengths by arbitrary real factors (while keeping the angles fixed) becomes difficult. So I'd be interested in answers both for the Euclidean and for the hyperbolic cases, especially if they require different arguments or have different answers. -If an answer can be found in the published literature, references would be helpful. - -REPLY [4 votes]: In hyperbolic, spherical and Euclidean geometry the answer is no. -For Euclidean polytopes it is a result of B. Jessen (proved here). Namely, let $\Omega^1_{\mathbb{R}/\mathbb{Q}}$ be a $\mathbb{Q}-$vector space of Kähler differentials. Then the image of Dehn invariant equals to the kernel of a surjective map -$$ -\mathbb{R}\otimes \mathbb{R}/2 \pi \mathbb{Z} \longrightarrow \Omega^1_{\mathbb{R}/\mathbb{Q}} -$$ -sending $l \otimes \alpha$ to $l \dfrac{d \sin(\alpha)}{\cos(\alpha)}.$ -For hyperbolic and spherical polytopes similar results were proven by Dupont and Sah. The role of Kähler differentials is played here by Milnor $K-$group $K_2(\mathbb{C})$. For instance for hyperbolic case the image of Dehn invariant equals to the kernel of a map -$$ -\mathbb{R}\otimes \mathbb{R}/2 \pi \mathbb{Z} \longrightarrow K_2(\mathbb{C}) -$$ -sending $l \otimes \alpha$ to $\{e^{l},e^{i \alpha}\}.$ -A. Goncharov proposed a conjectural higher dimensional generalization of these results, see here.<|endoftext|> -TITLE: Any finite simplicial complex has the homotopy type of an algebraic variety -QUESTION [8 upvotes]: A construction that I've seen at least a couple times in Deligne's work is that any finite simplicial complex has the homotopy type of a pretty natural algebraic variety. This appears, for instance, in 6. of his 1974 ICM address and in his paper with Sullivan on vanishing of Chern classes of flat bundles. (So for instance, the construction is robust enough to "transfer" a topological vector bundle on the simplicial complex to an algebraic vector bundle on the variety.) -Unfortunately I don't really understand his construction and was hoping somebody could explain it to me. -My French is awful, but here's my attempt to translate what he says. Suppose you have a finite set S and a set $\mathcal{S}$ of subsets of S, corresponding to simplices, with the usual property that it's closed under subsets. Consider the space $\mathbb{R}^S$, and for $\sigma \subset S$, let $|\sigma|$ be the subspace spanned by the basis vector corresponding to $s \in \sigma$. Let -$$|S| = \bigcup_{\sigma \in \mathcal{S}} |\sigma|$$ -Then the corresponding algebraic variety is the same thing construction $\mathbb{C}^S$. -I don't understand how to go from this to the desired result. Isn't $|S|$ obviously contractible (onto the origin)? - -REPLY [7 votes]: the geometric picture is - -a simplex extends naturally to its real affine space which deformation retracts back to the simplex. -a real affine extends to a complex affine space which deformation retracts back -to the real affine space. -now apply these two steps to any simplicial complex -finite ones yield complex varieties defined over Q which deformation retract to any finite simplicial complex.<|endoftext|> -TITLE: Central binomial coefficients deprived of $2$'s: not radicals? -QUESTION [5 upvotes]: In the paper, P Erdos, R Graham, I Ruzsa, E Straus, On the prime factors of $\binom{2n}n$, Math. Comp., 29:83–92, 1975, it was conjectured that the central binomials are never square-free for $n>4$. The proof was given in A Granville, O Ramare, Explicit bounds on exponential sums and the scarcity -of squarefree binomial coefficients, Mathematika, 43:73–107, 1996. -Cute fact: $\frac12\binom{2n}n\in\mathbb{Z}$, and it is odd iff $n=2^k$. So, remove powers of $2$ to write $\frac1{2^n}\binom{2n}n:=\frac{U_n}{V_n}$ as reduced fractions, and ask: - -Question. Is this true? There exists $n*\in\mathbb{N}$ such that - $U_n$ is never square-free whenever $n>n*$. - -REPLY [14 votes]: The real thrust of the work of Granville and Ramare was to get explicit bounds, so that one could get a complete resolution of the Erdos problem. Earlier work of Sarkozy already gave asymptotic results that are quite a bit sharper. Thus, Sarkozy showed that if one writes -$$ -\binom{2n}{n} = (s(n))^2 q(n), -$$ -where $q(n)$ is square-free, then -$$ -\exp((c-\epsilon)\sqrt{n}) \le s(n)^2 \le \exp((c+\epsilon)\sqrt{n}) -$$ -for large $n$. Here -$$ -c= \sqrt{2} \sum_{k=1}^{\infty} \Big( \frac{1}{\sqrt{2k-1}} -\frac{1}{\sqrt{2k}}\Big) -$$ -is a positive constant. Note that the power of $2$ dividing $\binom{2n}{n}$ is no more than $\log_2 (2n)$, and so the square-factor in Sarkozy's result certainly doesn't just come from powers of $2$.<|endoftext|> -TITLE: Were 3-manifolds with $\sec>0$ known to be space forms before Ricci flow? -QUESTION [13 upvotes]: It is well known that R. Hamilton (JDG 1982) used Ricci flow to show that a closed $3$-manifold with positive Ricci curvature must be diffeomorphic to a spherical space form $S^3/\Gamma$, since such metrics evolve to a limit which has constant curvature. -Together with some colleagues, I was under the suspicion that it was previously known that closed $3$-manifolds with positive sectional curvature satisfied this conclusion, but cannot seem to find such an earlier proof independent of Ricci flow. More precisely, I suspected it was possible to prove this statement by establishing that the Heegaard genus of such a manifold must be $\leq2$. So here is my: - -Question. Is there a proof that if a closed manifold $M^3$ has $\sec>0$ then $M^3\cong S^3/\Gamma$ using Heegaard splittings instead of Ricci flow? - -REPLY [12 votes]: No, this was not known before Hamilton's paper. I think a Ricci-flow-free argument would still be of interest, so if you know how to do it, by all means...<|endoftext|> -TITLE: Equivariant cohomology vs. invariant cohomology vs. cohomology of quotient space -QUESTION [16 upvotes]: Given a space $X$ and an action of a group $G$ on $X$, the $G$-invariant cochains with coefficients in an Abelian group $A$ define a sub-cocomplex $\mathcal{C}^{\bullet}_G$ of the cocomplex $\mathcal{C}^{\bullet}$ of cochains with coefficients in $A$. I will call the cohomology $H^{\bullet}(\mathcal{C}_G)$ the "invariant cohomology" of $X$. What is the relationship between the follwing three objects: (i) the invariant cohomology, (ii) the usual equivariant cohomology $\mathcal{H}_G^{\bullet}(X, A)$, and (iii) the cohomology of the quotient space $H^{\bullet}(X/G,A)$? Note that I am most interested in the case where $A$ is a finite Abelian group. -I think that there are always homomorphisms -$$H^n(X/G,A) \to H^n(\mathcal{C}_G) \to \mathcal{H}_G^n(X,A).$$ -The first map comes from pulling back the projection $X \to X/G$, and the second can be seen from the interpretation of equivariant cohomology as the total cohomology of the double cocomplex of group cochains of $G$ valued in $\mathcal{C}^{\bullet}$. -But how does one characterize the kernel and image of these maps? Is there a simple statement if $X$ is contractible, say? -[Edit: regarding the relationship between $H^n(X/G,A)$ and $H^n(\mathcal{C}_G)$, please see the comments on Mark Grant's answer. The upshot is that with simplicial cochains, $H^n(\mathcal{C}_G)$ seems to depend on the choice of triangulation. But, if one defines $\mathcal{C}_G$ in a suitably triangulation-independent way, i.e. either singular cochains, or simplicial cochains on the abstract simplicial complex containing all simplex embeddings into X, then $H^n(\mathcal{C}_G)$ and $H^n(X/G,A)$ are almost certainly not the same.] - -REPLY [4 votes]: If $G$ is a finite group acting by simplicial automorphisms on a simplicial complex $X$, then -$$H^{\ast}(X/G;\mathbb{Q}) = (H^{\ast}(X;\mathbb{Q}))^{G} = H^{\ast}((C^{\ast}(X;\mathbb{Q}))^G),$$ -where the superscripts indicate that we are taking the invariants with respect to $G$. For a proof, see Proposition 1.1 (and its proof) in my note "The action on homology of finite groups of automorphisms of surfaces and graphs", available on my webpage here. It is definitely necessary to take $\mathbb{Q}$ coefficients here (the referenced proof will make it clear why this is the case).<|endoftext|> -TITLE: Non-trivial problems about the trivial group -QUESTION [6 upvotes]: Is there any non-trivial problem (maybe open problem) about the trivial group? -I asked already a question about the Laws characterizing the trivial group. There is a description of such laws. As another example, one can ask about the first order sentences characterizing the trivial group. I am not sure if these sort of problems are trivial or easy. Is there any important (open) question about the trivial group? - -REPLY [3 votes]: In Group trisections and 4-smooth manifolds, the authors proved that the smooth 4-dimensional Poincaré conjecture is equivalent to the following (purely group theoretical) statement about the trivial group: Every $(3k, k)$–trisection of the trivial group is stably equivalent to the trivial trisection of the trivial group.<|endoftext|> -TITLE: Topological Hochschild homology and Hochschild homology of dg algebras -QUESTION [8 upvotes]: Topological Hochschild homology is a generalization of Hochschild homology from rings to $E_\infty$-ring spectra. On the other hand, there is a natural way to extend the notion of Hochschild homology to dg rings (either by explicitly writing down the bar complex, which is now a double complex, or by defining it as a derived tensor product and using semi-free resolutions). -On the other hand, if $k$ is a $\mathbb{Q}$-algebra, then $E_\infty$-ring spectra over $Hk$ and dg $k$-algebras are equivalent. Under this equivalence, does topological Hochschild homology coincide with dg-Hochschild homology? Is this written down anywhere? - -REPLY [4 votes]: The notion of Hochschild homology can be defined abstractly in any suitable homotopical context in which a tensor product exist (say, in any presentably symmetric monoidal $\infty$-category). Given an associative algebra object $A$ in such a context, its Hochschild homology is the (suitably defined) tensor product of $A$ with itself over $A^{\rm op} \otimes A$. For example, dg-Hochschild homology is the one obtained by working in the $\infty$-category of chain-complexes (and their tensor product), while topological Hochschild homology is the one associated to spectra (and their smash product). Sometimes one comes across objects which might a-priori be interpreted as belonging to two different $\infty$-categories. For example, if we have an ordinary ring, then we may think of it as an object in either chain-complexes or spectra (via the associated Eilenberg-MacLane spectrum), and consequently define both its Hochschild homology and its topological Hochschild homology, which may be different. -When $A$ is a $\mathbb{Q}$-algebra its associated Eilenberg-MacLane spectrum is rational (i.e., the map $HA \to H\mathbb{Q} \wedge HA$ is an equivalence). The $\infty$-category of rational spectra is equivalent to the $\infty$-category of chain-complexes over $\mathbb{Q}$. Moreover, this equivalence is symmetric monoidal and identifies the smash product on the spectra side with the tensor product on the chain-complex side. As a result, the topological Hochschild homology of such an Eilenberg-MacLane spectrum (defined using smash product of spectra) will coincide with its Hochschild homology when considered as a dg-algebra over $\mathbb{Q}$. You can also decide that you add an action of some $\mathbb{Q}$-algebra $k$ on both sides, but this will not matter much: as soon as the two interpretations yield $\infty$-categories which are symmetrically monoidal equivalent, they will produce the same Hochschild homology, essentially by definition.<|endoftext|> -TITLE: "sinc-ing" integral -QUESTION [7 upvotes]: Let $a_1,\dots,a_n, b$ be positive real numbers. - -*Question.** Is this true? - $$\int_{-\infty}^{\infty}\frac{\sin(bx+a_1x+\cdots+a_nx)}{x}\prod_{j=1}^n\frac{\sin(a_jx)}{a_jx}\,\,dx=\pi.$$ - My most immediate quest is "why is it independent of $b$, in particular?" - -REPLY [11 votes]: A more general result is due to C. Störmer (Acta Mathematica December 1895, Volume 19, Issue 1, pp 341–350)<|endoftext|> -TITLE: Almost complex structures on a 4-ball that are not tamed -QUESTION [5 upvotes]: Recall that an almost complex structure $J$ on a manifold $M^{2n}$ is called tamed if there exists a symplectic form $\omega$ on $M^{2n}$ such that $\omega(v,Jv)>0$ for any non-zero tangent vector $v$. -Question. Is there an almost complex structure $J$ on a closed ball $B^4$, such that any $C^{\infty}$-small perturbation of $J$ is not tamed by any symplectic form? (I assume that $J$ behaves nicely on the boundary of $B^4$, in particular it is smooth there). -Remark. Clearly, if such $J$ exists on $B^4$, it would exists on any $4$-manifold admitting an almost complex structure. -This question is a follow-up to the following one, where a global obstruction for "tamebility" was found for $\mathbb CP^2$ -Almost complex structures on $\mathbb CP^2$ that are not tamed - -REPLY [6 votes]: The following construction provides plenty of examples of non-tamed almost complex structures: -Consider an almost complex structure $J$ on $B^4$ for which the contact hyperplanes of the overtwisted contact structure on $S^3=\partial B^4$ (in the same homotopy class as the standard tight contact structure) become $J$-complex. If there was a taming symplectic form, we would have constructed a so-called weak symplectic filling of an overtwisted contact manifold. However, This is not possible by a result of Eliashberg and Gromov. -Finally, observe that the contact condition is open, and by Gray's stability also a perturbation of $J$ would be tangent to $\partial B^4$ along some tangent hyperplane distribution being an overtwisted contact structure.<|endoftext|> -TITLE: Kirby's torus trick -QUESTION [27 upvotes]: My basic question is: What is Kirby's torus trick and why did it solve so many problems? -I can get a glimmer of it from looking at Kirby's original paper, "Stable Homeomorphisms and the Annulus Conjecture," and its mathscinet review. However, it is a little bit unclear to me what exactly the torus trick was. It seems that there are two important ideas: the first is that by lifting along ever higher covers one may make obstructions to surgery vanish, and the second is that one may pull back differential structures to tori by immersing them in $\mathbb{R}^n$ and forming diagrams like: -$$ -\require{AMScd} -\begin{CD} -{T^n-D^n}@>{\mathrm{id}}>> \widetilde{T^n-D^n}\\ -@VVV @VVV \\ -\mathbb{R}^n @>{g}>> \mathbb{R}^n -\end{CD} -$$ -I looked for a reference that would summarize the situation but I was unable to find one. If there some paper that gives a summary of the trick as well as the context to understand it, I would be grateful to know it. I was hoping to find a succinct summary on Wikipedia but the torus trick link on Kirby's page is sadly red. If someone could give an example of the sort of problem that the torus trick is good at solving that would also be great. - -REPLY [2 votes]: I recently read an interview with Sergei Novikov where he claimed that he introduced "toric neighborhoods" in his proof of the topological invariance of Pontrjagin classes. He draws a comparison to Grothendieck's invention of "generalized open covers" (now known as Grothendieck topologies), and remarks that the same kind of construction was used by Kirby et al. in later developments.<|endoftext|> -TITLE: multiplicity one for restriction of representations from $GU$ to unitary group -QUESTION [6 upvotes]: Let $E/F$ be a quadratic field extension of p-adic fields. Let $V$ be a (skew-)Hermitian space and $U(V)$ be the unitary group. Let $GU(V)$ be the similitude unitary group. Given an irreducible smooth representation $\pi$ of $GU(V)$, do we know that the restriction $\pi|_{U(V)}$ has multiplicity one? -For the pair $(GL(n),SL(n))$ similar results are proved by Tadic. For the pair $(GSp, Sp)$, similar results are proved by Adler-Prasad. I am wondering if the unitary group version is true or not. -Thanks. - -REPLY [2 votes]: In case you didn't already see it, this question has now been answered in the affirmative by Adler and Prasad in Theorem 12a here: $[$1$]$ -$[$1$]$ Jeffrey D. Adler, Dipendra Prasad. Multiplicity upon restriction to the derived subgroup, 2018. (arXiv link)<|endoftext|> -TITLE: Furstenberg decomposition for non-compact spaces -QUESTION [5 upvotes]: Given a topological group $G$, a $G$-space is a topological space $X$ equipped with an action of $G$, such that the map $(g,x) \mapsto g.x$ is continuous. The action is distal if no non-diagonal orbit of $G$ on $X \times X$ has an accumulation point on the diagonal, and minimal if every orbit is dense. -A theorem of Furstenberg (The Structure of Distal Flows, 1963) says that if $G$ is a locally compact group and $X$ is a compact metrizable $G$-space such that the action of $G$ is minimal and distal, then $X$ is built out of isometric extensions: that is, there is a (potentially transfinite) sequence $(X_{\beta})_{\beta \le \alpha}$ of compact $G$-spaces and surjective morphisms $X_{\beta+1} \rightarrow X_{\beta}$ so that $X_0$ is the one-point space, $X_{\alpha} = X$, $X_{\beta}$ is the inverse limit of $(X_{\gamma})_{\gamma < \beta}$ when $\beta$ is a limit ordinal, and for each successor ordinal $\beta+1$ there is a $G$-invariant real-valued function on $X_{\beta+1} \times X_{\beta+1}$ that restricts to a compatible metric on each fibre of $X_{\beta+1} \rightarrow X_{\beta}$. (Conversely, any $G$-space built out isometric extensions in this way is distal; this direction is much easier and doesn't need any special hypotheses.) -Subsequent authors have generalized this result by removing various hypotheses and obtaining a similar conclusion about the decomposition of distal minimal $G$-spaces; sometimes the $G$-invariant metrics need to be replaced with some version of equicontinuity, but this is just an inevitable consequence of allowing $G$-spaces that are not first-countable. One constant though seems to be that $X$ needs to be compact. Are there any similar results if $X$ is instead some 'nice' topological space, e.g. a Polish space or a locally compact Hausdorff space, but not necessarily compact? Alternatively, are there some instructive examples of distal minimal $G$-spaces that don't break into equicontinuous pieces? -I am mainly interested in the case when $X = G/K$ for some closed subgroup $K$ of a locally compact or Polish group $G$, if that makes life easier (so in particular, the action is transitive). - -REPLY [4 votes]: Here is a counter example: -take $G=\text{SL}_2(\mathbb{R})$ and $X=\mathbb{R}^2-\{0,0\}$. -It is easy to check that the action is distal, and that there is no non-trivial factor carrying an invariant metric (the only proper factor is $\mathbb{P}^1(\mathbb{R})$). - -More generally (I think), taking any simple Lie group $G$ and $X=G/U$ for a unipotent subgroup $U$, you will get a distal action with no isometric factors. -This follows from the fact that $U$ has closed orbits on $G/U$ and Theorem 6.2 in https://arxiv.org/pdf/1408.4217.pdf. - -However, the following answers positively the case $X=G/K$ where $G$ is locally compact and $K -TITLE: Computationally challenging integer sequences -QUESTION [9 upvotes]: I wonder what are the examples of integer sequences, where only few elements are known and the researchers are still actively looking for the new terms. I think this discussion might be a good reference for those who would like to contribute to mathematical community by doing large scale calculations, extending the numerical data and providing computational evidence towards known conjectures. -The examples should include only those sequences where it is potentially feasible to compute one of the unknown terms if one has an optimized algorithm and enough computing power. In other words, the problem should be more computational rather than theoretical. -Perhaps, the most famous example are Mersenne primes (A001348), where only the first 45 consecutive terms are known (even though 49 Mersenne primes are known): -$3, 7, 31, 127, 2047, 8191, 131071, 524287, 8388607, 536870911, 2147483647, \ldots$ -The members of GIMPS (Great Internet Mersenne Prime Search) are actively searching for the new terms, and according to the website are currently trying to prove that $M_{37156667}$ is the 46th Mersenne prime. -Another interesting example is the number of arrangements of $n$ circes in the affine plane (A250001), where only the first 5 terms are known: -$1, 1, 3, 14, 173.$ -All of these elements got computed by Jon Wild, and on the OEIS webpage for this sequence he mentioned that the 5th element (the sequence starts with the 0th term) is probably equal to 16942. -I'd be interested to see other curious examples. - -REPLY [4 votes]: The Hales-Jewett numbers $c_{n,k}$ are defined, essentially, to be the largest possible size of a subset of $\mathbb Z_k^n$ free of $k$-term arithmetic progressions with the difference in $\{0,1\}^n$ and satisfying the additional restriction that the support of the initial term of the progression is disjoint from the support of its difference. A significant part of the Polymath project Density Hales-Jewett and Moser Numbers consists in determining the numbers $c_{n,3}$. We know the first seven Hales-Jewett numbers (with $k=3$): - $$ c_{0,3}=1,\ c_{1,3}=2,\ c_{2,3}=6,\ c_{3,3}=18,\ c_{4,3}=52,\ c_{5,3}=150,\ c_{6,3}=450; $$ -see Theorem 1.4 of the aforementioned project, or OEIS sequence A156989. -What if we drop the support disjointness restriction and define, say, $\mu_n$ to be the largest possible size of a subset of $\mathbb F_3^n$ free of three-term arithmetic progressions with the difference in $\{0,1\}^n$? It is not particularly difficult to see that - $$ \mu_0=1,\ \mu_1=2,\ \mu_2=6,\ \mu_3=14,\ \text{and}\ 36\le\mu_4\le 40. $$ -Virtually nothing is known about the order of growth of the numbers $\mu_n$, and to develop some intuition, it would be extremely helpful to compute several more of them. - -Added March 13, 2017 -Robert Israel reports $\mu_4=36$ and $\mu_5\ge 102$, using cplex. Moreover, computations seem to suggest that, indeed, $\mu_5=102$ may hold true.<|endoftext|> -TITLE: "Economic" Eilenberg-MacLane topological abelian groups -QUESTION [8 upvotes]: This might be regarded as a sequel to my previous "Economic" CW-structure for Eilenberg-MacLane spaces? However the content seems to be quite different. -I believe it is easy to prove that for any topological abelian group $A$ there is a model of $BA$ (that is, a space $X$ with $\Omega X$ homotopy equivalent to $A$) which is a topological abelian group; in particular, for any (discrete) abelian group $\pi$ and any $n\ge0$ there is a model for $K(\pi,n)$ which is a topological abelian group. -This latter can be actually quite easily seen using the Dold-Kan correspondence: take the chain complex with $\pi$ in degree $n$ and zero everywhere else; take the corresponding simplicial abelian group; take its geometric realization. Since the latter preserves products, it will be a topological abelian group. -However I somehow feel that this is not the most economic way to do it; in any case I do not quite "see" the result. In fact the only case when I know a satisfactory description (for me) is $K(\mathbb Z/2\mathbb Z,1)$ built from the infinite-dimensional sphere. It follows e. g. from the wonderful paper "Using the generic interval" by Gavin Wraith that $S^\infty$ has a topological $\mathbb F_2$-vector space structure. Since it is contractible, its quotient by any two element subgroup is a $K(\mathbb Z/2\mathbb Z,1)$ which is a topological elementary abelian 2-group. -I don't know if $S^\infty$ has any other topological abelian group structure which would turn the infinite lens spaces into subgroup quotients so as to obtain a $K(\mathbb Z/p\mathbb Z,1)$ topological abelian group. Is something like this known? -Infinite symmetric power of $S^n$ is a $K(\mathbb Z,n)$ (related discussion here on MO is in Why the Dold-Thom theorem?); it is a commutative topological monoid. Can it be made into a group? Is its group of quotients again a $K(\mathbb Z,n)$? -Also here on MO there is the question about H-space structure on infinite projective spaces where something similar is done but honestly speaking I could not understand from answers there whether there is a topological abelian group structure on, say, $\mathbb C\mathrm P^\infty$ or not. There, they also mention the page by John Baez about Classifying Spaces Made Easy but again, I could not quite find answers to my questions there. -Other than that, I've seen Stephan Stolz using the projective unitary group of an infinite-dimensional separable Hilbert space as a model for $K(\mathbb Z,2)$ (in "A conjecture concerning positive Ricci curvature and the Witten genus", Math. Ann. 304(1) 1996, page 795) but that is very-very nonabelian. -Not counting the circle as $K(\mathbb Z,1)$, and lots of nice $K(\pi,1)$s for (mostly) nonabelian $\pi$, these are basically the only ones that I know. -Is there any systematic construction of $K(\pi,n)$ topological abelian groups which would be minimal in the sense that they do not have contractible [closed] subgroups or contractible [continuous] quotients? Are there any restrictions on the resulting groups? For example, what kind of torsion they should have if $\pi$ is cyclic? Obviously the case when $\pi$ is finitely generated reduces to the investigation of cyclic $\pi$'s. What about, say, $\mathbb Q$? Are there any nice topological abelian $K(\mathbb Q,n)$ groups known? - -REPLY [2 votes]: For a discrete abelian group $\pi$ the following simplicial abelian group models for $K(\pi,1)$ are isomorphic, I believe: -Simplicial abelian group whose normalized chain complex is $0\leftarrow \pi\leftarrow 0\leftarrow 0\leftarrow \dots$. -Nerve of category having one object and having morphisms $\pi$ (with obvious addition law since $\pi$ is abelian). -Universal example of simplicial abelian group $A$ equipped with a based map of simplicial sets from $S^1=\Delta^1/\partial$ to the simplicial set $Hom(\pi,A)$. -When realized as a topological abelian group, this becomes the universal example of a topological abelian group $A$ equipped with a continuous based map from the topological space $S^1$ to the space of homomorphisms from $\pi$ to $A$. -This latter space can also be described using configurations as in the previous answer. -In each of these constructions of $B\pi$, the universal cover $E\pi$ has its own corresponding description. -I believe that when $\pi$ has order two then $E\pi$ is homeomorphic to $S^\infty$, but I don't know if this abelian group structure on $S^\infty$ is the one referred to in the question. -$E\pi$ as a space, or even as a simplicial set, does not depend on the group structure in $\pi$; if you have a set $\pi$ then you can make $E\pi$ using the nerve of the category that has one object for each element of $\pi$ and one morphism between any two objects. If $\pi$ is nonempty then this is contractible, and if $\pi$ is a group then it gets an obvious free group action. It seems likely that if it is homeomorphic to $S^\infty$ when $\pi$ has two elements then the same is true when it has more than two (but finitely many) elements. -These don't seem to have the kind of strong minimality property that you are asking about: when $\pi$ has order two then there are plenty of contractible closed subgroups of $B\pi$ (for which the quotient is isomorphic to $B\pi$). But maybe that property is too much to wish for.<|endoftext|> -TITLE: Lie groups and differential equations -QUESTION [6 upvotes]: I'm trying to understand Lie groups method for solving differential equations and I have some basic questions. -For instance, Let $\Sigma = 0$ be a differential equation of order $n$ and let $X^{(n)}$ be the n-th prolongation of the vector field $X$. Then $e^{\epsilon X}$ is a symmetry if $X^{(n)} \Sigma = 0$ whenever $\Sigma = 0$. These two simultaneous problems gives rise to determining equations which are usually solved by means of a polynomial ansatz. -So here are my questions: - -How can one know the degree of the polynomial one should use? -Does one always find more symmetry generators by increasing the degree of the polynomial? Or maybe there is a maximum bound? -All symmetry generators found in this way necessarily close a Lie algebra? - -Thanks very much in advance for any insights! - -REPLY [3 votes]: All of the questions are answered immediately by looking at the symmetries of the Cauchy--Riemann equations in complex analysis of one complex variable, or the Laplace equation on the 3-sphere. - -You can't know in advance what degrees to try: all complex polynomial vector fields prolong to preserve the Cauchy--Riemann equations. -You do not always get more symmetry generators by increasing the degree. Typically, most differential equations one encounters have only a finite dimensional symmetry Lie algebra. The Laplace equation on the 3-sphere has symmetry group the conformal group, a finite dimensional Lie group, and the symmetry vector fields are polynomial in the standard Ptolemaic projection coordinates. -The Lie bracket of any two symmetry vector fields is also a symmetry vector field, but might not be one you have already encountered, because it might be of a higher degree of polynomial than you have checked for. For example, with the Cauchy--Riemann equations, two generic complex polynomial vector fields in one complex variable, of degree 3 or more have higher degree bracket. Of course, there is no guarantee that the symmetries must be polynomial, or even that there must be a choice of coordinates in which all symmetry vector fields are polynomial; but Lie bracket of polynomial vector fields yields a polynomial vector field. If the symmetry Lie algebra (the Lie algebra of the symmetry pseudogroup, not necessarily of the global symmetry group) is infinite dimensional (as, for example, with the Cauchy--Riemann equations on a complex manifold), then you will never close up to find a Lie algebra, unless you happen by accident to have all of your symmetry vector fields land inside some finite dimensional subalgebra. For example, you might find the subalgebra of polynomial vector fields of degree at most 2 inside the Lie algebra of holomorphic vector fields, and this might incorrectly lead you to believe that you have found all of the symmetries.<|endoftext|> -TITLE: Number theoretic interpretation of the integral $\int_{-\infty}^\infty\frac{dx}{\left(e^x+e^{-x}+e^{ix\sqrt{3}}\right)^2}=\frac{1}{3}$? -QUESTION [22 upvotes]: Is there any explanation based on algebraic number theory that the integral -$$ -\int_{-\infty}^\infty\frac{dx}{\left(e^x+e^{-x}+e^{ix\sqrt{3}}\right)^2}=\frac{1}{3}\tag{1} -$$ -has a closed form? Analytic proof of this integral is given in this MSE post, however this proof does not explain why a similar looking integral -$$ -\int_{-\infty}^\infty\frac{e^{ix\sqrt{3}}\ dx}{\left(e^x+e^{-x}+e^{ix\sqrt{3}}\right)^3}=\frac{\sqrt{3}}{8\pi}\int_{0}^\infty\frac{dx}{\left(1+\frac{x^3}{1^3}\right)\left(1+\frac{x^3}{2^3}\right)\left(1+\frac{x^3}{3^3}\right)\ldots} -$$ -probably does not have a closed form. Is it possible that $(1)$ is related to Eisenstein integers? -Alternative formulation of the integral $(1)$ is -$$ -\int_0^\infty\frac{dt}{(1+t+t^{\,\alpha})^2}=\frac23, \quad \alpha=\frac{1+i\sqrt3}{2}. \tag{1a} -$$ - -REPLY [7 votes]: The following formula gives a parametric extension of $(1)$ for $|a|$ sufficiently small -\begin{align} -\small\int_{-\infty}^\infty\frac{dx}{\left(e^x+e^{-x}+e^{a+ix\sqrt{3}}\right)^2}+e^a\int_{-\infty}^\infty\frac{dx}{\left(e^{a+x}+e^{-x}+e^{ix\sqrt{3}}\right)^2}+e^a\int_{-\infty}^\infty\frac{dx}{\left(e^{a+x}+e^{-x}+e^{-ix\sqrt{3}}\right)^2}=1 -\end{align} -This means that $(1)$ is not an isolated result. -In view of this one might be very sceptical that any number theoretic interpretation of the integral exists.<|endoftext|> -TITLE: Solutions of $x^d=1$ in the symmetric group -QUESTION [9 upvotes]: L Moser and M Wyman, On solutions of $x^d = 1$ in symmetric groups, Canad. J. Math., 7 (1955), pages 159-168, explored asymptotic behavior of the cardinality of such permutations: -$$f_d(n):=\#\{\pi\in\mathfrak{S}_n:\, \pi^d=1\}.$$ -In particular, $f_2(n)$ counts the number of involutions in the symmetric group $\mathfrak{S}_n$. - -Question 1. I recall (but forgot where) $f_d(n)$ enumerate pseudoknots in RNA molecular folding. Can you explain this or point me to a readable reference? - -Also, - -Question 2. Does this generating function hold true? - $$\sum_{n\geq0}f_d(n)\frac{x^n}{n!}=e^{\sum_{c\,\vert\,d}\frac1cx^c}.$$ - -REPLY [4 votes]: For Q1, there is a quite nicely written 2014 preprint by Cheda and Gupta.<|endoftext|> -TITLE: nonabelian reciprocity law -QUESTION [33 upvotes]: I heard the following relation in a talk by Peter Scholze. Could someone explain "in a simple way" what is the precise relation between the polynomial $x^4-7x^2-3x+1 $ and the integral homology of the orbit space $\mathbb{H}^3/\Gamma$ where: - -$\mathbb{H}^3$ is the hyperbolic 3-space. -$\Gamma\subset SL_{2}(\mathbb{Z}[i])$ such that for any $(\gamma_{r,s})_{r,s\in\{1,2\}}=\gamma\in \Gamma$ we have $\gamma_{1,2}\equiv 0 \textrm{ mod } 183$ and $\gamma_{2,2}\equiv 1 \textrm{ mod } 3$ - -REPLY [27 votes]: Will Sawin's answer is perfectly correct, but I wanted to add some further perspective. -What Peter Scholze has proven is a profound generalization of (one half) of Class Field Theory. The particular example you state is simply a special case of a much more general result. -To set the stage, let me recall some of the set-up from class field theory. Let $F$ be a number field. For simplicity of exposition, suppose that the class group of $F$ is equal to one. Let $\mathfrak{q}$ be a prime of $F$, and suppose that $\mathfrak{q}$ has norm $q$. Now ask the question: does $F$ admit an abelian extension of degree $p$ which is only ramified at the prime $\mathfrak{q}$? -Class Field Theory provides an answer to this question. The answer is that one should consider a certain quotient of the Idele Class group $F^{\times} \backslash \mathbf{A}^{\times}_F$ depending on $\mathfrak{q}$, namely, the quotient -$$X_{\mathfrak{q}}:=F^{\times} \backslash \mathbf{A}^{\times}/U_{\mathfrak{q}},$$ -where $U = \prod U_v$ is a compact open of the finite adeles such that $U_v = \mathcal{O}^{\times}_{F,v}$ -for all primes $v$ except for $\mathfrak{q}$, and for $v = \mathfrak{q}$, $U_v$ is subgroup of $\mathcal{O}^{\times}_{F,\mathfrak{q}}$ consisting of elements congruent to $1 \mod \mathfrak{q}$. This quotient space is, geometrically, a disconnected union of a product of circles. What does class field theory say? It says that the zeroth homology group: -$$H_0(X_{\mathfrak{q}},\mathbf{Z})$$ -completely determines (and is determined by) the abelian extensions of $F$ which are unramified everywhere away from $\mathfrak{q}$ and have degree prime to $q$. In particular, we have a geometric object whose homology relates directly to Galois Representations. Now it turns out that one can give a quite precise description of the connected components of $X_{\mathfrak{q}}$ in terms of the quotient of $(\mathcal{O}_F/\mathfrak{q})^{\times}$ by the global units, which leads to an explicit computational way to determine the existence of abelian extensions. Here's an explicit example. -Let $F = \mathbf{Q}(\sqrt{2})$, and $\epsilon = \sqrt{2} - 1$. Let $\mathfrak{q}$ be a prime of $\mathcal{O}_F$ whose norm is prime and congruent to $1 \mod 3$. Then there exists an abelian extension of $F$ of degree $3$ unramified outside $\mathfrak{q}$ if and only if $\epsilon$ is a cube modulo $\mathfrak{q}$. -OK, now let's turn to $2$-dimensional representations. Again, let's fix a field $F$ and a prime $\mathfrak{q}$. This time, consider the space -$$X_{\mathfrak{q}} = \mathrm{GL}_2(F) \backslash \mathrm{GL}_2(\mathbf{A})/U_{\mathfrak{q}},$$ -where now $U_{\mathfrak{q}}$ is a compact open subgroup whose level reflects the choice of prime $\mathfrak{q}$. It turns out that the homology of this space is related to the existence of two dimensional Galois representations. It's usually traditional (convenient) to also take the quotient by a further compact subgroup $K$ of $\mathrm{GL}_2(\mathbf{R})$ as well, and also the $\mathbf{Q}$-split part of the center. -So what kind of Galois representations do these homology classes relate to, at least conjecturally? Well, one has to be quite careful in specifying the precise degrees, which depend somewhat delicately on the signature of the field $F$. But let's consider the case when $F = \mathbf{Q}$, in which case the objects above are none other than the classical modular curves. In this case, the cohomology of modular curves corresponds, via the Eichler-Shimura isomorphism, to classical modular forms of weight two. And then, by the combined efforts of many people (Shimura, Deligne, Langlands, etc), we expect that these then give rise (taking into account the Hecke operators) to Galois representations: -$$\rho:\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \rightarrow \mathrm{GL}_2(\overline{\mathbf{Q}}_p).$$ -What is surprising at first is that these Galois representaitons have infinite image, and determining (even conjecturally) precisely what the image should be requires the entire theory of integral p-adic Hodge Theory, local-Langlands, and so on. The actual construction of these Galois representations crucially uses the fact that the objects $X_{\mathfrak{q}}$ we have constructed above are actually algebraic varieties over number fields (in this case curves), so we can use the powerful methods of etale cohomology developed by Grothendieck and his school to construct Galois representations. This is usually viewed as one half of the problem of "Reciprocity" in the Langlands program: going from automorphic representations (in this case in their avatar as cohomology classes, which requires the representations to be of a special kind) to Galois Representations. One should view this as the analog in class field theory of constructing a non-trivial abelian extension (say the Hilbert class field) starting with the class group. Naturally, there is another half of the problem which is the converse, and here the profound work of Wiles is relevant. -To get back to Scholze's talk, however, requires a slight detour. Already for the case $F = \mathbf{Q}$ above, one could consider the cohomology not with coefficients in characteristic zero, but in characteristic $p$. In this case, there is a generalization of Reciprocity which associates (in a precise bijective way) such eigenclasses under Hecke operators to representations: -$$\rho:\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \rightarrow \mathrm{GL}_2(\overline{\mathbf{F}}_p).$$ -(again with some technical conditions on what representations can occur). This amounts to a famous conjecture of Serre, proved spectacularly by Khare and Wintenberger, surely one of the great theorems in the subject since Wiles. In this case, the arguments crucially use the following observation: If one takes the mod-p cohomology of an algebraic curve (or orientable topological surface), then such classes always lift to characteristic zero. By Serre's conjecture, this implies a correpsonding statement on the Galois side which is incredibly profound. However, at least one has tools to study this quesiton, since there is a direct link between mod-p torsion classes in $H_1$ and classes in characteristic zero. -Now let us take $F$ to be an imaginary quadratic field, say $\mathbf{Q}(i)$. In this case, the corresponding $X_{\mathfrak{q}}$ turn out to be a disjoint union of spaces of the shape -$$\Gamma \backslash \mathbf{H}^3,$$ -where $\mathbf{H}^3$ is hyperbolic three space, and $\Gamma$ is a congruence subgroup of $\mathrm{GL}_2(\mathbf{Z}[i])$. This means (since the cover is contratible) that the homology groups of this space are just the homology groups of $\Gamma$. Once again, Langlands reciprocity predicts, from any cohomology class which is a Hecke eigenform, a corresponding Galois representation: -$$\rho:\mathrm{Gal}(\overline{\mathbf{Q}}/F) \rightarrow \mathrm{GL}_2(\overline{\mathbf{Q}}_p)$$ -with particular properties. Now it's much harder to see why this might be true, since the hyperbolic three manifold no longer has any association with algebraic geometry. One method is to use the fact that characteristic zero classes in homology can often be computed by de Rham cohomology, and hence by harmonic differentials, and hence by automorphic forms. Then one might hope to move the automorphic forms around between different groups (using Langlands base change or variations thereof), and move to a situation where one has access to geometry. This was done (in this case) by Harris-Soudry-Taylor, who found the Galois representations. -The REAL issue, however, is when one now considers mod-p cohomology classes. There is no reason why (and it is often false) a mod-p cohomology class of $\Gamma$ (or on the corrresponding arithmetic three manifold) should lift to characteristic zero. Hence this is not directly related to Langlands reciprocity. That is we HAVE GONE BEYOND THE LANGLANDS PROGRAM AS WE KNOW IT. However, an amazing generalization of this reciprocity (due to Ash and others) says that the something analogous happens in this case too, namely, given a mod-p torsion eigenclass in cohomology, one obtains a mod-p Galois representation. This is an outrageous conjecture, which was completely open until Peter Scholze stunningly proved it (in many cases) in 2013. -It's interesting to ask about the precise relationship between cohomology and Galois representations, which was your (great!) question. Will Sawin has explained what is going on in this case. But it is worth noting that the recipe (going from eigenalues of Hecke operators to the trace of Frobenius elements) is a completely general feature of these mod-p Langlands reciprocity conjectures. To recall the precise statment: -$$\mathrm{Trace}(\rho(\mathrm{Frob}_{v})) = a_v \mod p,$$ -where $a_v$ is the eigenvalue of $T_v$. In this case, the prime $p$ si equal to three, and so representation of interest has the shape: -$$\mathrm{Gal}(\overline{\mathbf{Q}}/F) \rightarrow \mathrm{GL}_2(\overline{\mathbf{F}}_3).$$ -The composite of this map to $\mathrm{PGL}_2(\overline{\mathbf{F}}_3)$ turns out the land in $\mathrm{PGL}_2(\mathbf{F}_3) = S_4$ and turns out to have image $A_4$, and moreover is the natural representation of the Galois group of the given polynomial. -Let's now use this to understand solutions to the polynomial $x^4−7x^2−3x+1$ modulo $q$, for a prime $q = N(\mathfrak{q})$ of the Gaussian integers. Certainly, since the polynomial has rational coefficients, it has a solution modulo $q$ if and only if it has a solution modulo $\mathfrak{q}$, since $\mathbf{Z}/q = \mathbf{Z}[i]/\mathfrak{q}$. It turns out that one can tell if any polynomial of degree $n$ has a root modulo $q$ in terms of the corresponding Frobenius element considered as a conjugacy class inside $S_n$, where the Galois group acts on the roots in the obvious way. In this case, the Galois group is $A_4$ acting on the roots in the obvious way to identify the group with $A_4$. The group $A_4$ only has four conjugacy classes. In general, a polynomial doesn't have a root if and only if the corresponding element of $S_n$ has no fixed points. In $A_4$, this means it is conjugate to $(12)(34)$. There are four roots if Frobenius is trivial, and one root of Frobenius is conjugate to $(123)$ or its inverse. In the first, second, and third cases, the corresponding elements of $A_4 = \mathrm{SL}_2(\mathbf{F}_3)$ are -$$\left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right), -\quad -\left(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right), -\quad -\left(\begin{matrix} 0 & 1 \\ -1 & 1 \end{matrix} \right).$$ -Any lift of the first matrix to $\mathrm{GL}$ has trace zero, and any lift of the other two matrices do not, thus the trace of $\rho(\mathrm{Frob}_v)$ will be zero modulo three exactly when the quartic has a rational solution modulo $q$. But now, the magic happens, and Scholze's theorem associates these traces to the Hecke eigenvalues, and thus, remarkably, one can deduce that the Hecke operator $T_{\mathfrak{q}}$ on this eigenclass is non-trivial modulo $3$ exactly when the quartic has a solution modulo $q$. -Of course, one should really think of this example not in isolation but as part of the vast web of interlacing conjectures in the modern Langlands program. Still, it's nice to have explicit examples! -(There are other interesting features of this example, namely that the $A_4$ extension comes from the restriction of an even representation of $\mathbf{Q}$, but we do not discuss those here.)<|endoftext|> -TITLE: What is the limit set of a hyperbolic lattice? -QUESTION [5 upvotes]: My claim is as follows: - -Let $\Gamma$ be a discrete subgroup of $\operatorname{Isom}(\Bbb{H}^{n})$, the isometries of hyperbolic $n$-space. If $\Gamma$ is a lattice in $\operatorname{Isom}(\Bbb{H}^n)$ then the limit set of $\Gamma$ is $\partial \Bbb{H}^{n}$. - -I'm inclined to believe that the statement above is true purely on intuition gleaned from $\operatorname{PSL}(2,\Bbb{Z})$ in $\operatorname{PSL}(2,\Bbb{R})$ and that it (or probably a discussion of a more general statement) should be in the literature somewhere; however, I have yet to find it through Google or in my go-to print resources (Ratcliffe, Bearon, and Margulis). -Is anyone able to provide insight into the validity of this statement and/or point to potential resources? It would be greatly appreciated. Thank you. - -REPLY [4 votes]: This question is probably not really research level given that the answer exists in many textbooks. Nevertheless, it may be useful to have different viewpoints. -An elementary geometric argument follows from Chapter 8 of Thurston's notes. In fact, the argument shows that if $\mathbb{H}^n/\Gamma$ does not contain a half-space, the limit set $L_\Gamma=S^{n-1}_\infty$. -Thurston notes in Prop. 8.2.3 that the action of $\Gamma$ on $D_\Gamma=S_\infty-L_\Gamma$ (which is an open subset) is properly discontinuous. The proof is by taking the convex hull of $L_\Gamma$, and noting that there is a retract from $\mathbb{H}^n\cup S^{n-1}_\infty$ to this set, which send $D_\Gamma$ to the boundary of the convex hull. But the action of $\Gamma$ on the boundary of the convex hull is discrete (being in the interior of $\mathbb{H}^n$), hence the action on $D_\Gamma$ is. -Now observe that there is a ball in $D_\Gamma$ which is mapped disjointly from itself by all elements of $\Gamma$, from the proper discontinuity of the action (this requires a bit of argument, but follows from the existence of a free orbit of the action of $\Gamma$ on $\mathbb{H}^n$ and hence on $D_\Gamma$). Hence there is an embedded half-space in $\mathbb{H}^n/\Gamma$, so it is infinite volume.<|endoftext|> -TITLE: Existence of Minkowski units in totally real cyclic fields of prime degree -QUESTION [6 upvotes]: Does anyone know of any papers which give results on the existence of Minkowski units in totally real cyclic fields of prime degree? F. Marko has some results for composite degree in his paper "On the existence of Minkowski units in totally real cyclic fields" but I'm more interested in the prime degree case. See https://www.emis.de/journals/JTNB/2005-1/pages195-206.pdf - -REPLY [3 votes]: See this short note by Brumer and the book Elementary and Analytic Theory of Algebraic Numbers by Narkiewicz, in particular Section 3.3.<|endoftext|> -TITLE: "Explicit" perturbations of Morse-Bott functions -QUESTION [7 upvotes]: There are explicit perturbations of Morse-Bott functions $f:X\to\mathbb{R}$ used in the literature (ex: Austin-Braam, Banyaga-Hurtubise, Bourgeois) to help solve various problems (ex: building Morse homology, and constructing nondegenerate contact forms). Namely, Morse functions $f_i$ are chosen on the critical submanifolds $C_i\subset crit(f)$ and then $0<\epsilon<<1$ is chosen so that $f+\epsilon\sum_i\rho_if_i$ is Morse (whose critical points have become $crit(f_i)$ on each $C_i$). Here $\rho_i$ is a bump function which has support on a tubular neighborhood of $C_i$ and takes value 1 in a smaller tubular neighborhood. -I am questioning how "rigid" this approach is, i.e. if I can replace the above description with something similar, to achieve a Morse function whose critical points live within the original $\operatorname{crit}(f)$. From the description, the $f_i$ are extended to be constant in the normal direction of $C_i\subset X$ (before being cut-off to zero). Is this "constancy condition" crucial? -Can I instead extend $f_i$ to a small tubular neighborhood in any fashion (there is at least one way by Whitney's extension theorem) and then cut it off to zero? This extension would have no requirement to be constant in the normal directions to $C_i$, and I can still choose $\epsilon$ arbitrarily small and have the critical manifolds $C_i$ be perturbed into $crit(f_i)$. In other words, the original approach pulls back $f_i$ to the normal bundle $N_{C_i}$ of $C_i$ so that locally the perturbation is $(c,x)\mapsto f_i(c)$, and I'm looking for other functions $(c,x)\mapsto \tilde{f}_i(c,x)$ satisfying $\tilde{f}_i(c,0)=f_i(c)$. -The more I think about this though, the more I understand why the original description is used (so that $\nabla f\perp\nabla f_i$ away from $C_i$, helping to ensure no new critical points appear). - -REPLY [5 votes]: Here is a local model for a Bott-style Morse function on $\mathbb R^n$: -$$f(x_1, \cdots, x_n) = a + x_1^2 + \cdots + x_k^2 - x_{k+1}^2 - \cdots - x_{k+j}^2$$ -The critical-point set is $x_1=x_2=\cdots = x_{k+j} = 0$, a linear subspace of dimension $n-k-j$. -Let $\beta$ be a bump-function in the variables $x_1, \cdots, x_{k+j}$ centred at the origin, then if I understand the suggestion you are making, the idea is to replace $f$ with -$$g(x_1, \cdots, x_n) = a + x_1^2 + \cdots + x_k^2 - x_{k+1}^2 - \cdots - x_{k+j}^2 + \beta(x_1,\cdots,x_{k+j})( \pm x_{k+j+1}^2 + \cdots + \pm x_n^2 )$$ -So the function is unchanged outside of a neighbourhood of the critical submanifold. The critical submanifold of this function is just a single point (the origin). The flow of gradient preserves the original critical-point set $x_1 = x_2 = \cdots = x_{k+j} = 0$. -I take it you want to understand the flowlines in the support of $\beta$. The way I've set things up, there's no magic. No flowlines stay in the support of $\beta$, (using both forward and backward time) other than the original critical-point set $x_1 = \cdots = x_{k+j} = 0$. -Does that help at all?<|endoftext|> -TITLE: Relative version of de Rham cohomology with local coefficients -QUESTION [8 upvotes]: Given a vector bundle $E \to M$ with connection $\nabla$, we get a twisted de Rham sequence using the exterior covariant derivative: -$$\mathcal{E} \xrightarrow{d^\nabla=\nabla} \Omega^1_M \otimes_{\Omega^0_M} \mathcal{E} \xrightarrow{d^\nabla} \Omega^2_M \otimes_{\Omega^0_M} \mathcal{E} \xrightarrow{d^\nabla} \cdots$$ -Here, I am using $\mathcal{E}$ to denote the sheaf of smooth sections of $E$, and $\Omega_M$ the sheaf of smooth differential forms on $M$. -We say that the connection $\nabla$ is flat if $(d^\nabla)^2 = 0$, and in this case we get an actual complex of sheaves. In this situation, the sheaf $\mathcal{L}$ of parallel sections is a local system and we can use this complex (a soft resolution) to compute its sheaf cohomology. -As described in Kashiwara and Schapira, or in Liviu Nicolaescu's answer here, given a closed submanifold $i: Z \hookrightarrow M$, and $j: M \setminus Z \hookrightarrow M$ the inclusion of its complement, we get a short exact sequence of sheaves -$$0 \to j_!j^{-1}\mathcal{L} \to \mathcal{L} \to i_*i^{-1}\mathcal{L} \to 0.$$ -The natural maps here are the counit $j_!j^{-1}\mathcal{L} \to \mathcal{L}$ of the $j_! \dashv j^{-1}$ adjunction, and the unit $\mathcal{L} \to i_*i^{-1}\mathcal{L}$ of the $i^{-1} \dashv i_*$ adjunction, respectively. -Note that the pullback bundle $i^* E \to Z$ inherits the flat connection from $E \to M$, and this gives rise to a complex of sheaves on $Z$ -$$\mathcal{i^*E} \xrightarrow{d^\nabla=\nabla} \Omega^1_Z \otimes_{\Omega^0_Z} i^*\mathcal{E} \xrightarrow{d^\nabla} \Omega^2_Z \otimes_{\Omega^0_Z} i^*\mathcal{E} \xrightarrow{d^\nabla} \cdots$$ -Note that $i^*= \Omega^0_Z \otimes_{i^{-1} \Omega^0_M} i^{-1}$, so it is not the same as the inverse image functor above. Is it correct to say that this complex gives a soft resolution of the sheaf $i^{-1} \mathcal{L}$? If so, how can I show that this is true? -This problem has come up in my research (in differential geometry) but my department doesn't really have anyone who works with sheaves. I would really appreciate some help with this! - -REPLY [2 votes]: I had a look in Ramanan's Global Calculus, and sure enough he gives the necessary details: -Let $\mathbb{R}_M$ denote the constant sheaf corresponding to $\mathbb{R}$ on $M$. -Note that $\Omega^k_M\otimes_{\mathbb{R}_M} \mathcal{L} = \Omega^k_M\otimes_{\Omega^0_M} \mathcal{E},$ and hence the connection is recovered from the exterior derivative by the formula: -$$\nabla = d \otimes_{\mathbb{R}_M} 1: \mathcal{E} \to \Omega^1_M \otimes_{\Omega^0_M} \mathcal{E}.$$ -So in particular, tensoring the de Rham resolution of $\mathbb{R}_M$ with $\mathcal{L}$ recovers the twisted de Rham resolution of $\mathcal{L}$. -With regards to the resolution of $i^{-1} \mathcal{L}$ on $Z$, note that we have the following canonical isomorphisms: -\begin{align} -\Omega^k_Z \otimes_{\Omega^0_Z} i^* \mathcal{E} &= \Omega^k_Z \otimes_{\Omega^0_Z} (\Omega^0_Z \otimes_{i^{-1}\Omega^0_M} i^{-1} \mathcal{E}) \\ -&\cong \Omega^k_Z \otimes_{i^{-1}\Omega^0_M} i^{-1} \mathcal{E} \\ -&\cong \Omega^k_Z \otimes_{i^{-1}\Omega^0_M} i^{-1} (\Omega^0_M\otimes_{\mathbb{R}_M} \mathcal{L}) \\ -&\cong \Omega^k_Z \otimes_{i^{-1}\Omega^0_M} i^{-1}\Omega^0_M \otimes_{\mathbb{R}_Z} i^{-1} \mathcal{L} \\ -&\cong\Omega^k_Z \otimes_{\mathbb{R}_Z} i^{-1} \mathcal{L}. -\end{align} -Hence, tensoring the de Rham resolution of $\mathbb{R}_Z$ with $i^{-1} \mathcal{L}$ gives the usual de Rham resolution of $i^{-1} \mathcal{L}$, and this computes the twisted de Rham cohomology on $Z$. -The same idea works to get the resolution of $j^{-1} \mathcal{L}$ on $M \setminus Z$. Then, applying the "extension by zero" functor $j_!$, we get a resolution for $j_!j^{-1} \mathcal{L}$, which computes the relative twisted de Rham cohomology.<|endoftext|> -TITLE: Galois theory for products of fields (aka finite etale extensions) -QUESTION [8 upvotes]: Let $F$ be a field. By a Galois algebra over $F$ I mean a finite etale extension, that is, a product $K = K_1 \times \cdots \times K_r$ of finite (separable) field extensions, of total degree $[K : F] = n$, equipped with a subgroup $G \subseteq \operatorname{Aut}_F K$ of $n$ linearly independent automorphisms. For example, $\mathbb{C} \times \mathbb{C}$ equipped with the set of four automorphisms $(z,w) \mapsto \{(z,w), (\bar{w},z), (\bar{z},\bar{w}), (w,\bar{z})\}$ is a cyclic Galois algebra over $\mathbb{R}$ of degree $4$. -I'd like to find a reference (or counterexample) to the following simple assertions: - -$K$ is a Galois algebra iff it is a product $K = K_1^r$ of copies of a field that is Galois over $F$, and the $G$-action permutes the $n$ ring maps from $K$ to $K_1$ simply transitively; -If $H \leq G$ is a subgroup, then $K/K^H$ is also Galois under the natural $H$-action (more precisely, for each field factor of $K^H$, the portion of $K$ above it is Galois under the $H$-action); -Also, if $H$ is normal, then $K^H/F$ is Galois under the natural $G/H$-action; -If $E$ is any extension field of $F$, then $L \otimes E$ is Galois over $E$, with the $G$-action extended linearly; -If $K$ and $K'$ are Galois algebras over $F$ with Galois groups $G$ and $G'$ respectively, then $K \otimes K'$ is Galois under the natural action of $G \times G'$. - -A construction due to Bhargava embeds any finite etale extension of degree $n$ (in characteristic $0$) into an $S_n$-Galois algebra. -It seems that the theory of such algebras is similar in depth to, but not a trivial consequence of, Galois theory of fields (e.g. for part 2, how do we compute $[K^H : F]$?) We no longer have that every subalgebra of $K$ is the fixed field of some subgroup; but on the other hand, the elegant properties 4 and 5 are missing from the classical presentation (because the tensor product of fields need not be a field). -It could be that this is all a special case of Grothendieck's theory of finite etale covers of schemes, but I wouldn't like to dredge up Grothendieckian formalism while discussing mainly about elementary properties of number fields, with a bit of class field theory. - -REPLY [8 votes]: Let $K$ be a separable algebraic closure of $k$. Then the functor sending an etale $k$-algebra $A$ to $Hom(A,K)$ is an equivalence to the category of finite sets with a continuous action of the Galois group of $k$ (baby form of Grothendieck's theory). There is an elementary exposition of this in Milne's notes on Fields and Galois theory. I think you can answer all your questions by looking at the sets with Galois action.<|endoftext|> -TITLE: Categorical mapping class group action -QUESTION [11 upvotes]: Let $\Sigma$ be a closed genus $g$ surface. Assume that $\mathcal{C}$ is a smooth and proper dg- (or $A_\infty$-) category which admits a faithful action -$$ MCG(\Sigma) \to Auteq(\mathcal{C})$$ -by the mapping class group by auto-equivalences. - -What good properties about $\mathcal{C}$ does that imply? (or -: why can representation theory tell me about the category or its auto-equivalences) -Are there any known cases where such an action arises which do not come from symplectic topology/mirror symmetry? The cases I know of so far come from Seidel-Thomas spherical twists or Ivan's Floer cohomology and pencils of quadrics paper. I assume there must be some examples from geometric representation theory or categorification and alike... -What happens if I replace the mapping class group with the Torelli group $\mathcal{I}_g$ (or more generally, level $k$ in the Johnson filtration)? - -REPLY [8 votes]: [This is an elaboration of parts of Mark Penney's answer] -A natural source of categorical actions of the mapping class group is the category assigned by any 4d TFT to a surface. Such categories are often written as Fukaya categories - eg Donaldson theory and Seiberg-Witten theory will attach Fukaya categories of moduli of G-bundles or symmetric products of the curve - but there are other examples, coming e.g. from factorization homology constructions as Mark wrote. -Here's what seems like the most natural example: the MCG acts on moduli spaces of G local systems ("character varieties" - or really character derived stacks) for G a complex reductive group say. So it acts on any category of sheaves on these moduli spaces. One natural answer is to look at quasicoherent sheaves - i.e. to any $\Sigma$ we look at $QC(Loc_G\Sigma)$. This is neither smooth nor proper as dg category, but it's faithful and interesting.. It's also the output of factorization homology, when the input is the braided (in fact symmetric) monoidal category $Rep(G)$ of representations of $G$. A more interesting answer coming from [Betti version of] geometric Langlands is to replace $QC$ by $IndCoh$ (possibly with fixed singular support). This is supposed to indeed be the value of an interesting 4d TFT on $\Sigma$ - namely the Kapustin-Witten B-twist of $\mathcal N=4$ super Yang-Mills. [An even more interesting (but conjectural) answer is the A-side of geometric Langlands.] -The character varieties have a symplectic form preserved by $MCG(\Sigma)$, so one can try to quantize them and get more interesting categorical actions. That's what is done e.g. in my work with Brochier and Jordan that Mark referenced, using factorization homology.<|endoftext|> -TITLE: Dual space of $L^2(\mathbb{R},L^1(0,1))$? -QUESTION [11 upvotes]: I was wondering what the dual space of $L^2(\mathbb{R},L^1(0,1))$ is? (equipped with Lebesgue measures) -Formally, one would suspect that it is just $L^2(\mathbb{R},L^{\infty}(0,1))$. But this may be a bit too quick, as there many references that suggest that this formally replacing each space by its dual is only possible in the reflexive case. -Here, is already a good answer at Math.stackexchange, but it only works for finite measure spaces: click me. - -REPLY [2 votes]: Your question has been partially answered in that it has been shown that the naive version is false. However, there is a concrete representation of the dual which might be of interest. The result becomes more transparent in a more general setting which might also be of interest to you. Consider the space $L^p(\mu,E)$ where $\mu$ is a finite measure, $1\leq p<\infty$ and $E$ is a Banach space. Then its dual is naturally identifiable with $L^q(\mu,E')$ if and only if $E'$ has RNP ($q$ the conjugate of $p$). This is treated in detail in the beginning of chapter IV of "Vector measures" by Diestel and Uhl. However, there is a representation which works for any Banach space. I have not found this result explicitly in the literature but the basic idea -is due to L. Schwartz (Sem. d'anal. fonctionelle (1974-75), Exp. 4---readily available online). I will state the result for the case of separable $E$: The dual is the space $L^q_{w^\ast}(\mu,E')$, where the subscript $w\ast$ refers to the fact that we are using functions which are weak star rather than norm measurable. Thus in your case, the dual is $L^2_{w\ast}(\mu,L^\infty)$. (As pointed out above there are standard methods to extend such results to the case of $\sigma$-finite measures, in particular the real line with Lebesgue measure).<|endoftext|> -TITLE: Minimal cover v.s random reals -QUESTION [9 upvotes]: The following set theoretical question is inspired by a question from recursion theory: - -Question: Is there an $L$-random real $r$ which is a minimal cover over another real $x$? - -Where a minimal cover $r$ over $x$ means that $x\in L[r] \wedge r\not\in L[x]$ but there is not real $z$ so that $x\in L[z]\wedge z\in L[r]\wedge r\not\in L[z]\wedge z\not\in L[x]$. - -REPLY [6 votes]: Suppose $r$ is random over $L$, $x \in 2^{\omega} \cap L[r]$ and $r \notin L[x]$. For $y \in 2^{\omega}$, let $y_0, y_1 \in 2^{\omega}$ denote the even and odd parts of $y$ - So $y = y_0 \oplus y_1$. Let $B$ denote the random algebra and $A$, the complete subalgebra of $B$ generated by $x$. Let $G_A$ be an $A$-generic filter over $L$ such that $L[x] = L[G_A]$. Then $L[r]$ is obtained by adding a $B/G_A$-generic filter over $L[x]$. Since this is same as adding a random real over $L[x]$, we can choose a real $y \in L[r]$ such that $y$ is random over $L[x]$ and $L[x][y] = L[r]$. Let $z = x \oplus y_0$. Then $L[x] \subsetneq L[z] \subsetneq L[r]$.<|endoftext|> -TITLE: Forcing in Constructive Set Theories -QUESTION [8 upvotes]: I searched on the internet, but I could not find anything useful about applications of forcing in constructive set theories. -Are there any developments of forcing in CZF or IZF? -Thanks in advance. - -REPLY [8 votes]: See -Forcing for IZF in sheaf toposes -Also -Toposes from Forcing for Intuitionistic ZF with Atoms. - -Edit: -Maybe more references: -Heyting-valued models for intuitionistic set theory -The book "Intuitionistic logic, model theory and forcing" by Fitting. -see also the following web-page where some references are also given: -Constructive Set Theory: Forcing, Large Sets, and Mathematics -Topological forcing semantics with settling.<|endoftext|> -TITLE: hooks and contents: Part I -QUESTION [6 upvotes]: For a cell $\square$ in the Young diagram of a partition $\lambda$, let $h_{\square}$ and $c_{\square}$ denote the hook length and content of $\square$, respectively. -R Stanley proved the following in Theorem 2.2: -$$\sum_{n\geq0}x^n\sum_{\lambda\vdash n}\prod_{\square\in\lambda}\frac{t+c_{\square}^2}{h_{\square}^2}=(1-x)^{-t}.$$ -In the same spirit, I ask: - -Question. if $\mathcal{O}(\pi)$ and $\mathcal{E}(\pi)$ stand for the number of odd and even cycles in $\pi$, respectively, then is this averaging formula true? - $$\frac1{n!}\sum_{\pi\in\mathfrak{S}_n}t^{\mathcal{O}(\pi)+2\mathcal{E}(\pi)}=\sum_{\lambda\vdash n}\prod_{\square\in\lambda}\frac{t+c_{\square}}{h_{\square}}.$$ - -REPLY [13 votes]: Specialize the Cycle Index Formula to get -$$\prod_{n=1}^\infty \exp \bigl( \frac{a_i}{i}z^i \bigr) = \sum_{n=0}^\infty \frac{z^n}{n!} \sum_{\pi \in \mathfrak{S}_n} a_1^{\mathrm{cyc}_1(\pi)} a_2^{\mathrm{cyc}_2(\pi)} \ldots, $$ -by setting $a_i = t^2$ if $i$ is even and $a_i = t$ if $i$ is odd. We get -$$\sum_{n=1}^\infty \frac{z^n}{n!} \sum_{\pi \in \mathfrak{S}_n} t^{\mathcal{O}(\pi)+2\mathcal{E}(\pi)} = \frac{1}{(1-z^2)^{t^2/2}} \frac{(1+z)^{t/2}}{(1-z)^{t/2}}. $$ -It is well known (see Corollary 7.21.4 in Stanley, Enumerative Combinatorics II) that -$$\prod_{\Box \in \lambda} \frac{d+c_\Box}{h_\Box} = s_\lambda(1^d)$$ -is the number $|\mathrm{SSYT}_d(\lambda)|$ of semistandard Young tableaux of shape $\lambda$ with entries from $\{1,\ldots,d\}$, or equivalently, the dimension of the irreducible polynomial representation of $\mathrm{GL}_d(\mathbb{C})$ labelled by the partition $\lambda$. -Therefore it is equivalent to show that if $d \in \mathbb{N}$ then -$$ \frac{1}{(1-z^2)^{d^2/2}} \frac{(1+z)^{d/2}}{(1-z)^{d/2}} = \sum_{n=0}^\infty z^n \sum_{\lambda \in \mathrm{Par}(n) } |\mathrm{SSYT}_d(\lambda)| \tag{$\star$} $$ -Now, as suggested by Gjerji Zaimi in a comment on my first answer, we use Littlewood's Identity. It states that -$$ \sum s_\lambda(x_1,\ldots,x_d) = \prod_{1 \le i < j \le d} \frac{1}{1-x_ix_j} \prod_{1 \le k \le d} \frac{1}{1-x_k} $$ -where the sum is over all partitions $\lambda$. (For an online source, see (6) in this paper by Betea and Wheeler, and multiply by $\prod_{k=1}^d \frac{1}{1-x_kz} = \sum_{n=0}^\infty h_n(x_1,\ldots,x_d)z^n$ using Young's rule.) Specialize Littlewood's Identity by setting $x_i = z$ for all $i \in \{1,\ldots, d\}$ and sum over $n$ to get -$$ \sum_{n=0}^\infty \sum_{\lambda \in \mathrm{Par}(n)} s_\lambda(1^d) z^n = \frac{1}{(1-z^2)^{(d^2-d)/2}(1-z)^d}. $$ -The right-hand side agrees with the left-hand side of ($\star$), so we are done.<|endoftext|> -TITLE: Simultaneous triangularisation of an exterior power of a set of matrices -QUESTION [6 upvotes]: I'm working on some research problems relating to random matrix products, and this is taking me into areas of mathematics I've not previously studied: Lie groups, representation theory, and real algebraic groups. In this area I encounter a lot of questions which I don't have the tools for, but which are probably easy for people who do have the correct tools. I previously posted this question on Stack Exchange, but it sank like a stone: - - -Question 1. Let $X \subset GL_d(\mathbb{R})$ be nonempty, let $1 -TITLE: DGLA or $L_{\infty}$-algebra controlling the deformation of Einstein metrics and instantons -QUESTION [13 upvotes]: As proposed by Quillen, Drinfeld, and Deligne and other important mathematicians, there is supposed to be a philosophy that, at least over a field of characteristic zero, assigns to every "deformation problem" a differential graded Lie algebra or $L_{\infty}$-algebra that controls it. -I've seen this idea realized in various situations, like for example the deformation theory of a compact complex manifold, which is "controlled" by the Kodaira-Spencer DGLA, or the deformation theory of Dirac structures in exact Courant algebroids. However, from my naive point of view as an outsider, I see this set of techniques completely disconnected from a different type of moduli problems of more "analytic" character. I refer for example to the moduli of Einstein metrics or the moduli of instantons in Donaldson's theory. It looks like the DGLA-philosophy has played virtually no role in the study of the moduli spaces of Einstein metrics and instantons. Why is this so? Does the DGLA-principle still applies to these problems but it does not add anything interesting? What is the DGLA controlling the deformation theory of these problems? It looks like the more analysis requires a moduli space problem, the less of a relevant role is played by the DGLA-philosophy, which seems to be more "algebraically inclined". I wonder because sometimes it looks like the moduli-literature is very polarized in different communities using different techniques to study moduli problems, so I would like to know to which extent the methods of one community apply to the problem considered by a different community. -Thanks. - -REPLY [9 votes]: With Domenico's clear explanation, I can actually write down more or less explicitly the DGLA describing the deformations of Einstein metrics. -First, some notation. Let $\bar{g}_{ab}$ denote a given (background) Einstein metric, with corresponding Levi-Civita connection $\bar{\nabla}_a$, Riemann tensor $\bar{R}_{abc}{}^d$, and Ricci tensor $\bar{R}_{ac} = \bar{R}_{abc}{}^b$. It is given that $\bar{\nabla}_a \bar{g}_{bc} = 0$ and $\bar{R}_{ac} = k \bar{g}_{ac}$ for a fixed constant $k$. -Let $\nabla_a$ denote an arbitrary symmetric (torsion free) affine connection, which differs from the background Levi-Civita connection as $\nabla_a v^b = \bar{\nabla}_a v^b + C^b_{ac} v^c$. Let me call $C^b_{ac} = C^b_{(ac)}$ the corresponding Christoffel tensor (or connection coefficients). The equation for an Einstein metric $g_{ab} = \bar{g}_{ab} + h_{ab}$ (with the same constant $k$) can be written in the form -\begin{align*} - \nabla_a g_{bc} - &= \bar{\nabla}_a h_{bc} - - C^d_{ab} \bar{g}_{dc} - C^d_{ac} \bar{g}_{bd} - - C^d_{ab} h_{dc} - C^d_{ac} h_{bd} , - \\ - R_{ac}[C] - k h_{ac} - &= -kh_{ac} - \bar{\nabla}_a C^b_{bc} + \bar{\nabla}_b C^b_{ac} - + C^b_{ac} C^d_{db} - C^b_{ad} C^d_{cb} , -\end{align*} -where $R_{ac}[C] = R_{abc}{}^b[C]$ is the usual Ricci contraction of the curvature tensor $R_{abc}{}^d[C]$ of $\nabla_a$. The first of the above equations is the $g$-compatibility condition for $\nabla_a$. Solving it for the Christoffel tensors identifies $\nabla_a$ with the Levi-Civita connection of $g_{ab}$ and plugging that solution into the second equation gives the equation $R_{abc}[g] - k g_{ac} = 0$ for an Einstein metric, due to the identity $R_{ab}[g] = \bar{R}_{ab} + R_{ab}[C]$. -Now the equations are precisely in the form that Domenico described, $f[h,C] + Q[h,C;h,C] = 0$, with $f$ linear and $Q$ quadratic -\begin{align*} - \begin{pmatrix} - f_{abc}[h,C] \\ f_{ac}[h,C] - \end{pmatrix} - &= \begin{pmatrix} - \bar{\nabla}_a h_{bc} - - C^d_{ab} \bar{g}_{dc} - C^d_{ac} \bar{g}_{bd} \\ - -kh_{ac} - \bar{\nabla}_a C^b_{bc} + \bar{\nabla}_b C^b_{ac} - \end{pmatrix} -\\ - \begin{pmatrix} - Q_{abc}[h,C;h',C'] \\ Q_{ac}[h,C;h',C'] - \end{pmatrix} - &= \frac{1}{2} \begin{pmatrix} - - C^d_{ab} h'_{dc} - C^d_{ac} h'_{bd} - - C'^d_{ab} h_{dc} - C'^d_{ac} h_{bd} \\ - + C^b_{ac} C'^d_{db} - C^b_{ad} C'^d_{cb} - + C'^b_{ac} C^d_{db} - C'^b_{ad} C^d_{cb} - \end{pmatrix} -\end{align*} -It remains to describe how infinitesimal symmetries (diffeomorphisms) act on the $h_{ab}$ and $C^b_{ac}$ tensor fields, as well as on $f$ and $Q$. They are generated by vector fields $u^a$. They act on tensors via the usual Lie derivative, which we find convenient to express via $\bar{\nabla}_a$, so that $\mathcal{L}_u X^a = u^b \bar{\nabla}_b X^a - X^b \bar{\nabla}_b u^a$ and $\mathcal{L}_u Y_a = u^b \bar{\nabla}_b Y_a + Y_b \bar{\nabla}_a u^b$, and they act on each other via the usual Lie bracket $[u,u'] = \mathcal{L}_u u' = -\mathcal{L}_{u'} u$. But special attention must be paid to the identities -\begin{align*} - \mathcal{L}_u \bar{\nabla}_b X^a - \bar{\nabla}_b \mathcal{L}_u X^a - & = u^c \bar{\nabla}_c \bar{\nabla}_b X^a - - (\bar{\nabla}_b X^c) \bar{\nabla}_c u^a - + (\bar{\nabla}_c X^a) \bar{\nabla}_b u^c \\ - & \quad {} - -\bar{\nabla}_b (u^c \bar{\nabla}_c X^a - X^c \bar{\nabla}_c u^a) \\ - &= u^c \bar{R}_{bcd}{}^{a} X^d + X^c \bar{\nabla}_b \bar{\nabla}_c u^a - = (\bar{\nabla}_{(b} \bar{\nabla}_{c)} u^a - u^d \bar{R}_{d(bc)}{}^a) X^c , \\ -% &= u^c \bar{R}_{bcd}{}^{a} X^d - X^c \bar{R}_{bcd}{}^a u^d + X^c \bar{\nabla}_c \bar{\nabla}_b u^a \\ -% &= X^c \bar{R}_{cdb}{}^a u^d + X^c \bar{\nabla}_c \bar{\nabla}_b u^a , \\ - \mathcal{L}_u \bar{\nabla}_b Y_a - \bar{\nabla}_b \mathcal{L}_u Y_a - & = u^c \bar{\nabla}_c \bar{\nabla}_b Y_a - + (\bar{\nabla}_b Y_c) \bar{\nabla}_a u^c - + (\bar{\nabla}_c Y_a) \bar{\nabla}_b u^c \\ - & \quad {} - -\bar{\nabla}_b (u^c \bar{\nabla}_c Y_a + Y_c \bar{\nabla}_a u^c) \\ - &= -u^c \bar{R}_{bca}{}^{d} Y_d - Y_c \bar{\nabla}_b \bar{\nabla}_a u^c - = (\bar{\nabla}_{(b} \bar{\nabla}_{a)} u^c - u^d \bar{R}_{d(ba)}{}^c) Y_c , -\end{align*} -which identify the action of $[\mathcal{L}_u, \bar{\nabla}_b]$ as a derivation on the algebra of tensors. This means that infinitesimal diffeomorphisms generate the infinitesimal transformation $(h,C) \mapsto (h,C) + \epsilon K[u;h,C] + O(\epsilon^2)$, where -\begin{equation*} - \begin{pmatrix} - K_{ab}[u;h,C] \\ K_{ab}^c[u;h,C] - \end{pmatrix} - = \begin{pmatrix} - \bar{g}_{ac} \bar{\nabla}_b u^c + \bar{g}_{cb} \bar{\nabla}_a u^c \\ - \bar{\nabla}_{(a} \bar{\nabla}_{b)} u^c - u^d \bar{R}_{d(ab)}{}^c - \end{pmatrix} . -\end{equation*} -Finally, we can put these formulas together in the definition of a DLGA $(L,d,[-,-])$. $L$ itself will break down into a sum of sections of certain tensor bundles. For simplicity, I will write $T$ to denote the space of sections of the bundle of vectors, $S^2T^*$ for symmetric covariant 2-tensors, etc. The breakdown by degree is -\begin{gather*} - \begin{array}{c|ccccc} - & 0 && 1 && 2 \\ - \hline - L & T &\to& S^2 T^*\oplus S^2T^*\otimes T &\to& S^2T^* \otimes T^* \oplus S^2 T^* \\ - d & & K & & f & - \end{array} , \\ - \begin{array}{c|ccc} - [-,-] & 0 & 1 & 2 \\ - \hline - 0 & [u,u'] & K[u;h',C'] & \mathcal{L}_u \\ - 1 & -K[u';h,C] & 2Q[h,C;h',C'] & 0 \\ - 2 & -\mathcal{L}_{u'} & 0 & 0 - \end{array} -\end{gather*} -I believe that this DGLA could be extended by one more degree to take the Bianchi identities into account. But I will stop here. -There are of course other ways to present the same DGLA and one can find explicit attempts in the literature of writing it down. Here's one that uses a somewhat different presentation: - -Michael Reiterer, Eugene Trubowitz The graded Lie algebra of general relativity arXiv:1412.5561<|endoftext|> -TITLE: Instanton Moduli Space on ALE Spaces -QUESTION [10 upvotes]: I asked this on MathStackExchange and was instructed it would be better here. -I've recently been learning about moduli spaces of instantons on $\mathbb{C}^{2}=\mathbb{R}^{4}$. From what I can gather, one can consider the framed moduli space of torsion-free sheaves on $\mathbb{P}^{2}$ of rank $N$ and second Chern class $k$, which we denote $\mathcal{M}(k,N)$. I believe in the rank one case, we can identify this moduli space with the symmetric product of $\mathbb{C}^{2}$, which of course can be crepantly resolved to the Hilbert scheme. In other words, -$$\text{Hilb}^{k}(\mathbb{C}^{2}) \to \text{Sym}^{k}(\mathbb{C}^{2}) = \mathcal{M}(k,1)$$ -From what I can gather, this is what's known as the "instanton moduli space on $\mathbb{C}^{2}$." There is then this whole "geometric engineering" story by Vafa, Hollowood, et. al. where they consider either the $\chi_{y}$ genus of these moduli spaces or the elliptic genus $\text{Ell}_{y,q}$ and construct the instanton partition function: -$$ \sum_{k} p^{k} \chi_{y} (\text{Hilb}^{k}(\mathbb{C}^{2})) \,\,\,\,\,\, \text{or} \,\,\,\,\,\, \sum_{k} p^{k} \text{Ell}_{y,q} (\text{Hilb}^{k}(\mathbb{C}^{2}))$$ -One can then show that these partition functions are very remarkably equal to partition functions in topological string theory on certain Calabi-Yau varieties. -So really I'm curious about replacing $\mathbb{C}^{2}$ with the ALE spaces, specifically the $A_{N}$ resolutions of the singularities $\mathbb{C}^{2}/\Gamma$ where $\Gamma$ is a finite subgroup of $SU(2)$. The above story with the Hilbert schemes was only for the rank one case, $N=1$ so it's very tempting to hope that maybe the higher rank moduli spaces $\mathcal{M}(k,N)$ might be related to the $A_{N}$ resolutions somehow? I was hoping someone could help me understand what the moduli space of instantons on ALE spaces looks like, and whether there are nice partition functions like the ones above arising from such a space. I know there is physics literature here (like the Vafa-Witten https://arxiv.org/pdf/hep-th/9408074.pdf) but I'm having serious issues understanding the physics! Does considering the Hilbert scheme of points on the $A_{N}$ resolutions provide anything of physical relevance, or do we need something more complicated perhaps? - -REPLY [4 votes]: For every $k$, $M$ and $N$ positive integers, one can consider the moduli space $\mathcal{M}_M(k,N)$ of $U(N)$ instantons of instanton number (=second Chern class) $k$ on the resolution of the $A_{M-1}$ surface singularity. The case $M=1$ corresponds to instantons on $\mathbb{C}^2$. In particular we have at least three parameters: $k$, $M$, $N$ and things can become confusing if one mixes them or tries to obtain non-trivial relations between them. -On how the moduli space $\mathcal{M}_M(k,N)$ looks like: a fairly concrete description is due to Nakajima ( http://math.mit.edu/~ptingley/QuantumGroupsSpring2011/Nakajima94.pdf ) (generalizing the ADHM construction for $M=1$) by hyperkähler quotient construction, more precisely as Nakajima quiver varieties attached to the $\hat{A}_{M-1}$ Dynkin quiver. This quiver construction has a natural physics interpretation: it is the natural description of the moduli space "from the point of view of the instantons" (to make sense of that, one has to be able to separate the instantons from the four dimensional geometry, which is done naturally in string theory in the $D(-1)-D3$ brane system). -For any smooth complex surface $S$, there is a formula due to Göttsche giving the generating series of Euler characteristics of Hilbert schemes of points on $S$: it is the answer for $S=\mathbb{C}^2$ (the generating function of partitions) to the power $\chi(S)$ where $\chi(S)$ is the topological Euler characteristic of $S$. -In physics, the idea of geometric engineering is to consider a 10-dimensional string theory on a space time of the form $X_4 \times X_6$ and to study the effective gauge theory living on the 4-dimensional space $X_4$ in terms of the geometry of the 6-dimensional space $X_6$. In this context, there are at least two ways to introduce ALE spaces: one could take $X_4$ to be an ALE space or/and one could take some $X_6$ whose geometry contains an ALE space. All these possibilities are interesting and one should not confuse them. -For example, one can take $X_4=\mathbb{C}^2$ and $X_6$ a non-trivial fibration in ALE spaces of type $A_{M-1}$ over the complex projective line $\mathbb{P}^1$. -The resulting gauge theory on $\mathbb{C}^2$ has gauge group $SU(N)$ and geometric engineering will give a relation between the Nekrasov partition function of this gauge theory, some generating function of equivariant integrals over moduli spaces of $SU(N)$ instantons over $\mathbb{C}^2$, and topological string on $X_6$. But because $X_6$ is a fibration in $A_{M-1}$ ALE space, topological string on $X_6$ is in fact related to the Hilbert schemes of points on the ALE space (see https://arxiv.org/abs/0802.2737 , this is a key step in the proof of the MNOP conjecture relating Gromov-Witten and Donaldson-Thomas theories for 3-folds).<|endoftext|> -TITLE: Rational curves on elliptic surfaces -QUESTION [8 upvotes]: I need a reference for the following assertion (if it's true). Let $X$ be a minimal elliptic surface over the field of complex numbers. Assume that its Kodaira dimension $\kappa(X)=1$. Then $X$ does not contain rational curves. Thanks! - -REPLY [7 votes]: I don't think this is true. -Take a general pencil of cubics in $\mathbb{P}^2$, and blow-up the 9 fixed points to get an elliptic fibration $f:S\rightarrow \mathbb{P}^1$ which admits a section (at least 9 in fact). Pull back by a degree $n\geq 3$ covering $\mathbb{P}^1\rightarrow \mathbb{P}^1$ branched along two points $p_1,p_2$ such that $\ f^{-1}(p_i)\ $ is smooth. You get a new fibration $f':S'\rightarrow \mathbb{P}^1$ with $\kappa (S')=1$, $S'$ minimal; the sections of $f$ pull back to sections of $f'$ which are smooth rational curves.<|endoftext|> -TITLE: Drawing trees on small number of lines in 2D and 3D -QUESTION [7 upvotes]: Problem. Given a tree do we need fewer lines in 3D than in 2D in order to draw it straightline and crossing-free? - -(Asked 01.10.2016 by Alexander Wolff on page 20 of Volume 1 of the Lviv Scottish Book). -Clarification. For a given finite simple graph $G$ by $\rho^1_2(G)$ (resp. $\rho^1_3(G)$) we denote the minimal number of straight lines needed to cover all edges in a straight-line crossing-free drawing of the graph $G$ in 2D (resp. 3D). We are asking for a tree $T$ such that $\rho^1_3(T)<\rho^1_2(T)$. -The problem was originally asked 11.06.2016 by Oleg Verbitsky. - -REPLY [2 votes]: I'm not 100% sure how to prove this, but here is my candidate for an embedding in 2D that minimizes the number of lines used: -For each node in the tree if it has an even number of edges, $2e$ then embed pairs of edges on the same line (i.e. put the edge $\frac{\pi}{e}$ radians apart) so that the node uses $e$ lines in total; if the node has an odd number of edges, $2e + 1$ then do the same, but have one edge be on its own line (put it at some angle not used by the other edges) to use $e+1$ lines in total. This process is possible by making the embedding spread out enough that there is no crossing. -I believe that you cannot do better than this because an edge can be on the same line of at most one incident edge. Furthermore I believe you cannot do better in 3D as well for the same reason.<|endoftext|> -TITLE: Periodic tilings of the plane by regular polygons -QUESTION [9 upvotes]: Let $A$ be a tiling of $\mathbb{R}^{2}$ using regular polygons. Assume that the tiling is edge-to-edge. Assume also that there are two directions of periodicity, so that $\mathbf{u},\mathbf{v}\in \mathbb{R}^{2}$ are linearly independent vectors, and $A+\mathbf{u}=A+\mathbf{v}=A$. -Question: Must there always exist orthogonal directions of periodicity? That is, must there always exist non-zero vectors $\mathbf{u},\mathbf{v}\in \mathbb{R}^{2}$ such that $\mathbf{u}\cdot \mathbf{v}=0$, and $A+\mathbf{u}=A+\mathbf{v}=A$? -Note that the assumption of regularity is necessary, since we can tile the plane with identical parallelograms in such a way that there are no orthogonal directions of periodicity. It is also necessary that we have an edge-to-edge tiling, since otherwise we can construct an example with identical squares that does not have orthogonal directions of periodicity. -Does this follow from some feature of the wallpaper groups? - -REPLY [10 votes]: I claim that the tiling https://upload.wikimedia.org/wikipedia/commons/6/66/5-uniform_310.svg does not admit a orthogonal period. -The basis for the period lattice is given by the two vectors $$ v_1 = \begin{pmatrix} 3 + \sqrt{3} \\ -1 \end{pmatrix}, v_2 = \begin{pmatrix} 1/2 \\ (2+\sqrt{3})/2 \end{pmatrix}. $$ So if there is an orthogonal period, then there has to be integers $a_1, a_2, b_1, b_2$ such that $\langle a_1 v_1 + a_2 v_2, b_1 v_1 + b_2 v_2 \rangle = 0$ and $(a_1, a_2), (b_1, b_2) \neq (0,0)$. Expanding out, we can write this as -$$(13 + 6\sqrt{3}) a_1 b_1 + \frac{1}{2} (a_1 b_2 + a_2 b_1) + (2 + \sqrt{3}) a_2 b_2 = 0. $$ -Since $\sqrt{3}$ is irrational, we then obtain -\begin{align*} 0 &= 6 a_1 b_1 + a_2 b_2, \\ 0 &= 26 a_1 b_1 + (a_1 b_2 + a_2 b_1) + 4 a_2 b_2. \end{align*} -Multiplying $-25 + \sqrt{7}$ to the first equation and multiplying $6$ to the second equation and adding them up gives -\begin{align*} 0 &=(6 + 6 \sqrt{7}) a_1 b_1 + 6(a_1 b_2 + a_2 b_1) + (-1 + \sqrt{7}) a_2 b_2 \\ &= (-1 + \sqrt{7}) ((1 + \sqrt{7}) a_1 + a_2) ((1 + \sqrt{7}) b_1 + b_2). \end{align*} -Then again by irrationality of $\sqrt{7}$, either $a_1 = a_2 = 0$ or $b_1 = b_2 = 0$.<|endoftext|> -TITLE: Finite-variable fragments of $\Delta_0$-formulas -QUESTION [6 upvotes]: Consider sets definable in the usual structure of arithmetic $(\mathbb{N},0,1,+,\times)$ by $\Delta_0$-formulas, i.e., formulas with bounded quantifiers. The quantifier alternation hierarchy has been studied,* which stratifies $\Delta_0$ formulas by the number of nested alternations of bounded quantifiers they contain. -My question is, is anything known about the finite variable hierarchy, in which formulas are stratified by the number of (distinct) variables that occur? In particular, - -Does it collapse or is it strict? (Perhaps under some complexity-theoretic assumptions?) -One expects that since syntactically one can write formulas with arbitrary high quantifier alternation using only finitely many variables, or arbitrarily many variables using only a finite amount of quantifier alternation, if we restrict ourselves to formulas from a finite level in either hierarchy, the other one still does not collapse (Assuming neither collapse absolutely.). Is this true? - -*See, e.g., Keith Harrow, "The bounded arithmetic hierarchy," http://www.sciencedirect.com/science/article/pii/S0019995878902577 - -REPLY [2 votes]: The bounded variable hierarchy collapses to a fixed finite level, because of the existence of definable pairing functions. -In more detail, I will use the Cantor pairing function -$$\def\p#1{\langle#1\rangle}\def\N{\mathbb N}\let\eq\leftrightarrow\p{x,y}=\frac{(x+y)(x+y+1)}2+x,$$ -which is a bijection $\N^2\to\N$. Let $l(z)$, $r(z)$ denote the corresponding projections, so that -$$l(\p{x,y})=x,\qquad r(\p{x,y})=y,\qquad\p{l(z),r(z)}=z.$$ -Since $\p{x,y}$ is a polynomial with rational coefficients, $\p{x,y}=z$ is equivalent to a quantifier-free (hence $3$-variable) formula, namely -$$\p{x,y}=z\eq z+z=(x+y)\cdot(x+y+1)+x,$$ -and consequently the functions $l$ and $r$ are definable by $\Delta_0$ formulas using $3$ distinct variables: -$$l(z)=x\eq\exists y\le z\,\p{x,y}=z,$$ -and similarly for $r$. - -Proposition: Every $\Delta_0$ formula in one free variable is equivalent to a $\Delta_0$ formula using only $3$ distinct variables. - -Proof: -It will be convenient to preprocess the formulas first. Let us call a bounded quantifier $\exists y\le t(\vec x)\,\theta(\vec x,y)$ safe if -$$\bigl(\exists y\le t(\vec x)\,\theta(\vec x,y)\bigr)\eq\bigl(\exists y\,\theta(\vec x,y)\bigr),$$ -in which case we can replace the bound $t(\vec x)$ with any larger bound without affecting the truth of the formula. Similarly for universal quantifiers. By replacing $\exists y\le t(\vec x)\,\theta$ with $\exists y\le t(\vec x)\,(y\le t(\vec x)\land\theta)$, and unwinding terms with the help of extra existential quantifiers, it is easy to see that any $\Delta_0$ formula is equivalent to a $\Delta_0$ formula $\theta(\vec x)$ such that - -all quantifiers in $\theta$ are safe, -the index $i$ of any variable $x_i$ quantified in any subformula $\xi$ of $\theta$ is higher than the indices of all free variables of $\xi$, and -all atomic subformulas of $\theta$ (except quantifier bounds) are of the forms $x_i=x_j$, $x_i+x_j=x_k$, or $x_i\cdot x_j=x_k$. - -Let us call such formulas special. - -Claim: For any special formula $\theta(x_0,\dots,x_n)$, there is a formula $\xi(z)$ using only $3$ distinct variables such that - $$\xi(z)\eq\theta\bigl(r(l^n(z)),\dots,r(l(z)),r(z)\bigr).$$ - -Note that the Claim implies the Proposition: if $\xi(z)\eq\theta(r(z))$ can be written using $3$ variables, then so can -$$\theta(x)\eq\exists z\le x^2\,\bigl(z=\p{0,x}\land\xi(z)\bigr).$$ -We will prove the Claim by induction on the complexity of $\theta$. -The induction steps for Boolean connectives are trivial. The step for bounded quantifiers is also easy: if $\theta(x_0,\dots,x_n)$ is $\exists x_{n+1}\le t(\vec x)\,\theta'(x_0,\dots,x_{n+1})$, where the quantifier is safe, and $\xi'$ is a $3$-variable formula equivalent to $\theta'(r(l^{n+1}(z)),\dots,r(l(z)),r(z))$, then -$$\begin{align*} -\theta(r(l^n(z)),\dots,r(z))&\eq\exists x_{n+1}\,\xi'(\p{z,x_{n+1}})\\ -&\eq\exists w\,(l(w)=z\land\xi'(w))\\ -&\eq\exists w\le s(z)\,(l(w)=z\land\xi'(w)) -\end{align*}$$ -for a suitable term $s(z)$. -It remains to prove the Claim for special atomic formulas. First, notice that since $l$ and $r$ are definable using $3$ variables, the same holds for any finite composition of these functions: for example, -$$x=l^{n+1}(z)\eq\exists y\le z\,(x=l(y)\land y=l^n(z)),$$ -where we proceed to expand the inner formulas. This immediately takes care of atomic formulas $x_i=x_j$, as -$$r(l^i(z))=r(l^j(z))\eq\exists x\le z\,(x=r(l^i(z))\land x=r(l^j(z))).$$ -For the formulas $x_i+x_j=x_k$ (and similarly $x_i\cdot x_j=x_k$), we can proceed as follows: -$$r(l^i(z))+r(l^j(z))=r(l^k(z))\eq\exists w\le t(z)\,(l(l(w))=r(l^i(z))\land r(l(w))=r(l^j(z))\land r(w)=r(l^k(z))\land\exists x,z\le w\,(w=\p{\p{x,z},x+z})).$$ -(Note that we recycled the $z$ variable.) Here, $l(l(w))=r(l^i(z))$ can be written as $\exists x\le w\,(x=l(l(w))\land x=r(l^i(z)))$, which we know how to handle, and similarly for the other two conjuncts. Finally, -$$w=\p{\p{x,z},x+z}$$ -can be written as an identity using no further variables, as $\p{\p{x,z},x+z}$ is a polynomial with rational coefficients in $x$ and $z$ (and $x+z$). QED - -Corollary: Every $\Delta_0$ formula in $n$ free variables is equivalent to a $\Delta_0$ formula using $\max\{3,n+1\}$ distinct variables. - -Comments: - -We only used elementary properties of the pairing function, hence the equivalence is not just true in $\N$, it is provable in a very weak theory: IOpen is certainly enough (probably $\mathrm{PA}^-$ will suffice with a little care). -The result still holds for bounded formulas in a richer language with extra binary functions or relations. Even more generally, if we expand the language with $k$-ary functions and relations with $k>2$, the result holds with $k+1$ in place of $3$. -The transformation more-or-less preserves quantifier complexity: for any $i\ge1$, an $E_i$ formula in $n$ free variables is equivalent to an $E_i$ formula using $\max\{3,n+1\}$ distinct variables, and similarly for $U_i$. However, we need to make sure the $E_i$ and $U_i$ classes are defined in such a way as to explicitly allow conjunctions and disjunctions; of course, we cannot reduce the number of distinct variables in prenex formulas.<|endoftext|> -TITLE: The importance of relations between automorphic forms and arithmetic functions -QUESTION [7 upvotes]: As I understand things, one of the classical reasons to care about modular forms was their relation to interesting arithmetic functions/counting questions, i.e. on sums of squares and partitions. When I read Diamond and Shurman’s book([Diamond&Shurman]A First Course in Modular Forms), this point of view was briefly mentioned as an interesting application, but most of the book was focused on their role in the Modularity Theorem. I have now been working on Gelbart’s book on automorphic forms([Gelbart]Automorphic Forms on Adele Groups), which feels like it is moving even further off in this direction. -This point of view is certainly interesting and there’s obviously a lot of important mathematics here, but it leaves me wondering what happens to this when we move from modular forms to automorphic forms. Are there any interesting arithmetic functions leading to automorphic forms over groups besides $SL(2,\mathbb{Z})$? If not, what makes this particular group so special? - -REPLY [7 votes]: Let me just highlight one aspect that I find particularly interesting. -Without any doubt, one the most basic arithmetic functions is $\tau_k(n)$, which counts the number of ways $n$ can be written as a product of $k$ factors. It is the $k$-fold convolution of the constant $1$ function with itself, so it is multiplicative and its Dirichlet series is an $L$-function of degree $k$: -$$ \sum_n \frac{\tau_k(n)}{n^s} = \prod_p\frac{1}{(1-p^{-s})^k},\qquad \Re s>1. $$ -This $L$-function is apparently $\zeta(s)^k$, so it has a meromorphic continuation, functional equation, and the (analogue of the) Riemann hypothesis seems to be true for it. -A natural question is if $\tau_k(n)$ fits into some broader family of arithmetic functions which perhaps allows us to understand $\tau_k(n)$ better and gives insight into the mentioned properties of its $L$-function. A natural candidate for such arithmetic functions is the convolution of $k$ Dirichlet characters. Then the resulting $L$-function is the product of $k$ Dirichlet $L$-functions, which equally seems to satisfy the (analogue of the) Riemann hypothesis. It turns out that there is geometry behind these more general coefficients: they are Hecke eigenvalues of special Eisenstein series on the group $\mathrm{GL}_k$. But these Eisenstein series do not form a spectrally complete family: they are missing the more general Eisenstein series and the cusp forms. In these missing cases the Euler denominators $(1-p^{-s})^k$ are more general degree $k$ polynomials of $p^{-s}$, and the closer we are to a cusp form, the less we can canonically decompose them into lower degree polynomials. The $L$-functions of general Eisenstein series on $\mathrm{GL}_k$ decompose into a product of $L$-functions of cusp forms on $\mathrm{GL}_j$ with $j -TITLE: Two divergent series conspiring? -QUESTION [11 upvotes]: Consider the sequence $a_n=2^{2n}\binom{2n}n^{-1}$. Stirling's approximation shows that $a_n\sim \sqrt{\pi n}$, thus -$$\sum_{n\geq0}\frac{\pi}{2a_n}\qquad \text{and} \qquad -\sum_{n\geq0}\frac{a_n}{2n+1}$$ -are both divergent series. However, their difference should converge with terms of order $\sim\frac1{n^{3/2}}$. - -Question. In fact, is this true? - $$\sum_{n=0}^{\infty}\left(\frac{\pi}{2a_n}-\frac{a_n}{2n+1}\right)=1.$$ - -REPLY [29 votes]: We have - $$ f(x):=\sum_{n\geq 0}\frac{x^{2n}}{a_n} = - \frac{1}{\sqrt{1-x^2}} $$ -and - $$ g(x):=\sum_{n\geq 0} \frac{a_n}{2n+1}x^{2n} = - \frac{\sin^{-1}x} {x\sqrt{1-x^2}}. $$ -It is routine to compute that - $$ \lim_{x\to 1-}\left(\frac 12\pi f(x)-g(x)\right)=1 $$ -and then apply Abel's theorem.<|endoftext|> -TITLE: Is there a trace inequality for the product of a sequence of hermitian postive definite matrices? -QUESTION [6 upvotes]: Let $A$ and $B$ be two Hermitian matrices with positive eigenvalues. -Let $k>0$ be a integer. -Let $P=(P_1,P_2,\dots,P_{2k})$ be a sequence of $k$ $A$s and $k$ $B$s in any given order. -Do we have -${\rm tr}\,\prod^{2k}_{i=1} P_i \leq {\rm tr}\,A^k B^k$ ? - -REPLY [5 votes]: A counterexample was apparently first found by Plevnik (2016) (pdf): -$$ -A = \begin{bmatrix}76&0&0\\0&0&0\\0&0&1\end{bmatrix} -\qquad\text{and}\qquad -B = \begin{bmatrix}20&-14&13\\-14&2880&3100\\13&3100&3380\end{bmatrix} -$$ -give -$$ -\operatorname{tr} A^4BAB^4 = 7608677695167720100 > 7566365725138281700 = \operatorname{tr} A^5B^5. -$$ -To get one with $A$ positive definite as requested, replace entry $A_{22}$ by $0.01$: the two traces become -$$ -7584680876077508226.18611992\ \ > \ \ 7566365725573314229.03610008. -$$<|endoftext|> -TITLE: Proof of a theorem about the size of the number of sign changes of Hecke eigenvalues -QUESTION [5 upvotes]: In their paper Sign changes of Hecke eigenvalues, Matomaki and Radziwill showed that (Theorem 1.2 of the paper) for a large enough $x$ , the number of sign changes of sign changes of in the non-vanishing sequence of Hecke eigenvalues $(\lambda_f(n))_{n\leq x}$ is -$\asymp \frac{x}{k(x)},$ where $$k(x)=\prod_{\substack{p \leq x\\ -\lambda_f(p)=0}} \left(1+\frac{1}{p}\right).$$ In the proof of this result, they used the fact that there is a positive proportion of $x$ in $[X,2X]$ such that -$$\left|\sum_{x\leq n \leq x+h k(x)} \operatorname{sgn}(\lambda_f(n))w_{n}\right|< -\sum_{x\leq n \leq x+h k(x)} w'_{n}$$ once $h$ is large enough but satisfying some condition. They say that since $w'_n \leq w_n$ for every $n$ then for every $x$ satisfying the upper bound then a sign change of $\lambda_f(n)$ must occur in the interval $[x, x+h k(x)]$. From which they say that there are $\gg \frac{X}{k(X)}$ sign changes. -I have two questions: -1) I could not understand how they get their deduction about the lower bound of sign changes. -2) Why they used the same proof to deduce an upper bound although they did not mention it! -Did I misunderstand something? - -REPLY [5 votes]: The proof of Theorem 1.2 relies on Propositions 3.4 and 3.5. In particular, if $h$ is sufficiently large but fixed, the bound you quote is true for a positive proportion of $x\in\mathbb{N}\cap[X,2X]$. Call such an $x$ nice. -Consider a maximal family $F$ of nice points $x\in[X,2X]$ such that any two of them differ by more than $hk(2X)$. Then any nice $x\in[X,2X]$ is within distance $hk(2X)$ from some $x\in F$, hence the cardinality of $F$ satisfies $hk(2X)|F|\gg X$, i.e., $|F|\gg X/k(2X)$. -Shrink $F$ slighly to $G:=F\cap[X,2X-hk(2X)]$. Using that $k(2X)\ll\log X$, it is clear that -$$|G|\gg X/k(2X)\gg X/k(X).$$ Now $[x,x+hk(x)]$ with $x\in G$ are pairwise disjoint subintervals of $[X,2X]$, because $k(x)\leq k(2X)$, and in each such subinterval $\lambda_f(n)$ changes sign. So the number of sign changes of $\lambda_f(n)$ in $[X,2X]$ is at least $|G|$ which is $\gg X/k(X)$. -Regarding your second question, note that the number of sign changes of $\lambda_f(n)$ in $[X,2X]$ cannot exceed the number of $n\in[X,2X]$ with $\lambda_f(n)\neq 0$. However, the number of such $n$'s is $\ll X/k(2X)\leq X/k(X)$ by part (i) of Lemma 3.1 in the paper, so we are done.<|endoftext|> -TITLE: Maximal number of triple intersection points of $n$ circles -QUESTION [5 upvotes]: It is easy to show that $n$ (mutually different) circles on the plane can have maximum $n(n-1)$ intersection points. In our optimal graph drawing research we have encountered a counterpart of this result. For a natural number $n$ let $c_3(n)$ denotes the maximum number of triple intersection points of $n$ (mutually different) circles on the plane. (A point is a triple intersection point if in belongs to at least three circles of the family). This problem looks natural and simple, so it should be already known. Thus I tried to google for it. Surprisingly, I found nothing, hence I decided to ask here for a reference. -We can easily show that $c_3(n)\le n(n-1)/3$, by counting intersection points (there are at most $n(n-1)$ of them) taking into account their multiplicity (each triple intersection point needs at least there of them) and maybe the value of $c_3(n)$ is asymptotically close to this upper bound. But for our applications we are mainly interested in upper bounds for small values of $n\le 10$. For instance, the following pictures combined with the above upper bound shows that $c_3(5)=6$, $8\le c_3(6)\le 10$, and $12\le c_3(7)\le 14$. - -REPLY [2 votes]: The following construction shows that $c_3(n)\ge n^2/4-1$. -Put $k=\lceil \frac{n+1}2\rceil$ and draw a regular $k$-gon $K$ with side $1$. -Initially, we place $k$ black unit circles centered in the -corners of $K$. This generates $k - 1$ circular layers of $k$ double -intersection points (see, for instance, the last picture at the question). -Since $k-1\ge n-k$, we can cover $n-k$ layers by red circles, -achieving at least $k$ triple intersection points on each layer -and $k(n-k)\ge n^2/4-1$ of them in total.<|endoftext|> -TITLE: Solids with constant surface area during "erosion" -QUESTION [10 upvotes]: Imagine a drug, a pill that you swallow, which is designed to dissolve in your -stomach at a constant rate. It must be shaped such that the surface area -remains constant when the volume is "eroded" uniformly over its surface. -Two dimensions -Define "erosion" of a distance $\delta$ from a shape as removing a slither of width $\delta$ around the whole perimeter of the shape. This is a parallel curve, at least initially. (Note that a re-entrant corner will become rounded after erosion. An alternative defintion would have the corner preserved, which makes the question slightly easier but is harder to justify.) -Does a shape exist whose perimeter remains constant after is eroded (at least for erosion of a distance $\delta$, for all $\delta <= D$, for some $D > 0$)? -I have an unsatisfying solution: - - An annulus, assuming internal erosion is allowed, because the change - of perimeter of the inner and outer circles cancel out until the area - is zero. This works - mathematically but not for how I posed the question. So I would be - interested in solutions that don't have holes. - - But for an annulus with a channel of zero width connecting the inner - and outer circle, after erosion the perimeter will have been reduced - by about twice the width of the channel. So I would conjecture that - it can't be done without holes. - - (This solution doesn't generalise to spheres in 3D although holes - might still help.) - -Three dimensions -Define "erosion" of a distance $\delta$ from a solid as shaving a width $\delta$ from the whole surface of the solid. -Does a solid exist whose perimeter remains constant after is eroded (at least for erosion of a distance $\delta$, for all $\delta <= D$, for some $D > 0$)? - -REPLY [10 votes]: It seems that such pill exists. -Take a ball and drill a hole through it, so you get a solid torus; -we assume it has smooth boundary $\Sigma$. -By Gauss--Bonnet formula, we gave -$$\int\limits_\Sigma G=0,$$ -where $G$ denotes Gauss curvature. -Denote by $H$ the mean curvature of $\Sigma$; -it is mostly very negative in the surface of the hole. -It is easy to make a hole such that -$$\int\limits_\Sigma H >0,$$ -but, if the hole wriggles badly, then -$$\int\limits_\Sigma H\approx -\infty.$$ -It follows that for a reasonably wriggling hole, we get -$$\int\limits_\Sigma G=\int\limits_\Sigma H=0.$$ -Let $\Sigma_r$ be the $r$-equidistant surface from $\Sigma$. -Then by Weyl's formula -$$\textrm{area}\,\Sigma_r=\textrm{area}\,\Sigma+ r\cdot\int\limits_\Sigma H +r^2 \cdot\int\limits_\Sigma G$$ -for sufficiently small $r$. -Therefore the identities above imply that the area of equidistant surfaces stays constant for a short time.<|endoftext|> -TITLE: In what sense do Jones' original subfactors come from quantum SU(2) -QUESTION [8 upvotes]: In his paper Index for subfactors [Invent. Math., vol. 72 (1983), pp. 1-26], Vaughan Jones proved his remarkable index rigidity theorem, i.e., the fact that the possible index values for a (type II$_1$) subfactor are precisely those in the set -$$ -\{4\cos(\pi/n)^{2}\,:\,n\geq 3\}\cup [4,+\infty]. -$$ -In particular, he constructed a subfactor with index $\alpha$ for each value $\alpha$ in the "discrete series" $\{4\cos(\pi/n)^{2}\,:\,n\geq 3\}$. -In section 2.4 of the report https://www.birs.ca/workshops/2014/14w5083/report14w5083.pdf it is stated, with reference to these subfactors, that - -The subfactors arising from the discrete series in his [Vaughan Jones'] - article come from SU$_q$(2) at a root of unity. - -I would like to know precisely what is meant by this statement. - -REPLY [2 votes]: The fusion category $\mathcal{C}_\ell$ of unitary highest weight projective representations of level $\ell$ of the loop group $LSU(2)$ is equivalent to ${\rm Rep}({\rm SU}_q(2))$ with $q = e^{\frac{i \pi}{\ell + 2}}$ (see this paper, first paragraph p5). -Now, for any simple object $\rho$ of $\mathcal{C}_\ell$ (characterized by its spin $i \le \ell/2$), there is a Jones-Wassermann subfactor (see this Jones' survey Section 6, or also this answer): $$ \rho (L_I)'' \subseteq \rho (L_{I^c})'$$ of index $\frac{sin^{2}(p\pi/m)}{sin^{2}(\pi/m)}$ with $m=\ell + 2$ and $p=2i+1$. -At spin $1/2$, it is exactly the Jones's original subfactor of index $4cos^2(\frac{\pi}{m})$.<|endoftext|> -TITLE: Application of toric varieties for problems that do not mention them -QUESTION [14 upvotes]: I wonder whether there are problems whose statement do not mention toric varieties (nor simple polytopes, vanishing sets of binomials, etc.), but whose proof nicely and essentially uses them? -To give an idea, two simple random examples -- not for toric varieties, but for other concepts: - -a one-line computation of $H^*(K/T)$, the cohomology of flag variety, using equivariant cohomology, in an answer MSO:21670 by Allen Knutson; -the hard Lefschetz theorem, that for a Kähler $X$, the map is $\omega^{n-i}: H^i(X) \to H^{2n-i}(X)$ is an isomorphism; the now standard proof by Chern uses representation theory of $sl_2$. - -REPLY [7 votes]: There are lots of applications of toric varieties to singularities, e.g., the proof of the weak factorization theorem in characteristic zero. (Indeed, the name of the linked paper is "Torification and factorization of birational maps.") The weak factorization theorem states roughly that a birational map between (possibly singular) varieties can be factored into blow-ups and blow-downs.<|endoftext|> -TITLE: Most general maximum principle for non-integrable almost complex structures -QUESTION [8 upvotes]: Let $H\subseteq\mathbb C^n$ be a smooth co-oriented real codimension one hypersurface. If $H$ is weakly pseudo-convex, then holomorphic maps $u:\Delta\to\mathbb C^n$ ($\Delta$ denotes the unit disk) satisfy the following maximum principle with respect to $H$: - -If $u(0)\in H$ and $u$ maps a neighborhood of $0$ to the "inside" of $H$, then in fact $u$ maps a neighborhood of $0$ to $H$. - -Moreover, my understanding is that this result is sharp: if $H$ is not weakly pseudo-convex, then there are maps $u$ violating the maximum principle. -Question: Is there a sharp maximum principle for pseudo-holomorphic maps $u:\Delta\to(X^{2n},J)$ where $J$ is a (not necessarily integrable) almost complex structure and $H\subseteq X$ is as before? -I know of various sufficient conditions on $H$ which imply a maximum principle, however they generally require rather special assumptions on the local geometry of $J$ (such as the existence of a Liouville $1$-form) which seem far from the most general possible. It would be very nice to know what the "right" assumptions are here. - -REPLY [5 votes]: You may already know this, and this is not the most general statement, probably, but it does work for all almost complex structures (integrable or not): -First, some notation: Using the almost complex structure $J$ on $X$, split the exterior derivative into graded pieces $d^{s,t}:\Omega^{p,q}(X)\to\Omega^{p+s,q+t}(X)$, where $(s,t)\in\{(2,-1),(1,0),(0,1),(-1,2)\}$. Of course, $d^{2,-1}$ and $d^{-1,2}$ are tensorial (i.e., zeroth order linear differential operators) and vanish when $J$ is integrable. Maintain the standard abbreviations $d^{1,0}=\partial$ and $d^{0,1}=\overline\partial$. [Note that, because pseudo-holomorphic mappings preserve bi-degree (and, of course, 'commute' with the exterior derivative), these operators commute with pullback via pseudo-holomorphic maps.] -Given a (say, smooth) function $f:X\to\mathbb{R}$ define the Levi form of $f$ to be -$$ -{\mathsf L}(f) = i\,\partial\overline{\partial}\,f \in \Omega^{1,1}(X). -$$ -Then ${\mathsf L}(f)$ is a real-valued $(1,1)$-form on $X$. One says that $f$ is plurisubharmonic on $X$ if ${\mathsf L}(f)\ge0$ (in the sense that it is nonnegative on all (oriented) $1$-dimensional complex subspaces of $TX$). -Proposition: Let $f:X\to\mathbb{R}$ be plurisubharmonic. If $u:\Delta\to X$ is $J$-holomorphic and $f\bigl(u(z)\bigr)\le f\bigl(u(0)\bigr)$ for all $z\in\Delta$, then $f\bigl(u(z)\bigr) =f\bigl(u(0)\bigr)$ for all $z\in \Delta$. -The relation with the OP's question is this: If $X\subset\mathbb{C}^n$ and $c$ is a regular value of $f:X\to\mathbb{R}$, where $f$ is plurisubharmonic, then $H_c=f^{-1}(c)$ is weakly pseudo-convex. -It's not clear to me that a smooth weakly pseudo-convex hypersurface in $X$ is locally a level set of a plurisubharmonic function $f$, but it seems plausible. (See the comment below, which asserts that this is true in the integrable case, by a result of Diederich-Fornæss.) -The proof of the above proposition follows from the (usual) maximum principle on $\Delta$ and the formula -$$ -\frac{\partial^2(f{\circ}u)}{\partial z\,\partial \overline z} -= \frac1i \,u^*\bigl({\mathsf L}(f)\bigr)\left(\frac{\partial}{\partial z},\frac{\partial}{\partial \overline{z}}\right) \ge0, -$$ -which is a consequence of the fact that the Levi form operator commutes with pullback under pseudo-holomorphic maps. (Note, in particular, that the Nijnhuis tensor does not affect the formula.)<|endoftext|> -TITLE: Analytic properties of Eisenstein series -QUESTION [7 upvotes]: Let $\Gamma$ be a discrete subgroup of $SL_2(\mathbb{R})$ which has a cusp at $\infty.$ suppose that $\mu(\Gamma\setminus\mathbb{H})<\infty,$ consider the Eisenstein series :$$E(z,s,\Gamma)=\sum_{\gamma\in\Gamma_\infty\setminus\Gamma}\dfrac{y^s}{|cz+d|^{2s}}$$ -what is the analytic properties of $E(z,s,\Gamma)$ ? - -REPLY [6 votes]: The main properties of these Eisenstein series (meromorphic continuation, functional equation, poles and residues) are discussed and derived in Chapter 6 of Iwaniec: Spectral methods of automorphic forms (2nd edition, AMS, 2002). For their role in the spectral decomposition of $L^2(\Gamma\backslash\mathbb{H})$ see Chapter 7 of the same book. Further analytic properties in the arithmetic case (e.g. results about their value distribution) can be found in recent papers and theses (e.g. Luo-Sarnak, Spinu, Young, Huang-Xu).<|endoftext|> -TITLE: trace and involution permutations: Part I -QUESTION [13 upvotes]: Let $\operatorname{Inv}(\mathfrak{S}_n):=\{\pi\in\mathfrak{S}_n: \pi^2=1\}$ be the set of involutions in the symmetric group $\mathfrak{S}_n$. Denote $I_n:=\#\operatorname{Inv}(\mathfrak{S}_n)$. Let $\operatorname{tr}(\pi)$ be the number of fixed points of a permutation $\pi$. We know -$$I_n=\sum_{k\geq0}\frac{n!}{2^kk!(n-2k)!}.$$ -Some experimental evidence convinces me that it is possible to express $I_{n+m}$ in terms of a sum of polynomials in $\operatorname{tr}(\pi)$ over $\operatorname{Inv}(\mathfrak{S}_n)$. We may attempt a modest special case. - -Question. Is this true? - $$I_{n+2}=\sum_{\pi\in \operatorname{Inv}(\mathfrak{S}_n)}\left(\operatorname{tr}(\pi)^2+\operatorname{tr}(\pi)+2\right).$$ - -Note. $\operatorname{tr}(\pi)^2$ is understood as $\operatorname{tr}(\pi)\operatorname{tr}(\pi)$. - -REPLY [20 votes]: Let us identify an involution $\sigma$ in $\mathfrak{S}_n$ with a set partition $\Pi_\sigma$ of $[n] := \{1,2,...,n\}$ into nonempty blocks of size at most two in the obvious way: we have $\{a,b\} \in \Pi_\sigma$ if and only if $\sigma(a)=b$. This is obviously a bijection between involutions and such set partitions; and the fixed points of $\sigma$ correspond exactly to the singletons of $\Pi_\sigma$. -Now to get a set partition $\Pi'$ of $[n+2]$ with block sizes at most two from such a set partition $\Pi_\sigma$ of $[n]$ corresponding to the involution $\sigma \in \mathfrak{S}_n$, we just have to decide what blocks the elements $n+1$ and $n+2$ belong to: - -they could each be singletons (one way); -they could be in a size-two block together (one way); -$n+1$ could form a block of size two with something that was a singleton in $\Pi_\sigma$, while $n+2$ is a singleton ($\mathrm{tr}(\sigma)$ many ways); -$n+2$ could form a block of size two with something that was a singleton in $\Pi_\sigma$, while $n+1$ is a singleton ($\mathrm{tr}(\sigma)$ many ways); -$n+1$ could form a block of size two with something that was a singleton in $\Pi_\sigma$, and $n+2$ could form a block of size two with a different singleton in $\Pi_\sigma$ ($\mathrm{tr}(\sigma)(\mathrm{tr}(\sigma)-1)$ many ways). - -Adding these all together, we get $1+1+\mathrm{tr}(\sigma)+\mathrm{tr}(\sigma)+\mathrm{tr}(\sigma)(\mathrm{tr}(\sigma)-1)=\mathrm{tr}(\sigma)^2+\mathrm{tr}(\sigma)+2$ such set partitions for each involution $\sigma \in \mathfrak{S}_n$, proving your claimed identity. -EDIT: -Here is the formula for the more general case of $I_{n+m}$: -$$ I_{n+m} = \sum_{\sigma \in \mathrm{Inv}(\mathfrak{S}_n)} \sum_{j=0}^{m} \binom{m}{j} \cdot I_{m-j} \cdot \frac{\mathrm{tr}(\sigma)!}{(\mathrm{tr}(\sigma)-j)!}.$$ -It can be proved in the same manner: for each involution $\sigma \in \mathfrak{S}_n$, we choose $j$ of the elements of $\{n+1,n+2,\ldots,n+m\}$ to join with singletons in $\Pi_{\sigma}$, and hence choose an ordered subset of $j$ singletons from $\Pi_{\sigma}$ to pair with these elements, and then we put an involution on the remaining $m-j$ elements of $\{n+1,n+2,\ldots,n+m\}$ in $I_{m-j}$ ways (with the convention $I_0 =1$). -EDIT 2: -Maybe one final repackaging of this result is worth recording. -Set $I_n(t) := \sum_{\sigma \in \mathrm{Inv}(\mathfrak{S}_n)} t^{\mathrm{tr}(\sigma)}$ for all $n \geq 0$ (with $I_0(t)=1$). Then for all $m \geq 0$, -$$ I_{n+m}(t) = \left(\sum_{k=0}^{m}\binom{m}{k} \, I_{m-k}(t) \, \frac{d^k}{dt^k} \right) \, I_{n}(t).$$ -In particular, -$$ I_{n+1}(t) = \left(t+\frac{d}{dt}\right) I_{n}(t);$$ -hence we also have, -$$ I_{n+m}(t) = \left(t+\frac{d}{dt}\right)^m I_{n}(t).$$<|endoftext|> -TITLE: Existence of well-ordering of epsilon_0 in weak theories -QUESTION [6 upvotes]: In a discussion on a youtube video on the hydra game I jokingly mentioned how everyone was assuming that $\varepsilon_0$ was well-ordered. This lead to a bit of disagreement (in a nice way!) about the existence of the well-ordering of $\varepsilon_0$, assuming that $\varepsilon_0$ exists in whatever weak theory one was working in. The person with whom I was discussing seemed to think it obvious that if $\varepsilon_0$ existed, then it was an ordinal, hence well-ordered. I am very skeptical, to say the least. To clear things up, I said I would ask here: - -Is it possible, in some suitably weak theory, to define (by a code, or outright) the object $\varepsilon_0$ whose elements (or 'elements') are the usual countable ordinals given by Cantor normal form using only smaller ordinals, but the well-ordering on $\varepsilon_0$ is not available in the theory? What is a good reference for the answer, in either case? - -To me this seems like it should be true, but I can't pin down what the required weak theory is. Some form of second-order arithmetic seems reasonable to try, but I don't want one that implies the consistency of (first-order) PA, since then clearly $\varepsilon_0$ is well-ordered. - -REPLY [10 votes]: The existence of $\epsilon_0$ and its order is not a problem, its well-foundedness is. -Cantor normal forms (recursively expanded) of ordinals below $\epsilon_0$ can be written as strings over a finite alphabet, and in that form can be manipulated in weak fragments of arithmetic, say in $I\Delta_0+\mathrm{EXP}$. (Even much weaker theories would suffice, such as a theory of polynomial-time functions.) -That is, $I\Delta_0+\mathrm{EXP}$ can define $\epsilon_0$ (as a “definable class” of Cantor normal forms) in a natural way by a low-complexity ($\Delta_0(\exp)$) formula, and it can also define the order $\alpha<\beta$ on its elements, and basic operations like $\alpha+\beta$, $\alpha\cdot\beta$, and $\alpha^\beta$. The theory can prove elementary properties of these operations such as associativity, and in particular, it can prove that $<$ is a linear order. When interpreted in the standard model of arithmetic $\mathbb N$, these definitions give a structure isomorphic to the actual $(\epsilon_0,<,+,\cdot,x^y)$ from the outside world. -In a “second-order” theory of arithmetic like $\mathrm{RCA}_0^*$, we can even turn the above definition of $\epsilon_0$ into an actual object of the theory, rather than just a definable class. -However, what these relatively weak theories cannot prove is that $<$ is a well order, i.e., that it is well founded. More precisely, well foundedness is a second-order property, and thus can be properly stated only in second-order theories: $\forall X\,[\forall\alpha\in\epsilon_0\,((\forall\beta<\alpha\,\beta\in X)\to\alpha\in X)\to\forall\alpha\in\epsilon_0\,\alpha\in X]$. In first-order theories of arithmetic, it is approximated by a transfinite induction schema; $I\Delta_0+\mathrm{EXP}$, or even $\mathrm{PA}$, cannot prove the $\epsilon_0$-induction schema even for formulas of low complexity ($\Delta_0$).<|endoftext|> -TITLE: Normal closures of finitely generated subgroups of a free group -QUESTION [11 upvotes]: Is it true that for every finitelty generated subgroup $H$ of infinite index in a free - group $F$ on the two letters $\{x,y\}$, there exists a finite index - subgroup $K$ of $H$, such that the normal subgroup $N$ of $F$ - generated by $K$ is of infinite index in $F$ ? - -The normal subgroup $N$ of $F$ generated by $K$ is given by -$$N = \langle \bigcup_{g \in F} g^{-1}Kg \rangle.$$ - -REPLY [12 votes]: This is true, though the proof uses some heavy machinery! Theorem A.1 of Agol--Groves--Manning's appendix to Agol's proof of the Virtual Haken conjecture states: - -Let $G$ be a hyperbolic group, let $H\leq G$ be a quasi-convex virtually special subgroup. For any $g\in G-H$, there is a hyperbolic group $\mathcal{G}$ and a homomorphism $\phi:G\to\mathcal{G}$ such that $\phi(g)\notin\phi(H)$ and $\phi(H)$ is finite. - -Furthermore, if you examine the proof carefully, you obtain the following addendum. - -If $H$ has infinite index in $G$ then $\mathcal{G}$ can be taken to be infinite. - -These statements use Groves--Manning/Osin's combinatorial Dehn filling technology, and Wise's Malnormal Special Quotient Theorem. We don't need to remember all the definitions, except to recall that finitely generated free groups are hyperbolic and virtually special, and that all finitely generated subgroups of free groups are quasiconvex. -Therefore, taking $F=G$, we obtain an infinite, hyperbolic quotient $\phi:F\to\mathcal{G}$ such that $\phi(H)$ is finite. Taking $K=\ker\phi|_H$, we see that $K$ has finite index in $H$ and that $\ker\phi$ contains the normal closure of $K$. In particular, the normal closure of $K$ has infinite index in $F$.<|endoftext|> -TITLE: Does the Doob-Dynkin lemma hold for any measurable space that separates points? -QUESTION [13 upvotes]: This is cross-posted from math.se, where I got no responses. -Let's say that a measurable space $(Z, \mathcal Z)$ has the "Doob-Dynkin property" iff for any set $X$, measurable space $(Y, \mathcal Y)$ and function $f:X\to Y$, a function $g:X\to Z$ is $f^{-1}(\mathcal Y)$ measurable if and only if it's a measurable function of $f$. This, with $\mathbb R$ playing the role of $Z$, is the Doob-Dynkin lemma as I was taught it. -It's easy to show that not every measure space has the Doob-Dynkin property. But let's say a separating measure space is one in which, given $x\ne y$, there exists a measurable set containing $x$ but not $y$. - -Does every separating measure space have the Doob-Dynkin property? - -I think I have a simple proof that this is the case when $f$ is surjective. Can this result be proven for a non-surjective $f$, or is there a counter-example? - -Here's an outline of my proof. -First - with the above notation, if $g$ is $f^{-1}(\mathcal Y)$ measurable, and $Z$ separates points, then $g$ is a function (not necessarily measurable) of $f$. Proof: we only need to show that $f(x)=f(y)\implies g(x)=g(y)$, so let $x$ and $y$ be such that $f(x)=f(y)$. Then every set in $f^{-1}(\mathcal Y)$ either contains $x$ and $y$, or contains neither. If $g(x)\ne g(y)$, then let $A$ be some set of $\mathcal Z$ containing $g(x)$ but not $g(y)$. Then $g^{-1}(A)$ contains $x$ but not $y$ and so is not in $f^{-1}(\mathcal Y)$, thus $g$ is no measurable. Thus $g$ is a function of $f$, say $g=h\circ f$. -If $f$ is surjective then $h$ (which is unique when $f$ is surjective) is measurable. Proof: let $A\in \mathcal Z$. Since $g$ is $f^{-1}(\mathcal Y)$ measurable, $g^{-1}(A)=f^{-1}(B)$ for some $B\in\mathcal Y$. One can check that $h^{-1}(A)=B$. - -REPLY [9 votes]: A characterization of spaces with the Doob-Dynkin property was given by N. Pintacuda in his paper Sul lemma di misurabilità di Doob (1989). Unfortunately, the paper is in Italian and it doesn't show up on Google. -L. Pratelli gave another proof of Pintacuda's result in his paper Sur le lemme de mesurabilité de Doob (1990). You can find this paper (in French) here. His main result is the following (the translation from French is my own, and I changed the notation and terminology to match yours): - -Theorem: A measurable space $(Z, \mathcal{Z})$ has the Doob-Dynkin property if and only if it is separated and injective. - -A measure space $(Z, \mathcal{Z})$ is said injective if for any space $(Y, \mathcal{Y})$ and any subset $A \subset Y$ (measurable or not), any measurable map from $A$ (with the subspace sigma algebra) to $Z$ can be extended to a measurable map from $Y$ to $Z$. -The proof of the "if" part is essentially the one you gave, and the injectivity assumption is an ad hoc way to account for the fact that $f$ might not be surjective. You first prove that there exists a map $h: f(X) \to Z$ such that $g = h \circ f$. The author shows that this map is measurable if $f(X)$ is equipped with the subspace sigma algebra. By the injectivity of $Z$, it extends to a measurable map from $Y$ to $Z$. When $f$ is assumed surjective, the proof reduces to the one you gave. -Let's say that a space $Z$ has the weak Doob-Dynkin property if it has the Doob-Dynkin property with respect to surjective maps $f$. By the discussion above, a separated space has the weak Doob-Dynkin property. In fact, I believe that the converse also holds, i.e. a space has the weak Doob-Dynkin property if and only if it is separated. -Here is a sketch of a proof, which is essentially a minor modification of Proposition (1.1) in Pratelli's paper. Suppose $(Z, \mathcal{Z})$ has the weak Doob-Dynkin property. Take $X= Z$ and $g: Z \to Z$ the identity. Following Pratelli, consider the space $\tilde{Y}$ of maps $\mathcal{Z} \to \{ 0, 1\}$ and the map $\tilde{f}: Z \to \tilde{Y}$ that sends $x \in Z$ to the dirac measure $\delta_x: \mathcal{Z} \to \{0, 1 \}$. Equip $\tilde{Y}$ with the sigma algebra $\mathcal{A}$ generated by sets of the form -$$ -A_{U,V} = \{ h: \mathcal{Z} \to \{ 0, 1 \} \; : \; h(U) \in V \} -$$ -where $U$ ranges over $\mathcal{Z}$ and $V$ ranges over the power set of $\{0, 1 \}$. Now, let $Y \subset \tilde{Y}$ be the image of $\tilde{f}$, $f: Z \to Y$ be the restriction and $\mathcal{A}_Y$ be the subspace sigma algebra on $Y$. Then $f$ is surjective and one can check that it is measurable and strict, i.e. $f^{-1}(\mathcal{A}_Y) = \mathcal{Z}$ (this is Pratelli's terminology). -By the Doob-Dynkin property, there is a measurable map $h: Y \to Z$ with $id_Z = h \circ f$. But this implies that $f$ injective. It is an easy check that this is equivalent to $Z$ being separated. -[1] MR1008597 Pintacuda, Nicolò. On Doob's measurability lemma. (Italian. English summary) Boll. Un. Mat. Ital. A (7) 3 (1989), no. 2, 237–241. -[2] MR1071531 -Pratelli, Luca. -Sur le lemme de mesurabilité de Doob. (French) [On Doob's measurability lemma] Séminaire de Probabilités, XXIV, 1988/89, 46–51, -Lecture Notes in Math., 1426, Springer, Berlin, 1990.<|endoftext|> -TITLE: Pro-representability of deformation functor associated to a DG Lie algebra -QUESTION [6 upvotes]: Edit : There are several satisfying proofs in the case each $L^i$ is finite-dimensional. It is proven (for example, Hinich DG coalgebras as formal stacks) that for $A$ : local Artin ring then $\operatorname{Def}_L(A)=\operatorname{Hom}(A^*, H^0(\mathscr{C}(L)))$. So when everything is finite-dimensional, just use basic linear algebra and duality to get $\operatorname{Hom}(H^0(\mathscr{C}(L))^*, A)$. How could one reduce to that case ? If one can not, what happens ? How to obtain pro-representability ? - -I work with deformation theory and DG Lie algebras, motivation coming from Goldman-Millson theory. Here DG Lie algebras are always assumed to have $L^i=0$ for $i<0$ and $H^i(L)$ finite-dimensional. -Context : -To such a DG Lie algebra $L$ over a field $\mathbf{k}$ of characteristic zero is associated a functor $\operatorname{Def}_L$ on local Artin rings, sending $(A,\mathfrak{m}_A)$ so the set of solutions of the Maurer-Cartan equation $dx+\frac{1}{2}[x,x]=0$ in $L^1\otimes\mathfrak{m}_A$ modulo the exponential action of $L^0\otimes\mathfrak{m}_A$ by gauge transformations. It is known by classical methods in deformation theory that $\operatorname{Def}_L$ is pro-representable if $H^0(L)=0$ (Manetti, Deformation theory via differential graded Lie algebras, section 4 on the Kuranishi map). -On the other hand I'm reading about more modern deformation theories (Kontsevich, Hinich, Manetti...) They describe a functor which has several different names: Quillen $\mathscr{C}$ functor, bar construction, Chevalley-Eilenberg complex... obtained as follows : take the symmetric algebra on $L$ shifted by $1$, $\operatorname{Sym}(L[1])$, and turn it into a DG coalgebra $\mathscr{C}(L)$ (the codifferential has one part coming from the differential in $L$ and one part coming from the Lie bracket). Its dual $\mathscr{C}(L)^*$ is a DG complete local algebra. -Question : -Now I would like the following theorem to be true but I never find it stated so explicitly and each time I try to extract it from the existing litterature (Kontsevich, Hinich, Manetti) there seems to be some subtle points (about duality algebras-coalgebras, finite-dimensionality, or algebras up to quasi-isomorphism) that I may not understand. - -Let $L$ be a DG Lie algebra ($L^i=0$ for $i<0$ and $H^i(L)$ finite-dimensional). If $H^0(L)=0$ then $\operatorname{Def}_L$ is pro-represented by $H^0(\mathscr{C}(L))^*$. - -Is this theorem true ? I would like some help to understand it (I work at the most "concrete" possible level of deformation theory, with the least possible amount of $\infty$ / simplicial / derived techniques). I'm suprised I never see this so stated nor used. It means, that $\operatorname{Def}_L$ is pro-represented by some very explicit and functorial object, which is much better than choosing a non-canonical splitting of $L^1$ as is done in Manetti's lecture notes. - -REPLY [3 votes]: Edit: The answer below only works for the case where all the $L^i$'s are finite dimensional. -The statement is true. In fact, under the assumptions of the question, it is also true that ${\scr C}(L)^*$ pro-represents the associated derived moduli problem for commutative artinian dg-algebras. Recall that the derived deformation problem determined by $L$ associates to a commutative artinian dg-algebra $A$ the derived mapping space ${\rm Map}_{{\rm dg-Lie}}({\scr D}A,L)$, where ${\scr D}A$ is the dg-lie algebra Koszul dual to $A$. On the other hand, the commutative dg-algebra ${\scr C}(L)^*$ is the Koszul dual of $L$. Furthermore, under the given assumptions $L$ is also the Koszul dual of ${\scr C}^*(L)$ (see Proposition 13.3.1.1 of http://www.math.harvard.edu/~lurie/papers/SAG-rootfile.pdf). One then obtains a natural equivalence ${\rm Map}_{{\rm dg-Lie}}({\scr D}A,L) \simeq {\rm Map}_{{\rm dg-Lie}}({\scr D}A,{\scr D}{\scr C}(L)^*) \simeq {\rm Map}_{{\rm dg-com}}({\scr C}(L)^*,A)$. To get the pro-representability in the classical setting use the fact if $A$ is concentrated in degree 0 then ${\rm Map}_{{\rm dg-com}}({\scr C}(L)^*,A) \simeq {\rm Map}_{{\rm com}}(H_0({\scr C}(L)^*),A)$. Note that if the cohomology groups $H^i(L)$ are not finite dimensional but $H^0(L)$ is still 0 then then the moduli problem associated to $L$ is still pro-representable (see Proposition 13.3.3.1 of http://www.math.harvard.edu/~lurie/papers/SAG-rootfile.pdf), but just not usually by ${\scr C}(L)^*$, since in this case $L$ is generally not the Koszul dual of ${\scr C}(L)^*$.<|endoftext|> -TITLE: Can we realize the smooth metric of an Alexandrov space with nonnegative curvature by a Riemannian structure? -QUESTION [7 upvotes]: We know that a smooth Riemannian manifold with nonnegative curvature is an Alexandrov space (with induced metric) of nonnegative curvature. -What about the converse? That is, given a smooth metric d on a smooth manifold M such that M is an Alexandrov space with nonnegative curvature, can we find a smooth Riemannian structure g on M so that d is induced by g ? -Otsu and Shioya showed partial results in the paper The Riemannian structure of Alexandrov spaces. Has there been any other progress? And are there other references? - -REPLY [7 votes]: Yes, smooth distance functions plus Alexandrov means Riemannian, -but you should make all the definitions precise. -After Otsu and Shioya, there was a paper of Perelman "DC structure on Alexandrov space with curvature bounded below". The key "new" ingredient is an application of the construction from the Riemann's lecture which reconstructs the metric tensor from sufficiently many distance functions.<|endoftext|> -TITLE: Tell me an algebraic integer that isn't an eigenvalue of the sum of two permutations -QUESTION [36 upvotes]: Can you tell me an algebraic integer, with all archimedean absolute values less than 2, which is not an eigenvalue of $\pi_1 + \pi_2$ for any two permutation matrices $\pi_1,\pi_2$? -Is it conceivable that every algebraic integer satisfying the archimedean condition can be so expressed? - -REPLY [44 votes]: Yes: if $\alpha$ is an algebraic integer which obeys $|\alpha| < 2$ for all archimedean norms $|\ |$ then $\alpha$ is an eigenvalue of a sum of two permutations matrices. -I remark that this is not true when $|\alpha|=2$, for example, $(1+\sqrt{-15})/2$ is an algebraic integer with absolute value $2$ which is not twice a root of unity. -Notation: Let $A = \mathbb{Z}[\alpha]$ and let $V = A \otimes \mathbb{R}$. Since $\alpha$ is an algebraic integer, $A$ is a discrete full rank sublattice of $V$. Each archimedean norm $| \ |_v$ extends to a continuous homogenous function $V \to \mathbb{R}_{\geq 0}$ and we define $|\ |$ on $V$ by -$$|x|^2 = \sum_v |x|_v^2$$ -where the sum is over archimedean places, this is a positive definite norm on $V$. Let $c = \max_v |\alpha|_v$, so $c<2$. Note that $|\alpha x| \leq c |x|$ for any $x \in V$. -All references to "distance", "radius", etc on $V$ are with respect to the norm $|\ |$. Write $B_R$ for the closed ball of radius $R$ around $0$. -Step $1$: There is a nonnegative integer matrix $C$ whose rows sum to $2$ with eigenvalue $\alpha$. -Choose $M$ large enough that a ball of radius $M$, centered anywhere in $V$, contains a point of $A$. In other words, choose $M$ larger than the covering radius of the lattice $A$. -Choose $R$ large enough that $cR/2+M < R$. -I claim that, for any $z \in B_R \cap A$, there are $z_1$ and $z_2 \in B_R \cap A$ such that $\alpha z = z_1 + z_2$. Proof: Let $z_1$ be the nearest point of $A$ to $\alpha z/2$, and let $z_2 = \alpha z - z_1$. Then $|z_1 - \alpha z/2| = |z_2 - \alpha z/2| \leq M$ so $|z_1|$, $|z_2| \leq |\alpha z/2| + M \leq cR/2 + M \leq R$. We have shown that $z_1$ and $z_2$ are in $B_R \cap A$ as desired. -Let $z_1$, $z_2$, ..., $z_N$ be the points of $B_R \cap A$ and choose a way to write each $\alpha z_i$ as $z_{j_1}+z_{j_2}$. Then the matrix $C$ which has $1$'s in positions $(i,j_1)$ and $(i,j_2)$ (and a $2$ if $j_1=j_2$) and $0$'s elsewhere has eigenvector $(z_1 \ z_2 \ \cdots \ z_N)^T$ with eigenvalue $\alpha$ and all rows sum to $2$. -Before heading into step $2$, it is convenient to modify this argument slightly. Take $M$ large enough that any ball of radius $M$ contains at least $3$ points of $A$. In this way, we can ensure that, for any $z \in B_R \cap A$ we can write $\alpha z = z_1 + z_2$ with $z_1$, $z_2 \in B_R \cap A \setminus \{ 0 \}$. -We can then take $z_1$, $z_2$, $\dots$, $z_N$ be the points of $B_R \cap A \setminus \{ 0 \}$. It will be convenient at the next step to make sure none of the entries of our eigenvector are $0$. -Step $2$: We may assume that there is no way to permute the rows and columns of $C$ to give it the block structure $\left( \begin{smallmatrix} \ast & 0 \\ \ast & \ast \end{smallmatrix} \right)$. In other words, $C$ is irreducible in the sense of the Perron-Frobenius theorem. -Suppose we could permute the rows and columns, so $C = \left( \begin{smallmatrix} C_{11} & 0 \\ C_{21} & C_{22} \end{smallmatrix} \right)$ and $C \vec{z} = \alpha \vec{z}$ where we can write the eigenvector $\vec{z}$ as $\left( \begin{smallmatrix} \vec{z}_1 \\ \vec{z}_2 \end{smallmatrix} \right)$. Then $C_{11}\vec{z}_1 = \alpha \vec{z}_1$ We arranged above that none of the components of $\vec{z}$ is $0$, so $\vec{z}_1 \neq 0$. We see $C_{11}$ is a smaller matrix with eigenvalue $\alpha$ and row sums $2$, and we may consider it instead. -Step $3$: Making the column sums $2$. So now $C$ has row sums $2$, meaning that $(1 \ 1 \ \cdots \ 1)^T$ is a right eigenvector with eigenvalue $2$. Let $(d_1 \ d_2\ \cdots \ d_n)$ be the corresponding left eigenvector. By Perron-Frobenius, the $d_i$ are all positive. Since the corresponding eigenvalue is rational, the $d_i$ can be taken to be integers. -Build a new matrix $D$ of size $\sum d_i \times \sum d_i$, broken into $d_i \times d_j$ blocks. We will arrange that it has eigenvalue $\alpha$ with corresponding eigenvector $\vec{w}:=(z_1 \ z_1 \cdots z_1 z_2 z_2 \cdots z_2 \cdots )^T$ where $z_i$ is repeated $d_i$ times. -Within the $d_i \times d_j$ block, we will place $C_{ij}$ ones in each row. This is enough to force $D \vec{w} = \alpha \vec{w}$ and to force the row sums to be $2$. -Moreover, within the $d_j$ columns in the $j$-th block, there will be a total of $\sum_i C_{ij} d_i$ ones. We have $\sum_i C_{ij} d_i = 2 d_j$ by the choice of $d$ as a left eigenvector. -So we can place $2$ of them in each column and we win. -By the Birkhoff-von Neumann theorem, a nonnegative integer matrix with row and column sums $2$ is a sum of two permutation matrices. QED - -Here is an illustration of the trick at the end. Let -$$C = \begin{bmatrix} 1&1&0\\1&0&1\\2&0&0 \end{bmatrix}.$$ - The left eigenvector of $2$ is $(4,2,1)$. -Let -$$D = -\left[ -\begin{array}{|cccc|cc|c|} -\hline -1& 0 & 0&0 & 1& 0 &0\\ -1& 0 & 0&0 & 1& 0 &0\\ -0& 1 & 0&0 & 0& 1 &0\\ -0& 1 & 0&0 & 0& 1 &0\\ -\hline -0& 0 & 1&0 &0& 0 &1\\ -0& 0 & 1&0 &0& 0 &1\\ -\hline -0& 0 & 0&2 &0& 0 &0\\ -\hline -\end{array} \right]$$ -Then $D$ has the same eigenvalues as $C$, plus a bunch of $0$ eigenvalues.<|endoftext|> -TITLE: Classification of (complex algebraic) vector bundles on punctured affine space -QUESTION [5 upvotes]: The Quillen-Suslin theorem asserts that there are no nontrivial vector bundles over the affine space $\mathbb{A}^{n+1}$, $n\geq 0$. -Let's work over the complex numbers. What can be said about vector bundles on the punctured affine space $X_n=\mathbb{A}^{n+1}\smallsetminus\{0\}$? -According to this paper, there seem to be room for nontrivial vector bundles. -Let $\mathbb{C}^{*}$ act on $X_n$ by the action $\lambda.(x_0,\dots,x_n):=(\lambda x_0,\lambda x_1,\dots, \lambda x_n)$ whose quotient is $\mathbb{P}^n$. -Notice that equivariant v.b. on $X_n$ are in bijection -via pullback- with v.b. on $\mathbb{P}^n$, and the latter form already a rich moduli problem on its own. In this question we concentrate on the specificity of $X_n$ - -1. Is there some sort of classification of v.b. on $X_n$, taking as a starting base -say- the "classification" of stable v.b. on $\mathbb{P}^n$ given by the corresponding moduli spaces? -What about particular ranks, for example the case of line bundles? -2. Are there vector bundles on $X_n$ that are not pullbacks of v.b. on $\mathbb{P}^n$, that is, v.b. on $X_n$ that do not admit an equivariant structure? - -REPLY [3 votes]: I spoke to my colleague Song Sun, and he reminded me of a discussion that he and I had about Question 2 some time ago. For $n\geq 2$, there are many examples of locally free sheaves on $X_{n} = \mathbb{A}^{n+1}\setminus\{0\}$ that admit no equivariant structure. Denote by $S$ the polynomial ring $\mathbb{C}[x_0,\dots,x_n]$. Denote by $S_+$ the maximal ideal $\langle x_0,\dots,x_n \rangle$. Let $\underline{f}=(f_0,\dots,f_n)$ be a regular sequence of elements in $S_+$. -Associated to this regular sequence there is a Koszul complex $(K_\bullet(\underline{f}),d_\bullet)$ of $S$-modules where $K_0(\underline{f})$ equals $S$, where $K_1(\underline{f})$ equals $S^{\oplus(n+1)}$, where $K_r(\underline{f})$ is the $r^{\text{th}}$ exterior power of $K_1(\underline{f})$, and where $d_r:K_r(\underline{f}) \to K_{r-1}(\underline{f})$ is the unique sequence of $R$-module homomorphisms such that $d_1(g_1,\dots,g_n) = f_1g_1 + \dots + f_ng_n$ and such that $(K_{\bullet}(\underline{f}),d_\bullet)$ is a differential graded $R$-algebra. To be precise, $d_{r-1}\circ d_r$ equals the zero homomorphism, and $d_{r+s}(\alpha \wedge \beta) = d_r(\alpha)\wedge \beta + (-1)^r\alpha\wedge d_s(\beta)$ for every $r,s\geq 0$, for every $\alpha\in K_r(\underline{f})$, and for every $\beta\in K_s(\underline{f})$. -For every integer $r\geq 1$, denote by $Z_r(\underline{f})$ the kernel of $d_r$. Since $\underline{f}$ is a regular sequence, this is the same as the image of $d_{r+1}$. Thus, also define $Z_0(\underline{f})$ to be the image of $d_1$, i.e., the ideal $I$ generated by $\underline{f}$. In particular, $Z_{n+1}(\underline{f})$ is the zero module, and $Z_n(\underline{f})$ is $K_{n+1}(\underline{f})$. -Fact 1. For all $r$ with $1\leq r \leq n$, $Z_r(\underline{f})$ is a reflexive $S$-module. -Proof. By construction $K_r(\underline{f})$ is a free $S$-module, hence reflexive, and $K_{r-1}(\underline{f})$ is torsion-free (even free). The kernel of every $S$-module homomorphism from a reflexive module to a torsion-free module is reflexive. QED -Fact 2. For $M=Z_{n-1}(\underline{f})$, the first Fitting ideal $\text{Fitt}_1(M)$ equals the ideal $I$ generated by $(f_0,\dots,f_n)$. -Proof. The Fitting ideal can be computed from any finite presentation of $M$. For $Z_{n-1}(\underline{f})$, one such presentation is $d_{n+1}:K_{n+1}(\underline{f}) \to K_n(\underline{f})$. By self-duality of the Koszul complex, the Fitting ideal of $d_{n+1}$ is $I$. QED -Since the Fitting ideal is intrinsic, if $M$ is equivariant, then $I$ is a homogeneous ideal. However, there are many $S_+$-primary ideals $I$ that are generated by a regular sequence, yet are not homogeneous ideals. For instance, one example is $$\underline{f}=(x_0,x_1,\dots,x_{n-2},x_{n-1}-x_n^2,x_n^3).$$ For $n\geq 2$, for such $\underline{f}$, the module $M=Z_{n-1}(\underline{f})$ is reflexive, the restriction of $\widetilde{M}$ to $\mathbb{A}^{n+1}\setminus\{0\}$ is locally free (visibly it is locally free on each $D(f_i)$ for $i=0,\dots,n$). Yet $\widetilde{M}$ is not equivariant since $I$ is not a homogeneous ideal. -Edit. Song Sun asked the following variant of the question. For a reflexive $S$-module that is locally free on $\mathbb{A}^{n+1}\setminus\{0\}$ and whose Fitting ideals are all homogeneous, is the module equivariant? Also, note that every module as constructed above (also allowing other syzygy modules of the complexes) has rank $\geq n$. So here is a second variant: is every reflexive $S$-module that is locally free on $\mathbb{A}^{n+1}\setminus\{0\}$ equivariant provided that the rank is $\leq n-1$?<|endoftext|> -TITLE: Does there exists a Fukaya category with no objects -QUESTION [6 upvotes]: ... and really without even the possibility of having objects, so it's not a matter of just finding the "correct" flavour of Fukaya category to use. -Question: Does there exist interesting symplectic manifolds $(M,\omega)$ without: - -Any unobstructed Lagrangians (and therefore, with a $Ob(Fuk(M,\omega)) = \varnothing$)? -Any (smooth) Lagrangians? -Any (immersed) Lagrangians? - -REPLY [3 votes]: Let's assume that $M$ is an $n$-dimensional exact symplectic manifold and we are only interested in Fukaya categories of closed exact Lagrangian submanifolds. Then for any subcritical Weinstein manifold, $\mathcal{F}(M)$ is trivial. This follows from the fact that there is a well-defined open-closed string map -$\mathit{HH}_\ast(\mathcal{F}(M),\mathcal{F}(M))\rightarrow\mathit{SH}^{\ast+n}(M)$ -and $\mathit{SH}^\ast(M)=0$, which is in fact nothing else but a modernization of a classical argument of Viterbo, which shows that any closed exact Lagrangian submanifold will contribute non-trivially to symplectic cohomology. -The above result can be generalized in both directions. Namely you can assume $M$ to be a subflexible Weinstein manifold rather than subcritical. For example, there are certain exotic cotangent bundles of spheres constructed by Maydanskiy and Seidel: https://arxiv.org/abs/0906.2230. On the other hand, you can also relax the assumption that the Lagrangian submanifolds under consideration are exact. For example, Seidel and Smith proved that for any Liouville 4-manifold $M$ which contains a weakly unobstructed Lagrangian torus $T$ with $\mathit{HF}^\ast(T,T)\neq0$, $\mathit{SH}^\ast(M)\neq0$. An application of this result is given in the paper of Lekili-Maydanskiy: https://arxiv.org/abs/1202.5625. They proved that for certain rational homology balls $B_{p,q}$, one has $\mathit{SH}^\ast(B_{p,q})\neq0$ but there is no exact Lagrangian submanifold in $B_{p,q}$. Note that this is an illustration of the fact that the above open-closed map is far from an isomorphism, it is an isomorphism only after certain non-compact exact Lagrangian submanifolds of $M$ are taken into consideration. -In the study of homological mirror symmetry, one usually defines the Fukaya category of an exact symplectic manifold by using only exact Lagrangian submanifolds. However, in view of the examples of Lekili-Maydanskiy, this does not seem to be a good definition. Since in this case the mirrors of $B_{p,q}$ are just certain finite coverings of Milnor fibers of $(A_{p-1})$ singularities with superpotantials, which admit non-trivial triangulated categories of matrix factorizations. It is believed by many people that a correct definition of $\mathcal{F}(M)$ should also involve certain immersed Lagrangian submanifolds as its objects. A good supporting evidence of this point of view seems to be the paper of Evans-Smith: https://arxiv.org/abs/1606.08656. Based on this viewpoint, I believe that there is always an immersed exact Lagrangian submanifold whose Floer cohomology is non-trivial in any Liouville domain with $\mathit{SH}^\ast(M)\neq0$. -Another typical example which is discussed in Seidel's lecture notes on categorical dynamics is $\mathbb{C}^2$ with a conic $\{xy=1\}$ removed. This manifold is algebraically and symplectically self-mirror. In this case, $M$ contains an exact Lagrangian torus $T$, and an immersed Lagrangian sphere $S\subset M$ can also be explicitly constructed. Computation shows that -$\mathit{HF}^\ast(S)\cong\mathit{HF}^\ast(T)$ -Recall that the SYZ mirror construction of $M$ starts from a Lagrangian torus fibration on $M$ which has a unique singular fiber. Because of this, it's easy to see the algebraic counts of holomorphic discs only has to go through one wall and therefore the mirror $M^\vee\cong M$ consists of two $(\mathbb{C}^\ast)^2$ charts patched together using the wall-crossing formula of algebraic counts of holomorphic discs bounded by Lagrangian fibers on two different chambers. Under mirror symmetry, $T$ corresponds to the skyscraper sheaf of a point on one of the $(\mathbb{C}^\ast)^2$ chart, but $S$ corresponds to the skyscraper sheaf of the origin. -Another supporting evidence of my conjecture is the work of Alston, you can find many computations of Floer cohomologies of immersed Lagrangian spheres in his paper: https://arxiv.org/abs/1311.2327. -By the way, if you don't assume non-triviality of Floer cohomology or non-displaceability, then your second and third questions do not make sense.<|endoftext|> -TITLE: Convergence of Radon Nikodym derivatives -QUESTION [6 upvotes]: I apologise in advance if my question is too basic. -Some notation: - -$(X,\cal{X})$ denotes a measurable metric space -where $X$ is a metric space and -$\cal{X}$ is the associated Borel sigma algebra. -$B(X)$ is the space of all bounded continuous -functions defined on $X$. - -Let $\{\mu_n\}$ and $\{\nu_n\}$ be sequences of -probability measures on the above measurable space $(X, \mathcal{X})$. -Assume -that -each $\mu_n$ is absolutely continuous with respect to -$\nu_n$, with an density $h_n\in B(X)$. -Suppose that $\mu_n\to \mu$, $\nu_n\to \nu$ -in the weak star topology and $h_n$ converges -to a bounded continuous function $h$. -Question: - I would like to know if $\mu\ll\nu$. If so, is $h$ the density? If not, is there some condition in order -to have $\mu\ll\nu$? -Other information that can be useful - is that each $\nu_n$ and $\mu_n$ has support in a -compact subset $K_n\subset X$ which -increase to $X$, i.e, $X=\bigcup K_n$. -Edit: $h_n$ converges to $h$ uniformly in compacts sets - -REPLY [3 votes]: I'm going to assume that your space is locally compact (as well as $\sigma$-compact), so that $X$ is the union of a sequence of compact sets where each lies in the interior of the next. -In this case, the answer to your question is yes. -Fix any $g \in B(X)$ with $g \geq 0$. We need $\int_X g \, d\mu_n \to \int_X gh \, d\nu$. -Fix $\varepsilon>0$. Let $M_g:=\sup_{x\in X} g(x)$ and likewise $M_h:=\sup_{x\in X} h(x)$. Let $K \subset X$ be a compact set with $K^\circ$ sufficiently large that -$$ \mu(X \setminus K^\circ) < \frac{\varepsilon}{4M_g} \hspace{4mm} \textrm{and} \hspace{4mm} \nu(X \setminus K^\circ) < \frac{\varepsilon}{4M_gM_h}. $$ -Let $N \in \mathbb{N}$ be such that for all $n \geq N$, -$$ \mu_n(X \setminus K^\circ) < \frac{\varepsilon}{4M_g} \ , \hspace{4mm} \max_{x \in K} |h_n(x)-h(x)| < \frac{\varepsilon}{4M_g} \ , \hspace{4mm} \left| \int_{K^\circ} gh \, d\nu_n - \int_{K^\circ} gh \, d\nu \right| < \frac{\varepsilon}{4}. $$ -The third statement is possible since $\nu(\partial K)<\frac{\varepsilon}{4M_gM_h}$ and so $\int_{\partial K} gh \, d\nu < \frac{\varepsilon}{4}$. (This reasoning can be seen by adapting the argument for 3,4$\Rightarrow$5 on p3 of here.) -Then for all $n \geq N$, we have that -\begin{align*} -\Bigg| \int_X g \, d\mu_n & - \int_X gh \, d\nu \Bigg| \\ -&\leq \ \left| \int_{K^\circ} gh_n \, d\nu_n - \int_{K^\circ} gh \, d\nu \right| \ + \ \int_{X \setminus K^\circ} g \, d\mu_n \ + \ \int_{X \setminus K^\circ} gh \, d\nu \\ -&< \ \left| \int_{K^\circ} gh_n \, d\nu_n - \int_{K^\circ} gh \, d\nu \right| \ + \ \frac{\varepsilon}{4M_g}M_g \ + \ \frac{\varepsilon}{4M_gM_h}M_gM_h \\ -&\leq \ \left| \int_{K^\circ} g.\!(h_n - h) \, d\nu_n \right| \ + \ \left| \int_{K^\circ} gh \, d\nu_n - \int_{K^\circ} gh \, d\nu \right| \ + \ \frac{\varepsilon}{2} \\ -&< \ M_g.\max_{x \in K} |h_n(x)-h(x)| \ + \ \frac{3\varepsilon}{4} \\ -&< \ \varepsilon. -\end{align*}<|endoftext|> -TITLE: Every PD group is $\pi_1$ of an aspherical manifold -QUESTION [6 upvotes]: It is conjectured that for a discrete, finitely presented group $G$ such that $BG$ satisfies Poincaré duality, there actually exists a closed manifold $M$ which is homotopy equivalent to $BG$. -This is somehow pointing in the opposite direction as Borel's conjecture, which implies that the homeomorphism type of such a manifold $M$ is uniquely determined. -Who conjectured this first? Is it also due to Borel, or was it Wall, or somebody else? - -REPLY [6 votes]: The reference for the first appearance of the conjecture (still without the condition that the PD group has to be a priori finitely presented) seems to be http://www.worldcat.org/title/homological-group-theory-proceedings-of-a-symposium-held-at-durham-in-september-1977-on-homological-and-combinatorial-techniques-in-group-theory/oclc/6022486 from 1977. -You find a survey on the state of the subject around 2000 in Davis: "Poincaré duality groups".<|endoftext|> -TITLE: Yoneda Lemma for internal presheaves -QUESTION [8 upvotes]: I'm looking for a reference explaining under what conditions the internal Yoneda lemma holds; in particular, I am wondering if it is known what properties of the ($2$-)category of categories are responsible for the following. -Take a (small) category $\mathcal C$, i.e. a category internal to the category of sets. Promote it to a category internal to the category of (small) categories via the arrow category $\mathcal C^\to$. Then (split) fibrations over $\mathcal C$ are a reflective subcategory of the category of internal presheaves on $\mathcal C$, and furthermore are precisely the internal presheaves for which the Yoneda lemma holds. - -REPLY [10 votes]: The Yoneda lemma holds in any finitely complete $2$-category $\mathscr{K}$, such as the $2$-category $\text{Cat}(\mathscr{C})$ of internal categories in a finitely complete category $\mathscr{C}$. -The statement is that for any object $B$ of $\mathscr{K}$, any morphism $b \colon 1 \to B$, and any discrete fibration [i.e. internal presheaf in the case $\mathscr{K} = \text{Cat}(\mathscr{C}$)] $p \colon E \to B$, there is a bijection between hom-sets in the slice category $\mathscr{K}/B$ -$$ \text{Hom}(d\colon B/b \to B,p\colon E\to B )\cong \text{Hom}(b \colon 1 \to B, p\colon E \to B)$$ -induced by the canonical morphism from $(b \colon 1 \to B)$ to the projection $(d\colon B/b \to B)$. -This is due to Ross Street: a more general version of the internal Yoneda lemma (for general morphisms $b \colon X \to B$ and two-sided discrete fibrations) was proved in the paper - -Street, Ross. Fibrations and Yoneda's lemma in a $2$-category. Category Seminar (Proc. Sem., Sydney, 1972/1973), pp. 104--133. Lecture Notes in Math., Vol. 420, Springer, Berlin, 1974. - -For further generalizations see Street's paper Categories in categories, and size matters, which moreover contains a proof of the following fact. For any internal category $C$ in a finitely complete category $\mathscr{C}$, there is a isomorphism between the category of split fibrations over $C$ in the $2$-category $\mathscr{K} = \text{Cat}(\mathscr{C})$, and the category of discrete fibrations (i.e. internal presheaves) over $\text{sq}C$ in the $2$-category $\text{Cat}(\mathscr{K}) = \text{Dbl}(\mathscr{C})$. Here $\text{sq}C$ denotes the internal category in $\mathscr{K}$ whose "object of objects" is $C$ and whose "object of morphisms" is $C^\mathbf{2}$. I believe this addresses your last question.<|endoftext|> -TITLE: On a generalization of the classical Cauchy's functional equation -QUESTION [5 upvotes]: I start with some known preliminaries on the problem: -Classical result. The one-dimensional Cauchy functional equation -$$ -\forall x,y \in \mathbb{R}, \,\,\,f(x+y)=f(x)+f(y) -$$ -with $f:\mathbb{R}\to \mathbb{R}$ is only solved by the trivial solutions $f(x)=cx$, for some $c \in \mathbb{R}$, if $f$ satisfies for some additional conditions, e.g., continuity. -Classical result with restricted domain. Now let $\mathbb{R}^+:=(0,\infty)$. It is clear from the proof of the above classical result that if $f:\mathbb{R}^+\to \mathbb{R}^+$ is a continuous function such that -$$ -\forall x,y \in \mathbb{R}^+, \,\,\,f(x+y)=f(x)+f(y) \, , -$$ -then there exists $c \in \mathbb{R}^+$ such that $f(x)=cx$ for all $x$. -Multidimensional Cauchy functional equation. It is also well known that if $f:\mathbb{R}^2\to \mathbb{R}$ is a continuous function such that -$$ -\forall x,y \in \mathbb{R}^2, \,\,\,f(x+y)=f(x)+f(y), -$$ -then there exist $A,B \in \mathbb{R}$ such that $f(x,y)=Ax+By$ for all $(x,y) \in \mathbb{R}^2$. -I know that the following generalization holds true as well. In particular, I already know how to prove it, by using a variant of the classical proof. In the following, a cone $C\subseteq \mathbb{R}^2$ is a set for which $\alpha x+\beta y \in C$ whenever $\alpha,\beta \in \mathbb{R}^+$ and $x,y \in C$. - -Fact. Let $C\subseteq \mathbb{R}^2$ be a non-empty cone and $f:C \to \mathbb{R}$ be a continuous function such that - $$ -\forall x,y \in C, \,\,\,f(x+y)=f(x)+f(y). -$$ - Then there exist $A,B \in \mathbb{R}$ such that $f(x,y)=Ax+By$ for all $(x,y) \in C$. - -Is it a known result? In such case, does anyone have a reference for this result? - -REPLY [3 votes]: As I have just learned from Janusz Matkowski, this follows from Theorems 5.5.2 and 18.2.1 in Kuczma's book (the same mentioned in my comments to the OP), after ruling out the trivial case when the cone $C$ is a line or a half-line. In particular, Theorem 5.5.2 is about the (unrestricted) Cauchy functional equation in $\mathbf R^n$, and shows that the only continuous solutions are precisely the functions of the form $\mathbf R^n \to \mathbf R: (x_1, \ldots, x_n) \mapsto \sum_{i=1}^n c_i x_i$ with $c_1, \ldots, c_n \in \mathbf R$, while Theorem 18.2.1 reads as follows: - -Let $G$ and $H$ be abelian groups (written additively), and let $S$ be a subsemigroup of $G$ such that $G = S - S := \{x-y: x, y \in S\}$. If $g : S \to H$ is a (semigroup) homomorphism, then there exists a unique - homomorphism $f : G \to H$ whose restriction to $S$ is $g$. - -Each of these results is somehow a piece of folklore (for people -working primarily on functional equations and related topics), but I think it is not harmful to have a reference.<|endoftext|> -TITLE: Approximating dense subspaces of Fréchet spaces -QUESTION [6 upvotes]: If $H$, $H_0$ are two separable Hilbert spaces and $H$ is continuously and densly embedded in $H_0$, it is possible to construct a sequence of linear operators -$$ P_n : H_0 \to H $$ -such that for all $x \in H_0$ one has convergence $P_n x \to x$ in the $H_0$-norm. -The motivation is to generalize the idea of smoothing operators. For example, $H^2(\mathbb R)$ is densely embedded in $L^2(\mathbb R)$ and it is well known that one can approximate each $L^2$-function by an $H^2$-function. In this case $P_n$ could be a convolution with a smooth kernel with width $1/n$. -Can this be generalized to Fréchet spaces or even beyond? If $H$ and $H_0$ are Fréchet spaces rather than Hilbert spaces, is it always possible to construct a sequence of bounded operators $P_n$ as above? -In the Hilbert case one way to construct $P_n$ is by considering an unbounded, self-adjoint operator $A : D(A) \to H_0$, with $D(A)=H$, that represents the inner product via -$$ \langle v,w \rangle_H = \langle Av, Aw \rangle_{H_0}\,. $$ -If $\{ P_\Omega\}$ is the spectral measure accociated to $A$, then we can use $P_{[-n,n]}$ as our sequence of operators. It is less clear to me, what to with Fréchet spaces. - -REPLY [3 votes]: Thanks to Jochen, Matthew and Bill, this is a detailed proof for Fréchet spaces. -Proposition. Let $E$ be a separable Fréchet space with the bounded approximation property, $F$ a topological vector space, continuously and densely embedded in $E$. Then there exists a sequence of continuous linear maps $P_n : E \to F$, such that -$$ -\forall x \in E\,:\, P_n x \to x \text{ in }E\,. -$$ -Proof. -Let $(x_n)_{n \in \mathbb N}$ be a countable, dense sequence in $E$ and $(\|\cdot\|_n)_{n \in \mathbb N}$ an increasing fundamental system of seminorms. We assumed that $E$ has the bounded approximation property, hence there exists an equicontinuous sequence of linear maps $T_n : E \to E$ with finite rank that converge to $\operatorname{Id}_E$, uniformly on compact sets. By passing to a subsequence we can assume that -$$ -\| T_n x_j - x_j \|_n \leq \frac 1n \text{ for }j \leq n\,. -$$ -Due to equicontinuity there exists for each $m$, an $N_m \in \mathbb N$ and $C_m>0$ such that -$$ -\forall n \in \mathbb N\,,\; \forall x \in E\,:\, \| T_n x \|_m \leq C_m \| x \|_{N_m}\,. -$$ -For each $n$, the space $T_n(E)$ is finite dimensional. Let $n'=n'(n)$ be such that $\|\cdot\|_{n'}$ is a norm on $T_n(E)$. We can construct a map $S_n : T_n(E) \to F$ with -$$ -\| S_ny - y \|_{n} \leq \frac 1n \| y \|_{n'}\,, -$$ -for all $y \in T_n(E)$. -To see that this is possible choose a basis $y_1, \dots, y_m$ of $T_n(E)$ and note that it is sufficient to define $S_n(y_i) \in F$, such that $\| S_n(y_i) - y_i \|_{n}$ is small enough. This is possible, because $F$ is dense in $E$. Define $P_n = S_n T_n$. -We have to show convergence $P_n x \to x$. Fix $x \in E$ and a seminorm $\|\cdot\|_m$. For $n$ and $k$ satisfying $m ,\,N_m,\,N_{m'} \leq n$ and $k \leq n$ we have -\begin{align*} -\| P_n x &- x \|_m \leq \|S_n T_n(x-x_k) -T_n(x-x_k) \|_m + \| T_n(x-x_k) \|_m + \\ -&\qquad\qquad\qquad -+ \|S_n T_n x_k - T_n x_k \|_m +\| T_n x_k - x_k\|_m + \| x_k - x\|_m \\ -&\leq \frac 1n \| T_n(x-x_k) \|_{m'} + C_m \| x - x_k \|_{N_m} + \frac 1n \| T_n x_k \|_{m'} + \frac 1n + \|x_k - x \|_m \\ -&\leq -\frac {C_{m'}}{n} \| x-x_k \|_{N_{m'}} + C_m \| x - x_k \|_{N_m} -+ \frac {C_{m'}}n \left( \| x\|_{N_{m'}} + \| x-x_k \|_{N_{m'}} \right) + \\ -&\qquad\qquad + \frac 1n + \|x_k - x \|_m \,. -\end{align*} -We see that by choosing $n$ large enough and $\|x - x_k\|_n$ small enough we can achieve convergence.<|endoftext|> -TITLE: Continuing generalization of the Simson line -QUESTION [6 upvotes]: In 2014, I found a nice result in plane geometry, the result is a generalization of the Simson line theorem, and there are nine proofs for this result were published in [1]-[7]. Continuing, I find a new generalization of the old result as follows: -Problem: Let $ABC$ be a triangle, let $P$ be a point in the circumcircle, the circumcenter is $O$. Let $Q$ be the point in the plane. The circles $(APQ), (BPQ), (CPQ)$ meet $OQ$ again at $A', B', C'$ respectively. Let $A_1, B_1, C_1$ be the the projections of $A', B', C'$ onto $BC, CA, AB$ respectively. Then $A_1, B_1, C_1$ are collinear, and the new line through a fixed point on the Nine point circle when $Q$ be moved on the given line (or $P$ be moved in the circumcircle). When $Q$ in infinity we get the old result (the result I found in 2014). - -My question, could you give a proof for this problem? - - -Note: Check the result is true with applet by click here -Note: You can see some problem around this configuration in -** References:** -[1]-Nguyen Van Linh, Another synthetic proof of Dao's generalization of the Simson line theorem, Forum Geometricorum, 16 (2016) 57--61. -[2]-Nguyen Le Phuoc and Nguyen Chuong Chi (2016). 100.24 A synthetic proof of Dao's generalisation of the Simson line theorem. The Mathematical Gazette, 100, pp 341-345. doi:10.1017/mag.2016.77. -[3]-Leo Giugiuc, A proof of Dao’s generalization of the Simson line theorem, tạp chí Global Journal of Advanced Research on Classical and Modern Geometries, ISSN: 2284-5569, Vol.5, (2016), Issue 1, page 30-32 -[4]-Tran Thanh Lam, Another synthetic proof of Dao's generalization of the Simson line theorem and its converse, Global Journal of Advanced Research on Classical and Modern Geometries, ISSN: 2284-5569, Vol.5, (2016), Issue 2, page 89-92 -[5]-Ngo Quang Duong, A generalization of the Simson line theorem, to appear in Forum Geometricorum. -[6]-Three other proofs by Telv Cohl, Luis Gonzalez, Tran Quang Huy A Generalization of Simson Line -[7]-Another proof https://www.artofproblemsolving.com/community/c6h1075523p5181203 - -REPLY [4 votes]: Let $CC'$ meet a circle $\omega=(ABC)$ in a point $S\ne C'$. Then $\angle (CP,CS)=\angle (CP,CC')=\angle (QP,QC')=(QP,QO)$. Thus $AA'$, $BB'$ pass through the same point $S$. The following argument is not synthetic, but it explains what is this fixed point on an Euler circle and what is another point in which $A_1B_1C_1$ meets Euler circle. Thus it hopefully may help with a synthetic argument too. -Consider the complex coordinates for which $\omega=\{z:|z|=1\}$, $A,B,C$ correspond to complex numbers $a,b,c$, $OQ$ to a real line, $S$ to $s$. Then $C'$ corresponds to $c'=(c+s)/(1+cs)$ (this is a formula for central projection from $\omega$ to a real line from the point $s$, as may be checked for three points $1,-1,-s$). Next, a projection of $z$ to a line between $a,b$ is $(z-\bar{z}ab+a+b)/2$, as may be checked for points $a,b,0$. So, $C_1$ corresponds to $c_1=(c'(1-ab)+a+b)/2$. Denote $c_2=2c_1-(a+b+c)$. Note that $z\rightarrow 2z-(a+b+c)$ is a homothety which sends Euler circle of $ABC$ to $\omega$. Thus for points $a_2,b_2,c_2$ we should prove that they are collinear and the line passes through a point on $\omega$ not depending on $s$. We get $c_2=c'(1-ab)-c$ and I claim that $c_2$ lies on a line between $s$ and $-abc$. Indeed, the direction between $s$ and $-abc$ is a direction of $s+abc$. The direction between $s$ and $c_2$ is a direction of $c_2-s=-c'(ab+cs)$, that is, direction of $ab+cs$, but the ratio of $s+abc$ and $ab+cs$ is indeed real.<|endoftext|> -TITLE: groupring morphisms and bialgebra -QUESTION [6 upvotes]: Let $G_{1}$ and $G_{2}$ be two groups. Suppose that we have a morphism $\mathbb{Z}[G_{1}]\rightarrow \mathbb{Z}[G_{2}] $ of bialgebras is it true that this morphism comes from a morphism of groups $G_{1}\rightarrow G_{2}$ ? -In case when the answer is "no", is it true that if $\mathbb{Z}[G_{1}]\rightarrow \mathbb{Z}[G_{2}] $ is an isomorphism of rings then there exists an isomorphism of bialgebras $\mathbb{Z}[G_{1}]\rightarrow \mathbb{Z}[G_{2}] $ ? - -REPLY [5 votes]: I will show that any bialgebra homomorphism $\mathbb QM_1\to \mathbb QM_2$ of monoid algebras is induced by a monoid homomorphism $M_1\to M_2$. This will imply what the OP wants. -An element $g$ of a bialgebra is called group-like if $\Delta(g)=g\otimes g$ and $\eta(g)=1$ where $\eta$ is the counit. It is well known that the group-like elements are linearly independent and form a monoid (cf. Lemma 2.1 http://www.math.wisc.edu/~passman/balgebra.pdf). -If $M$ is a monoid then the elements of $M$ are group-like in the monoid algebra and from the above linear independence they are the only group-like elements. -Since bialgebra morphisms preserve grouplikes it follows any bialgebra morphism of monoid algebras is inducted by a monoid homomorphism. Hence any bialgebra morphism of group algebras is induced by a group homomorphism. - -REPLY [4 votes]: Your first claim is true even if you substitute $\mathbb{Z}$ with any integral domain $\Bbbk$. Actually what is true is that we have a bijection -$$\text{Bialg}_\Bbbk(\Bbbk[G],B)\cong\text{Mon}(G,\mathcal{G}(B))$$ -where $\mathcal{G}(B)$ denotes the monoid of group-like elements in $B$ and $B$ is a $\Bbbk$-algebra via a ring homomorphism $\gamma:\Bbbk\to B$. -Notice that if $f:\Bbbk[G]\to B$ is a morphism of bialgebras then the relations -\begin{gather} -\Delta_B(f(g))=(f\otimes_\Bbbk f)(\Delta_{\Bbbk[G]}(g))=f(g)\otimes_\Bbbk f(g), \\ -\varepsilon_B(f(g))=\varepsilon_{\Bbbk[G]}(g)=1_{\Bbbk}, -\end{gather} -imply that $f(g)$ is group-like in $B$ for every $g\in G$. Thus we may (co)restrict $f$ to $f':G\to \mathcal{G}(B)$, which gives the assignment from left to right. -Conversely, every morphism of monoids $f:G\to \mathcal{G}(B)$ can be extended in a unique way to a morphism of $\Bbbk$-algebras $F:\Bbbk[G]\to B$ by letting -$$F\left(\sum_{g\in G}k_gg\right)=\sum_{g\in G}\gamma(k_g)f(g)$$ -and this turns out to be a morphism of bialgebras. -If you take $B=\Bbbk[H]$ for $H$ another group, then you may check that $\mathcal{G}(\Bbbk[H])=H$ (here you should need the integral domain hypothesis). -About the question you asked in the comments, every $\Bbbk$-bialgebra morphism $ f:A\to B$ between Hopf algebras preserves the antipodes as both $fS_A$ and $S_Bf$ are convolution inverses of $f$ in $\text{Hom}_\Bbbk(A,B)$ (see also Sweedler, Hopf algebras, Lemma 4.0.4). -To complete Benjamin answer to your last question, in this paper you may find an example of two non-isomorphic groups whose group algebras are instead isomorphic.<|endoftext|> -TITLE: Etymology of "exterior" in "exterior calculus" -QUESTION [20 upvotes]: What is the origin of the term "exterior" in "exterior calculus"? How does this term relate to "interior products" and "inner products", if it does at all? - -REPLY [35 votes]: I think it's well known to have been introduced by Grassmann. He explains the word choice in Die lineale Ausdehnungslehre (1844, pp. x-xi): - -I have shown how one can understand as product of two segments the parallelogram (...); this product is distinguished in that factors can only be interchanged with a sign change, while at the same time the product of two parallel segments vanishes. This concept stood in contrast to another (...) Namely when I projected one segment perpendicularly onto another, then the arithmetic product of this projection with the segment projected upon represented likewise a product of the two, insofar as the distributive relation with addition also held. But that product was of a whole other kind than the first, as its factors were interchanged without sign change, and the product of two perpendicular segments vanished. I named the former product exterior, the latter interior, reflecting that the former was nonzero only when involving independent directions, the latter only when involving a shared i.e. partly common one. - -and again in the second edition (1862, p. 51): - -I chose the name exterior multiplication to emphasize that the product is only nonzero when one factor lies entirely outside the span of the other. Exterior multiplication stands in contrast with the interior one (chap. 4).<|endoftext|> -TITLE: Are there only finitely many distinct cubic walk-regular graphs that are neither vertex-transitive nor distance-regular? -QUESTION [7 upvotes]: The class of walk-regular graphs contains the vertex-transitive graphs and the distance-regular graphs. However, there are walk-regular graphs that are neither vertex-transitive nor distance-regular. In particular, I believe that I found a first example of a cubic walk-regular graph that is neither vertex-transitive nor distance-regular on 20 vertices: its Graph6 code is SsP@@?OC?S@C@_@C?K?A_?AG?D??C_??]. -My question here is: are there only finitely many cubic walk-regular graphs (up to isomorphism) that are neither vertex-transitive nor distance-regular (or is it e.g. possible to construct an infinite family of graphs of this kind)? -There are e.g. only finitely many distinct cubic distance-regular graphs - could it e.g. also be true that there are only finitely many distinct non-vertex-transitive cubic walk-regular graphs? - -REPLY [6 votes]: What follows is a proof that semisymmetric graphs are walk-regular. -Say vertices $u$ and $v$ in a graph $X$ are cospectral if the graphs $X\setminus u$ and $X\setminus v$ are cospectral. If $X$ is semisymmetric, then the vertices in each colour class are cospectral. Two vertices $u$ and $v$ in a bipartite graph are cospectral if, for each non-negative integer $k$, the number of closed walks of length $2k$ starting at $u$ equals the number starting at $v$. [This is proved, for example, in Section 8.13 in one of my favourite texts on algebraic graph theory. :-)] -The adjacency matrix $A$ of any bipartite graph can be written in the form -\[ - A =\begin{pmatrix}0&B\\ B^T&0\end{pmatrix} -\] -and hence -\[ - A^{2k} = \begin{pmatrix}(BB^T)^k&0\\ 0&(B^TB)^k&\end{pmatrix} -\] -Since $X$ is semisymmetric, the diagonal entries of $(BB^T)^k$ are all equal, as are the diagonal entries of $(B^TB)^k$. -By the cyclic symmetry of trace, we see that $(BB^T)^k$ and $(B^TB)^k$ have the same trace. Therefore the diagonal entries of these two matrices are equal, and it follows that $X$ is walk regular.<|endoftext|> -TITLE: Do successive maximum permutations pick latin squares uniformly? -QUESTION [7 upvotes]: Suppose we start with a $n\times n$ matrix with entries sampled independently and uniformly at random from $[0,1]$. The weight of a set of entries will simply be the sum of those entries. A permutation refers to a set of $n$ entries, no two on the same row or column. -Pick a permutation whose corresponding entries have the largest weight, and place a label "1" in all those entries. Proceed to erase the values of those entries. At step k, pick a permutation from the remaining entries with largest weight and place a label "k" in all the selected entries. -This procedure ends with an $n\times n$ latin square. Do all latin squares appear with the same probability? - -REPLY [3 votes]: Studying the results given in YBerman's answer gave me the following. Consider these two possibilities after drawing two symbols (in order) for $n=4$: -$$ -\begin{align*} -A &= \begin{pmatrix} -1 & 2 & . & . \\ -2 & 1 & . & . \\ -. & . & 1 & 2 \\ -. & . & 2 & 1 -\end{pmatrix}, & -B &= \begin{pmatrix} -1 & 2 & . & . \\ -. & 1 & 2 & . \\ -. & . & 1 & 2 \\ -2 & . & . & 1 -\end{pmatrix} -\end{align*} -$$ -(Up to permutation of rows and columns, which doesn't change the probability distribution, any square with two symbols is equivalent to one of these.) -The square $A$ can be completed to $4$ different Latin squares, and $B$ can be completed to $2$ different Latin squares. So the probability of $A$ should be twice the probability of $B$. However, out of 10,000,000 trials, $A$ occurred 35,581 times and $B$ occurred 51,481 times. -So "almost surely" this is a counterexample, but I don't know if there's a simple proof for that.<|endoftext|> -TITLE: When do we get $CON(ZF)$ in transfinite progressions of consistency statements? -QUESTION [6 upvotes]: Given the work of Turing and Feferman all arithmetical truths can be isolated through a transfinite progression of theories like $T_0=PA$, $T_{\beta+1}=T_β \ plus \ CON(T_\beta)$ and $T\lambda=\cup T\mu(\mu\prec\lambda)$ - when $\lambda$ is a limit ordinal - through all the recursive ordinals. What is the smallest ordinal $\sigma$ such that $T_\sigma$ proves CON(ZF)? How do such ordinals for arithmetical consistency statements align with proof theoretical ordinals? -Edit: My question does not ask for the proof theoretic ordinal of ZF. -Update: Phillip Welch gives a very readable account of such things as I hint to in comments concerning Feferman's work in an answer to a question here: -Pi1-sentence independent of ZF, ZF+Con(ZF), ZF+Con(ZF)+Con(ZF+Con(ZF)), etc.? -Update 2: My question was badly prepared, as evidenced also by the previous update and the comments in discussion. Noah Schweber kindly suggested that I unaccept his reply until more is clarified concerning my question as related to the Feferman style process I had in mind, and which through a detour into Shoenfield's recursive omega rule (non-constructively) captures all arithmetical truths. I would be surprised if Turing like collapses down to $\omega+1$ could occur in Feferman style processes. - -REPLY [8 votes]: Note that the progression $T_\alpha$ really isn't defined for ordinals but rather ordinal notations. Once we realize this, there is a disappointing answer to your question: for any true $\Pi^0_1$ sentence $\varphi$ (of which a consistency statement is an example), there is a notation $n$ for $\omega+1$ such that $T_n$ proves $\varphi$. -See this answer by Francois Dorais for more details. -This phenomenon breaks the initial hope of assigning an interesting ordinal to a theory $S$ measuring the difficulty of proving $S$'s consistency via iterated consistency statements. However, we can fix things by working below some fixed notation for a "large enough" ordinal: e.g. the ordinal $\epsilon_0$ has, in addition to really stupid notations, very natural notations, and we can work below such a notation to develop the fast-growing hierarchy. -So if we fix a notation $n$, it may be that some notation $m<_\mathcal{O}n$ for a smaller ordinal satisfies "$T_m$ proves $Con(ZF)$"; and if $n$ is "nice", this $m$ might be really interesting! Unfortunately this is putting the cart before the horse: in order to find such an $n$, we basically already need to know everything relevant about $ZFC$, including (at least something close to) its proof-theoretic ordinal.<|endoftext|> -TITLE: Irreducible elements in endomorphism rings -QUESTION [5 upvotes]: Let $(G, +)$ be a commutative group. The endomorphism set $\text{End}(G)$ of all group endomorphisms $f:G\to G$ is a ring, where $+$ is taken pointwise and the multiplication is the composition of endomorphisms. -Suppose $f\in \text{End}(G)$ is not bijective (i.e. not a unit in the ring) and not a zero-divisor. Can it be written as a product of irreducible elements? -(We call $g\in\text{End}(G)$ reducible if there are non-units $h_1,h_2\in\text{End}(G)$ such that $g = h_1\circ h_2$; and we call $g$ irreducible otherwise.) - -REPLY [7 votes]: Take $G = \mathbb{Q}/\mathbb{Z}$; then $\text{End}(G)$ is the profinite integers -$$\widehat{\mathbb{Z}} \cong \prod_p \mathbb{Z}_p$$ -where $\mathbb{Z}_p$ is the $p$-adic integers. The element $\prod_p p$ is neither a unit nor a zero divisor in this ring, and it cannot be written as a product of irreducible elements. This is because the only irreducible elements $x = \prod_p x_p$ are those of the form $x_q = q$ for a fixed prime $q$ and $x_p = 1$ otherwise, and unit multiples of these (there are lots of units), and so the elements that can be written as a product of irreducibles are precisely those where $x_p$ is a unit for all but finitely many $p$, and nonzero otherwise. - -REPLY [7 votes]: If $R$ is a commutative unital ring, then $R \cong_{\sf Ring} {\rm End}_{{\sf Mod}_R}(R_R)$, with the elements of $R$ acting on $R$ by left multiplication. Now, let $R$ be any non-atomic integral domain and note that the multiplicative monoid of ${\rm End}_{{\sf Mod}_R}(R_R)$ is a divisor-closed submonoid of the multiplicative monoid of ${\rm End}_{{\sf Grp}}(R_R)$. Lastly, recall that a commutative unital ring is atomic (i.e., every non-unit, non-zero element is a product of some atoms) iff so is its multiplicative monoid, and that a monoid $H$ with zero is atomic only if so are all the divisor-closed submonoids of $H$ (we say that a submonoid $M$ of $H$ is divisor-closed if $x \mid_H y$ and $y \in M$ imply $x \in M$).<|endoftext|> -TITLE: Morse theory in infinite dimensions -QUESTION [12 upvotes]: It seems that people often talk of "doing Morse theory" on loop spaces in two quite different contexts. -Case 1: When one does Morse theory on a loop space $\Omega(M; p,q)$ using the energy functional as the Morse function, then critical points are geodesics and the index is the index is computed via the second variation formula. As explained in Milnor's classic book, one can recover the homology of the loop space by this technique. -Case 2: On the other hand, when one does Morse theory on the space of (unbased) contractible loops of a symplectic manifold endowed with a Hamiltonian H by using the symplectic action functional, then Floer famously showed that (under suitable hypotheses) one recovers the singular homology of the underlying symplectic manifold. -It seems to me that Case 1 can be interpreted as an infinite dimensional generalization of finite dimensional Morse theory in a very straightforward way. However, Case 2 seems to be of a different nature since the Morse function is not computing the homology of the space on which this function is defined. -Is there an intuitive way to explain this difference? Given a functional on an infinite dimensional space, can we reasonably "guess" what kind of homology theory we will get by "doing Morse theory" with this functional? - -REPLY [16 votes]: The first case has finite indices and parabolic gradient flow; the second infinite (co)indices and elliptic gradient flow. -In more detail, the Morse theory of the energy functional $E$ on $X:=\Omega(M;p,q)$ has the following behavior: -1) For generic metrics it's a Morse function (and for other special metrics of interest, it's Morse-Bott with finite-dimensional critical manifolds). -2) Morse indices of critical manifolds are finite. -3) The downward gradient flow amounts to a parabolic PDE; the flow exists for all positive times. The unstable manifold at a point in a critical manifold is a cell whose dimension is the index. -4) There's reasonable control over the limiting behavior of such flowlines (Palais-Smale condition). -So we can build an increasing sequence of finite-dimensional subspaces $X_k$ of $X$, where $X_k$ is the union of the critical manifolds of index $\leq k$ and their descending manifolds. Reasonably enough, the union $\bigcup X_k$ has the homotopy type of $X$. -The symplectic action functional (actually closed 1-form) on the free loopspace $LM$ of a compact symplectic manifold $M$ (or on its component $(LM)_0$ of nullhomotopic loops) shares property 1), but in other respects is very different: -2') All Morse indices and co-indices are infinite. -3') The downward gradient "flow" amounts to an elliptic PDE. It isn't a flow; you can't flow from an arbitrary initial loop. -4') There are well-behaved spaces of gradient flowlines from one critical point $x_-$ to another $x_+$; these are (virtual) manifolds whose dimension at a flowline $u$, the Fredholm index of the linearization at $u$ of the flowline equation, is finite. -These properties are shared by other Floer theories. You can try to build some kind of a homotopy type from this flow (a "Floer homotopy type") as envisioned by Cohen-Jones-Segal, though the technical difficulties are considerably greater than in setting up Floer homology. But there's no reason for it to look anything like $LM$.