diff --git "a/stack-exchange/math_overflow/shard_27.txt" "b/stack-exchange/math_overflow/shard_27.txt" deleted file mode 100644--- "a/stack-exchange/math_overflow/shard_27.txt" +++ /dev/null @@ -1,2563 +0,0 @@ -TITLE: Explicit expression for recursive sums -QUESTION [10 upvotes]: Let $t_1,t_2,\dots,t_k$ be non-negative integers. Can the following sum -$$f_k(t_1,t_2,\dots,t_k):=\sum_{j_1=0}^{t_1} \sum_{j_2=0}^{t_2+j_1} \sum_{j_3=0}^{t_2+j_2} \dots \sum_{j_k=0}^{t_k+j_{k-1}} 1$$ -be explicitly expressed as a polynomial in $t_1,t_2,\dots,t_k$ or via known combinatorial entities? -We surely have a recurrence formula: -$$f_{k+1}(t_1,t_2,\dots,t_{k+1}) = \sum_{j=0}^{t_1} f_k(j+t_2,\dots,t_{k+1}),$$ -which does not seem to easily unroll. -Just in case, first few terms are -\begin{split} -f_0 &= 1,\\ -f_1(t_1) &= 1+t_1,\\ -f_2(t_1,t_2) &= (1+t_1)(1+t_2) + \frac{t_1(1+t_1)}2,\\ -f_3(t_1,t_2,t_3) &= \left[ (1+t_1)(1+t_2) + \frac{t_1(1+t_1)}2 \right](1+t_3) + \frac{t_2(1+t_2)}2 + \frac{3t_2^2 + 6t_2 + 2}6t_1 + \frac{1+t_2}2t_1^2 + \frac16t_1^3. -\end{split} - -UPDATED. Billy Joe found that -$$f_k(n,d,d,\dots, d) = \frac{n+1}k \binom{n+k(d+1)}{k-1}.$$ -In particular, at $f_k(1,1,\dots,1)$ gives $(k+1)$-st Catalan number. - -REPLY [9 votes]: $f_k(t_1,\dots,t_k)$ is counting the number of integer points in the -"Pitman-Stanley polytope" $\Pi_k(t_1,\dots,t_k)$ defined here. The -notation $N(\Pi_k(\mathbf{t}))$ is used in this paper, which has the -determinantal formula given by Hugh Denoncourt, as well as a -combinatorial formula. The combinatorial formula is equivalent to -$$ f_k(t_1,\dots,t_k)=\sum_{\mathbf{h}\in K_k} \binom{t_1+h_1}{h_1} - \prod_{i=2}^k \binom{t_i+h_i-1}{h_i}, $$ -where -$$ K_k := \{\mathbf{h}\in\mathbb{N}^k\colon \sum_{i=1}^j h_i\geq - j\ \mathrm{for\ all}\ 1\leq j\leq k-1\ \mathrm{and}\ \sum_{i=1}^k - h_i=k \}. $$ -The set $K_k$ has a Catalan number $C_k$ of elements.<|endoftext|> -TITLE: Derivations on the continuous functions of a manifold -QUESTION [6 upvotes]: For a manifold $M$ a vector field is a derivation of the algebra $C^{\infty}(M)$ of smooth functions on $M$. What happens if look instead as derivations on the continuous functions of a manifold. I guess we get fewer derivations . . . but I'm not sure how one might prove this. - -REPLY [10 votes]: More is true: if $X$ is a topological manifold, then in fact $\operatorname{Der}(C(X)) = 0$, where $C(X)$ denotes the $\mathbb{R}$-algebra of $\mathbb{R}$-valued continuous functions on $X$. In particular, this is so for smooth manifolds $M$. -Here is one proof: https://ncatlab.org/nlab/show/derivation#DerOfContFuncts<|endoftext|> -TITLE: Specific application of Cauchy-Schwarz and Large Sieve -QUESTION [5 upvotes]: Im reading a paper by Matomaki here, and the following is stated (I'm paraphrasing): -"By the Cauchy-Schwarz inequality and the large sieve, we have -$$\sum_{q \leq Q}\frac{q}{\phi(q)}\sum_{\substack{\chi\text{(mod $q$)}\\\text{primtiive}}} \big{|}\sum_{a \in \mathcal{A}}\chi(a)\sum_{b \in \mathcal{B}}\chi(b) \big{|} \leq (Q^2 + N)(AB)^{1/2}$$ -where $Q,N$ are positive integers and $\mathcal{A}, \mathcal{B}\subseteq \{1,...,N\}$ and $|\mathcal{A}| = A$ and $|\mathcal{B}| = B$." -Now I am not so concerned with the application of the large sieve, but I am a little confused about how she applied Cauchy-Schwarz. Of course the large sieve she is referring to states -$$\sum_{q \leq Q}\frac{q}{\phi(q)}\sum_{\substack{\chi\text{(mod $q$)}\\\text{primtiive}}} \big{|}\sum_{n \leq N }a_n\chi(n)\big{|}^2 \leq (Q^2 + N)\sum_{n \leq N}|a_n|^2.$$ -But I am unsure of how she used Cauchy-Schwarz, especially with multiplicative characters. Does anyone have any thoughts? - -REPLY [11 votes]: The answer of Ofir Gorodetsky is perfectly fine, but one can also apply the Cauchy-Schwarz inequality for $L^2$ spaces directly. -Indeed, let us consider the $L^2$ space of functions on the set of pairs $(q,\chi)$, where the measure of the pair $(q,\chi)$ is $q/\phi(q)$. Then, for any complex-valued functions $f$ and $g$ on the set of these pairs, the inequality says that -$$\sum_{\substack{q \leq Q\\\text{$\chi$ mod $q$}\\\text{primitive}}}\frac{q}{\phi(q)}|f(q,\chi)g(q,\chi)|\leq\Biggl(\sum_{\substack{q \leq Q\\\text{$\chi$ mod $q$}\\\text{primitive}}}\frac{q}{\phi(q)}|f(q,\chi)|^2\Biggr)^{1/2}\Biggl(\sum_{\substack{q \leq Q\\\text{$\chi$ mod $q$}\\\text{primitive}}}\frac{q}{\phi(q)}|g(q,\chi)|^2\Biggr)^{1/2}.$$ -Applying this for -$$f(q,\chi):=\sum_{a\in\mathcal{A}}\chi(a) -\qquad\text{and}\qquad -g(q,\chi):=\sum_{b\in\mathcal{B}}\chi(b)$$ -yields the claimed result (using the large sieve inequality). -The point is that the result is purely formal from the Cauchy-Schwarz inequality. Ofir Gorodetsky's response adapts the usual proof of the Cauchy-Schwarz inequality to this situation.<|endoftext|> -TITLE: Does permission always work? -QUESTION [7 upvotes]: Suppose $g$ is a total computable injective function and $f$ is a total computable function satisfying $$g(x) k$, then it's not the case that $\Phi(X) = \bigcup_n f(n)$ and $\Psi(\bigcup_n f(n)) = X$. For ease of notation, let $Z = \bigcup_n f(n)$. -For this module, we choose large $m_1 > m_0$ and enumerate the axiom $m_0 \not \in X$ with use $m_1 \not \in ran(g)$. We require that all future definitions of $g$ must use values greater than $m_1$, and we wait until we see $f$ converge on all $y \le k$ and all $y$ where we have already defined $g(y) < m_0$. We wait further until we see an expansionary stage: some $s$, $r_0$ and $r_1$ with $\Phi_s(X_s\upharpoonright r_0) = Z_s\upharpoonright r_1$ and $\Psi(Z_s\upharpoonright r_1) = X_s\upharpoonright m_0+1$. We pick an $m_2 > r_1$, we use our next definition of $g$ to enumerate $m_1$ into $ran(g)$, and we enumerate the axiom $m_0 \not \in X$ with use $m_2 \not \in ran(g)$. We then require that there be no enumerations into $X$ below $r_0$ and no further definitions of $g$ with values below $r_1$. -Now, we wait until $f$ converges on all $y$ where we have already defined $g(y) < r_1$. If $f$ uses any of these values to enumerate a new element into $Z$ below $r_1$, we then have $\Phi_s(X_s\upharpoonright r_0)$ incompatible with $Z$, so we win by maintaining the restraint on $X$ below $r_0$. -If $f$ does not take this opportunity to enumerate new elements into $Z$ below $r_1$, then $f$ will have lost the opportunity to ever again enumerate further elements into $Z$ below $r_1$ (assuming that our choice of $k$ was correct and we maintain the restraint on $g$). So $Z\upharpoonright r_1 = Z_s\upharpoonright r_1$. Now we enumerate $m_0$ into $X$ and $m_2$ into $ran(g)$, giving us $\Phi(Z)(m_0) = \Phi_s(Z_s\upharpoonright r_1)(m_0) = 0 \neq X(m_0)$, so we have won. -Now just arrange these modules into a finite injury priority argument. - -REPLY [8 votes]: Dan's answer is very nice and should be the accepted answer. However, I thought it might be worth pointing out that there's a much easier counterexample to your first question (with the stricter requirement that $X = \bigcup_{n \in \mathbb{N}}f(n)$). Namely, we can let $g$ be any total computable injective function which enumerates a non-computable c.e. set and then just take $X$ to be the range of $g$ itself. -Here's why. Suppose $f$ is a total computable function with the properties listed in the question. Note that by modifying $f$ we can assume that for all $n > 0$, $g(n) < \min(f(n))$ (rather than all sufficiently large $n$). We can do this by resetting $f(n)$ to be $f(n) \cap \mathbb{N}_{> g(n)}$ and then setting $f(0)$ to consist of all missing values. -We can now compute $X$ (by computing the true stages of $g$) as follows. - -Set $n_0$ to be the minimum element of $f(0)$. Note that this must actually be the minimum element of $X$: due to the restriction on $f$ and $g$, $f$ cannot enumerate the least element of $X$ on any later stage. -Let $m_0$ be the point at which $g$ enumerates $n_0$ and set $n_1$ to be the second least element of $f(0)\cup f(1)\cup \ldots \cup f(m_0)$. Note that $n_1$ must be the second least element of $X$. -And so on. This results in a computable increasing enumeration $n_0, n_1, \ldots$ of $X$, hence $X$ is computable. - -Note that even if you change the requirement in your question to $g(n) \leq \min(f(n))$, this proof can easily be adapted by taking $X$ to be the range of $g - 1$.<|endoftext|> -TITLE: de Rham theorem for tempered distributions -QUESTION [5 upvotes]: I am wondering if the following statement holds. - -If $u\in \mathscr{S}'$ satisfies $\left< u,\Phi\right>=0$ for all $\Phi \in \mathscr{S}$ with $\mathrm{div}\, \Phi=0$, then there exists $p\in \mathscr{S}'$ such that $u=\nabla p$ in $\mathscr{S}'$. Here <,> denotes the dual pairing. - -It is well known that if we replace $\mathscr{S}$ with $\mathscr{D}$, then this is well-known theorem due to de Rham. -It is hard for me to find the analog version of tempered distribution. -Thank you very much for your time. - -REPLY [4 votes]: This works. As explained in my comment, we need only show that if $p\in\mathcal D'(\mathbb R^n)$ and $\nabla p\in\mathcal S'$ (vector valued), then $p\in\mathcal S'$. -The condition for a function $\varphi$ to be a divergence of a vector field is $\int\varphi=0$; see here. So if we fix a $\varphi_0\in\mathcal D$ with $\int\varphi_0=1$ and $\varphi\in\mathcal S$ is arbitrary, then we can write $\varphi=\textrm{div }\Phi + c\varphi_0$, for some $\Phi\in\mathcal S$, and with $c=\int\varphi$. -Let $a=(p,\varphi_0)$. We then obtain -$$ -(p,\varphi)=(p,\textrm{div }\Phi) + a \int\varphi = -(\nabla p, \Phi) + a\int\varphi . -$$ -This shows that $p$ is well defined and continuous on $\mathcal S$, so is a tempered distribution. (Strictly speaking, I was too lazy here to actually show the continuity explicitly and I more relied on the meta-principle that anything that has a natural definition as a functional on $\mathcal S$ is continuous. To do it properly, we'd have to establish that $\Phi$ can be made to depend continuously on $\varphi$ in the topology of $\mathcal S$.)<|endoftext|> -TITLE: Why does representing functors help solving Diophantine equations? -QUESTION [25 upvotes]: Here I read: - -Another insight of Grothendieck and his school was, how important it is to represent functors in algebraic geometry - regardless of what you want at the end. [as Mazur reports, Hendrik Lenstra was once sure that he did want to solve Diophantine equations and did not want to represent functors - and later he was amused that he represented functors to solve Diophantine equations.] - -My question is already in the title: Why does representing functors help solving Diophantine equations? -As a category theory enthusiast I am fascinated by the fact that sometimes category theory is helpful to prove concrete things (here: solving Diophantine equations). So I'd really want to know how that works, roughly. - -REPLY [8 votes]: Let me give an answer that pertains to the Diophantine equation that, according to David Speyer's answer, Lenstra was specifically talking about. -How does representing functors help solve the Diophantine equation $a^n + b^n=c^n$? -A solution to this Diophantine equation defines an elliptic curve $y^2 = x (x-a^n)(x-b^n)$ (Frey). The Galois group of the rational numbers acts on the $\overline{\mathbb Q}$-points of this curve, thus on the $\ell^m$-torsion points for each $m$. Taking an inverse limit as $m$ goes to $\infty$, we obtain a Galois action on a rank two free $\mathbb Z_\ell$-module (the Tate module). -We also obtain Galois actions on rank two free $\mathbb Z_\ell$-modules from modular forms, after Eichler-Shimura and Deligne. In fact, for many different rings $S$ we can obtain Galois actions on rank two free $S$-modules. -It turns out that the Galois representations arising from curves of the form $y^2 = x (x-a^n)(x-b^n)$ have very special properties on congruence mod $n$, properties which the Galois representations arising from modular forms cannot have (Ribet). So if we can show that every Galois representation arising from an elliptic curve also arises from a modular form, we can obtain a contradiction from any solution to the Diophantine equation. -Now here's where we introduce the functors to be represented. We consider the category of complete local rings $S$ of residue characteristic $\ell$, and the functor that sends each such ring to the set of isomorphism classes of rank two free $S$-modules with a Galois action that is congruent mod $\ell$ to some fixed Galois representation, satisfying some conditions known to hold for elliptic curves. We can consider another functor that sends the ring $S$ to only the set of isomorphism classes that arise from modular forms. -We'd like to prove the natural transformation from one of these functors to the other is an isomorphism, which is equivalent to the statement about every Galois representation arising from a modular form. In general, this is a hard problem. -But if we can represent these functors by complete local rings of residue characteristic $\ell$, then it becomes equivalent to proving that a map of local rings $R \to T$ is an isomorphism. That's a question which is much easier to tackle, because we can use all the commutative algebra theory of local rings. In particular, we can hope to prove a criterion for the map to be an isomorphism that depends on concrete properties of these rings which can be expressed in terms of the original functors. (The simplest case of this is that a map of smooth local rings is an isomorphism if and only if it's an isomorphism on the residue field and tangent space, which both have a functorial interpretation in terms of very simple rings.) We then reduce the problem to checking some new properties are satisfied by the functors. These properties turn out to be much more tractable than proving the isomorphism directly. In part, this is because they connect to areas where there is a pre-existing theory (related to $L$-functions, the class number formula, Iwasawa theory) that can be applied (though of course deep new ideas were needed as well). - -REPLY [6 votes]: Solving a Diophantine equation is the same thing as showing that the functor defined by a certain scheme $S$ of finite type over $\bf Z$ gives a non-empty set when evaluated at ${\rm Spec}({\bf Z})$. So it is about a functor in the first place. -Now when trying to solve that (extremely hard) problem, one will be led to study different and more tractable functors on the category of schemes of finite type over ${\rm Spec}({\bf Z})$. In order to say something about these new functors, one will wonder whether they too have geometric meaning, which roughly translates to asking whether they are representable in the same category. -As explained in Donu Arapura's answer, this is at work in Faltings's proof of the Mordell conjecture but an easier example is Buium's proof of the Manin-Mumford conjecture for curves. In this situation, one starts with a curve $C$ over a number field $K$ and one takes a prime $\mathfrak p$ of $K$, which is unramified over $\bf Z$ and where $C$ has good reduction. A property of a prime to $\mathfrak p$ torsion point of the Jacobian of $C$, which lies on $C$, is that it provides a point of the curve $C_{{\mathcal O}_K/{\mathfrak p}}$ (the curve over a finite field obtained by reduction), which lifts to an element of $C_{{\mathcal O}_K/{\mathfrak p}^2}({\mathcal O}_K/{\mathfrak p}^2)$, which has the property that it is divisible in the Jacobian by the prime number $p$ lying under $\mathfrak p$. Buium's proof shows that there can only be finitely many such points and to do this, he uses that fact that the set $C_{{\mathcal O}_K/{\mathfrak p}^2}({\mathcal O}_K/{\mathfrak p}^2)$ can be seen as the set of ${\mathcal O}_K/{\mathfrak p}$-points of a certain variety over ${\mathcal O}_K/{\mathfrak p}$, which represents a functor, which appears as an adjoint to a naturally representable functor. If one could not represent this adjoint, one couldn't understand the set $C_{{\mathcal O}_K/{\mathfrak p}^2}({\mathcal O}_K/{\mathfrak p}^2)$ geometrically and it would not be clear how to approach the statement proven by Buium. -So I would say that one solves Diophantine equations by representing functors, because a Diophantine equation is a question about a functor, which can be studied by replacing it by simpler ones. If you can show that the new ones are also representable then you will be led to another Diophantine problem and you can start all over again. This is exactly what happens in the example given above.<|endoftext|> -TITLE: Motivation of the fundamental theorem of covering spaces -QUESTION [6 upvotes]: The fundamental theorem of covering spaces states that for a nice topological space $X$, there is an equivalence of categories between covering spaces over $X$ and left $\pi_1(X)$-sets. "Grothendieck's Galois theory" states that is is even true for $X$ a connected scheme, if we replace $\pi_1(X)$ be its étale analogue, covering spaces by finite étale covers, and left $\pi_1(X)$-sets by finite continuous left $\pi_1(X)$-sets. -Certainly these are nice statements, because they both show that two a priori different stories ($G$-sets vs. coverings) turn out to be the same. However, I keep wondering: -QUESTION: What is the actual motivation people (could) came up with these statements? This could be answered by: What are some applications of these theorems? -A more down-to-earth formulation used in algebraic topology courses is the "local" version using posets instead of categories (isomorphism classes of covering spaces $\cong$ conjugacy classes of subgroups of $\pi_1(X)$), as discussed in Hatcher's book Theorem 1.38. I skimmed through this section of Hatcher's book but I can't find any concrete motivation except "here's a nice theorem / classification result: ...". - -REPLY [8 votes]: Many results in algebraic topology are proved using an argument along the following lines. Suppose such and such holds. Then there is a subgroup of the fundamental group with the following properties. Pass to the corresponding covering space and do something. Contradiction. For example, Miles Reid has a nice argument along these lines which shows that the etale fundamental group of a Godeaux surface has size at most 5. See theorem 0.2.1 of his paper Godeaux and Campedelli surfaces: https://homepages.warwick.ac.uk/~masda/surf/more/Godeaux.pdf<|endoftext|> -TITLE: Does the first Laplacian eigenfunction on a homogeneous space have a unique maximum? -QUESTION [8 upvotes]: For convex domains $\Omega \subset \mathbb{R}^n$ with Dirichlet boundary conditions, it's known that any first Laplacian eigenfunction is log-concave. In particular, it has a unique maximum. -These are functions $f$ satisfying $-\Delta f = \lambda_1 f$ for the smallest possible $\lambda_1 > 0$ and $f|_{\partial \Omega} = 0$. -Q: Is there such a concavity result for homogeneous spaces, ie. Riemannian manifolds with a transitive action of the isometry group? Does $f$ at least have a unique maximum? -I imagine that this might even be apparent from an explicit formula for these eigenfunctions, but I haven't been able to find one in general. -I can at least check that it holds for spherical harmonics. Actually for spheres I think it can be derived from the domain result by separation of variables (take $\Omega = \mathbb{R}^n$). Perhaps such a method can work for any homogeneous space with a nice enough embedding, such as as the boundary of a convex domain? -I'd appreciate any thoughts or relevant references! - -REPLY [10 votes]: The flat torus $\mathbb{T} = \mathbb{R}^2/\Lambda$ gives a counterexample: The first nontrivial eigenvalue is of the form $\lambda_1 = \xi_1^2+\xi_2^2$, where $\xi = (\xi_1,\xi_2)$ is a nonzero element of the dual lattice $\Lambda^*$ of smallest norm, and the correspnding eigenfunctions are of the form $f(x_1,x_2) = a\cos(\xi_1 x_1 + \xi_2 x_2 + b)$ for some constants $(a,b)$. This function has a whole circle of maxima. -In general, on a compact manifold, an eigenfunction with nonzero eigenvalue must change sign because its average value on the manifold must be zero. (Integration by parts.) -Oh, another example occurred to me that you might find more interesting: Let $M = \mathrm{SO}(3)$ with its standard biïnvariant Riemannian metric. The first nontrivial eigenvalue has multiplicity 9, and the corresponding eigenfunctions are the 9 entries $a_{ij}$ of the standard matrix embedding of $\mathrm{SO}(3)$ into the space of 3-by-3 matrices. Each $a_{ij}$ has its maximum and minimum values equal to $\pm 1$, but it attains each on a circle embedded in $\mathrm{SO}(3)$. Meanwhile $f = -a_{11}-a_{22}-a_{33}$ has a maximum value of $1$, attained on a copy of $\mathbb{RP}^2$ embedded in $\mathrm{SO}(3)$, and a minimum value of $-3$, attained only at $I_3\in\mathrm{SO}(3)$.<|endoftext|> -TITLE: Is it possible to solve sextic equations using the Fox H function? -QUESTION [5 upvotes]: Although the Kampé de Fériet function can solve the sextic equation, the details about it are shrouded in the fog of more than a century ago. -In contrast, we know more about the Fox H function, and we have even succeeded in implementing it on CAS. -I would like to know if the experts of the Fox H function have a more modern understanding of the solution of the sextic equation. - -REPLY [6 votes]: Trinomial sextic equations $z^6-z-t=0$ can be solved in terms of the Fox H function: -$$z_j=\tfrac{1}{5}t H_{2,3}^{1,2}\biggl(te^{j\frac{2 i \pi }{5}}\biggl| -\begin{array}{c} -(0,1),(0,6/5) \\ - (0,1),(-1,1),(0,1/5) \\ -\end{array} -\biggr)+e^{-j\frac{2 i \pi }{5}},\;\;j\in\{0,1,2,3,4,5\}.$$ -Mathematica input: -(t/5)*FoxH[{{{0,1},{0,6/5}},{}},{{{0,1}},{{-1,1},{0,1/5}}},t Exp[(2 Pi I)/5]^j]+Exp[(2 Pi I)/5]^-j - -This is a special case of a more general solution of trinomial equations of arbitrary order, $z^\alpha-\beta z+\gamma=0$, described in All the trinomial roots, their powers and logarithms from the Lambert series, Bell polynomials and Fox–Wright function (J. Math. Chem. 57 (2019) pp 59–106).<|endoftext|> -TITLE: Is the ring of power series with coefficients in a field free as a module over the polynomials subring? -QUESTION [5 upvotes]: Is the ring of power series with coefficients in a field free as a module over the polynomials subring? - -REPLY [12 votes]: No. Take some element $f$ of $K[x]$ which is not in the ideal $(x)$ and is not invertible. Then -$$K[[x]] \otimes_{K[x]} K[x]/(f) \cong K[[x]]/(f) = 0$$ -where the latter follows from the fact that since $f \notin (x)$, it becomes invertible in $K[[x]]$. -In particular, $K[[x]]$ is not faithfully flat over $K[x]$, so it, being a nonzero module, cannot be free over it.<|endoftext|> -TITLE: Are the non-free factors of Grushko decomposition of a finitely generated convex-cocompact (but not cocompact) subgroup of PSL$(2,\mathbb{R})$ finite? -QUESTION [6 upvotes]: See Grushko decomposition theorem. -Are the non-free factors of Grushko decomposition of a finitely generated convex–cocompact (but not cocompact) subgroup of $\operatorname{PSL}(2,\mathbb{R})$ finite? -In the cocompact case it is not true, since the group is not a free group and cannot be split into a non-trivial free product. -For convex–cocompact but not cocompact I know of particular examples with affirmative answer. Is it always the case? - -REPLY [7 votes]: For a discrete, noncocompact subgroup $\Gamma < \text{PSL}(2,\mathbb R)$, the quotient $\mathbb H^2 / \Gamma$ is a noncompact, 2-dimensional oriented orbifold, i.e. a noncompact surface with an orbifold locus consisting of cone singularities forming a closed, discrete subset. -Assuming in addition the finite type hypothesis, the underlying surface of the quotient orbifold is obtained from some closed oriented surface by removing a finite subset, and the orbifold locus is a finite set. -Such an orbifold has a spine (an orbifold deformation retract, thus having the same fundamental group) which is a finite graph of groups, whose vertices include all of the cone singularities and perhaps some other points with trivial vertex group, and whose edges all have trivial group. -The Grushko decomposition of the graph-of-groups fundamental gorup therefore has the form $A_1 * ... * A_K * F_n$ for some finite cyclic groups $A_1,...,A_K$ and some finite rank free group $F_n$. -So yes, the non-free factors of the Grushko decomposition are all finite.<|endoftext|> -TITLE: Complex vector bundles on compact complex manifolds -QUESTION [7 upvotes]: The complex vector bundles on complex projective space $\mathbb{CP}^n$ are explicitly classified for low dimensions. When $n\leq 3$, they are exactly the holomorphic vector bundles; when $n\geq 4$ we can use the Schwarzenberger condition together with the result of A. Thomas to classify complex vector bundles according to chern classes. -In general we have the Grassmann manifold $\text{Gr}(\mathbb{C}^{\infty})$ as the classfying space of complex vector bundles, but it is not so useful for computation (e.g. to describe the set of isomorphic classes $\text{Vect}^n(X)$ and topological $K$ group). -Now I wonder whether we have some known examples on the classification of complex vector bundles over other compact complex manfolds e.g. hypersurfaces. - -REPLY [11 votes]: This is an explanation of my comment above, namely: "Complex vector bundles over a CW complex of dimension $\leq 4$ are classified by their Chern classes and rank. Moreover, every possible choice of Chern classes and rank arises. In particular, we have a complete answer for complex manifolds $X$ with $\operatorname{dim}_{\mathbb{C}}X \leq 2$." -Let $B$ be a topological space. A Postnikov tower for $B$ consists of: - -a sequence of spaces $B_n$ with $\pi_i(B_n) = 0$ for $i > n$, -maps $\alpha_n : B \to B_n$ which induce isomorphisms on $\pi_n$ for $i \leq n$, and -fibrations $f_n : B_{n+1} \to B_n$ such that $f_n\circ\alpha_{n+1} = \alpha_n$. - -Note, from the long exact sequence in homotopy, we see that $f_n$ has fiber $K(\pi_{n+1}(B), n+1)$. If $\pi_1(B)$ acts trivially on $\pi_n(B)$ for every $n$, then one can take $f_n$ to be a principal fibrations for every $n$. The principal fibration $f_n$ is classified by $k_n \in H^{n+2}(B_n; \pi_{n+1}(B))$ called the $n^{\text{th}}$ $k$-invariant of $B$. -Consider the Postnikov tower of $BU(r)$, the classifying space for rank $r > 1$ complex vector bundles. Note that $\pi_1(BU(r)) \cong \pi_0(U(r)) = 0$, so all the fibrations will be principal and hence have associated $k$-invariants. As $\pi_2(BU(r)) \cong \pi_1(U(r)) = \mathbb{Z}$, the first non-trivial stage of the Postnikov tower is $BU(r)_2$ which is a $K(\mathbb{Z}, 2)$ and is therefore homotopy equivalent to $\mathbb{CP}^{\infty}$. Now note that $\pi_3(BU(r)) \cong \pi_2(U(r)) = 0$ so $BU(r)_3 = BU(r)_2$. At the next step however we have $\pi_4(BU(r)) \cong \pi_3(U(r)) \cong \mathbb{Z}$, so there is a principal fibration $f_3 : BU(r)_4 \to BU(r)_3$ with fiber $K(\mathbb{Z}, 4)$ classified by $k_3 \in H^5(BU(r)_3; \mathbb{Z}) = H^5(\mathbb{CP}^{\infty}; \mathbb{Z}) = 0$. As $k_3 = 0$, the principal fibration $f_3$ is trivial and hence $BU(r)_4 = BU(r)_3\times K(\mathbb{Z}, 4) = K(\mathbb{Z}, 2)\times K(\mathbb{Z}, 4)$. -The map $\alpha_4 : BU(r) \to BU(r)_4$ induces an isomorphism on $\pi_i$ for $i \leq 4$, so for any CW complex $X$ of dimension $\leq 4$, there is a bijection -$$[X, BU(r)] \to [X, BU(r)_4] = [X, K(\mathbb{Z}, 2)\times K(\mathbb{Z}, 4)] = H^2(X; \mathbb{Z})\times H^4(X; \mathbb{Z}).$$ -After composing with a self-homotopy equivalence of $K(\mathbb{Z}, 2)\times K(\mathbb{Z}, 4)$ if necessary, one can arrange for this map to correspond precisely to $(c_1, c_2) : \operatorname{Vect}^{\mathbb{C}}_r(X) \to H^2(X; \mathbb{Z})\times H^4(X; \mathbb{Z})$. So any rank $r > 1$ complex vector bundle $E \to X$ is determined up to isomorphism by $c_1(E)$ and $c_2(E)$. Moreover, for every $\alpha \in H^2(X; \mathbb{Z})$ and $\beta \in H^4(X; \mathbb{Z})$ and $r > 1$, there is a rank $r$ complex vector bundle $E \to X$, unique up to isomorphism, such that $c_1(E) = \alpha$ and $c_2(E) = \beta$. For $r = 1$, we instead obtain $BU(1)_2 = BU(1)$ which gives rise to the familiar bijection $c_1 : \operatorname{Vect}^{\mathbb{C}}_1(X) \to H^2(X; \mathbb{Z})$; note, there is no restriction on $X$ in this case. -Complex vector bundles of rank $r > 1$ over CW complexes of dimension $> 4$ are no longer determined by their Chern classes. For example, as $\pi_5(BU(2)) = \pi_4(U(2)) \cong \mathbb{Z}_2$, there is a non-trivial rank two complex vector bundle $E \to S^5$ which necessarily has $c_1(E) = 0$ and $c_2(E) = 0$. A construction of this vector bundle can be found here.<|endoftext|> -TITLE: Can a variety acquire points in a purely transcendental extension? -QUESTION [9 upvotes]: Let $X$ be an algebraic variety defined over a field $k$ of characteristic zero. Suppose $X(K)$ is non-empty for some extension $K/k$. Is it true that $X(L)$ is non-empty where $L$ is the algebraic closure of $k$ in $K$? An equivalent formulation is: suppose $X(k)$ is empty; does that force $X(K)$ to be empty for every purely transcendental extension $K/k$? -In my particular application $k$ is a number field with a real place and $K=\mathbb{R}$ and then claim is true since the theory of real closed fields eliminates quantifiers, but I'm wondering about the general case. Relatedly Chevalley's Theorem (elimination of quantifiers for algebraically closed fields) does produce $\bar{k}$-points, but I don't see why they have to be $\bar{k}\cap K$-points. - -REPLY [11 votes]: The following answers your second question. - - -Let $X$ be a finite type separated scheme over an infinite field $k$. -Let $K/k$ be a purely transcendental extension. -If $X(k)$ is empty, then $X(K)$ is empty. - - -Proof. We may assume that $K$ has finite transcendence degree over $k$. Then, any element of $X(K)$ extends to a morphism $U\to X$, where $U$ is a dense open of some $\mathbb{P}^n$. Since $k$ is infinite, the set $U(k)$ is dense (hence non-empty). This contradicts $X(k)$ being empty. -Your first question is answered by Laurent Moret-Bailly. - - -Let $X$ be a geometrically connected finite type scheme over $k$ (and thus non-empty). Let $K=K(X)$ be the function field of $X$. Then $X(K)$ is non-empty. However, there is no reason for $X(k)$ to be non-empty. Take $X$ to be a smooth proper conic over $\mathbb{Q}$ with no $\mathbb{Q}$-points, for example.<|endoftext|> -TITLE: Unrigorous British mathematics prior to G.H. Hardy -QUESTION [25 upvotes]: I was looking at a bio-movie of Ramanujan last night. Very poignant. -Also impressed by Jeremy Irons' portrayal of G.H. Hardy. -In G.H. Hardy's wiki page, we read: -. . . "Hardy cited as his most important influence his independent study of Cours d'analyse de l'École Polytechnique by the French mathematician Camille Jordan, through which he became acquainted with the more precise mathematics tradition in continental Europe." -and -. . . "Hardy is credited with reforming British mathematics by bringing rigour into it, which was previously a characteristic of French, Swiss and German mathematics. British mathematicians had remained largely in the tradition of applied mathematics, in thrall to the reputation of Isaac Newton (see Cambridge Mathematical Tripos). Hardy was more in tune with the cours d'analyse methods dominant in France, and aggressively promoted his conception of pure mathematics, in particular against the hydrodynamics that was an important part of Cambridge mathematics." -Are we to understand from this that up to the late 1800s, British mathematics used only partial or inductive proofs or what ? -On the face of it, this would have been quite a state of affairs. -What exactly - in general or by a specific example - did Hardy bring to mathematics by way of rigour that had previously been absent ? -If someone introduced a new and sketchily proven theorem in the days of Hardy's childhood - and we are talking about Victorian times here (...) - then surely all the mean old men of the profession would have been disapproving of it and would obstruct its publication ? - -REPLY [8 votes]: The quote in question contrasts Hardy’s rigor with “the hydrodynamics that was an important part of Cambridge mathematics.” -So to understand Hardy’s role, it makes sense to look at that hydrodynamics, e.g. Alfred Basset’s “Treatise on Hydrodynamics, with Numerous Examples”, conveniently available online at the Internet Archive. (Since Basset studied math at Cambridge, did his math afterwards without any professorship, and published multiple editions of this book with Cambridge University Press, I think it’s fair to call this “Cambridge math”; one could probably find this or similar books listed in Cambridge math syllabi too.) -Browsing through the book, I do not see a lot of proofs; I see calculations from mathematical hypotheses deemed appropriate for various physical problems. -So Hardy brought his rigor to pure mathematics; what came before him in England was less about pure mathematics done unrigorously, and more about applied mathematics done under different constraints.<|endoftext|> -TITLE: Can you perturb an inscribed polytope so all its edges grow? -QUESTION [12 upvotes]: Consider the family of convex simplicial polytopes with vertices in the unit sphere of $\mathbb{R}^n$ which have the origin as an interior point. -My question is the following: - -Let $P, P'$ be two non-congruent combinatorially identical polytopes from the above family, with -vertices of each polytope labelled so that the face lattices of the -two polytopes are identical. Is it possible that $\|a-b\|\leq\|a'-b'\|$ whenever -edge $\{a,b\}$ in $P$ corresponds to edge $\{a',b'\}$ in $P'$? -In other words, can you perturb one such polytope to another making all the edges grow? - -This is impossible in $\mathbb{R}^2$: Take an inscribed polygon from the family and perturb it so that it remains in the family and has the same combinatorial structure. If all the edges grow, then all the central angles grow, which would make those central angles add up to a value larger than $2\pi$. Therefore in a perturbed polygon, if some edges grow then other edges must contract. -Does this idea transfer to dimension $n=3$, or more generally to $n\geq3$? And, if yes, is there a reference? - -REPLY [4 votes]: OK, let me address the case of a simplex. -In fact, it follows from the `dual Kneser--Poulsen'conjecture, as stated, e.g., in this nice paper. A good thing is that the simplex has only $n+1$ vertices, and in such partial case, according to that paper, the conjecture has been established by Gromov. To reach the desired result, apply the conjecture to the balls of unit radius centered at the vertices of both simplices. (One still needs to check that the intersection strictly changes, but that is doable). -Anyway, here is a direct proof. -Let $u_0,\dots,u_n$ be the vertices of the first simplex (hence unit vectors), and let $v_0,\dots,v_n$ be the vertices of the second one, with $\|u_i-u_j\|\leq\|v_i-v_j\|$ for all $i$ and $j$, where at least one inequality is strict. -Choose the positive $\alpha_i$ such that -$$ - \sum_i\alpha_i=1 \quad\text{and}\quad \sum_i \alpha_iu_i=0; -$$ -these are the barycentric coordinates of $0$ in the first simplex. Take the point $p$ with the same barycentric coordinates in the second, i.e., -$$ - p=\sum_i \alpha_iv_i. -$$ -Clearly, $p$ lies inside the second simplex. -Notice that $\langle v_i,v_j\rangle\leq\langle u_i,u_j\rangle$. Therefore, for all $j$ we have -$$ - \langle v_j,v_j-p\rangle - =-\sum_{i\neq j}\alpha_i\langle v_j,v_i\rangle - +(1-\alpha_j)\langle v_j,v_j\rangle - \geq -\sum_{i\neq j}\alpha_i\langle u_j,u_i\rangle - +(1-\alpha_j)\langle u_j,u_j\rangle - =\langle u_j,u_j\rangle=1, -$$ -where sometimes an inequality is strict. Hence $\|v_j-p\|\geq 1$, with sometimes strict inequality; in particular, $p\neq 0$. -This shows that all the $v_i$ lie on the unit sphere centered at $0$, but outside the (open) unit ball centered at $p$. This shows that they all (along with $0$( are on the same side of the hyperplane equidistant from $0$ and $p$. Hence the second simplex cannot contain $p$ --- a contradiction.<|endoftext|> -TITLE: Field extensions over which algebraic varieties cannot acquire points -QUESTION [7 upvotes]: The following fact (slightly reworded here) is proven in this answer: - -If $K$ is a purely transcendental extension of an infinite¹ field $k$, then whenever a separated scheme $X$ of finite type over $k$ has points over $K$, it already has points over $k$ (viꝫ., $X(k) = \varnothing$ implies $X(K) = \varnothing$). - -Let us say that the field extension $K$ of $k$ is “astigmagenic”² when the conclusion of this statement (the part that follows the “then”) holds. Thus, the above states that purely transcendental extensions of infinite fields are astigmagenic. But they are not the only ones: if $k$ is algebraically closed, then any extension of $k$ is astigmagenic (and it need not be purely transcendental: consider the field of functions of an algebraic variety that isn't rational). -Question: Can we characterize these “astigmagenic” field extensions in a simpler way? (E.g., do the finite type ones coincide with rational function fields of algebraic varieties for which the rational points are Zariski-dense?) -Note: I used “separated scheme of finite type” in the definition because this is what the linked answer does. But we can also try variations around this: e.g., maybe say that $K$ is weakly astigmagenic when $X(k) = \varnothing$ implies $X(K) = \varnothing$ for $X$ a geometrically integral separated scheme of finite type. It is very unclear to me how much this changes the condition, so I'm interested in the relations between the various variations: feel free to answer with whatever variation seems to make the question most natural or interesting. - -As pointed out by Laurent Moret-Bailly in a comment, the word “infinite” was missing from the answer. - -Pardon my Greek. This is supposed to mean “that does not create points”. - -REPLY [10 votes]: Let $K$ be a finite type field extension of $k$ which corresponds to a rational function field of an algebraic variety $V$ for which the rational points are not Zariski dense. -Let $U$ be the complement in $V$ of the Zariski closure of the rational points. -Then $U$ has no $k$-point, but has a $K$-point (the generic point), so $K/k$ is not astigmagenic. -Since every finite type extension is the rational function field of a variety, it follows that a finite type extension is astigmagenic if and only if it is the rational function field of an algebraic variety for which the rational points are Zariski-dense. (Well, that proves "only if", and "if" follows from the same proof as for purely transcendental - given a $K$-point, we obtain a rational map from the variety, hence from an open subset of the variety, and then look at the image of a rational point).