<|endoftext|> -TITLE: Prove that matrix is positive definite -QUESTION [11 upvotes]: I faced a hard question in kernel methods theory, which I can't answer for about one week. Initially it was formulated in terms of positive valued functions, but it could be reformulated easier: -Let $\{x_1, \dotsc, x_n\}$ and $\{y_1, \dotsc, y_n\}$ be two sets of real positive numbers. Prove that matrix $A$ is positive definite where $A_{i, j} = \min(x_i\cdot y_j, x_j\cdot y_i)$. Equivalently, I want to prove that for arbitrary $a_1, \dotsc, a_n$ the following inequality holds: -$$ -\sum_{i,j=1}^n a_i a_j \min(x_i\cdot y_j, x_j\cdot y_i)\ge 0. -$$ -I've run out of ideas on how it could be proved, so any advice is welcome. - -REPLY [5 votes]: This is a very similar proof but based on $\mathbb{E}B_{z_{i}}B_{z_{j}}=\min(z_{i},z_{j})$ where $B_{z}$ is the standard brownian motion starting at zero: -$ -0\leq \mathbb{E}\left(\sum_{j} a_{j} x_{j} B_{\frac{y_{j}}{x_{j}}} \right)^{2}=\sum_{i,j}a_{i}a_{j}\min(x_{i}y_{j},x_{j}y_{i}) -$<|endoftext|> -TITLE: What is known about the functor $G\mapsto k[G]^\times/k^\times$? -QUESTION [6 upvotes]: Let $k$ be a ring (resp. profinite ring), $G$ a group (resp. profinite group), and $k[G]$ the group algebra (resp. completed group algebra). -For any such $G$, we may associate to it the group of units $k[G]^\times$ of $k[G]$, and this association is clearly functorial. Has this functor been studied at all? -For example, given an exact sequence -$$1\rightarrow G\rightarrow G'\rightarrow G''\rightarrow 1$$ -we get a sequence -$$1\rightarrow k[G]^\times/k^\times\rightarrow k[G']^\times/k^\times\rightarrow k[G'']^\times/k^\times\rightarrow 1$$ -which is probably not exact, but at least the map $k[G]^\times/k^\times\rightarrow k[G']^\times/k^\times$ is injective, and so one might hope that the functor $G\mapsto k[G]^\times$ is left exact. -If it's left-exact, is this actually exact? If not, is anything known about its right-derived functors? -I suppose this is best posed under the assumption that $G$ is abelian, though I'm also very interested in the nonabelian case (or at least whatever still makes sense there in terms of cohomology) - -REPLY [6 votes]: Forget about exactness; this functor isn't additive, which pretty much tanks any hope of doing homological algebra to it even if you restrict to abelian groups. -Here's a special case that's easy to understand. Suppose $G$ is finite and $k$ has characteristic not dividing $|G|$. Then $k[G]$ is semisimple, so we have a decomposition -$$k[G] \cong \prod_i M_{n_i}(D_i)$$ -where the product is indexed over the irreducible representations $V_i$ of $G$ over $k$, with $\text{End}_G(V_i) = D_i$ division algebras and $n_i = \dim_{D_i} V_i$. Hence -$$k[G]^{\times} \cong \prod_i GL_{n_i}(D_i)$$ -and quotienting by $k^{\times}$ just removes the trivial representation. The functoriality with respect to morphisms $f : G \to H$ comes from pulling back irreps of $H$ to $G$ and seeing how they decompose. -$k[G]^{\times}$ naturally occurs as the automorphism group of the forgetful functor from $G$-representations to $k$-vector spaces (regarded just as a functor, not as a monoidal functor).<|endoftext|> -TITLE: the character tables of irreducible representations of $SL(3,Z_q)$ -QUESTION [5 upvotes]: The following paper gives a classification of the character tables of irreducible representations of $SL(3,GF(q))$ where $q$ is a power of a prime number, and $ GF(q)$ a finite field of $q$ elements. -WILLIAM A. SIMPSON AND J. SUTHERLAND FRAME -Can. J. Math., Vol. XXV, No. 3,1973, pp. 486-494 -THE CHARACTER TABLES FOR -SL(3, q ), SU(3, *•), PSL(3, q), PSU(3, q *) -Here I would to ask "do we have a classification of the character tables of irreducible representations of $SL(3,Z_q)$, where $Z_q=Z/qZ$?" - -REPLY [8 votes]: As was already mentioned, the answer to the question asked is "no, we currently do not have a good classification". Here I wish to describe a successful and interesting recent line of research which does not aim at giving such a classification, rather merely at counting how many representations we do have. -To put things in context, let me first make the following observations: - -the set of irreducible representations of a product of groups is in bijection with the product of the sets of irreducible representations of the groups. -$\text{SL}_n$ over a product of rings (commutative with 1) is isomorphic to the product of $\text{SL}_n$ over the rings. -for a natural $m$, $\mathbb{Z}/m\mathbb{Z}$ is the product of the rings $\mathbb{Z}/q\mathbb{Z}$ where $q$ is a prime power. - -Thus, the study of the rep theory of $\text{SL}_n(\mathbb{Z}/m\mathbb{Z})$ is naturally reduced to the study of the rep theories of the groups $\text{SL}_n(\mathbb{Z}/p^k\mathbb{Z})$ for prime $p$. Fixing $p$ and varying $k$, these are grouped together as the (continuous) representation theory of $\text{SL}_n(\mathbb{Z}_p)$, where $\mathbb{Z}_p$ is the ring of $p$-adic integers. -For $n=2$ the irreducible representations of $\text{SL}_n(\mathbb{Z}_p)$ are indeed classified, as appeared in the answere by Jim Humphreys. -To this answer I wish to add the reference to Uri Onn's paper in which he classifies all irreducible representations over an arbitrary discrete valuation ring. See also a comment below by A Stasinski. -For $n\geq 3$ the representation theory of $\text{SL}_n(\mathbb{Z}_p)$ is much more complicated and a classification is currently out of reach. However we do know that $\text{SL}_n(\mathbb{Z}_p)$ has only finitely many irreducible representations at each dimension $d$. Denoting this number by $r_d$ we are interested in the study of the sequence $(r_d)_{d=1}^\infty$. On this problem there have been in recent years a remarkable progress on which I wish to report, and this is why I am writing this answer. -For $n=3$ it is shown in by Avni-Klopsch-Onn-Voll -that the sequence $r_d$ grows similarly to $d^{3/2}$. For a precise statement see their Duke paper. -Recently Aizenbud-Avni got a uniform statement for all $n$, that is $r_d$ is $O(d^{22})$. -For this see their Inventiones paper. -Whether such a uniform bound exists was asked as a question in Larsen-Lubotzky. -I would be happy to tell more about the techniques of these remarkable papers, but this answer is already getting too long. You will have to read through the sources. -Let me just mention that in order to study the sequence $(r_d)$ we set the representation zeta function $\zeta(s)=\sum r_d\cdot d^{-s}$. A general result of Jaikin-Zapirain tells us that this function is rational in the variable $p^{-s}$. The representation growth is then related to the study of the poles of this zeta function. Let me also mention that there is a global theory here, which comes from the representation theory of $\text{SL}_n(\mathbb{Z})$. -Edit: As remarked by Alexansder Stasinski (thanks!), for $n=3$ (but not for higher $n$'s) the representation zeta function is explicitly computed in the Avni-Klopsch-Onn-Voll paper alluded to above (see Theorem E) and one should be able get out of it the exact numbers of irreducible representation of $\text{SL}_3(\mathbb{Z}/q\mathbb{Z})$ at each dimension, which is closer to answering the OP question. One can also hope to get a full answer using Kirillov orbit method. However, things get messy for higher $n$'s.<|endoftext|> -TITLE: Determinant of a matrix having diagonal and subdiagonal entries zero -QUESTION [8 upvotes]: I have found the determinant of the following matrix of order $n\ge3$ using some elemetary operations -$$\begin{bmatrix}0& 1 & 1& \dots & 1 \\ 0 & 0 & 1& \ddots & 1\\ 1 & 0 & 0 & \ddots & 1 \\ \vdots & \ddots & \ddots & \ddots & 1 \\ 1 & 1 & 1 & \dots & 0 \end{bmatrix}_{n\times n},$$ that is, a matrix which have diagonal, subdiagonal entries zero and rest of entries are equal to 1. Its determinant comes out to be equal to $$(-1)^n\Big(\lfloor\frac{n-2}{2}\rfloor\Big),$$ where, $\lfloor.\rfloor$ is the floor function. Now, I want to find determinant of the following matrix -$$ -\begin{bmatrix}0& -1 & 1& \dots & 1 \\ 0 & 0 & -1& \ddots & 1\\ 1 & 0 & 0 & \ddots & 1 \\ \vdots & \ddots & \ddots & \ddots & -1\\ 1 & 1 & 1 & \dots & 0 \end{bmatrix}_{n\times n},$$ that is, a matrix having diagonal and subdiagonal entries zero. Superdiagonal elements are -1 and rest of the entries are equal to 1. - -REPLY [3 votes]: I will try to give a different explicit solution. From the problem formulation of answer 1, $$\det(A+uu^T)=(1+u^TA^{-1}u).\det(A).$$ From here, we need to solve some recursive expressions, in order to calculate the determinant and inverse of $A$. I solve these recursive expressions using roots of their characteristic equations. For the determinant of $A$, recursive expression is $$f_n=-f_{n-1}-2f_{n-2},\ \ \ \ \ f_0=1,\ f_{-1}=0.$$ Roots of the resulting characteristic equation $x^2+x+2=0,$ are $$ s_1=\frac{-1}{2}+ \frac{\iota \sqrt{7}}{2},\ \ s_2=\frac{-1}{2}- \frac{\iota \sqrt{7}}{2}.$$ Hence, $$\det(A)=f_n=r_1.s_1^n+r_2.s_2^n,$$ where, using initial conditions $$r_1=\frac{1}{2} +\frac{\iota }{2\sqrt{7}},\ \ r_2=\frac{1}{2} -\frac{\iota }{2\sqrt{7}}.$$ -Now, to calculate $A^{-1}$ we need to solve following recursive expressions $$g_i=-g_{i-1}-2g_{i-1}, \ \text{for} \ \ i=2, 3\dots,n,\ \ g_0=1,\ g_1=-1$$ $$ h_i=-h_{i+1}-2h_{i+2}, \ \text{for} \ \ i=n-1,\dots,1, \ \ h_{n+1}=1,\ h_{n}=-1.$$ Similar to $f_n$, solving these recursive expressions we get $$g_i=r_{1}.s_1^i+r_{2}.s_2^i.$$ And, $$h_i=r_{h1}.s_1^{n-i}+r_{h2}.s_2^{n-i},$$ where, $$r_{h1}=\frac{-1}{2}-\frac{3}{2\iota\sqrt(7)},\ \ \ r_{h2}=\frac{-1}{2}+\frac{3}{2\iota\sqrt(7)}.$$ Entries of $A^{-1}$ are clearly given by $g_i, h_i$ (here). -$$A^{-1}_{ij}=\begin{cases}\frac{2^{j-i}g_{i-1}h_{j+1}}{g_n} - & \mbox{if $i\le j$} -\\ \frac{g_{j-1}h_{i+1}}{g_n} &\mbox{if $j< i$} -\end{cases}.$$<|endoftext|> -TITLE: Etiquette of publishing folklore results -QUESTION [27 upvotes]: I am wondering what is the etiquette of publishing a "folklore" result? -Though special cases of the result are well-known, the proof is not readily available in any reference text or paper I've seen. I will include it in the background of my thesis, and I could stick it onto another paper which I am working on (in which I use the folklore result). However, the paper I am working on is in a very niche area, whereas the proof of this result is useful to the bulk of researchers in my field. -My options are: - -Include it in my current paper, which is in a niche subfield -Put the result into a (very) short paper and submit it to some journal -Post the proof on arXiv, and not submit it to a journal -Post the theorem and proof on nLab - -What are your thoughts? Are there any journals that accept this sort of "folklore" publication? Or am I better off just sticking the proof in my current paper or posting the proof on arXiv or nLab? - -REPLY [4 votes]: Or, you can do as they (Phelps et al) did at U Washington, and publish numerous results under a pseudonym (Rainwater in this case) ....<|endoftext|> -TITLE: Automorphisms and isometries of the quaternions -QUESTION [7 upvotes]: Each automorphism of the quaternion algebra is inner and it is an orthogonal mapping with the determinant 1. -Let $f: \mathbb H \rightarrow \mathbb H$ be in $SO(4)$. Does there exists a quaternion $q$ with $\|q\|=1$ and an automorphism $g$ of $\mathbb H$ such that $f(x)=qg(x)$ for $x\in \mathbb H$ ? - -REPLY [14 votes]: In other words, what you are asking is whether every $f\colon\mathbb{H}\to\mathbb{H}$ in $\mathit{SO}_4$ takes the form $x\mapsto \bar u x v$ where $u,v$ are unit quaternions (the connection with your notation is that then $f$ is the composition of the inner automorphism $g\colon x\mapsto \bar v x v$ with left-multiplication by $q = \bar u v$). This is a well-known fact: see, e.g., Conway & Smith, On Quaternions and Octonions (A. K. Peters 2003), §4.1. -Furthermore, this can be seen as an isomorphism $\mathit{Spin}_3\times\mathit{Spin}_3 \to \mathit{Spin}_4$ taking a pair $(u,v)$ of unit quaternions (the group of unit quaternions is isomorphic to $\mathit{Spin}_3$ acting by conjugation) to $x\mapsto \bar u x v$. - -REPLY [13 votes]: This is a standard result in representation theory: -Let the quadratic form on $\mathbb{H}$ be $\langle x,x\rangle> = x\bar x$ (which is positive definite). (Note that I am considering $\mathbb{R}$ as a subset of $\mathbb{H}$, in fact, the center of $\mathbb{H}$.) Then (because $\mathbb{H}$ is associative), for any two unit quaternions $p,q\in \mathbb{H}$ (i.e., $p\bar p = q\bar q = 1$), the linear map $f_{p,q}:\mathbb{H}\to\mathbb{H}$ -$$ -f_{p,q}(x) = px\bar q -$$ -preserves the inner product: $f_{p,q}(x)\overline{f_{p,q}(x)}=x\bar x$ for all $x\in\mathbb{H}$. Moreover, again by associativity and the fact that $\overline{xy} = \bar y \bar x$, -$$ -f_{p_1,q_1}\circ f_{p_2,q_2} = f_{p_1p_2,q_1q_2}, -$$ -so $f$ defines a homomorphism -$$ -f:S^3\times S^3\longrightarrow \mathrm{SO}(4), -$$ -where $S^3 = \{p\in\mathbb{H}\ |\ p\bar p = 1\}$. -It is not hard to show that the kernel of $f$ is $\{(\pm1,\pm1)\}\simeq\mathbb{Z}_2$, and that $f$ is surjective. -This is the standard proof that $\mathrm{Spin}(4)$, the nontrivial double cover of $\mathrm{SO}(4)$, is $S^3\times S^3$. -Finally, $\mathrm{Aut}(\mathbb{H})$ is the subgroup consisting of elements in $\mathrm{SO}(4)$ that are of the form $f_{p,p}$.<|endoftext|> -TITLE: Bijection implies isomorphism for algebraic varieties -QUESTION [10 upvotes]: Let $f:X\to Y$ be a morphism of algebraic varieties over $\mathbb C$. Assume that -a) $f$ is bijective on $\mathbb C$-points -b) $X$ is connected -c) $Y$ is normal. -Does it imply that $f$ is an isomorphism? If not, are there stronger reasonable conditions under which it is true? - -REPLY [11 votes]: I am just expanding the comment above. The statement is local on $Y$, so assume that $Y$ is connected. Since $Y$ is normal, this is equivalent to assuming that $Y$ is irreducible. Grothendieck's formulation of Zariski's Main Theorem is EGA $\textrm{III}_2$, Théorème 4.4.3, p. 136. According to that theorem, there exists a factorization of $f$, $f=\overline{f}\circ i$ where $i:X\hookrightarrow\overline{X}$ is a dense open immersion, and where $\overline{f}:\overline{X}\to Y$ is a finite morphism. Identify $X$ with the image of $i$. -Since $f$ is bijective on points, there exists a unique irreducible component $\overline{X}_0$ of $\overline{X}$ that dominates $Y$. The restriction of $\overline{f}$ to $\overline{X}_0$ is a finite, surjective morphism that is generically one-to-one. Since the characteristic is $0$, this morphism is birational (in positive characteristic, it might be purely inseparable but not birational). Thus, by the classical form of Zariski's Main Theorem, $\overline{f}$ restricts to an isomorphism from $\overline{X}_0$ to $Y$, cf. Mumford's Red Book of Varieties and Schemes, p. 288. -By way of contradiction, assume that $\overline{X}_0$ is a proper subset of $\overline{X}$. Denote by $Z \subset \overline{X}$ the union of all irreducible components different from $\overline{X}_0$. Since $X$ is a dense open in $\overline{X}$, also $X\cap Z$ is a dense open in $Z$. Thus, also $(X\cap Z) \setminus (\overline{X}_0\cap Z)$ is a dense open subset of $Z$. Denote by $W$ the image in $Y$ of this open subset. Since $f$ is injective, the constructible subset $W$ is disjoint from the dense open $V=f(X\cap \overline{X}_0)$ of $Y$. Thus, the closure of $W$ is disjoint from $V$. Since $\overline{f}$ is finite, this closure equals the image of $Z$. Thus, $X\cap \overline{X}_0$ is disjoint from $\overline{X}_0\cap Z$, i.e., $X\cap \overline{X}_0\cap Z$ is empty. Thus, the open subsets $\overline{X}_0\setminus (\overline{X}_0\cap Z)$ and $Z\setminus (Z\cap \overline{X}_0)$ of $\overline{X}$ pullback to disjoint open subsets of $X$ that cover $X$. Since $X$ is connected, this is a contradiction. -Therefore, $\overline{X}_0$ equals all of $\overline{X}$. Since $\overline{X}_0\to Y$ is an isomorphism, it follows that $f$ is an open immersion. Since $f$ is also surjective, $f$ is an isomorphism.<|endoftext|> -TITLE: Reference request quantum SU(3) -QUESTION [5 upvotes]: Woronowicz shows that the C*-algebras of quantum $SU(2)$ are isomorphic (only as C*-algebras, forgetting the quantum group structure). Are there similar results for quantum $SU(n)$ for $n \geq 3$? - -REPLY [2 votes]: For SU(3), there's [1]. The general case is not fully worked out yet, I guess. -[1] Nagy, Gabriel; A rigidity property for quantum SU(3) groups. Advances in geometry, 297–336, Progr. Math., 172, Birkhäuser Boston, Boston, MA, 1999.<|endoftext|> -TITLE: The largest Wasserstein distance to uniform distribution among all probability distributions with uniform marginals -QUESTION [7 upvotes]: I am looking for the largest Wasserstein distance to the uniform distribution among all probability distributions with uniform marginals. -More specifically, let $\Xi=\{1,2,\ldots,N\}^2$, and let $\nu$ be a uniform distribution on $\Xi$, namely, $\nu_{ij}:=\nu(\{(i,j)\})=\frac{1}{N^2}$, for all $1\leq i,j\leq N$. - Consider the problem - \begin{equation} - \max_{\mu\geq0} \bigg \{ W(\mu,\nu):\ \sum_{i}\mu_{ij} = \frac{1}{N}, \forall j,\ \sum_{j} \mu_{ij}=\frac{1}{N}, \forall i \bigg\}, \tag{1} -\end{equation} -where $W_1(\mu,\nu)$ is the Wasserstein distance between probability distribution $\mu:=\{\mu_{ij}\}_{1\leq i,j\leq N}$ and $\nu$, defined as - $$ - W_1(\mu,\nu) := - \min_{\pi\geq0} \bigg\{\sum_{1\leq i,j,i',j' \leq N} ||(i,j)-(i',j')||_1 \pi_{(i,j),(i',j')}:\ \sum_{i,j} \pi_{(i,j),(i',j')} = \nu_{i'j'},\forall i',j',\ \sum_{i',j'} \pi_{(i,j),(i',j')} = \mu_{ij},\ \forall i,j\bigg\}. - $$ -My conjecture is that the maximizer of problem $(1)$ is given by $\mu_{ij}=\frac{1}{N}{1}_{\{i=j\}}$, or $\mu_{ij}=\frac{1}{N}{1}_{\{i+j=N+1\}}$, namely, the comonotonic/countermonotonic distribution. But how to prove/disprove it? Also, if it is true, could the result be extended to the multivariate case, namely, $\Xi=\{1,2,\ldots,N\}^K$ for $K>2$, or be extended for norms other than $\ell_1$-norm, or other $W_p$ distance ($p>1$)? -Update: Steve provides an affirmative answer for $\ell_1$-norm with $W_1$ distance in the case of $K=2$, and I provide a proof for $\ell_2$ norm with $W_2$ distance for all $K\geq2$. I am wondering if we can get some other result, such as $\ell_1$-norm with $K>2$, or $\ell_\infty$-norm. - -REPLY [2 votes]: I think I have an answer for the case p = 1, K = 2. I write "I think" because my computation does not coincide with the example values for $N=4$ posted earlier by OP in a comment, but I really cannot find any error in my proof, so I wanted to share it. -As already mentioned, we only need to consider permutation measures of the form $\mu = \sum_{i=1}^N \frac{1}{N} \delta_{i,\sigma(i)}$, since these are the extremal points of the set we optimize over. -For any such $\mu$, we can define a coupling $\pi$ by $\pi_{(i,\sigma(i)),(i,j)} = 1/N^2$ for $i,j \in \{1,...,N\}$ and $\pi_{(i,j),(i',j')} = 0$, if $i,j,i',j'$ are not of the form as before. That this is an admissible coupling in the definition of $W_1$ is trivial. -The corresponding "value" is -$$ -\sum_{1\leq i,j,i',j' \leq N} ||(i,j)-(i',j')||_1 \pi_{(i,j),(i',j')} = \sum_{1 \leq i,j' \leq N} |\sigma(i) - j'| \frac{1}{N^2}. -$$ -We further show that this value is minimal if $\mu$ is the monotonic or comonotonic measure, which will yield the claim. I only show this for the monotonic case, $\mu = \frac{1}{N}\sum_{i=1}^N \delta_{i,i}$. Then for any admissible coupling $\hat{\pi}$, by the constraints in the calculation of the Wasserstein distances, we see -\begin{align*} -\sum_{1\leq i,j,i',j' \leq N} ||(i,j)-(i',j')||_1 \hat{\pi}_{(i,j),(i',j')} =& \sum_{1\leq i',j' \leq N} \sum_{1 \leq i \leq N} ||(i,i)-(i',j')||_1 \hat{\pi}_{(i,i),(i',j')} \\ \geq& \sum_{1\leq i',j' \leq N} \sum_{1 \leq i \leq N} ||(i',i')-(i',j')||_1 \hat{\pi}_{(i,i),(i',j')}\\ =&\sum_{1\leq i',j' \leq N} |i'-j'| \frac{1}{N^2}, -\end{align*} -where the inequality is elementwise and the last term corresponds to the value for the coupling $\pi$. -Note that this only shows that monotonic/comonotonic measures are maximizers, but not that these are the only maximizers.<|endoftext|> -TITLE: "unexpected" residue formula for $\Gamma^3(s)/(\Gamma(3s)(e^{2\pi is}-1)) $ -QUESTION [10 upvotes]: There is a related problem in my current work: to find the residue of the following function at any negative integer $s=-n$: -$$f(s)=\frac{\Gamma^3(s)}{\Gamma(3s)(e^{2\pi is}-1)}$$ -It seems to be a tedious calculation. As far as I know, the first few terms of Laurent series of $\Gamma(s)$ around $s=-n$ are given by -$$\Gamma(s)=\frac{(-1)^n}{n!}\left(\frac 1{s+n}+\psi(n+1)+\frac 16(3\psi(n+1)^2+\pi^2-3\psi'(n+1))(s+n))+\frac16(\psi(n+1)^3+(\pi^2-3\psi'(n+1))\psi(n+1)+\psi''(n+1))(s+n)^2+O((s+n)^3)\right)$$ -where $\psi(s):=\frac{\Gamma'(s)}{\Gamma(s)}$ is the digamma function. -So we might expect that the residue formula of $f(s)$ at $s=-n$ involves the values of digamma function and/or its derivatives. However, numerical values indicate that the formula for $Res (f,-n)$ might only involve $\pi$, $i$, and rational numbers, which is quite unexpected. Here are a few examples: -$$Res (f,-3)=7527 + \frac{4299 i}{4 \pi} + 2100i\pi$$ -$$Res (f,-10)=\frac {1065144125784453}{40} - \frac {17136782690536253 i}{11200 \pi} + 6938745989175 i \pi$$ -I have verified some other integers as well. Can we find any reason for this pattern or any counterexample? It also seems that $f(s)$ is very special, for instance, if we multiply another $\Gamma(s)$ to the numerator, i.e. $g(s)=\frac{\Gamma^4(s)}{\Gamma(3s)(e^{2\pi is}-1)}$, then the formula for $Res(g,-n)$ does not have such property. - -REPLY [9 votes]: $\def\Res{\operatorname*{Res}} -\def\G{\Gamma} -\def\e{\varepsilon} -\def\p{\pi} -\def\ZZ{\mathbb{Z}} -\def\QQ{\mathbb{Q}} -\def\NN{\mathbb{N}} -\def\j{\psi} -\def\z{\zeta} -\def\To{\rightarrow} -\def\f{\varphi} -\def\g{\gamma} -\def\a{\alpha} -\def\b{\beta} -\def\i{\mathrm{i}} -\def\bell{\mathcal{B}} -\def\bern{B} -\newcommand\set[1]{\{#1\}}$After a straightforward but tedious calculation one finds that the residue is given by -\begin{align*} -\frac{(3 n)! i}{4 \pi (n!)^3} - &\Big(8 \pi ^2 + - 18\pi i (H_n-H_{3 n}) -27 (H_n-H_{3 n})^2 \\ - &\quad +9 \big(\psi^{(1)}(n+1)-3 \psi ^{(1)}(3 n+1)\big)\Big) -\end{align*} -Now note that -$H_n, H_{3n}\in\mathbb{Q}$ -and -$\psi^{(1)}(m+1) = \pi^2/6 - \underbrace{\sum_{k=1}^m 1/k^2}_{\in\mathbb{Q}}$ for $m\in\mathbb{Z}^+$. -Thus, -$$\Res_{s=-n}f(s) = a + bi/\pi + ci\pi,\quad -\textrm{where }a,b,c\in\mathbb{Q},$$ -as claimed. -Addendum -Below we consider a wider class of residues of the form -\begin{align*} -\Res_{s=-n} -\frac{\G(s)^m}{\G(m s)}\a(s), -\end{align*} -where $m\in\NN$ and $\a$ has an isolated pole of order $p$ at $s=-n$. -That is, we examine -\begin{align*} -\Res_{\e=0} -\frac{\G(-n+\e)^m}{\G(-m n+m\e)} \frac{\b(\e)}{\e^p}, -\end{align*} -where $\b$ is analytic at $\e=0$. -We find the general form of such residues. -This is an extension of the types of residues suggested by @Wolfgang in the comments. -In the posted question, $m=3$, $p=1$, and $\b(\e) = \e/(e^{2\p \i\e}-1)$. -A routine calculation results in -\begin{align*} -\frac{\G(-n+\e)^m}{\G(-m n+m\e)}\frac{\b(\e)}{\e^p} - &= - \frac{1}{\e^{m+p-1}} - m\b(\e) - \left(\frac{\p\e}{\sin\p\e}\right)^m - \frac{\sin\p m\e}{\p m\e} - \frac{\G(mn+1-m\e)}{\G(n+1-\e)^m} -\end{align*} -where $s = -n+\e$. -Thus, -\begin{align*} -\Res_{s=-n} -\frac{\G(s)^m}{\G(m s)}\a(s) - &= - \Res_{\e=0} - \frac{1}{\e^{m+p-1}} m\b(\e)f(\e)g(\e), -\end{align*} -where -\begin{align*} -f(\e) &= - \left(\frac{\p\e}{\sin\p\e}\right)^m - \frac{\sin\p m\e}{\p m\e} \\ -g(\e) &= \frac{\G(m n+1-m\e)}{\G(n+1-\e)^m}. -\end{align*} -We find -\begin{align*} -\Res_{s=-n} -\frac{\G(s)^m}{\G(m s)}\a(s) -&= - m - \sum_{i+j+k=m+p-2} - \frac{1}{i!j!k!} - \b^{(i)}(0) - f^{(j)}(0) - g^{(k)}(0). - \tag{1} -\end{align*} -For the original problem $\b^{(i)}(0) = a_i (\i\p)^{i-1}$, where $a_i\in\QQ$. -In fact, since $\b(\e)$ in this case is essentially the generating function for the Bernoulli numbers $B_i^-$ we find -\begin{align*} -\b^{(i)}(0) = (2\p\i)^{i-1}B_i^-. - \tag{2} -\end{align*} -The first few derivatives of $f$ are given as -\begin{align*} -f(0) &\equiv\lim_{\e\To0}f(\e) = 1 \\ -f'(0) &= 0 \\ -f''(0) &= -\frac{1}{3} (m-1) m \pi^2 \\ -f'''(0) &= 0. -\end{align*} -Note that $f^{(2i+1)}(0) = 0$ generally. -One can show that, as suggested above, -\begin{align*} -f^{(2i)}(0) &= b_{2i}\p^{2i}, -\end{align*} -where $b_{2i}\in\QQ$. -In fact, an explicit form may be found for these coefficients, -\begin{align*} -f^{(2i)}(0) &= (-1)^i \p^{2i} \bell_{2i}(x_1,\ldots,x_{2i}), - \tag{3} -\end{align*} -where $\bell_i(\cdots)$ is the $i$th complete Bell polynomial, -\begin{align*} -x_j &= \begin{cases} -\frac{m}{j} 2^j (m^{j-1}-1) B_j, & \textrm{$j$ even} \\ -0, & \textrm{else}, -\end{cases} \tag{4} -\end{align*} -and where $\bern_j$ is the $j$th Bernoulli number. -(Since $j$ is even, there is no distinction between $\bern^-_j$ and $\bern^+_j$.) -The first few complete Bell polynomials are -\begin{align*} -\bell_0 &= 1 \\ -\bell_1(t_1) &= t_1 \\ -\bell_2(t_1,t_2) &= t_1^2+t_2 \\ -\bell_3(t_1,t_2,t_3) &= t_1^3 + 3t_1 t_2+t_3. -\end{align*} -Note that -$$g(0) = \frac{(m n)!}{n!^m} .$$ -The first few derivatives of $g$ are given as -\begin{align*} -\frac{g'(0)}{g(0)} &= -m(H_{m n} - H_n) \\ -\frac{g''(0)}{g(0)} &= - m^2 (H_{m n} - H_n)^2 - + m^2 \j^{(1)}(m n+1)- m\j^{(1)}(n+1) \\ -\frac{g'''(0)}{g(0)} &= - \Bigg(m^3 (H_{m n} - H_n)^3 \\ - & \hspace{7ex} + 3 m (H_{m n} - H_n) - \left(m^2 \j^{(1)}(m n+1)-m\j^{(1)}(n+1)\right) \\ - & \hspace{7ex} + m^3 \j^{(2)}(m n+1) - m\j^{(2)}(n+1)\Bigg), -\end{align*} -where -$H_n$ is the $n$th harmonic number -and where -$\j^{(i)}$ is the polygamma function. -Generally we find -\begin{align*} -\frac{g^{(j)}(0)}{g(0)} - &= (-1)^j \bell(y_1,\ldots,y_j), - \tag{5} -\end{align*} -where -\begin{align*} -y_i &= m^{i}\j^{(i-1)}(m n+1)-m\j^{(i-1)}(n+1). -\end{align*} -Note that -\begin{align*} -y_i &= \begin{cases} -m(H_{m n} - H_{n}), & i=1 \\ -(-1)^i (i-1)!\Big( -(m^i-m)\z(i) - m^i H^{(i)}_{m n} + m H^{(i)}_n -\Big) & \textrm{else.} -\end{cases} - \tag{6} -\end{align*} -Thus, residues involving $y_3$ or higher will contain terms involving $\z(3)$. -We find -\begin{align*} -\Res_{s=-n} -\frac{\G(s)^m}{\G(m s)}\a(s) -&= - m - \frac{(m n)!}{n!^m} - \sum_{{0\le i,2j,k\le m+p-2}\atop{i+2j+k=m+p-2}} - \frac{1}{i!(2j)!k!} - (-1)^{j+k}\p^{2j} - \b^{(i)}(0) \\ - &\hspace{25ex}\times - \bell_{2j}(x_1,\ldots,x_{2j}) - \bell_k(y_1,\ldots,y_k) \\ -&= \sum_{{0\le i,2j,k\le m+p-2}\atop{i+2j+k=m+p-2}} - q(i,2j,k) - \b^{(i)}(0) - \p^{2j} \\ - & \hspace{16ex}\times - \bell_k\Big(q_1,q_2\z(2)+q'_2, - \ldots,q_k\z(k)+q_k'\Big), - \tag{7} -\end{align*} -where $q(i,2j,k),q_i,q_i'\in\QQ$ are explicitly calculable. -Thus, the residues involve sums of rational multiples of $\b^{(i)}(0)$ and products of powers of $\z(2),\z(3),\ldots$. -The residues do not involve the Euler-Mascheroni constant unless $\b^{(i)}(0)$ does. -For $\a$ in the question statement we find -\begin{align*} -\Res_{s=-n} -\frac{\G(s)^m}{\G(m s)(e^{2\p\i\e}-1)} -&= \sum_{{0\le i,2j,k\le m-1}\atop{i+2j+k=m-1}} - q'(i,2j,k) - \i^{i-1} - \p^{2j+i-1} \\ - & \hspace{14ex}\times - \bell_k\Big(q_1,q_2\z(2)+q'_2, - \ldots,q_k\z(k)+q_k'\Big), - \tag{8} -\end{align*} -where $q'(i,2j,k)\in\QQ$. -For $m=3$, one can quickly verify that this sum results in a residue of the form $a + b\i/\p + c\i\p$, where $a,b,c\in\QQ$. -(Schematically, the sum will be of the form -$ \{2,0,0\}+\{0,2,0\}+\{1,0,1\}+\{0,0,2\}$.)<|endoftext|> -TITLE: Strictly finer bornological topology on Hilbert space -QUESTION [7 upvotes]: Question: Let $E$ be a Hilbert space. Can there exist a strictly finer bornological topology on $E$? -The background to my question is as follows. I am looking at locally complete, locally convex spaces $E$ with bounded inner products $\gamma$. Then the topology induced by $\gamma$ is weaker than the bornological topology. I am interested in ways to recognise that $(E,\gamma)$ is a Hilbert space. Consider the following three properties - -$\gamma$ induces the bornological topology on $E$. -$\check \gamma : E \to E'$ is surjective; $E'$ are the bounded linear functionals on $E$. -$(E,\gamma)$ is a Hilbert space. - -I am able to show that (1) and (2) are equivalent and that either one implies (3). If I additionally assume that $E$ is webbed, then I can show that (3) implies (1) using a closed graph theorem. The closed graph theorem in question is from (13.3.4, Jarchow, 1981): if $E$ is ultrabornological and $F$ is webbed, then every closed linear map $E \to F$ is continuous. -Is the assumption, that $E$ is webbed, necessary? If yes, are there explicit counterexamples satisfying (3), but not (1)? - -REPLY [9 votes]: Every (real or complex) vector space $E$ can be endowed with its finest locally convex topology $\tau_{flc}$ where every seminorm is continuous and (equivalently) every absolutely convex absorbing set is a $0$-neighborhood. -This topology can be described as the locally convex inductive limit of all finite dimensional subspaces (with their unique Hausdorff locally convex topologies). Since bornologicity is stable with respect to inductive limits the finest locally convex topology is thus bornological (also ultrabornological). Using e.g. a Hamel basis of $E$ you find linear functionals which are discontinuous with respect to the Hilbert space topology. As they are clearly continuous with respect to $\tau_{flc}$ we get that the latter is strictly finer than the Hilbert topology. -The identity $(E,\|\cdot\|) \to (E,\tau_{flc})$ has closed graph (since it is continuous in the other direction) but is not continuous. This shows that being webbed cannot be dropped from the closed graph theorem and that $(E,\tau_{lfc})$ is not webbed.<|endoftext|> -TITLE: Existence of a faithful irreducible representation using Möbius function -QUESTION [16 upvotes]: Let $G$ be a finite group, $L(G)$ its subgroup lattice and $\mu$ the Möbius function. -Consider the Euler totient of $G$ defined as follows: -$$ \varphi(G) = \sum_{H \le G}\mu(H,G) |H| $$ -Let $X=\{M_1, \dots, M_n \}$ be the set of maximal subgroups of $G$. By applying the Crosscut Theorem with $X$ (see this comment of Richard Stanley) and next the inclusion–exclusion principle, we get that: -$$ \varphi(G) = |G \setminus \bigcup_{i=1}^n M_i| $$ In other words, $\varphi(G)$ is the number of elements $g \in G$ such that $\langle g \rangle = G$. It follows that -$$ \varphi(G) \neq 0 \Leftrightarrow G \text{ cyclic}$$ -Note that $\varphi(\mathbb{Z}/n) = \varphi(n)$ the usual Euler's totient function. -Now, consider the dual Euler totient of $G$ defined as follows: -$$ \hat{\varphi}(G) = \sum_{H \le G}\mu(1,H) |G:H| $$ -Question: $ \hat{\varphi}(G) \neq 0 \Leftrightarrow G$ has a faithful irreducible complex representation? -Remark: We will see below that $(\Rightarrow)$ is true. So the question reduces to $(\Leftarrow)$. -It is true for the finite simple group $G$ of order $<10000$: -$$ \begin{array}{c|c|c|c|c|c} -G & |G| & \hat{\varphi}(G) \newline - \hline - A_5 & 60 & 8 & \newline - \hline - PSL(2,7) & 168 & 228 & \newline - \hline - A_6 & 360 & 8748 & \newline - \hline - PSL(2,8) & 504 & 19056 & \newline - \hline -PSL(2,11) & 660 & 24932 & \newline - \hline -PSL(2,13) & 1092 & 105684 & \newline - \hline - PSL(2,17) & 2448 & 389496 & \newline - \hline - A_7 & 2520 & 188136 & \newline - \hline - PSL(2,19)& 3420 & 1148028 & \newline - \hline - PSL(2,16)& 4080 & 1935584 & \newline - \hline - PSL(3,3)& 5616 & 395496 & \newline - \hline - PSU(3,3)& 6048 & 507168 & \newline - \hline - PSL(2,23)& 6072 & 2234784 & \newline - \hline - PSL(2,25)& 7800 & 5391800 & \newline - \hline - M_{11} & 7920 & 1044192 & \newline - \hline - PSL(2,27)& 9828 & 7778916 & \newline -\end{array}$$ -Any idea about the meaning of these numbers? - -Proof of $(\Rightarrow)$ -Consider the relative version $$ \hat{\varphi}(H,G) = \sum_{K \in [H,G]}\mu(H,K) |G:K|.$$ Warning: $-\hat{\varphi}(H,G)$ is not the Möbius invariant of the bounded coset poset $\hat{C}(H,G)$ because $$-\mu(\hat{C}(H,G)) = \sum_{K \in [H,G]}\mu(K,G) |G:K| $$ and $\mu(K,G) \neq \mu(H,K)$ in general. -Now if $[H,G]$ is boolean of rank $n$ then $\mu(K,G) = (-1)^n \mu(H,K)$; moreover (independently) by Theorem 3.21 of this paper, if $ \hat{\varphi}(H,G) \neq 0$ then there is an irreducible complex representation $V$ of $G$ such that $G_{(V^H)} = H$. Next, using a dual reformulation of the Crosscut Theorem with $X$ the set of atoms, we can extend the proof of Theorem 3.21 to any interval $[H,G]$ (i.e. without assuming it to be boolean). Finally by taking $H = 1$, we get that for any finite group $G$, if $\hat{\varphi}(G) \neq 0$ then there is an irreducible complex representation $V$ such that $G_{(V)} = 1$, which means that $V$ is faithful. - -REPLY [6 votes]: No, the modular maximal-cyclic group $M_4(2)$, of order $16$, has a faithful irreducible complex representation (f.i.c.r.) of dimension $2$, whereas $\hat{\varphi}(M_4(2)) = 0$. -Let $B$ be the subgroup generated by the minimal subgroups of $G$. By the Crosscut Theorem, if $H \in [1,G] \setminus [1,B]$ then $\mu(1,H) = 0$. Then $$\hat{\varphi}(G) = |G:B| \hat{\varphi}(B)$$ So that $\hat{\varphi}(G) \neq 0 $ if and only if $\hat{\varphi}(B) \neq 0$. But we can't expect that the existence of a f.i.c.r. for $G$ implies the existence of a f.i.c.r. for $B$ (whereas the converse is true). - In fact, $G = M_2(4)$ has a f.i.c.r. but not its $B = C_2^2$ -Improved question: Is $(\Leftarrow)$ true if $B=G$? -[note that $B$ is normal (see here), so $B=G$ for $G$ simple] -Answer: No, there are exactly two counter-examples of order $\le 100$: $D_8 \rtimes C_2^2$ and $D_8 \rtimes S_3$. They are of order $64$ and $96$ resp., and both check $B=G$, $\hat{\varphi}(G)=0$ and have a f.i.c.r. of dimension $4$. -We can still specialized the question to $G$ simple. - -For answering the comment of Benjamin Steinberg, we will prove an alternative result involving the normal subgroup lattice, and giving a formula shorter than in Pálfy's paper (available here). -Let $G$ be a finite group and $\mathcal{N}(G)$ its normal subgroup lattice. Let $\mu_{\mathcal{N}}$ be the Möbius function for $\mathcal{N}(G)$. Consider the dual normal Euler totient: -$$ \hat{\varphi}_{\mathcal{N}}(G) = \sum_{H \in \mathcal{N} (G)}\mu_{\mathcal{N}}(1,H) |G:H| $$ Let $V_1, \dots, V_r$ be equivalent class representatives of the irreducible complex representations of $G$. -Theorem: $\hat{\varphi}_{\mathcal{N}}(G) = \sum_{V_i \text{ faithful}} \dim(V_i)^2$. -Corollary: $G$ has a faithful irreducible complex representation iff $\hat{\varphi}_{\mathcal{N}}(G) \neq 0$. - -Proof of the theorem: -Let the fixed-point subspace $V_i^H=\{v \in V_i \ | \ hv=v ,\ \forall h \in H \}$ -Claim 1: $|G:H| = \sum_i\dim(V_i) \dim(V_i^H)$ -Claim 2: $H$ is a normal subgroup iff $\forall i \ V_i^H = \{0\}$ or $V_i$. -Claim 3: $V_i$ is not faithful iff there is an atom $H$ of $\mathcal{N}(G)$ such that $V_i^H = V_i$. -Consider the set $L=\{ \bigoplus_i V_i^H \ | \ H \in \mathcal{N}(G) \}$ whose lattice structure is the reverse of $\mathcal{N}(G)$. -By applying the Crosscut Theorem and the inclusion-exclusion principle, together with the claims above, the result follows. $\square$ -Reference request: does this theorem exist in the literature?<|endoftext|> -TITLE: Can the constant rank theorem for smooth manifolds be generalized to nonconstant rank? -QUESTION [22 upvotes]: The constant rank theorem says that -if $f\colon M→N$ is a smooth map whose rank equals some fixed $k≥0$ at any point of $M$, then, locally with respect to $M$ and $N$, the map $f$ assumes the easiest possible form: -$f(x_1,…,x_k,…,x_m)=(x_1,…,x_k,0,…,0)$. -I am interested in a more general situation: the rank of $f$ is at most $k$ -at any point of $M$ and can actually drop below $k$ at some points. -The classification of local forms for such maps seems to be too difficult. -However, in my case one can replace the map f with any map homotopic to it -via a homotopy $M×[0,1]→N$ that is itself a smooth map of rank at most $k$. -Thus, one can inquire about the possible local forms of smooth maps -of rank at most $k$ up to smooth homotopy of rank at most $k$. -I am mostly interested in the case $k < \operatorname{dim} M$. -Such a problem could presumably be solved using the techniques of catastrophe theory, -but my knowledge of this area is virtually nonexistent. -Is it possible to write down a finite list of local forms of smooth maps -of rank at most $k$ considered up to a smooth homotopy of rank at most $k$? -This is motivated by the following question, -which arises in connection to holonomy and parallel transport: -given some $k -TITLE: Not all manifolds can be triangulated: In which dimensions? -QUESTION [35 upvotes]: I know that Ciprian Manolescu has settled the triangulation conjecture in the negative: Not all manifolds can be triangulated. I've only read secondary literature on this result, which did not detail in which dimensions it is known -that there exist untriangulable manifolds. My understanding is that Freedman and Casson showed there exist counterexamples in dimension $4$. - -Q. Does Manolescu's proof provide counterexamples in all dimensions $\ge 5$? - -REPLY [64 votes]: In dimensions up to three, every manifold is triangulable (this is classical). In dimension 4, there are simply connected non-triangulable manifolds (such as the E8 manifold); in fact, a closed 4-manifold is triangulable if and only if it's smoothable. (Pick a triangulation; the links of vertices are always both homology and homotopy spheres, thus by the Poincare conjecture are $S^3$, so this gives a PL structure, and thus is smoothable. This is probably true more generally than closed manifolds, but I'm feeling a little paranoid.) -The crucial papers pre-Manolescu are, firstly, Galewski-Stern and Matumoto's thesis (which I don't have a link to): these prove that every closed manifold if dimension $n \geq 5$ are triangulable if and only if there is a homology 3-sphere $Y$ such that $Y \# Y$ bounds a homology ball, and the Rokhlin invariant $\mu(Y) = 1 \in \Bbb Z/2\Bbb Z$. This is what Manolescu disproved (there is no such 3-manifold); the Galewski-Stern paper clarifies what dimensions their theorem is proved for after you start dropping assumptions about compactness and the boundary. -But importantly for your question is Galewski and Stern's sequel paper. Their theorem 2.1 (plus Manolescu's result) implies that there are non-triangulable manifolds in every dimension $n \geq 5$. However, all orientable 5-dimensional manifolds are triangulable. In dimensions at least 6, though, you can use their construction to produce non-triangulable orientable manifolds. -EDIT: I can't describe this better than Ciprian's own Lectures on the Triangulation Conjecture; Chapter 2 outlines the geography of triangulable manifolds, giving progressively more detail about the older results until he gives a sketch of his construction.<|endoftext|> -TITLE: When $\frac{a^2}{b+c}＋\frac{b^2}{a+c}＋\frac{c^2}{a+b}$ is integer and $a,b,c$ are coprime natural numbers, is there a solution except (183,77,13)? -QUESTION [18 upvotes]: Given $a,b,c\in \Bbb{N}$ such that $\{a,b,c\}$ are coprime natural numbers and $a,b,c>1$. When -$$\frac{a^2}{b+c}＋\frac{b^2}{c+a}＋\frac{c^2}{a+b}\in\mathbb Z\,?$$ -I know the solution $\{183,77,13\}$. Is there any other solution? - -REPLY [40 votes]: Yes, there is another solution. The next one I found is a bit big, namely -$$ a = 15349474555424019, b = 35633837601183731, c = 105699057106239769. $$ -This solution also satisfies the property that -$$ \frac{a^{2}}{b+c} + \frac{b^{2}}{a+c} + \frac{c^{2}}{a+b} = \frac{31}{21} (a+b+c), -$$ -which was true of $a = 13$, $b = 77$ and $c = 183$. If you clear denominators in the equation above, you find that both sides are multiples of $a+b+c$, and dividing out gives a plane cubic. -One wishes to search for rational points on this plane cubic, which by scaling can be assumed to be relatively prime integers. We also need points where $a, b, c > 0$ and $a+b+c$ is a multiple of $21$. This plane cubic is isomorphic to the elliptic curve -$$ E : y^2 + xy + y = x^3 - 8507979x + 9343104706, $$ -and $E(\mathbb{Q})$ has rank $3$ and the torsion subgroup is $\mathbb{Z}/6\mathbb{Z}$. Let $E(\mathbb{Q}) = \langle T, P_{1}, P_{2}, P_{3} \rangle$ where $6T = 0$. I searched for points of the form $c_{1}T + c_{2}P_{1} + c_{3}P_{2} + c_{4}P_{3}$ where $0 \leq c_{1} \leq 5$ and $|c_{2}|, |c_{3}|, |c_{4}| \leq 5$. Of these $7986$ points, there are $1464$ where $a$, $b$ and $c$ are all positive, and these yield $11$ solutions (where $a < b < c$), the smallest of which is $(13,77,183)$. The second smallest is the one I gave above. It should be straightforward to prove that one can get infinitely many solutions in this way. -EDIT: I searched for solutions that satisfy -$$ \frac{a^{2}}{b+c} + \frac{b^{2}}{a+c} + \frac{c^{2}}{a+b} = t(a+b+c) $$ -for all rational numbers $t = d/e$ with $d, e \leq 32$. This yields a number of smaller solutions. In particular, for $t = 9/5$ the rank of the curve is $2$ and one finds -$$ a = 248, b = 2755, c = 7227. $$ -For $t = 21/17$ the rank is $2$ and one finds -$$ a = 35947, b = 196987, c = 401897. $$ -Interestingly, $t = 31/21$ is the only case where the rank is $\geq 3$.<|endoftext|> -TITLE: Are two octonion algebras with different multiplications isomorphic? -QUESTION [6 upvotes]: Some authors, e.g. Baez, Ward, defined multiplication of octonions by formula -$ -(a,b) \cdot^B (c,d)=(ac-db^*, cb+a^*d) \textrm{ for } a,b,c,d\in \mathbb H, -$ -some others, e.g. Springer & Veldkamp, N. Jacobson, by -$ -(a,b)\cdot^S(c,d)=(ac-d^*b, da+bc^*) \textrm{ for } a,b,c,d\in \mathbb H -$ -($a^*$ denotes the conjugate quaternion to $a$). -Which of these multiplications is better? -Are algebras $\mathbb O=\mathbb H \times \mathbb H$ with multiplications $\cdot^B$ and $\cdot^S$ (and the standard addition and multiplications by real numbers) isomorphic? - -REPLY [9 votes]: Yes: the isomorphism is given by $(a,b) \mapsto (a,b^*)$. - -REPLY [5 votes]: The "S" formula you give corresponds to writing octonions in the form $q+r\ell$ where $q,r\in\mathbb{H}$ and $\ell$ is a unit octonion orthogonal to $\mathbb{H}$: we have -$$(q+r\ell)(q'+r'\ell) = (qq'-r'^*r)+(r'q+rq'^*)\ell$$ -which is your "S" formula, written with a more readable notation. The "B" formula you give corresponds to writing octonions in the form $q+\ell r$: we have -$$(q+\ell r)(q'+\ell r') = (qq'-r'r^*)+\ell(q^*r'+q'r)$$ -They can be deduced from one another by using the fact that $q + \ell r = q + r^*\ell$ (a consequence of either formula) which gives the isomorphism you ask for: it's an easy exercise to check that either formula gives the other by applying this isomorphism (e.g., $q(\ell r') = q(r'^*\ell) = (r'^* q)\ell = \ell(q^* r')$ applying the "S" formula for the middle equality). -As to which is best, it really depends whether you prefer to write $\mathbb{O}$ as $\mathbb{H} \oplus \mathbb{H}\ell$ or $\mathbb{H} \oplus \ell\mathbb{H}$. Personally prefer the former (so, "S"), but that's maybe just because I'm right-handed. -Incidentally, as for how to remember the formula, I think it's better to keep it split as (1) $q(r'\ell)=(r'q)\ell$, (2) $(r\ell)q' = (rq'^*)\ell$ and (3) $(r\ell)(r'\ell) = -r'^*r$ (obviously the purely quaternionic part doesn't need any remembering): the main difficulty is to remember the order of the factors. It's possible to recover them by remembering that $\ell$ anti-commutes with purely imaginary quaternions and anti-associates with a pair of orthogonal imaginary quaternions, but I think a better mnemonic is that for a unit quaternion $w$, the map fixing $\mathbb{H}$ and taking $r\ell$ to $(wr)\ell$ should be an automorphism of the octonions: this tells us, for example, that $r'$ should be on the left in the formula (1), because we need $q((wr')\ell)$ to be $(wr'q)\ell$, and it similarly constrains (2) and (3) so as to make mistakes impossible.<|endoftext|> -TITLE: yet another determinant and inverse of a matrix -QUESTION [11 upvotes]: This problem is some variation of another MO question. Consider the matrix -$$M_n:=\begin{bmatrix}-c& a & a& \dots & a \\ b & c & a& \ddots & a\\ b & b & -c & \ddots & a \\ \vdots & \ddots & \ddots & \ddots & a \\ b & b & b & \dots & (-1)^nc \end{bmatrix}_{n\times n},$$ -that is, a matrix with diagonal $(-1)^ic$, below diagonal entries $b$ and above diagonal entries $a$. - -Question 1. Is there a nice closed formula for the determinant $\det(M_n)$? This has been solved. -Question 2. Is there a nice expression for the inverse $M_n^{-1}$? This one awaits an answer. - -REPLY [7 votes]: Let me continue on from the notation and reasoning of Gjergji Zaimi's answer. By the Sherman-Morrison formula: -$$(A_n - a v^T v)^{-1} = A^{-1} + a \frac{ A^{-1} v v^T A^{-1} }{1 - a v^T A^{-1} v } =A^{-1} + a \frac{ A^{-1} v v^T A^{-1} }{\frac{ (a-b) P_n(a)}{a P_n(b) - bP_n(a)} } $$ -$$(A_n - b v^T v)^{-1} = A^{-1} + b \frac{ A^{-1} v v^T A^{-1} }{1 - b v^T A^{-1} v } =A^{-1} + b \frac{ A^{-1} v v^T A^{-1} }{\frac{ (a-b) P_n(b)}{a P_n(b) - bP_n(a)} } $$ -where in each case the denominator is calculated as a ratio of determinants using the determinant formulas in Gjergji Zaimi's answer. -Hence -$$b P_n(a) (A_n - a v^T v)^{-1} - aP_n(b) (A_n - b v^T v)^{-1} = (a P_n(b) - b P_n(A) ) A_n^{-1} + 0$$ -as when we take a suitable linear combination, the second terms drop out, and -$$A_n^{-1} = \frac{ b P_n(a) (A_n - a v^T v)^{-1} - aP_n(b) (A_n - b v^T v)^{-1}}{ bP_n(a) - aP_n(b) }$$ -Now $A_n - a v^T v$ and $A_n - b v^T v$, and their inverses, are lower-triangular and upper-triangular matrices respectively. I think these shouldn't be too hard to calculate explicitly.<|endoftext|> -TITLE: A necessary condition for differential entropy to be finite -QUESTION [7 upvotes]: An ensemble corresponding to a probability distribution usually has finite free energy so it is not a great loss of generality to assume that the ensemble also has finite energy in following discussions. -It is known that when the probability distribution $\mu$(assuming it is also dominated by Lebesgue measure and its density $f_{\mu}(x)$ for simplicity.) has a finite (compact) support then the classical differential entropy $$En(\mu) = \int_{\mathbb{R}} -\log[f_{\mu}(x)]f_{\mu}dx$$ -can be bounded by Bekenstein bound (Wiki) and hence finite. -However, when the support of $\mu$ is not finite (compact), there exists counter example that the entropy could be infinite (Math.SE). -Question: -(1)Is finite support of $\mu$ also a necessary condition to make sure the $\mu$ has a finite entropy? -Update: It is clear that a $\mu$ with infinite support can have finite entropy. Thanks Anthony Quas for pointing out. -(2)Is there a characterization(sufficient and necessary condition) of probability distributions/statistical ensembles with finite entropy? -(3)Also, it is known that Bekenstein bound may also be applied(with volumes defined only for atoms) to entropy defined for $\sigma$-algebras(MO.post), so can we translate the characterization in (2) onto $\sigma$-algebras? -(4)From physicists' viewpoint, what will a finite entropy system looks like? - -REPLY [5 votes]: A sufficient condition would be that the tails of $f$ decay as $O(x^{-1-\epsilon})$, which is most distributions you might encounter over $\mathbb{R}$. -That being said, the differential entropy of a continuous pdf isn't really a meaningful physical, or information theoretical, quantity. The equation isn't homogeneous: changing the units in which you measure $x$ changes the differential entropy (dimensional analysis can lead to a surprising amount of mathematical insights, even outside of physics). -What does make sense is the KL-divergence of one distribution with respect to another. For a pdf with a compact support, you're implicitly looking at the divergence with respect to the uniform distribution. However, there is no "uniform" distribution over the real line and thus the concept isn't meaningful.<|endoftext|> -TITLE: Dividing squares by sums -QUESTION [15 upvotes]: This question is out of curiosity and came to me thinking about another MO question which is linked below. - -Question: Do there exist positive integers $a,b,c$ such that $\gcd(a,b,c) =1 $ and each of $\frac{a^2}{b+c},\frac{b^2}{a+c},$ and $\frac{c^2}{a+b}$ - are also integers? - -My question was inspired by this MO question where MAEA2 asked about coprime positive integer solutions to: -$$\begin{equation}\frac{a^2}{b+c} + \frac{b^2}{a+c} + \frac{c^2}{a+b} \in \mathbb{Z}\tag{1}\end{equation}$$ -My naive self started looking for solutions to (1) via my question above, but could not find any. Jeremy Rouse has given an excellent answer using elliptic curves, but none of the points produced satisfy my question. -Note if we ask for each of $\frac{a^n}{b+c},\frac{b^n}{a+c},$ and $\frac{c^n}{a+b}$ to be integers there is clearly no solution for $n = 1,$ and for $n \geq 3$ we can take $a = 3, b = 5,$ and $c = 22$ since -$$\begin{align*} -\frac{3^n}{5+22} &= \frac{3^n}{3^3}\\ -\frac{5^n}{3+22} &= \frac{5^n}{5^2}\\ -\frac{22^n}{3+5} &= \frac{2^n 11^n}{2^3}. -\end{align*}.$$ -Also if we remove the positive condition or the $\gcd$ condition we can find solutions like $(1,-2,3)$ or $(2,2,2)$ as well as many others. - -REPLY [20 votes]: First, notice that such $a$, $b$, $c$ must be pairwise coprime (e.g., if prime $p\mid \gcd(a,b)$, then $(a+b)\mid c^2$ implies $p\mid c$, a contradiction to $\gcd(a,b,c)=1$). -As divisors of pairwise coprime numbers, $a+b$, $a+c$, $b+c$ are also pairwise coprime. -Now, since $(a+b)\mid c^2$, $(a+c)\mid b^2$, $(b+c)\mid c^2$, each of $a+b$, $a+c$, $b+c$ divides -$$D = (a+b)^2 + (a+c)^2 + (b+c)^2 - a^2 - b^2 - c^2.$$ -Then their product $(a+b)(a+c)(b+c)$ must also divide $D$. -Without loss of generality, assume that $a\leq b\leq c$ and so $a+b\leq a+c\leq b+c$. Then -$(a+b)(a+c)(b+c) \leq D < 3(b+c)^2$ and hence -$$(a+b)c < (a+b)(a+c) < 3(b+c) \leq 6c,$$ -implying that $a+b < 6$. -So, there is a finite number of cases to consider. It is easy to check that none of them gives a solution. Namely, for any fixed $a,b$, possible $c$ must belong to the finite set: -$$\{ d-b\ :\ d\mid a^2\} \cap \{ d-a\ :\ d\mid b^2\}.$$<|endoftext|> -TITLE: Counting Hilbert polynomials of projective varieties -QUESTION [11 upvotes]: EDIT. Fix $n,d,k\in\mathbb{N}$. Let us consider the set $\mathcal{P}_{n,d,k}$ of polynomials $P$ in one variable for which there exits a closed irreducible subvariety $X_P\subset \mathbb{C}\mathbb{P}^n$ -(which may be assumed to be smooth if necessary) of dimension $k$ and degree $d$ whose Hilbert polynomial is equal to $P$. -Question. Is it true that the set $\mathcal{P}_{n,d,k}$ is finite? - -REPLY [13 votes]: Edit. I edited the answer below so that it also applies to geometrically reduced schemes of degree $d$ and pure dimension $k$. Also, the argument shows that there is a single finite set $\mathcal{P}_{n,d,k}$ for all fields simultaneously, i.e., $\mathcal{P}_{n,d,k}$ is independent of the characteristic. I also edited the proof to make it simpler (in particular, there is no direct reference to existence of flattening stratifications). -Thank you for editing the question. That question is quite different from my first interpretation. -For geometrically reduced varieties of pure dimension $k$ and degree $d$, the set $\mathcal{P}_{n,d,k}$ is finite. In some sense, I believe this is already in Grothendieck's Bourbaki seminars constructing the Hilbert schemes. Bogomolov reminded me of the following argument a few years ago, together with an associated degree bound on the singular locus. I have seen a version of this also in an article by Mumford. -MR0282975 (44 #209) Reviewed -Mumford, David -Varieties defined by quadratic equations. 1970 -Questions on Algebraic Varieties (C.I.M.E., III Ciclo, Varenna, 1969) pp. 29–100 -Edizioni Cremonese, Rome -14.01 -For every field $\kappa$, for every closed subscheme $X\subset \mathbb{P}^n_\kappa$, denote by $\mathcal{I}_X$ the ideal sheaf of $X.$ For every integer $d$, denote by $I_{X,d}\subset H^0(\mathbb{P}^n_\kappa,\mathcal{O}(d))$ the $\kappa$-subspace $H^0(\mathbb{P}^n_\kappa,\mathcal{I}_X(d)).$ There is a natural map of coherent sheaves $I_{X,d}\otimes_\kappa \mathcal{O}(-d) \to \mathcal{O}_{\mathbb{P}^n_\kappa}$. Denote by $\mathcal{I}_{X,d}$ the image of this map. By construction $\mathcal{I}_{X,d}$ is an ideal sheaf that is contained in $\mathcal{I}_X$. Denote by $X_d$ the zero scheme of $\mathcal{I}_{X,d}$; this contains $X$ as a closed subscheme. -Proposition. Over every algebraically closed field $\kappa$, for every geometrically reduced, closed subscheme $X\subset \mathbb{P}^n_\kappa$ of dimension $k$ and degree $d$, the zero scheme $X_d$ equals $X$ set-theoretically. Moreover, on the smooth locus $X^o$ of $X$, $X_d$ equals $X$. -Proof. If $k$ equals $n-1$, then we are done: every reduced, degree $d$ subscheme of $\mathbb{P}^{k+1}_{\mathbb{C}}$ of pure dimension $k$ is a degree $d$ hypersurface, so $\mathcal{I}_{X,d}$ equals $\mathcal{I}_X$. -Since $X$ is geometrically reduced, the smooth locus $X^o\subset X$ is a dense open subscheme. For every $x\in X^o$, the union over all $y\in X\setminus{x}$ of the secant line $\text{span}(x,y)$ is a constructible subset of dimension $k+1$. Similarly, the "projective tangent space" to $X$ at $x$ is a linear subvariety of $\mathbb{P}^n_{\mathbb{C}}$ of dimension $k$. Finally, for a fixed closed point $z\in \mathbb{P}^n_{\mathbb{C}}\setminus X$, the union over all $y\in X$ of $\text{span}(z,y)$ is a closed subset of dimension $k+1$. There exists a linear subvariety $\Lambda\subset \mathbb{P}^n_{\mathbb{C}}$ of dimension $n-k-2$ that is disjoint from all of these constructible subsets of dimension $\leq k+1$. Thus, the $n-k-1$-plane $\text{span}(\Lambda,x)$, intersects $X$ at only $x$, and the intersection is (tangentially) transverse at this point. Moreover, for every $y\in X$, $\text{span}(\Lambda,y)$ does not contain $z$, or else $\Lambda$ would contain $\text{span}(z,y)$. -Thus, the linear projection "away" from $\Lambda$, $$\pi:\mathbb{P}^n_{\mathbb{C}}\setminus \Lambda \to \mathbb{P}^{k+1}_{\mathbb{C}},$$ restricts on $X$ to a morphism that is injective and unramified on $\pi^{-1}(U)$ for $U$ some Zariski open neighborhood of $\pi(x)$. The image $\pi(X)$ is a degree $d$ hypersurface, $\text{Zero}(G_{\Lambda,x})$ for some degree $d$ polynomial $G_{\Lambda,x}$ (even without these transversality hypotheses, there is a degree $d$ polynomial naturally associated to the image using "det", "Div" and "Chow" as in Knudsen-Mumford and Fogarty). Consider $F_{\Lambda,x}$, the pullback by $\pi$ of $G_{\Lambda,x}$ as a degree $d$ polynomial on $\mathbb{P}^n_{\mathbb{C}}.$ The zero scheme of $F_{\Lambda,x}$ contains $X$, but it does not contain $z$. -Varying $\Lambda$, the common zero locus of these polynomials is a closed subscheme that set-theoretically equals $X$ (since for every $z$, there is a $F_{\Lambda,x}$ that does not vanish at $z$), and that scheme contains a Zariski open subset that equals a Zariski open neighborhood of $x$ in $X$. Thus, up to varying both $\Lambda$ and $x$, the common closed subscheme set-theoretically equals $X$, and the scheme contains $X^o$ as a dense open subscheme (with its correct reduced scheme structure). This closed subscheme contains $X_d$, so the same is true for $X_d$. QED -Denote by $P_n(t)\in \mathbb{Q}[t]$ the numerical polynomial such that for every integer $r\geq -n$, $P_n(r)$ equals $\binom{n+r}{n}$. The dimension $m$ of $I_{X,d}$ is $\leq P_n(d)$. -Proposition. For every integer $m$ with $n-k\leq m \leq P_n(d)$, there is a finite set $\mathcal{P}_{n,d,k,m}$ of Hilbert polynomials such that for every algebraically closed field $\kappa$ and for every reduced closed subscheme $X\subset \mathbb{P}^n_\kappa$ of degree $d$, of pure dimension $k$, and with $\text{dim}_\kappa I_{X,d} = m$, the Hilbert polynomial of $X$ is in $\mathcal{P}_{n,d,k,m}$. -Proof. Denote by $G_m$ the Grassmannian over $\text{Spec}(\mathbb{Z})$ parameterizing rank $m$, locally free quotients of the free $\mathbb{Z}$-module $H^0(\mathbb{P}^n_{\mathbb{Z}},\mathcal{O}(d))^\vee.$ Denote by $\mathcal{Y} \subset G_m\times_{\text{Spec}(\mathbb{Z})} \mathbb{P}^n_{\mathbb{Z}}$ the closed subscheme cut out by the universal family of $m$-dimensional linear systems of degree $d$ polynomials. A locally closed subscheme $S$ of $G_m$ is transversal if (i) the restriction $\mathcal{Y}_S=S\times_{G_m} \mathcal{Y}$ is surjective and $S$-flat of pure relative dimension $k$, (ii) the smooth locus $\mathcal{Y}^o_S$ of $\rho_S:\mathcal{Y}_S\to S$ is set-theoretically dense in $\mathcal{Y}_S$, (iii) for the closure $\overline{\mathcal{Y}}^o_S \subset \mathcal{Y}_S$, $\overline{\mathcal{Y}}^o_S$ is $S$-flat, and (iv) for every geometric point of $S$, the fiber of $\mathcal{Y}^o_S$ over that point is schematically dense in the fiber over that point of $\overline{\mathcal{Y}}^o_S$. A locally closed subscheme $S$ of $G_m$ is irrelevant if either (a) $\mathcal{Y}_S$ is empty, (b) $\rho_S:\mathcal{Y}_S\to S$ is surjective and flat, but every geometric fiber has an irreducible component of dimension $ -TITLE: $f$ real-rooted forbid truncated $\frac1f$ to be so? -QUESTION [13 upvotes]: Let $f(x)$ be a polynomial in the ring $\mathbb{R}[x]$, the roots are all real and $f(0)=1$. Write the Taylor series of $1/f(x)$ around the origin as -$$\frac1{f(x)}=\sum_{k=0}^{\infty}a_kx^k,$$ -and denote $P_n(x)=\sum_{k=0}^na_kx^k$. - -Question. Is it true that the polynomial $P_{2n}(x)$ has no real roots at all? - -A prototypical example: If we choose $f(x)=1-x$ then $P_{2n}(x)=\sum_{k=0}^{2n}x^k=\frac{1-x^{2n+1}}{1-x}$ has all roots $r$ on the unit circle and $r\neq\pm1$. - -REPLY [8 votes]: Here is an alternative resolution motivated by Fedor's construction in his first line of argument. -Based on $f(0)=1$, it is clear that $\frac1{f(x)}=1+a_1x+a_2x^2+\cdots$ and now write -$$f(x)P_{2n}(x)=f(x)(1+a_1x+a_2x^2+\cdots+a_{2n}x^{2n})= -1-a_{2n+1}x^{2n+1}+\cdots.$$ -The RHS, call it $Q(x)$, is a non-constant polynomial. -More notably, observe that $Q(x)$ exhibits $2n$ (an even number of) vanishing consecutive coefficients. An application of Descartes' Rule of signs reveals $Q(x)$ has at least $2n$ non-real roots. This result is known as de Gua's Rule, which follows from Descartes' or Fourier-Budan Theorem. See the reference here, on page 28, Corollary 2. -Since $f(x)$ has only real roots, these non-real roots must come from the factor $P_{2n}(x)$ of $Q(x)$.<|endoftext|> -TITLE: Who proved the upper bound for the autocorrelation of higher order divisor functions? -QUESTION [7 upvotes]: Who first published a proof that -$$\sum_{n\leq x}d_{k}(n)d_k(n+h)=O(x(\log x)^{2k-2})$$ -for fixed $k$ and $h$ please? I am struggling to find a reference. Thank you. - -REPLY [9 votes]: I'm not sure if it was the first, but the bounds would follow from the main theorem in -Nair, Mohan; Tenenbaum, G\'erald, Short sums of certain arithmetic functions, Acta Math. 180, No.1, 119-144 (1998). ZBL0917.11048. -See also the more uniform bounds (in your case, this would give uniformity in $h$, after taking the singular series into account) obtained in -Henriot, Kevin, Nair-Tenenbaum bounds uniform with respect to the discriminant, Math. Proc. Camb. Philos. Soc. 152, No. 3, 405-424 (2012); erratum ibid. 157, No. 2, 375-377 (2014). ZBL1255.11048. -The result also follows from the properties of the pseudorandom majorants for multiplicative functions such as $d_k$ that were constructed by Matthiesen, see e.g. https://arxiv.org/abs/1606.04482<|endoftext|> -TITLE: Can ETCC/ETCS talk about 'size issues'? -QUESTION [8 upvotes]: In material set theories (like ZFC), one can prove that there is no set of all sets. Can one prove a similar statement in ETCS? This exact statement "there is no set x such that y in x for every set y" is not expressible in ETCS because ETCS is a structural set theory. But however is there a way to nevertheless talk about size issues in ETCS? -Also, in ETCC, which axiomatizes the category of categories, can one there prove/formulate a statement like "There is no category that contains every category"? - -REPLY [8 votes]: We can talk about a family of sets $Y_x$ parameterized by the elements $x \in X$ of another set using a bundle of sets, namely a map -$$Y = \bigsqcup_{x \in X} Y_x \mapsto X$$ -whose fibers are the family of interest. This is, for example, how ETCS talks about the axiom of choice. So we can phrase the nonexistence of a "bundle of all sets" as the following claim. - -Claim: There is no map $f : Y \to X$ such that, for every set $S$, there exists $x \in X$ such that $S \cong f^{-1}(x)$. - -(Note that $x \in X$ is the same as saying $x : 1 \to X$ and $f^{-1}(x)$ is the pullback of the diagram $1 \xrightarrow{x} X \xleftarrow{f} Y$; this really is an entirely categorical statement.) -What I want to say from here is that we should pick $S = 2^Y$ and then appeal to Cantor's theorem, or more categorically the Lawvere fixed point theorem. But what we need is a surjection $Y \to 2^Y$ and what we get is an injection $2^Y \to Y$, and I'm not sure how much of ETCS is necessary to make everything work out.<|endoftext|> -TITLE: One-ended finitely presented subgroups of hyperbolic groups -QUESTION [6 upvotes]: In Hyperbolic groups (page 82), Gromov claims that, by a standard application of Thurston's method of geodesic (hyperbolic) simplices, it can be prove that a hyperbolic group contains finitely many pairwise non conjugate subgroups isomorphic to a fixed one-ended finitely presented group. -I took a look on the usual references dealing with hyperbolic groups, but I didn't find any details. Do you know references about this result and/or Thurston's method? - -REPLY [9 votes]: There is a proof of this theorem due to Thomas Delzant: -T. Delzant, L’image d’un groupe dans un groupe hyperbolique, Comment. Math. Helv. -70 (1995), no. 2, 267–284. -There is also a version for relatively hyperbolic groups du to Francois Dahmani: -Accidental Parabolics and Relatively Hyperbolic Groups, Israel Journal of Mathematics, Volume 153, Issue 1, pp 93–127<|endoftext|> -TITLE: Extensions of local vector fields to whole manifold -QUESTION [5 upvotes]: Let $M$ be a smooth manifold (with boundary). Suppose I have a smooth vector field $T$ defined on the complement of a compact subset $K$ of $M$ and I wish to extend $T$ to the whole of $M$. What are the obstructions to doing so? Does $M$ need to be compact? Does $K$ need to be a submanifold? -I have a feeling that an argument involving a partition of unity of $K$ would suffice for extending $T$ into $K$, but have not been able to find a decent reference. - -REPLY [7 votes]: The sheaf of smooth functions on a manifold is fine and hence soft, so we can extend sections on closed subsets to global sections. However, it is not generally flabby (flasque): local sections on open subsets do not in general extend to global sections. Note that only discrete manifolds can have a flabby sheaf of smooth functions, which is the case because every subset is open and hence closed. -Modules over the sheaf of smooth functions are also soft, but not in general flabby (except for the zero module, for trivial reasons). Since the sheaf of vector fields is such a module, we wouldn't usually expect vector fields on open subsets to extend to global sections. -For example, if you have any sort of open (proper) subset which is diffeomorphic to an open ball, just take a smooth function that blows up as you go to the boundary of the ball. Then you can rescale any vector field that does not decay to zero at the boundary to a vector field which fails to globalize. -H. H. Rugh's example in his answer is also very good because it shows that even without blow ups, a smooth vector field on an open set (which in his case is diffeomorphic to an open disk) can fail to extend globally due to the topology of the sphere. -As mentioned by Ben McKay in the comments, you can have a look at Charles Fefferman's paper proving a sharp version of the Whitney extension theorem. It's available here.<|endoftext|> -TITLE: Pell-type equations with no integer solutions -QUESTION [11 upvotes]: Consider a Pell-type equation of the form: -$$x^2-ay^2=b$$ -where $a,b\in\mathbb{Z}$. Does there exist a method to exclude the existence of integer solutions to such an equation? - -REPLY [11 votes]: For a specific problem, there is a finite check to decide whether any solutions exist to $x^2 - a y^2 = b.$ Find the minimum integers $u,v > 0$ such that $u^2 - a v^2 = 1.$ Any solution to your problem, say with $x,y$ positive, creates an infinite sequence of solutions by -$$ (x,y) \mapsto (ux + av y, \; vx + u y). $$ This will increase the entries. In the other direction, that of decreasing one or both entries, -$$ (x,y) \mapsto (ux - av y, \; -vx + u y). $$ -Repeating this mapping gets us to a solution with both $x,y> 0$ but one of the entries in the left neighbor nonpositive, either -$$ ux - ay \leq 0 \; \; \; \mbox{OR} \; \; \; -vx + uy \leq 0. $$ -If you draw some picture, including the hyperbola $x^2 - a y^2 = b,$ you see how one or the other of $ux \leq ay$ or $uy \leq vx$ gives a bounded arc of the hyperbola. If there are no integer solutions in that arc, there are no solutions at all. -More generally, one may draw the Conway topograph for an indefinite form $a x^2 + bxy + c y^2.$ His "climbing lemma" shows how we need investigate only a finite region of the diagram to decide whether there is a (primitive) solution to $a x^2 + bxy + c y^2= n.$ -$$ x^2 - 2 y^2 = 7 \cdot 17 \cdot 23 $$ - -jagy@phobeusjunior:~$ ./Pell_Target_Fundamental - Automorphism matrix: - 3 4 - 2 3 - Automorphism backwards: - 3 -4 - -2 3 - - 3^2 - 2 2^2 = 1 - - x^2 - 2 y^2 = 2737 - -Sun Mar 12 14:48:31 PDT 2017 - -x: 53 y: 6 ratio: 8.83333 SEED KEEP +- -x: 55 y: 12 ratio: 4.58333 SEED KEEP +- -x: 57 y: 16 ratio: 3.5625 SEED KEEP +- -x: 73 y: 36 ratio: 2.02778 SEED KEEP +- -x: 75 y: 38 ratio: 1.97368 SEED BACK ONE STEP 73 , -36 -x: 107 y: 66 ratio: 1.62121 SEED BACK ONE STEP 57 , -16 -x: 117 y: 74 ratio: 1.58108 SEED BACK ONE STEP 55 , -12 -x: 135 y: 88 ratio: 1.53409 SEED BACK ONE STEP 53 , -6 -x: 183 y: 124 ratio: 1.47581 -x: 213 y: 146 ratio: 1.4589 -x: 235 y: 162 ratio: 1.45062 -x: 363 y: 254 ratio: 1.42913 -x: 377 y: 264 ratio: 1.42803 -x: 585 y: 412 ratio: 1.4199 -x: 647 y: 456 ratio: 1.41886 -x: 757 y: 534 ratio: 1.4176 -x: 1045 y: 738 ratio: 1.41599 -x: 1223 y: 864 ratio: 1.41551 -x: 1353 y: 956 ratio: 1.41527 -x: 2105 y: 1488 ratio: 1.41465 -x: 2187 y: 1546 ratio: 1.41462 -x: 3403 y: 2406 ratio: 1.41438 -x: 3765 y: 2662 ratio: 1.41435 -x: 4407 y: 3116 ratio: 1.41431 -x: 6087 y: 4304 ratio: 1.41427 -x: 7125 y: 5038 ratio: 1.41425 -x: 7883 y: 5574 ratio: 1.41424 -x: 12267 y: 8674 ratio: 1.41423 -x: 12745 y: 9012 ratio: 1.41423 -x: 19833 y: 14024 ratio: 1.41422 -x: 21943 y: 15516 ratio: 1.41422 -x: 25685 y: 18162 ratio: 1.41422 -x: 35477 y: 25086 ratio: 1.41422 -x: 41527 y: 29364 ratio: 1.41421 -x: 45945 y: 32488 ratio: 1.41421 -x: 71497 y: 50556 ratio: 1.41421 -x: 74283 y: 52526 ratio: 1.41421 -x: 115595 y: 81738 ratio: 1.41421 -x: 127893 y: 90434 ratio: 1.41421 -x: 149703 y: 105856 ratio: 1.41421 -x: 206775 y: 146212 ratio: 1.41421 -x: 242037 y: 171146 ratio: 1.41421 -x: 267787 y: 189354 ratio: 1.41421 -x: 416715 y: 294662 ratio: 1.41421 -x: 432953 y: 306144 ratio: 1.41421 -x: 673737 y: 476404 ratio: 1.41421 -x: 745415 y: 527088 ratio: 1.41421 -x: 872533 y: 616974 ratio: 1.41421 -x: 1205173 y: 852186 ratio: 1.41421 -x: 1410695 y: 997512 ratio: 1.41421 -x: 1560777 y: 1103636 ratio: 1.41421 -x: 2428793 y: 1717416 ratio: 1.41421 -x: 2523435 y: 1784338 ratio: 1.41421 -x: 3926827 y: 2776686 ratio: 1.41421 -x: 4344597 y: 3072094 ratio: 1.41421 -x: 5085495 y: 3595988 ratio: 1.41421 -x: 7024263 y: 4966904 ratio: 1.41421 -x: 8222133 y: 5813926 ratio: 1.41421 -x: 9096875 y: 6432462 ratio: 1.41421 -x: 14156043 y: 10009834 ratio: 1.41421 -x: 14707657 y: 10399884 ratio: 1.41421 - -Sun Mar 12 14:48:51 PDT 2017 - - x^2 - 2 y^2 = 2737 - -$$ x^2 - 2 y^2 = - 7 \cdot 17 \cdot 23 $$ - -jagy@phobeusjunior:~$ ./Pell_Target_Fundamental - Automorphism matrix: - 3 4 - 2 3 - Automorphism backwards: - 3 -4 - -2 3 - - 3^2 - 2 2^2 = 1 - - x^2 - 2 y^2 = -2737 - -Sun Mar 12 14:50:52 PDT 2017 - -x: 1 y: 37 ratio: 0.027027 SEED KEEP +- -x: 25 y: 41 ratio: 0.609756 SEED KEEP +- -x: 31 y: 43 ratio: 0.72093 SEED KEEP +- -x: 41 y: 47 ratio: 0.87234 SEED KEEP +- -x: 65 y: 59 ratio: 1.10169 SEED BACK ONE STEP -41 , 47 -x: 79 y: 67 ratio: 1.1791 SEED BACK ONE STEP -31 , 43 -x: 89 y: 73 ratio: 1.21918 SEED BACK ONE STEP -25 , 41 -x: 145 y: 109 ratio: 1.33028 SEED BACK ONE STEP -1 , 37 -x: 151 y: 113 ratio: 1.33628 -x: 239 y: 173 ratio: 1.3815 -x: 265 y: 191 ratio: 1.38743 -x: 311 y: 223 ratio: 1.39462 -x: 431 y: 307 ratio: 1.40391 -x: 505 y: 359 ratio: 1.40669 -x: 559 y: 397 ratio: 1.40806 -x: 871 y: 617 ratio: 1.41167 -x: 905 y: 641 ratio: 1.41186 -x: 1409 y: 997 ratio: 1.41324 -x: 1559 y: 1103 ratio: 1.41342 -x: 1825 y: 1291 ratio: 1.41363 -x: 2521 y: 1783 ratio: 1.41391 -x: 2951 y: 2087 ratio: 1.41399 -x: 3265 y: 2309 ratio: 1.41403 -x: 5081 y: 3593 ratio: 1.41414 -x: 5279 y: 3733 ratio: 1.41414 -x: 8215 y: 5809 ratio: 1.41418 -x: 9089 y: 6427 ratio: 1.41419 -x: 10639 y: 7523 ratio: 1.4142 -x: 14695 y: 10391 ratio: 1.4142 -x: 17201 y: 12163 ratio: 1.41421 -x: 19031 y: 13457 ratio: 1.41421 -x: 29615 y: 20941 ratio: 1.41421 -x: 30769 y: 21757 ratio: 1.41421 -x: 47881 y: 33857 ratio: 1.41421 -x: 52975 y: 37459 ratio: 1.41421 -x: 62009 y: 43847 ratio: 1.41421 -x: 85649 y: 60563 ratio: 1.41421 -x: 100255 y: 70891 ratio: 1.41421 -x: 110921 y: 78433 ratio: 1.41421 -x: 172609 y: 122053 ratio: 1.41421 -x: 179335 y: 126809 ratio: 1.41421 -x: 279071 y: 197333 ratio: 1.41421 -x: 308761 y: 218327 ratio: 1.41421 -x: 361415 y: 255559 ratio: 1.41421 -x: 499199 y: 352987 ratio: 1.41421 -x: 584329 y: 413183 ratio: 1.41421 -x: 646495 y: 457141 ratio: 1.41421 -x: 1006039 y: 711377 ratio: 1.41421 -x: 1045241 y: 739097 ratio: 1.41421 -x: 1626545 y: 1150141 ratio: 1.41421 -x: 1799591 y: 1272503 ratio: 1.41421 -x: 2106481 y: 1489507 ratio: 1.41421 -x: 2909545 y: 2057359 ratio: 1.41421 -x: 3405719 y: 2408207 ratio: 1.41421 -x: 3768049 y: 2664413 ratio: 1.41421 -x: 5863625 y: 4146209 ratio: 1.41421 -x: 6092111 y: 4307773 ratio: 1.41421 -x: 9480199 y: 6703513 ratio: 1.41421 -x: 10488785 y: 7416691 ratio: 1.41421 -x: 12277471 y: 8681483 ratio: 1.41421 -x: 16958071 y: 11991167 ratio: 1.41421 - -Sun Mar 12 14:51:24 PDT 2017 - - x^2 - 2 y^2 = -2737 - -$$ x^2 - 2 y^2 = -15 $$ -no integer solutions; prohibited by local conditions.<|endoftext|> -TITLE: How bad does a ring have to be for a failure of "going-in-between"? -QUESTION [12 upvotes]: Let $A\subset B$ be an integral extension of commutative unital rings. -Let $\mathfrak{p}_0\subset\mathfrak{p}_1\subset\mathfrak{p}_2$ be a saturated chain of primes in $A$ of length $2$. -Suppose $\mathfrak{q}_0,\mathfrak{q}_2$ lie over $\mathfrak{p}_0,\mathfrak{p}_2$, and $\mathfrak{q}_0\subset\mathfrak{q}_2$. -Is there necessarily a $\mathfrak{q}_1$ satisfying $\mathfrak{q}_0\subset\mathfrak{q}_1\subset\mathfrak{q}_2$ and lying over $\mathfrak{p}_1$? -It seems to me the answer is clearly yes if the rings $A,B$ are sufficiently geometric (edit: even this is no longer clear to me, see addendum), e.g. finitely generated algebras over an algebraically closed field, since in this case, if there is no $\mathfrak{q}_1$, then $\mathfrak{q}_0\subset\mathfrak{q}_2$ is saturated, and then $V(\mathfrak{q}_2)$ is a codimension one subvariety of $V(\mathfrak{q}_0)$, and $\operatorname{Spec}B\rightarrow\operatorname{Spec}A$ is a dimension-preserving map, so what is $V(\mathfrak{p}_1)$'s dimension? -But in general, it's not obvious to me. It seems to require that going-down holds in the integral extension of domains $A/\mathfrak{p}_0\subset B/\mathfrak{q}_0$, and because the going-down theorem requires an extra assumption of integral closure, shouldn't this fail sometimes? -So my question is: - -How bad do $A,B$ have to be for $\mathfrak{q}_1$ to fail to exist? Can it happen for noetherian rings? Cohen-Macaulay rings? What's the "least pathological" example? - -NB: This is crossposted from math.SE, where it hasn't gotten any answers after 2 weeks and a bounty. (Edit: it is no longer really the same question, see addendum.) -Addendum: A propos of an exchange in the comments with Jason Starr (partially deleted now), even in the geometric case it's no longer clear to me that the answer is yes. When I originally posted, I mistook the fact that a prime between $\mathfrak{q}_0$ and $\mathfrak{q}_2$ would necessarily lie over some prime strictly between $\mathfrak{p}_0$ and $\mathfrak{p}_2$ to imply it would lie over $\mathfrak{p}_1$. (Jason pointed out this mistake.) I corrected this by explicitly adding the requirement that $\mathfrak{q}_1$ lie over $\mathfrak{p}_1$, but now it is no longer true that the failure of $\mathfrak{q}_1$ to exist implies that $\mathfrak{q}_0\subset\mathfrak{q}_2$ is saturated. Thus, perhaps $V(\mathfrak{q}_2)$ is codimension 2 in $V(\mathfrak{q}_0)$, it's just that none of the codimension one subvarieties of $V(\mathfrak{q}_0)$ happen to lie over $V(\mathfrak{p}_1)$. It's not at all clear to me that this can't happen. -Fred Rohrer mentions a highly relevant paper by L.J. Ratliff in the comments. A cursory scan of this paper seems to indicate that the answer might be no most of the time. Ratliff states that the condition I'm asking for here is not even guaranteed if $A$ is a complete regular local ring (bottom of p. 778), though it's not completely clear from context whether he's requiring $B$ to be integral in this case. What's an example? -One last point is that these developments distinguish this question from the one I asked at math.SE, where I asked, "In what generality does it hold that if $\mathfrak{q}_0\subset\mathfrak{q}_2$ is saturated, and $\mathfrak{q}_i$ lies over $\mathfrak{p}_i$ in $A$, then $\mathfrak{p}_0\subset\mathfrak{p}_2$ is saturated?" This turns out (if I am thinking straight) to be the question addressed by Ratliff's paper, but I am asking for something stronger at present. - -REPLY [2 votes]: This is a partial answer, giving a two dimensional Noetherian counterexample. It starts from your observation that one has to consider the case where $A/\mathfrak{p}_0$ is not integrally closed. Geometrically the construction is as follows: take a normal affine surface $Y$, a curve $C$ on $Y$, a point $P_1 \in C$ and a point $P_2 \in Y \setminus C$. Construct a non-normal surface $X$ by identifying $P_1$ and $P_2$; let $P'$ be the image of $P_1$ and $P_2$ on $X$ and $C'$ be the image of $C$ on $X$. Then the chains $X \supset C' \ni P'$ on $X$ and $Y \ni P_2$ on $Y$ give a counterexample. -For an explicit example, take $Y = k^2$, where $k$ is any field, $C = x$-axis, $P_1 = (0,0)$, $P_2 = (0,1)$. So $B = k[x,y]$ and $A$ is the set of all polynomials $f \in B$ such that $f(0,0)= f(0,1)$. $B$ is integral over $A$, since e.g. $x \in B$ and $y(y-1) \in B$.<|endoftext|> -TITLE: A matrix norm inequality II -QUESTION [5 upvotes]: Let $\|\cdot\|$ be the spectral norm, i.e., largest singular value. The condition number of an invertible complex matrix $X$ is defined as $\kappa(X):=\|X\|\|X^{-1}\|$. -I am able to prove -Proposition Let $A, B$ be $n\times n$ positive definite matrices. If $X$ is an $n\times n$ invertible matrix such that $AXB$ is Hermitian, then \begin{eqnarray*} - \|X^{-1}AXB\|\le \kappa(X) \|AB\|. - \end{eqnarray*} -In particular, if moreover $X$ in the above proposition is unitary, then $\|X^{-1}AXB\|\le \|AB\|$. -I wonder if $\kappa(X)$ in the above proposition can always be replaced with $1$. - -REPLY [7 votes]: No the norm of the left side can be very large. -For example $\left\| X^{-1}AXB \right\| $ is an unbounded function in $(x,y)$ where $A,X,B$ are the following matrices: -Put $A=B= \begin{pmatrix} 1&0\\0&2 \end{pmatrix}$ and $X^{-1}=\begin{pmatrix} -x^3&y\\y&x \end{pmatrix}$ -where $(x,y) \in \mathbb{R}^{2} \setminus \{0\}$. -In this example $A,B$ are positive matrices and $AXB$ is equal to its transpose so is a Hermitian matrix. -This shows that the inequality $\left\| X^{-1}AXB \right\| \leq \left\| A B \right\| $ is not necessarily true. Because the left side is unbounded function in $(x,y)$ but the right side is a constant function. -The left side is unbounded because of the following argument: -Put $C=X^{-1}AXB$ then $C_{21}=xy/x^{4}+y^2$. Then $|C_{21}|$ tends to $+\infty$ when $y=x^2$ and $x \to 0$. So $\parallel C \parallel_{\infty} = Max |C_{ij}|$ goes to infinity. But the later norm is equivalent to the spectral norm.<|endoftext|> -TITLE: Fixed points of cardinal logarithm -QUESTION [7 upvotes]: For any cardinal $\kappa$ set $$\log(\kappa) = \min\{\mu\in \kappa\cup\{\kappa\}: 2^\mu \geq \kappa\}.$$ -Clearly $\log(\omega) = \omega$ and in $\textsf{GCH}$ we have $\log(\aleph_\omega) = \aleph_\omega$. -Is it consistent that $\log(\kappa)<\kappa$ for all uncountable cardinals $\kappa$? - -REPLY [12 votes]: No. Because strong limits cardinals exist in $\sf ZFC$. These are the $\beth_\alpha$ for a non-successor $\alpha$.<|endoftext|> -TITLE: Euclidean tangent cone implies Riemannian manifold -QUESTION [13 upvotes]: It is known that given a Riemannian manifold, then the tangent cone (as a metric space) at any point $p$ is isometric to the tangent space at $p$, with the metric given by the metric tensor. -Is there a converse, or an additional condition to have a converse, in the following form? - -Given a metric space $X$ such that the tangent cone at any point is a Euclidean space of dimension $n$, and (possibly additional condition), then $X$ is a Riemannian manifold of dimension $n$. - -REPLY [12 votes]: To provide some context the subsets of a Euclidean space that can be approximated by affine planes on every scale are known as Reifenberg-flat sets after E. R. Reifenberg who proved in the 1960s that such sets are bi-Holder to a Euclidean space. There is a substantial literature on the subject (search on "Reifenberg-flat"). -Now regarding your specific question: T. Colding and A. Naber construct in Lower Ricci Curvature, Branching, and Bi-Lipschitz Structure of Uniform Reifenberg Spaces a metric space $Y$ such that - -The tangent cones of $Y$ are all isometric to $\mathbb R^n$ (which by an earlier work of J. Cheeger and Colding implies that $Y$ is bi-Holder homeomorphic to $\mathbb R^n$). -Every bounded set of $Y$ is bi-Lipschitz embeddable in some Euclidean space. -There is no homeomorphism of $Y$ onto $\mathbb R^n$ such that the pullback geometry is induced by some $C^{0,\beta}$ Riemannian metric where $0<\beta<1$. - -To show (3) they prove that geodesic in $Y$ branch in the sense of Theorem 1.3 of the linked paper. -Moreover, $Y$ occurs ``in nature'' as a pointed Gromov-Hausdorff limit of a noncollapsing sequence of Riemannian $n$-manifolds with a common lower bound on Ricci curvature. -I am not sure whether the metric on $Y$ can be induced by a $C^0$ Riemannian metric but in any case branching of geodesics is not the property a decent $C^0$ metric should be proud of.<|endoftext|> -TITLE: Generalized Rayleigh-quotient gradient flow on Grassmannian -QUESTION [7 upvotes]: The following theorem appears without proof in : -Helmke, Uwe, and John B. Moore. Optimization and dynamical systems. Springer Science & Business Media, 2012. -Let $A$ be a symmetric $n\times n$ real matrix. Define the Stiefel manifold as -$St(k,n)=\{X\in \mathbb{R}^{n\times k}|X^TX=I\}$. -Then, we consider the following equation : -$\dot{X}=(I-XX^T)AX$, where $X\in St(k,n)$ -It is a matrix ODE which is invariant under right multiplication by $O(k)$. Hence it can be considered as an ODE on the Grassmannian. -Helmke and Moore state that it almost-surely converges to an $A$-invariant subspace spanned by a dominant $k$-dimensional eigenbasis of $A$. -I am looking for proof of this theorem. Does anyone know any suitable references ? - -REPLY [5 votes]: On the Stiefel manifold, consider the function $f(X)=\operatorname{tr}(X^TAX)$. It evolves under the flow of the given vector field as -$$\frac d{dt}f=2\operatorname{tr}(X^TA(1-XX^T)AX)\;.$$ -Because $1-XX^T$ describes the projection onto the orthogonal complement of the span $V_X$ of the columns of $X$, we have -\begin{gather*}\frac d{dt}f\ge 0\;,\\\frac d{dt}f=0\iff \dot X=0\iff\text{$V_X$ is $A$-invariant.}\end{gather*} -Hence, $X$ is a generalised gradient field for $f$ (it might in fact be half the actual gradient, but I have not checked). -Now it is easy to see that $f$ induces a Morse-Bott function on the Grassmannian (Morse if the eigenvalues of $A$ are distinct). The only stable fixpoints therefore correspond to the maximum, and the points that don't flow into one of the maxima form a stratified subset of lower dimension.<|endoftext|> -TITLE: Prokhorov's theorem in non separable metric spaces -QUESTION [13 upvotes]: Recently, working in some calculations I needed to use the Prokhorov's theorem -about compactness for probability measures. However, a friend warned me that -I had not the hypotesis of separability required by the theorem. -After some searching, over the books which I have reach, this is the version of the theorem that I found: -Let $ {\displaystyle (S,\rho )} $ be a separable metric space. Let -$ {\displaystyle {\mathcal {P}}(S)}$ denote the collection of all probability measures defined on ${\displaystyle S}$ (with its Borel σ-algebra). -Theorem (Prokhorov). - A collection ${\displaystyle K\subset {\mathcal {P}}(S)}$ of probability measures is tight if and only if the closure of ${\displaystyle K}$ is sequentially compact in the space ${\displaystyle {\mathcal {P}}(S)}$ equipped with the topology of weak convergence. -However, in a internet research I found the following document, -Prokhorov Theorem -. -where no separability is required to get one of the directions of the theorem, here a description of the result contained in this notes: -Theorem. Let $S$ be a metric space. - If collection ${\displaystyle K\subset {\mathcal {P}}(S)}$ of probability measures is tight then the closure of ${\displaystyle K}$ is sequentially compact in the space ${\displaystyle {\mathcal {P}}(S)}$ equipped with the topology of weak convergence. Conversely if $S$ is -separable and complete, then each relatively compact set is -tight. -Question: Is this version correct? If so, can someone provide me -some reference? - -REPLY [13 votes]: It is correct, see Theorem 8.6.7 in volume 2 of Bogachev's "Measure Theory" monograph. See also his Theorem 8.6.8 for a version of the second statement which covers the case of a non-separable space.<|endoftext|> -TITLE: map from 6-vertex model to domino tiling -QUESTION [7 upvotes]: I am trying to find a correspondence between 6-vertex model and an Aztec Diamond tiling. Here are the building blocks of the 8-vertex model: - -There seems to be more than one correspondence. I found at least two: - -Ferrari, Spohn Domino tilings and the six-vertex model at its free fermion point -Zinn-Justin Six-Vertex Model with Domain Wall Boundary Conditions and One-Matrix Model - -And there is likely even more. My goal had been to try to understand if there was equivalence between domino tilings and alternating sign matrix, with 6-vertex as an intermediate object. I wound up just totally confusing myself. - -ising model -alternating sign matrix -6-vertex model -domino tiling - - -Now ASM and Domino tilings in the same space have different numbers of tilings. ELKP shows how to obtain two different Alternating Sign Matrices from the domino tilings. -Every paper seems to give a different answer. My main question is whether the combinatorics models: - -domino, alternating sign - -are the same as the statistical mechanics models? Here I list a few candidates - -ising, 6-vertex, XXZ etc - -REPLY [7 votes]: In short: unlike the mapping between alternating-sign matrices and the configurations of the six-vertex model with domain-wall boundary conditions, the mapping between the six-vertex model and domino tilings is not one-to-one. This accounts for the fact that the relevant special case of the six-vertex model (with $a=b=c/\sqrt{2}$) corresponds to the 2-enumeration of alternating-sign matrices (ASMs). -On the statistical-mechanical side we're dealing with the special case $a=b -TITLE: Factorization of a Markov chain as the product of smaller chains -QUESTION [5 upvotes]: Consider a discrete-time Markov chain on $n$ states $S = \{1,2,\ldots,n\}$, with transition dynamics -$$ \mathbf{x}^{(t+1)} = \mathbf{P} \mathbf{x}^{(t)},$$ -where $\mathbf{P}$ is the transition matrix, a stochastic matrix (nonnegative entries and unit column sums). We are interested in the possibility of decomposing this chain into a product of 2 chains on two disjoint subject sets of states $A,B \subseteq S$ (of approx same size), -with dynamics $\mathbf{P}_A$ and $\mathbf{P}_B$ resp. so that -$$\mathbf{x}_A^{(t+1)} = \mathbf{P}_A \mathbf{x}_A^{(t)},\text{ and }\mathbf{x}_B^{(t+1)} = \mathbf{P}_B \mathbf{x}_B^{(t)}, $$ -where $\mathbf{P}_A$ and $\mathbf{P}_B$ are stochastic matrices on $\#A$ and $\#B$ states resp. When we can produce such a factorization, we shall say that the original chain is splittable (index $r = 1$), otherwise it is irreducible (index $r = 0$). -Question: I wish to extend / relax this split index $r$ to a general scenario and get a float $r(\mathbf{P}) \in [0, 1]$ which measures how well / bad the original chain can be approximated can be factored into smaller chains. Is there anything in literature (maybe linear algebra ?, graph theory ?) that can shed light on this ? - -REPLY [3 votes]: What you need is a definition of graph connectivity for undirected graphs. One can try to generalize the idea of Fiedler vector from directed graphs: if the second-to-last singular value of $I-P$ is zero, then the matrix is "splittable", otherwise it is not. For instance, you can use the ratio $\sigma_{n-1}/\sigma_1$ as a measure of connectivity. -If you have already a candidate approximate splitting into two blocks $\begin{bmatrix}P_{11} & P_{12} \\ P_{21} & P_{22}\end{bmatrix}$, another simple performance measure is $\frac{\|P_{12}\|+\|P_{21}\|}{\|P_{11}\|+\|P_{12}\|+\|P_{21}\|+\|P_{22}\|}$, which also makes sense as "distance from the closest splittable matrix".<|endoftext|> -TITLE: Upper shadow of a union-closed family -QUESTION [8 upvotes]: Is the following claim true? -Let $\mathcal{F} \subseteq 2^{[n]}$ be a union-closed family. That is, if $A,B \in \mathcal{F}$, then $A \cup B \in \mathcal{F}$. Then -$$|\partial^+\mathcal{F}\setminus \mathcal{F}|\leq 2^{n-1},$$ -where $\partial^+ \mathcal{F}$ is the upper shadow of $\mathcal{F}$. -Edit: To be explicit about what is the upper shadow of $\mathcal{F}$: -$\partial^+ \mathcal{F}=\{A\cup \{x\}|A \in \mathcal{F}, x\in [n]\setminus A\}$ -Edit: Ok, so the above statement turned out to be true. What about the following generalization? -Similar to the above definition of the upper shadow, we can also define, for any $k$, the $k$th upper-shadow of a family $\mathcal{F}\subseteq 2^{[n]}$ as follows: -$\partial_k^+\mathcal{F}:=\{A \cup K|A \in \mathcal{F}\wedge K\subseteq [n]\setminus A \wedge |K|=k\}$. -Denote $\partial_{\leq k}^+\mathcal{F}:=\bigcup_{i=1}^k\partial_i^+\mathcal{F}$. -Now, Let $\mathcal{F}\subseteq 2^{[n]}$ be a union-closed family. Is it true that for any $k \in [n]$, $|\partial_{\leq k}^+\mathcal{F}\setminus \mathcal{F} -|\leq 2^n-2^{n-k}$? We know it's true for $k=1$, but I wonder if it's true in general. If true, this would be tight for any $k$, since one can take the family $\mathcal{F}=2^{[n-k]}$ as a subfamily of $2^{[n]}$. -Also, it would be interesting to find bounds on $|\partial_k^+\mathcal{F}\setminus \mathcal{F}|$.I can prove that for any $k$ this quantity is at most $2^{n-1}$, and I know that for $k=1$ this is tight, but for larger $k$ perhaps the bound is lower. For instance, for $k=2$ the best I can do is $\frac{3}{8}2^n$, by taking $\mathcal{F}=2^{[n-3]}$. - -REPLY [3 votes]: I ended up proving that this is indeed the case. Theorem $1.3$ in this paper: -https://arxiv.org/abs/1708.01434<|endoftext|> -TITLE: Finite covers of hyperbolic surfaces and the `second systole´ -QUESTION [12 upvotes]: We are interested in the following ´relative´ version of residual finiteness for fundamental groups of surfaces. Similar discussions where given in this question: injectivity radius of hyperbolic surface (in particular with this answer). -The question is the following: -Given a closed hyperbolic surface $S$ with a simple closed geodesic of length less than $\ell$ and a value $K>0$, does there exist a finite cover $p: \hat S \to S$ such that $\hat S$ has a unique simple closed geodesic of length less than $\ell$ and every other simple closed geodesic has length $\geq K$. -(Notice that such a cover cannot be a normal covering as there cannot be an isometry of $\hat S$ sending $\gamma$ (the curve of length $\leq \ell$) to other preimages of $p(\gamma)$. ) - -REPLY [5 votes]: [17/4/17: edited to correct proof.] -This is true. First, we need a lemma which builds a related cover. Throughout, $\alpha$ is a simple closed geodesic of length $\ell$, and $\beta_1,\ldots,\beta_n$ are the finitely many (not necessarily simple) closed geodesics on $S$ of length at most $K$ which are not equal to $\alpha$. Note that we may assume that $\alpha$ is non-separating, by passing to a double cover if necessary. - -Lemma 1: Let $S$ be a closed, hyperbolic, orientable surface and $\alpha$ a simple closed curve. For every $K>0$ there exists a finite-sheeted cover $\widetilde{S}\to S$ so that $\alpha$ lifts to $\widetilde{S}$ and every simple closed curve in $\widetilde{S}$ of length less than $K$ is a preimage of $\alpha$. - -Proof: -Let $S_0$ be the result of cutting $S$ along $\alpha$. -Consider the result of killing $\alpha^k$, for sufficiently large $k$. There are various ways of thinking about this; my preferred way is to think of this as an orbispace $\Sigma$, obtained as follows. First, construct an orbifold $\Sigma_0$ from $S_0$ by replacing the two boundary components with cone points of degree $k$; then glue the two cone points together to obtain $\Sigma$. -Note that $\Sigma_0$ is still a hyperbolic orbifold; in particular, its fundamental group is Fuchsian and therefore residually finite. Furthermore, it's a classical fact that the fundamental group of a graph of residually finite groups with finite edge groups is itself residually finite; in particular, $\pi_1\Sigma=\pi_1S/\langle\langle\alpha^k\rangle\rangle$ is residually finite. -We next consider the images $\bar{\beta}_i$ of the $\beta_i$ in the orbispace $\Sigma$. For sufficiently large $k$, the images $\bar{\beta}_i$ still have infinite order in $\pi_1\Sigma$. This follows from the combinatorial Dehn filling machinery of Osin/Groves--Manning, but one can certainly give a more elementary proof in this context. -Since $\pi_1\Sigma$ is residually finite, we may find a finite quotient -$$ -\pi_1S\stackrel{\pi}{\to}\pi_1\Sigma\stackrel{\eta}{\to} Q -$$ -so that $\eta(\bar{\beta}_i)$ has order greater than $k$, for all $i$, and $\eta(\bar{\alpha})$ has order $k$. -Finally, let $q=\eta\circ\pi$ and let $\widetilde{S}\to S$ be the covering space corresponding to the (usually not normal) subgroup $H=\langle \alpha\rangle\ker q$, which has finite index in $\pi_1S$. Since $\alpha\in H$, we see that $\alpha$ lifts to $\widetilde{S}$. On the other hand, any closed geodesic $\tilde{\gamma}$ on $\widetilde{S}$ of length less than $K$ which doesn't map to a power of $\alpha$ maps to one of the $\beta_i$, which leads to a contradiction since the $\beta_i$ unwrap when they lift to $\widetilde{S}$. QED -At this point, we are only concerned about the components of the preimages of $\alpha$ on $\widetilde{S}$; let $\tilde{\alpha}_0,\ldots,\tilde{\alpha}_m$ denote these, where without loss of generality $\tilde{\alpha}_0$ is a lift of $\alpha$. Note that they form a disjoint set of non-separating curves on $\widetilde{S}$, and the complementary regions are covers of $S_0$, hence have genus at least one. - -Lemma 2: Fix an integer $N>0$. There is a finite-sheeted cover $\widehat{S}\to\widetilde{S}$ so that $\tilde{\alpha}_0$ has a unique lift to $\widehat{S}$, and the remaining components of the preimages of the $\tilde{\alpha}_i$ all unwrap at least $N$ times. - -Proof: By pinching a non-separating curve that separates a punctured torus containing $\tilde{\alpha}_0$ from the other $\tilde{\alpha}_i$, and collapsing the resulting torus to a circle, we obtain a surjection -$$ -\pi_1\widetilde{S}\to\langle\tilde{\alpha}_0\rangle*\pi_1S' -$$ -where $S'$ is a closed surface and the set $\{\tilde{\alpha}_i\mid 1\leq i\leq n\}$ maps to a disjoint collection of non-separating simple closed curves on $\pi_1S'$. Since the $\tilde{\alpha}_i$ are all non-zero in $H_1(S')$, it follows that we may find a surjection -$$ -\pi_1\widetilde{S}\to F_2=\langle\tilde{\alpha}_0\rangle*\langle b\rangle -$$ -so that the other $\tilde{\alpha}_i$ all map to non-zero powers of $b$. -We now pick a large prime $p$, and map $F_2\to \mathbb{F}^\times_p\ltimes \mathbb{F}_p$ in such a way that $\tilde{\alpha}_0$ maps to a generator of the Frobenius complement $\mathbb{F}^\times_p$, and $b$ maps to a generator of the Frobenius kernel $\mathbb{F}_p$. This gives a surjection -$$ -r: \pi_1\widetilde{S}\to \mathbb{F}^\times_p\ltimes \mathbb{F}_p -$$ -with the property that, for every $g\in\pi_1\widetilde{S}$, $\langle r(\tilde{\alpha}_0)\rangle\cap \langle r(\tilde{\alpha}_0^g)\rangle=1$ unless $r(g)\in \langle r(\tilde{\alpha}_0)\rangle$ (see here). For large enough $p$, we also have that every $r(\tilde{\alpha}_i)$ has order greater than $N$, for every $i$. -We now define $\widehat{S}\to\widetilde{S}$ to be the cover corresponding to the subgroup $\langle \tilde{\alpha}_0\rangle\ker r=r^{-1}(\mathbb{F}_p^\times)$. Clearly $\tilde{\alpha}_0$ lifts to $\widehat{S}$. -On the other hand, covering space theory shows that any other component of the preimage of $\tilde{\alpha}_i$ in $\widehat{S}$ corresponds to a non-trivial double coset $\langle r(\tilde{\alpha}_0)\rangle r(g) \langle r(\tilde{\alpha}_i)\rangle$, and the degree of unwrapping corresponds to the minimal power of $r(g\tilde{\alpha}_ig^{-1})$ contained in $\langle r(\tilde{\alpha}_0)\rangle$. By construction, this degree of unwrapping is at least $N$. QED -Choosing $\ell N>K$, we immediately obtain the cover we were looking for. - -Theorem: For any $K>0$ there is a finite-sheeted cover $\widehat{S}\to S$ so that $\alpha$ lifts to $S$ and every other closed curve on $\widehat{S}$ is of length greater than $K$.<|endoftext|> -TITLE: Hamiltonian, Lagrangian and Newton formalism of mechanics -QUESTION [24 upvotes]: If my thinking is wrong please let me know. I have little knowledge on beyond-college physics. -For research purposes, I read a few introductions to these three formalisms of classical mechanics [1,2,part of 5](Hamiltonian, Lagrangian and Newton formalism). -It seems to me that Newton formalisms is incapable of describing a quantum system(The uncertainty principle is not well addressed) while both Hamiltonian and Lagrangian are capable of describing a quantum system.So the Hamiltonian and the Lagrangian is simply a more general framework including Newton. -As Ben and Tobias pointed out in their answers, these three formalisms are equivalent their relationship are not simply inclusion but complementary. There are situations one of three systems that are particularly suitable to use. -For my purposes, I see there often seems to be one-to-one correspondence in-between Lagrangian construction and Hamiltonian construction for dynamic systems(OR in the simpler case the derived differential equations w.r.t. a chosen coordinate frame) and (sympletic) geometry when the concern is the dynamics on the manifold, say [3,4]. -A curious question in my mind is that if these two physical formalisms are equivalent, then why only Hamiltonian is studied in most cases(say geometric analysis and sympletic geometry)? - -(1)What is its(The Hamiltonian formalism) superiority over Lagrangian from mathematical perspective? - Does it lead to a richer structure or more natural structure?(by structure I mean manifold structure over which the system is defined) -(2)Moreover, is there any example that is easily formalized in - Hamiltonian formalism but too complicated/unnatural to formalize in - Lagrangian? - -Any comments or reference is appreciated! Please add some reference in your answer to support your claims, thanks a lot! -Reference -[1]http://www.macs.hw.ac.uk/~simonm/mechanics.pdf -[2]http://image.diku.dk/ganz/Lectures/Lagrange.pdf -[3]Boundary conditions and the relationship between Hamiltonian and Lagrangian Floer theories -[4]Koon, Wang Sang, and Jerrold E. Marsden. "The Hamiltonian and Lagrangian approaches to the dynamics of nonholonomic systems." Reports on Mathematical Physics 40.1 (1997): 21-62. -[5]Meyer, Kenneth, Glen Hall, and Dan Offin. Introduction to Hamiltonian dynamical systems and the N-body problem. Vol. 90. Springer Science & Business Media, 2008. - -Motivation of OP -And as for my motivation, I primarily want to figure out why [Mumford&Michor] proposed Hamiltonian approach in (which looks not quite natural to me at first since they are just laying down a framework for dynamics on $Cur(\mathbb{R}^2)$.) -[Mumford&Michor]Michor, Peter W., and David Mumford. "An overview of the Riemannian metrics on spaces of curves using the Hamiltonian approach." Applied and Computational Harmonic Analysis 23.1 (2007): 74-113. -From these nice answers below, I feel that the most convincing reason why Hamiltonian approach is preferred in [Mumford&Michor] is that -(1)the space of curves $Cur$ involves $\mathrm{Diff}(C)$ and Hamiltonian formalism is a convenient formalism to incorporate these transformations. -(2)And the infinitesimal generators of $\mathrm{Diff}(C)$ can be used to describe the velocity field along the curves in $Cur$. -(3)Combined with Tobias' answer, [Mumford&Michor] also said symmetry is a reason for Hamiltonian. Now I understand the sentence better. - -...The Hamiltonian approach also provides a mechanism for converting - symmetries of the underlying Riemannian manifold into conserved - quantities, the momenta.[Mumford&Michor] - -(Unless the authors disagree :) -Thanks again for everyone's input, I learnt a lot from you and willing to learn more! - -REPLY [18 votes]: The three formalisms of classical mechanics, i.e. the Newtonian, the Lagrangian (analytical mechanics) and the Hamiltonian (canonical formalism) are generally not equivalent to each other -at least not in some strict sense of the word "equivalent" which could be considered valid for the totality of physical systems. But in fact they are, for lots of systems of interest. In order to be more specific, let me try to sketch their logical interconnections: -$\blacktriangleright$ If we start from Newtonian mechanics and the assumptions: - -there are holonomic constraints only, i.e. constraints of the form -$f(x,y,z,t)=0$, which are independent of the velocity. -arbitrary virtual displacements are assumed to be in directions that are orthogonal to the constraint forces, i.e. the constraint forces do no work (this is equivalent to saying that the constraint forces are normal to the hypersurface determined by the constraints at a given instant $t=t_0$). - -                             - - -          - -If the constraints are scleronomic then the actual displacements $d\mathbf{s}$ are virtual displacements $\delta\mathbf{s}$. - -          - -($\Sigma$ is the hypersurface determined by the constraints $f(x,y,z)=0$, $\ R \ $ the constraint force, $F$ - -          - -is the resultant non-constraint force and $\mathbf{s}$ the position vector). - -then Newton's $2^{nd}$ law is equivalent to the D'Alembert's principle which in turn implies the Euler-Lagrange's equations of motion i.e. the analytical mechanics formalism. -(Note that, systems sattisfying the second of the above assumptions are sometimes called "mechanical" or "pure mechanical" systems in the literature. -On the other hand, one does not have to go far in order to find systems violating one or both of the above assumptions: rolling without slipping is a common system with non-holonomic constraints and generally systems with resistance forces -various friction forces for example- violate the second of the above assumptions). -The converse implication, that is starting from the Euler-Lagrange equations and deriving D'Alembert's principle and thus Newton's $2^{nd}$ law is relatively straightforward. -$\blacktriangleright$ If we start from a Lagrangian and its Euler-Lagrange equations of motion, under the assumptions: - -there are holonomic constraints only -the Lagrangian function is "standard" or "regular" in the sense that $\det\Big|\frac{\partial^{2}L}{\partial\dot{q}_{i}\partial\dot{q}_{j}}\Big|\neq 0$, i.e.the Hessian of the Lagrangian w.r.t. to the generalized velocities is non-degenerate (see: When does a Lagrangian dynamical system have an equivalent Hamiltonian description? for more details on this point) - -then we can derive the Hamiltonian and the canonical formalism of Hamilton's equations through a Legendre transformation. Note that Legendre's transformation, transforms functions on a vector space to functions on the dual space. In this case, it transforms the Lagrangian function (on the tangent bundle of the configuration space manifold) to the Hamiltonian function (on the cotangent bundle of the configuration space manifold). -Summarizing the above discussion: -$$ -\small{ -\left\{ -\begin{array}{c} -\text{Newtonian} \\ \text{mechanics} -\end{array} -\right\} -\underset{\begin{array}{c} -\text{pure} \\ \text{mechanical} \\ \text{system} -\end{array}}{\overset{\begin{array}{c} -\text{holonomic} \\ \text{constraints} -\end{array}}{\mathbf{\leftrightsquigarrow}}} -\left\{ -\begin{array}{c} -\text{d'Alembert's} \\ \text{principle} -\end{array} -\right\} -\leftrightsquigarrow -\left\{ -\begin{array}{c} -\text{Lagrangian} \\ \text{mechanics} -\end{array} -\right\} -\underset{\det\Big|\frac{\partial^{2}L}{\partial\dot{q}_{i}\partial\dot{q}_{j}}\Big|\neq 0}{\overset{\begin{array}{c} -\text{holonomic} \\ \text{constraints} -\end{array}}{\mathbf{\leftrightsquigarrow}}} -\left\{ -\begin{array}{c} -\text{Hamiltonian} \\ \text{mechanics} -\end{array} -\right\} -} -$$ -Remarks: -(a). in the framework of Newtonian mechanics, forces may depend on positions and velocities but not on accelerations and -(b). throughout the preceding disussion, we consider all non-constraint forces to be derivable from generalized scalar potentials depending on coordinates and -at most- linearly on velocities: $V=U(q_i,t)+A_j(q_i,t)\dot{q}_j$. Such systems are more general than conservative systems and fall into the class of monogenic systems. -Now, regarding your first question: I don't think it would be suitable to speak about "superiority" or "richer structure". From a technical aspect, Euler-Lagrange's equations are a system of $n$, second order, ODE and the Lagrangian function involved "lives" on the tangent bundle while The Hamiltonian formulation comprises of a system of $2n$, first order, ODE, involving the Hamiltonian function which "lives" on the cotangent bundle (of the configuration space manifold). -If we consider the passage to quantum mechanics, then both formalisms are suitable to handle the elementary aspects of the quantisation problem (at both levels, first and second quantization as well): Hamilton's equations have almost the same typical form in quantum mechanics with their classical counterparts, although their interpretation is quite different in the quantum case -but this is an apparently different story. The road to quantisation through the hamiltonian formalism is generally refered to as canonical quantisation while the road through Lagrangian formalism is known as path integral quantization. -Regarding your second question, and since you are asking for references, this article appears to discuss examples of classical Hamiltonian systems possessing no Lagrangian formulation. On the other hand, the Hamiltonian and the Lagrangian formulation of geometrical optics -as has already been mentioned in another answer- is a well known classical system possessing no meaningful description at the level of Newton's laws. -From one point of view, one may say that the Hamiltonian formalism is "wider" in the sense that it includes several systems of ODE not being directly or obviously related to classical mechanics (an example is the lagrangian/hamiltonian description of Maxwell's laws of the electromagnetic field) or to physics at all. In practise, the hamiltonian is frequently considered to be some abstract function while the Lagrangian is more intimately related to the concepts of kinetic and potential energy of some system. I think that you might also find some interest in the article Classical Mechanics Is Lagrangian; It Is Not Hamiltonian by Erik Curiel. -However, from another point of view, Newtonian formulation is "wider", in the sense that if one is to consider, frictions, dissipative forces, non-holonomic constraints etc then -although there are various formal generalizations of Lagrange's and Hamilton's equations in order to deal with such systems- usually one turns to the fundamentals that is Newton's laws. -So, concluding, i agree that there are no simple proper subset relationships between these three formulations of classical mechanics. -Some further References: - -Mathematical methods of classical Mechanics, by V.I. Arnold, and -A Treatise on the Analytical Dynamics of Particles and Rigid Bodies, by E.T. Whittaker, - -are invaluable -imo- sources, delving into the foundations, not trying to hide the pitfalls under the carpet, with an eye on details (either in the computations or the arguments). - -This: https://physics.stackexchange.com/q/89035/130499 question in physics.stackexchange has a natural overlap with the OP and has lots of answers. You might be interested in taking a look there as well. - -P.S.: One last thing: the above disussion and the diagram provided, aimed at emphasizing the logical interconnections of these three formalisms and the usual assumptions upon which one is being derived from or imply the other(s). However, it should be noted that both the Euler-Lagrange's and the Hamilton's equations, can be derived -each one of them seperately and independently of Newton's laws- in an "axiomatic" (or should I say "ad hoc" ?) manner through suitable variational principles: these are the Hamilton's principle (from which euler-lagrange's eqs are derived) and the modified Hamilton's principle (from which the canonical eqs of hamiltonian mechanics can be derived). They both determine the corresponding equations of motion and thus the evolution of the system, through the demand that the trajectories are such that the corresponding action-functional $S$ gets a stationary value (its $\delta$-variation becomes zero): -$$\delta S=0$$ -In the former case $S=\int_{t_1}^{t_2} Ldt$ and the principle is applied in the configuration space, while in the later case $S=\int_{t_1}^{t_2} (\dot{q}p-H)dt$ and the corresponding principle is applied in the phase space. -It is actually this possibility of independent foundation of these formalisms that inspired/enabled the extension of these formalism to non-mechanical systems such as classical fields.<|endoftext|> -TITLE: Reference for Feynman-Kac -QUESTION [6 upvotes]: I would like to have a reference with more in deep explanation of Feynman-Kac than in Evan's ‎An Introduction to Stochastic Differential Equations and, if possible, example of solution for equations like Schrödinger and others. - -REPLY [3 votes]: A rigorous treatment of the Feynman-Kac formula is given in the chapter 3 of the book "Quantum Physics: A Functional Integral Point of View" by J. Glimm and A. Jaffe: http://www.springer.com/in/book/9780387964775<|endoftext|> -TITLE: Real-Closed Fields and Intermediate Value Theorem -QUESTION [9 upvotes]: Artin and Schreier (1926) showed that a real-closed ordered field satisfies the intermediate value theorem for polynomials of a single variable. By the early 1980s authors such as Max Dickmann and Gregory Cherlin, working on the theory of real-closed rings, routinely appealed without proof or reference to the fact that: -For an ordered field $K$, $K$ is real-closed iff $K$ satisfies the intermediate value theorem for polynomials (of a single variable) over $K$. -Since that time numerous proofs of the equivalence have appeared without reference to earlier proofs and some authors, such as van den Dries in his book on o-minimality, simply define a real-closed ordered field as one that satisfies the intermediate value theorem for polynomials (of a single variable). Was the equivalence simply a "folk theorem" by the 1980s or are there earlier published proofs? If the latter, where might one find the earliest such published proof? -Edit 1: I have found a proof of the equivalence in P. Cohn's, Algebra, Volume II, Section 7.4, 1977. I'd still be interested in learning if there are earlier proofs. -Edit 2: As it turns out, already in Modern Algebra by Seth Warner, 1965, pp. 492-494, a real-closed ordered field is essentially defined as an ordered field satisfying the intermediate value theorem for polynomials (in one variable) and the equivalence between that characterization and the more familiar ones is established. I'm beginning to suspect that the equivalence was recognized quite early on for the reason expressed by Emil. - -REPLY [4 votes]: After a bit of searching, it seems reasonable to believe that the explicit recognition of the equivalence goes back to Tarski's great work: The Completeness of Elementary Algebra and Geometry, in The Collected Papers of Alfred Tarski, Vol. IV, Givant, S. R., and McKenzie, R. N., eds. Birkhäuser (1986). (See page 312). The bulk of the paper was written in 1930, three years after Artin and Schreier published their paper, but it wasn't completed and submitted for publication until 1939. Due to the war it didn't go into print until 1967 and the following revised version (that emphasized a decision procedure rather than completeness) was subsequently published: A Decision Method for Elementary Algebra and Geometry 1948/1951. In the latter paper the intermediate value theorem is replaced by a related assertion (See Note 9). -It is interesting to note that in those texts that explicitly say they are closely following the treatment of Artin and Schreier, including van der Waerden's Modern Algebra and Nathan Jacobson's Lectures in Abstract Algebra, there is no hint of the equivalence. The earliest explicit statement of the equivalence I have found in the more recent literature is in the classic text of Seth Warner (1965) mentioned in the second edit of my question. It is also implicit in Nathan Jacobson's Basic Algebra, Vol. 1, 1974 and explicit in P. Cohn's, Algebra, Volume II, Section 7.4, 1977. -Thus, in answer to the question I originally raised, the result appears to have become standard by the 1980s when model theorists and others freely appealed to it.<|endoftext|> -TITLE: Base change and the octahedron axiom -QUESTION [5 upvotes]: I am trying to understand "de Cataldo, Migliorini. The perverse filtration and the Lefschetz hyperplane theorem. Annals of Mathematics, 171(2010), 2089-2113." My question is about one detail in the paper. Here is my reproduction of the situation. -Let $\Sigma$ be a stratification of $\mathbb{P}^N$ adapted to a bounded complex $K$ with $\Sigma$-constructible cohomology. Let $\Sigma$ be also adapted to the embedding $Y \subseteq \mathbb{P}^N$ of an affine variety $Y$. Let $\Lambda \subset \mathbb{P}^N$ be a hyperplane, $H = \Lambda \cap Y$ and $\bar{H} = \Lambda \cap \bar{Y}$. Set $\bar{U} = \bar{Y} \setminus \bar{H}$ and $U = Y \setminus H$. Consider the cartesian diagram -$$\begin{array}[c]{ccccc} -H&{\xrightarrow{i}}&Y&{\xleftarrow{j}}&U\\ -\downarrow\scriptstyle{J}&&\downarrow\scriptstyle{J} && \downarrow\scriptstyle{J}\\ -\bar{H}&{\xrightarrow{i}}&\bar{Y}&{\xleftarrow{j}}&\bar{U} -\end{array}$$ -Now the two authors stated (page 2101, following the same diagram) "By the octahedron axiom, the map $J_!j_*j^*K \to j_*J_!j^*K$ is an isomorphism if and only if the natural base change map $i^*J_*K \to J_*i^*K$ is an isomorphism." -My question is: what is exactly used about the octahedron axiom for the statement? -Thanks! - -REPLY [2 votes]: I'm not sure what's going on, but here's a guess: -Firstly, the base change map goes the other way: $i^*J_*\mathcal{F}\rightarrow J_*i^*\mathcal{F}$. Say this is an isomorphism for the Verdier dual of $K$ (they end up saying that the condition they impose makes the base change morphism an isomorphism independent of $K$, so this is ok). -Then taking the Verdier dual of this isomorphism we get an isomorphism $J_!i^!K\rightarrow i^!J_!K$. Next note that the similarly defined base change map $J_!j^*K\rightarrow j^*J_!K$ gives an isomorphism, since $j$ is an open embedding and $j^*=j^!$. -The distinguished triangles to plug into the octahedron axiom are then -$J_!K\rightarrow J_!j_*j^*K\rightarrow J_!i_*i^!K[1]\rightarrow$ -and -$J_!K\rightarrow j_*j^*J_!K\rightarrow i_*i^!J_!K[1]\rightarrow$ -and -$J_!j_*j^*K\xrightarrow{f}j_*j^*J_!K\rightarrow cone(f)\rightarrow$ -By the axiom, there is a distinguished triangle -$J_!i_*i^!K[1]\xrightarrow{g} i_*i^!J_!K[1]\rightarrow cone(f)\rightarrow$ -and so if $g$ is an isomorphism, $cone(f)=0$ and $f$ is an isomorphism.<|endoftext|> -TITLE: rational effective implies effective? -QUESTION [8 upvotes]: Let $X$ be a weak del pezzo surface, I Wonder whether the following statment is true: -Let $L$ be a line bundle on $X$, then $h^0(L)=0$ implies $h^0(nL)=0$ for all $n\geq 1$. - -REPLY [2 votes]: Jun Yan found a counter example for weak del pezzo surface. -Let $X$ be weak del pezzo surface of degree $4$, Let $X=X_{4,4A_1}$, irreducible $(-2)$-curves are -$E_1-E_2,L_{123}=L-E_1-E_2-E_3,E_4-E_5,L_{345}=L-E_3-E_4-E_5$, take sum of all these curves, we -get:$2(L_{235})$ which is an effective divisor. But $L_{235}$ itself is not -effective divisor. Which means that rationally effective divisor is not -effective. And actually, there are a series such examples -But as Chen Jiang pointed out in the comment, the statement is true for del pezzo surface.<|endoftext|> -TITLE: Characterization of group characters -QUESTION [7 upvotes]: I wonder how to characterize the characters of a (say, finite) group $G$ as special class functions, in particular for the case $G=S_n$ (symmetric group). The answer to this is presumably well known to people working in group theory, so it is more of a reference request. -Any character $\chi$ is "positive" in the sense that given $g_1,...,g_k\in G$, the matrix with coefficients $\chi(g_i^{-1}g_j)$ is positive. In case of $G=S_n$, characters are also known to be integer-valued class functions. Are there any further properties of characters that distinguish them from general class functions, or even a characterization of characters as class functions with a list of properties? -Of course, the characters are exactly those class functions which are linear combinations of the irreducible characters with non-negative integer coefficients, but that is not what I have in mind here. - -REPLY [10 votes]: The best answer I can think of is Brauer's characterization of (generalized) characters: Recall that a generalized character is a difference of two characters. A Brauer elementary group is a group that is the direct product of a $p$-group and a cyclic group. Then Brauer's theorem states: - -A class function $\chi$ of a finite group $G$ is a generalized character if and only if its restriction $\chi_E$ to each Brauer elementary subgroup $E$ is a generalized character. - -This reduces the problem to certain subgroups of restricted structure. Since Brauer elementary groups are monomial, it is equivalent that $[\chi_E, \lambda ] \in \mathbb{Z}$ (inner product for class functions) for all linear characters $\lambda$ of all Brauer elementary subgroups $E\leq G$. (Thanks to Geoff Robinson for pointing this out in comment.) -Unfortunately, the word "generalized" can not be omitted from the theorem. We get that a class function $\chi$ is an irreducible character if in addition $\chi(1)>0$ and $[\chi,\chi]=1$. -The result is treated, for example, in the books by Isaacs, by Huppert (Character Theory...), or by Serre (Linear Representations of Finite Groups).<|endoftext|> -TITLE: In search of a combinatorial reasoning for a vanishing sum -QUESTION [8 upvotes]: Assume $s, j \in\mathbb{N}$. Define the set -$$\mathcal{A}_{j,s}:=\{(n_1,n_2,\dots,n_j)\in\mathbb{Z}_{\geq0}^j\vert \, -n_1+2n_2+\cdots+jn_j=j, \, n_1+n_2+\cdots+n_j=s\}.$$ - -Question. Is there a combinatorial argument for this? - $$\sum_{s=0}^j(-1)^{j-s}\sum_{\mathcal{A}_{j,s}}\binom{s}{n_1,\dots,n_j}=0.$$ - I've an algebraic proof. - -Update. Here is the proof that I mentioned. Below, $[x^k]F(x)$ denotes the coefficient of $x^k$ in $F(x)$. -This Multinomial Theorem with the specializations $a_i=x^i,\, k=j$ and $n=s$ offers -$$(x+\cdots+x^j)^s=\sum_{n_1+\cdots+n_j=s}\binom{s}{n_1,\dots,n_j}x^{n_1+2n_2+\cdots+jn_j} -=\sum_{j} x^j\sum_{\mathcal{A}_{j,s}}\binom{s}{n_1,\dots,n_j}.$$ -It follows that the coefficient of $x^j$ in $(x+\cdots+x^j)^s$ equals $\sum_{\mathcal{A}_{j,s}}\binom{s}{n_1,\dots,n_j}$. On the other hand, $(x+\cdots+x^j)^s=x^s(1-x^j)^s(1-x)^{-s}$. Consequently, the coefficient of $x^j$ is: -\begin{align} [x^j]\,\,\,x^s(1-x^j)^s(1-x)^{-s}&=[x^j]\,\,\,x^s\sum_a(-1)^s\binom{s}ax^{aj}\sum_b\binom{s+b-1}bx^b \\ -&=[x^j]\,\,\,x^s\sum_b\binom{s+b-1}bx^b=\binom{j-1}{j-s}=\binom{j-1}{s-1}. \end{align} -We proved $\sum_{\mathcal{A}_{j,s}}\binom{s}{n_1,\dots,n_k}=\binom{j-1}{s-1}$. -Binomial Theorem gives: $\sum_{s=1}^jy^s\binom{j-1}{s-1}=(1+y)^{j-1}$. We conclude that -$$\sum_{s=0}^j(-1)^{j-s}\sum_{\mathcal{A}_{j,s}}\binom{s}{n_1,\dots,n_k}=\sum_{s=0}^j(-1)^{j-s}\binom{j-1}{s-1}=0.$$ - -REPLY [16 votes]: For fixed $s,j>0$, $\sum_{\mathcal{A}_{j,s}}\binom{s}{n_1,\dots,n_j}$ enumerates the compositions of $j$ into $s$ positive parts by ordering the corresponding partitions. That is, we have -$$\sum_{\mathcal{A}_{j,s}}\binom{s}{n_1,\dots,n_j}=\binom{j-1}{s-1}$$ -and the identity in question follows trivially.<|endoftext|> -TITLE: Divisibility of a binomial sequence -QUESTION [8 upvotes]: Convincing numerical evidence prompts me to ask: - -Question. Is $\sum_{k=0}^n\sum_{j=0}^k\binom{k}j^2\binom{2j}j(2j+1)^2$ divisible by $(n+1)^2$? - -REPLY [3 votes]: The answer is yes, and the proof can be found in V. J. W. Guo, J.-C. Liu, Proof of some conjectures of Z.-W. Sun on the divisibility of certain double sums, Int. J. Number Theory 12 (2016), 615-623. An arXiv version is also available.<|endoftext|> -TITLE: Explicit generating acyclic cofibrations and right properness of a model category -QUESTION [7 upvotes]: Let $\mathcal{C}$ be a cofibrantly-generated model category. My impression is that the following two conditions are highly correlated: - -$\mathcal{C}$ is right proper. -There is an explicitly-describable set of generating acyclic cofibrations for $\mathcal{C}$. - -(Of course, "explicitly-describable" is vague, but let's at least stipulate that "all acyclic cofibrations between small objects" (the sort of description one gets from Jeff Smith's recognition theorem) is not an explicit description per se.) -For example, the Quillen model structure satisfies both (1) and (2) (witness the horn inclusions), while the Joyal model structure satisfies neither (1) nor (2). Taking a Reedy model structure or projectively-inducing a model structure along an adjunction -- operations that preserve property (2) -- also preserve property (1). In fact, I don't know a single example of a model category $\mathcal{C}$ satisfying (1) but not (2) or (2) but not (1)! This leads to a vague question: -"Question" A: Does $(1) \Leftrightarrow (2)$ hold in some sense? -Here's a more precise, and seemingly stronger, formulation that I haven't been able to rule out. In lieu of explicit generating acyclic cofibrations, one often works with what Simpson calls a pseudo-generating set: a set of morphisms $S$ such that - - -if $Y$ is fibrant (including the case where $Y$ is terminal), then $X \to Y$ is a fibration iff it has the right lifting property with respect to the morphisms of $S$. - - -Cisinski's theory (nicely generalized by Olschok) often makes it easy to get one's hands on a pseudo-generating set even when a generating set is hard to describe. For example, the set $\{\Lambda^k[n] \to \Delta[n]\}_{n \in \mathbb{N},0 < k < n} \cup \{\Delta[0] \to I\}$ (where $I$ is the walking isomorphism) is a pseudo-generating set, but not a generating set, for the Joyal model structure. And Cisinski theory easily shows that the horn inclusions form a pseudo-generating set for the Quillen model structure. But in order to see that they are actually a generating set, one needs a nice functor like $Ex^\infty$; and such a nice functor automatically entails that one's model category is right proper. Somehow I suspect that the horn inclusions can't be so special, and I'm led to consider the condition - -Every pseudo-generating set in $\mathcal{C}$ is an actual set of generating acyclic cofibrations. - -and to ask -Question B: Does $(1) \Leftrightarrow (3)$ hold? -even though I don't even know whether (3) holds for any $\mathcal{C}$ (unless every object is fibrant)! So I might as well also ask: -Question C: Is there an example of a model category $\mathcal{C}$ where not every object is fibrant where (3) holds? - -REPLY [6 votes]: If $\mathcal C$ is the category of simplicial presheaves on a small category $A$ equipped with the injective model structure, it is proper, but it is very unlikely that we will get "explicit generating (trivial) cofibrations" without further assumptions (for instance on the indexing category $A$, such as being elegant in the sense of Rezk and Bergner). And it is hard to define the adjective "explicit" mathematically with the level of generality you are aiming at... -That said, here are classes of examples which gives a hint about the freedom we have, and in particular, about the fact that having explicit generators and right properness are not directly related. -First, the left Bousfield localizations of the Kan-Quillen model structure on the category of simplicial sets which are proper are precisely the nullifications. This gives quite a lot of freedom to mess around. -Second, there are proper model structures with explicit generators but no nice functors such as the $Ex^\infty$-functor: for instance cubical sets. -To come back to the general case, it is a good thing to remember what properness is about: right properness means that the formation of slices is compatible with weak equivalences (i.e. pulling back along any weak equivalence $X\to Y$ induces a right Quillen equivalence from $\mathcal C/Y$ to $\mathcal C/X$). In particular, the canonical model structures induced on slices $\mathcal C$ are rather relevant when it comes to express whether the model structure $\mathcal C$ is right proper or not. -If ever we have "explicit" pseudo-generators, in practice, we can define yet another model structure on each slice $\mathcal C/X$, which is fully characterized through the following: the cofibrations are the maps which are cofibrations in $\mathcal C$, while the fibrant objects are the maps $E\to X$ which have the right lifting property with respect to the pseudo-generating set of trivial cofibrations. The usual sliced model structure on $\mathcal C/X$ is thus a left Bousfield localization of the latter, but there is no reason that they agree. In fact, they agree for all $X$ if and only if the pseudo-generators actually are generators of trivial cofibrations. -Here is an enlightening example. Horn inclusions of the form $\Lambda^n_k\to\Delta^n$ for $n\geq 1$ and $0\geq k< n$ do form a pseudo-generating set for the usual Kan-Quillen model structure. Moreover, on slices, the model structure induced by these pseudo generators are fully documented: they are the "contravariant model structures" modelling presheaves on quasi-categories. In particular, they are not always proper and they coincide with the usual sliced model structures only when we slice over simplicial sets whose fundamental category (obtained through the left adjoint of the nerve) is a groupoid. In particular, the fact that these pseudo-generators fail to be generators is not an accident nor a failure: it is needed because we do not want all presheaves to be locally constants!<|endoftext|> -TITLE: Non smallness of the set of anafunctors without AC? -QUESTION [9 upvotes]: Trying to construct a model category constructively is difficult. One often mention the fact that without the axiom of choice one cannot prove that the localization of the category of small categories at weak equivalence (i.e. functor which are fully faithful and essentially surjective) is locally small (ie. have small hom set) as an argument for the non existence of a model structure on cat that have these weak equivalences. -But actually proving that this cannot be proved is not easy:it can be shown that very weak choice principle like WISC or Makkai "small cardinality selection axiom" are enough to implies that this localization is locally small and we don't have that many model where those axioms fails. -My question is: Are we actually able to prove that the local smallness of this localization cannot be proved in ZF ? -I will now give some details on this, for non category theory people: - -Without the axiom of choice, Makkai has introduced the notion of 'anafunctor' which generalize the notion of functors in order to describe explicitely this localization. -In fancy words, his definition can be formulated as follow: The category of categories is a Brown category of fibrant objects, with fibrations being the isofibrations and weak equivalences being the fully faithful and essentially surjective functors (trivial fibrations are hence the functors which are fully faithful and surjective on objects). -Anafunctors $X \rightarrow Y$ are then spans $X \leftarrow Z \rightarrow Y$ where $Z \rightarrow X$ is a trivial fibrations, and we know by the work of Brown that homotopy class of such spans compute the localization of the category of categories at 'weak equivalences'. -In a more down to earth approach, and following Makkai's paper: -The idea of Makkai is that an anafunctor is like a functor but where the image of an object is not uniquely defined, but, following classical categorical philosophie, only well defined up to unique isomorphism. -More precisely: -If $X$ and $Y$ are (small) categories, an anafunctor from $X$ to $Y$ is the data of: --A set $|F|$ and a surjective map $\pi: |F| \rightarrow |X|$ (where $|X|$ denotes the set of objects of $|X|$). --On consider $F$ as a category over $X$ such that $\pi$ is extended into a fully faithful functor $\pi:F \rightarrow X$ (i.e. $Hom_F(a,b):=Hom_X(\pi(a),\pi(b))$. --One has a functor $f$ from $F$ to $Y$. -This fits the idea explained above as follow: if you have an object $x \in X$, you compute its image by $F$ by taking any $z \in |F|$ such that $\pi(z)=x$ and taking the image of $z$ by $f$. If one chose of a different $z'$ then there is a unique isomorphism between $z$ and $z'$ which is send to the identity of $x$ and this induce a canonical isomorphism between $f(z)$ and $f(z')$. -I refer to the paper of Makkai linked above for how anafunctor are composed as it is not relevant to the question... -The only thing we need to know here is when two anafunctors are isomorphic (equivalent): Let $F$ and $G$ be two anafunctors from $X$ to $Y$. One can construct a category $F \times_{X} G$ whose set of objects is: -$$ |F| \times_{|X|} |G| = \{ (x,y), x \in |F|, y \in |G|,, \text{ whose images in X are equal}\}$$ -and whose morphisms are the morphism between the projection to $X$. One then has two functors from $F \times_X G$ to $Y$ given respectively by $f$ and $g$ composed with the projection to $F$ and $G$. -An isomorphism of anafunctors is given by an isomorphism between these two functors from $F \times_X G$ to $Y$. More explicitly: -for each $x \in |F|, y \in |G|$ having the same image in X, one has an isomorphism $\theta_{x,y}: f(x) \rightarrow g(y)$ in $Y$, such that for every $a:x \rightarrow x'$ an arrow in $F$, and $b: y \rightarrow y' $ an arrow in $G$ such that $a$ and $b$ have the same image in $X$ then one has a commutative square: -$$\theta_{x',y'} \circ f(a) = g(b) \circ \theta_{x,y}$$ -Then the localization of the category of small categories at weak equivalences has the 'set' of isomorphisms class of anafunctors between $X$ and $Y$ has morphism from $X$ and $Y$, hence the claim we want to disprove is: for all small category $X$ and $Y$ there is only a set of isomorphism class of anafunctor from $X$ to $Y$. -More precisely: one wants to have a set of anafunctors such that any anafunctor is isomorphic to one in this chosen set. - -Let me mention a special case that will probably be easier to understand and which I think is equivalent to the general case: -If $X$ is a discrete category (a set, seen as a category with only identity arrow) and $Y=BG$ is a category which have only one object with a group $G$ of endomorphism. -An anafunctor from $X$ to $Y$ is the data of a set $\pi:E \twoheadrightarrow X$ together with for each $a,b \in E$ such that $\pi(a)=\pi(b)$ an element $g_{a,b} \in G$ such that $g_{a,b} =g_{b,a}^{-1}$ , $g_{a,a}=1_G$ and $g_{a,b} g_{b,c} = g_{a,c}$ (indeed $g_{a,b}$ is the image of the unique morphism from $a$ to $b$ corresponding to the identity of $ \pi(a)$). -Two such anafunctors $(E,g)$ and $(E',g')$ are isomorphic if and only if there exists a function $t_{a,a'}$ which maps any pairs $(a \in E,a'\in E')$ having the same image in $X$ to $t_{a,a'} \in G$ such that $g_{a,b} t_{b,b'}= t_{a,b'}$ and $t_{a,a'} g'_{a',b'} = t_{a,b'}$. -Moreover, Makkai (still in the paper linked above) has a theory of "saturated anafunctor" which construct for each such anafunctor an isomorphic "saturated anafunctor" which in this case are anafunctors of the form: -$\pi : E \twoheadrightarrow X$ is a surjection and $E$ caries an action of $G$ such that for each $a,b \in E$ having the same image in $X$ there is a unique $g \in G$ such that $g.b= a$. -One then define $g_{a,b}$ has being this unique $g$. -two such anafunctor are isomorphic if and only if they are isomorphic as set over $X$ endowed with a $G$ action. -Here again, what we want to prove more precesely is that there exists a set $F$ of anafunctors such that any anafunctor is isomorphic to one in this set. -Such anafunctor are generally called principale $G$-bundle over $X$, or $G$-torsor over $X$, and their isomorphism class form Giraud's definition of the non-abelian cohomology $H^1(X,G)$ of the discrete space $X$. -Andreas Blass has show that the triviality of all the $H^1(X,G)$ is equivalent to the axiom of choice, but here we just want to prove that they are sets. - -REPLY [7 votes]: This is a supplementing answer to aws' answer. In this answer I sketch the a class-symmetric extension in which there is a proper class of Russell sets, and there is no set of Russell cardinals. -(Recall that a Russell set is a set which can be partitioned into countably many pairs, such that no infinite family of those pairs admits a choice function.) -We work in $\sf ZFC+GCH$, where $\sf GCH$ is assumed for simplicity. $\DeclareMathOperator{\dom}{dom} -\DeclareMathOperator{\sym}{sym} -\DeclareMathOperator{\fix}{fix} -\newcommand{\PP}{\Bbb P} -\newcommand{\cG}{\mathcal G} -\newcommand{\HS}{\mathsf{HS}} -\newcommand{\ZF}{\mathsf{ZF}} -\newcommand{\cF}{\mathcal F} -\newcommand{\id}{\operatorname{id}} -\newcommand{\tup}[1]{\langle #1\rangle}$ -If $\PP$ is a forcing, and $\{\dot x_i\mid i\in I\}$ is a set (or class) or $\PP$-names, we define $\{\dot x_i\mid i\in I\}^\bullet$ to be the $\PP$-name $\{\tup{1_\PP,\dot x_i}\mid i\in I\}$. This extends to ordered pairs, and naturally to sequences. -1 Localized failure -Fix a regular cardinal $\kappa$. Let $\PP_\kappa$ be the following presentation of adding a Cohen subset to $\kappa$. We mimic the construction of Cohen's second model. -A condition in $\PP_\kappa$ is a partial function $p\colon\omega\times 2\times\kappa\to2$, such that $|\dom p|<\kappa$. -We define the following names, for $(n,i,m)\in\omega\times2\times\omega$: - -$\dot x_{n,i,m}=\{\tup{p,\check\alpha}\mid p(n,i,m,\alpha)=1\}$, -$\dot X_{n,i}=\{\dot x_{n,i,m}\mid m\in\omega\}^\bullet$, and -$\dot S_n=\{\dot X_{n,i}\mid i<2\}^\bullet$. - -Next we define the automorphism group $\cG_\kappa$, to be permutations $\pi$ of $\omega\times2\times\omega$ moving only finitely many points and satisfying that if $\pi(n,i,m)=(n',i',m')$, then $n=n'$; and for all $n$, either $i=i'$ or $i'=1-i$. It will be clearer to understand, once we see how these act on the names defined above: - -If $p\in\PP_\kappa$, then $\pi p(\pi(n,i,m),\alpha)=p(n,i,m,\alpha)$. This is the standard way this action is defined. It extends to $\PP_\kappa$-names recursively: $\pi\dot x=\{\tup{\pi p,\pi\dot y}\mid\tup{p,\dot y}\in\dot x\}$. -$\pi\dot x_{n,i,m}=\dot x_{\pi(n,i,m)}$, -$\pi\dot X_{n,i}=\dot X_{n,i'}$ if there are some $m$ and $m'$ such that $\pi(n,i,m)=\pi(n,i',m')$, and -$\pi\dot S_n=\dot S_n$. - -In other words, for every pair $A_n$, we decide whether or not we switch the elements in that pair, and separately, we may permute the elements of each set in the pair. -Finally, we take $\cF_\kappa$ to be the normal filter of subgroups generated by fixing finitely many points. Namely, if $E\subseteq\omega\times2\times\omega$ is finite, $\fix(E)$ is the group of all automorphisms which fix all the points in $E$ pointwise; and $\cF_\kappa$ is generated by $\fix(E)$ for $E$ finite. -We say that a name is symmetric if $\{\pi\mid \pi\dot x=\dot x\}\in\cF_\kappa$; and hereditarily symmetric if also every name which appears inside $\dot x$ is hereditarily symmetric. -We denote by $\HS$ the class of hereditarily symmetric names. If $G$ is a generic filter, then $\HS^G=\{\dot x^G\mid\dot x\in\HS\}$ is a model of $\ZF$. Let $M$ denote $\HS^G$, and we omit the dots to indicate the interpretations of the names, e.g. $\dot A_n^G$ is $A_n$ and so on. -Standard arguments show that: - -Each $\dot x_{n,i,m}$ is symmetric (and thus hereditarily symmetric), -Each $\dot X_{n,i}$ is symmetric (and thus ...), -The sequence $\tup{\dot A_n\mid n<\omega}^\bullet$ is hereditarily symmetric. -The name $\dot A=\{\dot X_{n,i}\mid(n,i)\in\omega\times 2\}^\bullet$ is hereditarily symmetric. - -So all these will then be in $M$. Moreover, - -For all $n$ and $i$, $X_{n,i}$, in $M$ is a Dedekind-finite set. -$A$ is a Russell set in $M$. Specifically, $\{A_n\mid n<\omega\}$ is a partition witnessing that. - -2 Global failure -Let $D$ be a class of regular cardinals. For example, all regular cardinals. For every $\kappa$ in $D$, note that $\PP_\kappa$ is $\kappa$-closed. Let $\PP$ be the Easton product of the $\PP_\kappa$'s. We then define an Easton support product of the groups and filters. -The class of hereditarily symmetric names of this system will include all the Russell sets we added by each $\PP_\kappa$. So it remains to show that there is no set of Russell sets which catch all of them up to equi-cardinality. -But then we get this quite easily. Any set of Russell sets will be added by some condition, and then some large enough $\kappa$ will add a new Russell set. -You can find details of a similar construction here: - -Karagila, Asaf, Embedding orders into the cardinals with $\mathsf {DC}_{\kappa} $, Fundam. Math. 226, No. 2, 143-156 (2014). ZBL1341.03068.<|endoftext|> -TITLE: Bounding the area of a convex body bounded in a sphere -QUESTION [10 upvotes]: I have a question which I believe to be pretty basic. -Let $\Gamma$ be some convex body, bounded inside a $L_2$ sphere of radius 1 $B(0,1)$. -Is it true that the surface area of $\Gamma$ is smaller than the surface area of the sphere? -I'm guessing that the answer involves finding a continuous deformation from $\Gamma$ to the sphere for which the area is monotonous, but I'm incapable of finding it - -REPLY [19 votes]: The nearest-point projection from the sphere to the convex set decreases distances.<|endoftext|> -TITLE: Asymptotic expansion of $\sum\limits_{n=1}^{\infty} \frac{x^{2n+1}}{n!{\sqrt{n}} }$ -QUESTION [15 upvotes]: I've been trying to find an asymptotic expansion of the following series -$$C(x) = \sum\limits_{n=1}^{\infty} \frac{x^{2n+1}}{n!{\sqrt{n}} }$$ -and -$$L(x) = \sum\limits_{n=1}^{\infty} \frac{x^{2n+1}}{n(n!{\sqrt{n}}) }$$ -around $+\infty$, in the from -$$\exp(x^2)\Big(1+\frac{a_1}{x}+\frac{a_2}{x^2} + .. +\frac{a_k}{x^k}\Big) + O\Big(\frac{\exp(x^2)}{x^{k+1}}\Big)$$ -where $x$ is a positive real number. As far as I progressed, I obtained only -$$C(x) = \exp(x^2) + \frac{\exp(x^2)}{x} + O\Big(\frac{\exp(x^2)}{x}\Big).$$ -I tried to use ideas from https://math.stackexchange.com/questions/484367/upper-bound-for-an-infinite-series-with-a-square-root?rq=1, https://math.stackexchange.com/questions/115410/whats-the-sum-of-sum-limits-k-1-infty-fractkkk, https://math.stackexchange.com/questions/378024/infinite-series-involving-sqrtn?noredirect=1&lq=1, but I was unable to make them work in my case. -Any suggestions would be greatly appreciated! -(If someone has a solid culture in this kind of things, is there are any specific names for $C(x) $ and $L(x) $ ?). -PS: -This question was asked on the math.SE but was closed as duplicate of https://math.stackexchange.com/questions/2117742/lim-x-rightarrow-infty-sqrtxe-x-left-sum-k%ef%bc%9d1-infty-fracxk/2123100#2123100. However, the latter question provides only the first term of the asymptotic expansion and does not address sufficiently the problem considered here. - -REPLY [8 votes]: This is another, totally different (and correct !) approach for answering the question. It is simply too long for a comment. So I decided to write it in a new answer. (Although that might look odd, but my previous answer, though accepted, is wrong.) -First define -$$ -I_{\mu}(y) = y^{1/2} \sum_{n=1}^{\infty} \ \frac{y^{n}}{n! \ n^{\mu}}, -$$ -and observe that the OP's functions are -$$ -C(x)= I_{\frac{1}{2}}(x^{2}) -$$ -and -$$ -L(x)=I_{\frac{3}{2}}(x^{2}) -$$ -Let $m$ be the integer part of $\mu$ and $\xi$ the fractional part, $0<\xi<1$. -Replace $n^{-\mu}$ in the definition of $I_{\mu}(y)$ by a ratio of $\Gamma$-functions times an asymptotic series in $n$ using this formula and for the coefficients of the asymptotic series in $n$ using the Norlund polynomials, $B_k^{(\alpha)}(x)$ found here. They are available in Mathematica via $\mathtt{NorlundB[k,\alpha,x]}$. Concretely, -$$ -n^{-\xi}\sim \frac{\Gamma(n)}{\Gamma(n+\xi)}\sum_{k=0}^{\infty}n^{-k} {\xi \choose k} B_k^{(1+\xi)}(\xi). -$$ -Now exchange the summation of the asymptotics in $n$ (with, say, summation index $k$) and the summation over $n$, which I assume light heartedly to be possible. The (now inner) sum over $n$ results in generalized hypergeometric functions of the form -$$ -_{m+k+2}F_{m+k+2}\left(\left. {1,\dots,1}\atop{2,\dots,2,1+\xi} \right\vert y\right), -$$ -with $m+k+2$ 1s in the upper line and $k+m+1$ 2s in the lower line. To get there one has to shift the summation index such that summation starts at $n=0$. Then insert $n+1=\frac{(2)_{n}}{(1)_{n}}$, with the usual Pochhammer symbols used. The defining formula for generalized hypergeometric functions results. -Using the asymptotic expansion of the generalized hypergeometric function for $y\rightarrow \infty$ given in a paper by Volkmer and Wood (downloadable from here) and after some simplifications one arrives at the asymptotic formula -$$ -I_{m+\xi}(y)=e^{y}\ y^{\frac{1}{2}-m-\xi}\ \frac{\xi\ \sin \pi\xi}{\pi}\sum_{k=0}^{\infty}(-1)^{k+1} y^{-k} \frac{\Gamma(k-\xi)}{k!}\ B_{k}^{(1+\xi)}(\xi) \\ -\sum_{s=0}^{\infty} y^{-s} \left\{ {\sum_{(s_{1},\dots,s_{m+k+1})}} \frac{\Gamma(\xi + s_{m+k+1})}{s_{m+k+1}!}\prod_{j=1}^{m+k+1} \frac{\Gamma\left(j+\sum_{i=1}^{j}s_{i}\right)}{\Gamma\left(j+\sum_{i=1}^{j-1}s_{i}\right)}\right\}. -$$ -$(s_{1},\dots ,s_{m+k+1})$ under the sum sign indicates summation over all (ordered) partitions of $s$ into $m+k+1$ non-negative integers, $s_{1},\dots ,s_{m+k+1}$. Order matters here, i.e., $(1,0)$ is different from $(0,1)$. -Numerical calculations of the coefficients (with highest possible precision on my laptop) show excellent match (5 or more digits) with the formula, even for exotic indices, like $\mu=\pi$ and orders up to $y^{-6}$. -For the OP's functions I get: -$$ -C(x) e^{-x^{2}} = 1 + \frac{3}{8} x^{-2} + \frac{65}{128} x^{-4} + \frac{1225}{1024} x^{-6} + \frac{131691}{32768} x^{-8} + O(x^{-10}) , -$$ -$$ -L(x) x^{2} e^{-x^{2}} = 1 + \frac{15}{8} x^{-2} + \frac{665}{128} x^{-4} + \frac{19845}{1024} x^{-6} + \frac{2989371}{32768} x^{-8} + O(x^{-10}) . -$$ -All calculations were done with Mathematica 11.<|endoftext|> -TITLE: Can we infer an isomorphism of quivers from an isomorphism of their corresponding path algebras? -QUESTION [7 upvotes]: Given a pair $\Delta, \Gamma$ of quivers and a field $K$ one can construct the corresponding path algebras $K\Delta, K\Gamma$. I came upon a paper claiming (section 3, 2nd paragraph) that an isomorphism of $K\Delta \cong K\Gamma$ of the path algebras induces an isomorphism of the quivers $\Delta \cong \Gamma$. However, no proof is given for this claim, and so I am looking for a reference giving a proof and stating explicitly the required conditions on the quivers or the exact meaning of a quiver isomorphism in this context. - -REPLY [2 votes]: Here's one way to recover $Q$ from $KQ$ when $Q$ is finite but not necessarily acyclic. -Consider the one-dimensional $KQ$-modules. Each is associated to some vertex of $Q$, and two such modules $M,N$ are associated to the same vertex if and only if there is some free algebra quotient $F=K\langle x_1,\dots,x_n\rangle$ of $KQ$ such that the actions of $KQ$ on $M$ and $N$ both factor through $F$. -So the vertex set of $Q$ can be recovered from $KQ$. Picking one one-dimensional module $M_v$ for each vertex $v$, the number of arrows from vertex $v$ to vertex $w$ is the dimension of $\text{Ext}^1(M_v,M_w)$.<|endoftext|> -TITLE: Can a perturbation of a matrix product always be represented as product of perturbations of its factor matrices? -QUESTION [6 upvotes]: Given $A=BC$ where $A\in\mathbb{R}^{m\times n}$ and for some $B\in\mathbb{R}^{m\times k}, C\in\mathbb{R}^{k\times n}$. We assume that $k>=\min(m,n)$ so that this decomposition always exists for any matrix $A\in\mathbb{R}^{m\times n}.$ -Can we prove that any perturbation $\bar{A}$ of $A$ can be represented as the product of two perturbations of $B$ and $C$ ? -Intuitively i think it should be possible but i cannot prove/disprove it. If not possible then what conditions are needed? - -REPLY [3 votes]: In numerical analysis lingo, you are more or less asking if matrix multiplication is backward stable. The answer seems to be no: see Section 3.5 of Higham, Accuracy and stability of numerical algorithms. I am unable to locate an explicit counterexample quickly, though.<|endoftext|> -TITLE: $E_n(\ell^\infty)=SL_n(\ell^\infty)$? -QUESTION [8 upvotes]: Let $R$ be a commutative unital ring $R$ with unit element $1$. -For $n\in \mathbb{N}=\{1,2,3,\cdots\}$, let $SL_n(R)$ be the group of all $n\times n$ matrices with entries from $R$ having determinant $1$. -An element $E\in SL_n(R)$ is called an elementary matrix if $E=I+a E_{ij}$, with $i\neq j$, where - -$a\in R$, -$I$ denotes the identity matrix, and -$E_{ij}$ the matrix with entry in $i$th row and $j$th column equal to $a$ and all other entries $0$. - -The subgroup of $SL_n(R)$ generated by the elementary matrices is denoted by $E_n(R)$. -We know that when $R=\mathbb{C}$, then for all $n$, $SL_n(\mathbb{C})=E_n(\mathbb{C})$. -Now let $R=\ell^\infty$, -the algebra of all complex valued bounded sequences with pointwise operations. -Question: Is it true that $SL_n(\ell^\infty)=E_n(\ell^\infty)$ for all $n$? - -REPLY [5 votes]: The answer is yes. The identity $SL_n(\ell^{\infty}) = E_n(\ell^{\infty})$ holds for every $n \ge 2$ and $\mathbb{C}$ can be replaced by any normed field. This can be inferred using the following lemma and the two observations below. - - -Lemma. Let $R = \prod_i R_i$ be a direct product of unital and commutative rings $R_i$. Let $n \ge 2$ and $A = (A_i )_i\in SL_n(R) \simeq \prod_i SL_n(R_i)$. If for every $i$ the matrix $A_i$ is the product of at most $k$ elementary matrices over $R_i$, then $A$ is the product of at most $k(n^2 - n)^k$ elementary matrices over $R$. In particular, we have $A \in E_n(R)$. -Proof. Consider each matrix $A_i$ as a word $w_i$ over an alphabet of elementary matrix types (there are $n(n -1)$ types). Since there are at most $(n^2 - n)^k$ such words $w_i$, they all fit in a word of length at most $k(n^2 - n)^k$. - - -Let $A \in SL_n(\mathbb{C})$. Then we have: - -The largest complex modulus of a coefficient of $A$ is greater or equal to $1/\sqrt[n]{n!}$ (expend $\det(A)$ and use the triangle inequality). -If the coefficients of $A$ are bounded above by $C > 1$, then $A$ is the product of $\nu(n)$ elementary matrices with coefficients bounded above by $2\sqrt[n]{n!} C$ where $\nu(n)$ depends on $n$ only (use the above observation as a basis for an induction on $n$). - -Combining the last observation with the lemma yields the result. -Let us detail the case of $SL_2(\ell^{\infty})$. - - -Proof for $n = 2$. - Multiplying $A = (a_{ij})_{1 \le i, j \le 2}$ by at most four elementary matrices whose coefficients are bounded above by $1$, we obtain a matrix $A' = (a'_{ij})_{1 \le i, j \le 2}$ whose coefficients are those of $A$, up to some permutation and some sign changes, and such that $\vert a'_{11} \vert$ is maximal. Using $a'_{11}$ as a pivot, we can turn $A$ into $\begin{pmatrix} a'_{11} & 0 \\ 0 & 1/a'_{11} \end{pmatrix}$ by means of two elementary matrices whose coefficients are bounded above by $1$. Using four more matrices whose coefficients are bounded above by $2\sqrt{2}C$, such as those used in Whitehead's lemma, we obtain the identity matrix. All in all, we used at most ten elementary matrices to perform our reduction and the coefficients of all these matrices are bounded above by $2\sqrt{2}C$ (hence we can set $\nu(2) = 10$).<|endoftext|> -TITLE: Does Laver forcing add cofinal branches to $\omega_1$-trees? -QUESTION [5 upvotes]: Given an $\omega_1$-tree $T$ in the ground model, can Laver forcing add a cofinal branch to $T$? Assume GCH in the ground model. -Definitions: -An $\omega_1$-tree is a well-founded tree of height $\omega_1$ with all levels countable. A cofinal branch of $T$ is a maximal chain of type $\omega_1$. Laver forcing $\mathbb{L}$ consists of subtrees $T$ of $\omega^{<\omega}$ where, for some $x\in T$, every $y<_T x$ has only one child and every $y\geq_T x$ has $\aleph_0$ children. $q\leq_{\mathbb{L}} p$ iff $q\subset p$. -Related results: -Silver proved in 1971 that a countably closed forcing cannot add a cofinal branch to $T$: given $p_\varnothing$ forcing a new cofinal branch $B$, construct conditions $(p_s:s\in 2^{<\omega+1})$ where $p_t\leq p_s$ for $s\subset t$ but for $s\perp t$ nodes $b_s\perp b_t$ of $T$ are respectively forced into $B$ by $p_s$ and $p_t$. This contradicts countability of any level $T_\delta$ chosen high enough that $b_s\in T_{<\delta}$ for all finite $s$. -I can modify the above argument to show that Sacks forcing (perfect subtrees of $2^{<\omega}$) does not add a cofinal branch to $T$. Given $s\in 2^n$ and $p_s$, I find, for each $y\in p_s$ minimal with respect to having $n$ splitting nodes below it, conditions $$q_{sy0},q_{sy1}\leq p_s|y=\{x\in p_s: x\not\perp y\}$$ that respectively force nodes $c_{sy0},c_{sy1}$ into $B$ such that $c_{sy0}\perp c_{sz1}$ for all $y,z$. Then $p_{si}=\bigcup_y q_{syi}$ preserves the first $n$ 'levels' of splitting nodes of $p_s$, ensuring that $\{p_{f|n}:n<\omega\}$ has a common extension $q_f$ for each $f\in 2^{\omega}$. Finally, choose $\delta$ sufficiently high and then extend each $q_f$ to $p_f$ deciding $B$ at level $\delta$. If $f\not=g$, then $p_f$ and $p_g$ disagree about $B$ at level $\delta$. -The above fusion argument also applies to Miller forcing; just replace 'splitting' with 'infinitely splitting.' However, for Laver forcing, $p_{si}=\bigcup_y q_{syi}$ above may fail to be a condition because the 'trunks' of the $q_{syi}$ may be too long. - -REPLY [4 votes]: No. -In our paper https://arxiv.org/abs/1409.4596 with Jindra Zapletal on Y-stuff we prove that Laver forcing is Y-proper. Consequently it has the $\omega_1$-approximation property (i.e., does not add fresh sets of size $\omega_1$ in Joel's terminology) and hence adds no new branches to trees. -PS: Thanks to Wolfgang for pointing me to this question.<|endoftext|> -TITLE: A property of nearby cycles functor -QUESTION [6 upvotes]: Let $f\colon X\to Y$ be a flat morphism of irreducible projective algebraic varieties over $\mathbb{C}$ (or any other algebraically closed field of characteristic 0). Assume that $Y$ is smooth, and the generic fiber of $f$ is smooth (these two assumptions seem to be important). -Let $U\subset Y$ be a Zariski open non-empty subset such that $f\colon f^{-1}(U)\to U$ is smooth. Fix a closed point $y\in Y$. -Let us denote by $S$ the "disk", more precisely $S$ is the spectrum of henselization of $\mathbb{C}[t]$ localized at the ideal generated by $t$. Let $\eta$ be the generic point of $S$. -For any morphism $\nu\colon S\to Y$ such that $\eta$ is mapped to $U$ and the closed point of $S$ to $y$, consider the fibered product $S\times_Y X\to S$. Notice that the generic fiber of this morphism is smooth over $\eta$. Consider the nearby cycle functor of the constant sheaf $\underline{\mathbb{\mathbb{Q}_l}}$. It is a perverse sheaf on $f^{-1}(y)$ which we will denote by $\mathcal{F}_\nu$ to emphasize dependence on the morphism $\nu$. - -QUESTION. Is it true that for all choices of $\nu$ as above the perverse sheaves $\mathcal{F}_\nu$ are isomorphic to each other? - -REPLY [2 votes]: The answer is no! -Start with any $g:Z \to S$, with smooth generic fiber. Write $Z_0$ for the closed fiber; in order for your conditions to be eventually satisfied this should be assumed smooth as well. Then set $X = Z\times S$, $Y = S\times S$, $f = g\times id.$ Taking the base change of $f$ with the two coordinate-axis embeddings $S \to S\times S$ gives two different schemes over $S$, namely $Z$ and $Z_0\times S$. In the latter case your nearby cycles sheaf will just be the constant perverse sheaf on $Z_0$; hence we will be done if we can find a starting $Z$ whose nearby cycles is not the constant perverse sheaf. For this there are many examples, but the easiest is when $Z_0$ contains an entire component of $Z$.<|endoftext|> -TITLE: Splitting of polynomials over rational function fields -QUESTION [12 upvotes]: Let $K$ be a number field, and let $P(t,X)$ be a monic polynomial in $X$ with coefficients in $K(t)$. -I would like to understand the set $T$ consisting of those $t_0 \in K$ such that the polynomial $P(t_0,X)$ is (totally) split over $K$. -Q1. Is there an algorithm to decide whether $T$ is empty, finite or infinite? -Q2. In the case $T$ is infinite, does there exist a non-constant rational function $f \in K(u)$ such that $t_0 \in T$ if and only if $t_0 = f(u_0)$ for some $u_0 \in K$? -Q3. In the case $T$ is infinite, is there an algorithm to find a non-constant rational function $f \in K(u)$ such that $P(f(u),X)$ splits over $K(u)$? -Here are some easy examples. -Example 1. Take $K=\mathbf{Q}$ and $P=X^2-t$. Obviously $T$ is the set of rational squares, and $f(u)=u^2$ satisfies the conditions in Q2 and Q3. -Example 2. Take $K$ arbitrary and $P=X^3-t$. If $t_0 \in K^\times$ and $P(t_0,X)$ splits over $K$, then $K$ must contain the cube roots of unity, and in this case the rational function $f(u)=u^3$ satisfies the conditions in Q2 and Q3. -My motivation for this question is that I'm trying to determine explicit equations for universal elliptic curves. For example, let $E_1(4)$ (resp. $E(4)$) be the universal elliptic curve over the modular curve $Y_1(4)$ (resp. $Y(4)$). A Weierstrass equation for $E_1(4)$ is given by $y^2+xy+ty=x^3+tx^2$ where $t$ is a generator of the function field of $Y_1(4)$ and the universal point of order 4 is given by $(0,0)$. Now determining $E(4)$ amounts to find those $t$ such that the $4$-division polynomial of $E_1(4)$ splits completely over $\mathbf{Q}(i)$. -So here is a concrete question: take $K=\mathbf{Q}(i)$ and $P(t,X)=X^4 + \frac12 X^3 + \frac32 tX^2 + 2t^2 X + t^3$. Can you find an explicit rational function $f \in K(u)$ such that $P(f(u),X)$ splits over $K(u)$? - -REPLY [15 votes]: Your question 1 is open and is equivalent to the problem of determining the rank of an elliptic curve over a number field. (In most "practical cases" it should be solvable.) In particular, we don't have an algorithm to determine if an elliptic curve $E/K$ has positive rank, if $E : y^2 = x^{3} + Ax + B$, this is asking precisely for the values of $t$ for which $P(t,X) = X^{2} - t^{3} - At - B$ splits. In general, if you let $L$ be the Galois closure of $K(t)[X]/(p(t,X))$ over $K(t)$, then $L$ is the function field of a curve, and you are asking about the $K$-rational points of this curve. Faltings's theorem says that if the genus of this curve is $> 1$, then there are finitely many $K$-rational points. The genus $1$ cases are a bit messier, and the genus zero case should be fairly simple. -Given the explanation above, you can probably see that the answer to question 2 is no. In particular, because $E : y^{2} = x^{3} - 2x$ is an elliptic curve and has positive rank, there are infinitely many $t$ so that $P(t,X) = X^{2} - (t^{3} - 2t)$ splits over $\mathbb{Q}$. However, there is no $f(u)$ so that $P(f(u),X)$ splits over $\mathbb{Q}(u)$. This would imply that there is a rational curve on $E$, which would force $E$ to be rational (which it's not because it has genus $1$). -You might also like to know that you can find a model for the univeral elliptic curve over $X(4)$ here. For me, $Y(4)$ is the modular curve consisting of pairs $(E,i)$, where $i : E[4] \to \mathbb{Z}/4\mathbb{Z} \times \mu_{4}$ is a Galois-equivariant map. To get the $E(4)$ you want, take this curve and consider the base change to $\mathbb{Q}(i)$. [ In my paper with David Zureick-Brown here, we explain how to compute models of these universal elliptic curves. ] -In regard to your last question about finding $f(u)$ so that $P(f(u),X)$ splits, the relevant curve is simply $P(t,X) = 0$. This has genus zero with a non-singular rational point, and thus can be parametrized. This yields -$$ f(u) = \frac{u^{3} - 3u^{2} + 4u - 2}{2u^{4}}. $$<|endoftext|> -TITLE: Explicit formulas for Carnot-Carathéodory distances on Carnot groups -QUESTION [13 upvotes]: Let $G$ be a Carnot group (aka stratified group), so that $G$ is a connected and simply connected finite-dimensonal Lie group, whose Lie algebra $\mathfrak{g}$ admits a decomposition $\mathfrak{g} = V_1 \oplus \dots \oplus V_n$ where $[V_1, V_k] = V_{k+1}$. Let $\mathcal{H} \subset TG$ be the left-invariant "horizontal distribution" which agrees with $V_1$ at the identity. Then $\mathcal{H}$ is bracket generating. If we fix an inner product $\langle \cdot, \cdot \rangle$ on $V_1$, we obtain a left-invariant sub-Riemannian metric $g$ on $G$. -Let $d$ be the corresponding Carnot–Carathéodory distance on $G$. That is, the distance $d(x,y)$ is the length of the shortest horizontal path joining $x$ to $y$. -If $G = \mathbb{H}^{2m+1}$ is a Heisenberg group, i.e. $n=2$ and $\dim V_2 = 1$, then an "explicit" formula for $d$ is known [1], for any inner product on $V_1$. (As Richard Montgomery points out, it is not explicit in the strongest sense because it is in terms of a solution of a transcendental equation. But that sort of thing is explicit enough for my purposes.) -There are also "explicit" formulas known in case $(G, \langle \cdot,\cdot \rangle)$ is H-type in the sense of Kaplan [2]; see [3,4,5]. - -Are there any other (classes of) Carnot groups for which we know how to "explicitly" compute the distance $d$? - - -[1]. Beals, Richard W.; Gaveau, Bernard; Greiner, Peter C., Hamilton-Jacobi theory and the heat kernel on Heisenberg groups, J. Math. Pures Appl., IX. Sér. 79, No.7, 633-689 (2000). ZBL0959.35035. -[2]. Kaplan, Aroldo, Fundamental solutions for a class of hypoelliptic PDE generated by composition of quadratic forms, Trans. Am. Math. Soc. 258, 147-153 (1980). ZBL0393.35015. -[3]. -Korányi, Adam, Geometric properties of Heisenberg-type groups, Adv. Math. 56, 28-38 (1985). ZBL0589.53053. -[4] Rigot, Séverine, Counter example to the Besicovitch covering property for some Carnot groups equipped with their Carnot-Carathéodory metric, Math. Z. 248, No. 4, 827-848 (2004). ZBL1082.53030. -[5]. Tan, Kang-Hai; Yang, Xiao-Ping, Characterisation of the sub-Riemannian isometry groups of $H$-type groups, Bull. Aust. Math. Soc. 70, No. 1, 87-100 (2004). ZBL1070.53013. - -REPLY [4 votes]: As pointed out by Richard Montgomery, there is no formula expressed in terms of elementary functions for the Carnot-Caratheodory distance between any two points, even in the case of the Heisenberg group. In the answer provided below we will see an explicit formula involving an inverse of an elementary function which is Corollary 3.2 in [2]. It is perhaps easily seen to be equivalent to (1.40) in [1] (see the answer of Richard Montgomery), but I haven't checked it. -The Heisenberg group $\mathbb{H}^n$ is $\mathbb{C}^n\times\mathbb{R}=\mathbb{R}^{2n+1}$ given the structure of a Lie group with multiplication -\begin{split} -&(z,t)*(z',t')= -\Big(z+z',t+t'+2 {\rm Im}\, \sum_{j=1}^nz_j\overline{z_j'}\Big)\\ -&= -\Big(x_1+x_1', \dots , x_n + x_n' , y_1+y_1', \dots, y_n+y_n', t+t'+2 \sum_{j=1}^n (x_j' y_j - x_j y_j')\Big). -\end{split} -A basis of a Lie algebra is given by -$$ -X_j = \frac{\partial}{\partial x_j} + 2y_j \frac{\partial}{\partial t}, -\quad -Y_j= \frac{\partial}{\partial y_j} - 2x_j \frac{\partial}{\partial t}, -\quad -T=\frac{\partial}{\partial t}, -\quad -j=1,2,\ldots,n. -$$ -The Heisenberg group is equipped with the horizontal distribution -$$ -H\mathbb{H}^n = \mathrm{span} \{ X_1,\dots, X_n, Y_1,\dots,Y_n \}. -$$ -The horizontal distribution is equipped with a Riemannian metric $g$ so that the basis $\{ X_1,\dots, X_n, Y_1,\dots,Y_n \}$ of $H\mathbb{H}^n$ is orthonormal. -Horizontal curves are the curves that are tangent to the horizontal distribution and we compute the length of such a curve with respect to the metric $g$ on -$H\mathbb{H}^n$. The Carnot-Caratheodory distance $d_{cc}(p,q)$ between $p,q\in\mathbb{H}^n$ is the infimum of lengths of horizontal curves connecting $p$ and $q$. -Define $H:(-1,1) \to \mathbb{R}$ as -$$ -H(s) = \frac{2 \pi}{1 - \cos(2 \pi s)} \left( s-\frac{\sin(2 \pi s)}{2 \pi} \right)= -\frac{\frac{2\pi s}{3!}-\frac{(2\pi s)^3}{5!}+\frac{(2\pi s)^5}{7!}-\ldots}{\frac{1}{2!}-\frac{(2\pi s)^2}{4!}+\frac{(2\pi s)^4}{6!}-\ldots}\, . -$$ -$H$ is a real analytic diffeomorphism of $(-1,1)$ onto $\mathbb{R}$ so the inverse $H^{-1}:\mathbb{R}\to (-1,1)$ is a real analytinc diffeomorphism too. - -Theorem. (Corollary 3.2 in [2]). For $z\neq 0$, the Carnot-Caratheodory distance between the origin $(0,0)$ and $(z,t)$, equals - $$ - d_{cc}((0,0),(z,t)) = t \sin(\pi H^{-1}(t|z|^{-2}))|z|^{-1} + - |z|\cos(\pi H^{-1}(t|z|^{-2})). -$$ - -Using a fact that left translations on the Heisenberg group are isometries, we can obtain a formula for the distance between any two points. It also follows that the Carnot-Caratheodory distance is a real analytic function away from the center of the group. -[1]. R. W. Beals,B. Gaveau, P. C. Greiner, Peter C., Hamilton-Jacobi theory and the heat kernel on Heisenberg groups, J. Math. Pures Appl., IX. Sér. 79, No.7, 633-689 (2000). ZBL0959.35035. -[2] P. Hajłasz, S. Zimmerman, Geodesics in the Heisenberg group. Anal. Geom. Metr. Spaces 3 (2015), 325–337.<|endoftext|> -TITLE: Is co-restriction in Galois cohomology in fact the norm map via Kummer isomorphism? -QUESTION [7 upvotes]: Let $\mathrm{F}$ be a field that contains a root of unity of order $p$, where $p$ is a prime number. Fix an element $a$ such that $a \in \mathrm{F}$ and $\sqrt[p]{a} \notin \mathrm{F}$. Consider the absolute Galois group of $\mathrm{F}$, denoted by $G_F$, and its open subgroup $G_{F[\sqrt[p]{a}]}$ of index $p$ (which is the absolute Galois group of the extension $F[\sqrt[p]{a}]$). There is a well-known map between the following Galois cohomology groups, the co-restriction: -$$Cor:H^1(G_{F[\sqrt[p]{a}]}, \mathbb{Z}/p) \rightarrow H^1(G_F, \mathbb{Z}/p)$$ -In addition, there is a map between the abelian groups $F[\sqrt[p]{a}]^*/{F[\sqrt[p]{a}]^*}^p$ and $F^*/{F^*}^p$ which is the field norm: -$$ N_{F[\sqrt[p]{a}] / F} : F[\sqrt[p]{a}]^*/{F[\sqrt[p]{a}]^*}^p \rightarrow F^*/{F^*}^p $$ -Kummer theory yields the isomorphisms $H^1(G_F, \mathbb{Z}/p) \cong F^*/{F^*}^p$ and $H^1(G_{F[\sqrt[p]{a}]}, \mathbb{Z}/p) \cong F[\sqrt[p]{a}]^*/{F[\sqrt[p]{a}]^*}^p$, so the natural question to be asked is whether the co-restriction map and the norm map are actually the same thing, via these identifications? In other words, does the corresponding diagram commute? I couldn't find any references for this. - -REPLY [7 votes]: The corestriction map on cohomology is indeed the norm in degree zero (see Tate's notes on Galois cohomology for example). By a dimension shifting argument, it then easily follows that the corestriction in degree one also corresponds to taking norms (under your isomorphisms).<|endoftext|> -TITLE: The Analog of Borel Subgroup in a Compact Real Form -QUESTION [6 upvotes]: I recently learned that there is a one-to-one correspondence between isomorphism classes of complex reductive groups and isomorphism classes of compact connected real Lie groups given by taking a compact real form in one direction and complexification in the other direction. I am happy with many parallels such as the existence of maximal tori, the definition of a Weyl groups, the classification of finite dimensional irreducible representations by dominant integral weights and so on. (Please correct me if I get any of the statements above wrong since I am just pulling them out from my memory.) -One question that bothers me is the following. In the complex reductive case, given a fixed maximal torus one can choose a Borel subgroup containing that torus, which is equivalent to choosing a collection of simple roots in the root system; is there an analog of this in the compact connected real Lie group case? (I don't know how to properly define a root system for a compact connected real Lie group without passing to the corresponding complex reductive group either.) Thanks a lot. - -REPLY [3 votes]: It sounds like you want a datum to associate to a compact Lie group and chosen torus, that's functorially the same as a choice of Borel containing that torus if one were to complexify, without using the word "complexify" anywhere. -If that's right, I suggest "choice of connected component of $(\mathfrak t^*$ with all points removed that have nontrivial $W$-stabilizer)".<|endoftext|> -TITLE: Where does $2\sqrt{d-1}$ come from in Ramanujan graphs? -QUESTION [8 upvotes]: Ramanujan graphs are the best spectral expanders: $\lambda_2 \le 2\sqrt{d-1}$. I'm looking for some intuition for this value $2\sqrt{d-1}$. -Friedman showed that every random $d$-regular graph achieves $\lambda_2 = 2\sqrt{d-1}+\epsilon$ and Alon-Boppana shows that every sufficiently large $d$-regular graph must have $\lambda_2 \ge 2\sqrt{d-1} - o(1)$. Unfortunately both these results seem quite involved. Is there an easy way to see that the answer should be around $2\sqrt{d-1}$, or at very least $\Theta(\sqrt{d})$? Intuition for either direction would be appreciated. - -REPLY [8 votes]: If you take an infinite regular tree of degree $d$ and fix one vertex $v$, then the number of closed walks of length $2k$ (there are none of odd length) starting at $v$ grows like $4^k(d-1)^k$ as $k\to\infty$. To see this heuristically, note that you have to move towards $v$ half the time and away from $v$ half the time. There are $2^{2k}$ choices for which steps you move away from $v$ and for each such step there are $d-1$ choices for which branch to take. As David says, this is related to the question because Ramanujan graphs are locally tree-like.<|endoftext|> -TITLE: Primitive integral solutions to $x^2+y^3=z^2$ -QUESTION [7 upvotes]: The Diophantine equation -$$x^2+y^3=z^2$$ -has solutions $(\pm 1,2,\pm 3)$ and $(\pm 13,3,\pm 14)$. -Brown [Int. Math. Res. Not. IMRN 2012, no. 2, 423–436; MR2876388] states that "there are infinitely many parameterized solutions with $xyz \neq 0$ and $z^2 \neq 1$ when $n \leq 5$" to the equation -$$x^2 +y^3 = z^n,$$ -but does not give a reference. I am looking for a reference that catalogs all primitive solutions. - -REPLY [4 votes]: A warm-up: -$$ \left(\frac{s^n-t^n}2\right)^2 +\ (s\cdot t)^n\ =\ - \left(\frac{s^n+t^n}2\right)^2 $$ -provides integer solutions for $\,\ s\equiv t \mod 2 $. -(Sorry, I couldn't help it). -  - -REMARK   More generally, going in the abc direction: -$$ \left(\frac{s^m-t^n}2\right)^2 +\ s^m\cdot t^n\ =\ - \left(\frac{s^m+t^n}2\right)^2 $$ -for $\ s\equiv t\equiv 1 \mod 2,\ $ and $\ \gcd(s\ t) = 1$.<|endoftext|> -TITLE: Brouwer's Theorem in the free topos? -QUESTION [12 upvotes]: In Introduction to Higher-Order Categorical Logic, Lambek & Scott remark that Brouwer's Theorem (all functions $\mathbb{R}\to\mathbb{R}$ are continuous) holds in the free topos $\mathcal{T}$. -EDIT: L&S do not say that it holds in the free topos, they are incorrectly quoted as saying this in the nLab. -Let me present a naïve syllogism which is bothering me: - -The free topos is the initial object in the category of toposes & logical morphisms / nLab -Logical morphisms preserve truth of sentences in the internal language of toposes. / nLab -Brouwer's theorem is false in the topos of sets $\mathbf{Set}$. -By (1) there is a logical morphism $\mathcal{T}\to\mathbf{Set}$ whence by (2) Brouwer's theorem should hold in $\mathbf{Set}$, which is not the case. - -Therefore, I think I have misinterpreted something. Is the claimed theorem (attributed to Joyal by Lambek & Scott) really that Brouwer's Theorem holds internally to $\mathcal{T}$ (i.e. as a statement in the internal language of $\mathcal{T}$)? -Or could it be that it holds externally of the functions defined in $\mathcal{T}$ or something? The latter case sounds plausible, because we can certainly show that the definable functions in extensional type theory are continuous, but we cannot do this internally for certain important reasons (I believe Beeson's Foundations of Constructive Mathematics explains why—later, Escardó and Xu have established this also for intensional type theory). -Thanks to anyone who can help demystify this for me. - -REPLY [6 votes]: Dear All: you must be precise on what you mean by Brouwer's theorem. The free topos is closed under many rules, but unlike the realizability topos, it is seldom closed under the internal implicative statement. Thus the free topos is closed under Church's rule (see the original paper by Lambek and me, Intuitionist type theory and the free topos, JPAA 19 (1980) doi:10.1016/0022-4049(80)90102-4), but obviously not the internal Church's theorem. The statement about Brouwer's principle (an original beautiful sketch of a proof, due to Joyal, but never published) is in LS, p. viii (Introduction). Here is the version you would want: using an appropriate version of the reals, if $\vdash f\colon \mathbb{R}\to \mathbb{R}$ (is provable in higher order int. logic) then $\vdash$ "$f$ is continuous". And, by the way, of course this is formalizable in the internal logic. For a reasonable proof, similar continuity rules were formalized and proved by Susumu Hayashi for second order logics and type theories in the 1970's. There are some versions of this also in Troelstra's Metamathematical Investigation of Intuitionistic Arithmetic and Analysis SLNM 344 (doi:10.1007/BFb0066739) (the "bible" of such formalizations). Phil Scott<|endoftext|> -TITLE: Is an $O(n^{d-1})$ bound known for the maximum number of edges in an ordered $n$-vertex hypergraph avoiding a fixed $d$-permutation hypergraph? -QUESTION [6 upvotes]: By a $d$-permutation hypergraph, I mean, for some fixed integer $k$, a $d$-uniform hypergraph on $[dk]$ with $k$ disjoint edges such that every edge has exactly one vertex from each of $\{1,\ldots,k\}$, $\{k+1,\ldots,2k\}$, $\ldots$, and $\{(d-1)k+1,\ldots,dk\}$. (So each vertex is contained in exactly one edge.) -By containment I mean that $G$ contains $H$ if we have an increasing injection $V(H)\to V(G)$ and an injection $E(H)\to E(G)$ that are compatible (that is, if $E$ is sent to $E'$, the vertices of $E$ are sent to (some of) the vertices of $E'$). In case it isn't clear, $n$ is allowed to vary and $d$ is fixed in this problem. -This, if true, generalizes two results of Klazar and Marcus in https://arxiv.org/abs/math/0507164; Section 2 proves the statement when $d=2$ and Section 3 proves (or nearly proves proves) the case where the containing hypergraph is restricted to be $d$-uniform. - -REPLY [3 votes]: Yes, the $O(n^{d-1})$ bound holds in this case as well, with a proof that is similar to the proof of the $d$-uniform case. -I would guess that this also follows from the proof of Klazar and Marcus with some easy modification, but I don't know it as well as our own proof, which you can find in Section 3 here: -https://arxiv.org/pdf/1408.4093.pdf. -First, make a $d$-uniform hypergraph $G'$ from $G$ by picking a set of size $d$ from each hyperedge. -The choice is arbitrary, except that we try to maximize the number of edges of $G'$. -In particular, if a hyperedge $e$ of $G$ is mapped to a hyperedge of $G'$ into which another edge of $G$ is also mapped, then all $d$-element subsets of $e$ are also in $G'$. -If this happens for a hyperedge $e$ of size at least $kd$, we are done, as we can find $H$ inside $e$. -Because of this, we can suppose that $G$ is uniform. -The bound trivially holds for $(d-1)$-uniform hypergraphs. -The rest of the proof is partly by induction, partly by repeating the argument for $d$-uniform hypergraphs. -Proof for $t$-uniform hypergrahs. -If $G$ has $n$ vertices, then we divide these vertices into intervals of length $s$, where $s$ is a constant that is large enough compared to $k,d$ and $t$. -Denote by $E_0$ the edges of $G$ that have at least two vertices in the same interval. -$|E_0|\le C_{t-1}n^{d-1}$ by induction. -(This is the only part where we use induction on $t$.) -Partition the rest of the edges of $G$ into $(\frac ns)^{t}$ blocks depending on which intervals the vertices belong to. -Denote the hyperedges of block $b$ by $E_b$. -Define $Proj_J$ as the projection that deletes the $j^{th}$ element of a hyperedge for every $j\in J$. -A block $b$ is $J$-wide if $Proj_J E_b$ contains $Proj_i H$ for every $i\in \{1,\ldots,d\}$. -If $b$ is not $J$-wide for any $J\subset\{1,\ldots, t\}$ with $|J|=t-d+1$, then we call it thin. -We will prove by induction (on $d$) and the Loomis-Whitney inequality that $|E_b|=o(s^{d-1})$ if $b$ is thin. -This will imply by contracting the intervals, the total number of edges in thin blocks is $o_s(n^{d-1})$. -The Loomis-Whitney inequality says that $|E_b|^{t-1}\le \Pi_{j} |Proj_j E_b|$. -To obtain $|E_b|=o(s^{d-1})$, it is enough to show for each $j$ that $|Proj_j E_b|=O(s^{d-2})$. -Otherwise, $Proj_j E_b$ would contain the hypergraph $Proj_i H$ for every $i$. -From this we want to show for some $J$ that $Proj_J E_b$ would also contain them. -And now, it seems that I'll have to cheat a little; this statement would be trivial if by induction we were proving instead the stronger statement that there is a $|J|=t-d$ such that $Proj_J G$ contains a copy of $H$. -But then we also have to verify this stronger statement in the earlier parts of this proof. -The estimate for $E_0$ still holds, as if we had a lot of $j^{th}$ and $(j+1)^{st}$ vertices in the same interval, that would lead to induction as before. -I don't know how to fix the argument for hyperedges that are bigger than $kd$, but luckily that can be avoided; we only need to raise the bound of $kd$ to something that also depends on $t$. -The only thing left to estimate is the number of edges in $J$-wide blocks. -Group the blocks into $I$-blockcolumns where $I\subset\{1,\ldots, t\}$ with $|I|=d-1$; every block belongs to exactly $\binom{t}{d-1}$ blockcolumns. -An $I$-blockcolumn is determined by $d-1$ intervals, and it contains every block that is defined by these $d-1$ intervals (and $t-d+1$ other intervals). -If an $I$-blockcolumn contains more than $\big((k-1)^{t-d+1}-1\big)\big(s(s-1)\cdot (s-k+1)\big)^{d(d-1)}$ $J$-wide blocks, then by the pigeonhole principle it contains $(k-1)^{t-d+1}$ blocks such that for $J=\{1,\ldots,t\}\setminus I$ the hypergraphs $Proj_J E_b$ contain the same copy of $Proj_i H$ for each $i\in\{1,\ldots,d\}$. -(These copies are all inside the $d-1$ intervals defining the $I$-blockcolumn.) -Using the pigeonhole principle again, for some $j\in J$ there are at least $k$ different intervals that contain the $j^{th}$ vertices of these blocks. -But then from any $k$ blocks that use these $k$ intervals and have the same copy of the respective $Proj_i H$, we can build a copy of $H$ in $G$. -Therefore, the number of $i$-wide blocks is small and the rest is calculation.