<|endoftext|> -TITLE: Deformation of (locally) ringed spaces and of their abelian categories of modules -QUESTION [7 upvotes]: I am interested in the general theory of deformations locally ringed spaces in the same language of the deformation theory of schemes/varieties that is already widely available. I am aware for example of what is written down in the stacks project (for example Tag 08UX) but I'm failing to find a more through treatment specially one that would put in contrast the differences with the special case of schemes/varieties. In my literature review, however, I stumbled into the following quote in -Lowen, Wendy; van den Bergh, Michel, Deformation theory of abelian categories, Trans. Am. Math. Soc. 358, No. 12, 5441-5483 (2006). ZBL1113.13009. - -These results confirm the fundamental insight of Gerstenhaber and Schack [6, -8] that one should define the deformations of a ringed space $(X, \mathcal{O}_X)$ not as the deformations of $\mathcal{O}_X$ as a sheaf of k-algebras, but rather as the deformations of the k-linear category $\mathfrak{u}$ (or of the “diagram” $(\mathcal{B}, \mathcal{O}_{\mathcal{B}})$ in case $X\in \mathcal{B}$). These “virtual” deformations are nothing but the deformations of the abelian category $Mod(\mathcal{O}_{X})$. - -The cited category $\mathfrak{u}$ is an appropriately defined category which is used to show that the deformations of the category of presheaves of modules over an appropriate basis $\mathcal{B}$ are equivalent to deformations of sheaves of modules over $\mathcal{O}_{X}$ -I have skimmed through the literature including the cited papers and while I have found some indication of the exact meaning of this claim I am still confused. My interpretation is that the claim is either strictly noncommutative in nature ( so passing through a reconstruction theorem ), or deformations of this abelian category directly give the 'correct' deformation theory of the space in some sense ( for example the relationship with Hochschild cohomology in Lowen, Wendy; van den Bergh, Michel, Hochschild cohomology of Abelian categories and ringed spaces, Adv. Math. 198, No. 1, 172-221 (2005). ZBL1095.13013.) -My first question would then be, could somebody clarify what exactly is the quote saying? -My second question is then, if the space is a (sufficiently nice) scheme then am I to understand that the deformation theory of the category of quasi-coherent sheaves controls the deformations of the scheme as I would find it written in classical texts? -Thanks in advance -The cited papers on the quote are, respectively: -Gerstenhaber, M.; Schack, S. D., On the deformation of algebra morphisms and diagrams, Trans. Am. Math. Soc. 279, 1-50 (1983). ZBL0544.18005. -and, -Gerstenhaber, Murray; Schack, Samuel D., The cohomology of presheaves of algebras. I: Presheaves over a partially ordered set, Trans. Am. Math. Soc. 310, No. 1, 135-165 (1988). ZBL0706.16021. - -REPLY [3 votes]: The answer to your second question is "no", I think. -Let's assume that sufficiently nice means that it is a smooth algebraic variety over a field of characteristic $0$. Then, as written in Severin Barmeier's answer, first order deformations of the abelian category of modules are given by $HH^2(X)\simeq H^0(X,\wedge^2 T_X)\oplus H^1(X,T_X)\oplus H^2(X,\mathcal O_X)$ [note that this isomorphism is a consequence of the Hochschild-Kostant-Rosenberg iso]. Now, only $H^1(X,T_X)$ amounts for classical first order deformations of $X$ as a scheme. $H^0(X,\wedge^2 T_X)$ provides first order (possibly noncommutative) deformations of the product of $\mathcal O_X$, while $H^2(X,\mathcal O_X)$ provides first order deformations of the glueing data of $\mathcal O_X$ (these are gerby/stacky deformations of $\mathcal O_X$). -All together, these three parts provide first order deformations of $\mathcal O_X$ as an algebroid stack, in the sense of Kontsevich (see Appendix A of this paper, as well as papers by Yekutieli or Kashiwara-Schapira). -As for your first question, the claims says that deforming the structure sheaf $\mathcal O_X$ as a sheaf of (possibly noncommutative) algebras is not enough to get all deformations of $Mod(\mathcal O_X)$. Indeed, for first order deformation this would only give $H^0(X,\wedge^2 T_X)\oplus H^1(X,T_X)$, and one would miss the part $H^2(X,\mathcal O_X)$. To get all deformations of $Mod(\mathcal O_X)$, the claim says that one shall deform the $k$-linear category $\mathfrak{u}$ (which is essentially the same as deformating $\mathcal O_X$ as an algebroid stack in the sense of Kontsevich- under the assumption that $X$ has a basis $\mathcal B$ so that for every $U\in\mathcal B$, $H^1(U,\mathcal O_U)=H^2(U,\mathcal O_U)=0$).<|endoftext|> -TITLE: What is $TP(\mathbb{Z}_p)$? -QUESTION [8 upvotes]: Let $TP$ be periodic topological cyclic homology. What is $\pi_* TP(\mathbb{Z}_p)$? -(i) I know that $\pi_* TP(\mathbb{F}_p) \cong \mathbb{Z}_p[v^{\pm 1}]$ with $v$ in degree $-2$ by IV.4.8 of Nikolaus-Scholze. -(ii) I know that $\pi_n THH(\mathbb{Z}_p)$ is $\mathbb{Z}_p$ for $n=0$, $\mathbb{Z}_p/m$ for $n=2m-1 \geq 0$, and zero otherwise. I tried to write down the Tate spectral sequence for $\pi_* TP(\mathbb{Z}_p)$ but it seems very complicated. I think Tsalidis and Bokstedt-Madsen calculated the homotopy groups of $TP(\mathbb{Z}_p)/p$ but I don't know how to get $\pi_* TP(\mathbb{Z}_p)$ from this. -(iii) A related question is what is the Breuil-Kisin twisted prismatic cohomology $\Delta_{\mathbb{Z}_p}\{i\}$? - -REPLY [4 votes]: The calculation of $\pi_* TP(\mathbb{F}_p) = \pi_* THH(\mathbb{F}_p)^{tS^1} = \pi_* \widehat{\mathbb{H}}(S^1, THH(\mathbb{F}_p))$ (the notation has changed over the years) was first published by Hesselholt-Madsen (Topology, 1997). The calculation of $\pi_* TP(\mathbb{Z})/p$ for odd primes $p$ was published by B"okstedt-Madsen (conference proceedings, 1994 and 1995) and, independently, by Tsalidis (Amer J. Math, 1997). They also calculated $\pi_* TC(\mathbb{Z})/p$. Comparison with known spectra related to topological $K$-theory allowed a determination of $\pi_* TC(\mathbb{Z})_p$, see Rognes (Math. Proc. Camb. Philos Soc., 1993, Corollary 3). I also made the corresponding calculations of $\pi_* TP(\mathbb{Z})/2$, $\pi_* TC(\mathbb{Z})/2$ and $\pi_* TC(\mathbb{Z})_2$ (Journal of Pure and Applied Algebra, 1999). There is some discussion of the additive extensions in the Tate spectral sequence for $\pi_* TP(\mathbb{Z})_2$ in Theorem 1.9 on page 231 of one of those JPAA papers, but the general picture appears to be complicated. My former PhD student Knut Berg determined (ca. 2013) the continuous mod $2$ homology of $TP(\mathbb{Z})$ as an $A_*$-comodule algebra, where $A_*$ is the dual Steenrod algebra, as well as the Adams $E_2$-term and $d_2$-differentials. This gives some other information about $\pi_* TP(\mathbb{Z})_2$ than the Tate spectral sequence, but a complete picture of the later differential pattern is missing.<|endoftext|> -TITLE: Stone-Čech compactification -QUESTION [5 upvotes]: Is every hyperstonean space a Stone-Čech compactification of a discrete space? -Is there a closed subset of Stone-Čech boundary that is extremally disconnected? - -REPLY [5 votes]: As for Q1, the answer is no. As remarked by Narutaka, the hyperstonean cover of $[0,1]$ does not have isolated points (Corollary 2.22) and all points of a discrete space $\Gamma$ are isolated in $\beta \Gamma$. Hyperstonean spaces can be way more wilder; all relevant information can be found in the book Banach spaces of continuous functions as dual spaces by Dales, Dashiell, Lau, and Strauss. -As for Q2, if you mean $\beta \mathbb N \setminus \mathbb N$, then yes, there are lots of copies of $\beta \mathbb N$ therein.<|endoftext|> -TITLE: Has this number-theoretic constant been studied? -QUESTION [11 upvotes]: Unless I made a mistake, the expected value of the largest exponent in the prime factorization of random positive integer (defined in the appropriate way) is $$\eta := \sum_{n=1}^\infty \Big(1-\zeta(n)^{-1}\Big)$$ with the convention $\zeta(1)^{-1} = 0$ (for aesthetics). I was just wondering whether this constant $\eta$ has a name, whether it's been studied, etc.. - -REPLY [32 votes]: If you calculate it to a few decimals, you find -$$ -1.705211140105\ldots -$$ -which is enough to locate it in the OEIS. -It's Niven's constant: MathWorld, Wikipedia, OEIS. -As mentioned by GH from MO in the comments, it was in fact proven by Niven in 1969 that the average largest exponent tends to $\eta$. -Since the question is essentially about finding literature related to a given numerical constant, I should probably mention this answer by myself on the other site, exhibiting some other methods (Steven Finch's book Mathematical Constants & how to google the decimals effectively).<|endoftext|> -TITLE: Can one glue De Rham cohomology classes on a differential manifolds? -QUESTION [15 upvotes]: Let $M$ be a differential manifold and $\mathcal H^k$ the presheaf of real vector spaces associating to the open subset $U\subset M$ the $k$-th de Rham cohomology vector space: $\mathcal H^k(U)=H^k_{DR}(U)$. Is this presheaf a sheaf? -Of course not! Indeed, given any non-zero cohomology class $0\neq[\omega]\in \mathcal H^k(U)$ represented by the closed $k$-form $\omega\in \Omega^k_M(U)$ there exists (by Poincaré's Lemma) a covering $(U_i)_{i\in I}$ of $U$ by open subsets $U_i\subset U$ such that $[\omega]\vert U_i=[\omega\vert U_i]=0\in \mathcal H^k(U_i)$, and thus the first axiom for a presheaf to be a sheaf is violated. -But what about the second axiom? -My question: -Suppose we are given a differential manifold M, a covering $(U_\lambda)_{\lambda \in \Lambda}$of $M$ by open subsets $U_\lambda \subset M$, closed differential $k-$forms $\omega_\lambda \in \Omega^k_M(U_\lambda)$ satisfying $[\omega_\lambda]\vert U_\lambda \cap U_\mu=[\omega_\mu]\vert U_\lambda \cap U_\mu\in \mathcal H^k(U_\lambda\cap U\mu)$ for all $\lambda,\mu \in \Lambda$. -Does there then exist a closed differential form $\omega\in \Omega^k(M)$ such that we have for the restrictions in cohomology: $[\omega]\vert U_\lambda=[\omega _\lambda]\in \mathcal H^k(U_\lambda)$ for all $\lambda\in \Lambda$ ? -Remarks - -This is an extremely naïve question which, to my embarrassment, I cannot solve. -I have extensively browsed the literature and consulted some of my friends, all brilliant geometers (albeit not differential topologists), but they didn't know the answer offhand. -For what it's worth, I would guess (but not conjecture!) that such glueing is impossible. -If the covering of $X$ has only two opens then we can glue. -This follows immediately from Mayer-Vietoris's long exact sequence -$$\cdots \to \mathcal H^k(M) \to \mathcal H^k(U_1) \oplus \mathcal H^k(U_2) \to \mathcal H^k(U_1\cap U_2)\to \cdots$$ - -Update -My brilliant friends didn't answer offhand but a few hours later, unsurprisingly, they came back to me with splendid counterexamples! See below. - -REPLY [3 votes]: Here is the great answer given by another of my brilliant friends: -Let $X$ be $\mathbb C, U_0$ be the open complement in $X$ of the closed disk $\bar D=\{z\in \mathbb C\vert \vert z\vert \leq1 \}$ -and add a few open discs $U_1,\cdots, U_n$ of radius $\lt 1$ covering $\bar D$ in order to obtain an open covering $U_0,U_1,\cdots U_n$ of $X$. -Now let $[0]\neq[\omega _0]\in \mathcal H^1(U_0)\cong \mathbb Z$ be a nonzero cohomology clas and define (no choice here!) $0=[\omega_i]\in \mathcal H^1(U_i)=0$. -The compatibility conditions are trivially satisfied since all intersections $U_i\cap U_j (i\neq j)$ are contractible, so that $\mathcal H^1(U_i\cap U_j )=0$. -Nevertheless we can't glue our cohomology classes $[\omega_i]$ to a global cohomology class $[\omega] \in \mathcal H^1(X)$ since the only global cohomology class on $X$ is $0\in \mathcal H^1(X)=0$, which does not restrict to $[\omega _0]\neq 0\in \mathcal H^1(U_0)$. -Remark -Here too the answer is entirely due to my geometer friend.<|endoftext|> -TITLE: Symmetries of contractable subsets of $\Bbb R^n$ -QUESTION [8 upvotes]: Let $K\subset\Bbb R^n$ be a non-empty compact subset of $\Bbb R^n$. A symmetry of $K$ is an isometry of $\Bbb R^n$ that fixes $K$ set-wise. Since $K$ is compact, there is always a point $x\in\Bbb R^n$ fixed by all symmetries. - -Question: Can the following three things be true at the same time: - -$K$ is contractible. -$x$ is the only point fixed by all symmetries of $K$. -$x\not\in K$. - - -The following image shows that such a $K$ exists if we ignore one of the properties: - -REPLY [6 votes]: I posted a refined/generalized question Does a compact contractible metric space have a point that is fixed by all isometries? and received an answer that contained all essential ingredients to provide an affirmative answer to this question. The related question Homeomorphic to the disk implies existence of fixed point common to all isometries? has an answer with further very interesting details. -There are two ingredients: - -there exists a group $G$ that acts without fixed points on the $n$-dimensional ball (for an appropriate $n>5$). -the ball can be smoothly embedded into some Euclidean space $\Bbb R^m$ so that the action of $G$ is now by linear transformations and only fixes the origin (Mostow's embedding theorem). - -If we choose $K$ to be this embedded ball and $x$ to be the origin, then this configuration satisfies all the conditions from my question. -References to the claims 1 and 2 can be found in the linked answers, except for the assertion that the linear transformations only fix the origin, which I found in the original paper "Equivariant Embeddings in Euclidean Space" by Mostow.<|endoftext|> -TITLE: A curious $q$-series identity on a truncated Euler function -QUESTION [7 upvotes]: Recall that a $q$-Pochhammer symbol is defined as -$$ -(x)_n = (x;q)_n := \prod_{l=0}^{n-1}(1-q^l x). -$$ -I found the following curious $q$-series identity that seems to hold for any $n\geq 0$: -$$ -(-1)^{n+1}q^{\frac{(n+1)(3n+2)}{2}}\sum_{j\geq 0}q^{j}(q^{j+1})_{n}(q^{j+2n+2})_{\infty} \overset{?}{=} \sum_{\substack{k\in \mathbb{Z}\\ |k| > n}}(-1)^{k}q^{\frac{k(3k-1)}{2}}. -$$ -Note, the right-hand side is a truncated version of the Euler function -$$ -\phi(q) := (q)_{\infty} = \sum_{k\in \mathbb{Z}}(-1)^{k}q^{\frac{k(3k-1)}{2}}. -$$ -How can we prove the above identity? Any suggestions/ideas would be greatly appreciated! - -REPLY [9 votes]: Let ${n\choose k}_q=\frac{(q)_n}{(q)_k(q)_{n-k}}$ denote a $q$-binomial coefficient. We start with the following version of $q$-Vandermonde convolution identity: -$$ -(x-y)(x-qy)\ldots(x-q^{n-1}y)\\=\sum_{k=0}^n(-1)^{k}q^{k(k-1)/2}{n\choose k}_q(y)_{n-k}(xq^{1-k})_k\quad\quad(\diamondsuit) -$$ -A short proof of $(\diamondsuit)$ may be obtained by checking it when $x=q^i$, $y=q^{-j}$ for non-negative integers $i,j$ such that $i+j\leqslant n$. In these points the values of both sides of $(\diamondsuit)$ are equal to 0, unless $i+j=n$ and we take the summand with $k=i$ in the right hand side. At this point, the identity is staightforward. It remains to observe that this collection of points is enough for reconstruction of the polynomial of degree at most $n$. -We apply $(\diamondsuit)$ for $y=q^{n+1}$, $x=tq^{2n+1}$ and multiply both sides by $-q^{2n+1}$. This gives, denoting for $k=0,1,\ldots,n$ -$$ -c_k:=(-1)^{k+1}{n\choose k}_qq^{k(k+1)/2}(q^{n+1})_{n-k}, -$$ -the following identity: -$$ -(-1)^{n+1}q^{(n+1)(3n+2)/2}(qt)_n\\=c_0q^{2n+1}+c_1q^{2n}(q^{2n+1}t)_1+c_2q^{2n-1}(q^{2n}t)_2+\ldots+c_{n}q^{n+1}(q^{n+2}t)_n. \quad \quad (\heartsuit)$$ -Denote now $s_n=(q^n)_\infty=\prod_{k\geqslant n}(1-q^k)$. Note that $$\sum_{k\geqslant n}q^ks_{k+1}=\sum_{k\geqslant n}(s_{k+1}-s_k)=1-s_n.\quad\quad(\clubsuit)$$ The idea is to reduce your sum to such telescoping sums. -Applying $(\heartsuit)$ for $t=q^j$ and multiplying by $q^j$ and using $(\clubsuit)$, your sum reads as -$$ -S=\sum_{i=0}^{n}c_i(1-s_{2n+1-i}). -$$ -I claim that $$-\sum_{i=0}^n c_is_{2n+1-i}=s_1=\sum_{k\in \mathbb{Z}} (-1)^kq^{k(3k-1)/2}\quad\quad(\spadesuit)$$ (the last equality is Euler's Pentagonal theorem). Thus $S-s_1=\sum_{i=0}^n c_i$, which is easy to see to be a polynomial in $q$ of degree $n(3n+1)/2$. On the other hand, $S$ is divisible by $q^{(n+1)(3n+2)/2}$ (that is seen from its very definition), thus this polynomial must be equal to $-s_1 \pmod {q^{(n+1)(3n+2)/2}}=-\sum_{|k|\leqslant n} (-1)^kq^{k(3k-1)/2}$, and we get your formula for $S$. -It remains to prove $(\spadesuit)$. Since $s_{2n+1-i}(q^{n+1})_{n-i}=s_{n+1}$, $(\spadesuit)$ reads as -$$ -s_{n+1}\sum_{i=0}^n (-1)^{i+1}{n\choose i}_q q^{i(i+1)/2}=-s_1, -$$ -or, equivalently, -$$ -\sum_{i=0}^n (-1)^{i}{n\choose i}_q q^{i(i+1)/2}=(q)_{n}, -$$ -which is a partial case of the $q$-binomial theorem $(x)_n=\sum_{i=0}^n (-1)^{i}{n\choose i}_q q^{i(i-1)/2}x^i$ for $x=q$.<|endoftext|> -TITLE: Is every folk cofibration of strict $\omega$-categories a monomorphism? -QUESTION [5 upvotes]: In the folk model structure on the category $sCat_\omega$ of strict $\omega$-categories, the cofibrations are generated by the boundary inclusions $\{\partial \mathbb G_n \to \mathbb G_n \mid n \in \mathbb N\}$, where $\mathbb G_n$ is the $n$-globe and $\partial \mathbb G_n$ is its boundary. -Question: Is every folk cofibration a monomorphism in $sCat_\omega$? -Folk cofibrations (or at least the cellular ones) are also called "relative computads / polygraphs". -I'm pretty sure the answer is yes, but I've been unable to find a reference. -Note that as mentioned here, monomorphisms in $sCat_\omega$ are not stable under cobase change in general. So an affirmative answer doesn't follow immediately from knowing that the maps $\partial \mathbb G_n \to \mathbb G_n$ are monomorphisms. - -REPLY [5 votes]: I just thought (or maybe remember) a neat proof of this fact. It involve ideas I worked on a few years ago but never published - but that's short enough so that I can explain the key ideas on MO. Let me know if you need it : I might write up the details of this proof in a short paper if you want to have a reference for it... -Fix $T$ a topos (It will be globular set) and $M$ a cartesian monad on $T$ (it will be the free $\infty$-category monad - so that the category of $M$-algebra is the category of strict $\infty$-categories). -Let $\Omega$ be the sub-object classifier of $T$. Then I claim: -Theorem 1 : $\Omega$ has a $M$-algebra structure. Moreover as a $M$-algebra it classifies the "cartesian sub-objects" of $M$-algebras. -By a cartesian subobject we mean a subalgebra $X \subset A$ such that the inclusion morphism $A \to M$ is a cartesian morphism of $M$-algebra. That is, the square witnessing it is a morphism of algebras is a pullback square. -Proof : consider $E$ the category whose objects are monomorphisms in $T$ and whose morphisms are cartesian squares. $M$ acts on this category by sending $A \subset B$ to $M(A) \subset M(B)$, and because its unit and multiplication are cartesian transformation, they give a monad structure to $M$ acting on $E$. -Now the terminal object of $E$ is $1 \to \Omega$, it automatically inherits a $M$-algebra structure and is terminal amongst $M$-algebras in $E$. -One then easily check that a $M$-algebra in $E$ is exactly a $M$-algebra with a cartesian sub-algebra and hence the theorem follows. -Remark : I had initially noticed this in the context of $\infty$-category a few years ago using the object classified instead of the subobject classifier. I used that to give a general proof that in an $\infty$-categorical context, if one defines "computads" for such a cartesian (or rather PRA) monad on a presheaf $\infty$-category then the category of computads is always a presheaf $\infty$-category. I gave a couple of talks about it in 2019 but never published the results because it seemed unclear what it was good for at the time. But it can be used to give a simple proof of the fact you mentioned. -Theorem 2 : If one defines cofibration in the category of $M$-algebra to be the left class of the weak factorization system cofibrantly generated by the $M(A) \to M(B)$ for $A \subset B$ an inclusion in $T$. Then all cofibration are cartesian monomorphism in the category of $M$-algebra. -In particular, the cofibrations of the Folk model structure are cartesian monomorphisms. -Proof : Using that $\Omega$ classified cartesian monomorphism it is easy to check that cartesian monomorphism are stable under pushout and transfinite composition. For example if $U \to V$ is a cartesian mono, classified by $\chi : V \to \Omega$, and $U \to X$ is any map, then one can define a map $X \coprod_U V \to \Omega$ that factor through to $1 \to \Omega$ on $X$ and is $\chi$ on $V$. One can then check* that this map classifies $X$, and hence that $X \to X \coprod_A B$ is a cartesian mono. -Finally, the $M(A) \to M(B)$ are cartesian monomorphism, so this conclude the proof. -(*) : ok here I'm hiding a part of the proof. To do this, One need to use that if $T_i$ is a diagram of $\infty$-category with a map $colim T_i \to \Omega$ then the subobject it classifies is the colimit of the subobject classified by the $T_i \to \Omega$. This is showed by generalizing theorem 1 to show that for every $\infty$-category $X$, the partial classifier $\tilde{X}$ is equiped with a $M$-algebra structure such that $M$-maps $B \to \tilde{X}$ correspodns to cartesian subobjects $A \subset B$ together with a $M$-map $A \to X$. This is proved exactly as Theorem 1 using the category $E$ whose object are inclusion $A \subset B$ together with a map $A \to X$ ($X$ being fixed), whose terminal object is $X \subset \tilde{X}$. -Once this is done, one can use it to show that the subobject classified by $colim T_i \to \Omega$ as the correct universal property to be the colimit of the object classified by the $T_i \to \Omega$ and really conclude the proof.<|endoftext|> -TITLE: Are the symmetric spaces $\operatorname{SU}(n)/{\operatorname{SO}(n)}$ always nontrivial in the bordism rings for $n>2$? -QUESTION [12 upvotes]: In my recent research, I need to know if the symmetric spaces $\operatorname{SU}(n)/{\operatorname{SO}(n)}$ are always nontrivial in the unoriented and oriented bordism rings for $n>2$. (For the motivation, by Benny Cheng, the cones over them are special Lagrangians in $\mathbb{C}^{n^2+n}$. I want to know if the cones are topologically smoothable. The $n=3$ case comes from Haskins & Pacini.) -I know that for $n=3$ we are dealing with the Wu manifold, and there are many references here. For $n>3$, I tried to find references about the Stiefel-Whitney numbers of these, which by Thom, is enough for the unoriented case. However, I've only been able to locate the cohomology ring of these in Topology of Lie Groups I & II by Mimura and Toda. I don't know much algebraic topology, but I think they did not state what the Stiefel–Whitney classes are. - -REPLY [15 votes]: There is a fibration $SU(n) \overset p\to SU(n)/SO(n) \overset j\to BSO(n)$, where the $j$ is the classifying map of $p$, viewed as (the projection of) a principal $SO(n)$-bundle. The Stiefel–Whitney classes for your Wu-esque manifolds are the $j^*$-images of the universal Stiefel–Whitney classes; this is in Borel and Hirzebruch's "Characteristic classes and homogeneous spaces I." -To find a nonzero Stiefel–Whitney number is to find a product of these classes of total degree $\dim SU(n)/SO(n)$. The computation in Mimura–Toda shows that the cohomology over $\mathbb F_2$ is an exterior algebra on one generator each of degrees $2$ through $n$, and that these are the images of the universal Stiefel–Whitney classes other than $w_1$. The product of these generators thus does represent the fundamental class.<|endoftext|> -TITLE: Does a compact contractible metric space have a point that is fixed by all isometries? -QUESTION [10 upvotes]: Let $(X,d)$ be a compact and contractible metric space. Let $\operatorname{Isom}(X)=\{\phi\colon X\to X\}$ be its group of isometries. - -Question: Is there a point $x\in X$ fixed by all $\phi\in\operatorname{Isom}(X)$? - -I am happy to assume some additional niceness conditions for $X$, enough to ensure that $X$ satisfies some fixed-point theorem, guaranteeing that every continuous map $\phi\colon X\to X$ has a fixed point (e.g. triangulable, locally contractible; see A version of Brower's fixed point theorem for contractible sets? for details). -The emphasize is therefore on whether all isometries have a common fixed point. -This post is a refinement/generalization of Symmetries of contractable subsets of $\Bbb R^n$. - -REPLY [15 votes]: There are finite groups that act smoothly on a disk without a global fixed point. You can arrange the metric to be isometric, e.g. via the Mostow-Palais embedding theorem, which equivariantly and smoothly embeds the disk into a Euclidean space where the group acts orthogonally, and taking the metric induced by Euclidean one. The full isometry group is potentially bigger, so it cannot fix a point.<|endoftext|> -TITLE: Can we state the Riemann Hypothesis part of the Weil conjectures directly in terms of the count of points? -QUESTION [22 upvotes]: For algebraic curves we can state the Riemann hypothesis part of the Weil conjectures directly as a formula for the number of points on the curves, sidestepping the zeta function. Namely, given a smooth algebraic curve $X$ of genus $g$ defined over $\mathbb{F}_q$ elements, let $N_n(X)$ be its number of points over $\mathbb{F}_{q^n}$. Then -$$ N_n(X) = q^n - \alpha_1^n - \cdots - \alpha_{2g}^n + 1 $$ -where all the numbers $\alpha_i \in \mathbb{C}$ have $|\alpha_i| = \sqrt{q}$. -Can we give a roughly similar formula for $N_n(X)$ when $X$ is a smooth higher-dimensional variety, say of dimension $d$? I know we can write -$$ N_n(X) = \sum_{k = 0}^{2d} (-1)^k \mathrm{tr}\left(\mathrm{Frob}^n \colon H^k(X) \to H^k(X) \right) $$ -where $H^k$ is etale cohomology, but I want to go a bit further, use the Riemann Hypothesis, and see numbers of magnitude $q^{k/2}$ showing up, presumably the eigenvalues of the Frobenius. -In fact what I'm really hoping for is a sum over pure motives: I hear pure motives defined using numerical equivalence give a semisimple category, so we can decompose $X$ uniquely as a sum of simple pure motives, and I'm hoping each one of these contributes a term to $N_n(X)$. If so, does each motive of dimension $k$ give a term whose absolute value grows like $q^{nk/2}$? -(By the way, I know we can state the Riemann hypothesis part of the Weil conjectures as a bound on the number of points, which is great, but I'm not interested in that here.) - -REPLY [26 votes]: (1) We have $$ N_n(X) = \sum_{k = 0}^{2d} (-1)^k \mathrm{tr}\left(\mathrm{Frob}^n \colon H^k(X) \to H^k(X) \right) = \sum_{k = 0}^{2d} (-1)^k \sum_{i=1}^{ h^k(X)} \lambda_{k,i}^n$$ -where $\lambda_{k,1},\dots,\lambda_{k,h^k(X)}$ are the eigenvalues of $\mathrm{Frob}$ on $H^k(X)$, counted with multiplicity. -Furthermore, we have $|\lambda_{k,i}| =q^{k/2}$ so $\left| \lambda_{k,i}^n\right| = q^{nk/2}$ as long as $X$ is a smooth projective variety (or even smooth proper). (If $X$ is not proper, we need to use compactly-supported cohomology to obtain the formula for the number of points, and then we only obtain an upper bound for the size of the eigenvalue.) -Technically speaking, the correct statement (see Lemma 1.7 of La Conjecture de Weil I by Pierre Deligne) is that the eigenvalues are algebraic numbers all whose Galois conjugates in $\mathbb C$ have size $q^{k/2}$. This deals with the issue that the eigenvalues are naturally $\ell$-adic integers and thus aren't naturally complex numbers, by considering the complex numbers that are Galois conjugates in the sense of satisfying the same minimal polynomial. -This notational difficulty is removed if we avoid étale cohomology and merely state the Riemann hypothesis in the form that -$$ N_n(X) = \sum_{k = 0}^{2d} (-1)^k \sum_{i=1}^{ h^k(X)} \alpha_{k,i}^n$$ -where $\alpha_{k,i}$ are complex numbers of absolute value $q^{k/2}$ and $h^i(X)$ is the $i$th Betti number of the de Rham cohomology of a characteristic $0$ lift. -This form is actually totally equivalent to the original formulation of Weil. We recover Weil's statement, in terms of the zeta function, by multiplying both sides by $U^n$, dividing by $n$, summing over $n$, and exponentiating. (In fact, this is exactly how Weil proves his conjecture for Fermat hypersurfaces in the paper where he originally formulates it, Number of Solutions fo Equations in Finite Fields.) We can recover this statement from Weil's by taking the logarithm of both sides, extracting the coefficient of $U^n$, and multiplying by $n$. -Of course, Weil could not give the statement in terms of étale cohomology because it hadn't been invented yet! -(2) This works in at least one notion of motives. -I think this is easiest to do if we use homological equivalence ($\ell$-adic, crystalline, whichever) as our equivalence relation for motives. -The idea here is that motives are supposed to have, for each cohomology theory, a realization functor that sends the motive to its cohomology groups, which on varieties $X$ recovers the cohomology groups of $X$, and is compatible with direct sums. -So if we express $X$ as a sum of (indecomposable, to avoid worrying about semisimplicity) motives, we can write the (signed) trace of $\mathrm{Frob}^n$ on the cohomology of $X$ as the sum of the (signed) trace of $\mathrm{Frob}^n$ on the cohomology of the motives. -To make the contribution of each motive of size $q^{kn/2}$, we just need each indecomposable motive to have cohomology only in a particular degree. -If we define motives using correspondences up to homological equivalence, say for $\ell$-adic homology, then the $\ell$-adic realization functor exists by construction. -Furthermore, because the eigenvalues of Frobenius on each cohomology group are distinct, we can define by linear algebra a polynomial in Frobenius, with rational coefficients, which acts by $1$ on the $k$th cohomology group and $0$ on each other cohomology group (the Künneth type standard conjecture). If we work with cycles up to homological equivalence, this cycle is idempotent, so defines a splitting of the motive of $X$. -Using these splittings, we can split $X$ into motives each of which has cohomology only in a single degree, as desired. -For Chow motives with respect to which other equivalence relations this can be made to work, as well as for which other notions of motives, I will leave to experts.<|endoftext|> -TITLE: The convex hull of a manifold whose cobordism class is trivial -QUESTION [6 upvotes]: Let $M$ be a compact orientable $n$ dimensional manifold. Assume that $M$ has trivial cobordism class. -Is there an embedding of $M$ in some Euclidean space $\mathbb{R}^m$ such that the convex hull of $M$ is a $n+1$ dimensional manifold whose boundary is $M$? -Here the image of $M$ under the embedding is denoted again by $M$. -Note: One can pose the same question in the following geometric manner: -Let $M$ be a compact Riemannian manifold with trivial cobordism class. Is there an isometric embedding of $M$ in some Euclidean space such that the convex hull of $M$ is a manifold whose boundary is $M$? - -REPLY [12 votes]: Implicit in the other responses is the fact that if $M$ bounds a convex manifold $W$, then $W$ is contractible and so M has the homology of a sphere. So any null-cobordant manifold that is not a homology sphere is a counterexample. Eg take $M = X \# -X$ where $X$ is any orientable manifold with non-trivial homology.<|endoftext|> -TITLE: Bernstein's corollary for the case of half space -QUESTION [5 upvotes]: The early seminal result of Bernstein in 1914 for $n=2$ is the well-known Bernstein theorem: - -The only entire solutions to the minimal surface equation in $\mathbb R^3$ are the affine functions -$$u(x,y)=ax+by+c,$$ -where $a, b, c\in\mathbb{R}$. - -Actually, Bernstein obtained his result as an application of the so called Bernstein’s geometric theorem: - -If the Gauss curvature of the graph of $u\in C^\infty(\mathbb R^2)$ in $\mathbb R^3$ satisfies $K\leq 0$ everywhere and $K<0$ at some point, then $u$ cannot be bounded. - -As a corollary, Bernstein proved a very general Liouville theorem: - -Suppose $u$ is a solution to the elliptic equation -$$\sum_{i,j=1}^2 a_{ij}u_{ij}=0\quad\text{in }\mathbb R^2$$ -such that -$$|u(x)|=o(|x|) \text{ as }|x|\to+\infty.$$ -Then $u$ is a constant. - -Note that in the above Liouville theorem, the equation doesn't need to be uniformly elliptic, hence it is a very powerful result. What I want to know if this result has a half space version, which is like harmonic functions. More precisely, I want to obtain the following propersition: - -Suppose $u\geq 0$ is a solution to the elliptic equation -$$\left\{\begin{aligned}\sum_{i,j=1}^2 a_{ij}u_{ij}&=0\quad\text{in }\mathbb R^2_+,\\ -u(x,0)&=0\quad \text{on }\mathbb R.\end{aligned}\right.$$ -Then $u$ is a linear function of form -$$u(x,y)=Ay,\quad A\geq 0.$$ - -Note that in the question, $a_{ij}$ could be degenerate or sigular at $\infty$. -This question is motivated by seeing Mooney's notes: The Monge-Ampère equations. He used partial Legendre transform to investigate the Liouville theorem for Monge-Ampère equation in half space, and one of steps in his proof used the similar propersition for harmonic functions, and which can be proved by boundary Harnack inequality and odd extension of $u$. But it is failed for the case without uniform ellipticity. -I have searched on the internet, and I didn't find any references about this propersition. Is this propersiton true? And if there are some references that I missed? Thanks in advance. - -REPLY [5 votes]: Here is a counterexample: let -$$u(x,y) = e^{-x^2}\sinh(y).$$ -Then -$$\det D^2u = -2e^{-2x^2}(\sinh^2(y) + 2x^2) < 0 \text{ ơn } \mathbb{R}^2 \backslash \{0\},$$ -and the equation -$$u_{xx} + (2-4x^2)u_{yy} = 0$$ -is uniformly elliptic in a neighborhood of the origin.<|endoftext|> -TITLE: Polynomial values are powers of two -QUESTION [14 upvotes]: The initial question comes from Komal in 1999. -Namely it asks to show that for infinitely many $n$ there is a polynomial $f\in\mathbb{Q}[X]$ of degree $n$ such that $f(0),f(1),\dotsc,f(n+1)$ are distinct powers of $2$. This is equivalent to finding $(t_0,\dotsc,t_n)$ different positive integers such that $\displaystyle \sum_{i=0}^{n} (-1)^{n-i}\dbinom{n+1}{i} 2^{t_i}$ is a power of $2$. -We could ask something more: -Is it true that there exists $c_n$ and we can bound it in terms of $n$ such that for all $f\in\mathbb{Q}[X]$ of degree $n$ we have that $f(0),f(1),\dotsc, f(c_n)$ cannot all be powers of $2$? The existence of $c_n$ and that it is bounded in terms of $n$ follows from a strengthened conjectural version of Falting's theorem for curves of the type $y^m=f(x)$. Can we say something unconjecturally about this? For $f(0),f(1),f(2),\dotsc, f(n)$ distinct powers of $2$ we can even construct $f\in\mathbb{Z}[X]$ thus $c_n\geq n+1$ always. - -REPLY [15 votes]: I'll prove a stronger statement. -Let $S$ be a finite set of primes. I claim there is a $c_{n,S}$ such that a polynomial $f$ with rational coefficients cannot take only values that are $S$-units on $\{1,\dots, c_{n,S}\}$. -Indeed, we can take $$c_{n, S} =(2n+1) \prod_{p \in S} p^{ \lfloor \log_p(n)\rfloor + 1}. $$ -This is a polynomial in $n$ of degree $|S|+1$. In particular, in the original case $S =\{2\}$, this is quadratic in $n$. -Proof: -We can write $f(x) = a \prod_{i=1}^n (x-\alpha_i)$ for $\alpha_i \in \overline{\mathbb Q}$. -If $p^k> n$ then there exists $x_p \in (\mathbb Z/p^k)$ such that $x_p$ is not congruent mod $p^k$ to $\alpha_i$ for any $i$. If $y$ is congruent mod $p^k$ to $x_p$, it follows that $$ v_p(f(y)) = v_p(a) +\sum_{i=1}^n v_p(y -\alpha_i) = v_p(a) +\sum_{i=1}^n v_p(x_p -\alpha_i)$$ since $y-\alpha_i = (y-x_p) + (x_p-\alpha_i)$ and the first term is divisible by $p^k$ while the second is not, so the $p$-adic valuation is independent of $y$ in this congruence class. -We can take $k = \lfloor \log_p(n)\rfloor + 1$. -By the Chinese remainder theorem, the number of $a \in \{1, \dots, (2n+1) \prod_{p \in S} p^{ \lfloor \log_p(n)\rfloor + 1}\}$ such that $a$ is congruent to $x_p$ modulo $p^{\lfloor \log_p(n)\rfloor + 1}$ is $2n+1$. -All $2n+1$ values in this arithmetic progression have the same $p$-adic valuation for all $p \in S$. If they are all $S$-units, then it follows they are all equal to the same value, up to $\pm 1$. But since $f$ is nonconstant, only $n$ can take the same value so only $2n$ can take the same value up to $\pm 1$, a contradiction. -So it is not possible to have all $c_{n,S}$ values $S$-units. -It is not too hard to see a similar argument works for $S$-units in an arbitrary number field. - -REPLY [13 votes]: Yes, such $c_n$ is bounded by something effective. Below is a cubic bound, which probably may be improved. (Update: see $n^2\log n$ upper bound by Will in the comments.) -Assume that $f(x)$ is a power of 2 for all integer $x$ on $[0,c_n]$. -Note that $[0,c_n]$ is partitioned onto at most $n$ segments, onto each of which $f$ is monotone. Thus there exist $N:=(c_n+1)/n$ consecutive integers onto which the values of $f(x)$ are, say, increasing powers of 2: $2^{m_1}<2^{m_2}<\ldots<2^{m_N}$. Assume that $N>(n+1)^2$. Denote $p_j=m_{1+j(n+1)}$ for $j=0,1,\ldots,n+1$. The numbers $2^{p_j}$ are the values of a polynomial of degree $n$ along $n+2$ elements of an arithmetic progression. Thus $$2^{p_{n+1}}-{n+1\choose 1}2^{p_n}+{n+1\choose 2}2^{p_{n-1}}-\ldots=0.$$ -But the first summand is greater than the sum of all others. -Let me also prove that for distinct powers of 2 the bound is linear. -Assume that $f(0),\ldots,f(m)$ are distinct owers of 2. Let $A\subset \{0,1,\ldots,m\}$ be a subset of size $n+1$ with $n+1$ minimal values. Denote $t=\max_{a\in A} f(a)$. For $x\in \{0,1,\ldots,m\}$ we get by Lagrange interpolation -$$ -|f(x)|=\left|\sum_{a\in A} f(a)\prod_{b\in A\setminus \{a\}} \frac{x-b}{a-b}\right|\leqslant t2^nm^n/n!\leqslant t(2em/n)^n -$$ -(I bounded all $f(a)$ as $t$, all $x-b$ as $m$, and the sum of reciprocals of absolute values of the denominators is obviously minimal when $A$ consists of $n+1$ consecutive numbers. In the latter case this reciprocals are equal to ${n\choose i}/n!$ for $i=0,\ldots,n$, thus the bound). -On the other hand, we have $f(x)\geqslant 2^{m-n}t$. Therefore $2^{n(m/n-1)}\leqslant f(x)/t\leqslant (2em/n)^n$ and $2^{m/n-1}\leqslant 2e m/n$, thus $m/n$ is bounded from above.<|endoftext|> -TITLE: Behavior of $|f'(z)|/(1+|f(z)|^2)$ as $|z| \rightarrow \infty$? -QUESTION [13 upvotes]: Let $f(z)$ be an entire holomorphic function in $\mathbb{C}$, and consider the real-valued function -$$g_f(z)=\frac{|f'(z)|}{1+|f(z)|^2}.$$ -If $f(z)$ is a polynomial, then it is easy to prove that $\lim_{|z|\rightarrow \infty}g_f(z)=0$. -When $f$ is transcendental, say $f(z)=e^z$, then $g_f(z)=\frac{e^x}{1+e^{2x}}$, which goes to zero when $Re(z)$ is going $\infty$, but remains a constant when $z$ is moving on any vertical line. So far I have not found any example such that the limit goes to zero when $f$ is transcendental. My question is, can we rigorously prove that there is no transcendental entire function $f$ such that $$ -\lim_{|z|\rightarrow \infty}g_f(z)=0\; ? -$$ - -REPLY [15 votes]: A concrete example is given by $f(z)=\cos{\sqrt{z}}$. Then -$$ -g_f(z) = \frac{|\sin\sqrt{z}|}{2|\sqrt{z}|(1+|\cos^2{\sqrt{z}}|)} . -$$ -Obviously, this is small for large $|z|$ if $\sin{\sqrt{z}}$ is not large, and if $\sin\sqrt{z}$ is large, then $|\cos\sqrt{z}|$ is of the same order of magnitude (since $\sin^2 w+\cos^2 w=1$), and again $g_f$ is small.<|endoftext|> -TITLE: Diagonal maps, Goodwillie calculus, and $T(n)$ local homotopy theory -QUESTION [6 upvotes]: Here is a collection of facts that all seem true, but together seem to give a nonsensical solution: - -After $T(n)$-localization, all natural transformations $F \sim G$ between homogenous functors $F,G:Sp \rightarrow Sp$ of different degrees appear trivial, in the sense that $\operatorname{cofiber}(\sim) \simeq G \vee\Sigma F$. This follows from Kuhn's splitting. - -If one has compositions of functors $f \circ g, f' \circ g'$ and natural transformations $\alpha: f \rightarrow f',\beta: g \rightarrow g'$ such that either $\alpha$ or $\beta$ is trivial on derivatives, then $\alpha \circ \beta$ is trivial on derivatives. This follows from the chain rule of Arone and Ching that the derivatives can be computed as a (derived) composition product. - -The natural transformation $\Sigma^\infty \Omega^\infty \rightarrow \Sigma^\infty (\Omega^\infty(-))^{\wedge 2}$ given by $1 \circ \Delta \circ 1$ induced by the diagonal $\Delta$ in $Top_*$ is $0$ on derivatives. This is because $\Sigma^\infty \Delta: \Sigma^\infty \rightarrow \Sigma^\infty(-)^{\wedge 2}$ is a natural transformation between homogenous functors of different degrees. - -The Taylor tower $(\Sigma^\infty \Omega^\infty )^{\wedge n}$ converges on $0$-connected spectra. I believe this follows from the fact for $n=1$. - -Hence, for a connected infinite loop space $X$, the cofiber of $\Sigma ^\infty \Delta: \Sigma^ \infty X \rightarrow \Sigma^\infty (X \wedge X)$ is $T(n)$-locally equivalent to $\Sigma^\infty (X \wedge X) \vee \Sigma^\infty X$. - - -This seems to imply that $T(n)$-locally, the diagonal maps of infinite loop spaces appear trivial in that they can't be detected by the cofiber. -I am wondering if this conclusion is correct, or if I have an error. I originally asked this question rationally, but figured I'd ask it in the $T(n)$-local category since it also applies there. - -REPLY [5 votes]: This is an elaboration on my comment above. Let us consider the natural transformation induced by the diagonal -$$\Sigma^\infty\Delta\colon \Sigma^\infty X \to \Sigma^\infty X\wedge X.