<|endoftext|> -TITLE: Intersection number of infinite graphs -QUESTION [5 upvotes]: If ${\cal C}$ is a collection of subsets of a set $X$, we associate to ${\cal C}$ a graph $G_{\cal C} = (V,E)$ where $V = {\cal C}$ and $$E = \big\{\{A,B\}: A\neq B\in {\cal C} \land A\cap B \neq \emptyset\big\}.$$ -If $G$ is a simple, undirected graph, we define its intersection number $\iota(G)$ to be the smallest cardinal $\kappa$ such that there is a collection ${\cal C}$ of subsets of $\kappa$ such that $G_{\cal C} \cong G$. -For any cardinal $\kappa>0$ we define $\log(\kappa) = \min\{\mu\in\kappa\cup\{\kappa\}: 2^\mu \geq \kappa\}.$ -If $G=(V,E)$ is infinite, is it true that $\iota(G) \in \{\;\log(|V|), |V| \;\}$? - -REPLY [3 votes]: I think the answer is no, at least not without some additional assumptions. Suppose that $|\omega| < |\omega_1| < \mathfrak c$ and let $G$ be the union of the complete graph on $\mathfrak c$ and an independent set of size $|\omega_1|$. As you mention in the comments, the clique can be represented as an intersection graph of subsets of $\omega$, so we can represent $G$ using a ground set of size $|\omega_1| < \mathfrak c$. But we certainly need a ground set of size at least $|\omega_1| > |\omega| = \log \mathfrak c$.<|endoftext|> -TITLE: Generalisations of Weyl's construction of irreducible representations -QUESTION [8 upvotes]: For the moment we work over the complex numbers. -Suppose that $G = \mathrm{SL}(V)$, or $G = \mathrm{Sp}(V)$, or $G = \mathrm{SO}(V)$. -Weyl gave explicit constructions of irreducible representations of a given heighest weight in terms of the faithful representation $V$. This is done using Schur functors and possibly intersecting them with kernels of contractions maps. Roughly speaking, Weyl gives an algorithm to construct all irreducible representations, starting from the tautological representation. -My question is about a generalisation of such constructions to other base fields, and other `starting representations'. All representations are assumed to be algebraic. I am not exactly sure about what I am looking for. So I will give a suggestion for a definition, but it is rather clunky. Maybe there is already a crystal clear concept out there, that describes what I am looking for. -Let $K$ be a field of characteristic $0$. Let $G$ be a reductive group over $K$. Let $V$ be a representation of $G$ (defined over $K$), the `starting representation'. Now we can start building a category from $V$. -(Note, this a subcategory of $\mathrm{Rep}_{K}(G)$, but it is not by definition a full subcategory). -To this category $\mathscr{C}$ of `algorithmically constructible' representations we add: - -$V$ -tensor powers of objects in $\mathscr{C}$ -sums of objects in $\mathscr{C}$ -duals of objects in $\mathscr{C}$ -contraction maps $W^{\otimes n} \to W^{\otimes m}$, for $W \in \mathscr{C}$ -morphisms that come from the action of $\mathfrak{S}_{n}$ -on $W^{\otimes n}$, for $W \in \mathscr{C}$ -kernels and cokernels of morphisms in $\mathscr{C}$ - -Maybe there are other natural algorithmic steps that should be added to this list, but I have not thought of any. -These steps allow us to define Schur functors, -so we find $\mathbb{S}^{\lambda}(V) \in \mathscr{C}$. -It is a well-known fact that if $V$ is faithful, then every representation of $G$ is a subobject of some object in $\mathscr{C}$, because $G$ is reductive. However, $\mathscr{C}$ is not closed under taking subobjects (in particular it need not be Tannakian). -I am not per se interested in knowing which algorithm would construct a given representation $W$, but rather in whether such an algorithm exists. -Thus, let us say that $V$ is an `algorithmically generating' representation if the inclusion functor $\mathscr{C} \hookrightarrow \mathrm{Rep}_{K}(G)$ is essentially surjective. -Now we get to my questions. In answering these questions, feel free to make additional assumptions on $G$. (E.g., semisimple, or absolutely simple.) - - -Does every group admit an `algorithmically generating' representation? - - -My other question is more specific. Weyl looked at the classical groups, but (as far as I know) he omitted the spin groups. Over a general field, the spin representation need not be defined over $K$, but the (even) Clifford algebra is. Therefore: - - -Let $(V,q)$ be a quadratic space - (i.e., a $K$-vector space $V$ with a nondegenerate quadratic form $q$). - Is the spin representation of $\mathrm{Spin}(V,q)$ - on the even Clifford algebra $\mathrm{Cl}^{+}(V,q)$ - an `algorithmically generating' representation? - - -Finally: - - -Is there anything that I am forgetting in my definition - of the category $\mathcal{C}$ of `algorithmically constructible' representations? - Is there a known concept that I am overlooking? - -REPLY [6 votes]: The short answers to your questions are all No. -The reason is that Weyl's construction works when the centraliser algebra -of a tensor power of $V$ is a quotient of a Brauer algebra. This is similar -to Schur's construction of representations of $SL(n)$ which assumes that the centraliser algebra is a quotient of the group algebra of the symmetric group. -In general, if $V$ is self-dual, then there is a homomorphism from the Brauer -algebra to the centraliser algebra. The constructions you describe are equivalent to taking images of (primitive) idempotents in the Brauer algebra. -For other representations the homomorphism is not surjective and there will be images which are reducible.<|endoftext|> -TITLE: Euler product approximation of the Riemann zeta function -QUESTION [12 upvotes]: It is well-known that the Riemann zeta function can be approximated in the critical strip by a finite Dirichlet polynomial. More precisely, -$$ -\zeta(\sigma+it) = \sum_{n \leq X} n^{-\sigma-it} + \frac{X^{1-\sigma-it}}{\sigma+it-1} + O ( X^{-\sigma}) -$$ -uniformly for all $X$ satisfying $2 \pi X / C \geq |t|$, where $C$ is a fixed constant greater than 1. (This is Theorem 4.11 in Titchmarsh's book). In particular this holds for $\sigma = 1/2$. -Question: is there a similar (simple) approximation of zeta by a finite Euler product? - -REPLY [9 votes]: See Finite Euler products and the Riemann Hypothesis by S. M. Gonek (2007): - -We show that if the Riemann Hypothesis is true, then in a region - containing most of the right-half of the critical strip, the Riemann - zeta-function $\zeta$ is well approximated by short $X$-term truncations $\zeta_X$ of its Euler - product. Conversely, if the approximation by products is good in this - region, the zeta-function has at most finitely many zeros in it. - - -This approximation also provides insight into why the zeroes on the critical line "repel", in the sense of level repulsion in the Gaussian Unitary Ensemble.<|endoftext|> -TITLE: Induced group action of the left regular representation strongly continuous -QUESTION [5 upvotes]: Let $G$ be a compact group and let $\lambda: G \rightarrow \mathcal{U}(L^2(G))$ be the left regular representation, i.e. $\lambda_sf(t)=f(s^{-1}t)$. Why is the induced group action $\overline{\lambda}:G \rightarrow \mathrm{Aut}(\mathcal{K}(L^2(G)))$, $\overline{\lambda}_s(T)= \lambda_s T\lambda_{s^{-1}}$ strongly continuous? -Thank you. - -REPLY [2 votes]: Let $g_j\to g$ a convergent net in $G$ and let $T\in\mathcal K$. We need to show that $\lambda_{g_j}T\lambda_{g_j^{-1}}$ converges to $\lambda_g T\lambda_{g^{-1}}$ in the strong topology. Now $\lambda_{g_j}$ converges to $\lambda_g$ strongly, i.e., $\lambda_{g_j}f$ tends to $\lambda_gf$ in the $L^2$ norm. This is seen for continuous $f$ by uniform convergence and for general $f$ by $L^2$-approximation with continuous functions. Now write $$ -\lambda_{g_j}T\lambda_{g_j^{-1}}f-\lambda_{g}T\lambda_{g^{-1}}f -=(\lambda_{g_j}T\lambda_{g_j^{-1}}f-\lambda_{g_j}T\lambda_{g^{-1}}f) -+(\lambda_{g_j}T\lambda_{g^{-1}}f-\lambda_{g}T\lambda_{g^{-1}}f) -$$ -and use the triangle inequality as well as the fact that $\lambda_g$ preserves norms.<|endoftext|> -TITLE: Simultaneous Diophantine approximation of $\sqrt{2}$ and $\sqrt{2\pm \sqrt{3}}$ -QUESTION [10 upvotes]: By using the LLL algorithm, I tried to find the best simultaneous Diophantine approximation of the three numbers $\sqrt{2} $ and $ \sqrt{2 \pm \sqrt{3}} $. I was expecting that to get a precision of $\epsilon$, the common denominator $q$ should be on the order of $\epsilon^{-3}$, because I have three irrational numbers. This is based on a well-known theorem by Dirichlet. -However, it is actually $\epsilon^{-2}$. -The point is that, the extra $\sqrt{2}$ does not make the task harder. -I used the LatticeRudection in Mathematica to find the common q. It turns out that the $q$'s in the triple case coincide with the double case of $\{ \sqrt{2\pm \sqrt{3}} \}$ in many cases! -Can anyone explain this? -If I replace $\sqrt{2}$ by $\sqrt{3}$, it is indeed $\epsilon^{-3} $. -If I do not use $\sqrt{2 \pm \sqrt{3}}$, but the square roots of the prime numbers, indeed the Dirichlet expectation is right. - -REPLY [26 votes]: $$\sqrt{2+\sqrt{3}}-\sqrt{2- \sqrt{3}}=\sqrt{2}$$<|endoftext|> -TITLE: Non-trivial examples of Stably diffeomorphic 4-manifolds -QUESTION [8 upvotes]: I am looking for some non-trivial examples of (smooth) 4-mflds $M,N$ such that $M$ and $N$ are STABLY diffeomorphic. I.e. $$M\sharp_n (S^2\times S^2) \cong N \sharp_r (S^2\times S^2)$$ for $r,n$ not necessarily the same - -By non trivial I mean that $M$ and $N$ are not diffeomorphic or that $M \cong N\sharp S^ 2 \times S^2$ or similar situation to these two. -Clearly the two manifolds have to have the same fundamental group and the same signature at least. What I am looking for is a concrete/explicit example. -By results of Gompf we know that if $M$ is orientable carries two non-equivalent smooth structures ($M_1,M_2$) then $M_1$ and $M_2$ are stably diffeomorphic (but clearly non-diffeomorphic). I would like to have a less exotic example as main example in mind when speaking about stable diffeomorphism. -I found abstract criterions for certain families of manifolds but I'm unable to cook up an example. - -REPLY [2 votes]: Since the examples given above are smooth, simply connected, and homotopy equivalent, they are also homeomorphic. You might like the following examples. Let $L$ and $L'$ be 3-dimensional lens spaces that are homotopy equivalent but not homeomorphic. Then $L \times S^1$ and $L' \times S^1$ are simple homotopy equivalent and stably diffeomorphic, but not homeomorphic.<|endoftext|> -TITLE: Simultaneous diagonalization of self-adjoint operators on Hilbert space -QUESTION [7 upvotes]: Suppose I'm given a finite set of possibly unbounded commuting self-adjoint operators $T_i : \mathfrak H \supset \mathscr D(T_i)\to \mathfrak H, i = 1 , \dots , N$ on a Hilbert space (in the sense of commuting resolvents $\forall i, j: \exists z_i \in \rho (T_i) , z_j \in \rho (T_j):[R_{T_i} (z_i ) , R_{T_j} (z_j)] = [R_{T_i} (\overline{z_i} ) , R_{T_j} (z_j)] = 0$). In linear algebra this implies that these operators can be simultaneously diagonalized. -Is there any general rigorous notion of simultaneous diagonalization in this functional analysis setting? Intuitively I would expect something like a projection-valued measure -$$E: \mathscr B(\mathbb R^N ) \to \mathbb C^N \otimes \mathfrak L(\mathfrak H)$$ -supported on $\sigma (T_1) \times \dots \times \sigma (T_N)$, such that $E^{(T_i)} (A) = E(\mathbb R \times \dots \times \mathbb R \times \underbrace{A}_{i} \times \mathbb R \times \dots \times \mathbb R )$. -Or maybe a guarantee that there is a common measure space $(\Omega , \mu )$ and a common unitary transformation $U : \mathfrak H \to L^2(\Omega, \mathrm d\mu )$ such that -$$\forall i : \exists f_i \in \text{Meas}(\Omega, \mathrm d\mu): T_i = U^* M_{f_i} U$$ -Is there anything like that? - -REPLY [3 votes]: There is a very nice treatment of this question in the textbook by Konrad Schmüdgen "Unbounded Self-adjoint Operators on Hilbert Space", see in particular Theorem 5.23 p. 103. -While there is no problem constructing a unique product measure $\mu\otimes\nu$ for $\sigma$-finite measure spaces $(X,\mathcal{M},\mu)$ and $(Y,\mathcal{N},\nu)$ the analogous question for projection-valued measures encounters a snag unless $X$ and $Y$ are sufficiently nice like $\mathbb{R}^n$. See the remark after Theorem 4.10 in the same book and the reference to the article by Birman, Vershik and Solomjak.<|endoftext|> -TITLE: A question with simple and indecomposable modules -QUESTION [5 upvotes]: Assume $M$ is both noetherian and artinian and fix $S_0\subseteq M$ a simple submodule. How to prove that $S_0$ is contained in some indecomposable direct summand of $M$? - -REPLY [8 votes]: As shown by Jeremy Rickard's answer, $S := S_0$ is usually not contained in an indecomposable direct summand. The purpose of this answer is to show the weaker statement - -$S$ can be embedded into an indecomposable direct summand of $M$. - -Proof: WLOG assume $S \neq 0$. Since $M$ is artinian, it is a direct sum of indecomposable submodules $M_1,..., M_m$. Let $n \le m$ be minimal such that there is an embedding (i.e. an injective hom. of modules) $S \hookrightarrow \oplus_{i=1}^n M_i$. If $n=1$ we are done. If $n > 1$ consider the composition $$S \hookrightarrow \bigoplus_{i=1}^n M_i \twoheadrightarrow \bigoplus_{i=1}^{n-1}M_i$$ -If it's kernel is zero, $S$ embedds into $\oplus_{i=1}^{n-1}M_i$, in contradiction to the minimality of $n$. Hence, the kernel is non-zero -and by simplicity of $S$, it's $S$, i.e. the composition is the zero map. Hence -$$\text{im}(S \hookrightarrow \bigoplus_{i=1}^n M_i) \subseteq \ker(\bigoplus_{i=1}^n M_i \twoheadrightarrow \bigoplus_{i=1}^{n-1}M_i) = M_n$$ -Thus the composition $S \hookrightarrow \bigoplus_{i=1}^n M_i \twoheadrightarrow M_n$ is injective. QED. - -Edit: Simpler proof: From $M=\oplus_{i=1}^m M_i$ we have $Hom(S,M)\cong \oplus_i Hom(S,M_i)$ and since the LHS is non-zero (it has the inclusion map), there is $i$ such that $Hom(S,M_i) \neq 0$. Let $0 \neq f: S \to M_i$ be a hom. By simplicity of $S$ we conclude $\ker f = 0$, i.e. $f$ is an embedding.<|endoftext|> -TITLE: Why does this Moiré pattern look this way? -QUESTION [12 upvotes]: I was making some gifs of Mobius transformations in Matlab, and some strange patterns began to appear. I'm not sure if a deeper knowledge of the filetype/algorithm is needed to understand this phenomenon, but I thought that there could perhaps be a purely mathematical explanation. The image is obtained by coloring the complex plane like a checkerboard, and then inverting it by taking the reciprocal of the complex conjugate. Here is the math psuedocode for the image with a given zoom $k$: -$\mbox{checkerboard}:\mathbb C \to\{\mbox{black},\mbox{white}\}$ -$\mbox{checkerboard}(z):=\begin{cases} -\mbox{black} & \mbox{if }\lfloor\Im(z)\rfloor+\lfloor\Re(z)\rfloor\equiv 0\mod 2\\ -\mbox{white} & \mbox{if }\lfloor\Im(z)\rfloor+\lfloor\Re(z)\rfloor\equiv 1\mod 2 -\end{cases}$ -$\mbox{image} = \{z\in\mathbb C:|\Re(z)|,|\Im(z)|\leq 1\}$ -$\mbox{color}:\mbox{image}\to\{\mbox{black},\mbox{white}\}$ -$\mbox{color}(z):=\mbox{checkerboard}(k/\overline{z})$ -And here are the pictures for $k=1$, $k=50$, and $k=200$. The resolution of each picture is 1000x1000. - - - -EDIT: More specifically, why does the Moiré pattern 'sync up' with the resolution of the picture at certain points? Can the Moiré pattern be predicted? -EDIT (Partial answer): I posted this question on image processing stack exchange, and I got a decent answer for why the pattern syncs up at certain points. I would, however, love a more detailed mathematical explanation of why the pattern seems to behave differently at different such points. Here is a gif I made illustrating the interesting stuff going on when you zoom in: https://media.giphy.com/media/3og0IwUINwEQAYoUDK/source.gif - -REPLY [4 votes]: EXTENDED COMMENT -Can you specify what "sync up" means here? All of your drawings are chaotic near the origin since they are trying to pack an infinite amounts of information (the checkerboard) into a very small amount of Euclidean space (the pixels). The best you can do is approximate the Möbius transform (which has become rather chaotic) with the nearest integer value. -the Arnold Cat map is chaotic however if you have a finite amount of pixels it will become periodic with a period related to the Fibonacci numbers, as outlined in a paper of Curtis McMullen. -i think it's clear any good answer involves the word "aliasing" where if we undersample on wave can look like another. spectral theory in hyperbolic space is challenging but maybe there is a good picture here and a quantitative discussion -One might check for correlations with basis of Haar Wavelet, Discrete Fourier Transform or Walsh-Hadamard Transform. The basis is exactly the checkerboards of various sizes and scales. These inversion transforms behave like affine maps on a "curved" space. And we can try to work that out. - -Some of the splotches or "aliasing" I am seeing look like hyperbolic hexagons or octagons. - -So a reasonable goal would be to quantify where and how much the inverted checkerboard correlates with these patterns. Or how quickly they are converging to absolute chaos. - -REPLY [2 votes]: (The following is not meant to be a full answer, but a few clues towards one, elaborating what I wrote in a comment. It's a bit too long for a comment.) -There are two things that overlap: one is the grid of square pixels, or more precisely, the "sample grid", namely the points where the "color" function has been evaluated (perhaps at the center of the square pixels, perhaps on one corner, it doesn't really matter, but certainly only at one point per pixel, that is, crucially, no anti-aliasing has been applied). The other is the pattern grid, defined by your function, and which is a conformal transformation (namely, a Möbius transformation) of a regular square grid; being a conformal transformation of a regular square grid, it looks locally like a checkerboard of squares, whose size and orientation are determined by the complex derivative of the conformal transformation applied (that is, of the Möbius tansformation). -Now to better understand what is going on, make two simplifying hypotheses: (A) that the pattern grid is, in fact, a simple square grid (colored in checkerboard pattern), not just locally like one, and (B) while we're at it, that we're only in $1$ dimension. In other words, you have an alternation of white and black intervals of equal length $\ell$ and you evaluate it at a given sample distance $d$. Clearly, if $d$ is equal to $2\ell$, the samples will systematically miss one color and "resonate" with the color, so you get a uniform color; if $d$ is only very close to $2\ell$, you get large intervals of one and the other color with length something like $d\cdot\ell/|d-2\ell|$ (an easy computation if you make a drawing — which I may or may not have gotten right, but it should look like this): it's perhaps clearer to say that the spatial frequency $1/\ell$ of the intervals is shifted by $2/d$, a phenomenon sometimes known as the Nyquist frequency. -Now if we let go of simplifying hypotheses (A) and (B), the places where you see a clear pattern patch emerge are those for which the complex derivative of the conformal transformation applied is such that the square grid which locally coincides with your pattern correctly matches up with the sample grid. One obvious possibility is that the sides of the pattern squares align with the axes of the sample grid, which happens for four different angles, and in each case for just the right size (the edge of the squares being one half the sample edge size), hence essentially in $8$ places arranged in a regular octagon since the derivative of a Möbius transformation is a degree $2$ map; another obvious one is that the diagonals of the pattern squares align with the axes of the sample grid, which also happens for four different angles (and this time the edge of the squares should be $1/\sqrt{2}$ times the sample edge size), so in another $8$ places in another octagon whose axes are shifted $\pi/8$ with respect to the other one and $\sqrt[4]{2}$ times larger. This "explains" the most prominent $16$ patches and why they form two regular octagons (and even the ratio of their sizes). -I don't think I can provide an explanation as to the color at the center of the patches, however, and I'm not sure there's much to be said on this subject.<|endoftext|> -TITLE: Inductive Definitions in Category Theory -QUESTION [6 upvotes]: I'm trying to pin down a notion of inductive definability in category-theoretic terms. -The sorts of inductively defined sets (and classes) I'm most interested in are those that admit of induction and recursion. So far, I've noticed that (weakly?) initial algebras of an endofunctor do a pretty good job of this. -We can capture natural number objects (and therefore $\mathbb{N}$) and other inductive types this way. But what about things like the constructive ordinals (or number classes) or even segments of the cumulative hierarchy? -I want to exclude things like the classical continuum, but not through cardinality, since the cumulative hierarchy is certainly quite large. And polynomial functors don't seem to cut it since the powerset functor isn't polynomial and doesn't have a smallest fixed point (and thus not an initial algebra). Should I restrict the powerset functor a certain way to get one? Based on inaccessible cardinals? -Is there a unified characterization of 'inductive' in category theory that would apply to these sorts of cases? - -REPLY [6 votes]: It is indeed the case that initial algebras for functors are the category-theoretic manifestation of inductive definitions. There's a whole industry surrounding this idea. -You are more specifically asking about (pieces of) the cumulative hierarchy. On that topic you should look at algebraic set theory (AST). A good overview of the subject was written up by Benno van den Berg and Ieke Moerdijk in arXiv:0710.3066. Do not be put off by all the talk about constructive set theories. They're just generalizing the ordinary (classical) set theory to natural cateogry-theoretic settings where excluded middle does not hold. -In AST we study categories of classes. Such categories have a notion of small map. The intuition is that a map between classes is "small" if the inverse image of every point is a set (rather than a class). This is a generalisation of the distinction "classes are large, sets are small" because a class $C$ is a set, and only if, the unique map $C \to 1$ is small. Smallness is axiomatised by some fairly straightforward axioms. -Next, to get the cumulative hierarchy, we study ZF-algebras. A ZF-algebra is a (large) poset $(P, {\leq})$ which is closed under small suprema and is equipped with a unary operation $s : P \to P$. The initial ZF-algebra $(V, {\sqsubseteq}, s)$ can be seen as the cumulative hierarchy, where $\sqsubseteq$ corresponds to subset inclusion and $s$ to the singleton map $x \mapsto \{x\}$. See section 4 of the van den Berg & Moerdijk's notes. -It should be possible to cook up chunks of $V$ by taking variations on a theme. For instance, $V_\omega$ ought to arise if we require closure under finite suprema (I have not checked this.) -By the way, we can get $V$ as the initial algebra of the small powerset functor in the category of classes. Let $\mathcal{C}$ be the (large) category of all classes. Define $P : \mathcal{C} \to \mathcal{C}$ by $P(X) = \{S \mid S \subseteq X\}$, i.e., $P(X)$ is the class of all subsets of $X$. Then $V$ is the initial $P$-algebra, isn't it?<|endoftext|> -TITLE: Embedding Riemann surfaces into $\mathbb P^2$ -QUESTION [6 upvotes]: Suppose I am given a Riemann surface $\Sigma_g$ of genus $g$. What is known about the sufficient and necessary conditions needed on $\Sigma_g$ to have an embedding into $\mathbb P^2$? -If $\mathcal T_g$ is the Teichmuller space of genus $g$ Riemann surfaces, is there any way to intrinsically describe the subspace of plane curves? - -REPLY [5 votes]: If I am not mistaken, for the exceptional values of the genus, the locus of degree $d$ plane curves has (complex) dimension -$$ -\frac{(d+2)(d+1)}{2}-9 -$$ -sitting inside a moduli space of dimension -$$ -\frac{3(d-1)(d-2)}{2}-3 -$$ -so it becomes relatively small as $d$ increases. -This article by Landesman seems to be pertinent for your question.<|endoftext|> -TITLE: dyadically recursive matrices: Part I -QUESTION [5 upvotes]: Introduce the $2^{n-1}\times 2^{n-1}$ matrix $B_n$ recursively as follows: $B_1(b_1)=\begin{pmatrix} b_1\end{pmatrix}$ and -$$B_n(b_1,\dots,b_n)=\begin{pmatrix} B_{n-1}(b_1,\dots,b_{n-1})& b_nJ_{n-1}\\ b_nJ_{n-1}&B_{n-1}(b_1,\dots,b_{n-1}) -\end{pmatrix}.$$ -Here $J_n$ is a $2^{n-1}\times 2^{n-1}$ matrix with $1$'s on the antidiagonal and zeros elsewhere. -Example. For $n=2$ and $n=3$, we have -$$B_2(b_1,b_2)=\begin{pmatrix} b_1&b_2\\b_2&b_1\end{pmatrix} -\qquad\text{and} \qquad -B_3(b_1,b_2,b_3)=\begin{pmatrix} b_1&b_2&0&b_3\\b_2&b_1&b_3&0\\0&b_3&b_1&b_2\\b_3&0&b_2&b_1\end{pmatrix}.$$ - -Question. Is there a closed (or interesting) formula for the determinant $\det(B_n)$? - -REPLY [5 votes]: For a vector $v$ of length $2^{n-2}$, denote by $v'$ the "mirrored" vector $J_{n-1}v$. -If $v$ is an eigenvector of $B_{n-1}$ for the eigenvalue $\lambda$, then $B_n\binom v{\pm v'}=(\lambda\pm b_n)\binom v{\pm v'}$. So this gives you all the eigenvalues of $B_n$ as the sums $ b_1\pm\cdots\pm b_n$ (note that $ b_1$ never has a "$-$" sign) and the eigenvectors as a certain subset (respecting the dyadically recursive structure) of the set of all vectors $ (1,\pm1,...,\pm 1)$ where an even number of minus signs is used. -The determinant is: $\det(B_n)=\prod(b_1\pm b_2\pm\cdots\pm b_n)$. - -REPLY [4 votes]: Yes and No. On the one hand, you can prove by a simple induction that $B_n$ commutes with $J_n$. Now, $B_n$ appears to be given blockwise, where all the blocks commute with each other. This allows you to write that the determinant of $B_n$ is the ``determinant of determinants'' (this is false for non-commuting blocks): -$$\det B_n=(\det B_{n-1})^2-(\det b_nJ_{n-1})^2=(\det B_{n-1})^2-b_n^{2^{n-1}}.$$ -On the other hand, it seems difficult to exploit this relation to give a closed form for $\det B_n$. Recall that even the iteration $z\mapsto z^2-a$ has outstanding complexity, which yields Julia and Fatou sets.<|endoftext|> -TITLE: Solution to a Diophantine equation -QUESTION [7 upvotes]: Find all the non-trivial integer solutions to the equation -$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4.$$ - -REPLY [2 votes]: There is one idea. To search for the solution of the equation. -$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=q$$ -If we know any solution $(a,b,c)$ of this equation. Then it is possible to find another $(a_2, b_2, c_2)$. Make such a change. -$$y=(a+b+2c)(q(a+b)-c)-(a+b)^2-(a+c)(b+c)$$ -$$z=(a+2b+c)(2b-qa-(q-1)c)+(b+2a+c)(2a-qb-(q-1)c)$$ -Then the following solution can be found by the formula. -$$a_2=((5-4q)c-(q-2)(3b+a))y^3+((5-4q)b+4(1-q)c)zy^2+$$ -$$+(3c+(q-1)(a-b))yz^2-az^3$$ -$$b_2=((5-4q)c-(q-2)(3a+b))y^3+((5-4q)a+4(1-q)c)zy^2+$$ -$$+(3c+(q-1)(b-a))yz^2-bz^3$$ -$$c_2=2(q-2)cy^3+3(2-q)(a+b)zy^2+$$ -$$+((5-4q)(a+b)+2(1-q)c)yz^2+(2c-(q-1)(a+b))z^3$$ -I tried this formula to simplify, but nothing happens. Maybe someone will check in Maple?<|endoftext|> -TITLE: Homology $H_*(TOP, \mathbb{Z}_2)$ of the stable homeomorphism space -QUESTION [7 upvotes]: Let $TOP$ be the stable homeomorphism space, with $TOP(n) = Homeomorphisms(\mathbb{R}^n)$. What is known about its $\mathbb{Z}_2$-homology $H_*(TOP, \mathbb{Z}_2)$? In particular I am interested in the map $H_*(TOP, \mathbb{Z}_2) \rightarrow H_*(G, \mathbb{Z}_2)$, where $G$ is the stable space of homotopy automorphisms of spheres. -Can $H_*(TOP, \mathbb{Z}_2)$ be computed knowing $H_*(G, \mathbb{Z}_2)$, $H_*(G/TOP, \mathbb{Z}_2)$ or $H_*(BTOP, \mathbb{Z}_2)$? -My specific question: -Is the map $H_{4k+2}(TOP, \mathbb{Z}_2) \rightarrow H_{4k+2}(G, \mathbb{Z}_2)$ injective on the image of $\pi_{4k+2}(TOP)$ under the Hurewicz homomorphism? So for example, is it injective on primitives? Does somebody have an idea or a reference I could look at? -Even more specific (and also sufficient): Let $\kappa \in \pi_{14}^s = \pi_{14}G$ be the element with Kervaire invariant 0. It lifts to $\pi_{14}(TOP)$. Is its image under $h : \pi_{14}(TOP) \rightarrow H_{14}(TOP, \mathbb Z_2)$ zero? (By Lemma 4.3 of arxiv.org/abs/1504.06752v2, $h(\kappa)=0$ in $H_{14}G$.) - -REPLY [5 votes]: I first comment on the specific case on the kernel of $H_{14}TOP\to H_{14}G$ when restricted to the image of Hurewicz homomorphism $\pi_{14}TOP\to H_{14}TOP$ (all homology groups are $\mathbb{Z}/2$-homology). I think it is a consequence of Sullivan's decomposition that at the prime $2$ the space $G/TOP$ decomposes as a product of Eilenberg-Moore spaces $K(\mathbb{Z}/2,4n-2)$ and $K(\mathbb{Z},4n)$ (for instance see Madsen's paper projecteuclid.org/download/pdf_1/euclid.pjm/1102868638). -Serre's exact sequence of homotopy groups for the fibration -$$\Omega(G/TOP)\to TOP\to G\to G/TOP\to BTOP\to BG$$ -yields -$$\pi_{14}\Omega(G/TOP)\to\pi_{14}TOP\to\pi_{14}G\to\pi_{14}G/TOP$$ -which reads as -$$0\to\pi_{14}TOP\to\mathbb{Z}/2\{\sigma^2,\kappa\}\to\mathbb{Z}/2.$$ -For the Hurewicz map $h:\pi_{14}G\to H_{14}G$ it is known that $h(\sigma^2)\neq 0$ whereas $h(\kappa)=0$. -It is known that there is a Kervaire invariant one element if its Hurewicz image maps nontrivially under $H_{2^i-2}G\to H_{2^i-2}G/TOP$. As $\sigma^2$ is a Kervaire invariant one element, we then deduce that $\sigma^2$ maps nontrivially under $\pi_{14}G\to\pi_{14}G/TOP$. Therefore, it is $\kappa$ which is detected by the map $TOP\to G$. So, $h(\kappa)=0$ only tells you that -$$\mathbb{Z}/2\{\kappa\}\simeq\pi_{14}TOP\to H_{14}TOP\to H_{14}G$$ -is trivial. Now, I suspect that $\kappa$ maps trivially under -$$\pi_{14}TOP\to H_{14}TOP$$ -which is what you asked for. -EDIT(added on 28th of March) I suggest the following route to prove that $\kappa$ maps trivially under $\pi_{14}TOP\to H_{14}TOP$. Note that here $\kappa\in\pi_{14}TOP$ is any element which maps to $\kappa\in\pi_{14}G$ where we have some abuse of notation here. Also, note that $\kappa$ really lives in $\pi_{14}SG$ where $SG=Q_1S^0$ which is homotopy equivalent to $Q_0S^0$. Furthermore, note that the Hopf invariant one element $\nu\in\pi_3^s\simeq\pi_3G$ pulls back to $\pi_3TOP$ and I think we can show it is a unique pull back (this uniqueness helps in showing that a triple Toda bracket in $\pi_*TOP$ is defined). Now, following arguments of Lemma 4.3 of https://arxiv.org/pdf/1504.06752v2.pdf we may construct $\kappa$ as an unstable map as a triple Toda bracket associated to -$$S^{13}\stackrel{\beta}{\longrightarrow}\Gamma^6(\Sigma^4K)\stackrel{\Gamma^6\alpha}{\longrightarrow}\Gamma^6 S^3\stackrel{\nu}{\longrightarrow}Q_0S^0$$ -where $\Gamma^6=\Omega^6\Sigma^6$. The epimorphism $\pi_3TOP\to\pi_3G\simeq\pi_3SG$ then allows to construct a triple Toda bracket for an element in $\pi_{14}TOP$ which maps to $\kappa$ with trivial indeterminacy, hence representing a generator of $\pi_{14}TOP$. A composition of the form -$$S^{14}\stackrel{\beta^{\flat}}{\longrightarrow}C_{\Gamma^6\alpha}\stackrel{\nu_\sharp}{\longrightarrow}TOP$$ -represents the element constructed as this triple Toda bracket. Now, by the same arguments in Lemma 4.3 https://arxiv.org/pdf/1504.06752v2.pdf we have $(\beta^\flat)_*=0$ which shows that the element $\kappa\in\pi_{14}TOP$ acts trivially in homology which is the desired result.<|endoftext|> -TITLE: Two ways to look at a double cover of the projective line -QUESTION [5 upvotes]: Let $f:L\rightarrow \mathbb P^1_{\mathbb C}$ be the line bundle associated to the invertible sheaf $\mathcal O_{\mathbb P^1}(2)$, $\phi=(X_0-X_1)^3X_0\in H^0(\mathbb P^1, \mathcal O_{\mathbb P^1}(4))$ and $C\subset L$ defined by the equation $T^2=f^*\phi$ where $T$ is the tautological section of $f^*\mathcal O_{\mathbb P^1}(2)$. -The curve $C$ is a singular rational curve, a double cover of the projective line ramified at $P=[1:1]$ and $Q=[0:1]$. As $Q$ seen in $C$ is a smooth point of $C$, if I am not mistaken, using $\mathcal O_C(3Q)$, we can embed $C$ in $\mathbb P^2=|\mathcal O_C(3Q)|$ as a cuspidal cubic curve. -Then given two smooth distinct points $P,P'$ of $C$, in the embedding in $\mathbb P^2$, we can consider the line $\ell\subset \mathbb P^2$ determined by $(P,P')$ and associate to this pair a third point (residual in the intersection $C\cap \ell$ in $\mathbb P^2$) of $C$. -Are there some interpretations of this line and this operation (associating a third point) related directly to $f:L\rightarrow \mathbb P^1$ and the definition of $C\subset L$? - -REPLY [6 votes]: The smooth points on a cuspidal cubic curve in $\mathbb P^2$ can be given the structure of an algebraic group, just as is done for smooth cubic curves. If the tangent directions of the cuspidal point are defined over your ground field $K$, then you get a copy of $\mathbb G_m$. If not, the tangent directions generate a quadratic extension $L/K$, and the group you get is the $L/K$ twist of $\mathbb G_m$. See for example, The Arithmetic of Elliptic Curves, Proposition III.2.5 for the split case. As usual, the group law is determined by the condition that three points add to 0 if and only if they are colinear.<|endoftext|> -TITLE: A symplectic form on a symplectic vector bundle -QUESTION [6 upvotes]: Suppose $E \to B$ is a symplectic vector bundle, i.e. it possesses a fibrewise linear symplectic form $\omega_F$. Further, suppose $\omega_B$ is a symplectic form on $B$. -Question: is there a symplectic form $\omega$ on $E$, such that (a) on any fibre of $E$, it restricts to $\omega_F$, and (b) on the zero section it restricts to $\omega_B$? -Remark: A closed 2-form can be obtained on $E$ that satisfies the first condition -- this can be done by choosing a connection on the associated principal bundle, and using Weinstein's theorem (Theorem 6.17 in McDuff-Salamon). - -REPLY [9 votes]: Such a symplectic form does not exist in general. In my paper (theorem 8.2). Gerbes, 2-gerbes and symplectic fibrations, I have shown that the obstruction of the existence of such a symplectic form on $E$ can geometrically be represented by a $2$-gerbes. My paper is available at -http://lanl.arxiv.org/pdf/math/0504274 -Tsemo Aristide -Gerbes 2-gerbes and symplectic fibrations. -Rocky Mountain J. Math. Volume 38, Number 3 (2008), 727-777.<|endoftext|> -TITLE: Shortest path connecting two opposite points on a cube -QUESTION [38 upvotes]: Is it true, that a path connecting two opposite points (i.e. such that the segment joining them passes through the centre of mass of the cube) on the surface of the $d$-dimensional unit cube (with $d>1$) is not shorter than $2$? - -REPLY [4 votes]: The impressive and beautiful @AntonPetrunin solution is external. Let me present an internal argument. The goal is to prove that the Euclidean length $\ |p|\ $ of any (rectifiable) path $\ p:[-\frac 12;\frac 12]\rightarrow F^{d-1}\ $ is at least $\ 2,\ $ where - $\ p(-\frac 12) = -p(\frac 12),\ $ and $\ F^{d-1}\ $ is the boundary of -$\ [-\frac 12;\frac 12]^{\,d},\ $ for every integer $\ d\ge2.$ (My solution will provide a little more). -Let $\ p\ :=\ \triangle_{k=1}^d\, p_k, $ where $\ p_1\ \ldots\ p_d\ $ are the coordinate functions. Define the set of the essential coordinates: -$$ D\ :=\ \{\, k : |\max p_k| = 1\, \} $$ -and let $\ \delta:=|D|\ $ be the essential dimension; we see that $\ \delta\ge 2.\ $ Finally, define the essential projection -$$ q\ :=\ \triangle_{k\in D}\, p_k $$ -The range of the essential path $\ q\ $ is contained in $\ F^{\delta-1},\ $ where $\ F^{\delta-1}\ $ is defined as the boundary of cube - $\ [-\frac 12;\frac 12]^D.$ Of course $\ q(-\frac 12) = -q(\frac 12). $ Since -$$ |p|\ge |q| $$ -it is enough to prove that $\ |q|\ge 2$. Observe that -$\forall_{k\in D}\, var(p_k)\ge 1 $ hence -$$ \sum_{k\in D}\, var(p_k)\ \ge D $$ -Let us also consider the $\ell^1$-norm in Euclidean spaces, namely -$\ ||x|| = \sum |x_k|.\ $ Let $\ ||q||\ $ be the -$\ell^1$-length of $\ q.\ $ We see that -$$ ||q||\ =\ \sum_{k\in D}\, var(p_k)\ \ge D $$ -Let's apply the comparison between the two norms in any $n$-dimensional euclidean space: $\ |x|\ge \frac{||x||}{\sqrt{n}}\ $ for arbitrary -$\ n=1\ 2\ \ldots\ $. It follows that -$$\forall_{k\in D}\,\forall_{x\ y\in F^{\delta-1}: p_k(x)=p_k(y)}\quad -|x-y|\ \ge \frac{||x-y||}{\sqrt{\delta-1}} $$ -and $\ |q|\ \ge\ \frac{||q||}{\sqrt{\delta-1}} $ hence - $\ |p|\ge|q| \ge \frac D{\sqrt{\delta-1}}.\ $ Thus, -THEOREM   - $\ \forall_{d\ge 2}\ |p|\ \ge\ \frac D{\sqrt{\delta-1}} $ -  -COROLLARY   - $\ \forall_{d\ge 2}\ |p|\ \ge\ 2 $ - -REMARK   When considering above implication Thm => Cor, - only $\ D=3\ $ makes you stop for a moment (unless you are a relative of John von Neumann).<|endoftext|> -TITLE: Representation theory and elementary particles -QUESTION [35 upvotes]: I have been looking for a clear expository mathematical text on the relation between the theory of elementary particles and the representation theory of $U(1), SU(2), SU(3)$, I was very disappointed about The step which explains why an elementary particle should correspond to an irreducible representation! -I asked physicists to explain The step and I get more confused then before asking. I was wondering if someone could explain this passage with mathematical rigor and physical principles! And pretty sure I'm not the only one in this situation :) -Thanks. -Edit Thank you very much all for your answers, unfortunately I can't choose one! It was great with lot of gratitude to all of you, still I do not understand the key step probably I need to learn more physics to feel what is really happening. - -REPLY [4 votes]: The answers given so far are all beautiful, but lack one detail which I think should be clarified. I'll try to answer the question concerning The Step in a more intuitive way, but essentially following John Baez as quoted by Carlo Beenakker. -First of all, the definition of elementary particle is not entirely rigorous. In high energy physics, the phrase elementary particle is unfortunately used for truly elementary particles (these are stable particles, which do not decay into others) and quite a lot of unstable particles. -Let me start with the stable elementary particles. The very fact that they are called elementary means that all mathematical structures needed to describe their properties should be such that these cannot be decomposed into smaller entities. Otherwise, it would be extremely likely that the particle described with these mathematical structures could also be decomposed into "smaller" particles, it would not be stable, or it could at least be fragmented by external forces. Thus, if properties of particles are related to representations of (Lie) groups, stable elementary particles should be related to the smallest possible representations, i.e. the fundamental representations (which happen to be irreducible). -Now, all the other particles, which happen to be not stable, will ultimately decay into stable elementary particles. Thus, they should be related to representations which can be decomposed into fundamental ones. A particle can only decay into other particles, if the tensor product of all the representations related to the fragments contains the representation of the original particle. -High energy experiments produce quite a lot of states which physicists like to associate with particles or excited states of particles. Since the compact Lie groups, which typically appear in physics, have the property that all of their finite dimensional representations can be decomposed into direct sums of irreducible representations, it became customary to associate the word particle in the sense of building block of physical entities to irreducible representations. But as explained above, only stable elementary particles are naturally associated with particular representations, the fundamental ones. For the others, it is merely a convenient way to get some order into the zoo of high energy physics. As Baez writes, working with the irreps is a convenient choice of basis. -Finally, physicists nowadays use the word quasiparticle for a lot of physical entities, which behave like particles, i.e. like excitations of quantum fields of some effective quantum field theories, in particular in condensed matter systems. Again, the word is defined to refer to entities which have well defined representation theoretic properties with respect to all groups involved. -I know, this is a very sloppy answer. However, I found it important to shift the focus towards the important distinction between truly elementary stable particles and all the other "elementary" particles.<|endoftext|> -TITLE: Algorithm to decide whether two constructible numbers are equal? -QUESTION [6 upvotes]: The set of constructible numbers -https://en.wikipedia.org/wiki/Constructible_number -is the smallest field extension of $\mathbb{Q}$ that is closed under square root and complex conjugation. I am looking for an algorithm that decides if two constructible numbers are equal (or, what is the same, if a constructible number is zero). -As equality of rational numbers is trivial, this algorithm probably needs to reduce the complexity of the involved number in several steps. -Does such an algorithm exist? If yes, what does it look like? - -REPLY [7 votes]: Although this can be done using the complicated algorithms for general algebraic numbers, there’s a much simpler recursive algorithm for constructible numbers that I implemented in the Haskell constructible library. -A constructible field extension is either $\mathbb Q$ or $F{\left[\sqrt r\right]}$ for some simpler constructible field extension $F$ and $r ∈ F$ with $\sqrt r ∉ F$. We represent an element of $F{\left[\sqrt r\right]}$ as $a + b\sqrt r$ with $a, b ∈ F$. We have these obvious rules for $a, b, c, d ∈ F$: -$$\begin{gather*} -(a + b\sqrt r) + (c + d\sqrt r) = (a + c) + (b + d)\sqrt r, \\ --(a + b\sqrt r) = -a + (-b)\sqrt r, \\ -(a + b\sqrt r)(c + d\sqrt r) = (ac + bdr) + (ad + bc)\sqrt r, \\ -\frac{a + b\sqrt r}{c + d\sqrt r} = \frac{ac - bdr}{c^2 - d^2 r} + \frac{bc - ad}{c^2 - d^2 r}\sqrt r, \\ -a + b\sqrt r = c + d\sqrt r \iff a = c ∧ b = d. -\end{gather*}$$ -To compute the square root of $a + b\sqrt r ∈ F{\left[\sqrt r\right]}$: - -If $\sqrt{a^2 - b^2 r} ∈ F$ and $\sqrt{\frac{a + \sqrt{a^2 - b^2 r}}{2}} ∈ F$, then -$$\sqrt{a + b\sqrt r} = \sqrt{\frac{a + \sqrt{a^2 - b^2 r}}{2}} + \frac{b}{2\sqrt{\frac{a + \sqrt{a^2 - b^2 r}}{2}}}\sqrt r ∈ F{\left[\sqrt r\right]}.$$ -If $\sqrt{a^2 - b^2 r} ∈ F$ and $\sqrt{\frac{a + \sqrt{a^2 - b^2 r}}{2r}} ∈ F$, then -$$\sqrt{a + b\sqrt r} = \frac{b}{2\sqrt{\frac{a + \sqrt{a^2 - b^2 r}}{2r}}} + \sqrt{\frac{a + \sqrt{a^2 - b^2 r}}{2r}}\sqrt r ∈ F{\left[\sqrt r\right]}.$$ -Otherwise, $\sqrt{a + b\sqrt r} ∉ F{\left[\sqrt r\right]}$, so we represent it as -$$0 + 1\sqrt{a + b\sqrt r} ∈ F{\left[\sqrt r\right]}\left[\sqrt{a + b\sqrt r}\right].$$ - -In order to compute with numbers represented in different field extensions, we need to rewrite them in a common field extension first. To rewrite $a + b\sqrt r ∈ F{\left[\sqrt r\right]}$ and $c ∈ G$ in a common field extension, first rewrite $a, b, r ∈ F$ and $c ∈ G$ in a common field extension $H$. If $\sqrt r ∈ H$, then we have $a + b\sqrt r, c ∈ H$; otherwise we have $a + b\sqrt r, c + 0\sqrt r ∈ H{\left[\sqrt r\right]}$. -I implemented the constructible real numbers and built the constructible complex numbers generically on top of those, to enable ordering relations and to avoid having to think too hard about branch cuts.<|endoftext|> -TITLE: 24 vectors in Leech lattice having scalar product $\frac{1}{4}$ pairwise -QUESTION [10 upvotes]: Two vectors from Leech lattice - as defined on wikipedia - have scalar product $\pm 32,\pm 16, \pm 8$ or $0$. Do there exist 24 vectors having scalar product 8 pairwise ? When we consider unit vectors in Leech lattice then scalar product is $\frac{1}{4}$ pairwise. -Motivation for this question is the fact that for 2b involution there are 196560 involutions in Monster group 2a conjugacy class which generate extraspecial group $2^{1+24}$. Modulo its center elements in this group correspond to Leech vectors. I believe vectors having scalar product 8 correspond to not commuting elements. -What I was able to achieve is that for $\Lambda_{16}$ sublattice there are at most 9 such vectors. The arguments are following. This sublattice can be seen as union of 18 disjoint $E_8$ lattices (see Wilson definition with octonions which refer to earlier Dixon definition). Given two vectors having scalar product 8 ($\frac {1}{2}$ in Wilson version where Leech vector is of legth $\sqrt 2$) must lie in two different E8s. -For Baby monster 2b element there are 4600 elements in 2a commuting with it. In Leech lattice language for fixed vector there are 4600 vectors having scalar product 16 with it. -Regards, - -REPLY [2 votes]: Here are 24 vectors in classic Leech lattice having pairwise scalar product 8. I obtained it from $(-5, 1^{23})$ using this. -$\begin{bmatrix} --1& 1& 1& 1& 1&-1& 1&-1& 1& 1&-1&-1& 1& 1&-1&-1& 1&-1& 1&-1&-1&-1&-1&-3 \\ -\dots\\ - 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1&-3 -\end{bmatrix}$ -The dots $\dots$ should be replaced by 22 vectors obtained from the first one by permutation $(1,2,...,23)$. -As a consequence we can obtain following useful -Fact -For each element $p$ of order $23$ in $M_{23}$ there exist dodecad $d$ such that dodecads $d, pd, p^2d,...,p^{22}d$ intersect in six points pairwise.<|endoftext|> -TITLE: What does the matrix of a mapping class tell you about the 3-manifold? -QUESTION [8 upvotes]: Let $H$ be a handlebody with $\Sigma = \partial H$. Given an automorphism $f : \Sigma \to \Sigma$ we can glue to obtain a closed 3-manifold $M = H \cup_f H$ and in fact all such 3-manifolds are obtained this way. Since only the isotopy class of $f$ is necessary in order to specify the homeomorphism type of the resulting 3-manifold, we have a map from the mapping class group $\mathop{MCG}(\Sigma)$ to the set of homeomrphism types of closed 3-manifolds. By looking at the action on $H_1(\Sigma;\mathbb{Z})$ which has a symplectic structure via the cup product, we obtain a surjection -$$ -\psi: \mathop{MCG}(\Sigma) \to \mathop{Sp}(H_1(\Sigma; \mathbb{Z})) -$$ -What information about $M$ can be obtained from $\psi([f])$? - -REPLY [4 votes]: Well, from the characteristic polynomial of the matrix, thanks to Casson, one can figure out if the gluing map is pseudo-Anosov (if the characteristic polynomial is irreducible, non-cyclotomic, and does not have the form $P(x) = Q(x^k),$ for some $k>1.$). In principle, there should be a some criterion in terms of the matrix for low-bounding the translation distance in the curve complex, and if that is at least $3,$ and so the manifold will be hyperbolic, but I don't know what the criterion is.<|endoftext|> -TITLE: Is there a topological space X homeomorphic to the space of continuous functions from X to [0, 1]? -QUESTION [28 upvotes]: In general, we might ask when we can find interesting spaces $X, Y$ such that $X$ is homeomorphic to $[X, Y]$. By the Lawvere fixed point theorem $Y$ must have the fixed point property. Happily, $Y = [0, 1]$ does in fact have the fixed point property (e.g. by the intermediate value theorem), so this is not an obstruction. -I know that computer scientists have constructed spaces $X$ such that $X$ is homeomorphic to $X^X$ in domain theory, but I don't know enough about it to tell whether the techniques are relevant here. - -REPLY [30 votes]: There cannot be such a homeomorphism, because Lawvere's fixed point theorem would give us something too constructive: a continuous map $[0,1]^{[0,1]} \to [0,1]$ that assigns a fixed point to each continuous map $[0,1] \to [0,1]$. -Following the proof of the Lawvere fixed point theorem, given a homeomorphism $f: X \cong [0,1]^X$ and a map $g: [0,1] \to [0,1]$, we construct a map $h: X \to [0,1]$ by $h(x) = g(f(x)(x))$, and then we get an element $x^* = f^{-1}(h)$ of $X$. Then, -$$ f(x^*)(x^*) = h(x^*) = g(f(x^*)(x^*)), $$ -so we have found a fixed point of $g$. This construction is continuous in $g$, since we're working with the exponential topology on $[0,1]^{[0,1]}$, so $g \mapsto f(x^*)(x^*)$ gives a continuous fixed-point-finding map $[0,1]^{[0,1]} \to [0,1]$. -There is a lot of work on the nonconstructivity of Brouwer's fixed point theorem for different senses of constructivity. To give a quick visual proof that no continuous map of the type considered here exists, it is sufficient to consider a path through $[0,1]^{[0,1]}$ along which such a map cannot be continuously defined. Consider linearly deforming green to red to purple in the below diagram; a fixed point (i.e. a point where the curve crosses the identity depicted in blue) cannot be continuously chosen. - -It's interesting that this works because we have a homeomorphism $X \cong [0,1]^X$ rather than just a continuous surjection. A continuous surjection $X \to [0,1]^X$ would be enough to deduce the intermediate value theorem using Lawvere's fixed point theorem, but not in a constructive enough way to similarly rule out.<|endoftext|> -TITLE: Minkowski's Linear Forms Theorem With Complex Coefficients -QUESTION [8 upvotes]: Minkowski's Linear Forms Theorem is often stated about linear forms with real coefficients. However, in Narkiewicz's Elementary and Analytic Theory of Algebraic Numbers, the following generalization of Minkowski's Linear Forms Theorem is stated (the theorem is actually stated for a general lattice in real space, but I'm only interested in the case where the lattice is $\mathbb{Z}^N$): -Let $U = \{ L_j(x_1, \ldots, x_N) = \sum_{i=1}^N a_{ij} x_i \mid 1 \leq j \leq N \}$ be a system of $N$ linear forms with complex coefficients satisfying the following condition: if $L_j \in U$, then the conjugate form, $\overline{L_j} \in U$. Let $M = [a_{jk}]_{1\leq j,k \leq N}$. Let $\epsilon_1, \ldots, \epsilon_N \in \mathbb{R}^+$ such that $\prod_{i=1}^N \epsilon_i \geq | \det M |$ and $L_i = \overline{L_j}$ implies $\epsilon_i = \epsilon_j$. Then there exists a nonzero point $(x_1, \ldots, x_n) \in \mathbb{Z}^N$ such that $|L_j(x_1,\ldots,x_N)| \leq \epsilon_j$ for each $1 \leq j \leq N$ with strict inequality holding for all but one $j$. -However, instead of providing a proof, Narkiewicz refers to two books by Cassels, both of which only prove the case where the coefficients of the forms are real. The proof of the real version---which uses Minkowski's Convex Body Theorem on $M^{-1} ([-1,1]^N)$---does not easily generalize to the complex version. Does anyone have an idea of how the proof of the generalization might go? I've tried translating the problem to real-space (using the fact that $N$-dimensional complex space is isomorphic to $2N$-dimensional real space), but the set that I want to apply Minkowski's Convex Body Theorem to does not have large enough volume. -EDIT: A proof may appear in the 39th "chapter" in Minkowski's Geometrie der Zahlen (on page 113). However, I cannot read German. Does anyone know of an English translation of the work? - -REPLY [2 votes]: Credit where it's due to @so-calledfriendDon who found a source before I figured this out. However, I figure I'll post the proof because Google Books can be a finicky thing. -First, note that we can assume that all $\epsilon_j = 1$. If not, replace $L_j$ by $\frac{L_j}{\epsilon_j}$ and note that all of the hypotheses pertaining to forms and their conjugates still hold. -Suppose there are $r$ forms which have real coefficients and $s$ pairs of complex conjugate forms so that $r + 2s = N$. Without loss of generality, $L_1, \ldots, L_r$ have real coefficients and $\bar{L}_{r+j} = L_{r+s+j}$ for each $1 \leq j \leq s$. Then we can write $$L_{r+j} = \frac{R_j + i I_j}{\sqrt{2}} \\ L_{r+s+j} = \frac{R_j - i I_j}{\sqrt{2}}$$ for linear forms with real coefficients $R_j$ and $I_j$ and $1 \leq j \leq s$. Observe that if we look at the set of forms $L_j$ for $1 \leq j \leq r$ and $R_k, I_k$ for $1 \leq k \leq s$, we have $r + 2s = N$ linear forms in $N$ variables, where all the linear forms have real coefficients. In particular, using the fact that $\bar{a}_{r+j,k} = a_{r+s+j,k}$ for all $1 \leq j \leq s$ and $1 \leq k \leq N$ (this is simply restating that $\bar{L}_{r+j} = L_{r+s+j}$), we can explicitly describe $R_j$ and $I_j$: $$ R_j(x_1, \ldots, x_N) = \sqrt{2} \text{Re}[a_{r+j,1}]x_1 + \cdots + \sqrt{2} \text{Re}[a_{r+j,N}] x_N \\ I_j(x_1, \ldots, x_N) = \sqrt{2} \text{Im}[a_{r+j,1}]x_1 + \cdots + \sqrt{2} \text{Im}[a_{r+j,N}] x_N$$ -Let $A$ be the matrix whose entries are the coefficients of the forms $L_j$ for $1 \leq j \leq r$ and $R_k, I_k$ for $1 \leq k \leq s$. I won't go into the gory details, but using elementary properties of determinants, you can show that $|\det(A)| = |\det(M)| \leq 1$. Now you can apply the version of Minkowski's Linear Forms Theorem where the coefficients of the linear forms are real (proofs of this theorem are plentiful) to find a nonzero point $\mathbf{x} = (x_1, \ldots, x_N) \in \mathbb{Z}^N$ where $|L_1(\mathbf{x})| \leq 1$, $|L_j(\mathbf{x})| < 1$ for $2 \leq j \leq r$ and $|R_k(\mathbf{x})|, |I_k(\mathbf{x})| < 1$ for each $1 \leq k \leq s$. -But then we have, for each $1 \leq j \leq s$ $$|L_{r+j}(\mathbf{x})|^2 = \left|\frac{R_j(\mathbf{x}) + i I_j(\mathbf{x})}{\sqrt{2}} \right|^2 = \frac{R_j(\mathbf{x})^2 + I_j(\mathbf{x})^2}{2} < 1 \\ |L_{r+s+j}(\mathbf{x})|^2 = \left|\frac{R_j(\mathbf{x}) - i I_j(\mathbf{x})}{\sqrt{2}} \right|^2 = \frac{R_j(\mathbf{x})^2 + I_j(\mathbf{x})^2}{2} < 1$$. -This shows that $|L_1(\mathbf{x})| \leq 1$ and for each $2 \leq j \leq N$, we have $|L_j(\mathbf{x})| < 1$, which is exactly what we wanted.<|endoftext|> -TITLE: Strictly Fréchet spaces versus strongly Fréchet spaces -QUESTION [13 upvotes]: For a topological space $X$ and a point $x\in X$, consider the following definitions: - -(Gerlits and Nagy): $X$ is strictly Fréchet at $x\in X$ if for any sequence $(A_n)_{n\in\omega}$ such that $x\in\bigcap_{n\in\omega}\overline{A_n}$ there exists a sequence $(x_n)_{n\in\omega}\in\prod_{n\in\omega}A_n$ such that $x_n\to x$. -(Siwiec): $X$ is strongly Fréchet at $x\in X$ if for any decreasing sequence $(A_n)_{n\in\omega}$ such that $x\in \bigcap_{n\in\omega}\overline{A_n}$ there exists a sequence $(x_n)_{n\in\omega}\in\prod_{n\in\omega}A_n$ such that $x_n\to x$. - -Clearly, strictly Fréchet spaces are strongly Fréchet. Since these conditions are equivalent for spaces of the form $C_p(X)$, I was wondering if the converse holds in general. More precisely, - -If $X$ is strongly Fréchet at $x$, then $X$ is strictly Fréchet at $x$? - -REPLY [3 votes]: You can find an example of a strongly Fréchet non-strictly Fréchet space in the preprint "Selective game versions of countable tightness with bounded finite selections" by L. Aurichi, A. Bella and R. Dias (see example 2.10). Basically, it's the one-point compactification of an appropriate Mrówka-Isbell $\Psi$-space.<|endoftext|> -TITLE: Do curvature differences obstruct a.e orientation-preserving isometries? -QUESTION [14 upvotes]: Is there an example of a pair $M,N$ of connected, oriented equidimensional Riemannian manifolds with the following properties: - -$M$ is everywhere non-flat, $N$ is flat. -There exist a map $f:M \to N$ which is differentiable almost everywhere (a.e), and $df$ is an orientation-preserving isometry a.e. - -(An easier goal: Find a pair of manifolds which are not locally isometric, but which admit a map as in 2. We should probably restrict here to manifolds without boundary, since otherwise $M=[0,1],N=\mathbb{R},f(x)=x$ is an example. ) -Context: -The point is to see whether curvature differences obstruct existence of a.e orientation-preserving isometries. -If we omit the requirement on the orientation, then there is a lot of flexibility; -Gromov showed that for any metric $g$ on the unit $d$-dimensional disk $\mathbb{D}^d$ there is an a.e isometry $f:(\mathbb{D}^d,g) \to (\mathbb{R}^d,e)$. ($e$ is the Euclidean metric). - -Further comments: - -Gromov's a.e isometry cannot be orientation-preserving: - -It's $1$-Lipschitz, and hence in $W^{1,\infty}(M,N)$, and every map $f \in W^{1,\infty}(M,N)$ satisfying $df \in \text{SO}$ a.e is smooth. (Thus Gromov's map cannot be orientation-preserving or orientation-reversing on any open subset of the domain. It must "switch" orientations in an"infinite rate"). - -An a.e orientation preserving isometry does not need to be smooth: - -For an example take $M=[0,1],N=[0,2],f(x)=c(x)+x$ where $c$ is the Cantor function. Then $f'=1$ a.e. -This example can be used to show that there is an a.e orientation-preserving isometry from a circle of radius $1$ into a circle of radius $2$. (Of course, there is no smooth local isometry from the former into the latter). - -REPLY [2 votes]: I will add an explanation of Anton Petrunin's comment -Question : There is no continuous map $$f :B\rightarrow -\mathbb{E}^2$$ where $B$ is a geodesic ball in $S^2(1)$ of radius -$\varepsilon$ s.t. $df$ is -isometric a.e. -Proof : Assume that $f$ is a continuous map. - Note that ${\rm length}\ f(\partial B)={\rm length}\ \partial B$. In further, $f(B)$ is - enclosed by $f(\partial B)$. Since $f$ is a volume preserving map, by isoperimetric inequality, it is a - contradiction.<|endoftext|> -TITLE: Who first proved that a vanishing Riemann tensor is sufficient for the existence of Euclidean coordinates? -QUESTION [11 upvotes]: Riemann famously introduced the notion of what we now call a Riemannian manifold and introduced the Riemann curvature tensor $R_{ijk}{}^l$, showing that it is an obstruction the local existence of Euclidean coordinates. It is now well known that $R_{ijk}{}^l = 0$ is also a sufficient condition for the local existence of Euclidean coordinates. So, who first proved this sufficiency? -It's quite possible that the original paper with the proof might be in German, or some other non-English language. In that case, what would be an English language reference translating/summarizing the contents of the corresponding original article? -Edit: For the purpose of collecting links to translations. Riemann's famous Commentatio paper, where he introduced the curvature tensor, is included in full English translation in the Appendix to -Farwell, Ruth; Knee, Christopher, The missing link: Riemann's ``Commentatio'', differential geometry and tensor analysis, Hist. Math. 17, No.3, 223-255 (1990). ZBL0743.01017. - -REPLY [10 votes]: Presumably it was Riemann himself who proved the sufficiency of $R=0$ for the flatness. Spivak's Chapter 4 in Volume II of "A comprehensive introduction to differential geometry" is a good source on Riemann's work. It gives a translation of Riemann's inaugural lecture and of a part of his prize essay. At the end of the prize essay Riemann says (I am citing from Spivak): - -Given an acquaintance with the traditional methods, it is demonstrated without difficulty that these ... conditions when they are satisfied, suffice... - -meaning the vanishing of the components of the curvature tensor. Later in Chapter 4, Spivak gives a proof of this claim by a method that could have been considered "traditional" by Riemann. -EDIT: It is possible that the first published proof is contained in -Christoffel, E. B., On transformations of homogeneous differential forms of degree two., Borchardt J. LXX, 46-70 (1869). ZBL02.0128.03. [English translation included as Section 8 of Fagginger Auer, B. O. Christoffel revisited. MSc thesis (2011, Utrecht)] -Christoffel deals with equivalence of two metrics, not just flatness.<|endoftext|> -TITLE: Dynkin diagram of the centralizer of a semisimple element in a Levi subgroup -QUESTION [6 upvotes]: Let $G$ be a connected reductive group over an algebraically closed field and consider a semisimple element $s \in G$ and let $L$ be a Levi subgroup containing $s$. -My question is about the two ways we can look at $C_L(s)^\circ$. On the one hand, it is the connected centralizer of $s$ in the Levi subgroup $L$, on the other hand, it is a Levi subgroup in the connected centralizer $C_G(s)^\circ$. -The Dynkin diagram of $C_{G}(s)^\circ$ can be obtained from the extended Dynkin diagram of $G$ by removing nodes. By removing appropriate nodes from the Dynkin diagram of $C_{G}(s)^\circ$, we obtain the Dynkin diagram of the Levi subgroup $C_{L}(s)^\circ$ of $C_{G}(s)^\circ$. -On the other hand, the Dynkin diagram of $C_L(s)^\circ$ can be obtained from the extended Dynkin diagram of $L$ and the Dynkin diagram of $L$ can be obtained by removing nodes from the Dynkin diagram of $G$. -For example let $G$ have type $B_{10}$ and choose $s$ such that $C_{G}(s)^\circ$ has type $D_4 \times B_6$, then I can find a Levi subgroup of $C_{G}(s)^\circ$ of type $D_4 \times A_2 \times B_3$. This Levi is of the form $C_L(s)^\circ$ for some Levi $L$ of $G$. -However, even though there exists a Levi subgroup of $G$ of type $A_2 \times B_7$ and one of its subgroups of maximal rank has the desired type $A_2 \times D_4 \times B_3$, the bases of the corresponding root systems seem to be completely different to me. In fact, I am not even sure if that subgroup of type $A_2 \times D_4 \times B_3$ lies in $C_G(s)$ or not. -Also, it seems to me that the other Levi subgroups of $G$ do not even have a subgroup of the desired type. -Did I do anything wrong here? If not, why is it not possible for me to obtain the same basis of the Levi of $C_{G}(s)^\circ$ when considering it as a maximal rank subgroup of a Levi subgroup of $G$? - -REPLY [3 votes]: In the Bourbaki numbering ($\alpha_i = \epsilon_i - \epsilon_{i + 1}$ for $i < 10$ and $\alpha_{10} = \epsilon_{10}$), I believe that you can take $s = \alpha_1^\vee(-1)\alpha_3^\vee(-1)$ (with connected centraliser of type $D_4 \times B_6$, where the base for the $D_4$ piece is $\{\alpha_3, \alpha_2, \alpha_1, -\mu\}$ and that for the $B_6$ piece is $\{\alpha_5, \dotsc, \alpha_{10}\}$) and $L$ to be the centraliser of the image of $\alpha_5^\vee + 2\alpha_6^\vee + 3(\alpha_7^\vee + \dotsb + \alpha_{10}^\vee)$, which, as you predicted, is of type $A_2 \times B_7$. The base for the $A_2$ piece is $\{\alpha_5, \alpha_6\}$, and that for the $B_7$ piece is $\{\alpha_1, \alpha_2, \alpha_3, \alpha_4 + \dotsb + \alpha_7, \alpha_8, \alpha_9, \alpha_{10}\}$. -I found Carter's paper Conjugacy classes in the Weyl group (MR) very useful for understanding this kind of calculation with what he calls admissible diagrams (Section 4 in his paper). Section 4 in my paper On counting orbits in root systems gives some amateurish examples.<|endoftext|> -TITLE: Finite subgroups of Lie group over algebraic ring of integers -QUESTION [10 upvotes]: I have frequently seen results like: -There are 4 isomorphism types of finite subgroups of $SL_2(\mathbb{Z})$, namely $\mathbb{Z}_2,\mathbb{Z}_3,\mathbb{Z}_4,\mathbb{Z}_6$. -I wonder what is known of we replace $\mathbb{Z}$ by the ring of integers $\mathcal{O}_K$ of an algebraic number field $K$. I've found these groups discussed in several contexts, but I could not find my question discussed, although I expect of course it has been....any hint or reference would be greatly appreciated. Thanx! -More concretely I wonder whether the finite groups $SL_2(3)$ and $SL_2(5)$ do appear in this way as a finite subgroup, maybe for $SL_2(\mathcal{O}_K)$ for $K=\mathbb{Q}(e^{2\pi i /3}),\mathbb{Q}(e^{2\pi i /4}), \mathbb{Q}(e^{2\pi i /5}),\mathbb{Q}(e^{2\pi i /6})$. - -REPLY [5 votes]: I'll comment on the last question, with the group $G = {\rm SL}(2,5).$ All $2$-dimensional irreducible complex representations of $G$ are real valued, so all have Schur index dividing $2$ by a standard theorem ( Speiser?). Since $G$ has a quaternion subgroup of order $8$, the Schur index of such a representation is necessarily $2$. Since the ring of integers of each of the fields mentioned in the question is a PID ( by Minkowski for example, though that is overkill) , it suffices to check whether $G$ is isomorphic to a subgroup of ${\rm SL}(2,K)$ for each of the listed fields. However any field over which such a representation can be realised must certainly contain the field generated by its character. This field necessarily contains $\sqrt{5}$ as it contains $\omega + \overline{\omega}$ -for $\omega = e^{\frac{2 \pi i}{5}}.$ Hence the only possible choice for the listed $K$ we need to concern ourselves with is $K = \mathbb{Q}[e^{\frac{2 \pi i}{5}}],$ since the other listed fields do not contain $\sqrt{5}.$ -Now for this choice of $\omega$, we note that ${\rm SL}(2,K)$ contains a double cover $E$ of a dihedral group of order $10$, which is isomorphic to a Sylow $5$-normalizer of $G,$ namely $E = \langle \left(\begin{array}{clcr}0 & \omega\\-\overline{\omega} & 0\end{array}\right), -\left(\begin{array}{clcr}\omega & 0\\0 & \overline{\omega}\end{array}\right) \rangle.$ Identifying $E$ with a Sylow $5$-normalizer of $G,$ and inducing the character $\mu$ of this representation, we find that ${\rm Ind}_{E}^{G}(\mu)$ has irreducible constituents of degree $2,4$ and $6.$ By the theory of the Schur index and the discussion above we know that the representation affording the $2$-dimensional constituent $\chi$ may be afforded over $K$ ( and hence over the ring of integers of $K$, since that ring of integers is a PID). For the Schur index $m_{K}(\chi)$ divides $2$, but we have exhibited a $K$-representation of $G$ which contains the character $\chi$ with multiplicity one. (Note that $K$ is not the field generated by the character values of $\chi,$ but a degree $2$-extension of that field, so this is consistent with earlier statements).<|endoftext|> -TITLE: Proper family deformation retracts onto special fiber -QUESTION [12 upvotes]: If $\mathbf{D}$ is the complex unit disc with coordinate function $s$ and $X \to \mathbf{D}$ is a proper flat holomorphic family (and it is smooth outside of the fiber $s=0$), will the total family $X$ deformation retract onto the fiber above $s=0$? -I don't believe this will be true, but I cannot find a counterexample. My idea was that if $X$ is a family of elliptic curves degenerating to a singular cubic, then this would fail (but, I think it actually works in this case!). -Edit: in the comments, the consensus is that this should be true - but we do not have a proof yet. - -REPLY [6 votes]: Here is the reference: -Persson, Ulf, -On degenerations of algebraic surfaces, -Mem. Amer. Math. Soc. 11 (1977), no. 189. -Clemens, C. H. Degeneration of Kähler manifolds. Duke Math. J. 44 (1977), no. 2, 215-290. -I found the reference here -http://web.math.ucsb.edu/~drm/papers/clemens-schmid.pdf -also worth reading -Morrison, David R. -The Clemens-Schmid exact sequence and applications, -Topics in transcendental algebraic geometry (Princeton, N.J., 1981/1982), 101-119, -Ann. of Math. Stud., 106, Princeton Univ. Press, Princeton, NJ, 1984.<|endoftext|> -TITLE: Does almost every pair of elements in a compact Lie group generates the connected component? -QUESTION [12 upvotes]: It is known that almost every pair of elements in a connected compact Lie group (topologically) generates the group. -Obviously this isn't true for non-connected groups but - -Given a compact Lie group $G$, is it true that almost every pair of elements of $G$ generates a subgroup containing the connected component $G^\circ$ of 1? - -REPLY [15 votes]: The closed subgroups of $G$ not containing the identity component lie in countably many conjugacy classes of subgroups. So it is sufficient to show that for each closed subgroup $H$ not containing the identity component, the probability that Haar-random $g_1$ and $g_2$ both lie in some conjugate of $H$ vanishes. -Such pairs are parameterized by the manifold of triples $x \in G/H$, $g_1 \in x H x^{-1}, g_2 \in x H x^{-1}$, which is a manifold of dimension $(\dim G - \dim H) + 2 \dim H = \dim G + \dim H$. -The image of this manifold in $G \times G$ under the projection $(x,g_1,g_2)\mapsto (g_1,g_2)$ must have measure $0$, as it is the image of a smaller-dimensional manifold (as $H$ does not contain the identity we have $\dim H < \dim G$) under a smooth map (by Sard's theorem). -For $H$ a subgroup, the same argument just barely fails to show that the probability that a single $g$ is contained in a conjugate of $H$ vanishes. Indeed in this case both manifolds have dimension $\dim(G)$. This is convenient as that statement is false, because we could take $H$ to be a maximal torus, or alternately the subgroup generated by a reflection in an infinite dihedral group, as in Noam's example.<|endoftext|> -TITLE: On binary quadratic forms which are not proper subforms of another binary quadratic form -QUESTION [9 upvotes]: Let $f(x,y)$ be a binary quadratic form with co-prime integer coefficients. We say that $f$ is a proper subform of $g(x,y)$ if there exists an integer matrix $A = \left(\begin{smallmatrix} a_1 & a_2 \\ a_3 & a_4 \end{smallmatrix}\right)$ with $|\det A| > 1$ such that -$$\displaystyle f(x,y) = g(a_1 x + a_2 y, a_3 x + a_4 y).$$ -For example, the form $f(x,y) = 4x^2 + 4xy + 5y^2$ is a proper subform of $g(x,y) = x^2 + y^2$, since -$$\displaystyle 4x^2 + 4xy + 5y^2 = (2x + y)^2 + (2y)^2.$$ -If $f$ is a proper subform, then the discriminant of $f$ is divisible by $\det(A)^2$, so it is not square-free. My question is the converse: suppose that $\Delta(f)$ is divisible by an odd square $m^2$. Is $f$ a proper subform of another form $g$? - -REPLY [8 votes]: Yes. It is about equivalence over $SL_2 \mathbb Z.$ Your form $f$ (primitively) represents some value not divisible by the fixed prime $p.$ Indeed, from original coefficients $\langle a,b,c \rangle$ we know that at least one of $a,c, a+b+c$ is not divisible by $p.$ We may therefore demand $a \neq 0 \pmod p$ in $\langle a,b,c \rangle.$ -For an integer $t$ we have the equivalent form -$$ \langle a, \; \;b + 2 a t, \; \;c + b t + a t^2 \rangle. $$ -As $p$ is odd and $a \neq 0 \pmod p,$ there exists some $\delta$ such that -$b + 2 a \delta \equiv 0 \pmod p.$ -At this point, we now have $\langle a,b,c \rangle$ with $a \neq 0 \pmod p$ and $b \equiv 0 \pmod p.$ You have stated that $$ b^2 - 4 a c \equiv 0 \pmod {p^2}. $$ It follows that $c \equiv 0 \pmod {p^2}.$ -We may now rewrite $$\langle \alpha, \; \;\beta p, \; \;\gamma p^2 \rangle.$$ The descent is -$$\langle \alpha, \; \;\beta p, \; \;\gamma p^2 \rangle \mapsto \langle \alpha, \; \;\beta, \; \;\gamma \rangle.$$ -As far as a factor of $4,$ taking $\Delta = b^2 - 4 a c,$ we can descend when $\Delta \equiv 4 \pmod {16},$ as $x^2 +3y^2 \mapsto x^2 + xy+ y^2,$ and - $\Delta \equiv 0 \pmod {16},$ as $x^2 +20y^2 \mapsto x^2 + 5y^2.$ No descent for the others, for example $x^2 + y^2$ or $x^2 + 2 y^2.$<|endoftext|> -TITLE: Pairs of elementary Fourier transforms in $L^2$ -QUESTION [8 upvotes]: It is customary to teach Fourier transform on the real line by starting with functions from $L^1$, $L^2$ or the Schwartz space. It is not so easy to illustrate the theory by computing explicit pairs of transforms that are both elementary. The list given in the standard books is short: algebraic fractions, gaussian, exponentials, sinc and that's it. -I was surprised to learn that the Fourier transforms of $1/\cosh(x)$ and -$\hbox{atan}(x+1) - \hbox{atan}(x-1)$ can be computed explicitely and are of an elementary nature. -Are you aware of other $L^1$ or $L^2$ examples besides the classical ones? -I use the convention $\hat{f}(\xi) = \int_{-\infty}^\infty f(x) e^{-i\xi x} dx$, so that -$$\widehat{\hbox{atan}(x+1) - \hbox{atan}(x-1)} = 2\pi \ e^{-|x|} {\sin(x)\over x}$$ -$$\widehat{1\over \cosh(x)} = {\pi \over \cosh({\pi x\over 2})}$$ -EDIT1: I just noticed that $\hbox{atan}(x+1) - \hbox{atan}(x-1) = \hbox{atan}(2/x^2)$ and that the Fourier transform of this function is computed in the classical book of Oberhettinger. -Another explicit pair as suggested in the answer of C. Beenakker. -$$ -{\hbox{tanh}(x)\over x} = --2 \ln(\hbox{tanh}({\pi |\xi|\over 4})) -$$ -EDIT2: Nemo in this post gives many examples of Fourier transforms which are self-reciprocal and provides references. Some of them are elementary functions with explicit Fourier transforms. Note that the convention concerning the definition of the transform is different from the one used here. - -REPLY [10 votes]: The following example is interesting -$$\widehat{\sin(x)\over x \sqrt{|x|}} = - \sqrt{2\pi} \ \left\{ - { - \matrix{ - \scriptstyle - \sqrt{|\xi+1|} + \sqrt{|\xi-1|} & - \hbox{ if } |\xi| \leq 1, \cr - {2\over \sqrt{|\xi+1|} + \sqrt{|\xi-1|}} & - \hbox{ if } |\xi| \geq 1.\cr}} - \right.$$ -because it is in $L^1$ but not in $L^2$ and its Fourier transform is continuous but with infinite derivative at $\xi = \pm 1$.<|endoftext|> -TITLE: What is this symmetric simplex category, concretely? -QUESTION [9 upvotes]: Let $\Delta_+$ denote the category of finite ordinal numbers with monotonic maps (the subscript indicates that $0$ is included, so this is the augmented simplex category). This has a monoidal structure (given by the sum), which is not symmetric. But we can make it symmetric in a universal way, see here for the general procedure. Let us denote this symmetric monoidal category by $(\Delta_+)_{\mathrm{sym}}$. -Question. What is a more "concrete" symmetric monoidal category which is equivalent to $(\Delta_+)_{\mathrm{sym}}$? -Notice that this is not the symmetric monoidal category $\mathcal{F}$ of finite sets. Whereas $(\Delta_+)_{\mathrm{sym}}$ classifies algebra objects in symmetric monoidal categories, $\mathcal{F}$ classifies commutative algebra objects in symmetric monoidal categories. Hence, there will be a strong symmetric monoidal functor $(\Delta_+)_{\mathrm{sym}} \to \mathcal{F}$, which is essentially surjective, but not fully faithful. - -REPLY [11 votes]: $\Delta_+$ is the monoidal category generated from the associative operad, considered as a non-symmetric operad. Similarly, $(\Delta_+)_{{\rm sym}}$ is the symmetric monoidal category generated from the associative operad, this time considered as a symmetric operad. This category can then be explicitly described as the category whose objects are finite sets and such that the morphisms from $I$ to $J$ are given by maps $f:I \to J$ together with, for each $j \in J$, a choice of a linear order on $f^{-1}(j)$. The symmetric monoidal structure is given by disjoint union.<|endoftext|> -TITLE: If the diagonal of a positive operator is compact, is the operator itself compact? -QUESTION [8 upvotes]: Let $H$ be a separable Hilbert space with a fixed orthonormal basis $\{e_n\}_n$. For a bounded operator $T$ on $H$, the diagonal of $T$ is the unique operator $D_T$ on $H$ which is diagonal with respect to the above basis, and whose diagonal entries are given by $d_n=\langle T(e_n),e_n\rangle$. It is well know that if $T$ is positive and $D_T=0$, then necessarily $T=0$. - -Question: If $T$ is positive and $D_T$ is compact, is $T$ necessarily compact? - -REPLY [5 votes]: If T also is sparse, i.e. for some number N then for each i - is non-zero for at most N values of j’s, -(and for each j it is non-zero for at most N values of i’s) then the claim is true. -I have a proof, but I guess this is well known. My own literature search has been unsuccessful, so anyone having a reference will make me happy.<|endoftext|> -TITLE: Connected planar compact set with finite length is path connected -QUESTION [6 upvotes]: Here length means 1-Hausdorff measure. This seems to be known, what is the reference? Or very short proof? - -REPLY [7 votes]: See Exercise 3.5 in "The geometry of fractal sets" by K. J. Falconer. -It says that any such set is an image of rectifiable curve. -For a short proof, check "Rectifiable curve" in my collection. -P.S. The earliest reference I found: Theorem 2 in Continua of finite linear measure. I. by Eilenberg and Harrold (1943).<|endoftext|> -TITLE: Gauss, Jacobi, Kloosterman sums and representation theory in the $\mathbb F_1$-world -QUESTION [11 upvotes]: This question is inspired by Why are Bessel function and Kloosterman sum similar? - it developed in me desire to understand Kloosterman sums better. -There seems to be common knowledge that Gauss, Jacobi and Kloosterman sums are analogs of the, respectively, Gamma, Beta and Bessel functions, and I would like to understand this better. - -(Subquestion one: are there other similar analogies known?) - -For example, are there any "finite sum analogs" of hypergeometric functions in this context? -Now there is a whole branch of the theory of special functions - their common treatment as matrix elements of Lie group representations; most comprehensive treatment that I know is in several volumes by Klimyk and Vilenkin - -(subquestion two: are there other texts you would recommend?) - -For example, Gamma appears in representations $R_\lambda$ of the group of affine transformations of the real line on functions via $R_\lambda[aX+b]:\varphi(x)\mapsto e^{\lambda bx}\varphi(ax)$, and there is (in the end of the second volume) a parallel definition of Gamma and Beta functions for any locally compact totally disconnected non-discrete field. (Seems like they in particular obtain $p$-adic $\Gamma$, although I am not sure.) Bessel functions are obtained from representations of $SO(n)$ on functions on the unit sphere, although I could not get a clear picture of it for me in the book. -So I thought - maybe there is some still more general unified treatment which would give all of the above but in addition would encompass the case of the field with one element; or, in more conventional terms, would give Gamma, Beta, Bessel, etc. functions for simple Lie groups and in parallel would give, respectively, Gauss, Jacobi, Kloosterman, etc. sums for their respective Weyl groups? -Or maybe $\mathbb F_1$ is not related in any way and one has simply to ask for a unified treatment including representations of linear groups over finite fields into above picture. Still in that case a question remains what would(could) one get in parallel from representations of Weyl groups. -And, since I am asking anyway - what about $q$-analogs? It seems that applying the above to $q$-deformations of representations of universal enveloping algebras one may get things like $q$-Gamma and $q$-hypergeometric functions; are there any finite field or $\mathbb F_1$-versions of these $q$-analogs known? - -REPLY [3 votes]: The analogy is obtained by placing the special function in a larger context of not-so-special functions. For instance I could compare the analytic function -$$\int_0^\infty e^{t^3x - t^2 x^2 + t x^4- x^8} x^{1/3} dx$$ -which presumably has no special significance, to the exponential sum -$$ \sum_{x\in \mathbb F_q} \psi(t^3x-t^2x^2+ t x^4-x^8) \chi(x)$$ -for $\psi$ an additive character and $\chi$ a multiplicative character of order $3$. -This transformation is a fairly straightforward process. The only subtlety is that we may have multiple choices for the contour of integration. Really one should compare the finite field function to a whole family of analytic functions. -Katz studied many examples in his book Exponential Sums and Differential Equations, including the hypergeometric ones. -I'm not familiar with the representation-theoretic interpretation of special functions, but it should generalize as long as the representation theory perspective gives us integrals that we can transform to exponential sums in this way.<|endoftext|> -TITLE: Weak equivalence over forcing notions -QUESTION [6 upvotes]: We know there are several definitions about forcing equivalence which imply that two forcings notions can be equivalent or not. In general we like to know the similarity between generic extensions by different forcings. Among them, we are familiar with complete embedding, densely embedding and so on. As a matter of fact If $\mathbb P$ and $\mathbb Q$ are two forcing notions, let's call them equivalent if their Boolean completion are isomorphic as Boolean algebras, then if it is the case we have V[G]=V[H], for any $\mathbb P$- generic filter $G$ and $\mathbb{Q}$- generic filter $H$. -I am interested to understand similarity between satisfaction of generic extension by different forcing notions forcing notions, let's unpack this: - - -Definition: - Fix $V\models ZFC$, Let $P$ and $Q$ be two forcing notions in $V$, we say $\mathbb P$ is weakly equivalent to $\mathbb Q$, if $V[G]\equiv V[H]$(i.e $V[G]$ is elementary equivalnet to $V[H]$), for all $G$ and $H$ which are $\mathbb {P}$- generic and $\mathbb Q$- genric filters over V, respectively. - - -Remark 1: Weakly equivalence probably is not definable in $V$. - -Question: Is there any characterization of the above-mentioned notion based on Boolean algebras or order structure of forcings? Or at least sufficient conditions for being weakly equivalent and there is nontrivial example i.e two foricngs which are not equivalent but weakly equivalent. - -Remark 2: By a theorem of Golshani and Mitchell, there is a model of $ZFC$, such that $Coll(\omega, \kappa)$ is weakly equivalent to $Coll(\omega,\lambda)$, for all $\kappa$ and $\lambda$ infinite. Thus consistently there are forcing notions which are weakly equivalent, but not equivalent. -Remark 3: Note that in Golshani-Mitchell model, trivial forcing is also weakly equivalent to collapse forcing, so it seems the situation is complicated. - -REPLY [5 votes]: Let me provide the alternative kind of example. -Theorem. If ZFC is consistent, then there is a model of ZFC in which any two weakly equivalent forcing notions are forcing equivalent. -Proof. Consider a pointwise definable -model $M$ that remains definable in all its set-forcing extensions, -such as a pointwise definable model of $V=L$. Suppose that $\newcommand\P{\mathbb{P}}\P\equiv\newcommand\Q{\mathbb{Q}}\Q$ are weakly equivalent. I claim that $\P$ and $\Q$ are forcing equivalent, in the sense -that they give rise to the same forcing extensions. -To see this, suppose that $G\subset\P$ and $H\subset\Q$ are $M$-generic. The theory of $M[G]$ includes the assertion that -the universe was obtained by forcing over $M$ with the poset $\P$, -and this is expressible since $M$ is definable in $M[G]$ and $\P$ is definable in $M$. Thus, since $M[H]$ has the same theory, it follows that this statement is also true in $M[H]$, and so $M[H]=M[G']$ for some $M$-generic -filter $G'\subset\P$. So every forcing extension by $\Q$ can be realized as a forcing extension by $\P$ and vice versa, and so the forcing notions are equivalent. QED -Let me remark further that your notion of equivalence applies only to forcing notions -$\newcommand\P{\mathbb{P}}\P$ with the property that all conditions -force the same sentences, since in particular every forcing -extension by $\P$ itself must have the same theory. Let us call -these the sententially homogeneous forcing notions. These are the forcing notions $\P$ with $\P\equiv\P$. Note that $\P\equiv\Q$ implies $\P\equiv\P$ and $\Q\equiv\Q$. -In the model of my proof, two forcing notions are weakly equivalent $\P\equiv\Q$ just in case they are each sententially homogeneous and $\P$ and $\Q$ are forcing equivalent. So it is relatively consistent with ZFC that weak equivalence is the restriction of a definable relation to the class of forcing notions to which it applies, the sententially homogeneous forcing notions. Meanwhile, I suspect that the question of whether a forcing notion is sententially homogeneous is not generally expressible in the language of set theory. -Aside. You said that two forcing notions are equivalent if and only -if their Boolean algebras are isomorphic, but that is too strong, -since a large lottery sum of a partial order with itself is forcing -equivalent, in the sense that it gives rise to exactly the same forcing extensions, but the Boolean algebra is much larger (and has a bad chain condition and so on). The solution is to restrict the forcing to -the cone below a condition. You want to say that two notions are -equivalent, if they give rise to exactly the same forcing -extensions, and this can be expressed in terms of their Boolean -algebras by the assertion: below any respective conditions in their -Boolean algebras, one can find still deeper isomorphic lower cones. See my -related blog post, Common forcing extension via different forcing -notions.<|endoftext|> -TITLE: sum-sets in a finite field -QUESTION [8 upvotes]: Let $\mathbb{F}_p$ be a finite field, $A=\{a_1,\dots,a_k\}\subset\mathbb{F}_p^*$ a $k$-element set, for $k -TITLE: Does generic projection into $\mathbb{R}^3$ preserve real-algebraic-curve-ness? -QUESTION [7 upvotes]: I'm interested in the topological properties of certain real algebraic curves in high-dimensional spaces. I want to visualize these curves (say, like this), and so I'm pursuing dimensionality reduction into $\mathbb{R}^3$. (Considering the proof of Theorem 3.1 in this paper, I expect a generic projection of this sort to successfully embed the curve.) -Is the generic projection of a real algebraic curve into $\mathbb{R}^3$ again a real algebraic curve? If so, is there an efficient procedure to derive the corresponding polynomials? -This question is related, but my setting should be fundamentally different, since the failure of Tarski–Seidenberg with algebraic sets appears to stem from a failure to embed. - -REPLY [5 votes]: The following is pretty much the standard argument due, I think, to Whitney (from the proof of his embedding theorem in the "stable range"). Suppose that $V\subset {\mathbb C}^N$ is an affine complex-algebraic subset defined over the real numbers (I.e. by polynomials with real coefficients), where $dim(V)=m$ and $2m+1< N$. Consider the map -$V\times V\times {\mathbb C}\to {\mathbb C}^N$, $(x,y,t)\mapsto t(x-y)$. Due to our dimension assumptions, this map is not surjective. For the same dimension reasons, its image $W$ (a complex algebraic subvariety) does not contain ${\mathbb R}^N$ and, moreover, is "small" in any meaningful sense (its dimension is $ -TITLE: "Big" groups $G$ with trivial $Out(G)$ -QUESTION [13 upvotes]: We are looking for examples of groups $G$ such that $G$ is "big", but $Out(G)$ is trivial. By "big" we mean things like virtually free, or large, or Golod-Shafarevich. However, we would like our groups to be residually-finite. -Edit: Henry, Igor and Lee thank you for all your help. Eventually, we only needed finite $Out(G)$ with $G$ having a trivial center. Thus, $A*B$ was the example we used. It is included in "Large normal subgroup growth and large characteristic subgroup growth" (sorry for the self-promotion, but I thought people might be interested in our motivation and the results of their help). - -REPLY [9 votes]: A countably-infinite family of groups is given at the end of Section 6 of a paper of A. D. Logan1: for each natural number $n>1$, the two-generator-, one-relator group -$$G_n := \langle a, b; (a^{−2}ba^4ba^{−3}ba^5b)^n\rangle$$ -has trivial outer automorphism group. -EDIT This has nothing to do with the above, but I don't want to add a third answer :) My discussion with Henry in the comments to the OP, might suggest that $Out(PGL(2, \mathbb{Z}))$ might be trivial. Apparently, this was proved by Hua and Reiner in 1951-1952, and then disproved(!) by Joan Dyer in 1978, where she constructed an outer automorphism (known ever since as the Dyer automorphism, so $Out(PGL(2, \mathbb{Z})) = C_2;$ the "so" is not trivial but true. -For completeness, her automorphism $\mathcal{D}$ is described as follows. The generators of $PGL(2, \mathbb{Z})$ are: -$$S = \pm \begin{pmatrix}0 &1\\ -1 & 0\end{pmatrix}, T=\pm \begin{pmatrix}1 &1\\ 0 & 1\end{pmatrix},B=\pm \begin{pmatrix}0 &1\\ 1 & 0\end{pmatrix}.$$ -Then the automorphism sends $S, T, B$ to $SB, TB, B.$ -1A. D. Logan: The outer automorphism groups of two-generator, one-relator groups with torsion. Proc. Amer. Math. Soc. 144 (2016) 4135-4150<|endoftext|> -TITLE: Do chain homotopic maps between dg-algebras induce "same" maps on dg-modules -QUESTION [8 upvotes]: Let $A$ and $B$ be two dg-algebras over a field $k$. Let $f, g: A\to B$ be two maps between dg-algebras. We call $f$ and $g$ chain homotopic if there exists a degree $-1$ map $h: A\to B$ such that $f-g=d_Bh+hd_A$. -Now let $M$ be a right $A$ dg-module. For a given map $f: A\to B$, we could obtain a right $B$ dg-module by tensor product: $M\mapsto M\otimes_AB$. To avoid confusion we denote it by $M\otimes^f_AB$. - -My question is: if $f, g: A\to B$ are chain homotopic, then is it true that $M\otimes^f_AB$ and $M\otimes^g_AB$ are quasi-isomorphic or further isomorphic up to homotopy? If not, is there any counter-example? - -REPLY [4 votes]: If I understand your question correctly, the answer is no. -Here is a counterexample. Let $A=k\langle 1,y\rangle$ with zero differential and trivial unital multiplication (put $y$ in degree $1$ for concreteness). Let $B=k[x]\langle 1, y\rangle$ with commutative multiplication and differential induced by $d(x) = y$. Then the inclusion $f:A\to B$ and the map $g:A\to B$ which sends $y$ to zero are both dg algebra maps, and are chain homotopic by $h(y)=x$. -Now let $M=k$ where $y$ acts trivially. Then $M\otimes_A^f B \cong k[x]$ with zero differential while $M\otimes_A^g B\cong B$, which is contractible. -Here are two related explanations for why this fails: 1) chain homotopy is not a good notion of homotopy for algebras and 2) projective or flat chain complexes are not necessarily projective or flat algebras.<|endoftext|> -TITLE: Large gaps in Singer's difference sets -QUESTION [6 upvotes]: This question is related to the question I asked earlier. -For a natural number $n$, a set $D$ of integer numbers is called a $n$-cyclic difference set if each integer number $x\notin n\mathbb Z$ can be uniquely represented as the difference $x\equiv a-b \mod n$ -for a unique pair $(a,b)\in D\times D$. For example, $\{0,1,3,6\}$ is a $13$-cyclic difference set. -For every power $q=p^k$ of a prime number $p$, Singer (1938) constructed a $(1+q+q^2)$-cyclic difference set $D$ of order $|D|=q+1$. A (still unproved) Prime Power Conjecture says that a number $n$ admitting an $n$-cyclic difference set is of the form $n=1+q+q^2$ for some power $q$ of a prime number. The Singer's cyclic difference sets play an important role in Additive Combinatorics. -We are interested in evaluating the largest gap in an $n$-cyclic difference set. A pair $(a,b)$ of numbers is called a gap in a set $A\subset \mathbb Z$ if the intersection $[a,b]\cap A$ coinsides with the doubleton $\{a,b\}$. The number $|a-b|$ is called the length of the gap. For a subset $A\subset\mathbb Z$ let $gap(A)$ be the largest lenght of a gap in $A$. For example, $g(\{0,1,3,6\})=|6-3|=3$. -Problem 1. What is the largest possible gap in an $n$-cyclic difference set $D$? -A more precise question: -Problem 2. Is it true that for any positive constant $C$ there exists an $n$-cyclic difference set $D$ with $gap(D)>C\cdot \sqrt{n}$? -Remark 1. In fact, Problem 2 was implicitely asked by Leech in 1956 and discussed by Golay in 1972. So, maybe this problem has been already solved, in which case I would greatly appreciate a corresponding reference. -Remark 2. Answering my preceding question, @Lucia proved that for every $\varepsilon>0$ there exists a number $n_\varepsilon$ such that $gap(D)<\varepsilon \cdot n$ for any $n$-cyclic difference set $D$ with $n \ge n_\varepsilon$. So, gaps in cyclic difference sets cannot be very large. -I can prove that for any prime power $q=p^k$ each $(1+q+q^2)$-cyclic difference set $D$ has $gap(D)\ge q+\sqrt{q}-1$, but it is very far from $gap(D)>Cq$. - -REPLY [3 votes]: Firstly, in my previous answer I showed that the gaps are bounded by $O(n^{\frac 34} \log n)$ (and one can remove the $\log$ with a little more care). I expect that this is the best known bound on the gap. -Problem 2 is definitely an open problem. Suppose one can show that there is a gap of size $C\sqrt{n}$. By translating the perfect difference set, one can assume that the gap is at the end -- that is, make the residue classes $\mod n$ lie in $[0, n -C \sqrt{n}]$. But if we simply think of these now as integers, then it would follow that there is a Sidon set (or $B_2$-set) in $[1,N]$ (with $N=n-C\sqrt{n}$) having more than $\sqrt{N} + C/2$ elements. This an open problem going back to Erdos, who offered \$ 500 for a solution -- there is a survey/annotated bibiliography of O'Bryant from 2004 that you may find useful; see section 4.1 there.<|endoftext|> -TITLE: Example of an additive functor admitting no right derived functor -QUESTION [20 upvotes]: I asked the same question a week ago on Mathematics Stackexchange but got no answer. -What would be a simple example of an additive functor $F:\mathcal C\to\mathcal C'$ of abelian categories such that the right derived functor -$$ -RF:\text D(\mathcal C)\to\text D(\mathcal C') -$$ -does not exist? -My reference for the notions involved in this post is the book Categories and Sheaves by Kashiwara and Schapira. -Here is a reminder of the definition of a right derived functor $RF$ in the above setting: -Let $\text K(F):\text K(\mathcal C)\to\text K(\mathcal C')$ be the triangulated functor induced by $F$ between the homotopy categories, let $X$ be in $\text K(\mathcal C)$, assume that the colimit -$$ -\operatorname*{colim}_{X\to Y}\ \text K(F)(Y),\tag1 -$$ -where $X\to Y$ runs over all the quasi-isomorphisms out of $X$ in $\text K(\mathcal C)$, exists in $\text D(\mathcal C')$, and denote this colimit by $RF(X)$. The right derived functor $RF$ of $F$ is defined at $X\in\text D(\mathcal C)$ if, for any functor $G:\text D(\mathcal C')\to\mathcal A$, the colimit -$$ -\operatorname*{colim}_{X\to Y}\ G(\text K(F)(Y)) -$$ -exists in $\mathcal A$ and coincides with $G(RF(X))$. The right derived functor $RF$ of $F$ exists if $RF$ is defined at $X$ for all $X$ in $\text D(\mathcal C)$. -The ideal would be to have an example of a pair $(F,X)$, where $F:\mathcal C\to\mathcal C'$ is an additive functor of abelian categories and $X$ is an object of $\text D(\mathcal C)$, such that (1) does not exist in $\text D(\mathcal C')$. - -REPLY [15 votes]: Let ${\cal C}$ be the category of finite dimensional ${\bf Z}/2$-vector spaces equipped with a ${\bf Z}/2$ action, let ${\cal C'}$ be the category of finite dimensional ${\bf Z}/2$-vector spaces and let $F: {\cal C} \to {\cal C'}$ be the fixed subspace functor $V \mapsto V^{{\bf Z}/2}$. Consider the chain complex $X$ such that $X_n = {\bf Z}/2$ with trivial action in each degree $n \in {\bf Z}$ and with trivial differentials. Then the colimit $RF(X) = {\rm colim}_{X \to Y}Y^{{\bf Z}/2}$ does exist as a complex of vector spaces (since this category has all limits and colimits), and $RF(X)$ will be the derived functor there, but it will not be degreewise finite dimensional (and from here a simple argument shows that the colimit cannot exist in ${\cal D}({\cal C'})$). In fact, since ${\rm Ext^n_{{\cal C}}({\bf Z}/2,{\bf Z}/2)} \cong {\bf Z}/2$ for all $n \geq 0$ a spectral sequence argument shows that $RF(X)$ has infinite dimensional homologies in all degrees.<|endoftext|> -TITLE: Classification of geometric structures through character varieties -QUESTION [9 upvotes]: Under what general assumptions $(G,X)$-geometric structures on a manifold $M$ are classified by their holonomies, yielding an injection -$$\Psi: \{(G,X)\text{-structures on M}\} \to H(\pi_1(M),G)/G ?$$ -When is the image a connected component of $H(\pi_1(M),G)/G$ in the Euclidean (other?) topology? (Probably $(G,X)$-structures need to be "marked".) -I believe that is the case for hyperbolic, Euclidean, affine, projective structures on surfaces. -Is it true in general? If not, what goes wrong? Perhaps extra conditions are necessary, say $RP^2$-structures on surfaces need to be convex. - -REPLY [3 votes]: This question was pretty much answered in the comments, so I will just give some relevant references: - -Geometric structures on manifolds and varieties of representations, by W. Goldman, Geometry of group representations (Boulder, CO, 1987), 169–198, Contemp. Math., 74, Amer. Math. Soc., Providence, RI, 1988. -Locally homogeneous geometric manifolds, by W. Goldman, Proceedings of the 2010 International Congress of Mathematicians, Hyderabad, India (2010), 717-744, Hindustan Book Agency, New Delhi, India - -The above two references describe the general picture pretty well: the moduli space of marked $(G,X)$-structures on a smooth compact manifold is locally modeled on the character variety (in some sense). -In practice, regardless of the properties of the holonomy map, one often uses the structure of the character variety to determine the structure of the moduli space of $(G,X)$-structures. This is exemplified well in Section 4 of this: -Trace coordinates on Fricke spaces of some simple hyperbolic surfaces, by W. Goldman, Handbook of Teichmüller theory. Vol. II, 611-684, IRMA Lect. Math. Theor. Phys., 13, Eur. Math. Soc., Zürich, 2009 -W. Goldman has also made available his working book on the subject: Geometric structures on manifolds, which contains many detailed examples.<|endoftext|> -TITLE: is it possible to have two non-isomorphic non-regular graphs with the same adjacent spectrum and the same laplacian spectrum? -QUESTION [10 upvotes]: For two regular graphs $G$ and $H$, it is possible for them to share the same adjacent spectrum and the same laplacian spectrum. While, on the other hand, is it possible to have two non-regular graphs $G$ and $H$, which share the same adjacent spectrum and the same laplacian spectrum? Any comments or references would be greatly appreciated. - -REPLY [15 votes]: Yes, Brendan McKay showed that almost all trees have mates that are simultaneously cospectral in both adjacency and Laplacian spectra. And more. -http://users.cecs.anu.edu.au/~bdm/papers/SpectralTrees.pdf -Edit: I wondered briefly what the smallest pair of such trees would be, and a few minutes of Sage told me that there are two on 11 vertices. -And here they are (for some reason I am having difficulty with the image uploader):<|endoftext|> -TITLE: Vector bundles over $RP^{\infty}$ -QUESTION [8 upvotes]: There is a theorem due to Sato which says that any vector bundle over $\mathbb CP^{\infty}$ decomposes as a direct sum of line bundles (which is a generalization of Grothendieck's result over $\mathbb CP^1$). I am wondering if there is a similar theorem for vector bundles over $\mathbb RP^{\infty}$ or a criterion for splitting. In my case I have a rank $2n$ grassmann bundle over $\mathbb RP^{\infty}$. Does it split as a direct sum of line bundles ? Thanks. - -REPLY [12 votes]: You are essentially asking about the set $[B\mathbb{Z}/2,BSO(2n)]$. This bijects with the set of conjugacy classes of homomorphisms from $\mathbb{Z}/2$ to $SO(2n)$, or in other words real, oriented representations of $\mathbb{Z}/2$. Essentially the same thing works for $[BP,BG]$ whenever $P$ is a finite $p$-group and $G$ is a compact connected Lie group. In fact, one can describe the full homotopy type of the space $\text{Map}(BP,BG)$: it is the classifying space of the groupoid of homomorphisms $P\to G$, and conjugacies between them. In this strong form the theorem is due to Dwyer and Zabrodsky. There are probably earlier references for the special case that you need. There is an exposition with references at http://www.math.purdue.edu/~wilker/papers/bzpztobg.pdf. -I am not sure exactly what to say if you want $O(2n)$ instead of $SO(2n)$, but I expect that there is some kind of straightforward reduction.<|endoftext|> -TITLE: Reference for dualizable chain complexes -QUESTION [6 upvotes]: Let $k$ be a commutative ring. Let $Ch(k)$ denote the monoidal category of chain complexes. I need a reference, including a proof, for the following "folklore fact" (see for example here after Definition 1.2.1): - -A chain complex $A \in Ch(k)$ is dualizable if and only if it is bounded and each $A_n$ is a dualizable, i.e. finitely generated projective $k$-module. - -The dual is then $(A^*)_n = (A_{-n})^*$. - -REPLY [10 votes]: This is Proposition 1.6 of 'Duality, Trace and Transfer' by Dold and Puppe, http://www.maths.ed.ac.uk/~aar/papers/doldpup2.pdf.<|endoftext|> -TITLE: Binary codes with length and distance a power of 2 -QUESTION [5 upvotes]: Let $A_2(n,d)$ denote the maximum number of words in a binary code (not necessarily linear) with length $n$ and distance $d$. The value of this function is not known in general, though there are tables for small values of $n$ and $d$, e.g. http://www.win.tue.nl/~aeb/codes/binary-1.html. -Suppose $n = 2^k$ and $d = 2^j$ (with $j < k$) are both powers of $2$. Is there a known formula for $A_2(2^k,2^j)$? It seems like a strong enough assumption that a closed formula should exist. -I am also interested if there are other exact results that apply when $n$ and $d$ can be large. - -REPLY [4 votes]: I doubt that a general formula is known, even in this special case. -But the largest and smallest $2+2$ cases are easy: -for all $n$, whether of the form $2^k$ or not, we have: - - $A_2(n,1) = 2^n$ (trivial); - $A_2(n,2) = 2^{n-1}$ (partiy code); - $A_2(n,n) = 2$ (repetition code), and most interestingly: - if $n$ is even then $A_2(n,n/2) \leq 2n$, with equality at least when - $n = 2^k$ (extended Hamming code). - [Likewise $A_2(n-1, n/2) \leq n$, with the same equality condition.] - -When $n=2^k$, these are all examples of -Reed-Muller codes, -and there's one of minimal distance $2^j$ for each $j$; but in general -they're far from optimal.<|endoftext|> -TITLE: Limit formula for the second derivative -QUESTION [9 upvotes]: Suppose that $f$ is a real-valued function which is twice differentiable in the interval $(-1,1)$. Does the following hold?: -$$\lim_{h \to 0} \frac{f(h) - 2f(0) + f(-h)}{h^2} = f''(0)$$ -If $f''(x)$ is continuous this follows from Taylor's theorem with the -Lagrange form of the remainder. What if $f''(x)$ is not continuous? --- Thanks. - -REPLY [6 votes]: Just so an answer is put here rather than in the comments: -As noted in the original question, the desired limit formula is a well-known, or at least straightforward, consequence of Taylor's theorem with Lagrange remainder in cases where $f\in C^2$, i.e. if we have some continuity of the 2nd derivative in a neighbourhood of $0$. -As pointed out by Nate Eldredge in the comments: for functions which are merely twice differentiable at the origin, i.e. where the second derivative exists at $0$ but might not be continuous in a neighbourhood of $0$, one still has Taylor's theorem (of order $2$) with Peano remainder: -$$ -f(x) = f(0) + f'(0)x + \frac{1}{2}f''(0)x^2 + h(x)x^2 -$$ -where $h(x)\to 0$ as $x\to 0$. This is enough to obtain the desired limit formula. -Remark. I took the liberty of contacting the OP and he said: - -I am teaching the first real analysis course - for math majors. In class I - proved that if you know the value of a twice - differentiable function at three points you - know the value of the second derivative at - some point ... and I assumed you need continuity - of the second derivative to prove the - limit formula.<|endoftext|> -TITLE: Density of numbers whose prime factors belong to given arithmetic progressions -QUESTION [7 upvotes]: By a theorem of Landau, the number of integers $n\leq x$ whose prime divisors belong to only arithmetic progressions $a_1,\dots,a_r$ mod $q$, with $r\leq\varphi(q)$ and $a_i$ coprime to $q$ for each $i$, is -$$Cx (\log x)^{r/\varphi(q)-1} + O(x (\log x)^{r/\varphi(q)-2}),$$ -for some constant $C$ depending on $r$ and $q$. -This is an old result. I am sure much more is known about the distribution of such $n$, but I do not know the literature well enough, and would be happy if someone could provide suggestions regarding the state of the art (or other classical theorems) regarding results of this nature. Thanks! - -REPLY [5 votes]: This result has been generalised a fair bit. The main generalisation is that you can replace congruence conditions by so-called "Frobenian conditions", namely conditions of the type which arise in the Chebotarev density theorem. -There have also been some improvements in the error term, but substantial improvements are not really possible without assuming GRH. -Serre has written quite a bit about such topics. See for example Theorem 2.8 of the paper: -Serre - Divisibilité de certaines fonctions arithmétiques. -The associated zeta functions don't admit a meromorphic continuation to all of $\mathbb{C}$ in general; they have natural boundaries along the line $\mathrm{re}(s) = 0$. You can read more about this in the paper: -Hashimoto - Partial zeta functions. -(the author works in an even greater generality than I describe here).<|endoftext|> -TITLE: Did Bishop make those comments in his oral presentation? -QUESTION [6 upvotes]: The 1975 published version of a 1974 talk at a workshop by Errett Bishop contains the following comment: - -"A more recent attempt at mathematics by formal finesse is non-standard analysis. I gather that it has met with some degree of success, whether at the expense of giving significantly less meaningful proofs I do not know. My interest in non-standard analysis is that attempts are being made to introduce it into calculus courses. It is difficult to believe that debasement of meaning could be carried so far." - -This was published in -Bishop, E. "The crisis in contemporary mathematics." Proceedings of the American Academy Workshop on the Evolution of Modern Mathematics (Boston, Mass., 1974). Historia Math. 2 (1975), no. 4, 507--517. -Was anyone present at Bishop's lecture who can testify whether Bishop actually made those comments in his oral presentation? -Note 1. As per discussion in the comments: The reason I doubt Bishop actually said that in his oral presentation is because in the ensuing discussion, also published in Historia Math. along with the lecture, nobody challenged Bishop on his comments even though a number of logicians were present. It seems likely that, had he said that publicly, there would have been some reaction and ensuing discussion. -Note 2. This question is primarily concerned with what Bishop said, or more precisely did not say, at the workshop, but since in the comments editors have responded with remarks on the effectiveness of teaching calculus using infinitesimals I would mention that there are many approaches to teaching the calculus, some more effective than others, but one rarely finds people dismissing an approach right out of hand without looking at the details and whether it actually works in the classroom, etc.--that is, except when it comes to true infinitesimal calculus, where everything seems to be allowed. Such a situation, I would argue, is partly the result of an atmosphere of demonisation of Robinson's framework created by the likes of Bishop, Halmos, and Connes. For more details on this aspect of the story see the many published articles here. - -REPLY [7 votes]: We were able to obtain an audio file of Bishop's talk from the American Academy of Arts and Sciences. Our analysis will be published in Historia Mathematica and is available on the arxiv. -We examine the preparation and context of the paper "The Crisis in Contemporary Mathematics" by Errett Bishop, published 1975 in Historia Mathematica. Bishop tried to moderate the differences between Hilbert and Brouwer with respect to the interpretation of logical connectives and quantifiers. He also commented on Robinson's Non-standard Analysis, fearing that it might lead to what he referred to as 'a debasement of meaning.' The 'debasement' comment can already be found in a draft version of Bishop's lecture, but not in the audio file of the actual lecture of 1974. We elucidate the context of the 'debasement' comment and its relation to Bishop's position vis-a-vis the Law of Excluded Middle. -One can only speculate concerning the reasons that may have led Bishop to suppress the 'debasement' comment when faced with an actual audience on 9 august 1974, or what he meant exactly when he declared, at the exact spot of the omission, that "that is all I want to say about pure mathematics." However, a reader of the 1975 published version who may have been surprised or disappointed not to find any reaction to the 'debasement' comment on the part of the audience that included a number of logicians will now have an explanation for their silence.<|endoftext|> -TITLE: A generalization of the Borsuk Ulam theorem -QUESTION [7 upvotes]: Is there a compact $n$-dimensional manifold $M$ or, more generaly, a compact $n$-dimensional topological space $M$ with the following property? -"For every continuous map $f:M \to \mathbb{R}^{n}$ there are points $a,b,c \in M $ with $f(a)=f(b)=f(c)$." -This is motivated by the following obvious consequence of the Borsuk-Ulam theorem: -"For every continuous map $f:S^n \to \mathbb{R}^n$ there are points $a, b \in S^n$ with $f(a)=f(b)$." - -REPLY [16 votes]: If M is allowed to be a simplicial complex, five $n$-simplices with a common $(n-1)$-dimensional face should do the job. If $M$ should be a manifold, take any closed non-orientable one. -The multiplicity of maps between manifolds can be studied with the help of characteristic classes, see this preprint of Roman Karasev and this preprint of Roman Karasev and Pavle Blagojevic. For example, any continuous map $\mathbb{R}\mathrm{P}^4 \to \mathbb{R}^4$ sends some four points in $\mathbb{R}\mathrm{P}^4$ to the same point in $\mathbb{R}^4$. -Another generalization of the Borsuk-Ulam theorem (more close to it in the spirit) deals with spaces with a group action (like $\mathbb{Z}_2$-action on the sphere) such that for any map to $\mathbb{R}^n$ some orbit is sent to a point. This is discussed in detail in Section 6 of -Matoušek, Jiří, Using the Borsuk-Ulam theorem. Lectures on topological methods in combinatorics and geometry. Written in cooperation with Anders Björner and Günter M. Ziegler, Universitext. Berlin: Springer (ISBN 978-3-540-00362-5/pbk; 978-3-540-76649-0/ebook). xii, 214~p. (2008). ZBL1234.05002.<|endoftext|> -TITLE: The need for nets in topology -QUESTION [6 upvotes]: I remember when I first heard about nets in topology (called also Moore-Smith sequences). I was told that most of useful topological properties which can be exressed in terms of sequences in the context of metric spaces can be generalized and expressed in the terms of nets: for example you can check continuity of mapping $f$ by checking whether $f(x_s) \to f(x)$ when $x_s \to x$ where $(x_s)_s$ is net. I was also warned about facts which no longer are true in the context of nets: for example if $x_s \to x$ then the set $\{x_s\}_s \cup \{x\}$ need not to be compact. I thought that this sort of things are due to the fact, that $(x_s)_s$ may be uncountable and the indexing set $S$ is only partially ordered instead of totally ordered. However even if you consider nets $(x_s)_s$ where $s$ runs over $\{0,1\} \times \mathbb{N} $ with lexycographic order (this is well ordered and countable set!) you can easily construct nets which are convergent and not bounded, with arbitary big closure etc. It raises the question: maybe we don't need to define nets as mappings $x:S \to X$ where $S$ is only partially ordered? -However in most of the proofs using nets, one usually uses the family of all possible neighborhoods of the given point (with reverse inclusion) as the indexing set which is partially ordered. Obviously in the context of metric spaces each point has a countable system of neigborhoods consisting of open balls with radius $\frac{1}{n}$ centered at that point. Therefore we can use standard sequences instead of general nets. Moreover, instead of writing $x_{B(x,\frac1n)}$ we simply write $x_n$. This system of neigborhoods has the property of being totally ordered. -Let us now consider the following example: $X=\beta \mathbb{N}$ the Stone-Cech compactification of the discrete countable space $\mathbb{N}$. This space is of cardinality $2^{\mathfrak{c}}$ but is also separable: $\mathbb{N}$ is the countable dense set in $X$. Therefore for each element $x \in X$ there is a subnet of $(n)_n$ which converges to $x$. Such a subnet is of the form $(n_U)_U$ where $U$ runs over the system of neigborhood of $x$. If for every $x$ the corresponding system of neighborhoods of $x$ could be totally ordered, we would get that $(n_U)_U$ is just $(n_k)_k$ (many terms are repeated) and thus there are only $\mathfrak{c}$ of possible subsequences of $(n)_n$ and therefore the closure of $\mathbb{N}$ cannot be of cardinality $2^{\mathfrak{c}}$. This contradiction shows that there are points $x$ such that their system of neighborhoods cannot be totally ordered. -EDIT: so to summarize things: we now from topology that $\overline{A}=\{ \lim_s a_s: (a_s)_s \ is \ a \ net \ in \ A \}$. -So my question is: - - -Whether the following is correct: "If we define nets in the topological space $X$ as maps $x:S \to X$ where $S$ must be totally ordered, then the above formula for the closure does not hold and the counterexample is every separable topological space of cardinality $2^{\mathfrak{c}}$". - - -And also - - -Is it correct that in any such space $X$ (separable, of cardinality $2^{\mathfrak{c}}$) there must be points with the property that their bases of neighborhoods cannot be totally ordered? - - -I hope that now everything is clear. Sorry for the previous inprecise formulation. - -REPLY [10 votes]: There seems to be a gap in your argument where, under the assumption that the system of neighborhoods could be totally ordered, you seem to use that this total order must be countable. To fill the gap, note that any total order has a cofinal well-ordered subset and that the neighborhoods correspond to the sets in $U$. So you'd have a well-ordered (by reverse inclusion) family of subsets of $\mathbb N$, and such a family is necessarily countable. -Your argument could be simplified a bit, using the known fact that no nonprincipal ultrafilter has a countable base. -Note also that it is consistent (for example it follows form the continuum hypothesis) that some non-principal ultrafilter on $\mathbb N$ has a base totally ordered by inclusion modulo finite, i.e., by the relation $A\subseteq^*B$ that means $A-B$ is finite. It's only total ordering by genuine inclusion that's impossible for a nonprincipal ultrafilter base.<|endoftext|> -TITLE: Is there a well-known notion of orientability for finite geometries? -QUESTION [11 upvotes]: I'm wondering if the notion of an orientable/non-orientable manifold has any reasonable extension that allows for a similar classification of finite geometries. -For example, the real projective plane is non-orientable. Does this somehow mean that a finite projective plane is "non-orientable" too? -I would also be interested in a definition that applies over a narrower domain, e.g. a notion of orientability for finite ordered geometries or something like that. - -REPLY [8 votes]: To do that you would need a notion of a non-orientable and an orientable linear transformation, i.e., essentially a notion of a "positive" and "negative" determinant, where "positive" determinants would form a subgroup not containing the element $-1$. This works for the field $\mathbb{Z}/p\mathbb{Z}$ if and only if the prime $p$ satisfies $p\equiv 3 (mod\; 4)$. -Thus denoting by $G\subset (\mathbb{Z}/p\mathbb{Z})^\ast$ the index-2 subgroup consisting of the quadratic residues modulo $p$, one obtains an "orientable double cover" $M=(F^3\setminus\{0\})/G$, where $F=\mathbb{Z}/p\mathbb{Z}$. Quotienting $M$ by the ("orientation-reversing") antipodal map, you get the projective plane over $F$.<|endoftext|> -TITLE: $c^\infty$ topology on $L(E, F)$ -QUESTION [5 upvotes]: In Kriegl/Michor's "Convenient Setting for Global Analysis", they put on the set $L(E, F)$ of bounded linear operators between locally convex spaces $E$, $F$ the subspace topology induced by the inclusion $L(E, F) \subset C^\infty(E, F)$, where smooth maps are those that map smooth curves in $E$ to smooth curves in $F$. -Question: Is this topology somehow related to one of the common topologies on $L(E, F)$, at least in special cases (i.e. when $E$, $F$ satisfy special assumptions)? -I am interested in this, because smooth maps from $M \rightarrow L(E, F)$ ($M$ a manifold) are needed to define vector bundles and vector bundle maps with locally convex spaces as fibers. - -REPLY [2 votes]: Since this was a bit long for a comment, I am posting it here as an answer (though it unfortunately does not answer your question per se). -It is not clear to me in general how the topologies are related to each other. In some case one can say something though: -Lemma 5.3. in the book states that a subset is bounded in $L(E, F)$ if and only if it is uniformly bounded on bounded subsets of E, i.e. the inclusion $L(E, F) \rightarrow C^\infty (E, F)$ is initial when we endow $L(E,F)$ with the topology of uniform convergence on bounded sets. Though this does nothing for you to establish that the topologies are the same one can read this as a statement on smoothness into the space $L(E,F)$ with respect to the uniform topology. As the concepts of differentiability discussed in the book do not depend on continuity but only on bornology, the above statement means that the smooth curves into (and thus the smooth mappings defined on) the spaces with both structures coincide. Hence to test the differentiability it makes no difference with respect to which structure you test. -Moreover, you can even get continuity if you assume that $E,F$ are Banach spaces and $M$ is a compact manifold. Then $C^\infty (M,L(E,F))$ is a complete metrisable space and being convenient smooth implies that the mapping is also continuous with respect to the original topologies (which then coinicde with the $c^\infty$-topology of the function space. -Remark: I suspect that for $E,F$ are finite dimensional, the induced topology on $L(E,F)$ should be the same as the operator norm topology. Though I was not able to show it yet, it suffices to prove that the topology turns $L(E,F)$ into a locally convex space (which is finite-dimensional in this case, whence it must be the standard topology by uniqueness of the norm topology)<|endoftext|> -TITLE: Locating a certain result on primes represented by a certain polynomial -QUESTION [5 upvotes]: In Theorem 2 of the paper "A polynomial divisor problem" by Friedlander and Iwaniec, Theorem 2 states that $$\sum_{a^6 + b^2\le x} \Lambda(a^6 + b^2)\sim cx^{2/3}$$ for some constant $c > 0$ (in the paper itself, they give a precise error term and make $c$ explicit). -They then say that "The proof of Theorem 2 will be given elsewhere." Where is this proof? - -REPLY [7 votes]: They prove it in section 14 ("An Application") of their paper - -John Friedlander & Henryk Iwaniec, "The Illusory Sieve" (2005) - -As pointed out by Lucia in the comments, the result is conditional on the existence of exceptional characters, an hypothesis which is generally not expected to hold.<|endoftext|> -TITLE: Limit of curvature near lightlike points -QUESTION [6 upvotes]: Let $\alpha \colon I \to \Bbb R^2_1$ be a regular curve and $t_0 \in I$ be such that $\alpha$ is lightlike at $t_0$, and not lightlike at $]t_0-r,t_0[$ for some $r>0$. Then, in that interval the curvature is given by $$k(t) = \frac{\det(\alpha'(t),\alpha''(t))}{\|\alpha'(t)\|^3}.$$I think that $\lim_{t \to t_0^-}|k(t)| = +\infty$, but I do not know how to prove it, nor I'm 100% sure if this is true. For every concrete example I tried until now, it was true. If suffices to check that $\det(\alpha'(t),\alpha''(t)) \not\to 0$, but I don't know how to estimate that determinant. -Help? - -REPLY [2 votes]: I think your conjecture is correct (assuming that "regular" means "smooth"), with one possible caveat that I'll mention below. -Since curvature is independent of parametrization, without loss of generality you can assume that -$$ \alpha(t) = (t, F(t)) $$ -for some function $F$. Then -$$ \alpha'(t) = (1, f(t)) $$ -where $f(t) = F'(t)$, and your hypothesis says that $f(t_0) = 1$ and $f(t) \neq 1$ for $t \in ]t_0 - r, t_0[$. -Here's the caveat: Suppose that the function $g(t) = f(t) - 1$ vanishes to finite order at $t_0$. Then we can write -$$ f(t) = 1 + (t-t_0)^n h(t) $$ -for some integer $n \geq 1$ and some smooth function $h$ with $h(t_0) \neq 0$. Then -$$ \frac{\det(\alpha'(t), \alpha''(t))}{\|\alpha'(t)||^3} = \frac{f'(t)}{(1 - f(t)^2)^{3/2}} = \frac{O((t-t_0)^{n-1})}{O((t-t_0)^{3n/2})} = O((t-t_0)^{-n/2-1}) $$ -as $t \to t_0$, so the limit is indeed infinite. -I suspect that the result is still true even if $g(t)$ vanishes to infinite order at $t_0$; e.g., it works if $g(t) = e^{-1/(t-t_0)^2}$. But I'm not sure sure how to give a rigorous proof in that case.<|endoftext|> -TITLE: composition factors of primitive components -QUESTION [5 upvotes]: A finite transitive permutation group $G$ can always be ``decomposed'' into primitive permutation groups, called its primitive components, although the decomposition is not unique. See Chapter 1 of Finite Permutation Groups by Wielandt. -Suppose $H$ is a primitive component of $G$, and $T$ is a composition factor of $H$ that is also a classical group $X(d,q)$. Here $X$ stands for $\mathrm{PSL}$, $\mathrm{PSU}$, etc. Then $T$ is a subquotient of a composition factor $T'$ of $G$. If $T'$ is an alternating group -$\mathrm{Alt}_k$, is there any lower bound on its degree $k$, in terms of $d$ and $q$? -If $T$ is a subgroup of $T'$ then $T$ has a faithful permutation representation of degree $k$. In this case I think a lower bound of order $q^{\Theta(d)}$ is known. See: -Cooperstein, Bruce N. "Minimal degree for a permutation representation of a classical group." Israel J. Math. 30 (1978), no. 3, 213-235. MR 506701 DOI: 10.1007/BF02761072. -I wonder if a similar bound is possible if $T$ is only a subquotient of $T'$. -A related question: for $k\in\mathbb{N}^+$, denote by $\Gamma_k$ the family of finite groups whose nonabelian composition factors are all isomorphic to subgroups of $\mathrm{Sym}_k$. If $G$ is a group in $\Gamma_k$, are its primitive components also in $\Gamma_k$ (or $\Gamma_{k'}$ for some $k'$ depending on $k$)? - -REPLY [3 votes]: Here is a proof of the claim I made in my comment: if $m(G)$ denotes the minimal degree of a faithful permutation representation of a finite group $G$ and $S$ is a composition factor of $G$, then $m(S) \le m(G)$. I think that answers your questions. (Note that, since $m(H) \le m(G)$ for any $H \le G$, this implies $d(S) \le d(G)$ for any composition factor $S$ of any subgroup of $G$.) -The claim is clear if $S$ is abelian, so assume that $S$ is nonabelian. As you observed yourself, if $N$ denotes the largest normal solvable subgroup of $G$, then $m(G/N) \le m(G)$. This also follows from Proposition 1.3 of -D. Easdown and C. E. Praeger, On minimal faithful permutation -representations of finite groups, Bull. Aust. Math. Soc. 38 (1988), -207-220. -So we may assume that $G$ has no nontrivial solvable normal subgroup. So the socle $K$ of $G$ (the group generated by its minimal normal subgroups) is a direct product $\times_{i=1}^k S_i$ of nonabelian simple groups $S_i$, which are permuted under conjugation by $G$. -et $L$ be the kernel of this conjugation action i.e. $L = \cap_{i=1}^k N_G(S_i)$. Then $L/K$ is isomorphic to a subgroup of $\times_{i=1}^k {\rm Out}(S_i)$. From the classification of fintie simple groups, we know that each ${\rm Out}(S_i)$ is solvable, and hece so is $L/K$. Also $G/L$ is isomorphic to a subgroup of $S_k$. -If $S$ is isomorphic to one of the $S_i$, ithen $m(S) = m(S_i) \le m(G)$ because $S_i \le G$. -Otherwise, $S$ is a composition factor of $G/L$ and $m(G/L) \le k$, so by induction we have $m(S) \le k$. -Now by Theorem 3.1 of the same paper by Easdown and Praeger, we have $m(G) = \sum_{i=1}^k m(S_i) \ge k$, so $m(S) \le m(G)$ as claimed. -Note also that $m(S)$ is known for all finite simple groups $S$. Unfortunately, many of the papers on this topic contain minor errors, but there is a hopefully correct table giving the results in the paper: -S.Guest, J.Morris, C.E.Praeger and P.Spiga. -On the maximum orders of elements of finite almost - simple groups and primitive permutation groups., which you can find here<|endoftext|> -TITLE: Large cardinal properties of $j(\kappa)$ -QUESTION [8 upvotes]: Let $\kappa$ be a measurable cardinal (or other large cardinal) and let $j:V\longrightarrow M$ be a witness. We know that $j(\kappa)$ has large cardinal properties in $M$, but what about $j(\kappa)$ in $V$? -Let me give a couple tentative nonstandard definitions: - -Call a measurable cardinal $\kappa$ weakly compact preserving, if -there is an elementary embedding $j$ witnessing measurability of $\kappa$, such that $j(\kappa)$ is weakly compact in $V$. - -Call a measurable cardinal $\kappa$ reflecting preserving if $V_\kappa\prec V_{j(\kappa)}$, again in $V$. - - -What I am really interested in is a weakly compact preserving–reflecting preserving measurable cardinal $\kappa$ such that there is a weakly compact cardinal $\lambda>\kappa$ such that $V_\lambda\prec V_{j(\lambda)}$. - -Question: What is the place of such a cardinal in the large cardinals hierarchy? - -REPLY [3 votes]: If $j$ is a superstrongness embedding, $j(\kappa)$ is a large cardinal: - -If $j : V \to M$ witnesses that $\kappa$ is superstrong (that is, $V_{j(\kappa)} \subset M$) then $j(\kappa)$ is worldly since $V_{\kappa} \prec V_{j(\kappa)}$ but not generally inaccessible, for by theorem 3.3 of Perlmutter 2013, a cardinal that is high-jump for strongness, which is implied by superstrongness, is superstrong with target equal to the clearance of the high-jump for strongness embedding and by lemma 3.1 of the same paper the clearance is singular. -If $j : V \to M$ witnesses that $\kappa$ is almost huge (that is, $\phantom{M}^{\lt j(\kappa)}M \subset M$), then $j(\kappa)$ is inaccessible since any short cofinal sequence would be a subset of $M$ of cardinality less than $\kappa$, and even $\Sigma_\omega$-Mahlo (defined by Bosch 2006, which seems to have become paywalled after I last read it) since $V_{\kappa} \prec V_{j(\kappa)}$ and $V_{\kappa} \vDash \text{"Ord is Mahlo"}$, but not generally Mahlo as the characterization in lemma 5.4 of Sato 2007, minus inaccessibility, is satisfied by a club of cardinals $\lambda \lt j(\kappa)$ in place of $j(\kappa)$ so if $j(\kappa)$ is Mahlo, there are stationarily many $\lambda \lt j(\kappa)$ that are almost-hugeness targets of critical point $\kappa$. -If $j : V_{\kappa+1} \to V_{j(\kappa)+1}$ witnesses that $\kappa$ is 1-extendible (by the coding argument of this Mathoverflow answer by Gabe Goldberg, 1-extendibility can equivalently be characterized by an elementary embedding $j : H_{\kappa^+} \to H_{j(\kappa)^+}$) then $\kappa$ is Woodin, locally measurable (defined by Holy and Lücke 2020) and thus strongly Ramsey (defined by Gitman 2008), and also $\Pi^1_n$-indescribable for every $n \lt \omega$ since those are properties of $H_{\kappa^+}$ so elementarity of $j$ applies. However, $j(\kappa)$ is not generally super-Ramsey (defined by the same paper by Gitman) or even super weakly Ramsey (defined by Holy and Schlicht 2018) since for any $k : M \to N$ as in the definition of super weakly Ramsey such that $j \in M$, $N \vDash \text{"$\kappa$ is 1-extendible with target $j(\kappa)$"}$ as $H_{j(\kappa)^+}^N=M \prec H_{j(\kappa)^+}$. They are probably not generally $\Pi^2_1$-indescribable as 1-extendibility targets are probably $\Delta^2_1$-definable since the satisfaction predicate for $V_{j(\kappa)+1}$ is $\Delta^2_1$-definable and being a 1-extendibility target involves no other third-order quantifiers, and probably not generally completely ineffable as that property has a characterization involving transitive models $M \prec H_{\beth_{j(\kappa)+1}^+}$ (Nielsen and Welch 2018, theorem 3.12). -If $\kappa$ is subcompact (that is, for every $A \subset H_{\kappa^+}$ there is a 1-extendibility embedding $j : (H_{\lambda^+}, B) \to (H_{\kappa^+}, A)$) then, in addition to the properties of 1-extendibility targets, $\kappa$ is fully Ramsey (defined by Holy and Schlicht 2018), for if there was a winning strategy $\sigma$ for player I in the Holy-Schlicht game (and whether a strategy is winning is definable over $H_{\kappa^+}$ since any play of the game is an element of $H_{\kappa^+}$), then, by subcompactness, there is an elementary embedding $j : (H_{\lambda^+}, \tau) \to (H_{\kappa^+}, \sigma)$ so that $\tau$ is a strategy witnessing that $\lambda$ isn't fully Ramsey, which is impossible since $\lambda$ is 1-extendible. However, $\kappa$ is not generally measurable for if $k : V \to M$ witnesses that $\kappa$ is measurable, $M \vDash \text{"$\kappa$ is subcompact"}$ because $M$ contains the true $H_{\kappa^+}$ and every $A \subset H_{\kappa^+}$ in M is in V, and not generally $\Pi^2_1$-indescribable as subcompactness is $\Pi^2_1$-definable. -If $\kappa$ is quasicompact (that is, for every $A \subset H_{\kappa^+}$ there is a 1-extendibility embedding $j : (H_{\kappa^+}, A) \to (H_{j(\kappa)^+}, j(A))$) then some of the targets of the witnessing embeddings are measurable, as can be seen by taking $A$ to be a measure witnessing that $\kappa$ is measurable. -If $j : V \to M$ witnesses that $\kappa$ is 2-fold 1-strong (that is, $V_{j(\kappa)+1} \subset M$; Sato 2007, definition 9.1) then $j(\kappa)$ is measurable since, if $U$ is a measure witnessing that $\kappa$ is measurable, $j(U)$ witnesses that $j(\kappa)$ is measurable as it measures all subsets of $j(\kappa)$. -If $j : V \to M$ witnesses that $\kappa$ is huge (that is, $\phantom{M}^{j(\kappa)}M \subset M$), then $j$ also witnesses that $\kappa$ is 2-fold 1-strong, so $j(\kappa)$ is measurable.<|endoftext|> -TITLE: How bad can the second derivative of a convex function be? -QUESTION [17 upvotes]: One can easily construct an example of a measurable function $f:(a,b)\to \mathbb{R}$ which satisfies the following property: -$$\label{p}\tag{P} -f\notin L^1(I),\ \mbox{for each interval}\ I\subset (a,b). -$$ -We know that a convex function $g:(a,b)\to \mathbb{R}$ is locally Lipschitz and its second derivative $g''$ exists a.e. $x\in (a,b)$. - -Can we find a increasing convex function $g$, such that $g''$ satisfies the property \ref{p}? - -If the answer is negative - -What can we say about the set of intervals $I$ for which $g''\notin L^1(I)$? - -Remark: I've asked a similar question on Math.stack with no answer. - -REPLY [16 votes]: The second derivative of a convex function, in the distributional sense, is a non-negative bounded measure. And conversely. If this measure $\mu$ contains a sum $\sum_na_n\delta_{x=x_n}$, where $(x_n)_n$ is dense and $a_n>0$, $\sum_na_n<\infty$, then $\mu|_I\not\in L^1(I)$ for every non-void open interval $I$. The corresponding $g$ is an example for the first question. -On the other hand $g''$ can be decomposed uniquely into a singular part $\mu_s$ and a part $\mu_a<\!<{\cal Leb}$. Because $\mu_s,\mu_a$ are non-negative, we have -$$\int_c^dd\mu_a\le|g''|<\infty,$$ -at least if $a -TITLE: How many generators does a direct product of alternating groups need? -QUESTION [12 upvotes]: P. Hall gave a formula for the number of generators of $G^n$ for any finite simple group $G$. One famous example is the fact that $A_5^{19}$ is 2-generated, but $A_5^{20}$ is not. The question of computing $d(G^n)$ has extensively been studied, starting with the work of Wiegold. However, what we need is a bound for the direct product of different factors. More precisely, for $k_1 -TITLE: Calabi-Yau manifolds and knot theory -QUESTION [10 upvotes]: In the paper "The Volume Conjecture and Topological Strings" it is said that the mirror Calabi-Yau threefold is given by -$X := \{ (x,y,u,v) \in \mathbb{C^* \times\mathbb{C^*} \times \mathbb{C} \times \mathbb{C}} $ : uv = A(x,y) }, -where A(x,y) is the A-polynomial for a given knot. -Let us take the A-polynomial for the trefoil knot: $A(x,y)=(y-1)(y+x^6)$ which gives us Calabi-Yau manifold defined by the equation: -$uv = (y-1)(y+x^6)$ -My questions are: -1) Why is this a Calabi-Yau manifold ? There are some very specific conditions to be satisfied, how do I check them or make sure that they are satisfied for manifolds defined this way? If I'm correct, these conditions would require for me to know for example a curvature form of the tangent bundle. I also don't know the metric so it would be hard to even compute something. -2) Is there any universal condition that can tell me if this is a Calabi-Yau manifold? Maybe something from knot theory? - -REPLY [11 votes]: $X$ is certainly Calabi-Yau in the algebraic sense: the canonical line bundle is trivialized by the holomorphic volume form -$\Omega=\frac{dx}{x} \wedge \frac{dy}{y} \wedge \frac{du}{u}$ -(this comes from the fact that $X$ is a conic bundle over $(\mathbb{C}^{*})^2$, degenerate over the curve $A=0$. The form $\Omega$ is constructed from the standard holomorphic volume form on $(\mathbb{C}^{*})^2$ and from the standard holomorphic volume on the $\mathbb{C}^{*}$ fiber). -In the compact Kähler case, this would be enough to insure the metric Calabi-Yau property, i.e. the existence of a Ricci flat metric, by Yau's theorem. But here, $X$ is not compact, so one has to be careful about question of completeness. My guess would be that in this particular example, there really exists a complete Ricci-flat metric but I don't know if/where it has been proved (if true). -In any case, what I have written is for any Laurent polynomial $A(x,y)$ and I don't think that the fact that $A$ comes from a knot should play a role (for this particular question). -EDIT (taking into account Vivek Shende's comment to the question): in my answer, I am assuming $A$ generic so that the curve $A=0$ is smooth and then $X$ is also smooth. But if $A=0$ is singular, as it is typical for the case coming from a knot (for a knot, there is always a factor $(y-1)$ if I remember correctly so the curve $A=0$ is always reducible), then $X=0$ will also be singular and so one has to be careful about what one means by Calabi-Yau).<|endoftext|> -TITLE: Sato-Tate conjecture when Fourier coefficients are complex numbers -QUESTION [7 upvotes]: Let $k\geq 1$ and let $f=\sum_{n\geq 1}a(n)q^{n}$, $a(n)\in\mathbb{R}$, be a normalised cuspidal Hecke eigenform of weight $2k$ for $\Gamma_{0}(N)$ without complex multiplication. So the result of Barnet-Lamb et al. tells us that the numbers $\frac{a(p)}{2p^{k-1/2}}$ are $\mu$-equidistributed in $[-1,1]$, where $\mu$ is the probability measure on the interval $[-1,1]$ defined by $\frac{2}{\pi}\sqrt{1-t^{2}}\,dt$, and $p$ runs through the primes not dividing $N$. -I am looking for any generalisation of Sato-Tate conjecture when Fourier coefficients $a(n)$ are complex numbers, $a(n)\in\mathbb{C}$? - -REPLY [6 votes]: First, note that if $f = \sum_{n \geq 1} a(n) q^{n}$ is a cuspidal Hecke eigenform lying in the new subspace of $S_{k}(\Gamma_{0}(N), \chi)$, where $\chi$ is a Dirichlet character modulo $n$, then one can determine the argument of $a(n)$ in terms of the values of $\chi$. In particular, for $\gcd(n,N) = 1$, then -$$ \overline{a(n)} = \chi(n) a(n). $$ -(A proof of this relation can be found in Iwaniec's "Topics in Classical Automorphic Forms" text.) -This is the reason that if $\chi$ is trivial, all the Fourier coefficient of $f$ are real. In general, if $\zeta$ is a root of unity and $\chi(p) = \zeta^{2}$, then $\frac{a(p)}{\zeta}$ is real. -A generalization of the Sato-Tate conjecture is stated and proved in the paper of Barnet-Lamb et. al. (which can be found online here). In particular, Theorem B of that paper states that if $\zeta$ is a root of unity with $\zeta^{2}$ in the image of $\chi$, then the numbers -$$ \left\{ \frac{a(p)}{2p^{(k-1)/2} \zeta} : p~\text{prime}, \chi(p) = \zeta^{2} \right\}$$ -are equidistributed in $[-1,1]$ with respect to $\frac{2}{\pi} \sqrt{1-t^{2}}$.<|endoftext|> -TITLE: Don Zagier's "Zetafunktionen und quadratische Körper" -QUESTION [5 upvotes]: Do you know of a text--preferably in English--whose treatment of the class number formula is based on (or follows closely) the one expounded by Zagier in sections II.8 (binary quadratic forms) and II.9 ($L(1, \chi)$ and the class number) of the aforementioned book? Unfortunately, my command of the German language is not so good yet and, to add insult to injury, it seems to me that Springer Verlag is still to commission a translation into English (or one of the Romance languages) of this notable book. -Let me thank you in advance for your attentive consideration of this question of mine. -BOUNTY! Jan-Christoph Schlage-Puchta suggests chapter 6 of H. Davenport's "Multiplicative Number Theory" as an alternative text for this topic. Though, I must confess that the fact that Davenport dedicates only two or three lines of the chapter to Lagrange's main result on reduction of binary quadratic forms makes one feel uneasy right from the start. So, if you are proficient in German and wish to provide a translation of pages 64-68 of Zagier's book, I will really appreciate your help and award you the bounty I am herewith offering on this question. - -REPLY [4 votes]: Zagier's proofs are compact, so I don't think there's any way to reduce further. 4 pages can be a lengthy translation but here is enough to to get started. - -Theorem 2 Let $f(x,y) = ax^2 + bxy + cy^2$ be a primitive quadratic form of discriminant $D$ (squarefree). There is a bijection between solutions to Pell's equation and the automorphism group of $f$ -$$ (t,u) \to \left( \begin{array}{cc} \frac{t - bu}{2} & - cu \\ au & \frac{t+bu}{2}\end{array} \right) $$ -This bijection is a group isomorhism by the composition rule: -$$ (t_1, u_1) \circ(t_2, u_2) = \left( \frac{t_1 t_2 + D u_1 u_2}{2}, \frac{t_1 u_2 + t_2 u_1}{2}\right)$$ -The group $U_f$ is finite if $D < 0$ and cyclic of order $w$ - -$\mathbb{Z}/6\mathbb{Z}$ for $D = - 3$ -$\mathbb{Z}/4\mathbb{Z}$ for $D = - 4$ -$\mathbb{Z}/2\mathbb{Z}$ for $D < - 4$ - -For positive $D > 0$ the group of solutions $U_f \simeq \mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ -Proof Zagier says the positive discriminant case follows from Pell's equation (it does). If you have a solution to Pell's equation (for any $f(x,y) = n$) , both group laws are pretty clear and that they map to one another. There is a map (injective homomorphism) from $U_f$ either to $\mathbb{C}^\ast$ or $\mathbb{R}^\ast$ given by -$$ (t,u) \mapsto \frac{t + u \sqrt{D} }{2} $$ -the case $D = - 3$ corresonds to $\mathbb{Z}[e^{2\pi i / 3}]$ and $D = - 4$ matches $\mathbb{Z}[i]$, with units $e^{2\pi i /3}$ or $i$. In all other cases $D < -4$ there are only the trivial units $\pm 1$. $\hspace{1.4in}\square$ - -Theorem 3 Let $D$ be a "fundamental discriminant" $n \neq 0$ be a whole number. Then the total number $R(n)$ of representation of $n$ by primitive forms of discriminant $D$ is given by -$$ R(n) = \sum_{m|n} \chi_D(m) $$ -where $m$ is a divisor of $n$ and $\chi_D(m)$ is a character. In particlar, $R(n)$ and all $R(n,f)$ are finite. -Proof There are no imprimitive forms, so we can leave outthe word "primitive" in the theorem. Let $R^\ast(n)$ be the number of inequivalent primitive representations of $n$ by (any) forms of discriminant $D$. Obviously -$$ R(n) = \sum_{g \geq 1, g^2 | n} R^\ast (n/g^2) $$ -since every representation is a multiple of a primitive one. The main step in the proof is the formula -$$ R^\ast(n) = \{ b \pmod {2n} \big| b^2 \equiv D \pmod{4n}\} $$ -Then he proceeds to explain the notion of a group action and variants of the orbit stabilizer formula -Let $G$ be a group $X$ and $Y$ be two sets and $S \subset X \times Y$. Then we can study the diagonal group action $(x,y) \mapsto (gx,gy)$ and observe the quotient sets $X/G$ as well as $S/G$. - -$Y_x = \{ y \in Y | (x,y) \in S \} $ -$G_x = \{ g \in G | gx = x \}$ -$|S/G| = \sum_{x \in X/G} |Y_x/G_x| $ -$|S/G| = \sum_{y \in Y/G} |X_y/G_y| $ - -The choice of group action by Zagier is set of invertible $2 \times 2$ matrices acting on pairs $X \times Y = $ -$$ \big\{\text{quadratic forms }ax^2 + bxy+cy^2 \text{ with }b^2 - 4ac = D \} \times \{ \text{pairs of integers }x,y \in \mathbb{Z}\big\} $$ -and $G = SL(2,\mathbb{Z})$ which is solutions to $ad-bc = 1$ for $a,b,c,d \in \mathbb{Z}$. Lastly the special set - -$S = S(n) =\{ (f,z) \in X \times Y : f(z) = n \} $ - -Then $X/G$ is the set of equivalence classes of discriminant $D$ , and by the group action formula: -$$ |S/G| = \sum_{[f]} R^\ast (n,f) = R^\ast(n) $$ -Every element of $Y$ is equivalent to $z = (1,0)$ (Exercise) for this element: - -$G_z = \left\{ \left(\begin{array}{cc} 1 & r \\ 0 & 1 \end{array} \right) : r \in \mathbb{Z}\right\} $ -$X_z = \{ nx^2 + bxy + \frac{b^2 - D}{4n} y^2: b \in \mathbb{Z}, b^2 \equiv D \pmod {4n} \}$ -these are related to the substitions $b \mapsto b + 2nr$ - -Zagier observes that $R^\ast(n)$ is multiplicative so that enough we solve for the primes -$$ R^\ast(p^r) = \# \big\{ b \pmod{p^r} \big| b^2 \equiv D \pmod{p^r} \big\} $$ -Then we can reconstruct $R(n)$ from the count of "primitive" representations of $n$, $R^\ast(n)$ - -$R(p^r) = \sum_{0 \leq s < r/2}2 + \sum_{s=r/2} 1 = r+1 = \sum_{0 \leq i \leq r} \chi_D(p^i) $ if $(\frac{D}{p})=+1$ -$R(p^r) = \sum_{0 \leq s < r/2}0 + \sum_{s=r/2} 1 = 0 \text{ or }1 = \sum_{0 \leq i \leq r} \chi_D(p^i) $ if $(\frac{D}{p})=-1$ -$R(p^r) = \sum_{0 \leq s < (r-1)/2}0 + \sum_{(r-1)/2 \leq s \leq r/2} 1 = 1 = \sum_{0 \leq i \leq r} \chi_D(p^i) $ if $(\frac{D}{p})=+1$ - -and Theorem 3 checks in all cases. $\hspace{2in}\square$ - -Corollary The average number of representations of $n$ is exactly the value of the L-function -$$ \lim_{N \to \infty} \left( \frac{1}{N}\sum_{n=1}^N R(n) \right) = L(1, \chi_D) $$ -The proof involves counting lattice points under a hyperbola. For a specific primitive $f$ the formula involves the fundamental unit. -Theorem 4 Let $\epsilon_0$ be fundamental unit of $f$. -$$ \lim_{N \to \infty} \left( \frac{1}{N}\sum_{n=1}^N R(n,f) \right) =$$ -$$\left\{ \begin{array}{cc} -\frac{2\pi}{w\sqrt{|D|}} & \text{ if }D < 0 \\ -\frac{\log \epsilon_0 }{w\sqrt{D}} & \text{ if }D > 0 \\ -\end{array}\right.$$<|endoftext|> -TITLE: Are there other semidirect product/crossed products in other areas -QUESTION [14 upvotes]: Suppose $(O, G, \alpha)$ is a triple where $O$ is some mathematical object, $G$ is a group and $\alpha : G \rightarrow Aut(O)$. Many different areas of mathematics study such triples. However, I only know of a couple of examples of a new object that encodes this system. -1) If $G$ is a group, $N$ a normal subgroup of $G$, $H$ a subgroup of $G$ and $\alpha(h)(n) = hnh^{-1}$, then one can encode $(N, H, \alpha)$ into the (outer) semidirect product $N \rtimes_\alpha H$. -2) If $O = \mathcal A$, a C$^*$-algebra, $G$ is a locally compact group and $\alpha : G \rightarrow Aut(\mathcal A)$ acts by $*$-automorphisms then $(\mathcal A, G, \alpha)$ is encoded as the crossed product algebra $\mathcal A \rtimes_\alpha G$. In very simple terms, this is accomplished by considering unitary representations of $G$. There is also a related construction in Statistical Mechanics called the covariance algebra. -My question then: are there any other semidirect/crossed product type constructions for such triples $(O, G, \alpha)$? - -REPLY [10 votes]: In any higher category $C$, given an object $X$ and an action of a group $G$ on it you can ask for the homotopy quotient $X_{hG}$ of $X$ by the action of $G$, which is defined by a homotopy-coherent version of the universal property of the usual quotient. This recovers the following constructions in the following higher categories: - -In the 2-category of groups (thinking of groups as one-object categories), this recovers the semidirect product, and actually a generalization of it describing general group extensions. -In the 2-category of rings (thinking of rings as one-object linear categories), this recovers the crossed product, and again a generalization of it including e.g. twisted group algebras. -In the $\infty$-category of spaces, this recovers the construction described in Mark Grant's answer. -In the $\infty$-category of chain complexes, this recovers a general form of group cohomology with nontrivial coefficients.<|endoftext|> -TITLE: tannakian description of vector bundles -QUESTION [11 upvotes]: Let $S$ be a scheme and $E$ be a rank $n$ vector bundle on $S$. $E$ corresponds to a $GL_n$ torsor $P$ via the definition -$$ -P = Isom_S(E,\mathbf{A}_S^n) -$$ -Tannakian theory tells us that $P$ corresponds to a fiber functor -$$ -\eta : Rep(GL_n) \to Vect_S. -$$ -(Recall that $\eta(V)$ is defined to be the pushout of $P$ via the representation $GL_n \to GL(V))$ -My questions is : is there an easy description of $\eta$ in terms of $E$ without going through $P$ ? - -REPLY [3 votes]: How is your vector bundle $E$ presented? (As a locally free sheaf? As it's total space? Or...) If you like to think in terms of transition functions, i.e., Čech cocycle $\underline{\sigma} = \{\sigma_{ij}\} \in H^1(S,GL_n)$ (of non-abelian cohomology group), then the representation $(\pi,V) \in Rep_k(GL_n)$ will be sent to the vector bundle corresponding to the Čech coycle $\{ \pi(\sigma_{ij}) \} \in H^1(S,GL_V)$. -To get an idea of what is going on in the previous answers: This is essentially "resolving" $S$ using a cover $\mathfrak{U}=\{U\}$ and the Čech nerve $C^\bullet(\mathfrak{U})$ and thinking of the stacks $BGL_n$ and $BGL_V$ as simplicial schemes via the Milnor bar constrcution. We get that the bundle $\eta(\pi,V)$ is classified by the composite -$C^\bullet(\mathfrak{U}) \to BGL_n^\bullet \to BGL_V^\bullet$ -where the first morphism (is the classifying map and) is precisely the data $\underline{\sigma}$ and the second is $B\pi$ (the functor $B$ applied to $\pi : GL_n \to GL_V$). -Is at least the first description closer to what you wanted?<|endoftext|> -TITLE: Are there any recent advances in formalizing the undecidability of $\mathit{CH}$? -QUESTION [6 upvotes]: The website Formalizing 100 Theorems by Freek Wiedijk contains a list of some theorems that were chosen at some point as good candidates for formalization (because of their complexity, their importance, etc.) This website seems to be updated very often. -Among the proofs not yet formalized is that of the independence of the Continuum Hypothesis from the axioms of set theory. - -What is the current state of the formalization of the independence of $\mathit{CH}$ from $\mathit{ZFC}$? - -I browsed this site for more information, and I found this recent question, as well as this one and this, and an answer in math.SE. But I couldn't find information directly concerned with my question. - -REPLY [7 votes]: Jesse Michael Han and Floris van Doorn recently formalized the independence of the continuum hypothesis in the Lean theorem prover. See the Flypitch project webpage for their papers and code.<|endoftext|> -TITLE: $n!$ divides a product: Part I -QUESTION [14 upvotes]: Question. The following is always an integer. Is it not? - $$\frac{(2^n-1)(2^n-2)(2^n-4)(2^n-8)\cdots(2^n-2^{n-1})}{n!}.$$ - -John Shareshian has supplied a cute proof. I'm encouraged to ask: - -Question. Can you give alternative proofs, even if they are not particularly as short? - -REPLY [6 votes]: Essentially the same argument: the ratio is a number of bases of the vector space $\mathbb{F}_2^n$. -A proof working also for composite values of $a=2$: -Let $p$ be a prime dividing $n!$. If $p$ divides $a$, then the product $\prod (a^n-a^k)$ is divisible by $p^{n-1}$, while $n!$ is not divisible by $p^n$. If $a$ and $p$ are coprime, it suffices to check that for any exponent $s$ the set $\{a^n-a^k\}$ contains at least as many numbers divisible by $p^s$ as the set $\{1,\dots,n\}$. It follows from the fact that a period of powers of a sequence $\{1,a,a^2,\dots\}$ modulo $p^s$ is less than $p^s$.<|endoftext|> -TITLE: What algebras are quotients of $\ell_1(\mathbf{N})$? -QUESTION [8 upvotes]: Every separable Banach space is a linear quotient of $\ell_1$, however not every separable Banach algebra is a Banach-algebra quotient of $\ell_1(G)$ for some group $G$ (these are the so called unitary Banach algebras). -Is every separable Banach algebra a quotient of $\ell_1(S)$ for some countable semigroup? -I'd be interested in knowing what could be the Banach-algebra quotients of $\ell_1(\mathbf{N}_0)$ and $\ell_1(\mathbf{Z})$. -Does either of them quotient onto $C(X)$ for some infinite compact metric space $X$? - -REPLY [8 votes]: (What follows is largely the result of digging around online, based on knowing a few more magic words than the OP.) -Answer to the first question (I think). -Let $V$ be a separable Banach space. The standard proof that $V$ is isometrically isomorphic to a linear quotient of $\ell_1$ works by choosing a countable set $X$ that is a dense subset of the unit sphere of $V$, defining $f:\ell_1(X) \to V$ in the obvious way, and then verifying two things: -(i) $f$ is surjective; -(ii) the natural map $f_1: \ell_1(X)/\ker(f) \to V$ is an isometry. -Suppose $V$ is actually a unital Banach algebra. Let $S$ be the multiplicative subsemigroup of $V$ generated by $X$; of course $S$ is countable. Although $S$ might not lie in the unit sphere of $V$, it does lie in the unit ball, and so we do get a norm-decreasing homomorphism $h: \ell_1(S) \to V$. -We may factorize $f$ as $h\circ \iota$ where $\iota: \ell_1(X)\to \ell_1(S)$ is the natural isometric embedding induced from $X\subset S$. Since $f$ is surjective, so is $h$. Moreover, since $f_1$ is an isometry: given $b$ in the unit ball of $V$ and $K>1$ we can find $\xi\in \ell_1(X)$ with $f(\xi)=b$ and $\Vert\xi\Vert_1 < K$. So $b=h(\iota(\xi))$ where $\Vert\iota(\xi)\Vert_1 < K$. This shows that $h$ is actually a quotient map of Banach spaces, as required. -Comment on the intermediate question. I am not sure right now what one can say about the Banach algebra quotients of these algebras in general. -Partial answer to the final question. Via the Gelfand transform we can regard $\ell_1({\bf Z}_+)$ as an algebra of continuous functions on the closed unit disc, and hence for any closed subset $E$ of the unit disc, there is a norm-decreasing homomorphism $r_E: \ell_1({\bf Z}_+)\to C(E)$ given by restriction of these functions. It is known that for some infinite closed $E\subset {\bf T}$ this restriction homomorphism $r_E$ can be surjective: according to remarks in Volume 2 of Hewitt&Ross, these were originally called Carleson sets, but Wik showed that this coincides with the more widely known notion of a Helson subset of ${\bf T}$, which corresponds to $r_E: \ell_1({\bf Z}) \to C(E)$. - -MR0125404 (23 #A2707) I. Wik, On linear dependence in closed sets. Ark. Mat. 4 1961 209–218. - -The condition that $r_E:\ell_1({\bf Z})\to C(E)$ be a quotient map of Banach spaces corresponds to the classical notion of $E$ being a subset of ${\bf T}$ with Helson constant one. Such sets are known to exist, but I confess I don't know anything about the proofs/constructions. Wik's result shows that Helson sets are Carleson sets, but it is not clear to me from his proof if "1-Helson sets" are automatically "1-Carleson". -So I can't answer to your question in the $\ell_1({\bf Z}_+)$ case, but the answer is positive in the $\ell_1({\bf Z})$-case.<|endoftext|> -TITLE: Ring of invariants -QUESTION [6 upvotes]: It is known that if you have a reductive group acting on a regular algebra, then the ring of invariants is Cohen-Macaulay. Is this still true for Gorenstein algebras? - -REPLY [7 votes]: That is not true. Let $k$ be a field. Let $R$ be the regular $k$-algebra, $R=k[x,y,z,w].$ Let $G = \text{Spec}\ k[s,s^{-1},t,t^{-1}]$ be the multiplicative $k$-group of rank $2$. Let $G$ act on $\text{Spec}(R)$ via the following coaction, $$m^*:k[x,y,z,w] \to k[s,s^{-1},t,t^{-1},x,y,z,w], $$ $$x\mapsto sx, \ \ y \mapsto s^{-1}y, \ \ z\mapsto tz, \ \ w\mapsto t^{-1}w.$$ The invariant subring is $k[u,v]$, where $u$ equals $xy$ and $v$ equals $zw$. -Now consider the ideal in $R$, $I=\langle x^2,yz\rangle$. This ideal is generated by a regular sequence. Thus the quotient algebra $R/I$ is Gorenstein, and even a complete intersection ring. Moreover, $I$ is $G$-stabilized. Thus, the action of $m$ on $\text{Spec}(R)$ restricts to an action of $m$ on the closed subscheme $\text{Spec}(R/I)$. -Since $G$ is linearly reductive over $k$, the invariant ring $(R/I)^G$ equals $R^G/I^G$. Finally, $I^G$ equals $\langle u^2,uv\rangle$. Thus the invariant ring $(R/I)^G$ is $k[u,v]/\langle u^2,uv \rangle$. This is not a Cohen-Macaulay ring; it is not even $S_1$.<|endoftext|> -TITLE: Self-covering spaces -QUESTION [15 upvotes]: Let $M$ be a connected Hausdorff second countable topological space. I will call $M$ self-covering if it is its own $n$-fold cover for some $n>1$. For instance, the circle is its own double cover by $e^{2\imath\varphi}\mapsto e^{\imath\varphi}$. -Questions: - -What would be a reasonable assumption on $M$ to ensure it is self-covering? Is every connected compact/closed topological manifold self-covering? -Assume that $M$ is a differentiable manifold, and let $\phi:M\to M$ be the local diffeomorphism providing the $n$-fold cover for $n>1$. Is it true that the modulus of the Jacobian $|\det J(d\phi(x))|<1$ everywhere aside from countably many isolated points $x$ where it can be 1? In other words, is a covering map a volume contraction? - -Please, note that I am not a topologist; more accessible language and references would be appreciated. - -REPLY [17 votes]: Here are some pretty examples of self covering manifolds: Suppose that $F$ is a manifold and $f \colon F \to F$ is a periodic homeomorphism of period $k$. We define $M_f = F \times I \,/\, f$ to be the mapping torus with monodromy $f$. That is, form the product $F \times I$ and identify the two ends via $f$. Note that $M_f$ is an $F$-bundle over the circle. -Now, the $(k+1)$-fold cover of $M_f$, obtained by "unwrapping the circle direction" is the mapping torus for $f^{k+1} = f$. Thus the $(k+1)$-fold cover is homeomorphic to $M_f$. -As a concrete example, the trefoil knot complement $X_T$ is a once-punctured torus bundle over the circle, with monodromy of order six. Thus $X_T$ seven-fold covers itself. (And also five-fold covers itself.) This trick works for any torus knot complement.<|endoftext|> -TITLE: Upsilon of an alternating knot -QUESTION [5 upvotes]: I have a couple of questions about how Oz-Stip-Sz computes the upsilon function invariant of an alternating knot in their upsilon ($\Upsilon$) paper here.1 This is theorem 1.14 (on the bottom of page 3); the proof is at the top of page 22. In the proof they state that for alternating knots one can find two sets of generators for $CFK^\infty$ with certain nice properties. These are the $x_i$'s and the $y_j$'s in the proof. My questions are: how do we find the $x_i$ and $y_j$ generators, and what is the role of the $y_j$ generators in determining the upsilon ($\Upsilon$) function? - -1Ozsvath, Peter, Andras Stipsicz, and Zoltán Szabó. "Concordance homomorphisms from knot Floer homology." arXiv:1407.1795 (2014). - -REPLY [4 votes]: The existence of generators $x_i$ and $y_i$ follows from the fact that alternating knots are thin (i.e. their knot Floer homology is supported on a diagonal $M-A=\text{constant}$) and (the filtered quasi-isomorphism class of) their $CFK^\infty$ is determined by the Alexander polynomial. This was proven originally by Ozsváth and Szabó in the paper: - -Peter S. Ozsváth and Zoltán Szabó, Heegaard Floer homology and alternating knots, Geom. Topol. 7 (2003), 225–254. - -Another, perhaps more modern, exposition is in Petkova's paper: - -Ina Petkova, Cables of thin knots and bordered Heegaard Floer homology, Quantum Topol. 4 (2013), no. 4, 377–409. - -As for the second question, I'm not sure I understand what you mean by "the role of $y_i$". In my view, they are just generators in the complex with a certain boundary. At a practical level, what they do is they identify each $x_i$ with $U^{i-j}x_j$ in $CFK^\infty$. -You might also want to take a look at Livingston's survey on $\Upsilon$, that has a more transparent reformulation of $\Upsilon$ in terms of certain "slanted filtrations" on $CFK^\infty$. - -Charles Livingston, Notes on the knot concordance invariant Upsilon, Algebr. & Geom. Topol. 17 (2017) 111–130.<|endoftext|> -TITLE: Non-free projective pearls (general and Abelian) -QUESTION [5 upvotes]: A pearl is an ordered pair $\ \mathbf P:=(G\,\ S),\ $ where $\ G\ $ is a group, and $\ S\ $ is a non-empty subset of G which does not contain the neutral element of $\ G\ $ (i.e. not 1 in the multiplicative notation, nor 0 in the additive notation). -If $\ \mathbf Q:=(H\,\ T)\ $ is another pearl then a morphism -$\ f:\mathbf P\rightarrow \mathbf Q\ $ is defined as a group homomorphism -$\ f:G\rightarrow H\ $ such that $\ f(S) \subseteq T$. -This defines the (general) category of pearls. Pearls for which their group is Abelian, form a full subcategory (with all their morphisms between them). -Let's define a free pearl $\ \mathbf P\ $ (as above) as one for which a subset $\ F\subseteq G\ $ is a set of the free generators of the group $\ G,\ $ and $\ S\subseteq F.\ $ Obviously Possibly, every free pearl is projective (in the category of pearls). -The definition of an abelian-free pearl (and the issue of projectivity) is similar as for the category of abelian pearls. -QUESTION   Can you provide examples of projective pearls which are not free? The same for the case of the category of abelian pearls. - -REMARK This requires paying close attention to pearl epimorphisms. - -                M O T I V A T I O N - -Categories of enhanced groups are among the most important. After groups, pearls are among the simplest among them. Thus it provides a view onto the next categories which already appear; -There are problems on the general categories (or a general class of them) when we would like to obtain a positive result or else a counter-example. Then pearls may provide a relatively simple testing ground and they may project onto other categories. -Truly, the notion of pearls appeals to me, I find them attractive. - - -EDIT   After reading @ArturoMagidin comments, I replaced my careless obviously by careful possibly. (I'd like to see Arturo's comments posted by him as one or more answers to make them more readable, please). - -REPLY [2 votes]: Pearls may be defined as first-order structures $\mathbf P = \langle P; *, {}^{-1},e,S\rangle$ in the language of one binary operation, one unary operation, one 0-ary operation, and one unary relation, axiomatized by - -identities saying that $\langle P; *, {}^{-1},e\rangle$ is a group, -the universal Horn sentence $\neg S(e)$, and -the existential sentence $\exists x S(x)$. - -Pearl morphisms have been defined so that they are exactly the structure-preserving maps. -Comment 1: Once you define the objects and morphisms of your category, and some underlying set functor is understood, you no longer have the right to define your free objects. If they exist, they are already determined. -Comment 2: The category of pearls, as defined in the problem, has no free objects. -Comment 3: If you delete Axiom 3, the one that asserts that $S$ interprets as a nonempty relation, then the category is a universal Horn class, so free objects exist. The free object over the set $X$ is the free group over $X$ with $S$ interpreted as the empty relation. These are exactly the projective objects in this category, too.<|endoftext|> -TITLE: n! divides a product: Part II -QUESTION [9 upvotes]: This is a follow up on another MO question. - -Question. For $n\geq2$, the following is always an integer. Is it not? - $$\frac{(2^n-2)(2^{n-1}-2)\cdots(2^3-2)(2^2-2)}{n!}.$$ - -REPLY [11 votes]: Here is a direct number theoretic proof. First of all note that -$$(2^n-2)(2^{n-1}-2)\cdots(2^2-2)=2^{n-1}(2^{n-1}-1)(2^{n-2}-1)\cdots(2^1-1).$$ -Now for any prime $p$ we have, by Legendre's formula and by Fermat's little theorem, -$$v_p(n!)\leq\left\lfloor\frac{n-1}{p-1}\right\rfloor\leq v_p\bigl(2^{n-1}(2^{n-1}-1)(2^{n-2}-1)\cdots(2^1-1)\bigr),$$ -where for the second inequality we argue separately for $p=2$ and for $p\geq 3$. The result follows. -P.S. The above proof was inspired by Cherng-tiao Perng's deleted response (which was not entirely correct).<|endoftext|> -TITLE: Thurston's 24 questions: All settled? -QUESTION [46 upvotes]: Thurston's 1982 article on three-dimensional manifolds1 ends with $24$ "open questions": - - -      $\cdots$ - - - -Two naive questions from an outsider: -(1) Have all $24$ now been resolved? -(2) If so, were they all resolved in his lifetime? - - -1Thurston, William P. "Three dimensional manifolds, Kleinian groups and hyperbolic geometry." Bull. Amer. Math. Soc, 6.3 (1982). - Also: In Proc. Sympos. Pure Math, vol. 39, pp. 87-111. 1983. - Citseer PDF download link. - - -Answered by Ian Agol, Andy Putman, and Igor Rivin. Ian: "Problems 1-18 have been completely answered....Problems 19-24 are more open-ended," and difficult to declare "all settled" (as emphasized by YCor). But, as Andy says, -"with the exception of problem 23." Back to Ian: -"One can imagine, however, a complete and satisfactory answer eventually to question 23." - -REPLY [52 votes]: A nice summary of the status of these problems may be found here: -Otal, Jean-Pierre, William P. Thurston: ``Three-dimensional manifolds, Kleinian groups and hyperbolic geometry", Jahresber. Dtsch. Math.-Ver. 116, No. 1, 3-20 (2014). ZBL1301.00035. -I would charaterize it this way: Problems 1-18 have been completely answered, although some of the answers are still unpublished (but exist as preprints and are still under submission). However, some of these questions (maybe 4., 7, and 8.) are less precisely formulated and somewhat open-ended, so one could argue whether or not they are answered completely. For problem 4., Hodgson's thesis addresses one part of the question: "Describe the limiting geometry -which occurs when hyperbolic Dehn surgery breaks down." I could list several other papers on this topic. -Problems 19-24 are more open-ended, and thus will likely never be completely satisfactorily addressed (I discussed some of these problems here). No progress has been made on problem 23 as far as I know (which was due to Milnor originally, not Thurston). One can imagine, -however, a complete and satisfactory answer eventually to question 23. -Otal does not comment much on Problem 24, which again is somewhat imprecise (what does "most" mean?). As Igor mentions, Joseph Maher has given a satisfactory answer. One could also argue that this was satisfactorily answered in Hempel's paper together with geometrization (Heegaard distance $>2$ implies hyperbolic). But there is other work on this question, such as giving a model for a hyperbolic manifold of bounded genus and bounded geometry (a lower bound on the injectivity radius) by Brock, Minsky, Namazi, and Souto. Moreover, these authors have a program to understand the unbounded geometry case as well (so it would eventually give a description of the geometry of all hyperbolic manifolds of bounded Heegaard genus in some sense). Thus, one could consider this problem to still be open for a variety of reasons. - -REPLY [23 votes]: They have all been resolved in some fashion with the exception of problem 23, which asserts that the volumes of all hyperbolic 3-manifolds are not rationally related. We know basically nothing about the diophantine properties of hyperbolic volumes.<|endoftext|> -TITLE: Diagram folding of simple Lie algebras -QUESTION [11 upvotes]: A non-simply laced simple root system can be constructed from the simply-laced root system by folding the Dynkin diagram and hence the corresponding non-simply-laced Lie algebra can be constructed by taking the fixed points of a non-trivial diagram automorphism (outer automorphism). Then how are their Grassmannians related ? Specifically, if $\sigma$ is an outer automorphism of $SL_{2n}$ and $P$ is a maximal parabolic, then is it true that $(G/P)^{\sigma}$ is a Grassamanian of $Sp_{2n}$ ? Is every Grassmannian of $Sp_{2n}$ obtained in this way ? - -REPLY [11 votes]: Most maximal parabolics of $SL_{2n}$ are not $\sigma$-invariant, not even up to conjugation. So $(G/P)^\sigma$ does not make sense. The correct statement is: Let $I\subseteq\{1,\ldots,2n-1\}$ be a symmetric subset, i.e., with $i\in I\Leftrightarrow 2n-i\in I$. Let $P_I\subseteq SL_{2n}$ be the corresponding parabolic. Then $J:=I\cap\{1,\ldots,n\}$ corresponds to a parabolic $P_J\subseteq Sp_{2n}$. Then $\sigma(P_I)=P_I$ and $(SL_{2n}/P_I)^\sigma=Sp_{2n}/P_J$. In particular, the $i$-th Grassmannian of $Sp_{2n}$ corresponds to the submaximal parabolic $P_I$ with $I=\{1,\ldots,2n-1\}\setminus\{i,2n-i\}$. -In geometric terms this corresponds to the following simple fact: the involution $\sigma$ induces an involution $U\mapsto U^\perp$ on subspaces of $\mathbb C^{2n}$. The $i$-th Grassmannian of $Sp_{2n}$ consists of $i$-dimensional isotropic subspaces $U$, i.e., with $U\subseteq U^\perp$. Thus, such a $U$ corresponds to a $\sigma$-invariant partial flag $U\subseteq V$ with $\dim U=i$ and $\dim V=2n-i$.<|endoftext|> -TITLE: A puzzle with some jumping frogs -QUESTION [31 upvotes]: (The following puzzle is ispired by this nice video of Gordon Hamilton on Numberphile) -In a pond there are $n$ leaves placed in a circle, for convenience they are numbered clockwise by $0,1,\ldots,n-1$. At the beginning, on each leaf there is a frog, so there are $n$ frogs. At each turn, the frogs can jump accordingly to the following rule: "If on leaf $j$ there are $k \geq 1$ frogs and if on leaf $(j + k) \bmod n$ there is at least one frog, then all the $k$ frogs on leaf $j$ can jump on leaf $(j + k) \bmod n$". -Is it true that the $n$ frogs can finally be all on the same leaf if and only if $n$ is a power of $2$? -It is quite easy to prove that if $n$ is a power of $2$ then there is a sequence of jumps that leads all the frogs on the same leaf. On the other hand, I checked by a brute force algorithm that no such sequence exists if $n \leq 14$ is not a power of $2$. -NOTE: I previously asked this question on MSE and I got non answers. -UPDATE 1: aorq checked that the answer is YES for all $n \leq 50$. -UPDATE 2: It might be worth trying to solve first this relaxed version of the problem: -To each solution of the puzzle associate a directed graph on the vertexes $0,1,\ldots,n-1$, where a directed edge $i \to j$ means a jump of all the frogs from leaft $i$ to leaf $j$. Say that the solution has $\ell$ leader frogs if its associated directed graph has exactly $\ell$ sources. -For which $\ell$ is it true that the $n$ frogs can finally be all on the same leaf using a series of jumps with $\ell$ leader frogs, if and only if $n$ is a power of $2$? -We know that $\ell = 1$ does the job, since with only one leader frog the problem is equivalent to "For which $n$ are the first $n$ triangular numbers distinct modulo $n$" and the answer is "only for $n$ a power of $2$". - -REPLY [3 votes]: Edit: As pertains to UPDATE 2, the post below gives references to settle the case of $\ell = 1$. - -This is a bit long for a comment, but I thought it worthwhile in case the following is non-obvious: - -It is quite easy to prove that if $n$ is a power of $2$ then there is a sequence of jumps that leads [to] all the frogs on the same leaf. - -(Maybe this really is "quite easy to prove," but I would not have known this to be the case at a glance without a familiarity with the references included below...) -If you begin with a frog and have it jump $1$ leaf, then have the newly formed duo jump $2$ leaves, then have the newly formed trio jump $3$ leaves, etc, then you move around the leaves by $+1, +2, +3, \ldots$ and, since you are jumping in a circle of $n$ leaves, the positions of frogs jumping together are at the triangular numbers modulo $n$. -And so the assertion elsewhere in this thread, which (in particular) covers the quotation-pull above, is that this will be sufficient to reach all the leaves if $n$ is a power of $2$. Equivalently, the set of the first $n$ triangular numbers modulo $n$ is $\{0, 1, \ldots, n-1\}$ if $n$ can be written as a power of $2$. -In fact, the last statement is true with if replaced by iff, and this was the subject of a very nice write-up in the New York Times' Numberplay column as Daniel Finkel’s Circle-Toss Game. There is another proof of this proposition by induction here, which includes a pointer to an earlier proof by D Knuth in: - -The Art of Computer Programming, Volume 3, Chapter 6.4. Excercise 20 in the second edition. - -Of course, with respect to the main question: - -Is it true that the $n$ frogs can finally be all on the same leaf if and only if $n$ is a power of $2$? - -just the "if" is covered: the strategy above (as proved in the three different references given) shows how to get all the frogs to the same leaf for $n$ being a power of $2$. The "only if" part is what remains difficult, since the OP is proposed in greater generality than the circle-toss game: instead of the frogs needing to jump together in the triangular number pattern already described, one has access to lots of other patterns. I have no new suggestions for the "only if" component, although perhaps seeing the different proofs mentioned above (I especially enjoy the "multi-ball argument" in the first link) could serve as some sort of inspiration.<|endoftext|> -TITLE: Dualizing the notion of topological space -QUESTION [71 upvotes]: $\require{AMScd}$ -Defining a topological space on a set $X$ is equivalent to designating certain subobjects of $X$ in ${\bf Set}$ (monomorphisms into $X$ up to equivalence) as open. The requirements on the open sets of a topological space $X$ are equivalent to requiring the following: - -$X \rightarrow X$ and $\emptyset \rightarrow X$ are open. -If $X_i \rightarrow X$ are open subobjects of $X$ for finite $i \in I$ then so is their product in the category of subobjects of $X$ -If $X_i \rightarrow X$ are open subobjects of $X$ for $i \in I$, then so is their coproduct in the category of subobjects of $X$. - -My question is about what happens when one dualizes this notion: -Let $X$ be a set and consider a subset of its quotient objects $\mathcal{S}$ such that: - -$X \rightarrow X \in \mathcal{S}$ and $X \rightarrow \{ * \} \in \mathcal{S}$. -If $X \rightarrow X_i \in \mathcal{S}$ for finite $i \in I$ then $\amalg_{i \in I} (X \rightarrow X_i) \in \mathcal{S}$, where the coproduct is taken in the category of quotient objects of $X$. -If $X \rightarrow X_i \in \mathcal{S}$ for $i \in I$, then $\prod (X \rightarrow X_i) \in \mathcal{S}$, where the product is taken in the category of quotient objects of $X$. - -Does this structure arise anywhere in practice? What is known about this notion of 'cotopological spaces'? Is there a place I can learn about them? - -Note: in the case of a set, the category of quotient objects is equivalent to the category of equivalence relations on the set in question. A pairwise coproduct of equivalence relations is then the equivalence relation generated by two equivalence relations. The product of a collection of equivalence relations is their intersection. Thus this notion of a 'cotopological space' is equivalent to putting a set of equivalence relations on a set $X$ closed under pairwise sum and arbitrary intersection, where the sum of two equivalence relations is the relation generated by them. We also require that collection includes the equivalence relation where all points are equivalent and the equivalence relation where no two distinct points are equivalent. - -Edits: - -Morphisms in the Category of Cotopological Spaces. - -Suppose $X$ and $Y$ are topological spaces with a set map $f: X \rightarrow Y$. We say $f$ is continuous if, for every open subobject $V \rightarrow Y$, the following pullback gives an open subobject $U \rightarrow X$: -\begin{CD} -X @>>> Y\\ -@AAA @AAA\\ -U @>>> V -\end{CD} -Analogously, suppose there is a set map $f : X \rightarrow Y$ of cotopological spaces $X$ and $Y$. We say $f$ is cocontinuous if for each open quotient object $X \rightarrow P$ the following pushout diagram forms a quotient object $Y \rightarrow Q$: -\begin{CD} -X @>>> Y\\ -@VVV @VVV\\ -P @>>> Q -\end{CD} -In terms of equivalence relations this translates to requiring that if $R$ is an open equivalence relation on $X$ then the relation generated by $R'$ where $x'R'y'$ if and only if $x' = f(x)$ and $y' = f(y)$ for $x, y \in X$ such that $xRy$ is open. - -Each metric space $(X, d)$ induces a cotopological space in the following way: - -Each open ball $B_{\epsilon}(x)$ induces an equivalence relation $R_{\epsilon} (x)$ where $y R_{\epsilon} (x) z \iff (z = y$ or $z, y \notin B_{\epsilon} (x))$. Form the set $T = \{ R_{\epsilon} (x) : \epsilon \in \mathbb{R}, x \in X \}$ and close it under intersection of equivalence relations. This forms a co-topological space. - -Each topological space $(X, T)$ induces a cotopological space in the following way: - -Let $B$ be a basis for $X$. Each open set $U \in B$ induces an equivalence relation $R(U)$ where $xR(U)y$ when $x = y$ or $x, y \notin U$. Form the set $S = \{ R(U) : U \in B\}$ and close it under intersection of equivalence relations. This forms a co-topological space. - -If we start instead with a topological space $(X, C)$ where $C$ is the set of closed sets on $X$, we end up with a space $(X, \mathcal{S})$ where $\mathcal{S}$ is a set of equivalence relations closed under finite intersections and arbitrary joins, where a join of equivalence relations is the smallest equivalence relation containing them. -Limits and colimits in the topology of Cotopological Spaces. - -Note: the forgetful functor $F : {\bf CoTop} \rightarrow {\bf Set}$ has a left and right adjoint and therefore preserves limits and colimits. Hence if $(X_i, \mathcal{S}_i) \cong \text{ colim } \Phi$ then $X_i \cong \text{ colim } F \circ \Phi$, and the same for limits. -Take cotopological spaces $(X_i, \mathcal{S}_i)_{i \in I}$. We define a cotopology $\mathcal{S}$ on $\amalg_{i \in I} X_i$ as follows: a set $R \subset \amalg_{i \in I} X_i \times \amalg_{i \in I} X_i$ is a relation in $\mathcal{S}$ if and only if there are $\{ R_i \}_{i \in I}$, with $R_i \in \mathcal{S}_i$, such that $x_i R y_j$ for $x_i \in X_i$, $y_j \in X_j$ if and only if $i = j$ and $x_i R_i y_i$. -Defining the product cotopology, "cofinal", and "coinitial" topologies for the case of direct and inverse limits is similarly straightforward. - -REPLY [18 votes]: I believe, this notion is closely related to the notion of the coarse structure which indeed had found many beautiful applications in geometric group theory and algebraic topology (including proofs of the Novikov conjecture for a lot of groups). For introduction, I'd recommend Lectures on coarse geometry by John Roe where in Chapter 2 he explains how to give a coarse structure to every metric space or a topological space (the latter upon fixing a compactification, which in your example should be Stone–Čech). -The idea of coarse geometry/topology is exactly dual to that of topology: one doesn't look at what happens at “small distances”, but rather at what happens at “large distances” in a space, and I believe, this is exactly what the definition of the cotopology also does (it defines what “large distances” mean). -However, the original definition of a coarse structure is done in a way as to define what are subsets with “not so large” distances, but I believe, it's a bit like defining topology by open subsets or closure operation. I'll try to sketch below the translation between the two notions. -Definition. A coarse structure on a set $X$ is a collection $\mathcal E$ of subsets of $X \times X$ called controlled sets, so that $\mathcal E$ contains the identity relation, is closed under taking subsets, inverses (transposes), and finite unions, and is closed under composition of relations. -The main intuitive point in the connection can be already seen in the examples: the metric cotopological structure given by $R_r(x)$ should correspond to the family of controlled sets generated by $\{(x,y)\mid d(x,y) -TITLE: Numerology with Ramanujan's pi formula -QUESTION [14 upvotes]: Given Ramanujan's famous $\frac1{\pi}$ formula $$\frac 1\pi=\frac {2\sqrt2}{99^2}\sum_{k=0}^\infty\frac {(4k)!