$$ -The natural transformation $\Sigma^\infty \Delta$ induces a map between the Goodwillie derivatives of the functors. Let us denote this induced map by -$\partial_*\Sigma^\infty\Delta\colon \partial_* \Sigma^\infty\to \partial_* \Sigma^\infty \mbox{Sq}$. Here Sq denotes the functor Sq$(X)=X\wedge X$. The question is what is $\partial_*\Sigma^\infty\Delta$, and in particular if it can be non-trivial. -The functor $\Sigma^\infty$ is homogeneous linear, and $\Sigma^\infty$Sq is homogeneous quadratic. So we have the following description of the derivatives as symmetric sequences -$$\partial_*\Sigma^\infty = (\Sigma^\infty S^0, *, *, \ldots, )$$ -$$\partial_* \Sigma^\infty\mbox{Sq} = (*, \Sigma^\infty{\Sigma_2}_+, *, *, \ldots, ).$$ -The space of maps from $\partial_*\Sigma^\infty$ to $\partial_* \Sigma^\infty\mbox{Sq}$ in the $\infty$-category of symmetric sequences is easily seen to be contractible. So it seems that $\partial_*\Sigma^\infty \Delta$ can only be the trivial map. However, we really are interested in the mapping space in the category of right modules over $\partial_*Id$. Let's see what this means. Let $O=\{O(n)\}$ be an operad in spectra. We consider reduced operads, which means that $O(0)=*$ and $O(1)=\Sigma^\infty S^0$. All our modules are truncated at $2$, so only $O(1)$ and $O(2)$ are relevant. Let $M(1), M(2)$ and $N(1), N(2)$ be (truncated) right modules over $O$. This means that we have $\Sigma_2$-equivariant maps $M(1)\wedge O(2) \to M(2)$ and $N(1)\wedge O(2) \to N(2)$. The mapping spectrum of $O$-module maps from $M$ to $N$ is equivalent to the homotopy pullback of the following diagram -$\require{AMScd}$ -$$\begin{CD} - @. \mbox{Spectra}(M(2), N(2))^{h\Sigma_2}\\ -@. @VVV\\ -\mbox{Spectra}(M(1), N(1)) @>>> \mbox{Spectra}(M(1)\wedge O(2), N(2))^{h\Sigma_2} -\end{CD}$$ -By contrast, the space of maps from $M$ to $N$ in the category of symmetric sequences is the product -$$\mbox{Spectra}(M(1), N(1))\times \mbox{Spectra}(M(2), N(2))^{h\Sigma_2}.$$ -There is an obvious forgetful map from the former mapping spectrum to the latter, which sends the middle term to a point. -In the case of interest, $O=\partial_*Id$, so $O(1)=\Sigma^\infty S^0$ and $O(2)\simeq\Sigma^{-1}\Sigma^\infty S^0$, where $\Sigma_2$ acts trivially on $O(2)$. The modules are $(M(1), M(2))=(S^0, *)$ and $(N(1), N(2)) = (*, \Sigma^\infty {\Sigma_2}_+)$. So the spectrum of $O$-module maps from $M$ to $N$ is the following pullback -$$\mbox{Spectra}(\Sigma^\infty S^0, *)\to \mbox{Spectra}(\Sigma^{-1}\Sigma^\infty S^0, \Sigma^\infty {\Sigma_2}_+)^{h\Sigma_2} \leftarrow \mbox{Spectra}(*, \Sigma^\infty{\Sigma_2}_+)^{h\Sigma_2}.$$ -This homotopy pullback is easily seen to be equivalent to $\Sigma^\infty S^0$. -It remains to show that the transformation $\Sigma^\infty\Delta$ actually induces a non-trivial map on derivatives. Here is one way to argue this. It is an exercise in Yoneda that the spectrum of natural transformations from $\Sigma^\infty X$ to $\Sigma^\infty X\wedge X$ is equivalent to the sphere spectrum, with the diagonal corresponding to a generator. There is a natural map of mapping spectra $$\mbox{Nat}(\Sigma^\infty X, \Sigma^\infty X\wedge X)\to \partial_*Id\!-\! mod(\partial_*\Sigma^\infty X, \partial_*\Sigma^\infty X\wedge X).$$ Both the source and the target are equivalent to $\Sigma^\infty S^0$. One can show that for polynomial functors whose derivatives are finite free spectra (or more generally, whose derivatives have vanishing Tate homology) this map is an equivalence.<|endoftext|> -TITLE: Is spherical trigonometry a dead research area? -QUESTION [36 upvotes]: When I was an undergrad, the field of spherical trigonometry was cited as a once-popular area of math that has since died. Is this true? Are the results from spherical trigonometry relevant for contemporary research? - -REPLY [2 votes]: Other answers give examples of advanced research, but I thought I'd point out a modest result from my somewhat recreational investigations in the area: - -If $W$ is area of the "hypotenuse-face" of a spherical tetrahedron with right "leg-faces" of area $X$, $Y$, $Z$, then -$$\cos\tfrac12W = \cos\tfrac12X\cos\tfrac12Y\cos\tfrac12Z+\sin\tfrac12X\sin\tfrac12Y\sin\tfrac12Z \tag{$\star$}$$ - -This is a counterpart of de Gua's theorem (aka, the "Pythagorean Theorem for Euclidean tetrahedra"): -$$W^2 = X^2 + Y^2+Z^2$$ -Of course, a hyperbolic counterpart exists as well; just append "h"s to all the trig functions in $(\star)$, and change "$+$" to "$-$". There are even associated Laws of Cosines, but I won't go into that here. -I will give a shout-out to an ancient, still-open question of mine: "Pythagorean theorem for right-corner hyperbolic simplices? -", which specifically seeks the $4$-dimensional hyperbolic counterpart of $(\star)$. (Since spherical and hyperbolic relations are likely comparable, this also counts as an open question in spherical geometry.)<|endoftext|> -TITLE: Why mpmath computes $\sum_{n=2}^\infty (-1)^n\log(n)=\log\left(\frac1 2 \sqrt{2} \sqrt{\pi}\right)$ -QUESTION [7 upvotes]: Working with precision 500 decimal digits, mpmath in sage computes: -$$\sum_{n=2}^\infty (-1)^n\log(n)=\log\left(\frac1 2 \sqrt{2} \sqrt{\pi}\right).\tag{1}\label{1}$$ -We believe the LHS of \eqref{1} diverges, so this isn't true. - -Q1. Are there theoretical reasons for mpmath to compute \eqref{1}? - -online code -Added Despite the interesting answers, I am ready bet mpmath -doesn't do any analytic stuff not related to summation, it -works purely numerically and the function is treated as black box, -returning real number. - -REPLY [7 votes]: As shown in my previous answer, the value of the sum that you see is -$$\lim_{t\uparrow1}\sum_{n=2}^\infty (-t)^n \ln n. $$ -Here is a "manual" way to show this. Writing -\begin{equation} - \ln n=\int_0^\infty dz\,\frac{e^{-z}-e^{-nz}}z, -\end{equation} -for $t\uparrow1$ we have -\begin{equation} -\begin{aligned} -&\sum_{n=2}^\infty (-t)^n \ln n \\ - &=\sum_{n=2}^\infty (-t)^n \int_0^\infty dz\,\frac{e^{-z}-e^{-nz}}z \\ - &=\int_0^\infty \frac{dz}z\sum_{n=2}^\infty (-t)^n (e^{-z}-e^{-nz}) \\ - &=\frac{t^2}{1+t}\int_0^\infty \frac{dz}z\,e^{-z}\frac{1-e^{-z}}{1+t e^{-z}} \\ - &\to\frac12\int_0^\infty \frac{dz}z\,e^{-z}\frac{1-e^{-z}}{1+e^{-z}} \\ - &=-\int_0^1 \frac{dx}{\ln x}\,\frac{1-x}{1+x} - = \frac12\,\ln\frac\pi2, -\end{aligned} -\end{equation} -by (say) Formula 4.267.1 on p. 545 in I.S. Gradshteyn and I.M. Ryzhik, Table of Integrals, Series, and Products, Seventh Edition, 2007.<|endoftext|> -TITLE: Examples of five-adjoint systems -QUESTION [9 upvotes]: I'm looking into Lawvere's formulation of unities of opposites and opposites of unities, and for this I would be interested in systems of five (or more) adjoint functors $X\stackrel{\stackrel{\longrightarrow}{\longleftarrow}}{\stackrel{\longrightarrow}{\stackrel{\longleftarrow}{\longrightarrow}}}Y$, ideally within the domains of topos theory and algebraic geometry, but really any examples would be interesting. I know they are very rare, but I feel like I have come across one at least once. Anyone? - -REPLY [9 votes]: As noted by Simon Henry, the nLab gives many examples of adjoint chains. You can get an infinitely long adjoint chains from any ambidextrous adjunction. For an example, let $G$ be a finite group and $H$ a subgroup of $G$. Let $\mathrm{Rep}(G)$ be the category of finite-dimensional representations of $G$ over your favorite field, and similarly for $\mathrm{Rep}(H)$. Then restriction gives a functor -$$ \mathrm{restriction} : \mathrm{Rep}(G) \to \mathrm{Rep}(H) $$ -and let -$$ \mathrm{induction} : \mathrm{Rep}(H) \to \mathrm{Rep}(G) $$ -be its left adjoint. Induction is also the right adjoint of restriction, so we get an infinitely long adjoint chain -$$ \cdots \dashv \mathrm{induction} \dashv \mathrm{restriction} \dashv \mathrm{induction} \dashv \cdots $$ -For this reason it might be more exciting to get adjoint strings of length 5, or more generally length $n$, that can't be extended further. The nLab also gives plenty of those. -Amazingly, Rosebrugh and Wood showed that any category with an adjoint string of length 5 -$$ U \dashv V \dashv W \dashv X \dashv Y $$ -where $Y$ is its Yoneda embedding must be equivalent to the category of sets. (More precisely, the category of small sets, where we assume the axiom of universes.) - -Robert Rosebrugh and R. J. Wood, An adjoint characterization of the category of sets, Proceedings of the American Mathematical Society 122, no. 2 (1994): 409–413. - -REPLY [7 votes]: You can build long chains of adjoints by taking functor categories and using Kan extensions. I will give an example. -Write $\underline{n}$ for the set $\{1, \ldots, n\}$ considered as a discrete category. As $\underline{1}$ is the terminal category, there is a unique functor $f \colon \underline{2} \to \underline{1}$. -Precomposing with $f$ gives a functor $f^* \colon [\underline{1}, \mathrm{Set}] \to [\underline{2}, \mathrm{Set}]$. As a functor $\underline{n} \to \mathrm{Set}$ is just an $n$-tuple of sets, we have that $f^*(X) = (X, X)$ so $f^*$ is the diagonal functor. The functor $f^*$ has adjoints on both sides, which are the Kan extensions. The left Kan $f_!$ provides a method of combining two sets $f_!(X,Y)$ so that a map $f_!(X,Y) \to Z$ is the same data as a pair of maps $(X \to Z, Y \to Z)$. In other words, $f_!(X,Y) = X \sqcup Y$ is the disjoint union. Similarly, the right Kan $f_*(Y,Z)$ provides a method of combining $Y$ and $Z$ so that $X \to f_*(Y,Z)$ is the same data as a pair of maps $(X \to Y, X \to Z)$. In other words, $f_*(Y,Z) = Y \times Z$. So far, we have a two-step adjunction of known useful functors -$$ -f_! \dashv f^* \dashv f_* -$$ -Precomposing with $f^*$ gives a functor $(f^*)^* \colon [[\underline{2},\mathrm{Set}], \mathrm{Set}] \to [[\underline{1},\mathrm{Set}], \mathrm{Set}]$. As a functor $[\underline{2},\mathrm{Set}]\to \mathrm{Set}$ is an operation $\odot \colon \mathrm{Set} \times \mathrm{Set} \to \mathrm{Set}$ that takes two sets and returns a set, we have that $(f^*)^*(\odot)$ is the "squaring" functor $(X) \mapsto X \odot X$. -Certainly we have Kan extensions -$$ -(f^*)_! \dashv (f^*)^* \dashv (f^*)_* -$$ -However, if $L \dashv R$ is an adjunction, we always have $L_! \cong R^*$ and $L^* \cong R_*$. Consequently, we can build the five-functor chain -$$ -(f_*)_! \dashv (f_*)^* = (f^*)_! \dashv (f^*)^* \dashv (f^*)_* = (f_!)^* \dashv (f_!)_* -$$ -For example, the functor $(f^*)_!$ takes a unary operation $\Gamma \colon \mathrm{Set} \to \mathrm{Set}$, considers it as a functor defined on the diagonal $\mathrm{Set} \overset{\Delta}{\to} \mathrm{Set} \times \mathrm{Set}$, and freely generates from here a binary operation. By the formula $L_! \dashv R^*$, we can calculate this rather concretely as sending $\Gamma$ to the binary operation $X \otimes_\Gamma Y = \Gamma(X \times Y)$. A natural transformation of bifunctors $\otimes_\Gamma \implies \odot$ will somehow be the same information as a natural transformation $\Gamma \implies (X \mapsto X \odot X)$. -We could evidently continue this pattern to build even longer and less-understandable chains.<|endoftext|> -TITLE: Max decoupling inequality -QUESTION [7 upvotes]: Let $X_1,\ldots,X_n$ be $\{0,1\}$-valued random variables drawn from some joint distribution. Let $\tilde X_1,\ldots,\tilde X_n$ be their independent version: $\mathbb{E}X_i=\mathbb{E}\tilde X_i$ for each $1\le i\le n$ and the $\tilde X_i$ are mutually independent (as well as being independent of the $X_i$'s). I would like to compare -$M:=\mathbb{E}\max_{1\le i\le n}X_i$ -and -$\tilde M:=\mathbb{E}\max_{1\le i\le n}\tilde X_i$. -It's obvious that $M$ can be made arbitrarily small, while $\tilde M$ is arbitrarily close to $1$, so in general, there can be no bound on $\tilde M/M$. But it seems that in the reverse situation, one should be able to say something. Is some bound of the form -$$ M \le c\tilde M $$ -known, where $c$ is an absolute constant? If such a thing is impossible (what's the counterexample?), what's the best dependence on $n$ in the bound? - -REPLY [10 votes]: Yes, this inequality holds with -$$c:=\frac e{e-1}.$$ -Indeed, let $A_i:=\{X_i=1\}$. Then -$$M=P\Big(\bigcup_i A_i\Big)\le\min(s,1)\le c(1-e^{-s}),$$ -where $s:=\sum_i P(A_i)$. On the other hand, -$$\tilde M=1-\prod_i(1-P(A_i)) -\ge1-\prod_i\exp(-P(A_i))=1-e^{-s}.$$ -So, $M\le c\tilde M$.<|endoftext|> -TITLE: Connection between Grothendieck's homotopy hypothesis and Lie's second and third theorems? -QUESTION [8 upvotes]: I'm not an expert on homotopy theory, but I speculated about this in my thesis, so I figured I'd ask about it here. As I understand it, the homotopy hypothesis says that $\infty$-groupoids, with $\infty$-categorical equivalences as weak equivalences, are equivalent to topological spaces, with weak equivalences being weak homotopy equivalences. -Now for simplicity, let me focus on the version of the homotopy hypothesis that says that $1$-groupoids, with categorial equivalences as weak equivalences, are equivalent to topological spaces, where weak equivalences are weak $1$-equivalences, ie. maps which induce isomorphisms on $\pi_0\,,\pi_1\,.$ -Now on the other hand, Lie's second and third theorems imply that there is an equivalence of categories between simply connected Lie groups and Lie algebras. These theorems generalize to Lie groupoids and Lie algebroids, where simply connected becomes source simply connected (well, in order for Lie's third theorem to hold completely one needs to use smooth spaces which are generalizations of manifolds, but we can always replace "Lie algebroids" with "integrable Lie algebroids", in any case). -Right now these two results may not seem related, for two reasons: - -There is no Lie algebroid present on the topological spaces. However, if we work with manifolds instead (or some appropriate class of infinite dimensional manifolds), then $M$ comes with a natural Lie algebroid, namely $TM\,.$ Therefore, there is a canonical Lie algebroid present. - -The groupoids in the homotopy hypothesis are discrete, so there is no source simply connected condition. However, the source simply connected integration of $TM$ is the fundamental groupoid $\Pi_1(M)\,,$ and this is Morita equivalent to $\pi_1(M)\,.$ Therefore $\Pi_1(M)\,,$ with the topology associated with the smooth structure, is equivalent to $\Pi_1(M)$ with the discrete topology, and I believe this is the correct groupoid to compare $M$ with in the homotopy hypothesis for 1-types. So in a sense, the source simply connected condition is naturally present. - - -Now, if we assume that we that we have a notion of weak equivalence of Lie algebroids which implies that a morphism $TM\to TN$ is a weak equivalence if the induced map $M\to N$ is a weak $1$-equivalence, then we seem to get a connection between Lie's theorems and the homotopy hypothesis, ie. the homotopy hypothesis for (smooth) 1-types seems to be implied by a version of Lie's theorems (this wouldn't exactly be Lie's theorems since Lie's theorems use isomorphisms as weak equivalences, but we can use other weak equivalences instead). -I can go into more detail, but has this connection been written about elsewhere in the literature, or is there any reason to doubt this connection? - -REPLY [4 votes]: There are analogues of Lie's theorems in homotopy theory, primarily for rational and $p$-adic homotopy types, as well as Lie ∞-groupoids, which can be seen as smooth homotopy types. -In the rational case, this was figured out by Quillen -in his 1969 paper Rational homotopy theory, -which establishes a Quillen equivalence between the model categories -of rational simply connected spaces and rational reduced differential graded Lie algebras. -This was later extended to non-simply connected spaces, see the book Rational Homotopy Theory II by Félix, Halperin, Thomas. -There is also a correspondence in the $p$-adic case, see the survey by Heuts Lie algebra models for unstable homotopy theory. -Finally, one should also mention Lie ∞-integration, which establishes a correspondence between Lie ∞-algebroids and Lie ∞-groupoids.<|endoftext|> -TITLE: Which varieties are sums of tensor powers of the Lefschetz motive? -QUESTION [5 upvotes]: Any smooth projective variety $X$ gives an object $h(X)$ in the category of pure Chow motives. If $X$ is a generalized flag variety, i.e. a quotient $G/P$ where $G$ is semisimple linear algebraic group and $P$ is a parabolic subgroup, I believe $h(X)$ is a direct sum of tensor powers of the Lefschetz motive, because $X$ can be decomposed into Schubert varieties which are copies of $\mathbb{A}^n$ for various $n$. -If this is correct, I'd like to know: which other smooth projective varieties give pure Chow motives that are direct sums of tensor powers of the Lefschetz motive? -My intuition is that any variety with something like a "Schubert decomposition" — roughly, a well-behaved way of expressing it as a disjoint union of copies of $\mathbb{A}^n$'s — will have this property. I feel there should be plenty. But I don't actually know any varieties with this property, apart from flag varieties! -Any variety $X$ of dimension $d$ over $\mathbb{F}_p$ having this property will have an associated polynomial $N_X$ of degree $d$ with natural number coefficients: -$$ N_X(q) = \sum_{n = 0}^d a_n q^n $$ -such that $X$ has $N_X(q)$ points over $\mathbb{F}_q$ when $q$ is any power of $p$. -Is the converse true? Is a smooth projective variety $X$ over $\mathbb{F}_p$ with a polynomial $N_X$ having this property always a direct sum of tensor powers of the Lefschetz motive? - -REPLY [4 votes]: One class of examples is already indicated in the comments, and the question itself. I thought it would be good to include this in an official answer. - -Proposition. Let $X$ be smooth projective variety over a field $k$, such that there exists a chain of closed sets -$$X=X_n\supset X_{n-1}\supset \ldots X_{-1}=\emptyset$$ -such that $X_i-X_{i-1}$ is a union of affine spaces or split tori (products of $\mathbb{G}_m$'s). Then the Chow motive is of Tate type, i.e. sum of powers of the Lefschetz motive. - - -Cor. The conclusion holds for flag varieties, and projective toric varieties. - -The proposition follows from the distinguished triangles -$$ M_{gm}^c(X_{i-1})\to M_{gm}^c(X_{i})\to M_{gm}^c(X_{i}-X_{i-1})$$ -where $M_{gm}^c$ is the motive with compact support -in Voevodsky's category (cf. Voevodsky Triangulated categories of motives over a field and Mazza, Voevodsky, Weibel Lectures on motivic cohomology), induction, and the identification of $M_{gm}^c(X)$ with the Chow motive of $X$ (up to translation and twist). -If you allow quasi projective varieties, then complements of hyperplane arrangements are also mixed Tate. -A few additional comments. As Ben Wieland and Will Sawin have pointed out, you get more examples by blowing an example you have of Tate type along a subvariety with the same property. I can think of a few more examples off the top of my head, such that the Hilbert scheme of points on a rational surface. On the other hand, I hope it's clear that examples would have to be quite special. -Over $\mathbb{C}$, the Hodge numbers $h^{pq}$ would vanish for $p\not=q$, so -most varieties would not satisfy this condition.<|endoftext|> -TITLE: What is the quotient group $D^*/{F^*(1+P_D)}$ for a quaternion division algebra $D$ over a local field $F$? -QUESTION [8 upvotes]: Let $F$ be the non-archimedean local field $\mathbb{Q}_p$ for some prime $p$ and $D$ be a quaternion division algebra over $F$. Let $\mathcal{O}_D$ and $\mathcal{P}_D$ denote the ring of integers of $D$ and its unique maximal ideal (respectively). Then, what is the finite group -$$ \frac{D^*}{F^*(1+ \mathcal{P}_D)} = {?}$$ where $D^*=D-\{0\}$ and $F^*=F-\{0\}$ are multiplicative groups. -Consider the reduced norm map $N_\text{rd}:D \rightarrow F$, then $N_\text{rd}(D^*)=F^*$ and if $D^1$ denotes the reduced norm one elements of $D$, then we have an exact sequence -$$1 \rightarrow D^1 \rightarrow D^* \rightarrow F^* \rightarrow 1$$ but we have $D^1 \cap F^*=\{\pm 1\}$. We know from Carl Riehm's article that $$ \frac{D^1}{(1+ \mathcal{P}_D)} \cong {_N}(\mathbb{F}_{p^2})= \text{Finite cyclic group of order } (p+1).$$ -Here ${_N}(\mathbb{F}_{p^2})$ is the subgroup of $\mathbb{F}_{p^2}$ consisting of norm 1 elements. -Question: Similarly, can we write $ \frac{D^*}{F^*(1+ \mathcal{P}_D)}$ in terms of finite fields? -Any comments or suggestions will be extremely helpful. - -REPLY [4 votes]: To give an alternative answer, let us first recall the article "Construction of Locally Compact Near-Fields from $p$-Adic -Division Algebras" by Detlef Groger: -Fix a prime element $\pi_F$ of $F$. Then $D$ is generated as a non-commutative $F$-algebra by an unramified extension $E/F$ of degree 2 and an element $\pi$ with $\pi^2=\pi_F$. we consider $\varpi$ for a (fixed) primitive $(p^2 -1)^\text{th}$ root of unity in $E$ and $ \varpi_F=\varpi^\frac{p^2 -1}{p-1} \in F$. -Suppose $U_F$, $U_D$ denote the groups of units in $F$ and $D$ with $U^1_F := 1 + P_F$ and $U^1_D := 1 + P_D$. -Put $C = \langle \varpi, \pi \rangle$ and $C_F=C \cap F^* = \langle \varpi_F, \pi_F \rangle$. Then, $C$ is a complement of $U^1_D$ in the semidirect product $D^*=U^1_D \rtimes C$, whereas the product $F^*=U^1_F \times C_F$ is direct product. Therefore, $F^* U^1_D= U^1_D \times C_F$ and, -$$\frac{D^*}{F^*(1+P_D)} \cong C/C_F \cong \langle\overline{\varpi}\rangle \rtimes \langle\overline{\pi}\rangle \cong Z_{p+1} \rtimes Z_2 .$$<|endoftext|> -TITLE: What are the indecomposable modules over $\mathbb{F}_2(C_2\times C_2)$? -QUESTION [5 upvotes]: Let $C_2$ be the cyclic group of order $2$ and $\mathbb{F}_2$ the field with $2$ elements. Consider the group algebra $A:= \mathbb{F}_2 (C_2\times C_2)$. It is well-known that $A$ has infinite representation type. Is there a classification of the finite dimensional indecomposable $A$-modules (and the Auslander-Reiten quiver of $\text{mod}\,A$) in this case? - -REPLY [7 votes]: A complete description of the indecomposable modules for $C_2\times C_2$, including the Auslander-Reiten quiver is available in David Benson's book Modular Representation Theory: New Trends and Methods, Springer 1984, pp.176-181. This is over $\bar{\mathbb{F}}_2$; the result over arbitrary fields is given in the following few pages.<|endoftext|> -TITLE: Is the ideal product presheaf a sheaf? Do we have any reasons to believe it will be / it won't? -QUESTION [9 upvotes]: My question is about one of those several concepts in algebraic geometry who everybody uses but nobody defines or introduces properly. -Given a ringed space $(X,\mathcal{O}_X)$ and ideal sheaves $\mathcal{I},\mathcal{J}\subset\mathcal{O}_X$, we define the ideal product presheaf $\mathcal{I}\cdot_p\mathcal{J}$ as the ideal presheaf -$$ -U\mapsto(\mathcal{I}\cdot_p\mathcal{J})(U)=\mathcal{I}(U)\mathcal{J}(U)\subset\mathcal{O}_X(U). -$$ -My question is: is this presheaf a sheaf? Or is it necessary to sheafify to obtain the correct definition of the ideal product sheaf? -I was quite a while trying to look for a counterexample myself and I couldn't find any. After asking as well to the lecturer in the graduate algebraic geometry course I am following, he told me he could not find a counterexample. I've already asked this question here on MSE, but nobody has answered yet. There I explain the approaches I've tried. The problem is that I don't even have any probability argument to believe the answer to be positive or negative. I don't see why the presheaf shouldn't be a sheaf, but a positive proof seems to be unlikely (as I explain it in the MSE post). So if at least I receive an answer of the like "I don't think it's a sheaf" from an expert on the field I think I would be somewhat content. -From my experience, I think there is not that many people in MSE interested on scheme-theoretic algebraic geometry, so I am reasking the question here on mathoverflow hoping there's more people here that could give any comments. - -REPLY [8 votes]: So here is a counterexample which is qcqs: -Take $X$ the affine line with double origin $a_1$ and $a_2$, then take $I_1$ and $I_2$ the ideal of functions vanishing each at one of the origins respectively. -In this case $I_1\cdot_p I_2$ evaluted at $X-a_i$ is $(t_i)\subset k[t_i]=\mathcal{O}(X-a_i)$, but evaluated at $X$ it is $(t^2)\subset k[t]= \mathcal{O}(X)$, whereas the sheaf property would imply that it is $(t_1)\times_{k[t_1,t_1^{-1}]} (t_2)\cong (t)\subset k[t]$.<|endoftext|> -TITLE: Can we prescribe the $L^2$ norm of the scalar curvature on a four-manifold? -QUESTION [5 upvotes]: As mentioned by Willie Wong, I modified to the following version: -Let $M$ be a closed smooth $4$ manifold. -Q -Suppose that $c>0$ is any positive number, can we find a Riemannian metric $g$ on $M$, such that the $\int_MScal^2_gdv_g=c$, where $Scal_g$ denotes the scalar curvature of $g$? If not, for any small $\epsilon>0$, can we find a metric $g_\epsilon$ such that $|\int_MScal^2_{g_\epsilon}dv_{g_\epsilon}-c|<\epsilon$? -PS I do not know whether the question is trivial or not. Any reference is welcome. - -REPLY [11 votes]: This is not always possible. -Let $M$ be a compact smooth manifold of dimension $n$. Consider the Einstein-Hilbert functional $\mathcal{E}$ given by -$$\mathcal{E}(g) = \dfrac{\displaystyle\int_Ms_g d\mu_g}{\operatorname{Vol}(M, g)^{\frac{n-2}{n}}}.$$ -If $\mathcal{C}$ is a conformal class, then by using the conformal Laplacian and Hölder's inequality, one can show that $\mathcal{E}|_{\mathcal{C}}$ is bounded below. So we can define the Yamabe constant of $\mathcal{C}$ to be the finite quantity $Y(M, \mathcal{C}) = \inf_{g\in\mathcal{C}}\mathcal{E}(g)$. A result of Aubin shows that $Y(M, \mathcal{C}) \leq Y(S^n, [g_{\text{round}}])$, so we can define the Yamabe invariant of $M$ to be the finite quantity -$$Y(M) = \sup_{\mathcal{C}}\mathcal{E}(g) = \sup_{\mathcal{C}}\inf_{g\in\mathcal{C}}\mathcal{E}(g).$$ -The following result is Proposition 1 from Kodaira Dimension and the Yamabe Problem by LeBrun. - -Let $M$ be a smooth compact $n$-manifold, $n \geq 3$. Then $$\inf_{g}\int_M|s_g|^{n/2} d\mu_g= \begin{cases}0 & \text{if}\ Y(M) \geq 0\\ |Y(M)|^{n/2} & \text{if}\ Y(M) < 0.\end{cases}$$ Here the infimum on the left hand side is taken over all smooth Riemannian metrics $g$ on $M$. - -In particular, if $n = 4$ and $Y(M) < 0$, then for all $c$ with $c < Y(M)^2$, there is no metric $g$ with $\displaystyle\int_Ms_g^2 d\mu_g = c$. -The question now becomes: is there an example of a compact smooth four-dimensional manifold with $Y(M) < 0$? The answer is yes as is shown in Theorem 2 of the same paper: - -Let $M$ be the underlying $4$-manifold of a complex surface with Kodaira dimension $2$. Then $Y(M) < 0$. Moreover, if $X$ is the minimal model of $M$, then $$Y(M) = Y(X) = -4\pi\sqrt{2c_1(X)^2}.$$ - -For complex surfaces, $c_1(X)^2 = 2\chi(X) + 3\sigma(X)$, so we can actually compute the Yamabe invariant for the manifolds in the above theorem precisely. -Example: Let $X$ be the product of two complex curves of genus $2$. Then $X$ has Kodaira dimension $2$ and is minimal. Moreover, it has Euler characteristic $4$ and signature $0$, so $c_1^2(X) = 8$. Therefore -$$Y(X) = -4\pi\sqrt{2c_1(X)^2} = -4\pi\sqrt{16} = -16\pi.$$ -The underlying smooth $4$-manifold is $\Sigma_2\times \Sigma_2$. So for any $c < (-16\pi)^2 = 256\pi^2$, we see there is no metric $g$ on $\Sigma_2\times\Sigma_2$ with $\displaystyle\int_{\Sigma_2\times\Sigma_2}s_g^2 d\mu_g = c$.<|endoftext|> -TITLE: Accelerating convergence for some double sums -QUESTION [13 upvotes]: I am interested in the following general double sums, for integers $a\geq 1$ and $b\geq 2$, -$$Z(a,b) = \sum_{k,\ell \geq 0} \frac{2k+3}{\binom{k+2}{2}^a} \frac{2\ell+3}{(\binom{k+2}{2}+\binom{\ell+2}{2})^b},$$ -which are converging very slowly. For these sums, there is also an alternative expression as an iterated integral in dimension $a+b$, similar to multiple zeta values. -I would like more particularly to find $Z(2,2)$ and $Z(1,3)$ with precision as large as possible. Using the naive summation, I could only obtain 4 decimal digits, namely $Z(2,2) \simeq 4.7058$ and $Z(1,3) \simeq 1.6470$. -It is known that $2 Z(2,2) + 4 Z(1,3) = 16$. - -What would be a smart way to accelerate the convergence, in general and in the special case using maybe the previous formula ? - -REPLY [10 votes]: This is easy to do with PARI/GP. -Here is my code -p(n) = binomial(n+2,2); -Y(k,b) = sumnum(l=0, (2*l+3)/(p(k)+p(l))^b,sumtable); -Z(a,b) = sumnum(k=0, (2*k+3)/p(k)^a*Y(k,b)); -default(realprecision,57); -sumtable = sumnuminit(); -print(2*Z(2,2)+4*Z(1,3)) -/* 16.0000000000000000000000000000000000000000000000000000000 */ - -It takes 1.361 CPU seconds for 57 decimal places. The -documentation -has some information about the methods being used and there -are several summation functions other than sumnum. For -example, I originally used sumpos but sumnum is faster. -Thanks to Henri Cohen for his comment to use sumnuminit -to speed up the calculation of Y(k,b). Thanks to Jorge -Zuniga for his comment that replacing sumnum with summonien -is much faster. And especially thanks to the developers of -PARI/GP who provided these fast numerical summation functions. -Some details are in the arXiv paper -Gaussian Summation: An Exponentially Converging Summation Scheme -by Hartmut Monien. (published in 2010 in -Mathematics of Computation).<|endoftext|> -TITLE: Hamiltonian path in bike-lock graph with $1$ known digit -QUESTION [6 upvotes]: Motivation. My youngest son has a bike lock with dials, and he forgot the unlocking combination completely, except that he remembered that digit $0$ appeared somewhere in the combination. So it was my task to go through all the zillions (but, fortunately, finitely many) possible combinations. Which led to the following problem. -Formal statement. Let $n \geq 2$ be an integer, so we have $n = \{0, \ldots, n-1\}$. For any integer $k>1$ let $$V^0_k = \{x \in n^k: (\exists j\in k)x(j) = 0\}, $$ and let two distinct elements $a\neq b \in V^0_k$ form an edge iff there is $j\in k$ such that - -$a(i) = b(i)$ for all $i\in n\setminus\{j\}$, and -$\{a(j), b(j)\} = \{x, x+1\}$ for some $x\in n-1$, or $\{a(j), b(j)\} = \{0, n-1\}$. - -Denote the set of edges by $E_k$. -For what positive integers $k, n$ does the graph $(V^0_k, E_k)$ have a Hamiltonian path? And, if there is a Hamiltonian path, can also a Hamiltonian cycle be found? (The second question doesn't need to be answered for acceptance.) - -REPLY [2 votes]: Partial answer to the cycle version. -Case $k=2$, any $n$: The graph has no Hamiltonian cycle. If $n \ge 3$, the graph consists of two $n$-cycles that intersect at a single point $(0,0)$. If $n=2$, the graph is just the chain $(0,1),(0,0),(1,0)$. - -Case $k \ge 3$ and $n$ even: The graph has no Hamiltonian cycle. This is because the graph is bipartite and $|V|$ is odd, and you cannot have an odd-length cycle with two alternating colors. - -To see that it is bipartite, color each vertex blue if the sum of -coordinates is even, and red if the sum of coordinates is odd; every -edge in the graph moves just one coordinate, either between $i$ and -$i+1$ (different parities), or between $n-1$ (odd) and $0$ (even), so -it flips the color. -To see that $|V|$ is odd, note that $V = A - \setminus B$, where $A = \{0,1,\ldots,n-1\}^k$, that is all $k$-digit -strings, and $B = \{1,2,\ldots,n-1\}^k$, that is all $k$-digit strings -without zeros. Clearly $|V|=|A|-|B|$, with $|A|$ even and $|B|$ -odd, so $|V|$ is odd. - - -Case $k \ge 3$ and $n$ odd: Computations suggest that there is a Hamiltonian cycle, but this is inconclusive. I have verified by straightforward SageMath code with $k=3$ up to $n=11$, and with $(k,n) \in \{(4,3),(4,5),(5,3)\}$. -Now the bipartity argument does not apply, because $0$ and $n-1$ have same parity, so the wraparound edges make the graph non-bipartite. (The wraparound possibility is essential; if we leave out the wraparound edges, there is again no Hamiltonian cycle, because the graph is again bipartite and has odd number of vertices.) -Here is an example solution with $k=3$, $n=5$. Green lines indicate wraparound between $0$ and $4$. I don't see any glaring structure in the machine-generated solution, so it is difficult to generalize from that. -000-040-140-240-340-300-310-410-420-320- -220-210-110-100-104-103-003-403-404-304- -303-302-301-201-200-204-203-202-102-101- -001-041-031-030-020-120-130-230-330-430- -440-400-401-402-002-012-013-023-033-043- -042-032-022-021-011-010-014-024-034-044- -004 - - -Update. Here is a manually designed solution for $k=3$, $n=7$. From the picture it is probably clear how it generalizes to any odd $n$. Interestingly, we only need to do one wraparound.<|endoftext|> -TITLE: Strictness of two operations on proarrow equipments -QUESTION [5 upvotes]: There are several equivalent definitions of a profunctor between categories $C$ and $D$. I'm interested in the following two: - -A functor $C\times D^o \to \text{Set}$ -A co-continuous functor between presheaf categories $\hat C \to \hat D$ - -These are equivalent by the free co-completion property of the Yoneda embedding. -The advantage of definition 2 is that the composition of profunctors is strictly associative/unital (just functor composition) whereas composition using definition 1 is only weakly associative/unital. -However it seems to me there is a corresponding advantage to definition 1. If we are interested in the pro-arrow equipment of categories/functors/profunctors/natural transformations, then we also use the operation of restriction of a profunctor $R : C \not\rightarrow D$ along functors $F : C' \to C$ and $G : D' \to D$ giving us a profunctor $R(F,G) : C' \not\rightarrow D'$. By definition 1, this operation is strictly associative in that -$$R(F\circ F',G\circ G') = R(F,G)(F',G')$$ -$$R(\text{id},\text{id}) = R$$ -However, I don't see a way to define this operation for definition (2) that is strictly associative. -So my questions are - -Is there a way to define restriction of co-continuous functors that is strictly associative? -Either way, is there a theorem showing that every pro-arrow equipment with weakly associative composition and restriction is equivalent to one where both composition and restriction are strictly associative/unital? - -REPLY [2 votes]: I believe the answer to (2) is yes. -First, apply the strictification theorem for bicategories twice, to make composition of arrows and proarrows both strictly associative. Thus, when our equipment is regarded as a double category, we have a strict double category. (We could also probably apply some coherence theorem for double categories directly, as Kevin Arlin suggested in a comment.) -Second, recall that the restriction of a proarrow $R:C \nrightarrow D$ along arrows $f:C'\to C$ and $g:D'\to D$ can be defined/constructed as $f_\bullet \odot R \odot g^\bullet$, where $f_\bullet$ is a companion of $f$, and $g^\bullet$ is a conjoint of $g$. Thus, if companions and conjoints can be chosed strictly functorially, the strict associativity of composition of proarrows will give strict functoriality of restriction. -Now redefine an "arrow" $f:C'\to C$ to consist of an arrow in the original (strictified) sense together with a chosen companion and conjoint. Since a composite of companions (resp. conjoints) is a companion (resp. conjoint) of the composite, and companions and conjoints are essentially unique, this is equivalent to the original 2-category of arrows. But now it admits a strictly functorial choice of companions and conjoints. -If we apply this construction to cocontinuous functors, we get a notion of "strictly functorial restriction" for them, but at the cost of fattening up the notion of "functor". We don't have to fatten it up quite as much, since with cocontinuous functors we automatically get a strictly functorial choice of conjoints, namely precomposition, but we do have to equip each functor with a chosen left Kan extension to presheaves to be the companions. It's possible there is some trick that can do better than this, but I don't know what it is offhand.<|endoftext|> -TITLE: Preservation of projective stationarity -QUESTION [7 upvotes]: A set $S \subseteq [\kappa]^\omega$ is called projective stationary if for every stationary $A \subseteq \omega_1$, and every algebra $F : \kappa^{<\omega}\to\kappa$, there is $z\in S$ such that $z$ is closed under $F$ and $z \cap \omega_1\in A$. -One can show that projective stationarity is preserved by ccc forcing. -Questions: - -Is projective stationarity preserved by proper forcing? How about semi-proper forcing? -Is the projective stationarity of the full $[\kappa]^\omega$ preserved by proper forcing? Semiproper? - -REPLY [7 votes]: It is possible that $\sigma$-closed forcing destroys projective stationarity: -Suppose $\mathcal A$ is a maximal antichain in $\mathrm{NS}_{\omega_1}^+$. Feng-Jech have shown that -$$\mathcal S=\{N\in [H_{\omega_2}]^\omega\mid N\prec H_{\omega_2}\wedge \exists S\in\mathcal A\cap N\ \omega_1\cap N\in S\}$$ -is projective stationary. Suppose $V[G]$ is a ($\omega_1$-preserving) generic extension of $V$ so that $\mathcal A$ is no longer maximal, i.e. there is a stationary $T$ with $S\cap T$ nonstationary for all $S\in\mathcal A$. I claim that $\mathcal S$ is no longer projective stationary in $V[G]$. Otherwise, we could find a countable $M\prec (H_{\omega_2})^{V[G]}$ with $T\in M$ so that for $N=M\cap (H_{\omega_2})^V$: - -$N\in \mathcal S$ -$\omega_1\cap N\in T$ - -By 1., there is $S\in\mathcal A\cap N$ so that $\omega_1\cap N\in S$. But $M$ knows that $S\cap T$ is nonstationary, so there is a club $C\in M$ with $C\cap S\cap T=\emptyset$, however $\omega_1\cap N\in C\cap S\cap T$, contradiction. -Now if $\mathrm{NS}_{\omega_1}$ is not saturated,such an extension can be realized by $\sigma$-closed forcing: -Let $\mathcal A$ be a maximal antichain in $\mathrm{NS}_{\omega_1}^+$ of size $>\omega_1$. The forcing $\mathbb P$ consists of conditions $(s, f)$ where - -$s\in 2^{{<}\omega_1}$ -$f:\mathcal A\rightarrow \omega_1$ is a partial function with countable domain -For $\gamma \in \mathrm{dom}(s)$ and $\gamma\in T\in\mathrm{dom}(f)$: If $\gamma\geq f(T)$ then $s(\gamma)=0$. - -with the obvious order. The idea is that the first components generically build a characteristic function for a new stationary set $T$ and the second component makes sure that $S\cap T$ is bounded for any $S\in\mathcal A$. $\mathbb P$ is clearly $\sigma$-closed. The only non-trivial thing to check is that $T$ is really stationary in the extension, and here is where we need that $\mathcal A$ is large. Suppose $p$ forces that $\dot C$ is a club. Build a continuous elementary chain $\langle X_i\mid i<\omega_1\rangle$ of countable elementary substructures of $H_\theta$, so that $X_0$ contains all the relevant information. As $\mathcal A$ is large, $\triangledown (\mathcal A\cap\bigcup_{i<\omega_1} X_i)$ is costationary. This allows us to find $\alpha=\omega_1^{X_\alpha}$ with -$$\alpha\notin\bigcup(\mathcal A\cap X_\alpha)$$ -Now build a $X_\alpha$-generic sequence through $\mathbb P\cap X_\alpha$ starting with $p$. If $q=(s, f)$ is the limit of that sequence then $q\Vdash\check\alpha\in \dot C$ and $\mathrm{dom}(f)=\mathcal A\cap X_\alpha$. We can now extend $q$ to $q'=(s\cup\{(\alpha, 1)\}, f)$. This is a condition by our choice of $\alpha$ and $q'\Vdash \dot C\cap \dot T\neq\emptyset$ where $\dot T$ is the canonical name for $T$. -Regarding question 2, it is consistent (from a measurable) that there is a semiproper forcing that makes $([\omega_2]^\omega)^V$ even nonstationary. The forcing is adding a Cohen real and then shooting a club through the complement of $([\omega_2]^\omega)^V$. Cox-Sakai proved that the semiproperness of this forcing is equivalent to a form of the Strong Chang Conjecture. Any semiproper forcing changing the cofinality of $\omega_2$ to $\omega$ would do the trick as well, of course. -I do not whether or not proper forcing can destroy the projective stationarity of the full $[\kappa]^\omega$.