}{k!^4}\frac {26390k+1103}{396^{4k}}$$ -which is a level 2 Ramanujan-Sato series. It can also be expressed as -$$\frac{1}{\pi} =\frac{192 \sqrt 2}{(396^2)^{3/2}} \sum_{k=0}^\infty \tbinom{2k}{k}\tbinom{2k}{k}\tbinom{4k}{2k}\frac{2\cdot58\cdot15015k+72798}{(396^4)^k}$$ -where $\binom{n}{k}$ is the binomial coefficient. In this form, its affinity is clear to the following level 8 Ramanujan-Sato series, -$$\begin{aligned} -\frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2+4)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_2(k)\,\frac{58\cdot15015k+(72798-37/4)}{(396^2+4)^k}\\ -\frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2+8)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_3(k)\,\frac{58\cdot15015k+(72798-37/2)}{(396^2+8)^k}\\ -\frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2+16)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_4(k)\,\frac{58\cdot15015k+(72798-37)}{(396^2+16)^k}\\ -\frac{1}{\pi}&\overset{\color{red}?}=\frac{192\sqrt{2}}{(396^2+32)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_5(k)\,\frac{58\cdot15015k+(72798-2\cdot37)}{(396^2+32)^k}\\ -\frac{1}{\pi}&\overset{\color{red}?}=\frac{192\sqrt{2}}{(396^2+64)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_6(k)\,\frac{58\cdot15015k+(72798-4\cdot37)}{(396^2+64)^k}\end{aligned}$$ -and integer sequences $s_n(k)$ starting with $k=0$, -$$\begin{aligned} -s_2(k)&=\sum_{j=0}^k\tbinom{k}{2j}\tbinom{2j}{j}\tbinom{2j}{j}=1, 1, 5, 13, 61, 221,\dots\\ -s_3(k)&=\sum_{j=0}^k\tbinom{k}{2j}\tbinom{2k-4j}{k-2j}\tbinom{2j}{j}=1, 2, 8, 32, 148, 712,\dots\\ -s_4(k)&=\sum_{j=0}^k\tbinom{k}{j}\tbinom{2k-2j}{k-j}\tbinom{2j}{j}=1, 4, 20, 112, 676, 4304,\dots\\ -s_5(k)&=1, 8, 68, 608, 5668, 54688, 542864,\dots\\ -s_6(k)&=1, 16, 260, 4288, 71716, 1215296, 20848016,\dots -\end{aligned}$$ - -Q: Are all the terms of $s_5(k)$ and $s_6(k)$ integers as well, and does it have a closed-form? - -P.S. I have already checked the OEIS. - -REPLY [3 votes]: (Per S. Cooper's request.) - -I. Table relating level $1$ with level $9$. - -The general form apparently is, -$$\frac{1}{\pi}=\frac{12}{(C)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k} \tbinom{3k}{k}\tbinom{6k}{3k} \frac{\color{red}3A\,k+B}{(-C^3)^k}$$ -and, -$$\frac{1}{\pi}=\frac{12}{(C-4\color{blue}\alpha)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k \color{blue}\alpha^{k-3j} \tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j} \frac{A\,k+B+N\color{blue}\alpha/3}{(-C+4\color{blue}\alpha)^k}$$ -for general real $\color{blue}\alpha$ and where, -$$\begin{array}{|c|c|c|c|c|} -\hline -d&A&B&C&N\\ -\hline -11&154/9&5&32&4/3\\ -19&114&25&96&4\\ -43 &5418 &789 &960 &24\\ -67 &87234 &10177 &5280 &76\\ -163 &181713378 &13591409 &640320 &1448\\ -\hline -\end{array}$$ -The variables $A,B,C$ are known to have closed-form expressions in terms of $d$. Presumably $N$ should have as well. -P.S. Note also the equivalent forms, -$$s_k(\color{blue}\alpha)=\sum_{j=0}^k \color{blue}\alpha^{k-3j} \tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j}=\sum_{j=0}^k \color{blue}\alpha^{k-3j} \tbinom{k}{j}\tbinom{k-j}{j}\tbinom{k-2j}{j}$$ -where the latter form is used in H. Chan and S. Cooper's paper "Rational analogues of Ramanujan's series for 1/π". The case $\alpha=-3$, -$$s_k(-3) = 1, -3, 9, -21, 9, 297, -2421$$ -is one of the six sporadic sequences studied by Zagier and Cooper. - -II. Table relating level $2$ with level $8$. - -The general form apparently is, -$$\frac{1}{\pi}=\frac{192\sqrt{2}}{(C)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k} \tbinom{2k}{k}\tbinom{4k}{2k} \frac{\color{red}2A\,k+B}{(C^2)^k}$$ -and, -$$\frac{1}{\pi}=\frac{192\sqrt{2}}{(C+4\color{blue}\alpha)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k \color{blue}\alpha^{k-2j} \tbinom{k}{2j}\tbinom{2j}{j}\tbinom{2j}{j} \frac{A\,k+B-M\color{blue}\alpha/4}{(C+4\color{blue}\alpha)^k}$$ -for general real $\color{blue}\alpha$ and where, -$$\begin{array}{|c|c|c|c|c|} -\hline -d&A&B&C&M\\ -\hline -6&\sqrt2&\sqrt2/4&(4\sqrt3)^2&\sqrt2/12\\ -10&10&2&12^2&1/3\\ -18 &70\sqrt6 &21\sqrt6/2 &28^2 &\sqrt6/2\\ -22 &385\sqrt2 &209\sqrt2/4 &(12\sqrt{11})^2 &17\sqrt2/12\\ -58 &870870&72798&396^2&37\\ -\hline -\end{array}$$<|endoftext|> -TITLE: Possible generalization to Kirszbraun's theorem for $\mathbb{R}^2$ -QUESTION [5 upvotes]: A key lemma to Kirszbraun's theorem for $\mathbb{R}^2$ states the following: -Given any two finite collections of points $x_1,\dots,x_n$ and $x_1',\dots,x_n'$ in $\mathbb{R}^2$ such that $|x_i'x_j'|\le |x_ix_j|$ for all $i,j=1,\dots,n$ and any $x\in \mathbb{R}^2$, it's always possible to find $x'$ in the convex hull of $x_1',\dots,x_n'$ such that $|x'x_k'|\le |xx_k|$ for all $k=1,\dots,n$. -Intuitively, it seems to me that this result could be generalized in the following way: -Given any two finite collections $x_1,\dots,x_n$ and $x_1',\dots,x_n'$ with $1 -TITLE: Sheaves of complexes and complexes of sheaves -QUESTION [17 upvotes]: Let A be an abelian category, and X a topological space. -There are two ways one could try to construct some oo-category of sheaves on X from this data: - -Consider the category $Sh(X,A)$ of sheaves on X with values in A. This is abelian, so we can consider its derived category; even better, if it has enough injectives, we can enhance this to have an oo-categorical structure for which the derived category is simply the homotopy category. We call $dgSh(X,A)$ the oo-category of complexes of sheaves. -Consider the category $Sh(X, dgA)$ of sheaves on X with values in complexes in A. That is, one can make out of A an oo-category dgA whose objects are complexes of A modules. Thus it makes sense to ask for those contravariant (oo-)functors $Open(X)^{op}\to dgA$ which send open covers to Cech objects realizing global sections as homotopy limits. - -When are $dgSh(X,A)$ and $Sh(X,dgA)$ equivalent? -I realize I am sweeping issues of boundedness out the picture when constructing dg of an abelian category, for which I apologize; if there are subtleties here, please do tell me. There are also ways to make dgA by favoring projectives, or favoring injectives; I also ignore these issues. - -REPLY [13 votes]: I'm going to restrict the discussion to Grothendieck abelian categories, because I'm not sure what can be said more generally. -The main reference for what follows is Appendix C in Lurie's book Spectral Algebraic Geometry. -The derived ∞-category is a stable ∞-category, but it is the stabilization of a prestable ∞-category which is more fundamental. A Grothendieck prestable ∞-category is the connective part of a t-structure on a stable presentable ∞-category where connective objects are closed under filtered colimits. If $C$ is (Grothendieck) prestable, the subcategory $C^\heartsuit$ of discrete objects is (Grothendieck) abelian. -Prestable ∞-categories are linear analogs of ∞-toposes (insert "Grothendieck" in front of everything): - -topos $\leftrightarrow$ abelian category -$n$-topos $\leftrightarrow$ abelian $n$-category -∞-topos $\leftrightarrow$ prestable ∞-category -hypercomplete ∞-topos $\leftrightarrow$ separated prestable ∞-category -Postnikov complete ∞-topos $\leftrightarrow$ complete prestable ∞-category - -The procedure for passing from the LHS to the RHS is tensoring with the ∞-category of connective spectra. There are also vertical adjunctions between these: Grothendieck abelian $n$-categories form a coreflective subcategory of Grothendieck prestable ∞-categories, and the separated and complete ones form reflective subcategories of the latter (the same happens on the topos side if you take left exact left adjoint functors as morphisms of ∞-toposes). -According to this picture, given a Grothendieck abelian $A$, there exist three versions of the nonnegative derived ∞-category of $A$: -$$ -D^\vee_{\geq 0}(A) \to D_{\geq 0}(A) \to D^\wedge_{\geq 0}(A). -$$ -The first is simply the universal one, that is, $A\mapsto D^\vee_{\geq 0}(A)$ is left adjoint to the functor sending a Grothendieck prestable ∞-category to its heart. The second is the universal separated one, and the third is the universal complete one. A useful characterization of the middle one is the following: $D_{\geq 0}(A)$ is the unique separated Grothendieck prestable ∞-category with heart $A$ which is 0-complicial, meaning that every object admits a $\pi_0$-epimorphism from a discrete object. -Stabilizing these prestable ∞-categories, we get stable ∞-categories with t-structures -$$ -D^\vee(A) \to D(A) \to D^\wedge(A) -$$ -with heart $A$. It turns out the classical derived category of $A$ is the homotopy category of $D(A)$, which explains the notation. The ∞-categories $D^\vee(A)$ and $D(A)$ agree for example if $A$ is compactly generated and every compact object has finite projective dimension. For $A$ the category of abelian groups, all three agree. -If $X$ is a 1-localic ∞-topos and $C$ is a 0-complicial Grothendieck prestable ∞-category, then $Shv(X,C)=X\otimes C$ is also 0-complicial. -In particular, the canonical functor -$$D^\vee_{\geq 0} (Shv(X, A)) = D^\vee_{\geq 0}(X\otimes A) \to X\otimes D^\vee_{\geq 0}(A) = Shv(X,D^\vee_{\geq 0}(A))$$ -is between $0$-complicial ∞-categories and restricts to an equivalence on the hearts, so it always induces an equivalence between $D_{\geq 0}(Shv(X,A))$ and the separation of $Shv(X,D^\vee_{\geq 0}(A))$, which is the same as the separation of $Shv(X,D_{\geq 0}(A))$. -So the general answer to your question is: $D(Shv(X,A))$ is a full subcategory of $Shv(X,D(A))$ and they agree if and only if the t-structure on $Shv(X,D(A))$ is separated. But I don't see an easy way to check that in general. There are two extreme cases that are easy: - -If $X$ is hypercomplete and $A=Ind(A_0)$ where $A_0$ has enough projectives (this reduces to the case $A=Ab$). -If $A$ is arbitrary and $X$ has a conservative family of limit-preserving points (this reduces to the case where $X$ is a point). - -For example, a concrete sufficient condition for $D(Shv(X,A))=Shv(X,D(A))$ is: $X$ is a paracompact space of finite covering dimension or a Noetherian space of finite Krull dimension, and $A=Ind(A_0)$ where $A_0$ is a small abelian category with enough projectives. -Remark. If we use $D^\vee$ instead of $D$, it is always true that $Shv(X,D^\vee(A))=D^\vee(Shv(X,A))$ for $X$ a 1-topos. This shows for instance that $Shv(X,D(Ab))$ can be recovered from the abelian category $Shv(X,Ab)$, even if it's not the derived ∞-category.<|endoftext|> -TITLE: Lifting Strict Comonoids and Comodules to Quasicategories -QUESTION [8 upvotes]: $\newcommand{\M}{\mathcal{M}}$ -Suppose I have a monoidal simplicial model category in which every object is cofibrant $(\M,\otimes,\mathbb{1})$ and I want to look at its underlying monoidal quasicategory, which I'll write as $N(\M)$, the simplicial nerve of $\M$. One way to do this is the following construction (following Variant 4.1.3.17 of Lurie's Higher Algebra): -First take the sub-simplicial monoidal category of fibrant and cofibrant objects of $\M$, denoted $\M^\circ$. This is still a monoidal simplicial category. Now produce its so-called "category of operators," i.e. the free semi-cartesian monoidal category on its underlying simplicial multicategory. To be explicit, do the following: -Let $Multi(\M^\circ)$ be the simplicial multicategory whose objects are the objects of $M^\circ$ and whose "multimapping" objects are $$Mul(\{m_1,\ldots,m_k\},m)=\coprod_{\alpha\in\Sigma_k} Hom_{\M^\circ}(m_{\alpha(1)}\otimes\cdots\otimes m_{\alpha(k)},m).$$ Now produce the category of operators of this multicategory, which I'll write as $(\M^\circ)^\otimes$. This category has as objects pairs $(\langle k\rangle,\{m_1,\ldots,m_k\})$ where $\langle n\rangle$ is the finite pointed set $\{\ast,1,\ldots,k\}$ and $\{m_1,\ldots,m_k\}$ is a finite list of objects of $Multi(\M^\circ)$ (hence a finite list of objects of $\M^\circ$). This category has mapping objects given by $$Hom_{(\M^\circ)^\otimes}((\langle k\rangle,\{m_1,\ldots,m_k\}),(\langle j\rangle\{l_1,\ldots,l_j\}))=\coprod_{f:\langle k\rangle\to\langle j\rangle}\prod_{1\leq r\leq j}Mul(\{m_i\}_{i\in f^{-1}(r)},l_r).$$ -This construction probably seems a bit unwieldy, but the point is that this final category $(\M^\circ)^\otimes$ admits a forgetful functor $(\M^\circ)^\otimes\to Ass^\otimes$, the category of operators of the associative operad. In this case, since all of our mapping objects are Kan complexes (this follows from the fact that we took bifibrant objects and that cofibrant objects are closed under tensor product), we get that $N((M^\circ)^\otimes)\to N(Ass^\otimes)$ defines a monoidal structure on the quasicategory $N(M^\circ)$. -Now, my question is about the following: Suppose my simplicial monoidal model category $\M$ has a tensor product that preserves fibrant objects in addition to cofibrant objects. Then take the opposite category of $\M^\circ$, which we'll write $(\M^\circ)^{op}$, which is still a simplicial monoidal category. Moreover, since the tensor product preserves fibrancy, the mapping objects of $((\M^\circ)^{op})^\otimes$ are Kan complexes and so we still have a structure map of quasicategories $N(((\M^\circ)^{op})^\otimes)\to N(Ass^\otimes)$ defining a monoidal structure on $N((\M^\circ)^{op})$. -My question is the following: is the monoidal structure so defined on $N((\M^\circ)^{op})$ equivalent to the monoidal structure on $N(M^\circ)^{op}$ obtained by using the Grothendieck construction to get a functor $N(Ass^\otimes)\to Cat_\infty$, composing with $op$ to get a functor $N(Ass^\otimes)\to Cat_\infty\overset{op}\to Cat_\infty$ and then reversing the Grothendieck construction? -----(EDIT)---- -Perhaps a bit of background. The main point of this whole thing here is to take a simplicial monoidal model category and lift strict comonoids and comodules therein to its associated monoidal quasicategory (in the sense of Lurie). These can be relatively easily defined to be strict monoids and modules in the opposite category, but one runs into difficulties using Lurie's set-up to access the monoidal structure on the nerve of the opposite category. When the tensor product preserves fibrancy it seems like the above construction might be a work around. At the very least, it lets you define comodules and comonoids in SOME monoidal quasicategory, but it's not clear that this is equivalent to the canonically defined (up to contractible space of choices) opposite monoidal quasicategory. If anyone has some other idea for lifting comonoids and comodules, I'd be very interested to know about it. - -REPLY [3 votes]: The answer to this question is yes, and it's the main result of this paper. -One thing to point out is that, even in the case that the tensor product of $\mathcal{M}$ preserves fibrant objects (so that the homotopy types of the mapping objects in the multicategory associated to $\mathcal{M}^\circ$ are well-behaved), we don't necessarily get a strictly monoidal simplicial category, which is the input described in the above cited paper. However, that's not a problem as we can apply standard machinery to replace any monoidal simplicially enriched category (i.e. a quasimonoid in the 2-category of simplicially enriched categories) by a strict monoid which has the same underlying quasicategory.<|endoftext|> -TITLE: Is there a "Langlands philosophy" reason for the fact that the L-function of the Jacobi theta function is (almost) the Riemann zeta function? -QUESTION [26 upvotes]: The Jacobi theta function $\theta(z) = 1 + 2 \sum_{n = 1}^\infty q^{n^2}$, with $q = e^{\pi i z}$ is a (twisted) modular form with weight $1/2$. It has an associated $L$-function $L(\theta, s) = \sum_{m =1}^\infty \frac{1}{(m^2)^s} = \sum_{m=1}^\infty \frac{1}{m^{2s}} = \zeta(2s)$, which stands roughly in the same relation to $\theta(z)$ as the $L$-function of a modular form $f$ stands to $f$. (They are both essentially Mellin transforms). -(Of course, there are a few issues here - $\theta$ is only a twisted modular form, the Fourier expansion for $\theta$ is in terms of $e^{\pi i z}$ instead of $e^{2\pi i z}$, there is the factor of $2$ appearing in the argument, etc.) -In the case of a weight $2k$ modular form $f$ (at least in the case that $f$ is a Hecke eigenform), there is an associated Galois representation with the same $L$-function by the Eichler-Shimura construction, and the Langlands conjectures for $GL_2(\mathbb{Q})$ predict that every Galois representation of a certain type arises this way. -Is there a similar explanation for the case of $\theta$? Of course, the $\zeta$ function is the $L$-function for the trivial one-dimensional Galois representation, but modular forms should be related to $2$-dimensional Galois representations... -(I've only thought much about the Langlands philosophy quite recently, so I'm probably missing some very basic point!) -EDIT: Thanks to GH from MO for pointing out that the trivial one-dimensional representation corresponds via (a trivial case of) class field theory to the $GL(1)$-automorphic representation defined by the trivial Hecke character. - -REPLY [15 votes]: Good question. I don't understand fully what's happening, but here is an idea. -Let $f=\sum a_n q^n$ be a modular form of weight $k+1/2$, nebentypus $\chi$. Assuming $k \geq 1$, the Shimura correspondence attaches to $f$ a -modular form of integral weight $2k$, $g = \sum b_n q^n$ such that -$$L(g,s) = L(\chi',s-k+1) \sum \frac{a_{n^2}}{n^s},$$ -where $\chi'(n)=\chi(n) \left(\frac{-1}{n} \right)^{k}$. -The Shimura correspondence is well understood in terms of the Langlands program, thanks to famous work of Waldspurger, namely functoriality from the metaplectic group. [I took this statement of the Shimura correspondence from Ono's book, the web of modularity] -Now for some reason (convergence, I suppose) the Shimura correspondence does not work as stated in weight $1/2$ (that is $k=0$) but let suppose it does and apply it boldly to $\Theta$ (neglecting the $q^{1/24}$-missing factor), which has trivial nebentypus. The character $\chi'$ would be trivial in this case. The function $g$ would be a weight $0$ modular form, that is a constant. The Galois representation attached to the constant modular form $g=1$ is the sum of the trivial character and the cyclotomic character inverted, and its $L$-function is thus $\zeta(s) \zeta(s+1)$. -So the displayed formula becomes -$$\zeta(s)\zeta(s+1) = \zeta(s+1) \sum a_{n^2}/n^{s}$$ -where $f=\theta(z)=\sum_{n \in \mathbb Z} q^{n^2} = \sum_{n \geq 0} a_n q^n$. Hence we get that $\sum \frac{a_{n^2}}{n^s}$ is the $\zeta(s)$,or that the Mellin transform of $\Theta$ is $\zeta(2s)$. This in some sense "explains" why it is so in terms of the Langlands program.<|endoftext|> -TITLE: On the Fourier-Mukai transform in families -QUESTION [5 upvotes]: Let $X$ be an abelian variety over $\mathbb C$ and $\hat X$ its dual. Let $S$ be a scheme over $\mathbb C$, and $$0\to E\to F\to G\to 0$$ a short exact sequence of coherent sheaves on $X\times S$, all flat over $S$. In my setting $E$ and $F$ are $S$-families of torsion-free sheaves on $X$, and the quotient $G$ is finite over $S$. I wonder if it is legal to apply the Fourier-Mukai transform to the above exact sequence, and if doing so one gets an exact triangle $$\Phi(E)\to \Phi(F)\to \Phi(G)\overset{+1}{\to} \qquad\textrm{in } D^b(\hat X\times S).$$ In other words, the hope is that with the above assumptions on the families of sheaves one could use the functoriality of FM transform to construct an actual morphism between the relevant moduli spaces of sheaves/complexes on $X$ and $\hat X$, and that in addition this would give an exact triangle on $X\times S$, which is certainly the case if $S=\textrm{Spec }\mathbb C$. Does any of this make sense? -Thanks! - -REPLY [4 votes]: $\,\!$Hi Andrea. It's not just legal to apply the Fourier-Mukai transform here, it is highly encouraged. For an abelian scheme $A \to S$ the Fourier transform defines an exact equivalence of triangulated categories $D^b(A) \to D^b(\widehat A)$, where $\widehat A$ is the dual abelian scheme over $S$. This applies in particular to your case when your abelian scheme is a trivial fibration $X \times S \to S$.<|endoftext|> -TITLE: symmetric models and HOD -QUESTION [5 upvotes]: Cohen's first non-AC model, the one with a Dedekind-finite infinite set $A$ of reals, can be defined in two ways, 1st, as $HOD(A)$ in a bigger generic model of ZFC (the one obtained as a generic extension of $L$ via the countable product of the Cohen forcing), and 2nd, via symmetric names. Jech notes in one of his books that both methods yield exactly same models. I wonder is there any consistent proof somewhere as a source of reference? - -REPLY [7 votes]: You might want to look at Grigorieff's paper which shows that every symmetric model is of the form $\mathrm{HOD}(V\cup X)^{V[G]}$, where $X\in V[G]$. -In the case of Cohen's model, it is easy to argue that $\mathrm{HOD}(A)$ is a submodel of the symmetric extension. From Grigorieff's paper you can extract (although not without difficulty) the needed information to get that this is exactly the symmetric model. - -Grigorieff, Serge, Intermediate submodels and generic extensions in set theory, Ann. Math. (2) 101, 447-490 (1975). ZBL0308.02060. - -(Essentially the idea is that you have a set of generators, and the groups which stabilize them form a basis for the filter of subgroups; then the ordinal is used to "choose a name" and give some sort of "partial interpretation". So in the case of Cohen's model, this would be fixing finitely many reals.) -I can offer another route, though, since we start over $L$ and the Cohen forcing is homogeneous, you might as well consider this as $L(A)$ instead. That would be the Halpern-Levy model, which one can show even "more by hand" that it is equivalent to the symmetric extension. Simply because every $x$ in $L(A)$ is definable from some finitely many elements of $A$, and $A$ itself. And that just tells you what is the support and the name in the symmetric extension. -This is [very] implicit in Jech's proof of the fact the model is linearly ordered in his Axiom of Choice book (Ch. 5), and in many papers of Gordon P. Monro, and others from that era (e.g. Felgner's book and more).<|endoftext|> -TITLE: The Möbius number of the nonabelian finite simple groups -QUESTION [13 upvotes]: Let $L$ be a finite lattice with minimum $\hat{0}$ and maximum $\hat{1}$. The Möbius function $\mu$ for $L$ is defined recursively by: for $\forall a,b \in L$ with $a -TITLE: Topology from the viewpoint of the filter endofunctor -QUESTION [10 upvotes]: Question. Are there any references that develop general topology from the viewpoint of a functor $$\Phi : \mathbf{Rel} \rightarrow \mathbf{Rel}$$ that assigns to every set $X$ the set $\Phi(X)$ of filters on $X$? (I'm not quite sure how to view $X \mapsto \Phi(X)$ as a functor on $\mathbf{Rel}$, but I'll bet it can indeed be viewed as such.) -If the answer is "no", any attempts to do this as an answer will be thoroughly appreciated. - -Anyway, here's what I have in mind: - -A topological space should be a coalgebra for an appropriate comonad structure on $\Phi$; we think of the relevant map $X \rightarrow \Phi(X)$ as assigning to each point of $X$ its neighbourhood filter. -A convergence space should be a "relational algebra" for an appropriate "relational monad" structure on $\Phi$; we think of the relevant map $\Phi(X) \nrightarrow X$ as assigning to each filter its set of limit points. - -Some further thoughts: -Given a function $N : A \rightarrow \Phi(B)$, we get a corresponding relation $\mathrm{lim}_N : \Phi(B) \nrightarrow A$ defined by $a \in \mathrm{lim}_N(F) \leftrightarrow N(a) \subseteq F$. I suppose it makes sense to call $N$ Hausdorff iff this relation is deterministic. And it should be straightforward to show that each topological space $X \rightarrow \Phi(X)$ induces a convergence space $\Phi(X) \nrightarrow X$, as defined above. -Let me talk about limits a bit. Suppose we're trying to make sense of the limit $\lim_{x \rightarrow 0}\frac{x}{x}.$ There's the standard Euclidean neighbourhood function on $\mathbb{R}$; lets call it $$E : \mathbb{R} \rightarrow \Phi(\mathbb{R}).$$ -This induces a notion of convergence of filters, $$\mathrm{lim}_E : \Phi(\mathbb{R}) \nrightarrow \mathbb{R}.$$ -And, I think it makes sense to interpret the limit of interest as -$$\lim_{x \rightarrow 0}\frac{x}{x} = \left(\mathrm{lim}_E \circ \Phi\left(x \mapsto \frac{x}{x}\right) \circ E\right)(0)$$ -Or maybe we need to work with punctured neighbourhoods or something. Honestly, I've never really understood this stuff. Anyway, we should be able to interpret left and right limits similarly. For instance, let $E^+$ denote the neighbourhood function corresponding to the lower limit topology on $\mathbb{R}$. Then: -$$\lim_{x \rightarrow 0^+}\frac{x}{x} = \left(\mathrm{lim}_E \circ \Phi\left(x \mapsto \frac{x}{x}\right) \circ E\right)(0)$$ -Again, I'm not quite sure about puncturedness. But you get the idea; limits are defined by conjugating the function of interest by a convergence structure on one side and a topological structure on the other, and then evaluating at the point of interest. -Following this line of thought, I think a reasonable definition of continuous functions between topological spaces $(X,N_X) \rightarrow (Y,N_Y)$ would be: $$f \subseteq \mathrm{lim}_{N_Y} \circ \Phi(f) \circ N_X,$$ or something like that. This should agree with the other possible definitions, namely, the definition that only mentions neighbourhoods, and the definition that only mentions convergence. - -REPLY [9 votes]: Almost this approach has been initiated by Barr in "Relational algebras" (1970). Recent monograph with lots of references on the subject is "Monoidal topology" by Hofmann, Seal and Tholen. -The difference is that topological spaces are viewed as lax algebras rather than coalgebras, and ultrafilters rather than filters are used. -Browsing the nLab entry on it that Todd Trimble linked to in the comment below, I saw that I forgot to mention the famous Manes theorem that started it all -- the category of compact Hausdorff spaces is isomorphic to the category of algebras over the ultrafilter monad.<|endoftext|> -TITLE: Ergodicity and mixing of geodesic and horocyclic flows -QUESTION [5 upvotes]: I am reading a theorem about the ergodicity and mixing of geodesic and horocyclic flows on unit tangent bundle of a compact hyperbolic surface. I find that there are two ways (be listed below) to approach that theorem. -First way: -Step 1. Define Liouville measure on unit tangent bundle which is the measure product of Lebesgue measure on $\mathbb{S}_{1}$ and hyperbolic measure on surface. By a evenly cover mapping, the hyperbolic measure on surface comes from hyperbolic area in upper plane model. -Step 2. Prove that geodesic flows and horocyclic flows preserves that measure. -Step 3. Use representation theory and operator analysis to prove the ergodicity and mixing of that flows. -Second way: -Step 0. Identify unit tangent bundle with the quotient space $\Gamma\setminus SL(2,\mathbb{R})$, where $\Gamma$ is a symmetric discrete subgroup of $SL(2,\mathbb{R})$. -Step 1. Define the measure of SL(2,R) be measure product of its subgroups K, A, N. -Step 2. Work on SL(2,R) which is the covering space of the quotient space. Prove that geodesic flows and horocyclic flows (be written in SL(2,R) - coordinates) preserves the measure in step 1. -Step 3. Use representation theory and operator analysis to prove the ergodicity and mixing of that flows. -My question: - -Are these above steps exact ? -Is there another way to approach that theorem ? - -REPLY [9 votes]: The algebraic approach is just one of many ways to deal with the geodesic and horocycle flows. Also your quick summary does not really pay tribute to the many ways to deal with it using "representation theory and operator analysis". -The geodesic flow can be defined physically, associated to curves with zero acceleration (Newton), as critical points of some action (Lagrange), or obtained from the energy functional (Hamilton, Jacobi). This corresponds to the free movement of a particle on a compact metric space of constant curvature. The metric itself can be defined following Gauss by building isothermal coordinates on the surface, which itself can be done in several ways, one of which consisting in solving some parabolic EDP, and then any self-respecting analysts will want to do it in many interesting ways. -There is a geometric definition using the Poincaré disk or the Poincaré upper half plane or some other model in hyperbolic geometry, either by taking a quotient or starting from the surface itself and using some uniformization procedure of the universal cover à la Hadamard, or maybe the classification of simply connected Riemann surfaces which admits several proofs by different methods, the original one by Riemann dealing with electromagnetism, the Laplacian and harmonic functions. Here again, the metric and the geodesic flow can be defined infinitesimally, at finite range or using a boundary at infinity. -It can be defined algebraically, as you alluded to, using $SL_2({\bf R})$, $PSU_{1,1}({\bf R})$ or the group of translation and homotheties of the real line or some other groups I can't remember, each of them being identified with the isometry group of the geometric models previously discussed. The geodesic flow is then defined as a matrix action, or maybe just as a one-dimensional Lie subgroup using its infinitesimal generator. -Of course there is also a symbolic representation of the geodesic flow. A very interesting example is given by the (non-compact) modular surface and the decomposition of real numbers in continuous fractions. There are several ways to build such models from a geometric or algebraic representation, cutting sequences, closed geodesics, etc. -And now we come to the properties of the geodesic and horocyclic flows which are numerous and in no way restricted to ergodicity and mixing. -E. Hopf himself gave several proofs of the ergodicity of the geodesic flow, the first one using holomorphic functions, harmonic analysis and -the Harnack principle. He is more often remembered for a dynamic proof known as the Hopf argument which uses the hyperbolic nature of the flow. -Hedlund proved the mixing property circa 1939. D.V. Anosov generalized these results to the variable curvature case and invented Anosov flows at the same time. Interestingly, the proof of mixing by Anosov and Sinai uses the symplectic nature of the flow, together with a bit of cohomology, and the then all new entropy theory. -The algebraic approach starts with Gelfand, then Mautner. The unique ergodicity of the horocyclic flow is due to Furstenberg using technics from harmonic analysis/representation theory again, and a parallel can be made with rotations of the circle. This is followed by a dynamic proof of Marcus, and then the symbolic work of Bowen, Marcus. A famous generalization to unipotent actions on Lie group is due to Ratner, 1984, and builds on the dynamic approach of Anosov and Sinai using entropy theory, Lyapunov exponents and more. The work of Howe and Moore, 1977, study the correlations through representation theory, which is perhaps the work you are alluding to. Moore 1987 also gives the exponential mixing of the geodesic flow. This is now extended to symplectic Anosov flows using improved symbolic technics by Dolgopyat or anisotropic Banach spaces and spectral gaps by Liverani and a number of coauthors. -I have only scratched the surface here. I could cite at least ten books with different proofs of the mixing of the geodesic flow, dozens authors, a hundred papers and there are proofs published almost every years, since a century.<|endoftext|> -TITLE: History of the pullback corner notation -QUESTION [15 upvotes]: Where/when did the convention originate of marking pullback (and/or pushout) squares by that little right-angle symbol in the corner? - -The earliest instance I’ve been able to find is in Paul Taylor’s diagrams package, from ≤1994, as mentioned in e.g. the changelog notes for v3.81 at http://www.paultaylor.eu/diagrams/ . But it seems more likely that this was to meet the demand for a notation that was already established, rather than being the origin? But looking at various well-known category theory textbooks from before 2000 (Mac Lane Categories for the Working Mathematician; Mac Lane and Moerdijk Sheaves in Geometry and Logic; Borceux Handbook of Categorical Algebra; Johnstone Topos Theory), none of them seem to use it, as far as I can find. - -REPLY [20 votes]: In an email to me dated 17 February 1992, Peter Freyd said: - -I was using a different notation in 1974 in lectures at Montreal. A high school teacher named Butler suggested the right-angle. It was an improvement. I have used it since. - -When the diagram gets too complicated for the pullbacks to be rectangles, such as in the final chapter of my book Practical Foundations of Mathematics, I strongly recommend making them at least parallelograms. Then it is clear that pullback is acting as a functor that transforms one part of the diagram to another. In particular, in a type-theoretic setting pullback is substitution; whilst this has been known for a long time, Section 8.2 of the book actually proves it. -William Butler (had) proved some important results about monads, which you will find in "Toposes, Triples and Theories" by Barr and Wells (free TAC reprints copy). -Other very smart categorists who left academia to become schoolteachers include Christian Mikkelsen (who was the first to derive colimits from limits in an elementary topos) and Sjoerd Crans (who did weak higher dimensional category theory).<|endoftext|> -TITLE: Are there such things as non-trivial entire semigroups? -QUESTION [7 upvotes]: I'll state the theorem I am posing up front, and then explain why I think this theorem appears to be true. I am asking if anyone can prove it, or knows references to where it is proved. Please, forgive me if it is trivial, or if a counter example is trivial too. - -If $\phi(s,z) : \mathbb{C}_{\Re(s)>0} \times \mathbb{C} \to -\mathbb{C}$ is holomorphic in both variables, and - $$\phi(s_0,\phi(s_1,z)) = \phi(s_0 + s_1,z)$$ then necessarily $\frac{\partial^2}{\partial z^2} \phi(s,z) = 0$, so that $$\phi(s,z) = -e^{qs}(z-z_0) + z_0$$ for $q,z_0 \in \mathbb{C}$ - or $$\phi(s,z) = z+cs$$ for $c \in \mathbb{C}$ - -I ask this question because -A) I've never encountered a counter example in an extensive study of semigroups. -B) I can prove on a whole bunch of occasions if $\phi(a,z) = f(z)$; for $\Re(a) > 0$, for specific $f$; implies $\phi(s,z)$ cannot exist. -Examples include $f = \sin,\cos,\exp,\exp(p(z)), p(\exp(z))$ where $p$ is an arbitrary polynomial. If $f$ has a super attracting fixed point, no such $\phi$ exists. If $f^{\circ n}$ has fixed points $f$ doesn't have, then no such $\phi$ exists. In all these cases, no such solution exists for the orbits of $f$ either. I'm wondering if this is a universal trait. That, in some sense, the theorem above can be a kind of Liouville theorem. -Let's say that $\phi$ satisfies $f$, if $\phi(a,z) = f(z)$ for some $\Re(a) > 0$. To highlight the similarity to Liouville's theorem, the original theorem can be restated as - -If $f:\mathbb{C} \to \mathbb{C}$ and some semigroup $\phi$ satisfies - $f$, then $f= mz+b$ for some $m,b \in \mathbb{C}$. - -Supposing the pair $f(\cdot)$ and $\phi(s,\cdot)$ don't have to map $\mathbb{C}$ to itself, that $\mathbb{C}$ is weakened to the unit disk $\mathbb{D}$, and $f(0) = 0$ with $|f'(0)| \neq 0,1$, then there always exists a $\phi$ satisfying $f$. It seems though, as soon as we lift from $\mathbb{D}$ (or any simply connected domain biholomorphic to $\mathbb{D}$) to $\mathbb{C}$ it fails. -I can show the result in a restricted form, which I also think is interesting - -If $p:\mathbb{C} \to \mathbb{C}$ is a polynomial, and there exists - a semigroup $\phi$ satisfying $p$ then $p(z) = mz+b$ for some $m,b \in \mathbb{C}$ - -My main avenue of approach for deriving a contradiction has been to consider the Weierstrass product. It follows that if $\phi(s_0,z_n) = z_n$ then $\phi(s,z_n) = z_n$, so $\phi$ has a countable list of fixed points invariant on our choice of $s$, allowing us to say -$$\phi(s,z) = z + e^{g(s,z)}\prod_{n=0}^{\infty}(1-\frac{z}{z_n})e^{-p_n(z/z_n)}$$ -where $p_n(z) = \sum_{j=1}^n \frac{z^j}{j}$. I've been fiddling around with this as it seems like a smart idea. Since $\lim_{s\to 0}\phi(s,z) = z$, and we have a semigroup property, it seems reasonable to think we can show that for all $\epsilon >0$ there exists a $\delta > 0$ such that $|\phi(\delta,z)-z| < C_\delta e^{|z|^{\epsilon}}$ then by Hadamard -$$\sum_{n=0}^\infty \frac{1}{|z_n|} < \infty$$ -which greatly reduces the candidates for $f$ that can be satisfied by some $\phi$—$f$'s fixed points have to be sufficiently spaced out, so to speak. -Overall, I'm lost on the general case, and seem to be only able to prove it in specific circumstances. I think it says something rather profound about multiplication and addition, that, for another reason, they are incredibly special. Help, comments, suggestions, edits, anything is welcome, thanks. - -REPLY [6 votes]: Set $f_t(z)=\phi(t,z)$. -Notice that entire functions $z\mapsto f_t(z)$ all commute with each other. -I. N. Baker proved that if $f$ is a non-affine entire function with a repelling fixed point -then the set of entire functions with commute with it is countable. -MR0147650 -Baker, Irvine Noel -Permutable entire functions. -Math. Z. 79 1962 243–249. -On the other hand, in a later paper he proved that for every non-affine entire function some iterate has a repelling fixed point. -MR0226009 -Baker, I. N. -Repulsive fixpoints of entire functions. -Math. Z. 104 1968 252–256. -Combination of these two theorems settles your question. Indeed, start -with $f_1$. Then $f_n$ with some positive integer $n$ has a repelling fixed point. Then the set of entire functions which commute with $f_n$ is at most -countable, contradiction. -Once we know that all $f_t$ are affine with respect to $z$, the rest is easy: -commutative sub-semigroups of the affine group are just what you listed. -EDIT. Second theorem of Baker was based on the Ahlfors Islands Theorem, which was considered "heavy machinery" at that time. Since then both Baker's proof and the Ahlfors Islands theorem were very much simplified. For an elementary proof of the Baker theorem see, for example, -MR1782673 Berteloot, François; Duval, Julien Une démonstration directe de la densité des cycles répulsifs dans l'ensemble de Julia. Complex analysis and geometry (Paris, 1997), 221–222, Progr. Math., 188, Birkhäuser, Basel, 2000.<|endoftext|> -TITLE: author of a paradoxical decomposition of the interval -QUESTION [7 upvotes]: I am looking for the original author and the date of publication of the following result. -Theorem -There exist subsets $E_i\subset [0,1)$, $i\in {\bf Z}$, pairwise disjoints and real numbers $a_i$ such that -$$ -[0,1) = \coprod_{i\in {\bf Z}} E_i, \quad [0,2) = \coprod_{i\in {\bf Z}} a_i + E_i. -$$ -I am pretty sure that this is either Borel or Lebesgue. I remember having seen the result in the collected works of one of these two mathematicians, who advertised it as a reason to reject the axiom of choice. - -REPLY [16 votes]: Your quoted theorem (actually, a generalization where the second set is $[0,n)$ or even $[0,\infty)$) appears as Lemme 32, p. 267 of the Banach-Tarski paper (1924). -This is on the way to their third main result (see Introduction and Thm 35, p. 270): In $\mathbf R^n$ ($n\geqslant 1$) any two subsets $A$, $B$ with nonempty interior are "equivalent by countable decomposition", i.e. $A=\coprod E_i$ and $B=\coprod g_i(E_i)$ for some disjoint $E_i$ and displacements $g_i$. -The only antecedent they quote is by Sierpiński (1918, p. 142): "Let us remark that, by using Mr. Zermelo's axiom, one could decompose a square into countably many sets with which one could then compose (by a suitable translation of each of these sets) a square larger than the given one." -Note. What you saw may well have been Borel's note Les paradoxes de l'axiome du choix (1947). By an argument which is only superficially different, he shows there that Zermelo's axiom implies "euclidean equality of the whole and the part, in a finite domain" (namely $[0,1]$), and concludes: "It seems preferable to me not to admit the axiom". He doesn't cite Banach, Tarski, or Sierpiński.<|endoftext|> -TITLE: Is every singular foliation induced by a Lie algebroid? -QUESTION [16 upvotes]: Let $M$ be a smooth manifold. -A smooth distribution $D$ on $M$ is the union of a family $\{D_p \leq T_p M : p\in M\}$ of vector spaces such that there is a family $\mathcal C $ of smooth vector fields on $M$ satisfying $D_p = \text{span}\{X_p : X\in \mathcal C \} $ for every $p \in M$. -Remark that we do not ask the dimension of the fiber $D_p $ to be constant: we call a distribution regular if the dimension is constant, and singular if it is not. -We call a distribution $D$ integrable if for every point $p \in M$ there is a submanifold $S\subseteq M$ which is tangent to $D$ and satisfies $T_q S = D_q $ for every $q \in S$. In this case, it can be proved that the maximal connected integral manifolds of the distribution form a partition of $M$ into weakly embedded submanifolds of $M$, which we call the foliation associated to $D$. -Typical examples of integrable distributions are given by Lie algebroids: if $A\to M$ is a Lie algebroid over $M$ with anchor map $\rho : A\to TM$, then the image of $\rho $ is an integrable distribution on $M$. For example, every Poisson manifold has an integrable, possibly singular distribution given by the image of the Poisson bivector field $\Pi : T^*M\to TM $, and the induced foliation is precisely the symplectic foliation of the Poisson manifold. -The integrability problem for regular distributions is solved by the Frobenious theorem: a regular distribution $D$ is integrable if and only if it's involutive. A singular version of the Frobenious theorem can be stated in the following way: a (possibly singular) distribution $D$ is integrable if and only if there is a family of vector fields $\mathcal C$ which span $D$ pointwise, such that the flow of every element of $\mathcal C$ preserves $D$ (see Theorem 3.5.10 of this book for a more precise statement and a proof). -A sufficient condition for the integrability of a singular distribution $D$ is the following: there exists a module $\mathcal C$ of compactly supported vector fields spanning $D$ which is locally finitely generated and involutive. Some people calls such an object a Stefan-Sussman foliation. -I have two related questions: -1) Is it true that every integrable distribution is spanned by a module $\mathcal C$ of compactly supported vector fields which is locally finitely generated and involutive? -2) Is it true that every integrable distribution is the image of the anchor map of some Lie algebroid? -Clearly, (2) implies (1). There is people which believe that (2) is true, and I would like to know if this question is still open. -Thank you! - -REPLY [5 votes]: For Stefan-Sussmann singular foliations, the answer is negative: See Prop. 1.3 in the following paper, for the construction of an explicit counterexample: -http://users.uoa.gr/~iandroul/AZsmooth_08DEc2011.pdf -Second, regarding the holonomy groupoid of a singular foliation: In fact, it turns out that it is a diffeological space, so one can do differential geometry with it. In particular, one can differentiate it in the sense of diffeological spaces (but this can be done explicitely) and obtain a Lie algebroid in a generalized sense. This Lie algebroid is nothing else than the original module $\cF$ of vector fields which defined the foliation in hand. -Of course this module $\cF$ will not be projective, unless the foliation is "almost regular" (this is the case studied by Debord). So the Serre-Swan theorem does not apply in general, which means that $\cF$ cannot be realized as the module of sections of some honest vector bundle. However, there does exist a "singular" bundle around -- its fiber at a point $x$ is the quotient of $\cF$ by the maximal ideal at x...<|endoftext|> -TITLE: Injectivity of the Fourier transform on $L^1$ without inversion -QUESTION [5 upvotes]: Is there a proof of the injectivity of the Fourier transform on $L^1({\bf R})$ that does not rely on an inversion formula? -The proofs I have seen in the literature ultimately rely either on the Fourier inversion formula for $L^2$ or Schwartz functions, on the Plancherel formula in $L^2$ which is more or less equivalent to the inversion formula, or on some inversion formula in $L^1$ using the Fejer kernel. - -REPLY [2 votes]: Following the comment of Denis Chaperon de Lauzières, I had a look at the book of Rudin, which indeed leads to a proof of the injectivity based on the Stone-Weierstrass theorem and does not rely on the $L^2$ theory. -One needs an integrable function $k$ with compactly supported transform. W. Rudin uses the inversion formula here but this can be obtained by a direct explicit computation if we are on ${\bf R}$, e.g. with $k(x) = {\sin^2 x\over x^2}$. Then, using Fubini theorem, -$$\langle e^{i\xi x}, \overline{\hat{k}} f \rangle = \langle \hat{k}, e^{-i\xi x} f \rangle = \langle k, \hat{f}(.-\xi)\rangle = 0.$$ -From the Stone-Weierstrass theorem (wrt the compact-open topology), we see that $\langle g, \overline{\hat{k}} f \rangle = 0$ for all continuous $g$ and by density, $\overline{\hat{k}} f = 0$. We get the result by dilating the support of $\hat{k}$. -Answering Alexandre Eremenko's comment, this gives a way to solve quickly the heat equation without needing to set up the whole $L^2$ theory. Indeed, denoting the heat kernel by $K_t$, and assuming we know how to compute its transform, getting to the heat equation in the phase space -$$\hat{u} = \widehat{K_t}\widehat{u_0} = \widehat{K_t * u_0}$$ -only uses the most basic properties of the Fourier transform on $L^1$ and injectivity gives the result. The inverse formulas are not needed here.<|endoftext|> -TITLE: About a Ramanujan-Sato formula of level 10, a recurrence, and $\zeta(5)$? -QUESTION [7 upvotes]: This is a long shot, but I am curious where it leads. First, recall the Dedekind eta function $\eta(\tau)$. - -I. Level 6 - -Define, -$$\begin{aligned} -j_{6A}(\tau) &= \Big(\sqrt{j_{6B}(\tau)} - \frac{1}{\sqrt{j_{6B}(\tau)}}\Big)^2 -\\ -j_{6B}(\tau) &= \Big(\tfrac{\eta(2\tau)\,\eta(3\tau)}{\eta(\tau)\,\eta(6\tau)}\Big)^{12}\end{aligned}$$ -then, -$$\sum_{k=0}^\infty \tbinom{2k}{k}\sum_{j=0}^k\tbinom{k}{j}^3\,\frac1{\big(j_{6A}(\tau)\big)^{k+1/2}}=\sum_{k=0}^\infty \color{blue}{\sum_{j=0}^k\tbinom{k}{j}^2\tbinom{k+j}{j}^2}\,\frac1{\big(j_{6B}(\tau)\big)^{k+1/2}}\tag1$$ -where the blue integer sequence $\alpha_k=1,5,73,1445,\dots$ are the Apery numbers. These numbers have the known $3$-term recurrence relation, -$$0=k^3\alpha_k-(2k-1)(17k^2-17k+5)\alpha_{k-1}+(k-1)^3\alpha_{k-2}$$ -We can use its polynomial coefficient to generate another integer sequence, -$$v_k = (2k-1)(17k^2-17k+5)= 5,117,535,1463,\dots$$ -which appears in the cfrac of $\zeta(3)$, -$$\zeta(3)=\cfrac{6}{5 - \cfrac{1^6}{117 - \cfrac{2^6}{ 535- \cfrac{3^6}{1463-\ddots } }}}$$ -and employed (among other means) by Apery to prove the irrationality of $\zeta(3)$. - -II. Level 10 - -Similarly, define, -$$\begin{aligned} -j_{10A}(\tau) &= \Big(\sqrt{j_{10D}(\tau)} - \frac{1}{\sqrt{j_{10D}(\tau)}}\Big)^2\\ -j_{10D}(\tau) &= \Big(\tfrac{\eta(2\tau)\,\eta(5\tau)}{\eta(\tau)\,\eta(10\tau)}\Big)^{6}\end{aligned}$$ -then, -$$\sum_{k=0}^\infty \sum_{j=0}^k\tbinom{k}{j}^4\,\frac1{\big(j_{10A}(\tau)\big)^{k+1/2}}=\sum_{k=0}^\infty \color{blue}{\beta_k}\,\frac1{\big(j_{10D}(\tau)\big)^{k+1/2}}\tag2$$ -where, -$\small \beta_k = 1, 3, 25, 267, 3249, 42795, 594145, 8563035, 126905185, 1921833075, 29609682273, 462653241939, 7313942412825, 116770179560211, 1880087947627377, 30492738838690395,\dots$ -Unfortunately, I don't have a closed-form for $\beta_k$ but one can find arbitrarily many terms. - -Questions: - - -What is the recurrence relation for $\beta_k$? -This is a long shot: Does its polynomial coefficient somehow appear in the cfrac of $\zeta(5)$? - -REPLY [9 votes]: Maple found this recurrence: -(n^4+6*n^3+12*n^2+10*n+3)*beta(n)+(-20*n^4-152*n^3-420*n^2-508*n-229)*beta(n+1)+(38*n^4+380*n^3+1416*n^2+2330*n+1431)*beta(n+2)+(-20*n^4-248*n^3-1140*n^2-2292*n-1689)*beta(n+3)+(n^4+14*n^3+72*n^2+160*n+128)*beta(n+4), beta(0) = 1, beta(1) = 3, beta(2) = 25, beta(3) = 267 -\begin{align} - 0 = \;&\left( {n}^{4}+6\,{n}^{3}+12\,{n}^{2}+10\,n+3 \right) \beta_n -\\& + \left( -20\,{n}^{4}-152\,{n}^{3}-420\,{n}^{2}-508 -\,n-229 \right) \beta_{n+1} -\\& + \left( 38\,{n}^{4}+380\,{n} -^{3}+1416\,{n}^{2}+2330\,n+1431 \right) \beta_{n+2} -\\& + - \left( -20\,{n}^{4}-248\,{n}^{3}-1140\,{n}^{2}-2292\,n-1689 \right) -\beta_{n+3} -\\& + \left( {n}^{4}+14\,{n}^{3}+72\,{n}^{2}+160 -\,n+128 \right) \beta_{n+4}, -\end{align} -with $\beta_0=1,\beta_1 =3,\beta_2 =25,\beta_3 =267$. Equivalently, by shifting indices -\begin{align}0=&\;(k+1)(k-1)^3\,\beta_{k-2}\\ -&+ (-20k^4 + 8k^3 + 12k^2 - 12k + 3)\,\beta_{k-1}\\ -&+ (38k^4 + 76k^3 + 48k^2 + 10k + 3)\,\beta_{k}\\ -&+ (-20k^4 - 88k^3 - 132k^2 - 68k - 1)\,\beta_{k+1}\\ -&+ k(k+2)^3\,\beta_{k+2}\end{align} -P.S. to Vladimir: For more terms that I needed, I used only -$$ -j_{10A} = j_{10D} + \frac{1}{j_{10D}} - 2 -$$<|endoftext|> -TITLE: Most general version for the Gauss-Bonnet theorem for polygons -QUESTION [7 upvotes]: Suppose $M$ is a 2-dimensional smooth Riemannian manifold and $P\subset M$ is an open and connected subset with compact closure and a piecewise geodesic boundary. -My question is: What further conditions must $P$ (and $M$) satisfy such that the Gauss-Bonnet theorem is fulfilled for $P$? -I have found a lot of different versions of the Gauss-Bonnet theorem. All of them demand for instance that $M$ is orientable. Sometimes the boundary of $\partial P$ is supposed to be simple, closed and without cusps. But can't we do better than that? Is there a good survey which treats also more general cases? -Best wishes - -REPLY [8 votes]: There are no other conditions, and in fact a more general statement is true. -The standard reference is the survey of Reshetnyak, Two-dimensional surfaces of bounded curvature, -in the book: -MR1263963 -Geometry. IV. -Nonregular Riemannian geometry. A translation of Geometry, 4. Translation by E. Primrose. Encyclopaedia of Mathematical Sciences, 70. Springer-Verlag, Berlin, 1993. -The original source is the book of Aleksandrov and Zalgaller, Two-dimensional manifolds of bounded curvature, Trudy Mat. Inst. Steklov. 63 1962. There is an English translation. -Gauss Bonnet is stated and proved there in much larger generality: for "Aleksandrov surfaces of bounded curvature".<|endoftext|> -TITLE: L-series of difference of two cusp forms -QUESTION [5 upvotes]: Let $f=\sum_{n\geq 1} a_nq^n$ and $g=\sum_{n\geq 1} b_nq^n$ be two distinct cusp eigenforms of same weight $k\geq 2$ with real Fourier coefficients. We normalize the coefficients by defining: $\widetilde {a_n}:=\frac{a_n}{n^{(k-1)/2}}$ and similarly $\widetilde{b_n}:=\frac{b_n}{n^{(k-1)/2}}$. I was wondering if there is a quick way to see that $$\sum_{p-prime}\frac {\widetilde{a_p}-\widetilde{b_p}}{p^s}$$ is bounded as $s\to 1^{+}$. - -REPLY [6 votes]: Let $L(s,f)$ denote the $L$-function of a Hecke eigenform $f$, normalised so that the critical line is $\Re(s) = 1/2$. Then for $\Re(s) > 1$, -\[L(s,f) = \prod_p \frac{1}{1 - \lambda_f(p) p^{-s} + p^{-2s}} = \prod_p \frac{1}{(1 - \alpha_f(p) p^{-s}) (1 - \beta_f(p) p^{-s})},\] -where $\lambda_f(p)$ is the normalised $p$-th Hecke eigenvalue and $\alpha_f(p) + \beta_f(p) = \lambda_f(p)$, $\alpha_f(p) \beta_f(p) = 1$; by Deligne's proof of the Ramanujan conjecture, $|\alpha_f(p)| = |\beta_f(p)| = 1$. It follows that for $\Re(s) > 1$, -\[\log L(s,f) = \sum_{p} \frac{\lambda_f(p)}{p^s} + \sum_{p} \sum_{k = 2}^{\infty} \frac{\alpha_f(p)^k + \beta_f(p)^k}{kp^{ks}},\] -and the second term is uniformly bounded as $s \searrow 1$ (in fact, this only requires the weaker bound $|\alpha_f(p)|, |\beta_f(p)| \leq p^{1/2 - \delta}$ for some $\delta > 0$). Moreover, the $L$-function $L(s,f)$ is nonvanishing and has no poles on the line $\Re(s) = 1$ (by the usual proof of the prime number theorem; see chapter 5 of Iwaniec and Kowalski), so $\log L(s,f) \to \log L(1,f)$ as $s \searrow 1$. -In particular, it follows that $\sum_{p} \frac{\lambda_f(p)}{p^s}$ is bounded as $s \searrow 1$, and the same is obviously true for $\sum_{p} \frac{\lambda_f(p) - \lambda_g(p)}{p^s} = \sum_{p} \frac{\lambda_f(p)}{p^s} - \sum_{p} \frac{\lambda_g(p)}{p^s}$. - -Similarly, let $L(s,\mathrm{sym}^2 f)$ denote the symmetric square $L$-function, so that for $\Re(s) > 1$, -\[L(s,f) = \prod_p \frac{1}{1 - \lambda_f(p^2) p^{-s} + \lambda_f(p^2) p^{-2s} - p^{-3s}} = \prod_p \frac{1}{(1 - \alpha_f(p)^2 p^{-s}) (1 - p^{-s}) - (1 - \beta_f(p)^2 p^{-s})}.\] -Then for $\Re(s) > 1$, -\[\log L(s,\mathrm{sym}^2 f) = \sum_{p} \frac{\lambda_f(p^2)}{p^s} + \sum_{p} \sum_{k = 2}^{\infty} \frac{\alpha_f(p)^{2k} + 1 + \beta_f(p)^{2k}}{kp^{ks}}.\] -Again, $L(s,\mathrm{sym}^2 f)$ is nonvanishing and has no poles on the line $\Re(s) = 1$ (though when the nebentypus is nonprincipal, there may be a pole; see this question). So the same argument goes through (though this time, we need bounds towards the generalised Ramanujan conjecture of the form $|\alpha_f(p)|, |\beta_f(p)| \leq p^{1/4 - \delta}$ for some $\delta > 0$, which is certainly known).<|endoftext|> -TITLE: Evolving curves by Alexander Polden -QUESTION [6 upvotes]: I am writing a piece on curve shortening flow and lots of my sources have referenced Alexander Polden's honours thesis 'Evolving Curves' from the Australian National University. I have tried to find this paper to no avail, even through ANU's thesis catalogue here which returns nothing. I was wondering whether anyone could provide me access to this paper (I don't know if there are any copyright issues with such theses) - I would very much like to read it and most likely cite it. Thanks. - -REPLY [6 votes]: Not exactly what you are seeking, but apparently a 21-page piece of it. In the paper by Gerhard Huisken and Alexander Polden, "Geometric evolution equations for hypersurfaces." Calculus of Variations and Geometric Evolution Problems. Springer Berlin Heidelberg, 1999. 45-84, they say: - - - -That last section is about 21 pages long. - -PDF download: Huisken-Polden paper. -Springer link to collection: Calculus of Variations and Geometric Evolution Problems.<|endoftext|> -TITLE: Artin map restricted to base field -QUESTION [5 upvotes]: Let $M/L/K$ be a tower of local fields such that $M/L$ is abelian with Galois group $G$. The Artin map $\psi_{M/L}$ restricted to $K^\times$ is a continuous map to $G$ and thus corresponds to some abelian extension $T/K$ with an embedding $\operatorname{Gal}(T/K) \hookrightarrow G$. Challenge: Find $T$. (I am primarily interested in the mixed characteristic case, i.e. $K$ extends $\mathbb{Z}_p.$) -I conjecture the following Galois-theoretic description. Let $\tilde M$ be the normal closure of $M$. Then there is an embedding $G_1 = \operatorname{Gal}(\tilde M/K) \hookrightarrow S_n \wr G$ (where $n = [L:K]$). Let $H$ be the intersection of $G_1$ with the subgroup -$$ - \{(\sigma,g_1,\ldots,g_n) \in S_n \wr G : \sum g_i = 0 \}. -$$ -Then $T = \tilde M^H$ is the $T$ we seek, and $\operatorname{Gal}(T/K) \cong G_1/H$ has a natural embedding into $G$. -Using Kummer theory, I was able to prove this when $\mu_m \subseteq K$, where $m$ is the exponent of $G$, or when $m$ is squarefree. Also the global analogue, where $M/L/K$ is a tower of number fields and we restrict $\psi_{M/L}$ to ideals of $K$, yields readily to manipulation of Frobenius elements. But there we are aided by having to consider only ideals composed of unramified primes. Indeed, the local unramified case is not hard. - -REPLY [2 votes]: I've finally found the answer to my question by perusing Serre's Local Fields, Ch.XIII, specifically Propositions 10-12, which contain the functorial properties of the Artin symbol used below. -We describe $\left.\phi_{M/L}\right|_K$ in terms of $\phi_{\tilde M / K}$: -\begin{alignat*}{5} - \text{Gal}(\tilde M / K)^{\text{ab}} &\xrightarrow{\text{Ver}} - \text{Gal}(\tilde M / L)^{\text{ab}} &&\to \text{Gal}(M/L) \\ - \phi_{\tilde M / K}(x) & \longmapsto \phi_{\tilde M / L}(x) && \mapsto \phi_{M/L}(x). -\end{alignat*} -But these maps between Galois groups can be described as the suitable restrictions of the maps of abstract groups -$$ - (G \wr S_n)^{\text{ab}} \xrightarrow{\text{Ver}} \left(G \times (G \wr S_{n-1})\right)^{\text{ab}} \xrightarrow{\pi_1} G. -$$ -Thus $T$ arises from a certain map $G \wr S_n \to G$ which, upon computation, is none other than -$$ - (g_1,\ldots,g_n,\sigma) \mapsto \sum g_i -$$ -as desired.<|endoftext|> -TITLE: Bijective proof of formula for rooted binary forests -QUESTION [8 upvotes]: For $n\ge 1$, let $f(n)$ be the number of rooted complete (unordered) binary trees with $n$ leaves labeled from $1$ to $n$ ("complete binary" means that every vertex has either $0$ or $2$ children and "unordered" means that the we do not specify which child is the left child or the right child). Then it is well known (e.g., Example 5.2.6 of Stanley's Enumerative Combinatorics 2) that the exponential generating function of $f(n)$ is given by -$$F(x) := \sum_{n\ge1} f(n)\, {x^n\over n!} = 1 - \sqrt{1-2x}.$$ -Now fix a positive integer $r$ and for $n\ge r$, let $f(n,r)$ be the number of rooted complete binary forests with $n$ leaves labeled $1$ to $n$, and $r$ roots labeled $n+1$ to $n+r$. By generatingfunctionology (e.g., Proposition 5.1.3 of Stanley's Enumerative Combinatorics 2), -$$\sum_{n\ge r} f(n,r)\, {x^n\over n!} = F(x)^r = (1 - \sqrt{1-2x})^r.$$ -This equation yields a formula for $f(n,r)$, and in fact we have: -Theorem. $$f(n,r) = {r(2n-r-1)!\over 2^{n-r}(n-r)!}.$$ -The current proof I have of the Theorem simply notes that the generating function $F(x)^r$ coincides with a generating function for the Catalan tree, for which the coefficients are known to obey the above formula. My question is: - -Is there a bijective proof of the Theorem? - -REPLY [6 votes]: I think the following might do the trick: -Erdös, Péter L., A new bijection on rooted forests, Discrete Math. 111, No.1-3, 179-188 (1993). ZBL0785.05049.<|endoftext|> -TITLE: Polynomial related to lognormal moments -QUESTION [5 upvotes]: Consider the polynomial: -$$p(x) = \sum_{k=0}^{r}(-1)^{r-k} {r \choose k} x^{k(k-1) / 2}$$ -I want to show that -$$p(x) = (x - 1)^{\lceil r/2 \rceil} \, q(x)$$ -That is, $(x - 1)^{\lceil r/2 \rceil}$ is a factor of $p(x)$. Even better, find a formula for the quotient polynomial $q(x)$. -This problem arises when trying to compute the central moments of the log-normal distribution. Raw moments of this distribution are given by $M_r = \left< x^r \right> = e^{r\mu + \frac{1}{2} r^2 \sigma^2}$. So central moments are -$$C_r = \left< (x - M_1)^r \right> = \sum_{k=0}^{r} {r \choose k} M_k \, (-M_1)^{r-k} \\ = -e^{r(\mu + \frac{1}{2}\sigma^2)} \sum_{k=0}^{r} (-1)^{r-k} {r \choose k} e^{\frac{1}{2}k(k-1)\sigma^2}$$ -You will recognize the sum in this expression as the polynomial in the problem above. Using Mathematica or a similar program, it's easy to test that the alleged property holds for any particular $r$ you care to test. For numerical work, it's valuable to be able to express $C_r$ in factored form because there are catastrophic cancellations in the original form. -Presumably this comes down to some binomial identity with which I am unfamiliar. - -REPLY [7 votes]: Here is a combinatorial proof. Let -$$t(x)=p(x+1)=\sum_{k=0}^r (-1)^{r-k}\binom rk (1+x)^{\binom k2}.$$ -We want to show that $t(x)$ is divisible by $x^{\left\lceil r/2\right\rceil}$. -The coefficient of $x^j$ in $t(x)$ is the number of graphs with vertex set $\{1,2,\dots, r\}$, $j$ edges, and no isolated vertices. This can be proved easily using inclusion-exclusion or properties of exponential generating functions. The coefficients of these polynomials, with this combinatorial interpretation, can be found in the OEIS as sequence A054548 or A276639. -Since a graph with $r$ vertices and no isolated vertices must have at least $\left\lceil r/2\right\rceil$ edges, $t(x)$ is divisible by $x^{\left\lceil r/2\right\rceil}$. -It's interesting to note that the cumulants of the log-normal distribution are related to the inversion enumerator for labeled trees. -Additional comment: -Here's a more detailed explanation of the combinatorial interpretation. Let -$$ -u_n(x) = \sum_G x^{e(G)}, -$$ -where the sum is over all graphs $G$ with vertex set $[n]:=\{1,2,\dots,n\}$ and $e(G)$ is the number of edges of $G$, -and let -$$ -t_n(x) = \sum_H x^{e(H)}, -$$ -where the sum is over all graphs $H$ with vertex set $[n]$ and no isolated vertices. -Then -$$u_n(x) = \sum_{k=0}^n \binom nk t_k(x)$$ -since any graph with vertex set $[n]$, and with $n-k$ isolated vertices, can be specified by choosing a $k$-subset $S$ of $[n]$, constructing a graph without isolated vertices with vertex set $S$, and leaving the elements of $[n]\setminus S$ as isolated vertices. - This may be inverted to give -$$t_n(x) = \sum_{k=0}^n (-1)^{n-k}\binom nk u_k(x).$$ -But $u_n(x) = (1+x)^{\binom n2}$, since a graph with vertex set $[n]$ may be specified by including or not including each of the $\binom n2$ possible edges. Therefore -$$t_n(x) = \sum_{k=0}^n (-1)^{n-k}\binom nk (1+x)^{\binom n2}.$$<|endoftext|> -TITLE: Closed manifolds with the fixed point property -QUESTION [13 upvotes]: The real projective plane ${\bf P}^2({\bf R})$ is the only closed surface with the fixed point property: all continuous maps from the plane to itself has a fixed point. The projective spaces ${\bf P}^{2k}({\bf R})$ and ${\bf P}^{2k}({\bf C})$ also have this property; this follows from the Lefschetz fixed point formula. There seems to be few examples of manifolds with this property though. I think that odd dimensional orientable manifolds always have a vector field without zeros, and a map without fixed points can be obtained by integrating such a vector field. -Are there any other examples of compact manifolds without boundary with the fixed point property? Please explain briefly how the fixed point property is obtained. -Is there some hope to get a complete list of such manifolds? - -REPLY [6 votes]: The Lefschetz argument you mentioned, of course, also works on ${\bf P}^{2k}({\bf H})$ and ${\bf P}^2({\bf O})$. -What is a bit more surprising, is that ${\bf P}^k({\bf H})$ has the fixed point for any $k > 1$. (Of course, ${\bf P}^1({\bf H})\cong S^4$ does not have the fixed point property: the antipodal map has no fixed points). -This is proven in Hatcher's Algebraic Topology book (The end of example 4L.4, pg 493 in my edition) as an application of Steenrod powers with $p=3$.<|endoftext|> -TITLE: Parallel transport on Riemannian symmetric spaces -QUESTION [10 upvotes]: What would be a reference for the following property of symmetric spaces? -Given a smooth curve $\gamma : [0,1] \rightarrow M$ on a symmetric space $(M,g)$, there exists an isometry $\varphi : M \rightarrow M$ taking $\gamma(0)$ to $\gamma(1)$ and such that the parallel transport map -$$ -P_\gamma : T_{\gamma(0)}M \longrightarrow T_{\gamma(1)}M -$$ -equals the differential of $\varphi$ at the point $\gamma(0)$. -I guess one can get this from two particular cases: (1) the curve $\gamma$ is closed and we are looking at holonomy vs. isotropy groups, and (2) the curve $\gamma$ is a geodesic segment and is the projection of a translation of one-parameter subgroup of isometries, but I would love just to be able to cite something. -I looked up in Helgason and it does not seem to be there, at least in this guise. - -REPLY [5 votes]: As for the reference see Kobayashi-Nomizu Vol I page 262 Corollary 7.6. -Besides Petrunin's well-known argument the claim is also an obvious corollary of Cartan's Theorem 2.1 page 157 of Do Carmo's "Riemannian Geometry"<|endoftext|> -TITLE: When will the supporting hyperplane of a convex set coincide with the tangent? -QUESTION [8 upvotes]: Due to the supporting hyperplane theorem, a convex set $C$ in a separable topological space has supporting hyperplance at each of its boundary points. The theorem only guarantees its existence, now I want to discuss the uniqueness of the supporting hyperplane. -A special situation where uniqueness holds is when the supporting hyperplane at $p\in\partial C$ coincides with the tangent pf $\partial C$ at $p$. For example, when the $\partial C$ is piecewise linear, then the supporting hyperplanes must coincide with tangent planes except at vertices of $C$. -There are sufficient conditions, say $C$ being strictly convex, which guarantees the uniqueness of the supporting hyperplane. But strict convexity does not lead to the conclusion that the supporting hyperplane coincide with tangents. Moreover. since the supporting hyperplane theorem is no more than Hahn-Banach theorem, I was wondering - -(i)Is there a necessary condition (on $\partial C$, say some sort of - algebraic regularity; conditions on $C$ are also fine but less interesting) about the - boundary of convex set $C$ to make the supporting hyperplane unique? -(ii)Is there a necessary condition (on $\partial C$, say some sort of algebraic regularity) about the - boundary of convex set $C$ such that the supporting hyperplane of $C$ - at any point of $p\in\partial C$ coincide with the tangent plane of - $\partial C$ at $p$? - -For my purpose, I only want to know the case $C\subset\mathbb{R}^n$ with the usual topology. -($C$ does not have to be closed convex set but only convex, closedness is too strong for my purpose.) -Additionally, how will the situation changes if $C\subset H$ for a general (infinite dimensional) Hilbert space? Or even more general, in a Banach space $B$. - -REPLY [5 votes]: I think it goes like this: Assume $C$ has nonempty interior. Rotating the convex set if needed, the boundary $\partial C$ near any given point $p \in \partial C$ can be written as the graph of a convex function. More or less by definition, a hyperplane is a supporting hyperplane at $p$ if and only if it corresponds to an element of the subdifferential of $f$. Moreover, $f$ is differentiable at $p$ if and only if the subdifferential contains only one element. In that case, the unique supporting hyperplane is the tangent hyperplane of $\partial C$ at $p$. -In summary, a supporting hyperplane is the tangent hyperplane at $p \in\partial C$ if and only if $f$ is differentiable at $p$ if and only if the subdifferential of $f$ at $p$ contains only one element if and only if there is only one supporting hyperplane at $p$. -This, I believe, follows by results stated and proved in the book Convex Analysis by Rockafellar. -EDIT: As HenryL points out, this holds only if the supporting hyperplane actually touches the boundary. If $C$ is noncompact, then it is possible this does not happen.<|endoftext|> -TITLE: Is there a gap between the Hausdorff and the lower Minkowski dimensions? -QUESTION [5 upvotes]: Does there exist a subset $A\subseteq\mathbb{R}^n$, for some $n$, and numbers $hm$? - -REPLY [6 votes]: I believe what you are asking is whether the Hausdorff dimension can be strictly less than the lower packing dimension (also called the lower modified box dimension). I'm pretty sure the strongest formulation, $h=0$ and $m=n,$ is possible, but I don't know of a reference. However, I think examples showing that $h < m$ is possible can be found in Pertti Mattila's 1995 book Geometry of Sets and Measures in Euclidean Spaces and/or in Falconer's 1990 book Fractal Geometry (based on what I wrote in this 13 January 2001 sci.math post), but I don't have either of these books with me now. -The following diagram shows the obvious ordering of the 5 dimensions --- $\dim _{H},$ $\underline{\dim }_{P},$ $\overline{\dim }_{P},$ $\underline{\dim }_{B},$ $\overline{\dim }_{B}$ --- for any fixed choice of a subset of ${\mathbb R}^{n},$ where larger numerical values lie to the right. (I got the diagram by playing around with the LaTeX code from [1], a file that I happened to have on the computer I'm using right now.) -$$\begin{array}{ccccccc} -& & & & \overline{\dim }_{P} & & \\ -& & & \nearrow & & \searrow & & & & \\ -\dim _H & \longrightarrow & \underline{\dim }_{P} & & & & -\overline{\dim }_B \\ -\ & & & \searrow & -\underline{\dim }_B & \nearrow \\ -& & & & & & -\end{array}$$ -What I'm fairly certain is true, but I don't know a reference, is that given any 5 real numbers $\alpha,$ $\beta,$ $\gamma,$ $\delta,$ $\epsilon$ such that -$$0 \;\; \leq \;\; \alpha \;\; \leq \;\; \beta \;\; \leq \;\; \min\{\gamma, \, \delta \} \;\; \leq \;\; \max\{\gamma,\, \delta \} \;\; \leq \;\; \epsilon \;\; \leq \;\; n $$ -then there exists a subset $E$ of ${\mathbb R}^n$ such that $\;\dim _{H}(E) = \alpha \;$ and $\;\underline{\dim }_{P}(E) = \beta \;$ and $\;\overline{\dim }_{P}(E) = \gamma \;$ and $\;\underline{\dim }_{B}(E) = \delta \;$ and $\;\overline{\dim }_{B}(E) = \epsilon.$ -I thought this was proved in [2] (for the case $n=1),$ but this paper does not include the dimension function $\underline{\dim }_{P}.$ Surely the result can be found somewhere, maybe buried in someone's undergraduate honors thesis or Masters thesis. Searching online, I found several items that discuss the dimension function $\underline{\dim }_{P}$ (such as [3]), but I did not find anything that gives a result like Spear's paper [2] for all 5 of the dimension functions I've brought up. -[1] Dave L. Renfro, A porosity description of the typical continuous graph, Real Analysis Exchange 22 #1 (1996-1997), 70-73. -[2] Donald W. Spear, Sets with different dimensions in $[0,1]$, Real Analysis Exchange 24 #1 (1998-1999), 373-389. -[3] Ying Xiong and Min Wu, Category and dimensions for cut-out sets, Journal of Mathematical Analysis and Applications 358 #1 (1 October 2009), 125-135.<|endoftext|> -TITLE: Maps to $K(\pi,1)$ spaces from manifolds with $S^1$-action -QUESTION [8 upvotes]: Suppose $M$ is a connected smooth manifold with a smooth $S^1$-action that fixes a point in $M$. Let $X$ be a $K(\pi,1)$-space and let $\varphi: M\to X$ be a continuous map. -Question. How to prove that $\varphi$ can be homotoped to a map $\varphi': M\to X$ that sends each orbit of the $S^1$ action to a point in $X$? -I would be grateful for a short proof or a reference. -Remark. I can prove the statement when $X$ is a negatively curved manifold, but I am confident that the statement holds as well when $X$ is $K(\pi,1)$. - -REPLY [3 votes]: I would like to explain why the answer to the question is negative in general, following the suggestion of Tom Godwillie (however see PS for a positive answer if $dimX<\infty$) -Let us note first that there is an $S^1$-action on $\mathbb RP^2=M$ such that the quotient is a the segment. -Now, apply the idea of Tom Goodwille. Namely take $\Gamma=\mathbb Z_2$, and consider the inclusion $\mathbb RP^2\to \mathbb RP^{\infty}=K(\mathbb Z_2,1)$ that induces an isomorphism on $\pi_1$. Clearly, this map is not contractible, so it gives a counterexample to the claim in the question. -PS. However, using the remark of Misha it is also possible to prove that the answer to the question is positive in the case when the group $\pi=\Gamma=\pi_1(X)$ has no torsion elements. In particular, it holds when $X$ is finite-dimensional. -Sketch proof. Denote $M/S^1$ by $Y$ and let $p$ be the projection map. I claim that the kernel of the homomorphism $\phi^M_Y:\pi_1(M)\to \pi_1(Y)$ induced by $p$ is generated by torsion elements. Since $\Gamma$ has no torsion elements, it follows that any homomorphism $\phi^M_X:\pi_1(M)\to \Gamma$ factors through a homomorphism $\phi^M_Y$, i.e. there exists $\phi^Y_X:\pi_1(Y)\to \pi_1(X)$ such that $$\phi^M_X=\phi^Y_X\circ\phi^M_Y.$$ -Now, since $X$ is aspherical, the homomorphism $\phi^Y_X$ is induced by a map $\psi: Y\to X$. Finally note that the original map $\varphi: M\to X$ and the composed map $\psi\circ p$ are homotopic, since they induce the same homomorphism on $\pi_1(M)$.<|endoftext|> -TITLE: p-adic valuation for multinomial coefficients -QUESTION [9 upvotes]: Kummer's formula -https://en.wikipedia.org/w/index.php?title=Kummer%27s_theorem&oldid=745783657 -says that -$$ - \text{ord}_p \binom{n}{k} -$$ -is the number of carries required when adding the base-$p$ expansions -of $k$ and $n-k$. Is there a similar formula for the $p$-adic -valuation of a multinomial coefficient -$$ - \binom{n}{k_1,\ldots,k_r} := \frac{n!}{k_1!\cdots k_r!} ? -$$ -If so, is there a good reference (free online for preference, but failing that, -in a book)? -There is a related question -Reference needed for Lucas' Theorem for multinomial coefficients modulo a prime, -but it involves the value of the multinomial coefficient modulo $p$, not the $p$-adic valuation. - -REPLY [15 votes]: Denote the sum of the digits of $n$ in base $b$ by $S(n)$. Then the number of carries when adding $k_1+k_2$ is -$$\frac{1}{b-1}\big(S(k_1)+S(k_2)-S(k_1+k_2)\big).$$ -This shows that the number of carries when successively adding $(((k_1+k_2)+k_3)+\cdots +k_r)$ is -$$\frac{1}{b-1}\left(\sum_{i=1}^r S(k_i)-S\left(\sum_{i=1}^r k_i\right)\right),$$ -and this last expression is clearly independent of the order in which they are added. The formula for the multinomial coefficient can thus be written as -$$\operatorname{ord}_p \binom{n}{k_1,\ldots,k_r}=\frac{\sum_{i=1}^rS(k_i)-S(n)}{p-1}.$$<|endoftext|> -TITLE: Lefschetz hyperplane section theorem for intersection homology -QUESTION [5 upvotes]: Let $X$ be a smooth, projective variety and $Y \subset X$ be a hyperplane section (possibly singular) of $X$. Suppose that the dimension of $X$ is $n$. Is it true that for any $k -TITLE: Homology Sphere Embedding into $\mathbb R^4$ -QUESTION [6 upvotes]: Let $Y$ be an oriented closed $3$-manifold, with trivial homology group, i.e. integer homological sphere. -Q: If $Y$ can be embedded into $\mathbb R^4$, is there any example, that such a $Y$ admits a negative scalar curvature? - -REPLY [6 votes]: L Z Gao and S T Yau showed in Invent Math 85 (1986) 637-652 that any compact 3 manifold admits a metric of negative Ricci curvature .J Lohkamp proved that any manifold of dimension 3 or higher admits a complete metric with Ricci curvature pinched between two negative constants .See Annals of Math 140 (1994) 653-683<|endoftext|> -TITLE: Commutation of endomorphisms of abelian varieties -QUESTION [7 upvotes]: Let $A$ be an abelian variety over an algebraically closed field $k$. -Let $\phi:A\to A$ be an étale isogeny (over $k$). Suppose that the set $\cup_{r\geq 0}({\rm ker}\,\phi^{\circ r})(k)$ is -Zariski dense in $A$. Here $\phi^{\circ r}:=\phi\circ\dots\circ\phi$ ($r$-times). -Now let $\lambda:A\to A$ be an endomorphism and suppose that for all $r\geq 0$, we -have $$\lambda(({\rm ker}\,\phi^{\circ r})(k))\subseteq ({\rm ker}\,\phi^{\circ r})(k)\,\,\,\,(*).$$ -Does it follow that we have $$\phi^{-1}\circ\lambda\circ \phi=\lambda\circ c$$ in ${\rm End}(A)_{\bf Q}$ for some $c\in{\rm End}(A)_{\bf Q}$ which commutes with $\phi$? Or more generally, what relation in ${\rm End}(A)_{\bf Q}$ between $\lambda$ and $\phi$ does $(*)$ imply? -One may of course ask the same question for any isogeny (not just étale) and -formulate a more general condition involving finite group schemes but the case above is the case I am interested in. - -REPLY [3 votes]: Remark. The condition that $\bigcup_{r \geq 0} \ker(\phi^r)(k)$ is Zariski dense is not needed. Indeed, if it is not satisfied, replace $\phi$ by $[\ell] \circ \phi$ for $\ell$ invertible in $k$. Since $([\ell] \circ \phi)^r = [\ell^r] \circ \phi^r$, the kernel contains all $\ell^r$-torsion, which is Zariski dense as $r \to \infty$. -Moreover, $[\ell] \circ \phi$ satisfies $(*)$ if $\phi$ does. Indeed, if $x \in \ker([\ell^r] \circ \phi^r)$, then -$$([\ell^r] \circ \phi^r) (\lambda(x)) = \phi^r(\lambda([\ell^r]x)) = 0,$$ -since $[\ell^r]x \in \ker \phi^r$ and by condition $(*)$ for $\phi$. Thus, $\lambda(x) \in \ker([\ell^r] \circ \phi^r)$, which proves condition $(*)$ for $[\ell] \circ \phi$. -Finally, $\phi^{-1} \circ \lambda \circ \phi$ is unaffected by this replacement, since scalars are central. -Negative answer to your question. Let $A = E \times E$ be the square of an elliptic curve. Let $\phi \colon (x,y) \mapsto (y,x)$ be the swap isomorphism, and let $\lambda \colon (x,y) \mapsto (x,0)$ be the projection. Condition $(*)$ is now trivially satisfied (there is no kernel). -But $\phi^{-1} \circ \lambda \circ \phi$ is now the other coordinate projection $(x,y) \mapsto (0,y)$. This is not of the form $\lambda \circ c$ for any $c$, for example since the images don't agree (even after inverting scalars). Nor is it of the form $c \circ \lambda$, because the kernels don't agree. -Is there anything we can say? The best I can do is the following: condition $(*)$ implies, by the fundamental theorem on homomorphisms¹, that there exist maps $\lambda_r \colon A \to A$ such that the diagrams -$$\begin{array}{ccc}A & \stackrel\lambda\longrightarrow & A \\ {\tiny{\phi^r}}\downarrow\ \ & & \ \ \downarrow\tiny{\phi^r} \\ A & \stackrel{\lambda_r}\longrightarrow & A \end{array}$$ -commute. In fact, this is equivalent to condition $(*)$: the converse follows from the diagram. -But as we saw above, the relation between $\lambda_1$ and $\lambda$ is not something simple like $\lambda_1 = \lambda \circ c$ or $\lambda_1 = c \circ \lambda$ for some $c$. - -¹One should say some words about set-theoretic inclusion and scheme-theoretic inclusion in $(*)$. This should be fine because $\ker(\phi^r)$ is an étale $k$-scheme, and $k$ is algebraically closed.<|endoftext|> -TITLE: Behavior of a Baker-Campbell-Hausdorff problem at infinity -QUESTION [9 upvotes]: The Baker-Campbell-Hausdorff problem is to obtain $\log(e^{X}e^{Y})$ where $X,Y$ are appropriate operators. The Dynkin series $$\log(e^{tX}e^{tY})=t(X+Y)+\frac{t^2}{2}[X,Y]+o(t^3)$$ gives an expansion in powers of a perturbation parameter $t$, with higher coefficients expressed in terms of successively more complicated commutators of $X,Y$. The parameter $t$ has been introduced to control the convergence of the series; so long as the operators $X,Y$ are bounded the Dynkin series has a finite radius of convergence (theorem 5 of Suzuki 1997). A more involved approach would be to replace $e^{t X}\to e^{s X}$; the appropriate Dynkin series would then presumably be in powers of $s,t$ and have finite radius of convergence in each variable. -Now suppose we instead consider the product $e^X e^{tY}$ for $t\gg 1$. My intuition had been that there would be an asymptotic series $$\log(e^{X} e^{tY})=tY+X+o(t^{-1}).$$ But this is clearly not the standard Dynkin series: powers of $t$ now only arise from the operator $Y$, and moreover $t=\infty$ will always fall outside the finite radius of convergence. Hence my question: - -Is there a known asymptotic series for $\log(e^Xe^{t Y})$ at large $t$? - -References: - - Suzuki, M. Commun.Math. Phys. (1977) 57: 193. doi:10.1007/BF01614161. - -REPLY [3 votes]: It may be helpful to consider an example that can be solved exactly; Using the Special-case closed form of the Baker-Campbell-Hausdorff formula one finds that if the commutator $[X,Y]$ evaluates to -$$[X,Y]=uX +vY +cI$$ -then the desired logarithm of the product of matrix exponentials equals -$$\log(e^X e^{tY})=tY+X+f(u,v,t)(uX +vY +cI)$$ -$$f(u,v,t)=\frac{(ut-v)e^{ut+v}-ute^{ut}+ve^{v}}{uv(e^{ut}-e^{v})}$$ -The large-$t$ limit can now be read off once the sign of $u$ is known: -$$\lim_{t\rightarrow\infty}f(u,v,t)=\begin{cases} --1/u&\text{if}\;u<0\\ -(t/v)(e^v-1)&\text{if}\;u>0\\ -(t/v)[v-1+v(e^v-1)^{-1}]&\text{if}\;u=0 -\end{cases}$$ -There are no terms greater than order $t$ in the large-$t$ limit for this class of commutators. (I'm actually a bit puzzled how $t^2$ terms and higher might appear at all, an explicit example would help me a lot.)<|endoftext|> -TITLE: Ternary associative multiplication -QUESTION [9 upvotes]: In this answer Brian M. Scott describes the following generalization of a binary associative multiplication to a ternary one: it is a function $$[\cdot,\cdot,\cdot] : G\times G \times G \to G$$ such that $$[[a,b,c],d,e] = [a, [b,d,e], [c,d,e]]$$ The reasoning is roughly that for ordinary binary associativity we have a left multiplication function $a:G \mapsto f_a : G \to G$, such that $f_a \circ f_b = f_{f_a(b)}$, while for a ternary associative operation we have a "left multiplication" $a:G \mapsto f_a: G \times G \to G$ such that $$f_a \circ (f_b, f_c) = f_{f_a (b,c)}$$ -It may also probably make sense to demand different symmetrizations of this relation, like $$[a,b,[c,d,e]] = [[a,b,c],[a,b,d],e]$$ and I'm also no entirely sure if the arguments on the right are correct, so we could consider various relations of the form $$[[a,b,c],d,e] = [a, [b,d,?], [c,?,e]]$$ The reasoning here is that if we would try to define "ternary algebras", then it would make sense to have all relations given by polylinear operations (which isn't true in the example above). I believe a natural generalization to finite-dimensional ternary algebras should have relations of the form $$[[a,b,c],d,e] = \frac{1}{\mathrm{dim}}\mathrm{Tr}_w [a, [b,d,w], [c,w,e]]$$ An example of such a ternary algebra would be the space of rank 3 tensors $a_{ijk}$ with the bracket given by $$[a,b,c]_{ijk} = \sum_{pqs} a_{piq} b_{qjs} c_{skp}$$ Thus it is possible that in the case of ternary groups there should also be some averaging over all elements in place of $?$. -My question is: have such structures been studied? If so, then what are their names and relevant references? Searching on the google for "ternary operations" or "ternary algebras" doesn't yield any useful results. -EDIT: I am not asking about ternary multiplications in general, most references that I found focus on a very naive approach which is equivalent to a binary associative multiplication in most interesting cases and isn't very far in general. I am interested specifically in the associativity condition of signature $[-, -, [ -, -, -]] = [[-,-,-], [-,-,-],- ]$, which has 2 multiplications on the left, but 3 on the right. Most articles discuss signatures of the form $[-,-,[-,-,-]] = [[-,-,-],-,-]$. -One can also try to generalize the above definitions to $n$-ary algebras for arbitrary $n$, but the precise relations are even less clear. - -REPLY [10 votes]: Perhaps this has already been said in some form, but the identity -$$[[a,b,c],d,e] = [a, [b,d,e], [c,d,e]]$$ -is exactly the $(2,2,2)$-associative law of clone theory. I think that this is the most natural way this identity arises. -A clone is a closed set of operations. Starting with a collection of finitary operations on some set, close them under composition, group them together according to their arity, and one obtains a multisorted structure called a clone: $\langle C_1, C_2, \ldots; \{\textrm{comp}_n^m\}, \{p^m_i\}\rangle$ where $C_i$ is the set of $i$-ary operations in the collection, $\textrm{comp}_n^m$ is a composition operation that applies an element of $C_m$ to $m$ elements of $C_n$ to obtain an element of $C_n$. (For example, if $f(x,y)\in C_2$ and $g_1(x,y,z), g_2(x,y,z)\in C_3$, then $\textrm{comp}^2_3(f,g_1,g_2)=f(g_1(x,y,z),g_2(x,y,z))=f(g_1,g_2)(x,y,z)\in C_3$.) -There is a Cayley representation theorem for clones, starting from axioms -that are similar to the axioms for monoids: axioms of the identity and axioms of associativity. The $(m,n,k)$-associative law is -$$ -[[f,g_1,\ldots,g_m],h_1,\ldots,h_n]=[f,[g_1,h_1,\ldots,h_n],\ldots,[g_m,h_1,\ldots,h_n]], -$$ -where $f\in C_m, g_i\in C_n, h_j\in C_k$ and $[f,g_1,\ldots,g_m]$ denotes $f(g_1,\ldots,g_m)$. -As I said above, the identity of this question is the $(2,2,2)$-associative law. -The Cayley representation is sufficient to show that models of -$$[[a,b,c],d,e] = [a, [b,d,e], [c,d,e]]$$ -are exactly those structures that are isomorphic to binary components of clones -under the operation of $\text{comp}_2^2(f,g,h)=:[f,g,h]$ PROVIDED you have the identity elements to complete the Cayley proof. These identity elements are the elements $p_1^2(x,y)=x$ and $p_2^2(x,y)=y$. So, if you have any structure with a ternary operation $[f,g,h]$ satisfying the identity of this problem, and you also have ``constants'' $p_1^2$ and $p_2^2$ such that the (universally quantified) identities $[p_1^2,f,g]=f$, $[p_2^2,f,g]=g$, and $[h,p_1^2,p_2^2]=h$ hold, then your structure is isomorphic to the binary component of some clone of finitary operations on some set. - -You can read about clones on the wikipedia page for clones, linked above. The $(m,n,k)$-associative law is the 6th bullet point in the subsection ``Abstract clones''. You might also examine -W. Taylor, -Characterizing Mal’cev conditions, -Algebra Universalis 3 (1973), 351-397.<|endoftext|> -TITLE: Bounding the minimum singular value of a block triangular matrix -QUESTION [7 upvotes]: Question: -What is the sharpest known lower bound for the minimum singular value of the block triangular matrix -$$M:=\begin{bmatrix} -A & B \\ 0 & D -\end{bmatrix}$$ -in terms of the properties of its constituent matrices? -Some known bounds: -Since the minimum singular value of $M$ is one over the norm of $M^{-1}$, we can equivalently look for upper bounds on $M^{-1}$, which has the following block structure: -$$M^{-1} = \begin{bmatrix} -A^{-1} & -A^{-1}BD^{-1} \\ -0 & D^{-1} -\end{bmatrix}.$$ -Applying the triangle inequality yields the simple bound: -$$\left\Vert M^{-1} \right\Vert \le \left\Vert D^{-1} \right\Vert + \left\Vert A^{-1}BD^{-1} \right\Vert + \left\Vert A^{-1} \right\Vert,$$ -A slightly more involved argument discussed here yields the (usually sharper) bounds: -\begin{align} -\left\Vert M^{-1} \right\Vert &\le \sqrt{\left\Vert A^{-1}\right\Vert^2\left(1 + \left\Vert BD^{-1} \right\Vert^2 \right) + \left\Vert D^{-1} \right\Vert^2} \\ -\left\Vert M^{-1} \right\Vert &\le \sqrt{\left\Vert D^{-1}\right\Vert^2\left(1 + \left\Vert A^{-1}B \right\Vert^2 \right) + \left\Vert A^{-1} \right\Vert^2}. -\end{align} -Conjecture: -The obvious conjecture is the following symmetrized version: -$$\left\Vert M^{-1} \right\Vert \overset{?}{\le} \sqrt{\left\Vert D^{-1} \right\Vert^2 + \left\Vert A^{-1}BD^{-1} \right\Vert ^2 + \left\Vert A^{-1} \right\Vert ^2},$$ -which I have been unable to prove. -Notes: - -This question was asked on math.stackexchange where it got many upvotes but no answers. -Presumably this has been systematically studied somewhere, but I have just been unable to find the right paper. -Perhaps there is no one "best" bound. In that case, it would still be useful to know what different bounds are out there, and what the tradeoffs are between them. - -REPLY [7 votes]: It suffices to state the bound for $M$, as also noted in C. Remling's comment. -In particular, recall that $\|M\|^2 = \|M^*M\|$, whereby we have the bound -\begin{equation*} -\|M\|^2 = \left\|\begin{bmatrix} -A^*A + B^*B & B^*D\\ -D^*B & D^*D -\end{bmatrix}\right\| \le \|A^*A + B^*B + D^*D\|, -\end{equation*} -where the final inequality holds as the block matrix involved is PSD. -One can even prove related bounds if norms other than the operator norm are desired. See for instance this interesting paper by K. Audenaert for bounds involving Schatten-$p$ norms (though it leads to somewhat different inequalities); while Bourin's paper provides such inequalities for all symmetric (i.e., unitarily invariant) norms. - -REPLY [6 votes]: Yes, this works. As I suggested in my comment above, we can rephrase the claim as $\|M\|^2 \le \|A\|^2 + \|B\|^2 + \|D\|^2$, and this we can check by direct calculation: Apply $M$ to $v=(x,y)^t$. Then -$$ -\|Mv\|^2 =\|Ax+By\|^2 + \|Dy\|^2 -$$ -and -$$ \|Ax+By\|^2 \le (\|A\| \|x\|+\|B\|\|y\|)^2 \le (\|A\|^2 + \|B\|^2)(\|x\|^2+\|y\|^2) , -$$ -(by the Cauchy-Schwarz inequality on $\mathbb R^2$ in the last step) as desired.<|endoftext|> -TITLE: Intersection of two quadrics that have a common inscribed sphere -QUESTION [9 upvotes]: This is related to a question I asked here on math.stackexchange. It didn't receive an answer there (except for my answer), and my question here is a generalization of that one, anyway. -Suppose I have two quadric surfaces $Q1$ and $Q2$ in plain ordinary 3D space, $\mathbb{R}^3$, and they have a common inscribed tangent sphere $S$, as shown in the picture. - -In general, the intersection of two quadric surfaces is a nasty curve of degree 4. But in our case, because of the common tangency, I suspect that the intersection of $Q1$ and $Q2$ is just a pair of ellipses. Numerical experiments seem to suggest this, anyway. My question: is my conjecture true, and, if so, what is the proof? -My tags might well be wrong, so please feel free to edit. - -REPLY [3 votes]: In fact, there is a far more general result. In Salmon's "Analytic Geometry of Three Dimensions", 4th edition, page 117, we find the following: - -Two quadrics having plane contact with the same third quadric intersect each other in plane curves. Proof: Obviously $U-L^2$ and $U-M^2$ have the planes $L-M$ and $L+M$ for their planes of intersection. - -This is essentially the same proof given in Ivan's answer.<|endoftext|> -TITLE: Prime distribution in CM field -QUESTION [6 upvotes]: For any prescribed infinite set of primes $P\subseteq \mathbb{Z}$, is there a general heuristic to finding a field satisfying certain algebraic or geometric constraints such that these primes all split? -I am interested, in particular, in obtaining a CM field (quadratic imaginary extension of either $\mathbb{Q}$ or, more generally, of a totally real field) in which infinitely many primes in a prescribed infinite set of primes $P$ split, and for my application, I can moreover assume that $P$ has density $0$. -I do not need to write down the explicit CM field (and it also does not need to be unique), as long as I know I can find a CM field for any degree $d$ over $\mathbb{Q}$. My background is largely in geometry, so I do not know if a question of this form is routine or difficult. - -REPLY [4 votes]: I'm not 100% sure I understand your question correctly, so let me first explain what I understand your problem to be. First, by a prime "splitting" in a field $K$, I understand you to mean that $p$ splits completely. Second, by "CM field," I assume you do not allow the field to be totally real (some people's definition of CM fields allows this). Finally, I understand your problem to be this: Given an infinite set $P$ of primes, does there exist a CM field $K$ such that infinitely many (but not necessarily all) primes $p$ in $P$ split in $K$. The answer to this question is that it is not always possible, so one will require more information about the particular set $P$. -Lemma: Let $K_1, K_2, \ldots K_n$ be CM fields. Then there exist infinitely many primes $p$ which do not split in any of the fields $K_i$. -Proof: A prime $p$ splits completely in a field $K$ if and only if it splits completely in the Galois closure of $K$, and thus we may assume the $K_i$ are all Galois. The Galois closure of a CM field is CM. Let $G$ denote the Galois group of the compositum of the $K_i$. By the Ceborarev Density Theorem, there exists infinitely many primes $p$ for which the Frobenius conjugacy class of $p$ is the conjugacy class of (any) complex conjugation in $G$. The image of complex conjugation in $G$ in the quotient $\mathrm{Gal}(K_i/\mathbf{Q})$ is also complex conjugation, which is non-trivial, because $K_i$ is a CM field. Hence such a prime $p$ does not split completely in $K_i$. $\square$ -We now construct an infinite set $P$ of primes as follows. -Enumerate the (countably infinite) CM fields as $K_1, K_2, \ldots$. Then choose $p_n \in P$ to be any prime not in $\{p_1,\ldots,p_{n-1}\}$ that does not split in any of the fields $K_1, \ldots, K_n$. That this is possible follows from the previous lemma. By construction, for any CM field $K$, only finitely many primes in $P$ split in $K$. -You say that "I can moreover assume that $P$ has density $0$." This is a slightly strange thing to say, because this assumption makes the problem more difficult. If you assume that $P$ does have positive density, then you can find a CM field such that infinitely many primes in $P$ split in an imaginary quadratic field $K$. Indeed, for a prime $p$ not to split in any of the fields -$$\mathbf{Q}(\sqrt{-1}), \mathbf{Q}(\sqrt{-2}), \mathbf{Q}(\sqrt{-3}), \ldots \mathbf{Q}(\sqrt{-n}),$$ -it must be the case that the quadratic symbol $(-k/p) = -1$ for all $k \le n$. The density of such primes tends to zero as $n$ tends to infinity. Hence, taking $n$ large enough so that the density of such primes is smaller than the density of $P$, it follows that a positive proportion of primes in $P$ splits in at least one of the fields above. -If one increases the set of CM fields to the set of all non-trivial finite extensions of $\mathbf{Q}$ (or even just all such fields which are either CM or totally real), then one can take $K$ to be either $\mathbf{Q}(\sqrt{-2})$, $\mathbf{Q}(\sqrt{-3})$, or $\mathbf{Q}(\sqrt{6})$. This is because the identity $(-2/p)(-3/p) = (6/p)$ means that every prime $p$ splits in at least one of these fields.<|endoftext|> -TITLE: Examples of discrete time martingales -QUESTION [5 upvotes]: In probability, a martingale is given by a sequence of integrable - random variables $(S_n)$ and an increasing sequence of - $\sigma$-algebras ${\cal F}_n$ such that - $S_n$ is ${\cal F}_n$-measurable and - $E(S_{n+1} \mid {\cal F}_n) = S_{n}$. -This is an important notion because there are many results concerning - convergence of martingales sequences, e.g. if it is bounded in $L^2$ - then it converges in $L^2$ norm and $a.e.$ -If $X_i$ is a sequence of i.i.d. random variables and - ${\cal F}_n = \sigma(X_i, i\leq n)$, then the following sequences - are martingales: - -$S_n - E(S_n)$, -$ \exp(S_n)/E(\exp(S_n))$, -$(S_n)^2-E(S_n^2)$, - -These are used in the theory of random walks to compute e.g. the mean time before reaching a given state. -Are there any other interesting examples of discrete time martingales? - -REPLY [4 votes]: If you're interested in a somewhat off-beat example of a discrete martingale, you could look at -Rafe Jones, Iterated Galois towers, their associated martingales, and -the $p$-adic Mandelbrot set, Compos. Math. 143 (2007), 1108-1126. -This is really a result in discrete dynamics over finite fields, and Jones uses tools from number theory (Galois theory, chebotarev density theorem) to show that a certain system constructed by iterating polynomials over finite fields is a martingale, and he then uses the convergence of martingales to deduce his final result. -(For those who don't have access to Compositio, there's a pre-print version on Jones's website: http://www.people.carleton.edu/~rfjones/Preprints/Jones_Iterated_Towers_Homepage.pdf)<|endoftext|> -TITLE: For which rings R is SL_n(R) generated by its n-1 fundamental copies of SL_2(R)? -QUESTION [15 upvotes]: By "fundamental copies" of $SL_2(R)$ in $SL_n(R)$, I mean those embedded along the diagonal (for instance, if $n=3$, those are the upper left and lower right corner copies of $SL_2(R)$ embedded in $SL_3(R)$). -When $R$ is a field or a Euclidean ring, $SL_n(R)$ is generated by elementary matrices. However, this need not be the case in general. My question is, whether one can nevertheless generate $SL_n(R)$ by its fundamental copies of $SL_2(R)$, even if $SL_n(R)$ is bigger than its elementary subgroup. I have the impression that the answer is at least "yes" when $R$ is a principal ideal domain or, more generally, a Bezout domain. -A similar question holds for the affine Kac-Moody group $SL_2(R[t,t^{-1}])$: when is it generated by its two fundamental copies of $SL_2(R)$, where the two embeddings of $SL_2(R)$ are given as follows: -$\begin{pmatrix}a &b\\ c &d \end{pmatrix}\mapsto \begin{pmatrix}a &b\\ c &d \end{pmatrix}$ and $\begin{pmatrix}a &b\\ c &d \end{pmatrix}\mapsto \begin{pmatrix}d &ct^{-1}\\ bt &a \end{pmatrix}$ - -REPLY [7 votes]: L. Vaserstein obtain a generation result in a very large generality, which encompasses all finitely generated commutative algebras over fields or PIDs: - -Theorem (Vaserstein, 2006, unpublished): for every commutative noetherian ring that is finitely generated (by $k$ elements) over a subring of Krull dimension $\le 1$ and every $r\ge 3$, every matrix decomposes as a product of one matrix in the upper $\mathrm{SL}_2$ and a bounded (lazy bound: $\le 28k^2r^2$) number of elementary matrices. - -(Note that obviously every elementary matrix is a product of a bounded (bound depending only on $r$) number of (elementary) matrices in one of the diagonal $\mathrm{SL}_2$.) -For somewhat absurd reasons, this paper is unpublished. I did in 2007 a detailed check of the proof, including a careful reading of his long 1976 paper with Suslin on which the paper strongly relies. Vaserstein's paper can be found on his web page: http://www.math.psu.edu/vstein/pm2.pdf; the above theorem follows from his Corollary 2. -Note that in case of finitely generated algebra over a field/PID, this is considerably more general than what is cited in Andrei's answer: the case of Bézout rings is strictly contained in the case of Krull dimension 1, and the case of stable $\le 2$ does not much go beyond Krull dimension 2: it covers the case of Krull dimension 1 and also maybe a few rings of Krull dimension 2. -(Input welcome! I do not know if there's a commutative algebra that is finitely generated over a PID, has stable rank $\le 2$ and Krull dimension $\ge 2$. There is a MO post resulting in the conclusion that $\mathbf{Z}[x]$ has stable rank 3. According to this paper by Gabel (Pacific J. Math. 1975), polynomial rings over fields with $s\ge 4$ indeterminates have stable rank $\ge 3$, and for subfields of the real field this already holds when $s\ge 2$.) - -Added: Vaserstein proved his result soon after Shalom proved and gave talks on a Kazhdan Property T result relying on the fact that for a ring $R$ of stable rank $d$, $\mathrm{SL}_r(R)$ for $r\ge d+1$ is boundedly generated by elementary matrices along with its upper left copy of $\mathrm{SL}_{d+1}(R)$. For $d=2$ this is of course the easy result on stable rank 2 retrieved later (link in Andrei's post). This is briefly mentioned in Shalom's ICM 2006 paper, with a reference to a paper which never appeared, and explained in detail in 2006/07 pdf notes of mine based on his talks. Combining Shalom and Vaserstein's result lead to the result that the subgroup generated by elementary matrices in $\mathrm{SL}_r(R)$ has Property T for every finitely generated commutative ring and $r\ge 3$. What happened a bit later is that the Property T result was obtained by other means, not using bounded generation, by Ershov and Jaikin-Zapirain (Inv. Math. 2010), in a wider generality ($R$ not necessarily commutative). This superseded the Property T result, but not Vaserstein's beautiful bounded generation result, which obviously is of interest independently of Kazhdan's Property T.<|endoftext|> -TITLE: Sum of independent random variables -QUESTION [15 upvotes]: We know that the sum of two independent normal random variables is again a normal random variable. But is the reverse right? If $X$ and $Y$ are independent random variables satisfying $X+Y$~$N(\mu,\sigma^2)$ for some $\mu$ and $\sigma$, can we conclude that both $X$ and $Y$ obey normal distribution? or under some conditions added on $p_X$ and $p_Y$ (the density functions of $X$ and $Y$)? - -REPLY [7 votes]: Yes, and the same holds for Poisson, and for mixtures of Gauss and Poisson. -All these are special cases of the general question: if $X_j$ are independent -and we know the distribution of their sum, what can be said about the distributions of the $X_j$. This general question is addressed in the book -Linnik, Ostrovskii, Decomposition of random variables and vectors, AMS 1977 -(translation from the Russian).<|endoftext|> -TITLE: Expanding in Fibonacci powers -QUESTION [7 upvotes]: Let $F_n$ denote the all-familiar Fibonacci numbers, with $F_0=0, F_1=1, F_2=1$, etc. -There is a plethora of properties for these numbers involving their sums, products, convolutions and so on. Here, we are concerned with expanding linear (polynomial) factors. Consider -$$P_n(x):=\prod_{k=2}^n(1-x^{F_k}).$$ - -Question. What is a closed formula for the number of monomials, $N(P_n)$, in the expansion of $P_n(x)$? - -Examples. $P_2(x)=-x+1, P_3(x)=x^3-x^2-x+1$ and hence $N(P_2)=2, N(P_3)=4$. - -REPLY [4 votes]: Here is the direct proof without using the result of Ardila. -We will represent a sum of distinct Fibonacci numbers $m=\sum_{j=2}^n b_{j-1} F_j$, where $b_j\in\{0,1\}$, as a bit string $b_1b_2\ldots b_{n-1}$. In the case of Zeckendorf's representation, in the corresponding bit string every pair of 1s is interspaced with at least one 0. -We remark that an arbitrary bit string $s$ can be converted into Zeckendorf's representation by scanning $s$ from right to left and replacing every substring $110$ with $001$ (and start scanning from the right-most end again). This way an $(n-1)$-bit string may be turned into an $n$-bit string. Conversely, we can "unroll" Zeckendorf's representation into another sum of distinct Fibonacci numbers by scanning the string from left to right and replacing some substring $001$ with $110$ (and starting scanning from the left-most end again). -The coefficient $c_m$ of $x^m$ in $P_n(x)$ is the difference between $(n-1)$-bit representations of $m$ with even and odd number of 1s. Furthermore, $c_m$ can be nonzero only if $m\leq F_{n+2}-2$. To compute $c_m$, we consider Zeckendorf's representation $Z_m$ of $m$ in the form of $n$-bit string. -The general structure of $Z_m$ is -$$Z_m = 0^{p_1} 1 0^{p_2} \dots 0^{p_t} 1 0^{p_{t+1}},$$ -where $t\geq 0$, $p_1\geq 0$, $p_{t+1}\geq 0$, $p_i\geq 1$ for $1 -TITLE: Rank of a combinatorial matrix -QUESTION [8 upvotes]: For any $2n$ with $n\in\mathbb{N}$, we conjecture the matrix $A\in\mathcal{M}_{3n\times 2n}$ to have rank $\lceil{\frac{5n-1}{3}}\rceil$ where $$A=\left(\begin{array}{cccccc} -0 & & & & & \\ -1 & 2 & & \\ -1 & 2 & 0 & & \\ -\vdots & \vdots & & \;\ddots \\ -1 & \binom{2n-2}{1} & \cdots & \binom{2n-2}{2n-3} & 0 & \\ -1 & \binom{2n-1}{1} & \cdots & \binom{2n-1}{2n-3} & \binom{2n-1}{2n-2} & 2\\ -\hline\\ -1 & & & & & -1\\ - & \ddots & & & \cdot^{\large{\cdot^{\Large{\cdot}}}} &\\ - & & 1 & -1 & & -\end{array} -\right).$$ -That is, the kernel of $A$ is the solution to the following set of equations. For $k=1,\dots,2n$, we set, for $k$ odd, $$x_{1}+\binom{k-1}{1}x_2+\dots+\binom{k-1}{k-2}x_{k-1}=0$$ and, for $k$ even, $$x_{1}+\binom{k-1}{1}x_2+\dots+\binom{k-1}{k-2}x_{k-1}+2x_k=0.$$ -Additionally, for $i=1,\dots,n$, we take $$x_i-x_{2n+1-i}=0.$$ -Computationally (with SageMath), we have found this to be true for up to at least $n=150$. We have obtained these equations while studying tensor valuations on lattice polygons. - -REPLY [7 votes]: If $(c_0,\dots,c_{2n-1})$ is a row of $A$, we look at a polynomial $c_0-c_1s+\dots-c_{2n-1}s^{2n-1}$. We have to find the dimension of the subspace $\Sigma$ formed by such polynomials in the space $\Pi$ of all polynomials of degree at most $2n-1$. The polynomials are $(1-s)^{k-1}-s^{k-1}$ for $k=1,\dots,2n$ and $\pm(s^{i-1}+s^{2n-i})$, $i=1,\dots,n$. Denote $$A={\text{span}}\,\left((1-s)^{m}-s^{m},0\leqslant m\leqslant 2n-1\right),\\B={\text{span}}\,\left(s^{i-1}+s^{2n-i},1\leqslant i\leqslant n\right).$$ -We have to find $\dim(A+ B)=\dim A+\dim B-\dim A\cap B$. Note that $A$ is the subspace of polynomials $f(s)\in \Pi$ satisfying $f(s)=-f(1-s):=\alpha(f)$ (this is clear if we denote $s=1/2+x$: then $A$ consists of all polynomials in $\Pi$ which are odd in the variable $x$.) Thus $\dim A=n$. Obviously $\dim B=n$, and $B$ is the space of all polynomials in $\Pi$ satisfying $f(s)=s^{2n-1}f(1/s):=\beta(f)$. So, $A\cap B$ is a set of polynomials in $\Pi$ which are invariant under both involutions $\alpha(f),\beta(f)$. It is straightforward (but crucial) that $(\alpha \beta)^3=\rm id$, so the group generated by $\alpha$, $\beta$ consists of 6 elements $\rm{id},\alpha,\beta,\alpha\beta,\alpha\beta\alpha,\beta\alpha$. -So, we have to prove that the space $C=A\cap B$ of the polynomials in $\Pi$ invariant under the action of this group $G\cong S_3$ has a dimension $2n-\lceil(5n-1)/3\rceil=\lfloor(n+1)/3\rfloor$. In other words, the multiplicity of the trivial 1-dimensional representation in our representation $\lambda$ of $S_3$ must be equal to $\lfloor(n+1)/3\rfloor$. This multiplicity is a scalar product of the characters of $\lambda$ and the trivial representation, that is, the sum of traces of all the six above linear operators divided by 6. The trace of $\rm id$ equals $2n$, the traces of involutions $\alpha,\beta,\alpha\beta\alpha$ are equal to $0$ (for $\beta$ or $\alpha$ this is clear in the standard basis, and $\alpha\beta\alpha$ is conjugate to $\beta$), and it remains to prove that the trace $t$ of $\alpha\beta$ (and of $\beta\alpha$ too) equals $3\lfloor(n+1)/3\rfloor-n$. There should be some clever way, but in any case we may do it in the standard basis $\{1,s,\dots,s^{2n-1}\}$. Indeed, $$(\beta\alpha)s^i=-s^{2n-1-i}(s-1)^i,$$ -so we should sum up the coefficients of $s^i$ in $-s^{2n-1-i}(s-1)^i$, or, the same, sum up the constant terms of $-s^{2n-1-2i}(s-1)^i$, or, the same, to take the constant term of $$-s^{2n-1}\sum_{i=0}^{2n-1}\left(\frac{s-1}{s^2}\right)^i=-s^{1-2n}\frac{s^{4n}-(s-1)^{2n}}{s^2-s+1},$$ -or, the same, to take the coefficient of $s^{2n-1}$ in $\frac{(s-1)^{2n}}{1-s+s^2}$. For doing this we note that $(s-1)^{2n}-(-s)^n$ is divisible by $1-s+s^2$, and the ratio has coefficient 0 of $s^{2n-1}$. While the coefficient of $s^n/(1-s+s^2)=(s^n+s^{n+1})/(1+s^3)=(s^n+s^{n+1})(1-s^3+s^6-\dots)$ is easy to find. Checking separately remainders of $n$ modulo 3 we verify that it is indeed $(-1)^n(3\lfloor(n+1)/3\rfloor-n)$.<|endoftext|> -TITLE: Are there known and interesting or easy examples of groups $G$ which admit $\mathbb Z$ as a distorted normal subgroup? -QUESTION [6 upvotes]: A finitely generated subgroup $H$ of $G$ is said to be undistorted if any word metric on $H$ and any metric on $H$ induced by a word metric of $G$ are roughly equivalent (i.e., they differ by a multiplicative and additive constant; in other words, $H \hookrightarrow G$ is a quasi-isometric embedding). It is said to be distorted otherwise. -Baumslag–Solitar groups $G = \langle a, b \,|\, aba^{-1} = b^n \rangle$ provide easy example of groups $G$ with distorted, infinite cyclic, non-normal subgroups $H$. I was wondering how difficult it is to find examples of groups with distorted, infinite cyclic, normal subgroups. - -REPLY [7 votes]: As explained by A. Sisto here on p.20 and p.21, "The subgroup generated by -$z$ -in the Heisenberg group -$$〈 -x,y,z -| -[ -x,y -] = -z, -[ -x,z -] = [ -y,z -] = 1 -〉$$ -is isomorphic to -$\mathbf Z$ -and distorted." As $\langle z \rangle$ is the center of the Heisenberg group and hence normal, this gives one example.<|endoftext|> -TITLE: Maximum $2$-D bootstrap percolation time for $n$ points on an $n\times n$ grid -QUESTION [7 upvotes]: I hesitate to ask this question here, but since it remained unanswered after a bounty on MSE, I ask it here with some reservation. -Is the maximum bootstrap percolation time for $n$ points on an $n\times n$ grid $\big{|}\left \lceil{(n^2-3)/2}\right \rceil + n - 1 \big{|}$ for $1\leq n\leq 8$, and $n(n+3)/2-7$ for $n\geq 9$? -Below are some possible starting positions for $1\leq n\leq 12$: - -and a possible construction method for $n\geq 10$ (based on the starting positions of $n-2$): - -$\hspace{2em}$ - -In Mathematica, this might be constructed as follows: -aa = Uncompress@"1:eJzVlsEOgjAQRLuAgvyF/+PJT/BAwskD/n/UNhGG7mwpqNFLw8K8zuxaCcfL9dy1zrmheiynfrh1wqv+WXUFCIpI0LtI5W8FugS61GlFGu6/FvrQkxWYVMSE6GdOwY4pJKkYXXaQCqp5KgJp0YI7k8kyWfZuPtseGoJKbYiQNEcIw7SKg72vZGjXZfC91TAVqPhURquaWGUD9Ei9zYGOdqMDSz7poYEhN0lcM0UqDayk4rwrvT7Wl5lQlCvTy/znB+owpbBa2qWy0VJqCyox+rOBybzVX4k1YbuarWeNafYCwU+SNmMjLQjyOehmXmL+X/IL4b+Q3zoNqsw+iAqffkmNyx0cTkRo"; -a[9] = aa[[9]]; a[10] = aa[[10]]; -a[n_] := If[n < 9, aa[[n]], With[{t = Length@#[[1]] + 2}, Flatten[{ReplacePart[Array[0 &, t], # -> 1] & /@ {1, t + 1, t, 1, t + 1, t - 1, 1}, Drop[Flatten[{Take[#, 2], #}, 1] &@(PadLeft[PadRight[#, t - 1], t] & /@ #), 7]}, 1] -] &@a[n - 2]]; - -or, a non-recurrence solution: -a[n_] := If[n < 9, aa[[n]], Partition[ReplacePart[ConstantArray[0, n^2], Thread[# -> 1]] &@ -With[{v = Join[{1, 3 #1, 1 + 3 #1, -1 + 6 #1, 1 + 6 #1}, LinearRecurrence[{0, 2, 0, -1}, {-2 + 8 #1, 2 + 8 #1, -3 + 10 #1, 3 + 10 #1}, # - 9], {(-3 - 8*#1 + 4*#1^2 + (-1)^#1*(-5 + 2*#1))/4, ((-1)^#1*(1 + (-1)^#1*(-1 - 6*#1 + 4*#1^2)))/4, ((-1)^#1*(1 + (-1)^#1*(-13 - 2*#1 + 4*#1^2)))/4, ((-1)^#1*(-1 + (-1)^#1*(9 - 2*#1 + 4*#1^2)))/4}] &@n}, -If[EvenQ@n, v, ReplacePart[v, (Length@v - 4) -> v[[Length@v - 4]] + 1]]], n]]; - -eg -Manipulate[With[{b = Most@FixedPointList[ -CellularAutomaton[{1018, {2, {{0, 2, 0}, {2, 1, 2}, {0, 2, 0}}}, {1, 1}}, {#, 0}][[1, 2 ;; -2, 2 ;; -2]] &, a[n]]}, -ArrayPlot[b[[length]], Mesh -> True]], -{length, 1, If[n < 9, {1, 2, 5, 10, 15, 22, 29, 38}[[n]], n (n + 3)/2 - 7], 1, Appearance -> "Open"}, -{{n, 10}, 1, 20, 1, Appearance -> "Open"}] - -The above is smaller than the lower bound shown in this paper of $13 n^2/18-14 n/9-5/3$, but a quick search for all permutations at $n=5$ shows that the maximum percolation time requires $>n$ initial points. -Does the above construction result in the maximum percolation time for $n$ initial startpoints? -Sets containing $>n$ initial points -In addition, I am looking through the paper by Fabricio Benevides and Michał Przykucki on maximum bootstrap percolation time and I am having trouble finding an example (or seeing how there could be a set of points) that takes a greater time to complete than the one given in their example of a set for a $12\times 12$ grid on page $20$: - -the following pattern is valid for every multiple of $12$ and requiring $4n/3-1$ initial points, takes $ n(17 n- 10)/24$ moves to complete: - -manipu[n_, m_] := -Manipulate[With[{b = Most@FixedPointList[ -CellularAutomaton[{1018, {2, {{0, 2, 0}, {2, 1, 2}, {0, 2, 0}}}, {1, 1}}, -{#, 0}][[1, 2 ;; -2, 2 ;; -2]] &, n]}, -ArrayPlot[b[[length]], Mesh -> True]], {length, 1, m, 1, Appearance -> "Open"}]; - -m12[n_] := -With[{y = Length@#}, manipu[#, y (17 y - 10)/24]] &@ With[{t = 12 n}, -Flatten[{Take[Flatten[{PadRight[{1}, t], PadLeft[{1}, t], -Array[0 &, t]} & /@ Range@Ceiling[t/6], 1], t/2], -Reverse@(CenterArray[Join[{0, 0, 1}, Array[0 &, #], {1}], t] & /@ -Range[8, t, 4]), {CenterArray[{0, 1, 0, 0, 0, 1}, t]}, -{CenterArray[{1, 0, 1}, t]}, CenterArray[Join[{1}, -Array[0 &, #], {1}], t] & /@ Range[6, t, 4]}, 1]]; - -m12[3] - -This differes from their minimum percolation time: the set following the pattern given in their example takes $ 17 n^2/24 +O(n)$, yet they state the lower bound is $13n^2/18+O(n)$. It is close, $(\lim{n\rightarrow\infty (17 n^2/24)/(13n^2/18)=51/52})$, but I can't see how to construct a set of initial points that meets their lower bound. What am I missing? - -REPLY [7 votes]: Thanks for your interest in our paper! A friend just messaged me linking to this thread. First, when the starting set has exactly $n$ infected sites, the maximum time is the integer nearest to $(5n^2 -2n)/8$ - see this paper. -Basically, to achieve this, you first infect a $2 \times \tfrac{n}{2}$ rectangle at the bottom of your square (in time $O(n)$), then grow it "from the corners" like in the pictures at the beginning of your question until the lower half of the square is infected (that takes time $\tfrac{\tfrac{n}{2}+n}{2} \tfrac{n}{2}+O(n) = \tfrac{3n^2}{8}+O(n)$), and you finish by infecting the rest horizontally, two rows at a time (that takes time $\tfrac12 \tfrac{n}2 n +O(n)= \tfrac{n^2}{4}+O(n)$). So as you see, in total this construction takes time $\tfrac{5n^2}{8}+O(n)$. -Now, regarding initial sets of arbitrary size, consider what happens when the value of $s$ in what we call the Mighty Crab is around $\tfrac{n}{3}$ (see Figure 12 in the paper you link to, which is also the one you copied here). If the only difference in your solution is that you hardcoded it to be $s=5$ then this should hopefully do the trick. Please let me know if anything is still unclear.<|endoftext|> -TITLE: Elementary subgroups of surface groups -QUESTION [13 upvotes]: From Sela's proof of Tarski's conjecture we know that the surface groups (i.e. fundamental group of a closed surface of genus $\geq 2$) and free (non-abelian) groups have the same first order theory. We also know that the inclusion of a free non-abelian group into another as a free factor is an elementary embedding. Now any subgroup $H$ of a surface group $G$ is either a surface group (if the index is finite) or a free group (if the index is infinite). Hence the elementary theory of $H$ is equivalent to that of $G$. On the other hand not every inclusion of a subgroup of the surface group is an elementary embedding. A necessary condition is provided in Theorem 1.4 of this paper. -Q) Is there a sufficient condition for the inclusion to be an elementary embedding? -Q) Can we completely classify the subgroups of surface groups for which the inclusion is an elementary embedding? -I am a newcomer in this area and I am currently reading Sela's proof of the above theorems therefore any suggestion about papers or references regarding the problems will be extremely helpful. Thanks in advance. - -REPLY [15 votes]: Perin's theorem 1.2 gives a necessary condition that is also sufficient in the case of a (closed) surface group (for general torsion-free hyperbolic group a -slight modification is needed). A subgroup H is elementary embedded in the fundamental group of a closed surface group S (of genus at least 2), if and only if H is a non-abelian free factor in the fundamental group of a (proper) subsurafce M of S, and there exists a retraction from S onto M. The proof of the necessary part appears in Perin's thesis (theorem 1.2), and the sufficient part follows using my own argument that the elementary core of a torsion-free hyperbolic group is elementary embedded in the hyperbolic group.<|endoftext|> -TITLE: Consequences of the Birch and Swinnerton-Dyer Conjecture? -QUESTION [17 upvotes]: Before asking my short question I had made some research. Unfortunately I did not find a good reference with some examples. My question is the following - -What are the consequences of the Birch and Swinnerton-Dyer Conjecture ? - -I have read the official statement of the conjecture in Clay Institute website and followed a very nice talk by Manjul Bhargava... -Just to be clear enough, I'm not interested on the explanation of the conjecture, I would like to know what are the consequences. -How can I make this question C.W. ? - -REPLY [21 votes]: There is a theorem of Michael Stoll that there are no $c\in\mathbb Q$ such that the polynomial $x^2+c$ admits a periodic 6-cycle starting at some $a\in\mathbb Q$, but the theorem is contingent on the Birch-Swinnerton-Dyer conjecture being true for the Jacobian $J$ of a certain curve $C$ appearing in Stoll's paper. The reason it's needed is because he needs to know the rank of $J(\mathbb Q)$, but the genus of $C$ is too large for him to do an explicit descent. So he's "just" using the rank part of BSwD. Here's the reference: -MR2465796 Stoll, Michael Rational 6-cycles under iteration of quadratic polynomials. LMS J. Comput. Math. 11 (2008), 367–380. - -REPLY [12 votes]: Implicit in the BSD conjecture are two other basic conjectures about elliptic curves: the Parity Conjecture and the finiteness of the Tate-Shafarevich group. Most applications I know follow from the rank part, but refined BSD will allow formulas for the order of the Tate-Shafarevich group for instance. -Two specific well-known elementary applications are: - -Tunnell's solution to the Congruent Number Problem -Rodriguez-Villegas and Zagier's solution to which primes are sums of two rational cubes - -These are both contingent on BSD for some cases. Here are a couple other consequences: - -There exists an elliptic curve/modular form of analytic rank 4. (Note the existence of an elliptic curve of analytic rank at least 3 is used to solve the Gauss class number 1 problem.) -(example) Let $E$ be an elliptic curve of conductor 17, and $-d$ a negative prime discriminant. Then the $-d$-th quadratic twist of $E$ has infinitely many rational points if and only if the Legendre symbol $(-d/17) = +1$. - -It would also provide an analytic way to attack Goldfeld's conjecture on average ranks of elliptic curves.<|endoftext|> -TITLE: Variance of Multi-Dimensional Ornstein Uhlenbeck process -QUESTION [5 upvotes]: I am trying to compute the asymptotic variance of OU process -$$ -d X_t = - H X_t dt + S dW_t -$$ -where $X_t$ takes value in $R^d$. $H$ and $S$ are $d\times d$ matrices that does not have $HS = SH$ in general. How to compute its variance at $t$? -Things are classical for one dimensional case as -$$ -Var (X_t) = \sigma^2 / \mu (1-e^{-\mu t}) -$$ - -REPLY [4 votes]: the large-$t$ limit $\rho$ of the variance of $X_t$ is given by -$$\rho=\int_0^\infty \exp(-tH)SS^{\dagger}\exp(-tH^{\dagger})dt$$ -where the superscript $\dagger$ indicates the transpose (or the conjugate transpose if the matrix is complex); this result holds also if $S$ is $d\times d'$ with $d'\leq d$; see, for example, page 11 of these notes, which also show how to evaluate the integral by Schur decomposition of $H$.<|endoftext|> -TITLE: Functoriality of the formality quasi-isomorphism of E-polydifferential operators -QUESTION [5 upvotes]: Given a smooth manifold $M$ and a Lie algebroid $E\rightarrow M$ we can consider the $E$-polydifferential operators $D_E$ and the $E$-polyvectorfields -$T_E$ as -$$D_E:=\bigoplus_{k=-1}^\infty\mathcal{U}(E)^{\otimes k+1}\hspace{0.3cm}\mbox{and}\hspace{0.3cm}T_E=\bigoplus_{k=-1}^\infty\Gamma\left(M;\bigwedge^{k+1}E\right) $$ -where $\mathcal{U}(E)$ denotes the universal enveloping Hopf algebroid of the Lie-Rinehart pair $(\Gamma(E),C^\infty(M))$ and the tensor products are over $C^\infty(M)$ on the left hand side. One now finds the usual DGLA structures given by the analogues of the Gerstenhaber bracket (expressed in terms of the coproduct) and the Hochschild differential (expressed as Gerstenhaber bracket with a certain degree 1 element) on the left and the Schouten-Nijenhuis beacket on the right. Calaque Showed in http://link.springer.com/article/10.1007/s00220-005-1350-5 -that one can construct a Kontsevich type formality $L_\infty$-quasi-isomorphism $$T_E\longrightarrow D_E$$ by using the Fedosov type techniques developed by Dolgushev in his thesis. -Reccently I had been studying this situation and this made me wonder about the functorial properties of these maps. Namely, consider the functors -$$T,D\colon LA(M)\longrightarrow L_\infty[W^{-1}]$$ -from the category of Lie algebroids over $M$ to the category of $L_\infty$-algebras localized at the quasi-isomorphisms. -Q: Is there a natural equivalence $T\rightarrow D$? -The work by Calaque shows that certainly $T_E$ is isomorphic to $D_E$ for all $E\rightarrow M$, but, after searching and thinking for a while, I could not really obtain any results about this kind of functoriality of these isomorphisms. -One thing that seems necessary (as Stefan pointed out in the comments) is that we consider the category of pairs $(E\rightarrow M,\nabla_E)$ where $\nabla_E$ is a torsion-free $E$-connection, since such connections are essential in constructing Calaque's quasi-isomorphism. Morphisms in this category would then have to somehow preserve these connections though. This seems to warrant considering only inclusions (considering only isomorphisms doesn't serve any purpose it seems to me). -In case there is some "stupid" reason that this question either makes no sense or is not worth investigating I would also welcome an explanation of this stupid reason, of course. - -REPLY [3 votes]: The construction is functorial with respect to algebraic morphisms of Lie algebroids (as opposed to geometric ones): see for instance my paper with Van de Bergh https://arxiv.org/pdf/0708.2725.pdf (it encompasses the globalization techniques that are used in the paper you cite). -As Stefan pointed out, the choice of a connection is rather important for the Dolgushev-Fedosov resolutions. In the paper with Van den Bergh we use much bigger resolutions that are independant of such a choice (they are somehow universal). A torsion free connection allows one to linearize the jet bundle. The resolution we construct in the paper with Van den Bergh involves an algebra that is universal among the ones that can linearize the jet bundle (and as such it somehow carries a universal connection). This in particular shows that the homotopy class of the $L_\infty$-morphism that you get from my paper does not depend on the choice of the connection involved in the construction. -Note that the algebraic morphisms from the paper with Van den Bergh are called comorphisms by several people (see e.g. https://arxiv.org/pdf/1210.4443.pdf). -Let me also observe that if you fix the base and only consider morphisms that are the identity on the base, then the categoy of Lie algebroids with comorphisms/algebraic morphisms is just the opposite category to the category of Lie algebroids with morphisms/geometric morphisms. -So, combining all these comments, I would say that the answer to your question is yes. -EDIT JULY 25 2017: -1. the functoriality only works for isomorphisms of Lie algebroids (see comments below), and more generally for étale morphisms (like open embeddings in the tangent Lie algebroid case). -2. there are probably other situations where it works. As suggested in the question, one could expect it to work for inclusions of Lie algebroids.<|endoftext|> -TITLE: Algebraization of Bayesian networks? -QUESTION [6 upvotes]: The algebraization of classical propositional logic is Boolean algebra. -Bayesian networks are a generalization of classical propositional logic with probability truth-values. -What is the corresponding algebraic form of Bayesian networks? Or at least outline how such an algebra would look like? I seem to be at a loss as to how to begin... - -REPLY [10 votes]: A bayesian network corresponds to an independence (algebraic) variety and hence to a polynomial ring over $\mathbb{R}$. -You can start from - -Garcia, Luis David, Michael Stillman, and Bernd Sturmfels. "Algebraic - geometry of Bayesian networks." Journal of Symbolic Computation 39.3-4 - (2005): 331-355.<|endoftext|> -TITLE: Can an intersection of ideals in a Noetherian ring be replaced by a countable intersection? -QUESTION [13 upvotes]: Let $(R,\frak m)$ be a Noetherian local ring, and let $X$ be a set of ideals in $R$. Assume $\bigcap_{I \in X} I = 0$. Is there some sequence $\{I_n\}_{n \in \mathbb N}$, with $I_n \in X$ for all $n$, such that $\bigcap_n I_n =0$? -The above is true if $R$ is complete, by a slight alteration of the argument in Chevalley's Theorem (lemma 7 in Chevalley, On the theory of local rings, Ann. Math. 44 (4), 1943). I want to know whether completeness is necessary. - -REPLY [12 votes]: Yes it's true. -First, in a (commutative) noetherian ring $A$, every chain of ideals is well-ordered by reverse inclusion. The supremum of ordinal types of such chains is denoted $o(A)$. -A particular case of Theorem 2.12 in [Bass71] is that for $A$ of finite Krull dimension $d$, we have $o(A)\le\omega^d$; hence $o(A)<\omega^\omega$ ($\omega^\omega$ is the countable ordinal $\sup_{d<\omega}\omega^d$). This holds in particular if $A$ is noetherian local. -Now if $A$ is a ring such that there is some family of ideals $(I_x)$ whose intersection is not achieved by any countable subfamily, then it is quite immediate that there is a descending chain of ordinal type $\omega_1$: indeed, by induction one defines for $\alpha<\omega_1$ $x_{\alpha}$ such that $I_{x_\alpha}$ does not contain $\bigcap_{\beta<\alpha}I_{x_\beta}$, and hence for $J_\alpha=\bigcap_{\beta<\alpha}I_{x_\alpha}$, we obtain $(J_{\alpha})_{\alpha<\omega_1}$ as the desired chain. -But actually Bass' main result (his Theorem 1.1) says that any chain of submodules of a f.g. module over an arbitrary noetherian ring, is countable. This shows that the result holds for arbitrary (commutative) noetherian rings and not only local ones. -The case of countable Krull dimension (encompassing the case of noetherian local rings, and finitely generated algebra over fields) is quite easy to understand using ordinal length, as in Gulliksen [Gull73]. -[Bass71] H. Bass. Descending chains and the Krull ordinal of commutative Noetherian rings, Journal of Pure and Applied Algebra -Volume 1, Issue 4, December 1971, Pages 347-360 (Sciencedirect link -poor scan) -[Gull73] T. Gulliksen. A theory of length for Noetherian modules. Journal of Pure and Applied Algebra -Volume 3, Issue 2, June 1973, Pages 159-170. (Sciencedirect link)<|endoftext|> -TITLE: Homotopy theory of non-test categories? -QUESTION [8 upvotes]: Let $\mathcal{C}$ be a category with pullbacks, and consider the functor $i_\mathcal{C}: \mathcal{C}^\mathrm{op} \to \mathsf{Cat}$, $X \mapsto \mathcal{C}/X$. We can equip $\mathcal{C}$ with a class $\mathcal{W}$ of "weak equivalences" by setting $\mathcal{W} = i_\mathcal{C}^{-1}(\mathcal{W}_\mathsf{Cat})$, where $\mathcal{W}_\mathsf{Cat}$ is the class of functors that induce weak equivalences of nerves. Cisinski shows that if $\mathcal{C}$ is a (Grothendieck) topos, then there is a model structure on $\mathcal{C}$ with weak equivalences $\mathcal{W}$ and cofibrations the monomorphisms. Moreover, Cisinski proves a conjecture of Grothendieck, saying that if $\mathcal{C}$ is the category of presheaves on a so-called test category, then the homotopy theory of $(\mathcal{C},\mathcal{W})$ is equivalent to the homotopy theory of spaces (for example, the model structure is Quillen equivalent to the usual Quillen model structure on simplicial sets). -But what about Grothendieck toposes which are not presheaf categories on test categories? Cisinski's theory gives us a way to associate a homotopy theory to any Grothendieck topos -- what is that homotopy theory? -I'm basically wondering what is known about the homotopy theory of $(\mathcal{C}, \mathcal{W})$. For example, is it related to the etale homotopy type of $\mathcal{C}$? I might guess that it is some sort of slice category over the etale homotopy type. - -REPLY [3 votes]: Here is an example. -Consider the truncated category $\Delta_{\le n+1}$. Then presheaves on this with the model structure presents $n$-types. -EDIT: -Andrea points out that the relevant Quillen equivalences+proofs can be found in Corollary 9.2.7 of Cisinski's book Les préfaisceaux comme modèles des types d'homotopie.<|endoftext|> -TITLE: Transformation converting power series to Bernoulli polynomial series -QUESTION [7 upvotes]: I wonder, can anyone describe an expression or formula of a transform that converts -$$\sum_{k=0}^\infty \frac{a_k x^k}{k!}$$ -into -$$\sum_{k=0}^\infty \frac{a_k B_k(x)}{k!}$$ -where $B_k(x)$ are Bernoulli polynomials. -For instance, -$$x^2\, \to\, x^2-x+1/6$$ -$$e^x \,\to \, \frac{e^x}{e-1}$$ -etc. - -REPLY [6 votes]: Another way, somewhat related with the above answers, is the $p$-adic Volkenborn integral. You can find this, for example, in Schikhof's or in Alain Robert's books on $p$-adic calculus, or Henri Cohen vol. 2 of his books on number theory. This approach is useful because of the relation of Bernoulli numbers and L-functions: one can easily define good and elementary $p$-adic zeta functions using the Volkenborn integral (actually, this was Kubota and Leopoldt's original approach). -Let $\mathbb{Z}_p$ be the ring of $p$-adic integers and let $\mathbb{C}_p$ be the topological completion of an algebraic closure of the field of fractions $\mathbb{Q}_p$ of $\mathbb{Z}_p$ (a nice and large field for doing $p$-adic analysis). Let $f:\mathbb{Z}_p\to\mathbb{C}_p$ be an analytic function, that is, $f$ is of the form -$$f(x)=\sum_{n\ge0}a_n\frac{x^n}{n!},\qquad a_n\in\mathbb{C}_p,\quad -\frac{a_n}{n!}\to0.$$ -(We suppose $f$ analytic for simplicity and because of what you are asking). Then the Volkenborn integral of $f$ is defined by the following $p$-adic limit: -$$\int_{\mathbb{Z}_p}f(t)dt=\lim_{m\to\infty}p^{-m}\sum_{k=0}^{p^m-1}f(k).$$ -Then, one has the following relation with Bernoulli numbers and polynomials: -$$\int_{\mathbb{Z}_p}t^ndt=B_n$$ -and -$$\int_{\mathbb{Z}_p}(x+t)^ndt=B_n(x).$$ -This Volkenborn integral is a continuous linear operator on a Banach space of functions (see the books mentioned above). Hence, with $f$ as above, one obtains: -$$\int_{\mathbb{Z}_p}f(t)dt=\sum_{n\ge0}a_n\frac{B_n}{n!}$$ -and -$$\int_{\mathbb{Z}_p}f(x+t)dt=\sum_{n\ge0}a_n\frac{B_n(x)}{n!}.$$ -Hope this helps. -Note: This integral is a special case of "$p$-adic distributions", which are one of the main tools that are now used to define $p$-adic zeta functions attached to arthmetic objects. See, for example, Washington or Lang books on cyclotomic fields for a nice introduction. -PS: For a nice "general zeta functions" interpretation of your question, see Lemma 2.4 in this article by Friedman and Pereira https://arxiv.org/abs/1105.2603 It was published in the IJNT, but the arxiv version is the same as the published version.<|endoftext|> -TITLE: Proof that a local fibration is a fibration, in May -QUESTION [40 upvotes]: I was reading "A Concise Course in Algebraic Topology" by J.P.May (page 52) and found the proof of the following theorem incomprehensible: -Let $p:E\rightarrow B$ be a map and let $\mathcal{O} $ be a numerable open cover of B. Then $p$ is a fibration if and only if $p:p^{-1} (U) \rightarrow U$ is a fibration for every $U\in \mathcal O$. -The only if part is clear. To check the if part, May tries to patch together path lifting functions of $p|p^{-1} (U)$ to construct a path lifting function. To do this, he defines some complicated functions, which I could not easily understand. -He starts by choosing maps $\lambda_U : B\rightarrow I$ such that $\lambda _U ^{-1} (0,1]=U$. For each finite ordered subset $T=\{U_1 , \cdots , U_n \}$ of $\mathcal O$, he defines $c(T)=n$, and $\lambda _T : B^I \rightarrow I$ as $$\lambda_T (\beta ) = \inf \{(\lambda_{U_i} \circ \beta ) (t) : (i-1)/n \le t \le i/n , 1\le i\le n \}$$ -and let $W_T = \lambda _T ^{-1} (0,1]$. I understand that $\{W_T\}$ is an open cover of $B^I$, and that $\{W_T : c(T)0\} \subset W_T.$$ -I understand that $\{V_T\}$ is a locally finite open cover of $B^I$. -Now, he chooses path lifting functions $$s_U : p^{-1} (U) \times _p U^I \rightarrow p^{-1} (U) ^I $$ for each $U\in \mathcal O $, and he constructs a global path lifting function by the following steps. -First, choose a total order on the sets of finite ordered subsets of $\mathcal O $. For a given $T=\{U_1, \cdots , U_n \}$, he defines $s_T (e,\beta [u,v]):[u,v]\rightarrow E $ for each $\beta \in V_T$, $u,v\in [0,1]$, $e\in p^{-1} (\beta (u))$ canonically, by lifting $\beta[u,v]$ by letting its starting point be $e$. ($\beta [u,v]$ is defined as the restriction of $\beta$ to $[u,v]$.) -Now here is the part where I don't understand at all. For each $e,\beta$ such that $\beta(0)=e$, he defines $s(e,\beta)$ to be the concatenation of the parts $s_{T_j} (e_{j-1} , \beta [u_{j-1} ,u_j ])$, $1\le j \le q$ where the $T_i $ run through the set of all $T$ such that $\beta \in V_T$, and $$u_0 =0, \ u_j = \sum _{i=1} ^j \gamma _{T_i } (\beta ).$$ -He asserts that $s$ is well defined and continuous, thus a path lifting function for $p$. -I could not understand at all why the definition of $s$ makes sense. For this to make sense, I believe that $u_q$ should be $1$, which I cannot get at all. For me, this definition seems completely random. Can someone please kindly explain why this definition makes sense? - -REPLY [86 votes]: May has carelessly made a typo in transcribing a proof in one of his books to another, -and it has taken an impressive high school student in Korea to catch it. In his earlier book ``Classifying -Spaces and Fibrations" --- the title is a joke: "classifying" here has two meanings --- -(http://www.math.uchicago.edu/~may/BOOKS/Classifying.pdf) -there is a generalization of the cited result in "Concise". In its proof, but using -the notation of "Concise", May remembers to normalize by setting -$u_j = \sum_{i=1}^{j}\gamma_{T_i}(\beta) / \sum_{i=1}^{q}\gamma_{T_i}(\beta).$ -With this correction (and correction of another typo, $s(e,\beta)= e$ should read -$s(e,\beta)(0) = e$) the proof is correct, albeit miserably hard to read, way too concise. May should apologize to his readers.<|endoftext|> -TITLE: Non-isomorphic measurable spaces -QUESTION [13 upvotes]: Suppose that $X$ and $Y$ are measurable spaces with the property: there are measurable bijections $f:X \to Y$ and $g:Y \to X$. Is it possible to find non-isomorphic spaces $X,Y$ with this property? One can state similar question in any catgeory and for example in the topological category it is possible. - -REPLY [7 votes]: (This is an edited example, simpler than the first version.) -Yes, there are such spaces $X$ and $Y$. -For $n=0,1,2$, let $A_n$ be a set of cardinality $\aleph_2$ with the $\sigma$-algebra of subsets of cardinality $\leq\aleph_n$ and their complements. Let $X$ be the sum (=disjoint union) of the spaces $A_n$, $n=0,2$. Let $Y$ be the sum of $X$ and $A_1$. Let $f\colon X\to Y$ be a bijection that is identity on $A_0$ and maps $A_2$ onto $A_1 \cup A_2$. Let $g\colon Y\to X$ be a bijection that maps $A_0 \cup A_1$ onto $A_0$ and is identity on $A_2$. -To prove that $X$ and $Y$ are not isomorphic, let $h\colon Y\to X$ be any bijection. Since $h(A_1)=(h(A_1)\cap A_0)\cup(h(A_1)\cap A_2)$, there is $k\in\{0,2\}$ such that the cardinality of $h(A_1)\cap A_k$ is $\aleph_2$. If $k=0$ then $h^{-1}$ is not measurable, and if $k=2$ then $h$ is not measurable. That proves that $X$ and $Y$ are not isomorphic.<|endoftext|> -TITLE: Expression of morphisms in motivic homotopy categories in terms of Nisnevich cohomology? -QUESTION [5 upvotes]: For a perfect field $k$ there is a collection of stable motivic homotopy categories equipped with the corresponding Morel's (homotopy) $t$-structures: $SH^{S^1}(k)$, $SH(k)$, $DA(k)$, and also modules over the motivic cobordism and motivic cohomology spectra (the latter are essentially Voevodsky motives); these categories also contain motivic spectra for all smooth varieties over $k$ that I will denote by $\Sigma^{\infty}(-_+)$. For an object $N$ of the heart of the corresponding homotopy $t$-structure, a smooth $X/k$, and $n\ge 0$ I would like to compute the morphism group $Hom(\Sigma^{\infty}(X_+),N[n])$ (in each of the aforementioned categories). -Now, to the object $N$ of the heart one can associate the presheaf $S_N$ sending a smooth $Y/k$ into $Hom(\Sigma^{\infty}(X_+),N)$; actually, $S_N$ is a Nisnevich sheaf. -I believe that $Hom(\Sigma^{\infty}(X_+),N[n])$ equals $H^n_{Nis}(X,S_N)$. Is there any reference for this fact (for any of the aforementioned motivic categories)? A natural way to prove it would be to start with the stable homotopy category $SH_{Nis}^{S^1}(k)$ of simplicial Nisnevich sheaves (of sets on smooth varieties over $k$) and to study adjoint functors connecting this category with the "motivic categories"; yet I don't have fine enough references for this argument. - -REPLY [4 votes]: $S^1$-spectra -Let me first show it when the category is $SH^{S^1}(k)$, that is the (∞-)category of $\mathbb{A}^1$-invariant sheaves of spectra. Then the heart of the t-structure is precisely the category of strictly $\mathbb{A}^1$-invariant sheaves of abelian groups. The embedding is given by the Eilenberg-MacLane functor, sending a sheaf of abelian groups $A$ to the Nisnevich sheafification $HA$ of $U\mapsto H(A(U))$ (as we will see strictly $\mathbb{A}^1$-invariant means precisely that $HA$ is $\mathbb{A}^1$-invariant). I will show the following more general result -Theorem: Let $A$ be a sheaf of abelian groups on a site. Then for all $X$ there is a natural isomorphism for all $i\in\mathbb{Z}$ -$$\pi_{-i}\Gamma(X,HA)\cong H^i(X;A)\,.$$ -Proof: First let us show the statement when $A$ is an injective sheaf of abelian groups. Then I claim that $HA=H\circ A$, that is that $U\mapsto H(A(U))$ is already a sheaf. But this is the well-known statement that the Čech cohomology of injective sheaves is trivial. Now to show the result for all sheaves it suffices to notice that both sides are $\delta$-functors that are annihilated by injective sheaves. So if we show that they coincide when $i=0$ we are done. But this is easily done by noticing that the functor $H$ sends short exact sequences of abelian sheaves into fiber sequences of sheaves of spectra, and using an injective resolution for $A$. $\square$ -Since $\mathrm{Hom}(\Sigma^∞_+X,HA[i])=\pi_i\Gamma(X,HA)$ this proves the result for the case of $S^1$-spectra. -Motivic spectra -Now let us extend this result to the category $SH(k)$ of motivic spectra. Recall that a motivic spectrum is a sequence $E=\{E_n\}_{n\ge0}$ of $S^1$-spectra together with fiber sequences -$$E_n(X)\to E_{n+1}(X\times\mathbb{G}_m)\to E_{n+1}(X)$$ -The t-structure on motivic spectra is then given by saying that $E$ is connective iff all $E_n$'s are connective. There is a right adjoint t-exact forgetful functor to $S^1$-spectra sending $E$ to $E_0$. They fit in a diagram of adjunctions -$$H(k)\rightleftarrows SH^{S^1}(k)\rightleftarrows SH(k) $$ -where in the first adjunction the right adjoint map is just postcomposition with $\Omega^∞$. -The heart of this category is the category of homotopy modules, that is sequences $A={A_i}_{i\ge0}$ of strictly $\mathbb{A}^1$-invariant Nisnevich sheaves of abelian groups together with short exact sequences -$$ 0\to A_i(X)\to A_{i+1}(X\times\mathbb{G_m})\to A_{i+1}(X)\to 0$$ -If $A$ is a homotopy module I will denote the corresponding motivic spectrum $\{HA_i\}_{i\ge 0}$ with $HA$. Since the functor $SH(k)\to SH^{S^1}(k)$ is t-exact, there is a clear forgetful functor from homotopy modules to strictly $\mathbb{A}^1$-invariant sheaves of abelian groups sending $A$ to $A_0$. I claim the following -$$\textrm{Map}(\Sigma^∞_TX_+,HA)=\Gamma(X,HA_0)$$ -If this is true, clearly the thesis follows from the previous result. But this follows immediately from the diagram of adjunctions I sketched earlier, since the right hand side is just $\mathrm{Map}(\Sigma^∞_{S^1}X_+,HA_0)$.<|endoftext|> -TITLE: What is this concept related to group actions? -QUESTION [8 upvotes]: Inspired by Group Equivariant Convolutional Networks by Taco Cohen and Max Welling I have been thinking about the following construction and I wonder where else it has been described and studied. -Suppose $G$ is a group and $X$ and $Y$ are sets with a $G$-action. Let $f : X \to Y$, not necessarily $G$-equivariant. Then we can cook up augmentations of $X$, $Y$ and $f$ ($X_G$, $Y_G$ and $f_G$ respectively) such that $f_G$ is $G$-equivariant. -The recipe is to define $X_G = G \times X$ with $G$-action $g(h, x) = (gh, gx)$ (resp. for $Y_G$) and $f_G : X_G \to Y_G$ by $f_G(g, x) = (g, gf(g^{-1}x))$. -Then we have that - -$f_G(g(h, x)) = f_G(gh, gx)$ -$= (gh, ghf((gh)^{-1}gx)$ -$= (gh, ghf(h^{-1}g^{-1}gx))$ -$= (gh, ghf(h^{-1}x))$ -$= g(h, hf(h^{-1}x))$ -$= gf_G(h, x)$ - -i.e. $f_G$ is $G$-equivariant. -It seems like this should be a well-known construction but I just can't quite grasp its implications at the moment. Is there somewhere I can read more about it? - -REPLY [8 votes]: In fact $X_G=G\times X$ with the action $g(h,x)=(gh,gx)$ is $G$-equivariantly isomorphic to $G\times|X|=G\times X$ with the action $g(h,x)=(gh,x)$. The isomorphism $i_X:X_G\to G\times|X|$ is given by $i_X(h,x)=(h,h^{-1}x)$, with the inverse $i_X^{-1}(h,x')=(h,hx')$. -Now observe that for any set $S$, the $G$-set $G\times S$ with the action $g(h,s)=(gh,s)$ is the free $G$-set on $S$; thus any map $f:S\to Y$ to a $G$-set $Y$ extends uniquely to a $G$-equivariant map $\bar f:G\times S\to Y$ with $\bar f(1,s)=f(s)$, namely $\bar f(h,s)=hf(s)$. -It follows that for any map $f:X\to Y$ between $G$-sets, there is a unique $G$-equivariant map $\tilde f:X_G\to Y$ with $\tilde f(1,x)=f(x)$, namely, $\tilde f=\bar f\circ i_X$, i. e. $\tilde f(h,x)=hf(h^{-1}x)$: -$$ -X_G\xrightarrow{i_X}G\times|X|\xrightarrow{\bar f}Y -$$ -In the above notation your $f_G$ is $(\pi_G,\tilde f)$, where $\pi_G:X_G=G\times X\to G$ is the projection (which is obviously equivariant).<|endoftext|> -TITLE: How many quit mathematics because they (are afraid that they) can not find a permanent job? -QUESTION [37 upvotes]: I think it is an important question, which frequently appears in discussions. Do you know any surveys which approach it? What are the reasons for people with completed PhD in mathematics to quit it? Definitely, there are many reasons, but how many of them would like to do research in math, but switch to something else because of uncertainty in the future? -Anyway, what are your expectations of such a survey? I heard many estimates, from 0% to 50%. -By quitting mathematics I mean, that a person stops doing research. So, switching to finance, programming, only teaching without active research means "quitting mathematics" in most cases. -I don't know, we can define "doing mathematics" by writing at least one research article in five years or so. Or working on research positions in academia or Microsoft Research and similar. Any surveys are welcome, in any country and time period. - -REPLY [11 votes]: This is more of a suggestion than an answer, but perhaps the following objectively measurable statistic would come close to addressing the question: - -The number of people in the final year of their current position who applied to at least (say) five academic positions but received no offers from academic institutions. - -The first part of this description indicates some kind of desire for an academic position, while the latter part suggests an involuntary exit from the academic world. Note that "final year" could be the final year of a Ph.D., a postdoc, or a tenure track. -Of course this does not exactly match the original question, for several reasons. - -It does not capture those who were (in Nikita's terminology) "afraid" that they would not get an academic position and therefore elected not to apply. -It does not distinguish between "academic position" and "position in which one does mathematical research," neither of which implies the other. -It does not allow for the possibility that someone works at a non-academic position for some time and then returns to an academic position later. - -But it seems close. I don't know that anyone has conducted a survey that tries to measure exactly this number, but perhaps some group that already conducts related surveys could be persuaded to add this to their existing questionnaires.<|endoftext|> -TITLE: The space of Whittaker functionals is at most one-dimensional -QUESTION [9 upvotes]: Let $\mathbf G$ be a connected, reductive group over a local field $F$, and let $(\pi,V)$ be a smooth, irreducible, admissible representation of $G = \mathbf G(F)$. Assume there exists a Borel subgroup $\mathbf B$ defined over $F$. Let $\mathbf T$ be a maximal torus of $\mathbf B$ which is defined over $F$, and let $\mathbf U = \mathscr R_u(\mathbf B)$, $U = \mathbf U(F)$. The choice of a nontrivial unitary character of $F$ and an "$F$-splitting" gives us a nontrivial unitary character $\chi: U \rightarrow S^1$. A linear functional $\lambda: V \rightarrow \mathbb{C}$ is called a $\chi$-Whittaker functional if $\lambda(\pi(u)v) = \chi(u) \lambda(v)$ for all $u \in U, v \in V$. -It is a well known result in the representation theory of reductive groups that the space of Whittaker functionals is at most one dimensional. I have seen a reference for a proof in the case $\mathbf G = \textrm{GL}_n$. Is there a reference for the general case? - -REPLY [8 votes]: This is proved in Shalika's multiplicity one paper: - -Shalika, J. A. - The multiplicity one theorem for $GL_n$. - Ann. of Math. (2) 100 (1974), 171–193. - -While Shalika starts off assuming $G=GL_n$ (he wants to prove the existence of Whittaker models as well), he notes in the introduction that the proof that the dimension is at most 1 works for quasi-split groups, and verifies what he needs in the appendix for quasi-split groups. -I remember parts of Shalika's paper being hard to read (I think the later parts, not where he proves dimension at most 1, but I'm no longer sure). -Anyway, if you are also interested in another reference for a somewhat weaker result, Rodier extends the Gelfand-Kazhdan method for $GL_n$ to split groups over nonarchimedean fields here: - -Rodier, François - Whittaker models for admissible representations of reductive p-adic split groups. Harmonic analysis on homogeneous spaces (Proc. Sympos. Pure Math., Vol. XXVI, Williams Coll., Williamstown, Mass., 1972), pp. 425–430. Amer. Math. Soc., Providence, R.I., 1973.<|endoftext|> -TITLE: Details about plethysm -QUESTION [7 upvotes]: I'm currently working on plethysm, i.e. the character of the composition $S^\lambda(S^\mu(V))$ of the Schur functors $S^\lambda$ and $S^\mu$ on a complex vector space $V$. We note this character $s_\lambda[s_\mu]$. One must know that the character of $S^\mu(V)$ is the Schur function $s_\mu$. -We want to find the irreductible representation of $S^\lambda(S^\mu(V))$, or equivalently, to write $s_\lambda[s_\mu]$ in a sum of Schur function. There is very little that is known about this, and this operation is rarely mentionned in the litterature I found. -I have worked with plethysm mostly in the context of a "math research traineeship", so I skipped some details of all this theory to emphasize on the use of the computer in pure maths. -I have 3 questions about all this and I don't find references: -1) I know how to compute plethysm in terms of power sum symmetric functions, but why is it defined this way? -2) What is known or conjectured about operations on plethysms that are Schur-positive (i.e. having only nonnegative coefficients when expressed in terms of Schur functions), like the Foulkes' conjecure, that said that $\forall a,b \in \mathbb{N}, \ a \leq b, \ h_b[h_a] - h_a[h_b]$ is Schur-positive, where $h_n = s_{(n)}$, the schur function indexed with only one part? -3) Where can I find clear versions (and with the use of modern notations) of the Thrall's proofs of $h_2[h_n]$, $h_n[h_2]$ and $h_3[h_n]$? -Thanks in advance for short explanations or for references! - -REPLY [8 votes]: Of the three plethysms in 3), $h_n[h_2]$ is the simplest. A "modern" proof can be found for instance in Example A2.9 (page 449) of Enumerative Combinatorics, vol. 2. This was known to D. E. Littlewood well before Thrall. It may be even older though not stated in the language of plethysm. -For $h_2[h_n]$, here is one argument. Note that $h_2=\frac 12(p_1^2+p_2)$. Hence - $$ h_2[h_n] =\frac 12(h_n^2+h_n(x_1^2,x_2^2,\dots)). $$ - Now - \begin{eqnarray*} \sum_{n\geq 0}h_n(x_1^2,x_2^2,\dots) & = & \frac{1} - {\prod(1-x_i^2)}\\ & = & \frac{1}{\prod(1-x_i)(1+x_i)}\\ -& = & \left( \sum_{j\geq 0} h_j\right)\left( \sum_{k\geq 0} - (-1)^k h_k\right), \end{eqnarray*} - whence - $$ h_2[h_n] =\frac 12\left(h_n^2+\sum_{k=0}^{2n}(-1)^k h_k - h_{2n-k}\right). $$ - Expand $h_n^2$ and $h_kh_{2n-k}$ into Schur functions by - Pieri's rule and collect terms to get - $$ h_2[h_n]=\sum_{k=0}^{\lfloor n/2\rfloor}s_{(2n-2k,2k)}. $$ -A reference for $h_3[h_n]$ is S. P. O. Plunkett, Canad. J. Math. 24 (1972), 541--552.<|endoftext|> -TITLE: The angles subtended in a TSP tour -QUESTION [12 upvotes]: If I sample a large number of uniform points in the unit square and take a traveling salesman tour of them, is there anything at all that can be said/is known about the distribution of angles at each point? The two pictures below show a TSP tour of 1000 points, together with a histogram of the angles. -It is intuitive that we expect a lot of angles to be near $\pi$, since an angle close to $0$ (or to $2\pi$) suggests that we're "doubling back" on a route and possibly being inefficient. - -REPLY [2 votes]: The distribution has a stunning similarity to the density function of random Delaunay angles mentioned on page 4 of The Expected Extremes in a Delaunay Triangulation - -An explanation might be, that optimal tours go along Delaunay edges for a major part of their course (that heuristic made me search for properties of Delaunay triangulations in the first place).<|endoftext|> -TITLE: Is this proof of union closed set conjecture verified? -QUESTION [13 upvotes]: I have always found union closed conjecture very interesting. There is an arxiv article claiming the solve the problem. My background is not appropriate to read the proof. Is this proof verified or wrong? Any comment would be appreciated. -The article was submitted in 2015 and it is not too long(about ten pages). Thus, I expect that it should be verified until now. - -REPLY [23 votes]: There is a recent thesis by André da Cruz Carvalho on this conjecture. It discusses the paper by Vladimir Blinovsky and shows why his proof does not work (page 16). NB: I have not read it carefully, and this refers to the first version of Blinovsky's paper.<|endoftext|> -TITLE: Sum of the norm of polynomials -QUESTION [8 upvotes]: Let $\bar D$ denote the closed unit disc in the complex plane. -Consider the function $f:\bar D\longrightarrow \mathbb{C}$, defined as $f(z)=z$ for all $z\in \bar D$. -Let $n\in \mathbb{N}$. For $1\leq i\leq n$, let $p_i:\bar D\longrightarrow \mathbb{C}$ be monic complex polynomials on $\bar{D}$, such that $p_i$’s have no roots in $\bar{D}$ and $c_i\in\mathbb{C}$ such that $$f=\sum_{i=1}^{n}c_ip_i.$$ Can we say that for any such representation of $f$, the value $$K=|c_1|\|p_1\|_\infty+|c_2|\|p_2\|_\infty+\cdot+|c_n|\|p_n\|_\infty\geq 2?$$ -$\|.\|_\infty$ denotes the uniform norm, defined as $\|p\|_\infty=sup\{|p(t)|: t\in \bar D\}$ for any complex polynomial $p$ on $\bar{D}$. - -REPLY [6 votes]: The question basically boils down to how large the coefficient at $z$ of a polynomial $p(z)$ that has no zeroes in the circle can be compared to the uniform norm of the polynomial. Indeed, one direction is clear. For the other direction, notice that if we have a polynomial $P$ of degree $n$ with the coefficient $1$ at $z$ and without zeroes, then for every $N>n$ we can write -$$ -z=\frac 1N\sum_{\zeta:\zeta^N=1}\zeta^{-1}P(\zeta z) -$$ -(the restriction that $p_j$ are monic is totally pointless because only the products $c_jp_j$ matter in the whole problem setup). -Now, let's try to figure it out in this new formulation. Nothing is special about polynomials here, we can always just take a long enough Taylor series of a function analytic in a slightly larger disk and we can always reduce the radius a bit and expand afterwards. Also, any function $F$ without zeroes in the disk is just $e^g$. So the question becomes something like that: given a function $g=az+\dots$ in the unit disk, what is the least left half-plane we can squeeze its image into. The answer (by the Schwarz lemma) is given by any conformal mapping of the disk to a left half-plane preserving the origin and having the first Taylor coefficient of some fixed absolute value $A$, which is, say, $g(z)=\frac {Az}{z+1}$ corresponding to the left half plane $\Re z\le\frac A2$. Thus, the best constant is $\min_A\frac{e^{\frac A2}}A=\frac e2\approx 1.36$, which is still noticeably above $1$, but definitely not $2$.<|endoftext|> -TITLE: Status of the basis exchange condition for symplectic matroids -QUESTION [13 upvotes]: Let $J_n := \{1,2,3,\ldots,n,1^*,2^*,\ldots,n^*\}$ with the involution $x\mapsto x^*$ exchanging $i$ and $i^*$ for $1\leq i\leq n$. The following is supposed to be standard, but to avoid any doubt as to the definition (since I am asking about possibly equivalent conditions): a symplectic matroid of rank $k$ on $J_n$ means a set $\mathscr{B}$ of $k$-element subsets of $J_n$, all of which satisfy $B \cap B^* = \varnothing$ (such subsets are called "admissible"), and such that for every total order $\preceq$ on $J_n$ for which $x \preceq y$ implies $y^* \preceq x^*$ for all $x,y\in J_n$ (such orders are also called "admissible") there is a $\preceq$-greatest $B \in \mathscr{B}$, where two admissible $k$-element subsets $A,B$ of $J_n$ are compared by letting $A \preceq B$ mean $a_i \preceq b_i$ for all $i$, where $A = \{a_1\preceq \cdots \preceq a_k\}$ and $B = \{b_1\preceq \cdots \preceq b_k\}$. This is essentially the definition found in Borovik, Gelfand & White, Coxeter Matroids (Birkhäuser 2003, Progress in Mathematics 216), §3.1.3. -The same book states, in §3.8, that more direct definitions of symplectic matroids, analogous to the definition of ordinary matroids using their sets of basis in the form of the basis exchange property, are not known (except in the particular case of Lagrangian matroids). They refer to a paper by Tim Chow, "Symplectic Matroids, Independent Sets, and Signed Graphs" (Discrete Math. 263 (2003), 35–45), which proves the equivalence of symplectic matroids with a complicated exchange property on independent sets, and proposes a conjectural equivalent property in terms of bases. -This is very odd, because in the older paper by I. M. Gelfand & V. V. Serganova (И. М. Гельфанд & В. В. Серганова, "Комбинаторные геометрии и страты тора на однородных компактных многообразиях", Успехи Мат. Наук 42 (1987), 107–134; English translation: "Combinatorial geometries and torus strata on homogeneous compact manifolds", Russian Math. Surveys 42 (1987), 133–168), where the concept of symplectic and (what are now called) Coxeter matroids first originated, the definition given of symplectic matroids is as a set $\mathscr{B}$ of admissible $k$-element subsets of $J_n$ satisfying the conjunction of the following two conditions: -(†1) for any $A,B\in\mathscr{B}$ and $a\in A\setminus B$ there is $b\not\in A$ such that either $(A\setminus\{a\})\cup\{b\} \in \mathscr{B}$ or $(A\setminus\{a,b^*\})\cup\{a^*,b\} \in \mathscr{B}$; -(†2) for any $A,B\in\mathscr{B}$ and $b\in B\setminus (A\cup A^*)$ there is $a\in A$ such that $(A\setminus\{a\})\cup\{b\} \in \mathscr{B}$. -The proposition 4 of §8.2 in the Gelfand-Serganova paper (p. 127 in the original, p. 158 in the translation) asserts that the conjunction of these conditions (which I'll call (†)) is equivalent to the definition of a symplectic matroid using the maximality property (i.e., the one given above). They do not bother to give a proof, and the obviousnes of the fact escapes me. -So, what is going on here? I can imagine several things: - -Either (†) is indeed equivalent to the condition of being a symplectic matroid (in the sense given at the top of this post), but for some reason it is not considered a satisfactory "basis exchange property". In this case, I would like to know why, and also where I can find the proof which Gelfand and Serganova omit. -Or the Gelfand-Serganova paper is wrong in stating that (†) is equivalent to being a symplectic matroid (in the sense given at the top of this post), and for some reason no later paper bothers to actually point out this mistake (this is particularly strange since the paper by Borovik, Gelfand and White, "Symplectic Matroids", J. Algebraic Combin. 8 (1998), 235–252, points out a different mistake in the Gelfand-Serganova, but makes no comment whatsoever concerning this characterization of symplectic matroids). In this case, I would like to know a counterexample to the equivalence (and also, if at least one implication holds between (†) and "being a symplectic matroid"). - -Finally, what is the status of Chow's proposed conjecture, or more generally, of the characterization of symplectic matroids using an exchange-like condition on bases? - -REPLY [5 votes]: I recall being told (by Neil White I think, but he might have been reporting an observation of Borovik) that the Gelfand–Serganova exchange condition is wrong. Unfortunately I no longer recall the counterexample, and I agree with you that this is an unfortunate lacuna in the literature. I would suggest contacting Borovik. Of course if you can straighten out the matter yourself and give a clear report, that would be even better. -I am not aware of any progress on my conjecture or of basis exchange for Coxeter matroids, but on the other hand people have sometimes proved a conjecture of mine without notifying me, so don't take this as the final word. - -EDIT: Zhexiu Tu has obtained a circuit axiomatization of symplectic matroids. This doesn't directly answer your question either, but I thought I would mention it here, since it is the only recent progress on the theory of symplectic matroids that I am aware of.<|endoftext|> -TITLE: Isolated degenerate fixed points -QUESTION [6 upvotes]: Here is a rather naïve question: to apply the classical holomorphic Lefschetz theorem, you often work with biholomorphisms of a complex compact manifold whose fixed points are non degenerate (i.e. $1$ is not an eigenvalue of the differential) even if work of O'Brian http://www.jstor.org/stable/1998512 allows to get rid of the degeneracy assumption and compute explicitly the multiplicities at the fixed points, provided they remain isolated. -However, I have encountered no interesting examples where some fixed point are isolated and degenerate. For instance, this condition implies (using Bochner's linearization) that the automorphism is not of finite order. -Ideally I would like to construct such an exemple with the manifold $X$ satisfying the additional assumptions $h^0(X, \, T_X)=h^1(X, \, T_X)=0$. A finite product of $\mathbb{P}^2$ blown up at 4 generic points seems a reasonable candidate, but I'm not sure this is the simplest path to start. - -REPLY [3 votes]: I think the surface constructed in Lemma 4 of https://arxiv.org/abs/1609.06391 should do what you want. No doubt there is a simpler example, so hopefully someone else will chime in. -It's a rational surface $S$, constructed by blowing up $15$ (carefully chosen) points in $\mathbb P^2$. There is a smooth rational curve $C \subset S$ and an automorphism $\phi : S \to S$ which fixes $C$ and restricts to $C$ as the automorphism $z \mapsto z+1$ in suitable coordinates. In particular, there is a unique fixed point $p$ on $C$, given by $\infty$ in these coordinates. This map acts trivially on the tangent space $T_p C \subset T_p S$, which gives the $1$-eigenvector you are looking for. -I think this should have all the properties you want. One should check there is not some curve of fixed points through $p$, but I am pretty sure of it. (Note that $C$ is not a curve of fixed points: although it's mapped to itself, the points on $C$ move around, and only $p$ is fixed.) The automorphism is indeed of positive entropy, hence not finite order, as you note.<|endoftext|> -TITLE: how to pass from algebraic power series to the analytic ones -QUESTION [5 upvotes]: Fix a field of zero characteristic, $k$, e.g. $\Bbb{R}$ or $\Bbb{C}$. Suppose $k$ is normed (and complete for its norm). Consider the ring extensions: $k[x_1,..,x_n]\subset \ k \ \subset k\{x_1,..,x_n\}$. -The algebraic power series are in some sense "controlled" by the ring polynomials. (e.g. because they satisfy polynomial equations). Is there some way "to control" the analytic power series by the algebraic ones? -e.g. given a morphism of two "large" rings, $R\stackrel{\phi}{\to}S$, each containing subrings of algebraic/analytic power series, suppose I know how $\phi$ acts on polynomials (and thus on the algebraic power series). Is there any way to determine how $\phi$ acts on the analytic power series? -(Here $\phi$ is just an algebraic morphism, without any assumption of continuity in the classical topology, coming from the topology on $k$.) -In the extreme case: suppose an automorphism $\phi\circlearrowright R$ is identity on the subring of algebraic power series. Is $\phi$ identity on the analytic power series? At least, does $\phi$ send analytic to analytic? -(Here $R$ can be non-Noetherian, e.g. $C^\infty(\Bbb{R}^1,0)$.) - -REPLY [3 votes]: I don't think that analytic power series can be controlled by algebraic power series in an algebraic way, because they are often transcendental over the former. - -Suppose an automorphism $\phi$ of $R$ is identity on the subring of algebraic power series. Is $\phi$ identity on the analytic power series? - -The answer is no. -For a counter-example let $k$ be any field of characteristic zero and let $K \subseteq k((X))$ be the field of elements that are algebraic over $k(X)$. By the accepted answer of What's an example of a transcendental power series?, the exponental function $\text{exp}$ is transcendental over $k(X)$ and by transitivity of algebraic extensions, $\text{exp}$ is also transcendental over $K$. Extend the identity on $K$ to the automorphism $$\phi: K(\text{exp}) \to K(\text{exp}),\,\,\text{exp}\mapsto \text{exp} + X$$ -(the inverse sends $\text{exp}\mapsto \text{exp} - X)$. -Let $R$ be an algebraic closure of $k((X))$. Then $\phi$ extends to an automorphism of $R$ (cf. https://en.wikipedia.org/wiki/Transcendence_degree, "Applications"). This $\phi$ is the identity on algebraic power series but not on analytic power series. - -At least, does $\phi$ send analytic to analytic ? - -I think the answer is no, but I have no example. To produce an example one could look for a non-analytic power series $f$ such that $\text{exp}, f$ are algebraically independent over $K$. Then modify the $\phi$ above by mapping $\text{exp} \mapsto f,\,\, f \mapsto \text{exp}$. -Added Apr 11, 2017: The answer to the second question is also no: -Let $k = \mathbb{C}$. As indicated above, it suffices to find a non-analytic power series $f$, that is algebraically independet over $K(\text{exp})$. -By the Newton-Puiseux Theorem (see http://www.emis.de/journals/UIAM/actamath/PDF/38-279-282.pdf), the field $P_a$ of analytic Puiseux series over $\mathbb{C}$ is algebraically closed. In particular, each non-analytic power series is transcendental over $P_a \supseteq K(\text{exp})$. So, we can take, for example, $f := \sum_{n\ge 1} n^n X^n$.<|endoftext|> -TITLE: Reverse Toponogov triangle comparison -QUESTION [8 upvotes]: See the wiki page https://en.wikipedia.org/wiki/Toponogov%27s_theorem -One consequence of the Toponogov comparison Theorem is that if the sectional curvature of a manifold $M$ is pinched below by a number $\delta$. Let $pqr$ be a geodesic triangle, i.e. a triangle whose sides are geodesics, in $M$, such that the geodesic $pq$ is minimal. Let $p′q′r′$ be a geodesic triangle in the model space $M_\delta$, i.e. the simply connected space of constant curvature $\delta$, such that the length of sides $p′q′$ and $p′r′$is equal to that of $pq$ and $pr$ respectively and the angle at $p′$ is equal to that at $p$. -Then $d(q,r) \le d(q',r').$ -The wiki then claim that "When the sectional curvature is bounded from above, a corollary to the Rauch comparison theorem yields an analogous statement, but with the reverse inequality." -Is this true? If so, can someone please provide a reference to this? It seems a bit weird to me since the statement of Rauch Theorem is symmetric, if this reverse statement is a corollary of Rauch Theorem, why isn't Toponogov Theorem also a corollary? -Any comments and references are welcome. - -REPLY [6 votes]: Rauch's comparison local result, while Toponogov's comparison is global. -For the upper curvature bound an analog of Toponogov's comparison holds only locally and indeed it follows from Rauch's comparison. There is a gloabal version for upper comparison. For the curvature bound $\kappa\le 0$ it has an addition assumption that space is simply connected. If $\kappa=1$ then one has to assume that any closed curve shorter than $2\cdot\pi$ can be contracted in the class of closed curves shorter than $2\cdot\pi$. (The case $\kappa>0$ can be reduced to $\kappa=1$ by rescaling.)<|endoftext|> -TITLE: Classification of non-Hausdorff topological vector spaces -QUESTION [13 upvotes]: It is well-known that up to topological isomorphism there is exactly one Hausdorff topological vector space (say, over $\mathbb{C}$) of a given dimension $n$, namely $\mathbb{C}^n$ with the euclidean topology. What happens if we drop the Hausdorff condition, can we still classify all topological vector spaces? For example, there is one $\mathbb{C}^n_{triv}$ with the trivial (indiscrete) topology, and more generally we have $\mathbb{C}^k_{triv} \times \mathbb{C}^{n-k}$ for $0 \leq k \leq n$. Are these all? -Here is how one might start. If $V$ is a topological vector space of dimension $n$, then we can consider its subspace $K=\overline{\{0\}}$, which has some dimension $k$. The quotient $V/K$ is a topological vector space of dimension $n-k$. But it is also Hausdorff, so that it is isomorphic to $\mathbb{C}^{n-k}$. One checks that $K$ has the trivial topology (if $C \subseteq K$ is closed and $p \in C$, then $C-p \subseteq K$ is closed and contains $0$, so that $C-p=K$ and hence $C=K+p=K$.) But it is not clear a priori if this quotient splits. - -REPLY [5 votes]: The quotient actually splits, without any assumption on dimensions of the topological vector space $X$. -Let $Y$ be any algebraic complement subspace of $K:=\overline{\{0\}}$, that is the restriction to $Y\times K$ of the addition map is a bijective (linear) map $s:Y\times K\to X$. I claim it is a homeomorphism, hence a TVS isomorphism. -As you said, $K$ has the trivial (aka indiscrete) topology, that is, $K$ is the only nbd of $0$ in $K$. Also, recall that a TVS $X$ ($T_0$ or not) always possesses a basis of closed nbd's of $0$. But if $V$ is a closed nbd of $0$ in $X$, $V=V+K$, since the closure of any $x$ is $x+K$. Then also $s((V\cap Y) \times K)=(V\cap Y) + K=V+K=V$, which means that $s$ is a homeomorphism. -In conclusion, $X$ splits in the TVS direct sum of the (Hausdorff, dense in $X$) subspace $Y$ and the (indiscrete, closed in $X$) subspace $K$. As a consequence, the restriction of the quotient map $\pi:X\to X/K$ to $Y$ is also an isomorphism $\pi_{|Y}:Y\to X/K$.<|endoftext|> -TITLE: Where does the really nice '8-dimensional' description of the $E_7$ root system come from? -QUESTION [13 upvotes]: The Wikipedia page on $E_7$ tells me: - -Even though the roots span a 7-dimensional space, it is more symmetric and convenient to represent them as vectors lying in a 7-dimensional subspace of an 8-dimensional vector space. - The roots are all the 8×7 permutations of (1,−1,0,0,0,0,0,0) and all the $\begin{pmatrix}8\\4\end{pmatrix}$ permutations of (½,½,½,½,−½,−½,−½,−½) - Note that the 7-dimensional subspace is the subspace where the sum of all the eight coordinates is zero. There are 126 roots. - -This presentation is indeed more symmetrical than anything 7-dimensional I could have come up with myself. But more importantly it makes computations really easy because, magically, the Killing form is just the restriction of the ordinary inner product on the ambient 8-dimensional space to the 7-dimensional subspace where the roots live. (Wikipedia does not state this explicitly but it can be easily checked from the list of simple roots corresponding to the nodes of the Dynkin diagram that Wikipedia is kind enough to list.) -The question in the title ('where does this come from?') is really two questions: -1) A reference request: if I use this presentation of the root system to simplify my computations, who do I give credit to? -2) A more 'philosophical' question: where does the nice presentation in terms of a bigger space 'come from', mathematically? -The idea of realizing a degree $n$ root system in the $n$-dimensional 'sum of coordinates equals zero'-subspace of an $(n+1)$-dimensional space is of course very familiar: it is how we normally describe the root systems of type $A_n$. -But in the $A_n$-case the appearance of the extra dimension seems very natural. Thinking about the root system as coming from the Lie algebra $\mathfrak{g} = \mathfrak{sl}_{n+1}$, we can either argue that the $(n+1)$-dimensional space 'surrounding' the $n$-dimensional Cartan subalgebra is `really' the Cartan of the central extension $\mathfrak{gl}_{n+1}$ of $\mathfrak{g}$ or simply accept that everything there is to understand about $\mathfrak{g}$ can be seen inside its 'defining' representation, which happens to be $(n+1)$-dimensional. (I say it a bit sloppy but you hopefully get what I mean.) -It seems that neither of these explanations is available in the $\mathfrak{g} = \mathfrak{e}_7$ case. It definitely does not have a non-trivial 8-dimensional representation (defining or otherwise) and I also never heard about any interesting central extensions. -So is there another explanation of this '$A_n$-like' behavior of $E_7$? The best I could come up with is that the nice 8-dimensional presentation of the $E_7$-root system is 'inherited' from the $A_7$-root system sitting inside of it. But as far as explanations go this feels like cheating since I only found out that there is an $A_7$ sitting inside $E_7$ by looking at the very description of the root system I am trying to explain! - -REPLY [6 votes]: [edited to include more examples of the construction and fix a typo] -This can also be seen purely in terms of lattices or quadratic forms. -As you note, for any $n$ the vectors in the slice $\sum_{i=0}^n x_i = 0$ of -${\bf Z}^{n+1}$ constitute the root lattice $A_n$; so we'll use -the case $n=7$ of the following general picture. -The dual lattice $A_n^*$ consists of the intersection of this slice with the -sublattice of $(n+1)^{-1} {\bf Z}^{n+1}$ with all $x_i \equiv x_j \bmod\bf Z$; -the map taking $(x_0,\ldots,x_n) \in A_n^*$ to the common value of -$((n+1) x_i) \bmod n+1$ descends to an isomorphism -$A_n^*/A_n \cong {\bf Z}/(n+1){\bf Z}$. Any intermediate lattice -then consists of the intersection of $A_n^*$ with $d^{-1} {\bf Z}^{n+1}$ -for some factor $d$ of $n+1$; this lattice is integral iff $d^2 | n+1$, -in which case it is even unless $n$ and $(n+1)/d^2$ are both odd. -In particular, taking $n=7$ and $d=2$ we obtain an even lattice -of discriminant $8/2^2 = 2$ that contains $A_7$ with index $2$; -this must be the lattice $E_7$ (for instance because we can count -${8 \choose 4} = 70$ new roots which together with the $8 \cdot 7 = 56$ -roots of $A_7$ give a total of $126$). -The root lattice $E_8$ can be constructed similarly with $(n,d) = (8,3)$ -as a lattice intermediate between $A_8^{\phantom*}$ and $A_8^*$, -with index $3$ in both directions. -Other examples: $(n,d) = (2,3)$ gives a lattice isometric with ${\bf Z}^3$; -for $(n,d) = (15,4)$ we get the unique unimodular lattice of rank $15$ -with no vectors of norm $1$; for $(17,3)$, one of the two indecomposable -even lattices of rank $17$ and discriminant $2$ (the other one has the -same theta series, and contains $D_{10} \oplus E_7$ with index $2$); -and for $(24,5)$, the Niemeier lattice with root system $A_{24}$. -(If you like coding theory or the Fano plane, you can also use a -symmetrical picture of $E_7$ in ${\bf R}^7$ that consists of -all vectors $2^{-1/2} a$ with $a \in {\bf Z}^7$ such that $a \bmod 2$ is -an element of the $[7,3,4]$ code, or equivalently either zero or the -complement of a line in the Fano plane. NB $[7,3,4]$ is the dual Hamming code; -using the Hamming code -$[7,4,3]$ itself yields $E_7^*$, while the extended [8,4,4] Hamming code -yields $E_8$.) - -REPLY [3 votes]: The span of the roots of E7 is the pentacontihexapentacosiheptacontihexaexon, aka Gosset polytope 2_13 = x3o3o3o *c3o3o3o. That one allows for a complemental compound decomposition into the small petated hexadecaexon = expanded octaexon = x3o3o3o3o3o3x and the hexadecaexon = trirectified octaexon = o3o3o3x3o3o3o. -In fact, the first one has 56 vertices, the latter one has 70 vertices. If these are compounded in the given orientation wrt. A7 (which most symmetrically can be represented within cartesian coordinates of one dimension plus), then the convex hull of these 126 vertices is nothing than that Gosset polytope 2_13. -You'll find that also on https://bendwavy.org/klitzing/incmats/laq.htm<|endoftext|> -TITLE: Monotone symplectic manifolds with Hamiltonian actions are Kähler? -QUESTION [5 upvotes]: I am wondering if the following is true: -Let $(M,\omega)$ be a compact symplectic manifold which is also monotone, i.e. $c_1(TM)=\lambda [\omega]$. -Moreover assume that it admits a Hamiltonian circle action with isolated fixed points. This forces $\lambda$ to be positive, hence we can rescale the symplectic form to satisfy -$$ c_1(TM)=[\omega].$$ -Is there any result already in the literature that says that in this case $(M,\omega)$ admits a Kähler structure (whose integrable $J$ is compatible with $\omega$)? -This structure is not required to be invariant under the action. - -REPLY [3 votes]: Recall that a compact symplectic manifold $(M, \omega)$ is called Fano if $c_1(M)=[\omega]\in H^2(M, \mathbb{R})$. As is stated in the question body, compact monotone symplectic manifold admitting a Hamiltonian $S^1$-action with isolated fixed points provide examples of symplectic Fano manifolds (possibly after rescaling). -In real dimension 4, it's known that a compact symplectic Fano manifold is diffeomorphic to a del Pezzo surface (Ohta&Ono). Since any two cohomologous symplectic forms on a del Pezzo surface are equivalent and symplectic cone of a del Pezzo surface coincides with its Kaehler cone, we see that any compact symplectic Fano manifold $(M, \omega)$ of dimension 4 admits an integrable a.c.s compatible with $\omega$ (just pullback the a.c.s. from the corresponding del Pezzo surface). Therefore, the answer to your question in real dimension 4 is yes. -In real dimension 6, there is a conjecture stating that a compact symplectic Fano manifold admitting a Hamiltonian $S^1$-action is diffeomorphic to a Fano 3-fold. It has been actually proved that such a manifold is necessarily simply-connected and satisfies $c_1c_2=24$ (for symplectic Chern classes). The conjecture, as far as I know, is still open in full generality. -The picture in higher dimensions is probably even more complicated (for instance, starting from real dimension 12 there are examples of not simply-connected symplectic Fano manifolds; of course, such symplectic manifolds can not admit compatible integrable a.c.s. since they would have a trivial fundamental group then).