<|endoftext|> -TITLE: How hard must "no high-degree irreducibles" proofs be? -QUESTION [10 upvotes]: Let $\mathsf{RCF}$ be the usual theory of real closed fields and for $n>2$ let $\theta_n$ be the sentence "No degree-$n$ polynomial is irreducible." Since $\mathsf{RCF}$ is complete, for every $n$ we have $\mathsf{RCF}\vdash\theta_n$. I'm curious how hard it must be to prove $\theta_n$, as a function of $n$. -One way to formalize this is the following. Let $\mathsf{RCF}_k$ be the (finitely axiomatizable!) subtheory of $\mathsf{RCF}$ gotten by restricting attention to polynomials of degree $\le k$. Then we can ask: - -What is the growth rate of the function $$\mathfrak{F}:\mathbb{N}_{>2}\rightarrow\mathbb{N}:n\mapsto\min\{k:\mathsf{RCF}_k\vdash\theta_n\}$$ sending each degree above $2$ to the highest degree we need to consider? - -I'm also interested in the qualitative question, "As $n$ increases, does $\theta_n$ get meaningfully harder to prove or does the same basic strategy work for all $n$?," but I'm a bit skeptical that that's sufficiently clear to have a good answer. I'd also be interested in questions about proof length (measured either in symbols or in steps), but there I'm skeptical that there is much that's easy to say (questions about proof length tend to get hairy, in my admittedly-limited experience). - -To clarify, the axiomatization I have in mind is the conjunction of - -the field axioms, - -the assertion that the relation "$b-a$ has a square root" is a compatible ordering, and - -for each $n$ the assertion that every degree-$n$ polynomial satisfies the intermediate value theorem. - - -Only this last clause is affected by the $\mathsf{RCF}\leadsto\mathsf{RCF}_k$ truncation. - -REPLY [7 votes]: I don’t know how to make use of the full power of the intermediate value theorem, hence I will work instead with a weaker axiomatization: let $\def\rcf{\mathrm{RCF}}\rcf'_k$ denote the first two groups of axioms from the question (i.e., $\rcf_1$) + the assertion that every polynomial of odd degree $d\le k$ has a root, and let -$$\def\fF{\mathfrak F}\fF'(n)=\min\{k:\rcf'_k\vdash\theta_n\},\quad n>2.$$ -Since $\rcf_k\vdash\rcf'_k$, we have $\fF(n)\le\fF'(n)$. (On the other hand, $\bigwedge_{n=3}^k\theta_n$ implies $\rcf_k$ over $\rcf_1$, hence $\rcf'_{\max\{\fF'(n):3\le n\le k\}}\vdash\rcf_k$. I don’t know if one can do better.) -I claim that if $n$ is not a power of $2$, then -$$\fF(n)\le\fF'(n)\le\binom n{n_2},\quad\text{where }n_2=2^{v_2(n)}$$ -(i.e., $n_2$ is the highest power of $2$ that divides $n$). Note that $\binom n{n_2}$ is odd (see e.g. Kummer’s theorem). -To see this, let $R\models\rcf'_k$ for $k=\binom n{n_2}$, and assume for contradiction that $f\in R[x]$ of degree $n$ is irreducible. Let $\{\alpha_i:i\in[n]\}$ list the roots of $f$ in a fixed splitting field of $f$ over $R$. -For any symmetric polynomial $s\in R[x_1,\dots,x_{n_2}]$, the coefficients of -$$p_s(x)=\prod_{i_1<\dots -TITLE: On the Artin-Rees Lemma for non-commutative rings -QUESTION [6 upvotes]: Consider a commutative noetherian ring $A$ with an ideal $I\subset A$. The Artin-Rees lemma implies that for f.g. modules $N\subset M$, the $I$-adic topology on $N$ agrees with the subspace topology coming from the $I$-adic topology on $M$. I wonder how much this can be generalized to the case where $A$ is not necessarily commutative. For example, let $\mathcal{R}$ be the valuation ring of a complete, discretely valued non-archimedean field $K$, and let $\pi$ be a uniformizer of $\mathcal{R}$. Assume you have an almost-commutative filtered algebra $A$ over $\mathcal{R}$ which is two-sided noetherian, and complete with respect to the $\pi$-adic topology. Are there any conditions on $A$ which assure that the Artin-Rees lemma holds for the $\pi$-adic topology? Notice that the previous statement covers, for example, -the case where $A$ is the universal enveloping algebra of a finitely generated Lie algebra over $\mathcal{R}$, and many other objects which have a PBW-type theorem. - -REPLY [3 votes]: There is some discussion of this in Rowen's "Ring Theory", volume I, Section 3.5, with additional references therein. -Exercise 19 on p. 462 in op. cit. states that a polycentral ideal $I$ of a noetherian ring $A$ has the Artin-Rees property, i.e., for every f.g. left module $M$ and a f.g. submodule $N\subseteq M$, there is $i\geq 1$ with $N\cap I^i M\subseteq IN$. (Applying this to $I^nN$ in place of $N$ shows that the $I$-adic topology on $N$ conicides with the topology induced from the $I$-adic topology of $M$.) -Here, an ideal $I$ is called polycentral if there is a chain of ideals $I=I_0\supseteq I_1\supseteq\dots\supseteq I_{t+1}=0$ such that $I_i = I_{i+1} +\sum_{r=1}^{s(i)} a_r A$ with $a_1,\dots a_{s(i)}\in A$ central modulo $I_{i+1}$. In particular, any ideal generated by central elements is polycentral. -In the context of your question, $I=\pi A$ is polycentral since it is generated by a central element, so it has the Artin-Rees property. -Actually, it is noted in op. cit. that the usual proof of the Artin-Rees lemma in the commutative case carries over in the non-commutative case for ideals generated by central elements.<|endoftext|> -TITLE: Reference request : table of quantum Clebsch-Gordan coefficient -QUESTION [5 upvotes]: From a quick Google search, one can find a table of the first Clebsch-Gordan coefficient. For example this table. Those are used to pass between the tensor product bases and the bases as sum of irreducible of the tensor product of two irreducibles representations of $U(\frak{sl_2})$. -I was wondering if there is such a table somewhere for the quantum Clebsch-Gordan coefficient, that is those occuring when you replace $U(\frak{sl_2})$ with $U_q(\frak{sl_2})$. - -REPLY [4 votes]: For rank-two quantum groups the Clebsch-Gordan coefficients are tabulated in arXiv:1004.5456 by Ardonne and Slingerland. This also includes mathematica notebooks to perform these and similar calculations. -More tables are in Hegde & Ramadevi.<|endoftext|> -TITLE: Did Euler know (unconsciously) to integrate by differentiating? -QUESTION [15 upvotes]: Considering a method to find the anti-derivative of an (sufficiently smooth) real function by differentiating published some years ago (equation (48) in Kempf et al., New Dirac Delta function based methods with -applications to perturbative expansions in quantum -field theory): -\begin{equation} -\int^x f(x')\,dx' = \lim_{y \to 0} f\left(\frac{\partial}{\partial y}\right) \frac{\mathrm{e}^{xy}-1}{y} +C, -\end{equation} -I'm wondering whether Euler in his very imaginative calculations (to say the least) did use some techniques (in special cases) that amount to this formula. -Any hints are welcome. - -REPLY [21 votes]: I don't know about that particular integral, but Euler certainly knew about integrating by differentiating. He wrote about it in his Exposition de quelques paradoxes dans le calcul integral (1758). A recent summary of that work can be found in -A. Fabian and H.D. Nguyen, -Paradoxical Euler: integrating by differentiating, -The Mathematical Gazette 97 (2013), no. 538, 61-74.<|endoftext|> -TITLE: A ribbon presentation for a torus knot -QUESTION [7 upvotes]: Let $K$ be a knot in $S^3$. It is well-known that the knot $K \# -\overline{K}$ is always ribbon. -The following picture describes the connected sum of the left-handed torus knot $T(3,4)$ and the right-handed torus knot $T(3,4)$. In Rolfsen's notation, $T(3,4) = 8_{19}$, for several descriptions see [1] and [2]. -I would like to find the ribbon move(s) for this composite knot but I could not elaborate. -Is there an easy way to see this or any trick to figure out the necessary ribbon moves? - -REPLY [3 votes]: If you want to find a set of ribbon moves, then you need to end up with an unlink with several components; therefore, you need to attack crossings! -This is your diagram: - -Then I add two bands along twists of given torus knots: - -Next, we have a link with three components: - -If you apply Reidemeister moves, you may eventually find an unlink with three components: - -There is no specific trick for any torus knot (my opinion) but you can control the number of additional ribbon moves. -Assume that $p$ and $q$ are relatively coprime integers with $p < q$. Then for the corresponding ribbon knot $T(p,q) \# \overline{T(p,q)} $, the number of ribbon bands is $p-1$.<|endoftext|> -TITLE: Some questions from the paper by Scholze-Weinstein -QUESTION [6 upvotes]: The following is from the paper by Scholze-Weinstein on moduli of $p$ divisible groups. -My question is from a part of Lemma 4.1.7: If $R$ is a semiperfect ring, then the canonical map $W(R^{\flat}) \rightarrow A_{\text{cris}}(R)$ is injective. -Here, they have claimed that elements of $W_{PD,n}$, as defined can be written uniquely as a sum $\Sigma _{i \in \mathbb{Z}} [r_i]p^i$. This is not clear to me. -Secondly, from remark 4.3.9, I have two questions: -That, for the perfect case, we can assume $R$ is local, then $T_0 ^+$ defined as $W(R) \otimes O_C$, and $T_0 = T_0[1/p]$ is connected. -Secondly, we can choose $\tilde{T_0}$ a direct limit of faithfully flat finite etale algebras such that $\text{Spec} \tilde{T_0}$ is connected without any etale covers. -EDIT: I am adding some other questions that I have from the paper and two questions that I have regarding Prof. Scholze's answer: -In Lemma 4.3.4 where it is proven that there is an isomorphism of $A_{\text{cris}}(R)$ algebras : -$ A_{\text{cris}}(S) \cong A_{\text{cris}}(S^{\prime}) \hat{\otimes}_{W(R^{\flat})} A_{\text{cris}}(R) $ there's a crucial step where it's been claimed that giving divided powers on $I^{\prime} + J\tilde{S^{\prime}}$ is equivalent to giving divided powers on $I^{\prime}$ and $J \tilde{S^{\prime}}$ separately. For the forward direction, why can we say that divided powers on $I^{\prime} + J\tilde{S^{\prime}}$ restrict to divided powers on $I^{\prime}$ and $J \tilde{S^{\prime}}$. -Secondly, in proposition 4.3.6, knowing that for any $x$, the sequence $0 \rightarrow \widehat{\mathcal{F}_{1,x}} \rightarrow \widehat{\mathcal{F}_{2,x}} \rightarrow \widehat{\mathcal{F}_{3,x}} \rightarrow 0$ has cohomologies killed by $p^{1/p-1} + \epsilon \forall \epsilon>0$ we want to say that $0 \rightarrow \mathcal{F}_{1}/p^n \rightarrow \mathcal{F}_{2}/p^n \rightarrow \mathcal{F}_{3}/p^n \rightarrow 0$ has cohomologies killed by $p^{2/p-1} + \epsilon$, one way to do this would be to apply UCT to the sequence w.r.t. the change of coefficients to $T^{+}/p^n$, but then we'd have to know that $\widehat{\mathcal{F}_{i,x}}/p^n = {\mathcal{F}_{i,x}}/p^n$. Can we do this over the non-noetherian ring we are working? -Regarding the answer below -Why is $R \otimes _{k} \overline{\mathbb{F}_p}$ connected? -And, finally, how can we get a countable directed system of finite etale algebras? (as countability is important in the proof of lemma 4.3.10.) - -REPLY [7 votes]: In Lemma 4.1.7, we actually assume that $R$ is f-semiperfect (i.e. a quotient of a perfect ring by a finitely generated ideal); I doubt the result is true without this assumption. -Note that $W_{PD}$ is a subring of $W(R^\flat)[1/p]$, and the latter is exactly the set of power series $\sum_{i\gg -\infty} [r_i] p^i$ with $r_i\in R^\flat$. So elements of $W_{PD}$ have unique expressions of the given sort. To get to $W_{PD,n}$, one kills all those elements where all $r_i\in \Phi^n(J)$. Thus, one gets similar unique representations of elements of $W_{PD,n}$, where now all $r_i\in R^\flat/\Phi^n(J)$. -For the next question: There's actually a small lapsus here: One should take the tensor product not over $\mathbb Z_p$, but over $W(k)$ where $k$ is the algebraic closure of $\mathbb F_p$ in $R$; also, the tensor product was meant to be ($p$-adically) completed. Then one argues as follows: As $T_0$ is integral perfectoid, it is almost integrally closed in $T_0[1/p]$. In fact, it is integrally closed, as any almost element of $T_0$ is already in $T_0$. (This reduces to the similar assertion in $O_C$, using that $T_0$ is a completed direct sum of copies of $O_C$ (as $W(R)$ is a completed direct sum of $\mathbb Z_p$'s).) Thus, any idempotent already lies in $T_0$. By $p$-adic completeness of $T_0$, the idempotents are then the same for $T_0$ and for $T_0/p$, or its reduced quotient, which is $R\otimes_k \overline{\mathbb F}_p$. But this has no nontrivial idempotents, as $k$ is algebraically closed in $R$. -The last thing is something very general: For any ring $A$ without nontrivial idempotents, one can fix a geometric point of $\mathrm{Spec}(A)$, and take the direct limit over all finite étale $A$-algebras $A'$ with a lift of this geometric base point. This gives such an algebra $\tilde{A}$, which is a direct limit of faithfully flat finite étale $A$-algebras, and such that any further finite étale cover of $\tilde{A}$ splits (as it can be approximated over some $A'$, but then occurs itself in this direct limit).<|endoftext|> -TITLE: On Cramér's theorem about roots of Zeta function -QUESTION [6 upvotes]: Cramér proved the following theorem (see the announcement in [1] and [2]): -Consider the following function: - -$$V(z)=\sum_k e^{\rho_kz}$$ - -Where $\rho_k$ runs through non trivial zeta zeros with $Im(\rho_k) > 0$ -Cramér proved $V(z)$ converges for $Im(z) > 0$ and has a singularity at the origin of the type $\frac{\log(z)}{(1-e^{-z})}$ -by which it means that the function - -$$F(z) = 2πiV(z) -\frac{\log(z)}{(1-e^{-z})}$$ - -has a meromorphic continuation to all $\Bbb C$, with simple poles at the points $\pm πin$ where $n$ ranges over the integers, and at the points $\pm\log(p^m)$ where $p^m$ ranges over the -prime powers. -I have following questions - - -I'm wondering if $V(z)$ has alternate explicit expression ? -(Simpler) reference where I can study about this ? ( Other than Cramér’s paper itself) - - -References -[1] Harald Cramér, "Sur les zéros de la fonction $\zeta(s)$" (French), Comptes Rendus Hebdomadaires des Séances de l’Académie des Sciences, Paris, tome 168 (Janvier-Juin), 539-541 (1919), -JFM 47.0289.02. -[2] Harald Cramér, "Studien über die Nullstellen der Riemannschen Zetafunktion" (German), Math. Zeitschr. 4, 104-130 (1919), JFM 47.0289.03. - -REPLY [8 votes]: Q1: There is a functional relation for $V(z)$, but no "explicit expression" I know of. -Q2: Cramér's paper is from 1919, a modern and more extensive treatment is in On Cramér's theorem for general Euler products with functional equation (1993).<|endoftext|> -TITLE: All saddles in the unit ball have area $<2\pi$? -QUESTION [19 upvotes]: Let $M$ be the saddle surface in $\mathbb R^3$ defined by $x^2-y^2+z=0$. For any $r\geq 0$ and $(x_0,y_0,z_0)\in\mathbb R^3$, let $rM+(x_0,y_0,z_0)$ denotes the surface obtained by scaling $M$ by $r$ and then translating by $(x_0,y_0,z_0)$. (Note that by $rM$, with $r=0$, we mean $\lim_{r\to 0} rM$, which is two perpendicularly intersecting plane.) And let $B$ be the unit ball. -Is it true that -$$\mathrm{area}\left[(rM+(x_0,y_0,z_0))\cap B\right]\leq 2\pi,$$ -with equality holds if and only if $r=0$ and $(x_0,y_0,z_0)=(0,0,0)$ (which gives two intersecting equatorial disks)? -Edit: Using the first variation formula, I can show some partial results: - -For any fixed $z_0$, the area of $(r(M+(0,0,z_0)))\cap B$ decreases as $r$ increases, for all $r\in(0,+\infty)$. -$(r,x_0,y_0,z_0)=(0,0,0,0)$ is a strict local maximum. - -REPLY [11 votes]: It is actually next to trivial if you choose the right parameterization (and rather puzzling if you don't, so it can make a decent take-home exam problem in multivariate calculus). -I'll use the line cover $x(s,t)=(s+t,s-t,4st)$. The area element is then -$2\sqrt{1+8s^2+8t^2}\,ds\,dt< 2(\sqrt{1+8s^2}+\sqrt{1+8t^2})\,ds\,dt$. -Now for a ball $B$ of radius $R$ centered at $(u,v,w)$ and for fixed $s$, we have the line in $t$ whose moving speed is $\sqrt{2+16s^2}=\sqrt 2\sqrt{1+8s^2}$ and whose square distance from the center of the ball is at least -$$ -\min_t[(s+t-u)^2+(s-t-v)^2]=2(s-\tfrac{u+v}2)^2=2(s-s_0)^2. -$$ -Thus, integrating in $t$ first, we have -$$ -\int_{s,t:x(s,t)\in B}2\sqrt{1+8s^2}\,dt\,ds\le \int_{s\in\mathbb R}2\sqrt 2\sqrt{[R^2-2(s-s_0)^2]_+}\,ds=\pi R^2 -$$ -(time = line length/speed; length = $2\sqrt{R^2-\text{(line distance to the center)}^2}$). -The other integral is done in exactly the same way, only you need to integrate with respect to $s$ first. Hence, the area is $<2\pi R^2$, which is equivalent to the requested bound after scaling.<|endoftext|> -TITLE: Proposition A.2.6.15 in HTT -QUESTION [9 upvotes]: This is a cross-post of a question in MSE. - -I am reading Lurie's Higher Topos Theory and I need some help to understand a part of the proof of Proposition A.2.6.15. (A.2.6.13 in the published version) This proposition is used repeatedly in the book to construct various model structures. -In the proposition, we work with a locally presentable category $\mathbf{A}$, a class $W$ of morphisms in $\mathcal{C}$, and a set $C_0$ of morphisms in $\mathbf{A}$, subject to the following conditions: - -The class $W$ contains all isomorphisms, has the two out of three property, is close under filtered colimits, and has a set $W_0$ such that every morphism in $W$ is a filtered colimit of morphisms in $W_0$. - -Given cocartesian squares - - -$$\require{AMScd} -\begin{CD} -X @>>> X' @>{g}>>X''\\ -@V{f}VV @VVV @VVV\\ -Y @>>> Y' @>>{h}> Y'', -\end{CD} -$$ -if $f\in C_0$ and $g\in W,$ then $h\in W$. - -Every map in $\mathbf{A}$ having the right lifting property with respect to $C_0$ lies in $W$. - -We then want to show that the condition (2) remains valid if we repalce $C_0$ by the weakly saturated class generated by $C_0.$ Lurie proves this by arguing that the class $P$ of morphisms for which (2) remains true when $C_0$ is replaced by $P$ is weakly saturated, i.e., closed under pushouts, transfinite compositions, and retracts. -I understand that $P$ is closed under pushouts and transfinite compositions: Closure under pushout is obvious. Closure under transfinite transfinite composition is a consequence of the fact that $W$ is closed under filtered colimits. But I don't see why $P$ is closed under retracts. Can anyone help me figure this out? Any help is appreciated. Thanks in advance! - -REPLY [10 votes]: Retracts of weak equivalences are weak equivalences. -Now if $f'$ is a retract of $f$ and you start with such a diagram with $f'$ on the left, you can create a new diagram with $f$, the same $X', X''$ but new $Y',Y''$, determined by cocartesianness of the two squares. -The claim is that the original diagram is a retract of this new one. This follows essentially because retracts of cocartesian squares are cocartesian.<|endoftext|> -TITLE: Are generalized symmetric groups maximal finite groups (in a certain sense)? -QUESTION [5 upvotes]: Let $S(m,n)$ be the generalized symmetric group which is a wreath product of the cyclic group of order $m$, denoted here by $\mathbb{Z}_m$, and the symmetric group $S_n$. A standard unitary representation of $S(m,n)$ is given by the semi-direct product of the $n\times n$ permutation matrices and $n \times n$ diagonal matrices with $m$-th roots of unity entries. (These matrices are sometimes called generalized permutation matrices.) -For finite $m$ and $n$ all such unitary matrices form a finite subgroup of $U(n)$ of size $n! m^n$. I'll denote this group of unitary matrices as $M(m,n)$. One way to see that $M(m,n)$ is not a maximal finite subgroup of $U(n)$ is to note that $S(2m,n)$ is a finite group which has $S(m,n)$ as a subgroup. I'm interested in a slightly different question. -Let $U'\in U(n)$ be a unitary matrix which is not in $M(m,n)$ for any $m$ (in other words, $U'$ is not a generalized permutation matrix). Is the group generated by $M(m,n)$ (for some fixed $m$) and $U'$ always infinite? - -REPLY [4 votes]: I think the answer is "yes" when $m >6$. By arguments along the lines of Frobenius, Schur and Blichfeldt, if we set $G = \langle M(m,n), U^{\prime} \rangle $ and assume that $G$ is finite, then the non-scalar elements of $G$ whose eigenvalues all lies on an arc of length less than $\frac{\pi}{3}$ on the unit circle generate an Abelian normal subgroup of $G$, say $A$. When $m >6$, this Abelian group has rank $n$, and may be assumed to consist of diagonal matrices. In that case, $C_{G}(A)$ also consists of diagonal matrices (using that $A$ has rank $n$). By Clifford's theorem, we may conclude that the given representation of $A$ is now induced from a $1$-dimensional representation of a subgroup of $G$ of index $n$. In other words, $G$ now consists of monomial matrices. -I suppose, to be more precise, this argument shows that if $m > 6$, any finite unitary overgroup of $M(m,n)$ is conjugate (via a unitary matrix) to a finite group of monomial matrices ( what you call "generalized permutation matrices" are what I am calling monomial matrices).