<|endoftext|> -TITLE: Interactions (functors) between equivariant sheaves for different groups? -QUESTION [8 upvotes]: Let $G$ be a finite group and $k$ a field (alg. closed char 0 for simplicity). -To every $G$ set $X$ we can assign the category of $G$-equivariant sheaves of $k$-vector spaces $Sh_G(X)$. It is essentially obvious that all standard operations on sheaves (pull,push,shriek push,tensor etc.) elevate to the level of $G$-equivariant sheaves on $G$-sets and with $G$-equivariant maps. -So as long as we remain in a context where everything has an action of $G$ things look very similar (almost identical) to the non-equivariant setting. -There are however other natural operations one can do with equivariant sheaves. -For instance let $X$ be a $G$-set and $p: X \to Y=X/G$ be the quotient map. In this case we have a natural functor which assigns to an equivariant sheaf $\mathcal{F}$ the $G$-invariant sections in the pushforward: -$$ \mathcal{F} \mapsto (p_*\mathcal{F})^G \in Sh(Y)$$ -Alternatively one could replace invariants with coinvariants. -For a different example consider a $G$ set $X$ and a subset $S \in X$. Suppose $Stab_G(S)=H$. There's a natural functor $Sh_G(X) \to Sh_H(S)$ which is the usual pullback at the level of sheaves but remembers the $H$-equivariant structure. I think there should be a natural functor: -$$Sh_H(S) \to Sh_G(X)$$ -But at this point i'm confused as to how to define it. In the particular case where $X$ is an orbit and $S=\{x\}$ is a point these functors should give a natural equivalence. - -In a general I'd like to understand when and how do equivariant categories for different groups interact. Slightly more specifically: let $X$ be a $G$-set and $Y$ an $H$-set. And - suppose $G$ and $H$ have some morphism between them (either $G \to H$ - or $H \to G$) And suppose we have a map of sets $X \to Y$ (or the other way round) which respects this morphism. -What kind of natural functors are there between the categories - $Sh_G(X)$ and $Sh_H(Y)$? - -EDIT: I phrased everything in the simple context of sets for clarity. I will prefer a natural answer that would generalize easily to any relevant context. Hopefully with some intuitive/geometric/hands-on interpretation of the the functors involved beyond formal existence arguments. - -REPLY [4 votes]: In light of Marc Hoyois's observation that $Sh_G(X)$ is the same as presheaves on some easy category, I will answer the more general question: - -Given a functor $f \colon \mathcal{C} \to \mathcal{D}$, how can I explicitly compute the three functors $f_! \dashv f^* \dashv f_*$? - - -Background -If $F \colon \mathcal{D} \to \mathbf{Vec}_k$ is a functor, then precomposing with $f$ gives another functor $\mathcal{C} \overset{f}{\to} \mathcal{D} \overset{F}{\to} \mathbf{Vec}_k$ which we write as $f^* F$ because it comes from "pulling $F$ back" along $f$. In the case where $f$ is an injection, then $f^*$ restricts the action from $\mathcal{D}$ to $\mathcal{C}$. If $f$ is a surjection, then $f^*$ inflates the action up from $\mathcal{D}$ to $\mathcal{C}$. -Observe that $f^*$ is always an exact functor, since all it does is reindex the linear maps we already have. As a general rule, that means we should look for a left adjoint (since $f^*$ is left exact) and a right adjoint (since $f^*$ is right exact). They exist! (Not only do they exist, but they are called Kan extensions; see nlab for details. I learned Kan extensions from Emily Riehl's excellent textbook.) The three functors satisfy -$$ -\mathrm{Hom}(f_! E, F) \simeq \mathrm{Hom}(E, f^* F) \;\;\;\;\;\; \mathrm{Hom}(f^* F, G) \simeq \mathrm{Hom}(F, f_* F) -$$ -which seems useful but maybe not explicit enough for everyday use. -First examples -Let $\bullet$ be the terminal category so that a functor $V \colon \bullet \to \mathbf{Vec}_k$ is just a $k$-vector space. Our first two examples take one of our categories to be $\bullet$. -Example 1: $\mathcal{C} = \bullet$ -A functor $\bullet \to \mathcal{D}$ is nothing but an object of $\mathcal{D}$. So suppose $d \in \mathcal{D}$ is some object, and write $i_d \; \colon \, \bullet \to \mathcal{D}$ for its inclusion. -Plainly, $(i_d)^*$ is evaluation of $F$ at $d$ so that $(i_d)^* F = Fd$. Let us set about computing $(i_d)_! E$. Since $(i_d)_!$ is additive, the key example is $E = k$, a one-dimensional vector space. The main property of $(i_d)_! k$ is that homming from it is the same as evaluating at $d$: -$$ -\mathrm{Hom}(\,(i_d)_! k \, , F) \simeq \mathrm{Hom}( k , (i_d)^* F) \simeq \mathrm{Hom}( k , Fd) \simeq Fd. -$$ -In other words, we want the representable functor $\mathrm{Hom}(\,(i_d)_! k \, , \, - \,)$ to be evaluation at $d$. But now we recall the Yoneda lemma, which says that evaluation at $d$ is already represented by an extremely explicit functor $Y^d$! And now, by Yoneda's lemma again, we get an isomorphism between representing objects $(i_d)_! k \simeq Y^d$. -Of course, the functor $Y^d$ is itself representable: $Y^d = k \mathrm{Hom}_{\mathcal{D}}(d, -)$. It sends an object of $\mathcal{D}$ to the free $k$-vector space on the set of morphisms from $d$. -From now on, we think of $Y^d$ as "freely generated by a vector in degree $d \in \mathcal{D}$" since a map from it is the same as a choice of vector in degree $d$. -(Of course, $d$ need not be a number or anything like that; "degree $d$" is a shorthand where we think of a functor $F \colon \mathcal{D} \to \mathbf{Vec}_k$ as a $k$-vector space that's graded by the objects of $\mathcal{D}$, and with structure maps coming from the arrows of $\mathcal{D}$.) -Example 2: $\mathcal{D} = \bullet$ -In this case, there is only the terminal functor $t\, \colon \mathcal{C} \to \bullet$. Since we made no choices, we can expect $t_!$ and $t_*$ to be familiar and important. -If $E \colon \mathcal{C} \to \mathbf{Vec}_k$, then the defining property of $t_! E$ is the same as the defining property of $\mathrm{colim}_{\mathcal{C}} E$, and similarly for $t_* E$ and $\mathrm{lim}_{\mathcal{C}} E$. In brief, -$$ -t_! \, = \mathrm{colim}_{\mathcal{C}} \;\;\;\;\;\;\;\;\; t_* \, = \mathrm{lim}_{\mathcal{C}}. -$$ -Or, thought of another way, $t_! = H_0(\mathcal{C}; -)$ and $t_* = H^0(\mathcal{C}; -)$. This notation reminds us that we also want to compute the derived functors $L_p t_!$ and $R^q t_*$. -Maps between the free modules $F^d$ and "matrices over $\mathcal{D}$" -Let $x, y \in \mathcal{D}$ be objects. By the main property the free modules, we may compute the maps from one free module to another -\begin{align} -\mathrm{Hom}(Y^y, Y^x) &= Y^x y \\ -&= k \mathrm{Hom}_{\mathcal{D}}(x, y). -\end{align} -So a formal $k$-linear combination of morphisms $x \to y$ is the same as a map $Y^y \to Y^x$. The order switches! -(This is the same switch that makes $\mathrm{End}_R(R_R) \simeq R^{op}$; only the left action of $R$ on $R_R$ commutes with the right $R$-module structure. Similarly, only maps between the free modules of $\mathcal{D}$ that are induced by precomposition commute with the (postcomposition) action of $\mathcal{D}$. ) -In the same way that a matrix over a ring gives a map between free modules for the ring, we want the same bookkeeping for maps between free modules for $\mathcal{D}$. In light of our calculation of $\mathrm{Hom}(Y^y, Y^x)$, a map $Y^{y_1} \oplus \cdots \oplus Y^{y_n} \to Y^{x_1} \oplus \cdots \oplus Y^{x_m}$ is naturally given by a "matrix over $\mathcal{D}$" -$$ -A = \left( -\begin{array}{ccc} - a_{11} & \cdots & a_{1n} \\ - \vdots & \ddots & \vdots \\ - a_{m1} & \cdots & a_{mn} \\ -\end{array} -\right) -$$ -where we think of the rows as labeled by the objects $x_1, \ldots, x_m$ and the columns as labeled by $y_1, \ldots, y_n$. We need the row and column labels because the entries are drawn from vector spaces that depend on the row and column label! The rule: $a_{ij}$ is a $k$-linear combination of $\mathcal{D}$-morphisms $x_i \to y_j$. -It is an amusing exercise to multiply two such matrices if the column labels of the first matrix match the row labels of the second. You will find that the morphisms keep themselves sorted out. Two morphisms that you need to multiply will be composable, and two that need to be added will have matching source and target. -(I introduced the terminology of "matrices over $\mathcal{D}$" in my thesis, but the general idea surely dates back to at least the early 70s, for example to Mitchell's Rings with several objects) -Defining a functor $F \colon \mathcal{D} \to \mathbf{Vec}_k$ by a presentation matrix -If the elements of some ring $R$ can be represented by alphanumeric strings, then one way to type an $R$-module into a computer is by a presentation matrix. The syntax of such a matrix is routine. The semantics come from thinking of the cokernel of the map between free modules that the matrix represents. -Similarly, a matrix over $\mathcal{D}$ gives a presentation for a functor $\mathcal{D} \to \mathbf{Vec}_k$. In detail, if we let $\natural \colon k \mathrm{Hom}_{\mathcal{D}}(x, y) \to \mathrm{Hom}(Y^y, Y^x)$ be the natural isomorphism coming from Yoneda's lemma, then any matrix $A$ over $\mathcal{D}$ with row labels $x_1, \ldots, x_m$ and column labels $y_1, \ldots, y_n$ gives rise to a natural transformation -$$ -A^{\natural} \, \colon \, Y^{y_1} \oplus \cdots \oplus Y^{y_n} \to Y^{x_1} \oplus \cdots \oplus Y^{x_m} -$$ -coming from applying $\natural$ to every entry. Then we say that $A$ gives a presentation matrix for a functor $F \colon \mathcal{D} \to \mathrm{Vec}_k$ if there's a natural isomorphism $F \simeq \mathrm{coker}(\, A^{\natural} \,)$. -Which functors $F$ may be so written? Well, the finitely presented ones. But note: if the category $\mathcal{D}$ is finite (meaning finitely many morphisms), then a functor is finitely presented if and only if it lands in the finite dimensional part of $\mathbf{Vec}_k$. -To build a presentation matrix is then not so hard. -Pick a few vectors $v_i \in F x_i$ for various $x_i \in \mathcal{D}$ so that pushing these vectors around using the arrows of $\mathcal{D}$ is enough to span every $k$-vector space $Fd$ for $d \in \mathcal{D}$. (Using our finiteness assumptions, we could naively just pick a basis for every $Fd$, but this would not be efficient!) These vectors are the "generators" of $F$, and they give rise to a surjection -$$ -Y^{x_1} \oplus \cdots \oplus Y^{x_m} \twoheadrightarrow F -$$ -using the Yoneda lemma to interpret the vectors $v_i \in Fx_i \simeq \mathrm{Hom}(Y^{x_i}, F)$ as natural transformations. -Of course, this surjection has a kernel $K$. Our finiteness assumption tells us that we may similarly pick generators for the kernel. Thinking in terms of the original functor $F$, these will be relations on the generators $v_i$. For example, there may be some morphism $\varphi \colon x_1 \to d$ and another $\psi \colon x_2 \to d$ so that the induced linear maps $F\varphi \, \colon Fx_1 \to Fd$ and $F\psi \, \colon Fx_2 \to Fd$ send the vectors $v_1, v_2$ to the same place. In this case, there would be a column in the presentation matrix -$$ -\left( -\begin{array}{c} - \varphi \\ - -\psi \\ - 0 \\ - \vdots \\ - 0 \\ -\end{array} -\right) -$$ -and the column label would be $d$. More generally, we could have a linear combination that draws on several generators $v_i$ in which case the column would be potentially more complicated. Note, however, that an element of $Kd$ is the exact same information as a matrix over $\mathcal{D}$ with row labels $x_1, \ldots, x_m$ and a single column with label $d$. So a degree-$d$ vector in the kernel gives rise to a column in the presentation matrix with label $d$. -In this way, each generator of $F$ gets a row in the presentation matrix, and each generator of $K$ gets a column. The row label is the degree of the generator, and the column label is the degree of the relation. -Computing $L_p f_!$ and $R^q f_*$ on a finitely presented functor with presentation matrix $A$ -The method we just described of finding a presentation matrix for a functor may be extended leftward in an iterative fashion, so as to give as many steps in a resolution as required. If $E \colon \mathcal{C} \to \mathbf{Vec}_k$ is a finitely presented functor with presentation matrix $C_0$, then extend the resolution $p+1$ steps in preparation for computing $L_p f_! E$ obtaining matrices over $\mathcal{C}$ called $C_1, \ldots, C_{p}$ so that $C_i C_{i+1} = 0$. By construction, the induced chain complex of free functors -$$ -* \overset{C_p^{\natural}}{\longrightarrow} * \overset{C_{p-1}^{\natural}}{\longrightarrow} \;\;\;\;\; \cdots \;\;\;\;\; \overset{C_1^{\natural}}{\longrightarrow} * \overset{C_{0}^{\natural}}{\longrightarrow} * \longrightarrow 0 -$$ -is exact except at the very end where the homology is $F$. -Ok, we are ready now. Apply the functor $f$ to the morphisms appearing in the entries of $C_i$. The result is a sequence of matrices $fC_0, fC_1, \ldots, fC_p$, still with $(fC_i)(fC_{i+1}) = 0$ because $f$ is a functor. Each $fC_i$ is now a matrix over $\mathcal{D}$, and so we get a chain complex of free functors $\mathcal{D} \to \mathbf{Vec}_k$ -$$ -* \overset{(fC_p)^{\natural}}{\longrightarrow} * \overset{(fC_{p-1})^{\natural}}{\longrightarrow} \;\;\;\;\; \cdots \;\;\;\;\; \overset{(fC_1)^{\natural}}{\longrightarrow} * \overset{(fC_{0})^{\natural}}{\longrightarrow} * \longrightarrow 0. -$$ -This sequence may have homology! Indeed, the top homology is $L_p f_! E$, the thing we were trying to compute. In particular, $f_! E$ is the cokernel of $(fC_0)^{\natural}$. -In order to compute $R^q f_*$, just take dual vector spaces to build $E^{\vee} \colon \mathcal{C}^{op} \to \mathbf{Vec}_k$. Then $(R^q f_*) E \simeq ((L_q f^{op}_!) (E^{\vee}))^{\vee}$. -Computing homology of $F$ -As an example, suppose we wish to compute the homology $H_0 (\mathcal{D}; F)$, which is to say, the colimit of $F$. Well, write a presentation matrix over $\mathcal{D}$ for $F$, replace each morphism that appears in the matrix with the scalar $1 \in k$, and take the cokernel of that matrix. That's the colimit! -More generally, to get the homology $H_p(\mathcal{D};F)$, the "higher colimits", just write a resolution for $F$ using matrices over $\mathcal{D}$, and replace every morphism in every matrix with the number $1$. This gives back a chain complex over $k$ whose homology computes $H_p$! And if you want cohomology, you can take $k$-linear duals and compute using the opposite category in exactly the same way.<|endoftext|> -TITLE: distinguishing E(K)/E_0(K) groups of order 4 -QUESTION [7 upvotes]: Let $K$ be a local field, complete with respect to a discrete valuation $v$ and let $E/K$ be an elliptic curve. -We also let $E_0(K)$ is the set of points with nonsingular reduction. -It is known that $E(K)/E_0(K)$ is of order at most 4 if $E$ does not have split -multiplicative reduction over $K$ (see Theorem VII.6.4 of Silverman's "The Arithmetic of Elliptic Curves"). -My question concerns the case when the order of this group is exactly 4. -How can we tell if this group is ${\mathbb Z} / {2\mathbb Z} \times {\mathbb Z} / {2\mathbb Z}$ or ${\mathbb Z} / {4\mathbb Z}$? -I looked at Tate's Algorithm, as described in Section IV.9 of Silverman's "Advanced Topics in the Arithmetic of Elliptic Curves", and although that tells us how to tell when the order is 4, and Table 4.1 states what the group is when the residue field, $k$, is algebraically closed, it does not tell us how to determine the actual group itself. -Any references, ideas, etc would be very welcome. - -REPLY [8 votes]: This is not a complete answer, since there are some subtleties in residue characteristic $2$, but in all other cases there is a simple answer. -In residue characteristic $p>2$, you can simply look at the $2$-torsion of $E(K)$. If (a) it either has only one point of order $2$, or it has three, but one is contained in $E_0(K)$, then $E(K)$ is cyclic of order four. The reason is that, if $p>2$, then $E(K)/E_0(K)$ is a direct summand of $E(K)$. [Indeed, since the reduction is additive, this follows from the fact that $E_0(K)/E_1(K)$ is of order $p$, so the short exact sequence relating $E(K)/E_1(K)$ to the two smaller quotients splits; therefore any element of order $2$ in $E(K)/E_0(K)$ lifts to a unique element of order $2$ in $E(K)/E_1(K)$, which lifts to $E(K)$ by Hensel.] -Otherwise (b) we must have that $E(K)$ has three points of order $2$, none of which are in $E_0(K)$, so then $E(K)/E_0(K)$ must necessarily be isomorphic to the Klein four-group. -One brief thing that one can say about the case $p=2$ is that if $E(K)$ has three points of order $2$ which are all of bad reduction, or $E(K)$ has only one point of order $2$, then you're again immediately done. The tricky case remains the one where $p=2$, $\# E(K)[2] = 4$ and $\# E_0(K)[2] = 2$. In that case I can't exclude the possibility that $E(K)/E_0(K)$ could be the Klein-four group, but only one of its elements of order $2$ lifts to an element of order $2$ in $E(K)/E_1(K)$, the other two lifting to elements of order $4$. (But I have no idea whether of not this case actually occurs.)<|endoftext|> -TITLE: Fixed points of an involution -QUESTION [9 upvotes]: Let $V=\mathbb C^{2n}$ with the standard basis $\{e_1,e_2, \cdots , e_{2n}\}$ and let $\sigma$ be the involution $e_i \mapsto -e_{2n+1-i}$. This induces an involution of the Grassmannian $G(n,2n)$ of $n$ dimensional subspaces of $\mathbb C^{2n}$. Then what are the fixed points of this involution ? Does it have a nice structure as a projective variety ? - -REPLY [9 votes]: In general, the fixed points of any map of the Grassmannian induced by a diagonalizable linear transformation is just a union of products of Grassmannians in the eigenspaces. Any subspace $V$ fixed by a diagonalizable transformation $A$ is the sum of the intersections of $V$ with the eigenspaces of $A$ (since the projection to each eigenspace is a polynomial in $A$). If we fix the dimension of each of these intersections, we get a map to the product of Grassmannians of the eigenspaces, which is obviously an isomorphism. -In this case, $\mathbb{R}^{2n}$ is the sum of the 1 and -1 eigenspaces, both having dimension $n$. Thus, the fixed points are a disjoint union of $Gr(k,n)\times Gr(n-k,n)$ for all $0\leq k\leq n$.<|endoftext|> -TITLE: No more absolute simples than $p$-regular conjucacy classes: elementary proof? -QUESTION [6 upvotes]: I am wondering whether anyone knows an elementary (no number fields, no Brauer characters; ideally, not even passing to algebraic extensions) proof of the following fact: - -Let $G$ be a finite group. Let $F$ be a field, and let $p = \operatorname{char} F$. We say that an element $g \in G$ is $p$-regular if the order of $g$ in $G$ is coprime to $\begin{cases} p, & \text{ if } p > 0 ; \\ 1, & \text{ if } p = 0 \end{cases}$. (Thus, for $p = 0$, each element of $G$ is $p$-regular.) We say that a conjugacy class $C$ of $G$ is $p$-regular if and only if each element of $C$ is $p$-regular (or, equivalently, at least one element of $C$ is $p$-regular). -A simple $FG$-module $U$ is said to be absolutely simple if $\operatorname{End}_{FG} U \cong F$ as $F$-algebras. -Theorem 1. The number of pairwise non-isomorphic absolutely simple $F$-modules is at most the number of $p$-regular conjugacy classes of $G$. - -Theorem 1 is attributed to Brauer in Mark Wildon's Representation theory of the symmetric group, where it is used to verify that the simple $F S_n$-modules constructed as quotients of Specht modules (for $p$-regular partitions) comprise the whole list of absolutely simple $F S_n$-modules. Of course, when $p = 0$, Theorem 1 reduces to the well-known fact that the number of pairwise non-isomorphic absolutely simple $F$-modules is at most the number of conjugacy classes of $G$; this is not hard to prove elementarily (by arguing that their characters are orthonormal and therefore linearly independent). But how would one prove Theorem 1 in the general case? -(This is, of course, inspired by math.stackexchange question #2212663; but it is meant to be independent from it.) - -Question 2. What if we replace "absolutely simple" by "simple"? - -REPLY [2 votes]: Berman proved the following general result. If $F$ is a field of characteristic $p$ and $G$ is a finite group let $n$ be the lcm of the orders of the $p$-regular elements of $G$. Let $\zeta$ be a primitive $n^{th}$ root of unity in an algebraic closure of $F$. The Galois group of $F(\zeta)$ over $F$ can be identified with a subgroup $T$ of $(\mathbb Z/n)^*$. Then call two $p$-regular elements $g,h\in G$ $F$-conjugate if $xgx^{-1}=h^j$ with $j\in T$ abusing notation in the obvious way. This is an equivalence relation on $G$. Then the number of simple $FG$-modules is the number if $F$-conjugacy classes of $p$-regular elements. -Proofs can be found in Curtis and Reiner for characteristic 0 and in http://www.ams.org/journals/proc/1964-015-05/S0002-9939-1964-0168665-8/S0002-9939-1964-0168665-8.pdf for positive characteristic.<|endoftext|> -TITLE: Why do people study curve shortening flows? -QUESTION [19 upvotes]: I've recently been studying Riemannian geometry with goal of studying and doing research in Ricci flow, however, I've been noticing that a lot of work in Riemannian geometry seems to be done in curvature shortening flows. Now, most people are familiar with the power of Ricci flow – besides being essential in proving the Poincaré Conjecture and the Sphere Theorem, Ricci flow is in general very useful for uniformizing metrics on 3-dimensional manifolds, and is additionally finding new uses in higher dimensions. -But, why do people care about curve shortening flow? Curves seem so simple that there would be relatively little to study (besides well known aspects such as the fundamental group, or holonomy). Why do people study curve shortening flow? - -REPLY [8 votes]: You absolutely must consult the Wikipedia article on CSF. -Some interesting applications: - -Isoperimetric inequality in the place (noted in the answers here already) -Closed geodesic theorems (noted in the answers here already) -Determination of isoperimetric regions in paraboloids of revolution: A new isoperimetric comparison theorem for surfaces of variable curvature -Used in the proof of the Poincaré conjecture! Finite extinction time for the solutions to the Ricci flow on certain three-manifolds -Evolution of interfaces in variety of applications: Applications of CSF -In particular, CSF of networks models two-dimensional, multi-phase systems (original Mullins problem): Evolution of networks with multiple junctions<|endoftext|> -TITLE: Rationality of conic bundles -QUESTION [5 upvotes]: Let $\pi:X\rightarrow\mathbb{P}^2$ be a $3$-fold conic bundle, and let $\Delta\subset\mathbb{P}^2$ be its discriminant. Assume that both $X$ and $\Delta$ are smooth and that $deg(\Delta)\geq 6$. -Can we make hypotheses on $X$ and $\Delta$ ensuring that $X$ is rational or unirational ? - -REPLY [6 votes]: By Corollary 5.6.1 here -https://arxiv.org/pdf/1712.05564.pdf -if $\text{deg}(\Delta)\leq 4$ then $X$ is rational. -If $\text{deg}(\Delta) = 5$ then $X$ could be rational or not depending on whether the double cover $\widetilde{\Delta}\rightarrow\Delta$ is defined by an even or an odd theta characteristic. For instace, by blowing-up a line in a smooth cubic $3$-fold in $\mathbb{P}^4$ we get a conic bundle, with discriminant of degree five, that is unirational but not rational. -By Theorem 9.1 of the same paper if $\text{deg}(\Delta)\geq 6$ then $X$ is not rational. However, by Theorem 7 here -https://arxiv.org/pdf/1712.05564.pdf -or Corollary 1.2 here -https://arxiv.org/pdf/1403.7055.pdf -if $\text{deg}(\Delta)\leq 8$ then $X$ is unirational.<|endoftext|> -TITLE: Is Homeo($M,D^n$) torsion-free? -QUESTION [9 upvotes]: Let $M$ is a connected smooth $n$-manifold, and suppose that $f$ is a self-homeomorphism of $M$ that has finite order, i.e. $f^k = \text{id}$ for some $k\geq 1$. Suppose moreover that $f$ fixes a non-empty open set point-wise. Does it follow that $f = \text{id}$? In other words, is $\text{Homeo}(M,D^n)$ torsion-free? -If homeomorphism is replaced by diffeomorphism, this is easily seen to be true: create a Riemannian metric such that $f$ is an isometry (by averaging), then use the exponential map at one of the point whose tangent space is fixed by $f$ to deduce that $f = \text{id}$. -It is also true as stated for $M = S^n$; this can be proved using Smith theory. -But what about the general case? - -REPLY [8 votes]: Yes, this is due to Newman (1931): A theorem on periodic transformations of spaces. And Smith reproves it in Part III of his papers (on transformations of finite period).<|endoftext|> -TITLE: Maximize the determinant of Boolean combinations of positive definite matrices -QUESTION [5 upvotes]: I have the following optimization problem. -$$\begin{array}{ll} \text{maximize} & \det \left(\sum^n_{i=1}z_i W_i \right)\\ \text{subject to} & \sum_{i=1}^n z_i = N\\ & z_i -\in \{0,1\}\end{array}$$ -where - -all $W_i$ are given; they are constant, symmetric, and positive definite matrices. -$N$ is also given and strictly less than $n$ (typically much less than $n$---for example, if $n = 200$, then $5 \leq N \leq 20$). - -1 A sub-optimal or near-optimal solution is acceptable for my problem; -2 the maximum size of $W_i$ can be around $100 \times 100$. -Here are the questions. -1 Is there any existed analytically strict solver/algorithm to handle this except random search type (that is, of derivative-free type) algorithm (which I have already tried, but found to be too slow)? -2 Is there a reformulation technique to re-cast it as convex as possible? -3 Is there a reformulation technique to re-cast it as smooth as possible? -One trick is to relax the $z_i$ to be continuous variables with values in $[0,1]$ (let's call it relaxed-ver1; but even if we go through this, the relaxed-ver1 sill involves a sum of a series of matrices weighted by the decision variables $z_i$. -Now, I can write down the gradient of the objective function w.r.t $z_i$, that is, $\frac{\partial W}{\partial z_i} $ (I will type it here later); but the Hessian involves a matrix derivative for the adjunct (or adjugate) matrix adj(W), so I will just stop here: -$$\frac{\partial \text{adj}(W)}{\partial z_i},$$ -where $W = \sum^n_{i=1} z_i W_i$. - -Another possibility is to change the objective functions to some "similar type". For example, I have already thought about using (1) quadratic forms; (2) traces (e.g., $\text{trace}\sum^n_{i=1}z_i W_i$); (3) minimum eigenvalues. But so far, no additional progress. - -REPLY [2 votes]: Given $n$ symmetric positive definite matrices $\mathrm W_1, \mathrm W_2, \dots, \mathrm W_n \in \mathbb R^{m \times m}$ and $s \in \mathbb N$, where $s < n$, -$$\begin{array}{ll} \text{maximize} & \det \left( \displaystyle\sum_{i=1}^n z_i \mathrm W_i \right)\\ \text{subject to} & \displaystyle\sum_{i=1}^n z_i = s\\ & \mathrm z \in \{0,1\}^n\end{array}$$ -where the objective function to be maximized is concave. Were it not for the Boolean constraints, we would have a convex optimization problem. Let us find bounds on the maximum. - -A naive lower bound -Since the matrices are positive definite and $z_i \geq 0$, we have -$$\det \left( \displaystyle\sum_{i=1}^n z_i \mathrm W_i \right) \geq \displaystyle\sum_{i=1}^n \det \left( z_i \mathrm W_i \right) = \displaystyle\sum_{i=1}^n z_i^m \det \left( \mathrm W_i \right) = \displaystyle\sum_{i=1}^n z_i \det \left( \mathrm W_i \right)$$ -Let $c_i := \det \left( \mathrm W_i \right)$. The following binary integer program (IP) -$$\begin{array}{ll} \text{maximize} & \displaystyle\sum_{i=1}^n c_i z_i\\ \text{subject to} & \displaystyle\sum_{i=1}^n z_i = s\\ & \mathrm z \in \{0,1\}^n\end{array}$$ -provides a lower bound on the maximum of the original optimization problem. This lower bound may be too loose, however. - -An upper bound -Replacing the (non-convex) Boolean constraints $z_i \in \{0,1\}$ with the (convex) inequality constraints $z_i \in [0,1]$, the following convex relaxation of the original optimization problem -$$\begin{array}{ll} \text{maximize} & \det \left( \displaystyle\sum_{i=1}^n z_i \mathrm W_i \right)\\ \text{subject to} & \displaystyle\sum_{i=1}^n z_i = s\\ & \mathrm z \in [0,1]^n\end{array}$$ -provides an upper bound on the maximum. In [0], Joshi & Boyd used Newton's method to solve the following approximation of the relaxed problem -$$\begin{array}{ll} \text{maximize} & \log \det \left( \displaystyle\sum_{i=1}^n z_i \mathrm W_i \right) + \gamma \displaystyle\sum_{i=1}^n \left( \log (z_i) + \log (1 - z_i) \right)\\ \text{subject to} & \displaystyle\sum_{i=1}^n z_i = s\end{array}$$ -where $\gamma > 0$. Note that the latter is devoid of inequality constraints. - -Reference -[0] Siddharth Joshi, Stephen Boyd, Sensor Selection via Convex Optimization, IEEE Transactions on Signal Processing, Vol. 57, No. 2, pages 451-462, February 2009.<|endoftext|> -TITLE: Infinite dimensional simple algebras -QUESTION [7 upvotes]: Are there examples of non-commutative finitely-presented infinite-dimensional simple algebras? -I am looking for examples and the only example I know is the Weyl algebra. - -REPLY [4 votes]: First, a couple of constructions. If $R$ is simple and finitely presented (f.p.) then the matrix ring $M_n(R)$ has the same properties. Moreover, $\operatorname{G.rk}(M_n(R))=n\operatorname{G.rk}(R)$, where $\operatorname{G.rk}$ is the Goldie rank. Thus if $R$ has finite Goldie rank (e.g. if $R$ is noncommutative domain, so that $\operatorname{G.rk}(R)=1$), all rings $M_n(R)$ are f.p., simple and mutually non-isomorphic. Likewise, if $S$ is a multiplicative Ore set and $R$ is simple, then so is the Ore localization $S^{-1}R$. If, moreover, $S$ is finitely generated and $R$ is f.p., then $S^{-1}R$ is simple and f.p. Frequently, one can show that $S^{-1}R$ is not isomorphic to $R$ by considering their units. Both constructions can be applied to the Weyl algebra $A_n$ (and iterated), yielding plenty of examples of non-isomorphic f.p. simple infinite-dimensional algebras. You should check standard sources on noncommutative ring theory, such as McConnell and Robson, for this theory and other standard examples, as well as for the filtered algebra argument mentioned below. -An important family of examples of simple (and more generally, primitive) f.p. rings dear to my heart arises from universal enveloping algebras $U(\frak{g})$ of semisimple Lie algebras $\frak{g}$: namely, the quotient of $U(\frak{g})$ by a maximal (resp. primitive) ideal is a simple algebra; if the ideal is completely prime, one even gets a noncommutative domain. These simple quotients are generically infinite-dimensional and cannot be isomorphic to Weyl algebras. Note that since $A=U(\frak{g})$ is a filtered algebra whose associated graded algebra is a free polynomial ring (namely, the symmetric algebra of $\frak{g}$), it follows by standard filtered techniques that every left (or right) ideal of $A$ is finitely generated — the only assumption here is that $\frak{g}$ is finite-dimensional. Thus, every quotient $A/I$ is finitely presented. The upshot of the theory of primitive ideals (assuming the ground field is the complex numbers) is that maximal ideals in $U(\frak{g})$ are parametrized by $W$-orbits on $\frak{h}^{*}$ and the quotient algebra is finite-dimensional if and only if the corresponding orbit $W\lambda$ is integral and non-singular (i.e. $\lambda$ has these properties). In particular, most of these simple quotients are f.p. simple infinite-dimensional algebras, as desired. The somewhat involved technical details, which require knowledge of semisimple Lie algebras and their representation theory, and more explicit examples are described below the fold. This theory is exposed in the books of Dixmier and Jantzen on the enveloping algebras (and in subsequent original papers), for those interested in the proofs. - -Primitive ideals in $U(\frak{g})$ for a complex semisimple Lie algebra $\frak{g}$ have been completely classified in the work of Dixmier, Duflo, Borho, Jantzen, Joseph, Lusztig, Vogan, Barbasch, and others. Each primitive ideal $I$ has "infinitesimal character", determined by the maximal ideal $I\cap Z(\frak{g})$ and there is a unique maximal ideal in each infinitesimal character ($Z(\frak{g})$ denotes the center of $U(\frak{g})$). -In more detail, maximal ideals of $U(\frak{g})$ have the form $I(\lambda)=\operatorname{Ann} L(\lambda-\rho)$, where $\lambda\in\frak{h}^*$ is maximal in its $W$-orbit (the orbit corresponds to the infinitesimal character), $L(\lambda)$ is the simple highest weight module with highest weight $\lambda$, and the linear dual of the Cartan subalgebra $\frak{h}^*$ is partially ordered by $\lambda\geq \mu \iff \lambda-\mu$ is a non-negative linear combination of simple roots $\{\alpha_i, 1\leq i\leq l\}$, where $l=\operatorname{rank}{\frak{g}}$ is the rank of $\frak{g}$. The ideal $I(\lambda)$ has finite codimension if and only if the corresponding simple module is finite-dimensional, i.e. the highest weight $\lambda-\rho$ is dominant integral (i.e., $(\lambda,\alpha_i)$ is a positive integer for all $1\leq i\leq l$); otherwise, $U({\frak{g}})/I(\lambda)$ is an infinite-dimensional primitive ring. -Here are some more specific examples. At one extreme (generic case), if $(\lambda,\alpha_i)\notin\Bbb{Z}\setminus 0$ for all $1\leq i\leq l$, the Verma module $M(\lambda-\rho)$ is a simple infinite-dimensional module whose annihilator is a maximal ideal. A theorem of Dixmier asserts that this annihilator is generated by a maximal ideal of $Z(\frak{g})$, and $Z(\frak{g})$ is known to be a polynomial algebra in $l=\operatorname{rank}{\frak{g}}$ generators, making the quotient manifestly finitely presented. The ideals just described have maximal Gelfand-Kirillov (GK) dimension $\dim{\frak{g}}-l$. At the other extreme are the Joseph ideal for types other than $A$, which is a unique maximal completely prime ideal of minimal positive GK dimension $2l$ (and its analogues in type $A$ that form a one-parameter family). Rather than going through the rather involved construction, let me give the most elementary example: for ${\frak{g}}={\frak{sp}}_{2n}$, the quotient is the even Weyl algebra $A_{n}^{\Bbb{Z}_2}$ (the nontrivial element of cyclic group of order 2 acts by $-1$ on the Weyl algebra; it is easy to see that this action preserves the algebra structure). Thus, the even Weyl algebra is an infinite-dimensional simple ring of GK dimension $2n$, in fact, a non-commutative domain, that is not isomorphic to a Weyl algebra. -The preceding theory can be made fairly explicit in type $A$, including the description of generators for $I(\lambda)$ in several important cases (see e.g. https://arxiv.org/abs/0802.1952, where some examples are worked out, but without spelling out the maximality condition).<|endoftext|> -TITLE: Is Kripke Platek theory finitely axiomatizable? -QUESTION [13 upvotes]: I know that closure under the Gödel operations is equivalent to $\Delta_0$-separation (plus extensionality, union, pair, foundation). This is finitely axiomatizable. But when we add $\Delta_0$-collection, to obtain KP, is it still finitely axiomatizable? And in this case, how to prove it? - -REPLY [10 votes]: The answer to the question depends on how "foundation" is formulated, i.e., as a single axiom or as a full scheme (one which makes sure that every nonempty parametrically definable class has a minimal element, equivalently: the principle of $\in$-induction holds). Emil Jeřábek 's answer pertains to the latter formulation, but not the former. -In the old days, beginning with Barwise, KP was defined so as to include the full scheme of foundation, perhaps because of the fact that KP was designed to study the class of well-founded models of a well-behaved fragment of ZF. But this has changed somewhat, for example in Adrian Mathias' paper The Strength of MacLane Set Theory (Annals of Pure and Applied Logic, 2001) KP only includes $\Pi_1$-Foundation, asserting that every nonempty $\Pi_1$-definable class (parameters allowed) has a minimal element (see p.111 of the above hyperlinked text). -It is well-known that KP as defined by Mathias is finitely axiomatizable. The proof uses the fact that there is a definable truth definition for $\Pi_1$-formulae in KP.<|endoftext|> -TITLE: Finite and infinite 4-regular vertex-transitive graphs with identical $R$-balls -QUESTION [9 upvotes]: Consider the set of all 4-regular connected vertex-transitive graphs. -By compactness, for every integer $R \ge 0$, there is an integer $N_R$, so that for every 4-regular vertex-transitive graph of radius larger than $N_R$, there is an infinite vertex-transitive graph with identical $R$-balls. -Do we know any bounds on $N_R$? - -REPLY [3 votes]: This is not an answer to the question, but it is a bit long for a comment. -My aim here is to note that, if one works with Cayley graphs with edges oriented and labeled by the corresponding generators $S=\{a,b\}$, then it is known that the analogous number $N'_R$ grows at least exponentially fast with respect to $R$. (One can replace $4$ by some fixed integer in "$4$-regular", or equivalently $S$ by some larger fixed finite set.) -The basic observation is the following: if $G$ is a group with finite generating set $S$, then its Cayley graph has the same labeled $R$-ball as some infinite Cayley graph if and only if the group $\langle S\mid A\rangle$ if infinite, where $A$ are the group words of length less than $2R$ that are trivial in $G$. -Indeed, $\langle S\mid A\rangle$ is the largest group with the same $R$-balls as $G$, because every relation in $G$ that one can see in a ball of radius $R$ has length $\leq 2R$. -This observation leads to the following equivalent definition of $N'_R$. Consider all presentations $\langle S\mid A\rangle$ for $A$ a set of group words of length less than $2R$ with respect to $S$. We get finitely many group presentations, some of infinite groups and some of finite groups. Then $N'_R$ is the maximal radius of those finite groups. -It is known that there are finite groups with relations of length $\leq R$ and radius $\geq e^{cR}$. For example the finite nonabelian simple groups, see the paper Presentations of finite simple groups: a quantitative approach (or arXiv link) by Guralnik, Kantor, Kassabov and Lubotzky. One concludes that $N'_R \geq e^{cR}$ for labeled Cayley graphs. -For unlabeled Cayley graphs, I wonder whether the finite simple groups can be used to show that $N_R$ also grows at least exponentially. My gut feeling is that $N_R$ grows much faster.<|endoftext|> -TITLE: Functional equation Dedekind zeta function -QUESTION [8 upvotes]: I'd like to know to what point is it possible to generalize -this method - for obtaining the functional equation for the Dedekind zeta function $\zeta_K(s)$ of a number field ? - -Let $\mathfrak{C}$ be the ideal class group -$$\zeta_K(s) =\sum_{I \subset \mathcal{O}_K} N(I)^{-s} = \sum_{C_j \in \mathfrak{C}} \ \ \sum_{I \subset \mathcal{O}_K,I \,\sim \, C_j} N(I)^{-s}$$ -Assuming $\mathcal{O}_K^\times$ is finite, each ideal $C_j \in \mathfrak{C}$ being a rank $n=[K:\mathbb{Q}]$ free $\mathbb{Z}$-module with basis $b_{j,1}, \ldots, b_{j,n}$, letting $C_j C_j^{-1} = (d_j)$ : -$$\zeta_{K,C_j}(s)=\sum_{I \subset \mathcal{O}_K,I \,\sim \, C_j} N(I)^{-s} = \frac{1}{|O_K^\times|} \sum_{a \in C_j^{-1}} N(\frac{a}{d_j} C_j)^{-s}$$ $$=\frac{1}{|O_K^\times|}\frac{N(C_j)^{-s}}{N(d_j)^{-s}}\sum_{m \in \mathbb{Z}^n \setminus (0)} N(\sum_{l} m_l b_{j,l})^{-s}$$ -$$\Gamma(s) \zeta_{K,C_j}(s) = \frac{1}{|O_K^\times|} -\frac{N(C_j)^{-s}}{N(d_j)^{-s}}\int_0^\infty x^{s-1} \sum_{m \in \mathbb{Z}^n \setminus (0)} e^{-x\ N(\sum_{l} m_l b_{j,l})}dx$$ -Therefore it reduces to finding a Poisson summation formula for $\displaystyle\Theta_j(x) = \sum_{m \in \mathbb{Z}^n} e^{-x \, N(\sum_{l} m_l b_{j,l})}$. - -I did it for $\mathbb{Q}(\sqrt{-5})$ whose ideal class group has two elements, obtaining that $$\Lambda(s)= 5^{s/2}\pi^{-s}\Gamma(s)4 \zeta_{\mathbb{Q}(\sqrt{-5})}(s)=\Lambda(1-s) $$ I guess this method works at least for any imaginary quadratic field ? -If $K$ is imaginary quadratic, $\mathcal{O}_K = \mathbb{Z}[w]$ and any ideal has the form $(k,l+ow)$ then due to the property of $2\times 2$ matrices (see how $\scriptstyle\begin{pmatrix} 2 & 1 \\ 0 & \sqrt{5}\end{pmatrix}$ is treated in the link), I'll get a functional equation $\Lambda_{K,C_j}(s) = \lambda_j^{s/2} \pi^{-s}\Gamma(s)\zeta_{K,C_k}(s)=\Lambda_{K,C_j}(1-s)$, but I don't know how to show $\lambda_j $ is the same for each ideal class $C_j$. -What about the other cases ($\mathcal{O}_K^\times$ infinite, cubic extension, non-monogenic field...) ? - -REPLY [4 votes]: In any event, this is essentially Hecke's original argument, the only other truly independent known proof being that of Tate. The original paper (in german) is, I believe, - -Erich Hecke "Über die Zetafunktion beliebiger algebraischer Zahlkörper" (1917) - -The method works for arbitrary number fields and, with some extra work, for any twist by a Hecke character. -English expositions of the proof can be found at least in Neukirch and Lang's textbooks.<|endoftext|> -TITLE: GIT quotients for linear representations of $SL(2,\mathbb C)$ -QUESTION [7 upvotes]: Let $V$ be the standard two-dimensional representation of $SL(2,\mathbb C)$ and let ${\rm Sym}^2V$ be its symmetric square. Let $n$ be a positive integer and consider the following two representations 1) $V^{\oplus n}$ and 2) $({\rm Sym}^2V)^{\oplus n}$ of $SL(2,\mathbb C)$. -Question 1) Is there some explicit description of the GIT quotient $V^{\oplus n}// SL(2,\mathbb C)$? In particular, is it true that the ring of invariant polynomials is generated by $\frac{n(n-1)}{2}$ quadratic polynomials $vol(v_i,v_j)$ $i\ne j$, where $v=(v_1,\cdots ,v_n)=v\in V^{\oplus n}$ and $vol$ is a volume preserved by $SL(2,\mathbb C)$? -Question 2). Is there some explicit description of the GIT quotient $({\rm Sym}^2V)^{\oplus n}// SL(2,\mathbb C)$? In particular, is it true that the ring of invariant polynomials is generated by $\frac{n(n+1)}{2}$ quadratic polynomials $(v_i,v_j)$ , where $v=(v_1,\cdots ,v_n)=v\in ({\rm Sym}^2V)^{\oplus n}$ and $(.,.)$ is a symmetric bilinear form on $({\rm Sym}^2V)$ preserved by $SL(2,\mathbb C)$? - -REPLY [4 votes]: Both questions are extensively dealt with in Weyl's book "Classical invariant theory" who investigated the invariant of classical groups on multiple copies of their defining representations. Determining a set of generators is called a "First Fundamental Theorem" (FFT) while the relations are given in a "Second Fundamental Theorem". -Question 1: This is best regarded as the action of $Sp(2n)$ on $V=\mathbb C^{2n}$ for $n=1$. Then, indeed, the ring of invariants is generated by all pairings $f_{ij}:=\omega(v_i,v_j)$. These are neatly organized in a $2n\times 2n$ skew-symmetric matrix. The relations are generated by all "principal" Pfaffian minors of size $(2n+2)\times (2n+2)$. For $n=2$ these are quadratic polynomials in 3 variables called the "Plücker relations". The quotient consists of all skew-symmetric matrices of rank $\le 2$ (which is, of course, the affine cone over a Grassmannian). -Question 2: In this case one is dealing with the group $SO(n)$ acting on $\mathbb C^n$ for $n=3$. Here things are a bit more complicated since $SO(n)$ is not strictly a classical group. The better problem is to look at the group $O(n)$ instead. In this case, the invariants are indeed generated by all pairings $p_{ij}=(v_i,v_j)$ which can be organized into a $n\times n$ symmetric matrix. The relations are generated by all $(n+1)\times(n+1)$-principal minors. In your case, the quotient would be the set of symmetric matrices of rank $\le3$. Since you are dealing with $SO(n)$ instead of $O(n)$ things are more complicated. In this case there are additional generating invariants namely all determinants of the form $\det(v_{i_1},\ldots,v_{i_n})$. For $n=3$ the quotient is the subset of $S^2\mathbb C^n\oplus\wedge^3\mathbb C^n$ given be two sets of relations: the rank conditions and the condition that the square of the determinant can be expressend as a Gram matrix $\det((v_{i_\mu},v_{i_\nu}))$.<|endoftext|> -TITLE: How quickly can the derivative of an everywhere differentiable function change sign? -QUESTION [30 upvotes]: Let $f : [a,b] \to \Bbb R$ be everywhere differentiable with $f'(a) = 1$ and $f'(b) =-1$. -By Darboux theorem, we know that $f'([a,b])$ is an interval containing $[-1,1]$. In particular, the set $\{x \in [a,b]: |f'(x)| < 1\}$ is uncountable. But how small can it be? Or to be more formal: - -Can $\{x \in [a,b]: |f'(x)| < 1\}$ have measure zero? - -I guess not, because I have never heard of such a counterexample. But I don't see how to prove it. - -REPLY [16 votes]: Fact 1 (Goldowsky-Tonelli): Let $F:(a, b) \to \mathbb{R}$ be continuous and have finite derivative everywhere. Suppose $F' \geq 0$ almost everywhere. Then $F$ is monotonically increasing. -For a proof of this, see Saks, Theory of the integral, Chapter 6, page 206. -Suppose $X = \{x \in [a, b]: -1 < f'(x) < 1\}$ has zero measure. Let $Y = \{x \in [a, b]: f'(x) \leq -1\}$ and $Z = \{x \in [a, b]: f'(x) \geq 1\}$ -Claim 1: Every point of $X$ is a limit point of $Y$ and a limit point of $Z$. -Proof: Suppose for example $x \in X$ is not a limit point of $Y$ - the other case is similar. Let $I$ be an open interval around $x$ disjoint with $Y$. Then at almost every $y \in I$, $f'(y) \geq 1$. Using Fact 1, it follows that the function $y \mapsto f(y) - y$ is monotonically increasing on $I$ and hence for every $y \in I$, $f'(y) \geq 1$ which is impossible as $x \in Y \cap X$. -Claim 2: The set of points of continuity of $f' \upharpoonright \overline{X}$ is dense in $\overline{X}$ (the closure of $X$). -Proof: Well known (using Baire category theorem). -Now let $I$ be any open interval around $x \in X$. Then the supremum of $f' \upharpoonright I$ is at least $1$ and its infimum is at most $-1$ by Claim 1. By Darboux theorem, it follows that $f' \upharpoonright I$ and hence also $f' \upharpoonright (I \cap X)$ takes every value in $(-1, 1)$. So $f'$ is everywhere discontinuous on $\overline{X}$ which contradicts Claim 2.<|endoftext|> -TITLE: Is the category Idem filtered? -QUESTION [18 upvotes]: I have recently been reading Lurie's "Higher Topos Theory", and come upon what I believe to be an erronous claim. However, the author goes to some pain as a result of that claim, and the error seems to affect multiple other statements. Also, I am relatively unfamiliar with arguments about sizes of categories, which seem to be relevant here. As a result, I am hesitant to trust my understanding. -Specifically, the author defines the category $Idem$ in definition 4.4.5.2 (it is also defined in nlab). To my understanding, it is the nerve of the ordinary category with one object and two morphisms, $id$, $e$ with $e\circ e=e$. -Above proposition 4.4.5.15, the author claims that this category is not filtered. This does not seem true to me: for any simplicial set $K$ and a map $K\rightarrow Idem$, it is possible to extend it to a cone point under $K$ via $e$. -This issue resurfaces again later. Indeed, in proposition 5.3.1.15, it is claimed that a sufficiently small filtered category has a final object. To my understanding, $Idem$ must be a counterexample to that theorem. Otherwise, consider a very small $\infty$-category $\mathcal{C}$, and a much larger cardinal $\kappa$. The ind-category $Ind_\kappa(\mathcal{C})$ should be the idempotent completion of $\mathcal{C}$. However, it is also given by the $\kappa$-right-exact functors on $\mathcal{C}$, which are the $\kappa$-filtered right fibrations over $\mathcal{C}$. By choosing $\kappa$ sufficiently large (and perhaps enlarging the universe), we may assume that such fibrations are $\kappa$-small, and thus have a final object and are representable. This would imply that $Ind_\kappa(\mathcal{C})=\mathcal{C}$, a contradiction. This suggests that $Idem$ should, in a sense, be the only counterexample. -In fact, the above argument could be made even more concrete: let $\mathcal{C}=Idem$. Its idempotent completion $Idem^+$ contains an extra object $x$, which defines a right fibration over $Idem$ via $Idem\times_{Idem^+}Idem^+_{/x}$. This is easily seen to be equivalent to the trivial right fibration $Idem\rightarrow Idem$. Since this fibration belongs to $Ind_\kappa(Idem)$ for all $\kappa$, it should be $\kappa$-filtered. -Finally, I believe that I also know where the error is in the proof of 5.3.1.15. The author claims that the existence of a retract $\mathcal{C}^\triangleright\rightarrow\mathcal{C}$ implies that $\mathcal{C}$ has a final object. This is false, again as can be demonstrated by $Idem$. -I am uncertain how to reconcile any of this with the author's claims, and would gladly appreciate any help! - -REPLY [3 votes]: As noted in the comments, this was a mistake in the book, and is now corrected in the online version.<|endoftext|> -TITLE: Surprisingly short or elegant proofs using Lie theory -QUESTION [71 upvotes]: Today, I was listening to someone give an exhausting proof of the fundamental theorem of algebra when I recalled that there was a short proof using Lie theory: - -A finite extension $K$ of $\mathbb{C}$ forms a finite-dimensional vector space over $\mathbb{C}$, so the group of units $K^\times$ would be of the form $\mathbb{C}^n\setminus\{0\}$, which is simply connected for $n>1$. Since the operation on $K^\times$ is essentially just multiplication of polynomials over $\mathbb{C}$, it must be a Lie group. In sum, if $n>1$, then $K^\times$ is a simply connected abelian Lie group, thus isomorphic (as a Lie group) to $\mathbb{C}^n$, which is absurd (since $\mathbb{C}^n$ is torsion-free). Thus, $n=1$. - -What other examples are there of theorems which yield such short or elegant proofs by appealing to Lie theory? -To clarify the criteria: I'm looking for (nontrivial) theorems that are usually stated in terms outside of Lie theory (e.g. the fundamental theorem of algebra) that can be proven in a particularly short or elegant way using Lie groups or Lie algebras. - -REPLY [4 votes]: I originally intended to write this as a comment but it turned out to be too long. The following is a proof of the fundamental theorem of algebra similar to OP's but which does not rely on the correspondence between connected, simply-connected Lie groups and Lie algebras but on somewhat easier results on Lie theory and some topological facts. -Suppose $K$ is a non-trivial field extension of $\Bbb{R}$ of degree $n$. Then, $K^\times$ is a connected, abelian real Lie group of dimension $n$. Commutativity implies that the exponential map $\mathrm{exp}:\Bbb{R}^n\to K^\times$ is a group morphism, and by connectedness we get that it is surjective and thus a covering map. The kernel $\Gamma$ of the exponential map is then a discrete subgroup of $\Bbb{R}^n$, isomorphic to $\Bbb{Z}^d$, and so $K^\times \simeq \Bbb{R}^{n-d}\times (S^1)^d$. -We know that $d\geq 1$ since there is torsion in $K^\times$, and in fact we must have $d=1$, for otherwise there would be $n^d$ $n$-th roots of unity in $K$. Therefore $K^\times \simeq \Bbb{R}^{n-1}\times S^1$ but also $K^\times \simeq \Bbb{R}^n\setminus\{0\}$, and we get that $n=2$ for otherwise $K^\times$ would be simply-connected.<|endoftext|> -TITLE: Is the asymptotic for $\sum_n (\mu(n)/\sigma(n)) \log(x/n)$ also an upper bound? -QUESTION [14 upvotes]: For $x\geq 1$, -$$\rho(x) = \sum_{n\leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n}.$$ -As $x\to\infty$, this sum tends to $\zeta(2) = \pi^2/6$. Is it in fact the case that $\rho(x)\leq \zeta(2)$ for all $x\geq 1$? (Computations confirm this for $x\leq 65000000$; this takes just a minute or two, but I'm using interval arithmetic, and precision issues preclude me from going further.) How would one go about proving such an inequality? -(We can derive an asymptotic expression for $\rho(x)$ from an asymptotic expression for $\check{m}(x) = \sum_{n\leq x} (\mu(n)/n) \log x/n$ (bounded by Balazard and Ramaré); the hard part is showing that the error term is negative.) - -REPLY [17 votes]: We first observe that as $\frac{\mu(p^k)}{\sigma(p^k)}$ is equal to $1$ if $k = 0$, $-\frac{1}{p + 1}$ if $k = 1$, and $0$ otherwise, its Dirichlet series is -\[\sum_{n = 1}^{\infty} \frac{\mu(n)}{\sigma(n) n^s} = \prod_p \left(1 - \frac{1}{(p + 1) p^s}\right) = \frac{R(s)}{\zeta(s + 1)},\] -where -\[R(s) = \prod_p \left(1 + \frac{1}{(p + 1)(p^{s + 1} - 1)}\right).\] -Note that for $\sigma > -1$, the terms in the Euler product for $R(\sigma)$ are of the form $1 + a_p(\sigma)$ with $a_p(\sigma) = \frac{1}{(p + 1)(p^{\sigma + 1} - 1)}$. Since $\sum_p a_p(\sigma)$ converges for $\sigma > -1$, it follows that $R(s)$ is absolutely convergent for $\Re(s) > -1$ and defines a holomorphic function in that region. Moreover, $R(0) = \zeta(2) = \frac{\pi^2}{6}$. -Let $\Theta$ denote the supremum of the real parts of the zeroes of $\zeta(s)$. Suppose in order to obtain a contradiction that -\[\sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6} < x^{-1 + \Theta - \varepsilon}\] -for all $x > x_{\varepsilon}$. Then Landau's lemma implies that if $\sigma_c$ is the infimum of $\sigma \in \mathbb{R}$ for which -\[\int_{1}^{\infty} \left(x^{-1 + \Theta - \varepsilon} - \sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} + \frac{\pi^2}{6}\right) x^{-\sigma} \, \frac{dx}{x}\] -is convergent, then -\[\int_{1}^{\infty} \left(x^{-1 + \Theta - \varepsilon} - \sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} + \frac{\pi^2}{6}\right) x^{-s} \, \frac{dx}{x}\] -is holomorphic in the right half-plane $\Re(s) > \sigma_c$ but not at the point $\sigma_c \in \mathbb{R}$. On the other hand, this integral is equal to -\[\frac{1}{s + 1 - \Theta + \varepsilon} - \frac{R(s)}{s^2 \zeta(s + 1)} + \frac{\zeta(2)}{s}\] -for $\Re(s) > 0$ and hence for $\Re(s) > \sigma_c$ by analytic continuation. However, this expression has a pole at $s = -1 + \Theta - \varepsilon$ and no other poles on the real line segment $\sigma > -1 + \Theta - \varepsilon$, yet by the definition of $\Theta$, there are poles in the strip $-1 + \Theta - \varepsilon < \Re(s) \leq -1 + \Theta$. Thus a contradiction is obtained, and so it follows that -\[\sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6} = \Omega_+\left(x^{-1 + \Theta - \varepsilon}\right).\] -The same method shows that -\[\sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6} = \Omega_-\left(x^{-1 + \Theta - \varepsilon}\right),\] -so that -\[\sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6}\] -changes sign infinitely often. With a little extra work, we can show that -\[\sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6} = \Omega_{\pm}\left(x^{-1/2}\right).\] - -All of this raises the question: - -Why does there seem to be a bias? - -In other problems like disproving Pólya's conjecture, the bias stems from a term of the same order as the oscillatory terms coming from the zeroes of $\zeta(s)$. But this is not the case in this setting; as Greg mentions in the comments, if we additionally assume the Riemann hypothesis and the Linear Independence hypothesis, then we can show (modulo some details about the growth of $R(s)$ on the line $\Re(s) = -1/2$) that the set -\[\left\{x \in [1,\infty) \colon \sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6} < 0\right\}\] -has a limiting logarithmic density that is equal to $1/2$. -So where does the bias come from? I believe it's due to a lower order term. As $s$ tends to $-1$, $1/\zeta(s+1)$ tends to $-2$, whereas $R(s)$ is positive and blows up as $s \to -1$. In fact, it blows up in two ways: one way from the prime $p = 2$, and the other from the product over the remaining primes. This suggests that $R(s) \sim C/(s + 1)^2$ as $s \to -1$ for some positive constant $C > 0$. So this will give a lower order term of size $-\frac{C' \log x}{x}$ in the asymptotic expansion of $\sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6}$, where $C' > 0$ is some positive constant.<|endoftext|> -TITLE: A non vanishing vector field on $S^3$ with a periodic attractor -QUESTION [5 upvotes]: Is there a non vanishing real analytic vector field $X$ on $S^3$ such that $X$ has an attractor periodic orbit(An asymptotically stable periodic orbit) ? What about the smooth case? - -REPLY [2 votes]: $S^3$ has the structure of a Lie group. Consider a left invariant vector field $X$. (Check that you can get closed orbits!) In a tubular neighborhood of a small amount of time around a closed orbit $o$, the vector field looks like a constant field on $\mathbb{R}^3\simeq D^2 \times I$. Consider the functionr $f = \beta\cdot d(\cdot,o)^2$, where $\beta$ is a smooth compactly supported bump function. The choice of unit vector $\nu$ pointing towards the central orbit $o$ from within this neighborhood is clear. Now set $X' = X + \epsilon\cdot f\cdot \nu$ for some choice of $\epsilon$. -For $X'$ we have the old orbit $o$, unperturbed. Moreover, nearby orbits will veer towards $o$. Thus now $o$ is a stable attractor. One checks that with $\epsilon$ small enough, $X'$ is still nonvanishing.<|endoftext|> -TITLE: Counting monomials in cyclotomic polynomials -QUESTION [9 upvotes]: Let $\Phi_n(x)$ denote the $n$-th cyclotomic polynomial. There are numerous properties and utilities of these polynomials. My interest is more basic and in the spirit of - -Tewodros Amdeberhan and Richard P. Stanley, Polynomial Coefficient Enumeration, preprint (2008) arXiv:0811.3652 (pdf). - -Namely, how many non-zero terms $N(\Phi_n(x))$ does $\Phi_n(x)$ have? I suspect this might be hard. So, let's consider a modest problem. - -Question. If $p, q, r$ are distinct primes, is there an explicit (or semi-explicit) formula for - $$N(\Phi_{pqr}(x))\,\,?$$ - -UPDATE. If $r=2$, direct calculation shows that $\Phi_{2pq}(-x)=\Phi_{pq}(x)$. Therefore, the result of Carlitz (see Igor's link below) implies the following very special case of my question: -$$N(\Phi_{2pq}(x))=N(\Phi_{pq}(x))=\frac{2(p-u)(uq+1)}p-1.$$ -UPDATE. I am still waiting for some answers to the question. - -REPLY [8 votes]: In a 1966 Monthly paper -Carlitz, Leonard, The number of terms in the cyclotomic polynomial $F_{pq}(x)$, Am. Math. Mon. 73, 979-981 (1966). ZBL0146.26704. -Carlitz shows that -$$ N(\Phi_{pq}(x)) = \frac{(p-u)(u q + 1)}{p} + \frac{u(pq - uq - 1)}{p},$$ where $q>p$ and $u = -1/q \mod p.$ Perhaps his methods can be extended to more primes, who can say.<|endoftext|> -TITLE: When is differential geometry on moduli spaces possible (and productive)? -QUESTION [5 upvotes]: I will start out by saying I am not well-informed about moduli theory in the slightest. However, it is known that some moduli spaces (of algebro-geometric objects) have severe pathologies (Ex: Murphy's Law with regards to Hilbert schemes), while others are very nice, such as $\overline{\mathcal{M}}_{1,1}$ which can be identified with the famous modular surface, and on such things differential geometry can be done. This leads me to ask (whether anyone has already asked): -1) When does a moduli space (parametrizing algebro-geometric objects) admit natural Riemannian structure(s) besides possibly on some singular locus? -2) In this case, do geodesics correspond to useful information (nice degenerations between objects perhaps)? - -REPLY [5 votes]: You need that the objects are not varying `wildly'. In algebraic geometry, this may mean that they can always be fit into flat families. Also you might find that the moduli space is broken up into many components: this is the case if considering all Riemann surfaces (of arbitrary genus) or all sheaves on a scheme. (Actually the same underlying issue here is that the Hilbert polynomial is misbehaving.) If this is happening, restrict your class of objects a bit. -If you are in a nice situation, say with a single connected component of moduli, then one tries to exhibit the base objects as local patches of the moduli space. If some of the objects you are parameterizing happen to be very symmetric, i.e., admit larger automorphism groups than their friends nearby, then the structure of the moduli space at this point will have a `stacky' behaviour. Two ways of dealing with this: - -Rigidify: For example, on curves (Riemann surfaces), one can assume as part of the data sets of points $S = \{p_0, \ldots, p_n\}$. Or, perhaps, as is done in number theory, choose a basis of the torsion group $E[p]$, for an elliptic curve $E$ -Admit defeat and work with stacks. Particularly nice stacks are $DM$-stacks and arose specifically for dealing with moduli of curves. If you are interested in the Riemannian angle, then it should be pointed out that putting Riemannian metrics on orbifolds (a type of stack, without a lot of AG baggage if you want) is totally a thing. - -So for your question: - -When does a moduli space admit a Riemannian structure? Ans: This happens when the objects of interest happen to always live in families $E \to B$ where the base $B$ has some natural Riemannian structure. -In this case, do geodesics correspond to useful information? Ans: This is really up to how you set up moduli problem. And there will in general be multiple ways of assigning meaningful metrics, e.g., there is an abundance of meaningful metrics on $M_g$ (not all Riemannian). What do geodesics mean? Maybe quickest way to deform one object into the other given some constraints.<|endoftext|> -TITLE: A book on elliptic curves using scheme theory? -QUESTION [16 upvotes]: I'm interested in learning some stuff about elliptic curves. I've been learning scheme theory, and I'm interested in seeing these tools "in action". It seems that the standard introduction to elliptic curves is Silverman's book, which doesn't make use of schemes at all. So I'm curious, is there an introduction to elliptic curves from the point of view of schemes? - -REPLY [13 votes]: Not exactly a book, but there are course notes on abelian varieties from a course that Brian Conrad taught a few years ago. It is definitely from the perspective of scheme theory/functor of points.<|endoftext|> -TITLE: t-Stochastic Neighbor Embedding vs Topological Data Analysis -QUESTION [5 upvotes]: The shortest form of this question is: - -How much TDA can be done with tSNE? - -Specifically, I'm referring to the application of TDA to clustering data, so, think along the lines of Ayasdi's implementation: - -My understanding is that TDA constructs simplical complexes on a continuum of scales, to then find persistent components. This is an oversimplification because there is also work done by the Mapper algorithm to recognize persistent homologies. -Whereas, tSNE is a 2-D stochastic embedding, which assumes two separate distributions: a gaussian distribution that generates neighbors in high dimensions, and a Cauchy distribution in 2 dimensions, and then constructs an embedding that preserves distances as best as possible between the original space and the embedded space. The analogy to persistent components can be made in our choice of the tSNE perplexity parameter, which specifies the width of each Gaussian distribution, and can therefore give us a simplical complex where edges connect pairs of points that generate each other with some minimal probability. -I would think then that tSNE is equivalent to TDA if we also compute persistent homologies? - -REPLY [10 votes]: I am a co-founder of Ayasdi and contributed to the the original research at Stanford. -The screen shot that you pasted is the output of the Mapper algorithm. A couple of notes of clarification: - -Mapper produces a graph as an output, where nodes are groups of data points and edges between them signify non-empty intersection i.e. some points can be in more than one node and whenever that happens, Mapper connects all such nodes. -The output of t-SNE (and other dimensionality reduction methods such as PCA, MDS etc..) is a lower dimensional representation. -Mapper uses as its input results of dimensionality reduction (e.g. Mapper is happy to consume the results of T-SNE as an input). - -The above is to say that comparing the results of mapper and dimensionality reduction methods does not make sense - they produce different objects. -Why would you want to use Mapper when you have awesome dimensionality reduction? - -Dimensionality reduction methods suffer from projection loss - points well separated in high dimensional space, might appear to be in the same neighborhood in the lower dimensional projection. Mapper protects against this by clustering points in high dimensional space after determining a grouping from the lower dimensional projection. -The output of mapper is very convenient - it is easy to work with graphs and build ML pipelines on top of them. - -Some other notes/pointers: - -LargeVis - since you are interested in efficient dimensionality reduction. -Your idea on constructing simplicial complexes - there are many ways of constructing complexes (e.g. vietoris-rips), and yours is interesting as well, but if you want a complex, simpler methods are generally better.<|endoftext|> -TITLE: n-Engel groups as "homotopy associative" groups -QUESTION [7 upvotes]: Maybe the question (and particularly title) is somewhat silly. Anyway, one can observe that variety of 2-Engel groups may be alternatively described as variety of groups which have skew associative commutator operation: $\mathfrak B ([x, y, y]) \equiv \mathfrak B ([[x, y], z][x, [y, z]])$. - -Can we define n-Engel groups through "higher associators" identities? -Concretely, is multiplicative pentagonator identity of the form $$[x, y, z, w]^{\epsilon}[x, [y, z], w]^{\epsilon}[x, [y, z, w]]^{\epsilon}[x, [y, [z, w]]]^{\epsilon}[[x, y], [z, w]]^{\epsilon}$$ with some choice of $\epsilon 's = \pm 1$ defining for 3-Engel groups? (it obviously implies 3-Engelness for any choice of exponents). - -If not, maybe those free pentagon groups will share some local properties with free 3-Engel; every 3-Engel group is locally nilpotent. - -Are pentagon groups nilpotent? -Are their 4-generator subgroups nilpotent of fixed degree? For example, groups where operations [x, y] and [y, x] distribute over each other are 3-locally metabelian but may fail to be metabelian with minimal counterexample being something like 2-Sylow subgroup of $S_{14}$. - -I think (but it's too lousy for actual statement) that possibility of associator presentations is related to gaps (or their absence) between conditions in this chain of implications: - -(a) $\langle x\rangle^G$ are $n$-nilpotent $\Rightarrow$ (b) $\langle - x\rangle^G$ are $n$-Engel $\Rightarrow$ (c) $G$ is $(n+1)$-Engel - -(c) implies (a) for $n \le 2$ but up to my knowledge equivalence is unknown for higher $n$'s. (see edit) - -What is known today about normal closures of elements in n-Engel groups? - -Edit: N. Gupta and F. Levin constructed in [1] infinitely generated counterexample where (c) $\not \Rightarrow$ (a) for $n = \text{odd prime} + 1$: take countably generated free 2-nilpotent exponent $p$ group $H$ and let $G$ be the semidirect product $\mathbb F_p[H] \rtimes H$ via regular representation. Then $G$ is $p+2$-Engel, but normal closure of $(1, x)$ is not nilpotent for $x$ a generator of $H$. -In same paper they provide intricate example of finitely generated 4-Engel group where normal closures of elements have nilpotence degree 4. -// I've also reformulated and listed subquestions for comprehensibility. -[1] N. Gupta and F. Levin, On soluble Engel groups and Lie algebras, Arch. Math. 34 (1980) - -REPLY [2 votes]: For Quesion 5: -What is known today about normal closures of elements in n-Engel groups? -see the following papers: -Traustason, Gunnar, Locally nilpotent 4-Engel groups are Fitting groups. -J. Algebra 270, No. 1, 7-27 (2003). -Vaughan-Lee, Michael, On 4-Engel groups. LMS J. Comput. Math. 10, 341-353 (2007). -In the latter paper it is proved that a group is a 4-Engel group if and only if the normal closure of every element is a 3-Engel group.<|endoftext|> -TITLE: Parallel transport as algebra isomorphism -QUESTION [6 upvotes]: Assume that there is an smooth structure of the matrix algebra $M_{n}(\mathbb{R})$ on fibers of the tangent bundle of a $n^2$ dimensional manifold. - -Is there a Riemannian metric on $M$ such that all operator of parallel transports would be an algebra isomorphism? - -REPLY [15 votes]: It is a classic theorem in linear algebra that any ($\mathbb{R}$-linear) automorphism $\phi$ of the the ring $M_n(\mathbb{R})$ is inner, i.e., of the form $\phi(x) = axa^{-1}$ for some invertible $a\in M_n(\mathbb{R})$. In particular, the group of automorphisms of the algebra is $\mathrm{PGL}(n,\mathbb{R})$, a simple group of dimension $n^2{-}1$, whose maximal compact subgroups are all conjugate to $\mathrm{PO}(n) = \mathrm{O}(n)/\{\pm I_n\}$. -To avoid trivial cases, assume that $n>1$ and, at least to start with, assume also that $M$ be simply connected. If $M^{n^2}$ carried a Riemannian metric $g$ whose Levi-Civita connection $\nabla$ preserved an $M_n(\mathbb{R})$ algebra structure on $TM$, then, because its holonomy would be compact, there would be an underlying $\nabla$-parallel $\mathrm{PO}(n)$-structure on $M$. -Now, the representation of $\mathrm{PO}(n)$ on $M_n(\mathbb{R})$ is reducible, being the sum of three terms: -$$ -M_n(\mathbb{R}) = \mathbb{R}{\cdot}I_n\ \oplus\ A_n(\mathbb{R})\ \oplus\ S_n(\mathbb{R}), -$$ -where $A_n(\mathbb{R})\subset M_n(\mathbb{R})$ consists of the skew-symmetric $n$-by-$n$ matrices, while $S_n(\mathbb{R})\subset M_n(\mathbb{R})$ consists of the traceless symmetric $n$-by-$n$ matrices. -When $n>2$, the representations of $\mathrm{PO}(n)$ on both $A_n(\mathbb{R})$ and $S_n(\mathbb{R})$ are irreducible and (almost) faithful. By the Bianchi identities, the two corresponding subbundles of $TM$ will be $\nabla$-parallel and, hence by the reducibility of the holonomy, the metric on $M$ will locally split as a product into three factors The (almost) faithfulness of two of the representations guarantees (by the second Bianchi identity), that the curvature of $\nabla$ must vanish identically. Hence, the only case that occurs is the locally flat structure. One can, of course, do things with open sets and discrete quotients, but, locally, the flat case is the only case. -When $n=2$, things are different. The group $\mathrm{PO}(2)$ acts on $A_2(\mathbb{R})\simeq \mathbb{R}$ as $\{\pm1\}\simeq \mathbb{Z}_2$, and it acts on $S_2(\mathbb{R})\simeq \mathbb{R}^2$ as $\mathrm{O}(2)$. Since $M$ is simply connected, it follows that, at least locally, $(M,g)$ can be written as a metric product -$$ -M = \mathbb{R}\times \mathbb{R}\times \Sigma, -$$ -where $(\Sigma,h)$ is an oriented Riemannian surface, which, hence, has an associated orthogonal complex structure $J$ and area form $\omega$. Then one can define an $M_2(\mathbb{R})$ structure on $T_pM$ = $T_x\mathbb{R}\times T_y\mathbb{R}\times T_z\Sigma = \mathbb{R}\oplus\mathbb{R}\oplus T_z\Sigma$ by the rule -$$ -(a,b,v)(a',b',v') = (aa'{-}bb'{+}h(v,v'),\ ab'{+}a'b{+}\omega(v,v'),\ av'{+}a'v{+}b\,Jv' {+} b'\,Jv). -$$ -Thus, the general solution essentially depends on a Riemannian metric in dimension $2$, which is one function of two variables, up to diffeomorphism. -If one is willing to consider pseudo-Riemannian metrics instead of only Riemannian ones, there are other solutions. For example, when $n=2$, $\mathrm{PGL}(2,\mathbb{R})$ acts preserving the irreducible splitting -$$ -M_2(\mathbb{R}) = \mathbb{R}{\cdot}I_2\ \oplus Z_2(\mathbb{R}) -$$ -where $Z_2(\mathbb{R})\simeq \mathbb{R}^3$ is the space of $2$-by-$2$ traceless matrices. The action of $\mathrm{PGL}(2,\mathbb{R})$ on $Z_2(\mathbb{R})$ preserves a Lorentzian inner product $\langle,\rangle$ that satisfies $\langle v,v\rangle = \det(v)$, so that $v^2 = -\langle v,v\rangle I_2$ and the 'outer' or 'skew' product $[v,w] = vw-wv$. -Consequently, if $(M^4,g)$ is a simply-connected Lorentzian $4$-manifold whose holonomy preserves a $M_2(\mathbb{R})$ algebra structure on $TM$, then $M$ can be written locally as a metric product -$$ -M^4 = \mathbb{R} \times \Sigma^3 -$$ -where $(\Sigma^3,h)$ is an oriented Lorentzian $3$-manifold and $g$-parallel algebra structure on $TM= T\mathbb{R}\oplus T\Sigma$ can be written in the form -$$ -(a,v)(a',v') = (aa'-h(v,v'),\ av' + a'v + v{\times}v') -$$ -where $v{\times}v' = \ast_h(v\wedge v')$ and where $\ast_h:\Lambda^2(T\Sigma)\to T\Sigma$ is the Hodge star associated to the Lorentzian metric $h$ and the orientation. This gives a family of (Lorentzian) solutions depending on $3$ functions of $3$ variables locally. -Finally, when $n>2$, there is the natural bi-invariant $M_n(\mathbb{R})$ structure on $M=\mathrm{GL}(n,\mathbb{R})$ itself, regarded as a pseudo-Riemannian symmetric space, together its dual symmetric space $M^* = \mathrm{GL}(n,\mathbb{C})/\mathrm{GL}(n,\mathbb{R})$. By Berger's classification of non-symmetric holonomies (or by simply computing the space of curvatures of this particular $\mathrm{PGL}(n,\mathbb{R})$-structure and seeing that the only possibility is the locally symmetric one), these are, locally, the only examples when $n>2$.<|endoftext|> -TITLE: Sets of sets: near-pencils and projective planes -QUESTION [7 upvotes]: Suppose $n$ is a positive integer. Let ${\cal C}$ be a set of subsets of $X:=\{1,\ldots,n\}$ with the following properties: - -all members of ${\cal C}$ contain at least $2$ elements, and $X\notin {\cal C}$; -$A\neq B\in {\cal C}$ implies $|A\cap B| = 1$; and -$|{\cal C}| = n$. - -Doe this imply that at least one of the following statements is true? -(S1) All members of ${\cal C}$ have equal cardinality (in that case ${\cal C}$ is called a "projective plane"); -(S2) There is $A\in{\cal C}$ such that $|A| = n-1$ (in that case ${\cal C}$ is called a "near pencil"). - -REPLY [9 votes]: Yes, it is true and known. See Bourbaki Theory of sets, excersises to the chapter III (ordered sets), $\S 5$ (properties of integers). It is ex. 14 in English edition of 1968. -The proof goes as follows. At first, we denote $|\cal C|=m$ and do not assume for a moment that $m=n$, but prove that $m\leqslant n$. Assume that some element $y$ belongs to a unique set $A\in \cal C$. If $|A|=2$, all sets in $\cal C$ should contain $A\setminus \{y\}$ and have no other common elements, this gives even $m\leqslant n-1$. If $|A|\geqslant 3$, we may remove $y$ from $X$ and from $A$ and (say, using induction) we again get $m\leqslant n-1$. So, we may suppose that each element $y$ belongs to at least two sets from $\cal C$. -Take an element $x\in X$ and a set $A\in \cal C$, denote $f(x)$ the number of sets containing $x$. Note that if $x\notin A$, then $f(x)\leqslant |A|$, else by pigeonhole principle there exist two sets $A_1,A_2\in \cal C$ containing $x$ for which $C_1\cap A=C_2\cap A$, and hence $|C_1\cap C_2|\geqslant 2$, a contradiction. Choose a set $A$ such that $|A|:=t$ is minimal. For any $a\in A$ choose a set $B_a\in \cal C$ containing $a$ and different from $A$. If $A=\{1,2,\dots,t\}$, we have $i\notin B_{i+1}$ (indicies are taken modulo $t$). Therefore $\sum_{i=1}^t f(i)\leqslant \sum_{i=1}^t |B_i|$. If now $X=\{1,\dots,n\}$, we have $f(i)\leqslant |A|$ for $i=t+1,\dots,n$. Summing up we get $\sum_{i\in X} f(i)\leqslant \sum_{i=1}^t |B_i|+(n-t)|A|$. But $\sum_{i\in X} f(i)=\sum_{B\in \cal C} |B|$, so we get that the sum of sizes of certain $m-t$ sets from $\cal C$ does not exceed $(n-t)|A|$. By minimality of $|A|$ this is not possible when $m>n$. If $m=n$, it is possible only if all sets except $B_1,\dots,B_t$ have size equal to $t$ and each element outside $A$ is contained exactly in $t$ sets from $\cal C$. Consider two cases. -1) $t=2$, $A=\{1,2\}$. If $1$ is contained in $f(1)=k+1$ sets from $\cal C$, then 2 is contained in $n-k$ sets from $\cal C$, and for all pairs of sets $B\ni 1$, $C\ni 2$ from $\cal C\setminus \{A\}$ the intersections $B\cap C$ are different, thus $n\geqslant 2+k(n-1-k)$, $k=1$ or $k=n-2$. If, say, $k=1$, then the set containing 1 different from $A$ contains $n-1$ elements and we get a near-pencil. -2) $t>2$. Since we have equality, we get $f(1)=|B_2|\geqslant t\geqslant 3$, so we may replace $B_1$ to a different set containing 1. This gives $|B_1|=t$. Analogously, all sets have cardinality $t$.<|endoftext|> -TITLE: A surjective morphism of abelian varieties induces an epimorphism on the torsion subgroups -QUESTION [5 upvotes]: Let $f:A\to B$ be a surjective morphism of abelian varieties (over an algebraically closed field). -Why is it true that $f$ induces an epimorphism on the points of finite order $A_{\mathrm{tors}}\to B_{\mathrm{tors}}$? - -REPLY [5 votes]: Clearly the image of a torsion point by a group homomorphism is again a torsion point, so it only remains to prove the surjectivity of the restriction $f \colon A_{\mathrm{tors}} \to B_{\mathrm{tors}}$. -By using Stein factorization, we see that $f$ factors as a surjective homomorphism having an abelian variety as kernel and an isogeny, so it sufficies to prove the result in these two situations. -$\boldsymbol{(i)}$ Assume first that the kernel of the surjective homomorphism $f \colon A \to B$ is an abelian subvariety. Let $b \in B_{\mathrm{tors}}$ be a point of order $n$, namely $nb=0$. Then if $a \in A$ is any preimage of $b$ via $f$, we have $$f(na)=nf(a) =nb=0,$$ hence $na \in \ker f$. -By assumption $\ker f$ is an abelian subvariety of $A$, in particular it is a divisible group; so we can find $a' \in \ker f$ such that $na = na'$, that is $n(a-a')=0.$ So $a-a' \in A_{\mathrm{tors}}$ and moreover $$f(a-a')=f(a)-f(a') = b-0= b,$$ -i.e. $a-a'$ is a preimage of $b$ that lies in $A_{\mathrm{tors}}$. This shows the desired surjectivity in this case. -${}$ -$\boldsymbol{(ii)}$ Assume now that $f \colon A \to B$ is an isogeny; then $\ker f$ is finite, in particular it is a torsion subgroup of $A$. Choosing $b$ and $a$ as above, we see that $$f^{-1}(b)=a+ \ker f.$$ -Now $na \in \ker f$ and the fact that $\ker f$ is a torsion subgroup imply that $a$ is a torsion element, too. So in this case all the preimages of $b$ via $f$ are torsion elements of $A$, in particular the surjectivity of $f \colon A_{\mathrm{tors}} \to B_{\mathrm{tors}}$ follows also in this case.<|endoftext|> -TITLE: Index Theorem for the Twisted Dirac Operator -QUESTION [5 upvotes]: In the Mirror Symmetry monograph (http://www.claymath.org/library/monographs/cmim01c.pdf), on page 297, the index theorem is used for a two-dimensional twisted Dirac operator. Below equation 13.37, it is claimed that the number of $\psi_-$ zero modes is equal to the number of $\overline{\psi}_+$ zero modes, and the number of $\overline{\psi}_-$ zero modes is equal to the number of ${\psi}_+$ zero modes. How does one show that this is true? -In addition, on page 811, a similar problem is considered for a two-dimensional surface with boundary. Here it is claimed in equation 39.213 that the index of the twisted Dirac operator is -\begin{equation} -\textrm{Index }\mathcal{D}=\#[(\psi_-,\overline{\psi}_+)\textrm{ zero modes}]-\#[(\overline{\psi}_-,{\psi}_+)\textrm{ zero modes}]. -\end{equation} -However, as far as I understand there is no well-defined index for $\mathcal{D}$, but only well-defined indices for its chiral or antichiral parts $D$ and $\overline{D}$, where -\begin{equation} -\mathcal{D}=\bigg(\begin{array}{cc} - 0 & D \\ - \overline{D} & 0 \\ - \end{array} \bigg). -\end{equation} - What exactly does equation 39.213 mean? Would it be correct to interpret $\textrm{Index }\mathcal{D}$ as $\textrm{Index }\overline{D}$? - -REPLY [7 votes]: First, we need a spin structure to define the spinor bundle. The index theorem does not care which one we take, so we may take even spinors to be $(0,0)$-forms and odd spinors to be $(0,1)$ forms. Both bundles are trivial on $T^2$, so we may take functions as spinors of both parities. Then the positive part of the Dirac operator is $\not\!\partial_+=\partial_{\bar z}$, and the negative is its adjoint, which becomes $\not\!\partial_-=\partial_z$ on functions. -Now twist with $E$ and the connection $A$ to get twisted Dirac operators $\not\!\partial_+^E=D_{\bar z}$ and $\not\!\partial_-^E=D_z$ in their notation. Equation (13.37) follows from Riemann-Roch (or Atiyah-Singer, if you prefer). -Taking adjoint bundles is the same as doing a complex conjugation if the operators are compatible with a Hermitian metric, so it maps $E\otimes S_\pm$ to $E^*\otimes S_\mp$, preserving the connections. -In particular, it maps $\psi_\pm$ zero modes to $\bar\psi_\mp$ zero modes and vice versa. -For your second question, regard the operator $\mathcal D$ as the formal difference of the operators $\not\!\partial^E$ and $\not\!\partial^{E^*}$. This operator only has an index if we specify suitable boundary conditions; take those of (39.212). Then the index (39.213) is indeed well-defined. The easiest way to see this is to consider a twisted Dirac operator on the double of $\Sigma$, and that is explained in some detail after (39.213). Because the double has no more boundary, you can use Atiyah-Singer (or Riemann-Roch, if you prefer) to see what the index is.<|endoftext|> -TITLE: Proof of Fisher's inequality in combinatorial terms -QUESTION [6 upvotes]: Suppose $n$ is a positive integer. Let ${\cal C}$ be a set of subsets of $X:=\{1,\ldots,n\}$ with the following properties: - -all members of ${\cal C}$ contain at least $2$ elements, and $X\notin {\cal C}$; and -$A\neq B\in {\cal C}$ implies $|A\cap B| = 1$. - -A version of Fisher's inequality states that $|{\cal C}| \leq n$. There are short proofs relying on Linear Algebra. Is there a purely combinatorial proof of this statement? - -REPLY [7 votes]: A combinatorial proof of a more general inequality is given by Douglas Woodall. -One line proof Fisher's inequality is given by Renaud Palisse -Palisse, Renaud, A short proof of Fisher's inequality, Discrete Math. 111, No.1-3, 421-422 (1993). ZBL0788.04003. -Woodall, Douglas R., A note on Fisher's inequality, J. Comb. Theory, Ser. A 77, No.1, 171-176, TA962729 (1997). ZBL0878.05011.<|endoftext|> -TITLE: A step in the proof on the uniqueness of mass -QUESTION [6 upvotes]: I am reading the survey paper "The Yamebe Problem" by Lee and Parker. In section 9, Theorem 9.6 in P.78, it was proved that the mass is well defined in the sense that $m(g)$ depends only on the metric $g$. But there is one step in the proof which I cannot understand: in the proof of Theorem 9.6, they said that "The first key observation is that the radial distance functions $\rho=|z|$ and $\tilde{\rho}=|\tilde{z}|$ are related by $C^{-1}\tilde{\rho}\leq \rho\leq C\tilde{\rho}$ for some positive constant." I wonder how one can conclude this. I guess it follows from the previous sentence, which says -"$\tilde{z}^i=z^i+\varphi^i$, where $\varphi^i\in C^{2,\alpha}_{-\tau+1}(N_\infty)$". But I am not exactly sure. I also tried to read the original paper "The mass of an asymptotically flat manifold" by Bartnik. But it seems to me that it was not mentioned explicitly. I guess there are many people in this site who have read these papers. I would appreciate if I can get help from these people. - -REPLY [10 votes]: Yes it follows from the previous sentence. Since $\tau > (n-2)/2 \geq 0$ by assumption you have that $|\varphi^i| \leq C \rho$ from the definition of the norm. This implies that $\tilde{\rho} \leq C \rho$ for some possibly different $C$ by triangle inequality. The condition is symmetric between $\rho$ and $\tilde{\rho}$, so the other inequality follows also immediately.<|endoftext|> -TITLE: Infinite groups with oligomorphic conjugation action -QUESTION [16 upvotes]: The action of a group $G$ on a set $X$ is called oligomorphic if the diagonal action on $X^n$ has finitely many orbits for each $n$. - -Question: Is there an infinite (maybe even finitely generated) group $G$ such that the conjugation action of $G$ on itself is oligomorphic? - -Edit: In view of the comments, I would already be happy to see an example of an infinite group $G$ whose conjugation action on $G^2$ or $G^3$ has finitely many orbits. - -REPLY [8 votes]: Another argument to show that a conjugation-oligomorphic group G is finite is as follows. Let $(s_i,t_i)$ be a set of representatives of the $G$-action on $G\times G$ and let $X$ be the subgroup generated by these representatives. Note that whenever we consider a subgroup generated by $X$ and two more elements $a,b$, $Y=\langle X,a,b\rangle$, then $Y$ is a subgroup of $\langle X,t\rangle$ where $t$ is such that it conjugates $(a,b)$ to some $(a_i,b_i)$. -Continuing this process, i.e. passing from $\langle X,a,b\rangle$ to $ \langle X,t\rangle$ to $\langle X,t,u\rangle$ to $\langle X, v\rangle$ (where $v$ is such that it conjugates $(t,u)$ into $X$)... we get an increasing chain of $(m+1)$-generated subgroups. As $G$ is uniformly locally finite (see the comments or (3) in YCor's answer), $G$ must be finite.<|endoftext|> -TITLE: real analyticity, Fourier coefficients -QUESTION [5 upvotes]: Question. Suppose $f$ is periodic in $[0,2\pi]$. What conditions on the Fourier coefficients of $f$ would guarantee real analyticity of $f$? Please provide me with a reference. - -REPLY [10 votes]: A function $f:\mathbb{R}\to\mathbb{C}$ periodic by $2\pi$ is real analytic if and only if it extends holomorphically to $\{z\in\mathbb{C}:\ |\Im z|0$, because the interval $[0,2\pi]$ is compact. The latter condition is easily seen to be equivalent to the exponential decay of the Fourier coefficients: $a_n\ll e^{-c|n|}$ for come $c>0$. (Necessity of this condition follows from Cauchy's theorem about integrals of holomorphic functions. Sufficiency of the condition follows from Morera's theorem and Cauchy's theorem about the analyticity of holomorphic functions.) - -REPLY [6 votes]: The condition is $|c_n|\le A e^{-B|n|}$ for some positive constants $A,B$ and all $n$. This is not difficult to prove directly (a reference might be exercise I.4.4 in Katznelson's Introduction to Harmonic Analysis, but it is given as an exercise...).<|endoftext|> -TITLE: Torsion in the Atiyah–Hirzebruch spectral sequence of a classifying space -QUESTION [20 upvotes]: Let $G$ be a compact, connected Lie group. There is an Atiyah–Hirzebruch spectral sequence -$$H^*(BG;K^*) \implies K^*(BG)$$ -connecting $H^*BG$, which generally contains torsion, with $K^*BG \cong \widehat R(G)$, which does not. -Is it known whether the torsion situation "only improves"? -More precisely, let $P_r \subsetneq \mathbb N$ be the set of torsion primes of the $r$th page: - -Is it always the case that $P_{r+1} \subseteq P_r$? -Is it at least true that $P_r \subseteq P_2$? - -REPLY [27 votes]: Of course, in any spectral sequence $E_{r+1}$ is a subquotient of $E_r$ (the kernel of $d_r$ divided by the image of $d_r$). And in general new torsion can appear in the sense of torsion elements in $ker/im$ that are not represented by torsion elements in $ker$. -But this cannot happen when the spectral sequence is rationally trivial, that is, when the image of $d_r$ is in the torsion subgroup. So in that case the torsion subgroup of $E_{r+1}$ is a subquotient of that of $E_r$. -This holds for the Atiyah-Hirzebruch spectral sequence of any simply connected space and any cohomology theory $k^\ast$. The reason is that $k^\ast\otimes \mathbb Q$ is a product of ordinary cohomology theories. -So, yes the torsion situation only improves. -Edit: I overstated things a bit. Tyler Lawson makes a point that I had not appreciated: at the level of generality of my answer, tensoring the AHSS with $\mathbb Q$ does not yield the AHSS for another cohomology theory. The argument sketched above applies to the homology AHSS of any simply connected space, or spectrum, but in the case of cohomology the conclusion is false in general. Think of the case when $k$ has only two nontrivial coefficient groups $k^0(\ast)=A$ and $k^n(\ast)=B$, $n>0$. The one nontrivial differential $d_{n+1}$ is a stable cohomology operation $H^\bullet(X;B)\to H^{\bullet +n+1}(X;A)$, corresponding to an element of $H^{n+1}(HB;A)\cong Hom(H_{n+1}(HB),A)\oplus Ext(H_{n}(HB),A)$. This group need not be a torsion group, even though $H_n(HB)$ and $H_{n+1}(HB)$ are torsion groups for any abelian group $B$. But it's OK if $B$ is finitely generated.<|endoftext|> -TITLE: Variation of the Green function with respect to the metric -QUESTION [9 upvotes]: Consider a (closed) Riemann surface and let $G(x,y)$ be the Green function of the Laplace-Beltrami operator. We can informally identify $G$ with the two-point correlation function for the Gaussian random field: -$$G(x,y)=\left<\phi(x)\phi(y)\right>=\frac{1}{Z} \int \mathcal{D}\phi\;\phi(x)\phi(y) \exp\left(-\frac{1}{2}\int \left| \nabla\phi(z)\right|^2 dV_g(z)\right).$$ -This is easy to vary with respect to the metric $g_{\mu\nu}$: -$$\delta G(x,y)=-\frac{1}{2}\int dV_g(z)\;\delta g^{ij}(z)\left<\phi(x)\phi(y) \nabla_i\phi(z)\nabla_j\phi(z)\right>_c,$$ -where $\left<\right>_c$ is the connected correlation function (the disconnected diagram $x\leftrightarrow y,z\leftrightarrow z$ is cancelled by the variation of the partition function). We can use Wick's theorem to compute this, getting -$$\boxed{\frac{\delta G(x,y)}{\delta g^{ij}(z)}=-\nabla_{[i} G(x,z)\nabla_{j]} G(z,y)}$$ -(derivatives w.r.t. $z$). This formula looks wickedly similar to the Hadamard variation formula for the variation of the boundary of the domain in flat space. Yet I haven't been able to find any mentions of this in the mathematical literature. -Furthermore, if we define the regularized Green's function (a.k.a. the Robin function) by -$$G^R(x)=\lim_{y\to x} \left(G(x,y)-\frac{1}{2\pi}\ln d(x,y)\right),$$ -where $d(x,y)$ is the local geodesic distance, then I'm conjecturing the following variational formula: -$$\frac{\delta G^R(x)}{\delta g^{ij}(z)}=-\nabla_{i} G(x,z)\nabla_{j} G(x,z)-\frac{1}{4\pi}\nabla_i \nabla_j G(x,z).$$ -The second term is motivated by the well-known formula for conformal variations (where it is $\frac{1}{4\pi}\delta_x(z)$) and seems to be necessary to cancel the second order pole in this variation. Edit: this guess turned out to be wrong, see my answer below. - -REPLY [3 votes]: It seems that the naive derivation from the path integral only picks up the term coming from quasiconformal variations. Combining the known result for the quasiconformal variation with the much more well known conformal variation, I arrived at the expression -$$\frac{\delta G(x,y)}{\delta g^{\mu\nu}(z)}=(-\nabla_{(\mu}G(z,x)\nabla_{\nu)}G(z,y)+\frac{1}{2}g_{\mu\nu}(z)\nabla_\rho G(z,x)\nabla^\rho G(z,y))+\frac{1}{2V}(G(x,z)+G(z,y))g_{\mu\nu}(z).$$ -The traceless part in the brackets represents the quasiconformal variation (which in complex conformal coordinates reduces to just $8\partial_z G(z,x)\partial_z G(z,y)$, as found in the literature). The last "trace" term is the conformal variation, which is not probed by the path integral because it only shows up at finite volume. -I believe this formula to be correct based on the independence of conformal and quasiconformal variations. The resulting tensor is also divergence-free away from $x$ and $y$, as required by general covariance. The more difficult part is proving the following conjecture for the Robin function: -$$\frac{\delta G^{R}\left(x\right)}{\delta g^{\mu\nu}\left(z\right)}=-\nabla_{(\mu}G\left(z,x\right)\nabla_{\nu)}G^R\left(z\right)+\frac{1}{2}g_{\mu\nu}(z)\nabla_\rho G(z,x)\nabla^\rho G^R(z)+\frac{1}{V}G\left(z,x\right)g_{\mu\nu}\left(z\right)-\frac{1}{4\pi}\nabla_{\mu}\nabla_{\nu}G\left(x,z\right)-\frac{\delta(x,z)}{4\pi}g_{\mu\nu}(z)+\frac{1}{8\pi V}g_{\mu\nu}(z)+\frac{1}{2V}f_{\mu\nu},$$ -where $f_{\mu\nu}$ is an unknown symmetric traceless tensor defined by -$$\nabla^\mu f_{\mu\nu}=\nabla_\nu G^R,$$ -as required by the divergence-free condition on the metric variation of $G^R$. -This result is based on the exact formula -$$G^R(x)=\frac{1}{4\pi}\int G\left(x,y\right)R\left(y\right)\mathrm{d}V(y)+\frac{1}{V}\zeta^{R}\left(1\right)-c,$$ -where $\zeta^{R}(s)=\zeta(s)-\frac{V}{4\pi}\frac{1}{s-1}$ and $\zeta$ is the spectral zeta function of the positive Laplace-Beltrami operator, $c=\frac{1}{2\pi}\left(\gamma-\ln2\right)$ (reference below). -Steiner, Jean, A geometrical mass and its extremal properties for metrics on $S^2$, Duke Math. J. 129, No. 1, 63-86 (2005). ZBL1144.53055.<|endoftext|> -TITLE: Complex varieties with non-torsion homotopy groups -QUESTION [10 upvotes]: Is there some kind of classification of (connected) smooth complex varieties such that every homotopy group of the manifold of complex points is torsion-free? Any reference on this topic will be most welcome. - -REPLY [11 votes]: Following Mark Grant's comment, referencing David Chataur's answer here: McGibbon and Neisendorfer proved that a finite-dimensional 1-connected space with any nonzero reduced homology has infinitely many homotopy groups with torsion. In particular, this applies to the universal cover $\tilde{X}$ of the space $X$ we care about. So $\tilde{X}$ must be acyclic (and is simply-connected), and is thus contractible by Hurewicz's theorem. Therefore $X$ is aspherical. - -REPLY [2 votes]: The answer is likely to be a bit messy, since every projective curve is a $K(\pi, 1),$ and so are cartesian products and (iterated) fibrations of such (classifying which of these are algebraic is, presumably, still open, even for surface bundles over surfaces). I assume, in my ignorance, that complements of hyperplane arrangements are not of this form, which makes me think that the question is hopeless.<|endoftext|> -TITLE: Homotopy limit and Bousfield localization -QUESTION [6 upvotes]: Suppose $E$ is a spectrum and $p$ is a prime. We can then $(H\mathbb{Z}/p)_*$-localize to obtain $L_{H\mathbb{Z}/p}E$. Is it true that the natural map $L_{H\mathbb{Z}/p}E\rightarrow\text{holim}_n(L_{H\mathbb{Z}/p}E)/p^n$ is an $(H\mathbb{Z}/p)_*$-equivalence? - -REPLY [6 votes]: Yes, $H\mathbb{F}_p$-localizations are $p$-complete. We have the cofiber sequence -$$ -\Sigma^{-1}\mathbb{S}/p^\infty \xrightarrow{j} \mathbb{S}\to \mathbb{S}[p^{-1}] \to \mathbb{S}/p^\infty -$$ -where $\mathbb{S}/p^\infty\approx \mathrm{colim}_n \mathbb{S}/p^n$, from which it is not hard to see that the $p$-completion map $X\to \mathrm{holim}_n X/p^n=X^\wedge_p$ is equivalent to -$$ -F(\mathbb{S},X) \xrightarrow{j^*} F(\Sigma^{-1}\mathbb{S}/p^\infty, X). -$$ -In particular, $X\to X^\wedge_p$ is an equivalence iff $F(\mathbb{S}[p^{-1}], X)=0$. -We have that $H\mathbb{F}_p\wedge \mathbb{S}[p^{-1}]=0$, so $F(\mathbb{S}[p^{-1}], L_{H\mathbb{F}_p}X)=0$.<|endoftext|> -TITLE: Deformation invariance of Fano varieties -QUESTION [11 upvotes]: Let $f:X \to T$ be a flat, projective morphism of noetherian schemes with $T$ an irreducible curve. Suppose that there exists a point $0 \in T$ such that the fiber $f^{-1}(0)$ is Fano. - -Q. Is it true that in this case, for all $t$ near $0$, the fiber $f^{-1}(t)$ is Fano? - -N.B. If necessary, one can assume that $f$ is smooth, but if such a statement holds without this assumption, it would be more interesting. - -REPLY [15 votes]: The answer is yes, in fact the following result holds. - -Theorem. Let $f \colon X \to T$ be a flat deformation of a Fano variety $X_0:=f^{-1}(0)$ having at most terminal, $\mathbb{Q}$-factorial singularities. -Then $X_t:=f^{-1}(t)$ is a Fano variety with at most terminal, $\mathbb{Q}$-factorial singularities, for all $t$ in a neighborhood of $0$ in $T$. - -For a proof, see Corollary 3.2 and Proposition 3.8 in -De Fernex, Tommaso; Hacon, Christopher D., Deformations of canonical pairs and Fano varieties, J. Reine Angew. Math. 651, 97-126 (2011). ZBL1220.14026.<|endoftext|> -TITLE: About the category of all small diagrams -QUESTION [8 upvotes]: Let $\mathcal{K}$ be a category. I denote by $\mathcal{D}\mathcal{K}$ the category of all small diagrams over $\mathcal{K}$: an object is a functor $F:I\to \mathcal{K}$ from a small category $I$ to $\mathcal{K}$ and a morphism from $F:I\to \mathcal{K}$ to $G:J\to \mathcal{K}$ is a functor $\phi:I\to J$ together with a natural transformation $\mu:F\Rightarrow G\circ \phi$. If $\mathcal{K}$ is complete and cocomplete, then $\mathcal{D}\mathcal{K}$ is complete and cocomplete as well. I have two basic questions about $\mathcal{D}\mathcal{K}$: - -When $\mathcal{K}$ is cartesian closed, is $\mathcal{D}\mathcal{K}$ cartesian closed ? -Does it exist any paper/book gathering what is known about this category which is certainly not new ? In particular, does this construction have a name ? - -REPLY [5 votes]: If $\mathcal{K}$ is assumed both complete and cartesian closed, then $\mathcal{DK}$ is also complete and cartesian closed. Since completeness is not in question for the OP, I'll skip over that part and focus on cartesian closure, although we will need to recall the structure of finite cartesian products. -So: if $g: J \to \mathcal{K}$ and $g': J' \to \mathcal{K}$ are two objects of $\mathcal{DK}$, then their product is the composite $J \times J' \stackrel{g \times g'}{\to} \mathcal{K} \times \mathcal{K} \stackrel{\text{prod}}{\to} \mathcal{K}$. I'll denote this as $g \cdot g': J \times J' \to \mathcal{K}$. If $p: L \to \mathcal{K}$ is another object, I claim the exponential $(p: L \to \mathcal{K})^{(g': J' \to \mathcal{K})}$ is the functor $p^{g'}: L^{J'} \to \mathcal{K}$ defined by a formula given by an end of exponentials in $\mathcal{K}$, -$$p^{g'}(F) := \int_{j': J'} p(F(j'))^{g'(j')}.$$ -This is a straightforward calculation: morphisms from $(g: J \to \mathcal{K}) \times (g': J' \to \mathcal{K})$ to $p: L \to \mathcal{K}$, say $(l: J \times J' \to L, \beta: g \cdot g' \Rightarrow pl)$, are clearly in natural bijection with pairs $(\hat{l}: J \to L^{J'}, \hat{\beta}: g \Rightarrow p^{g'} \circ \hat{l})$ where $\hat{l}(j) := l(j, -)$ and $\hat{\beta}$ is the natural transformation whose component at $j \in Ob(J)$ is naturally associated with the family $\beta_{j, j'}$ (indexed over $j'$) according to the evident natural bijection between morphisms indicated below: -$$\frac{g(j) \times g'(j') \stackrel{\beta_{j, j'}}{\to} pl(j, j')}{g(j) \underset{\hat{\beta}_j}{\to} \int_{j'} pl(j, j')^{g'(j')}}$$ -(holding $j$ fixed and currying the family $\beta_{j, j'}$ which is natural in $j'$ yields a family dinatural in $j'$, so that we get an induced map to the end which is a universal wedge for that dinatural family). -(Perhaps this somewhat pedestrian calculation is a special case of some more abstract consideration, possibly involving a form of Artin gluing. For example, if we follow Karthik's comment and specialize to where the categories $I, J$ are discrete and $\mathcal{K}$ is say $\text{Set}$ or even a topos $E$, then $Fam(E)$ is the Artin gluing along $\Delta: \text{Set} \to E$. And indeed, if I recall correctly, Carboni and Johnstone do discuss Artin gluings in the doctrine of cartesian closed categories, so maybe something like this can be made to work. However, I am doubtful that one could dispense with completeness of $\mathcal{K}$, as the end calculation seems to mandate this.)<|endoftext|> -TITLE: Complex symmetric Matrices over the the field of Laurent series -QUESTION [7 upvotes]: Let $K=\mathbb C((z))$ be the field of Laurent series in the variable $z$, and consider the involution on $K$ that sends $f(z)$ to $f(-z)$. A complex symmetric matrix of size $r$ over $K$ is a matrix $A(z)\in M_r(K)$ such that $$^tA(z)=A(-z),$$ where $^tA(z)$ is the transpose (not conjugate transpose) matrix of $A(z)$. - -Is it true that any complex symmetric matrix can be written in the form $^tB(z)\cdot B(-z)$? - -REPLY [3 votes]: The answer is yes -- this follows from the general theory of reduction of hermitian matrices. -The matrix $A$ such that $^tA(z) = A(-z)$ is an hermitian matrix in the terminology of Bourbaki, Algèbre, Chap. 9, $\S$ 3, n°1. The reduction theory ($\S$ 6, n°1, Cor. 2 in loc.cit.) tells you that there exists an invertible matrix $P \in \mathrm{GL}_r(K)$ such that $A(z)= {}^tP(z) D(z) P(-z)$ with $D(z)=\operatorname{diag}(f_1(z), \ldots , f_m(z),0, \ldots, 0)$, where $f_i$ are elements of $K$ such that $f_i(z)=f_i(-z)$ and $m$ is the rank of $A$. -It remains to check that any $f \in K$ fixed under the involution can be written $f(z)=g(z) g(-z)$ for some $g \in K$. Note that $f \in \mathbb{C}((z^2))$ and we may assume without loss of generality that $f(z)=1+\sum_{n\geq 1} a_n z^{2n}$. Using the Taylor expansion of the square root, we get $f(z) =g(z^2)^2$ for some $g \in K$, and thus $f(z)=g(z^2)g((-z)^2)$ as desired. -The proof of the reduction result is entirely similar to the proof for classical hermitian matrices. Letting $\phi$ be the hermitian form associated to $A$, the starting point is to find a vector $x \in K^r$ such that $\phi(x,x) \neq 0$, and then to proceed by induction on $r$. To find $x$, choose vectors $x_0,y_0 \in K^r$ such that $\phi(x_0,y_0) \neq 0$, then at least one of the vectors $x_0,y_0,x_0+y_0,x_0+zy_0$ will work. This provides an algorithm (but I don't claim it is optimal).<|endoftext|> -TITLE: Problem on distances in a polygon -QUESTION [6 upvotes]: In $\mathbb{R}^2$ consider a square (call it $S$) and three triangles (one acute $T_2$ and two obtuse $T_1$ and $T_3$) such that each triangle shares one different side with the square and the triangles and the square are disposed exactly as in the following picture. - -Define $P:=S\cup T_1\cup T_2\cup T_3$. -Call $x_i$ the vertex of $T_i$ opposed to the side of $T_i$ shared with the square $S$. Choose any point $p$ inside the square $S$ such that all three segments $\overline{px_i}$ are entirely contained in $P$. -Now move all the vertexes of $P$ in a continuous way in such a manner that all the following lengths are not increased: - -the lengths of all the sides of the square and of the three triangles -the lengths of the two diagonals of the square and the lengths of the segments from each $x_i$ to the vertexes of the square which are contained in $P$. - -In the following picture I've drawn in blue all the segments whose lengths are not increased moving the vertexes of $P$: - -Call $P':=S'\cup T_1'\cup T_2'\cup T_3'$ the polygon obtained in such a manner and $x_i'$ the vertexes of the new triangles. -For any $p'\in S'$ define $d(x_i',p')$ in the following way: - -If $\overline{x_i'p'}\subset P'$, then $d(p',x_i'):=|p'x_i'|$ -If $\overline{x_i'p'}\not\subset P'$, then $d(p',x_i'):=$ minimum of $|v'p'|+|v'x_i'|$ for $v'$ which varies between the vertexes of $S'\cap T_i$ - -I'm wondering if it's always possible to find $p'\in S'$ such that $d(p',x_i')\le |px_i|$ for $i=1,2,3$. -So my question is of course how to prove the existence of $p'$. I'm trying to consider all possible cases in which the vertexes of $P$ could move (given the bonds of the lengths), but it's quite complicated. Do you thing there's a better way to proceed? - -REPLY [4 votes]: Let me prove a bit more general statement. - -Let $P=[v_1\dots v_n]$ and $P'=[v_1'\dots v_n']$ be two solid polygons such that if $[vw]$ is a side of $P$ or a diagonal which lies in $P$ completely then $|v-w|\ge |v'-w'|$. Then there is a short map $P\to P'$ which sends any vertex $v$ of $P$ to the corresponding vertex $v'$ of $P'$. - -Note that if there is one diagonal $[vw]$ such that $|v-w|=|v'-w'|$ then we can cut $P$ in two polygons and reduce the question to a simpler case. -First note that we can assume that $|v-w|=|v'-w'|$ for any side of $[vw]$ of $P$. -If this is note the case choose a vertex $x'$ so that the triangle $[x'v'w']$ lies in $P'$. -Put joints in these vertexes and increase $|v'-w'|$ up to $|v-w|$ so that the rest of diagonals stay the same or increase. The new poygon $P''$ admits a short map to $P'$ so we can make each side to be the same as in $P$ one by one and reduce the question to this case. -Note that the convex angles get smaller in $P'$. -Therefore one concave angle, say $a$, should go up, in particular it stays concave. -Choose 4 vertexes: $a$, its neighbors $x$ and $y$ and a vertex $b$ visible from $a$. -Put joints in these vertexes. -We can move the polygon straighting the angle at $a$, keeping the distances $|a-x|$, $|a-y|$, $|b-x|$ and $|b-y|$ and nondecreasing the rest of diagonals. -Repeating if necessary, we will get one more diagonal fixed. -P.S. Doing these operations, you can end in a generalized solid polygon --- a flat disc with polygonal boundary, when it is mapped to the plane you can get overlaps.<|endoftext|> -TITLE: Standard Monomial basis for other types -QUESTION [8 upvotes]: For the algebraic group $SL_n$ (type $A_{n-1}$) and for a dominant weight $\lambda$ the standard monomials are indexed by the semi-standard young tableaux of shape $\lambda$ and they form a basis for the representation $V_{\lambda}^*$. For other types of simple algebraic groups do we have a description of standard monomials in terms of tableau ? If so, then how are the semi-standard young tableaux described ? A small example illustrating the basis elements say for $B_4$ or $D_5$ in terms of tableau will be highly appreciated. - -REPLY [11 votes]: Standard monomial theory has been extended to all classical groups by Lakshmibai, Seshadri and others in the series of papers "Geometry of $G/P$ I-IX". -A very concise description of standard tableaux in this setting can be found in the appendix of "Littelmann, Peter: A generalization of the Littlewood-Richardson rule. J. Algebra 130 (1990), no. 2, 328–368". -Note, that the standard monomial approach was later generalized er even superseded by Littelmann's path model and Kashiwara's crystals. These work even for symmetrizable Kac-Moody algebras.<|endoftext|> -TITLE: Gap principle for a diophantine inequality -QUESTION [7 upvotes]: Let $d$ be a positive, non-square integer, and let $B > 1$ be a real number. Consider the inequality -\begin{equation} |x^2 - dy^2| \leq B. \end{equation} -This inequality has infinitely many solutions in integers $x$ and $y$ (for example, there are already infinitely many solutions to the Pell equation $x^2 - dy^2 = 1$). Let $T > 1$ be a positive number, and consider the subset of solutions to the above inequality satisfying -$$\displaystyle |x - y\sqrt{d}| < B^{1/2} T^{-1}$$ -and -$$\displaystyle |x + y \sqrt{d}| > B^{1/2} T.$$ -Suppose there exists $k > 1$ such that $1/2k < B^{1/2}T^{-1} < 1/k$. Are distinct integral solutions $(x,y), (u,v)$ satisfying the above inequalities necessarily bounded away from each other? If so, what are the best general bounds in terms of $B,T,k$ and $d$? One can assume that there exists $0 < \alpha \leq 1$ such that $d \leq B^\alpha$. -Any help would be appreciated. - -REPLY [2 votes]: Conjecture If $(s,t)$ and $(u,v)=(s+i,t+j)$ are in $\{{(x,y) \mid |x^2-dy^2| \le B\}}$ then, under the right conditions, $$i \ge \frac{s}{2B}-\frac{s}{8B^2}.$$ (Near) equality occurs when $s^2-dt^2=\mp B,$ $u^2-dv^2=\pm B$ and $i^2-dj^2=\pm 1.$ - -By the right conditions I mean something vaguely like $d\sqrt{d} \lt B$ and $B\sqrt{B} -TITLE: Can a index 2 subgroup of $\pm\Gamma(n)\le \text{SL}_2(\mathbb{Z})$ be noncongruence? -QUESTION [10 upvotes]: One way of interpreting the question might be: Is the property of being congruence a topological property? Ie, is it detected at the level of Riemann surfaces $\mathcal{H}/\Gamma$? -My motivation is that I'm looking for an efficient algorithm to test if a given finite index subgroup of $SL_2(\mathbb{Z})$ is congruence. -The best algorithm I'm aware of is Hsu's algorithm, but this only tests if a finite index subgroup of $\text{PSL}_2(\mathbb{Z})$ is congruence. -Given a finite index $\Gamma\le\text{SL}_2(\mathbb{Z})$, the image of $\Gamma$ in $\text{PSL}_2(\mathbb{Z})$ is congruence if and only if $\pm\Gamma$ is congruence. Here, $\pm\Gamma$ is the subgroup generated by $\Gamma$ and $-I$. -Let $\ell$ be the "Wolhfart level" of $\Gamma$ - that is, $\ell$ is the least common multiple of the cusp widths of $\mathcal{H}/\Gamma$. Then, a classical theorem of Wolhfart/Klein says that $\pm\Gamma$ is congruence if and only if it contains $\Gamma(\ell)$, equivalently if it contains $\pm\Gamma(\ell)$ (this theorem is actually quoted incorrectly in many sources, where the $\pm$ in $\pm\Gamma$ is ignored!!) -The first question is: Is testing for congruence in $\text{PSL}_2(\mathbb{Z})$ enough? Ie, if $\pm\Gamma$ is congruence, then must $\Gamma$ be congruence? -Of course this is only relevant if $-I\notin\Gamma$, in which case $\pm\Gamma = \Gamma\times\{\pm I\}$, and $\pm\Gamma(\ell) = \Gamma(\ell)\times\{\pm I\}$. Intersecting $\Gamma$ with $\pm\Gamma(\ell)$, there are two cases: -A) If $\ell = 2$, then if $\pm\Gamma$ is congruence, it must contain $\Gamma(2) = \pm\Gamma(2)$, so $\Gamma\cap\Gamma(2)$ has index 2 in $\Gamma(2)$, with quotient an 2-torsion abelian group, but the maximal 2-torsion abelian quotient of $\Gamma(2)$ is $\Gamma(2)/\Gamma(4) = C_2\times C_2\times C_2$, hence $\Gamma$ contains $\Gamma(4)$ and is congruence. -In particular, this shows that the Sanov subgroup - the index 12 subgroup of $\text{SL}_2(\mathbb{Z})$ generated by -$$\begin{bmatrix} 1&2\\0&1 -\end{bmatrix} \quad\text{and}\quad\begin{bmatrix} 1&0\\2&1 -\end{bmatrix}$$ -is congruence, even though the cusp widths are all 2 and it doesn't contain $\Gamma(2)$. -B) If $\ell\ge 3$, then again $\Gamma\cap\pm\Gamma(\ell)$ has index 2 inside $\pm\Gamma(\ell)$, and certainly any index 2 subgroup of $\pm\Gamma(\ell)$ can appear, and so the question reduces to: -Reformulation of question: For any $\ell\ge 3$, is every index 2 subgroup of $\pm\Gamma(\ell)$ a congruence subgroup? -or equivalently: For any $\ell\ge 3$, is the composite $\Gamma(\ell)'\Gamma(\ell)^2$ congruence? (the first is the commutator subgroup, the second is the subgroup of squares) -I don't even know of a reasonable way to computationally test this, since without Wolhfart/Klein's result, one might a priori have to test non-containment of infinitely many $\Gamma(n)$'s to prove that some $\Gamma$ is noncongruence. -Of course, if the answer is negative, then naturally one might ask: -Is there an efficient algorithm to test whether a given finite index subgroup of $\text{SL}_2(\mathbb{Z})$ is congruence? - -REPLY [15 votes]: For the first question: it can happen that $\pm \Gamma$ is congruence but $\Gamma$ is not; there is a beautiful paper on this phenomenon, with lots of examples, by Kiming, Schütt and Verril here. -For the second question, the existence of an efficient algorithm: I set this once as a fourth-year research project to a Warwick undergrad, Tom Hamilton. Tom successfully generalised Hsu's algorithm to subgroups of SL2 instead of PSL2. You can read our paper on this here; the algorithm is implemented in recent versions of Sage. - -In fact, in Sage there is a function which, given a finite-index subgroup of SL2Z containing -1, will enumerate all index 2 subgroups of that group not containing -1 (i.e. all liftings of $\Gamma / \{\pm 1\}$ to SL2Z). Using this, one can show (for instance) that there is an index 2 subgroup of $\pm \Gamma(7)$ which is non-congruence. It's generated by the following matrices: -( -[-1 -7] [-48 7] [209 -56] [113 -35] [ 55 -21] [-120 49] -[ 0 -1], [ -7 1], [ 56 -15], [ 42 -13], [ 21 -8], [ -49 20], - -[-15 7] [ 239 -140] [113 -70] [-232 161] [-181 133] [-8 7] -[-28 13], [ 70 -41], [ 21 -13], [ -49 34], [ -49 36], [-7 6], - -[ 76 -105] [-169 238] [ 43 -63] [ 309 -490] [ 134 -217] -[ 21 -29], [ -49 69], [ 28 -41], [ 70 -111], [ 21 -34], - -[-281 476] [-230 399] [-15 28] [-97 231] [ 218 -525] -[ -49 83], [ -49 85], [ -7 13], [-21 50], [ 49 -118], - -[ 279 -763] [ 22 -63] [-118 399] [-29 112] [ 139 -609] [-36 175] -[ 49 -134], [ 7 -20], [ -21 71], [ -7 27], [ 21 -92], [ -7 34], - -[-43 252] -[ -7 41] -)<|endoftext|> -TITLE: Understanding Vaughan's Identity -QUESTION [19 upvotes]: Vaughan's identity https://proofwiki.org/wiki/Vaughan%27s_Identity is a very useful identity in analytic number theory. The identity expresses the von-Mangoldt function $\Lambda(n)$ as a sum of several sums. -I have seen a couple of applications of Vaughan's identity and I can follow the proof fine but I don't seem to have the intuition on why it works or why the identity is how it is. At the moment the identity seems like expressing $\Lambda(n)$ as a sum of random sums to me...(which I am very sure is not the case!). I would greatly appreciate some explanation on why Vaughan's identity is how it is. Thank you very much! - -REPLY [19 votes]: The point of Vaughan's identity is to express $\Lambda(n)$ (on some range, e.g. $n \in [X,2X]$) into two types of sums that are reasonably tractable: "Type I components" $\sum_{d|n: d \leq D} a_d$ where $D$ is fairly small (in particular, significantly smaller than $X$), or "Type II components" $\sum_{n = d_1 d_2: d_1 \geq D_1, d_2 \geq D_2} a_{d_1} b_{d_2}$ where $D_1, D_2$ are fairly large. This decomposes sums such as $\sum_n \Lambda(n) f(n)$ into "Type I sums" $\sum_{d \leq D} a_d \sum_m f(dm)$ and "Type II sums" $\sum_{d_1 \geq D_1} \sum_{d_2 \geq D_2} a_{d_1} b_{d_2} f(d_1 d_2)$, the first of which can often be dealt with through upper bound estimates on the magnitude of the inner sums $\sum_m f(dm)$, and the latter can be dealt with through bilinear sum methods (often based ultimately on using the Cauchy-Schwarz inequality to eliminate the pesky weights $a_{d_1}, b_{d_2}$). -Because the logarithm function $L(n) = \log n$ is so slowly varying, it behaves like $1$, and so we also consider expressions such as $\sum_{d|n: d \leq D} a_d \log \frac{n}{d}$ to be Type I. -In terms of Dirichlet convolutions, the task is to decompose $\Lambda$ into some combination of Type I components $a_< * 1$ or $a_< * L$, where $a_<$ is supported on small numbers, and Type II components $a_> * b_>$, where $a_>, b_>$ are supported on large numbers. The Vaughan identity is not the only identity that achieves this purpose, but it is amongst the simplest such identity, and is already sufficient for many applications. -Now, we do have the basic identity $\Lambda = \mu * L$, which looks sort of like a Type I component, except that the Mobius function $\mu$ is not restricted to the small numbers. Nevertheless, one can try to truncate this identity by performing the splitting -$$ \Lambda = \mu_< * L + \mu_> * L$$ -where $\mu_<, \mu_>$ are the restrictions of $\mu$ to small and large numbers respectively (let us ignore for now exactly where to make the cut between the two types of numbers). The first component is of Type I; we just need to figure out what to do with the second component. -At this point we introduce another basic identity, $L = \Lambda * 1$, hence $\mu_> * L = \mu_> * \Lambda * 1$. This begins to look a bit like a Type II component, except that only one of the factors is restricted to be large. So we perform another truncation, this time on the $\Lambda$ factor: -$$ \mu_> * L = \mu_> * \Lambda_> * 1 + \mu_> * \Lambda_< * 1.$$ -The first factor is of Type II (after grouping, say, $\mu_> * 1$, into a single factor supported on large numbers). So we're left with understanding $\mu_> * \Lambda_< * 1$. Here we use a final basic identity, $\mu * 1 = \delta$ (where $\delta$ is the Kronecker delta). This implies that $\mu * \Lambda_< * 1 = \Lambda_<$, which will vanish on the desired range $[X,2X]$ if we truncated $\Lambda$ properly. So we can flip the truncation on $\mu$: -$$ \mu_> * \Lambda_< * 1 = - \mu_< * \Lambda_< * 1.$$ -But this is a Type I component if we group $\mu_< * \Lambda_<$ into a single function supported on (reasonably) small numbers. -Incidentally, analytic prime number theory would be a lot easier if we could somehow eliminate the Type II components, and have a Vaughan-type identity that expresses $\Lambda$ solely in terms of Type I components, which are usually a lot easier to deal with. Sadly, this is not the case (except in the presence of a Siegel zero, but that's another long story...). The easiest way to see this heuristically is by assuming the Mobius pseudorandomness conjecture, which among other things will imply that all Type I components will have small correlation with the Mobius function. On the other hand, the von Mangoldt function has a very large correlation with the Mobius function, as the latter is almost always $-1$ on the support of the former. So one cannot efficiently decompose von Mangoldt solely into Type I components, and one needs something like a Type II component somewhere in the decomposition also. (But sometimes one can work with other components that look like truncated versions of the divisor functions $d_2 = 1*1$, $d_3 = 1*1*1$, $d_4 = 1*1*1*1$, etc., which in some applications are preferable to Type II sums. The Heath-Brown identity is particularly well suited for producing components of this form to replace some or all of the Type II sums). - -REPLY [14 votes]: The analytic version of Vaughan's identity is -$$ -\frac{\zeta'}{\zeta} = F+\zeta'G-FG\zeta + \left(\frac{\zeta'}{\zeta}-F\right)(1-\zeta G). -$$ -Here the last factor to the right is the most complicated, so to simplify the right hand side one should take $F$ as an approximation to $\frac{\zeta'}{\zeta}$ and $G$ to be an approximation to $\frac{1}{\zeta}$. The simplest approximation one could think of is a truncation of the Dirichlet series, which then yields Vaughan's identity. -Expressions as above were known before in the theory of density estimates for $\zeta$, going back to the work of Hardy and Littlewood. The simplest version is just one Dirichlet polynomial $M$, which approximates $\zeta^{-1}$. Then $M\zeta$ is on average smaller than $\zeta$, and one can get better bounds for the zeroes of $M\zeta$ then for $\zeta$. Since the roots of $\zeta$ are among the roots of $M\zeta$, one obtains bounds for the roots of $\zeta$. As far as I know, Vaughan had found the analytic identity above and applied it to density estimates before he noticed, that it can be turned into a useful elementary identity for its coefficients.<|endoftext|> -TITLE: On the causal structure of spacetimes: piecewise $C^1$, $C^k$ or $C^\infty$? -QUESTION [8 upvotes]: This is a more technical question but it seems that there is some confusion in the literature on the choice of curves used to define the causal relations in time-oriented Lorentz manifolds: the infinitesimal causal relation is the choice of a forward light cone at every point, i.e. a time orientation. Now the (global) causal relation would take two points $p$ and $q$ and call $q$ in the future of $p$ if there is a causal curve from $p$ to $q$. -Of course you have different futures depending on whether you take timelike or just causal curves, their interplay is of crucial importance in understanding the global geometry of Lorentz manifolds. -Now my question is what types of curves one actually uses: -One option is to stay with $C^\infty$-curves, simple but sometimes difficult in technical aspects as one would like to have arguments with broken geodesics etc. -The other option is to use piecewise $C^\infty$ curves, where the jumps of the velocities at break points stay in the same (forward) light cone. This should make the above disadvantage mostly disappear. -The option I have also seen in the literature (O'Neill...) is to take piecewise $C^1$-curves, again with the same condition at the break points. Now this is probably the most general version where one has still a meaning of what is "future-directed". However, many important(!) arguments in exploring the causal structure use variational formulas for arclength etc which can handle finitely many break points but require higher differentiability, say $C^2$ or even $C^3$. -So does the above choice matter? If so, what is the good convention? -EDIT: For convenience, here the more precise definitions: a tangent vector is called timelike if it is in the open interior of the lightcone and causal if it is inside the closed lightcone. It is called lightlike if it is in the boundary. Correspondingly, one has timelike, lightlike and causal curves, by using this for the tangent vector to the curve. -The set $J^+(p)$ is the causal future (we assume to have a time orientation) of the point $p$ which can be reached by causal curves, while $I^+(p)$ is the timelike future of the point $p$, i.e. those points which can be reached by timelike curves. While the condition "timelike" seems to be rather robust (tangent vectors inside the open cone) for causal this is more touchy. Nevertheless, for normally hyperbolic pdes on the Lorentz manifold, i.e. those with principal symbol given by the metric, it is the causal future $J^+$ which controls the propagation of singularities etc. -To give a flavour of known results (see e.g. O'Neill): for a causal piecewise $C^3$-curve $\gamma$ which can not be reparametrized into a lightlike geodesic one finds a variation $\gamma_s$ such that for all $s > 0$ the curve $\gamma_s$ is timelike and has the same starting and end point. This result is important for the transitivity of the causal relations. But the proof uses very much $C^3$. - -REPLY [4 votes]: Let me expand on Igor Khavkine's answer, especially: - -"The answer is that it doesn't matter, as long as the metric itself is sufficiently regular." - -In our recent paper: The future is not always open we clarified all these issues with the regularity of the curves vs. the regularity of the metric (and showed that there are some pathologies in low regularity). Maybe this is also of interest to you.<|endoftext|> -TITLE: Going beyond the surreal numbers -QUESTION [7 upvotes]: Denote the class of surreal numbers No. We can create new "number", like the gap $\infty=\{\infty^L|\infty^R\}$, defined by $\infty^L=\{x:\exists n\in\mathbb N,xn\}$. $\infty\notin$No because $\infty^L$ and $\infty^R$ are not sets, but classes. Apparently we can construct new numbers in the form $\{L|R\}$, where $L$ and $R$ are subclasses of No such that no $x\in L$ is $\geq$ any $y\in R$. Then we can use these new numbers to construct more numbers, and continue to expand No (possible indefinitely?). Is there any reason we do not do so? - -REPLY [17 votes]: Relentlessly filling cuts is of course the main construction idea of the surreal numbers---at every ordinal birthday, one fills all the cuts that exist in the previously-born surreals. Your proposal is to continue filling cuts after all ordinal birthdays are completed. -All such cuts will have cofinality Ord on one side or the other, and so each such cut will take a proper class to represent it. So the first thing to say is that there will be certain set-theoretic foundational difficulties with undertaking the construction. For example, this is not straight-forwardly a ZFC construction, but you could proceed in GBC for a step or so. To proceed much further, you will need stronger second-order set-theoretic axioms, such as the axiom ETR of elementary transfinite recursion, which allows one to undertake recursions on proper class well-founded relations whose rank exceeds Ord. -But another perspective is that what you are proposing is just the situation that arises when one has a smaller universe $V_\kappa$ extended to the full universe $V$. The surreal numbers $\text{No}^{V_\kappa}$ as constructed up to $\kappa$ are the surreal numbers of the universe $V_\kappa$, which proceeded in $\kappa$ many birthdays, but the surreal numbers of the full universe $V$ simply carried on with birthday $\kappa$, filling the cuts that you describe, and then $\kappa+1$ and so on through the ordinals of $V$. -One can imagine a hierarchy of universes -$$V_\kappa\prec V_\lambda\prec V_\theta\prec\cdots$$ -and the surreal numbers of each of them are related to the next in just this way. -So your construction is realized by considering the surreal numbers as constructed in various universes and realizing how each next universe continues the cut-filling construction of the surreals of the smaller universes.<|endoftext|> -TITLE: On the equivalence of a pair of binary quadratic forms -QUESTION [8 upvotes]: Let $f,g, u,v \in \mathbb{Z}[x,y]$ be binary quadratic forms with co-prime coefficients. We say that the pair $(f,g)$ and $(u,v)$ are $\operatorname{GL}_2(\mathbb{Z})$-equivalent if there exists $T = \left(\begin{smallmatrix} t_1 & t_2 \\ t_3 & t_4 \end{smallmatrix} \right) \in \operatorname{GL}_2(\mathbb{Z})$ such that -$$\displaystyle f(t_1 x + t_2 y, t_3 x + t_4 y) = u(x,y).$$ -and -$$\displaystyle g(t_1 x + t_2 y, t_3 x + t_4 y) = v(x,y).$$ -Note that if the pair $(f,g)$ is equivalent to $(u,v)$, then $f$ is equivalent to $u$ and $g$ equivalent to $v$ as individual binary quadratic forms, but the converse need not be true, since it is not clear that there exists a common matrix $T$ that sends $f$ to $u$ and $g$ to $v$. -Aside from the discriminants $\Delta(f), \Delta(g)$ of the individual forms $f$ and $g$, the pair $(f,g)$ has one more invariant which can be taken to be -$$\displaystyle \Delta(f,g) = 2 f_2 g_0 - f_1 g_1 + 2 f_0 g_2,$$ -where $f(x,y) = f_2 x^2 + f_1 xy + f_0 y^2$ and $g(x,y) = g_2 x^2 + g_1 xy + g_0 y^2$. -Denote by $h(d_1, d_2, d_3)$ to be the class number of $\operatorname{GL}_2(\mathbb{Z})$-equivalence classes of pairs of binary quadratic forms $(f,g)$ with $\Delta(f) = d_1, \Delta(g) = d_2$, and $\Delta(f,g) = d_3$. Are there any estimates for $h(d_1, d_2, d_3)$ in terms of $d_1$ and $d_2$, or average estimates with $0 < \max\{|d_1|, |d_2|,|d_3|\} \leq X$? -I have consulted an old paper by Dickson (https://www.jstor.org/stable/2370100?seq=1#page_scan_tab_contents) but he only treated the easier case of $\operatorname{GL}_2(\mathbb{Q})$-equivalence. -Any help or reference would be appreciated! - -REPLY [4 votes]: The exact class number formula was given by Jorge Morales in the following paper: -Jorge Morales. The classification of pairs of binary quadratic forms. Acta Arith., 59(2):105–121, 1991. -In particular, he showed that -$$\displaystyle h(d_1, d_2, d_3) = m \sum_{n | 4(d_3^2 - d_1 d_2)} \left(\frac{d_1}{n}\right),$$ -where $m = 1$ if $d_3^2 - d_1 d_1 \geq 1$ and $m = 2$ if $d_3^2 - d_1 d_1 < 0$, and $\left(\frac{\cdot}{n}\right)$ is the Jacobi symbol. -The case when computing the class number $h(d_1, d_2, d_3)$ of pairs $(f,g)$ with $\Delta(f) = d_1, \Delta(g) = d_2, \Delta(f,g) = d_3$ and $f,g$ are both positive definite was resolved by an earlier paper by K. Hardy and K. Williams.<|endoftext|> -TITLE: Direct combinatorial proof that $2^{2k} = \sum \binom{2i}{i}\binom{2j}{j}$? -QUESTION [21 upvotes]: In a purely algebraic way, I've just stumbled onto the fact that -$$2^{2k} = \sum_{i+j=k} \binom{2i}{i}\binom{2j}{j},$$ -i.e. that the self-convolution of the sequence $\binom{2k}{k}$ is the sequence $2^{2k}$. This is the type of identity for which I would expect there to be a beautiful, simple, direct, and probably well-known combinatorial argument. I just spent some time looking for it but couldn't find it. - -Is there a direct combinatorial proof of the above? - -Geometric remark: such an argument would partition the vertices of an even-dimensional hypercube into some sort of combinatorially meaningful classes of size $\binom{2i}{i}\binom{2j}{j}$. For example, for $k=2$, the vertices of the $4$-cube would be partitioned into classes of size $6=1\cdot 6$, $4=2\cdot 2$, and $6=6\cdot 1$. - -REPLY [3 votes]: An elementary proof. -From -\begin{equation} -\arcsin x=\sum_{\ell=0}^{\infty}\frac{1}{2^{2\ell}}\binom{2\ell}{\ell}\frac{x^{2\ell+1}}{2\ell+1}, \quad |x|<1, -\end{equation} -it follows that -\begin{equation*} -\frac{1}{2}\arcsin(2x)=\sum_{k=0}^{\infty}\frac{1}{2k+1}\binom{2k}{k}x^{2k+1},\quad |x|<\frac{1}{2} -\end{equation*} -and, by differentiation, -\begin{equation*} -\frac{1}{\sqrt{1-4x^2}\,}=\sum_{k=0}^{\infty}\binom{2k}{k}x^{2k},\quad |x|<\frac{1}{2}. -\end{equation*} -Squaring on both sides, using the geometric series expansion, and utilzing the Cauchy product of the multiplication of two power series yield -\begin{equation*} -\sum_{k=0}^\infty2^{2k}x^{2k}=\frac{1}{1-4x^2}=\Biggl[\sum_{k=0}^{\infty}\binom{2k}{k}x^{2k}\Biggr]^2 -=\sum_{k=0}^\infty\Biggl[\sum_{i=0}^k\binom{2i}{i}\binom{2(k-i)}{k-i}\Biggr]x^{2k}, \quad |x|<\frac{1}{2}. -\end{equation*} -Equating the coefficients of $x^{2k}$ results in -$$ -\sum_{i=0}^k\binom{2i}{i}\binom{2(k-i)}{k-i} -=\sum_{i+j=k}\binom{2i}{i}\binom{2j}{j} -=2^{2k}, \quad k\ge0. -$$ -The idea of this elementary proof comes from the following paper -Feng Qi, Chao-Ping Chen, and Dongkyu Lim, Several identities containing central binomial coefficients and derived from series expansions of powers of the arcsine function, Results in Nonlinear Analysis 4 (2021), no. 1, 57--64; available online at https://doi.org/10.53006/rna.867047. -This elementary proof is simpler than the proof of Theorem 3.4 on page 232 in the following paper -Necdet Batır, Hakan Küçük, and Sezer Sorgun, Convolution identities involving the central binomial coefficients and Catalan numbers, Transactions on Combinatorics 10 (2021), no. 4, 225--238; available online at https://dx.doi.org/10.22108/toc.2021.127505.1821.<|endoftext|> -TITLE: Arithmetic progressions in finitely generated groups -QUESTION [7 upvotes]: Suppose $\Gamma$ is a finitely generated countable discrete torsion free group with a generating set $S$. Let $l$ be the word length function given by $S$. -Let $F_n=\{s\in\Gamma| l(s)\leq n\}$. -Assume that $\Lambda$ is a subset of $\Gamma$ such that $$\limsup_{n\to\infty} \frac{|\Lambda\cap F_n|}{|F_n|}>0.$$ -Question: For every positive integer $k$, do there exist $b$ and $a$ in $\Gamma$ such that $\{b^{j}a\}_{j=0}^{k-1}\subseteq\Lambda$? -When $\{b^{j}a\}_{j=0}^{k-1}$has $k$ distinct elements, we call it a left arithmetic progression of length $k$ in $\Gamma$. -Remark: One would avoid the trivial case that $b=e_\Gamma$. In fact, let $\Gamma=\mathbb{Z}$ with $S=\{1\}$, the answer to the above question is affirmative by Szemeredi's theorem, which says that a subset of $\mathbb{Z}$ with positive upper density contains arbitrarily long arithmetic progressions. - -REPLY [2 votes]: Your definition of "arithmetic progression" in a group is wrong. It should be as follows. Let $w(x,a_1,...,a_n)$ be a word. Then an arithmetic progression is the sequence $w(a_1,a_1,...,a_n), ..., w(a_n,a_1,...,a_n)$ for some a_1,...,a_n in the group (see https://en.wikipedia.org/wiki/Hales%E2%80%93Jewett_theorem). In this formulation the result is true as proved by Furstenberg and Katznelson (it was a \$100 problem from a list by Ron Graham, as far as I remember, http://www.math.ucsd.edu/~ronspubs/08_06_old_and_new.pdf).<|endoftext|> -TITLE: Current status of Grothendieck's homotopy hypothesis and Whitehead's algebraic homotopy programme -QUESTION [30 upvotes]: (Disclaimer: I'm no expert in homotopy theory nor in higher categories!) If I understand it correctly, Grothendieck's homotopy hypothesis states that there should be an equivalence (of $(n+1)$-categories) between "homotopy $n$-types" and $n$-groupoids. Where, by "homotopy $n$-types" is probably meant the $(\infty,n+1)$-category that has (nice) topological spaces with vanishing homotopy groups above the $n$-th as objects, and higher morphisms given by homotopies and homotopies-between-homotopies etc. And by "$n$-groupoid" is probably understood $(\infty,n)$-groupoid. -Edit: the homotopy types probably are defined to be some localization of the thing I stated above? - -To which extent has the homotopy hypothesis been proved? By "proved" I mean precise statements and rigorous proofs, not just "philosophical" evidence; and not "tautological" solutions in which homotopy types are defined to be $\infty$-groupoids in the first place. - -All this fits in the context of Whitehead's algebraic homotopy programme - -Which is the present status of that programme, both in the sense of formalization and of proof? -How can any advance in the programme be made at all if Grothendieck's conjecture is not fully proven first? - -REPLY [10 votes]: I think this question is an interesting one, and the two approaches should be compared. I have more knowledge of Whitehead's programme, and have been involved with others in the development of aspects of that programme. A report on the background of part that work is in the paper Modelling and Computing Homotopy Types: I, to appear in 2017 in a special issue of Indagationes Mathematicae in honor of L.E.J. Brouwer. A notable feature of this work is that it deals not with "bare" spaces but with "Topological Data", and this somewhat reflects Grothendieck's view expressed in "Esquisse d'un Programme" Section 5. Also among the aims of Whitehead's work was to introduce invariants which allowed specific calculation and even if possible enumeration. -I should explain that having from 1965 to 1974 tried to define for a space $X$ a strict homotopy double groupoid which could satisfy a van Kampen type Theorem, and so enable specific calculations, it was a considerable relief to find with Philip Higgins that this could be done in a natural and intuitive way for a pair $(X,A,c)$ of pointed spaces by mapping the square $I^2$ into $X$ so that the edges of $I^2$ mapped to $A$ and the vertices mapped to $c$. This gave a vast generalisation of a tricky theorem on free crossed modules, proved in Section 16 of the 1949 paper "Combinatorial Homotopy II" (CHII). -Further this definition of homotopy double groupoid generalised, with considerable more work, to all dimensions using filtered spaces, so continuing Whitehead's programme in CHII. -The algebraic data used here of strict cubical $\omega$-groupoids, and the equivalent crossed complexes, do not model all homotopy types, and indeed contain in essence only "linear" information, i.e. no Whitehead products, for example. However these models do contain more information than chain complexes with a group of operators, conforming with an observation of Whitehead in CHII. -However a meeting with J.-L. Loday in 1981 in Strasbourg started our link with his work on what he called $n$-cat-groups, and we later agreed to call cat$^n$-groups, and which are $n$-fold groupoids in which one direction is a group. Loday had proved in 1982 modelled pointed homotopy $(n+1)$-types. We conjectured then and eventually proved a van Kampen type theorem in the context of $n$-cubes of spaces; this work was eventually accepted for the journal Topology, and a companion paper was accepted for Proc. London Math Soc, both appearing in 1987. The latter paper proved an $n$-adic Hurewicz Theorem, the triadic version of which was a conjecture of Loday in 1981. -Actually Grothendieck objected to the pointed condition, and did not recognise what had been achieved - he always wanted the most general conditions! However one aspect of the work with Loday, a nonabelian tensor product of groups which act on each other, has been well taken up by group theorists, and a current bibliography has 158 items dating from 1952. -H.-J. Baues in several books has continued aspects of Whitehead's programme, but he does not apply the Brown-Loday work, and his models do not satisfy the Criteria set out in Section 1 of the cited paper "Modelling and Computing Homotopy Types:I". However paper II of that has been delayed for various health reasons. -This paper, in a volume in memory of J.F.Adams, refers to work of G.J.Ellis and R. Steiner which applies the Brown-Loday work to solve an old problem in homotopy theory, on which Whitehead had written, to determine the value of the critical (i.e. first non vanishing) group of an $(n+1)$-ad. -Looking again at Esquisses d'un Progamme, it seems that programme has currently little relation to Whitehead's; but a 1983 letter from Grothendieck to the writer, reprinted as Problem 16.1.29 of Nonabelian Algebraic Topology, may suggest that despite his interest in the 2-d van Kampen theorem, the lack of progress with the issues he there raises indicates a limitation of this writer. -Grothendieck was very interested at one point in the idea I once conveyed to him that $n$-fold groupoids model homotopy $n$-types. But this is not true as stated even for $n=2$, since $2$-fold groupoids can be much more complicated than crossed modules over groupoids, which are a good model of homotopy $2$-types, and are equivalent to a special kind of double groupoid. -Added 18/04/2017: A final remark is that in my work with Higgins the cubical aspect is essential, and analogous results have not been obtained by simplicial methods. However the work with Loday does use also advanced simplicial methods for the proof of the main result. In both cases, the cubical methods are used to express multiple compositions, which are tricky simplicially.<|endoftext|> -TITLE: Are there indecomposable unsolvable four and five dimensional Lie algebras? -QUESTION [6 upvotes]: In many places one can find an information that real Lie algebras are classified up to dimension 5. The only reference containing a classification I was able to find is Invariants of real low dimensional Lie algebras. The list presented there contains indecomposable Lie algebras of dimension up to five, yet every for and five dimensional algebra presented there is solvable. Is it true that there are no indecomposable, unsolvable algebras in that dimensions or the article very implicitly deals in fact only with solvable algebras. -An answer or a solid reference for a classification will be very appreciated. - -REPLY [2 votes]: The classification of indecomposable non-solvable Lie algebras in dimension $\le 7$ over a field of characteristic zero is a simple exercise. -First, the semisimple ones have dimension 3 or 6: in dimension 3, $\mathfrak{so}(q)$ for quadratic forms in 3 variables, in dimension 6, the same over a quadratic extension, or products of two simple ones of dimension 3. -Next we use a Levi decomposition $\mathfrak{s}\ltimes\mathfrak{r}$. If $\mathfrak{s}$ has dimension 6, $\mathfrak{r}$ has dimension $\le 1$ and is a direct factor. -If $\mathfrak{s}$ has dimension 3, one has to discuss on $\mathfrak{r}$. If $\mathfrak{r}$ is abelian, it comes with a representation of $\mathfrak{r}$, which splits into irreducibles. If a 1-dimensional rep occurs, then it is a direct factor, so in the indecomposable cases the only possibilities are 2,3,4 and 2+2. (Note that if $\mathfrak{s}$ has a nontrivial 2-dimensional rep, then it is isomorphic to $\mathfrak{sl}_2$; the only 3-dimensional irreducible is the adjoint rep. In dim 4, I'm not sure in general fields, but in the real case both the split form and the non-split form have a unique 4-dimensional irreducible rep, which in the case of $\mathfrak{su}_2$ is not absolutely irreducible). -If $\mathfrak{r}$ is nilpotent and not abelian then it is (by the classification of nilpotent Lie algebras of dimension $\le 4$, either $\mathfrak{h}_3$ (Heisenberg), or $\mathfrak{h}_3\times\mathfrak{a}_1$ (its product with a 1-dimensional abelian) or $\mathfrak{f}_4$ (filiform of dimension 4). In the last case, the derivation algebra is solvable and hence $\mathfrak{r}$ is a direct factor. In the first two cases, either $\mathfrak{r}$ is a direct factor, or we have the following: modulo the center, we have an irreducible 2-dimensional $\mathfrak{s}$-module (hence $\mathfrak{s}$ is $\mathfrak{sl}_2$); then this lifts to an irreducible 2-dimensional submodule in $\mathfrak{r}$, and we see we have the (unique) nontrivial semidirect product $\mathfrak{sl}_2\ltimes\mathfrak{h}_3$, or its direct product with $\mathfrak{a}_1$. -The last case is when $\mathfrak{r}$ is not nilpotent. In a solvable Lie algebra $\mathfrak{r}$ with nilpotent radical, it is easy to check that $\dim(\mathfrak{n})\ge\dim(\mathfrak{r})/2$, with equality only for powers of the 2-dimensional nonabelian Lie algebra. The latter has a solvable derivation algebra and hence only occurs as direct factor. Otherwise in our case the pair $(\dim(\mathfrak{r}),\dim(\mathfrak{n}))$ is $(4,3)$ or $(3,2)$. Since in these cases $\mathfrak{n}$ has codimension 1, it has a $\mathfrak{s}$-invariant complement. So the Lie algebra has the form $(\mathfrak{s}\times\mathfrak{a}_1)\ltimes\mathfrak{n}$, where $\mathfrak{a}_1$ does not act on $\mathfrak{n}$ in a nilpotent way, and acts in the centralizer of the $\mathfrak{s}$-action. Excluding the case with abelian direct factors, we find $\mathfrak{gl}_2\ltimes\mathfrak{h}_3$, - $(\mathfrak{s}\times\mathfrak{a}_1)\ltimes\mathfrak{v}$ where - $\mathfrak{v}$ is an irreducible of dimension 2 or 3 and - $\mathfrak{a}_1$ acts by scalar multiplication. -So the list (dimension in parentheses): - -(3) $\mathfrak{s}$ simple of dimension 3 ($\mathfrak{so}$ of some 3-dimensional quadratic form, unique up to scalar multiplication and equivalence: in the real case, $\mathfrak{sl}_2=\mathfrak{so}(2,1)$ or $\mathfrak{su}_2=\mathfrak{so}(3)$) -(6) $\mathfrak{s}$ simple of dimension 6 (idem over a quadratic extension of the ground field; in the real case: $\mathfrak{sl}_2(\mathbf{C})=\mathfrak{so}_3(\mathbf{C})$) -(5) $\mathfrak{sl}_2(K)\ltimes K^2$ -(7) $\mathfrak{sl}_2(K)\ltimes (K^2\oplus K^2)$ -(6) $\mathfrak{sl}_2\ltimes\mathfrak{h}_3$ -(6) $\mathfrak{gl}_2(K)\ltimes K^2$ -(7) $\mathfrak{gl}_2\ltimes\mathfrak{h}_3$ -(6) $\mathfrak{s}\ltimes V_\mathfrak{s}$ where $\mathfrak{s}$ is simple of dimension 3 and $V_\mathfrak{s}$ is the adjoint representation of $\mathfrak{s}$ -(7) $(\mathfrak{s}\times\mathfrak{a}_1)\ltimes V_\mathfrak{s}$ where $\mathfrak{s}$ is 3-dimensional simple, $\mathfrak{a}_1$ acts by scalar multiplication -(7) $\mathfrak{s}\ltimes V$, $V$ irreducible of dimension 4. To be more precise, in the real case, the only case beyond the split case $\mathfrak{sl}_2(K)\ltimes\mathrm{Sym}^3(K^2)$ is $\mathfrak{su}_2\ltimes\mathbf{C}^2$.<|endoftext|> -TITLE: Arithmetic problem for bicolored graphs -QUESTION [13 upvotes]: A bicolored graph is a graph each vertex of which has been assigned one of two colors such that each edge connects vertices of different colors. A bipartite graph is a graph $G$ which admits such a coloring. Given $j$ white and $i$ black vertices, there are $2^{ji}$ ways to join vertices of different colors. Thus the number of (labeled) bicolored graphs on n vertices is -$$b_n=\sum_{i+j=n}\binom{n}i2^{ij}.$$ - -Question. Is it true that $b_n$ is never divisible by $5$? If so, any proof? - -Update. One combinatorial proof and two number-theoretic/algebraic proofs are supplied. - -REPLY [12 votes]: Here is an illustration of Ira Gessel's comment about a number-theoretic proof. -Lemma. For $n\geq1$ and an odd prime $p$, the sequence $b_n$ mod $p$ is periodic of period $p-1$. -Proof. Let $n=m(p-1)+r$, where $1\leq r\leq p-1$ and $m\geq0$. Proceed with a successive application of Fermat's, Lucas' and Chu-Vandermonde's theorems so that -\begin{align*} -b_n&=\sum_{j=1}^{p-1}\sum_{k=0}^m\binom{m(p-1)+r}{k(p-1)+j}2^{[k(p-1)+j][m(p-1)+r-k(p-1)-j]} \\ -&\equiv_p\sum_{j=1}^{p-1}\sum_{k=0}^m\binom{m(p-1)+r}{k(p-1)+j}2^{j(r-j)} \\ -&\equiv_p\sum_{j=1}^{p-1}2^{j(r-j)}\sum_{k\geq0}\binom{m}k\binom{r-m}{j-k}\\ -&\equiv_p\sum_{j=1}^{p-1}2^{j(r-j)}\binom{r}j\\ -&=\sum_{j=1}^r\binom{r}j2^{j(r-j)}=b_r. -\end{align*} -The proof is complete. $\square$ -Example. The first 4 values of $b_n$ mod 5 are $[2,2,1,1]$, as Wojowu mentioned. By the Lemma, we conclude that $\nu_5(b_n)=0$, that is, $b_n$ is never divisible by $5$.<|endoftext|> -TITLE: Does projection of 3D points reduce distances by exactly 1/3? -QUESTION [31 upvotes]: Let $P$ be a set of $n$ random points uniformly distributed inside -a unit-radius sphere centered on the origin. -Orthogonally project $P$ to a random plane through the origin; -call the projected points $P_{\bot}$. -Let $A$ be the distance matrix for $P$, and $B$ the distance matrix for $P_{\bot}$, -where an $n \times n$ distance matrix records the Euclidean distance -between pairs of points. So these matrices are symmetric across a diagonal of zeros. -Finally, define the distance between matrices $A$ and $B$ as -$$ -d(A, B) = \frac{\sqrt{\sum_{i=1}^n \sum_{j=1}^n (a_{ij} - b_{ij})^2}}{n} \;. -$$ -(This matrix distance likely has a name, but I don't know it.) -Simulations suggest that $d(A,B)$ is close to $\frac{1}{3}$, independent of $n$. -For example, for $n{=}200$, $50$ random trials led to $0.332$. - -Q1. Is it true $d(A,B) = \frac{1}{3}$ exactly? -Q2. If so, is there some intuitive way to see this without calculation? - - -          - - -          - -$n=200$ points $P$ (blue) in $\mathbb{R}^3$ projected to $P_{\bot}$ (red). - - -Update 1. -Following @Henry.L's suggestion in the comments, for a sphere of radius -$r{=}2$, I find $d(A,B) \approx \frac{2}{3}$. -This suggests the matrix distance might be $\frac{r}{3}$. -Update 2. A histogram of $d(A,B)$ for $n{=}100$, $5000$ random trials: - -REPLY [23 votes]: No. In the large $n$ limit, this is equivalent to asking whether the expected value of $(a_{ij}-b_{ij})^2$ is $(1/3)^2$, but in fact the expected value is $2-\frac{3\pi}{5} \approx .115$. -We can compute the expectation as follows: $a_{ij}^2 = r_i^2 + r_j^2 + 2 r_i r_j \cos \theta_{ij}$, where $\theta_{ij}$ is the angle between the vectors from the origin to points $i$ and $j$. The variable $\theta_{ij}$ is independent of $r_i,r_j$ and $\cos \theta_{ij}$ has expected value zero. So $$E[a_{ij}^2]=2E[r_i^2]=2\frac{\int_0^1 r^2 \cdot r^2\,dr}{\int_0^1 r^2 \,dr} = \frac{6}{5}$$ -We have $b_{ij}=a_{ij} \sin \phi_{ij}$, where $\phi_{ij}$ is the angle between the vector from $i$ to $j$ and the vertical. Furthermore, $\phi_{ij}$ is independent of $a_{ij}$. So $E[(a_{ij}-b_{ij})^2]=E[a_{ij}^2] E[(1-\sin \phi_{ij})^2]$. We can compute -$$ E[(1-\sin \phi_{ij})^2]=\frac{\int_0^{\pi} (1-\sin \phi)^2 \sin \phi\, d\phi}{\int_0^{\pi} \sin \phi\, d\phi}=\frac{5}{3}-\frac{\pi}{2}$$ -Putting these together gives $E[(a_{ij}-b_{ij})^2]=2-\frac{3\pi}{5}$.<|endoftext|> -TITLE: Computation of a homotopy colimit of pro-spectra -QUESTION [7 upvotes]: Suppose $E\simeq\text{hocolim}_iE_i$ is a filtered homotopy colimit. Suppose $X=\{X_j\}_j$ is a pro-space (assume some finiteness conditions on the spaces or spectra if you have to...for example, assume $X_i$ are simplicial sets with finitely many simplicies in each degree). Is it true that in the $\infty$-category of pro-spectra, we have $\text{hocolim}^{pro}_i\{E_i\wedge X_j\}_j\simeq \{\text{hocolim}_i(E_i\wedge X_j)\}_j$? More generally, is it true that if $E_i$ are pro-spectra with some mild finiteness condition and all indexed by the same cofiltered diagram, then I can compute their homotopy colimit degreewise? - -REPLY [7 votes]: This is not true, unfortunately. Here's an example. -Let $E_i$ be the directed system -$$ -\DeclareMathOperator{\hocolim}{hocolim} -\DeclareMathOperator{\holim}{holim} -S^0 \to S^0 \vee S^1 \to S^0 \vee S^1 \vee S^2 \to \dots -$$ -of wedge inclusions, and let $X_j$ be the inverse system -$$ -S^0 \leftarrow S^0 \vee S^1 \leftarrow S^0 \vee S^1 \vee S^2 \leftarrow \dots -$$ -of projections. Both are systems of objects of finite type, and both the homotopy limit and homotopy colimit are also of finite type. -We also have a natural isomorphism -$$ -Map(E_i \wedge X_j, F) \cong \bigoplus_{k=0}^i \bigoplus_{l = 0}^j Map(S^{i+j}, F). -$$ -Let $F$ be any spectrum, viewed as a constant pro-object. Then we can show that the two spectra you describe are different by calculating homotopy classes of maps out to $F$: -$$ -\begin{align*} -Map_{pro}(\{\hocolim_i(E_i\wedge X_j)\}_j, F) &= \hocolim_j \holim_i Map(E_i \wedge X_j, F)\\ -Map_{pro}(\hocolim^{pro}_i\{E_i\wedge X_j\}_j, F) &= \holim_i \hocolim_j Map(E_i \wedge X_j, F) -\end{align*} -$$ -Thus, we have isomorphisms -$$ -\begin{align*} -[\{\hocolim_i(E_i\wedge X_j)\}_j, F]_{pro} &= \bigoplus_l \prod_k \pi_{k+l}(F)\\ -[\hocolim^{pro}_i\{E_i\wedge X_j\}_j, F]_{pro} &= \prod_k \bigoplus_l \pi_{k+l}(F) -\end{align*} -$$ -In particular, the natural map from the top to the bottom is not an isomorphism unless the homotopy groups of $F$ are zero above a certain degree. -As you can see, finiteness properties on the $E_i$ or $X_j$ don't buy you what you need. If you are taking a hocolim over a finite diagram instead (pushouts, finite wedges, things built from those by finitely many steps), then the fact that the category of pro-spectra is stable allows you to move the homotopy colimit inside the inverse system.<|endoftext|> -TITLE: Is there an asymptotic formula that describes the correlation of multiplicative inverses in Farey sequences? -QUESTION [11 upvotes]: Added 14/04/17: So I am placing a bounty on this question because I am very interested in knowing about strategies for calculating the asymptotic behaviour of this sum. I have calculated $S(X)$ for $X<10^8$ and I am fairly confident that we have $S(X)\sim X/2$. Alas this doesn't prove anything yet though. Can this be proved? Due to the lack of answers, I suppose this is a hard problem, so I would also accept answers that can make this clear in some way. For instance, if solutions to this sort of problem depend on the solution of some open problem or the resolution of a conjecture which lies deep, then an explanation would be welcome. - -It is not clear to me what the level of difficulty of this problem is but it appears to involve questions about notions of randomness in number theory, and so it seems like a reasonable MO question. -The motivation here comes from considering the Dirichlet series $$\sum_1^{\infty}\frac{d(n)}{n^s}\sum_{q|n-1:q\leq X}1,$$ and looking at the value of (the analytic continuation of-) this function at $s=0$. Such Dirichlet series are related to questions about the autocorrelation of the divisor function, for which the asymptotics were worked out by Ingham and Estermann in the 1920's and 1930's. -The objective is to obtain an asymptotic formula for the sum -$$S(X)=\sum_{q\leq X}\frac{1}{\phi(q)}\sum_{\chi}L^2(0,\chi),$$ -and it can easily be shown that determining this is equivalent to knowing something interesting about Farey fractions. -It is fairly elementary that -\begin{eqnarray}\label{} -S(X)=\sum_{q\leq X}\frac{1}{\phi(q)}\sum_{\chi} \left( \sum_{a\leq q} \chi(a)\zeta\left(0 ,\frac{a}{q}\right) \right)^2 -\end{eqnarray} -where $\zeta(s,a)$ is the Hurwitz zeta function so, denoting by $a^*$ the multiplicative inverse of $a$ modulo $q$, expanding out the square and using the orthogonality relations for the Dirichlet characters one obtains -$$S(X)=\sum_{q\leq X}\sum_{(a,q)=1}\left(\frac{a}{q}-\frac{1}{2}\right)\left(\frac{a^*}{q}-\frac{1}{2}\right).$$ -Noting that the mean value of the non-trivial Farey fractions ($0$ and $1$ being trivial) is $1/2$ for all $X$, at this point one probably realises that $S(X)/X^2$ is proportional (by a factor of $3/\pi^2$) to the correlation of the non-trivial Farey fractions of order $X$ with their multiplicative inverses modulo their denominators. -Since one would expect the values of $a/q$ and $a^*/q$ to be independent as $q\rightarrow\infty$, one would expect that $$S(X)=O(X^{1+\epsilon})$$ and so I would like to pose the following question: - -Can an asymptotic formula for $S(X)$ be determined, or just a non-trivial upper or lower bound, via the Farey fractions or the Dirichlet L-functions? - -Potentially, one way to proceed is to multiply out the product and use the symmetry about $1/2$ to obtain -$$S(X)=-\frac{1}{2}+\sum_{q\leq X}\sum_{(a,q)=1}\left(\frac{aa^{*}}{q^2}-\frac{1}{4}\right).$$ -Using the fact that $aa^{*}-k(a)q=1$, where $k(a)$ is the greatest number of integer multiples of $q$ less than $aa^{*}$, and $k(a)/q$ is another non-trivial Farey fraction, one obtains -$$S(X)=-\frac{1}{2}+\sum_{q\leq X}\sum_{(a,q)=1}\left(\frac{k(a)}{q}-\frac{1}{4}\right) +O(\log X).$$ -However, the mapping $a\rightarrow k(a)$ is not an automorphism of the multiplicative group of integers modulo $q$, and the behaviour of $k$ seems to be quite complex. - -Added 14/04/17: I should also mention that it appears that even the strongest conjectures on bounds on sums of Kloosterman sums (e.g. of Selberg type) seem to be of little utility here. - -REPLY [4 votes]: Finding asymptotic for $S(X)$ amounts to estimation of $\sum_{(a,q)=1} aa^*$ for any given $q$. -From the rearrangement inequality, it follows that -$$\frac{1}{2}q^2\varphi(q) - T = \sum_{(a,q)=1} a(q-a) \le \sum_{(a,q)=1} aa^* \le \sum_{(a,q)=1} a^2 = T,$$ -where -$$T = \sum_{(a,q)=1} a^2 = \sum_{d\mid q} \mu(d) d^2 \frac{q/d(q/d+1)(2q/d+1)}{6} = \frac{1}3 q^2\varphi(q) + O(q^2).$$ -Hence, -$$\frac{1}6 q^2\varphi(q) \lesssim \sum_{(a,q)=1} aa^* \lesssim\frac{1}3 q^2\varphi(q).$$ -UPDATE. Notice that the assumption of independent $a$ and $a^*$ essentially suggests that $\sum_{(a,q)=1} aa^* \sim\frac{1}4 q^2\varphi(q)$, which deviates from the asymptotic bounds above by constant factors. However, these asymptotic bounds are not tight enough to obtain the anticipated asymptotic formula for $S(X)$.<|endoftext|> -TITLE: Avoiding equal distances -QUESTION [7 upvotes]: Is the following consistent? -There exists $X \subseteq [0, 1]$, such that $X$ does not have measure zero and for every $Y \subseteq X$, if $Y$ does not have measure zero, then there are $y_1 < y_2 < y_3 < y_4$ in $Y$ such that $y_2 - y_1 = y_4 - y_3$. -Edit: Under CH, there is no such $X$. This follows from a result of Erdos and Kakutani that says that CH is equivalent to the following statement: -The real line $\mathbb{R}$ can be partitioned into countably many rationally independent sets. -The category analogue has a positive answer. -Maybe I should add a comment to Pietro Majer's response: -It is a simple consequence of Lebesgue density theorem that every non null measurable $Y$ must contain equal distances. There are non trivial problems about avoiding patterns in positive measure sets (Erdos similarity problem) but this one is quite different. - -REPLY [2 votes]: Any measurable $X$ with positive measure does it, e.g. $X:=[0,1]$. Indeed, given any measurable $Y\subset\mathbb{R}$ with positive measure, let $a\in\mathbb{R}$ be such that both $Y_-:=Y\cap (-\infty,a)$ and $Y_+:=Y\cap (a,+\infty)$ have positive measure. By Steinhaus theorem, $Y_--Y_-$ and $Y_+-Y_+$ are nbds of $0$, so in particular there are $y_1 -TITLE: Theta functions, re-expressed -QUESTION [17 upvotes]: Recall the classical $\theta(q):=\prod_{k=1}^{\infty}(1-q^k)$ and -define the sequences $a_n$ and $b_n$ by -$$\frac{\theta^3(q)}{\theta(q^3)}=\sum_{n=0}^{\infty}a_nq^n \qquad \text{and} \qquad -F(q):=\sum_{i,j\in\Bbb{Z}}q^{i^2+ij+j^2}=\sum_{n=0}^{\infty}b_nq^n.$$ -Edit. In accord with Noam's commentary, we may replace $\theta$ by $\eta$. - -Question. Is the following true? If so, any proof? - -$$b_n=\begin{cases} -\,\,\,\,\,\,\, a_n\qquad \text{if $a_n\geq0$} \\ -2a_n \qquad \text{if $a_n<0$}. \end{cases}$$ - -REPLY [22 votes]: Yes, it is true. -This generating function $\sum_n a_n q^n$ -turns out to be the same as $(3F(q^3)-F(q))/2$: -they coincide through the $q^{100}$ term, which is more than enough -to prove equality between modular forms of weight $1$ -for a congruence group of such low index in ${\rm SL}_2({\bf Z})$. -Your conjecture then follows from $b_{3n} = b_n$ for all $n$ -(for which an explicit bijection is $(i,j) \mapsto (i+2j,i-j)$). -P.S. I think the generating function for the $a_n$ would be "classically" called -not $\theta^3(q) / \theta(q^3)$ but $\eta^3(q) / \eta(q^3)$, -where $\eta(q) = q^{1/24} \prod_{k=1}^\infty (1-q^k)$.<|endoftext|> -TITLE: Is the Gaussian Correlation Inequality universal? -QUESTION [8 upvotes]: T. Royen proved the Gaussian correlation inequality in the context of Gamma distributions back in 2014, which was since popularized by Latala and Matlak. The properties of Gaussian integration seem heavily exploited in the original proof: even the reduction from convex sets to cylindrical ones rely on the fact that projections of Gaussian vectors are again Gaussian. Has anyone attempted to prove the result for more general distributions beyond multivariate Gamma, or are there obvious counterexamples? - -REPLY [3 votes]: I'm not that familiar with this area but there do exist some results for more general probability distributions but more restricted sets. For example, as explained by Li and Shao, the Gaussian correlation inequality may be reformulated as saying that if $(X_1, \ldots, X_n)$ is a centered, Gaussian random vector, then -$$\mathbb{P}\left(\max_{1\le i\le n} |X_i| \le 1\right) \ge -\mathbb{P}\left(\max_{1\le i\le k} |X_i| \le 1\right) -\mathbb{P}\left(\max_{k+1\le i\le n} |X_i| \le 1\right)$$ -for all $1\le k < n$. The case $k=1$ was proved by Khatri and Šidák (independently), and for this special case, Das Gupta et al. have proved a generalization to elliptically contoured distributions. Das Gupta et al. also give some counterexamples to overly strong generalizations of the conjecture. - -REPLY [2 votes]: Unlike Chow's answer, I do not think the results for elliptically contoured distributions is in the same spirit as GCI because they are controlling the bound of extreme values, which is more like results from U-statistics instead of the generality of GCI. -I think Royden's thinking is basically following Renyi's theorem [5](Or Cramer-Wold if you like) and consequential work in this direction is ongoing using Renyi's divergence applied on convex bodies[4]. - -...even the reduction from convex sets to cylindrical ones rely on the - fact that projections of Gaussian vectors are again Gaussian. - -According to the technique that Royden used, it relies heavily on the fact that the Gamma family is reproducing[1] (OR projection closed, which does not generalize to many other families). The key arguments in his proof, as pointed out by Latala and Matlak[2], is the repeatative use of rectangular sets and the projected images onto these sets. -So I am doubtful that the GCI can be generalize further to other families beyond Gamma. At least I do not believe that these set of techniques can be generalized directly for otherwise Latala and Matlak must have already done.:) There is also another discussion about the application of GCI [3]. -Reference -[1]Teicher, Henry. "On the convolution of distributions." The Annals of Mathematical Statistics (1954): 775-778. https://projecteuclid.org/euclid.aoms/1177728664 -[2]Latała, Rafał, and Dariusz Matlak. "Royen's proof of the Gaussian correlation inequality." arXiv preprint arXiv:1512.08776 (2015). https://arxiv.org/abs/1512.08776 -[3]https://stats.stackexchange.com/questions/270639/consequences-of-the-gaussian-correlation-inequality-for-computing-joint-confiden -[4]Kumar, M. A., & Sason, I. (2016). Projection Theorems for the Rényi Divergence on $\ alpha $-Convex Sets. IEEE Transactions on Information Theory, 62(9), 4924-4935. -[5]Renyi, Alfréd. "On projections of probability distributions." Acta Mathematica Hungarica 3.3 (1952): 131-142.<|endoftext|> -TITLE: Irreducible representations occuring in $\mathrm{Ind}_G^{S_{|G|}}1$ for $G$ finite group -QUESTION [16 upvotes]: Let $G$ be a finite group with $|G|=n$, let $S_G=S_n$ be the group of $n!$ permutations of the set $G$. Then $G$ is a subgroup of $S_G$ via left-translation (i.e. $g\in G$ corresponds to the permutation $h \mapsto gh$). -My question: What can be said about the decomposition into irreducibles of the $S_G$-representation $\mathrm{Ind}_G^{S_G} 1$, where $1$ is the trivial $G$-representation? Is there some combinatorial way to identify the partitions $\lambda$ of $n$ corresponding to the irreducible $S_G$ representations $V_\lambda$ that appear in this induced representation? -Using Frobenius reciprocity, the representation $V_\lambda$ occurs iff restricted to $G$ it contains a copy of the trivial representation. The question can also be expressed by asking for which $\lambda$ we have -$$\frac{1}{n} \sum_{g \in G} \chi_\lambda(h \mapsto gh)\neq 0,$$ -where $\chi_\lambda$ is the character of $V_\lambda$. As the cycle-type of $h \mapsto gh$ is uniquely determined by $\mathrm{ord}(g)$, this shows that the answer of the question above only depends on the multiset $\{\mathrm{ord}(g): g \in G\}$. - -REPLY [4 votes]: Let me collect two partial results that I learned since asking the question. I would still be very happy to have a more general answer of course. -1) For $G=\mathbb{Z}/n\mathbb{Z}$ the question can be explicitly answered (as described in the comments above). Indeed, combining Theorem 8.9 (case $i=0$) and Theorem 8.8 of the book Free Lie Algebras by Reutenauer, the multiplicity with which the representation $V_\lambda$ appears in $\mathrm{Ind}_{G}^{S_n} 1$ is the number of standard tableaux of shape $\lambda$ and major index congruent to $0$ mod $n$. For convenience, let me recall that in a standard tableau $T$, a descent is an index $i \in \{1, \ldots, n-1\}$ such that $i+1$ is in a lower row than $i$ in $T$ and the major index is the sum of all descents for $T$. -2) For general finite groups $G$, the only useful result I know is that the restriction of $\mathrm{Ind}_{G}^{S_n} 1$ to $S_{n-1}$ is isomorphic to the regular representation of $S_{n-1}$ by Mackey's formula. Indeed, note that since $G$ acts simply transitively on itself by left-translation, we have $S_n=G \cdot S_{n-1}$. Since any nontrivial element of $G$ has no fixed point as a permutation of $G$, we have $G \cap S_{n-1}=\{()\}$. Then by Mackey's formula -$$\left(\mathrm{Ind}_{G}^{S_n} 1 \right)|_{S_{n-1}} = \mathrm{Ind}_{\{()\}}^{S_{n-1}}1 = \mathrm{Reg}_{S_{n-1}}.$$ -Translated to partitions, this means the following: take all partitions $\lambda$ with the multiplicity in which $V_\lambda$ appears in $\mathrm{Ind}_{G}^{S_n} 1$. From this form the multiset of all partitions $\mu$ obtainable by removing a block of the partitions $\lambda$. Then the collection of $\mu$ obtained this way is the set of all partitions of $n-1$ with the multiplicity given by the dimension of $V_{\mu}$. In a certain sense this says that a lot of partitions $\lambda$ appear in $\mathrm{Ind}_{G}^{S_n} 1$. Indeed, any partition $\mu$ of $n-1$ must be obtainable from such a $\lambda$ by removing one block.<|endoftext|> -TITLE: When $C (X) $ is zero dimensional -QUESTION [9 upvotes]: Let $X $ be a Tychonoff topological (completely rgular) space and $C (X) $ be the ring of all real valued functions over $X $. When is the krull dimension of $C (X) $ zero? - -REPLY [8 votes]: I had written this as a comment, but since the discussion is now a bit confused, it is best to write it as an answer. -The completely regular spaces $X$ such that the ring $C(X)$ is zero-dimensional (i.e., every prime ideal of $C(X)$ is maximal) are known as the "P-spaces" (in the sense of Gillman and Henriksen). The book Rings of Continuous Functions by Gillman and Jerison (Springer 1960, GTM 43) describes a number of properties about them: specifically in exercise 4J and theorem 14.29 (and various other places listed after the latter theorem). -Among the equivalent properties, P-spaces are those in which every function which vanishes at a point $p\in X$ vanishes in a neighborhood of $p$, of in which every $G_\delta$ (countable intersection of open sets) is open. -These spaces look in many ways like discrete spaces, but they are not necessarily discrete: Gillman and Jerison give examples (exercises 4N and 13P) examples of nondiscrete P-spaces. -(Edit.) Here is a simple but interesting example of a non-discrete P-space: consider the set $Q$ of functions $x\colon \omega_1 \to \{0,1\}$ which are eventually $0$ (i.e. there is $\alpha<\omega_1$ such that $x(\xi)=0$ for $\xi\geq\alpha$) and order them lexicographically (i.e., $x$ and $y$ are compared as $x(\xi)$ and $y(\xi)$ for the smallest $\xi$ for which $x(\xi)\neq y(\xi)$). Put the order topoplogy on $Q$. Then $Q$ has no isolated point, but it is still a P-space (Gillman & Jerison, theorem 13.20 + exercise 13.P(1)).<|endoftext|> -TITLE: Classification of pointed Hopf algebras up to gauge equivalence -QUESTION [5 upvotes]: The classification of finite-dimensional pointed Hopf algebras over an algebraically closed field of characteristic zero and whose group of group-like elements is abelian is very much completed. However, from a more categorical point of view it would make sense to classify such Hopf algebras up to gauge equivalence,i.e. two (quasi-)Hopf algebras are called gauge equivalent if their representation categories are tensor equivalent. -Are people interested in the latter problem? Are there results in this direction? -Edit: In this paper by Wakui, he classified all $8$-dimensional Hopf-algebras up to gauge equivalence. In fact the representation ring alone distinguishes between these Hopf-algebras. Clearly people are interested in this. Are there further results? - -REPLY [2 votes]: There's been quite a lot of interest in classifying fusion categories of (low) dimensions, and by reconstruction this effectively classifies the represented objects up to gauge equivalence. A few Arxiv preprints on this: - -Classification of fusion categories of dimension $pq$ -On the classification of certain fusion categories -Morita equivalence methods in classification of fusion categories -On the Classification of Pointed Fusion Categories up to weak Morita Equivalence - -You can search the Arxiv for results specifically about classifications of certain Hopf algebras, of which there are many papers. -Gauge equivalence of $H$ and $K$ means there is a "twist"—a 2-cycle-esque object that modifies the comultiplication, but for which there is generally no structured homology theory—$F$ of $K$ such that $H\cong K^F$. The twists of the so-called "big quantum groups" are fully classified to my understanding (based on hearsay; I don't think it's been published), and the twists of the small quantum groups $u_q(\mathfrak{g})$ (which remain Hopf algebras, at least) are conjecturally classified by certain Belavin-Drinfeld triples on the Dynkin diagram of $\mathfrak{g}$ and certain twisted automorphisms. -There are also certain conjectures (well, they were conjectures a few years ago, anyway; possibly I'm out of the loop, but I think these were fairly deep and unlikely to have been resolved) in the theory of vertex operator algebras which would imply the existence of braided equivalences between the representation categories of certain twisted group doubles. Extensive work has been done to (successfully) classify such equivalences, mostly by Naidu and/or Nikshych. -So, yes, in short, a lot of people are interested in your latter question, though I only know of two cases where the pointed property is relevant: - -The small quantum groups are pointed, and these are of immense and active research interest on a wide range of topics for a wide range of reasons and applications. As linked above, a very recent paper of Negron concerns the determination of the gauge equivalence class of small quantum groups. -A semisimple pointed Hopf algebra has the property that all simple objects of $\text{Rep}(H)$ are invertible. And the classification of pointed fusion categories is basically known (see the first set of links), and the most commonly encountered ones (in my experience, at least) are the $G$-graded (f.d.) vector spaces with associativity $\omega$: $\text{Vec}_G^\omega$. So the non-semisimple case would be more interesting; or at least more difficult. - -It seems the pointed property might be more ubiquitous in combinatorial Hopf algebras, but these are outside my expertise. Otherwise, the quasitriangular case (rather than the pointed one) has been of substantial interest because braided fusion categories appear throughout a vast range of mathematical physics.<|endoftext|> -TITLE: There are no "holes" in the Bruhat decomposition of parabolic cell $Pw_1P$ -QUESTION [7 upvotes]: Let $G$ be a split reductive algebraic group (over a local field if you like), $B$ be a fixed Borel subgroup, and $P$ be a fixed standard parabolic subgroup. Let $W$ be the Weyl group of $G$. For $w\in W$, denote $C(w)=BwB$. -Given Weyl elements $w>w'>w_1$ (Bruhat order), if $C(w)\subset Pw_1P$, do we know that $C(w')\subset Pw_1P$? It's like to ask whether there are holes in $Pw_1P$. On the other hand, this amounts to asking if $Pw_1P\cap \Omega_{w_1}$ is closed in $\Omega_{w_1}$, where $$\Omega_{w_1}:=\coprod_{w\ge w_1}C(w).$$ -Any comments and references are welcome. Thanks in advance. - -REPLY [3 votes]: Your condition $C(w)\subseteq Pw_1P$ implies $PwP=Pw_1P\Rightarrow\overline{PwP}=\overline{Pw_1P}$. -Moreover $w\ge w'\Rightarrow\overline{BwB}\supseteq\overline{Bw'B}\Rightarrow\overline{PwP}\supseteq \overline{Pw'P}$. -Similarly $w'\ge w_1\Rightarrow\overline{Bw'B}\supseteq\overline{Bw_1B}\Rightarrow\overline{Pw'P}\supseteq \overline{Pw_1P}$. -Together, one gets $\overline{Pw'P}=\overline{Pw_1p}$. -But then $Pw'P=Pw_1P$ and in particular $C(w')\subseteq Pw_1P$. -Remark: One can show stronger that the set of $w$ with $C(w)\subseteq Pw_1P$ is the double coset $W_Iw_1W_I$ and that this double coset forms a Bruhat interval, i.e., there are $w_{min},w_{max}\in D$ such that $w\in D\Leftrightarrow w_{min}\le w\le w_{max}$. After a very cursory search I found it without proof in the paper "Parabolic double cosets in Coxeter groups" by Billey et al. Proposition 2.<|endoftext|> -TITLE: Is there a relationship between the moduli space of spatial polygons and the moduli space of labeled points? -QUESTION [5 upvotes]: It is well known that the set of all polygons with consecutive side lengths $l_1, \dots, l_n$ in $\mathbb{R}^3$, considered up to rigid motions, is a compact complex manifold. Of course, I am assuming, for the sake of simplicity that there are no "straight line" polygons i.e., the length vector $L:= (l_1,\dots, l_n)$ is generic. I will denote this space by $M_L$. -Consider configurations of $n$ not necessarily distinct points $p_i\in \mathbb{P}^1$. Each point $p_i$ is assigned the weight $l_i$. A configuration is called stable if the sum of weights of equal points is less than the total weight $\sum l_j$. The moduli space of stable points on $\mathbb{P}^1$ is defined as the stable configurations modulo projective automorphisms. -It was shown by Kapovich and Millson (and also by Klyachko) that $M_L$ is isomorphic to the moduli space of stable points on $\mathbb{P}^1$. -Before I ask my questions let me recall the definition of one more moduli space. -The moduli space of $n$ distinct labeled points on $\mathbb{P}^1$ is obtained from $(\mathbb{P}^1)^n$ by removing the diagonals $z_i = z_j$ for $i\neq j$ and factoring out by the $PSL(2, \mathbb{C})$ action. This space, denoted $M_{0,n}$, has a smooth compactification denoted $M_n$. The way I understand the construction of $M_n$ is as follows: first realize that $M_{0,n}$ is the complement of the braid arrangement in $\mathbb{P}^{n-1}$. Then one reaches $M_n$ by successively blowing along certain intersections. -Now my questions: - -For any (generic) choice of length vector one can observe that $M_{0, n} \subset M_L$ as the set of all those polygons for which no to directions are same (i.e., stable configurations where are all points are distinct). In view of this, can one say that $M_L$ is a smooth compactification of $M_{0,n}$? -Is there some nice map from $M_n$ to $M_L$ (or in the other direction) which is an isomorphism on $M_{0,n}$? I have read the following description of $M_n$ in several papers: '' when two points in $M_{0,n}$ come close too each other instead of colliding a new $\mathbb{P}^1$ pops out and these points move on it". I don't understand what this means mathematically but this pictures suggests a map. Given a length vector $L$ the essential information is the collection of all so-called short subsets. I guess this data should help us identify which "popped out" $\mathbb{P}^1$s to contract and obtain a stable configuration (or the other way round?). I know what I have written here is pretty vague but I have a strong feeling that somebody must have thought about this earlier. -The complement $M_L\setminus M_{0,n}$ is certainly not a normal crossing divisor. So, if one were to blow up along the subspaces of colliding points -what would one get? Does this even make sense? - -Motivation: For the past few months I have been working on some problems related to the moduli space of planar polygons. One day I realized that the planar polygons modulo $O(2)$ contain $M_{0, n}(\mathbb{R})$ as an open subset (which is just a disjoint union of $(n-1)!/2$ open discs). This observation lead me to above questions. -I should also confess that I have no expertise in algebraic geometry. I am completely new to the world of moduli space of stable points and labeled points. - -REPLY [5 votes]: Yes. -I assume that your $M_n$ is what is more usually denoted $\overline{M_{0,n}}$. Then the answer is yes, there is a natural map $\overline{M_{0,n}} \twoheadrightarrow M_L$, for each $L$. Specifically, let $\gamma$ be a tree of $\mathbb P^1$s glued along nodes and with $n$ labeled points (not at the nodes). Then at each node, measure the total weight on either side. By your assumption, on one side it'll be smaller than the other. Contract all those $\mathbb P^1$s on that smaller side to a point, collapsing the labeled points together, and repeat until there is only one $\mathbb P^1$. - -The more general statement here is that the polygon spaces are GIT quotients, while $\overline{M_{0,n}}$ is a Chow quotient (in each case, of the $2$-Grassmannian by a torus). The relevant paper is Kapranov's "Chow quotients of Grassmannians", which I'm not finding online (although -Keel-Tevelev's followup is available). - -You're right that blowing up a divisor does nothing if it's normal crossings, but more generally the condition is that it be "Cartier", i.e. locally given by a single equation. Here the local picture is of $n$ points in $\mathbb C$, and the single equation is the Vandermonde $\prod_{i -TITLE: Bijection from the plane to itself that sends circles to squares -QUESTION [43 upvotes]: Let me apologize in advance as this is possibly an extremely stupid question: can one prove or disprove the existence of a bijection from the plane to itself, such that the image of any circle becomes a square? Or, more generally, are there any shapes other than a square such that a bijection does exist? (obviously, a linear map sends a circle to an ellipse of fixed dimensions and orientation) - -REPLY [66 votes]: There is no such bijection. -To see this, imagine four circles all tangent to some line at some point $p$, but all of different radii, so that any two of them intersect only at the point $p$. (E.g., any four circles from this picture.) Under your hypothetical bijection, these four circles would map to four squares, any two of which have exactly one point in common, the same point for any two of them. You can easily convince yourself that no collection of four squares has this property. - -REPLY [8 votes]: The problem is that two distinct squares can have more than two common points (easy to make an example), and under such a bijection these squares would have to go to two distinct circles with more than two common points - an impossibility.<|endoftext|> -TITLE: Strange convergence of Euler's series for $\zeta(2)$ -QUESTION [32 upvotes]: Using Maple to compare $\pi^2$ and the partial sums of $6\sum_{n=0}^{\infty}\frac{1}{n^2}$ I have noticed something that appears strange. -For instance, let $S_{k}=6\sum_{n=0}^{k}\frac{1}{n^2}$ be the kth partial sum, for the first 50 digits we, have -9.8636074000893588188343481429332935379126206789621 ($S_{1000}$) -9.8696044010893586188344909998761511353136994072408 ($\pi^2$ ) -Notice that the first three digits coincide (this is of course expected), but we have a strange coincidence of digits that are quite surprising. -A table with the difference between the digits is given below (read from right to left!) -[-7, 2, 2, 7, 1, 7, 2, -9, -7, 0, -1, 0, 6, 2, 0, 4, 2, 4, 1, -3, -4, 1, -7, -5, -8, 8, -5, -1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, -1, 0, 0, 3, 0, 0, -6, 0, 0, 0], -The zeros indicate that the digits coincide. -Below the tables for -$k=10^4$: -[-7, 1, 4, 7, -1, 1, -3, -4, -1, 3, 4, 1, -1, -4, -1, 3, 4, 1, 7, -5, -1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 3, 0, 0, 0, -6, 0, 0, 0, 0] -$k=10^5$: -[-4, 9, 2, 6, 1, 7, 5, -2, -7, -5, 2, 7, 5, -2, 0, 0, 0, 0, 0, 0, 0, 0, 0, -8, -9, 1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 3, 0, 0, 0, 0, 4, -1, 0, 0, 0, 0] -$k=10^6$: -[-1, 3, 4, 1, -3, 6, -2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, -7, 1, 0, 0, 0, 0, 4, 9, -1, 0, 0, 0, 0] -Notice that in the table above the initial (from right to left) expected sequence of zeros has length 4, but we can see a sequences with 11 and 12 consecutive zeros! -Can someone explain this surprising abundance of coinciding digits? -I have checked that a similar phenomenon occurs for other values of Riemann's $\zeta$ function. For instance: -If we compare $\zeta(3)$ and the partial sums of $\sum_{n=1}^{\infty} \frac{1}{n^3}$ we have, for 100 digits and $k=10^{11}$ the following table -[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -5, -2, 7, -4, -3, -3, -3, -3, 7, 6, -4, -3, -3, 7, -4, 7, -4, -3, -3, -3, 2, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, -5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] -so we have the first 21 digits that coincide (ok, expected) and "much later" 21 that don't coincide followed by 19 that coincide. -Is this a well known phenomenon? - -REPLY [36 votes]: Yes, this is a well-known occurrence of Bernoulli numbers arising from Euler-MacLaurin summation applied to the zeta function. See the AMM article of Borwein-Borwein-Dilcher for details.<|endoftext|> -TITLE: Understanding a germ of a GIT quotient -QUESTION [7 upvotes]: Let $X$ be a smooth complex affine variety, let $G$ be a complex reductive group acting on $X$. Suppose that the stabilizer $G_x$ of a point $x\in X$ is reductive and connected. Let $\varphi: X\to X//G$ be the GIT quotient. I would like to understand the germ of $X//G$ at the point $\varphi(x)$, and in particular understand if the following is correct: -Guess. The analytic germ of $X//G$ at $\varphi(x)$ is isomorphic to the analytic germ at $0$ of the GIT quotient of $T_X(x)/T_{O_x}(x)$ by the linear action of $G_x$. -Question. Is the above guess correct? If yes, is there a reference for this statement? In particular, how to prove this statement in the case when $x$ is fixed by $G$? - -REPLY [10 votes]: For a fixed point your guess is right and one doesn't need Luna's slice theorem to prove it: Let $T$ be the tangent space in $x$ and let $\mathfrak m_x\subset\mathbb C[X]$ be the maximal ideal. Then the canonical surjective linear map $\mathfrak m_x\to\mathfrak m_x/\mathfrak m_x^2\cong T^*$ has a $G$-equivariant section giving rise to a $G$-linear map $T^*\hookrightarrow\mathbb C[X]$ and therefore to a morphism $X\to T$ which maps $x$ to $0$ and is étale in $x$. Now use the fact that invariants commute with completions (by complete reducibility) to see that $X//G\to T//G$ is étale in the image of $x$. As Jason indicated this implies an isomorphism of analytic neighborhoods. -It is true that the above argument is part of Luna's slice theorem but it is actually only the "easy" part. In fact, Luna's slice theorem is not just a statement about étale or formal neighborhoods but has a global aspect, as well. One consequence is, e.g., that the unstable set with respect to $x$ (i.e., the set of points having $x$ in their orbit closure) is globally isomorphic to the unstable set of $0$ in $T$. -Concerning your more general guess: One has to be very careful with Luna's slice theorem: it requires that the orbit of $x$ is closed. This is much stronger than $G_x$ being reductive. So, your guess is ok if $Gx$ is closed in $X$ (which is trivial if $x$ is a fixed point). -Otherwise, there is the following counterexample. Let $G=SL(2,\mathbb C)$, let $X$ be the space of $5$-forms, and $x=\xi\eta(\xi+\eta)^3\in X$. It is easily seen that $G_x=1$. So $V:=T_x(X)/T_{\mathcal O_x}(x)=\mathbb C^3$. -Let $N\subset X$ be the set of unstable $5$-forms. Hilbert has proved that these are those $5$-forms having a linear factor of multiplicity $\ge3$. So $x\in N$. Now, it is well known that $X//G$ is singular in the image of $0$. In particular, it is not isomorphic to $V//G_x=V$. -The problem of this example is that the codimension of $N$ in $X$ is $2$. Thus, $S\cap N$ is nontrivial and maps to $0$ where $S\cong V$ is a slice to the orbit of $x$. -Edit: Concerning the equivalence of étale and analytic singularities see this MO thread.<|endoftext|> -TITLE: Binomial Coefficient Identity, Double Series, Floor Function -QUESTION [6 upvotes]: I came across this (supposed) identity for natural numbers $m$ and $n$: $$ \sum_{i=0}^{n}{ \sum_{j=0}^{m}{ \left(-1 \right)^{\lfloor \frac{i}{2} \rfloor+j}2^{n-i}\binom{n-i+\lfloor \frac{i}{2} \rfloor}{j}\binom{n-i+\lfloor \frac{i}{2} \rfloor-j}{\lfloor \frac{i}{2} \rfloor}\binom{i}{m-j}}}= \left(-1 \right)^{m} \binom{2n+1}{2m+1} $$ -and would like to prove it. -Cross posting from MSE after getting no replies. -I am trying to keep a certain tone to my work so I am looking for a human, non-analytic, combinatorial or algebraic proof to the above. To be more specific, I'd like to avoid using generating functions, calculus, complex numbers, trigonometric functions, chabyshev polynomials and induction; every other technique would do. -Update #1: tested to be true for $n,m \leq100$ -Update #2: I would like to give the reason as to why I would like to avoid using the above techniques. I am trying to generlize this identity (12): $$ \left(\sin \left(nx \right) \right)^{2}= \left(\sin \left(x \right) \right)^{2} \left(\sum_{m=0}^{\frac{n-1}{2}}{\left(-1 \right)^{m} \binom{n}{2m+1}} \left( \cos^{2}x\right)^{\frac{n-1}{2}-m} \left( \sin^{2}x\right)^{m}\right)^{2}$$ -for an odd natural number $n$ and a number $x$ to a general field, using the framework of Rational Trigonometry and its Spread Polynomials. I want to show that the latter identity is essentially a polynomial or an "arithmetical" identity, meaning that it is independent of the framework of the classical trigonometric functions and the framework of complex numbers, and that the problem of proving said identity is essentially a (rather complicated) problem of arithmetic or of counting. In this endeavor, I would like to keep a "theme" of not invoking techniques which regard the classical trigonometric functions, complex numbers or calculus. The binomial coefficient identity I wanted to prove came up in the process of this work. - -REPLY [10 votes]: Notice that -$$\binom{n-i+\lfloor \frac{i}{2} \rfloor}{j}\binom{n-i+\lfloor \frac{i}{2} \rfloor-j}{\lfloor \frac{i}{2} \rfloor} = \binom{n-i}{j}\binom{n-i+\lfloor \frac{i}{2} \rfloor}{\lfloor \frac{i}{2} \rfloor}.$$ -This lets us easily sum up the terms depending on $j$: -$$\sum_{j=0}^m (-1)^j \binom{n-i}{j} \binom{i}{m-j} = [x^m]\ (1-x)^{n-i}(1+x)^i = [x^{2m}]\ (1-x^2)^{n-i}(1+x^2)^i.$$ -(We will see later why dealing with squares of $x$ is preferable.) -So, it remains to evaluate the following sum: -$$\sum_{i=0}^n (-1)^{\lfloor \frac{i}{2} \rfloor} (2-2x^2)^{n-i} (1+x^2)^i -\binom{n-i+\lfloor \frac{i}{2} \rfloor}{\lfloor \frac{i}{2} \rfloor}.$$ -Let $i=2s+t$ where $s=\lfloor \frac{i}{2} \rfloor$ and $t=i\bmod 2$. Then the sum becomes -$$\sum_{s\geq 0} \sum_{t=0}^1 (-1)^{s} (2-2x^2)^{n-2s-t} (1+x^2)^{2s+t} -\binom{n-s-t}{s}$$ -$$=\sum_{t=0}^1 \frac{(1+x^2)^{2n-t}}{(2-2x^2)^{n-t}} (-1)^{n-t} \sum_{s\geq 0} \binom{n-s-t}{s} \left(-\frac{4(1-x^2)^2}{(1+x^2)^2}\right)^{n-s-t}$$ -$$(\star)\quad = \sum_{t=0}^1 \frac{(1+x^2)^{2n-t}}{(2-2x^2)^{n-t}} (-1)^{n-t} \frac{(1+x^2)^2}{4(x-x^3)}\Im\left( \frac{2(1-x^2)}{(I-x)^2}\right)^{n-t+1}$$ -$$=\Im \sum_{t=0}^1 \frac{(1+x^2)^{2n-t+2}}{2x} (-1)^{n-t} (I-x)^{-2(n-t+1)}$$ -$$=-\Im \sum_{t=0}^1 \frac{(1-Ix)^{2n-t+2}(1+Ix)^t}{2x}$$ -$$=-\Im \frac{(1-Ix)^{2n+1}}{x},$$ -where $I$ is the imaginary unit ($I^2 = -1$) and $\Im X$ denotes the imaginary part of $X$. -Now we are ready to take the coefficient of $x^{2m}$: -$$[x^{2m}]\ -\Im \frac{(1-Ix)^{2n+1}}{x} = -\Im \binom{2n+1}{2m+1} (-I)^{2m+1} = (-1)^m\binom{2n+1}{2m+1}$$ -as expected. - -UPDATE. To clarify step $(\star)$: -$$\sum_{p\geq 0} \binom{p}{r-p} u^p = \sum_{p\geq 0} [z^{r-p}]\ (1+z)^p u^p = [z^r]\ \sum_{p\geq 0} (z(1+z)u)^p $$ -$$= [z^r]\ \frac{1}{1-z(1+z)u} =[z^r]\ \frac{1}{(z_1-z_2)z}\left(\frac{1}{1-z_1z}-\frac{1}{1-z_2z}\right) = \frac{z_1^{r+1}-z_2^{r+1}}{z_1-z_2},$$ -where $z_{1,2} = \frac{u\pm\sqrt{u^2+4u}}{2}$ are the reciprocals of the zeros of $f(z)=1-z(1+z)u$, so that $f(z)=(1-z_1z)(1-z_2z)$. When $u^2+4u=-v^2$ like in the case $(\star)$, we further have -$$\frac{z_1^{r+1}-z_2^{r+1}}{z_1-z_2} = \frac{2}{v}\Im \left(\frac{u+Iv}{2}\right)^{r+1}.$$<|endoftext|> -TITLE: Friable Numbers In Short Intervals: Density Estimates? -QUESTION [6 upvotes]: I am hoping for explicit numerical estimates like the following sample (with made up numbers, though it might be true): for every $n \gt 10^6$ and every $b$ with $b^2 \lt n \lt b^3$, the number of somewhat smooth numbers in the interval $(n,n+b)$ is at least $b/4$ and at most $3b/7$. Here a number $m$ is somewhat smooth if each of its prime factors is at most the square root of $m$. -The literature on Dickman's function and recent surveys on smooth numbers by Moree and Granville say a lot for larger intervals than what is considered above, and in more generality, but the error terms are not explicit, and I have yet to construct a version with explicit terms from above based on my searches. The actual value approaches $b(1 - \log 2)$ as $n$ grows, but I do not know how fast. Also, somewhat smooth corresponds to $u=2$ in the literature, where $u=\log x / \log y$ is a common parameter in counting $y$-smooth positive integers less than $x$, but I may need to adapt somewhat smooth to $u$-smooth for $u$ some real between $2$ and $3$, so a good answer would address these other values of $u$. If smaller values of $b$ can be used, that would be great, but I think I may only need $n \lt b^3$, and doubt I will need $n \lt b^4$. Note that I am emphasizing explicit and reliable over asymptotic; I do not need tight estimates or small error terms especially if they are not explicit, and I do not want a set of exceptional $n$ or $b$ unless they are all less than a finite explicit and small bound. -Can someone point to or derive such explicit estimates of this type? Computations suggest that the sample above not only is true, but holds for $n$ much smaller than $10^6$. -Gerhard "Quite More Than Somewhat Unsure" Paseman, 2017.04.10. - -REPLY [2 votes]: Given $u$, for every prime $p$ let $S_p = \{m \in (n,n+b)\colon p\mid m,\, p^u > m \}$. Then a lower bound on the quantity you want is $(b-1) - \sum_p \#S_p$. (Indeed, this is the exact value when $u<2$.) The smallest $p$ you have to consider is $n^{1/u}$, and an upper bound for $\#S_p$ is $\lceil\frac{b-1}p\rceil \le \frac{b-2}p+1$. You're therefore led to need to estimate $\sum_{n^{1/u} < p < n+b} 1$ and $\sum_{n^{1/u} < p < n+b} 1/p$. For both of these, the complete sums (where the lower bound is $1$) should appear explicitly in the work of Rosser–Schoenfeld, so just subtract two of those from each other to get the interval versions above.<|endoftext|> -TITLE: Non standard extension of real numbers via nonprincipal ultra filters -QUESTION [14 upvotes]: Assume That $U,V$ are two filters on the natural number $\mathbb{N}$. -We say that $U$ is equivalent to $V$ if there is a bijection $\phi: \mathbb{N} \to \mathbb{N}$ such that $\tilde{\phi}(U)=V$ where $\tilde{\phi}:P(\mathbb{N}) \to P(\mathbb{N})$ is the natural extension of $\phi$ to the power set $P(\mathbb{N})$. -Let $U,V$ be two non principal ultra filter on $\mathbb{N}$. -Let $\mathbb{R}^*_{U}$ and $\mathbb{R}^*_{V}$ be the corresponding nonstandard extension of real numbers associated with $U$ and $V$, respectively. -Assume that $\mathbb{R}^*_{U}$ and $\mathbb{R}^*_{V}$ are isomorphic as fields. Does this imply that $U$ and $V$ are equivalent filters? -My apology in advance, if the question is elementary. The question arose me about 17 years ago when I was trying to understand the application of non standard analysis to ordinary differential equations. - -REPLY [4 votes]: As far as I understand the question is still open under $\neg CH$, namely whether isomorphism of hyperreal fields implies equivalence of filters (up to permutation of index set). Perhaps one can try the following approach. -In a universe $V$ satisfying $CH$, we can take two inequivalent filters and obtain fields that are automatically isomorphic by the result of Erdos &Co mentioned in the MSE post linked in the comments above. Now the idea is to take a forcing extension $V^F$ satisfying $\neg CH$. -One can't transfer naively the construction of the hyperreal field to $V^F$ because the ultrafilter is not definable, but perhaps one can work with the definable hyperreal field of Kanovei and Shelah (which exploits a huge index set using all ultrafilters simultaneously, thereby defeating non-definability). Perhaps one can specify two variants of the Kanovei-Shelah construction whether the ideals are not equivalent but the quotient fields will be by the Erdos-type argument, and then take a forcing extension to exhibit a similar phenomenon under $\neg CH$.<|endoftext|> -TITLE: When the longest element of Weyl group is rational? -QUESTION [5 upvotes]: Let $\mathbb{G}$ be a connected reductive group over $\mathbb{F}_q$, and let $G$ be the base change to an algebraic closure of the base field. Denote by $F$ the associated geometric Frobenius. -Let $T$ be an $F$-rational (i.e. $FT=T$) maximal torus of $G$. As $T$ is $F$-rational, the Frobenius $F$ acts on the finite Weyl group $W(T)$. In general, the subgroup $W(T)^F$ consisting of $F$-fixed points is a proper subgroup, and depends on the choice of $T$. -How to determine when the longest element of $W(T)$ appears in $W(T)^F$? Is there a (maybe case by case) list available? - -As commented below, the longest element is with respect to a Borel subgroup $B$ containing $T$: Then the longest element is the element corresponding to the open piece in $G=\cup_{w\in W(T)}BwB$. - -REPLY [8 votes]: Let $B$ be a Borel subgroup containing $T$. As $F(B)$ and $B$ are both Borel subgroups containing $T$ there exists an element $n \in N_G(T)$ such that ${}^nF(B) = B$. Thus the Frobenius endomorphism $F' : G \to G$ defined by $F'(g) = nF(g)n^{-1}$ induces an automorphism $F' : W \to W$, where $W = N_G(T)/T$, and this stabilises the Coxeter generators $\mathbb{S} \subseteq W$ determined by $B$. Thus $F'$ is a length preserving automorphism so must fix the longest element (by uniqueness). Hence, $F$ fixes the longest element if and only if $\bar{n} \in C_W(w_0)$, where $\bar{n} \in W$ is the image of $n$. -One can ask how unique the element $n$ is. Assume $m \in N_G(T)$ also satisfies ${}^mF(B) = B$ then $mn^{-1} \in N_G(B) = B$ so $mn^{-1} \in N_B(T) = T$ thus $\bar{n} = \bar{m}$. Hence the condition that $F$ fixes the longest element does not depend upon the choice of element $n$. -Let's see how to compute this in practice. Choose an $F$-stable maximal torus and Borel subgroup $T_0 \leqslant B_0 \leqslant G$. There then exists an element $x \in G$ such that $(T,B) = ({}^xT_0,{}^xB_0)$. Note, by the above argument that such an element $x$ is unique up to right multiplication by elements of $T_0$. Let $(W_0,\mathbb{S}_0)$ and $(W,\mathbb{S})$ be the Coxeter system defined with respect to $(T_0,B_0)$ and $(T,B)$ respectively. Then conjugation by $x$ induces an isomorphism $W_0 \to W$ mapping $\mathbb{S}_0$ onto $\mathbb{S}$. Hence, if $w_0 \in W_0$ is the longest element then ${}^xw_0 \in W$ is the longest element. -Assume $x^{-1}F(x) = n \in N_G(T_0)$ then we have $$F(B) = F({}^xB_0) = {}^{xn}B_0 = {}^{xnx^{-1}}B.$$ Hence $xn^{-1}x^{-1} \in N_G(T) = {}^xN_G(T_0)$ is an element as above. We have $x\bar{n}^{-1}x^{-1} \in C_W({}^xw_0) = {}^xC_{W_0}(w_0)$ if and only if $\bar{n} \in C_{W_0}(w_0)$. Thus one can easily construct examples where the longest element is not fixed.<|endoftext|> -TITLE: On the topology induced by a Lorentzian metric -QUESTION [7 upvotes]: Let $(M,g)$ be a time-oriented smooth Lorentzian manifold, with Lorentzian metric $g$. In the following thread: -https://physics.stackexchange.com/questions/228669/why-pseudo-riemannian-metric-cannot-define-a-topology/228676?noredirect=1#comment730891_228676 -physicist @ValterMoretti makes the following claim: -"As a matter of fact, a (connected) Lorentzian smooth metric $g$ over the time-oriented smooth manifold $M$ does define a topology, the same already present on $M$." -I have not been able to find a proof of the statement above, unless some relatively strong assumptions are made on $(M,g)$ (for example imposing that $(M,g)$ is strongly causal). Can the statement be proved in general without further extra-assumptions? I was under the impression that in fact one of the key points of departure between Riemannian and Lorentzian geometry is the failure for a Lorentzian metric to define in general a topology equivalent to the pre-existing manifold topology, in contrast to what happens in Riemannian geometry. -Thanks. - -REPLY [8 votes]: In general, this topology is coarser than the original topology of the manifold, and, without further assumptions, strictly coarser. It coincides with the original one iff the Lorentz manifold is strongly causal, see e.g. Prop 3.11 in Beem-Ehrlich-Easley. To get counter-example consider the cylinder $\mathbb{S}^1 \times \mathbb{R}$ with time direction being $\mathbb{S}^1$, i.e. periodic, and the usual flat metric. Then you can connect any two points by a timelike curve, thus the only non-empty open diamond is the whole spacetime. In this case the induced topology is the in-discrete one.<|endoftext|> -TITLE: Is the following series consisting of equally distributed $\pm 1$ bounded? -QUESTION [14 upvotes]: Apologise in advance if this problem isn't research-level (I'm quite certain it isn't). It's just I found it quite intriguing because it turned out to be much more subtle than it appeared at my first glance. Also, this problem didn't get an answer so far (nor much attention) from MSE, so I decided to post it here in hopes I can get some valuable insights. Cheers! -Here goes the original post from MSE. - - -Let $b=\dfrac{\sqrt{5}-1}2$ and $a_n:=(-1)^{[nb]\bmod 2}$ where $[\cdot]$ denotes the floor function. Then is $\sum a_n$ bounded? - -As far as I know $nb\bmod 2$ is equidistributed in $[0,2]$ and so as $n$ gets large there will be about half of indices in $\{1,2,\cdots,n\}$ that correspond to $1$ and the other half to $-1$. This doesn't help much though because what it says is basically $\sum a_n\in o(n)$. By testing on computer softwares, it seems to be $O(\log n)$ (but of course any such "test" is worthless). But I don't know what else I can do. -Also, is there anything special about $b$ besides being an irrational real number (or, being the golden ratio)? What if I replace it by, say, $\pi$? - -PS: A comment suggested a probable connection with the 1D random walk, but I really can't see a possible way to formulate such a connection beyond the intuitive level. - -REPLY [16 votes]: The sequence $\sum a_n$ is unbounded. -This is a consequence of a general result from Kesten, -On a conjecture of Erdös and Szüsz related to uniform distribution mod 1, Acta Arithmetica (1966). The proof is not very long but quite computational, using properties of continued fractions. -Theorem -Let $\xi \in [0,1]$, $0\leq a < b \leq 1$, denote by $N(M, \xi, a,b)$ the number of integers less than $M$ such that $a \leq \{k\xi\} \leq b$. Then -$N(M,\xi, a,b) - M(b-a)$ is unbounded if $b-a$ is rational and $\xi$ is irrational. -You are interested in the $1/2$ discrepancy of an irrational rotation of the circle. Note that from the Denjoy-Koksma inequality, -there is a sequence $q_n$ such that -$\sum_{k= 0}^{q_n-1} a_k$ is bounded by $2$. The sequence exists for any irrational rotation of the circle and is given by the denominators of the partial fraction decomposition of the rotation number. For the golden ratio, these numbers are given by the Fibonacci numbers. -The golden ratio is a bit special because its decomposition in continued fraction is just $[1,1,1...]$ so it is badly approximated by the rational numbers. So the computation of Kesten should be simpler in that case. -EDIT: the theorem is also mentioned shortly in the book of Kuipers and Niederreiter, "uniform distribution of sequences". Some additional references can be found in the notes of chapter 2, section 3 p128.<|endoftext|> -TITLE: Chromatic number of Voronoi diagrams of lattices -QUESTION [11 upvotes]: Let $L$ be a Euclidean lattice. Define a graph whose vertex set is $L$ and where two points $x,y\in L$ are declared to be adjacent whenever the cells of $x$ and $y$ in the Voronoi diagram of $L$ have a facet in common, or equivalently¹, when $z := x-y$ and $-z$ are the only shortest vectors in the coset $z + 2L$ (such vectors $z$ are called "relevant"). Let $\chi(L)$ be the chromatic number of this graph (i.e., the minimal number of colors required to color the Voronoi diagram of $L$ such that two cells sharing a facet never have the same color). -Main question: Does this quantity $\chi(L)$ appear in the literature? Does it have a name? What are the standard facts about it? -Trivial observation: $\chi(L) \leq 2^d$ where $d$ is the dimension of $L$. (Proof: color $L$ by mapping $x\in L$ to its coset in $L/2L$.) -More specific question: What is the value of $\chi(L)$ for some standard lattices like the root lattices, their duals, and the Leech lattice? -Note: If $L$ is a root lattice, then the relevant vectors are precisely the roots. (In fact, this characterizes root lattices: Rajan & Shande, "A Characterization of Root Lattices", Discrete Math. 161 (1996), 309–314.) -Example observation: $\chi(A_n) \leq n+1$. (Proof: color $(x_0,\ldots,x_n)\in\mathbb{Z}^{n+1}$ such that $\sum_i x_i = 0$ by $\sum_i i x_i$ mod $n+1$.) This inequality is sharp as Fedor Petrov points out in a comment below. Even more trivially, $\chi(\mathbb{Z}^n) = 2$ (using a checkerboard coloring). - -Conway & Sloane, Sphere Packings, Lattices and Groups (Springer 3rd ed., 1999), theorem 10 of chapter 21. - -REPLY [5 votes]: It's been a bit long in the making, but Mathieu Dutour Sikirić, Philippe Moustrou, Frank Vallentin and I just uploaded to the arXiv a paper on “Coloring the Voronoi tessellation of lattices” which answers some aspects of the above question. In particular, we computed the following chromatic numbers: - -$n+1$ for the root lattice $A_n$ and its dual $A_n^*$, -the chromatic number of the half-cube graph for the root lattice $D_n$, and $4$ for its dual $D_n^*$, -$9,14,16$ for the root lattices $E_6,E_7,E_8$, and $16$ for their duals $E_6^*$ and $E_7^*$, -and $4096$ for the Leech lattice. - -We also show that the greatest chromatic number of an $n$-dimensional lattice grows exponentially with $n$.<|endoftext|> -TITLE: How is the "conformal prediction" conformal? -QUESTION [11 upvotes]: The question is clarified by Prof.V.Vovk. See his answer below for discussion. -Recently, early works of Gammerman, Vanpnik and Vovk[4] are rediscovered by Wasserman et.al[1] and proposed it as a promising candidate towards distribution-free inference coming along with confidence level guarantee. -Given the current literature, especially [2,3], the main improvement in the conformal inference(implemented in support vector machine) is providing a confidence region with certain credibility. For example, in the classical regression model we can obtain the pointwise prediction interval at each of observation sample points. The CP - -...Unlike traditional regression methods which produce point predictions, - Conformal Predictors output predictive regions that satisfy a given confidence - level.The regions produced by any Conformal Predictor are automatically valid, however their tightness and therefore usefulness depends on the nonconformity measure used by each CP....what we call in this - paper “Conformal Prediction” (CP).[2] -...We also obtain a measure of “credibility” which serves as an - indicator of the reliability of the data upon which we make our - prediction.[3] - -And in their introductory book, they explained their usage of the name "Conformal Prediction" by saying - -Most of this book is devoted to a particular method that we call - "conformal prediction". When we use this method, we predict that a new - object will have a label that makes it similar to the old examples in - some specified way, and we use the degree to which the specified type - of similarity holds within the old examples to estimate our - confidence in the prediction. Our conformal predictors are, in other - words, "confidence predictors".[5]pp.7-8 - -To be more precise and in response to @RHahn comment below, I do not think "conformal" is an arbitrary choice of word, since Vovk mentioned that - -"In 1963–1970 Andrei Kolmogorov suggested a different approach to - modelling uncertainty based on information theory; its purpose was to - provide a more direct link between the theory and applications of - probability. On-line compression models are a natural adaptation of - Kolmogorov's programme to the technique of conformal prediction. - Working Paper 8 introduces three on-line compression models: - exchangeability (equivalent to the iid model), Gaussian and - Markov."[6] - -Therefore I do believe there is a deeper motivation of the "conformal prediction" from complex analysis that I was not aware of. Thanks! -Therefore my question is, -(1)Given the name "Conformal prediction", is this method more or less associated with the concept of conformal mapping in (multivariate) complex analysis? -Does it mean that the old sample can be mapped locally conformally to the new samples? -Since most of CP are implemented using SVM, is this related to the shape of classifying hyperplane determined by the SVM? -(2)(More like a opinion-based question) How is conformal predictor different from the existing robust predictors while they both come with a guarantee that the true value will fall into the confidence region with high probability? -Reference -[1]Lei, Jing, et al. "Distribution-free predictive inference for regression." Journal of the American Statistical Association just-accepted (2017). -[2]Papadopoulos, Harris, Vladimir Vovk, and Alexander Gammerman. "Regression conformal prediction with nearest neighbours." Journal of Artificial Intelligence Research 40 (2011): 815-840. -[3]Saunders, Craig, Alexander Gammerman, and Volodya Vovk. "Transduction with confidence and credibility." Proceedings of the Sixteenth International Joint Conference on Artificial Intelligence (IJCAI'99). Vol. 2. 1999. -[4]Shafer, Glenn, and Vladimir Vovk. "A tutorial on conformal prediction." Journal of Machine Learning Research 9.Mar (2008): 371-421. -[5]Vovk, Vladimir, Alexander Gammerman, and Glenn Shafer. Algorithmic learning in a random world. Springer Science & Business Media, 2005. -[6]http://www.vovk.net/cp/index.html - -REPLY [12 votes]: Thanks for your interest. The term “conformal prediction” was suggested by Glenn Shafer, and at first I did not like it exactly for the reason that you mention: it has nothing (or very little) to do with conformal mappings in complex analysis. But then I discovered other meanings, even in maths; e.g., Wikipedia has five on its disambiguation page for “conformal”: - -Conformal film on a surface (same thickness) -Conformal fuel tanks on military aircraft -Conformal coating in electronics -Conformal hypergraph, in mathematics -Conformal software, in ASIC Software - -So the word did not look taken to me anymore. The expression that we had used before Glenn proposed “conformal prediction” was even worse (“transductive confidence machine”). -Thanks to Hengrui Luo for drawing my attention to this question. -As for question (2), the answer depends on which robust predictors you have in mind. The predictors with most similar properties are the ones in classical statistics (such as the standard prediction intervals in linear regression based on Student's t distribution); the main difference is that they are parametric. There is a predictive version of tolerance intervals in nonparametric statistics, but their treatment of objects (x parts of observations (x,y), where y are labels) is limited. Upper bounds on the probability of error given by standard PAC predictors are often too high to be useful.<|endoftext|> -TITLE: Non-density of invertible elements in $\ell_1(\mathbb{N}_0)$ -QUESTION [7 upvotes]: Consider the Banach algebra $\ell_1(\mathbb{N}_0)$ (with convolution / Cauchy product of series). I am looking for an elementary proof of the fact that the group of invertible elements in this algebra is not dense. -If one wishes to use some hammers, here is a way to do so. Call this algebra $A$; it is not too hard to show that $A$ is semi-simple. Thus the Gelfand transform $g\colon A\to C(\Phi_A)$ is injective and has dense range. However $\Phi_A = \overline{\mathbb D}$ and it is well known that invertible elements in $C(\overline{\mathbb D})$ are not dense. - -Q1. Is there a more elementary way to see this (preferably that would work for the real scalars too)? -Q2. Is there a characterisation of semigroups $S$ for which invertible elements in $\ell_1(S)$ are (not) dense? - -REPLY [6 votes]: Q1 can be answered with the following result. - -An element in the boundary of the invertible group of a unital Banach algebra is a topological divisor of zero. - -I must confess I remembered the statement but not the proof, but it can be shown using elementary arguments related to the fact that the invertible group is open. It should be in e.g. Allan's book on Banach algebras (since it is in the lecture notes which evolved into that book) and probably also in Bonsall and Duncan. If I have time in the next day or two I can include the argument here. -Regarding Q2: if $S$ is a (discrete) group with polynomial growth, then I think the invertible elements should be dense in $\ell^1(S)$, using the above criterion and Corollary 1.9 in this paper of Tessera the results of https://doi.org/10.1016/j.jfa.2010.07.014 (sorry for being sketchy right now, please remind me if I don't update this answer within the next couple of days) -I have a feeling that Q2 is open even when S is a discrete solvable group of exponential growth, but I have not kept up with progress in this area for a while. The case of commutative monoids might be tractable - -REPLY [4 votes]: For Q1: -Every $f=\sum a_nz^n\in A$ induces a continuous function $\hat{f}$ on the closed disk of radius $1$ and $f\mapsto \hat{f}$ is a continuous operator (for $\ell^1$-norm $\|f\|=\sum|a_n|$ on the left and sup norm $g=\sum_{|z|\le 1}|g(z)|$ on the right), which in addition is multiplicative. -The for $u(z)=z$, it is standard ($*$) from differential/algebraic topology that every $g$ close enough to $\hat{u}$ (the identity of the closed disk) in the supremum norm has a zero. It follows that $u$ (which is indeed $\delta_1=(0,1,0,\dots)$ is in the interior of the set of non-invertible elements of $A$. The same argument works for $z^n$, $n\ge 1$. -($*$) indeed if $g$ is close enough to the identity then it is homotopic to the identity, and if $g$ does not vanish, then we can deduce a homotopy from the identity of the circle to a constant map.<|endoftext|> -TITLE: Deligne's Canonical Extension in Algebraic Varieties? -QUESTION [7 upvotes]: Suppose $C/\mathbb{Q}$ is an algebraic curve (not necessarily complete) defined over $\mathbb{Q}$, and $p$ is a $\mathbb{Q}$ valued point of $C$. Suppose there is a algebraic fibration -\begin{equation} -\pi: Z \rightarrow C -\end{equation} -such that $Y=\pi^{-1}(p)$ is the only singular fiber, i.e. -\begin{equation} -\pi: X \rightarrow S -\end{equation} -where $X=Z \setminus Y$ and $S=C \setminus p$, is a smooth morphism between smooth varieties. Then by theorem 1 of the paper "On the differentiation of de Rham cohomology classes with respect parameters" by Katz and Oda, there is a canonical integrable connetion $\nabla_{GM}$ --Gauss-Manin connection--on the relative de Rham cohomology sheaf $R^q\,\pi_*(\Omega_{X/S}^*)$. -If we take extension of field and then analytification, we got maps -\begin{equation} -\pi: Z_{\mathbb{C}}^{an} \rightarrow C_{\mathbb{C}}^{an}\\ -\pi:X_{\mathbb{C}}^{an} \rightarrow S_{\mathbb{C}}^{an} -\end{equation} -The Gauss-Manin connection now is the the analytification of $\nabla_{GM}$, and let us denote it by $\nabla_{GM}^{an}$. For the purpose of this question, let me assume that $\nabla_{GM}^{an}$ has regular singularities at $p$. -Then from the works of Griffiths and Schmid on variations of Hodge structures, there exists Deligne's canonical extension of the locally free sheaf $R^q\pi_*(\Omega_{X/S}^{*an})$ on $S_{\mathbb{C}}^{an}$ to a locally free sheaf on $C_{\mathbb{C}}^{an}$. The process of extending it over $p$ could be described explicitly as follows. -Choose a local coordinate in a neighbourhood of $p$, say $t$ and $p$ is the point with $t=0$, i.e. origin. Choose muli-valued local sections $e=(e_1,e_2,\cdots,e_n)^{t}$ of $R^q\pi_*(\Omega_{X/S}^{*an})$ in a neighbourhood of $0$ which form a multi-valued local frame. Suppose the monodromy matrix around $t=0$ with respect to thess muli-valued sections is $T$, define $N= \log T$, the vector -\begin{equation} -\exp (-\frac{\log t}{2 \pi i} \,N)\,e -\end{equation} -is single valued and we use it as a local frame to do the extension. This is Deligne's canonical extension in analytical case. -My questions are -1, With the assumption of $\nabla_{GM}^{an}$ is regular, is $\nabla_{GM}$ regular. Actually I am not clear with the definition of regularity of $\nabla_{GM}$ when it is defined over $\mathbb{Q}$, could someone explain it? -2, Could $R^q\,\pi_*(\Omega_{X/S}^*)$ be canonically extended to a locally free sheaf on $C$? -3, If such an extension exists, is there an explicit way like the analytical case to describe this extension using monodromy matrix? -4, If such an extension exists, after $\otimes_{\mathbb{Q}} \mathbb{C}$ and analytification, is this extension compatible with the Deligne's canonical extension in the analytical case? -Any comments and references will be appreciated. - -REPLY [4 votes]: A short answer is that Deligne's definition of canonical extension (https://publications.ias.edu/sites/default/files/71_Localbehavior.pdf) is not analytic, and works fine over $\mathbb Q$. Rather than the exponential you define, he defines a new connection via an algebraic formula, which is constant, and whose horizontal sections map the sections you define. Then he defines the canonical extension to be generated by the horizontal sections. This can be done algebraically by finding horizontal sections in a formal power series ring, say. - -If more explanation is needed, my way of thinking of this is to consider everything in terms of the action of the operator $t \nabla$ on the sections of the vector bundle $R^q \pi_*$ over the field of formal Laurent series, $\mathbb Q((t))$ and $\mathbb C((t))$. The condition that $\nabla$ have regular singularities is, I believe, equivalent to the condition that $t \nabla$ can be put in Jordan normal form - i.e. an arbitrary section can be written as an infinite sum of sections that lie in a generalized eigenspace of $t \nabla$. -So the descent of this to $\mathbb Q((t))$ is the condition that an arbitrary section can be written as an infinite sum of sections in a finite-dimensional $t\nabla$-stable subspace. -In the case when the connection comes from algebraic geometry, one knows that all the eigenvalues of the monodromy, which are the same as the eigenvalues of $e^{ 2\pi i t\nabla}$, are roots of unity, so the eigenvalues of $t \nabla$ are rational numbers. -Hence one can simply define the canonical extension to be the sum of the subspaces generated by elements in generalized eigenspaces whose eigenvalue is a nonnegative rational number. (We want nonnegative because multiplying by $t$ adds $1$ to the eigenvalue.) -In fact, I think you want $N$ to be nilpotent, in which case all eigenvalues of $t\nabla$ are integers, and you can do the same thing with nonnegative integers. In fact you can take it to be the submodule generated by the subspace of sections in generalized eigenspaces with eigenvalue zero. (If a generator has eigenvalue a positive integer, we can lower it to zero by dividing by the corresponding power of $t$, so we may take generators to be those sections with eigenvalue $0$.) Using this description, you should be able to see it matches Deligne's characterization that the matrix of 1-forms defining the connection has logarithmic poles with nilpotent residues.<|endoftext|> -TITLE: Every 4-manifold has a $Spin^c$ Structure -QUESTION [10 upvotes]: I'm having trouble understanding the proof given in Morgan's The Seiberg-Witten Equations that every 4-manifold $X$ admits a $Spin^c$ structure (Lemma 3.1.2). One can easily see from the exact sequence: -\begin{equation*} -H^1(X;Spin^c) \to H^1(X; SO(n)) \oplus H^1(X;\mathbb{Z}) \xrightarrow{c_1+w_2} H^2(X;\mathbb{Z}_2) -\end{equation*} -that a $Spin^c$ structure will exist iff $w_2(TX)$ lifts to an integral class, which we can check using Bockstein homomorphisms. After that, I'm lost; I'm not sure if these are theorems, or whether they are supposed to be obvious: - -In what sense is every $\mathbb{Z}/2^k \mathbb{Z}$ 3-class represented by a mapping from a smooth $\mathbb{Z}/2^k \mathbb{Z}$-manifold into $X$? -Why are integral 2-classes that represent torsion elements necessarily represented by embedded oriented surfaces? - -EDIT: Since the proof from Morgan's book is quite short, I may as well write out the whole thing here: -"We need only see that $w_2(X)$ lifts to an integral class $c \in H^2(X;\mathbb{Z})$ in order to prove the existence of a $Spin^c$ lifting. But for any class $x \in H_2(X;\mathbb{Z}/2 \mathbb{Z})$ the value of $w_2(X)$ on $x$ is given as follows: represent $x$ as an embedded (possibly non-orientable) closed surface in $X$ and take the self-intersection of this surface modulo two. To see that $w_2(X)$ lifts to an integral class, we must see that its integral Bockstein $\delta w_2(X)$ is zero. But this torsion integral class is zero if and only if it evaluates trivially on every $\mathbb{Z}/2^k \mathbb{Z}$ class of dimension three. Any such class is represented by a mapping of a smooth $\mathbb{Z}/2^k \mathbb{Z}$-manifold into $X$. The value of $\delta w_2(X)$ on such a class is equal to the value of $w_2(X)$ on the Bockstein of this $\mathbb{Z}/2^k\mathbb{Z}$-manifold. Thus, we need only see that $w_2(X)$ vanishes on integral classes which represent torsion elements in $H_2(X;\mathbb{Z})$. But this is clear, any such class is represented by a smoothly embedded oriented surface with zero self-intersection". -I suppose what I'm really asking is which statements in this proof are non-trivial theorems about the topology of 4-manifolds, and which ones are supposed to be obvious? -Sorry, I don't seem to be able to comment, so I'll just say here: Ryan Budney, I hope this makes the question less vague, and Anton Fetisov, yes, there are other proofs of this fact that I do understand, but I'm specifically trying to understand this proof, because it seems very slick. - -REPLY [3 votes]: First let me stress the importance of Anton Fetisov's comment that $X$ must be orientable. Indeed, this seems to be used at nearly every step of the proof you cite. -For example, consider the claim "for any class $x\in H_2(X;\mathbb{Z}/2\mathbb{Z})$ the value of $w_2(X)$ on $x$ is given as follows: represent $x$ as an embedded (possibly non-orientable) closed surface in $X$ and take the self-intersection of this surface modulo two." This can fail if $X$ is non-orientable. For example, take $X=\mathbb{R}P^2\times\mathbb{R}P^2$, which has $w_2(X)=a^2\times 1 + a\times a + 1\times a^2$ where $a\in H^1(\mathbb{R}P^2;\mathbb{Z}/2\mathbb{Z})$ is the generator. Consider the homology class $x\in H_2(X;\mathbb{Z}/2\mathbb{Z})$ represented by the embedding $\iota:\mathbb{R}P^2\hookrightarrow \mathbb{R}P^2\times\mathbb{R}P^2$ of the first factor. Then -$$ -\langle w_2(X),x\rangle =\langle w_2(X),\iota_*[\mathbb{R}P^2]\rangle = \langle \iota^*w_2(X),[\mathbb{R}P^2]\rangle = \langle a^2,[\mathbb{R}P^2]\rangle \equiv 1, -$$ -whereas it is clear that the self-intersection is zero in this case. -Now to your actual questions: - -Here a $\mathbb{Z}/2^k\mathbb{Z}$-manifold means a manifold which is oriented with respect to singular homology with $\mathbb{Z}/2^k\mathbb{Z}$ coefficients. When $k=1$ this is all manifolds, and the claim reduces to Thom's positive answer to the mod $2$ Steenrod realizability problem. When $k>1$, a $\mathbb{Z}/2^k\mathbb{Z}$-manifold is an orientable manifold $M^m$ with a choice of fundamental class $[M]\in H_m(M;\mathbb{Z}/2^k\mathbb{Z})$. -By Thom's results, a homology class $a\in H_3(X;\mathbb{Z}/2^k\mathbb{Z})$ is represented by an embedded $\mathbb{Z}/2^k\mathbb{Z}$-manifold if and only if its Poincaré dual in $H^1(X;\mathbb{Z}/2^k\mathbb{Z})$ is induced from the universal Thom class $t_{\mathbb{Z}/2^k\mathbb{Z}}\in H^1(MSO(1);\mathbb{Z}/2^k\mathbb{Z})$ under some map $f: X\to MSO(1)$. Now note that $MSO(1)\simeq S^1$, and that the Thom class induces an isomorphism $$t^*_{\mathbb{Z}/2^k\mathbb{Z}}:H^1(MSO(1);\mathbb{Z}/2^k\mathbb{Z})\cong H^1(K(\mathbb{Z}/2^k\mathbb{Z},1);\mathbb{Z}/2^k\mathbb{Z}).$$ It follows that every $a\in H_3(X;\mathbb{Z}/2^k\mathbb{Z})$ is represented by an embedded $\mathbb{Z}/2^k\mathbb{Z}$-manifold. - -(Note that orientability of $X$ was used here, to get Poincaré duality with $\mathbb{Z}/2^k\mathbb{Z}$ coefficients.) - -Again by orientability and Thom's results, every class in $H_2(X;\mathbb{Z})\cong H^2(X;\mathbb{Z})$ is represented by an embedded oriented surface $\Sigma\hookrightarrow X$. The torsion assumption is just used to get the final clause, that $\Sigma$ has zero self-intersection. This is because the self-intersection is Poincaré dual to cup square, and the square of a torsion class in $H^2(X;\mathbb{Z})$ is a torsion class in $H^4(X;\mathbb{Z})\cong \mathbb{Z}$, therefore is zero.<|endoftext|> -TITLE: Is $H_{et}^1(X,F) = H^1(\pi_1^{et}(X), F(\bar{k}))$ true? -QUESTION [8 upvotes]: Let $X$ be a smooth geometrically connected scheme over a field $k$ of characteristic 0 (but not necessarily algebraically closed, I can take it to be a number field). Let $F$ be a finite algebraic group over $k$. - -Is the following statement true: $H_{et}^1(X,F) = H^1(\pi_1^{et}(X), F(\bar{k}))$? - -REPLY [10 votes]: This is true, yes. More generally, if $X$ is a scheme and $F$ is a locally constant étale sheaf of finite abelian groups on $X$, then -$$ -H^1_{et}(X,F) = H^1(\Pi_1^{et}(X), \tilde F), -$$ -where $\Pi_1^{et}(X)$ is the étale fundamental pro-groupoid of $X$ and $\tilde F$ a certain local system on $\Pi_1^{et}(X)$ corresponding to $F$. This follows from Proposition 5.9 in Friedlander's book on étale homotopy [the assumption there that $X$ is locally noetherian is not needed when $F$ is finite], together with the fact that $H^1$ of a (pro-)space is the same as $H^1$ of its 1-truncation. -When $X$ is connected and $x$ is a geometric point of $X$, $\Pi_1^{et}(X)\simeq B\pi_1^{et}(X,x)$, so that local systems on $\Pi_1^{et}(X)$ can be identified with $\pi_1^{et}(X,x)$-modules. Under this identification, the local system $\tilde F$ is just $F(x)$ with its $\pi_1^{et}(X,x)$-action.<|endoftext|> -TITLE: Laplace-like / cofactor expansion for Pfaffian -QUESTION [7 upvotes]: Wikipedia presents a recursive definition of the Pfaffian of a skew-symmetric matrix as $$ \operatorname{pf}(A)=\sum_{{j=1}\atop{j\neq i}}^{2n}(-1)^{i+j+1+\theta(i-j)}a_{ij}\operatorname{pf}(A_{\hat{\imath}\hat{\jmath}}),$$ where $\theta$ is the Heaviside step function. I have failed to find a proper reference for this result. Is it standard or does it require proper citation? What should one cite? - -REPLY [5 votes]: Your formula can be found in -Fulton, William, Pragacz, Piotr: Schubert varieties and degeneracy loci, Lecture Notes in Mathematics 1689, Springer. xi, 148 p. (1998). ZBL0913.14016, -see in particular equation (D.1) p. 116. - -REPLY [5 votes]: Here is how you get the Pfaffian. $\newcommand{\bR}{\mathbb{R}}$ Consider the symplectic $2$-form -$$ \omega = dx^1\wedge dx^2+\cdots +x^{2n-1}\wedge x^{2n} \in\Omega^2(\bR^{2n}). $$ -Note that -$$ \frac{1}{n!}\omega^{\wedge n}:=\frac{1}{n!}\underbrace{\omega\wedge \cdots \wedge \omega}_n=dx^1\wedge dx^2\wedge \cdots \wedge dx^{2n-1}\wedge dx^{2n} $$ -Suppose that $A=(a_{ij})$ is a skew-symmetric $2n\times 2n$ matrix. None that $\newcommand{\be}{\boldsymbol{e}}$ -$$a_{ij}=(\be_i, A\be_j), $$ -where $(\be_i)$ is the canonical basis of $\bR^{2n}$. Consider now the $2$-form -$$ \omega_A=-\sum_{i -TITLE: If $X\times X$ is rational, must $X$ also be rational? -QUESTION [14 upvotes]: Is there an example of a smooth projective variety $X$ such that $X$ is irrational, but $X\times X$ is rational? -For instance, is $X\times X$ irrational for a smooth cubic threefold $X$? - -REPLY [21 votes]: For the first question I am not that pessimistic. At least there are candidates as follows: Recall that $Z$ is stably rational if there is $n\ge0$ such that $Z\times\mathbf A^n$ is rational. Now suppose there is such a $Z$ such that the minimal $n$ is $\ge2$. I would be extremely surprised if that didn't exist. Then put $Y=Z\times\mathbf A^{n-2}$ and $X=Y\times\mathbf A^1$. By assumption, $X$ is not rational but $X\times\mathbf A^1$ is. Let $d=\dim X\ge1$. Then -$$ -X\times X\cong Y\times \mathbf A^1\times X\cong Y\times\mathbf A^{d+1}\cong X\times\mathbf A^1\times\mathbf A^{d-1}\cong\mathbf A^{d+1}\times\mathbf A^{d-1}\cong\mathbf A^{2d} -$$ -A candidate for $Y$ would be the non-rational $3$-fold constructed by Beauville et al. for which Shepherd-Barron proved that $Y\times\mathbf A^2$ is rational. I don't know whether $Y\times\mathbf A^1$ is rational or not.<|endoftext|> -TITLE: Rate of convergence of uniform order statistics to their expectations -QUESTION [5 upvotes]: This is a problem that I encountered in my research and have no clues to fully -resolve it. Basically, I need large (or moderate) deviation bounds on the -difference between an order statistic of independent and identically -distributed (i.i.d.) random variables on the compact interval $\left[ -0,1\right] $ and the expectation of this order statistic. -Let $X_{1}% -,...,X_{n}$, $n\in\mathbb{N}$ be i.i.d. and uniformly distributed on $\left[ -0,1\right] $. Let their order statistics be $X_{\left( 1\right) }\leq -X_{\left( 2\right) }\leq...\leq X_{\left( n\right) }$, where we can ignore -the zero probability event that any two order statistics are equal. Let $E$ -and $V$ denote respectively the expectation operator and variance operator. -Then each $X_{\left( r\right) }$ follows a Beta distribution, -$$ -E\left[ X_{\left( r\right) }\right] =\frac{r}{n+1},r=1,...,n -$$ -and -$$ -V\left[ X_{\left( r\right) }\right] =\frac{r\left( n-r+1\right) -}{\left( n+1\right) ^{2}\left( n+2\right) },r=1,...,n -$$ -This implies two things: -(a) when $r=o\left( n\right) $ where the small $o$ notation means that -$\lim_{n\rightarrow\infty}\frac{r}{n}=0$, for any $\varepsilon>0,$ -\begin{equation} -P\left( \left\vert X_{\left( r\right) }-\frac{r}{n+1}\right\vert ->\varepsilon\right) \leq\frac{o\left( 1\right) }{n -\varepsilon^{2}},\label{eq1}% -\end{equation} -where $o\left( 1\right) $ denotes a nonnegative sequence that converges to -$0$ as $n\rightarrow\infty$; in this case, $\varepsilon=n^{-\alpha}$ with $0 \le \alpha < \frac{1}{2}$ can be -set, such that -\begin{equation} -P\left( \left\vert X_{\left( r\right) }-\frac{r}{n+1}\right\vert >\frac -{1}{n^{\alpha}}\right) \le \frac{o(1)}{n^{1-2\alpha}}\rightarrow0,\label{eq3}% -\end{equation} -where the $o\left( 1\right) $ can be set to be no smaller in order than rate -$\frac{r}{n}$. -(b) when $r=O\left( n\right) $ where the big $O$ notation here means that -$\liminf_{n\rightarrow\infty}\frac{r}{n}>0$, for any $\varepsilon>0$, -\begin{equation} -P\left( \left\vert X_{\left( r\right) }-\frac{r}{n+1}\right\vert ->\varepsilon\right) \leq C\frac{1}{\left( n+2\right) \varepsilon^{2}% -}\label{eq2}% -\end{equation} -from some constant $C\leq2$; in this case $\varepsilon=o\left( \sqrt -{n}\right) $ can be set such that -\begin{equation} -P\left( \left\vert X_{\left( r\right) }-\frac{r}{n+1}\right\vert >\frac -{1}{n^{\alpha}}\right) \leq\frac{2}{n^{1-2\alpha}}\rightarrow0\label{eq4}% -\end{equation} -for any $0\leq\alpha<\frac{1}{2}$. -Observation: -A simple conclusion from the above discussion is that, regardless of the value -of $r$, we have that $X_{\left( r\right) }$ for each $r$ converges to -$E\left[ X_{\left( r\right) }\right] $ in probability as $n\rightarrow -\infty$. Further, we know the rate of convergence is $n^{-\alpha}$ for any $0 \le \alpha < 1/2$. My question is "are the deviation bounds given above -the best?" Very likely NOT. -Question: -Let $a_{n,r}$ be a positive sequence -that depends on $n$ and $r$ such that -$$ -\lim_{n\rightarrow\infty}a_{n,r}=0 -$$ -for each $r=1,...,n$, what is the best result available on -$$ -\beta_{n,r} \ge P\left( \left\vert X_{\left( r\right) }-\frac{r}{n+1}% -\right\vert >a_{n,r}\right) -$$ -where $\beta_{n,r} \to 0$ as $n \to \infty$? -By this I mean, what is the bound $\beta_{n,r}$ corresponding to the sequence $a_{n,r}$ that converges to $0$ at relatively and possibly the fastest speed? -Any pointers or hints would be greatly appreciated! Thanks! -Update 1: (see update 3 below) -Thanks for Henry's comment. I found this: https://projecteuclid.org/euclid.ecp/1465263184, Concentration inequalities for order statistics. But this paper is mainly about concentration of order statistics of i.i.d. standard Gaussian random variables. The second paragraph in the Introduction of this paper quotes without a proof a general concentration of measure phenomenon for i.i.d. standard Gaussian random variables. If someone can point out to me a reference on how this result is obtained, that will be great. Since I guess I can reverse engineer this result from its proof to get a result for i.i.d. standard uniform random variables. -Update 3 (April 14, 2017): -By Theorem 2 of Chung 1949 "An estimate concerning the Kolmogoroff limit distribution, Trans. Amer. Math. Soc. 67: 36–50", we see -$$ -P\left( \limsup\limits_{n\rightarrow\infty}\dfrac{n\sup\limits_{t\in\mathbb{R}% -}\left\vert \mathbb{S}_{n}\left( t\right) -S_{\ast}\left( t\right) \right\vert -}{\left( 2^{-1}n\log_{\left( 2\right) }n\right) ^{1/2}}=1\right) -=1 -$$ -for any continuous CDF $S_{\ast}$ on $\mathbb{R}$ with $\mathbb{S}_{n}\ $ as its empirical CDF (ECDF), where $\log_{(s)}$ means the natural logarithm composed by itself $s$ times. Therefore, these order statistics converges at a rate as per the iterated logrithm. -In other words, when $n$ is very large, the classic location of each $X_{(r)}$ is $r/n$, with asymptotic deviation upper bounded by -$$\tag{1} -\frac{\sqrt{2 \log_{(2)}n}}{\sqrt{n}} -$$ -Basically, we know where these order statistics are. -However, this raises an interesting question as follows. Let $u_1 = X_{(1)}$, $u_2 = X_{(2)}-X_{(1)}$, $\ldots$, $u_k = X_{(k)}-X_{(k-1)}$, $\ldots$ , $u_n = X_{(n)}-X_{(n-1)}$, and $u_{n+1}=1-X_{(n)}$ be the uniform spacings. Further, let -$$ -u^{\ast} = \max_{1 \le k \le n+1} u_{k} -$$ -be the maximal uniform spacing. - Then by Devroye (1981, 1982) "Laws of the iterated logarithm for order statistics of uniform spacings" and "A log log law for maximal uniform spacings", we know -$$ -P\left( \limsup\limits_{n\rightarrow\infty}\frac{n{u} -^{\ast}-\log n}{2\log_{\left( 2\right) }n}=1\right) =1, -$$ -Namely, the maximal spacing is asymptotically no larger than -$$\tag{2} -\frac{\log n + 2 \log_{(2)} n}{n} -$$ -Devroye (1981, 1982) also provide the law of the iterative logarithm for the minimal uniform spacing $u_{\ast} = \min_{1 \le k \le n+1} u_k$. Specifically, -$$ -P\left( \liminf\limits_{n\rightarrow\infty}\left( n -{u}_{\ast}-\log n +\log_{\left( 3\right) }n\right) =-\log2\right) =1 -$$ -This means that the miminal spacing asympotitcally is no smaller than -$$\tag{3} -\frac{-\log 2 + \log n - \log_{(3)}n}{n} -$$ -With this piece of information, we can approximately locate all $X_{(r)}$, by first locating $X_{(n)}$, then $X_{(n-1)}$ and so on. -Question #2: -I am curious on the following: -Look at the largest order statistic $X_{(n)}$ and the smallest $X_{(1)}$. Then each of them has two different rates of convergence, $(1)$ and $(2)$, respectively to $0$ and $1$. Since both rates are correct, we see that the extreme statistics converge much faster than non-extreme ones. -Why is the maximal uniform spacing so small in magnitude compared to the maximal oscillation in the empirical distribution? Which rate of convergence would you use to identify the locations of uinform order statistics? (I am aware of one quick and intuitive answer, which is "on average the individual spacing should be around $n^{-1}$ and for the empirical distribution central limit theorem plays a role to give $n^{-1/2}$". But this does not seems to be somewhat convincing.) -Question #3 (April 16, 2017): -For the convergence of the order statistics to their classic locations, the first rate is based on deviation of empirical distribution, whereas the second based on uniform spacing. -This prompts me to think: is there a rate of convergence for uniform order statistics that is better than $(1)$ if we know that $F(t)=t$, i.e., is the rate in Chung's 1949 paper optimal for convergence of empirical distribution of uniform random variables? (Probably not since that result was obtained for the space of functions of totally bounded variation.) -Update (April 21, 2017): answer to question #3 -The convergence rate of uniform order statistics to their expecations obtained from $(1)$ is NOT optimal for order statisic $X_{(k)}$ for which $k$ or $n-k$ is bounded. However, it is optimal for $X_{(k)}$ for which $k= c_n n$ with $$1 > \limsup c_n \ge \liminf c_n >0.$$ In other words, close to the edges 0 or 1, the optimal reate of convergence is proportional to $$\frac{\log n}{n},$$ whereas in other parts of the compact interval $[0,1]$ the optimal rate of convergence is $$\frac{\sqrt{2 \log_{(2)}n}}{\sqrt{n}}.$$ This can be seen from: -Lemma A2.1 on page 148 of "asymptotic expansions for the power of distribution free tests in the one-sample problem" by Albers, Bickel and Zwet. - -REPLY [3 votes]: For your first question, Rigollet has a series of notes(with minor typos) that dicusses basics of this kind of tail bounds. The result you mentioned in the "introduction" is actually the classic Hoeffding bound that mentioned by Rigollet in this set of notes. - -High Dimensional Statistics, Philippe Rigollet (2015) - http://www-math.mit.edu/~rigollet/PDFs/RigNotes15.pdf - -If you are concerned with the order statistics, you usually want to restrict yourself to a narrower class of distributions, say the classic paper [2]. -For a general distribution family, it is almost impossible to obtain a tail-bound on the order statistics due to the concentration of measures phenomenon. Another keyword you may want to look into is "U-statistics" because the (full) order statistic is an example of U-statistics, the following quote from [1] is the best description of the power of U-statistics - -Theoretically, for these U-statistics we can study the whole spectrum - of asymptotic problems which were investigated for independent - variables. As a matter of fact, it is necessary to control the nature - of dependence in order to obtain meaningful results. We restrict - ourselves to exchangeable and weakly exchangeable variables, rank - statistics, samplings from finite populations, weakly dependent random - variables, bootstrap-variables, and to order statistics.[1]p.15 - -Concentration bounds on U-statistics is a research subject that involves many false claims and hardcore techniques, so I think it is not too hard to find one but never too careful to use one. - -Update in response to update3. -Why is the maximal uniform spacing so small in magnitude compared to the maximal oscillation in the empirical distribution? -This is not something surprising. you can always have very large spacings but small Kolomogorov-Smirnov norm, which measures on the space of $\mathcal{M}(\mathbb{R})$ while the spacing is measuring $\mathbb{R}$. If you look into Luc's argument, you will see his comments on it. -For the simplest example, given a set of data $\{0,0.1,0.9\cdots,0.9\text{(n repeated 0.9)},1\}$ and $\{0,0.1\cdots,0.1\text{(n repeated 0.1)},0.9,1\}$, their empirical measures has completely different cdfs but they both have the same maximal spacing of $0.8$. as the number $n$ of repeated observation increase, the KS norm between these two cdfs can be arbitrarily close to 1. -Reference -[1]Korolyuk, Vladimir S., and Yu V. Borovskich. Theory of U-statistics. Vol. 273. Springer Science & Business Media, 2013. -[2] Gupta, S. Das, et al. "Inequalities on the probability content of convex regions for elliptically contoured distributions." Proceedings of the Sixth Berkeley Symposium on Mathematical Statistics and Probability (Univ. California, Berkeley, Calif., 1970/1971). Vol. 2.1972.<|endoftext|> -TITLE: Graphs with only disjoint perfect matchings -QUESTION [20 upvotes]: Let $G(V,E)$ be a graph. I am searching for graphs with only disjoint perfect matchings (i.e. every edge only appears in at most one of the perfect matchings). -Examples: - -Cyclic graph $C_n$ with even $n$, with $m=2$ disjoint perfect matchings. -Complete graph $K_4$, with $m=3$ disjoint perfect matchings. - -I have three questions: - -How are such graphs called? -Are there other examples than $C_n$ and $K_4$? -What is the maximum number $m$ of perfect matchings, if the graph has only completly disjoint perfect matchings? - -For question 3, it seems to me that $K_4$ with $m=3$ different, disjoint perfect matchings is the optimum, but I have no proof for that. -Every hint to an answer or to relevant literature would be very much appreciated! -Edit: I am interested in undirected graphs only for the moment. -Edit2: The answer to this question I have used in a recent article in Physical Review Letters, where I cite this MO question as reference [24]. See Figure 2 for a detailed variant of the application of Ilya's answer. Thanks Ilya! - -REPLY [21 votes]: $m=3$ is indeed the maximum, and $K_4$ is the only example for this value of $m$. -Two perfect matchings form a disjoint union of cycles. If there is more than one cycle, then you may swap one of them, obtaining a third matching on the same edges. So any two of the $m$ matchings form a Hamiltonian cycle. -Assume that $m\geq3$; consider a Hamiltonian cycle $v_1,\dots,v_{2n}$ formed by the first two matchings, and check how the third one looks like. -If some its edge $(v_i,v_j)$ subtends an arc of odd length (i.e. if $i-j$ is odd), then we may split the vertices outside this arc into pairs, and split the cycle $v_i,\dots,v_j$ into edges including $(v_i,v_j)$, obtaining a matching intersecting the third one but not coinciding with it. This should not be possible; thus each subtended arc is even. -Now take an edge $(v_i,v_j)$ subtending minimal such arc, and consider an edge $(v_{i+1},v_k)$ of the third matching (going otside this arc). Now split the cycle $v_i,v_j,v_{j+1},\dots,v_k,v_{i+1}$ into edges containing $(v_i,v_j)$, and split all the remaining vertices into edges of the Hamiltonian cycle (it is possible, according to the parity). If $2n>4$, you again get a fourth matching sharing edges with the third ones bot diferent from it. -Thus $2n\leq 4$ and $m\leq 3$. - -REPLY [12 votes]: It looks like Ilya Bogdanov has answered your question, but here is a pointer to related literature. -An edge that appears in at most one perfect matching is known as a forcing edge. Che and Chen have written a survey of the literature on forcing edges and related topics.<|endoftext|> -TITLE: Generalization of Legendre`s conjecture -QUESTION [6 upvotes]: Legendre`s conjecture states that there is always a prime between $n^2$ and $(n+1)^2$ for every natural $n$. -It is natural to create following generalization: - -Is it true that for every $\varepsilon \in (0,1]$ there exists natural number $n_0(\varepsilon)$ such that for every $n \geq n_0(\varepsilon)$ there is always a prime between $n^{1+\varepsilon}$ and $(n+1)^{1+\varepsilon}$? - -I do not know if some results from number theory do not allow this generalization to be true but if it is not known is it true or not what is known, if anything, about this generalization of Legendre`s conjecture? - -REPLY [6 votes]: As mentioned in the remarks, this conjecture is equivalent to the statement that if $p$ is a prime, then the next prime satisfies $p'\ll_\delta p+p^\delta$ for any $\delta>0$. The parameter $\delta$ corresponds to the OP's $\epsilon/(1+\epsilon)$. For $\delta>1/2$ (which corresponds to $\epsilon>1$) this follows from Legendre's conjecture and also from the Riemann Hypothesis, for $\delta>21/40$ (which corresponds to $\epsilon>21/19$) it follows unconditionally from the work of Baker-Harman-Pintz. For arbitrary $\delta>0$, the statement follows from Cramér's conjecture. The bottom line is: we don't know the truth yet.<|endoftext|> -TITLE: Asymptotic formula for sums of four squares? -QUESTION [8 upvotes]: Does there exist asymptotic formula for ways to write n as sum of four squares? Or can this be proved impossible? I can only find reference for sums of five squares. - -REPLY [5 votes]: Stanley Yao Xiao gave a perfect answer, but let me remark that $r_4(n)$ also equals $n$ times the usual singular series (familiar from the circle method) when $4\nmid n$. The difference with five or more squares is that the singular series does not vary between two positive constants: instead, it varies between $1$ and a constant times $\log\log n$. Note also that $r_4(n)$ is very small when $n$ is divisible by a large power of $2$. -For more details, see Heath-Brown: A new form of the circle method, and its application to quadratic forms (J. reine angew. Math. 481 (1996), 149-206), especially Theorem 4 and Corollary 1.<|endoftext|> -TITLE: Chow ring of Hilbert scheme of 4 points in $\mathbb{P}^2$ -QUESTION [11 upvotes]: What is known about the Chow ring of the Hilbert scheme of length 4 subschemes of $\mathbb{P}^2$? -I know there is work on cycles on Hilbert schemes in the literature, but I don't know what can be deduced about this special case. - -REPLY [11 votes]: In principle, the Chow rings of Hilbert schemes of length $d$ subschemes in $\mathbb{P}^2$ are known (though it may still be a nontrivial task to extract information from the known descriptions). Here are some literature references. (Note that some of these talk about integral cohomology or homology, but because of the Bialynicki-Birula cell structure the cycle class map from Chow ring to integral cohomology ring is an isomorphism.) -There is one description of the Chow rings of Hilbert schemes in terms of the representation theory of the infinite-dimensional Heisenberg algebra. This is due to Nakajima and Grojnowski (after work of many people, check out the references in the papers) - -H. Nakajima: Heisenberg algebra and Hilbert schemes of points on projective surfaces. Ann. of Math. 145 (1997), 379-388. link to arXiv paper -I. Grojnowski: Instantons and affine algebras I. The Hilbert scheme and vertex operators. Math. Res. Lett. 3 (1996), 275-291. link to arXiv paper - -There is another description based on computations with equivariant cohomology in the following paper (which also contains an explicit computation for the Hilbert scheme of 3 points on $\mathbb{P}^2$): - -L. Evain: The Chow ring of punctual Hilbert schemes on toric surfaces. Transformation Groups 12 (2007), 227--249. link to arXiv paper - -A different basis for the integral cohomology (given by explicit geometric configurations) was given in the following paper - -R. Mallavibarrena and I. Sols. Bases for the homology groups of the Hilbert scheme of points in the plane. Compositio Math. 74 (1990), 169-201. link to numdam - -Section 5 of this paper also contains some computations of intersection products in the Chow ring of the Hilbert scheme of 4 points. This may be the most relevant for the question, showing that computations of intersection products can be made explicit. -There are also some lecture notes on these results: - -H. Nakajima: Lectures on Hilbert schemes of points on surfaces. University Lecture Series 18. Amer. Math. Soc. 1999. -G. Ellingsrud and L. Göttsche: Hilbert schemes of points and Heisenberg algebras. link to ICTP website - - -Edit: Some more references. There is of course - -L. Göttsche. Hilbert schemes of zero-dimensional subschemes of smooth varieies. Lectures Notes in Math. 1572. Springer 1994. - -In the references of Göttsche's book I found the following which also provides a computation of the intersection pairing on ${\rm CH}^4$, the middle dimension of the Chow ring for ${\rm Hilb}^4(\mathbb{P}^2)$ (results are obtained by using specific geometric features, not by specialization from results for general ${\rm Hilb}^d$). - -D. Avritzer and I. Vainsencher. ${\rm Hilb}^4(\mathbb{P}^2)$. In: Enumerative Geometry (Sitges 1987), Lecture Notes in Math. 1436. Springer 1990, pp. 30-59.<|endoftext|> -TITLE: What does positivity of the first Pontryagin number of a vector bundle tell us? -QUESTION [9 upvotes]: Some context: -In the theory of compact, oriented Riemannian Einstein 4-manifolds, there is a a fundamental topological constraint that is implied by the Einstein equations. To wit, if $\chi$ and $\tau$ denote the Euler characteristic and signature of a fixed 4-manifold $M$, then $$2\chi\pm3\tau=\textstyle\frac{1}{8\pi^2}\displaystyle\int_M\,\big(\displaystyle\frac{R^2}{12}-\mathring{|Rc|}^2+|W^{\pm}|^2 \big)\,dV$$ -for $any$ Riemannian metric on $M$. As the traceless Ricci tensor $\mathring{Rc}$ vanishes identically for an Einstein metric, these 2 (well, 1 if $\tau=0$) topological invariants are non-negative if an Einstein metric exists on $M$. This is known as the Hitchin-Thorpe inequality. -One can calculate that in fact $2\chi \pm 3\tau=p_1(\Lambda^2_{\pm})$ by using the connection which is induced on $\Lambda^2_{\pm}$ by the Levi-Civita connection, so the presence of an Einstein metric on $M$ means that the first Pontryagin number of the bundles $\Lambda^2_{\pm}$ is non-negative. -My question is plainly this: Sticking with a 4-manifold $M$, what does it mean in a tangible, geometric sense, to say that a vector bundle over $M$ has non-negative/positive first Pontryagin number? Or perhaps more generally, what does the quantity $p_1$ in fact quantify? Does it say something about generic sections of the bundle (I vaguely recall that the Steifel-Whitney classes quantify something like this), or something else palpable? -I'd be perfectly happy if one were to restrict discussion to the bundles $\Lambda^2_{\pm}$ above or maybe general rank-3 bundles if either simplifies the discussion. - -REPLY [6 votes]: Suppose $M^4$ is a simply-connected 4-manifold, and $V$ is a 4-dimensional real vector bundle over $M$. Then $V$ is classified by a map $M \to BSO(4)$. Rationally, the cohomology of this space has the following form: -$$H^{\ast}(BSO(4);\mathbb Q) = \mathbb Q[p_1,e].$$ -Here $p_1$ is the first Pontrygion class, and $e$ is the Euler class (the square of this class is equal to the second Pontryagin class $p_2$). The obstruction for $V$ to admit a non-vanishing section that you mention above is given by $e$, not by $p_1$. -In fact we can change $V$ in such a way that $e$ becomes $0$, but $p_1$ does not change. If we do this, we can split off a line bundle, which corresponds to lifting the classifying map to a map $M \to BSO(3)$. Now there is (since $M$ is simply connected) no obstruction to split off another line bundle, we thus finally get that $V$ is classified by a map $V \to BSO(2) = \mathbb CP^{\infty}$, so in fact $V$ comes from a complex line bundle $\xi$. This complex line bundle has a Chern class $c_1(\xi) \in H^2(M),$ which is the obstruction to trivializing it. By definition, $p_1(V) = -c_1(\xi)^2 \in H^4(M)$. I hope this satisfies your request for understanding $p_1$ geometrically... -In fact, the difference between $e$ and $p_1$ is on e of the things that Milnor exploits in his famous paper "On manifolds homeomorphic to the 7-sphere" (I highly recommend reading this paper!), which was what he got the Fields medal for.<|endoftext|> -TITLE: Fibonacci series captures Euler $e=2.718\dots$ -QUESTION [35 upvotes]: The Fibonacci recurrence $F_n=F_{n-1}+F_{n-2}$ allows values for all indices $n\in\mathbb{Z}$. There is an almost endless list of properties of these numbers in all sorts of ways. The below question might even be known. Yet, if true, I like to ask for alternative proofs. - -Question. Does the following identity hold? We have - $$\frac{\sum_{k=0}^{\infty}\frac{F_{n+k}}{k!}}{\sum_{k=0}^{\infty}\frac{F_{n-k}}{k!}}=e,$$ - independent of $n\in\mathbb{Z}$. - -REPLY [18 votes]: More generally, -$$\sum_{k=0}^\infty F_{n+k} \frac{x^k}{k!} = e^x\sum_{k=0}^\infty F_{n-k}\frac{x^k}{k!},\tag{1}$$ -which is equivalent to Will Sawin's identity. -Similarly, -$$e^x\sum_{k=0}^\infty F_{n+k}\frac{x^k}{k!}= \sum_{k=0}^\infty F_{n+2k}\frac{x^k}{k!},\tag{2}$$ -and more generally, from the formula -$$(-1)^q F_{p-q}+F_{\!q}\,\phi^p = F_{\!p}\,\phi^q,$$ -where $\phi = (1+\sqrt5)/2$, and the analogous formula with $\phi$ replaced by $(1-\sqrt5)/2$, we get for any integers $p$, $q$, and $n$, -$$ -e^{(-1)^q F_{p-q}\ x}\sum_{k=0}^\infty F_{q}^k F_{n+pk}\frac{x^k}{k!} - = \sum_{k=0}^\infty F_p^k F_{n+qk} \frac{x^k}{k!}. -$$ -The cases $p=1, q=-1$ and $p=1,q=2$ give $(1)$ and $(2)$. -Related identities (also involving Lucas numbers) can be found in -L. Carlitz and H. H. Ferns, -Some Fibonacci and Lucas Identities, -Fibonacci Quarterly 8 (1970), 61–73.<|endoftext|> -TITLE: For a round-robin tournament, what is the favorite's least favorite size? -QUESTION [28 upvotes]: Suppose we have a round-robin tournament (i.e., each player plays exactly one game with each other player) with $n$ players, who are all equally skillful except for one player, the favorite, whose probability of winning a game against any other player is some fixed value $p > 1/2$. Assume that all games are independent, and also that no individual game ends in a tie. The winner of the tournament is the player who wins the most games; let us assume that if several players are tied for first place, then one of them is chosen (uniformly) at random to be the winner. Let $\pi(p,n)$ be the probability that the favorite wins the tournament. -Fact. For any fixed $p>1/2$, $\lim_{n\to\infty} \pi(p,n) = 1$. -Sketch of proof: The probability that a given ordinary player scores higher than the favorite goes to zero at a rate that is exponentially fast in $n$ (by, e.g., Hoeffding's inequality), but there are only $n$ competitors, so even a union bound suffices to show that the probability that any player scores higher than the favorite goes to zero exponentially fast. -In light of the above Fact, it may be slightly surprising that for a fixed $p$, especially for $p$ close to $1/2$, the value of $\pi(p,n)$ actually declines for a while (as $n$ increases), and I think it may even wobble around, before eventually climbing to 1. -Intuitively, what's happening for small $n$ is that the increase in the number of competitors is increasing the chances that one of them will do well and upset the favorite, and that this is initially a more important effect than the fact that the increase in the number of games is giving the favorite an opportunity to demonstrate a skill edge. -This has led me to consider the following question. - -Let $N(\epsilon)$ denote the value of $n$ that minimizes $\pi(1/2 + \epsilon, n)$. What can be said about $N(\epsilon)$ as $\epsilon\to0$? - -Presumably, $\lim_{\epsilon\to0} N(\epsilon) = \infty$, but approximately how fast? - -REPLY [11 votes]: I can show that $N(\epsilon)$ is equal to $\epsilon^{-2}$ up to a log factor on each side. -The strategy I'll use is to give an upper bound for $\pi(1/2+\epsilon,n)$. Optimizing it, we obtain an upper bound for $\pi(1/2+\epsilon,N(\epsilon))$. Then using lower bounds for $\pi(1/2+\epsilon,n)$ we can rule out certain values of $n$ as being $N(\epsilon)$. -For any $m \geq\epsilon n$, we have the upper bound $$ \pi(1/2+\epsilon,n) \leq \frac{(1+2\epsilon)^{n/2+m} (1-2\epsilon)^{n/2-m} }{n} + e^{ - 2 (m-\epsilon n)^2/n}$$ -Indeed by Hoeffding's inequality the second term is an upper bound on the probability of getting greater than $m$ wins, so it suffices to prove the first term is an upper bound on the probability of winning the tournament with at most $m$ wins. However, for each possible outcome of the tournament (specifying the winner of every match) involving at most $m$ wins for a certain player, the probability of getting it when that player is the a favorite divided by the probability of getting it if that player is the same as all the others is at most $(1+2 \epsilon)^m (1-2\epsilon)^{n-m}$ (just multiply out the probability). Hence the probability of winning the tournament with at most $m$ wins as the favorite is at most $(1+2 \epsilon)^m (1-2\epsilon)^{n-m}$ times the probability of winning the tournament as an average player and at most $m$ wins, which is at most the probability of winning the tournament as an average player, which is $< 1/n$. -We can further estimate $$ (1+2\epsilon)^{n/2+m} (1-2\epsilon)^{n/2-m} = e^{ 4 \epsilon m - \frac{(2\epsilon)^2}{2} n + O( \epsilon^3 n -) }$$ -So setting $m = \epsilon n + \sqrt{n \log n /2}$, the second term is $1/n$ and the first term is $$\frac{ e^{2 \epsilon^2 n + 2 \epsilon -\sqrt{2 n \log n}} }{n}$$ so choosing $n$ to be approximately $\epsilon^{-2}/ \log (\epsilon^{-1})$, the exponent is $O(1)$, so $\pi(1/2+\epsilon,n)=O(1/n) = O( \log (\epsilon^{-1})/ \epsilon^{-2})$. -Thus $\pi(1/2+\epsilon,N(\epsilon)) = O( \log (\epsilon^{-1})/ \epsilon^{-2})$. -Now using the lower bound $\pi(1/2+\epsilon, n) \geq 1/n$, we obtain $N(\epsilon) \geq C \epsilon^{-2} / \log (\epsilon^{-1})$ for some constant $C$. -To upper bound $N(\epsilon)$, we use a different lower bound. By Hoeffding's inequality, the probability that either some player scores above $(1/2 + \epsilon/2 n)$ or the favorite scores below $(1/2 + \epsilon/2 n)$ is at most $e^{ - \epsilon^2 n /2}$, so $\pi(1/2+\epsilon,n) \geq 1 -n e^{-\epsilon^2 n/2}$. In particular, because $\pi(1/2+\epsilon,N(\epsilon))=o(1)$ then $ \epsilon^2 N(\epsilon)/2 \geq \log N(\epsilon) - o(1)$, so $N(\epsilon)/\log N(\epsilon) \geq 2 \epsilon^{-2} (1-o(1))$ and hence $N(\epsilon) \geq 4 \epsilon^{-2} \log (\epsilon^{-1}) (1+o(1))$.<|endoftext|> -TITLE: Is every rational realized as the Euler characteristic of some manifold or orbifold? -QUESTION [23 upvotes]: Let me first ask the question for two-dimensional compact, connected manifolds and orbifolds. -Then, if the answer is No, one can remove various conditions on the dimension, -and allow non-compact examples and disconnected examples, to realize a (perhaps) wider range of rationals. -This came up after a class I'm teaching and I didn't know the answer. -Related: - -MO question "Euler characteristic of orbifolds." -Wikipedia table for 2-dim orbifolds - -REPLY [34 votes]: Products of 2-orbifolds with manifolds will do the trick. There are 2-orbifolds of Euler characteristic $1/n$ (take a quotient of $S^2$ by a rotation of order $2n$). Then take a product with a manifold of Euler characteristic $m \in \mathbb{Z}$ to get all rationals.<|endoftext|> -TITLE: If a right adjoint to the product functor exists, must it be the diagonal? -QUESTION [19 upvotes]: Let $C$ be a category with binary products. The product functor $\times : C^2 \to C$ is right adjoint to the diagonal $\Delta: C \to C^2$. If $C$ has biproducts, then $\times$ is also left adjoint to $\Delta$. But from the fact that $\times$ is left adjoint to some functor $R$, can we conclude that $R = \Delta$? -In the case of nullary products, the answer is yes: if the terminal object $1: C^0 \to C$ has a right adjoint $R$, there's only one functor $C \to C^0$, so of course $R$ coincides with the diagonal $\Delta$. So a terminal object is also initial as soon as the functor $1: C^0 \to C$ has a right adjoint. -To say that $\Delta$ is right adjoint to $\times$ is to say that $\times$ is also a coproduct functor. I believe that if the binary product is also a binary coproduct, this implies that the $C$ is canonically enriched in pointed sets, with the "0" points on the homsets providing the structure maps for the "identity matrix" natural isomorphism $\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}: \amalg \Rightarrow \times$. So in this case $C$ almost has biproducts (except that the 0 object might not actually be representable). -The picture I have in mind is the string of adjoints $\amalg \dashv \Delta \dashv \times$, and the question "when can this adjoint string be extended further?", which leads to the question "if the adjoint string extends further, is it always (as in familiar examples) periodic of period 2?". - -REPLY [9 votes]: Building on Garlef Wegart's work, the answer is yes in general. The idea is to show that if $C$ has a right adjoint to $\times$, then it is enriched in pointed sets, so that $C$ embeds fully faithfully in a category $C_\ast$ with a zero object and the easy argument can be applied. And as noted already, in this case $C$ will have biproducts with respect to a (necessarily unique) enrichment in pointed sets. -Suppose that $R= (R_1, R_2): C \to C \times C$ is right adjoint to $\times : C \times C \to C$. Then $\pi_1 : X \times Y \to X$ is natural in $X$ and $Y$. By adjointness, this corresponds to $\pi_{11}: X \to R_1 X$ and $\pi_{12}: Y \to R_2 X$ natural in $X$ and $Y$. Similarly, $\pi_2: Y \times X \to X$ gives us $\pi_{21} : Y \to R_1 X$ and $\pi_{22}: X \to R_2 X$. Thus we get $(\pi_{12}, \pi_{21}): Y \to R_2 X \times R_1 X$ natural in $X$ and $Y$, which we can compose (after swapping the factors) with the counit $R_1 X \times R_2 X \to X$ to obtain a map $Y \to X$, natural in $X$ and $Y$. -That is, $C$ admits a (necessarily unique) enrichment in pointed sets. So there is a canonical way to add a zero object, and the resulting inclusion $C \to C_\ast$ is fully faithful. Moreover, $C_\ast$ still has products, computed as in $C$ with $X \times 0 = 0 \times X = X$. One can extend $R$ by hand to a functor $R_\ast: C_\ast \to C_\ast \times C_\ast$ by setting $R_\ast(0) = (0,0)$, and then check by hand that $R_\ast$ is still right adjoint to $\times_{C_\ast}$. (Conceptually, zero objects are absolute for $\mathsf{Set}_\ast$-enrichment, and the forgetful functor from $\mathsf{Set}_\ast$-enriched categories to $\mathsf{Set}$-enriched categories creates products and adjunctions.) -Hence the argument I gave in the comments applies: $X \times Y = (X + 0) \times (0 + Y) = (X \times 0) + (0 \times Y) = X + Y$ in $C_\ast$, because $\times$, as a left adjoint, must preserve colimits (note that the coproduct $(X,0) + (0,Y) = (X,Y)$ holds automatically in $C_\ast$; we need not assume the existence of coproducts). Hence $\times_{C_\ast} = +_{C_\ast}$ and taking adjoints we have $R_\ast = \Delta_{C_\ast}$. By restricting along the fully faithful inclusion $C \to C_\ast$, we get $R= \Delta_C$ as desired.<|endoftext|> -TITLE: Computing the equivariant cohomology of a specific $(\mathbb{Z}/2\mathbb{Z})^2$-space -QUESTION [7 upvotes]: In the paper On the Castelnuovo-Mumford regularity of the cohomology ring of a group, Symonds describes the following space. -Let $G = (\mathbb{Z}/2\mathbb{Z})^2 = \{1,a,b,ab\}$ be an elementary abelian $2$-group, and let $Z = \{ z_1, \dots, z_6\}$ be a discrete space with six elements. Let $G$ act on $Z$ such that each rank one subgroup of $G$ is a stabilizer of some element of $Z$. If I'm not mistaken, this means that the action is for example given as follows: -$$a \cdot z_1 = b \cdot z_1 = z_2;\\ -a \cdot z_3 = z_4; \; b \cdot z_3 = z_3; \; b \cdot z_4 = z_4;\\ -b \cdot z_5 = z_6; \; a \cdot z_5 = z_5; \; a \cdot z_6 = z_6.$$ -Let $SZ$ be the suspension of $Z$, with the induced $G$-action. -Question. How to compute the equivariant cohomology $H^*_G(SZ)$ as an $H^*_G$-module (with coefficients $\mathbb{F}_2$)? -I believe I could compute the equivariant cohomology $H^*_G(Z)$, using the Mayer–Vietoris sequence for equivariant cohomology (it split as a disjoint union of three $G$-spaces), the Künneth theorem, and the known results about the equivariant cohomology of spaces with a free (resp. trivial) action. But unfortunately there's no suspension isomorphism for equivariant cohomology, as far as I can tell... -Of course I know the definition of equivariant cohomology; the classifying space $BG$ is $(\mathbb{RP}^\infty)^2$, and $H^*_G = \mathbb{F}_2[x,y]$ is a polynomial algebra on two variables of degree $1$. The total space of the universal bundle is $(S^\infty)^2$ with $a$ (resp. $b$) acting by the antipodal action on the first (resp. second) factor, and $H^*_G(SZ) = H^*(EG \times_G SZ)$. But this isn't exactly helpful, or at least I don't see how to compute the equivariant cohomology just from that. - -REPLY [6 votes]: The standard way to compute equivariant cohomology of a $G$-space $X$ is to use the spectral sequence of the fibration -$$X\to EG\times_G X\to BG,$$ -where the projection is induced by $X\to \ast$. With $\mathbb{F}_2$ coefficients this takes the form -$$ -E_2^{p,q} = H^p(BG; H^q(X;\mathbb{F}_2))\Rightarrow H^*_G(X;\mathbb{F}_2). -$$ -In your case, the space $X=SZ$ is connected and one-dimensional, so there are only two non-trivial rows: $q=0$ (which just gives $H^*(BG;\mathbb{F}_2) = H^*_G$) and $q=1$ (the cohomology of $G=\mathbb{Z}/2\times\mathbb{Z}/2$ with coefficients in the module $H^1(SZ;\mathbb{F}_2)$). Hence the $E_2$-term is computable. -There is only one possible non-trivial differential $$d_2:H^p(BG;H^1(SZ;\mathbb{F}_2))\to H^{p+2}(BG;H^0(SZ;\mathbb{F}_2))=H^{p+2}(BG;\mathbb{F}_2).$$ -However, the fact that $SZ$ has $G$-fixed points (the suspension points) implies that the above fibration admits a section, and so the induced map $H^*(BG;\mathbb{F}_2)\to H^*_G(SZ;\mathbb{F}_2)$ is injective. This agrees with the edge homomorphism of the spectral sequence, implying that $d_2$ is in fact trivial. Hence $E_2=E_\infty$. There are no extension problems, since we are using field coefficients, so this gives the equivairiant cohomology as a $\mathbb{F}_2$-module. -I believe it also gives that $H^*_G(SZ;\mathbb{F}_2)$ is free as an $H^*_G$-module.<|endoftext|> -TITLE: What does it mean that $[X]+[Y]=0$ in the Grothendieck ring of varieties? -QUESTION [11 upvotes]: This is a really basic question. If I have two non-isomorphic varieties $X$ and $Y$, is it possible that $[X]+[Y]=0$ in the Grothendieck ring? -If so, what does this mean geometrically? Obviously $[\emptyset]-[X]-[Y]$ is not one of the standard relations modded into the ring, so $[X]+[Y]$ has to be some non-trivial combination of such relations. I'm having trouble seeing how this could happen though. An example would be especially nice. - -REPLY [15 votes]: The element $[X]+[Y]$ of $K_0({\rm Var}_k)$ is the class of the disjoint union $X\cup Y$. So your question amounts to understand whether the class of a variety, say $X$, can vanish in $K_0({\rm Var}_k)$. -As is implicit in the questions, and in the other answers, the ring $K_0({\rm Var}_k)$ is hard to understand, and is essentially only understood thanks to a few geometrically motivated ``motivic measures'': - -If $k$ is finite, the counting measure $N\colon K_0({\rm Var}_k)\to\mathbf Z$; -The Euler characteristic using any cohomology theory with compact supports you can imagine; -If $k=\mathbf C$, the Hodge-Deligne polynomial, valued in $\mathbf Z[u,v]$; -Using Deligne's theory of weights on étale cohomology, one can define a Poincaré polynomial, valued in $\mathbf Z[t]$; -If $k$ is a field of characteristic $0$, Larsen and Lunts have defined an exotic — and useful — motivic measure with valued in the free abelian group of stable birational class of varieties. - -In any case, any of these motivic measures shows that the class of a non-empty variety does not vanish. They can also be useful to show that the classes of two varieties are not equal.<|endoftext|> -TITLE: Existence of a 2-labelled Hamiltonian Path decomposition of $K_{2n}$ -QUESTION [6 upvotes]: I am trying to see if, for the complete graph $K_{2n}$, there exists a labelling of the vertices with two labels $a$ and $b$ (each used exactly $n$ times), such that we can decompose the graph into $n$ hamiltonian paths that have the same labelling. -For example, if I take n=3, I numerate my vertices from $1$ to $6$, and associate respectively the labels $a$,$b$,$b$,$a$,$a$,$b$. Then, I take the three following paths : -$1-2-3-4-5-6$ -$4-6-2-5-1-3$ -$5-3-6-1-4-2$ -The three paths have the same labelling and they are edge disjoint so it works. -I was able to prove the existence for $n$ odd, using Walecki's construction. For $n$ even, I know Walecki's construction does not work as the labels of the two extremities of the paths have to be different, but as I understand it is not the only construction I can use, so it does not necessarily mean that the labelling does not exist. I am leaning towards the idea that it does not exist, with the example $n=2$, but I don't see any theoretical argument to prove that. Would some of you have any ideas ? Thank you in advance. - -REPLY [4 votes]: I think I have a proof that such a labelling cannot exist if $n$ is even. -Suppose we have a labelling $\ell : V(K_m) \to \{ a, b \}$ and a decomposition of $K_{m}$ into a family $\mathcal{P}$ of Hamiltonian paths. It is clear by counting edges that $p := |\mathcal{P}| = m/2$, so in particular $m$ must be even. Every path $P = v_1 \ldots v_m$ in $K = K_{m}$ has a trace $\ell(v_1) \ldots \ell(v_m) \in \{ a, b \}^m$. We assume that every path in $\mathcal{P}$ has the same trace, which we denote by $T$. -Let $A$ be the set of all vertices that receive label $a$ and $B$ the set of those that receive label $b$. Suppose that neither $A$ nor $B$ is empty. -Counting the edges incident to a fixed vertex, it is easy to see that every vertex is the end vertex of precisely one $P \in \mathcal{P}$. In particular, every $P \in \mathcal{P}$ has one end in $A$ and the other in $B$ and $|A| = |B| = p$ (without loss of generality, all paths start in $A$ and end in $B$). -Edited with a simplification: -Every $P \in \mathcal{P}$ has the same number -$$ -e_A := | \{ 1 \leq i < n : T_i = T_{i+1} = a \} | -$$ -of edges within $A$. Then, since every edge of $K[A]$ lies in precisely one $P \in \mathcal{P}$, we get -$$ -p(p-1)/2 = |E(K[A])| = pe_A -$$ -or $e_A = (p-1)/2$, from which it follows that $p$ must be odd.<|endoftext|> -TITLE: Can a row of five equilateral triangles tile a big equilateral triangle? -QUESTION [89 upvotes]: Can rotations and translations of this shape - -perfectly tile some equilateral triangle? - -I originally asked this on math.stackexchange where it was well received and we made some good progress. Here's what we learnt: - -Because the area of the triangle has to be a multiple of the area of the tile, the triangle must have side length divisible by $5$ (where $1$ is the length of the short edges of the tile). -The analogous tile made of three equilateral triangles can tile any equilateral triangle with side length divisible by three. -There is a computer program, Burr Tools, which was designed to solve this kind of problem. Josh B. used it to prove by exhaustive search that there is no solution when the side length of the triangle is $5$, $10$, $15$, $20$ or $25$. The case of a triangle with side length $30$ would take roughly ten CPU-years to check using this method. -Lee Mosher pointed me in the direction of Conway's theory of tiling groups. This theory can be used to show that if the tile can cover an equilateral triangle of side length $n$ then $a^nb^nc^n=e$ in the group $\left$. But sadly it turns out that we do have that $a^nb^nc^n=e$ in this group whenever $n$ divides by $5$. -In fact one can use the methods in this paper of Michael Reid to prove that this tile's homotopy group is the cyclic group with $5$ elements. I think this means that the only thing these group theoretic methods can tell us is a fact we already knew: that the side length must be divisible by $5$. -These group theoretic methods are also supposed to subsume all possible colouring arguments, which means that any proof based purely on colouring is probably futile. -The smallest area that can be left uncovered when trying to cover a triangle of side length $(1,\dots,20)$ is $($$1$$,\,$$4$$,\,$$4$$,\,$$1$$,\,$$5$$,\,$$6$$,\,$$4$$,\,$$4$$,\,$$6$$,\,$$5$$,\,$$6$$,\,$$4$$,\,$$4$$,\,$$6$$,\,$$5$$,\,$$6$$,\,$$4$$,\,$$4$$,\,$$6$$,\,$$5$$)$ small triangles. In particular it's surprising that when the area is $1\;\mathrm{mod}\;5$ one must sometimes leave six triangles uncovered rather than just one. -We can look for "near misses" in which all but $5$ of the small triangles are covered and in which $4$ of the missing small triangles could be covered by the same tile. There's essentially only one near miss for the triangle of side $5$, none for the triangle of side $10$ and six (1,2,3,4,5,6) for the triangle of side $15$. (All other near misses can be generated from these by rotation, reflection, and by reorienting the three tiles that go around the lonesome missing triangle.) This set of six near misses are very interesting since the positions of the single triangle and the place where it "should" go are very constrained. - -I'd also be interested in learning what kind of methods can be used to attack this sort of problem. Are there any high-level approaches other than the tiling groups? Or is a bare hands approach most likely to be successful? - -REPLY [31 votes]: Since nobody has posted it, here's the smallest triangle tilable by the 'straight pentiamond', ie a side-30 triangle. Simple backtracking program, takes 0.5 seconds to show no tilings of the side-20 triangle, 4:39 for the side 25. Took 12 minutes to find 120 tilings for the side-30 before I stopped it.<|endoftext|> -TITLE: An explicit isomorphism between the 1st Cech cohomology and the 1st hypercohomology -QUESTION [8 upvotes]: Let $\mathbf{X}$ be a Grothendieck topos and let $A$ be an abelian group in $\mathbf{X}$. -Verdier's Theorem allows one to describe $\mathrm{H}^n(\mathbf{X},A)$ in terms of hypercoverings, namely, as the colimit of $\mathrm{H}^n(U_\bullet,A)$ where $U_\bullet$ ranges over the category of hypercoverings (of the terminal object of $\mathbf{X}$). -One can further form the Cech cohomology $\check{\mathrm{H}}^n(\mathbf{X},A)$, which is the colomit of $\mathrm{H}^n(U_\bullet,A)$ as $U_\bullet$ ranges over hypercoverings with $U_\bullet=\mathrm{cosk}_0(U_\bullet)$. Let me call them Cech hypercoverings. (These are the hypercoverings with $U_n=U_0^{\times (n+1)}$ and the standard simplicial structure.) -A well-known spectral sequence relates the derived-functor cohomology with the Cech cohomology, implying in particular that $\mathrm{H}^1(\mathbf{X},A)$ is canonically isomorphic to $\check{\mathrm{H}}^1(\mathbf{X},A)$. I am not absolutely certain, but it seems correct that under the description provided by Verdier's Theorem, the isomorphism $\check{\mathrm{H}}^1(\mathbf{X},A)\to \mathrm{H}^1(\mathbf{X},A)$ is the obvious one, namely, if a Cech cohomology class is represented by a $1$-cocycle in $Z^1(U_\bullet,A)$, then its image in $\mathrm{H}^1(\mathbf{X},A)$ is the cohomology class represented by that $1$-cocycle. -The latter statement means that for a hypercovering $U_\bullet$ (not necessarily Cech) and any $1$-cocycle $\alpha\in Z^1(U_\bullet,A)$, one can find a Cech hypercovering $U'_\bullet$ and $\alpha'\in Z^1(U'_\bullet,A)$ representing the same cohomology class. -I am looking for a way to construct these $U'$ and $\alpha'$ directly. -The main problem is that $U_\bullet$ cannot be refined to a Cech hypercovering in general. -An alternative approach to the problem (which also applies to non-abelian $A$) is via the correspondence with $A$-torsors. It is a standard fact that $\check{H}^1(\mathbf{X},A)$ is in 1-1 correspondence with isomorphism classes of $A$-torsors. Explicitly, if $\alpha\in Z^1(U_\bullet,A)$ with $U_\bullet$ being a Cech hypercovering, then the $A$-torsor $P$ corresponding to $\alpha$ can be described by -$$ -P(V)=\{a\in A(U_0\times V)~:~ \alpha_V \cdot d_0^1a =d_1^1 a\} -$$ -where $d^1_i:A(U_0\times V)\to A(U_1\times V)=A(U_0\times U_0\times V)$ is induced by $d^1_i:U_1\to U_0$. -Suppose now that $U_\bullet$ is an arbitrary hypercovering (not necessarily Cech). The question will be resolved if one can show that $P$ constructed above is still an $A$-torsor. The difficult thing to check is that $P\neq \emptyset$. (However, when $U_\bullet$ is Cech, one can check that $\alpha\in P(U_0)$.) - -REPLY [2 votes]: Some hints in the literature led me to an answer, which I find a bit surprising: -One can take $U'_\bullet=\mathrm{cosk}_0(U_\bullet)$ and the $1$-cocycle $\alpha\in Z^1(U_\bullet,A)\subseteq A(U_1)$ descends uniquely to a $1$-cocycle in $\alpha'\in Z^1(\mathrm{cosk}_0(U_\bullet),A)\subseteq A(U_0\times U_0)$ along $(d_0,d_1):U_1\to U_0\times U_0$. In other words: -Proposition: For any hypercovering $U_\bullet$, the canonical map $Z^1(\mathrm{cosk}_0(U_\bullet),A)\to Z^1(U_\bullet,A)$ is an isomorphism. -Consequently, the map $\mathrm{H}^1(\mathrm{cosk}_0(U_\bullet),A)\to \mathrm{H}^1(U_\bullet,A)$ is an isomorphism. -This is a nontrivial statement so let me sketch the ad-hoc proof I have. -Step 1: We may assume $U_\bullet=\mathrm{cosk}_1(U_\bullet)$. -Indeed, since $U_\bullet$ is a hypercovering, the map $U_2\to \mathrm{cosk}_1(U_\bullet)_2$ is a covering, and this easily implies that $Z^1(\mathrm{cosk}_1(U_\bullet),A)\to Z^1(U_\bullet,A)$ is an isomorphism, so replace $U_\bullet$ with its $1$-coskeleton. -One consequence of this assumption is that $U_2=\underline{\mathrm{Hom}}_{\mathrm{Simp}}(\partial \Delta^2, U_\bullet)$, where $\partial \Delta^2$ is the boundary of the $2$-simplex, realized as a constant simplicial object (i.e. sheaf) in $\mathbf{X}$, and $\underline{\mathrm{Hom}}$ denote internal $\mathrm{Hom}$ in $\mathbf{X}$. -Step 2: Let $\alpha\in Z^1(U_\bullet,A)$. We claim that $a\in A(U_1)$ descends along $(d_0,d_1):U_1\to U_0\times U_0$ to some $\alpha'\in G(U_0\times U_0)$. -Let $V=U_1\times_{U_0\times U_0}U_1$ and let $\pi_1,\pi_2:V\to U_1$ denote the first and second projections. Let $S$ denote the simplicial object of $\mathbf{X}$ obtained by gluing two copies of $\Delta^1$ along their vertices. Then $V=\underline{\mathrm{Hom}}_{\mathrm{Simp}}(S, U_\bullet)$. -There is a simplicial map $\partial \Delta^2\to S$ which degenerate the edge $\{1,2\}$ into a vertex. This gives rise to a map $V\to U_2$, which in turn gives rise a map $A(U_2)\to A(V)$. Applying this map to the cocycle equation $d_0^2\alpha-d_1^2\alpha+d^2_2\alpha=0$ in $A(U_2)$ gives $0-\pi_2^*\alpha+\pi_1^*\alpha=0$ in $A(V)$, which means that $\alpha$ descends to $\alpha'\in G(U_0\times U_0)$. -Step 3: We finally claim that $\alpha'$ lies in $Z^1(\mathrm{cosk}_0(U_\bullet),A)$, which proves the surjectivity of $Z^1(\mathrm{cosk}_0(U_\bullet),A)\to Z^1(U_\bullet,A)$. The injectivity follows easily from the fact that $(d_0,d_1):U_1\to\mathrm{cosk}_0(U_\bullet)_1=U_0\times U_0$ is a covering. -To show the claim, it is enough to show that the canonical map $U_2\to \mathrm{cosk}_0(U_\bullet)_2=U_0\times U_0\times U_0$ is a covering. Indeed, if this holds, then the fact that the $1$-cocycle equation holds for $\alpha$ in $A(U_2)$ implies that it holds for $\alpha'$ in $A(U_0\times U_0\times U_0)$. Proving that $U_2\to U_0\times U_0\times U_0$ is a covering amounts to showing that any $3$ "vertices" in $U_0$ can be joined by a "triangle" in $U_2$ locally. But this follows from $U_2=\underline{\mathrm{Hom}}_{\mathrm{Simp}}(\partial \Delta^2, U_\bullet)$ and the fact that $(d_0,d_1):U_1\to U_0\times U_0$ is a covering. -I would value references for this proposition in the literature, if you know them.<|endoftext|> -TITLE: Unipotent algebraic group action on quasi-affine (vs. affine) variety? -QUESTION [8 upvotes]: This question arises from a comment by user nfdc23 on an unrelated recent MO question here. It concerns textbook treatments of what has been called the "Theorem of Kostant-Rosenlicht", stated as Theorem 2 in this 1961 paper by Rosenlicht. (Apparently Kostant had addressed a similar question earlier involving Lie algebras in characteristic 0, though this may be unpublished.) Rosenlicht was mainly concerned with quotient varieties related to the action of a unipotent algebraic group $G$ on a quasi-affine variety $V$ (an open set in an affine variety). By a somewhat complicated induction on the dimension of $G$ (which can just as well be assumed connected), he proved in Theorem 2 that the $G$-orbits in $V$ are all closed. As an immediate corollary, the conjugacy classes in $G$ itself are closed. -Each of the three textbooks called Linear Algebraic Groups includes some version of this theorem: Borel's 1969 Benjamin lecture notes (expanded to a second edition in the Springer GTM series in 1991) stated Rosenlicht's theorem as Proposition 4.10 but with a proof which works only when $V$ is affine. My own GTM book in 1975 included just an exercise (17.8) on the affine case, with an outline of the proof. Then Springer in 1981/1998 treated the affine case briefly in his Proposition 2.4.14. - -Is there a natural example of a quasi-affine but not affine variety $V$ to which Rosenlicht's theorem applies in an essential way? - -I've found it difficult to place the theorem in the study of algebraic groups and their actions. For example, it doesn't seem to play any role in the basic structure theory of reductive groups. -ADDED: The answers and comments by nsfdc23, YCor, and Friedrich Knop have been very helpful in clarifying the issues here. The implicit answer to my question seems to be that there are no examples of the sort I asked about. Indeed, there seems to be little distance between the affine and quasi-affine formulations. This is less obvious in Rosenlicht's paper, especially because the language he used isn't as flexible as the language of schemes. -Borel apparently saw how to simplify Rosenlicht's approach by invoking the action of $G$ on regular functions, where any finite dimensional $G$-invariant subspace has a fixed point. But he skipped over the shift from an affine to a quasi-affine variety. Perhaps this was due to impatience and also having too many other projects in mind. (I recall the kind of time pressure he faced in 1968 in splicing together two sets of lecture notes on arithmetic groups which differed a lot in notation. I found many minor corrections to be made but didn't share all of them in a timely way; so the published version still has uncorrected pages.) -I don't remember now why I stated my exercise just for affine varieties, since I was usually following Borel's lecture notes. Unfortunately it's too late to ask Borel or Springer how they arrived at their choices. Springer might for example have understood the extra argument needed to deal with the quasi-affine case and decided to avoid it. - -REPLY [6 votes]: Since the quasi-affine case is so easily reduced to the affine case, one doesn't really get much extra mileage out of it. -After checking my papers I am pretty sure that I never used the quasi-affine case seriously. To the contrary, in some cases I had to reduce to the affine case anyway because one is actually losing information by passing to the quasi-affine case. -For example, for affine varieties it is clear that any two closed orbits can be separated by a global function $f$ (i.e., $f$ is constant $0$ on one orbit and constant $1$ on the other). So, this also holds for unipotent groups acting on quasi-affine varieties. But this cannot be deduced just from all orbits being closed. Consider, e.g., $\mathbf G_m$ acting on $X=\mathbf A^2\setminus\{0\}$ by scalar multiplication. Then all orbits are closed but no two orbits can be separated. The reason is of course that in $\mathbf A^2$ none of the orbits stay closed. -So why bother stating theorems in the quasi-affine case, at all? I think, this is mostly to alert that one is dealing with a property which is passed on to open subsets. So one is allowed to remove all kinds of undesired points from your variety like singularities, fixed points etc.<|endoftext|> -TITLE: "Jacobian Conjecture" for $k[x_1,\ldots,x_n,x_1^{-1},\ldots,x_n^{-1}]$? -QUESTION [5 upvotes]: Is there exist a similar conjecture to the famous Jacobian Conjecture with $\mathbb{C}[x_1,\ldots,x_n,x_1^{-1},\ldots,x_n^{-1}]$ instead of -$\mathbb{C}[x_1,\ldots,x_n]$? -Namely, let $f$ be $\mathbb{C}$-algebra endomorphism of $\mathbb{C}[x_1,\ldots,x_n,x_1^{-1},\ldots,x_n^{-1}]$, denote -$f_i:= f(x_i)$, and further assume that the Jacobian of $\{f_1,\ldots,f_n\}$ is in $\mathbb{C}-\{0\}$. Is such $f$ an automorphism? -(I guess that first one should be familiar with the group of automorphisms of -$\mathbb{C}[x_1,\ldots,x_n,x_1^{-1},\ldots,x_n^{-1}]$, see this question). -Edit: I also wonder if there exists any nice connection between the Jacobian Conjecture and my above conjecture (which is not exactly phrased yet); for example, are the two conjectures equivalent? - -REPLY [15 votes]: Counterexample: the endomorphism of the product of two punctured lines (complement of the curve of equation $xy=0$ in the plane) given by $$(x,y)\mapsto f(x,y)=\left(\frac{x}{y},y^2\right)$$ -We have -$$\begin{pmatrix}\partial_1f_1 & \partial_2f_1\\ \partial_1f_2 & \partial_2f_2\end{pmatrix}=\begin{pmatrix}\frac{1}{y} & -\frac{x}{y^2}\\ 0 & 2y\end{pmatrix},$$ -so the Jacobian is constant equal to 1. -It's not an automorphism, since both $(1,1)$ and $(-1,-1)$ map to $(1,1)$.<|endoftext|> -TITLE: Applications of Crystalline Cohomology for Physics -QUESTION [8 upvotes]: I know this is a very vague question, so I may restrict it to quantum theories for their more category theoretic setting. Even if the concept of crystalline cohomology is very abstract, it has made me wonder if there could be any applications of it to physical theories. This question arises from the growing use of higher category theory in many quantum theories and their mathematical analogs, for example homological mirror symmetry. -Question: - Are there any applications of crystalline cohomology to physical theories? - -REPLY [11 votes]: A talk that explores the physics connection to crystalline cohomology in the context of string theory is Motives and Strings by Jan Stienstra, with a challenge for each community: - -Why look for a MOTIVE-STRING relation? -Computations in Type IIb string theory proceed by manipulating solutions of certain differential equations. During the computations there are many denominators. In the end these drop out and true integers remain. - Many differential equations in Type IIb string can be recognized as - Picard-Fuchs equations in De Rham cohomology of families of varieties. - The integrality statements can be recognized as consequences of theorems about the crystalline cohomology of families of ordinary varieties. -Challenge for Motive people: Crystalline cohomology deals with only - one prime $p$ at a time and puts out statements about $p$-adic - integrality. What mechanism synchronizes the primes and leads to true - integers? -Challenge for String people: Crystalline cohomology implies extra - symmetries in the differential equations. Where are these extra - symmetries in Nature?<|endoftext|> -TITLE: Fourier series of a continuous function converging to non-value of function? -QUESTION [5 upvotes]: I know of examples of periodic continuous functions whose Fourier series diverge on sets of measure zero. Is it possible for a Fourier series of a periodic continuous function to converge to something other than the function value, at some point? If so, can someone provide an example or reference? - -REPLY [9 votes]: If a series converges to a limit $L$, its Cesaro averages converge to the same limit $L$. -The Cesaro averages of a Fourier series are given by convolution with the Fejer kernel, and it is known that convolving an element of $C({\bf T})$ with the Fejer kernel always gives a sequence of trigonometric polynomials converging uniformly to the given function. - -So I think the answer to your question is negative: if $f$ is continuous, and if the partial sums of $\sum_{n\in{\bf Z}} \hat{f}(n)e^{2\pi int}$ converge to some $L$, then $L=f(t)$.<|endoftext|> -TITLE: To derive or not to derive, that is the question -QUESTION [5 upvotes]: What are concrete and abstract examples of problems (even whole programmes of inquiry) when one has a choice to use a "derived" theory (e.g., $\infty$-categories, DAG, HAG, $DRep_k(G)$, "higher" counting,...) or its non-derived variant (1-categorical arguments, basic algebraic geometry, "stupid" counting, $Rep_k(G)$ ...) but the derived variant will not work? -Motivation: when defining, say, some geometric theory, is it still useful to write up a 1-categorical treatment distinct from an $\infty$-categorical treatment? - -REPLY [8 votes]: The comments are long, so I will post this as an answer. One example that has both a classical and a derived aspect is the construction, in algebraic geometry, of the virtual fundamental class by Kai Behrend and Barbara Fantechi. In Kontsevich's paper, "Enumerating Rational Curves via Torus Actions", he proposed a derived approach. However, there are several difficult things to check to even make sense of the construction, and there are many other compatibilities in order to prove that the construction satisfies the Kontsevich-Manin axioms for Gromov-Witten invariants. These things need to be proved with "classical" arguments about cycles on algebraic varieties and cotangent complexes. What I find most beautiful about the paper of Behrend-Fantechi is that they also clarify longstanding constructions like the excess intersection formula with their approach.<|endoftext|> -TITLE: SO(3) action on (simply connected) 6 manifold with discrete fixed point -QUESTION [16 upvotes]: If a 6-dimensional orientable smooth manifold $M$ admits a smooth effective $SO(3)$ action with discrete fixed point set, what can we say about the topology of $M$? What if we assume that in addition M is a Riemannian manifold with nonnegative/positive sectional curvature, and the group action is isometric? -The question is motivated by Fuquan Fang's paper: Positively curved 6-manifolds with simple symmetry groups, in which he tried to classify all such 6-manifolds. But the author overlooked the possibility of finite isotropy groups, and his proof turned out to have a gap. Since finite isotropy groups occur for the $SO(3)$ action on $SU(3)/\mathbb{T}^2$, I am trying to look at this one particular case that he missed. -I am also aware that some people had worked on similar things on 4-manifolds with $\mathbb{S}^{1}$ symmetry, e.g. https://arxiv.org/pdf/1703.05464.pdf. Through studying local isotropy representation and applying some signature formula, the author was able to classify the fixed point data of circle actions on 4-manifolds with discrete fixed point when there are few fixed points. -I am wondering if one can do analogous things on 6-manifolds with $SO(3)$ symmetry and discrete fixed point set. The isotropy representation near an isolated fixed point on $M^6$ must be $\mathbb{R}^{3}\oplus \mathbb{R}^{3}$, on which SO(3) acts diagonally. Unfortunately, this is the only piece of related work I know so far. -I want to point out that according to an unpublished note of Fabio Simas, if we have an effective isometric action of $SO(3)$ on a positively curved 6-manifold $M^6$ with only isolated fixed points, then the number of fixed points is at most 3. This follows from a "q-extent" argument in comparison geometry. I'm really most interested in the case where M carries a metric with positive sectional curvature. In this case, if we assume M admits an $SO(3)$ action with isolated fixed points, then we know the orbit space $M/SO(3)$ is a 3-dim Alexandrov space. I want to find a way of reconstructing the manifold M from the structure of the quotient $M/SO(3)$ and information about stabilizer groups,and information of "gluing maps" which identify the boundaries of different "orbit types" in M. But so far I'm still trying to work it out. -I have 2 examples for such actions. One is the linear $SO(3)$-action on $\mathbb{S}^6$, induced from the 7-dimensional representation $\mathbb{R}^3\oplus \mathbb{R}^3\oplus \mathbb{R}$, where $SO(3)$ acts diagonally and trivially on the last factor $\mathbb{R}$. This action has 2 isolated fixed points. Another is linear $SO(3)$-action on $\mathbb{CP}^3$, induced from the 4-dimensional complex representation $\mathbb{C}^3\oplus \mathbb{C}$, where $SO(3)$ acts trivially on $\mathbb{C}$. This action has 1 isolated fixed point. I'm wondering if there exists an example with 3 isolated fixed points, but yet still positively curved. - -REPLY [3 votes]: So I am answering my own question. I have been thinking about this question and other related questions for months, and now I have some results. -For this question, if we have a positively curved 6-manifold $M$ with $SO(3)$-action with discrete fixed point set, then the orbit space $M/SO(3)$ is homeomorphic to a 3-ball. For a priori it is a simply connected compact 3-manifold with boundary, thus it is homeomorphic to 3-ball minus finitely many open disks. But it is also an Alexandrov space with positive curvature: now by the soul theorem, it has at most one boundary component; so it is either a 3-sphere or a 3-ball. Now by the slice representation of $SO(3)$ around the discrete fixed point, the orbit space has boundary and the boundary orbit types mainly consist of $SO(3)/SO(2)$. -Now I need a few extra assumptions. I need to assume that there are no exceptional orbits, or in other case, there are no finite non-trivial stabilizer groups. Under this assumption, I can show that the boundary 2-sphere of the orbit space has only 2 fixed points, and no other orbit types. The argument is as follows: -Take $S^1$-fixed point $M^{S^1}$ of this action, where $S^1$ is any maximal torus of $SO(3)$. The $S^1$-fixed point component $M^{S^1}_0$ above the boundary of the orbit space is a branched double cover of $S^2$, since in each $SO(3)/SO(2)$-orbit, the $SO(2)$-fixed point set is a set of 2 elements. But $M^{S^1}_0$ is also a 2-sphere since it is a totally geodesic submanifold of $M$, thus it also has positive curvature. By Riemann-Hurwicz formula, a branched double cover of $S^2$ over $S^2$ has 2 branched points. Thus we have 2 points on the boundary of the orbit space which are "more singular" than $SO(3)/SO(2)$. These could a priori be fixed points or $SO(3)/O(2)$. But from the representation of $O(2)$, the existence of $SO(3)/O(2)$ orbit will force the existence of finite non-trivial stabilizer group, violating my assumption. Thus the 2 branched points must be 2 fixed points. -Now our picture of the orbit space is clearer. It is a 3-ball, whose principal isotropy group in the interior is trivial, and the boundary orbit types are $SO(3)/SO(2)$ and fixed points. There are 2 fixed points. We have two cases: Case I: interior orbits are all $SO(3)$; Case II: there is one singular orbit $SO(3)/SO(2)$ in the interior of the orbit space. There can't be more singular orbits in the interior, because by the q-extent argument, the total number of interior singular orbits and boundary fixed points is at most 3. -I can solve Case I. In this case, $M^6$ is a suspension of a 5-dim $SO(3)$-space $N^5$ whose orbit space corresponds to the equator disk of $M/SO(3)$. In other words, $N^5/SO(3)$ is a 2-disk with two orbit types $SO(3)$ and $SO(3)/SO(2)$, and all the singular orbits lie on the boundary. According to the "second classification theorem" in Bredon's book , Ch.5, Theorem 6.1 and Corollary 6.2, there are 2 such $SO(3)$-spaces, and they are $S^5$ and $S^2\times S^3$. $M^6$ is suspension of one of these two, but only suspension of $S^5$ is a manifold. Thus $M^6$ is homeomorphic to $S^6$. Case I is done. -For Case II and more general cases where one allows exceptional orbits, I do not know what to do so far. I can not get a homeomorphism classfication in these cases, but still I can say something about the (co)homology groups, using tools like Mayer-Vietoris sequence.<|endoftext|> -TITLE: What is the upper bound of $R\underbrace{(3,3,3, \ldots,3)}_\text{$k$ times}$? -QUESTION [9 upvotes]: Just some context: $R\underbrace{(3,3,3, \ldots,3)}_\text{$k$ times}\leq n$, means that any colouring of a complete graph, $k_n$, on $n$ vertices or more with $k$ colours must contain a monochromatic triangle. -Claim: $R\underbrace{(3,3,3, \ldots,3)}_\text{$k$ times} \leq 3k!$ -Proof: -Consider the minimum number of monochromatic edges connected to an arbitrary vertex, $V$. At least $\big\lceil{\frac{3k!-1}{k}}\big\rceil$ edges will be monochromatic. $\big\lceil{\frac{3k!-1}{k}}\big\rceil$ =$\frac{3k!}{k}$ when $k>1$. Therefore, vertex $V$ must be connected to $3(k-1)!$ monochromatic edges. -By trying to avoid a monochromatic triangle of color $k_1$, we will be forced into creating a monochromatic triangle of color {$k_2$, $k_3$, $\ldots$, or $k_i$}. If we connected any of the vertices $\{1,2,3, \ldots, 3(k-1)!\}$ with a blue edge, a blue monochromatic triangle would be formed, to avoid this, the remaining $3(k-1)!$ vertices must be colour with the remaining $k-1$ colours. -We can then consider a subgraph of the remaining vertices $\{1,2,3,\ldots,3(k-1)!\}$ and colour it with the remaining $k-1$ colours. We can focus on a new vertex, $V'$, within the subgraph. -Vertex $V'$ will be connected to at least $\displaystyle{\frac{3(k-1)!}{k-1}}$ monochromatic edges. Which equals $3(k-2)!$ edges. -We can continue this process until we consider the subgraph of remaining vertices $3(k-(k-1))!$ and colour it with the remaining $k-(k-1)$ colours. This means that this final subgraph has 3 vertices and has to be coloured with 1 colour. Therefore, a monochromatic triangle is unavoidable. Therefore, $R\underbrace{(3,3,3, \ldots,3)}_\text{$k$ times} \leq 3k!$. -I managed to get it down to $3k!$, I was wondering if anyone had better approximations? - -REPLY [15 votes]: All questions about Ramsey numbers for small graphs should be first checked in Staszek Radziszowki's amazing frequently updated survey. On page 40 we find the upper bound $(e-\frac16)k!+1\approx 2.55 k!$, proved by Xu Xiaodong, Xie Zheng and Chen Zhi in a paper published in Chinese.<|endoftext|> -TITLE: The set of ergodic mesures being $G_\delta$: about a theorem of K. R. Parthasarathy -QUESTION [6 upvotes]: Something I do not understand. It is Theorem 2.1 of the article of K.R. Parthasarathy "On the category of ergodic measures, Illinois J. Math. 5 (1961), pages 648-656 Full text here: (projecteuclid link). -Does this theorem states, as a particular case, that for any self-homeomorphism of a compact metric space, the set of ergodic invariant measures is dense in the set of all invariant measures? -If not, please can someone explain me what my mistake is in trying to apply that theorem to deduce the above statement? -If yes, please can you explain how to approximate with ergodic measures the following invariant measure in the example below? -Example in the two-dimensional torus $T: (x,y) \rightarrow (x+y,y)$; -$y_1 \neq y_2$ are irrational. So $T$, if restricted to the circle $y= y_1$ and if restricted to the circle $y=y_2$, is irrational rotations. Hence the Lebesgue measures $\mu_1$ and $\mu_2$ along each of both circles, are ergodic. Take -$\mu = (\mu_1 + \mu_2)/2$. It is non-ergodic and invariant, and it seems to me that it can not be approached by ergodic measures. - -REPLY [2 votes]: Theorem 2.1 states that the set of ergodic measures is a $G_\delta$ set in the set of invariant probability measures. This means that this set is a countable intersection of open sets, this does not imply that the set is dense. Examples of $G_\delta$ sets that are not dense include the empty set, singletons etc. Theorem 3.1 in the article asserts density, but for shifts on product spaces, not for all homeomorphisms on compact spaces. -A simple example of a compact space on which the ergodic measures are not dense is given by $x\mapsto x^2$ on $[0,1]$ for which the invariant measures are of the form $p\delta_0 +(1-p)\delta_1$, $p\in [0,1]$, whereas the ergodic ones are just $\delta_0$ and $\delta_1$. -You are perhaps confused by the Baire Category Theorem that asserts that a countable intersection of dense open sets is dense, for example if the ambient space is a complete metric space.<|endoftext|> -TITLE: Generalizations of the Robbins lemma and Gaussian integration by parts -QUESTION [10 upvotes]: This is getting no attention, so I'll try this here: - -The Robbins lemma, named after Herbert Robbins, says that if $X\sim\operatorname{Poisson}(\lambda)$ and $g$ is a function for which $\operatorname{E}(|X g(X)|) < \infty,$ then $$\operatorname{E}(Xg(X)) = \lambda \operatorname{E}(g(X+1)).$$ -"Gaussian integration by parts" is an identity that says that under suitable assumptions about the function $g$, if $X\sim N(0,\sigma^2),$ then $$ \operatorname{E}(Xg(X)) = \sigma^2\operatorname{E} (g\,'(X)). $$ - -Both of these propositions are used in empirical Bayes methods. -Both of these are of the form $$ \operatorname{E}(Xg(X)) = \operatorname{var}(X) \cdot \operatorname{E} ((Tg)(X)) $$ -where $T$ is a linear operator on functions $g$. -QUESTION: Might there be, for each linear operator $T$, some probability distribution for which this holds? And might all of these be useful in empirical Bayes methods? -(P.S.) BETTER BUT LESS LOGICALLY PRECISE VERSION: Are both of these instances of some more general fact of interest? - -REPLY [6 votes]: If I allow a probability distribution $P(X)=\delta(X)$ with ${\rm var}\,X=0$ then this solves the relation for any $T$ acting on functions $g(X)$ without a singularity at $X=0$, so that would be a trivial answer. -Let me therefore exclude the delta function distribution. Then there seems -seems to be a counterexample: take for $T$ the identity, and try $g=1$ and $g=X$; this gives the two equations -$$E(X)=E(X^2)-E(X)^2,\;\;E(X^2)=E(X)E(X^2)-E(X)^3$$ -which have the only solution $E(X)=0$, $E(X^2)=0$, which is excluded. - -REPLY [4 votes]: Note that there always is a stupid answer. For any $T$ the Dirac delta at $0$ will do the trick. However there are operators $T$ for which there does not exist a measure with nonzero variance. -Try the operator -$$ Tg(x)=g'(x)-xg(x). $$ -Suppose that there is a probability measure $\mu$ associated to it such that $\newcommand{\bE}{\mathbb{E}}$ $\DeclareMathOperator{\var}{var}$ -$$ -\bE_\mu\big[ Xg(X)\big]= \var_\mu(X)\bE_\mu\big[ Tg(X)\big]. -$$ -$\newcommand{\si}{\sigma}$ - Set $\si^2:=\var_\mu[X]$. We deduce -$$(1+\si^2)\bE_\mu[Xg(X)\big]=\si^2\bE_\mu[g'(X)],\;\;\forall g. $$ -If we take $g(x)=1$ we deduce $\bE[X]=0$.If we take $g(x)=x$ we deduce -$$ (1+\si^2)\si^2=(1+\si^2)\bE_\mu[X^2]=\si^2\bE_\mu[1]=\si^2 $$ -which shows that $\si^2=0$ so $\mu$ must be the Dirac delta concentrated at the origin.<|endoftext|> -TITLE: Explicit description (=pictures!) of elements in $Mod_g[k]$? -QUESTION [6 upvotes]: This question is probably obvious to experts but I couldn't find the answer in the literature... -Background: Consider the mapping class group $Mod_g$ of the closed genus $g$ surface. There are many nice sets of generators (i.e. Humpreys famous $2g+1$ Dehn twists or Wajnryb's 2-element generating set etc). -If we consider the Torelli group $\mathcal{I}_g = Mod_g[1]$, we know it is finitely generated by bounding pair maps. D. Johnson proved that the next level in the Johnson filtration $Mod_g[2]$ (now called the Johnson kernel) is generated by Dehn twists around separating curves. -My question: - -Is there an explicit description of elements deeper in the mapping class group (i.e. in $Mod_g[k]$ with $k \geq 2$)? Or at least examples in picture form... -Is there any known characterization (even for special values of $g$ and $k$) of such elements? -Finally (unrealistically optimistic) - is there a known generating set? - -REPLY [6 votes]: Generators for the higher terms in the Johnson filtration are not known. -Probably the best way to find explicit elements is to use the fact that the Johnson filtration forms a central filtration, and thus the kth term of the lower central series of the Torelli group lies in the kth term of the Johnson filtration. You can thus get elements by taking iterated commutators of elements in the Torelli group (e.g. bounding pair maps or separating twists). Unfortunately, it is known that this will not give generating sets; indeed, Hain proved in -R. Hain, -Infinitesimal presentations of the Torelli groups, -J. Amer. Math. Soc. 10 (1997), no. 3, 597–651. -that the lower central series and Johnson filtrations of the Torelli group are not only different, but even define different topologies on the Torelli group. -There are some things that are known about generating sets for the Johnson filtration. Let me point out three of them. - -I. In my paper -A. Putman, Small generating sets for the Torelli group, -Geom. Topol. 16 (2012), no. 1, 111–125. -I construct a generating set for the Torelli group which is much smaller than Johnson's generating set (its cardinality is roughly $c g^3$ for some constant $c$, while Johnson's is exponential in the genus; the abelianization has rank on the order of $g^3$, so this is the best you can do). - -II. In our paper -T. Church & A. Putman, Generating the Johnson filtration, Geom. Topol. 19 (2015), no. 4, 2217–2255. -we prove that for all $k$, there exists some $G_k$ such that the kth term of the Johnson filtration is generated by elements supported on subsurfaces of genus at most $G_k$. - -III. Very recently Ershov-He proved that the Johnson kernel subgroup is finitely generated for $g \geq 5$. The was extended independently by Ershov-He and by Church-Putman to prove that the kth term of the Johnson filtration is finitely generated as long as the genus is sufficiently large. The relevant papers are -M. Ershov & S. He, -On finiteness properties of the Johnson filtrations, -preprint 2017 -and -T. Church & A. Putman, -Generating the Johnson filtration II: finite generation, -preprint 2017. -These can be downloaded from Ershov's webpage here and from my webpage here. The other papers by myself that I list above can also be downloaded from my webpage.<|endoftext|> -TITLE: Singularities of fibrations -QUESTION [5 upvotes]: Let $f:X\rightarrow \mathbb{P}^2$ be a fibration, here $X$ is a projective variety of dimension three. -Assume that there exixts a smooth curve $C\subset\mathbb{P}^2$ such that for any $p\in\mathbb{P}^2\setminus C$ the fiber $f^{-1}(p)$ is a smooth curve and when $p\in C$ then $f^{-1}(p) = A_p\cup B_p$ where $A_p, B_p$ are smooth curves intersecting transversally in a point $x_p = A_p\cap B_p$. -In this situation what could be said on the singularities of $X$? For instance deos this imply that $X$ is singular at most in the points of the form $x_p$ and that the locus of these points is a locus of at worst canonical singularities for $X$ ? - -REPLY [4 votes]: This is true. Actually, even better, these singularities will be terminal and Gorenstein, so as mild as it can get. Well, at least if we assume that you are working over an algebraically closed field, but otherwise you would have to be more careful about what exactly do you mean by these assumptions and questions, so I'll assume that that's what you meant. -Your assumptions imply that all the fibers are Gorenstein. Since $\mathbb P^2$ is smooth, this implies that then $X$ is Gorenstein. -Furthermore, your assumptions imply that all the fibers are $1$-dimensional, so $f$ is equidimensional, and hence it is flat by "miracle flatness". So, as Ariyan noted, then $f$ is smooth at every $x\in X$ such that $x\neq x_p$ and hence $X$ is smooth at all of those points. -This also implies that $X$ is normal (it is $S_2$ since it is Gorenstein and its singular locus has at most codimension $2$, so it is also $R_1$). -Now, looking at one of these $x_p$'s, we still have that $X$ is Gorenstein and we know that a complete intersection curve through this point is a simple node. As an exercise, try to prove directly (say via a direct local computation) that this implies that then $X$ is canonical (of index $1$) at these points. -If you get stuck, then use this argument: -Let $C_1$ and $C_2$ be two smooth curves in $\mathbb P^2$ that intersect transversally at $p$ and let $D_i=f^*C_i\subseteq X$ for $i=1,2$. (For the record, the $D_i$ are reduced divisors on $X$.) Now, $D_1\cap D_2 =f^{-1}(p)$, which is a nodal curve, so it has slc singularities. -Then by inversion of adjunction (applied twice) $(X,D_1+D_2)$ is log canonical. Now if you scrape away the $D_i$, this actually means that $X$ is terminal at the points of the form $x_p$. (Note that $(D_1, D_1\cap D_2)$ is also log canonical, so $D_1$ is canonical. It is locally isomorphic to a cone over a quadric.)<|endoftext|> -TITLE: Relationship between the Witt algebra and vector fields on the circle -QUESTION [10 upvotes]: I have seen some (apparently contradictory) claims by mathematicians and physicists regarding the existence of an (infinite dimensional) Lie group whose Lie algebra is the Virasoro algebra. -The Virasoro algebra is a central extension of the Witt algebra, which is slightly simpler to describe and for which the same confusion is present, so I will formulate my question in that setting. -We define the Witt algebra to be $\mathfrak g:=\mathbb R[L_n\colon n\in\mathbb Z]$ where $L_n$ is the operator on the Laurent polynomial ring $\mathbb R[z,z^{-1}]$ given by $L_np(z)=z^{1-n}p'(z)$. When equipped with the commutator of operators as Lie bracket, $\mathfrak g$ becomes an (infinite dimensional) Lie algebra. -The Witt algebra can also be described abstractly, as the Lie algebra generated by elements $(L_n)_{n\in\mathbb Z}$ satisfying the commutation relations -$$ -[L_m,L_n]=(m-n)L_{m+n},\qquad m,n\in\mathbb Z. -$$ -There is also a complex Witt algebra, $\mathfrak g_{\mathbb C}$, obtained by tensoring with $\mathbb C$. Equivalently, this is the algebra generated as above by operators acting on the Laurent polynomial ring $\mathbb C[z,z^{-1}]$. -Question 1. Are $\mathfrak g$ and $\mathfrak g_{\mathbb C}$ dense subsets of reasonable Banach algebras? -Regardless of the answer to the previous question, both $\mathfrak g$ and $\mathfrak g_{\mathbb C}$ can be realized as dense subsets of reasonable locally convex topological vector spaces of vector fields. -For a compact smooth manifold $M$, the group $\text{Diff}(M)$ of smooth diffeomorphisms can be given a Lie group structure such that the Lie algebra $\text{Vect}(M)$ consists of the smooth vector fields on $M$. -If I am understanding several claims by physicists correctly, taking $M=\mathbb S^1$ in the previous paragraph should yield a Lie group whose Lie algebra is closely related to either $\mathfrak g$ or $\mathfrak g_{\mathbb C}$. This leads me to the following questions. -Question 2. Does $\text{Vect}(\mathbb S^1)$ contain a Lie subalgebra isomorphic to $\mathfrak g$? -Question 3. What is the precise relationship between $\text{Vect}(\mathbb S^1)$ and $\mathfrak g_{\mathbb C}$? - -References: -The article Kac-Moody and Virasoro algebras in relation to quantum physics by Goddard and Olive claims that the diffeomorphism group of the circle has the infinitesimal symmetries of the Witt algebra. Schottenloher's book A Mathematical Introduction to Conformal Field Theory claims, in section 5.4, that there is no complex Virasoro group and also no complex Witt group. Remarks on infinite dimensional Lie groups by Milnor discusses infinite dimensional Lie groups modeled upon locally convex topological Lie algebras and, in particular, section 6 of that article is devoted to groups of smooth diffeomorphisms of spheres. -I have asked a related question on the physics stackexchange. - -REPLY [3 votes]: The following answers Question 2 and 3: - -For 2, $\frak{g}$ is a dense Lie subalgebra of Vect($S^1$). Tensoring with $\mathbb{C}$ gives you 3. - -Identify an infinitestimal diffeomorphism on the circle to a vector field on $S^1$. Visualize a vector field as a field of tangent vectors. Each tangent vector is a multiple of $\partial_\theta$, so a vector field can be described by $A(\theta)\partial_\theta$, where $A$ is a smooth function on $S^1$. Fourier theory lets you express $A$ in terms of its Fourier series, so the space of vector fields is generated by $\{e^{in\theta}\ | n\in \mathbb{Z}\}$. Define -$$ -L_n = -ie^{in\theta}\partial_\theta -$$ -It's easy to check that these satisfy the generator relations you gave: for all smooth functions $f$ on $S^1$, -$$ -[L_n,L_m]f = e^{i(n+m)\theta} i (n-m) f = (m-n) L_{n+m} f, -$$ -and thus -$$ -[L_n,L_m] = (m-n) L_{n+m}. -$$<|endoftext|> -TITLE: What are _all_ of the exactness properties enjoyed by stable $\infty$-categories? -QUESTION [19 upvotes]: Alternate formulation of the question (I think): What's a precise version of the statement: "In a stable $\infty$-category, finite limits and finite colimits coincide"? -Recall that a stable $\infty$-category is a type of finitely complete and cocomplete $\infty$-category characterized by certain exactness conditions. Namely, - -There is a zero object $0$, i.e. an object which is both initial and terminal. -Every pushout square is a pullback square and vice versa. - -Item (2) takes advantage of a peculiar symmetry of the "square" category $S = \downarrow^\to_\to \downarrow$; namely $S$ can either be regarded as $S' \ast \mathrm{pt}$ where $S' = \cdot \leftarrow \cdot \rightarrow \cdot$ is the universal pushout diagram, or $S$ can be regarded as $S = \mathrm{pt} \ast S''$ where $S'' = \cdot \rightarrow \cdot \leftarrow \cdot$ is the universal pullback diagram. Hence it makes sense to ask, for a given $S$-diagram, whether it is a pullback, a pushout, or both. Item (1) similarly takes advantages of the identities $\mathrm{pt} = \emptyset \ast \mathrm{pt} = \mathrm{pt} \ast \emptyset$. -But I can't shake the feeling that notion of a stable infinity category somehow "transcends" this funny fact about the geometry of points and squares. For one thing, one can use a different "combinatorial basis" to characterize the exactness properties of a stable $\infty$-category, namely: -1.' The category is (pre)additive (i.e. finite products and coproducts coincide) -2.' The loops / suspension adjunction is an equivalence. -True, (2') may be regarded as a special case of (2) -- but it may also be regarded as a statement about the (co)tensoring of the category in finite spaces. -Both of these ways of defining stability say that certain limits and colimits "coincide", and my sense is that in a stable $\infty$-category, all finite limits and colimits coincide -- insofar as this makes sense. -Question: -Is there a general notion of "a limit and colimit coinciding" which includes - -zero objects -biproducts (= products which are also coproducts) -squares which are both pullbacks and pushouts -suspensions which are also deloopings - -and if so, is it true that in in a stable $\infty$-category, finite limits and finite colimits coincide whenever this makes sense? -I would regard this as investigating a different sort of exactness to the exactness properties enjoyed by ($\infty$)-toposes. In the topos case, I think there are some good answers. For one, in a topos $C$, the functor $C^\mathrm{op} \to \mathsf{Cat}$, $X \mapsto C/X$ preserves limits. Foir another, a Grothendieck topos $C$ is what Street calls "lex total": there is a left exact left adjoint to the Yoneda embedding. It would be nice to have similar statements here which in some sense formulate a "maximal" list of exactness properties enjoyed by (presentable, perhaps) stable $\infty$-categories, rather than the "minimal" lists found in (1,2) and (1',2') above. - -REPLY [2 votes]: The fundamental stability axiom (pushouts are pullbacks and vice-versa) can be seen geometrically as a particular instance of Verider--Lurie duality for the circle. So, Lurie's formulation of Verdier duality (summarized below) for general spaces can be thought of as an answer to your question. -Let $X$ be a locally compact Hausdorff space. Given a presheaf $F$ on $X$ valued in a pointed $\infty$-category $C$ which sends $\varnothing\subseteq X$ to a zero object, we can define a "dual cosheaf" -$$\mathbb DF(U)=\operatorname{cofib}(F(X)\to F(X\setminus U))$$ -which is a precosheaf on $X$ valued in $C$ which sends $\varnothing$ to a zero object. I will ignore the detail that $X\setminus U$ isn't usually an open subset of $X$ (but see Lurie HTT 7.3.4 for why this isn't too bad). There is a natural map $F\to\mathbb D\mathbb DF$. -When $C$ is stable, $\mathbb D$ sends sheaves to cosheaves and the natural map $F\to\mathbb D\mathbb DF$ is an isomorphism (Lurie HA 5.5.5). It's likely (just by considering very simple compact Hausdorff spaces and very simple (co)sheaves) that the converse is true (if $\mathbb D$ sends sheaves to cosheaves then $C$ is stable). -Remark: It would indeed be great to replace compact Hausdorff spaces in the above discussion with finite topological spaces. Is there a duality result for finite topological spaces, or perhaps finite posets with their poset topology? -Remark: There should be a notion of "global stable $\infty$-category" and the initial such category (with marked object(s)) should be the global stable homotopy category (compatible families of $G$-spectra for all compact Lie groups $G$). Then I expect there is a form of the above discussion where orbispaces or global spaces or local quotient stacks replace compact Hausdorff spaces.<|endoftext|> -TITLE: Asymptotic behavior of integral with gamma functions -QUESTION [5 upvotes]: Consider the following function defined for complex numbers $z\in\mathbb{C}$ with $\Re(z)\geq \frac{1}{2}$: -$$F(z)=\frac{1}{5^{\Re(z)}}\int_0^\infty \left| \frac{\Gamma(z+ix)\Gamma(z-ix)}{\Gamma(z)^2} \cdot \exp\left( \frac{\pi}{2} x\right)\right| dx.$$ -I am wondering about the behavior of $F(z)$ for $\vert z \vert \rightarrow \infty$. - -REPLY [2 votes]: Too long for a comment: -By using the approach of @Carlo Beenakker I was able to get an asymptotic formula for $z=\frac{1}{2} + z_{0} e^{i \phi}$ with $z_{0}\rightarrow\infty$ and $-\frac{\pi}{2}<\phi<\frac{\pi}{2}$. Here it is: -$$ -F(z) \sim \left(\frac{2}{5}\right)^{\Re (z)} \sqrt{\pi z_0 \cos \phi} \exp\left\{z_0 \left[\phi \ \sin{\phi}+\cos{\phi} \ln{(\cos{\phi})}\right]\right\}. -$$ -Near the limits of the validity interval for $\phi$ the approximation breaks down. For $\phi = \pm\frac{\pi}{2}$ and $z_0 \rightarrow \infty$ I found -$$ -F(\frac{1}{2} \pm i z_{0})\sim \frac{2}{\pi\sqrt{5}}\left(2\ e^{z_{0}\frac{\pi}{2}}-1\right). -$$ -Edit: However, in the last formula a factor is missing. But for the asymptotic case $z=\frac{1}{2} + i z_{0}$ with $z_{0}\rightarrow \infty$ it is anyway easier to exploit the identity -$$ -\left|\Gamma\left(\frac{1}{2} + i y\right)\right|^2 = \frac{\pi}{\cosh (\pi y)}. -$$ -After inserting in the definition of $F(z)$ and some variable transformations on gets -$$ -\sqrt{\frac{2}{5}}\frac{\beta+\beta^{-1}}{2^{3/2}\pi}\int_{1}^{\infty} dz \ z^{-1/4} \left(z^2 +(\beta+\beta^{-1})z+1\right)^{-1/2} -$$ -with the abbreviation $\beta = e^{\pi z_{0}}$. Mathematica can solve this integral in terms of Gaussian Hypergeometric functions and an Appell function, $F_{1}$ (see for example here). This can be expanded for large $z_{0}$ -$$ -F\left(\frac{1}{2}+ i z_{0}\right) = \sqrt{\frac{2}{5}} \frac{\Gamma\left(\frac{1}{4}\right)^2}{(2 \pi)^{3/2}} e^{\frac{\pi}{2}z_{0}}+O(e^{-\frac{\pi}{2}z_{0}}) -$$<|endoftext|> -TITLE: Example of 2-locally finite group that is not locally finite -QUESTION [9 upvotes]: Define a group to be 2-locally finite if, for any two elements, the subgroup generated by them is finite. -Define a group to be locally finite if the subgroup generated by any finite subset is finite. -I want an example of a 2-locally finite group that is not locally finite. -This paper https://arxiv.org/pdf/1403.0331.pdf discusses a special case where 2-locally finite groups must be locally finite (namely, the case of groups with a planar lattice of subgroups). -More generally: for a nonnegative integer n, call a group n-locally finite if every subset of size at most n generates a finite group. -I'm interested more generally in examples of n-locally finite groups that are not (n + 1)-locally finite. Currently I know of solutions for: - -n = 0: This is trivial; any group that has elements of infinite order will do. -n = 1: This means a periodic group (every element has finite order) where there is an infinite subgroup generated by two elements. Examples here include the Grigorchuk group and some negative solutions to the Burnside problem (specifically, things like Tarski monsters); I've compiled this info at https://groupprops.subwiki.org/wiki/Periodic_not_implies_locally_finite - -REPLY [10 votes]: Golod (MR link; Some problems of Burnside type. (Russian) 1968 Proc. Internat. Congr. Math. (Moscow, 1966) pp. 284–289 Izdat. "Mir'', Moscow; English translation Amer. Math. Soc. Transl. (2) 70 (1968), 49) produced infinite $n$-generator groups in which all $(n-1)$-generated subgroups are finite, for all $n$. -From the MR review: (...) The main theorem follows: For each field $k$ and each integer $d\ge 2$, there is a $d$-generator infinite-dimensional $k$-algebra in which every $(d−1)$-generator subalgebra is nilpotent and $\bigcap_{n=1}^\infty A^n=(0)$. Corollary 1 states that for each $d\ge 2$ and each prime $p$ there is a $d$-generator infinite residually finite $p$-group G whose $(d−1)$-generator subgroups are finite. -(I haven't checked but I expect that $A$ is meant to be associative but not not unital and that $A^n$ denotes the subspace generated by products of $n$ elements, which is a two-sided ideal.)<|endoftext|> -TITLE: Is $\mathbb{CP}^2$ with a line collapsed a complex analytic space? -QUESTION [6 upvotes]: Consider the quotient space of $\mathbf{CP}^2$ obtained by collapsing a line (a $\mathbf{CP}^1$) to a point. Is this a complex analytic space (in a natural way)? - -REPLY [10 votes]: The answer is no, because of the following general result. - -Theorem (Grauert's contractibility criterion). Let $X$ be a smooth complex surface and let $E \subset X$ be a connected curve in $X$, with irreducible components $E_i$. Then there exists an analytic contraction $$\pi \colon X \to Y$$ - of $E$ to a point $p \in Y$, where $Y$ is a (possibly singular) complex surface, if and only if the intersection matrix $(E_i \cdot E_j)$ is negative defined. - -If $E \subset \mathbb{CP}^2$ is a line, then $E^2=1$ and so $E$ cannot be analytically contracted to a point. -For a reference, see [K. Matsuki, Introduction to the Mori program, Theorem 4-6-25 p.234].<|endoftext|> -TITLE: Stiefel-Whitney total class with prescribed zeros -QUESTION [8 upvotes]: First things first, I am aware of the existence of this topic. It's related, but old and my question hasn't been discussed there. So I hope it's not wrong to start a new topic. -I'm currently searching for a vector bundle $E\to M$ with $M$ a manifold (no conditions on the rank of $E$) such that $$ w(E) = 1 + w_2(E)+w_3(E) $$ -with $w_2(E),w_3(E)$ both non zero. -I've shown that : - -If $w_2(E)=0$ then $w_3(E)=0$ (Wu's formula), so we can't really simplify the question. -$M$ can not have a $\mathbf Z/2$ cohomology engendered by a single element of degree $1$. In fact one can not have $w_2(E)=x^2$ with $x$ of degree 1. -$E\to M$ can not be the tangent bundle of $M$ (where $M$ is in this case a smooth manifold of dimension $3$ or $4$). (Wu's formula for tangent bundle and results on spin structures in dimension 4) - -If you have any ideas, It would be much appreciated. Thanks! -edit: I'd like something less "trivial" than just the universal oriented vector bundle on something like an approximation of the grassmannian. The goal is to get something like a geometrical interpretation of such a total class. It's well understood that $w_1$ represents orientation and $w_2$ spin structure, but it's still a deep mystery to me the meaning of $w_3$. -edit2 : Mark Grant gave a first answer, and thanks to him. But it seems to me that it's not clear if such a manifold $M$ exists if we ask $M$ to be low dimensional : dimension $3$ or $4$ at most. Of course it get's uglier, mostly because we can't consider the tangent bundle as I pointed out before. - -REPLY [2 votes]: In your second edit, you ask whether there exists an example of such a bundle over a lower-dimensional manifold. -Four-dimensional example -Let $M = (\mathbb{RP}^2\times\mathbb{RP}^2)\#(S^1\times S^3)$. Note that -$$H^1(M; \mathbb{Z}_2) \cong H^1(\mathbb{RP}^2\times\mathbb{RP}^2; \mathbb{Z}_2)\oplus H^1(S^1\times S^3;\mathbb{Z}_2).$$ -Let $a$ and $b$ denote elements of $H^1(M; \mathbb{Z}_2)$ corresponding to generators of $H^1(\mathbb{RP}^2\times\mathbb{RP}^2; \mathbb{Z}_2)$, and let $c$ denote the element of $H^1(M; \mathbb{Z}_2)$ corresponding to the generator of $H^1(S^1\times S^3; \mathbb{Z}_2)$. -Consider the rank four vector bundle $E = L_a \oplus L_b \oplus L_c\oplus L_{a + b + c}$ where $L_x$ is the unique real line bundle over $M$ with $w_1(L_x) = x$; note that $L_{a+b+c} \cong L_a\otimes L_b\otimes L_c$. We have -\begin{align*} -w_1(E) =&\ w_1(L_a) + w_1(L_b) + w_1(L_c) + w_1(L_{a + b + c})\\ -=&\ a + b + c + (a + b + c) = 0\\ -&\\ -w_2(E) =&\ w_1(L_a)w_1(L_b) + w_1(L_a)w_1(L_c) + w_1(L_a)w_1(L_{a + b + c})\\ -&+ w_1(L_b)w_1(L_c) + w_1(L_b)w_1(L_{a + b + c}) + w_1(L_c)w_1(L_{a + b + c})\\ -=&\ ab + ac + a(a + b + c) + bc + b(a + b + c) + c(a + b + c)\\ -=&\ ab + a^2 + b^2 \neq 0\\ -&\\ -w_3(E) =&\ w_1(L_a)w_1(L_b)w_1(L_c) + w_1(L_a)w_1(L_b)w_1(L_{a + b + c})\\ -&+ w_1(L_a)w_1(L_c)w_1(L_{a + b + c}) + w_1(L_b)w_1(L_c)w_1(L_{a + b + c})\\ -=&\ abc + ab(a + b + c) + ac(a + b + c) + bc(a + b + c)\\ -=&\ a^2b + ab^2 \neq 0\\ -&\\ -w_4(E) =&\ w_1(L_a)w_1(L_b)w_1(L_c)w_1(L_{a + b + c})\\ -=&\ abc(a + b + c) = 0. -\end{align*} -So $E$ is a rank four vector bundle over a four-manifold $M$ with $w(E) = 1 + w_2(E) + w_3(E)$. -In fact, we can do better. As $H^4(M; \mathbb{Z}) \cong \mathbb{Z}_2$, reduction mod $2$ defines an isomorphism $H^4(M; \mathbb{Z}) \to H^4(M; \mathbb{Z}_2)$. Under this isomorphism, $e(E)$ is mapped to $w_4(E) = 0$, so $e(E) = 0$ and hence $E \cong F\oplus\varepsilon^1$. Note that $F \to M$ is a rank three vector bundle with $w(F) = 1 + w_2(F) + w_3(F)$. -Three-dimensional characterisation -Let $X$ be a three-dimensional CW complex. Recall that there is a bijection between isomorphism classes of orientable rank three bundles on $X$ and homotopy classes of maps $X \to BSO(3)$. As $X$ is three-dimensional, we can instead map to $BSO(3)[3]$, the third stage of the Postnikov tower for $BSO(3)$. As $\pi_1(BSO(3)) = 0$, $\pi_2(BSO(3)) = \mathbb{Z}_2$, and $\pi_3(BSO(3)) = 0$, we see that $BSO(3)[3]$ is a $K(\mathbb{Z}_2, 2)$. Moreover, as the map $BSO(3) \to BSO(3)[3]$ induces an isomorphism on $\pi_1$ and $\pi_2$, the map $H^2(BSO(3)[3]; \mathbb{Z}_2) \to H^2(BSO(3); \mathbb{Z}_2)$ is also an isomorphism. It follows that there is a bijection between orientable rank three bundles on $X$ and $H^2(X; \mathbb{Z}_2)$ given by the second Stiefel-Whitney class of the bundle. -Now suppose that $X$ is a connected three-dimensional manifold. In order for $w_3(E) \in H^3(X; \mathbb{Z}_2)$ to be non-zero, we need $X$ to be closed. Furthermore, if $X$ is closed, -$$w_3(E) = \operatorname{Sq}^1(w_2(E)) = \nu_1(X)w_2(E) = w_1(X)w_2(E)$$ -so $X$ must be non-orientable. By Poincaré duality, there is at least one $\alpha \in H^2(X; \mathbb{Z}_2)$ such that $w_1(X)\alpha \neq 0$. For each such $\alpha$, there is a unique $SO(3)$-bundle $E \to X$ with $w(E) = 1 + \alpha + w_1(X)\alpha$. -In conclusion, we have the following statement: - -Let $X$ be a connected, closed three-manifold. There is a real rank three vector bundle $E \to X$ with $w(E) = 1 + w_2(E) + w_3(E)$ if and only if $X$ is non-orientable. Moreover, on any non-orientable $X$, for every choice of $\alpha \in H^2(X; \mathbb{Z}_2)$ satisfying $w_1(X)\alpha\neq 0$, there is a unique real rank three bundle $E$ with $w(E) = 1 + \alpha + w_1(X)\alpha$.<|endoftext|> -TITLE: Artin Rings, Noetherian Rings, and the Axiom of Choice -QUESTION [17 upvotes]: It is a well-known fact that in ZF, the axiom of choice is equivalent to the statement that every commutative ring has a maximal ideal. On the other hand, for Noetherian rings, this is not necessary (either in the sense that it only requires dependent choice or that it requires no choice at all, depending on your definition of Noetherian). It can also be shown (in ZFC) that every Artin ring is Noetherian, and the proof of this fact in Atiyah-MacDonald essentially relies on the following two results: - -In an Artin ring, every prime ideal is maximal (proved without choice), and -The nilradical of a ring is the intersection of all of the prime ideals in a ring (proved with choice). - -Result 2. is equivalent to the statement that every Artin ring contains a prime ideal. To see this, note simply that given a non-nilpotent element $f$ in a ring $A$, there is a prime ideal of A not containing $f$ if and only if $A_f$ has a prime ideal. According to this answer, the fact that every commutative ring has a prime ideal is equivalent in ZF to the Boolean prime ideal theorem, so at least full choice is not necessary. Can this be weakened for Artin rings? Alternatively, is there another proof that Artin $\Rightarrow$ Noetherian that uses a weaker form of the axiom of choice (or no form of it)? Thank you. - -REPLY [15 votes]: Suppose $A$ is a nonzero artinian ring. Then the collection of all nonzero (possibly improper) ideals of $A$ has a minimal element, say $I\subseteq A$. Then $I$ has no nonzero proper submodules, and so is a simple $A$-module. In particular, it must be cyclic (generated by any nonzero element) and the kernel of a surjection $A\to I$ will be a maximal ideal of $A$. -(Assuming "artinian" means "every nonempty collection of ideals has a minimal element", this does not use Choice. Assuming "artinian" means "every descending sequence of ideals stabilizes", it uses only Dependent Choice.)<|endoftext|> -TITLE: De Bruijn tori in higher dimensions? -QUESTION [9 upvotes]: Q. Do there exist De Bruijn tori in dimension $d > 2$? - -A De Bruijn torus -is a two-dimensional generalization of a -De Bruijn sequence. -A De Bruijn sequence is, for two symbols, -a cyclical bit-string that contains all bit strings of length $n$ -as consecutive, left-to-right bits (with wrap-around). -For example, here is a sequence of $8$ bits that contains all $2^3$ bit -strings of length $3$: -$$ -\begin{matrix} -\color{red}{0} & \color{red}{0} & \color{red}{0} & 1 & 1 & 1 & 0 & 1\\ -0 & \color{red}{0} & \color{red}{0} & \color{red}{1} & 1 & 1 & 0 & 1\\ -\color{red}{0} & 0 & 0 & 1 & 1 & 1 & \color{red}{0} & \color{red}{1}\\ -0 & 0 & \color{red}{0} & \color{red}{1} & \color{red}{1} & 1 & 0 & 1\\ -\color{red}{0} & \color{red}{0} & 0 & 1 & 1 & 1 & 0 & \color{red}{1}\\ -0 & 0 & 0 & 1 & 1 & \color{red}{1} & \color{red}{0} & \color{red}{1}\\ -0 & 0 & 0 & 1 & \color{red}{1} & \color{red}{1} & \color{red}{0} & 1\\ -0 & 0 & 0 & \color{red}{1} & \color{red}{1} & \color{red}{1} & 0 & 1 -\end{matrix} -$$ -Here is a De Bruijn torus that includes all $2 \times 2$ bit-matrices -exactly once (from here): -$$ -\begin{matrix} -0 & 1 & 0 & 0 \\ -0 & 1 & 1 & 1 \\ -1 & 1 & 1 & 0 \\ -0 & 0 & 1 & 0 -\end{matrix} -$$ -A $4 \times 4$ De Bruijn torus has been explicitly constructed. -My question is: Is it known that there exist -De Bruijn tori in dimensions larger than $d=2$? -Perhaps for every dimension? -For example, a three-dimensional pattern of bits that includes -every $k \times k \times k$ bit-(hyper)matrix? - -REPLY [9 votes]: Yes. See "New constructions for de Bruijn tori" by Hurlbert and Isaak<|endoftext|> -TITLE: Newform of a cuspidal Automorphic Representation -QUESTION [8 upvotes]: I was going through these notes https://www.dpmms.cam.ac.uk/~ty245/2008_AGR_Fall/2008_agr_week1.pdf . There, Theorem 9.2 states that: If $\pi ^{\infty}$ is a cuspidal automorphic representation of $\text{GL}_2(\mathbb A^{\infty})$ (on $V$), then there exists $N \in \mathbb Z _{>0}$ with $V^{U_1(N)} \ne 0$ and for minimal such $N$, we have $\dim_{\mathbb C}V^{U_1(N)} = 1$. The cusp form $\varphi \in \mathcal A_k( U_1(N))$ generating (unique upto scalar) $V^{U_1(N)}$ is defined to be a "newform". -Later it is shown that this $\varphi$ gives a classicial cusp form w.r.t. $\Gamma_1(N)$. My question is: -Quetion: Does it follow that $\varphi$ is a newform according to the classical definition (i.e., the one involving Petersson inner product)? -I understand that $\varphi$ is a Hecke eigenform w.r.t. $T_n$ and $$ for $n>0$. So it follows that $\varphi$ is either a newform or an old form. If it is an old form one can associate a newform $f$ to it with some conductor $M \mid N$. I was trying to show that $f \in V$, which, then by minimality of $N$ would answer the question. -Thank you. - -REPLY [6 votes]: Yes, the classical version of this adelic newform is the newform in the sense of Atkin-Lehner, and vice versa. See Casselman: On some results of Atkin and Lehner (Math. Ann. 201 (1973), 301-314), especially Theorem 4 there. Another important reference is Miyake: On automorphic forms on GL_2, and Hecke operators (Ann. of Math. 94 (1971), 174-189.). See also Sections 5.3-5.4 in Goldfeld-Hundley: Automorphic Representations and L-Functions for the General Linear Group, Volume 1. -Added. To your last paragraph: the adelization of $f$ lies in $V$, because it has the same Hecke eigenvalues as $\varphi$. The multiplicity one theorem says (or implies) that adelic (almost) Hecke cusp forms with the same Hecke eigenvalues (outside any finite set of places) generate the same cuspidal representation.<|endoftext|> -TITLE: An elementary, short proof that the group of units of the ring of integers of a number field is finitely generated -QUESTION [17 upvotes]: Dirichlet's unit theorem states that (i) the group of units, $\mathscr{U}_K$, of the ring of integers of a number field $K$ is finitely generated, and (ii) the rank of $\mathscr{U}_K$ is equal to $r_1 + r_2 - 1$, where $r_1$ is the number of real embeddings and $r_2$ the number of conjugate pairs of complex embeddings of $K$. - -Q. What about a reference to an elementary, short proof of (i)? - -In principle, (i) is much weaker than (ii), so the question shouldn't be so implausible. The reason why I'm asking is that I seem to have an alternative, extravagant proof of a well-known result, but for the approach to be of any potential interest I need (a reference to) an elementary, short proof of (i). - -REPLY [5 votes]: I do not know if the following qualifies as "short" or "elementary": but it does not follow the usual pattern through Minkowski's Convex Body Theorem. Rather, it mimics the classical proof of Mordell–Weil: honestly, it copies it, but I write the details down to give a direct feeling of what comes into the game. -First of all, Hermite—Minkowski tells you that $\mathcal{O}_K^\times/(\mathcal{O}_K^\times)^2$ is finite, as follows. For every unit $u$, the extension $K(\sqrt{u})/K$ has discriminant whose norm equals a bounded power of $2$, because every integral element $r+s\sqrt{u}\in \mathcal{O}_{K(\sqrt{u})}$ (with $r,s\in K$) has minimal polynomial $f(X)=X^2-2rX+r^2-us^2$ with $2r,r^2-us^2\in\mathcal{O}_K$. This shows that $\mathcal{O}_{K(\sqrt{u})}/\mathcal{O}_K(\sqrt{u})$ is a finite group killed by $4$ and of rank bounded by $2[K\colon\mathbb{Q}]$. The discriminant of $K(\sqrt{u})/K$ is then $2\sqrt{u}\cdot[\mathcal{O}_{K(\sqrt{u})}\colon\mathcal{O}_K(\sqrt{u})]$ -and thus we get a bound $\operatorname{Disc}_{K(\sqrt{u})/\mathbb{Q})}\leq \operatorname{Disc}_{K/\mathbb{Q}}^2\cdot 2^{a(K)}$ for a constant $a(K)$ depending on $K$ only. It follows that the quotient $\mathcal{O}_K^\times/(\mathcal{O}_K^\times)^2$ classifies quadratic extensions of $K$ (this is easy: Kummer theory works over any field, in this case of characteristic different from $2$) of degree bounded by $2[K\colon\mathbb{Q}]$ and discriminant bounded by $\operatorname{Disc}_{K/\mathbb{Q}}^22^{a(K)}$, and this is a finite set of extensions by Hermite–Minkowski. One might say that Hermite–Minkowski is not trivial, which is true: but it is easier than the full proof of Dirichlet's Unit Theorem, in the sense that it comes well before it in almost all books I know, and Hermite's original proof consists of elementary, although quite tedious, algebra not involving any topology. -Now comes the ''Mordell–Weil" part: consider the usual height -$$ -H(u)=\sqrt[{[K\colon\mathbb{Q}]}]{\prod_{\sigma\in\mathcal{M}_K^\infty}\max\{1,\vert\sigma(u)\vert\}^{\varepsilon(\sigma)}}\qquad\text{ for all }u\in\mathcal{O}_K^\times -$$ -where the product is over all infinite places and $\varepsilon(\sigma)=1$ if $\sigma(K)\subseteq\mathbb{R}$, and $2$ otherwise. It is immediate to see that $H(u)\geq 1$ for all $u$, that $H(uv)\geq H(u)H(v)$ for all $u,v$ and that $H(u^{m})=H(u)^m$ for all $u$ and $m\geq 1$; and it is a classical result that the Northcott property holds, namely that for every given bound $B$ there are only finitely many elements in $\mathcal{O}_K^\times$ such that $H(u)\leq B$: this is elementary again, using simply that algebraic integers of bounded degree and bounded height are roots of polynomials in $\mathbb{Z}[X]$ with bounded coefficients and degree, hence a finite number of polynomials. A proof of this can be found on page 503 of Northcott's original paper - -Northcott, D. (1949). An inequality in the theory of arithmetic on algebraic varieties. Mathematical Proceedings of the Cambridge Philosophical Society, 45(4), 502-509. doi:10.1017/S0305004100025202 - -By the first part, we know that there are finitely many units $\eta_1,\dots,\eta_r$ representing the elements in $\mathcal{O}_K^\times/(\mathcal{O}_K^\times)^2$; we also fix now a bound $B$ so that by the Northcott property there are only finitely many units $v_1,\dots,v_s$ of height bounded by $B$. Pick any $u\in\mathcal{O}_K^\times$: for every $n\geq 1$ we can write -$$ -u=\eta_{i_0}u_1^2,\quad u_1=\eta_{i_1}u_2^2,\quad\dots \quad u_{n-1}=\eta_{i_{n-1}}u_{n}^2 -$$ -and thus -$$ -u=\Big(\prod_{j=0}^n\eta_{i_j}^{2^j}\Big)u_{n+1}^{2^n}:=\eta(u;n)\cdot u_{n}^{2^n}. -$$ -with $\eta(u;n)$ belonging to the subgroup generated by $\eta_1,\dots,\eta_r$; moreover, the above equation shows -$$ -H(u)\geq H(u_n^{2^n})=H(u_n)^{2^n}\Rightarrow H(u_n)\leq H(u)^{1/2^n}\overset{n\to+\infty}{\longrightarrow} 1\quad(u\text{ is fixed}). -$$ -Therefore, if we choose $n$ to be big enough, we get $u_n\in\{v_1,\dots,v_r\}$ and therefore $u$ belongs to the subgroup generated by $\{\eta_1,\dots,\eta_r,v_1,\dots,v_s\}$, showing that $\mathcal{O}_K^\times$ is finitely generated.<|endoftext|> -TITLE: Geodesics equation on Lie groups with left invariant metrics -QUESTION [7 upvotes]: First of all, I am so sorry if this question is not appropriate to be here. I tried to ask something similar on Math Stack Exchange but it didn't have much attention. Any comment and I delete the question. -I was reading the classical paper from Milnor entitled Curvature of Left Invariant Metrics on Lie Groups. The approach is to consider an orthonormal frame on the Lie algebra, since all geometric information is gained considering an inner product on it vector space, once we have the correspondence between left invariant metrics and inner products on the Lie algebra. -From this is easy to take information about Levi-Civita connection, Curvatures and etc. But my question is, what about the geodesics? Is there any formula that relates the frame on the Lie algebra and the geodesic equation? Is possible to describe all the geodesics just looking for the orthonormal frame on the Lie algebra? -If it is not the case, what is the general approach to obtain geodesics on Lie groups? The only way is trying to describe the general Levi-Civita connection? - -REPLY [7 votes]: Let $\nabla$ be the connection induced by the Levi-Civita connection. $\nabla$ is left invariant. It thus defines a bilinear product $b$ on ${\cal G}$ the Lie algebra of $G$. Let $c(t)$ be a geodesic. We can write $\dot c(t)=dL_{c(t)}(x(t))$ where $x(t)\in {\cal G}$. It you write the equation $\nabla_{\dot c(t)}\dot c(t)=0$, you obtain -$\dot x(t)+b(x(t),x(t))=0$.<|endoftext|> -TITLE: On the solutions of $f(x) = y^k$ with $f \in \mathbb{Z}[x]$, $k \in \mathbb{N}$ -QUESTION [7 upvotes]: I was wondering if it is true that the set of integer solutions of the equation -$$ f(x) = y^k $$ -is finite, where $f$ is an irreducible integer polynomial of degree $d \ge 2$ and $y \in \mathbb{Z}$, $k \in \mathbb{N}$ with $k \ge 3$. -That is, given an integer irreducible polynomial $f$ of degree greater than two, if the set $\{ (x, y, k) \in \mathbb{Z}^3 | k \ge 3, f(x) = y^k \}$ is finite. -I searched in the literature but could not find the answer, and trying to prove it myself has proved unsuccesful. -The MO question Polynomials which always assume perfect power values asks the question for the same equation but assumes that it holds for every integer number, so it isn't quite the solution to current question. -Thanks in advance for any help or counterexample. -Ok, due to me not paying lot of attention to how i've written it down, the question has been misread a lot, so I've rephrased the question. I hope it is more clear now. - -REPLY [14 votes]: To expand a bit on Peter Mueller's answer, Siegel's theorem isn't overkill, although this case is a bit easier. When $k=2$, these curves are classically called hyperelliptic curves, and for higher $k$ they were named superelliptic curves by Serge Lang. Here's a rough sketch of the proof of finiteness for $k\ge3$, but you can find a full proof in many books. Factor $f(x)$, so $y^k=c\prod(x-\alpha_i)$. Then in any solution, the quantities $x-\alpha_i$ are more-or-less relatively prime, so they are more-or-less $k$'th powers. More precisely, letting $K$ be the splitting field of $f(x)$ and $R_K$ its ring of integers, there is a finite set of ideals $\{\mathfrak a_1,\ldots,\mathfrak a_t\}$ so that every $(x-\alpha_i)R_K$ is one of the $\mathfrak a_i$ times the $k$'th power of an ideal. Using finiteness of class number, one finally get $x-\alpha_i=b_iz_i^k$, with $b_i$ chosen from a finite set and $z_i\in R_K$. In particular, we get -$$\alpha_2 - \alpha_1 = b_2z_2^k - b_1z_1^k.$$ -It follows (after further work) that $z_2/z_1$ is a very good approximation to $(b_1/b_1)^{1/k}$, and now an application of Roth's theorem (or one of the weaker variants, the one due to Thue will suffice) gives finiteness of solutions. However, this gives an ineffective bound for the largest solution, although it does give an effective bound for the number of solutions. You can use Baker's theorem on linear forms in logarithms to get an effective bound for the largest solution. -As you can see from this sketch, this is a hard theorem, since it relies on a bunch of fairly deep results (finiteness of class number, finite generation of the unit group, and Diophantine approximation to algebraic numbers). - -REPLY [6 votes]: For $d=k=2$ this isn't true. For instance take $f(x)=2x^2+1$, the Pellian equation $y^2-2x^2=1$ has infinitely many integral solutions. -If $d\ge3$ or $k\ge3$, then the curve given by $f(X)-Y^k=0$ has positive genus, and such curves have (by an old theorem of Siegel, a precursor of Falting's theorem) only finitely many integral points. -Maybe Siegel's theorem is overkill here. On the other hand, even the elliptic curve case $d=3$, $k=2$ isn't easy.<|endoftext|> -TITLE: Mumford-Tate conjecture cases with small $l$-adic monodromy groups -QUESTION [5 upvotes]: My question concerns the Mumford-Tate conjecture for abelian varieties over number fields. -Most proven cases (that I am familiar with) show that the l-adic monodromy group is as large as it can possibly get because of the conditions imposed and has the same rank as the Mumford-Tate group. A notable exception is the paper of Zhao on four-folds with endomorphism ring $\mathbb{Z}$ which starts out with abelian varieties such that the derived subgroup $G_l^{\mathrm{der}}$ of the $l$-adic monodromy group is a product of three copies of $SL_2$ and shows that the same is true for the Mumford-Tate group (as opposed to the latter being $GSp_{8}$). -Are there any other known results of this type? - -REPLY [3 votes]: This is not a clear answer, but let me attempt to clarify the question a little, and also explain why the problem, properly interpreted, is so difficult: -Deligne's theorem that Hodge classes are absolute Hodge shows that the identity component of the $\ell$-adic monodromy group is a subgroup of the Mumford-Tate group. Hence in some sense, every case of the Mumford-Tate conjecture will involve showing that the $\ell$-adic monodromy group is "as large as it can possibly get because of conditions imposed" on the Mumford-Tate group. -Focusing on the size of the $\ell$-adic monodromy group is a bit of a red herring, I think. -First, because the Mumford-Tate group is an upper bound, one can prove the conjecture under the assumption that the Mumford-Tate group is quite small, as Pink did in Theorem 5.15 of https://people.math.ethz.ch/~pinkri/ftp/AMT-v3.pdf. -Second, one can prove the conjecture also in the case that the $\ell$-adic monodromy group is as small as possible (i.e. abelian) because then by Falting's theorem the abelian variety has many endomorphisms and thus must be CM. -So to state things more precisely, the situation that is hard to rule out is when the $\ell$-adic monodromy group is less than the maximum possible size that commutes with the endomorphism algebra of the abelian variety, but the Mumford-Tate group is larger. -I am not aware of any other results of this type. -Basically, the reason it is so difficult is that we have very few general tools to relate the $\ell$-adic monodromy group and the Mumford-Tate group together, other than the ones already mentioned ( absolute Hodge cycles, Falting's theorem / Hodge conjecture in codimension $1$) and so progress has mostly revolved around proving in as many cases as possible that there is a unique $\ell$-adic monodromy group with a given endomorphism algebra.<|endoftext|> -TITLE: Do all countable $\omega$-standard models of ZF with an amorphous set have the same inclusion relation up to isomorphism? -QUESTION [17 upvotes]: In my recent paper with Makoto Kikuchi, - -J. D. Hamkins and M. Kikuchi, The inclusion relations of the countable models of set theory are all isomorphic. manuscript under review. (arχiv) - -we proved the following theorem. -Theorem. All countable models $\langle M,\in^M\rangle\models\text{ZFC}$, whether well-founded or not, have the same inclusion relation $\langle M,\subseteq^M\rangle$, up to isomorphism. -And the same is true for much weaker theories, such as KP and even finite set theory, provided that one excludes the $\omega$-standard models without any infinite sets and also the $\omega$-standard models with amorphous sets. -The proof proceeds by proving that for most of the models of set theory $\langle M,\in^M\rangle$, the corresponding inclusion relation $\langle M,\subseteq\rangle$ is an $\omega$-saturated model of what we have called set-theoretic mereology, which is the theory of an unbounded atomic relatively complemented distributed lattice. Those are the basic facts of $\subseteq$ in set theory and that is a complete, finitely axiomatizable, decidable theory. Since the theory is complete, and the resulting models are $\omega$-saturated, it follows by the back-and-forth method that all countable saturated models are isomorphic. -But the possibility of amorphous sets throws a wrench in the works. For most models of set theory, we prove that the inclusion relation is $\omega$-saturated; but when the model is $\omega$-standard and has an amorphous set, then it is not $\omega$-saturated, and so this is an irrirating obstacle to the general phenomenon. What we want to know is whether the ZF models provide just one or many different isomorphism types for the inclusion relation. -Question. Do all countable $\omega$-standard models of ZF with an amorphous set have isomorphic inclusion relations? -Kikuchi and I state in the paper that we believe that the answer to this question will come from an understanding of the Tarski/Ersov invariants combined with a knowledge of the models of ZF with amorphous sets. - -REPLY [2 votes]: No, except in trivial cases.$\newcommand{\ck}{\omega_1^{\mathrm{ck}}} -\newcommand{\nmereo}{n_{\mathrm{mereo}}} -\DeclareMathOperator{MT}{MT} -\DeclareMathOperator{rk}{rk} -\DeclareMathOperator{rks}{rks}$ If $\nmereo$ is the cardinality of the set of isomorphism classes of reducts of countable models of ZF to their inclusion relation, then $\nmereo\neq 2.$ -A suitable notion of rank is studied by Mendick and Truss [1]. A simpler permutation model and further connections are discussed in Truss [2, Section 4], which cites [3] for this algebraic definition. For each Boolean algebra $B$ define the following transfinite sequence of ideals. - -$I_0(B)$ is $\{0\}$ -$I_{\alpha+1}(B)$ is the ideal of elements $x\in B$ such that the equivalence class $x+I_\alpha(B)$ is a finite sum of atoms in $B/I_\alpha(B)$ -$I_\beta(B)=\bigcup_{\alpha<\beta}I_\alpha(B)$ at limit ordinals $\beta>0$ - -The rank $\rk(B)$ is $-1$ if $B$ is the zero ring, or else the least ordinal $\alpha$ such that $I_{\alpha+1}(B)=B,$ if it exists, and otherwise $\infty.$ The Mendick-Truss rank $MT(x)$ of a set $x$ is the rank of the powerset $\mathcal P(x)$ with its usual Boolean algebra structure. -(Beware that [3] uses the similar but different quantity $\delta(B)=\rk(B)+1,$ i.e. the Cantor-Bendixson rank of the Stone space. “Rank” is an overloaded term, but this use for Boolean algebras appears in at least one paper [4]. These ideals are what you get by applying Stone duality to Cantor-Bendixson derivatives. They're more appropriate for studying powersets than the Tarski-Ershov ideal of sums of atomic and atomless elements: every element of a powerset is atomic.) -We can look at the rank of powersets within a model or externally, and it is useful if the two notions agree. Let $|p|$ denote the order type of a code $p$ in Kleene’s $\mathcal O,$ to be concrete. Then for any $\omega$-standard model $M$ of ZFC, $|p|^M$ is isomorphic to $|p|.$ So statements about ranks below $\ck$ will be computed correctly: $M\Vdash \rk(B)=|p|$ if and only if $\rk(B)=|p|.$ The following argument only needs to compare ranks of at most $\omega.$ -For a model $M$ of ZFA, define $\rks(M)$ to be the least ordinal $\alpha$ such that the rank of $(\mathcal P(x)^M,\subseteq^M)$ is less than $\alpha$ for all $x\in M.$ So at least for models of ZF, $\rks(M)$ is an invariant of the inclusion reduct. We can do a similar computation for class models $M$ adequate for the Boolean algebra structure on powersets, though we might get $\infty$ if there is no such ordinal. This justifies the convenient approach of working within a model, instead of reasoning externally to a model. -Assuming that there is an $\omega$-standard countable model of ZFC, the two different inclusion reducts will come from: - -An $\omega$-standard countable model of ZF with $\rks(M)>\omega$ -An $\omega$-standard countable model of ZF with $\rks(M)=\omega$ - -We’ll make use of permutation models. So I should explain how to get a model of pure set theory instead of just ZFA. I think the best approach is to appeal to embedding theorems. See Note 103 in “Consequences of the Axiom of Choice” for the statements; the proofs only use the usual techniques of forcing and inner models, so they should work in the setting of countable $\omega$-standard models. For (1) it’s easy because we just need a property to hold of one powerset algebra, which is exactly the kind of thing the Jech-Sochor embedding theorem allows. For (2) we can use the fact that if a set $x$ admits a surjection to $\omega$ then its MT rank is $\infty$ [1, Lemma 1.3]. A statement about powersets of sets that do not admit a surjection to $\omega$ is “surjectively boundable” so Pincus’s embedding theorem applies. -For (1), apply Jech-Sochor with a permutation model containing a set of MT rank $>\omega.$ [1, Theorem 3.1] [2, Section 4]. -For (2), apply Pincus’s embedding theorem with the first Fraenkel model $\mathcal N1.$ -This has a set of urelements $A,$ with full permutation group $G.$ The action of $G$ extends by abuse of notation to the whole of $\mathcal N1.$ -Any $x$ breaks up as a disjoint union of non-empty sets -$$x\cong \bigcup_{i\in\alpha} x_i$$ -with a family of functions $f_i$ with $\operatorname{dom}(f_i)\subseteq A^{k_i}$ for some $k_i\in\omega$ and $\operatorname{rng}(f_i)=x_i.$ -I don’t need any stronger properties of this model. -This decomposition is standard, but I don’t know of a suitable reference for this particular case. We can work in the full universe $V(A)$ where choice holds. Each set $x\in\mathcal N1$ is supported by a finite sequence $\xi\in A^{<\omega},$ meaning that $x$ is fixed by the stabilizer group $G_\xi\subset G$ of automorphisms that fix $\xi.$ Enumerate the $G_\xi$-orbits of $x$ by $x_i,$ $i<\alpha.$ Pick $y_i\in x_i$ for each $i<\alpha.$ Pick $\eta_i\in A^{<\omega}$ such that $y_i$ is supported by $\eta_i.$ Define $f_i=\{(\pi \eta_i,\pi y_i) : \pi\in G_\xi\}.$ Then the family $(f_i)_{i<\alpha}$ is fixed by $G_\xi$ and each $f_i$ is a surjection to $x_i.$ -If $\alpha$ is infinite then there is easily seen to be a surjection $x\to\omega.$ In this case $\rk(x)=\infty$ [1, Lemma 1.3]. If $\alpha$ is finite then $\MT(x)$ is finite because $\MT(A)=1$ and the class of sets of finite MT rank is closed under finite unions, products, subsets, and quotients: the first two operations are covered by [1, Corollary 1.5] and [1, Theorem 1.9], and as [1, page 2] says "[...] It is also easy to see by transfinite induction that the image of any [MT] rank $\alpha$ set under a function has rank $\leq \alpha,$ and hence that a subset of a rank $\alpha$ set has rank $\leq \alpha$." -This shows $\rks(\mathcal N1)\leq \omega.$ And $\MT(A^k)=k$ by [1, Theorem 1.9], so $\rks(\mathcal N1)$ is exactly $\omega.$ - -I've written up some thoughts on trying to show $\nmereo\geq\aleph_0$ (assuming $\nmereo>1$), but didn't prove anything. You might want to try asking Mendick and Truss for their advice. - -[1] Mendick, G. S.; Truss, J. K., A notion of rank in set theory without choice, Arch. Math. Logic 42, No. 2, 165-178 (2003). ZBL1025.03047. -[2] J. K. Truss, The axiom of choice and model-theoretic structures, submitted. http://www1.maths.leeds.ac.uk/~pmtjkt/preprints.html -[3] Day, G. W., Superatomic Boolean algebras, Pac. J. Math. 23, 479-489 (1967). ZBL0161.01402. -[4] Bonnet, Robert; Rubin, Matatyahu, A thin-tall Boolean algebra which is isomorphic to each of its uncountable subalgebras, Topology Appl. 158, No. 13, 1503-1525 (2011). ZBL1229.54045.<|endoftext|> -TITLE: Inductive and reducible functions -QUESTION [9 upvotes]: The question was asked by a Computer Scientist and is closely related to parallel computing. But it is clearly of algebraic nature, so I decided to post it here. -Let $X$ be a set and $\bar X$ be the union of all Cartesian powers $X^n$. -Let $f$ be a function from $\bar X$ to $X$. We say that $f$ is inductive if -there exists a binary operation $F\colon X\times X \to X$ such that for -every $x_1,...,x_n \in X$, $n>1$, we have -$f(x_1,...,x_n)=F(f(x_1,...,x_{n-1}), x_n)$. If, in addition, $F$ is -associative and $(X,F)$ is a monoid (has the identity element), then $f$ -is called reducible. For example, if $X$ is the set of all natural -numbers, $f$ is the max function on $\bar X$, then we can take -$F(x,y)=\max(x,y)$. So $f$ is reducible. -We say that $(X,f)$ has an inductive (resp. reducible) extension if -there exists a set $Y$ containing $X$ with a function $g$ from $\bar Y$ to $Y$ -and a projection function $p: Y\to X$ such that $$p(g(x_1,...,x_n))=f(x_1,...,x_n)$$ for -all $x_1,...,x_n$ in $X$ and $(Y, g)$ is inductive (resp. reducible). -Question: 1) What is known about the set of functions $f$ with inductive -(reducible) extensions? 2) In particular, does every function $f$ admit an -inductive (reducible) extension? - -REPLY [2 votes]: I assume that $\bar X$ also contains $\varnothing$ as the Cartesian product of 0 copies of $X$. -Set $Y=\bar X$. If $y_i=(x_{i,1},\dots,x_{i,k_i})\in X^{k_i}$ with $i=1,\dots,\ell$, set -$$ - g(y_1,\dots,y_\ell)=(x_{1,1},\dots,x_{1,k_1},x_{2,1},\dots,x_{2,k_2},\dots)\in X^{k_1+k_2+\dots}\subset Y -$$ -(so $g$ is simply the concatenation). Clearly, $g$ is reducible (with $F$ being a binary concatenation; we get a monoid, since we added $\varnothing$). Finally, for $y=(x_1,\dots,x_k)$ define -$$ - p(y)=f(x_1,\dots,x_k) -$$ -(we can define $p(\varnothing)$ arbitrarily; on the other elements, $p$ does all the job). -Thus we obtain a reducible extension of $f$. -So, each function $f$ indeed has a reducible extension.<|endoftext|> -TITLE: Provoking involutions further -QUESTION [6 upvotes]: Let $\mathfrak{S}_n$ denote the permutation group, and $I_0(n)=\sum_{j\geq0}\binom{n}{2j}\frac{(2j)!}{2^jj!}$ stand for involutions see A000085 for more interpretations. There is also these numbers $I_1(n)=\sum_{j\geq0}\binom{n}jI_0(j)I_0(n-j)$ described in A000898 by several means. -Let me introduce the numbers $I_2(n)=\sum_{j\geq0}\binom{n}jI_0(j)^2I_0(n-j)^2$. I was able to verify the exponential generating function -$$\sum_{n\geq0}I_2(n)\frac{x^n}{n!}=\frac1{1-x^2}e^{\frac{2x}{1-x}}.$$ -However, it is desirable to know: - -Question. Is there a combinatorial meaning to the numbers $I_2(n)$? - -Remark. Of course, it is also interesting if one can provide any other context where $I_2(n)$ appears. - -REPLY [9 votes]: The generating function for involutions with respect to the number of fixed points is given by an evaluation of Hermite polynomials . The bilinear generating function of Hermite polynomials is given by "Mehler's formula": -$$\sum_{n\geq 0}\frac{t^n H_n(x)H_n(y)}{2^nn!}=\frac{1}{\sqrt{1-t^2}}\exp\left(\frac{t^2(x^2+y^2)-2txy}{t^2-1}\right).$$ -Foata wrote the paper "A combinatorial proof of the Mehler formula" where he gives a combinatorial interpretation and bijective proof of the formula. Mehler's formula (after an appropriate specialization) is technically related to the square root of your generating function. Your generating function also has a similar refinement and it's what Zeilberger calls "the heterosexual version of Mehler's formula". To see your identity in Zeilberger's set up you need to set $x=y=1$ and $s=t$ in his main identity. - -REPLY [6 votes]: Define a standard bitableau of size $n$ to be a pair $(P_1, P_2)$ of standard tableaux of total size $n$ such that each of the integers $1,\dotsc, n$ occurs exactly once in either tableau. -Then $I_2(n)$ is the number of pairs of standard bitableaux $((P_1, P_2), (Q_1, Q_2))$ of size $n$ such that $P_1$ has the same content as $Q_1$. In fact, the $j$th summand in the sum defining $I_2(n)$ is the number of such pairs where $P_1$ and $Q_1$ have $j$ cells, and the same content.<|endoftext|> -TITLE: Determinant of the "quantum" version of the group $\mathbb{Z}_n$ -QUESTION [5 upvotes]: Let $[0]_q:=0$ and $[n]_q:=\frac{1-q^n}{1-q}=1+q+\cdots+q^{n-1}$, for $n\geq1$. - -Question. Is there a closed formula (with proof) for the determinant of the matrix of $(i,j)$-entries - $$[i+j\bmod n]_q, \qquad i,j=1,2,\dots,n.$$ - -Remark. To bring in some context to the problem, this determinant is the specialization -$$x_{i+j\bmod n}\rightarrow [i+j\bmod n]_q$$ -in the group determinant of Frobenius for the (finite) additive group $\mathbb{Z}_n$. -EDIT. Sorry, I was meant to write $[0]_q=0$ not $[0]_q=1$. - -REPLY [4 votes]: Let me give a few details on Fedor's calculation. If we swap the columns with index $j$ and $n+1-j$ we get the matrix $M_n(q)$ with $(i,j)$ entry equal to $[1+i-j \mod n]_q$. Since we have swapped $\lfloor \frac{n}{2}\rfloor$ columns, the determinant of your original matrix is equal to -$$(-1)^{\lfloor \frac{n}{2}\rfloor}\det M_n(q)=(-1)^{\frac{n(n-1)}{2}}\det M_n(q).$$ -Now, $M_n(q)$ is a circulant matrix so it has eigenvectors $(1,\omega^k,\dots,\omega^{k(n-1)})$, with corresponding eigenvalue $[1]_q+[0]_q\omega^k+[n-1]_q\omega^{2k}+\cdots+[2]_q\omega^{(n-1)k}$, for all $0\le k\le n-1$ where $\omega=\exp(\frac{2\pi i}{n})$. -Next we observe that -$$[1]_q+[0]_q\omega^k+[n-1]_q\omega^{2k}+\cdots+[2]_q\omega^{(n-1)k}=\omega^{k}\left([0]_q+\frac{1+\frac{q}{\omega^k}+\cdots \frac{q^{n-1}}{\omega^{k(n-1)}}}{q-1}\right)=\omega^k\left([0]_q+\frac{q^n-1}{(q-1)(\frac{q}{\omega^k}-1)}\right)$$ -when $k\neq 0$. So we can calculate -$$\det M_n(q)=\omega^{\frac{n(n-1)}{2}}\left([0]_q+[1]_q+\cdots+[n-1]_q\right)\prod_{k=1}^{n-1}\frac{[0]_q(q-\omega^k)+\omega^k [n]_q}{q-\omega^k}$$ -if you take $[0]_q=0$ this gives -$$\det M_n(q)=[n]_q^{n-2}\frac{[n]_q-n}{q-1}$$ -and if you take $[0]_q=1$ it gives -$$\det M_n(q)=(-q)^{n-1}\left(1+\frac{[n]_q-n}{q-1}\right)\frac{1+(-1)^{n-1}[n-1]_q^n}{[n]_q(1+[n-1]_q)}.$$<|endoftext|> -TITLE: Convergence of Lagrange interpolation polynomials to entire functions -QUESTION [7 upvotes]: Consider an entire function $\ f:\mathbb C\rightarrow\mathbb C.\ $ Let $\ (a_n\in\mathbb C:n=0\ 1\ \ldots)\ $ be an infinite sequence, where $\ a_k\ne a_n\ $ whenever $\ k\ne n.\ $ Let $\ L_n\ $ be the the degree $\le n$ polynomial $\ L_n\ $ such that $\ L_n(a_k) = f(a_k)\ $ for every $\ k=0\ \ldots\ n.\ $ What are the results about convergence of $\ L_n\ $ to $\ f\ $, e.g. uniform convergence on compact subsets of $\ \mathbb C\,?$ - -The results may depend on the nature of the sequence $\ a_n\ $ or on a subclass of the entire functions to which $\ f\ $ belongs, or perhaps on both. (Myself, I don't know any such results, be them positive or negative). - -REPLY [10 votes]: There are too many results to survey them here. The principal books addressing this question are: -B. Levin, Distribution of zeros of of entire functions, -A. Gelfond, Calculus of finite differences, -J. M. Whittaker, Interpolatory function theory. -All these books exist in multiple editions, and can be found on Internet. -But the general form of the results is as you wrote: under certain restrictions -on the nodes and values, Lagrange polynomials (or Lagrange series) converge, -without any restrictions it diverges.<|endoftext|> -TITLE: A simple number theory confirmation -QUESTION [11 upvotes]: Suppose $a,b\in\Bbb N$ are odd coprime with $a,b>1$ then is it true that if all four of $$x_1a+x_2b,\mbox{ }x_2a-x_1b,\mbox{ }x_1\frac{(a+b)}2+x_2\frac{(a-b)}2,\mbox{ }x_2\frac{(a+b)}2-x_1\frac{(a-b)}2$$ are in $\Bbb Z$ for some $x_1,x_2\in\Bbb R$ then $x_1,x_2\in\Bbb Z$ should hold? - -REPLY [5 votes]: This is a simplification of Cehrng-tiao Perng's proof. As observed in that proof, it suffices to show that the Gaussian integers -$$ \gamma:=a+bi\qquad\text{and}\qquad \delta:=\frac{1+i}{2}(a-bi) $$ -are coprime. We know that some $\mathbb{Z}$-linear combination of $a$ and $b$ equals $1$, while also -$$\gamma+(1-i)\delta=2a\qquad\text{and}\qquad -i\gamma+(1+i)\delta=2b,$$ -hence some $\mathbb{Z}[i]$-linear combination of $\gamma$ and $\delta$ equals $2$. This shows that $\gcd(\gamma,\delta)$ divides $2$, hence the only possible common prime divisor of $\gamma$ and $\delta$ is $1+i$. However, -$$ \frac{\delta}{1+i}=\frac{a-bi}{2}\not\in\mathbb{Z}[i],$$ -because $a$ and $b$ are odd integers. Hence $\delta$ is not divisible by $1+i$, and we are done.<|endoftext|> -TITLE: Families of curves on compact complex surfaces and algebraicity -QUESTION [8 upvotes]: Let $S$ be a compact complex manifold of dimension $2$ and assume that there exists a two-dimensional family of curves on $S$. Is it true then that the algebraic dimension of $S$ is $2$, i.e. that $S$ is Moishezon (and hence projective)? -By a two-dimensional family of curves I mean a complex analytic subset $X \subset S \times T$ where $T$ is some compact complex space of dimension 2, and such that fibres $X_t$ are complex spaces of dimension 1 (possibly singular and reducible), and such that $X_t$ and $X_s$ are distinct for $t \neq s$. One can assume that the projection of $X$ on $T$ is flat or smooth, if it helps. - -REPLY [5 votes]: The answer is yes under the assumption that the general element of the mobile part of the family is irreducible, and a proof goes at follows. -Write your analytic family of curves as $$\{X_t\} = Z + \{M_t\},$$ -where $Z$ is the fixed part (i.e., the maximal effective divisor contained in any member of the family) and $\{M_t\}$ is a $2$-dimensional analytic family with at most isolated base points (the mobile part). -Then, calling $M$ the cohomology class of $M_t$, which is by definition independent on $t$, we claim that $M^2 >0$. In fact, given any irreducible $M_t$ and a general point $p$ on it, our assumption on the dimension implies that we can find an irreducible $M_s$, with $s \neq t$, such that $p \in M_s$, hence $p \in M_t \cap M_s$ and this proves our claim. -Now we are done, since a smooth, compact, complex surface is projective if and only if there exists on it a divisor with strictly positive self-intersection, see Barth-Peters-Van de Ven, Compact Complex Surfaces, Chapter 4, Theorem 5.2. -Remark. If the general element of the mobile part $\{M_t\}$ is not irreducible, the result is in general false. For instance, take a surface $S$ with algebraic dimension $1$, and consider the fibration $f \colon S \to \mathbb{P}^1$ given by the essentially unique meromorphic function on $S$ (if necessary, blow-up some point on $S$ in order to make it a holomorphic map). Writing $F_{u}$ for the fibre over $u \in \mathbb{P}^1$, the linear system $|F_0 + F_{\infty}|$ provides a $2$-dimensional family of curves on $S$, whose members are all disjoint unions of two fibres.<|endoftext|> -TITLE: Number of connected components of an Automorphism group -QUESTION [7 upvotes]: Let $X$ be a smooth quasi-projective irreducible variety over the field of complex numbers $\mathbb{C}$. We denote by $\mathrm{Aut}(X)$ the group of algebraic automorphisms of $X$. Moreover, for a variety $V$, we call a map $V \to \mathrm{Aut}(X)$ a morphism, if the induced map $V \times X \to X$ -is a morphism of varieties. The connected component of $\mathrm{Aut}(X)$ -of the neutral element $e \in \mathrm{Aut}(X)$ we define by -$$ -\mathrm{Aut}(X)^\circ = \left\{ g \in \mathrm{Aut}(X) \Big| -\begin{array}{l}  -\, \textrm{$\exists$ -an irreducible variety $V$ and a morphism} \\ -\textrm{$V \to \mathrm{Aut}(X)$ s.t. the image contains $g$ and $e$} -\end{array} \right\} \, . -$$ -This notion goes back to Ramanujam, see [Ram64]. Clearly, $\mathrm{Aut}(X)^\circ$ is a normal subgroup of $\mathrm{Aut}(X)$. -I am interested in the size of the group $Q(X) = \mathrm{Aut}(X) / \mathrm{Aut}(X)^\circ$. In case $X$ is projective, then $Q(X)$ is countable. Also in case $X$ is affine, -$Q(X)$ is countable. My question is, whether this is true in general, i.e. whether for all smooth irreducible quasi-projective variety $X$, the group $Q(X)$ -is countable. Every proof, counter-example or textbook reference would be perfect. -[Ram64] Ramanujam, C.P., A note on automorphism groups of algebraic varieties, Math. Ann. 156, 25-33 (1964). ZBL0121.16103. - -REPLY [3 votes]: Let $i:X\to \overline{X}$ be any dense open immersion of $X$ in a projective scheme. For every morphism $\phi:X\to X$, denote by $\overline{\Gamma}_\phi\subset \overline{X}\times_{\text{Spec}(\mathbb{C})}\overline{X}$ the closure of the graph of $\phi$. For every choice of ample invertible sheaf $\mathcal{L}$ on $\overline{X}\times_{\text{Spec}(\mathbb{C})}\overline{X}$, there is an associated Hilbert polynomial $P(t)\in \mathbb{Q}[t]$ of $\overline{\Gamma}_\phi$ with respect to $\mathcal{L}$. For every choice of $P(t)$, denote by $\text{Hilb}^{P(t)}$ the associated Hilbert scheme parameterizing closed subschemes $Z\subset \overline{X}\times_{\text{Spec}(\mathbb{C})}\overline{X}$ with Hilbert polynomial $P(t)$. There is an open subscheme $U$ that parameterizes those $Z$ such that both projections $Z \cap (X\times_{\text{Spec}(\mathbb{C})} X) \to X$ are isomorphisms. As an open subscheme of a projective scheme, $U$ has only finitely many irreducible components. Also there are only countably many possible numerical polynomials $P(t)$. Thus, the group $Q(X)$ is finite or countably infinite.<|endoftext|> -TITLE: Special cases of the Kimura-O’Sullivan conjecture, i.e., examples of finite dimensional motives? -QUESTION [7 upvotes]: In this 2005 paper, Kimura introduces a notion of finite dimensionality for Chow motives, defined in terms of vanishing of high symmetric and wedge powers. Toward the end of his paper, he conjectures that the Chow motive $h(X)$ of any smooth variety $X$ is finite dimensional [Conjecture 7.1]. The (more general?) Kimura-O’Sullivan conjecture states that every Chow motive is finite dimensional. -In his paper, Kimura proves that Chow motives of smooth projective curves are finite dimensional [Corollary 4.4] and that tensor products of finite dimensional motives are finite dimensional [Corollary 5.11]. What is the current state of knowledge about Chow motives satisfying the Kimura-O’Sullivan conjecture? More specifically: -Q1. What other specific Chow motives are known to be finite dimensional? -Q2. What other constructions from finite dimensional Chow motives are known to return finite dimensional Chow motives? -Q3. I would expect that for any finite surjective map $f:X\longrightarrow Y$ of smooth proper $k$-schemes over a fixed field $k$, if the Chow motive $h(Y)$ is finite dimensional, then the Chow motive $h(X)$ is also finite dimensional. Is there an obvious way to see this? Is this known? (If I understand correctly, the dual statement minus the finiteness assumption — the statement that if $h(X)$ is finite dimensional and $f$ is surjective then $h(Y)$ is finite dimensional — is Proposition 6.9 in Kimura's paper.) - -REPLY [8 votes]: Q1 + Q2: At present the Chow motives known to be finite dimensional are precisely those that are contained in the thick tensor subcategory generated by motives of abelian varieties. That is, the motives that can be obtained from motives of abelian varieties by tensorial operations, extensions, quotients and subobjects. -Q3. This is definitely not known at all! Any projective variety is a finite cover of a projective space, and $h(\mathbb P^n)$ is of course finite dimensional. -That said, there is recent dramatic progress in this area. According to a preprint of Ayoub ("Topologie feuilletée et conservativité des réalisations classiques en caractéristique nulle", available on his webpage), the conservativity conjecture in characteristic zero may be within reach (but see the disclaimers in his preprint). This would in particular imply finite dimensionality for all Chow motives which have only even or odd cohomology.<|endoftext|> -TITLE: fibers of birational contraction for complex manifolds - are they Moishezon? -QUESTION [5 upvotes]: Let $X$ be a smooth complex manifold and -$\phi:\; X \mapsto Y$ a proper holomorphic -map which is birational ("birational contraction"), -and $Z= \phi^{-1}(y)$ its fiber in a point $y$. -The variety $Y$ is not assumed to be smooth. -In this case I think that $Z$ is Moishezon. -I would be very grateful for a reference or a simple argument. -Misha - -REPLY [2 votes]: This is proved in Hironaka, Flattening theorem in complex-analytic geometry, Corollary 2.<|endoftext|> -TITLE: The completion of the space of finite groups -QUESTION [15 upvotes]: Edit: I revise the question based on the comment conversations -Let $\mathcal{F}$ be the set of all equivalence classes of finite groups under the "Isomorphism" equivalence relation. -We define a pseudo metric $d$ on $\mathcal{F}$ as follows: -$$d(G,H)= \inf \{Hd(\tilde {G}_{n},\tilde{H}_{n})\} $$ -where $\inf$ is taken over all arbitrary isomorphic copies $\tilde{G}_{n}$ and $\tilde{H}_{n}$ of $G$ and $H$ in $Gl(n,\mathbb{R})$, respectively, while $Hd$ is the Hausdorff distance in $GL(n,\mathbb{R})$ induced by its standard left invariant metric. -The definition of this metric is motivated by the Hausdorff Gromov metric on the space of compact Riemannian manifolds. - -Is $d$ a metric on $\mathcal{F}$? If the answer is yes, we denote by $\bar{\mathcal{F}}$ the completion of $\mathcal{F}$. What can be said about an object $Z$ in $\bar{\mathcal{F}}$? - -Can one consider the unit circle, in some reasonable sense, as an object in this completion? -Is there a natural group structure on every element $Z\in \bar{\mathcal{F}}$? Is there a natural topology on $Z$? -Is $\bar{\mathcal{F}}$ a compact space? - -REPLY [12 votes]: i don't think it's a metric. Take a large prime $p$. By embedding $\mathbb Z/p\mathbb Z$ and $\mathbb Z/(p^2+p)\mathbb Z$ in the circle $S^1 \subseteq GL(2,\mathbb R)$, one sees that the distance between them is at most $O(1/p)$. By embedding $\mathbb Z/p \mathbb Z \times \mathbb Z/p \mathbb Z$ and $\mathbb Z/p \mathbb Z \times \mathbb Z/(p+1)\mathbb Z$ in the torus $S^1 \times S^1 \subseteq GL(4,\mathbb R)$, one again sees that the distance between them is at most $O(1/p)$. -But of course $\mathbb Z/(p^2+p)$ and $\mathbb Z/p \mathbb Z \times \mathbb Z/(p+1)\mathbb Z$ are isomorphic, so one would be forced to conclude by the triangle inequality that the distance between $\mathbb Z/p\mathbb Z$ and $\mathbb Z/p \mathbb Z \times \mathbb Z/p\mathbb Z$ is $O(1/p)$. -But this is false. If they had that distance in some $GL(n,\mathbb R)$, then by pidgeonhole, $p$ different elements of $\mathbb Z/p \mathbb Z \times \mathbb Z/p\mathbb Z$ would have to be within $O(1/p)$ of some element of $\mathbb Z/p\mathbb Z$ and hence within $O(1/p)$ of each other. By left invariance, $p$ different elements of $\mathbb Z/p \mathbb Z \times \mathbb Z/p\mathbb Z$ would have to be within $O(1)$ of the identity. -But in any representation of $\mathbb Z/p\mathbb Z \times \mathbb Z/p\mathbb Z$, only $o(p)$ elements have eigenvalues within $o(1/\sqrt{p})$ of the identity, since we can write the representation as a sum of characters, the eigenvalues on each character must be $p$th roots of unity, and each element is determined by its eigenvalues on two independent characters. -So we just need to check that every element within $O(1/p)$ of the identity has eigenvalues within $o(1/\sqrt{p})$ of the identity. -In fact we can show more is true, and an element within $d$ of the identity matrix can't move any vector of length one by a distance of greater than $e^{d}-1$. Since $e^{O(1/p)}-1 = O(1/p) = o(1/\sqrt{p})$, we obtain the desired conclusion. To check this, differentiate $Mv$ with respect to $M$ and observe that its operator norm with respect to your metric is the operator norm of $M$, so if $f(x)$ is the maximum total distance moved a vector of length one by a matrix within $x$ of the identity, $df/dx \leq 1+f$ so $f(x) \leq e^x-1$.<|endoftext|> -TITLE: Algebraic fundamental group of a variety -QUESTION [8 upvotes]: I have a very explicit question. Consider a projective variety (a Fano 3-fold) in $\mathbb P^{10}$ defined by 3 quadrics and 32 cubic equations. I want to show that the algebraic fundamental group of the variety is {1}. Is there a way to show it? - -REPLY [5 votes]: I am just collecting my comments above as an answer. Over $\mathbb{C}$ (or any algebraically closed field of characteristic $0$), every log $\mathbb{Q}$-Fano variety is simply connected. A log $\mathbb{Q}$-Fano variety is a pair $(X,D)$ of a normal projective variety $X$ and an effective $\mathbb{Q}$-Cartier divisor $D$ on $X$ such that $(X,D)$ has at worst Kawamata log terminal singularities and $-(K_X+D)$ is nef and big. For such a pair $(X,D)$, for every desingularization $\nu:\widetilde{X}\to X$, $\widetilde{X}$ is rationally connected: this is a theorem of Qi Zhang. -MR2208131 (2006m:14021) -Zhang, Qi(1-MO) -Rational connectedness of log Q-Fano varieties. (English summary) -J. Reine Angew. Math. 590 (2006), 131–142. -14E30 (14J45) -https://arxiv.org/abs/math/0408301 -This was reproved by Hacon and McKernan -as part of their proof of the Shokurov conjecture. -For a smooth projective variety that is rationally connected in characteristic $0$, or more generally if it is separably rationally connected in any characteristic, the variety is algebraically simply connected. This was first proved over $\mathbb{C}$ by Campana (who also proves that the topological fundamental group of the underlying complex manifold is finite, so that the complex manifold is simply connected). In positive characteristic this was proved by Kollár. One nice reference is Section 3.4 of the following. -MR2074059 (2005g:14096) -Debarre, Olivier(F-STRAS-I) -Variétés rationnellement connexes (d'après T. Graber, J. Harris, J. Starr et A. J. de Jong). -Séminaire Bourbaki. Vol. 2001/2002. -Astérisque No. 290 (2003), Exp. No. 905, ix, 243–266. -14M20 (14D06) -https://eudml.org/doc/110309 -Since normal varieties are unibranch, the fundamental group of $X$ is the image of the fundamental group of $\widetilde{X}$ (in general the surjective group homomorphism can have nontrivial kernel -- e.g., for cones over plane curves of degree $d\geq 3$). Since $\widetilde{X}$ is simply connected, also $X$ is simply connected. -Of course it is also important to understand the fundamental group of the smooth locus of $X$. I believe the best results for the smooth locus of $X$ are due to Chenyang Xu, who proves that the algebraic fundamental group of the smooth locus is finite.<|endoftext|> -TITLE: Reference for result on partial sums of Taylor series -QUESTION [5 upvotes]: I remember seeing somewhere that whenever $f$ is a holomorphic function with radius of convergence at $z$, $0 -TITLE: An interesting integral expression for $\pi^n$? -QUESTION [54 upvotes]: I came on the following multiple integral while renormalizing elliptic multiple zeta values: -$$\int_0^1\cdots \int_0^1\int_1^\infty {{1}\over{t_n(t_{n-1}+t_n)\cdots (t_1+\cdots+t_n)}} dt_n\cdots dt_1.$$ -Only the variable $t_n$ goes from $1$ to $\infty$, the others all go from $0$ to $1$. Numerically, I seem to be getting a rational multiple of $\pi^n$. I would like to prove this. Has anyone ever seen an integral like this before? -Edit: After reworking out why I thought it would be a power of $\pi$, I now have a different integral, which numerically really does seem to give a rational multiple of $\zeta(n)$ for each $n$ (though I have only been able to go up to n=4 numerically). I want to integrate $1/(z_1\cdots z_n)$ over the part of the simplex $0\le z_n\le \cdots \le z_1\le 1$ in which $|z_i -z_{i+1}|>\varepsilon$, then let $\varepsilon\rightarrow 0$. The integral is $$\int_{n\epsilon}^{1-\varepsilon}\int_{(n-1)\varepsilon}^{z_1-\epsilon}\cdots \int_{\varepsilon}^{z_{n-1}-\varepsilon} {{1}\over{z_1\cdots z_n}} dz_n\cdots dz_1.$$ It actually diverges when $\varepsilon\rightarrow 0$, but it can be regularised like ordinary multizeta values by computing the integral as a power series in $ln(\varepsilon)$ and $\varepsilon$ and then taking its constant term. It's related to the previous integral by the variable change $z_1=\varepsilon(t_1+\cdots+t_n),\ldots,z_n=\varepsilon t_n$, but the new bounds of the integral form an $(n+1)$-angled polyhedron, not a cube. - -REPLY [4 votes]: As suggested by მამუკა ჯიბლაძე, for n=2,3 there is indeed an interesting connection of -$$I_n=\int\limits_0^1\cdots \int\limits_0^1 dt_1\cdots dt_{n-1}\int\limits_1^\infty dt_n \frac{1}{t_n(t_n+t_{n-1})\cdots(t_n+\cdots +t_1)}$$ -with the Beukers-type integrals. Using the Feynman parametrization -$$\frac{1}{A_1A_2\cdots A_n}=(n-1)!\int\limits_0^1\cdots \int\limits_0^1\frac{\delta\left(1-\sum\limits_{i=1}^n x_i\right)dx_1\cdots dx_n}{(x_1A_1+x_2A_2+\cdots +x_nA_n)^n},$$ -we get after the parametrization and subsequent trivial integration over $t_n$: -$$I_n=\frac{(n-1)!}{n-1}\int\limits_0^1\cdots \int\limits_0^1 \frac{dx_1\cdots dx_n dt_1\cdots dt_{n-1}\;\delta(1-x_1-\cdots-x_n)}{[1+t_{n-1}(x_{n-1}+\cdots x_1)+\cdots +t_1x_1]^{n-1}}.$$ -In particular, for n=2 we get immediately -$$I_2=\int\limits_0^1\int\limits_0^1 \frac{dx dt}{1+xt}=\frac{1}{2}\zeta(2),$$ -and for n=3 we get -$$I_3=\int\limits_0^1\cdots\int\limits_0^1 \frac{dx_1 dx_2 dx_3dt_1 dt_2\;\delta(1-x_1-x_2-x_3)}{[1+t_2(x_1+x_2)+t_1x_1]^2}=$$ $$\iint\limits_0^1\frac{dx_1dx_2dx_3\delta(1-\sum\limits_{i=1}^3 x_i)}{x_1(x_1+x_2)}\left [\ln{(1+x_1+x_2)}+\ln{(1+x_1)}-\ln{(1+2x_1+x_2)}\right]=$$ $$\iiint\limits_0^1\frac{dx_1dx_2dx_3\;\delta(1-\sum\limits_{i=1}^3 x_i)}{x_1(1-x_3)}\left [\ln{(2-x_3)}+\ln{(1+x_1)}-\ln{(2+x_1-x_3)}\right].$$ -Making $x=x_1,y=1-x_3,z=x_2$ change of variables in the integral, and -performing integration over $z$ thanks to the $\delta$-function, we get -$$I_3=\iint\limits_0^1\frac{dx dy}{xy}\left[\ln{(1+x)}+\ln{(1+y)}-\ln{(1+x+y)}\right]\theta(y-x)=$$ $$ -\frac{1}{2}\iint\limits_0^1\frac{dx dy}{xy}\left[\ln{(1+x)}+\ln{(1+y)}-\ln{(1+x+y)}\right], \tag{1}$$ -where in the last step we used the symmetry of the integrand and $\theta(y-x)+\theta(x-y)=1$. This result can be represented in the form -$$I_3=\frac{1}{2}\iiint\limits_0^1\frac{dx dy dz}{1+x+y+xyz}. \tag{2}$$ -Note the resemblance with Beukers integrals -$$ \iiint\limits_0^1\frac{dx dy dz}{1-z+xyz}=2\zeta(3),\;\; -\iiint\limits_0^1\frac{dx dy dz}{1+xyz}=\frac{3}{4}\iiint\limits_0^1\frac{dx dy dz}{1-xyz}=\frac{3}{4}\zeta(3).$$ -Analogously, for $I_4$ we get -$$I_4=\iiint\limits_0^1\frac{dx dy dz}{xyz}\left[\ln{(1+x)}+\ln{(1+y)}+ -\ln{(1+z)}+\ln{(1+x+y+z)}-\right . $$ $$ \left . \ln{(1+x+y)}- \ln{(1+x+z)}-\ln{(1+y+z)}\right]\theta(z-x)\theta(y-z).$$ -$\theta(z-x)\theta(y-z)$ term is a consequence of integration with the help of the $\delta$-function and reflects $x_2=z-x>0$ and $x_3=y-z>0$ conditions. After symmetrization, we finally get -$$I_4=\frac{1}{6}\iiint\limits_0^1\frac{dx dy dz}{xyz}\left[\ln{(1+x)}+\ln{(1+y)}+ -\ln{(1+z)}+\ln{(1+x+y+z)}-\right . $$ $$ \left . \ln{(1+x+y)}- \ln{(1+x+z)}-\ln{(1+y+z)}\right],$$ -and the analogy with Beukers-type integrals is less obvious, because it doesn't seem this can be represented in the form analogous to (2). -I don't know whether there are some further deeper connections to Beukers-type integrals, or to its generalizations (for example Vasilenko integrals - see section 8 in https://eudml.org/doc/249095). -P.S. From (1) we have -$$I_3=\frac{1}{2}\int\limits_0^1f(x)\;d\ln{x}=-\frac{1}{2}\int\limits_0^1 \ln{x}\,\frac{df(x)}{dx}\;dx,$$ -where -$$f(x)=\int\limits_0^1\frac{dy}{y}\left[\ln{(1+x)}+\ln{(1+y)}-\ln{(1+x+y)}\right],$$ -and -$$\frac{df(x)}{dx}=\int\limits_0^1 \frac{dy}{y}\left[\frac{1}{1+x}-\frac{1}{1+x+y}\right]=\frac{\ln{(2+x)}-\ln{(1+x)}}{1+x}.$$ -Therefore -$$I_3=-\frac{1}{2}\left[\int\limits_0^1 \frac{\ln{x}\ln{(2+x)}}{1+x}-\int\limits_0^1 \frac{\ln{x}\ln{(1+x)}}{1+x}\right]dx=-\frac{1}{2}\left[-\frac{13}{24}+\frac{1}{8}\right]\zeta(3)=\frac{5}{24}\zeta(3),$$ -because $$\int\limits_0^1 \frac{\ln{x}\ln{(2+x)}}{1+x}dx=-\frac{13}{24}\zeta(3),$$ and $$\int\limits_0^1 \frac{\ln{x}\ln{(1+x)}}{1+x}dx=-\frac{1}{8}\zeta(3).$$ -In the case of $I_4$, we analogously get -$$I_4=\frac{1}{6}\int\limits_0^1\int\limits_0^1\frac{\ln{x}\ln{y}}{(1+x+y)^2}\left[\ln{(2+x+y)}-\ln{(1+x+y)}+\frac{1}{2+x+y}\right],$$ -however further progress doesn't seem feasible.<|endoftext|> -TITLE: Why not $\mathit{KSO}$, $\mathit{KSpin}$, etc.? -QUESTION [24 upvotes]: If $X$ is a compact Hausdorff space, we can consider the Grothendieck ring of real vector bundles on $X$, -$\mathit{KO}^0(X)$, and this extends to a generalized cohomology theory represented by a ring spectrum -$\mathit{KO}$. Using complex vector bundles, we get another generalized cohomology theory, represented by -$\mathit{KU}$. Using quaternionic vector bundles, we get a third one, represented by $\mathit{KSp} \simeq -\Sigma^4\mathit{KO}$. -In the same way, one can define the Grothendieck ring of oriented vector bundles $\mathit{KSO}(X)$, spin vector -bundles $\mathit{KSpin}(X)$, and so on for any $G$-structure. Do these -constructions extend to spectra $\mathit{KSO}$, $\mathit{KSpin}$, and so on? If not, what issues arise? -The first concern I would imagine is a failure of Bott periodicity, but $\pi_k(\mathit{SO})$ and -$\pi_k(\mathit{Spin})$ agree with $\pi_k(O)$ for $k\ge 2$, so in at least these cases some sort of -construction might be possible. -An alternative approach sidestepping Bott periodicity would be to use algebraic $K$-theory: if $\mathsf{Vect}_{k}$ denotes the topological -symmetric monoidal category of $k$-vector spaces, then $K(\mathsf{Vect}_{\mathbb R}) = \mathit{ko}$ and -$K(\mathsf{Vect}_{\mathbb C}) = \mathit{ku}$. So if $\mathsf{Vect}^{\mathrm{or}}$ denotes the topological symmetric monoidal -category of oriented real vector spaces, would it be reasonable to define $\mathit{kso}:= K(\mathsf{Vect}^{\mathrm{or}})$? Is -this an interesting object? This could also generalize to $G$-structures. -Someone must have thought about this, but I can't find it written down anywhere. -Edit: Though I've accepted Denis' excellent answer, I would still be interested to learn of existing references which consider these kinds of cohomology theories/spectra. - -REPLY [3 votes]: Regarding preexisting references, I found a few examples of people using Grothendieck groups of oriented or spin vector bundles, but no cohomology theories or spectra. I'll leave them here in case anyone else is interested. -For $\mathit{KSpin}$: - -Li-Duan, “Spin characteristic classes and reduced $\mathit{KSpin}$ group of a low-dimensional complex” looks at the reduced Grothendieck group $\widetilde{\mathit{KSpin}}(X)$. -Sati, “An approach to M-theory via KSpin” applies Li-Duan's construction to string theory. -Karoubi, “Clifford modules and invariants of quadratic forms” defines a group $\mathit{KSpin}(X)$ in a different way. - -For $\mathit{KSO}$: - -Pejsachowicz, “Bifurcation of Fredholm maps II: the dimension of the set of bifurcation points.”<|endoftext|> -TITLE: Can an amenable group have a weak mixing unitary representation without almost invariant vectors? -QUESTION [5 upvotes]: Does there exist a finitely generated discrete amenable group $G$ that acts on a separable Hilbert space $\mathcal{H}$ by unitary transformations, and where (1) $\mathcal{H}$ has no finite dimensional $G$-invariant closed subspaces and (2) there does not exist a sequence $(v_n)_n$ of unit vectors in $\mathcal{H}$ such that $\lim_n \|g.v_n - v_n\|=0$ for all $g \in G$? -This is impossible if $\mathcal{H}$ is $L^2$ of an invariant probability measure. - -REPLY [5 votes]: If $G$ is amenable, then every weakly mixing representation $\pi$ has almost invariant finite-dimensional subspaces. This means that $\pi \otimes \bar \pi$ has almost invariant vectors. -Results like this can be found in -M.E.B. Bekka. Amenable unitary representations of locally compact groups. Invent. Math. 100 (1990), 383–401. -A concrete example of a weakly mixing representation without almost invariant vectors would be given by the Heisenberg group $H(\mathbb Z)$ acting on $\ell^2(\mathbb Z^2)$ with generator of the center acting by multiplication with some irrational $\theta \in S^1$. It is easy to see that there are no finite-dimensional subrepresentations and since the generator of the center acts by $\theta$ (and not by $1$), there cannot be any almost invariant vectors.<|endoftext|> -TITLE: Convex hull with genus information -QUESTION [5 upvotes]: Are there convexity generalizations that admit genus information? -For example in genus $1$ is there a way to think of this polyhedron as convex while this polyhedron as non-convex? Any two points can be joined by a line or a circle seems to work. -Is there a good definition that works in higher dimensions for which appropriate generalization of traditional convex geometry inequalities such as Brunn-Minkowski inequality can be given (A suitable notion of convex hull needs to be first defined). - -REPLY [6 votes]: Natural higher genus analogues of convex surfaces are usually considered to be surfaces which satisfy the "two piece property" or are "tight". -A closed surface in Euclidean space is said to have the two piece property or be tight if no plane cuts it into more than two pieces. So for instance tori of revolution are tight as are all convex surfaces, and the polyhedral torus referenced in the question. An equivalent formulation is that the "total positive curvature", or the integral of the Gauss curvature where it is positive, must be 4π, which is its lowest possible value. This implies that all points of positive curvature must lie on the boundary of the convex hull of the surface. -The study of these surfaces originates with Alexandrov, who showed that tight analytic surfaces are isometrically rigid. Later significant contributions were made by Chern and Lashoff (who studied total curvature), Nirenberg (Who tried to extend Alexandrov's rigidity theorem to smooth surfaces), Banchoff (who coined the term "two piece property"), Kuiper, and others. -A very good reference is the survey book Tight and Taut Subamnifolds, edited by Cecil and Chern.<|endoftext|> -TITLE: Induced map of an approximately inner automorphism on the multiplier algebra of $A\otimes\mathcal{K}$ -QUESTION [8 upvotes]: Suppose that $A$ is a separable, simple, non-unital C*-algebra. Let $\varphi$ be an approximately inner automorphism on $A\otimes\cal K$, meaning that there exists a sequence of unitaries $v_n$ in the multiplier algebra $\mathcal{M}(A\otimes\mathcal{K})$ with $v_n x v_n^* \stackrel{n\to\infty}{\longrightarrow} \varphi(x)$ for all $x\in A\otimes\cal K$. We also denote by $\varphi$ the induced automorphism on the multiplier algebra. -Let $e_{11}\in\cal K$ denote the obvious rank one projection. - -My question: Is $1\otimes e_{11}$ Murray-von-Neumann equivalent to $\varphi(1\otimes e_{11})$ in the multiplier algebra $\mathcal{M}(A\otimes\mathcal{K})$? - -It is obvious that this is true if $A$ is unital, but unclear (to me) otherwise. I don't know whether simplicity of $A$ is a red herring here, but that is the case that I am interested in for now. - -REPLY [3 votes]: If $A$ has stable rank one then this can be deduced from the following result: -If $B$ has stable rank one and $a,b\in B_+$ are Cuntz equivalent then -the right ideals $\overline{aB}$ and $\overline{bB}$ are isomorphic as Hilbert modules. This is Theorem 3 of -Coward, Kristofer T.; Elliott, George A.; Ivanescu, Cristian. The Cuntz semigroup as an invariant for $C^*$-algebras. J. Reine Angew. Math. 623 (2008), 161--193. -See also Proposition 1 of -Ciuperca, Alin; Elliott, George A.; Santiago, Luis. On inductive limits of type-I $C^*$-algebras with one-dimensional spectrum. Int. Math. Res. Not. IMRN 2011, no. 11, 2577--2615. -It's a little awkward to translate this into an answer to the question. Let me try. -First, this: Let $p,q\in M(B)$ be multiplier projections such that $pB\cong qB$ as Hilbert modules. Then $p$ is Murray-von Neumann equivalent to $q$. Proof: Let $v\colon pB\to qB$ be an isomorphism. Extend $v$ to $B$ setting it to 0 on $(1-p)B$. Then $v\in M(B)$ (regarded as the adjointable operators on $B$) and $v^*v=p$, $vv^*=q$. -Now let $B$ be separable and of stable rank one. Let $\phi$ be an approximately inner automorphism extended to $M(B)$. Let us show that any projection $p\in M(B)$ is Murray-von Neumann equivalent to $\phi(p)$. It suffices to show that $pB\cong \phi(p)B$. Let $c\in B$ be a strictly -positive element of $pBp$ (exists since $B$ is separable). Then $pB=\overline{cB}$. -Since $c^{1/n}\uparrow p$ strictly and $\phi$ is strictly continuous, we also have $\phi(p)B=\overline{\phi(c)B}$. But $c$ and $\phi(c)$ are approximately unitarily equivalent, so they are Cuntz equivalent. We can use the theorem recalled above. -To answer the question, we take $B=A\otimes \mathcal K$ and $p=1\otimes e_{11}$. -Notice that to answer the question we only need $A$ $\sigma$-unital rather than separable. -Remark: There exist examples of positive elements $a,b$ in an $A$ (of sr=2) that are approximately unitarily equivalent and yet $\overline{aA}$ and $\overline{bA}$ are not isomorphic as Hilbert modules. However, (1) $A$ is non-simple and they way the examples work they can't be tweaked to get this, (2) The unitary conjugations moving $a$ to $b$ don't seem to converge pointwise to an automorphism. So there is a chance that the question has a positive answer without sr1.<|endoftext|> -TITLE: Frequency of papers showing academic misconduct among the articles indexed by MathSciNet and Zentralblatt MATH -QUESTION [31 upvotes]: Among the papers indexed by MathSciNet and Zentralblatt MATH, -I occasionally have seen papers which consist essentially only -of text copied from elsewhere without proper attribution and without -adding any significant value. I would be interested whether anyone -has an idea what the frequency of such papers among those indexed in -the mentioned databases roughly is. -- -Are these extremely rare cases, or are such papers more common than one usually thinks, and perhaps even not easy to keep out of the databases if -one doesn't want to be too restrictive in which journals to cover? -- -Is there any data known on this? -Also, if one spots such a paper -- should one report this to the -authors or copyright holders of the pieces of text from which the -paper is composed, to the editorial board of the journal in which -the paper is published, or to MathSciNet / Zentralblatt MATH -- -or rather just ignore it? What is common practice in such case? - -REPLY [35 votes]: On behalf of MathSciNet / Mathematical Reviews, I concur with Olaf Teschke that we appreciate notification of such cases. We too sense that the number of cases has increased. The majority of the alerts we receive come from our reviewers. We sometimes have authors contact us to say that their papers have been republished by someone else. And we do receive a number of notifications from third parties. -If you wish to notify MathSciNet of such a case, you may write to mrexec@ams.org (as in MR Executive Editor). If you wish to become a reviewer, you may write to us at mathrev@ams.org. -As with zbMATH, when a case of duplicated text or duplicated papers comes up, we contact the editorial boards of both journals, inform the editors at zbMATH, and add a note to the listing of the paper. -Edward Dunne, -Executive Editor, -Mathematical Reviews<|endoftext|> -TITLE: Peter-Weyl vs. Schur-Weyl theorem -QUESTION [22 upvotes]: Let $V$ be a finite dimensional complex vector space. -According to the Peter-Weyl theorem there is a decomposition $\mathcal O(\mathrm{GL}(V)) \cong \bigoplus_\lambda V_\lambda \otimes V_\lambda^\ast$ of the algebraic coordinate ring of $\mathrm{GL}(V)$ into a direct sum indexed by partitions, where $V_\lambda$ denotes the representation of highest weight $\lambda$. -According to Schur-Weyl duality there is a decomposition $T(V) \cong \bigoplus_{\lambda} V_\lambda \otimes \sigma_\lambda$ of the tensor algebra on $V$, where $\sigma_\lambda$ now denotes the Specht module associated to a partition $\lambda$. -The two statements look very similar. Is there a direct relation between the commutative ring $\mathcal O(\mathrm{GL}(V))$ and the associative algebra $T(V)$? E.g. a map between them that behaves nicely w.r.t. the decompositions? - -REPLY [8 votes]: Allen's nice answer led me in a slightly different direction. Let me try to give another answer to the question. -Let's start from the Cauchy identities, -$$ \prod_{i,j \geq 0} (1-x_i y_j)^{-1} = \sum_\lambda s_\lambda(x)s_\lambda(y) $$ -which is an equality between bisymmetric functions in infinitely many variables $\{x_i\}$ and $\{y_i\}$. -Now recall that there is a correspondence between symmetric functions of degree $n$, representations of $S_n$, and polynomial functors $\mathrm{Vect}_{\mathbb C}\to\mathrm{Vect}_{\mathbb C}$ of degree $n$. Passing to the completion of the ring of symmetric functions wrt degree gives instead a correspondence between (completed) symmetric functions, sequences of representations of $S_n$ (i.e. "tensorial species") and analytic functors $\mathrm{Vect}_{\mathbb C}\to\mathrm{Vect}_{\mathbb C}$. -Using this we can give three different interpretations of the Cauchy identities: -(1) Consider both the $x$- and $y$-variables as corresponding to representations of the symmetric groups. The Cauchy identities become -$$ \bigoplus_{n \geq 0} \mathbb C[S_n] = \bigoplus_{\lambda} \sigma_\lambda \otimes \sigma_\lambda,$$ -i.e. the Peter-Weyl theorem for $S_n$. -(2) Consider the $x$-variables as corresponding to an analytic functor and the $y$-variables as corresponding to a sequence of representations. Then the left hand side becomes the analytic functor $V \mapsto T(V)$ and the right hand side becomes $V \mapsto \bigoplus_\lambda V_\lambda \otimes \sigma_\lambda$. -(3) Consider both $x$- and $y$-variables as corresponding to analytic functors. The left hand side becomes the analytic functor $(V,W) \mapsto \mathcal O(V\otimes W)$ and the right hand side becomes $(V,W) \mapsto \bigoplus_\lambda V_\lambda \otimes W_\lambda$. -Specializing to $W = V^\ast$ in (3) gives the coordinate ring of the matrix space as in Allen's answer. -The three interpretations of the Cauchy identities can be seen as equalities between sequences of representations of $S_n \times S_n$, sequences of polynomial functors of degree $n$ into $S_n$-representations, and analytic functors of two variables, respectively. But in all cases there is also an obvious multiplication on the left hand side: given by the inclusion $\mathbb C[S_n] \otimes \mathbb C[S_m] \to \mathbb C[S_{n+m}]$, the multiplication in the tensor algebra, and the multiplication in the coordinate ring, respectively. This is because in all three cases we have a commutative algebra object in the respective symmetric monoidal category, and the structure of commutative algebra object gets transferred along the different equivalences of categories. For example, a commutative algebra object in the category of tensorial species is what's usually called a twisted commutative algebra, so we are saying that the tensor algebra $T(V)$ is a twisted commutative algebra (even though the multiplication in $T(V)$ is certainly not commutative), and so on. So the multiplication in $T(V)$ is in a precise sense "the same" as the multiplication in the coordinate ring of $V \otimes W$! -PS - I certainly hope all the above is correct. But I am confused about the fact that what appears is $\sigma_\lambda \otimes \sigma_\lambda$ in case (1), rather than $\sigma_\lambda \otimes \sigma_\lambda^\ast$ which would be more expected. (Of course $\sigma_\lambda \cong \sigma_\lambda^\ast$, but I would still like the dual to be there!)<|endoftext|> -TITLE: Can Khovanov homology have arbitrarily large torsion? -QUESTION [11 upvotes]: Can Khovanov homology have arbitrarily large torsion? -That is, given $N\gg 0,$ does there exist $k>N$, a knot (diagram) $D$ and $i,j \in \mathbb{Z}$ such that $\operatorname{Kh}^{i,j}(D) = \mathbb{Z}/k\mathbb{Z}$? - -REPLY [17 votes]: This paper from earlier this year (Jan 18, to be precise) proves the existence of $\mathbb{Z}/n\mathbb{Z}$-torsion for $n\le 8$ and $\mathbb{Z}/2^s\mathbb{Z}$-torsion for $s\le23$. It also states at the beginning of Section 3.4: - -Until now, no knot or link with torsion larger than $\mathbb{Z}/8\mathbb{Z}$ was known. - -I believe this is the state of art. -Computations suggest that $T(p^k,p^k+1)$ should have $\mathbb{Z}/p^k\mathbb{Z}$-torsion for each $p$ prime and $k\ge 1$.<|endoftext|> -TITLE: Is there a four-manifold whose tangent bundle is an endomorphism bundle? -QUESTION [7 upvotes]: Is there a smooth four-manifold $M$ such that $TM \cong \operatorname{End}(E)$ for some rank $2$ bundle $E \to M$? - -If $M$ is parallelisable, then one can take $E$ to be the trivial rank $2$ bundle over $M$. I would like to know if there is a non-trivial example. -As sections of the bundle $\operatorname{End}(E)$ are vector bundle morphisms $E \to E$ (covering the identity map $M \to M$), the identity map $\operatorname{id}_E$ defines a nowhere-zero section of $\operatorname{End}(E)$. Therefore, such an $M$ must have Euler characteristic zero by the Poincaré-Hopf Theorem. -Using the real splitting principle, I was able to show that $w(\operatorname{End}(E)) = 1 + w_1(E)^2$. In particular, $M$ must be orientable and $w_2(M)$ is a square; such manifolds were asked about here, but no explicit four-manifold examples were given. -Choosing an orientation for $M$, I would like to compute its first Pontryagin class and hence the signature of $M$, but I have been unable to do so. Even if I had this information, I would have no idea how to proceed. -Added Later: Thanks to Igor Belegradek for his help. Using his advice, one can show that $p_1(\operatorname{End}(E)) = -4c_2(E_{\mathbb{C}}) = 4p_1(E)$ where $E_{\mathbb{C}} = E\otimes_{\mathbb{R}}\mathbb{C}$ is the complexification of $E$. If $E$ is orientable, it can be viewed as a complex line bundle, in which case $E_{\mathbb{C}} = E\oplus\overline{E}$ and the expression for $p_1(\operatorname{End}(E))$ reduces to $4c_1(E)^2$ as in my answer below. -This expression for $p_1$, together with the Dold-Whitney Theorem, might also lead to examples where $E$ is non-orientable. - -REPLY [2 votes]: Note, this answer was conceived before I understood Igor Belegradek's comments regarding the calculation of $p_1$. - -Suppose $E$ is orientable, i.e. $w_1(E) = 0$. Then $w_2(\operatorname{End}(E)) = w_1(E)^2 = 0$, so $M$ must be spin. -Choosing an orientation for $E$, we can view $E$ as a complex line bundle. Then -$$\operatorname{End}(E) = \operatorname{End}_{\mathbb{C}}(E)\oplus\overline{\operatorname{End}}_{\mathbb{C}}(E)$$ -where the terms of the decomposition are complex linear and complex antilinear endomorphisms respectively. If $J$ denotes the almost complex structure on $E$, then the decomposition is given by $L \mapsto \frac{1}{2}(L - JLJ) + \frac{1}{2}(L+JLJ)$. Note that $\operatorname{id}_E$ defines a nowhere-zero section of $\operatorname{End}_{\mathbb{C}}(E)$, so $\operatorname{End}_{\mathbb{C}}(E) \cong \varepsilon_{\mathbb{C}}^1$ (alternatively, $\operatorname{End}_{\mathbb{C}}(E) \cong E^*\otimes E \cong \varepsilon^1_{\mathbb{C}}$). On the other hand, a complex anti-linear endomorphism of $E$ can be viewed as a complex linear homomorphism $E \to \overline{E}$, so -$$\overline{\operatorname{End}}_{\mathbb{C}}(E) \cong \operatorname{Hom}_{\mathbb{C}}(E, \overline{E}) \cong E^*\otimes\overline{E} \cong \overline{E}^2.$$ -Therefore -\begin{align*} -p_1(\operatorname{End}(E)) &= p_1(\varepsilon_{\mathbb{C}}^1\oplus\overline{E}^2)\\ -&= p_1(\overline{E}^2)\\ -&= -c_2(\overline{E}^2\otimes_{\mathbb{R}}\mathbb{C})\\ -&= -c_2(\overline{E}^2\oplus E^2)\\ -&= -c_1(\overline{E}^2)c_1(E^2)\\ -&= -4c_1(\overline{E})c_1(E)\\ -&= 4c_1(E)^2. -\end{align*} -Now we can use the Dold-Whitney Theorem. - -Dold-Whitney Theorem: Oriented rank four real bundles on an closed, orientable four-manifold are uniquely determined by their second Stiefel-Whitney class $w_2$, Euler class $e$, and first Pontryagin class $p_1$. - -Under the assumption that $E$ is orientable, the bundle $\operatorname{End}(E)$ is an oriented rank four real bundle on an orientable four-manifold with $w_2(\operatorname{End}(E)) = 0$, $e(\operatorname{End}(E)) = 0$ and $p_1(\operatorname{End}(E)) = 4c_1(E)^2$. If we can find an orientable four-manifold $M$ which is spin, has $\chi(M) = 0$, and $p_1(M) = 4c_1(E)^2$ for some complex line bundle $E$ on $M$, then by the Dold-Whitney Theorem, $TM \cong \operatorname{End}(E)$. -Let $M$ be a closed, smooth, spin four-manifold with $\chi(M) = 0$. If $\tau(M) = 0$, then $p_1(M) = 0 = 4c_1(E)^2$ where $E$ is the trivial complex line bundle. If $\tau(M) \neq 0$, then the intersection form of $M$ is of the form $\pm 2mE_8\oplus nH$ for some non-negative integers $m, n$. By Donaldson's theorem on definite intersection forms, if $m \neq 0$, then $n \neq 0$. So if $\tau(M) \neq 0$, it must have a copy of the hyperbolic lattice $H$ in its intersection form. Therefore every even integer is of the form $c^2$ for some $c \in H^2(M; \mathbb{Z})$. By Rokhlin's theorem, $\tau(M)$ is a multiple of $16$, so $\frac{3}{4}\tau(M) = \frac{1}{4}p_1(M)$ is a multiple of 12 (in particular, even), so there is a class $c$ such that $c^2 = \frac{1}{4}p_1(M)$, i.e. $p_1(M) = 4c^2$. As every element of $H^2(M; \mathbb{Z})$ is the first Chern class of some complex line bundle, we see that $p_1(M) = 4c_1(E)^2$ for some complex line bundle $E$. Therefore, we have the following: - -Let $M$ be a closed, smooth, spin four-manifold with $\chi(M) = 0$. Then $TM \cong \operatorname{End}(E)$ for some $E$. - -Moreover, by Dold-Whitney, such an $M$ is parallelisable if and only if $\tau(M) = 0$. In particular, if $M$ is spin with $\chi(M) = 0$, $\tau(M) \neq 0$, we get a non-trivial example. -Here's one way to construct such examples. Let $X$ be a closed, smooth, simply connected, spin four-manifold and set $M = X\# k(S^1\times S^3)$ where $k = \frac{1}{2}\chi(X)$. Then $M$ is a spin manifold with $\chi(M) = 0$, so $TM \cong \operatorname{End}(E)$ for some $E$. As $\tau(M) = \tau(X)$, we obtain non-trivial examples whenever $\tau(X) \neq 0$. The simplest non-parallelisable manifold that arises from this construction is -$$M = K3\#12(S^1\times S^3).$$ -In this case $E$ is the unique complex line bundle with $c_1(E)^2 = -12$.<|endoftext|> -TITLE: Weird analogy between quadratic forms and formal systems -QUESTION [7 upvotes]: A fundamental connection between provability and consistency for formal systems is that, if $Q$ is a formal system and $A$ is a sentence in the language of $S$, then - -$Q$ proves $A$ if and only if $Q + \neg A$ is inconsistent - -(see for example Franzén's excellent Gödel's theorem: An Incomplete Guide to its Use and Abuse, p.19.) -In quadratic form theory (more precisely: for nondegenerate quadratic forms over a field of characteristic $\neq 2$), an equally fundamental connection between representability (a quadratic form $q$ is said to represent a scalar $a \in K$ if there is a nonzero vector $v$ in the vector space on which $q$ is defined s.t. $q(v) = a$) and isotropy (a quadratic form is isotropic if it represents $0$) is that - -$q$ represents $a$ if and only if $q \oplus \left<-a\right>$ is isotropic - -(see for example Lam's equally excellent Introduction to Quadratic Forms over Fields, p. 11). -My question is simply: what's up with that? -To be more specific: is there a general relevant framework which allows us to link formal systems and the sentences they prove on the one side, and quadratic forms and the scalars they represent on the other? -The two theories having quite distinct flavours, I would find such a framework really fascinating. -(I've asked the exact same question on MSE, without success.) - -REPLY [4 votes]: I like this question, and I think there is a general relevant framework which links the two cases, but it is so general that you may be disappointed. -Let me start by reformulating the two settings a little. First, if you factor out tautological equivalence, then the set of sentences of a theory forms a Boolean algebra. Given a set of sentences, the sentences provable from them can be regarded as those which are "generated" via formal derivations. -Second, every quadratic form can be diagonalized, i.e., put in the form $\sum a_ix_i^2$, and so can be represented by the multiset $\{a_1, \ldots, a_n\}$. The scalars represented by the form can be viewed as those which are "generated" from this multiset via taking linear combinations with coefficients which are squares. -So, very generally, in both cases we have a set with some algebraic structure (field or Boolean algebra) which includes a negation operation, and we have a notion of elements generated by a subset (or multisubset). -Now in both cases one implication is rather easy. If $Q$ proves $A$, then $Q + \neg A$ is inconsistent --- that is just because $Q + \neg A$ proves $A \wedge \neg A$ and $A \wedge \neg A$ implies $\perp$ (falsehood). If $q$ represents $a$ then $q \oplus \langle -a\rangle$ is isotropic --- this is just because $q \oplus \langle -a\rangle$ represents $a + (-a) = 0$. (Note that in the multiset representation $q \oplus \langle -a\rangle$ becomes $q \cup \{-a\}$.) -The reverse implications are more interesting. If $Q + \neg A$ is inconsistent, i.e., it proves $\perp$, then $Q$ proves $A$ --- this is a special case of the fact that $Q + B \vdash C$ implies $Q \vdash B \to C$. Because $Q + \neg A \vdash \perp$ implies $Q \vdash \neg A \to \perp$ and $\neg A \to \perp$ equals $\neg\neg A$, which under tautological equivalence is the same as $A$. The analogous statement for quadratic forms is that if $q \oplus \langle b\rangle$ represents $c$ then $q$ represents either $c$ or $cx^2 - b$ for some scalar $x$. So that if $q \oplus \langle -a\rangle$ represents $0$ then either $q$ represents $0$ and hence $a$ (it is a theorem that if nondegenerate $q$ represents $0$ then it represents everything) or else $q$ represents $0 -(-a) = a$.<|endoftext|> -TITLE: Exact sequence of $n$th powers of abelian groups -QUESTION [8 upvotes]: Let $A,B,C$ be finitely generated abelian groups. Assume that there is an exact sequence $$0 \to C \to A^n \to B^n \to 0,$$where $A^n = A \oplus \dotsc \oplus A$ as usual. It is not assumed that $A^n \to B^n$ is induced by some $A \to B$. - -Assume that $C$ can be generated by $ -TITLE: Meaning of the determinant of cohomology -QUESTION [14 upvotes]: The Arakelov intersection number on arithmetic surfaces is defined as an "extension" of the classical intersection number on algebraic surfaces. It was introduced to get a nice intersection theory that behaves well up to linear equivalence of divisors in the arithmetic case. In particular, we need some analytic data on the fibers at infinity and a new extended concept of divisors, namely Arakelov divisors. -Everything works well and we even have a correspondence between Arakelov divisors and metrized line bundles. The big problem is that, of course, we don't have any cohomology theory for such kind of line bundles, because the data at infinity is somewhat artificial. -The Faltings-Riemann-Roch theorem deals with the so called determinant of the cohomology introduced by Deligne. Formally, it is a way to associate to each coherent module on our surface $X$, a line bundle to the base scheme. -I think that it should be something similar to the concept of dimension of some cohomology group. But the problem is that I'm not able to understand what is the intuition behind this new tool. What information do we want to capture with the determinant of cohomology? What is the analogy with the geometric case? -Many thanks in advance - -REPLY [2 votes]: Let us consider the arithemetic surface case, which is already very difficult (see the recent work by Gerald Montplet, for example). In this case Faltings-Riemann-Roch established that -$$ -\chi(O_{X}(D))-\chi(O_{X})=\frac{1}{2}(D,D-K) -$$ -Therefore formally in order to evaluate $\chi(O_{X}(D))$ it is suffice to find $\chi(O_{X})$. -There are however at least two issues at here, one is that the intersection pairing $$ -(D, D-K) -$$ -in Arakelov theory depends on the choice of Green functions, which is dependent upon the choice of the Arakelov metric. If we use a different metric, then everything will have to be changed. The other issue is Faltings' construction using the technique of "reduction to the Jacobian" in an essential way and cannot be easily generalized to higher dimensions (see the MathScithnet review, for example). -The one-step "magic solution" of the issue essentially comes from the introduction of the Quillen metric, namely finding a metric on the determinant line bundle over $X_{\infty}$ that satisfies the following properties: -1) It should be smooth on the moduli space of genus $g$ complex algebraic curves. Namely, when we "slide" on the "huge" space parametrizing the whole space of genus $g$ complex algebraic curves, the metric on the determinant line bundle should be a smooth function. -2) It should be compatible with Serre duality. -3) It should not be really dependent upon the metric. Namely if we have two different metrics $d_1, d_2$ on $X$, then the metric on the determinant of cohomology should be independent of $s$ for $d_s=sd_1+(1-s)d_2$. In other words, it should be coming from a topological invariant. -4) It should be naturally generalizable to higher dimensions, not using special nice properties only available for Riemann surfaces. -In Ray-Singer's first paper, they proved (3) for analytic torsion (not for Quillen metric, which did not exist by then). For a survey paper on this, see Pavel Mnev's article. In Quillen's paper, he claimed (1) and Soule provided a detailed proof in his book. I think (2) comes from Deligne's introduction of Deligne pairing and (4) comes from Bismut-Gillet-Soule's work. I think this is partly why analytic torsion considered to be so important for Arakelov theory. I do not know any other analytic invariant that can satisfy (1)-(4) in the same time. If it exists, then it has potential to be the building block for a generalized Arakelov cohomology theory. -Part of the difficulty is that analytic torsion is a "secondary global invariant" that involves all positive time of the heat kernel (you have to deal with $Tr(\int^{\infty}_{0}t^{s}e^{-\Delta t}dt|'_{s=0}$). So any potential good candidate that can define $h^{0}_{Ar}, h^{1}_{Ar}$ would have to involve spectral theory of elliptic operators in an essential way. And it is not just the $\zeta$-function of the operator, but its derivative at $s=0$. Therefore the problem become exceedingly difficult and from what I read even computing $\chi(O_{X})$ for $\mathbb{P}^{N}$ in general is not easy. -I am really a beginner in the field and maybe for you discussing this with experts (Gillet, Soule, Bost, Montplet, Faltings, etc) will be helpful.<|endoftext|> -TITLE: Simple Proof that a Reductive Group is Unimodular? -QUESTION [5 upvotes]: Let $G$ be a connected, reductive group over a local field $k$ of characteristic zero. I thought of a simple proof that $G(k)$ is unimodular, but I realize it is almost certainly wrong: $G(k)$ is generated by its derived group and its center, and the modular character is trivial on both of these, Q.E.D. -I am relying on some results from algebraic groups over algebraically closed fields which I am not certain carry over to arbitrary fields. So what I wanted to know which of these results are false for arbitrary $k$ (all being true in the case $k = \overline{k}$) --$[G,G](k) = [G(k),G(k)]$ --$Z(G)(k) = Z(G(k))$ --$G(k)$ is generated by $[G,G](k)$ and $Z(G)(k)$ -Moreover, can the idea of this proof be modified to correctly show that $G(k)$ is modular? - -REPLY [10 votes]: The modular character $\Delta: G(k)\rightarrow {\mathbb R}_{>0}$ on $G(k)$ is trivial on the commutator subgroup and is trivial on a compact open subgroup and also on the centre. These three groups generate a subgroup of finite index since the abeliansation of $G(k)$ modulo centre is compact. Hence the modular character is trivial on an open subgroup of finite index, and is hence trivial. This works in arbitrary characteristic. -I am sure that I have seen this argument somewhere in the literature, but old age and failing memory prevent me from recalling the reference. - -REPLY [5 votes]: Since you assume that $k$ has characteristic zero, one can make use of the Lie algebra. If $G$ is an arbitrary Lie group over a locally compact field $k$ of characteristic zero, $G$ is unimodular if and only of the adjoint action of $G$ on its Lie algebra is by elements of determinant of modulus 1. In the case of $G=\mathbb{G}_k$, with $\mathbb{G}$ reductive connected, the Lie algebra has the form $\mathfrak{s}\oplus\mathfrak{z}$ with $\mathfrak{s}$ semisimple and $\mathfrak{z}$ central, and the adjoint $G$-action preserves this decomposition and is trivial on $\mathfrak{z}$. Moreover the automorphism group of $\mathbb{s}$ has a finite index subgroup acting with determinant 1 (because self-derivations of $\mathfrak{s}$ are inner and have trace zero because $\mathfrak{s}$ is perfect). This entails the unimodularity statement.<|endoftext|> -TITLE: Slick construction of Hochschild complex -QUESTION [5 upvotes]: Let $R$ be a $k$-algebra and $M$ be an $(R,R)$-bimodule. Let $[n] \mapsto M \otimes R^{\otimes n}$ be the simplicial $k$-module which defines the Hochschild homology $H_*(R,M)$. Is it possible to write down explicitly the $k$-linear map $f^* : M \otimes R^{\otimes n} \to M \otimes R^{\otimes m}$ associated to a monotonic map $f : [m] \to [n]$? Does this give a more slick way to construct the simplicial module? I imagine that once we have a formula, one can prove $(f \circ g)^* = g^* \circ f^*$ directly and does not have to verify all the simplicial identities. - -REPLY [3 votes]: The category Δ_a of finite linearly ordered sets with disjoint union is a monoidal category. -A monoid M in a monoidal category F is a strong monoidal functor Δ_a → F. -Discarding the monoidal structure on such a functor yields an augmented simplicial object, -which is (by definition) the bar construction of M. -The reversal functor Δ_a → Δ_a turns a monoid into its opposite. -The category Δ_e of pointed objects in Δ_a, where the basepoint is the smallest element, -is a right module over Δ_a. -A right module X over a monoid M in a monoidal category F can now be defined -as a strong right module over M in the sense of strong monoidal functors, -with structure maps X(x)⊗M(m) → X(x⊗m). -There is a canonical inclusion Δ_e → Δ_a (add a basepoint as the smallest element), -and restricting along it gives back M. -A left module is defined in the same way, but using Δ^e, where the basepoint is now the largest element. -We also have an inclusion Δ^e→Δ_a. -The reversal functor Δ^e → Δ_e turns a right module into a left module, -and vice versa. -Finally, M-N-bimodules can be defined using Δ_*, the category of pointed objects in Δ (or Δ_a). -We have two inclusions Δ_e → Δ_* and Δ^e → Δ_*, which recover M and N. -The reversal functor Δ_* → Δ_* sends an M-N-bimodule into an N^op-M^op-bimodule. -For M-M bimodules we can “glue” together the left and right action -and define an M-cyclic module as a functor C_* → F -whose restriction along the inclusion Δ_* → C_* yields an M-M-bimodule. -Here C_* denotes the pointed cyclic category (definition left as an exercise) -and Δ_* → C_* closes a finite linearly ordered set into a cycle. -(If the bimodule is M with its left and right M-action, -then cycles need not be pointed, and we recover the cyclic bar construction C → F of M this way.) -The underlying simplicial object of an M-cyclic module -can be now recovered by precomposing with the functors Δ → Δ_* → C_*. -All of this is easy to remember if one keeps in mind -that all these combinatorial categories -are small models for the monoidal Weiss sites of various topological manifolds with boundaries and defects: -the interval (0,1) for Δ_a, -[0,1) for Δ_e, -(0,1] for Δ^e, -(-1,1) with {0} pointed for Δ_*, -S^1_* for C_*, -S^1 for C.<|endoftext|> -TITLE: Automatic transfer of pointwise metric computations to bundle computations -QUESTION [7 upvotes]: $\newcommand{\M}{\mathcal{M}}$ -$\newcommand{\N}{\mathcal{N}}$ -$\newcommand{\deriv}[2]{\frac{d#1}{d#2}}$ -$\newcommand{\sAverage}[1]{\langle#1\rangle} $ -$\newcommand{\IP}[2]{\sAverage{#1,#2}}$ -$\newcommand{\Cof}{\text{Cof}}$ -$\newcommand{\Hom}{\operatorname{Hom}}$ -$\newcommand{\tr}{\operatorname{tr}}$ -$\newcommand{\TM}{\operatorname{T\M}}$ -$\newcommand{\TN}{\operatorname{T\N}}$ -$\newcommand{\TstarM}{\operatorname{T^*\M}}$ -There is a well-known folklore saying that "any linear algebraic construction/statement can be lifted to vector bundles" (e.g tensor products, direct sums, quotients etc). -I am interested in a metric version of this phenomena: - -Does every statement about inner-prodcut spaces admit a vector bundle analog? - -Specifically, I am interested in "derivations-type" results: -On various occasions, I need to compute derivatives of certain "geometric quantities" associated with bundle maps over a manifold. (examples are given below). -Often, I find it's easier to start with a finite dimensional analogous computation. The computation in the bundle context then becomes a routine adaptation of the original calculation, modulu some extra justifications (revolving around the compatiblity of connections with metrics). -Soft Question: Is there a way to "automate" this transfer? (I want to avoid repeating essentially the same calculation twice). In other words, is there a way to prove a "meta-theorem" which says that the result in the pointwise context carries over to the bundle context? -Main Example: Calculating the derivative of the determinant. -We want to prove the following: - -Theorem 1: Let $f:\M \to \N$ be a smooth map between $d$-dimensional oriented Riemannian manifolds. - Define $\Cof df= (-1)^{d-1} \star_{f^*TN}^{d-1} (\wedge^{d-1} df) \star_{TM}^1.$ - Then for all $V \in \Gamma(\TM)$ - $$ - d(\det df)(V)= \IP{\Cof df}{\nabla_V df}_{\TM,f^*{\TN}} . - $$ - - -We start first with the "pointwise" analogy, where the vector spaces are fixed and only the linear map is changing: - -Proposition: (The cofactor is the gradient of the determinant) - Let $V,W$ be oriented $d$-dimensional inner product spaces. Then $$d(\det)_A(B)=\tr\left( \Cof A^T B \right)=\IP{\Cof A}{B}_{V,W}.$$ - -Specific question: Can we deduce the theorem from the proposition? (without using the proof of the proposition, like I am doing below). -One obvious way to achieve this would be to view $p \to \det(df_p)$ as the determinant of a changing map between fixed vector spaces. This can be done by representing $df$ w.r.t orthonormal frames. However, one then needs to track the derivative of this matrix in terms of $V$ which looks cumbersome. (I would say that even if this approach would work, it is less aesthetic - an invariant way would be better). -Edit: -As pointed out by Deane Yang, there is a more general version of theorem $1$ which is the right "bundle-analog" of the finite-dim proposition: - -Theorem 2: Let $E$ and $F$ be rank $d$ oriented vector bundles over $\M$ with smooth metrics and compatible connections. Let $A:E \to F$ be a smooth bundle map. - Define $\Cof A= (-1)^{d-1} \star_{F}^{d-1} (\wedge^{d-1} A) \star_{E}^1.$ - Then for all $V \in \Gamma(\TM)$ - $$ - d(\det A)(V)= \IP{\Cof A}{\nabla_V A}_{E,F}. - $$ - -The proof of theorem $2$ is exactly the same as the proof of theorem 1 (see below) - we just replace $df \to A$ everywhere (that proof does not use the fact $df$ is the differential of a map, just the bundle-structures). -The question still remains- can we use the statement of the proposition -to deduce theorem $2$, without looking at the proof. -(This is not a trivial consequence of the proposition, where the two vector spaces, while different, are fixed). - -proof of the proposition: - Let $A_t$ be a smooth family of mappings in $\Hom(V,W)$: $A(0)=A,A'(0)=B$, and let $e_1,\dots,e_d$ be a positive orthonormal basis of $V$. -$$ - \det(A_t)= \star^d_W \circ \bigwedge^d A_t \circ \star^0_V(1)= \star^d_W \bigwedge^d A_t \big( e_1 \wedge \dots \wedge e_d \big)= \star^d_W \big( A_t e_1 \wedge \dots \wedge A_te_d \big) - $$ -Using the Leibniz rule we get: -$$ -\left. \deriv{\det A_t}{t} \right|_{t=0}= \star^d_W \left. \deriv{}{t} \right|_{t=0}\big( A_t e_1 \wedge \dots \wedge A_te_d \big) =$$ -$$ \star^d_W \sum_{i=1}^d \big( A e_1 \wedge \dots \wedge Be_i \wedge \dots \wedge Ae_d \big) = \sum_{i=1}^d \star^d_W (-1)^{i-1} \big( Be_i \wedge A e_1 \wedge \dots \wedge \widehat Ae_i \wedge \dots \wedge Ae_d \big) $$ -$$ =\sum_{i=1}^d \star^d_W (-1)^{i-1} \big( Be_i \wedge (-1)^{d-1} \star^1_W \star^{d-1}_W ( A e_1 \wedge \dots \wedge \widehat Ae_i \wedge \dots \wedge Ae_d) \big) = $$ - $$(-1)^{i-1} (-1)^{d-1} \sum_{i=1}^d \star^d_W \big( Be_i \wedge \star^1_W \star^{d-1}_W ( A e_1 \wedge \dots \wedge \widehat Ae_i \wedge \dots \wedge Ae_d) \big) = $$ - $$(-1)^{i-1} (-1)^{d-1} \sum_{i=1}^d \IP{Be_i}{ \star^{d-1}_W ( A e_1 \wedge \dots \wedge \widehat Ae_i \wedge \dots \wedge Ae_d)}_W = $$ - $$ (-1)^{d-1} \sum_{i=1}^d \IP{Be_i}{ \star^{d-1}_W \big( \bigwedge^{d-1} A ( \star_V^1 e_i )\big)}_W=\sum_{i=1}^d \IP{Be_i}{ \big( (-1)^{d-1} \star^{d-1}_W \bigwedge^{d-1} A \star_V^1\big) e_i }_W=$$ - $$\sum_{i=1}^d \IP{Be_i}{ \Cof A(e_i) }_W=\sum_{i=1}^d \IP{(\Cof A)^TBe_i}{ e_i }_V=\tr\left( \Cof A^TB \right)= \IP{\Cof A}{B}_{V,W}.$$ - -proof of Theorem $1$: - We want to imitate the proof above: -A positive orthonormal frame $e_1,\dots,e_d$ of $\TM$ will replace the basis for $V$, covariant differentiation will replace the time derivation, so $A \to df,A'(0)=B \to \nabla_Vdf$. There are two obstacles with using this analogy verbatim: - -The $e_i$ cannot be chosen to be parallel w.r.t $\nabla^{\TM}$ if $\M$ is not flat, while in the original setting, the $e_i$ were constant vectors (time-independent). -The derivative w.r.t time commuted with $\star_W$, since it was a fixed linear operator. This time we need to establish a stronger commutation property between Hodge duals and covariant differentiation. - -We shall see that a miracle will happen - metricity shall come to our aid. - $$ - \det(df)= \star^d_{f^*T\N} \circ \bigwedge^d df \circ \star^0_{\TM}(1)= \star^d_{f^*T\N} \big( df(e_1) \wedge \dots \wedge df(e_d) \big),$$ -So -$$ V\det df = V \star^d_{f^*T\N}\big( df(e_1) \wedge \dots \wedge df(e_d) \big) \stackrel{(1)}{=} $$ -$$ \star^d_{f^*T\N} \nabla_V \big( df(e_1) \wedge \dots \wedge df(e_d) \big)= - \star^d_{f^*T\N} \sum_{i=1}^d \big( df(e_1) \wedge \dots \wedge \nabla_V \big(df(e_i)\big) \wedge \dots \wedge df(e_d) \big) = -\star^d_{f^*T\N} \sum_{i=1}^d \big( df(e_1) \wedge \dots \wedge (\nabla_V df)e_i \wedge \dots \wedge df(e_d) \big) + $$ -$$ \star^d_{f^*T\N} \sum_{i=1}^d \big( df(e_1) \wedge \dots \wedge df(\nabla_{V}e_i) \wedge \dots \wedge df(e_d) \big) \stackrel{(2)}{=} $$ -$$ \IP{\Cof df}{\nabla_V df}_{\TM,f^*{\TN}}+ \star^d_{f^*T\N} \bigwedge^d df( \sum_{i=1}^d e_1 \wedge \dots \wedge \nabla_Ve_i \wedge \dots \wedge e_d)= -\IP{\Cof df}{\nabla_V df}_{\TM,f^*{\TN}}+ \star^d_{f^*T\N} \bigwedge^d df\big( \nabla_V (e_1 \wedge \dots \wedge e_i \wedge \dots \wedge e_d) \big)=\IP{\Cof df}{\nabla_V df}_{\TM,f^*{\TN}}. -$$ -Where equality $(1)$ follows since metric connections and Hodge duals commute, and equality $(2)$ is exactly the formal repeatition of the calculation in the pointwise setting (where $A \to df,B \to \nabla_Vdf$). -Admittedly, this repeatition is not huge, but I have other examples on my mind where the computations are much longer, so a general "transfer-principle" would be nice to have. - -REPLY [2 votes]: The following is the meta-theorem I have in mind (I can't swear that what I've written is 100% correct): -Given $G < \mathrm{GL}(n)$, let $\Phi: \mathbb{R}^n \rightarrow \mathbb{R}$ a smooth $G$-invariant function. Let $\Phi': \mathbb{R}^n \rightarrow (\mathbb{R}^n)^*$ denote the differential of $\Phi$, defined to be -$$ -\langle\Phi'(x),v\rangle = \left.\frac{d}{dt}\right|_{t=0}\Phi(x + tv). -$$ -Let $V$ be a rank $n$ vector bundle with structure group $G$ (using the same representation as above) and compatible connection $\nabla$. The function $\Phi$ induces naturally a function $\widehat\Phi: V \rightarrow \mathbb{R}$, which is equal to $\Phi$ on each fiber $V_x$. Similarly, $\Phi'$ induces a bundle map $\widehat\Phi': V \rightarrow V^*$. -Given any section $v$ of $V$, let $f(x) = \widehat\Phi(v(x))$. -Then given any tangent vector $t \in T_x\mathcal{M}$, -$$ -\langle df(x), t\rangle -= \langle \widehat\Phi'(v(x)), \nabla_tv(x)\rangle. -$$ -I think this (or something like it) can be proved by using a curve passing through $x$ and tangent to $t$ and a $G$-frame parallel translated along the curve and writing the section $v$ in terms of the frame.<|endoftext|> -TITLE: A thickening map on integer partitions -QUESTION [5 upvotes]: I am looking for the name of the following map $t(\mu)$, defined for integer partitions $\mu=\mu_1\geq\mu_2\geq\dots$: - -if $\mu$ is empty, return $\mu$. -if the first part $\mu_1$ of $\mu$ is at least the number of parts of $\mu$, return the partition $\mu_1, t(\mu_2,\dots)$ -otherwise, let $\lambda=\mu^t$ and return $\lambda_1, t(\lambda_2,\dots)$. - -One property of $t$ is that $t(\mu)$ dominates $\mu$, but I'm not sure whether this is the maps natural context. - -REPLY [2 votes]: The partitions produced by the map are called superdiagonal partitions in OEIS, so absent a better reference you could call it the superdiagonalising map.<|endoftext|> -TITLE: Fastest algorithm for counting perfect matchings in a general graph -QUESTION [8 upvotes]: Let $G(V,E)$ be a undirected graph. I am interested in the fastest known algorithm for counting the number of perfect matchings in $G(V,E)$ (which is known to be in $\#P$). In particular, what is the scaling depending on the number of vertices $|V|$ and edges $|E|$? -Randomized solutions can be found in polynomial time; for bipartite graphs it corresponds to calculation of the permanent (which can be solved by Ryser's formula in $O(2^n n^2)$). -But for general undirected graphs I was not able to find any algorithm or it's scalings. Thank you. - -REPLY [8 votes]: Have a look at: approximating Hafnians. In particular, there are a several works on computing Hafnians, I can add them also in case you are unable to find them starting from the cited reference -- which also includes a short but good discussion about the complexity of the problem; in polynomial space, a Ryser like algorithm seems to be known, and thus $O(n^22^n)$; for non-polynomial space, the linked paper provides a $\tilde{O}(2^{n/2})$ method.<|endoftext|> -TITLE: supermanifolds - elementary introduction? -QUESTION [16 upvotes]: I am looking for an elementary but mathematically precise introductory text on supermanifolds in a modern differential geometric setting. -Elementary in the sense that there is plenty of motivation for the concepts and methods, and that these are explained in some detail with simple examples featuring only few bosonic and one fermionic coordinates, for example $R^{1|1}$ or $OSp(1|1)$. - -REPLY [3 votes]: I would like to add Riemannian supergeometry by Oliver Goertsches. Abstract: - -Motivated by Zirnbauer in J Math Phys 37(10):4986–5018 (1996), we develop a theory of Riemannian supermanifolds up to a definition of Riemannian symmetric superspaces. Various fundamental concepts needed for the study of these spaces both from the Riemannian and the Lie theoretical viewpoint are introduced, e.g., geodesics, isometry groups and invariant metrics on Lie supergroups and homogeneous superspaces.<|endoftext|> -TITLE: How is this (Tannakian) de Rham fundamental group calculated? -QUESTION [11 upvotes]: $\newcommand{\dR}{\mathrm{dR}}$Edit: I originally asked this question on MSE, but migrated it to MO after a long period of inactivity and a recommendation from another user. - -Let $X$ be a complex elliptic curve and $e$ the identity element of $X$. Let $E^\times$ denote the punctured curve $X\backslash\left\{e\right\}$. In their 2007 paper Towards Multiple Elliptic Polylogarithms, Levin and Racinet compute the de Rham fundamental group $\pi_1 ^{\dR} (E^\times, e)$. To do this they prove that there is an equivalence of categories (Theorem 1) -$$R\mathsf{-NMod}\xrightarrow{\sim} \mathsf{NConn}(X;e)$$ -where: - -$R = \mathbb{C}\langle\langle \mathbf{t}, A\rangle\rangle$ is the Hopf algebra of noncommutative power series in two formal variables $\mathbf{t}$ and $A$. $R$ is also the completed universal enveloping algebra of the free Lie algebra $\mathbb{L}(\mathbf{t}, A)$ generated by $\mathbf{t}$ and $A$. The coproduct in $R$ is given by - -$$\Delta(\mathbf{t}) = \mathbf{t}\otimes 1 + 1\otimes\mathbf{t},\quad \Delta(A) = A\otimes 1 + 1\otimes A;$$ - -$R$-$\mathsf{NMod}$ is the category of (finitely generated) "nilpotent left $R$-modules", which I take to mean (as they define) finite-dimensional complex vector spaces $V$ with two nilpotent endomorphisms $\mathbf{t}, A\in\text{End}(V)$; -$\mathsf{NConn}(X;e)$ is the category of "nilpotent vector bundles" on $X$ with log-poles at $e$, which I take to mean unipotent vector bundles i.e. those $\mathcal{V}$ possessing a finite filtration by subbundles (with compatible connections) such that the quotients are all isomorphic to the trivial bundle with trivial connection $(\mathcal{O}_X, d)$. This category is often called $\mathsf{Un} (X;e)$ or something to that effect (e.g. in the papers of M. Kim). - -By definition the de Rham fundamental group $\pi_1^{\dR} (E^\times;b)$ is the Tannakian fundamental group of the category $\mathsf{Un}(X;e)$, with fiber functor $\omega$ sending a bundle to the stalk over a basepoint $b$ of $X$ (or for $e$, a suitable notion of "tangential basepoint"). By the above equivalence, it follows that there is an isomorphism between $\pi_1^{\dR} (E^\times;e)$ and the Tannakian fundamental group of $R$-$\mathsf{NMod}$ (with the forgetful functor as fiber functor). -In Corollary 2.2.7, Levin and Racinet claim that this Tannakian fundamental group is -$$\pi_1^{\dR} (E^\times; e)\cong \exp \mathbb{L}(\mathbf{t}, A).$$ -I don't understand how this is calculated as the Tannakian fundamental group of $R$-$\mathsf{NMod}$. Would anyone be able to help me or point to how this is calculated? Many thanks! - -REPLY [8 votes]: First, a correction, I believe "simultaneously nilpotent" in this context means that any sequence of $t$s and $A$s eventually multiplies to zero, or in other words there is a filtration such that both $t$ and $A$ send elements to elements of lower degree. This makes sense as we want to get the same filtration on the vector bundle. It's equivalent to say that every element of the Lie algebra, not just the generators, acts nil potently. -To calculate the Tannakian group, we can use the alternate description of the Tannakian fundamental group, that it is the unique pro-algebraic group whose category of representations (w/ Tannakian structure) is equivalent to your Tannakian category. -For any Lie algebra, there is an equivalence of categories between the representations of the Lie algebra and the representations of the universal enveloping algebra. -There is also an equivalence of categories between representations of a simply-connected Lie group and its Lie algebra. In the algebraic group setting, however, we have to be careful, because not all representations produced this way are algebraic. -Instead, note that the Lie algebra in question is an inverse limit of nilpotent Lie algebras (the quotients by the ideal generated by the $n$th order commutators, say). Hence its exponential is an inverse limit of unipotent Lie groups. Algebraic representations of a unipotent Lie group are exactly the representations of its Lie algebra sending every element to a nilpotent matrix. -To check this, note that any unipotent subgroup of $GL_n$ is contained in the upper triangular subgroup, and take the logarithm. Conversely, use the power series formula for exponentials, and note that it is algebraic when the object being exponentiated is nilpotent. -It follows that there is a series of equivalences of categories between nilpotent modules for the universal enveloping algebra, representations of the pro-Lie algebra that make every element nilpotent, and algebraic representations of the pro-unipotent pro-algebraic exponential group. Because these equivalences agree with the tensor product and the fiber functor, this must be the Tannakian group.<|endoftext|> -TITLE: Moment bounds on exponential martingale -QUESTION [6 upvotes]: Consider the exponential martingale used in the Girsanov transformation of -measure: -$$Z(t) = \exp\Big(\int_0^tXdW - \frac{1}{2}\int_0^t|X|^2ds\Big)$$ -so that $Z$ solves the sde $dZ = ZXdW$ where $W$ is a one dimensional -Brownian motion. Under certain conditions (e.g. Novikov) on $X,$ $Z$ is a martingale. Many things are known as well, like -$$W_t - \int_0^t X_sds$$ -is a Brownian motion under $dQ/dP = Z(t)$ where $P$ is the original -measure attached to $W.$ -I'm interested in moments of $Z$ given in terms of moments of $X.$ -Using the sde above, we can see that -$$Z_t^2 = 1 + 2\int_0^tZ_sdZ_s + \int_0^tZ_s^2X_s^2ds$$ -which shows -$$E Z_t^2 = 1 + \int_0^t E(Z_s^2X_s^2)ds$$ -where I was hoping to apply a Gronwall inequality to get a bound on -$E Z_t^2.$ It seems we are unable to do this unless we know $X_s$ is bounded and can apply an $L_1, L_\infty$ bound. -Does anyone have any reference or knowledge on moment bounds of this exponential martingale in terms of moment bounds of $X_s?$ - -REPLY [3 votes]: There are a number of ways to bound moments of $Z$ in terms of exponential moments of $X$. For some sharp results, see Theorem 1.5 of Kazamaki's book, "Continuous exponential martingales and BMO," as well as Remark 1.2 thereafter (page 8). -For a more pedestrian bound we need nothing more than Holder's inequality and the fact that a stochastic exponential always has expectation at most $1$. Let $p,q,q^* > 1$ with $1/q + 1/q^* =1$. Then -\begin{align*} -\mathbb{E}\left[Z_t^p\right] &= \mathbb{E}\left[\exp\left(p\int_0^tX_sdW_s - \frac{qp^2}{2}\int_0^t|X_s|^2ds\right)\exp\left(\frac{p(qp-1)}{2}\int_0^t|X_s|^2ds\right)\right] \\ - &\le \mathbb{E}\left[\exp\left(\frac{pq^*(qp-1)}{2}\int_0^t|X_s|^2ds\right)\right]^{1/q^*}. -\end{align*} -A simple exercise shows that $q^*(qp-1) = \frac{q(qp-1)}{q-1}$ is minimized by $q= 1 + \sqrt{1-\frac{1}{p}}$.<|endoftext|> -TITLE: Stable marriage with contracts: is it known? -QUESTION [9 upvotes]: Consider the following generalization of the classical Stable Marriage Problem. The rough idea is that instead of merely specifying who marries whom, a matching now chooses a set of "marriage contracts" (out of a given set, which may be large since a man and a woman can choose between several different contracts for marrying each other); correspondingly the men and the women rank not their potential partners but rather the potential contracts available to them (e.g., a given man $m$ can prefer marrying a woman $w_1$ via a contract $c_1$ to marrying a woman $w_2$ via a contract $c_2$, but at the same time prefer marrying $w_2$ via $c_2$ to marrying $w_1$ via a different contract $c_3$). In detail, the generalized problem is stated as follows: - -Contracted Stable Marriage Problem. Suppose that we have a population of $k$ men and $k$ women - (for some $k \in \mathbb{N}$). - Assume furthermore that a finite set $C$ of "contracts" is given. - Each contract involves exactly one man and exactly - one woman. - (Think of the contracts as marriage contracts, each prepared for some man and some woman, but not signed.) -Assume that, for each pair $\left(m, w\right)$ consisting of a man $m$ - and a woman $w$, there is at least one contract - that involves $m$ and $w$. -Suppose that each person has a preference list of all the - contracts that involve him/her; i.e., he/she ranks - all contracts that involve him/her in the order of - preferability. - (No ties are allowed.) -A matching shall mean a subset $K$ of $C$ such - that each man is involved in exactly one contract in $K$, - and such that each woman is involved in exactly one - contract in $K$. - Thus, visually speaking, a matching is a way to marry off - all $k$ men and all $k$ women to each other (in the - classical meaning of the word -- i.e., heterosexual and - monogamous) - by having them sign some of the contracts in $C$ - (of course, each person signs exactly one contract). -If $p$ is a person and $K$ is a matching, then the - unique contract $c \in K$ that involves $p$ will be called - the $K$-marriage contract of $p$. -If $K$ is a matching and $c \in C$ is a contract, then - the contract $c$ is said to be rogue (for $K$) if - -this contract $c$ is not in $K$, -the man involved in $c$ prefers $c$ to his - $K$-marriage contract, and -the woman involved in $c$ prefers $c$ to her - $K$-marriage contract. - -Thus, roughly speaking, a rogue contract is a contract - $c$ that has not been signed in the matching $K$, but that - would make both persons involved in $c$ happier than - whatever contracts they did sign in $K$. -A matching $K$ is called stable if there exist - no rogue contracts for $K$. -The problem now asks you to find a stable matching. - -Theorem. A stable matching always exists. -The proof of the above theorem is a fairly straightforward modification of the classical proof of the Gale-Shapley Stable Marriage Theorem (which is discussed in several places, the most readable one perhaps being Section 6.4 of Eric Lehman, F. Thomson Leighton, Albert R. Meyer, Mathematics for Computer Science, revised 16th April 2017). In generalizing it, I was motivated by an elegant solution of a graph-theoretical exercise suggested by Alex Postnikov (Exercise 3 on UMN Spring 2017 Math 5707 midterm #2; see its Second Solution for Postnikov's argument). -But I am left wondering: Is the above Theorem, and the problem it answers, known? I suspect it has further applications, which the (standard) Stable Marriage Theorem is a tad too narrow to handle, as the latter treats a marriage as uniquely determined by the two partners involved. -One thing the added generality seems well-equipped to model is "matching with dowry"; i.e., a marriage should come with a payment from one partner to the other, and the preferability of the marriage depends on this payment. (At least if the set of all possible payments is finite, this is easily obtained as a particular case of the Contracted Stable Marriage Problem by formulating a contract for each possible payment.) This rather unromantic modification appears suited for various real-life applications of the Stable Marriage Problem, given that few of them involve actual marriages. At least this case is probably known? - -REPLY [11 votes]: There is a humongous literature on "matching with contracts", starting with: - -Hatfield, John William, and Paul R. Milgrom. "Matching with - contracts." The American Economic Review 95.4 (2005): 913-935. - -The existence result of that paper is actually a special case of the earlier, even up to the basic proof approach (via Tarski's fixed point theorem). - -Fleiner, Tamás. "A fixed-point approach to stable matchings and some - applications." Mathematics of Operations Research 28.1 (2003): - 103-126. - -It was also shown in - -Echenique, Federico. "Contracts versus salaries in matching." The - American Economic Review 102.1 (2012): 594-601. - -that the model of Hatfield and Milgrom is equivalent to the model in - -Kelso Jr, Alexander S., and Vincent P. Crawford. "Job matching, - coalition formation, and gross substitutes." Econometrica: Journal of - the Econometric Society (1982): 1483-1504. - -There exists also generalizations that are genuinely new and the matching with contracts model is pretty much the state of the art of stable matching theory in economics. There is also a large and even older literature on matching with transferable utility or imperfect transferable utility, which covers the "dowry case". This material can even be found in the classic textbook - -Roth, Alvin E., and Marilda A. Oliveira Sotomayor. "Two-sided - matching, volume 18 of Econometric Society Monographs." (1990), - -where you can find further references.<|endoftext|> -TITLE: Seeing modules as ring extensions -QUESTION [7 upvotes]: In a paper about Poisson modules that I can't seem to find at the moment, I saw a description of Poisson modules that was rather different from most descriptions I'd seen; however, after considering it, it seemed to naturally extend to many other situations. The description was approximately: -Let $A$ be a Poisson algebra. Then a Poisson module over $A$ is a vector space $E$ and a Poisson algebra structure on $A \oplus E$ such that 1) it agrees with the original structure on $A$, and 2) $\{e_1, e_2\} = e_1 e_2 = 0$ for any $e_1, e_2 \in E$. -The general idea does seem to work for other situations than Poisson; it clearly works for commutative algebras, and works for Lie algebras in the same way as for Poisson algebras. -This leads to a few questions, then. 1) Is this a "good" way to think about algebras (of various forms) and modules? Are there any situations where it doesn't work? 2) The apparent thing that "separates" the "base" algebra structure from the module structure is that the module part is nilpotent (in fact, it "squares" to 0). This made me think back to non-reduced schemes, and made me think that the "right way" to think of a nonreduced scheme is a reduced scheme with extra module structures on top "localized at" the nonreduced points? 3) If so, is there a natural way to think about the "higher modules", since nilpotence can happen at higher powers than 2? 4) Are there any other natural ideas that result from this? -Edited to give a partial answer I thought of afterwards: This really isn't correct; the appropriate idea here is when $E$ is a bimodule. It just happens that in the cases I was thinking of, a module is naturally a bimodule. So if the type of algebra was "arbitrary not-necessarily-commutative algebra", then the above structure would not be equivalent to a module, but a bimodule. - -REPLY [2 votes]: You might be interested in the Deformation Theoretic content of this. For the commutative algebra version the construction is in -S. LICHTENBAUM AND M. SCHLESSINGER, The cotangent complex of a morphism, Trans. Amer. Math. Sot. 128 (1967), 41-70, and shortly afterwards Quillen used this to discuss cohomology in his paper: -D. QUILLEN, On the homology of commutative rings, Proc. Sympos. Pure Marh. 17 (1970), 65-87. Quillen ascribes the idea to Beck. Following up the cotangent complex idea takes one through Grothendieck's cofibred categories and Illusie's thesis, which may link up with your later questions. -The idea is also known as Idealisation as it turns a (bi)module over an algebra into an ideal of another algebra. -There is also a very general theory of split epimorphisms in semi-abelian categories (or similar) that describe this process in great generality. (I can give some references for that if needs be, just ask.) -I seem to recall that in non-commutative geometry this construction has been used but that is out of my comfort zone!<|endoftext|> -TITLE: Is there a residually finite non-elementary hyperbolic group whose profinite completion is boundedly generated? -QUESTION [8 upvotes]: Is there a residually finite hyperbolic group $G$ that is not virtually cyclic, such that there exists finitely many procyclic closed subgroups $C_1, \dots, C_n$ of the profinite completion $\hat{G}$ of $G$ satisfying the property that for every $g \in \hat{G}$ there exist $c_1 \in C_1, \dots, c_n \in C_n$ with $g = c_1 \cdots c_n$ ? -It is known that $G$ itself can't be boundedly generated. - -REPLY [9 votes]: I'm fairly certain that no example is known. Of course, it's a famous open problem whether every hyperbolic group is residually finite. This turns out to be equivalent to many other questions about the profinite topology on hyperbolic groups (see, for instance, https://arxiv.org/abs/0802.0709 and http://www.math.uiuc.edu/~kapovich/PAPERS/dani.dvi.gz ). It's very tempting to conjecture that the existence of a non-elementary hyperbolic group with boundedly generated profinite completion would imply the existence of a non-residually finite hyperbolic group. -Most of the hyperbolic groups that we know to be residually finite are known to be so because they are cubulable (i.e. the fundamental group of a compact, non-positively curved cube complex). Agol's theorem, which resolved the virtual Haken conjecture, implies that these groups are residually finite. In this case, we can say more. Any non-elementary, cubulable, hyperbolic group always has a finite-index subgroup that retracts to a non-abelian free group. It follows that the profinite completion virtually surjects a non-abelian profinite free group, and hence can't be boundedly generated. -In the other direction, one would like candidate positive examples; here, the first example to look at is a cocompact lattice $\Gamma$ in $Sp(n,1)$. The congruence subgroup problem is open for such lattices. This asks whether the profinite completion of $\Gamma$ is equal to the congruence completion, meaning the inverse limit of all the congruence quotients of $\Gamma$. If so, then $\Gamma$ is said to satisfy the congruence subgroup property (CSP). -The lattice $\Gamma$ is linear, and hence certainly residually finite. Nevertheless, Lubotzky observed that, if the CSP holds for $\Gamma$, then there is a non-residually finite hyperbolic group. And, indeed, the OP points out in comments that the congruence completion of $\Gamma$ is known to be boundedly generated. -In summary, we can say that bounded generated of the profinite completion is closely tied to the existence of a non-residually finite hyperbolic group. On the one hand, in cubulable groups, which are very "robustly" residually finite, the profinite completion is not boundedly generated. And in quaternionic lattices, the most plausible source of a non-residually finite hyperbolic group would also lead to a non-elementary, residually finite hyperbolic group with boundedly generated profinite completion.<|endoftext|> -TITLE: How can one understand the Eisenstein series E2 in terms of automorphic representation? -QUESTION [37 upvotes]: The weight 2, level 1 Eisenstein series $E_2(z)$ is a non-holomorphic automorphic form. It is defined as the analytic continuation to $s = 0$ of the series -$$ E_2(z, s) = \sum_{\substack{m, n \in \mathbf{Z} \\ (m, n) \ne (0,0)}} \frac{\operatorname{Im}(z)^s}{(mz + n)^2 |mz + n|^{2s}} $$ -which is convergent for $\operatorname{Re}(s) > 0$ (but not for $s = 0$). - Moreover for any prime p the function $E_2(z)-pE_2(pz)$ is holomorphic. -My question is: what's the Archimedean component of the automorphic representation corresponding to $E_2$? (If it it the holomorphic discrete series of weight two then seems the vector corresponding to $E_2$ should be holomorphic.) - -REPLY [6 votes]: (I took @LSpice's inquiry as encouragement to add some remarks to @GH from MO's good answer... But/and one of the issues that helped me overcome my skepticism about the utility of representation theory long ago was its clarification of exactly these issues about "Hecke summation", Maass-Shimura operators, and such. In particular, that it is not necessary to write out Fourier expansions using special functions, etc.) -First, as in the comments that helped @GH...'s answer get on track, in a typical (naive) indexing scheme, the $2k$-th holomorphic discrete series is not a subrepresentation of the $2k$-th principal series (also in a naive normalization), but just of the $k$-th. Also, there are choices about when-and-how to append/renormalize the $y^k$ to weight-$2k$ automorphic forms, to make them left $\Gamma$-invariant and right $K$-equivariant. Lots of chances to make a mess... -A more interesting issue than normalization is the fact that (the map...) formation of Eisenstein series (manifestly) does not commute with meromorphic continuation. Such a thing already occurs in looking at the spectral decomposition of the space of pseudo-Eisenstein series in terms of Eisenstein series. -For generic principal series $I_s$, the raising and lowering operators $R,L$ change the $K$-type by $\pm 2$, with the lowering operator having a coefficient of $s-k$ (here the normalization matters, obviously). So, up to irrelevant constants, with $E_{s,2k}$ being the Eisenstein series hit by $I_s$ with $K$-type $2k$, $LE_{s,2k}=(s-k)E_{s,2k-2}$. Thus, at $s=1$, if $E_{s,2k-2}$ has no pole, then $E_{s,2k}$ is annihilated by the lowering operator. -It is partly some sort of crazy luck that the lowering operator $L$, in coordinates, is essentially the Cauchy-Riemann operator, so annihilation by it implies holomorphy. -But/and of course with $2k=2$, the Eisenstein series $E_{s,0}$ has a pole at $s=1$. The residue is just a constant, not very exciting, but not $0$. Thus, in whatever normalization one operates, the application of the lowering operator (Cauchy-Riemann) to (the meromorphically continued) $E_{s,2}$ gives that residue, not $0$. -For Hilbert modular forms over (totally real) number fields of degree $>1$ over $\mathbb Q$, the lowering operators attached to the various archimedean factors of the group do not map $E_{s,(2,2,2,2,...)}$ to $E_{s,0}$, where there is a pole, so they annihilate $E_{s,(2,2,...)}$. -The weight-one Eisenstein series are another story... -(Edit: some "$s-1$"'s should have been "$s-k$"'s, but I was thinking about $k=1$... maybe repaired...)<|endoftext|> -TITLE: Reduction and fibre product -QUESTION [7 upvotes]: Given two analytic spaces $X$ and $Y$ over a third analytic space $S$, is it true that -$$ -(X\times_S Y)_{red}=X_{red}\times_{S_{red}}Y_{red}? -$$ -Any reference is welcome. - -REPLY [13 votes]: In general the answer is no, as shown by the following example. -Given a fibration $f \colon X \to S$, where $X$ and $S$ are smooth $k$-schemes, and setting $Y=\{s\} \hookrightarrow S$ with its reduced scheme structure, then $X_s:=X \times _S Y$ is the scheme-theoretic fibre of $f$ over $s$. -Of course such a fibre can be non-reduced even if $X$ and $S$ are both reduced, so in this case $$ -(X\times_S Y)_{red} \neq X_{red}\times_{S_{red}}Y_{red}=X \times _S Y, -$$ -because the first space is reduced, whereas the second is not. -For instance, take $$X= \mathrm{Spec} \, k[x, \, y, \, t]/(ty-x^2), \quad S=\mathrm{Spec}\, k[t]$$ -and let $f \colon X \to \mathbb{A}^1$ be the fibration in conics induced by the inclusion $$k[t] \to k[x,\, y,\, t]/(ty-x^2).$$ If $Y= \{0\} \hookrightarrow \mathbb{A}^1$ then $$X\times_\mathbb{A^1} Y= X_{red} \times_{\mathbb{A^1}_{red}} Y_{red}=k[x,\, y]/(x^2),$$ which is a double line.<|endoftext|> -TITLE: How to get a closed form for a possibly simple combinatorial sequence -QUESTION [7 upvotes]: I hope this question isn't too simple for MO. Combinatorics is not my forte. -I have two positive integers $n,k$ that define a resultant integer. I am running an experiment and I have a collection of sequences of integers as follows. For each $n$ I list the results for $k = 1,\dots 20$ although there are more integers that follow the same pattern. - -$n = 2$ gives 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 11. - -This is a simple sequence with an obvious pattern. - -$n = 3$ gives 1, 3, 5, 8, 11, 15, 19, 24, 29, 35, 41, 48, 55, 63, 71, 80, 89, 99, 109, 120 - -The differences between consecutive numbers is exactly the sequence for $n = 2$ (excluding the first number $1$ in the preceding sequence. - -$n = 4$ gives $1, 4, 9, 17, 28, 43, 62, 86, 115, 150, 191, 239, 294, 357, 428, 508, 597, 696, 805, 925$. - -Again the differences between consecutive numbers is exactly the sequence for $n = 3$ (excluding the first number $1$ in the preceding sequence). -And it continues this way for larger values of $n$. - -Assuming this is indeed the general rule, how could one find a closed - form formula for arbitrary positive $n ,k$ (with $n \geq 2$). - -REPLY [14 votes]: It's worth to consider the sequence for $n=1$: -$$1,1,0,1,0,1,0,\dots$$ -Let $s_k^n$ denote the $k$-th term of the $n$-th sequence. -In particular, $s_1^1=1$ and for $k>1$, $s_k^1$ equals 1 if $k$ is even and $0$ if $k$ is odd. -Then for $n>1$, we have -$$s_k^{n} = \sum_{i_{n-1}=1}^k s_{i_{n-1}}^{n-1} = \cdots = \sum_{i_{n-1}=1}^k \sum_{i_{n-2}=1}^{i_{n-1}} \cdots \sum_{i_1=1}^{i_2} s_{i_1}^1=\sum_{i_1,i_2,\dots,i_{n-1}\atop 1\leq i_1\leq i_2\leq \dots \leq i_{n-1}\leq k} s_{i_1}^1$$ -$$ = \sum_{i_1=1}^{k} s_{i_1}^1 \sum_{i_2,i_3,\dots,i_{n-1}\atop i_1\leq i_2\leq \dots \leq i_{n-1}\leq k} 1 = \sum_{i_1=1}^{k} s_{i_1}^1 \left(\!\!\binom{k-i_1+1}{n-2}\!\!\right) = \sum_{i_1=1}^{k} s_{i_1}^1 \binom{k-i_1+n-2}{n-2},$$ -where $\left(\!\!\binom{k-i_1+1}{n-2}\!\!\right) = \binom{k-i_1+n-2}{n-2}$ gives the number of combinations with repetitions $(i_2,i_3,\dots,i_{n-1})$ from the set $\{t_1,i_1+1,\dots,k\}$ of size $k-i_1+1$. -Recalling the formula for $s_{i_1}^1$, we have -$$s_k^n = \binom{k+n-3}{n-2} + \sum_{\ell=1}^{\lfloor k/2\rfloor} \binom{k+n-2-2\ell}{n-2},$$ -which holds for all $n>1$ and $k\geq 1$. -UPDATE. Noticing that $\binom{k+n-2-2\ell}{n-2}$ is the coefficient of $x^{n-2}$ in $(1-x)^{-(k+1-2\ell)}$, the formula above can be further reduced to the sum of $O(n)$ terms (which may be preferable when $k$ is large as compared to $n$): -$$s_k^n = \binom{k+n-3}{n-2} + \frac{(k+1)\bmod 2}{2^{n-1}} + \sum_{i=1}^{n-1} \binom{k-2+i}{i}\frac{1}{2^{n-i}}.$$<|endoftext|> -TITLE: Intuition behind centralizers of Langlands parameters -QUESTION [7 upvotes]: In the description of the Langlands correspondence for $\mathbb{Q}_p$, we consider admissible representations of $G(\mathbb{Q}_p)$ for $G$ a reductive group defined over $\mathbb{Q}_p$, and admissible Langlands parameters $\phi: W \to ^LG$, where $W$ denotes the absolute Weil group of $\mathbb{Q}_p$. In particular, to each irreducible admissible representation $\pi$ of $G(\mathbb{Q}_p)$, we (to some extent conjecturally) assign an $L$-parameter $\pi_{\phi}$. The map $\pi \mapsto \pi_{\phi}$ satisfies a number of compatibility conditions and is finite to one. The set theoretic fiber of an $L$-parameter is called an $L$-packet and denoted $\Pi_{\phi}$. Various descriptions of the contents of $\Pi_{\phi}$ have been suggested (in particular in the tempered case), typically involving some variant of the set of irreducible representations of the centralizer of the image of $\phi$, denoted $S_{\phi}$. -My question asks for an intuitive (or otherwise) explanation of why representations of this centralizer group should be involved in parametrising the contents of the corresponding $L$-packet. - -REPLY [7 votes]: The simplest non-trivial case for this is for $G = SL_2$. If you take an unramified principal series representation of $\tilde G = GL_2$ and restrict it to $G$, then it will almost always be irreducible; but there is a bad case where it's a direct sum of 2 things, which is when the ratio of the Satake parameters is $-1$, and these two summands form a non-singleton $L$-packet. This is all worked out very beautifully in Labesse--Langlands. -Now, since the Langlands parameter for $\tilde G$ is just a diagonal matrix, you can explicitly compute its centraliser in ${}^L \tilde G = GL_2(\mathbf{C})$ and in ${}^L G = PGL_2(\mathbf{C})$. [*] You will see that the former is always connected, but the latter has either 1 or 2 components, and it has 2 components exactly when the ratio of Satake params is $-1$. -This, at least, shows that the component group of the centraliser is giving the right answer in this simple case. -([*] I am lying a bit here: this is the dual group $\hat G$, not the $L$-group ${}^L G$ which is the semidirect product of $\hat G$ with the local Weil group. However, in our case the semidirect product is actually direct, because both groups are split, so we can ignore the second factor.)<|endoftext|> -TITLE: Galois representation and weight one Hilbert modular form -QUESTION [6 upvotes]: Let $f$ be a primitive weight one Hilbert modular form for the totally real field $F$ (the weight of $f$ is parallel). Assume that $f$ is $N$-new, $p$-stable and cuspidal. Let $\rho:G_F \rightarrow \mathrm{GL}_2(\mathbb{C})$ be the galois representation attached to $f$. -I have the following question : -Let $q$ be a prime ideal of the ring of integers of $F$ such that $q$ is prime to $p$ and $ q \mid N$: - -If $U_q (f)=a_q.f$ with $a_q \ne 0$. Is it equivalent to the fact that there exits a line of $\mathbb{C}^2$ fixed by the inertia group $I_q$ and on which $\mathrm{Frob}_q$ acts via $a_q$ ? -If $U_q(f)=0$. Is it equivalent to the fact that $\rho^{I_q}=0$ ? - -Since $f$ is $p$-stable and of weight one, $f$ is $p$-ordinary and we can put $f$ is a Hida family $F$ of level $N$, and we know that $U_q^s (U_q^2 - Tr \rho_F(\sigma) U_q + \det \rho_F(\sigma))=0$, where $\sigma$ is a lift of the Frobenius at $q$, $s$ is an integer such that $q^s \mid N $ and $q^{s+1} \nmid N$ and $\rho_F$ is the galois representation attached to the hida family $F$, and by density we can proof what I want. But is there a simple argument without using Hida family. - -REPLY [2 votes]: These are all simple instances of a much more powerful statement, which is local-global compatibility in the Langlands program: the Weil-Deligne representation obtained from $\rho |_{D_q}$ should correspond under the Langlands program to $\pi_{f, q}$. When the weights are all $\ge 2$, this goes back to Carayol. For parallel weight 1 forms I believe this is now entirely known, due to Jarvis and Newton. (There are a few cases that are not yet fully resolved for partial weight 1 forms, but these involve Steinberg primes and those cannot occur for parallel weight 1.)<|endoftext|> -TITLE: Quillen equivalence for under-categories -QUESTION [9 upvotes]: I believe the following statement is true under certain technical conditions. -Let $$L\colon C \rightleftharpoons D\colon R$$ be a Quillen equivalence. Suppose we are given objects $c\in C$, $d\in D$, and an equivalence $c\to R(d)$. (Perhaps we need to assume that $d$ is fibrant, but $c$ doesn't have to be cofibrant.) Then one has a Quillen equivalence of under-categories: $\hat L\colon c/C\rightleftharpoons d/D\colon \hat R$, where $\hat R$ is an obvious functor sending $d\to d_1$ to the composition $c\to R(d)\to R(d_1)$; and $\hat L$ sends $c\to c_1$ to -$$ -d\to \mathrm{colim}\bigl(d\leftarrow L(c)\rightarrow L(c_1)\bigr). -$$ -I wonder what are the conditions under which this is true? Any reference? In fact I am interested in the case C and D are categories of bimodules over operads. - -REPLY [2 votes]: Actually a while ago Benoit Fresse gave me an answer to this question. I figured I should post it here. -The condition on $C$ and $D$ is that they should be left proper and $R$ should preserve weak equivalences.<|endoftext|> -TITLE: Functors between simplicial sets and cubical sets with connections -QUESTION [8 upvotes]: Let $sSet = Set^{\Delta^{op}}$, $cSet = Set^{\square^{op}}$, $ccSet = Set^{\square_c^{op}}$ be the categories of simplicial sets, cubical sets, and cubical sets with connections, respectively. Then there is a left adjoint functor $cSet \to sSet$ such that a representable cubical set $\square_n$ is mapped to $\Delta[1]^n$. But I'm looking for a left adjoint functor going in the other direction. -We could try to define a functor $\Delta \to cSet$ in such a way that $\Delta[n]$ is mapped to a quotient of $\square_n$. For example, $\Delta[2]$ is mapped to the square in which one of the faces is degenerated. This does not work since we cannot define a map of cubical sets corresponding to one of the degeneracy maps $\Delta[2] \to \Delta[1]$. But it seems that we can do this if we assume that our cubical sets have connections. -So, the question is: can we construct a (nontrivial) functor $\Delta \to ccSet$? Is there such a construction in the literature? - -REPLY [2 votes]: Such an adjunction between simplicial sets and cubical sets with connections is constructed in a paper by Krzysztof Kapulkin, Zachery Lindsey and Liang Ze Wong, A co-reflection of cubical sets into simplicial sets with applications to model structures. Moreover, it is shown there that the left adjoint $\mathrm{sSet} \to \mathrm{ccSet}$ is full and fathfull and induces a Quillen equivalent model structure on $\mathrm{ccSet}$ for any cofibrantly generated model structure on $\mathrm{sSet}$ in which all cofibrations are monomorphisms.<|endoftext|> -TITLE: Supremum of a stochastic process -QUESTION [6 upvotes]: Let $(x_1,...,x_N)$ be points in $R^d$, and $\sigma=(\sigma_1,...,\sigma_N)$ are i.i.d. Rademacher variables (+1 or -1 with probability 0.5 each). (Or alternatively, $\sigma$ could be a standard Gaussian vector.) Also let $z$ in an infinite subset of $R^d$, and $\|\cdot\|_2$ will denote the Euclidean norm. -I am looking for a good upper estimate of: -$E_{\sigma} \sup_z \| \sum_{n=1}^N \sigma_n \frac{x_n-z}{\|x_n-z\|_2} \|_2$. -So far I only got the upper bound of $N$ (by triangle inequality), and the lower bound $\sqrt{N}$ by Jensen inequality, and I would like to know if there is a way to improve on the upper estimate or the lower estimate? -If not, are there conditions (on the domain of $z$) under which the upper bound would reduce to ${\mathcal O}(\sqrt{N})$? -Edit: The points $x_1,...,x_N$ are assumed to be in general position (in a ball of radius $R$). -Edit2: The domain of values of $z$ is an infinite set in $R^d$. Can be assumed to be bounded. -Edit3: The domain of $z$ is allowed to depend on $N$, $R$ (or on other assumed characteristics of the point set), but not on the individual points. -Edit4: I also want to avoid dependence on $d$. - -REPLY [5 votes]: First, if the subset $Z$ of $R^d$ of allowed values for $z$ is a singleton, then writing $e_n=(x_n-z)/\|x_n-z\|_2=(e_{n,j})_{j=1}^d$, we have -$$E_\sigma \left\|\sum\sigma_ne_n\right\|_2=E_\sigma\sqrt{\sum_j \left(\sum_n\sigma_ne_{n,j}\right)^2}\le\sqrt{E_\sigma\sum_j\left(\sum_n\sigma_ne_{n,j}\right)^2}$$ -$$%=\sqrt{\sum_jE_\sigma\left[\left(\sum_n\sigma_ne_{n,j}\right)^2\right]} -=\sqrt{\sum_j\mathrm{Var}_\sigma\left[\sum_n\sigma_ne_{n,j}\right]} -=\sqrt{\sum_j\sum_n\mathrm{Var}_\sigma\left[\sigma_ne_{n,j}\right]}$$ -$$=\sqrt{\sum_j\sum_ne_{n,j}^2}=\sqrt{\sum_n \|e_n\|_2^2}=\sqrt{N}.$$ -Second, if $Z$ is allowed to depend on $N$, then let $Z$ be of very small diameter $\delta>0$, relative to - -the distances $\|x_i-x_j\|_2$, -$N$, and -the modulus of continuity of the function $z\mapsto \left\| \sum \sigma_n e_n\right\|_2$, - -then we should be able to get -for a fixed $z_0\in Z$, and for all $\sigma$, -$$\sup_{z\in Z}\left\|\sum\sigma_ne_n\right\|_2 \le \left\|\sum\sigma_ne_n\text{ (for $z_0$)}\right\|_2 + \epsilon$$ -and hence -$$E_\sigma \left(\sup_z\left\|\sum\sigma_ne_n\right\|_2\right) \le E_\sigma\left( \left\|\sum\sigma_ne_n\text{ (for $z_0$)}\right\|_2 + \epsilon\right) = - \mathcal O(\sqrt{N}).$$<|endoftext|> -TITLE: Monadicity of Beck-Chevalley bifibrations via a distributive law -QUESTION [12 upvotes]: Fix a category $B$ with pullbacks. The category of bifibrations over $B$ satisfying the Beck-Chevalley condition appears to be monadic over $\mathsf{Cat}/B$. Is this discussed somewhere in the literature? -Let $\mathsf{Fib}(B)$ be the category of fibrations and Cartesian functors over $B$. It's well-known that the forgetful functor $\mathsf{Fib}(B) \to \mathsf{Cat}/B$ is monadic; the induced monad $T$ is given by a comma category construction $T: (E \overset{p}{\to} B) \mapsto (B \downarrow p \to B)$. In fact, this monad is colax-idempotent. Dually, the monad $S: (E \overset{p}{\to} B) \mapsto (B \downarrow p \to B)$ is lax-idempotent, with algebras $\mathsf{OpFib}(B)$ given by opfibrations. -Thus far we have required no hypotheses on $B$. Now if $B$ has pullbacks, there appears to be (I have not checked all the coherence conditions) a (pseudo-)distributive law $TS \Rightarrow ST$ which takes a diagram $p(e) \to b \leftarrow b'$ to its pullback. And it looks like an algebra for the composite monad $ST$ is a bifibration (a functor which is simultaneously a fibration and an opfibration) satisfying the Beck-Chevalley condition over all pullback squares. -(Dually, if $B$ has pushouts, then there will be another distributive law in the other direction whose algebras will be bifibrations satisfying a dual Beck-Chevalley condition over pushout squares, but I've never heard of such a thing in nature.) -Not only does it seem good to know that the Beck-Chevalley condition arises from a monad, but this seems like a really cool example of a pseudo-distributive law! In general, it seems like these are a pain to work with because of the number of coherence conditions (even with the simplifications afforded by the (co)lax idempotency). Maybe that's why I've never heard of this one. But pullbacks can be strictified -- perhaps the pseudodistributive law can also be strictified? - -REPLY [5 votes]: Yes, this pseudo-distributive law and the characterization of its algebras as bifibrations appear in section 1.5 of Tamara von Glehn's thesis.<|endoftext|> -TITLE: Can the projection onto a compact set always be taken to be measurable? -QUESTION [6 upvotes]: This may be a very basic question. -Let $X$ be a complete metric space and let $T$ be a compact subset of $X$. Say that a function $\pi: X \to T$ is a projection if -$$ -d(x, \pi(x)) = d(x, T) \quad \forall x \in X\,. -$$ -If $T$ is a closed convex subset of $\mathbb{R}^d$ then the (unique) projection $\pi$ onto $T$ is always continuous; of course this will not happen in general. I am interested in the following weaker question: -Does there always exist a projection that is Borel measurable? - -REPLY [6 votes]: Consider the closed multimap $\Pi:X\to 2^T$, whose graph is the closed set $\mathrm{graph}(\Pi):=\big\{(x,t)\in X\times T\, : d(x,t)=\min_{s\in T}d(x,s)\big\}\subset X\times T$. You want a measurable selection $\pi:X\to T$ of $\Pi$, that is $\mathrm{graph}(\pi)\subset\mathrm{graph}(\Pi)$. Since $K$ is a compact metric space, thus a Polish space, the classical Kuratowski–Ryll-Nardzewski measurable selection theorem does the job. -rmk. The Kuratowski–Ryll-Nardzewski measurable selection theorem admits an easy proof in the particular case of a closed, non-empty set valued multimap $\Pi:X\to2^T$ with $T$ a compact metric space. In this case (by the Alexandroff-Hausdorff theorem) $T$ is a continuous image $T=\kappa( C)$ of the Cantor set $C$ via some continuous surjective map $\kappa:C\to T$. The pre-image of the closed set $\mathrm{graph}(\Pi)$ via $\mathrm{id}_X\times\kappa:X\times C\to X\times T$ is therefore the graph of a closed multimap $X\to 2^C$, that is the multimap $x\mapsto \kappa^{-1}\Pi(x)\neq\emptyset;$ a lower semi-continuous, hence measurable selection of the latter, is $x\mapsto \min\kappa^{-1}\Pi(x) $, and a measurable selection of $\Pi$ is therefore -$$\pi(x):=\kappa\big(\min\kappa^{-1}\Pi(x)\big)\in\Pi(x).$$<|endoftext|> -TITLE: "Transcendental tilings": Do they exist? -QUESTION [13 upvotes]: Let $T$ be a tiling of the plane. -Fix an origin and shoot a ray $r$ from the origin. -Mark off points $p_i$ along $r$ separated by unit distance. -Compute from $r$ a binary number $0 < b(r) < 1$ that alternates $0$'s and $1$'s -for each marked point $p_i$ as the ray enters a new tile of $T$. -For example, the square tiling and illustrated ray below lead to -$$ -.00011011100010011011100 \ldots -$$ - -          - - -          - -Square tiles side length $=7/3$. Ray slope $= 1/\sqrt{2}$. - - -To avoid thin tiles, assume every tile includes a disk of diameter $> 1$ so -that more than one $p_i$ could land in a tile. -One needs a rule when $p_i$ is on the boundary of a tile to make -$b(r)$ well-defined, but I think that detail is not relevant to my question. -It is not difficult to find tilings and rays where $b(r)$ is rational, -irrational, or transcendental, for example, by selecting the slope appropriately -in the above example. - -Q1. Is there a tiling $T$ such that every $b(r)$, for all origins and rays $r$, is transcendental? - -If the answer to Q1 is No, the following two questions are superfluous: - -Q2. What is the class of all such transcendental tilings - (if I may coin a term)? -Q3. How does this class relate to the - aperiodic tilings? - -REPLY [7 votes]: Answering Q1, I believe there is a transcendental tiling. Let us begin with this tiling with congruent convex pentagons: -          - -Notice that one can rearrange any of the triples of pentagons that form the regular hexagon, by rotating any triple we want by 180 degrees. Thus we can have two kinds of triples: pointing up or down. Now, the entire tiling can be viewed as formed by triples, each filling a hexagon, and we can view the tiling by hexagons as the union of concentric "annuli" with disjoint interiors. Make it so that the annuli's thickness (not just diameter) is strictly increasing, if needed, increasing exponentially. Then arrange all triples in the same annulus to be pointing in the same direction, alternating the direction when passing from an annulus to the adjacent one. - - - - -(Image added by J.O'Rourke.) - - -Then, although I do not have a complete proof, I believe the tiling will be transcendental. My feeling is based on the example of a binary transcendental number formed by alternating blocks of zeros and ones of strictly increasing length.<|endoftext|> -TITLE: Monopole Floer Homology vs. Heegaard-Floer theory -QUESTION [11 upvotes]: I have a (possibly very naive) question: what is the relation between Monopole Floer Homology and Heegaard-Floer theory? (both known and conjectured) - -Is there some version of Atiyah-Floer conjecture that relates the two theories? -Is there some physical explanation to this phenomena? - -REPLY [6 votes]: Ozsvath and Szabo constructed HF as a topological interpretation of SWF, and they noticed many links between the two. Roughly speaking, their Euler characteristics are the same, and the analog of the Atiyah-Floer conjecture is thus HF = SWF. As mentioned in the comments, there is a 5-part paper that develops such an equivalence (though still in peer-review phase?). The construction passes through a version of Embedded Contact Homology, a la Taubes' proof SWF = ECH. -Some further motivation/explanation for the equivalence is that Taubes has shown that SW theory on symplectic 4-manifolds is related to $J$-holomorphic curve theory (counts of monopoles are equal to counts of surfaces). During their graduate career, Hutchings and Lee developed a 3-dimensional version of this story: SW theory on 3-manifolds is equal to counts of gradient trajectories of $S^1$-valued Morse functions on 3-manifolds. It's the same spirit as Taubes' proof: Sequences of solutions to (perturbed) SW equations will limit to objects that are localized around flowlines/curves. These equivalences extend down one more dimension, where Taubes started it all: Vortices on surfaces (2-dimensional versions of SW theory) are in bijection with symmetric products of the surfaces. I think that based on this history, i.e. similar in spirit to Taubes' and Hutchings-Lee's construction, Tim Perutz has formulated a precise "Atiyah-Floer conjecture" for HF and SW on 3-manifolds (at least he has described this in recent seminar talks that I attended). -As for your last question in the comments, there is something similar: Bordered Monopole Floer Theory developed by Jon Bloom and John Baldwin (I think still in progress). Their theory uses a finite "generating" set of bordered (i.e. parametrized surface-boundary) handlebodies $\lbrace (H_\alpha,\partial H_\alpha\cong\Sigma)\rbrace$. There is a pairing $\mathcal{C}(Y_1\cup_\Sigma Y_2)\simeq_\text{quasi}\mathcal{C}(Y_1)\tilde{\otimes}_{\mathcal{A}(\Sigma)}\mathcal{C}(Y_2)$ using the $A_\infty$-tensor product, where $\mathcal{A}(\Sigma):=\bigoplus_{\alpha\beta}\widehat{CM}(H_\alpha\cup_\Sigma -H_\beta)$ and $\mathcal{C}(Y):=\bigoplus_\alpha\widehat{CM}(Y\cup_\Sigma -H_\alpha)$.<|endoftext|> -TITLE: Are strict $\infty$-categories localized at weak equivalences a full subcategory of weak $\infty$-categories? -QUESTION [14 upvotes]: One has a nice "folk" model structure on strict $\infty$-categories due to Yves Lafont, Francois Metayer and Krzysztof Worytkiewicz whose notion of weak equivalences seem to be the notion of weak equivalences for weak $\infty$-category (I.e. a weaker notion than the existence of a strict inverse). -This produces a weak $(\infty,1)$-category of strict $\infty$-categories. My question is: is it expected to be a full subcategory of the category of weak $\infty$-category ? -(Note: I know that strict infinity categories with strict functors between them do not form a full subcategory of weak infinity categories, what I'm asking here is different, essentially because in the model structure mentioned above not all objects are cofibrant) -I'm actually not sure we have satisfying model for general weak $\infty$-category, and I might prefer to avoid the sort of problems mentioned in this answer, so I'll be happy with an answer dealing with $(\infty,n)$-categories defined for exemple as $n$-fold segal spaces, Rezk $\Theta_n$ spaces, Ara $n$-quasicategories or any other reasonable model. Also an answer focusing on $\infty$-groupoid or $(\infty,1)$-category would already be interesting. -Also, as a side question, assuming this is indeed fully faithful, is there any known result about which are the $\infty$-categories (or maybe $\infty$-groupoids) that are representable by strict $\infty$-categories ? -Edit: Let me clarify a few things which from what I read in the comments where unclear. -From the model structure of Lafont, Metayer, Worytkiewicz one obtains a notion of weak $\infty$-functor between strict $\infty$-category: as every objects in this model structure is fibrant a weak functor (or a weak anafunctor) from $X$ to $Y$ is a morphism $\widetilde{X} \rightarrow Y$ from some cofibrant replacement $\widetilde{X}$ of $X$, and notion of natural isomorphism of weak functor as morphism $\widetilde{X} \rightarrow PY$ where $PY$ is the path object for $Y$ in this model structure. -One can chose a functorial cofibrant replacement to have something more canonical, or even a comonadic one in order to obtain associative composition, but the choice of the cofibrant replacement does not have any effects on the question I'm asking, and it is possible to formulate it without choosing ones. -My question can be formulated as: does it defines the correct set of equivalence class of weak functor between strict $\infty$-categories if one see these as weak $\infty$-categories (and more generally, the correct space of morphism if one push things a little further). -Also note that I'm only interested in the 'canonical' way of sending strict $\infty$-categories to weak $\infty$-category, by just forgeting their strictness. -I know there is ways to send strict $\infty$-categories to weak $\infty$-groupoids or weak $(\infty,1)$-categories by formally (weakly) inverting all arrows or all $k$-arrow for $k>0$ , but then the image by this construction functor is no longer a strict $\infty$-category, and this construction has absolutely no chance to be fully faithful (it will like asking if the geometric realization functor from categories to the homotopy category of spaces is fully faithful). -The construction I'm refering to has good chances to be fully faithful (which is what I'm asking) but is clearly not essentially surjective even on $\infty$-groupoids: the groupoids in its image have for example trivial whitehead products $\pi_2 \times \pi_2 \rightarrow \pi_3$. The follow up question I asked is about knowing if we do have a good characterization of the image of this functor (for example, by the vanishing for all whitehead product in degree higher than $(2,2)$ or something like that). But please don't try to explain that there is construction which allow to represent all $\infty$-groupoid by strict $\infty$-category. - -REPLY [4 votes]: To record the jist of Charles' comment as an answer, the answer to the title question should decidedly be no if we take the $\infty$-category of strict $\infty$-categories to be that presented by the model structure of LaFont, Metayer, and Worytkiewicz (i.e. the morphisms are strict functors and the weak equivalences are created by the inclusion into weak $\infty$-categories). -If the inclusion of strict $\infty$-categories into weak $\infty$-categories were fully faithful, then the inclusion of strict $\infty$-groupoids into weak $\infty$-groupoids would be fully faithful. It would follow that the inclusion of pointed, 1-connected strict $\infty$-groupoids into pointed, 1-connected weak $\infty$-groupoids would also be fully faithful. But this is very far from the case -- pointed, 1-connected strict $\infty$-groupoids are equivalent as a 1-category to 1-connected crossed complexes, i.e. to 1-connected chain complexes. The inclusion functor is the usual forgetful functor. Moreover, the homotopy theory is also the same as the usual homotopy theory on chain complexes, as observed by Ara, so the we get the $\Omega^\infty$-functor restricted to 1-connected $H\mathbb Z$-modules, which is decidedly not fully faithful as a functor to pointed 1-connected spaces. -To spell this out a bit further, the essential image of this functor is the simply-connected products of Eilenberg-MacLane spaces. A map between two of these is an (unstable) cohomology operation. There are lots of these which don't come from maps of chain complexes. For instance, the "squaring" map $K(R,n) \to K(R,2n)$ for $R$ a commutative ring is one such (it can't come from a map of chain complexes for degree reasons). For that matter any element of the mod $p$ Steenrod algebra in other than the scalars or the Bockstein is a cohomology operation which again can't be represented by a chain complex map for degree reasons; these operations are even stable -- so they correspond to maps of spectra which are not maps of $H\mathbb Z$-modules. So our inclusion functor factors as $H\mathbb Z - Mod \to Spectra \to Spaces$, and neither functor in the composite is at all full. -As Harry points out, if we want to get fully faithfulness, then we can get more morphisms between strict $\infty$-groupoids by using pseudofunctors and such rather than strict functors. I'm not sure the inclusion becomes fully faithful with this modification (no pun intended!), although something like this does work in the 2-truncated case. For a nice overview of the possibilities in the 2-truncated case, there's this paper of Noohi.<|endoftext|> -TITLE: Probability of a given ordering of independent random variables -QUESTION [8 upvotes]: I ran into the following problem and I can't figure out a way of solving it... Suppose I have $N$ continuous random variable $X_1$,...,$X_N$ that are independent, but their law is different, and I know the cdf of each of them, i.e. the cdf of $X_i$ is $F_i(x)$. -I want to compute the probability of a given ordering of those random variable, for example -\begin{equation} -\mathcal{P}\left[X_12$. -Thank you - -REPLY [2 votes]: Kapouet, since my comments are long and exceed the number of allowed characters in a regular comment, i had to put it so (i.e., "an answer"). anyway, anyone can edit the comments and even delete them. Here are some thoughts: -(1) suppose there were a genealization with a simple form, what would you expect it to be? -(2) say, you have to do the brutal force intergation, then any systematic reduction of the integral will necessarily require some invariant property of the joint distribution of the random variables. recall the permutation invariance of the joint distribution of indepenent and identically distributed random variables. -(3) since the random variables are independent and the set is induced by succecive "cuts" (compare it with the case $N=2$), you can try mathematical induction to get a general formula without possibly computing the multiple intergal by brutal force. specifically, try a conditional argument on one or more of the random variables and combine induction. this way you are very likely to get some elegant and general formula -(4) finally, define the left inverse from $F_i$ and convert the problem to the case of random variables on the unit cube (if possible, say, when $F_i$ are absolutely continuous). then you proceed with (3) -Good luck! -Comments (4/22/2017): in my comments, i did not assume that the random variables are identically distributed; i was just referring to the i.i.d. case to illustrate that some invariant properties of the joint distribution or the set for which the probability is computed is needed for a systematic reduction of the integral. -This type of thing is usually hard to compute and one usually does not have an answer for this at hand unless he/she has done research along this line. i am confident that the strategy posted is the correct way to go! however, it is unlikely that someone else will compute the probability for the question owner since we are busy and sometimes very busy ....<|endoftext|> -TITLE: Spin structures on Sasakian manifolds and the Kähler analogy -QUESTION [12 upvotes]: A Sasakian manifold is often said to be the odd dimensional analogue of a Kähler manifold. -Now for a $2n$-dimensional Kähler manifold we know from Atiyah that it is spin exactly if the line bundle $\Omega^{(0,n)}$ admits a square root ${\cal S}$, and a choice of spin structure is equivalent to a choice of holomorphic structure ${\cal S}$. In this case the associated spinor bundle is the tensor product of ${\cal S}$ with the anti-holomorphic complex. Moreover, the associated Dirac operator is the tensor product of $\overline{\partial} + \overline{\partial}^*$ with the $\overline{\partial}$-operator corresponding to the choice of holomorphic structure. -So does the above analogy present any Sasakian versions of these spin geometry to complex geometry dictionary? - -REPLY [2 votes]: Every Sasakian manifold $M$ (of dimension $2k+1$) has a canonical $\mathrm{Spin}^c$ structure, because the cone $\overline{M}$ over $M$ is Kähler and thus has a canonical $\mathrm{Spin}^c$ structure which restricts to $\mathrm{Spin}^c$ structure on $M$. -If $M$ is Einstein, then the cone $\overline{M}$ is Ricci flat and that in turn implies that the auxiliary bundle $\Lambda^{k+1, 0} \overline{M}$ of the canonical $\mathrm{Spin}^c$ structure on the cone $\overline{M}$ is flat. Now $\mathrm{Spin}^c$ structures on simply connected manifolds with trivial auxiliary bundle is canonically identified with a spin strucutre. -For more details see Moroianu, Andrei: Parallel and Killing spinors on $Spin^c$ manifolds, Comm. Math. Phys. 187 (1997), no. 2, 417–427<|endoftext|> -TITLE: The closure of cyclic modules under direct sums and direct summands -QUESTION [5 upvotes]: Cohen and Kaplansky have proven that a commutative ring $R$ has the property - -C: Every $R$-module is a direct sum of cyclic $R$-modules. - -if and only if $R$ is an Artinian principal ideal ring. Can we classify those commutative rings $R$ with the following property? - -C': Every $R$-module is a direct summand of a direct sum of cyclic $R$-modules. - -Actually I am interested in the following property, which is more involved: - -C'': For every $R$-module there is a commutative $R$-algebra $R'$ satisfying $[\forall T \in \mathsf{Mod}(R):~R' \otimes_R T = 0 \Rightarrow T=0]$ (for example, a faithfully flat algebra) such that the $R'$-module $R' \otimes_R M$ is a direct summand of a direct sum of localizations of cyclic $R'$-modules. - -If $R$ is arbitrary, what can we say about this class of $R$-modules? - -REPLY [4 votes]: The introduction of -Warfield, R.B.jun, Rings whose modules have nice decompositions, Math. Z. 125, 187-192 (1972). ZBL0218.13012. -says - -It is also shown that if $R$ is a commutative ring and there is a cardinal number $\mathfrak{n}$ such that every $R$-module is a summand of a direct sum of modules with at most $\mathfrak{n}$ generators, then $R$ is an Artinian principal ideal ring. - -That seems to answer question C'.<|endoftext|> -TITLE: Why is free probability a generalization of probability theory? -QUESTION [25 upvotes]: Note: This question was already asked on Math.SE nearly a week and a half ago but did not receive any responses. To the best of my knowledge, free probability is an active topic of research, so I hope that the level of the question is appropriate for this website. If not, please let me know so I can delete the question myself. - -Question: One often sees statements to the effect that "free probability is a generalization of probability theory, which is commutative, to the non-commutative case". -But in what sense does classical probability theory only concern itself with commutative quantities? -If my understanding is correct, and probability theory also deals with non-commutative quantities, then in what sense is free probability a generalization of probability theory? - -Context: The simplest random variables are real-valued, and obviously real numbers have commutative multiplication. But random variables can take values in any measurable space (at least this is my understanding and also that of Professor Terry Tao), i.e. random variables can also be random vectors, random matrices, random functions, random measures, random sets, etc. The whole theory of stochastic processes is based on the study of random variables taking values in a space of functions. If the range of the functions of that space is the real numbers, then yes we have commutative multiplication, but I don't see how that's the case if we are e.g. talking about functions into a Riemannian manifold. -EDIT: To clarify what I mean by "classical probability theory", here is Professor Tao's definition of random variable, which is also my understanding of the term (in the most general sense): - -Let $R=(R, \mathcal{R})$ be a measurable space (i.e. a set $R$, equipped with a $\sigma$-algebra $\mathcal{R}$ of subsets of $R$). A random variable taking values in $R$ (or an $R$-valued random variable) is a measurable map $X$ from the sample space to $R$, i.e. a function $X: \Omega \to R$ such that $X^{-1}(S)$ is an event for every $S \in \mathcal{R}$. - -Then, barring that I am forgetting something obvious, classical probability theory is just the study of random variables (in the above sense). -/EDIT -To be fair though, I don't have a strong understanding of what free probability is. Reading Professor Tao's post about the subject either clarified or confused some things for me, I am not sure which. -In contrast to his other post, where he gives (what seems to me) a more general notion of random variable, in his post about free probability, Professor Tao states that there is a third step to probability theory after assigning a sample space, sigma algebra, and probability measure -- creating a commutative algebra of random variables, which supposedly allows one to define expectations. (1) How does one need a commutative algebra of random variables to define expectations? (2) Since when was defining a commutative algebra of random variables part of Kolmogorov's axiomatization of probability? -Later on his post about free probability, Professor Tao mentions that random scalar variables form a commutative algebra if we restrict to the collection of random variables for which all moments are finite. But doesn't classical probability theory study random variables with non-existent moments? Even in an elementary course I remember learning about the Cauchy distribution. -If so, wouldn't this make classical probability more general than free probability, rather than vice versa, since free probability isn't relevant to, e.g., the Cauchy distribution? -Professor Tao also mentions random matrices (specifically ones with entries which are random scalar variables with all moments finite, if I'm interpreting the tensor product correctly) as an example of a noncommutative algebra which is outside the domain of classical probability but within the scope of free probability. But as I mentioned before, aren't random matrices an object of classical probability theory? As well as random measures, or random sets, or other objects in a measurable space for which there is no notion of multiplication, commutative or non-commutative? -Attempt: Reading Professor Tao's post on free probability further, it seems like the idea might be that certain nice families of random variables can be described by commutative von Neumann algebras. Then free probability generalizes this by studying all von Neumann algebras, including non-commutative ones. The idea that certain nice families of random variables correspond to the (dual category of) commutative von Neumann algebras seems like it is explained in these two answers by Professor Dmitri Pavlov on MathOverflow (1)(2). -But, as Professor Pavlov explains in his answers, commutative von Neumann algebras only correspond to localizable measurable spaces, not arbitrary measurable spaces. While localizable measurable spaces seem like nice objects based on his description, there is one equivalent characterization of them which makes me suspect that they are not the most general objects studied in probability theory: any localizable measurable space "is the coproduct (disjoint union) of points and real lines". This doesn't seem to characterize objects like random functions or random measures or random sets (e.g. point processes), and maybe even not random vectors or matrices, so it does not seem like this is the full scope of classical probability theory. -Thus, if free probability only generalizes the study of localizable measurable spaces, I don't see how it could be considered a generalization of classical probability theory. By considering localizable measurable spaces in the more general framework of possibly non-commutative von Neumann algebras, it might expand the methods employed in probability theory by borrowing tools from functional analysis, but I don't see at all how it expands the scope of the subject. To me it seems like proponents of free probability and quantum probability might be mischaracterizing classical probability and measure theory. More likely I am misinterpreting their statements. -Related question. Professor Pavlov's comments on this article may be relevant. -I am clearly deeply misunderstanding at least one thing, probably several things, here, so any help in identifying where I go wrong would be greatly appreciated. - -REPLY [12 votes]: There have been many good answers to this question, but it might be that -the main point gets lost in too many details. So, as kind of expert on free -probability theory, let me try to give a short direct answer to the question -“Why is free probability a generalization of probability theory.” -There are two main ingredients in free probability theory: first, the general -notion of a non-commutative probability space and second, the more specific -notion of freeness (or free independence). -A non-commutative probability space consists of an algebra and a linear functional. -The algebra can (despite the use of “non-commutative”) also be commutative and -thus a classical probability space (encoded in the commutative algebra of random variables -and the functional given by taking the expectation with respect to the underlying probability -measure) is also an example of a non-commutative probability space. -However, in this generality non-commutative (as well as classical) probability spaces -are not too exciting. One needs more structure for interesting statements. In the classical -setting, the most basic additional structure is “independence”. In free probability the corresponding -structure is “free independence”. However, free independenc is NOT a generalization of -independence; it is an analogue. What independence means for classical (commuting) -random variables, free independence means for non-commuting variables. Apart from -trivial situations, there are no classical random variables which are free. Hence freeness -is not a kind of dependence for classical variables; it is a special relation for non-commuting -variables, which behaves in many respects like independence. -Hence the above question has two possible answers, depending on how it is interpreted. -Read as “Why is a non-commutative probability space a generalization of a classical -probability space?” the answer is just: because a commutative algebra is also allowed -as an example of a non-commutative algebra. -Read as “Why is free independence a generalization of classical independence?” the answer is: -this is actually not true, free independence is not a generalization, but an analogue of -classical independence.<|endoftext|> -TITLE: Is algebraic geometry constructive? -QUESTION [40 upvotes]: Notes: 1) I know next to nothing about algebraic geometry, although I am greatly interested in the field. 2) I realize that "constructive" might be a technical term, here I am using it only in an informal manner. -I hope that this question belongs on this site, since it is not strictly research-level. -As an autodidact (I have an ongoing formal education in physics, but the amount of math we learn here is abysmal, so most of my mathematics knowledge is self-taught), I have noted that algebraic geometry seems to be really impenetrable for somebody who has no formal education in the field, unlike, say, differential geometry or functional analysis, which are areas I can effectively learn on my own. -Pretty much every time I encounter AG-related stuff on sites such as this one or math.se, I see layers and layers of abstractation on top of one another to the point where it makes me wonder, is this field of mathematics constructive, in the sense that can it be used to actually calculate anything or have any use outside mathematics? -The point I am trying to make is that, using differential geometry as an example, is is constructive. No matter how abstractly do I define manifolds, tensor fields, differential forms, connections, etc, they are always resoluble into component functions in some local trivializations, with which one can actually calculate stuff. -Every time I used DG to calculate stellar equilibrium or cosmological evolution or geodesics of some model spacetime, I get actual, direct, palpable, realizable results in terms of real numbers. -I can use differential forms to calculate the volumes of geometric shapes, and every time I use a Lagrangian or Hamiltonian formalism to calculate trajectories for classical mechanical systems, I make use of differential geometry to obtain palpable results. -On top of that, I know that DG is useful outside physics too, I have heard of uses in economy, music theory etc. -I am curious if there is any real-world application of AG where one can use AG to obtain palpable results. I am not curious (for the purpose of this question) about uses to mathematics itself, I know they are numerous. But every time I try to read about AG I get lost in the infinitude of sheaves, stacks, schemes, functors and other highly abstract objects, which often seem so impossible to me to be resolved into calculable numbers. -The final point is, I would like to hear about some interesting applications of algebraic geometry outside mathematics, if there are any. - -REPLY [3 votes]: Slightly roughly, a smooth algebraic variety over $\mathbb{R}$ has an open covering by algebraic varieties of the form $U = \{ (x_1 , \ldots , x_n) \in \mathbb{R}^n \ | \ f_1 (x_1 , \ldots , x_n) = 0 , \ldots , f_m (x_1 , \ldots , x_n) = 0 \}$ where the $f_1 , \ldots , f_m$ are polynomials (with some condition on derivatives making this closed submanifold of $\mathbb{R}^n$ smooth). One can then describe the transition function and cocycle condition etc., so this is exactly like a smooth manifold, but in fact of a more restrictive nature (you can only construct from manifolds which are cut out by polynomials). -You can then cover $U$ by open subsets of $\mathbb{R}^{n - m}$, but this will be "not algebraic". An algebraic way of doing this further simplification is called "etale cover". -One then needs a series of Q&A's, to explain why you then see mountains of abstraction, so perhaps this is not the place. Basically and roughly, this is because you can ask for solutions of the same polynomial equations also over $\mathbb{C}$ and over $\mathbb{Z} / p \mathbb{Z}$ and in fact over a lot of other fields and rings, and it turns out that even if you were interested in only one ring, the "manifolds" that you get from others turn out to be related and interconnected, and contain useful information for your original pursuit. Thus, one would like to think of an algebraic manifold as being over all rings at once, and then there are various devices and theories which look quite abstract if one is not used to them, for dealing with this.<|endoftext|> -TITLE: When do the polynomial algebra and free algebra coincide in brave new algebra? -QUESTION [18 upvotes]: Given an $\mathbb E_\infty$-ring (highly structured commutative ring spectrum if you want) $R$, we have the free $R$-algebra (on one generation) $R\{t\}\simeq \bigoplus_{n\ge 0} R_{\mathrm h\Sigma_n}$ vs. the polynomial $R$-algebra $R[t]\simeq \bigoplus_{n\ge 0} R.$ -It is well-known that $R\{t\}\simeq R[t]$ when $R$ is rational, i.e. a $H\mathbb Q$-algebra. It is also very easy to verify that this is not the case in many other cases, for instance when $R= H\mathbb Z$ and $H\mathbb F_p$. -Q: Does $R\{t\}\simeq R[t]$ for a (connective?) $\mathbb E_\infty$-ring imply that $R$ is a $H\mathbb Q$-algebra? -In case not, I would be very happy with some example of non-rational ring spectra $R$ for which polynomial and free $R$-algebras agree. Also, in that case, what condition on an $\mathbb E_\infty$-ring $R$ does the condition that $R\{t\}\simeq R[t]$ imply? -In the special case of a discrete ring $R$ = Eilenberg-MacLane $\mathbb E_\infty$-ring $HR,$ this question reduces to one about group cohomology of symmetric groups. More precisely: -Q': Does the condition that $\mathrm H^i(\Sigma_n, R)\simeq 0$ for all $n$ and all $i\ge 1$ for a commutative ring $R$ imply that $R$ is a $\mathbb Q$-algebra? -This sounds like a possibly-very-easy piece of classical algebra, but to my shame, I do not know the answer even in this particular case. - -REPLY [7 votes]: Any morphism of $R$-algebras $\varphi : R\{t\} \to R[t]$ is determined up to homotopy by an element of $\pi_0(R[t]) \approx \pi_0(R)[t]$. If $\varphi$ is an equivalence, then this element must be the generator $t$, so we may as well assume $\varphi$ is the canonical map $\varepsilon_R : R\{t\} \to R[t]$. -Claim: -Let $R$ be an $E_\infty$-ring spectrum. Then the map $\varepsilon_R$ is invertible if and only if $R$ is a $\mathbf{Q}$-algebra. -Useful observations: -1) The map $\varepsilon_R$ is compatible with extensions of scalars. Therefore if $\varepsilon_R$ is invertible, then $\varepsilon_{R'}$ is invertible for any $R$-algebra $R'$. -2) $R$ is a $\mathbf{Q}$-algebra if and only if $\pi_0(R)$ is a $\mathbf{Q}$-algebra (see comments). -Sufficiency: -Suppose that $R$ is a $\mathbf{Q}$-algebra. The map $\varepsilon_R$ is obtained by extension of scalars along the connective cover $R_{\ge 0} \to R$, so we may assume $R$ is connective. The $\infty$-category of connective $E_\infty$-algebras over $\mathbf{Q}$ is equivalent to that of simplicial commutative $\mathbf{Q}$-algebras. Under this equivalence, the map $\varepsilon_R$ corresponds to the identity morphism $R[t] \to R[t]$ (where by abuse of notation $R$ also denotes the corresponding simplicial commutative $\mathbf{Q}$-algebra). -Necessity: -Suppose $\varepsilon_R$ is an equivalence. Since the map $\varepsilon_R$ is compatible with formation of connective covers, we may replace $R$ by its connective cover. We may also extend scalars along $R \to \pi_0(R)$ to assume $R$ is discrete. -Consider any residue field $R \to k$. -If $k$ is of characteristic $p>0$, then $\varepsilon_k$ cannot be invertible. Indeed, it is well-known that $\varepsilon_{\mathbf{F}_p}$ is not invertible, and $\varepsilon_k$ is the extension of scalars along the faithfully flat map $\mathbf{F}_p\{t\} \to k\{t\}$. -Thus every residue field $k$ must be of characteristic zero, so $R$ is a $\mathbf{Q}$-algebra.<|endoftext|> -TITLE: How much real determinacy can live in $L(\mathbb{R})$? -QUESTION [6 upvotes]: It's well-known that AD$_\mathbb{R}$ fails in $L(\mathbb{R})$, provably in ZFC. This is because: - -AD$^{L(\mathbb{R})}$ implies DC$^{L(\mathbb{R})}$. -Over ZF+DC, AD + "Every set of reals has a scale" is equivalent to AD$_\mathbb{R}$. This is due to Woodin; see Theorem 32.23 in Kanamori's book. -But in $L(\mathbb{R})$, properly $(\Pi^2_1)^{L(\mathbb{R})}$ sets of reals do not have scales. - -I'm curious how much of AD$_\mathbb{R}$ is consistent (relative to large cardinals) with ZF + $V=L(\mathbb{R})$ + AD. Specifically, it's clear that the limit should be around $\Sigma^2_1$, but I'm not sure exactly where it lies (I can't recall precisely the level-by-level connection between scales and strategies, and I can't find it in Moschovakis); it's not even clear to me that projective real determinacy is consistent with this theory. Additionally, I'm interested in the large cardinals beyond a proper class of Woodins required to get (roughly) this amount of real determinacy in $L(\mathbb{R})$. - -REPLY [3 votes]: I think the following may be relevant to your question: - -First, several bullet points in your question are nontrivial results. However, there is another way to see $\mathsf{AD}_\mathbb{R}$ does not holds in $L(\mathbb{R})$. -Using a two step game, you can show $\mathsf{AD}_\mathbb{R}$ can prove uniformization for all sets of reals. Consider the relation $R(x,y)$ if and only if $y$ is not ordinal definable from $x$. Assuming that $L(\mathbb{R}) \models \mathsf{AD}_\mathbb{R}$, $R$ would have a uniformization. Every set of reals in $L(\mathbb{R})$ is ordinal definable from some real. Use this real to get a contradiction by diagonalization. -Also note that if you put the discrete topology on $\mathbb{R}$, a two point game is a closed game. Hence $\mathsf{ZF + AD}$ can not prove the Gale-Stewart theorem (closed determinacy) for games on $\mathbb{R}$. Of course, this is because there is no well-ordering of $\mathbb{R}$. - -"The Extent of Scales in $L(\mathbb{R})$" Corollary 6 shows that $\mathsf{DET}_\mathbb{R}(\mathbf{\Pi}_1^1)$ can not hold in $L(\mathbb{R})$. Steel shows this by proving that $G^\mathbb{R}\Pi_1^1$ (the game quantifier class) is $\Sigma_1(L(\mathbb{R}, \{\mathbb{R}\})$. By $\mathsf{DET}_\mathbb{R}(\mathbf{\Pi}_1^1)$, $G^\mathbb{R} \Sigma_1^1$ is the dual of $G^\mathbb{R}\Pi_1^1$. Hence $G^\mathbb{R}\Sigma_1^1$ is $\Pi_1(L(\mathbb{R}, \{\mathbb{R}\})$. But you can directly show that $G^\mathbb{R}\Sigma_1^1$ is contained in $\Sigma_1(L(\mathbb{R}),\{\mathbb{R}\})$. Contradiction. -($\Sigma_1(L(\mathbb{R},\{\mathbb{R}\})$ is also $(\Sigma^2_1)^{L(\mathbb{R}}$. Using Wadges lemma and the not-ordinal-definable relation $R$ from above, Steels showed that if $\mathsf{AD}^{L(\mathbb{R})}$ holds, then a set has a scale in $L(\mathbb{R})$ if and only if it belongs to $(\mathbf{\Sigma}^2_1)^{L(\mathbb{R})}$. ) -The same paper of Steel also notes that assuming $\mathsf{AD}$, $\mathsf{DET}_\mathbb{R}(\mathbf{\Pi}_1^1)$ is equivalent to the existence of $\mathbb{R}^\sharp$. (At least in the paper, Steel states it is open whether the equivalence can be proved in just $\mathsf{ZF} + \mathsf{DC}$.) Assuming $\mathsf{AD}$, $\mathsf{DET}_\mathbb{R}(\mathbf{\Pi}_1^1)$ has stronger consistency strength than $\mathsf{AD}$.<|endoftext|> -TITLE: When are $\mathbb{R}^n$ and $\mathbb{R}^m$ essentially similar? -QUESTION [9 upvotes]: Here is a rather vague and subjective question: for which $n$ and $m$ are -$\mathbb{R}^n$ and $\mathbb{R}^m$ ``essentially similar''? The answer depends partly on what type of -mathematician is answering it. -Of course, in a certain technical sense, $\mathbb{R}^n$ and $\mathbb{R}^m$ are different for any $n\neq m$ since they -are not homeomorphic (or linearly isomorphic); thus they by definition differ in some describable topological (or algebraic) way. -But qualitatively all Euclidean spaces, especially ones of large dimension, seem to be ``similar.'' -The question is really: how large does $n$ have to be for Euclidean spaces of dimension $\geq n$ -all to behave essentially the same way? -I have included my own discussion of some possible answers in my Real Analysis manuscript at http://wolfweb.unr.edu/homepage/bruceb/Meas.pdf (Section XI.18.3). I invite comments on this discussion. - -REPLY [6 votes]: There are geometric and topological properties that significantly distinguish between Cartesian spaces of odd and even dimensions. Take, for example, the "sphere combing" property. The $(n-1)$-dimensional sphere (the boundary of a ball) in $\mathbb{R}^n$ admits a non-vanishing continuous tangent vector field if and only if $n$ is even. -Another striking difference is related to the existence of Hadamard matrices. It is known that an $n\times n$ Hadamard matrix exists only if $n=1,\ 2$, or is a multiple of $4$ (see https://en.wikipedia.org/wiki/Hadamard_matrix ), and Hadamard conjectured that there exists such a matrix for every $n$ divisible by $4$. The conjecture is still unsolved, but there exist many constructions producing an $n\times n$ Hadamard matrix for infinitely many values of $n$. Now, the geometric connection is this: Some $n+1$ vertices of the $n$-dimensional cube form a regular simplex if and only if there exists an $(n+1)\times(n+1)$ Hadamard matrix. -Thus, spaces $\mathbb{R}^n$ and $\mathbb{R}^{n+1}$ often differ essentially for infinitely many values of $n$, and I think many other examples of this nature can be readily presented.<|endoftext|> -TITLE: Question about Fourier coefficients of a newform at primes -QUESTION [5 upvotes]: For $q:=e^{2\pi i z},$ let $f(z)=\sum_{n\ge 1}\lambda(n)n^{(k-1)/2}q^n$ be a normalized newform of type $(k,\chi)$ and level $N$. For any prime $p,$ we have -$$\lambda(p)=2\cos(\theta_p)\;\;\;\text{for some}\;\;\; \theta_p\in[0,\pi]$$ -I wonder: Is $\theta_p\pmod{2\pi}$ take only finitely many values when $p$ run over primes number? - -REPLY [7 votes]: No, as this would contradict the Sato-Tate conjecture. For CM forms this conjecture is easier and has been long known, see e.g. this MO entry.<|endoftext|> -TITLE: Is the fundamental group of any compact hyperbolic 3-manifold embeddable into a p-adic group? -QUESTION [12 upvotes]: Is it true that for every compact hyperbolic $3$-manifold $M$ there exists a prime $p$, a finite field extension $K/\mathbb{Q}_p$, and an injective group homomorphism $$\tau \colon \pi_1(M) \to \mathrm{PSL}_2(\mathcal{O}_K) ?$$ -Here $\mathcal{O}_K$ is the ring of integers of $K$. -This question is obviously related to the existence of faithful low dimensional representation of hyperbolic groups (see the following survey by Aschenbrenner, Friedl, and Wilton). -An affirmative solution will give a (partial) confirmation of Conjecture 1 from this paper by Luo. -A paper by Calegari and Emerton also discusses such group homomorphisms $\tau$. -Note that the answer is obviously 'yes' if $M$ is arithmetic. - -REPLY [11 votes]: I think this follows easily from a couple of facts. The discrete faithful representation of an orientable hyperbolic 3-manifold leads to an embedding $\pi_1(M) \to PSL_2(S) \subset PSL_2(k)$, where $k$ is a number field, and $S$ is a finitely-generated subring (generated e.g. by the coefficients of matrices of generators). For most prime ideals $\mathcal{P} \subset \mathcal{O}_k$, $S$ will have non-negative valuation, and hence will lie in the valuation ring $R_{\mathcal{P}} \subset k_{\mathcal{P}}$ (the ring of $\mathcal{P}$-adic integers), hence $\pi_1(M) \hookrightarrow PSL_2(R_\mathcal{P})$. Such a field is a finite extension of $\mathbb{Q}_p$, where $\mathcal{P}|p$. (I'm attempting to follow the notation and results of Chapter 0.7 of MacLachlan-Reid). -It's possible I've misinterpreted your notation. I'm interpreting your $K=k_\mathcal{P}, \mathcal{O}_K=R_\mathcal{P}$ (the valuation ring) to translate to MacLachlan-Reid's notation. If $\mathcal{O}_K$ does not denote the valuation ring, then I think you should clarify what you mean. This seems to me to be the only reasonable interpretation with $K=\mathbb{Q}_p, \mathcal{O}_K=\mathbb{Z}_p$.<|endoftext|> -TITLE: How to braid a ribbon knot -QUESTION [6 upvotes]: Is there any algorithm known for braiding ribbon knots? More specifically I need to braid a generic ribbon knot presented as boundary of a ribbon surface= union of two 0-handles and one 1-handle. (Unfortunately MO did not allow me to add a picture). - -REPLY [6 votes]: If I understand your question correctly, the answer is in the paper - -Lee Rudolph, Braided surfaces and Seifert ribbons for closed braids, Comment. Math. Helvetici 58 (1983), 1–37. - -In fact, in Section 3 Rudolph proves that every ribbon surface is isotopic to one arising from a braid (in a very concrete way, explained in detail in the paper), which is a lot stronger than what you ask for.<|endoftext|> -TITLE: explicit description of the product map in K-theory -QUESTION [5 upvotes]: Let $A$ and $B$ be unital $C^*$-algebras. Let $u\in M_n(A)$ be a unitary representing an element $[u]\in K_1(A)$ and $p\in M_m(B)$ be a projection representing an element $[p]\in K_0(B)$. Then the unitary $$u\otimes p+1_A\otimes (1_B-p)$$ represets the product $[u]\times [p]\in K_1(A\otimes B)$ (see in Nigel Higson's and John Roes book "Analytic K-homology", proposition 4.8.3). This gives us a description of a product map $K_1(A)\times K_0(B)\to K_1(A\otimes B)$. -Now, let $A$ non-unital and $B$ unital, but $u$ and $p$ as above, where $u\in K_1(A)\cong K_1(A^+)$ with $A^+$ the unitization of $A$. -My question is: -Is it possible to give an explicit description for $K_1(A)\times K_0(B)\to K_1(A\otimes B)$ as above for this case to map $[u]$ and $[p]$ to $K_1(A\otimes B)$ naturally? -I guess that this is not possible. My first try was to consider $$u\otimes p+1_{A^+}\otimes (1_B-p),$$ which represents an element in $K_1(A^+\otimes B)$. However, since $A\otimes B$ is non-unital, the product map should map to $K_1(A\otimes B)\cong K_1((A\otimes B)^+)$, but $ K_1((A\otimes B)^+)$ -is not isomorphic to $K_1(A^+\otimes B)$. For $A$ nonunital and $B$ unital and nuclear, there is the following relation: there is a commutative diagram -$$\require{AMScd}\def\Z{\mathbb{Z}} -\begin{CD} -0 @>>> K_1(A)\otimes K_0(B) @>>> K_1(A^+)\otimes K_0(B) @>>>> K_1(\mathbb{C})\otimes K_0(B) @>>> 0 \\ -@. @VV V @VV\alpha V @VV\alpha V @. \\ -0 @>>>K_1(A\otimes B) @>>> K_1(A^+\otimes B) @>>> K_1(\mathbb{C}\otimes B) @>>> 0 \\ -\end{CD} -$$ -where $K_1(\mathbb{C})=0$, $K_1(\mathbb{C}\otimes B)\cong K_1(B)$ and $K_1(A)\otimes K_0(B) \to K_1(A^+)\otimes K_0(B)$ is an isomorphism. If $B$ is nonnuclear, then the sequence in this diagram is not exact at $K_1(A\otimes B)$. - -REPLY [3 votes]: I believe the correct formula is $$(u-1_{A^+}) \otimes p + 1_{(A \otimes B)^+}$$ which makes sense if you work this out in the following special case. Let $A$ be $C_0(0,1)$ and let $B$ be $C(\{0,1\})$, and take $u(s)=e^{2\pi is}$ and $p(t)=t$. In this case $$((u-1_{A^+}) \otimes p + 1_{(A \otimes B)^+})(s,t)= (e^{2\pi i s}-1)t + 1$$ which is constantly one on $(0,1)\times\{0\}$ and loops around the unit circle on $(0,1)\times\{1\}$. Check that this is plus-or-minus the correct answer for this special case. Now use the universal properties of $C_0(0,1)$ and $C(\{0,1\})$ to get maps from these to a generic $A_1$ and $B_1$ and use the fact that the product map is natural to show this is the correct formula in general.<|endoftext|> -TITLE: The spectral radius of a binary matrix -QUESTION [6 upvotes]: Let $\mathcal B_n$ denote the set of $n\times n$ matrices with entries in $\{0,1\}$. It follows from the spectral radius formula that if $M\in\mathcal B_n$ is not nilpotent, then $\rho(M)$, the spectral radius of $M$, is $\ge1$. -I need a condition for $\rho(M)>1$ in terms of the proportion of 1s. -QUESTION. Is there a $c\in(0,1)$ such that for any $n\ge1$ and any $M\in\mathcal B_n$ with the number of 0s less than $cn^2$ we have $\rho(M)>1$? -If so, what's the best known value of $c$? - -REPLY [12 votes]: Perron-Frobenius theory clarifies this quickly. Observe first of all that if $M$ is irreducible in the sense that it cannot be made a block upper triangular matrix by a permutation of the standard basis vectors, then $\rho(M)>1$. This follows because the Perron-Frobenius eigenvector $x$ has positive entries, and if, say, $x_1>x_2> \ldots >x_n$, then $\lambda=1$ would force $M_{1j}=0$ for $j\ge 2$, so $M$ was reducible. A similar argument works if $x_1 = \ldots = x_m>x_{m+1} \ge \ldots$. -So $\rho(M)\le 1$ forces $M$ to be block upper triangular after a permutation of basis vectors and moreover the diagonal blocks must be zero or $1\times 1$ matrices (the argument in the preceding paragraph implicitly used that the matrix has size $n\ge 2$). Clearly such a matrix has $\ge (n^2-n)/2$ zeros, so any $c<1/2$ works. Moreover, we have also obtained an example with $(n^2-n)/2$ zeros and $\rho(M)=1$, so this is optimal.<|endoftext|> -TITLE: Comparing the growth of $f\circ g$ and $g\circ f$ -QUESTION [9 upvotes]: I asked this Question on Math.StackExchange without success. Then I learned, that this might be the better place to ask. So, sorry for crossposting. I would agree on deleting my old question. - -Let $\mathbb R_0^+:=\{x\in\mathbb R\mid x\geq 0\}$. Further let $f:\mathbb R_0^+\rightarrow \mathbb R_0^+$ and $g:\mathbb R_0^+\rightarrow \mathbb R_0^+$ two strictly increasing continuous functions (this can be weakened if necessary). - -Is there anything that can be said about which of the functions $f\circ g$ and $g\circ f$ grows asymptotically faster (in some sense) with only assuming something about the asymptotic growth behavior of $g$ and $f$? - -I want to discuss this in a very general context. So I am open for all kinds of useful definitions of "grows faster" and "growth behavior". E.g. one can consider the usual $\mathcal O$-notation and ask for whether -$$ f\circ g\in\Omega(g\circ f)$$ -whenever $f\in\Omega (g)$ or $g\in\Omega(f)$. But neither seems to hold in general. For example, conider $$f(x)=\log(x)\quad\text{and}\quad g(x)=x^\alpha.$$ Which of $f\circ g$ and $g\circ f$ grows faster depends on $\alpha$, while $g\in\Omega(f)$ regardless of $\alpha$. Also possible: we can call $f$ growing faster than $g$ if $f(x)>g(x)$ for all sufficiently large $x$. -I was not abled to proof anything remotely useful for the connection between the "growth" of $f$ and $g$ and the connection between the "growth" of $f\circ g$ and $g\circ f$. So this is a soft question because I hope for input from no specific branch of math. - -Further useful assumptions might be - -$f$ and $g$ are convex functions. -$f(x)>x$ and $g(x)>x$. - - -I tried to prove - -If $f>g^n$ for all $n\in\mathbb N$ ($g^n$ means $g$ iterated $n$ times), it holds $g\circ f< f\circ g$. - -For example, use $g(x)=x^2$ and $f=\exp$. I have not succeeded for any definition of growth so far but it seems plausible to me, at least for convex functions $f$ and $g$ strictly greater than identity. - -REPLY [2 votes]: Suppose that $g$ satisfies $g(x)>x$ for all $x$, and also that $f$ and $g$ lie in a Hardy field, as suggested by Todd Trimble. (A Hardy Field, called an order of infinity by Hardy is a collection of germs of differentiable functions at $\infty$ that is closed under differentiation and composition, as well as the field operations. In particular, this means that for any Hardy field, $H$, and any $f\ne g\in H$, $f-g\in H\setminus\{0\}$, so that $1/(f-g)\in H$. In particular, $f-g$ is eventually positive, or eventually negative. That is, there is a total ordering on a Hardy field.) -Then I believe there is an extension Hardy field containing $f$, $g$ and a function, $h$ satisfying $h(g(x))=h(x)+1$. That is, $h(x)$ counts how many times you have to apply $g$ to 0 to get to $x$ (think of $\log^* x$). -Assuming the existence of such a Hardy field, there is now an affirmative answer to your question. If you conjugate $g$ by $h$, you obtain $\tilde g:=h\circ g\circ h^{-1}$ is $\tilde g(x)=x+1$. Conjugating $f$ by $h$ also to obtain $\tilde f$, your question is equivalent to asking whether $\tilde f(\tilde g(x))>\tilde g(\tilde f(x))$. That is, whether $\tilde f(x+1)>\tilde f(x)+1$. Notice that both $\tilde f$ and $\tilde g$ belong to the Hardy field. -The condition that $f>g^n$ for all $n$ means that $\tilde f(x)>x+n$ for large $x$, that is $\tilde f(x)-x\to\infty$. In a Hardy field, any function satisfying this satisfies $\tilde f'(x)>1$ for all large $x$. By the mean value theorem, this implies that $\tilde f(x+1)>\tilde f(x)+1$, as required.<|endoftext|> -TITLE: Sources for Alexandrov surfaces -QUESTION [7 upvotes]: There are two distinct notions in differential geometry associated -with A. D. Alexandrov: (1) Alexandrov spaces of courvature bounded -from below; (2) Alexandrov surfaces of bounded total curvature (more precisely integral of absolute value of Gaussian curvature is bounded). -(The confusion is reflected in our tag alexandrov-geometry.) Here -(1) was extensively studied in particular by Burago, Gromov, and -Perelman. -This question concerns the notion (2). These surfaces were -extensively studied by Reshetnyak e.g., -Reshetnyak, Yu. G. On the conformal representation of Alexandrov -surfaces. Papers on analysis, 287-304, Rep. Univ. Jyväskylä -Dep. Math. Stat., 83, Univ. Jyväskylä, Jyväskylä, 2001. -There is a nice survey article by Troyanov: -Marc Troyanov, Les surfaces à courbure intégrale bornée au sens -d'Alexandrov. https://arxiv.org/abs/0906.3407 -However, I haven't found anything like a complehensive or definitive -treatment of these surfaces, and am therefore looking for such a -reference. - -REPLY [6 votes]: I agree with the answer by Ivan Izmestiev, but let me add the original monograph -by Alexandrov and Zalgaller, -Intrinsic geometry of surfaces, English translation, -American Mathematical Society, Providence, R.I. 1967. -It is a different treatment of the subject in comparison with Reshetnyak, and it is also quite comprehensive. -Instead of potential theory of Reshetnyak, their main method is elementary geometry. Of the recent applications of this theory let me mention two -papers by M. Bonk: -MR2006006 -Bonk, Mario; Lang, Urs -Bi-Lipschitz parameterization of surfaces, -Math. Ann. 327 (2003), no. 1, 135–169, and -MR1804531 Bonk, M.; Eremenko, A. Covering properties of meromorphic functions, negative curvature and spherical geometry. Ann. of Math. (2) 152 (2000), no. 2, 551–592.<|endoftext|> -TITLE: Strong (Inverse of) Residue Theorem -QUESTION [6 upvotes]: Let $C$ be a compact Riemann surfaces of genus $g$. Let $p$ be a point, $\Delta$ a disc around $p$, and $\Delta^*$ the disc minus $p$. Let $\omega$ be a holomorphic one form defined on $\Delta^*$. -Given a meromorphic function $f$ on $C$, regular outside $p$, we can consider the residue of $f\omega$ at $p$. -If $\omega$ is the restriction of a regular form on $C\setminus p$ to $\Delta^*$, then this residue is zero by the classical Residue Theorem. -I read that also the converse is true. So, if the residue of $f\omega$ at $p$ is zero for every meromorphic function $f$ on $C$, regular outside $p$, then $\omega$ is the restriction of a regular form on $C\setminus p$ to $\Delta^*$. -In the book "Vertex algebras and Algebraic Curves" this goes under the name of strong residue theorem (9.2.9). I struggle to find a proof in the literature, this could be something relatively elementary to be found in some book on Riemann surfaces. It should follow from a variant of Serre duality. -I would like to have a proof and/or a reference about this result. I am interested both in the case $C$ smooth and $C$ nodal. The latter case I think makes really sense just when $C$ is irreducible, but I am not sure. -Thanks - -REPLY [4 votes]: Indeed, this follows from Serre duality: Take your favorite local holomorphic coordinate $z$ centered at $p$, and use this as the transition function $g_{0,\infty}=z$ of the line bundle $L(p)$ with global holomorphic section $s_p$. Consider the section $$\omega\otimes s_{-np}$$ and take a $C^\infty$ cut-off away from $p$ as follows: Let $\varphi$ be a function with support in $\Delta$ which is constantly 1 near $p$. This gives a global section $\tilde\omega_n=\varphi \omega\otimes s_{n-p}\in \Gamma(C\setminus\{p\},K\otimes L(-np))$ which is meromorphic near $p$. Apply the $\bar\partial$ operator to obtain a smooth section $$\bar\partial\tilde\omega_n$$ of $\bar K KL(-np)$ with support in an annulus $A\subset\Delta,$ $p\notin A$. -Consider a section $s\in H^0(C,L(np)).$ Then $$\int_C \bar\partial(\tilde\omega_n) s=\int_C(\bar\partial \varphi) \omega s_{-np} s=\int_{\partial A}\varphi \omega s_{-np} s=-\int_\gamma \omega s_{-np}s=-2\pi i Res_p(f\omega),$$ -where $\gamma$ is a small curve around $p$ along which $\varphi$ is 1, and $f$ is the meromorphic function $f=s_{-np}s.$ Therefore, by your assumption, the pairing of $\bar\partial\tilde\omega_n$ with any holomorphic section in $L(np)$ vanishes, and Serre duality -yields a smooth section $s$ of $KL(-np)$ such that $$\bar\partial\tilde \omega_n=\bar\partial s_n,$$ hence $$\tilde{\omega}_n-s_n$$ is a global meromorphic section of $KL(-np),$ or equivalently, a meromorphic 1-form $\omega_n$ with prescribed behavior up to order $n$ around $p.$ For $n$ big enough ($\geq 2g-2$), all $\omega_n$ are the same (e.g., by the easy part of the Serre duality) and hence coincide with $\omega $ on $\Delta\setminus\{p\}$.<|endoftext|> -TITLE: Undefinability of $\mathbb{Z}$ in the reals -QUESTION [14 upvotes]: It is a well-known fact that $\mathbb{Z}$ is not definable in the structure $\mathcal{R}=(\mathbb{R}, +, ., < , 0, 1)$. This follows from Tarski's quantifier elimination, and in fact, we can conclude that the structure -$\mathcal{R}$ is an o-minimal structure. -Another proof, suggested in the answer by Mikhail Katz, is to use the Godel's incompleteness theorem and the fact that the theory of the structure is complete. - -Question. Is there a more direct proof of the above undefinability result? - -I essentially mean a proof which does not use the above results of Tarski or Godel or its variants. -In general, what other different proofs of the above result exist? -Providing references is appreciated. - -In the paper A dichotomy for expansions of the real field a criteria is given for the undefinability of $\mathbb{Z}$ in expansions of the real field. A natural question is if we can use this criteria and prove the theorem directly? - -REPLY [5 votes]: This is not a real answer but rather an observation. The undefinability of $\mathbb{Z}$ follows from the fact that every infinite definable set in such a structure has uncountable cardinality. This property is strictly weaker than both o-minimality and quantifier elimination. Nevertheless, I do not know any proof of this fact that does not use neither of those. I guess this simply induces a nice sub-question of the original one.<|endoftext|> -TITLE: Galois invariant line bundles on a product of varieties -QUESTION [7 upvotes]: Let $k$ be a field with separable algebraic closure $k^{\rm s}$ and corresponding absolute Galois group $\varGamma={\rm Gal}(k^{\rm s}/\,k)$ and let $X$ and $Y$ be geometrically connected and geometrically reduced $k$-schemes locally of finite type. - -Question. Are there examples where the canonical map $(\,{\rm Pic}\,X^{\rm s})^{\varGamma}\oplus(\,{\rm Pic}\,Y^{\rm s})^{\varGamma}\to{\rm Pic}\,(X^{s}\times_{\,k^{\rm s}}Y^{\rm s})^{\varGamma}$ is an isomorphism but the map ${\rm Pic}\,X^{\rm s}\oplus{\rm Pic}\,Y^{\rm s}\hookrightarrow{\rm Pic}\,(X^{s}\times_{k^{\rm s}}Y^{\rm s})$ is not? - -It is known that, if $X$ and $Y$ are smooth, projective, geometrically irreducible and of finite type over $k$, then the cokernel of the latter map is naturally isomorphic to ${\rm Hom}({\rm Pic}_{ Y^{\rm s}}^{\vee},{\rm Pic}_{X^{\rm s}})$, where ${\rm Pic}_{X^{\rm s}}$ is the Picard variety of $X^{\rm s}$ (i.e., the largest reduced subscheme of the identity component of the Picard scheme of $X^{\rm s}$ over $k^{\rm s}$) and ${\rm Pic}_{Y^{\rm s}}^{\vee}$ is the dual of ${\rm Pic}_{Y^{\rm s}}$. Thus, while looking for such examples, one should assume that ${\rm Pic}_{Z}\neq 0$ for $Z=X^{s}$ and $Y^{s}$, in particular that $H^{1}(Z,\mathcal O_{Z})\neq 0$ for such $Z$ (a condition which excludes the rational varieties). - -REPLY [3 votes]: Take $X$ a non-CM elliptic curve over $k$ (over finite fields, a similar argument involving ordinary curves will work) and $Y$ its quadratic twist over some quadratic extension of $k$. -$X$ and $Y$ are smooth, projective, geometrically irreducible and of finite type over $k$, so as you note there is an exact sequence $$0 \to ( \operatorname{Pic} X^s \oplus \operatorname{Pic} Y^s \to \operatorname{Pic} (X^s \times Y^s) \to \operatorname{Hom} ( \operatorname{Pic}^0(Y^{\vee s}), \operatorname{Pic}^0(X^s)) \to 0$$ -I have written the Picard varieties as $ \operatorname{Pic}^0$ to distinguish them from Picard groups. -This implies, by the left exactness of $G \mapsto G^\Gamma$, an exact sequence. -$$0 \to (\operatorname{Pic} X^s)^\Gamma \oplus (\operatorname{Pic} Y^s)^\Gamma \to (\operatorname{Pic} (X^s \times Y^s))^\Gamma \to (\operatorname{Hom} ( \operatorname{Pic}^0(Y^{\vee s}), \operatorname{Pic}^0(X^s)))^\Gamma$$ -So it suffices to check that $\operatorname{Hom} ( \operatorname{Pic}^0(Y^{\vee s}), \operatorname{Pic}^0(X^s)) \neq 0$ but $(\operatorname{Hom} ( \operatorname{Pic}^0(Y^{\vee s}), \operatorname{Pic}^0(X^s)))^\Gamma=0$, so that the second exact sequence becomes -$$0 \to \operatorname{Pic} X^s)^\Gamma \oplus (\operatorname{Pic} Y^s)^\Gamma \to (\operatorname{Pic} (X^s \times Y^s))^\Gamma \to 0.$$ -To do this, observe that, because $X$ and $Y$ are elliptic curves, they are their own $\operatorname{Pic}^0$, and are also self-dual, so we are just working with $\operatorname{Hom}(Y^s, X^s)$. Because $Y^s$ and $X^s$ are isomorphic, and are a non-CM elliptic curve $\operatorname{Hom}(Y^s, X^s) = \mathbb Z$. The quadratic twisting means exactly that the Galois action on this is by the quadratic character of the corresponding field extension, and in particular the invariants are trivial, as desired.<|endoftext|> -TITLE: Compact Lie group inclusions that are trivial on all homotopy groups -QUESTION [10 upvotes]: Is there a classification of closed subgroups $H\le SO(n)$ such that the inclusion $H\to SO(n)$ is trivial on all homotopy groups? -This happen e.g. when the group $H$ is finite. Are there other examples? - -REPLY [12 votes]: Well, the identity component $H^0$ of $H$ would have to be abelian, since, otherwise, it would have a compact simple component, and the induced map on $\pi_3$ would then be nontrivial. -If $H^0$ has positive dimension and is abelian, then what you are asking is whether $\pi_1(H^0)\to\pi_1(\mathrm{SO}(n))\simeq \mathbb{Z}_2$ is trvial or not. This does happen, of course. For example, if $H^0=S^1$ and it represents the zero element in $\pi_1(SO(n))\simeq \mathbb{Z}_2$, then the induced map on homotopy is trivial in $\pi_1$ and hence in all higher cases as well. This first happens for $n=4$, as there is such an $S^1$ sitting in $\mathrm{SO}(4)$ (in fact, a countably distinct family of them). It also happens for all $n\ge 4$. -As far as classification goes: Given $n$, there will be a countable number of codimension $1$ subtori $T^{r-1}$ in a maximal torus $T^r\subset\mathrm{SO}(n)$ (where $r$, the rank is the greatest integer in $n/2$) such that the inclusion $T^{r-1}\hookrightarrow T^r$ induces an inclusion $\pi_1(T^{r-1})\subset \pi_1(T^r)$ so that all the elements of $\pi_1(T^{r-1})$ are zero in $\pi_1(\mathrm{SO}(n))$. Then the abelian groups that you want are the ones that are conjugate to a subgroup of such a sub-maximal torus.<|endoftext|> -TITLE: The Hopf invariant is an isomorphism for $\pi_3 (S^2)$ -QUESTION [6 upvotes]: Does any one have a reference to a proof that the Hopf invariant classifies the homotopy classes of maps from $\mathbb{S}^3$ to $\mathbb{S}^2$. -It is quite standard to find a proof that the Hopf invariant is a homotopy invariant. However, after searching MathSciNet for all the books covering the Hopf invariant, I could not find any proofs more recent than Hopf's original 1931 paper written in German. The best that I have found are some statements without any proof or reference. - -REPLY [7 votes]: Theorem. The Hopf invariant is a non-zero group homomorphism $\mathcal H:\pi_{4n-1}(\mathbb S^{2n})\to\mathbb Z$ and it is an isomorphism only when $n=1$. -For a proof that $\mathcal H$ is a group homomorphism, -see Proposition 4B.1 in [3]. -Hopf, in [4] (see Satz II and Satz II') proved that for any $n$, there is a map -$h:\mathbb S^{4n-1}\to\mathbb S^{2n}$ with $\mathcal H h\neq 0$ and hence the homomorphism -$\mathcal H:\pi_{4n-1}(\mathbb S^{2n})\to\mathbb Z$ is non-zero. -Since the Hopf invariant of the Hopf fibration $h:\mathbb S^3\to\mathbb S^2$ equals $1$ (Example 17.23 in [2]), -$\mathcal H:\pi_3(\mathbb S^2)\to\mathbb Z$ is an isomorphism. However, for $n\geq 2$ the Hopf invariant is never an isomorphism. -Indeed, Adams [1] proved that mappings with Hopf invariant equal $1$ exist only when $n=1,2$ and $4$, -so these are the only cases when one may suspect $\mathcal H$ to be an isomorphism, but -$\pi_7(\mathbb S^4)=\mathbb Z\times\mathbb Z_{12}$ and $\pi_{15}(\mathbb S^8)=\mathbb Z\times\mathbb Z_{120}$, so $\mathcal H$ cannot be an isomorphism. -[1] J. F. Adams, J., -On the non-existence of elements of Hopf invariant one. -Ann. of Math. 72 (1960), 20-104. -[2] R. Bott, L. W. Tu, L. W. -Differential forms in algebraic topology. -Graduate Texts in Mathematics, 82. Springer-Verlag, New York-Berlin, 1982. -[3] A. Hatcher, -Algebraic topology. -Cambridge University Press, Cambridge, 2002. -[4] H. Hopf, -Über die Abbildungen von Sphären auf Sphären niedrigerer Dimensionen. -Fundamenta Math. 25 (1935), 427-440.<|endoftext|> -TITLE: Anti-large cardinal principles -QUESTION [22 upvotes]: I'm interested in axioms that prevent the existence of large cardinals. More precisely: -(Informal definition) $\Phi$ is an anti-large-cardinal axiom iff $V \models \Phi \Rightarrow V \not \models \Psi$ for some garden-variety large cardinal axiom $\Psi$. -Probably the most famous are: -Variety 1 Axioms of constructibility (e.g. $V=L$, $V=L[x]$). -$V=L$ for example implies that there are no measurable cardinals in $V$. Often we don't need the full power of $L$ to get anti-large cardinal features though. For instance: -Variety 2 $\square$-principles -often prevent the existence of large cardinals via their anti-reflection features. For example $\square_\kappa$ prevents the existence of cardinals that reflect stationary sets to $\alpha < \kappa$ (and hence rule out cardinals like supercompacts). -Trivially, we might look at axioms of the following form: -Variety 3 $\neg \Phi$ or $\neg Con(\Phi)$ for some large cardinal axiom $\Phi$. -These obviously and boringly prevents the existence of a large cardinal in the model. Another possible candidate (formalisable in $NBG + \Sigma^1_1$ comprehension) is: -Variety 4 Inner model hypotheses. -For example, the vanilla inner model hypothesis that any parameter-free first-order sentence true in an inner model of an outer model of $V$ is already true in an inner model of $V$. This prevents the existence of inaccessibles in $V$. Other variants rule out cardinals not consistent with $L$. -Final example (if we're being super liberal about what we allow to be revised): -Variety 5 Choice axioms. -For instance $AC$ prevents the existence of Reinhardt (and super-Reinhardt, Berkeley etc.) cardinals in $V$ (assuming that these are, in fact, consistent with $ZF$, which is pretty non-trivial). -My question: -Are there other kinds of axiom with anti-large-cardinal features? Especially non-trivial ones (i.e. not like Variety 3). - -REPLY [5 votes]: The following is part of unpublished work by Toshimichi Usuba: -Definition. A cardinal $\kappa$ is hyper-huge if for every $\lambda > \kappa$ there is an elementary embedding -$$ -j \colon V \to M -$$ -into some inner model $M$ such that $\operatorname{crit}(j) = \kappa$, $\lambda < j(\kappa)$ and $^{j(\lambda)} M \subseteq M$. -Definition. An inner model $M$ of $V$ is a ground if there is some set forcing $\mathbb P \in M$ and some $\mathbb P$-generic filter $G$ over $M$ such that $M[G] = V$. -Theorem (Usuba). If there is a hyper-huge cardinal, then there are only set many grounds. -He also proved that $V$ has a $\subseteq$-minimal ground (which is known as the bedrock) iff there are only set many grounds. Thus the nonexistence of the bedrock is an anti-large cardinal axiom. -On the other hand, in Inner Model Theoretic Geology, Fuchs and Schindler provided an example for which the bedrock does not exist, namely: If $L[E]$ is a pure extender model without a strong cardinal satisfying some -'nice inner model theoretic properties' (*) then the bedrock of $L[E]$ does not exist. (**) -(*) $L[E]$ is assumed to be tame, fully iterable in $V$, internally not fully iterable as guided by $\mathcal{P}$-constructions and the extender sequence $E$ and its canonical extensions by $\operatorname{Col}(\omega, \theta)$, for cutpoints $\theta$ of $L[E]$, are ordinal definable in their respective models. -(**) Please note that they provide a much more detailed analysis of the class of grounds for this model.<|endoftext|> -TITLE: $\mathbb{Z}$-module structure of the subring generated by an algebraic number -QUESTION [14 upvotes]: Let $a$ and $b$ be algebraic numbers which are not necessarily algebraic integers. Is there some invariant that allows us to determine whether $\mathbb Z[a]$ and $\mathbb Z[b]$ are isomorphic as $\mathbb Z$-modules? - -Initial version of the question: Let $a$ be an algebraic number which is not necessarily an algebraic integer. What is the $\mathbb Z$-module structure of $\mathbb{Z}[a]$? - -REPLY [10 votes]: Let $a$ be an algebraic number, $K = \mathbb Q(a)$ the associated number field, $\mathcal O_K$ its ring of integers. Then the ring $\mathcal O_K[a]$ depends only on the set of places of $K$ at which $a$ is not integral. -$\mathcal O_K[a]$ is a good place to start for studying $\mathbb Z[a]$ because $\mathbb Z[a]$ is a finite index submodule of $\mathcal O_K[a]$. In particular, if $a \in \mathcal O_K$ then $\mathcal O_K[a]=\mathcal O_K$ is a finite free $\mathbb Z$-module, and hence $\mathbb Z[a]$ is as well, and the problem is trivial. -To say a little more about the structure of $\mathcal O_K[a]$, the key fact is that as a ring, it is $\mathcal O_K$ with some number of primes inverted. Hence as a $\mathbb Z$-module it is an extension of a sum of modules isomorphic to $\mathbb Q_p/\mathbb Z_p$ by the finite free $\mathbb Z$-module $\mathcal O_K$. Specifically, the multiplicity of $\mathbb Q_p/\mathbb Z_p$ is the total degree times ramification index of the primes lying over $p$ at which $a$ is not integral. Because $\mathbb Z[a]$ is a finite-index submodule of this, it is easy to see that it can also be expressed as a similar extension. -This should be enough information about the $\mathbb Z$-module structure for most practical purposes in number theory. However, as YCor points out, knowing the rank of $\mathcal O_K$ and the multiplicity of $\mathbb Q_p/\mathbb Z_p$ does not come close to uniquely determining $\mathcal O_K[a]$ or $\mathbb Z[a]$, except in some special cases. In particular, he raises the question of determining when $\mathbb Z[a]$ and $\mathbb Z[b]$ are isomorphic as $\mathbb Z$-modules. - -Generalizing the case where $a$ is integral, there is one case where the problem simplifies considerably. Suppose there is some proper subfield $L$ of $K$ and ring extension $\mathcal O_L'$ with $\mathcal O_L \subseteq \mathcal O_L' \subseteq \mathcal O_L$ such that $\mathcal O_K[a]= \mathcal O_L' \otimes_{\mathcal O_L} \mathcal O_K$. Then $\mathcal O_K[a]$ is a locally free module over $\mathcal O_L'$ (because $\mathcal O_K$ is a locally free module over $\mathcal O_L$) and $\mathbb Z[a]$ is a finite index submodule of $\mathcal O_K[a]$. This is only a small amount of extra data beyond $\mathcal O_L[a]$ and it seems impossible to recover much about $a$ from this data. -Thus, let us focus on the case where there does not exist such a proper subfield $L$ and ring $\mathcal O_L'$. Then, following YCor's suggestion, we can show that the ring of $\mathbb Z$-module endomorphisms of $\mathbb Z[a]$ is $\mathbb Z[a]$. In particular, this implies that (in this case) $\mathbb Z[a]$ and $\mathbb Z[b]$ are only isomorphic as $\mathbb Z$-modules if they are already isomorphic as rings. - -It suffices to show that the ring of $\mathbb Z$-module endomorphisms, tensored with $\mathbb Q$, is isomorphic to $K$. This is because the ring of endomorphisms would then be commutative, hence commute with $a$, hence consist of $\mathbb Z[a]$-module endomorphisms, which are clearly only $\mathbb Z[a]$. -Since $\mathbb Q\otimes_{\mathbb Z}\operatorname{End}_{\mathbb Z} Z[a]$ contains $K$ and is contained in $M_n(\mathbb Q)$, it is a semisimple algebra, so it must be a matrix algebra of rank $r$ over a division algebra of rank $d$ over some field $L$. The field $L$ commutes with $K$ and so is contained in $K$. The minimal faithful representation of such an algebra has dimension $r d^2$ over $L$, so the index of $L$ in $K$ is at most $rd^2$, and the maximal commutative subalgebra has dimension $rd$ over $L$, so the index is at least $rd^2$. Thus $d=1$ and the index of $L$ is $r$. Hence this algebra is the full centralizer of $L$ inside $M_n(\mathbb Q)$. In particular, if $L=K$ then the endomorphism algebra must be $K$, as desired, so it suffices to handle the case when $L$ is a proper subfield. To do this, we will show that the set of places of $K$ at which $a$ is not integral is the pullback of a set of places from $L$. Because $\mathcal O_K[a]$ is isomorphic to $\mathcal O_K$ with that set of places inverted, it is isomorphic to $\mathcal O_L$, with the corresponding set of places inverted, tensored over $\mathcal O_L$ with $\mathcal O_K$, contradicting the assumption on the nonexistence of $\mathcal O_L'$. -Choose a prime $p$ over which $a$ is not integral and tensor everything with $\mathbb Z_p$. The ring $\mathcal O_K[a] \otimes_{\mathbb Z} \mathbb Z_p$ is isomrphic to a product of local rings at places of $\mathcal O_K$ where $a$ is integral and local fields at places of $\mathcal O_K$ where $a$ is not integral. its endomorphisms preserve the subspace of $p$-divisible elemnts, which is the product of the local fields at the $p$-divislbe places. The endomorphisms contain the centralizer of $L$, and the only way everything in the centralizer of $L$ can preserve a certain $\mathbb Q_p$-subspace is if it is the kernel of some element of $L \otimes_{\mathbb Q} \mathbb Q_p,$ i.e. a product of the local fields at the places lying over some set of places of $L$ over $p$. Hence the set of places (over $p$) of $\mathcal O_K$ at which $a$ is not integral is the inverse image of a set of places of $\mathcal O_L$, as desired.<|endoftext|> -TITLE: Yoneda extensions in exact categories and their derived categories -QUESTION [6 upvotes]: If $\mathcal{A}$ is any abelian category, then for all objects $X,Y$ in $\mathcal{A}$ and for all integers $i \geq 0$, there is an natural isomorphism -$$\mathrm{Ext}_\mathcal{A}^i(X,Y) \simeq \mathrm{Hom}_{D(\mathcal{A})}(X,Y[i]),$$ -where the left-hand side denotes the Yoneda $\mathrm{Ext}$-group in $\mathcal{A}$ and the right-hand side denotes the morphisms between $X$ and the shift of $Y$ by $i$ in the derived category $D(\mathcal{A})$ of $\mathcal{A}$. -Note that this holds without any assumption on $\mathcal{A}$ (e.g. existence of enough injectives/projectives), cf. Verdier's thesis III.3.2. -Now, I am interested in an exact category $\mathcal{E}$ (in the sense of Quillen, i.e. a strictly full subcategory closed under taking extensions of an abelian category $\mathcal{A}$). -Exact categories encapsulate just enough of the formalism of abelian categories to define the Yoneda $\mathrm{Ext}$-groups (as sequences $E_1E_2...E_i$ of short exact sequences with compatible ends). -Moreover, under certain assumptions one can define the derived category $D(\mathcal{E})$ of $\mathcal{E}$. -In my situation, $\mathcal{E}$ has kernels so that $D(\mathcal{E})$ can be defined as in [Beilinson-Bernstein-Deligne, Faisceaux pervers, §1.1.4]. -In this setup, do we have a natural isomorphism -$$\mathrm{Ext}_\mathcal{E}^i(X,Y) \simeq \mathrm{Hom}_{D(\mathcal{E})}(X,Y[i])$$ -for all objects $X,Y$ in $\mathcal{E}$ and for all integers $i \geq 0$? - -REPLY [9 votes]: Firstly, for any Quillen exact category $\mathcal E$, one can define the derived category $D(\mathcal E)$, as well as its bounded versions $D^+(\mathcal E)$, $D^-(\mathcal E)$, and $D^b(\mathcal E)$. -Existence of kernels in $\mathcal E$ is irrelevant. The relevant conditions are idempotent-closedness (Karoubianness) and weak idempotent closedness. -The (bounded or unbounded) derived category of an exact category $\mathcal E$ can be defined without these conditions. But assuming one of them (depending on how bounded is the derived category which you want to define) simplifies some questions related to this definition. -References: - -A. Neeman, "The derived category of an exact category", Journ. of Algebra 135 (1990). -B. Keller, "Derived categories and their uses", Handbook of Algebra vol.1 (1995), p.671-701. -T. Buehler, "Exact categories", Expositiones Math. 28 (2010), Section 10. - -Secondly, the isomorphism -$$ - \operatorname{Ext}_{\mathcal E}^i(X,Y)\simeq - \operatorname{Hom}_{D(\mathcal E)}(X,Y[i]) -$$ -for all $X$, $Y\in\mathcal E$ and $i\ge0$ holds for all Quillen exact categories $\mathcal E$. -I would be glad to be informed of an earlier/better reference, but having glanced through the three papers cited above and not found this assertion explicitly formulated there, I can only suggest Proposition A.13 in my paper - -L. Positselski, Mixed Artin-Tate motives with finite coefficients, Moscow Math. Journ. 11 #2 (2011), Appendix A. - -In the arXiv version, https://arxiv.org/abs/1006.4343 , Appendix A, this is the only (unnumbered) Proposition in Section A.7. -The proof of this proposition in my paper is very short and omits all the details, but the assertion in question is indeed quite straightforward (particularly, if you already know how to prove this for abelian categories). This may also be the reason why this result is not even formulated in some other references. -Further discussion can be found in Section A.8.<|endoftext|> -TITLE: The uniqueness of a $K$-fixed vector in a spinor representation -QUESTION [5 upvotes]: Consider $G=SO(2n)$ and $K=U(n)$. $(G,K)$ is a symmetric pair. I'm interested in (zonal) spherical functions on $G/K$ which are matrix elements with respect to $K$-fixed vectors in irreducible representations of $G$. - -A classical result by E. Cartan says that if $(G, K)$ is a Riemannian symmetric pair then the algebra of integrable functions on $G$ which are bi-invariant under $K$ is commutative, that is, $(G, K)$ is a Gelfand pair. - -(source) -Then this book cited by Wikipedia says that for a Gelfand pair, the space of $K$-fixed vectors in every irreducible unitary representation of $G$ is at most one-dimensional. But an irreducible spinor representation of $SO(2n)$ has more than one orthogonal $U(n)$-fixed vectors. (I only know how to phrase it in physicists' language: The subgroup $U(n)$ preserves the number of fermions, so it fixes the state with no fermion and the fully filled state.) Is this because a spinor representation really is a representation of $Spin(2n)$ and not $SO(2n)$? Do I miss something here? -EDIT: Assume everything is over $\mathbb{C}$. - -REPLY [5 votes]: There is no contradiction. To use your physical language, the fully filled state is not $K$-fixed as a vector. Rather, the group ${\rm U}(n)$ acts on it by the determinantal character. The subtle point here is that in quantum mechanics, one considers projective representations, so the action by a character cannot be distinguished from the trivial one. The full decomposition of the (half-)spinor representation into $K$-types uses exterior algebra and is well known — see for example Goodman and Wallach's book. In particular, the spinor representation is multiplicity-free: in fact, every $K$-type that occurs has multiplicity 1, and not just the trivial one. This, of course, is a much stronger property. -Incidentally, for odd $n$, the vacuum (1, the state with fermion number $0$) and the fully filled state (the volume form, fermion number $n$) have different parity. So they appear in different halves of the spinor representation.<|endoftext|> -TITLE: Relation - Anabelian geometry and Tate conjecture -QUESTION [5 upvotes]: A lot has been said about the relation between Anabelian algebraic geometry and Mordell conjecture. - -I would like to know what is the relation between Anabelian algebraic geometry and Tate conjecture. - -Grothendieck said (Esquisse d'un programme): "c'est alors que se dégage la 'conjecture fondamentale de la géométrie algébrique anabélienne', proche des conjectures de Mordell et de Tate que vient de démontrer Faltings" - -REPLY [5 votes]: The Tate conjecture is the statement that $End(A,B)\otimes \mathbb{Q}_p$ is isomorphic to $End_G(T_p A,T_p B)\otimes \mathbb{Q}$ for abelian varieties $A,B$ over a number field $K$ with absolute Galois group $G$ and $T_pA$ the Tate module of $A$. A map $\pi_1(A) \to \pi_1(B)$ preserving the fundamental exact sequences $1 \to \pi_1({\bar A}) \to \pi_1(A) \to G \to 1$, etc, is a map $\pi_1({\bar A}) \to \pi_1({\bar B})$ preserving the Galois action and since $\pi_1({\bar A}) = \prod_p T_pA$, we can see that the Tate conjecture is pretty close to a statement that map between the $\pi_1$'s correspond to maps between the varieties. I think it's not quite the same because of the $\mathbb{Q}$ coefficients, but abelian varieties are not anabelian :-).<|endoftext|> -TITLE: A Converse of the Skorokhod Embedding Theorem -QUESTION [7 upvotes]: I am wondering whether the following "sort of converse" of Skorokhod's embedding theorem holds: -Suppose that $\{D_t\}_{t \geq 0}$ is a stochastic process with continuous paths, $D_0 = 0$, and suppose that the following is true: -For any mean zero, finite variance random variable $X$ there exists a stopping time $\tau$ with respect to the filtration $\mathcal{F}_t \stackrel{def}{=} \sigma\{D_s:0\leq s\leq t\}$, such that $X \stackrel{d}{=} D_\tau$ and $\mathbb{E}X^2 = \mathbb{E}\tau$. -My question is, does this imply that the process $\{D_t\}$ is a Brownian motion? -Even if not, under what condition(s) does it follow that $\{D_t\}$ is a Brownian motion? For example does strong Markov property or independent increment assumption of $\{D_t\}$ help? -Any help will be greatly appreciated. - -REPLY [3 votes]: Here is a thought: take a 50-50 mixture of Brownian motions with volatilities $\sigma_1^2, \sigma_2^2$ satisfying $\frac{1}{2}(\frac 1 {\sigma_1^2} + \frac 1 {\sigma_2^2}) = 1$. The mixer ought to be $F_0$ measurable since you can observe the volatility in any time interval $(0, \epsilon)$. Given $X$ use the stopping time $\tau_i$ you would use on either process by itself. Then as each satisfies $\sigma_i^2 E(\tau_i) = E(X^2)$ the package should satisfy $E(\tau) = .5*(E(\tau_1) + .5 E(\tau_2) = E(X^2)\frac{1}{2}(\frac 1 {\sigma_1^2} + \frac 1 {\sigma_2^2}) = E(X^2)$<|endoftext|> -TITLE: Fixed point set of smooth circle action -QUESTION [11 upvotes]: Suppose $M$ is a connected closed smooth $d$-dimensional manifold, and suppose $S^1 = SO(2)$ acts smoothly on $M$. Then the fixed point set $Y = M^{S^1}$ will be a submanifold of $M$ of even codimension, and we also have $\chi(Y) = \chi(M)$. -(For instance, there are different linear circle actions on $S^d$ for which the fixed point set will be $S^{d-2j}$; here $j$ can be any number between $0$ and $d/2$.) -My question is rather vague, I simply want to know what else we can say about $Y$. I am also very interested in seeing more examples where $Y$ is an interesting manifold. In particular I would like to know whether it can happen that $Y$ has components of different dimensions, and whether the number of components of $Y$ can be bounded in some way. Is there any result that tells us that the homotopy type of $Y$ cannot be much more complicated than that of $M$? - -REPLY [6 votes]: To answer your question: Yes $Y$ can have components in different dimensions: Consider for example the following action on $\mathbb{CP}^2$, where we identified $S^1\subset \mathbb{C}$: -$$ -w\cdot [z_0,z_1,z_2]=[w z_0,z_1,z_2]. -$$ -Then $[1,0,0]$ is an isolated fixed point, and also $[0,z_1,z_2]$ is a fixed $\mathbb{CP}^1$. To get more of such examples you should have a look at Delzant's classification of Hamiltonian Torus actions. Suitably choosing an $S^1$ sitting inside the torus will produce many more examples. -There is a large literature on $S^1$ actions. Even the case of a discrete fixed point set is very interesting. In the almost complex setting one could have a look at https://arxiv.org/pdf/1411.6458.pdf and references therein.<|endoftext|> -TITLE: What are the tangent $\infty$-categories to $\mathrm{Top}^\mathrm{op}$ and $E_\infty$-$\mathrm{Ring}^\mathrm{op}$? -QUESTION [15 upvotes]: Given an $\infty$-category $\mathcal{C}$ with finite limits and finite colimits, there are two ways to make it stable. One is to pass to the pointed objects and take the category of $\Omega$ objects $\mathrm{Sp}(\mathcal{C}_\ast)$. An $\Omega$ object is a sequence of pointed objects with equivalences $X_n = \Omega X_{n+1}$. Or dually, we pass to co-pointed objects and take $\mathrm{Sp}^\vee(\mathcal C^\ast) = \mathrm{Sp}((\mathcal{C}^\mathrm{op})_\ast)^\mathrm{op}$ the category of $\Sigma$-objects. In terms of the original cateogry $\mathcal{C}$, a $\Sigma$-object is a sequence with equivalences $X_n = \Sigma X_{n+1}$. (We could also mix and match (co)pointing and loops/suspension -- I'd be interested to hear about this too.) -This works in families, too: the tangent $\infty$-category $T\mathcal{C}$ of $\mathcal{C}$ is the category fibered over $\mathcal{C}$ whose fiber $T_C\mathcal{C}$ at $C \in \mathcal{C}$ is the stabilization $\mathrm{Sp}((\mathcal{C}_{/C})_\ast)$ of the slice. We could also take the dual tangent category $T^\vee \mathcal{C} = T(\mathcal{C}^\mathrm{op})^\mathrm{op}$ (I don't want to call it the cotangent category, since there's already something different called the cotangent complex). Note that the result of the pointing / co-pointing procedure is the same in both cases, yielding the category $\mathcal{C}_{C/-/C}$ of split objects over $C \in \mathcal{C}$. That is, $T_C \mathcal{C} = \mathrm{Sp}(\mathcal{C}_{C/-/C})$, while $T^\vee_C\mathcal{C} = \mathrm{Sp}^\vee(\mathcal{C}_{C/-/C})$. The difference likes in the spectrum procedure. -Now, the two foremost interesting examples of this construction are when $\mathcal{C} = \mathrm{Top}$, where $T\mathcal{C}$ is the category of parameterized spectra, and $\mathcal{C} = E_\infty$-$\mathrm{Ring}$ the category of $E_\infty$ ring spectra, where $T\mathcal{C}$ is the category whose objects are modules over $E_\infty$ rings, and morphisms are composites of ring homomorphisms and module homomorphisms (i.e. the Grothendieck construction applied to the functor $\mathrm{Mod}: E_\infty$-$\mathrm{Ring}^\mathrm{op} \to \mathrm{Cat}$ sending a ring to its category of modules). -The funny thing is, I'm more inclined to think of $\mathrm{Top}$ as being analogous to $E_\infty$-$\mathrm{Ring}^\mathrm{op}$ ("derived affine schemes") than to $E_\infty$-$\mathrm{Ring}$ for "geometric" purposes. So it seems strange to apply the same, non-self-dual construction to both categories. So I ask: -Questions: - -If $X$ is a space, what is the category $\mathrm{Sp}^\vee(\mathrm{Top}_{X/-/X})$? Is it somehow a category of modules? -If $R$ is an $E_\infty$ ring, what is the category $\mathrm{Sp}^\vee(R$-$\mathrm{Alg}_\mathrm{aug})$? Is it somehow a category of derived schemes parameterized over $\mathrm{Spec R}$? - -For (1), I should note that when $X$ is the empty space $\emptyset$, so that we're actually asking about the (opposite of the) stabilization of $\mathrm{Top}^\mathrm{op}$ as in my question title, the answer is that the stabilization is the terminal category, because $\emptyset$ is a strict initial object in $\mathrm{Top}$, and the first step of stabilization is to take pointed objects. But this also happens when one computes the stabilization of $E_\infty$-$\mathrm{Ring}$, so apparently this is no obstacle to the construction being interesting at other slices. The question is what a ``$\Sigma$-spectrum over $X$" looks like. When $X$ is a point, this is again trivial because a space which is an $n$-fold suspension for every $n$ is contractible. I think this will also happen whenever $X$ is simply-connected? But still, this may be interesting over other $X$'s. -For (2), one likewise needs to understand suspension in $R$-$\mathrm{Alg}_\mathrm{aug}$, where $R$ is an $E_\infty$ ring. The temptation is to say that this is given by topological Hochschild homology, but not so fast -- in augmented $R$-algebras, THH is the tensoring over $\mathrm{Top}_\ast$ with $S^1_+$, whereas suspension is tensoring with $S^1$ itself. So I don't know what the suspension of an augmented $E_\infty$-algebra is. this is entirely correct. -EDIT So a $\Sigma$-object in $R$-$\mathrm{Alg}_\mathrm{aug}$ is an augmented $R$-algebra $A_0$ equipped with the data of infinite "de-THH'ing": for each $n \in \mathbb{N}$, there is an $A_n$ with $THH_R(A_{n+1}) \simeq A_n$. Is anything known about this kind of thing? It would be amazing to have some analog of infinite loopspace theory to recognize when this can be done... It makes me think of iterated algebraic K-theory, and also about redshift... - -REPLY [6 votes]: As indicated in the original question, an object in $Sp^{\vee}(\mathcal C_*)$ would be a $\Sigma$-object: a sequence of objects with weak equivalences $X_k \simeq \Sigma X_{k+1}$, and this is not so interesting when $\mathcal C = Top$. -Ah ... but what if one changes the notion of weak equivalence? In particular, localize $Top$ like Bousfield likes to do: fix a prime $p$, and let $L_n^f$ be localization with respect to a map $\Sigma A \rightarrow *$, where $A$ is a finite complex whose suspension spectrum has type $n+1$ and whose mod $p$ homology has connectivity as small as possible. Then I think that the category of $\Sigma$ objects is roughly the same thing as the category of $K(n)$--local spectra (maybe subject to correction due to the telescope conjecture), with the correspondence being: given such a spectrum $X$, there is an $L_n^f\Sigma$-object with $k$th space $\Theta_n(\Sigma^{-k}X)$, where $\Theta_n$ is the `left adjoint' to the better known telescopic functor $\Phi_n: Top \rightarrow Spectra$. -[See my 2008 HHA paper for these constructions and references.] -I think that Brayton Gray made some use of this idea, when $n=1$, but I can't see anything about this on his publications listed on MathSciNet.<|endoftext|> -TITLE: Generating Random Curves with Fixed Length and Endpoint Distance -QUESTION [7 upvotes]: Are algorithms already known, that generate (arbitrarily good approximations of) random curves, w.l.o.g. with unit length, and joining endpoints $(0,0)$ and $(\alpha,0)$ with $\alpha \lt1$ given? - -The fixed distance between the endpoints is essential for the question, because otherwise a simple rescaling of an arbitrary curve would work. - -edit -In view of the comments and the answer of Bjørn Kjos-Hanssen, I see the need for some clarification: - -By random curve of unit length connecting $(0,0)$ and $(\alpha,0)$, I mean a random sample from the space of all such curves; that means, that the algorithm should be capable to approximate every such curve to arbitrary precision with a finite number of steps. -So "random" is not restricted to the appearance of the curve. -Being able to generate Brownian Bridges is not sufficient, because I would like the algorithm to be able to generate curves (ideally in any $\mathbb{R}^n$) and not only functions. - -So my apologies for not being precise enough. - -I have used the formulation "are algorithms already known", because I have found one, that seems to be able to produce all those curves. -I will provide details in a later edit. - -Here are the promised details: -the algorithm, that motivated this question is essentially based on realizing, that no point of the curve can lie outside the ellipse centered at $\left(\frac{\alpha}{2},0\right)$, foci $p$ at $\left(0,0\right)$ and $q$ at $\left(\alpha,0\right)$, for which the length of the semi-major axis equals $\frac{1}{2}$ and, $\sqrt{\left(\frac{1}{2}\right)^2-\left(\frac{\alpha}{2}\right)^2}$ for the semi-minor axis. -If the intermediate curve-point $r$ is chosen from the boundary of that ellipse, then the "length-stock" is used up and the algorithm terminates with a curve consisting of two line-segments and exact length $1$, joining $p$ and $q$ as demanded. -Otherwise the length-stock is split up and assigned to two newly generated line-segments and the original problem of finding a curve of fixed length with endpoints at fixed has to be solved recursively for both segments separately. - -Pseudo code: -$\text{expand}$(Point $p$, Point $q$, Length $\ell$, Curve curve) -$\quad$Point $r\in \lbrace x\in\mathbb{R}^n\ |\ \| r-p \| + \|q-r\|\ \le \ell\rbrace$; - $\quad$Length $\ell_{pr}$ := $\|r-p\|$; - $\quad$Length $\ell_{rq}$ := $\|q-r\|$; - $\quad$Length $\Delta\ell$ := $\ell-\left(\ell_{pr}+\ell_{rq}\right)$; - $\quad$Scalar $a\in\left[0,1\right]$ - -$\quad$if (a < threshold) - $\quad\quad$curve.append($r-p$); - $\quad$else - $\quad\quad\text{expand}$($p$,$\ r$,$\ \ell_{pr}$+$a$ * $\Delta\ell$); - -$\quad$if ($1-a$ < threshold) - $\quad\quad$curve.append($q-r$); - $\quad$else - $\quad\quad\text{expand}$($r$,$\ q$,$\ \ell_{rq}$+(1-$a$) * $\Delta\ell$); - - -Some remarks: -the pseudo code is aimed at full generality and also covers "degenerate" cases; those need to be ruled out by further checks. One such case is the collinearity of $p$, $\ q$ and $r$ with positive $\Delta\ell$. -Selecting $r$ from the mentioned elliptical regions with foci $p$ and $q$ can also be interpreted as chosing one of the intersection points of a circle around $p$ and $q$. That covers the algorithm of Matt F. as a special case. -The followup challenge is now to control further properties of the curve via taylored rules for selecting $r$ and distributing $\Delta\ell$ on each recursion level. -Or, play with the options to discover interesting curves and fractals. - -REPLY [2 votes]: Defining $z_t(s)=\int_0^s e^{itu(\sigma)}\ d\sigma$ transforms any function $u$ : $(0,1) \to \mathbb R$ into a one-parameter family of length $1$ curves $C_t=\{z_t(s):\ 0\leq s\leq1\}$ whose distance between end points should initially decrease from $1$, and shrink to $0$ as $t\to\infty$ (provided $u\neq0$ a.e.). -This may not qualify as an "algorithm" though, since the choice of $t$ to get the proper length remains implicit.<|endoftext|> -TITLE: How to construct a basis for the dual space of an infinite dimensional vector space? -QUESTION [16 upvotes]: Let $V$ be an infinite-dimensional vector space over a field $K$. Then it is known that $\dim V < \dim V^*$. More precisely, by a result attributed to Kaplansky and Erdos, we have $\dim V^* = |K|^{\dim V}$. -I have not seen an actual construction of a basis of $V^*$. My question is: given a basis $B$ of $V$, is there an explicit description of a basis of $V^*$ in terms of $B$? Can you do this at least in the case where $\dim V$ is countable? - -REPLY [30 votes]: It is consistent with the axioms of $\sf ZF$ that this is impossible. Specifically, if you consider $\Bbb R[x]$, then its dual space is just $\Bbb{R^N}$. And it is consistent with $\sf ZF$ that $\Bbb{R^N}$ does not have a Hamel basis. -(Under $\sf ZF+DC$, if all sets are Lebesgue measurable, or have the Baire property, then every group homomorphism between Polish groups is continuous. It follows that if $\Bbb{R^N}$ has a basis, then there is a discontinuous functional from $\Bbb{R^N}$ to $\Bbb R$ simply by cardinality arguments. And therefore such theories prove that $\Bbb{R^N}$ does not have a Hamel basis.) -It follows that there is no explicit way to specify how you get a basis of the dual space. You have to appeal to Zorn's lemma.<|endoftext|> -TITLE: Determinant of real Wishart matrix -QUESTION [6 upvotes]: Suppose $A$ is a real $N \times P$ matrix, $P \geq N$, with entries drawn independently according to $A_{ij} \sim \mathcal{N}(0,1)$. Then $W = A \, A^\top$ is a member of the real Wishart ensemble. What is the distribution of $\det W$? I'm particularly interested in the large $N$ behaviour. There might be a qualitative distinction between cases when $N / P(N) \rightarrow c$, with $c = 1$ or $c < 1$. In that case I'd be more interested in the former. Thanks! - -REPLY [6 votes]: The Distribution of the Determinant of a Complex Wishart Distributed Matrix proves that the determinant is distributed as the product of independent random variables with a chi-squared distribution, -$${\rm det}\,W\simeq\chi^2_P\chi^2_{P-1}\cdots\chi^2_{P-N+1}.$$ -(The title of the paper refers to the complex case, but the real ensemble is also considered towards the end.) -Results for the large-$N$ asymptotics can be found here. If we take $N,P\rightarrow\infty$ at fixed ratio $c=N/P<1$, the asymptotics is -$$\frac{\log{\rm det}\,(W/P)-\sum_{k=1}^N\log(1-k/P)}{\sqrt{-2\log(1-c)}}\rightarrow z,$$ -with $z$ normally distributed. For $c\rightarrow 0$ this means that $(2c)^{-1/2}\log{\rm det}\,(W/P)$ tends to a normal distribution. At the other extreme, for $c\rightarrow 1$ one has -$$\frac{\log{\rm det}\,(W/P)+P+\tfrac{1}{2}\log(P/2\pi)}{\sqrt{2\log P}}\rightarrow z,$$<|endoftext|> -TITLE: Graduate-level reference on temporal point processes -QUESTION [5 upvotes]: I am looking for a modern, graduate level, rigorous book on temporal point processes which also treats self-correcting point processes, and self-exciting point processes. -It would be even more interesting if the book was also pointing me to some research questions in the field. - -REPLY [5 votes]: You can briefly read a short introduction to decide whether you are interested in the theoretic side or application side of the research subject. Say Rasmussen's Notes. -If you are more interested in the theory, there is a two-volume standard text on theory of point processes. It includes a panoramic discussion on temporal/spatial process in Chap 14,15. - -Daley, Daryl J., and David Vere-Jones. An introduction to the theory - of point processes: volume I&II, 2ed, Springer, 2007. - -If you are more interested in the application, there is a very popular book ABG discusses a variety of applications of point processes in survival analysis. - -Aalen, Odd, Ornulf Borgan, and Hakon Gjessing. Survival and event - history analysis: a process point of view. Springer Science & Business - Media, 2008.<|endoftext|> -TITLE: The spectral radius of a binary matrix - polynomial growth? -QUESTION [5 upvotes]: (This is a follow-up to The spectral radius of a binary matrix) -Let $\mathcal B_n$ denote the set of $n\times n$ matrices with entries in $\{0,1\}$. -QUESTION. Is there a $\delta\in\bigl(0,\frac12\bigr), \alpha\in(0,1)$ and $b>0$ such that for any $n\ge1$ and any $M\in\mathcal B_n$ with the number of 0s less than $\delta n^2$ we have $\rho(M)\ge bn^\alpha$? -If so, what's the best known value of $\delta$? -The motivation for this question comes from symbolic dynamics (hence the tags). Namely, is it true that for any topological Markov chain given by a matrix with "sufficiently many" 1s its topological entropy is positive? - -REPLY [4 votes]: A paper that seems to directly address your question is the 1987 paper of Brualdi and Solheid, *On the minimal spectral radius of matrices of zeros and ones". That paper shows that if the number of 1's is $(\frac12+\delta)n^2$, then the spectral radius is essentially $2\delta n$. The paper also allows you to answer the question of how many 1's you do need if you want sublinear spectral radius. The extremal examples are the 0--1 matrices that are all 1's below the diagonal and have square blocks of 1's centred around the diagonal. That is, you have a number of completely connected components, all of the same size (with one remainder component), and each component had edges only to later components.<|endoftext|> -TITLE: Was Cauchy prescient? -QUESTION [6 upvotes]: Cauchy proved a sum theorem for series of continuous functions in 1821, and published another article on the subject in 1853. -Michael Segre, writing in Archive for History of Exact Sciences, claimed concerning Cauchy's sum theorem: -What is amazing here is Cauchy's attitude. He totally disregarded Fourier's counterexample and did not admit having made a mistake: not only did he "prove" his theorem, but he repeated it in a paper read to the Academie des Sciences as late as 1853. (page 233 in Segre, Michael. Peano's axioms in their historical context. Arch. Hist. Exact Sci. 48 (1994), no. 3-4, 201-342) -For his part, Umberto Bottazzini wrote: -The language of infinites and infinitesimals that Cauchy used here seemed ever more inadequate to treat the sophisticated and complex questions then being posed by analysis... The problems posed by the study of nature, such as those Fourier had faced, now reappeared everywhere in the most delicate questions of "pure" analysis and necessarily led to the elaboration of techniques of inquiry considerably more refined than those that had served French mathematicians at the beginning of the century. Infinitesimals were to disappear from mathematical practice in the face of Weierstrass' epsilon and delta notation (p. 208 in Bottazzini, Umberto. The higher calculus: a history of real and complex analysis from Euler to Weierstrass. Translated from the Italian by Warren Van Egmond. Springer-Verlag, New York, 1986) -These authors make Cauchy appear rather obstinate with regard to what is described by some historians as his famous "mistake". To a number of mathematicians who have studied Cauchy's work, such claims by historians seem surprising. Are we to accept them at face value? Is there more to the story than meets the eye? -An analysis of this question by my coauthors and myself is presented in this 2017 publication in Foundations of Science. Additional relevant material is referenced at this regularly updated site. What I am seeking are other possible responses to this question from people who have examined Cauchy's writings. -Note 1. I included in the article (on page 6) an extensive quotation from Cauchy that includes in particular his improbable substitution of $x=\frac{1}{n}$ in the remainder term $r_n$; see (new version of) article linked above. To a mathematician trained in the Weierstrassian framework this looks like a freshman calculus error. However, Robinson's framework enables an interpretation of this as evaluation at an infinitesimal point. Recall that the salient mathematical point here is that uniform convergence is expressible by a pointwise condition in the extended continuum. This is analogous to uniform continuity being expressible by a pointwise condition, namely S-continuity or microcontinuity (this last point is not strictly speaking related to the sum theorem but may help sort this out for those not closely familiar with the framework). -Note 2. For a related discussion of Cauchy see this MSE post. -Note 3. A detailed response to objections by Jesper Luetzen, Craig Fraser, and others appears in this 2017 publication in Mat. Stud. - -REPLY [18 votes]: After having read Katz' article, I must say I am not convinced and find that the standard interpretation, namely that of Cauchy making a mistake in 1821 and failing to acknowledging it or correcting it properly in 1853 is closer to the truth. In other words, even after reading your paper, I see nothing more than meets the eye. -One of your main point is the word "toujours" (always) which appears in the 1853 version of Cauchy's theorem, but not in the 1821 version. Quoting your paper, the 1821 version says - -When the various terms of series $u_0 +u_1 +u_2 + \dots +u_n + u_{n+1} + \dots$ are functions of the same variable $x$, continuous with respect to this variable in the neighborhood of a particular value for which the series converges, the sum s of the series is also a continuous function of $x$ in the neighborhood of this particular value. - -(I would have liked to see the French version, by the way). -The 1853 version is: - -Théorème 1. Si les différents termes de la série $$u_0,u_1,u_2,\dots,u_n,u_{n+1},\dots \ \ (1)$$ - sont des fonctions de la variable réelle $x$, continues, par rapport à cette variable, entre des limites données; si, d’ailleurs, la somme - $u_n +u_{n+1} + \dots + u_{n′−1}$ - devient toujours infiniment petite pour des valeurs infiniment grandes des nombres entiers $n$ et $n′ > n$, la série (1) sera convergente et la - somme $s$ de la série sera, entre les limites données, fonction continue de la variable $x$. - -You interpret "toujours" as meaning "for real (archimedean) and for infinitesimal values of the variable $x$". But I note that it is more natural to interpret it simply as meaning "for all real (archimedean) values of $x$". This interpretation would be enough to make the 1853 statement different, precisely with a stronger hypothesis, than the 1821 statement, for plainly the 1821 statement requires only the convergence of the series for a particular value $x_0$ (and the continuity of the $u_n$ on a neighborhood of $x_0$) to conclude the continuity of the sum $s$ at $x_0$. Thus we would have two statements of Cauchy's theorem, which both happen to be false. -The second important point of your argument is the discussion of Cauchy's treatment of a potential counter-example related lo Abel's objection in section 2.3. Cauchy claims that this is not a counter-example to his 1853 theorem because it fails some hypothesis. But here, since you give no quotation of Cauchy, it is impossible to know if Cauchy's arguments support your interpretation or are simply mistaken. - -REPLY [2 votes]: For convenience of our readers I provide a summary of the article linked in the question: - -Cauchy's sum theorem is a prototype of what is today - a basic result on the convergence of a series of functions in - undergraduate analysis. We seek to interpret Cauchy's proof, and - discuss the related epistemological questions involved in comparing distinct interpretive paradigms. Cauchy's proof is often interpreted - in the modern framework of a Weierstrassian paradigm. We analyze - Cauchy's proof closely and show that it finds closer proxies in - a different modern framework... - Interpretation of texts written in the nineteenth century, and the - meaning we give to technical terms, procedures, theories, and the like - are closely related to what we already know as well as our expectations - and assumptions. This paper provides evidence that a change in the - cultural-technical framework of a historian provides new explanations, - which are arguably more natural, and new insights into Cauchy’s work. - -Any serious interpretation of Cauchy's proof of his sum theorem has to take into account his argument involving the point generated by the sequence $(\frac{1}{n})$. I am not aware of any reasonable interpretation of such a point as a nonzero point of a standard Archimedean continuum.<|endoftext|> -TITLE: References for reasoning about the spectrum of a convex body? -QUESTION [14 upvotes]: By "spectrum of a convex body", I mean: start with a convex body $B$ in $\mathbb{R}^d$, then consider the corresponding $d \times d$ covariance matrix resulting from a uniform distribution over $B$ -- what can I say about the spectrum of this matrix? -For example, when the convex body $B$ is just the unit ball by logic like this its covariance matrix is diagonal and all eigenvalues are the same. My intuition is that this is because the unit ball is equally "fat" in all dimensions. -However, while I can similarly guess at the spectrum of other shapes based on thinness or fatness, I'm not sure how to approach the question rigorously. Again falling back on intuition, I would expect that the closer $B$ is to a sphere (e.g. if it both contains a sphere of radius $r$ and is contained in a sphere of radius $r'$) the more similar its spectrum (roughly). -Does anyone know of references for this problem, or similar variants? Thanks. - -REPLY [8 votes]: A very direct result giving characteristic function control for uniform measures on compact convex sets and hence spectrum is [Kulikova& Prokhorov]. -It sounds to me that all you want to study is the uniform distribution's spectrum on a convex body $K$. With some additional symmetries it is not surprising that [MO1] can lead to a strong claim, if you are to investigate this particular type of argument, then it is probably better to investigate uniform distributions on symmetric spaces instead of distributions on general convex bodies, like [Duenez]. -For a general convex body, there is little thing to say. The study in this direction is mostly about an isotropic convex body. If your problem is invariant under affine transformations then it does not lose generality by assuming isotropic. -The covariance matrix on an isotropic convex body can be studied by observing the behavior of inertia matrix [Aubrun1], seminal results due to R.Kannan， - L.Lovasz and J.Bourgain exist. [Brazitikos et al.] -There are a series of works by S.G.Bobkov and his collaborators that studied the concentration of mass in isotropic convex bodies. They formalized what they called "Central limit property" in [Bobkov&Koldobsky] around p.50 Theorem 2. So in high dimension $d$ the behavior of convariance matrix is asymptotically known. See [Paouris] Theorem 1.6 also.[Bobkov&Nazarov] showed how the geometry of unconditional measures will look like. Later [Aubrun2] extends Bai-Yin theorem to yield a sharper bound on the inertia matrix of uniform sampling. (For more details about Bai-Yin theorem please see part 1 of this [MO2] post.) -I have heard that there is a combinatoric approach to the problem of distributions on a general convex body but never have a chance to look into it. The book I heard about is [Kolchin et.al]. (Not sure but perhaps helpful.) -If you are interested in the boundary of a convex body and a distribution over it(say a random matrix with entries of uniform on the sphere $S^n$)instead of the convex body itself, a correct reference to look into is [Mardia&Peter] Chap 10. - -Reference -[Kulikova& Prokhorov]Kulikova, Anna A., and Yu V. Prokhorov. "Uniform distributions on convex sets: Inequality for characteristic functions." Theory of Probability & Its Applications 47.4 (2003): 700-701. -[Duenez]Duenez, Eduardo. "Random matrix ensembles associated to compact symmetric spaces." Communications in mathematical physics 244.1 (2004): 29-61. https://arxiv.org/pdf/math-ph/0111005.pdf -[Aubrun1]Aubrun, Guillaume. "Sampling convex bodies: a random matrix approach." Proceedings of the American Mathematical Society 135.5 (2007): 1293-1303. -[Aubrun2]Aubrun, Guillaume. "Random Points in the Unit Ball of ℓ n p." Positivity 10.4 (2006): 755-759. -[Bobkov&Nazarov]Bobkov, Sergey G., and Fedor L. Nazarov. "On convex bodies and log-concave probability measures with unconditional basis." Geometric aspects of functional analysis. Springer Berlin Heidelberg, 2003. 53-69. -[Bobkov&Koldobsky] Bobkov, Sergey G., and Alexander Koldobsky. "On the central limit property of convex bodies." Geometric aspects of functional analysis. Springer Berlin Heidelberg, 2003. 44-52. -[Kolchin et.al]Kolchin, Valentin Fedorovich, Boris Aleksandrovich Sevastyanov, and Vladimir Pavlovich Chistyakov. "Random allocations." (1978). -[Paouris]Paouris, Grigoris. "Concentration of mass on convex bodies." Geometric and Functional Analysis 16.5 (2006): 1021-1049. -[Mardia&Peter]Mardia, Kanti V., and Peter E. Jupp. Directional statistics. Vol. 494. John Wiley & Sons, 2009. -[Brazitikos et al.]Brazitikos, Silouanos, et al. Geometry of isotropic convex bodies. Vol. 196. Providence: American Mathematical Society, 2014.<|endoftext|> -TITLE: the sigma-p property -QUESTION [5 upvotes]: An operator ideal $\mathcal{I}$ possesses the $\sum_{p}$-property, say $10$, $1 -TITLE: Associativity of Steenrod's cup-i product -QUESTION [7 upvotes]: In the paper Products of Cocycles and Extensions of Mappings, -Steenrod introduced the cup-i product (and Steenrod square). I would like to ask if Steenrod's cup-i product associative or not? The paper did not discuss this property (it seems). - -REPLY [9 votes]: The cup-i products are not associative for $i > 0$. -For example, Steenrod's cup-1 product has the following description (taken mod 2 for expedience). For cocycles $f$ of degree $p$ and $g$ of degree $q$, and a simplex of degree $p+q-1$ with vertices $[v_0, v_1, \dots, v_{p+q-1}]$, the cup-1 product is defined by -$$ -(f \cup_1 g) [v_0 \dots v_{p+q-1}] = \sum_{i=0}^{p-1} f[v_0, \dots, v_i, v_{q+i}, v_{p+q-1}] \cdot g[v_i, \dots, v_{q+i}]. -$$ -In other words, you sum up over all ways to apply $g$ to a "middle" portion of the simplex. -If $f$ and $g$ have degree 2 and $h$ has degree 1, we find -$$ -\begin{align*} -((f \cup_1 g) \cup_1 h) [v_0, v_1, v_2, v_3] =& (f[v_0, v_2, v_3] g[v_0, v_1, v_2] + f[v_0, v_1, v_3] g[v_1, v_2, v_3]) \\&\cdot (h[v_0,v_1]+ h[v_1, v_2] + h[v_2, v_3]) \\ -(f \cup_1 (g \cup_1 h)) [v_0, v_1, v_2, v_3] =& f[v_0, v_2, v_3] g[v_0, v_1, v_2] \cdot (h[v_0, v_1] + h[v_1, v_2]) \\&+ f[v_0, v_1, v_3] g[v_1, v_2, v_3] \cdot (h[v_1, v_2] + h[v_2, v_3]) -\end{align*} -$$ -and the difference between the two is -$$ -f[v_0, v_2, v_3] g[v_0, v_1, v_2] h[v_2, v_3] + f[v_0, v_1, v_3] g[v_1, v_2, v_3] h[v_0, v_1]. -$$<|endoftext|> -TITLE: generalized elements in monoidal categories -QUESTION [7 upvotes]: In a category $\mathcal{C}$, a generalized element of an object $A$ means a morphism to $A$. It follows from Yoneda lemma that the object $A$ is determined by the collection of generalized of elements of $A$. -In a monoidal category $(\mathcal{C},\otimes)$, it seems more natural to consider generalized elements of a tensor product $A\otimes B$ in the form $a\otimes b$, meaning a tensor product of two morphisms, one to $A$, another to $B$. -So, my first questions are - -Is the tensor product $A\otimes B$ determined by generalized elements of the form $a\otimes b$? -Let $f,g\colon A\otimes B\to C$ be two morphisms. If for all generalized elements of the form $a\otimes b$, the compositions $f(a\otimes b)=g(a\otimes b)$, is it true that $f=g$? - -In nLab, it mentioned that generalized elements of the form $I\to A$, where $I$ is the tensor unit, is important in the theory of enriched category. However, the functor $A\mapsto\mathrm{Hom}(I,A)$ may not be full or faithful. Hence those generalized elements cannot determine the objects or morphisms in general. So - -Why generalized elements of the form $I\to A$ important in enriched category theory regardless its application to define underlying categories? - -REPLY [2 votes]: If I'm not wrong, the following is an answer to 2. -For any objects $A,B,C$ of $\mathcal{C}$, we have the following natural map -\begin{align} -\Phi\colon\mathrm{Hom}(A\otimes B,C) &\to \mathrm{Nat}(h_A\boxtimes h_B,h_C\circ\otimes) \\ -f &\mapsto f\circ- -\end{align} -where - -$h_A\boxtimes h_B$ denotes the functor $\mathcal{C}\times\mathcal{C}\to\mathrm{Set}$ mapping each pair of objects $S,T$ of $\mathcal{C}$ to the set $\mathrm{Hom}(S,A)\times\mathrm{Hom}(T,B)$; -$h_C\circ\otimes$ denotes the functor $\mathcal{C}\times\mathcal{C}\to\mathrm{Set}$ mapping each pair of objects $S,T$ of $\mathcal{C}$ to the set $\mathrm{Hom}(S\otimes T,C)$; -$f\circ-$ means mapping each pair $a\colon S\to A, b\colon T\to B$ to the composition $f(a\otimes b)$. - -On the other hand, by the definition of End, we have (each "$=$" means a canonical isomorphism) -\begin{align} -\mathrm{Nat}(h_A\boxtimes h_B,h_C\circ\otimes) -&= \int_{(S,T)\in\mathrm{ob}\mathcal{C}\times\mathcal{C}}\mathrm{Map}(\mathrm{Hom}(S,A)\times\mathrm{Hom}(T,B),\mathrm{Hom}(S\otimes T,C)) \\ -&= \int_{S\in\mathrm{ob}\mathcal{C}}\int_{T\in\mathrm{ob}\mathcal{C}} -\mathrm{Map}(\mathrm{Hom}(S,A)\times\mathrm{Hom}(T,B),\mathrm{Hom}(S\otimes T,C)) -& (\text{by Fubini theorem})\\ -&= \int_{T\in\mathrm{ob}\mathcal{C}}\int_{S\in\mathrm{ob}\mathcal{C}} -\mathrm{Map}(\mathrm{Hom}(S,A)\times\mathrm{Hom}(T,B),\mathrm{Hom}(S\otimes T,C)) -& (\text{since $\mathrm{Set}$ is symmetric monoidal})\\ -&= \int_{T\in\mathrm{ob}\mathcal{C}}\int_{S\in\mathrm{ob}\mathcal{C}} -\mathrm{Map}(\mathrm{Hom}(S,A),\mathrm{Map}(\mathrm{Hom}(T,B),\mathrm{Hom}(S\otimes T,C))) \\ -&= \int_{T\in\mathrm{ob}\mathcal{C}} -\mathrm{Nat}(h_A,\mathrm{Map}(\mathrm{Hom}(T,B),\mathrm{Hom}(-\otimes T,C))) \\ -&= \int_{T\in\mathrm{ob}\mathcal{C}} -\mathrm{Map}(\mathrm{Hom}(T,B),\mathrm{Hom}(A\otimes T,C)) -& (\text{by Yoneda lemma}) \\ -&= \mathrm{Nat}(h_B,\mathrm{Hom}(A\otimes -,C)) \\ -&= \mathrm{Hom}(A\otimes B,C)& (\text{by Yoneda lemma again}) -\end{align} -By trace the canonical isomorphism involved, the above gives the inverse of $\Phi$. The bijectivity of $\Phi$ implies 2 as desired.<|endoftext|> -TITLE: Semi-Simplicity of Mod-$\ell$ Galois Representations -QUESTION [6 upvotes]: I am currently studying mod-$\ell$ Galois representations of $CM$ elliptic curves. More precisely, let $\ell$ be a prime, $K$ a number field and $E/K$ a $CM$ elliptic curve, where all endomorphisms of $E$ are defined over $K$, then I am looking at the representation -$\rho_{E,\ell}: \mathrm{Gal}(\overline{K}/K) \to \mathrm{Aut}(E[\ell])$. -I was wondering if there is an effective/computable criterion to determine for which $ \ell's $ is this representation semi-simple. - -REPLY [5 votes]: Let us assume that $E$ has complex multiplication by a maximal order $\mathcal{O}$. Then $E[\ell]$ is a free rank $1$ module over $\mathcal{O}/\ell\mathcal{O}$. The image $G$ of $\rho_{E,\ell}$ is contained in the $\mathcal{O}$-automorphism of $E[\ell]$, which are isomorphic to $(\mathcal{O}/\ell\mathcal{O})^{\times}$. -If $\ell$ is unramified in $\mathcal{O}$, then $G$ is contained in a group of order either $\ell^2-1$ or $(\ell-1)^2$. Since $\ell$ will not divide the group order of $G$, the action on $E[\ell]$ will be semi-simple. -Instead for ramified primes $\ell$, it can happen that $E[\ell]$ is not semi-simple. For instance for the curve $y^2 = x^3 + 1$ over $K=\mathbb{Q}(\sqrt{-3})$ and $\ell=3$. Then $K(E[3])/K$ is an extension of degree $3$ and hence the Galois group acts on $E[3]$ via the matrix $\bigr(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\bigr)$. -I believe in the ramified case one has to check by hand if $E$ admits two isogenies of degree $\ell$ defined over $K$. Of course in practice one can determine $G$ completely for these $\ell$.<|endoftext|> -TITLE: Do all $\mathcal{N}=2$ Gauge Theories "Descend" from String Theory? -QUESTION [18 upvotes]: I asked this on PhysicsSE, but I think it also fits here as it's related to algebro-geometric connections to string and gauge theory. -I'm thinking about the beautiful story of "geometrical engineering" by Vafa, Hollowood, Iqbal (https://arxiv.org/pdf/hep-th/0310272.pdf) where various types of $\mathcal{N}=2$ SYM gauge theories on $\mathbb{R}^{4} = \mathbb{C}^{2}$ arise from considering string theory on certain local, toric Calabi-Yau threefolds. -More specifically, the topological string partition function (from Gromov-Witten or Donaldson-Thomas theory via the topological vertex) equals the Yang-Mills instanton partition function which is essentially the generating function of the elliptic genera of the instanton moduli space. (In various settings you replace elliptic genus with $\chi_{y}$ genus, $\chi_{0}$ genus, Euler characteristic, or something more fancy.) -From my rough understanding of the Yang-Mills side, we can generalize this in a few ways. Firstly, we can consider more general $\mathcal{N}=2$ quiver gauge theories where I think the field content of the physics is encoded into the vertices and morphisms of a quiver. And there are various chambers where one can define instanton partition functions in slightly different ways, though they are expected to agree in a non-obvious way. For a mathematical account see (https://arxiv.org/pdf/1410.2742.pdf) Of course, the second way to generalize is to consider not $\mathbb{R}^{4}$, but more general four dimensional manifolds like a K3 surface, a four-torus, or the ALE spaces arising from blowing up the singularities of $\mathbb{R}^{4}/\Gamma$. -My questions are the following: -Is it expected that this general class of $\mathcal{N}=2$ quiver gauge theories comes from string theory? In the sense that their partition functions may equal Gromov-Witten or Donaldson-Thomas theory on some Calabi-Yau threefold. If not, are there known examples or counter-examples? Specifically, I'm interested in instantons on the ALE spaces of the resolved $\mathbb{R}^{4}/\Gamma$. -In a related, but slightly different setting, Vafa and Witten (https://arxiv.org/pdf/hep-th/9408074.pdf) showed that the partition function of topologically twisted $\mathcal{N}=4$ SYM theory on these ALE manifolds give rise to a modular form. Now I know Gromov-Witten and Donaldson-Thomas partition functions often have modularity properties related to S-duality. So I'm wondering, does this Vafa-Witten partition function equal one of these string theory partition functions? Or are they related? - -REPLY [10 votes]: The answer is Yes, at least when the ALE space (more precisely the ALF space instead) is of type $A$. Very roughly, one consider Donaldson-Thomas invariants for the same noncompact Calabi-Yau space, but not of rank $1$, of higher ranks instead. (I do not know whether they correspond to Gromov-Witten invariants.) I do not know physics literature, but I wonder that it was probably known as the answer is simple. -Here is a little more explanation. -We consider the Coulomb branch of a framed affine quiver gauge theory as is defined in my paper 1601.03586 with Braverman, Finkelberg. The Coulomb branch in this case is expected to be the moduli spaces of instantons on an ALF space of type $A$, and is proved when the affine quiver is of type $A$ in 1606.02002. Here the type of the quiver determines the gauge group of instantons, and the type of the ALF space is the level of weights given by $V$, $W$. For example, if the quiver is of type $A$, it is the total sum of dimensions of $W$. When the level is $1$, it is the Taub-NUT space, $\mathbb R^4$ with non-flat hyper-Kaehler metric. -The Hilbert series of the Coulomb branch is nothing but the K-theoretic instanton partition function of pure gauge theory on the ALF space. When both of two weights determined by $V$, $W$ are dominant, one can replace the ALF space of the ALE space, as proved in 1606.02002 for type $A$. -On the other hand, as is explained in 1503.03676, its Hilbert series is a generating function of Donaldson-Thomas type invariants. More precisely, no stability condition is imposed in 1503.03676, but the ordinary Donaldson-Thomas invariants can be deduced from the Hilbert series by stuying Harder-Narashimhan stratification. This HN stratification argument is not written, but was explained in my talk at Oxford.<|endoftext|> -TITLE: What are the advantages of various "models" for the motivic stable homotopy category -QUESTION [15 upvotes]: People use several distinct models for the motivic stable homotopy category (so, there are some choices for the underlying category and a collection of available model structures). I would like to ask for help in "navigating" in this matter. So, which advantages does each of the model have? Does any "guide" for this subject exist? -In particular, is the category of symmetric (motivic) $T$-spectra endowed with the injective (="levelwise"?) model structure a "reasonable" monoidal model category? - -REPLY [10 votes]: The injective model structure is monoidal (satisfies the pushout product axiom), see Hornbostel's paper "Localizations in motivic homotopy theory", Thm 1.9 and Lemma 1.10. The projective model structure is also monoidal. See Hovey's "Spectra and symmetric spectra in general model categories", where he introduces the projective model structure. -Later, Hornbostel introduced the positive model structure, a modification of the projective model structure that, like the case for ordinary (non-motivic) symmetric spectra, avoids Gaunce Lewis's obstruction regarding commutative monoids. Specifically, this model structure is Quillen equivalent to the projective (and injective), but now the unit (sphere spectrum) is not cofibrant. Furthermore, there is now a transferred model structure on commutative ring spectra, which Lewis proved cannot exist for the projective model structure. For non-motivic spectra, the positive model structure was introduced in Mandell-May-Schwede-Shipley ("diagram spectra"). If you google, you can find old drafts online where I thought I'd done the positive analogue of Hovey's general-spectra paper, but Hornbostel beat me to it by several years. -There is also a positive flat model structure on symmetric spectra (see Shipley's paper "A convenient model category for commutative ring spectra") that has the extra property that cofibrant commutative monoids forget to cofibrant objects in the underlying category. This property is generally not true for the non-flat variant. I am not sure if this has been constructed for motivic symmetric spectra, but it surely must exist. An analogous program for equivariant spectra was carried out in the thesis of Martin Stolz. If this has been done for motivic, it would probably be in the paper of Pavlov and Scholbach. -If you have other questions not answered here, please let me know and I'll edit to try and answer. I agree it would be good to have a unified place with lists of pros/cons of the various structures, and that may as well be here. All these model structures are combinatorial, stable, and proper. What other properties do you care about?<|endoftext|> -TITLE: Tweetable Mathematics -QUESTION [108 upvotes]: Update: Please restrict your answers to "tweets" that give more than just the statement of the result, and give also the essence (or a useful hint) of the argument/novelty. - -I am looking for examples that the essence of a notable mathematical development fits a tweet (140 characters in English, no fancy formulas). - -Background and motivation: This question was motivated by some discussion with Lior Pachter and Dave Bacon (over Twitter). Going over my own papers and blog posts I found very few such examples. (I give one answer.) There were also few developments that I do not know how to tweet their essence but I feel that a good tweet can be made by a person with deep understanding of the matter. -I think that a good list of answers can serve as a good educational window for some developments and areas in mathematics and it won't be overly big. -At most 140 characters per answer, single link is permitted. Tweeting one's own result is encouraged. -Update: I learned quite a few things, and Noam's tweet that I accepted is mind-boggling. - -REPLY [2 votes]: By using the Combinatorial nullstellensatz and a related theorem of Alon- Tarsi, it can be proved that the list version to Brooks' theorem is true. See here<|endoftext|> -TITLE: Did Lagrange change his mind about infinitesimals? -QUESTION [16 upvotes]: Lagrange is famous for his attempt to found analysis algebraically using power series expansions, an approach that, as we know today, is limited to analytic functions. Lagrange is also known as the initiator of a contest at the French Academy at the end of the 18th century, for the best essay on justifying infinity and infinitesimals. L. Carnot did not win the contest but his essay eventually became very popular. The outcome of the contest was something of a disappointment to its organizers who did not find any of the submissions fully satisfactory. -What is less known is that toward the end of his career, and already in the 19th century, Lagrange fully embraced infinitesimals in the introduction to the second, 1811 edition of his book Analytical Mechanics, in the following terms: -Once one has duly captured the spirit of this system [i.e., infinitesimal calculus], and has convinced oneself of the correctness of its results by means of the geometric method of the prime and ultimate ratios, or by means of the analytic method of derivatives, one can then exploit the infinitely small as a reliable and convenient tool so as to shorten and simplify proofs. (translation mine) -Is there any work analyzing the actual use of infinitesimals by Lagrange in the body of the 1811 edition of his book? -I just noticed that in the article -Carpinteri, Alberto; Paggi, Marco. Lagrange and his Mécanique analytique: from Kantian noumenon to present applications. Meccanica 49 (2014), no. 1, 1-11 -the authors comment as follows on the second edition: -In the new formulation of mechanics proposed by Lagrange, the quantity of primary importance is no longer the force, but the work done by forces for infinitesimal arbitrary movements. (page 6) -The authors then go on to quote a passage from Lagrance: -Lagrange thus arrived at the formulation of the following expression ([11, 12], p. 21): "the principle of virtual velocities can be considered as very general by expressing it in the following manner: if any system, composed of any number of points or bodies, each of which solicited by any force, is in equilibrium; and if this system is given any small movement, in virtue of which each point travels by an infinitely small displacement, which shall express its virtual velocity - then the sum of the forces, each one multiplied by the displacement travelled (in its direction) from its point of application, shall always be equal to zero; considering the displacements travelled in the same direction of the forces as positive and small; and the travelled displacements in opposite direction as negative". (emphasis added) - -REPLY [8 votes]: Gert Schubring's Conflicts Between Generalization, Rigor, and Intuition, page 397 and following, gives a critical assessment of this issue:<|endoftext|> -TITLE: ccc after strongly proper forcing -QUESTION [8 upvotes]: Let $P, Q \in V$ be such that $P$ is strongly proper and $Q$ is ccc. Does $Q$ continue to be ccc after forcing with $P$? -Since strongly proper forcings do not add new branches to $\omega_1$-trees, they do preserve the ccc-ness of some partial orders: Suslin trees, the poset to specialise an Aronszajn tree. - -REPLY [10 votes]: Yes. The proof of Claim 3.8 in Neeman's paper on forcing with side conditions shows that: -(*) if $P$ is strongly proper and $Q$ is proper, $M$ is a countable elementary submodel, $p$ is an $(M,P)$ strong master condition, and $q$ is an $(M,Q)$ master condition, then $p$ forces that $\check{q}$ is an $(M[\dot{G}_P], Q)$-master condition. -(this is an abstraction of a lemma of Sy Friedman) -Now consider also that the following 3 properties of a poset $Q$ are equivalent (this is an old result, I don't recall where, but it's easy to prove): - -$Q$ is c.c.c. -$1_Q$ is a master condition for stationarily many countable models -$1_Q$ is a master condition for club-many countable models - -So back to your question: assume $P$ is strongly proper and $Q$ is c.c.c. Then there is a club $C$ of countable models $M$ such that $P$ is strongly proper w.r.t. $M$, and $1_Q$ is an $(M,Q)$ master condition (the latter uses equivalence of 1 with 3). Let $G_P$ be $(V,P)$-generic. Using a genericity argument, in $V[G_P]$ the set of $M \in C$ for which $G_P$ includes a strong master condition is stationary; let $S \in V[G_P]$ denote this stationary set. Also $S':= \{ M[G_P] \ : \ M \in S \}$ is stationary. Since $S \subseteq C$, then for every $M \in S$, $1_Q$ was an $(M,Q)$-master condition in the ground model. Then by (*), $V[G_P]$ believes that $1_Q$ is a master condition for every model in the stationary set $S'$. Hence by the equivalence of 1 and 2 above, $V[G_P]$ believes that $Q$ is c.c.c.<|endoftext|> -TITLE: Reference request: type C, D Catalan numbers -QUESTION [6 upvotes]: Catalan numbers are generalized to type B: https://oeis.org/A000984. -Are there some references about Catalan numbers of type C, D? Thank you very much. - -REPLY [11 votes]: The Catalan numbers you get in types other than A depend on which interpretation of the Catalan numbers you are generalizing. There are at least two possibilities: the number of anti-chains in the root poset (these are sometimes called non-nesting partitions) gives the so called Coxeter-Catalan number; these are well-studied, see e.g. "A uniform bijection between nonnesting and noncrossing partitions" by Armstrong, Stump, and Thomas. The number of Stembridge's "fully commutative" elements (in type A, this is the same as 321-avoiding permutations) of the Coxeter group is another generalization which gives different numbers, see "The enumeration of fully commutative elements of coxeter Groups" by Stembridge for explicit generating functions in all types.<|endoftext|> -TITLE: Is any CW complex with only finitely many nonzero homology groups homotopic to a finite dimensional CW complex? -QUESTION [10 upvotes]: My question is - -Is any CW complex with only finitely many nonzero homology groups homotopic to a finite dimensional CW complex? - -(My thoughts on this which might not be useful at all.) Since an infinite dimensional CW complex could have only finitely many nontrivial homology groups($S^\infty$ for example), it seems to me that the relation between the dimension and the zeroness of the CW complexes is not very strong. On the other hand, we know that Moore spaces are unique up to homotopy equivalence and any CW complex with prescribed homology groups can be construct by taking the wedge sum of Moore spaces. If this statement above is true, then it means every CW complex with only finitely many nonzero homology groups is essentially built up in this way... -Please note that I meant finite dimensional CW complex instead of finite CW complex. Otherwise the infinite dimensional discrete space will serve the purpose. Also, I really appreciate it if someone can point it out whether the statement can become true buy adding some small conditions(One of my professors said we need $X$ to be simply connected). - -REPLY [6 votes]: Supplementing Chris's answer, you could take $G$ to be any acyclic group of infinite cohomological dimension. Such groups exist, e.g. binate groups. See -Berrick, A.J., The acyclic group dichotomy., J. Algebra 326, No. 1, 47-58 (2011). ZBL1253.20055. -Then any model of $BG$ will have trivial homology but must be infinite-dimensional. Taking wedge sums or cartesian products of $BG$ with a finite complex will give plenty more counter-examples. -For the (positive) simply-connected statement mentioned by John Klein, search for "minimal CW structures".<|endoftext|> -TITLE: what is $\mathrm{Bun}(G)$? -QUESTION [13 upvotes]: I don't even know where to begin. There's a discussion of stacks and they talk about $\mathrm{Bun}(G)$. I don't know what it is, or what it's elements are or why it is important. Google and wikipedia don't really help since they pre-suppose. -One resource says $\mathrm{Bun}(G)$ is the moduli stack of $G$-bundles where $G$ is an affine algebraic group over a field $k$. - -the embedding $G \to GL_n$ induces a morphism of stacks $\mathrm{Bun}_G \to \mathrm{Bun}_{{GL}_n}$ -$\mathrm{Bun}_G$ is depends on the space $X$, $\mathrm{Bun}_G(X)$ is a groupoid -$\mathrm{Bun}_G$ is a functor, meaning that it is well behaved under maps of spaces. A map $Y \to X$ yields another "induced" map: -$$ \mathrm{Bun}_G(X) \to \mathrm{Bun}_G(Y) $$ - -I believe $X$ is a scheme; projective, connected smooth curve over a field $k$, not necessarily algebraically complete. $G$ is an affine algebraic group over $k$. -What are the étale or fppf topologies here? - -The only projective curves I can think of are Riemann surfaces, which are often smooth and connected. The field $k$ could be $\mathbb{C}$ but possibly $\mathbb{Q}$ or a number field. My idea of a "bundle" is something with local trivialization and around each loop you assign an element of $G$; gluing copies of $U_i \times G$ with transition maps. Over a number field, perhaps we lose the analytic topology. -The embedding theorem maybe says we can encode the gluing information using $n \times n$ matrices (a representation) and functorality says a map between schemes should "lift" to a map between bundles. -So these moduli stacks should be spaces of $n \times n$ matrices with various restrictions. They may themselves form a variety of some kind (certainly a scheme). -As you can see I am totally clueless. - -REPLY [33 votes]: I am surprised that nobody has mentioned (yet) that an essential point of the algebro-geometric notions of (algebraic) stack and algebraic space is to completely shift the burden of construction problems: one gives up on trying to make any kind of actual ringed space at all, and in fact the whole point is to create a kind of "geometry for functors". In particular, Bun$_G$ is demoted to the status of a definition (there's no "space" to be constructed: one just defines a certain groupoid-valued functor) and the real content is to show that it is "near enough" to representable functors that one can import to it many concepts from algebraic geometry. -It is easiest to explain this with an example. Let's contrast the approaches to the "Hilbert scheme" for a projective scheme $X$ over a field $k$ (to fix ideas). In the original approach of Grothendieck, he defined a functor $\underline{\rm{Hilb}}_{X/k}$ on the category $k$-schemes, namely -$\underline{\rm{Hilb}}_{X/k}(S)$ is the set of finitely presented closed subschemes $Y \subset X \times S$ such that the structure map $Y \rightarrow S$ is flat; this is a contravariant functor of $S$ via pullback (i.e., for any $f:S' \rightarrow S$ over $k$, the map -$$\underline{\rm{Hilb}}_{X/k}(f): \underline{\rm{Hilb}}_{X/k}(S) \rightarrow \underline{\rm{Hilb}}_{X/k}(S')$$ carries $Y \subset X \times S$ to $Y \times_S S' \subset (X \times S) \times_S S' = X \times S'$). -The real substance in this approach is to show $\underline{\rm{Hilb}}_{X/k}$ is a representable functor. This amounts to constructing a "universal structure": a distinguished $k$-scheme $H$ equipped with an $H$-flat finitely presented closed subscheme $Z \subset X \times H$ such that for any $k$-scheme $S$ and any $Y \in \underline{\rm{Hilb}}_{X/k}(S)$ whatsoever there exists a unique $k$-map $g:S \rightarrow H$ for which the $S$-flat closed subscheme -$$Z \times_{H,g} S \subset (X \times H) \times_{H, g} S = X \times S$$ -coincides with $Y$. In effect, $k$-maps to $H$ "classify" the functor $\underline{\rm{Hilb}}_{X/k}$. -But Grothendieck's representability result was proved via methods of projective geometry, and there were strong indications that one should seek a result like this more widely for proper $k$-schemes (beyond the projective case) yet nobody could find such a way to do this in the framework of schemes. Artin brilliantly solved the problem by widening the scope of algebraic geometry via his introduction of algebraic spaces, but it must be appreciated that in so doing one gives up on constructing a "representing space" in any sense akin to what Grothendieck does. In particular, algebraic spaces are not ringed spaces at all. Instead, they are Set-valued functors which are "near enough" to representable functors that it becomes possible to make sense of many concepts from algebraic geometry for these structures (and there is an associated topological space permitting one to speak of irreducibility, connectedness, etc.). -The crucial point is that in the Artin approach, the functor is the algebraic space. That is, in Artin's approach it is a complete misnomer to speak of "constructing the moduli space" for the Hilbert functor $\underline{\rm{Hilb}}_{X/k}$ for a proper $k$-scheme: the functor one has defined above literally is to be the algebraic space, so there is nothing to be constructed there. Instead, the real work is to build the representable functors "near enough" to this so that one can conclude that the functor one has already defined is indeed an algebraic space (and hence one can do meaningful geometry with it, after getting acclimated to the new setting). -Artin's breakthrough was to identify checkable criteria with which one could show (entailing some real work) that many functors of interest are "near enough" to representable functors to be algebraic spaces (and his PhD student Donald Knutson devoted his thesis to working out very many concepts of algebraic geometry in the wider setting of algebraic spaces, often entailing new kinds of proofs from what had been done for schemes). But even with $\underline{\rm{Hilb}}_{X/k}$ for projective $X$ there is a fundamental difference in the nature of the results: Grothendieck built the Hilbert scheme ${\rm{Hilb}}_{X/k}$ as a countably infinite disjoint union of quasi-projective schemes, so one knows the connected components are quasi-compact, whereas the Artin approach gives no information whatsoever on quasi-compactness properties for the algebraic space $\underline{\rm{Hilb}}_{X/k}$ (but of course it has the huge merit of being applicable far more widely, allowing to "do algebraic geometry" with many many more functors of interest; nonetheless, Grotendieck's quasi-compactness results for Hilbert schemes remain vital as a tool for proving quasi-compactness features of rather abstract algebraic spaces). -As another illustration, consider the functor $M_{g,n}$ of $n$-pointed smooth genus-$g$ curves. In the Mumford approach via GIT (say with $n$ big enough to get rid of non-trivial automorphisms), there is tremendous effort done to build a universal structure. In the Artin approach via stacks (which allows $n$ to be quite small too), the moduli problem is the stack: it is incorrect to speak of "constructing the moduli space" in this approach, since one has literally defined $M_{g,n}$ and there's all there is to it (granting that one is sufficiently fluent with descent theory and coherent duality to see at a glance that $M_{g,n}$ enjoys nice descent properties). But instead the real effort is to build appropriate "scheme charts" over $M_{g,n}$ to give precise meaning to a sense in which $M_{g.n}$ is "near enough" to representable functors that we can make sense of many concepts of algebraic geometry for this functor or for its groupoid-valued version when $n$ is small. -The situation with Bun$_G$ is similar: the groupoid assignment is the geometric object. Techniques of Artin provide a precise sense in which some schemes can be regarded as "smooth" over this gadget, with those used to define geometric concepts for Bun$_G$ in the same spirit as one uses open balls to define concepts for manifolds, but it is rather a misnomer to speak of "constructing Bun$_G$"! That is, one has to clearly distinguish the serious task of describing some "charts" on Bun$_G$ (with which to make computations and prove theorems) from the much more mundane matter of simply defining what Bun$_G$ is: it is the assignment to any $X$ of the groupoid of $G$-bundles on $X$, and the descent-like properties are easy to verify (so it is a "stack in groupoids"), and there is nothing more to do as far as "making the moduli space" is concerned. Of course, one cannot do anything serious with this until having carried out the real effort with Artin's criteria to affirm that this stack admits enough "scheme charts" to be an Artin stack and hence admit the rich array of meaningful concepts for it as in algebraic geometry (irreducible components, coherent sheaf theory, cohomology, smoothness properties, dimension, etc.) There is no gluing to be done: we define the global moduli problem at the outset. We have to do work to show this admits meaningful geometric concepts, and in practice there can be especially useful "scheme charts" or techniques or "local description" with which one can explore it. However, one really should not regard these explicit descriptions as "constructing Bun$_G$"; the construction is just the initial basic definition mentioned above. Another wrinkle is that in the algebro-geometric setting one has to use some serious Grothendieck topologies to make this all work, so it isn't as simple as with making manifolds by gluing open balls (likewise for the notion of $G$-bundle). -Of course, there are interesting and instructive analogies with constructions in homotopy theory, but to make coherent sense of actual algebro-geometric proofs involving Bun$_G$ one should recognize that this "space" is demoted to the status of a definition (in complete contrast with the setting for Hilbert schemes by Grothendieck, where he had to really build a representing scheme before geometry could be done) and the substantial effort is to build many kinds of interesting "scheme charts" over it with which to explore the geometric features of this stack. The framework of Artin stacks gives a systematic way to make sense of this process for many kinds of moduli problems encountered in practice. But at the end of the day stacks and their cousins are not ringed spaces (even though their structure can be explored using many auxiliary ringed spaces); e.g., even though there is an associated topological space with which to define various topological notions, in no sense is a map determined in terms of some kind of map of ringed spaces resting on those associated topological spaces (in contrast with the cast of schemes). The way one gets around the loss of contact with a specific ringed space is by a huge amount of descent theory. I think it is very important to always keep that in mind or else a lot of relevant issues in definitions and proofs will be rather confusing.<|endoftext|> -TITLE: On structures that are not submitted to compatibility conditions -QUESTION [7 upvotes]: In mathematics, it seems that 100 % of the time, we deal with objects that have a number of different structures that are "compatible" with each other. For example, -a Lie group is a manifold and a group such that these two structures are compatible in the sense that the group operations are smooth. -So my question is on a metamathematical level : - is it actually true that it's always the case ? -More precisely : are there some cases where the "right" object to study and define is an object with two (or more) structures that are not required to be made compatible, except that they live on the same set ? - -REPLY [13 votes]: Here's the most clear-cut example I can think of: - -A Galois representation consists of a field extension $K \subseteq L$, the group $G = Gal(L/K)$, and a representation $G \to GL(V)$ on a vector space $V$ with no compatibility condition between the field structure and the representation (although I have the sense that in more sophisticated treatments, one at least asks for compatibility between a profinite topology on $G$ and the representation.) - -I'd point out that oftentimes, the most obvious notion of "compatibility with structure" is too rigid, and one needs to loosen up the notion of compatibility, if not throw it out altogether. For example: - -There are lots of structures in differential geometry which are not functorial on the category of manifolds and smooth maps (or its opposite), so one needs to loosen up and study constructions that are functorial on the category of manifolds and diffeomorphisms, or perhaps local diffeomorphisms. For example, tensors of mixed variance. -In algebraic geometry, a scheme $X$ has a rich structure. When studying sheaves on $X$, one can ask the sheaf to to be compatible with all this structure and satisfy Zariski descent. But for many purposes this is too rigid, and instead one asks for compatibility with less structure, for example by asking for etale descent. -Similarly, in differential topology, one might be studying a smooth manifold, but might need to talk about constructions on the manifold that respect only the topological structure, or PL structure. -In algebra, the category $\mathrm{Vect}_k$ of vector spaces over a fixed field $k$ is naturally enriched in, well, itself. But there are interesting functors $\mathrm{Vect}_k \to \mathrm{Vect}_k$ which don't respect the enrichment -- for example Schur functors like $V \mapsto V \otimes V$ or $V \mapsto \Lambda^k V$ or $V\mapsto \mathrm{Sym}^k V$. Thus one ends up studying functors which are not compatible with all the structure they could be. -In algebraic topology, one can consider a nice cohomology theory (or $E_\infty$-ring spectrum) like, say ordinary cohomology with $\mathbb{Z}/2$ coefficients ($H\mathbb{Z}/2$). It turns out to be very important to study the Steenrod algebra, which is the algebra of stable natural transformations $H^{\bullet}(-,\mathbb{Z}/2) \to H^{\bullet+k}(-,\mathbb{Z}/2)$ (where $k$ can vary from operation to operation). In some sense, there's a missing compatibility condition, because one is talking about maps $H\mathbb{Z}/2 \to \Sigma^k H \mathbb{Z}/2$ which are not required to be $H\mathbb{Z}/2$-linear. But it's important to loosen things up: this actually ends up revealing more structure (the action of the Steenrod algebra) which is important to think about. -An example I'm fond of: Norbert Roby defined a polynomial law between modules $M$ and $N$ over a fixed ring $k$ to be a natural transformation $V(M \otimes U-) \Rightarrow V(N \otimes U- )$ where $U$ is the forgetful functor $R$-$\mathrm{Alg} \to k$-$\mathrm{Mod}$ and $V$ is the forgetful functor $k$-$\mathrm{Mod} \to \mathrm{Set}$ -- so the definition involves twice forgetting structure, deliberately leaving out compatibility requirements, and it turns out to yield a loosening of the notion of linear map which is just what one wants. -If $V,W$ are representations of a group $G$, then when studying homomorphisms from $V$ to $W$, one has basically two things one can look at: the space $Hom_G(V,W)$ of linear maps $V \to W$ which are compatible with the action of $G$, or rather the space $Hom(V,W)$ of all linear maps $V \to W$ with no compatibility. The latter is interesting, though, because $Hom(V,W)$ itself carries an action of $G$, and so can be analyzed as a $G$-representation. Similarly, between chain complexes $C,D$, one can consider the space of homomorphisms which respect the boundary maps, or instead the whole chain complex of homomorphisms which are not required to be compatible with the boundary maps. - -So in some sense there are a number of different ways to study interacting structure. It depends on what your goals are, and what the interesting examples do. - -REPLY [7 votes]: In -Shelah, Saharon; Simon, Pierre, Adding linear orders, J. Symb. Log. 77, No. 2, 717-725 (2012). ZBL1251.03038. -the authors show that if $<$ is an arbitrary linear order of an infinite vector space $V$ over $\mathbb{F}_P$, then $(V, <)$ has the independence property (IP). (Note that there are no natural compatibility conditions one could even ask for here.) The independence property is a measure of complexity of structures. This paper is part of a general project: given a structure $M$ without the independence property (NIP), classify the NIP expansions of $M$. -On a similar note, it is an open problem (that I heard from Erik Walsberg) whether there is a dense linear ordering $<_*$ of the natural numbers such that $(\mathbb{N}, <, <_*)$ has NIP. (Here $<$ is the usual ordering of the natural numbers.)<|endoftext|> -TITLE: Generalize Wu formula to integral cohomology classes -QUESTION [9 upvotes]: For $\mathbb{Z}_2$ cohomology classes, we have a very useful Wu formula: -In $d$-dimensional manifold and for a $n$-cocycle in $x_n \in H^n(M^d; \mathbb{Z}_2)$, we have $Sq^{d-n}(x_n)=u_{d-n}\cup x_n$, where $u_m$ is the $m^{th}$ Wu class, and $Sq^k$ is the Steenrod square. -I wonder if there is a similar formula for integral cohomology classes? For example, if there is a cohomological operation $P^4$, such that -$P^4(y_{d-4}) = p_1 \cup y_{d-4}$ where $p_1$ is the first Pontryagin class and $y_n \in H^n(M^d; \mathbb{Z})$? - -REPLY [5 votes]: The Wu classes are, or at least can be, defined by $Sq^{d-n}(x_n) = u_{d-n}\cup x_n$. The interesting part is not that the Wu classes exist, it is that they are related to Stiefel--Whitney classes, which is not at all a priori. Indeed, this implies that SW classes are homotopy-invariant objects (since $Sq^i$ are), which is not at all god-given. Most characteristic classes are not homotopy-invariant, and so will not participate in a Wu-type formula in terms of cohomology operations.<|endoftext|> -TITLE: Signature of a quadratic form -QUESTION [13 upvotes]: This may be a really dumb question, but here goes: is there any algorithm to compute the signature of a quadratic form (or a symmetric matrix, if you prefer) more efficient (asymptotically or otherwise) than actually computing the eigenvalues? - -REPLY [10 votes]: Gauss reduction gives you the answer. It writes, quite fast, the quadratic form $q$ as a sum -$$\sum_ja_j\ell_j(x)^2$$ -where the $\ell_j$'s are independent linear forms. The number of squares gives you the rank of $q$. The signs of the coefficients $a_j$ gives you the signature. I teach that in my undergraduate course in Algebra. -Remark that you cannot calculate the eigenvalues, at least in close form, because this is computing the roots of quite a general polynomial, and this is impossible in dimension $\ge5$.<|endoftext|> -TITLE: Arithmetic progressions in Van der Waerden's theorem -QUESTION [6 upvotes]: Recall that a syndetic subset of the integers is any $S\subseteq \mathbb{Z}$ with bounded gaps, i.e. there is some $k< \omega$ so that consecutive members of $S$ have distance at most $k$. One way to state Van der Waerden's theorem is that every syndetic set contains arbitrarily long arithmetic progressions (which I will call APs). -Call an AP of the form $\{a, a+d,...,a+(\ell-1)d\}$ a $d$-AP of length $\ell$. Now fix a syndetic $S\subseteq \mathbb{Z}$. For each $\ell < \omega$, let $d_\ell$ denote the least $d$ so that $S$ contains a $d$-AP of length $\ell$. My question is about the function $\ell\to d_\ell$. Is there a syndetic $S\subseteq \mathbb{Z}$ where $d_\ell\to \infty$? - -REPLY [6 votes]: The answer is yes, in fact the sequence $S\subset \mathbb{N}$ indexed by Thue-Morse sequence works, see this MO question. Note that this sequence is syndetic because it contains one of $\{2n, 2n+1\}$ for every $n$. You can extend to the negatives by reflecting and trivially only lose a factor of two on $\ell$.<|endoftext|> -TITLE: Is there an intrinsic definition of weak equivalence in Cat or RelCat? -QUESTION [7 upvotes]: It's known that Cat with the Thomason model structure serves as a model for $\infty\mathrm{Grpd}$, and that RelCat has a corresponding model structure that serves as a model for $\infty\mathrm{Cat}$. (with Cat embedding in RelCat as those relative categories where everything is a weak equivalence) -In Barwick and Kan's paper, they define a homotopy relation generated by the natural transformations whose components are weak equivalences. -While appealing, this is inadequate; if I understand anything at all, the infinite zigzag category $Z$ depicted as -$$ \ldots \leftarrow \bullet \to \bullet -\leftarrow \bullet \to \bullet \leftarrow \bullet \to \ldots $$ -where all arrows are weak equivalences is supposed to have geometric realization homeomorphic to $\mathbb{R}$, and thus have the homotopy type of a point... but $Z$ is not homotopy equivalent to the terminal category, since any homotopy from $1_Z$ can only take a fixed, finite number of steps, but arbitrarily large steps are needed to connect every object to a specified one. -Every exposition on the topic I have seen simply punts the question over to simplicial sets or bisimplicial sets or similar: that whether or not a map is a weak equivalence is determined by the map it induces on nerves. -Is there a description of weak equivalences that can be phrased entirely within Cat or RelCat without taking a detour through simplicial sets or topological spaces? - -REPLY [2 votes]: There is the following characterization. -Homotopy Limit Functors on Model Categories and Homotopical Categories (DKHS) gives, for any saturated relative category C with objects $x$ and $y$, a category $\mathbf{Gr}(\mathbf{C})^\mathbf{T}(x,y)$ of zigzags from $x$ to $y$. These categories assemble into a strict $2$-category $\mathbf{Gr}(\mathbf{C})^\mathbf{T}$ called its Grothendieck construction. -DKHS show that the Grothendieck construction is the correct $\mathbf{Cat}_{\mathrm{Thomason}}$-enriched model for C in the sense that taking nerves of the hom-categories of $\mathbf{Gr}(\mathbf{C})^\mathbf{T}$ gives a simplicially enriched category weakly equivalent to the Hammock localization $L^H\mathbf{C}$. -Thus, we have - -Let $F : \mathbf{C} \to \mathbf{D}$ be a functor between saturated relative categories. It is a weak equivalence if and only if: - -$\mathbf{Ho}(F) : \mathbf{Ho}(\mathbf{C}) \to \mathbf{Ho}(\mathbf{D})$ is an equivalence of ordinary categories -For every pair of objects $x,y$, it induces a weak equivalence $\mathbf{Gr}(\mathbf{C})^\mathbf{T}(x,y) \to \mathbf{Gr}(\mathbf{D})^\mathbf{T}(F(x), F(y))$ - - -Furthermore, on $\mathbf{Cat}_{\mathrm{Thomason}}$ (viewed as the subcategory of RelCat of categores where every arrow is a weak equivalence), we can carry out the definition of weak equivalence as isomorphisms on homotopy groups. - -Let $F : \mathbf{C} \to \mathbf{D}$ be a functor between categories. In the Thomason model structure, it is - -A $0$-equivalence if it induces a bijection on connected components -An $n$-equivalence if, for every pair of objects $x,y$, it induces $(n-1)$-equivalences $\mathbf{Gr}(\mathbf{C})^\mathbf{T}(x,y) \to \mathbf{Gr}(\mathbf{D})^\mathbf{T}(F(x), F(y))$ -A weak equivalence if it is an $n$-equivalence for every $n$ - - -If $x,y$ are in the same connected component, then $\mathbf{Gr}(\mathbf{C})^\mathbf{T}(x,x) \simeq \mathbf{Gr}(\mathbf{C})^\mathbf{T}(x,y)$, so the middle condition only needs checked on one endomorphism category per connected component.<|endoftext|> -TITLE: The connection between the Weil conjectures and Ramanujan's conjecture -QUESTION [6 upvotes]: I'm writing an essay about Ramanujan's conjecture and have some questions: -1 How is Ramanujan's conjecture connected with the Weil conjectures? -2 How could Ramanujan's conjecture be assumed true or deduced when Deligne proved the Weil conjecture? -3 How Is Ramanujan's conjecture connected or equivalent to the Riemann Hypothesis? -4 Are there any good articles on these subjects? -Thanks ann regards. - -REPLY [9 votes]: Basically, the coefficients of an holomorphic cusp form are related to the number of points on a certain smooth projective variety over $\mathbb{F}_p$, and the Weil-Riemann hypothesis gives the neccesary error term for the number of points. This is the content of - -Pierre Deligne, "Formes modulaires et représentations l-adiques" (1969) - -The Riemann hypothesis over finite fields was of course proved later by Deligne in his "Weil I" paper. -I'm not sure what the best place to learn these things is (I mean the Ramanujan conjecture, conditionally on the Weil conjecture; for the later there are plenty of resources, see here), other than Deligne's paper, but can find some information here on MO, - -Deligne's proof of Ramanujan's conjecture -Reference request for a proof of Ramanujan's tau conjecture - -As for you third question, there seems to be no relationship between Ramanujan-type conjectures and the location of zeros inside the critical strip, whether you are considering the L-funcion attached to a modular form, or any other L-function of arithmetic or geometric origin.<|endoftext|> -TITLE: Prime square offsets: Why is +7 more frequent than -7? -QUESTION [25 upvotes]: For a prime $p$, define $\delta(p)$ to be the smallest offset $d$ -from which $p$ differs from a square: -$p = r^2 \pm d$, for $d,r \in \mathbb{N}$. -For example, -\begin{eqnarray} -\delta(151) & = & +7 \;:\; 151 = 12^2 + 7 \\ -\delta(191) & = & -5 \;:\; 191 = 14^2 - 5 \\ -\delta(2748971) & = & +7 \;:\; 2748971= 1658^2 + 7 -\end{eqnarray} -For a particular $\delta=d$ value, define $\Delta(n,d)$ to be the number -of primes $p$ at most $n$ with $\delta(p) = +d$, minus the number with $\delta(p) = -d$. In other words, $\Delta$ records the cumulative prevalence of $+d$ offsets over $-d$. For example, $\Delta(139,5)=-2$ because there are two more $-5$'s than $+5$'s up to -$n=139$: -$$ -\delta(31)=-5 \;,\; \delta(41)=+5 \;,\; \delta(59)=-5\;,\; \delta(139) =-5 \;. -$$ -The figure below shows $\Delta(p,5)$ and $\Delta(p,7)$ out to the $200000$-th -prime $2750159$. -The offset $+7$ occurs $161$ times more than the offset $-7$, and -the reverse occurs for $|\delta|=5$: $-5$ is more common than $+5$. - -      - - - -Q. Is there a simple explanation for the different behaviors of offsets - $5$ and $7$? - -Obviously the question can be generalized to explaining the growth for -any $|\delta|$. -I previously asked a version of this question on MSE, -using somewhat different notation conventions -and with less focused questions. - -REPLY [42 votes]: Consider $n^2+7$ and $n^2-7$ modulo $3$. If these are to be prime they must be non-zero $\pmod 3$, and in the first case $n$ can be anything mod $3$, whereas in the second case $n$ must be $0 \pmod 3$. If you consider $n^2+5$ and $n^2-5$, you see that the pattern reverses. This is already a huge bias for one offset to be preferred over the other. -The Hardy-Littlewood conjectures make this precise. One expects that (for a number $k$ not minus a square) the number of primes of the form $n^2+k$ with $n\le N$ (say) is -$$ -\sim \frac 12 \prod_{p\ge 3} \Big(1 -\frac{(\frac{-k}{p})}{p-1} \Big) \frac{N}{\log N}, -$$ -where in the numerator of the product above is the Legendre symbol. The constants in front of the $N/\log N$ explain these biases. -You don't have to worry about a smaller offset than $\pm 5$ or $\pm 7$, since an application of the sieve shows that the numbers $n^2+a$ and $n^2+b$ (for fixed $a$, $b$) are both prime only $\le CN/(\log N)^2$ of the time for a constant $C$. - -REPLY [9 votes]: Modulo 6 the squares are 0,1,4,3,4,1 and the squares+7 (or -5) can only be 1,2,5,4,5,2, of which 3/6 can at all be prime. The squares-7 (or +5) are 5,0,3,2,3,0 of which only 1/6 can be prime. Obviously this not a proof, but there is clearly no first order surprise in the observed offsets. -UPDATE I checked offsets modulo different numbers (such as (mod 2*3*5*7*11*13*17*...)) and I found that mod(3*7*11*19*23) we get 64638/25806 for +7/-7 and 21945/73899 for +5/-5, while primes 2, 5, 13, 17, 29 don't affect the ratio. I imagine that exploring more primes will further refine the ratios, with primes $\equiv 3$ (mod 4) affecting them and others not. WHAT IS GOING ON HERE?<|endoftext|> -TITLE: Proof of the stable homeomorphism conjecture -QUESTION [6 upvotes]: I would like to get a demonstration of the stable homeomorphism conjecture (SHC$_n$), stating that any orientation-preserving homeomorphism $\mathbb{R}^n \rightarrow \mathbb{R}^n$ is stable in dimension $n$, for $n \neq 4$. -According to several sources, for example the upvoted answer to the post connectivity of the group of orientation-preserving homeomorphisms of the sphere , Robion Kirby proved it for $n > 4$ in the following paper http://www.maths.ed.ac.uk/~aar/papers/kirby.pdf . -("After that came R. Kirby (and L. Siebenmann) in 1968, who proved (using the results of surgery by Wall et al), that SHC$_n$ (hence AC$_n$) is true in all dimensions $n>4$.") -People often say in other papers that Kirby proves in fact in his paper not the SHC$_n$ for $n > 4$ but rather the annulus conjecture (AC$_n$) which would be equivalent to SHC$_n$, the equivalence being proved hypothetically in the paper of Brown and Gluck about "Stable structures on manifolds" (on JSTOR : https://www.jstor.org/stable/1970482?seq=1#fndtn-page_scan_tab_contents). -My two questions are the following : How can you justify precisely that : -1) Kirby proved AC$_n$ for $n > 4$ in his paper? -2) Brown and Gluck proved that AC$_n \Leftrightarrow$ SHC$_n$ for $n > 4$? -(I would like precise theorems in the papers and why they prove the results because after reading them a few times, I can't answer these two questions.) -PS : Nonetheless, if you have/know other research paths to prove SHC$_n$ for $n \neq 4$ don't hesitate, since it's my goal for all of this! - -REPLY [11 votes]: For your first question, yes Kirby did prove the n-dimensional annulus conjecture AC$_n$ for $n>4$ in the 1969 Annals paper that you cite. The only reason this might not be clear from reading the paper is that the original version of the paper proved AC$_n$ only assuming another statement that he calls HT$_n$ (namely the assertion that the PL structure on the $n$-torus is unique up to PL homeomorphism). But then Kirby added a note that appears in the published version of the paper, a note dated 1 December 1968, giving the argument that proves AC$_n$ for $n>4$ unconditionally. This argument uses a couple new results that had been proved by other people since the original version of the paper. Kirby's paper contains also a second added note dated 15 April 1969 which says that Siebenmann had just found an argument showing that HT$_n$ is in fact false for $n>4$. However, the first added note explains that the full strength of HT$_n$ is not needed, and that it suffices to assume the weaker version of HT$_n$ that asserts only that an arbitrary PL structure on the $n$-torus becomes standard after lifting it to a suitable finite-sheeted covering space. This weaker version had just been proved (for $n>4$) by Wall and Hsiang-Shaneson. -Actually Kirby does not deal directly with the annulus conjecture but rather with the stable homeomophism conjecture SHC$_n$. Thus the paper proves that the weaker version of HT$_n$ implies SHC$_n$ when $n>4$. Then Kirby quotes a 1964 result of Brown and Gluck that SHC$_n$ implies AC$_n$. -For your second question, Kirby states that Brown and Gluck also showed that AC$_k$ for all $k\leq n$ implies SHC$_n$. This of course is not the same as saying that AC$_n$ implies SHC$_n$ for a fixed $n$. However, Kirby's paper does not seem to use anything about AC implying SHC since he deals directly with SHC$_n$. This is fortunate since AC$_n$ for $n=4$ was only proved a number of years later by Quinn. -One other small comment: Kirby's note of 1 December 1968 contains a typographical error that might cause some confusion. He says that SHC$_n$ is a classical result for $n\neq 3$ when he must have meant $n\leq 3$.<|endoftext|> -TITLE: number of characters needed to separate conjugacy classes -QUESTION [11 upvotes]: It is very well-known that, if $G$ is a finite group, then the (complex) characters of $G$ separate the conjugacy classes; that is, if $g, h \in G$ are not conjugate, then there exists a character $\chi$ such that $\chi(g) \ne \chi(h)$. However, I have been wondering about the minimal number of characters needed to separate all the classes. -Consider the (very basic) example $G=C_n$, a cyclic group of order $n$. Then it is possible to find a single character $\chi$ which is injective on $G$, so the answer is $1$ in this case. -Looking at a few character tables using GAP, it seems the answer is often 1. -Is there any literature on this? Any reason why this number could be much easier to determine than I think? Is it always 1 ? I may very well have overlooked something simple. -Note that the answer would be different if you only allowed irreducible characters (then for $C_2\times C_2$, it is 2); it's also different if you allow virtual characters, I guess. -Thanks! -Pierre - -REPLY [2 votes]: Obviously Peter Müller's argument is perfectly fine. Here is a different way of doing it. -This algorithm yields a character that separates the conjugacy classes. It is terrible to compute with. -Index a set of representatives of the conjugacy classes $\{g_i\}$. Associate to every character $\chi$, the set of ordered pairs -$$D(\chi) = \left\{ (g_i,g_j): g_i \neq g_j, { \mathrm {but } }\ \chi(g_i) = \chi(g_j)\right\}.$$ -Now we show that if $D(\chi)$ is nonempty, there exists an irreducible character $\chi'$ and a positive integer $N$ (depending on $\chi$ and $\chi'$) such that $D(N\chi + \chi')$ is strictly contained in $D(\chi)$. -To see this, set $s $ to be the infimum of $|\chi(g_i) - \chi(g_j)|$ over all pairs of distinct representatives $(g_i,g_j)$ that are not in $D(\chi)$ -(this is an infimum over a finite set of nonzero numbers). Pick $(g,h) \in D(\chi)$; there exists an irreducible $\chi'$ such that $\chi'(g) \neq \chi'(h)$. Select $N$ a positive integer so large that $Ns > 2\chi'(1)$, and set $\psi = N\chi + \chi'$, a character. -If $\chi(g) \neq \chi(g')$, then $\psi (g) -\psi(g') = \left( N(\chi(g) - \chi(g')\right) + (\chi' (g) - \chi(g'))$; the first parenthesized term has absolute value at least $Ns$, while the second one has absolute value at most $2 \chi'(1)$. Hence $D(\psi)$ is contained in $D(\chi)$. -On the other hand, $(g,h)$ is not in $D(\psi)$, since $\psi(g) - \psi (h) = -\chi'(g) - \chi'(h)$. -This process (begin with any nontrivial character, and continue) must terminate, resulting in a character $\rho$ with empty $D(\rho)$, which is exactly what we want.<|endoftext|> -TITLE: Zeros of polynomials as a finite union of manifolds -QUESTION [5 upvotes]: Consider a polynomial in $d$ variables, $p:\mathbb{R}^{d}\rightarrow\mathbb{R}$. Denote by $\mathcal{C}$ its set of zeros, i.e. -$$\mathcal{C}=\{x\in\mathbb{R}^{d}\ |\ p(x)=0\}.$$ - -Q. Is it possible to find finitely many (not necessarily disjoint) manifolds $M_{1},\dots, M_{n}\subset\mathbb{R}^{d}$, with possibly different dimensions, such that - $$\mathcal{C}=\bigcup_{k=1}^{n}M_{k}?$$ - -My question arises in the context of degenerate real matrices, namely, the case where $\mathcal{C}$ consists on the set of symmetric real matrices with at least one repeated eigenvalue (in this context, $p(x)$ is the discriminant of the symmetric real matrix $x$). Any suggestion on how to approach this problem or a reference related to the problem will be greatly appreciated. - -REPLY [10 votes]: The answer is yes. -Your set $\mathcal{C}$ is by definition a (semi)algebraic subset of $\mathbb{R}^d$, and any such a subset has a Whitney stratification with finitely many (semi)algebraic strata. -See A. Dimca, Singularities and Topology of Hypersurfaces, ZBL07535700, -Chapter 1, Corollary 1.12 for more details. -Edit. It is also true that any closed analytic (or subanalytc) set of an analytic manifold admits a Whitney stratification whose strata are analytic submanifolds. In the complex setting, this is called a complex Whitney stratification. See -M. Tib$\check{\rm u}$ar, Polynomials and vanishing cycles, Cambridge Tracts in Mathematics 170 (2007), ZBL1126.32026, page 219 and the references given therein.<|endoftext|> -TITLE: Mathematical physics applications in present-day image processing -QUESTION [6 upvotes]: During the past few years several important areas of image processing and image classification or generation became dominated by convolutional neural networks. -I'm interested if there are any methods coming from mathematical physics context (like methods designed for solving ill-posed problems, spectral analysis, image deblurring and deringing), that outperform neural network-based approaches in 2017 for some common computer vision or image processing problem. Or methods that don't have any neural network-based rivals. Maybe in the field of biomedical image analysis (just a guess)? -The deeper and more specific the answer is, the better. - -REPLY [6 votes]: Neural networks are best-in-breed at solving a very specific kind of problem: computing a function $f \colon X \to Y$ given the values of $f$ on a large but finite subset $A \subseteq X$. Typically $Y$ (the space of "labels") is finite and small relative to $A$. Concrete example: $X$ is the space of images, $Y$ is a set of labels for images (e.g. "contains a car" or "is a picture of a sunset"), and $A$ is a large set of human-annotated images (the training data). -As it turns out, neural networks can learn $f$ so well that they can produce points in $f^{-1}(y)$'s near a prescribed $x \in X$ - that's why they are good at image / language generation. -This is certainly an important and broad class of problems (particularly for applications in industry), and in practice neural networks don't really have any serious rivals. But there are lots of other image processing problems for which they aren't really appropriate - compression, recovery, noise reduction, etc. I don't have a lot of expertise to draw from, but for far as I am aware the standard techniques from physics and engineering are close to the state of the art. -I'll conclude by remarking that there are some persuasive arguments that the practical effectiveness of neural networks is explained by principles in physics. There is theory which says that if $X$ is large (e.g. all images or all sequences of characters in an alphabet) then no machine learning algorithm can perform better than the most naive ones. So to understand why neural networks are so high-performing one must use something about the structure of the underlying data sources (photographs or human language) to restrict $X$. Here are some recent papers in this direction: -https://arxiv.org/abs/1608.08225 -https://arxiv.org/abs/1410.3831<|endoftext|> -TITLE: Negatively curved manifolds with many totally geodesic submanifolds -QUESTION [7 upvotes]: I'm curious about the following question and have not been able to find any literature on the topic: Suppose that $M$ is a closed negatively curved Riemannian manifold with a "large" quantity of totally geodesic submanifolds. Is there anything one can say about such a manifold? -To be more precise, consider such an $M$ (diffeomorphic to a complex hyperbolic manifold) carrying an almost complex structure $J$ which looks like a complex hyperbolic manifold in the following sense: for each $x \in M$ and each $v \in T_{x}M$ there is a totally geodesic surface $S$ tangent to the plane spanned by $v$ and $J(v)$ at $x$. It seems to be a natural question to ask if $M$ is itself complex hyperbolic (which would follow for example if $M$ was a Kahler manifold). I have no intuition as to whether this should be an easy or difficult problem so I'm curious (and thankful in advance) if anyone has some guidance/references. -Edit: I have made my question in the second paragraph more explicit below, since I think the original version may be ambiguous, -Let $X$ be a compact quotient of complex hyperbolic space $\mathbb{C} H^{n}$ ($n \geq 2$). Let $M$ be a Riemannian manifold obtained by taking a small smooth perturbation of the symmetric metric on $X$ (same underlying space, different metrics). Now suppose we are given an almost complex structure $J$ - not necessarily the standard one - such that for each $x \in M$ and each $v \in T_{x}M$ there is a totally geodesic subsurface $S$ of $M$ which is tangent to the plane spanned by $v$ and $J(v)$ inside of $T_{x}M$. Is $M$ isometric (up to rescaling) to $X$? -As far as I can tell from searching, Riemannian manifolds typically have very few higher dimensional totally geodesic submanifolds so the condition being imposed above is quite strong. -Edit 2: As pointed out in the comments, I actually want to assume these subsurfaces are also $J$-holomorphic, i.e., $TS \subset TM$ is preserved by $J$ for each surface $S$. - -REPLY [7 votes]: Here is a partial answer: Let $(M^{2n},g,J)$ be a compact Riemannian manifold endowed with a $g$-orthogonal almost complex structure $J$ with the property that, for every nonzero $v\in T_pM$ there exists a $J$-holomorphic curve $C\subset M$ passing through $p$ with tangent space spanned by $v$ and $Jv$ that is totally geodesic. Moreover, suppose that the scalar curvature of $g$ is negative. Then, up to a constant scale factor, $(M,g,J)$ is isomorphic to a compact quotient of $\mathbb{CH}^n$. -Note that the only additional hypotheses that I have added to the OP's problem are that the almost complex structure $J$ be $g$-orthogonal and that the totally geodesic surfaces should be $J$-holomorphic. -The argument goes as follows: -First, one shows by a local calculation that, if $(M^{2n},g,J)$ has the desired property of having sufficiently many totally $g$-geodesic surfaces that are $J$-holomorphic curves that there is one tangent to each $J$-complex line in $TM$, then $(M^{2n},g,J)$ must be nearly Kähler. -Second, a complete, simply-connected nearly Kähler manifold is known to be the product of a Kahler manifold and a strict nearly Kähler manifold. (See Proposition 2.1 of (Nearly Kähler geometry and Riemannian foliations, by Paul-Andi Nagy, arXiv:math/0203038v1). Meanwhile, by a theorem of Paul-Andi Nagy (On nearly Kähler geometry, arXiv:math/0110065), if $(N^{2n},g,J)$ is a complete strict nearly Kähler, then the scalar curvature of $g$ is constant and strictly positive. Since we are assuming that our given compact $(M^{2n},g,J)$ has negative curvature, it follows that its simply connected cover cannot have any strict nearly Kähler factor. Hence, $(M^{2n},g,J)$ must be Kähler. -Third, once one is in the Kähler category, it is easy to show that the existence of totally geodesic $J$-holomorphic curves tangent to every complex line implies that the metric has constant holomorphic sectional curvature, i.e., it is a complex 'space form'. Since the curvature is negative, it follows that, up to scale, it must be isometric to the standard Kähler structure on $\mathbb{CH}^n$. -Note: Without the negative curvature assumption, there do exist strictly nearly Kähler examples. E.g., the $6$-sphere with its standard $\mathrm{G}_2$-invariant metric and (non-integrable) almost complex structure has a totally geodesic $2$-sphere tangent to any given complex line in its tangent space.<|endoftext|> -TITLE: A question about the logarithmic complex and Morgan's paper -QUESTION [5 upvotes]: I have a question about the Morgan's paper ``The algebraic topology of smooth complex varieties''. Let $\Bbbk=\mathbb{C}$. Given a smooth non singular variety $X$ with normal crossing divisor $D$, the sheaf logarithmic forms $\mathcal{A}_{DR}^{\bullet}(\log(D))$ is defined as follows: for an open set $U$, a form in $\mathcal{A}_{DR}^{\bullet}(\operatorname{Log}(D))|_{U}$ can be written as -$w=\sum w^{J}\frac{dz_{1}}{z_{1}}\cdots \frac{dz_{j}}{z_{j}}$ -where $w^{J}$ is a smooth form on $U$. Let ${A}_{DR}^{\bullet}(\operatorname{Log}(D))$ be the differential graded algebra denoting the global sections of $\mathcal{A}^{\bullet}_{DR}(\operatorname{Log}(D))$. Let $A_{DR}^{\bullet}(X-D)$ be differential graded algebra of the ordinary complex smooth differential forms on $X-D$. Here my question: at page 154 theorem 3.3 it is written that -$(*)$"the inclusion ${A}_{DR}^{\bullet}(log(D))\hookrightarrow A_{DR}^{\bullet}(X-D) $ induces an isomorphism in cohomology" -I know that this statement is true in sheaf terms, i.e. the inclusion induces an isomorphism at the level of the Hypercohomology of the two sheaves, but in seems to me that the statements here are really about differential graded algebras. -My problem is the following: -Consider a torus $T=\mathbb{C}/\mathbb{Z}^{2}$ with coordinate $z=r+is$ and the obvious group action given by translations. Consider ${A}_{DR}^{\bullet}(T)$. Then the first cohomology group is generated by $dr$ and $ds$ and the second cohomology group is generated by $dr \, ds$. Let $D=0\subset T$ considered as a normal crossing divisor. The cohomology of ${A}_{DR}^{\bullet}(\log(D))$ is generated in degree $1$ by $dr$ and $ds$ and in degree $2$ we should have that $dr \, ds$ is exact. But $A_{DR}(\log(D))^{0}=A^{0}_{DR}(T)$ by definition and hence $drds$ is not exact. Where is my mistake? -Here is the link for the paper: http://www.numdam.org/item/PMIHES_1978__48__137_0 - -REPLY [5 votes]: For the first part of the question: what is well-known is that the cohomology of the open variety $X\setminus D$ can be computed as the hypercohomology of the holomorphic logarithmic complex $\Omega_X^\bullet(\log D)$. This is well explained in Voisin's book Hodge Theory and Complex Algebraic Geometry, corollary 8.19. But the above sheaf $\mathcal{A}^\bullet(\log D)$ is an acyclic resolution of $\Omega_X^\bullet(\log D)$ because it is a sheaf of modules overs $\mathcal{C}^\infty$ functions on $X$. This is stated in Deligne Hodge II, 3.2.3 b. Hence the hypercohomology of $\Omega_X^\bullet(\log D)$ can be computed as the cohomology of the global sections of $\mathcal{A}^\bullet(\log D)$. Again in your assertion $(*)$ that's because your two sheaves are acyclic and compute both the cohomology of $X\setminus D$ that the map is an isomorphism at the level of cohomology of the global sections. This is a general fact about holomogical algebra, see also Voisin's book corollary 8.9 and proposition 8.12. -For your second part. Your computation of $A^\bullet(\log D)$ is correct and agrees with the cohomology of the open manifold $T\setminus 0$ (which you seem to confuse with the cohomology of $T$), in particular $H^2(T \setminus 0)=0$. One way to see it is that $T\setminus 0$ deformation retracts onto a wedge sum of two circles; or invoke Poincaré duality for oriented non-compact manifolds, which implies that the top cohomology group is zero.<|endoftext|> -TITLE: Sequences with 3 letters -QUESTION [19 upvotes]: For a positive integer $n$ I would like to construct long sequences consisting of 0, 1 and 2's such that for any two subsequences consisting of $n$ consecutive elements the number of 0's , 1's or 2's are different. -Since there are $\binom{n+2}{2}$ different triples of nonnegative integers summing up to $n$ such a sequence has length at most $\binom{n+2}{2}+n-1$. For $n\leq 3$ it is possible to construct such sequences of this length: -$$\begin{eqnarray} -n=1&:& 012\\ -n=2&:& 0112200\\ -n=3&:& 011122200012 -\end{eqnarray}$$ -For $n\geq 4$ however I am unable to construct such sequences nor proving that there does not exist one. For given $n$ what is the maximal length of such a sequence? How can we construct it? I would be grateful for both the cases where $n$ is small as well as the asymptotics $n\rightarrow \infty $. - -REPLY [5 votes]: Here is a sketch of a proof that there are no such complete sequences for $n>4$. -Consider the graph where the vertices are the triples of nonnegative integers that sum to $n$ and construct -an edge between two vertices when one can get from one to the other by incrementing one element and decrementing another. -This graph looks like a triangle tiled by smaller triangles, in particular it is planar, so two paths cannot cross without intersecting. -A solution of the original problem induces a hamiltonian path on this graph. -Now look at the extreme vertices of the triangle $(0,0,n)$, $(0,n,0)$, and $(n,0,0)$. In the original problem, a path of length n that starts -or ends at one of these points must end or start at the opposite edge. As the induced path cannot cross itself, if the three vertices are all -on a path, they must occur adjacent to each other, so the path looks like wlog $*0^n1^n2^n*$. This means that two sides of the triangle are occupied so -neither the prefix nor the suffix can be longer than $(n-1)$, and if $n>4$ there are more than $2n-2$ elements that don't lie on the two edges. -ADDED: -I have translated this problem into a somewhat annoying javascript toy. This makes it somewhat easier to see the structure in Robert Israel's solutions. I am fairly convinced (but still cannot prove) that the size of the optimal solution grows quadratically, and I would not be surprised if the optimal solution covered all but a linear in $n$ number of combinations.<|endoftext|> -TITLE: For an intersecting family of $m$ sets there are at least $2m$ sets that are contained in at least one of them -QUESTION [9 upvotes]: Let $F$ be a finite family of non-empty sets such that any two of them intersect. Consider the set $F'$ consisting of all sets that are a subset of at least one element of $F$. Prove $|F'|\geq 2|F|$. -I conjectured this yesterday but have been completely unable to prove it. I tried doing things like inclusion-exclusion on the poset of the subsets but I was not able to solve it, I also tried things similar to Dilworth's theorem. -Since this problem is so natural I think that it should have been attempted before, does anyone have any references? - -REPLY [17 votes]: Assume that all the sets are contained in some master set of $n$ elements. Let $G$ be the set of super sets of at least one element of $F$. We apply the Harris-Kleitman inequality: -Lemma (Harris-Kleitman): For $F'$ a downward-closed and $G$ an upward-closed collection of subsets on a set with $n$ elements, $$|F'\cap G|\leq \frac{|F'| |G| }{2^n}.$$ -Proof. Induction on $n$. Let $G_0$ be the set of elements of $G$ that do not contain the first element, $G_1$ the set of elements that contain the first element, and so on. Then $$ -\begin{aligned} -|F'\cap G|&= |F'_0 \cap G_0|+|F'_1 \cap G_1| \\ -&\leq \frac{ |F'_0||G_0|}{2^{n-1}}+ \frac {|F'_1||G_1|}{2^{n-1}}\\ -&= \frac{ (|F'_0|-|F'_1|) (|G_0|-|G_1|)}{2^n}+ \frac{(|F'_0|+|F'_1|)(|G_0|+|G_1|)}{2^n} \\ -&\leq 0+ \frac{|F'||G|}{2^n} -\end{aligned} -$$ -QED -Now by the assumption on $F$, for any element of $G$, its complement is not in $G$, so $|G|\leq 2^{n-1}$, and thus -$$|F| \leq |F' \cap G| \leq \frac{|F'||G|}{2^n} \leq \frac{|F'|}{2}$$ as desired.<|endoftext|> -TITLE: Minimizing KL divergence: the asymmetry, when will the solution be the same? -QUESTION [6 upvotes]: The KL divergence between two distribution $p$ and $q$ is defined as -$$ -D( q \| p)\int q(x)\log \frac{q(x)}{p(x)} dx -$$ -and is known to be asymmetry: $D(q\|p)\neq D(p\|q)$. -If we fix $p$ and try to find a distribution $q$ among a class $E$ that minimize the KL distance, it is also known that minimizing $D( q \| p)$ will be different from minimizing $D( p \| q)$, e.g., https://benmoran.wordpress.com/2012/07/14/kullback-leibler-divergence-asymmetry/. -It is not clear which one to be optimized for a better approximation, although in many application we minimize $D(q\|p)$. -My question is that, when will the solution be the same -$$ -\underset{q\in E}{\operatorname{argmin}} D(q\|p) = \underset{q\in E}{\operatorname{argmin}} D(p\|q)? -$$ -For instance, if we take $E$ as the class of all Gaussian distribution, is there a condition on $p$ so that minimizing these two will lead to the same minimizer? - -REPLY [5 votes]: I don't have a definite answer, but here is something to continue with: -Formulate the optimization problems with constraints as -$$ -\mathrm{argmin}_{F(q)=0} D(q || p),\qquad \mathrm{argmin}_{F(q)=0} D(p||q) -$$ -and form the respective Lagrange functionals. Using that the derivatives of $D$ w.r.t. to the first and second components are, respectively, -$$ -\nabla_1 D(q||p) = \log(\tfrac{q}{p})+1\quad\text{and}\quad \nabla_2 D(p||q) = \tfrac{q}{p} -$$ you see that necessary conditions for optima $q^*$ and $q^{**}$, respectively, are -$$ -\log(\tfrac{q^*}{p})+1 + \nabla F(q^*)\lambda = 0\quad\text{and}\quad \tfrac{q^{**}}{p} + \nabla F(q^{**})\lambda = 0. -$$ -I would not expect that $q^*$ and $q^{**}$ are equal for any non-trivial constraint… -On the positive side, $\nabla_1 D(q||p)$ and $\nabla_2 D(q||p)$ agree up to first order at $p=q$, i.e. $$\nabla_1 D(q||p) = \nabla_2 D(q||p) + \mathcal{O}(\tfrac{q}{p})$$.<|endoftext|> -TITLE: Largest inscribed triangle with a given vertex -QUESTION [7 upvotes]: It is known from Blaschke and later Sas that any convex region $P$ with area $1$ admits an inscribed triangle with area at least $\frac{3\sqrt{3}}{4\pi}$. What if we require the triangle to have a given boundary point of P as vertex? The optimal constant should be between $\frac{1}{4}$ and $\frac{1}{3}$, but I can't tighten these bounds by too much. - -REPLY [5 votes]: The answer is indeed $1/\pi$, equality achieved in the case of $P$ a semicircle and the point its center. -For an arbitrary convex region $P$ and a point $X$ on its boundary, consider the convex hull of $P$ and $2X - P$, and call it $S$. Then any $S$ is clearly symmetric with respect to $X$. Any triangle in $P$ having $X$ as a vertex with area $a$, induces a parallelogram in $S$ with center $X$ and area $4a$. Also we have the trivial inequality $\operatorname{area}(S) \ge 2 \operatorname{area}(P)$. So it suffices to show that, given a centrally symmetric convex region $S$ with respect to $O$, there exists a parallelogram in $S$ with center $O$ and area at least $(2/\pi) \operatorname{area}(S)$. -This can be proved by essentially the same technique used by Sas. After scaling and rotation, let $S$ be contained in the unit disc, with $(1, 0), (-1, 0) \in S$. Parametrize the boundary of $S$ by $\pm (\cos \theta, f(\theta) \sin \theta)$, where $f : (0, \pi) \to [0, 1]$. Then the area of $S$ is -$$ \operatorname{area}(S) = 2 \int_{0}^{\pi} f(\theta) \sin^2 \theta.$$ -The area of the parallelogram with vertices $\pm (\cos \theta, f(\theta) \sin \theta)$ and $\pm (-\sin \theta, f(\theta + \pi/2) \cos \theta)$ is -$$\operatorname{area}(\mathrm{Par}_\theta) = 2 \Bigl( f(\theta) \sin^2 \theta + f\Bigl( \theta + \frac{\pi}{2} \Bigr) \sin^2 \Bigl( \theta + \frac{\pi}{2} \Bigr) \Bigr).$$ -Averaging this area over $0 \le \theta \le \pi / 2$ shows that there exists some $\theta$ such that -$$\operatorname{area}(\mathrm{Par}_\theta) \ge \frac{2}{\pi} \operatorname{area}(S). $$<|endoftext|> -TITLE: What is the consistency strength of "Every set is a member of a transitive model"? -QUESTION [10 upvotes]: Recall that $\kappa$ is a worldly cardinal if $V_\kappa$ is a model of $\sf ZFC$. While every worldly cardinal is a strong limit cardinal, it is not necessarily regular. The point being that the short cofinal sequence is not first-order definable, so Replacement is not violated. -In particular, the first worldly cardinal has countable cofinality, and in fact the first worldly cardinal which is a limit of worldly cardinal has countable cofinality (as do the worldly cardinals below it). -Consider the following statement "For every $x$ there is a transitive model $M$ such that $x\in M$". Clearly this statement follows from "There is a proper class of worldly cardinals". Does it also imply it, or at least is it equiconsistent with it? - -REPLY [11 votes]: The answer is no, because I claim that if $\kappa$ is worldly, then $V_\kappa$ thinks that every set is a member of a transitive model of ZFC. -To see this, note first that every worldly cardinal $\kappa$ is a beth-fixed point $\beth_\kappa=\kappa$ and furthermore $V_\kappa=H_\kappa$, the set of sets whose transitive closures have size less than $\kappa$. Now consider any $x\in V_\kappa$. By the Löwenheim-Skolem theorem, we can find an elementary substructure $X\prec V_\kappa$ with $x\subseteq X$ and $x\in X$, with $|X|=|\text{TC}(x)|<\kappa$. The transitive collapse $M$ of $X$ will be a model of ZFC containing $x$ as an element. And even though $\kappa$ is singular, and so perhaps $X$ is unbounded in $V_\kappa$, nevertheless we will have $M\in V_\kappa$ since it is small enough. For example, $M$ will have fewer than $\kappa$ many ordinals, and so $M\subseteq V_\beta$ for some $\beta<\kappa$ and hence $M\in V_{\beta+1}\subset V_\kappa$. -Incidentally, this axiom, that every set is an element of a transitive model of ZFC, has sometimes been put forth as an alternative to the Grothendieck universe axiom, asserting that there are unboundedly many inaccessible cardinals, since it captures much of the power of that axiom in its applications, by providing a robust small universe concept for any given set, while having considerably weaker large cardinal strength. But to use this axiom, one needs to make the move from Grothendieck universes to transitive models of ZFC, which do not necessarily compute the power set correctly and whose height may be singular, although this is not visible internally to the model. -The axiom has also been proposed as a natural arena in which to investigate the corresponding multiverse theory, a version of hyperversism, but with uncoutnable models. For this reason, I like the axiom very much. In some recent upcoming joint work with Øysten Linnebo, we analyze the modal logic of this collection of models, viewed as a Kripke model.<|endoftext|> -TITLE: What results are immediately generalised to higher dimensions, in light of Schoen and Yau's recent preprint? -QUESTION [46 upvotes]: Many problems in geometric analysis and general relativity have been established in dimensions $3\leq n\leq 7$, as the regularity theory for minimal hypersurfaces holds up to dimension 7*. In a recent preprint, Schoen and Yau show how the usual techniques can be generalised to arbitrary dimension. - -My question is: What known results can trivially said to be true in higher dimensions now, in light of this paper? Also, which related results with the same dimension restriction will be non-trivial to extend to higher dimensions? - -Apologies if this question is too open-ended (this is my first time posting here) – I'm hoping it will be considered something like a "community wiki", as I think such a list would be interesting to the community. -*More precisely, see Thompson's comment below - -REPLY [4 votes]: The progress of Schoen and Yau also allows to generalize many statements previously known only for spin manifolds to non-spin manifolds. See e.g. a preprint by Thomas Schick and Simone Cecchini, arXiv math.GT 1810.02116. In particular, they are now able to prove the following: -An enlargeable metric (complete or not) cannot have uniformly positive scalar curvature. Therefore, a closed enlargeable manifold cannot carry any metric of positive scalar curvature.<|endoftext|> -TITLE: Morphisms of $\mathbb E_l$-rings between $\mathbb E_k$-rings for $l -TITLE: Fourier transform that is almost a brick wall - but why? -QUESTION [8 upvotes]: Let $$g(x) := \sqrt{1+x^2},$$ and $$h(x) := g^{-3/2}(x) \exp(-i2\pi g(x)).$$ -I can observe that the Fourier transform $|H(f)|$ is almost flat if $|f|<1$, and $H(f)\approx 0, \; |f|>1$. -This observation is important for my work, but I cannot understand why it happens. I tried for quite some time to figure out why, but failed. It basically means that $h(x)$ is a $\sin(x)/x$ function. - -REPLY [6 votes]: The Fourier transform $H_p(f)$ of $h_p(x)=g^{-p}(x)\exp[-2\pi ig(x)]$, with $g(x)=\sqrt{1+x^2}$ has a closed form expression for $p=1$: -$$H_{1}(f)=\int_0^\infty h_{1}(x)\cos(2\pi f x)dx=K_0\left[2\pi\sqrt{f^2-1}\right],$$ -see page 17 of Erdelyi's "Tables of Integral Transforms" (Volume I). -The Fourier transform of $1/\sqrt g$ is also a Bessel function, -$$G(f)=\int_0^\infty g^{-1/2}(x)\cos(2\pi f x)dx=\frac{(\pi/f)^{1/4}}{\Gamma(\tfrac{1}{4})}K_{1/4}(2\pi f).$$ -The key thing to note at this point is that $G(f)$ is basically a broadened delta function. The function $H_{1}$ if real for $f>1$ and decays rapidly to zero. This is unaffected by the convolution with $G$. For small $f$ there is a plateau at $|K_0(2\pi i)|=0.4992$, not exactly $1/2$ but close. - -Plot of $|H_{1}(f)|$ (blue) and $|H_{3/2}(f)|$ (gold). - -Plot of $|H_{3/2}(f)|$ for $f<1$, to show that it is almost but not quite flat, and almost but not quite $1/2$ for $f\rightarrow 0$. The sharp peak at $f=1$ that was present in $|H_{1}(f)|$ has been greatly suppressed by the convolution with $G$.<|endoftext|> -TITLE: Do Betti numbers beyond the first have a "number of cuts" interpretation? -QUESTION [21 upvotes]: I have heard stated the following -Theorem. If $\Sigma$ is a (orientable) surface, then $\mathrm b_1(\Sigma)$ counts the maximum number of "circular cuts" (embedded circles $C_1,\ldots,C_m$) that you can make on $\Sigma$ without disconnecting it (i.e. with $\Sigma\smallsetminus (C_1\cup\ldots\cup C_m)$ still being connected). - -Is there a similar interpretation also for higher Betti numbers $\mathrm b_k(M)$ of manifolds in general? - -(I've been asked this by a non-mathematician, but I think it fits MO. If the topologists here think it's too trivial or well known, please move it to MSE). -Edit1: It wouldn't be bad to know if the above "theorem" is actually a theorem, and a reference to a rigorous proof. -Edit2: In the above, by a "generalization" I mean a statement of the form: $b=b_k(M)$ counts the maximum number of embedded orientable submanifolds $N_1,\ldots,N_b$ such that $M\setminus (N_1\cup\ldots\cup N_b)$ has clear topological property $\boldsymbol{\mathrm{P}}$. Maybe property $\boldsymbol{\mathrm{P}}$ could be about some $(k-1)$-connectedness condition? For the $b_1$ case, this would involve $0$-connectedness, i.e. just connectedness, and this is the case for $\dim M=2$ if the theorem I quoted above is true. - -REPLY [11 votes]: Here is one way that your observation about the first Betti number generalizes to higher-dimensional manifolds. -Suppose that $M$ a compact, orientable smooth manifold and $W$ is a regularly embedded, closed, orientable submanifold of codimension 1. Then if the number of path components of $W$ is greater than the first Betti number $b_1(M)$, then $M \setminus W$ has more path components than $M$. - -Here's the proof. Consider the long exact sequence -$$ -\dots \to H_1(M \setminus W) \to H_1(M) \to H_1(M, M \setminus W) \to H_0(M \setminus W) \to H_0(M) \to H_0(M, M \setminus W) \to 0. -$$ -Since $W$ is regularly embedded, it has a neighborhood diffeomorphic to the (1-dimensional) normal bundle $\nu$ of the embedding. By the excision axiom, there is an isomorphism -$$ -H_k(M, M \setminus W) \cong H_k(\nu, \nu \setminus W). -$$ -(Here we view $W$ as the same as the zero-section of $\nu$.) -Since $W$ and $M$ are both orientable, the normal bundle is orientable. Since the normal bundle is orientable of dimension 1, there is a Thom isomorphism -$$ -H_k(\nu, \nu \setminus W) \cong H_{k-1}(W). -$$ -Therefore, part of our exact sequence becomes -$$ -\dots \to H_1(M) \to H_0(W) \to H_0(M \setminus W) \to H_0(M) \to 0. -$$ -The rank of $H_0(W)$ is the number of path components of $W$, so if $W$ has more path components than $b_1(M)$ then the map $H_1(M) \to H_0(W)$ cannot possibly be surjective. If it is not surjective, then the image of $H_0(W)$ in $H_0(M \setminus W)$ is nontrivial. Since this image is the kernel of the next map, the map $H_0(M \setminus W) \to H_0(M)$ is not injective. But this map can only fail to be injective if there are more path components in $M \setminus W$ than there are in $M$. - -Here are some observations about this proof. - -The difference between the number of path components of $W$ and $b_1(M)$ actually gives us a lower bound on the number of new path components in $M \setminus W$, and if $b_1(M) = 0$ we get an exact count. -If we toss out the word "orientable" everywhere but replace homology with mod-2 homology everywhere, we get a similar bound using the mod-2 Betti number. This sees that we can remove the nonorientable submanifold $\Bbb{RP}^2 \subset \Bbb{RP}^3$ and still have a connected space, even though this can't happen for orientable submanifolds of $\Bbb{RP}^3$. -As given, the argument is only in one direction. It gives no guarantees that we can find an oriented submanifold $W$ with $b_1(M)$ connected components so that $M \setminus W$ is still connected. -If we replace codimension 1 in this argument with codimension $r$, we instead get an exact sequence -$$ -\dots \to H_r(M) \to H_0(W) \to H_{r-1}(M \setminus W) \to H_{r-1}(M) \to 0. -$$ -By the same argument, this allows us to say: if $b_r(M)$ is less than the number of connected components of $W$, the map $H_{r-1}(M \setminus W) \to H_{r-1}(M)$ cannot be injective. This is not quite as punchy as the statement for path components, but the de Rham theorem tells us that it does still indicate something about the existence of new differential forms on $M \setminus W$. -I don't think, in the original problem, that you wanted the submanifolds to be disjoint. If instead we have connected, closed, regularly embedded, orientable submanifolds $W_1,\dots,W_d$ of codimension $k$ which intersect transversely, one can show by a more complicated inductive argument with the (relative) Mayer-Vietoris sequence that $H_k(M, M \setminus \cup W_i)$ has rank at least $d$, and find that the map $H_{d-1}(M \setminus \cup W_i) \to H_{d-1}(M)$ has kernel of rank at least $d - b_k(M)$.<|endoftext|> -TITLE: Classification of the quotients of the ring Z/4 [X] -QUESTION [5 upvotes]: Is it possible to classify all cyclic $\mathbb{Z}/4$-algebras, i.e. the regular quotients of $\mathbb{Z}/4 [X]$? A typical example is $\mathbb{Z}/4 [X] / \langle X^n , 2 X^k \rangle$. For my purposes it is not necessarily important to decide which quotients are isomorphic (this is already unclear to me for $\mathbb{Z}/2[X]$). So an almost identical question is the following: -Is it possible to classify all ideals of $\mathbb{Z}/4 [X]$? Can we bound the number of generators? Or is this classification not feasible? -It is known that every $\mathbb{Z}/4$-module (also without any finiteness assumptions) is a direct sum of copies of $\mathbb{Z}/2$ and $\mathbb{Z}/4$. This should be useful. -Partial results are also helpful for me. Instead of $\mathbb{Z}/4$ it should be possible to take $R/p^2$ for any principal ideal domain $R$ and a prime element $p \in R$. -Edit. Here is how I understand YCor's proof that every ideal of $R/p^2 [X]$ can be generated by two elements; the proof works for every commutative ring $A$ with an element $p \in A$ such that $p^2=0$ and $A/pA$ is a principal ideal ring. Let $I \subseteq A$ be an ideal. Consider the $A$-module $I \cap p A$. Since it is killed by $p$, we may view it as an $A/pA$-module. But $A/pA$ is a principal ideal ring, and $pA$ is a cyclic $A/pA$-module. Hence, $I \cap pA$ is a cyclic $A/pA$-module, i.e. a cyclic $A$-module. On the other hand, $I/(I \cap pA) \cong (I + pA)/pA$ is an $A$-submodule of $A/pA$, and hence cyclic too. Hence, $I$ is generated by a generator of $I \cap pA$ together with any preimage of a generator of $(I+pA)/pA$. - -REPLY [6 votes]: Let $I$ be an ideal in $(\mathbf{Z}/4\mathbf{Z})[X]$. Let $J_I$ be its projection to $(\mathbf{Z}/2\mathbf{Z})[X]$, and write its intersection with $2(\mathbf{Z}/4\mathbf{Z})[X]$ as $2K_I$, for some ideal $K_I$ of $(\mathbf{Z}/2\mathbf{Z})[X]$ (precisely, first define $K'_I=\{P \in (\mathbf{Z}/4\mathbf{Z})[X]:2P\in I\}$ and then define $K_I$ as the projection of $K'_I$ modulo 2). -It's immediate that $J_I\subset K_I$. -Conversely for a given pair of ideals $(J,K)$ of $(\mathbf{Z}/2\mathbf{Z})[X]$ such that $J\subset K$, we can enumerate the $I$ such that $(J_I,K_I)=(J,K)$. First denote by $f\mapsto\bar{f}$ the lift $\mathbf{Z}/2\mathbf{Z}\to \mathbf{Z}/4\mathbf{Z}$ mapping $0\mapsto 0$, $1\mapsto 1$, and extend it coefficient-wise to polynomials. -Namely, let $f$ be the (unique) generator of $J$ and $g$ the generator of $K$ (so $g$ divides $f$). If $f=0$ (i.e., $J=0$), clearly $I$ is unique (and equal to the principal ideal $2K'_I$). -Now suppose $f\neq 0$ (so $g\neq 0$). Then for $h\in (\mathbf{Z}/2\mathbf{Z})[X]$ all $I_h=\langle \bar{f}+2\bar{h},2\bar{g}\rangle$ have $(J_{I_h},K_{I_h})=(J,K)$, and $I_h=I_{k}$ if and only $h-k\in\langle g\rangle$. Thus these ideals are parameterized by $(\mathbf{Z}/2\mathbf{Z})[X]/\langle g\rangle$, which is finite, and exhaust those ideals $I$ such that $(J_I,K_I)=(J,K)$. (In this case, $J_I=K_I$ iff $I$ is principal.) -I think this is a full description. -Example: the ideals $\langle X\rangle$ and $\langle X+2\rangle$ both have $(J,K)=(\langle X\rangle,\langle X\rangle)$ and these are the only ones. - -I should add a word on the structure of quotients. When $J_I\neq 0$, the quotient is finite (i.e., $I$ has finite index). The underlying additive group structure is isomorphic to $(\mathbf{Z}/4\mathbf{Z})^k\times (\mathbf{Z}/2\mathbf{Z})^\ell$ where $k$ is the degree of $f$ and $\ell+k$ is the degree of $g$ (some effort would yield something better, namely the decomposition of the finite length module $(\mathbf{Z}/4\mathbf{Z})[X]/I$ as sum of indecomposable modules). If $J_I=0$ and $K_I\neq 0$, the quotient is additely isomorphic to $(\mathbf{Z}/4\mathbf{Z})^k\times (\mathbf{Z}/2\mathbf{Z})^{(\aleph_0)}$ where $k$ is the degree of $g$, and of course if $I=0$ it is additively isomorphic to $(\mathbf{Z}/4\mathbf{Z})^{(\aleph_0)}$.<|endoftext|> -TITLE: Quiver representations and coherent sheaves -QUESTION [15 upvotes]: I've heard that under certain assumptions on an algebraic variety $X$ there exist a quiver $Q$ for which there is an equivalence $$D^b(\mathsf{Coh}(X))\simeq D^b(\mathsf{Rep}(Q))$$ between the corresponding bounded derived categories. My (naive) questions about this fact are the following: -1) Is there any intuitive reason for the existence on this relation between the geometry of $X$ and the combinatorics of $Q$? For instance, given some variety, is there a way to make a guess of what quiver could realize this equivalence? -2) Is there a generalization for this result when $X$ is a Deligne–Mumford stack? If this is the case, how is the structure of the stabilizers reflected in the quiver? - -REPLY [5 votes]: Let me try to answer in a slightly different way, with a few corrections (unless I'm the one that's wrong). It is a very general statement that, given a compact generator of a stable model or $\infty$-category, the categories of modules (appropriately defined) over the endomorphisms of that generator are equivalent to the original category. This goes back to Rickard's derived Morita equivalence, and the most general formulation is probably that of Schwede and Shipley-- it's also almost surely in Lurie's book. -So, in that sense, $Coh(X)$ is a bit of a red herring -- you don't need to know that your derived category comes from a variety, a scheme or whatever. You just need a compact generator. Once you have that, the question becomes, when do the endomorphisms of the generator look like the path algebra of a quiver. Specializing the the situation of a dg-category, some of the conditions are: -1) The dg-algebra of endorphisms must be quasi-equivalent to an ordinary algebra. In other words, the generator can't have any self-exts outside of degree zero. -2) The identity of the path algebra quiver decomposes into a set of idempotents associated with the nodes of the quiver. Associated to these idempotents are a set of projective modules. The direct sum of these projective modules is the algebra itself. Thus, the generator must decompose into a direct sum. It's the direct sum of these objects that is the generator; each is not a generator by itself. -3) Given two projective modules, $Hom(P_i,P_j)$ is isomorphic to the space of paths between those two nodes (not arrows). This need not be finite-dimensional if you are ok with your quiver having loops. However, a quiver also has a distinguished set of simple representations associated with the nodes that satisfy $Hom(P_i,S_j) = \mathbb{C}^{\delta_{ij}}$. These are just the one-dimensional representations associated with the nodes the quiver. -The dimension of the space of arrows between two nodes of the quiver are given by $Ext^1(S_i,S_j)$. So, if you want a finite number of arrows, that better be finite dimensional. In fact, you can recover the entire quiver algebra by looking at the self-ext dg/$A_\infty$-algebra. -Relating this to the comments, a full strong exceptional collection exactly gives you (1) and (2), but it's not a necessary condition -- you can get quivers for noncompact varieties, for example, a situation that is very common in studying D-branes at singularities in string theory. -One subtlety I've neglected so far is the difference between the unbounded derived category and the bounded derived category. I'm getting more out of my long-atrophied depth here, but I believe the equivalence from the unbounded category descends to the bounded one if perfect complexes are the same as compact objects. This is true, I think, for quasi-compact separated schemes. For stacks, you can see results in Ben-Zvi, Frances and Nadler.<|endoftext|> -TITLE: 1-Bridge Braids in Solid Tori (Berge-Gabai knots), dual knots, & knot exteriors -QUESTION [8 upvotes]: Let $K$ be a knot in a solid torus. Combining results of Berge and Gabai, we know that if $K$ admits a solid torus surgery, then $K$ is a 1-bridge braid. Using Gabai's result, we can figure out what the surgery slope is. If a 1-bridge braid admits a non-trivial solid torus surgery, the surgery slope is (almost) always unique (there is a unique 1-bridge braid which admits 2 non-trivial solid torus surgeries; most refer to it as K(7,2,4) in (w,b,t) notation). -If $B = K(7,2,4)$, we know that $B$ is equivalent to its dual, $B'$. -What is known about the relation between a 1-bridge braid and its dual in general? Namely: -Question #1: Given a 1-bridge braid $K$ that admits a solid torus surgery, when is it equivalent to its dual? -Question #2: Is it known when, given 1-bridge braids $K_1$ and $K_2$ such that $K_1 \not \simeq K_2$, their exteriors (in the solid torus $\mathbb{D}^2 \times S^1$) are homeomorphic? -I know Berge gives a necessary & sufficient condition for equivalence of knots, but I'm wondering if the answers to my above questions already exist in the literature in a more modern/non-G-pair-theoretic language. -Thanks! - -REPLY [4 votes]: For Question #1: -As you acknowledge, Berge already gives an answer to Question #1. Using his notation, this is summarized in Table 2 of his paper "The knots in $D^2 \times S^1$ which have nontrivial surgeries that yield $D^2 \times S^1$". Berge also already tells you how to translate from a G-pair to a standard braid form in his Lemma 2.2. So really, you just need to do the legwork to get it into a more common presentations of the dual. -Alternatively, a non-G-pair exhibition of the pairs of dual Berge knots is given in the setting of their tangle quotients in Baker-Buck, "The classification of rational subtangle replacements between rational tangles". Figures 3-6 show bandings (+isotopy) between tangles in plat form keeping track of the dual arc, starting from a tangle isotopic to the crossingless two cup tangle. -(In [Baker-Buck], since they’re dual, I had combined types III and V into one family, though I really shouldn’t have. With one exception, there’s no overlap of them (as noted in Table 2 of [Berge]). Referring back to my "Surgery descriptions and volumes of Berge knots II" one finds that in Figure 5 of [Baker-Buck]: Top row shows V to III, Bottom row shows III to V.) -For Question #2: -If the exteriors of $K_1$ and $K_2$ are both homeomorphic to the manifold $X$, then either - -(A) their meridians are on the same boundary component of $X$ in which case they are a pair of Berge-Gabai duals as in Question #1, or -(B) their meridians are on different boundary components of $X$ in which case this has been partially addressed by Wu in "The classification of Dehn fillings on the outer torus of a 1-bridge braid exterior which produce solid tori". - -In Theorem 3.6 of [Wu], Wu determines both the 1-bridge braids for which a filling of the outer torus boundary of the exterior produces a solid torus again and the slopes of these fillings. What’s still missing is the determination of the dual knot the resulting solid torus: Is it necessarily a 1-bridge braid? If so, which one? -Note that if you drop the assumption of 1-bridge braid, situation (B) becomes much more varied. You’re basically looking at two component links in lens spaces where each component is isotopic (individually) to a core of a Heegaard solid torus. For example in $S^3$ any two component link of unknots has this property.<|endoftext|> -TITLE: Rohklin’s formula and $\beta$ expansions -QUESTION [6 upvotes]: I've been reading K. Dajani and C. Kraaikamp's paper From greedy to lazy expansions and their driving dynamics. In the last page previous to the references, they mention that the entropy of the map $T_{\alpha, \beta}(x) = \beta x + \alpha$ is $\log(\beta)$. They mention this fact follows from the Rohklin’s formula for calculating the entropy. What I would like to know is a reference where this formula is stated and proved. -Thanks a lot in advance. - -REPLY [5 votes]: This is the formula -$$ -h_\mu(f) = \int \log |f'|\,d\mu -$$ -for the entropy of an absolutely continuous $f$-invariant measure $\mu$ on the unit interval which goes back to Rokhlin (http://www.ams.org/mathscinet-getitem?mr=143873), also see Ledrappier (http://www.ams.org/mathscinet-getitem?mr=627788).<|endoftext|> -TITLE: exponential functors on finite dimensional complex vector spaces -QUESTION [16 upvotes]: Denote by $Vect^{fin}_{\mathbb{C}}$ the category of finite dimensional complex vector spaces. We will call a pair $(F,\tau)$ that consists of a functor $F \colon Vect^{fin}_{\mathbb{C}} \to Vect^{fin}_{\mathbb{C}}$ and a natural isomorphism -$$ -\tau_{V,W} \colon F(V \oplus W) \to F(V) \otimes F(W) -$$ -an exponential functor if $\tau$ satisfies the following associativity condition: -$$ -(\tau_{V,W} \otimes id_{F(X)}) \circ \tau_{V \oplus W, X} = (id_{F(V)} \otimes \tau_{W,X}) \circ\tau_{V,W \oplus X} -$$ -An example of such a functor is the full exterior power $\Lambda^*$ together with the natural iso $\Lambda^*(V \oplus W) \cong \Lambda^*(V) \otimes \Lambda^*(W)$. Another example along the same lines is -$$ -F_X(V) = \Lambda^*(V \otimes X) -$$ -for a fixed finite dimensional complex vector space $X$. Since we insist on finite dimensional vector spaces as our target category for $F$, the full symmetric power is not an example. - - -Are there any other examples of exponential functors on $Vect^{fin}_{\mathbb{C}}$? Is there a classification of exponential functors? - -REPLY [7 votes]: In case anyone is still interested in this question: -There is a classification of polynomial exponential functors on the category $\mathcal{V}$ of finite-dimensional inner product spaces in terms of involutive $R$-matrices (i.e. involutive solutions to the Yang-Baxter equation). Basics about polynomial functors can be found in the book "Symmetric Functions and Hall Polynomials" by Macdonald (Chapter 1, Appendix A). -A functor $F \colon \mathcal{V} \to \mathcal{V}$ is polynomial, if for all linear maps $f_i \colon V \to W$ the result of $F(\lambda_1 f_1+ \dots + \lambda_n f_n)$ is a polynomial in the $\lambda_i$ with coefficients in $\hom(V,W)$. Each such functor has a direct sum decomposition into homogeneous components $F_n \colon \mathcal{V} \to \mathcal{V}$. Each homogeneous component has a linearisation $L_{F_n} \colon \mathcal{V}^n \to \mathcal{V}$ and $L_{F_n}(V_1, \dots, V_n)$ is defined to be the direct summand of $F_n(V_1 \oplus \dots \oplus V_n)$, on which $\lambda_1 id_{V_1} \oplus \dots \oplus \lambda_n id_{V_n}$ acts via multiplication by $\lambda_1 \cdot \lambda_2 \cdots \lambda_n$. -For an exponential functor $F$, the linearisation $L_{F_n}$ of its $n$-homogeneous summand $F_n$ is equivalent to $F_1^{\otimes n}$. Let $W = F_1(\mathbb{C})$ and $\tau \colon \mathbb{C}^2 \to \mathbb{C}^2$ be the isomorphism permuting the summands. Then $F_2(\tau) \colon F_2(\mathbb{C}^2) \to F_2(\mathbb{C}^2)$ restricts to a linear transformation $R \colon W^{\otimes 2} \to W^{\otimes 2}$ of the linearisation of $F_2$ that is involutive and satisfies the Yang-Baxter equation on $W^{\otimes 3}$. This can be extended to the statement that polynomial exponential functors up to natural equivalence of monoidal functors are in $1:1$-correspondence with certain equivalence classes of involutive $R$-matrices. -This is written up here. This result is used there to construct twists of complex $K$-theory localised at an integer over $SU(n)$, which was my original motivation for this question.<|endoftext|> -TITLE: Why is Quantum Field Theory so topological? -QUESTION [53 upvotes]: I understand that my question suffers from my lack of knowledge about the field, but as a mathematician without much knowledge of physics I have been wondering much about the following and I always felt a bit stupid to ask this in real: -Many parts of physics look mainly analytic to me, i.e. electrodynamics and fluid dynamics look like an application of vector analysis and PDEs, quantum mechanics seems to rely heavily on functional analysis. I also understand that there is a lot of structure hidden in those theories: The best example would be symplectic geometry and classical mechanics or topological phases in quantum mechanics. -Despite, it seems to me that this is a more modern development in the physics community and the core of those theories as they would be taught to undergraduates would still be mainly analytic. (Sorry if I am wrong about this assumptions but it is my feeling). -However, when it comes to Quantum Field Theory, I feel that very much revolves (especially from the math-community side) around topological and algebraic questions. There is for example a visible math-community with analysis background working on mathematical quantum mechanics, but I never noticed this community in Quantum Field Theory. -Please consider these thoughts of mine, to justify my question: Is there no analysis present in QFT or why do mathematicians concentrate so heavily on those other aspects? - -REPLY [29 votes]: The framing of your question is a bit ambiguous and perhaps there are two different questions -here depending on the context and interpretation. One could approach your question from the point of view of intrinsic scientific content and ask: why QFT seems to be intrinsically more related to topology than analysis? (Question A). But one can also -approach the question from the angle of how this is reflected in human activity (mathematicians doing mathematics) -in these subjects. -Namely, you could ask the question: why, among mathematicians interested in QFT, there are more topologists than -analysts? (Question B). -Here is a stab at answering these two very different questions. -Question A: -This is moot since it is based on a false premise. QFT is not only related to topology but also to analysis and, I even venture to say, to almost all of mathematics. The reason for this is that QFT or the problem of rigorously defining functional integrals is the logical and natural continuation of the development of calculus -as I explained in this MO answer. After the usual calculus sequence (I, II, III) concerning the finite-dimensional situation, it is natural to explore differentiation (Calc IV) and integration (Calc V) in infinite dimension. Although Calc IV can be traced back to the early work on the calculus of variations by Maupertuis, Euler and Lagrange, I think its mathematical development started in earnest with the work of Volterra. As for Calc V, Wiener's construction of Brownian motion would come to mind as -an important early milestone. The intrinsic programmatic content of Calc V -is exactly what the analysis of QFT is about. -As Robert mentioned, this area of mathematics already exists and is called constructive quantum field theory (CQFT), although -nowadays it is also called rigorous renormalization group theory. -To get an idea of what is going on in the field, have a look at the reports for the two recent Oberwolfach meetings: - -"Recent Mathematical Developments in Quantum Field Theory" -"The Renormalization Group". - -Another recent meeting around the new developments by Hairer and others in the strongly related field of stochastic quantization -is: - -"Rough Paths, Regularity Structures and Related Topics". - -The problem of defining a QFT functional integral is a well posed mathematical problem -(see this MO answer for details). -In a nutshell, one starts by putting UV and IR cutoffs as is familiar in the theory of -Schwartz distributions and one lets bare couplings vary with these cutoffs. The problem is to find the set of all weak -limits for the corresponding probability measures on Schwartz distributions. The main difficulty is to construct such weak limit points that are not Gaussian or free measures. One would also like to parametrize this collection of weak limits by a finite number of parameters called -renormalized couplings. The main tool to do this is the renormalization group (RG). -In this MO answer -I briefly explained what the RG is, but I did not give details about how the RG provides a strategy for solving -the above problem about weak limit points. For more explanations about this strategy see my article -"QFT, RG, and all that, for mathematicians, in eleven pages" -and my answer to the physics.stackexchange question Wilsonian definition of renormalizability. -What Robert said "I think there is a feeling that the "easy" questions have been answered, and much of what remains may be impossibly hard" -is not quite correct. There are plenty of doable problems to work on at present in CQFT other than $YM_4$. -For example one has analogous conjectures for the 2d Gross-Neveu model and the 2d $\sigma$-model. -These are not impossibly hard like $YM_4$ and they do not really require extraterrestrial "new ideas". -As in the millenium problem, what one has to do is a construction of the model without UV cutoffs and in infinite volume -together with a proof of mass gap. -Another interesting problem (the one I focused on in the above references) is to make contact with conformal field theory. -The good class of examples to study in this regard are three-dimensional -$N$ component phi-four models with fractional Laplacian -$(-\Delta)^{\alpha}$ in the kinetic term. -The cases $N=1$, $\alpha=\frac{3}{4}+\epsilon$ as well as $N$ large, $\alpha=1$ should be not be impossibly hard. -Other problems of current interest are: proving the operator product expansion and conformal invariance using the RG. -As for what I personally think is most important problem in constructive quantum field theory today, it is to develop a rigorous Wilsonian RG formalism for handling space-dependent couplings. -Question B: -This one is not moot. -It is a fact that there are more mathematicians working on the topological aspects of QFT rather than its analytical aspects. I think this state of affairs is simply due to the status quo, i.e., it's just the way things are. -With regards to the North American situation in particular, I think the main explanation is that if a graduate -student would like to work on the analysis of QFT, chances are there would simply be nobody in their department to teach them the subject -to the point of being research-ready. -I think there is nothing more to it, but this could change in the future.<|endoftext|> -TITLE: Are surjections $[n]\to [k]$ more common than injections $[k]\to [n]$? -QUESTION [20 upvotes]: Here's an interesting inequality involving binomial coefficient and Stirling numbers of the second kind that I believe holds for all $n,k$: -$$ k^n {n \choose k} \leq n^k {n \brace k} $$ -On the left-hand side we choose a $k$-element subset of the set $[n]=\{1,\ldots,n\}$, and then choose a function from $[n]$ into the chosen subset. On the right-hand side we choose a partition of $[n]$ into $k$ blocks, and then to each block we assign a label from $[n]$. The number of objects on the left-hand side seems to be at most the number of objects on the right-hand side. It is easy to show this for some special cases ($k=0,1,2,n-1,n$), but I don't know how to show it in general. -Another way of looking at the inequality is the following. Provided that $k$ and $n$ are either both zero or both positive, we can rearrange the inequality to get -$$ \frac{n^{\underline{k}}}{n^k} \leq \frac{{n \brace k} k!}{k^n} $$ -Here, $n^{\underline{k}} = n (n-1) \ldots (n-k+1)$ is the falling factorial power. Now the left-hand side is the ratio of injections to all functions $[k]\to[n]$, and the right-hand side is the ratio of surjections to all functions $[n]\to[k]$. Therefore the inequality intuitively expresses that surjections $[n]\to [k]$ are more common than injections $[k]\to[n]$. -Any ideas how to prove this? -UPDATE: As pointed out by Mark Wildon, this problem has been posted by American Mathematical Monthly with deadline June 30, 2017. It is not clear whether we should be discussing it before then. I tried to mark my attempt below as a spoiler, but the spoiler functionality doesn't seem to be working properly in combination with MathJax. So, if you don't want spoilers, I suggest you stop reading at this point. -SPOILER: -Here's a combinatorial approach that proves a weaker inequality: -$$ k^{n-k} {n \choose k} \leq n^k {n \brace k} $$ -We'll construct an injection from the objects on the left-hand side into the objects on the right-hand side. On the left-hand side we are given a $k$-element subset $K\subseteq [n]$, together with a function $f\colon [n]\setminus K \to K$. We form a partition $P=\{\{x\} \cup f^{-1}(x) \mid x\in K\}$ and a function $g\colon P\to [n]$ given by $g(\{x\} \cup f^{-1}(x))=x$. Function $g$ essentially picks a representative from each block, and serves to uniquely reconstruct the subset $K$ and the function $f\colon [n]\setminus K \to K$. -In the original inequality, on the left-hand side we have a function $f\colon [n]\to K$, meaning that in addition to a mapping $[n]\setminus K\to K$ we have a mapping $K\to K$. In order to construct an injection, our function $g\colon P\to [n]$ must now encode what the representative of each block is, and also how these representatives are mapped among themselves. It is not clear to me how to do this. -I've also tried to prove the inequality algebraically. There are many ways to expand the terms using various identities, but so far I always get stuck. - -REPLY [8 votes]: I have already accepted Mark Wildon's answer, but just for the reference, let me also post another proof that I found while the discussion was closed. The proof uses the following auxiliary notion that led me to the question in the first place. -Definition. Given a set $S\subseteq [n]$ with $k$ elements and a partition $P$ of $[n]$ with $k$ blocks, we say that $P$ splits $S$ if every block in $P$ contains exactly one element of $S$, that is, $B\cap S \neq \emptyset$ for every $B\in P$. -Proof. We show an equivalent inequality obtained by dividing both sides of the original inequality by $k^k$: -$$ k^{n-k} {n \choose k} \leq \frac{n^k}{k^k} {n \brace k} $$ -Let $M$ be a binary matrix whose rows are indexed by partitions of $[n]$ with $k$ blocks, columns are indexed by subsets of $[n]$ with $k$ elements, and such that an entry corresponding to partition $P$ and set $S$ is 1 if and only if $P$ splits $S$. -We count the number of ones in the matrix in two different ways. The number of ones in a column indexed by $S$ is the number of partitions that split $S$. Such a partition is uniquely determined by a map from $[n]\setminus S \to S$ that maps $x \in [n]\setminus S$ to $y\in S$ if $x$ and $y$ are in the same block of the partition. Hence the number of ones in the column is $k^{n-k}$, and the total number of ones in $M$ is $k^{n-k} {n \choose k}$. On the other hand, the number of ones in a row indexed by $P=\{B_1, \ldots, B_k\}$ is the number of sets split by $P$. Such a set is uniquely determined by a choice of one element from each block. Hence the number of ones in the row is $\lvert B_1 \rvert \cdots \lvert B_k \rvert$, and the total number of ones in $M$ is obtained by summing over all partitions with $k$ blocks. -From the identity obtained by double counting we get the desired claim by applying the AM-GM inequality: -\begin{align*} - k^{n-k} {n \choose k} - & = \sum_{P=\{B_1, \ldots, B_k\}} \lvert B_1 \rvert \cdots \lvert B_k \rvert \\ - & \leq \sum_{P=\{B_1, \ldots, B_k\}} \frac{(\lvert B_1 \rvert + \ldots + \lvert B_k \rvert)^k}{k^k} \\ - & = \frac{n^k}{k^k} {n \brace k} -\end{align*}<|endoftext|> -TITLE: Wielandt automorphism tower theorem -QUESTION [5 upvotes]: I wanted to know if anyone can point me to an (ideally freely available) english translation of the proof of Wielandt's Automorphism Tower Theorem (1939). -The theorem states the following: -Given a finite, centerless group $G$, its automorphism tower terminates (stabilizes) in finitely many steps. - -REPLY [4 votes]: The theorem appears as Theorem 9.10 in -Isaacs, I.Martin, Finite group theory. Graduate Studies in Mathematics 92. Providence, RI: American Mathematical Society (ISBN 978-0-8218-4344-4/hbk). xi, 350~p. (2008). DOI 10.1090/gsm/092. ZBL1169.20001, MR2426855. -The proof given there is very detailed (and well written). (But I think this book is not freely available.)<|endoftext|> -TITLE: Bull's-eye Riemann sum -QUESTION [15 upvotes]: Let $f:[a,b] \to \mathbb{R}^2$ be a continuous curve on the plane. - -Question: Are there numbers $a \leq x \leq c \leq y \le b$ such that $$(c-a)f(x)+(b-c)f(y) = \int_a^b f(t) \, dt \ ?$$ - -In other words, is there a Riemann sum with two terms that hits the bull's-eye? -EDIT: Prompted by a down-vote, maybe I should give some motivation: -It's not difficult to convince oneself (but not completely trivial to prove (*)) that the barycenter $\frac{1}{b-a}\int_a^b f(t) \, dt$ is a convex combination of two values of $f$, say $f(x)$ and $f(y)$ with $x0$. Spend time $\frac 23(1-\delta)$ at $(0,1)$. Then, in time $\delta/4$ travel the route $(0,1)\to(0,2)\to(-1,2)\to(-1,-2)$ so that the integral of the vertical coordinate is $0$. Then stay at $(-1,-2)$ for the time $\frac 16(1-\delta)$ and return to $(0,1)$ running the same movement backwards. Make a symmetric (with respect to the $y$ axis) round trip to the right. The resulting function has integral $(0,0)$ over the interval $[0,1]$. On the other hand, any two points on the associated curve that lie on the same line through the origin on the opposite sides of the origin can cancel each other only with coefficients $1/2$. However, the corresponding partition of $[0,1]$ puts them in the same interval. -This will certainly make an excellent extra credit question for multivariate calculus students. To make the life even more interesting, let us ask if any $n$ is sufficient on the plane.<|endoftext|> -TITLE: tensor products and intersections of subspaces -QUESTION [5 upvotes]: Given two real vector spaces $V$ and $U$ with subspaces $A, C \subset V$, $A\cap C \neq \{0\}$ and $B, D \subset U$, $B\cap D \neq \{0\}$, is it true that -$$(A\otimes B)\cap (C\otimes D) = (A\cap C)\otimes (B\cap D)$$ -and if so, is there a proof available in a paper or textbook? It is stated to be true in a comment to a previous question Tensor Products and Intersections but no proof is given. - -REPLY [8 votes]: See e.g. Lemma 1.4.5 in the book -S. Dascalescu - C. Nastasescu - S. Raianu: "Hopf Algebras. An Introduction", Pure and Applied Mathematics, 2001 -There the statement is proved under the assumption that $C\subseteq A$ and $D\subseteq B$: the general case easily follows from this.<|endoftext|> -TITLE: Holomorphic extension of an action by a compact Lie group on a complex homogeneous manifold -QUESTION [5 upvotes]: Let $G$ be a compact Lie group and let $M$ be a $G$-homogeneous manifold. Suppose that $M$ is endowed with a complex structure invariant by the action of $G$. Denote by $G_{\mathbb C}$ the complexification of $G$. -Are there known conditions so that the action of $G$ on $M$ can be extended to a holomorphic action of $G_{\mathbb C}$? - -REPLY [6 votes]: I think this should work for a connected group $G$. The Lie algebra action extends, by multiplying by $J$, to a complex Lie algebra action. By compactness of $M$, these vector fields are all complete. So some covering group of the complexification acts, for example the universal covering group $\tilde{G}_{\mathbb{C}}$ acts. Hochschild, The Structure of Complex Lie Groups, chapter XVII, §5, proves that if $G$ admits a faithful finite dimensional representation, then $G$ injects into its complexification. For example, all compact Lie groups admit faithful finite dimensional representations. The group $G$ is then the maximal compact subgroup of $G_{\mathbb{C}}$; Broecker and tom Dieck, Representations of Compact Lie Groups, p. 153, prop. 8.3. Every connected Lie group retracts to its maximal compact subgroup. In particular, the fundamental group of $G$ and of $G_{\mathbb{C}}$ are the same, represented by loops in $G$. -The fundamental group of $G_{\mathbb{C}}$ embeds into the universal covering group $\tilde{G}_{\mathbb{C}}$. We need to see which elements of it act trivially, so that we can see which quotient of $\tilde{G}_{\mathbb{C}}$ the action descends to. The fundamental group of the original compact group sits as a subgroup of $\tilde{G}$, but acts trivially on $M$. So the same is true for the fundamental group of $G_{\mathbb{C}}$: $\pi_1(G_{\mathbb{C}}) \subset \tilde{G}_{\mathbb{C}}$ acts trivially on $M$. Hence $G_{\mathbb{C}}$ acts on the complex homogeneous space $M$.<|endoftext|> -TITLE: Implicit equation between rational function and its derivative -QUESTION [6 upvotes]: Let $f$ be a complex rational function in one variable. How does one find a complex polynomial $P$ in two variables such that $P(f,f^\prime)=0$? -EDIT: as the answers explained there is a standard procedure using resultant to obtain $P$. see also https://en.wikipedia.org/wiki/Resultant#Algebraic_geometry . -However this process is computationally very heavy. I was wondering if in the case when $g=f^\prime$ we may know $P$ more explicitly, if we had some other, more efficient way to compute it. - -REPLY [3 votes]: The magic word is "implicitization". Consider the curve in $\mathbb{P}^2$ given by $C = (f(t), f^\prime(t)).$ (as pointed out by Robert Bryant in the comments). Now, we would like to find the equation satisfied by the curve. This can be done for any pair of rational functions (not just $f, f^\prime$). Namely, we write down the pair of equations: -$$x = \frac{p(t)}{q(t)} \Leftrightarrow x q(t) = p(t)$$ -$$y = \frac{r(t)}{s(t)} \Leftrightarrow y s(t) = r(t).$$ -We can now eliminate $t$ from these equations, but thinking of both of them as being defined over $\mathbb{C}[x, y],$ and setting the Sylvester matrix determinant resultant to zero. - -REPLY [3 votes]: As Igor noted, this is all quite standard, and I'm really not sure it belongs on MO, since I'm sure it would certainly receive a quick answer on MathSE. Anyway, in order to compute an equation for the image of the implicit curve $\bigl(f(t),g(t)\bigr)$, where $f(t)=f_1(t)/f_2(t)$ and $g(t)=g_1(t)/g_2(t)$ are rational functions, compute the $t$-resultant -$$ -\text{Resultant}_t\bigl( f_2(t)X-f_1(t), g_2(t)Y-g_1(t) \bigr). -$$ -More generally, this sort of question in answered by elimination theory, and as a practical matter, one uses Grobner bases to compute such quantities.<|endoftext|> -TITLE: What is the relationship between connective and nonconnective derived algebraic geometry? -QUESTION [32 upvotes]: "Derived algebraic geometry" usually means the study of geometry locally modeled on "$Spec R$" where $R$ is a connective $E_\infty$ ring spectrum (perhaps with further restrictions). Why "connective", though? -In my (limited) understanding, approaches to the subject like that of Toen and Vezzosi are motivated as an approach to studying things like intersection theory in ordinary algebraic geometry. The picture is that a connective $E_\infty$ ring $R$ (incarnated as a simplicial commutative ring, usually) is an "infinitesimal thickening" of the ordinary ring $\pi_0 R$. This picture breaks down if $R$ is not connective, motivating one to restrict attention to the connective case (moreover, I don't know of a way to model nonconnective ring spectra analogous to simplicial rings). -But another motivation comes from homotopy theory, $TMF$, and the moduli stack of elliptic curves, which is a nonconnective derived Deligne-Mumford stack. When the basic motivating objects are nonconnective, it leaves me puzzled that Lurie continues to focus primarily on the connective case in Spectral Algebraic Geometry. -I see basically two mutually exclusive possible reasons for this: - -The theory of nonconnective derived algebraic geometry is wild / ill-behaved / hard to understand, so one restricts attention to the connective cases which is more tractable. -The theory of nonconnective derived algebraic geometry is a straightforward extension of the theory of connective derived algebraic geometry; it is easy to study nonconnective objects in terms of connective covers, but the results are most naturally phrased in terms of the connective objects, so that's the way the theory is expressed. - -Question A. Which of (1) / (2) is closer to the truth? -Maybe as an illustrative test case, here are two statements pulled at random from SAG. Let $R$ be a connective $E_\infty$ ring, let $Mod_R$ denote its $\infty$-category of modules, and $Mod_R^{cn}$ its $\infty$-category of connective modules (both of which are symmetric monoidal), and let $M \in Mod_R$. - -$M$ is perfect(=compact in $Mod_R$) iff $M$ is dualizable in $Mod_R$. -$M$ is locally free (= retract of some $R^n$) iff $M$ is connective and moreover dualizable in $Mod_R^{cn}$. - -Question B. Do these statements have analogs when $R$ is nonconnective? If so, are they straightforward extensions of these statements from the connective case? -For Question B, feel free to substitute a better example of a statement if you like. - -REPLY [12 votes]: As Tyler pointed out, it is "too easy" to be representable in the non-connective world. This might sound good, but it comes at the cost of geometric intuition. It is related to the fact that negative homotopy groups of the cotangent complex arise from "stacky" phenomena, while in the non-connective setting it will be impossible to distinguish what comes from stackiness and what comes from non-connectiveness of the rings themselves. I will try to give an example of this below. -1) First, a slight reformulation of Tyler's example (just to show that this is a very general phenomenon). Let $X = Spec(A)$ be an affine scheme and $U \subset X$ a quasi-compact open subscheme. -Lemma: When considered as a nonconnective spectral scheme, $U$ is affine. -Proof: $U$ can be written as the vanishing locus of some perfect complex $F \in Perf(X)$. In other words, as a non-connective spectral stack, the functor of points of $U$ is as follows: a $T$-point $T \to U$ is a $T$-point $x : T \to X$ such that $x^*(F) = 0$. According to Prop. 1.2.10.1 in Toën–Vezzosi's HAG II, there exists a canonical epimorphism $A \to B$ of non-connective $E_\infty$-ring spectra such that $Spec(B)$ has the functor of points described. (This $B$ is discrete if and only if $U$ is actually affine as a classical scheme.) -2) Let $X$ be a (connective) spectral scheme and let $\mathcal{A}$ be a quasi-coherent $\mathcal{O}_X$-algebra. Consider the relative Zariski spectrum, the (connective) spectral stack $Spec_X(\mathcal{A})$ whose space of $T$-points is $Maps_{\mathcal{O}_T\text{-alg}}(x^*(\mathcal{A}), \mathcal{O}_T)$, for any $X$-scheme $x : T \to X$. In particular you can take $\mathcal{A} = Sym_{\mathcal{O}_X}(F)$ for any perfect complex $F$; let $V_X(F) := Spec_X(Sym_{\mathcal{O}_X}(F))$ denote the "generalized vector bundle" associated to $F$. -One can compute (see Theorem 5.2 in Antieau-Gepner) the relative cotangent complex of $V_X(F)$ at any point $s : T \to V_X(F)$, for an $X$-scheme $x : T \to X$, as $x^*(F)$. You can read off a lot of information about $V_X(F)$ from the cotangent complex. Namely, say $F$ is of tor-amplitude $[a,b]$ (I'm going to use homological grading). If $a \ge 0$, i.e. $F$ is of non-negative tor-amplitude (hence connective), then $V_X(F)$ is representable by a (connective) spectral scheme (which is affine over $X$): by Zariski descent, you can assume $X$ is affine, and then $V_X(F) = Spec(\Gamma(X, Sym_{\mathcal{O}_X}(F)))$. If $a \le 0$, then $V_X(F)$ is a spectral $(-a)$-Artin stack (this is what I meant about the cotangent complex controlling "stackiness"). If further $b \le 0$ then $V_X(F)$ is smooth. -That was the connective story. In the non-connective world, $V_X(F)$ will "automatically" become representable by a non-connective spectral scheme even when $F$ is non-connective. In other words, by passing to the non-connective world, we allowed ourselves to replace stacks by "schemes", but on the other hand we lost something significant: it is not clear anymore what information we can read from the cotangent complex about the geometry of the "scheme".<|endoftext|> -TITLE: Is 0.24681012141618202224... transcendental? -QUESTION [37 upvotes]: Is a number whose infinite decimal part is the sequence of even numbers, transcendental? How about a number whose infinite decimal part is the odd numbers? Would the odds be more difficult to prove since they contain almost the entire sequence of primes? - -REPLY [55 votes]: In point of fact, K. Mahler proved in this paper that, if $p(x)$ in a non-constant polynomial such that $p(n) \in \mathbb{N}$ for every $n\in \mathbb{N}$, then the number -$$0.p(1)p(2)p(3)p(4)\ldots,$$ -which is formed concatenating after the decimal point the values of $p(1), p(2), p(3), \ldots$ (in that order), is a transcendental and non-Liouville number. \ No newline at end of file