<|endoftext|> -TITLE: How to work out this elliptic function? -QUESTION [5 upvotes]: Let $f(x) = \sum\limits_{(n,m)\in\mathbb{Z}^2} \frac{1}{(x+ n + i m )^2}$ -If feel it should be $1/E(x)$ where $E$ is some elliptic function, like $sn^2$. But Wolfram Alpha is giving me some strange expression in terms of q-digamma functions. -But I would rather like to find it in terms of theta functions or elliptic functions. - -REPLY [8 votes]: This is a divergent series. But if one applies summation in the sense of Eisenstein, -$$\lim_{N\to\infty}\sum_{n=-N}^N\left(\lim_{M\to\infty}\sum_{m=-M}^M\right)$$ -then the sum is doubly periodic. Since the poles are at the lattice and residues are equal to $1$, it is equal $\wp(z)+C$. Looking at the Laurent expansion at $0$ we obtain $C=0$. So your sum is the Weierstrass function (if it is understood in the sense of Eisenstein). -Remark. The inner sum in parentheses is absolutely convergent. -Ref. A. Weil, Elliptic functions according to Eisenstein and Kronecker, Springer, 1976. -For Eisenstein summation, see also -Remmert, Classical topics in complex function theory, Springer 1998. He uses it to define trigonometric functions.<|endoftext|> -TITLE: Indecomposable contracting maps on the integers -QUESTION [15 upvotes]: $\def\ZZ{\mathbb{Z}}$Call a function $f : \ZZ \to \ZZ$ "contracting" if -$$|f(j) - f(i)| \leq |j-i|$$ -for all $i$, $j \in \ZZ$. The contracting functions form a monoid under composition; call it $C$. An element of a monoid is called a "unit" if it is invertible; the units of $C$ are the functions $x \mapsto \pm x + k$. An element of a monoid is called ``irreducible" if it is not a unit and cannot be factored as the composition of two non-units. - -Question 1: What are the irreducibles of $C$? - -To give two nonobvious examples, the maps $x \mapsto |x|$ and $x \mapsto \begin{cases} x & x \geq 0 \\ x+1 & x < 0 \end{cases}$ are both irreducible. -The problem which I actually want the answer to is a slight variant of $C$: Define $C_2$ to be the monoid of maps $f : \ZZ \to \ZZ$ which are contracting and obey $f(i) \equiv i \bmod 2$. So what I would really like to know is: - -Question 2: What are the irreducibles of $C_2$? - -If you prefer finite monoids, I am fine with you working with $\{ 0,1,2,\ldots,n \}$ instead of $\ZZ$ for either question. - -REPLY [9 votes]: I'll solve question 2, on $C_2$. -I will prove the irreducibles are those that only have one or two bends, verifying a prediction of Nate (and disproving a prediction of myself). -Call a "run" a maximal interval on which $f$ is linear. Clearly $f$ is linear on $[a,b]$ if and only if we either have $f(i)=i+1$ for all $i= a,\dots, b-1$, or $f(i) = i-1$ for all $i= a,\dots, b-1$. Then $[a,\dots, b]$ is maximal if, in addition, we have $f(a-1) = f(a)+1$ and $f(b+1) =f(b-1)$ in the first case or $f(a-1)=f(a)-1$ and $f(b+1)=f(b-1)$ in the second case. -I'll show that if $f$ is irreducible and $f$ has a run then $f$ has exactly one run. The number of runs is the number of bends minus one, so this is equivalent to Nate's claim. -The length $b-a$ of a run is a nonnegative integer, so if there is any run, there is a run of minimal length. If $[a,a+k]$ is a run of minimal length, then $f$ must be linear on $[a-k,a]$ and $[a+k,a+2k]$ as otherwise these intervals would contain a shorter run (the longest linear subinterval touching $[a,a+k]$). -Assume wlog the run of minimal length is increasing. Then $f( a-i ) = f(a)+i = f(a+i)$ for $0\leq i \leq k$ and $f(a+k-i) =f(a+k)-i = f(a+k+i)$ for $0 \leq i \leq k$. -So if we let $g(n)$ be given by the rule that $g(n) = n $ for $n \leq a$, $g(n) = 2a-n$ for $a\leq n \leq a+k$, and $g(n) = n-2k$ for $n\geq a+k$, and $h(n) = f(n)$ for $n \leq a$ and $h(n) = f(n+2k)$ for $n \geq a$, then $f =g \circ g$. -So if $f$ is irreducible, since $h$ is certainly not invertible (we have $k\geq 1$ since runs have length at least $1$), $g$ must be invertible, i.e. translation or reflection. So $f$ has the same number of runs as $h$, i.e., one.<|endoftext|> -TITLE: Original reference for generators and relations of 2-dimensional TQFT -QUESTION [5 upvotes]: What is the original reference where it was first proven that the generators and relations of the 2-dimensional cobordism category are those of commutative Frobenius algebras? -I've seen this article by Abrams being cited for it. But when I look into it I only find "Completeness of the relations follows easily by inspection" in proposition 12. I'm confused since I thought the completeness of relations was the only non-trivial part. The only place where I have seen something that looks like an actual proof is in this later article by Lauda and Pfeiffer in section 3.7, where they discuss 2-dimensional open-closed TQFT which obviously contains ordinary TQFT. - -REPLY [5 votes]: Generators and relations for the nonextended 2-dimensional bordism category already appear in Robbert Dijkgraaf's 1989 PhD dissertation, see Section 3.2.<|endoftext|> -TITLE: Can a non-Kähler complex manifold be rationally connected? -QUESTION [5 upvotes]: Let $X$ be a compact complex manifold. Suppose that $X$ is rationally connected in the sense that any two points lie in the image of a rational curve $\mathbb{CP}^1 \to X$. Are there any non-Kähler examples of such $X$? -If $X$ is Kähler and rationally connected, then $X$ is projective. So I might suspect that being rationally connected may force some algebraic structure on $X$, but I cannot find any reference for this question. - -REPLY [3 votes]: As Jason said already, there are many -examples of Moishezon manifolds which -are rationally connected. Indeed, any -manifold bimeromorphic to a rational -connected manifold is again rationally -connected, and there are many non-Kahler -Moishezon manifolds bimeromorphic to -$CP^n$ (or any other Fano manifold, -I suppose). -The better question would be whether all -rational connected manifolds are Moishezon. -The answer is, suprisingly, positive, -if you define rational connectedness -one way, and negative for another definition. -Recall that a rational curve is called -ample if its normal bundle -is $\bigoplus O(i_k)$ with all $i_k$ positive. -The first definition is -Defintion: -A compact complex manifold $M$ is called -rationally connected if there is an ample -curve with compact deformation space (Barlet -or Douady, does not matter). -In this case, any rationally connected manifold -is Moishezon, as shown by Campana -F. Campana, Reduction algebrique d'un morphisme -faiblement Kahlerien propre, Math. Ann. 256 (1980), 157--189. -Another definition is the same, but without the compactness -assumption. In this case you have manifolds which -are rationally connected, but not Moishezon, for example, -the twistor space of K3 surface or a torus. Curiously, -there is a rational curve passing through every -$n$ points of such a manifold; this is explained, -for example, in my paper -"Rational curves and special metrics on twistor spaces". -Personally, I think the first definition makes more sense.<|endoftext|> -TITLE: Alon-Füredi for homogeneous polynomials -QUESTION [9 upvotes]: A theorem of Alon and Füredi says that if $A$ and $B$ are finite, nonempty subsets of the field $\mathbb F$, and if a polynomial $P(x,y)\in\mathbb F[x,y]$ vanishes on all, but exactly one point of the grid $A\times B$, then -$\deg P\ge |A|+|B|-2$. -Can one improve this bound given that $P$ is homogeneous? - -What is the smallest possible degree of a homogeneous polynomial $P(x,y)$ vanishing on all, but exactly one point of the grid $A\times B$? - -The underlying field $\mathbb F$ can be assumed to have characteristic $2$. - -REPLY [3 votes]: In general homogeneity does not improve the bound. Take $A=\{1,q,q^2,\ldots,q^{a-1}\}$, $B=\{1,q,q^2,\ldots,q^{b-1}\}$, $f(x,y)=\prod_{i=-(b-1)}^{a-2}(x-q^iy)$. - -REPLY [3 votes]: If $0\notin B$, then $$F(x,y) = y^n \left(c_n \left(\frac{x}{y}\right)^n+\cdots+c_0\right).$$ Let $S=\{x/y:\ x\in A,y\in B\}.$ For $F$ to vanish on all the points of $A\times B$ except $1$, we must have that the one variable polynomial $$P(z)=c_n z^n+\cdots +c_0$$ vanishes on all the points of $S$ except $1$, and so $\deg F \geq |S|-1$. -In general, the Scherk-Kemperman Theorem states that $$|S|\geq |A|+|B|-\min_{c \in S}|A\cap cB|$$ (thank you Fedor for pointing this out) and equality is obtained if $A$ and $B$ are geometric progressions. Hence we will not have an improvement to the bound in the specific case where $A$ and $B$ are geometric progressions.<|endoftext|> -TITLE: Rational even polynomials maximally tangent to the unit circle -QUESTION [10 upvotes]: This question is motivated by College Mathematics Journal problem 1196, proposed by Ferenc Beleznay and Daniel Hwang. My solution to this problem (pre-publication version here) uses Chebyshev polynomials to show that for every $m \in \mathbb{N}$ there is an even polynomial $P_{2m}(x) \in \mathbb{Q}[x]$ of degree $2m$ such that the graph -$$\{(x, P_{2m}(x)) : -1 \le x \le 1\}$$ -is tangent to the unit circle $x^2+y^2=1$ at $2m-1$ points, and has two further points of intersection at $\pm 1$. Numerical calculations suggest that something stronger is true. - -Is is true that for every $m \in \mathbb{N}$ with $m \ge 2$ there is an even polynomial $Q_{2m}(x) \in \mathbb{Q}[x]$ of degree $2m$ such that the graph $\{(x, Q_{2m}(x)) : -1 \le x \le 1\}$ is tangent to the unit circle at $2m-3$ points, and has two further triple points of intersection? - -Examples. The quartic $Q_2(x) = 1 - \frac{(3x)^2}{2} + \frac{(3x)^4}{24}$ is tangent to the unit circle at $(x,y) = (0,1)$ and has triple points at $(x,y) = (\pm \frac{2\sqrt{2}}{3},-\frac{1}{3})$. The graphs below show the intersections for -\begin{align*} P_8(x) &= 1 - 25 x^2 + 104 x^4 - 144 x^6 + 64 x^8 \\ -Q_8(x) &= 1 - \frac{(7 x)^2}{2} + \frac{(7 x)^4}{24} - \frac{(7 x)^6}{864} + \frac{(7 x)^8}{96768} -\end{align*} - -Further remarks. - The polynomials $P_{2m}(x)$ in fact have coefficients in $\mathbb{Z}$. They meet the bound in Bezout's Theorem for the intersection multiplicity between the plane algebraic curves $y = P_{2m}(x)$ of degree $2m$ and $x^2 + y^2 = 1$ of degree $2$, since $(2m-1)\times 2 + 2 \times 1 = 4m$. If $Q_{2m}(x)$ exists then it also meets the bound, now with $(2m-3) \times 2 + 2 \times 3 = 4m$. - I've asked for a rational polynomial $Q_{2m}(x)$ because my numerical experiments suggest this is possible, and the coefficients of the polynomials I've found seem to have some interesting aritmetic properties, see the solutions below. (Does anyone recognise the denominators?) But I have not proved the existence of $Q_{2m}(x)$ working over the real numbers. -\begin{align*} -2:\quad & 1 - \frac{(3x)^2}{2} + \frac{(3x)^4}{24} \\ -3:\quad & 1 - \frac{(5x)^2}{2} + \frac{(5x)^4}{24} - \frac{(5x)^6}{1080} \\ -4:\quad & 1 - \frac{(7x)^2}{2} + \frac{(7x)^4}{24} - \frac{(7x)^6}{864} + \frac{(7x)^8}{96768} \\ -5:\quad & 1 - \frac{(9x)^2}{2} + \frac{(9x)^4}{24} - \frac{(9x)^6}{800} + \frac{(9x)^8}{64000} - \frac{(9x)^{10}}{14400000} \\ -6:\quad & 1 - \frac{(11x)^2}{2} + \frac{(11x)^4}{24} - \frac{7(11x)^6}{5400} + \frac{(11x)^8}{54000} - \frac{(11x)^{10}}{8100000} + \frac{(11x)^{12}}{3207600000} -\end{align*} -and for $m=7$ and $m=8$ there are -$$\begin{align*} 7: \quad & \scriptstyle 1 - \frac{(13x)^2}{2} + \frac{(13x)^4}{24} - \frac{(13x)^6}{756} + \frac{(13x)^8}{49392} - \frac{(13x)^{10}}{6223392} + \frac{(13x)^{12}}{1568294784} - \frac{(13x)^{14}}{999003777408}\\ 8: \quad & \scriptstyle 1 - \frac{(15x)^2}{2} + \frac{(15x)^4}{24} - \frac{3(15x)^6}{2240} + \frac{15(15x)^8}{702464} - \frac{11(15x)^{10}}{59006976} + \frac{(15x)^{12}}{1101463552} - \frac{(15x)^{14}}{431773712384} + \frac{(15x)^{16}}{414502763888640}.\end{align*}$$ -The prime factorizations of the denominators of the coefficients of $x^{2m}$ for $2 \le m \le 8$ are -$$2^3\cdot 3, \quad 2^3\cdot 3^3\cdot 5,\quad 2^9\cdot 3^3\cdot 7,\quad 2^9\cdot 3^2\cdot 5^5, \quad 2^7\cdot 3^6\cdot 5^5\cdot 11,\quad 2^7\cdot 3^6\cdot 7^7\cdot 13, \quad 2^{25}\cdot 3\cdot 5\cdot 7^7$$ -At Fedor Petrov's suggestion I computed the coordinates of the triple points. For $m \le 8$ the triple point with positive $x$-coordinate is -$$\bigl(\frac{\sqrt{(2m+1)^2-1}}{2m+1}, \frac{(-1)^{m-1}}{2m+1}\bigr).$$ -(Sign of $y$ coordinate corrected from $(-1)^m$.) There appears to be no simple closed form for the double points in my solutions. -The original problem also required odd polynomials $P_{2m-1}$ of degree $2m-1$ with analogous tangency properties. I conjecture that the analogous question for $Q_{2m-1}$ also has a positive answer. - -REPLY [4 votes]: Denote $\varepsilon=\frac1{2m-1}$ (I guess it should be minus in the denominator, not plus, please recheck). -We want to find a polynomial $h_{m-1}(t)$ of degree $m-1$ and a polynomial $p_m(t)$ of degree $m$ such that $$p_m^2(t)-t(t-1+\varepsilon^2)h_{m-1}^2(t)=1-t,\quad\quad\quad\,\,\,\,\,\,\,\,(1)$$ -$$h_{m-1}(1-\varepsilon^2)=0,\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(2)$$ -$$\text{all roots of}\, h_{m-1}\, \text{are distinct and belong to}\, (0,1)\quad (3).$$ -Indeed, if $\alpha$ is a root of $h_{m-1}$, then $(1)$ yields that the points $(\pm \sqrt{\alpha},p_m(\alpha))$ belong to a unit circle and to a graph of $y=p_m(x^2)$, and the tangency is at least double for all $\alpha$ and triple for $\alpha=\pm \sqrt{1-\varepsilon^2}$. Also $(0,p_m(1))$ is a point of tangency. -Now we look at (1) as Pell type equation $$\left(P-\sqrt{t(t-1+\varepsilon^2)}Q\right)\left(P+\sqrt{t(t-1+\varepsilon^2)}Q\right)=1-t.$$ There is a basic solution with 1 at right hand side: $P_0^2-t(t-1+\varepsilon^2)Q_0^2=1$ for $P_0=\alpha t-1$, $Q_0=\alpha$ with $\alpha:=\frac2{1-\varepsilon^2}$. A solution with $1-t$ in RHS is $(\beta t-1)^2-t(t-1+\varepsilon^2)\beta^2=1-t$, where $\beta(1\pm \varepsilon)=1$ (both signs are ok). So, we choose -$$ -p_m+\sqrt{t(t-1+\varepsilon^2)}h_{m-1}=\left(\beta t-1+\beta \sqrt{t(t-1+\varepsilon^2)}\right)\left(\alpha t-1+\alpha\sqrt{t(t-1+\varepsilon^2)}\right)^{m-1}. -$$ -For checking (2), we expand the brackets in RHS and choose only those which contain exactly one $\sqrt{t(t-1+\varepsilon^2)}$. In other words, $1-\varepsilon^2$ should be a root of $$(\alpha t-1)\beta+(m-1)\alpha(\beta t-1)=0.$$ -Well, this happens if $\beta=(1+\varepsilon)^{-1}$ (straightforward check). -It remains to check (3). For this, we note that when $t$ goes from 0 to $1-\varepsilon^2$, the point $\alpha t-1+\alpha\sqrt{t(t-1+\varepsilon^2)}$ goes along the half of the unit circle. Denote $t=\cos \tau$ for $0<\tau<\pi$. We need $m-2$ distinct points $\tau\in (0,\pi)$ for which the imaginary part of $$\frac{\alpha}{\beta}\left(p_m+\sqrt{t(t-1+\varepsilon^2)}h_{m-1}\right)=(\cos \tau+i\sin \tau)^{m-1}\left(\cos \tau+i\sin \tau-\right(\frac{\alpha}{\beta}-1\left)\right)$$ -equals to 0. As $\frac{\alpha}{\beta}-1=\frac{m}{m-1}$, this reads as $\sin m\tau=\frac{m}{m-1}\sin (m-1)\tau$, or $$\theta(\tau):=\frac{\sin (m-1)\tau}{\sin m\tau}=\frac{m-1}m.$$ -For each $k=1,2,\ldots,m-2$ consider an interval $(\frac{\pi k}m,\frac{\pi(k+1)}m)$. In this interval the function $\theta(\tau)$ is continuous, and the limit values at the endpoints are $+\infty$ and $-\infty$ in some order. Thus $\theta$ has at least one root in each of these intervals, totally $m-2$ roots on $(0,\pi)$, as we need.<|endoftext|> -TITLE: Classification of knots in solid torus -QUESTION [7 upvotes]: What is known about the classification of knots in a solid torus $S^1 \times D^2$? Is enumerating them a reasonable problem? Do we get a similar classification as for knots in $S^3$? Ideally there would be a simple description of the Seifert fibered knots in $S^1 \times D^2$ like for prime non-satellite knots in $S^3$ (they're exactly the torus knots). -The motivation is to better understand the classification of prime knots in $S^3$ as - -torus knots, -hyperbolic knots, or -nontrivial satellites, - -because satellites come from combining a knot in $S^3$ and a knot in $S^1 \times D^2$. -One issue is that we would want to exclude knots that are only knotted in a ball in $S^1 \times D^2$ because these aren't really new: they come from a knot in $S^3$, and they give composite knots in the satellite construction. Is there a relatively simple way to exclude these? -More formally: one class of knots to exclude are those obtained by taking a $(1,1)$-tangle in a $3$-ball, then closing up the ends so that the knot is not null-homologus in $S^1 \times D^2$. These are nontrivial knots in $S^1 \times D^2$ but they don't really use the topology of the solid torus in an interesting way, so we want to exclude them from our classification. - -REPLY [9 votes]: Up to Dehn twists, the class of knots in the solid torus is identical to the class of two-component links in the three-sphere, where the first component is an unknot. -For example, Seifert fibered knots in the solid torus give Seifert fibered links in the three-sphere. The base space is the orbifold $S^2(p,\infty,\infty)$. -There is a similar theory of “satellite” links. The knots you wish to exclude (knotted in a three-ball inside the solid torus) are a special case of these. When thought of as links, they are exactly the split links in the three-sphere (where again the first component is the unknot). -Finally, the knot complement in the solid torus is hyperbolic exactly when the link complement in the three-sphere is.<|endoftext|> -TITLE: Faster computation of p-adic log -QUESTION [8 upvotes]: As I see it, $p$-adic integers work very similar to formal power series over $x$ (e g. with regards to Hensel lifting). -When it comes to computing $\log P(x)$, one may use the formula -$$ -(\log P)' = \frac{P'}{P} -$$ -to compute the expansion of the logarithm of $P(x)$ with $P(0)=1$ as -$$ -\log P \equiv \int \frac{P'}{P} dx \pmod{x^n}. -$$ -This is the main trick to compute first $n$ coefficients of $\log P(x)$ in $O(M(n))$, where $M(n)$ is the maximum time needed to compute the product of two polynomials of degree at most $n$. -Is there any similar way to compute modulo $p^n$ the expansion of the $p$-adic logarithm of the $p$-adic integer $r$ such that $r \equiv 1 \pmod p$ in $O(M(n))$? -As I see it, the regular notion of polynomial derivatives can't be applied directly to $p$-adic integers, as $(uv)' = u'v+uv'$ woudln't hold. So, maybe there is a way to define some other reasonably invertible function $d(r)$ which is simple enough to compute and such that -$$ -d(uv) = u d(v) + v d(u) -$$ -for any $p$-adic numbers $u$ and $v$? - -REPLY [9 votes]: Fredrik Johansson gave a very good answer, but I feel like preprint's notation is a bit too dense and abstract (with finite extensions, uniformizers, ramifications and such), so I'll try to recap its content in a bit more plain manner here. It turned out to be somewhat lengthy, not fit for a comment, so I will write it as another answer. -For $x \equiv 0 \pmod p$ we define a sequence $y_1, y_2, \dots$ such that -$$ -1 - y_{k} \equiv (1-y_{k-1})(1+y_{k-1}) \pmod{p^{2^k}}. -$$ -with a starting number $y_1 \equiv x \pmod{p^2}$. It can be proven by induction that $y_k$ is divisible by $p^{2^{k-1}}$ and -$$ -(1-y_k)(1+y_k) \equiv 1 \pmod{p^{2^k}}. -$$ -For $m > \log_2 n$ it holds that $y_m$ is divisible by $p^{n}$, hence -$$ -1-y_m \equiv 1 \equiv (1-y_{m-1})(1+y_{m-1}) \pmod{p^n}. -$$ -Expanding it further, we get -$$ -1 \equiv (1-x)(1+y_1)(1+y_2)\dots(1+y_{m-1}) \pmod{p^n}. -$$ -Taking logarithms on both sides, we arrive at -$$ -\log(1-x) = -\sum\limits_{i=1}^{m-1} \log(1+y_i) \pmod{p^n}. -$$ -Now, $y_i$ is divisible by $p^{2^{i-1}}$ but itself is at most $p^{2^i}$. The logarithm of $1+y_i$ in such case can be computed from the direct expression -$$ -\log(1 + y_i) = \sum\limits_{k=1}^{\infty} \frac{(-1)^k y_i^k}{k}. -$$ -In this expression, only roughly first $\frac{n}{2^{i-1}}$ terms are of interest because $y_i$ is divisible by $p^{2^{i-1}}$ and they can be computed in $O(n \log^2 n)$ overall with divide and conquer algorithm using the fact that $y_i$ has only $2^{i-1}$ significant terms. -To be a bit more specific, to compute $\sum\limits_{k=1}^N \frac{y^k}{k_0+k}$, the following two sums are computed recursively: -$$ -S_1 = \sum\limits_{k=1}^{\lfloor N/2 \rfloor} \frac{y^k}{k_0+k} -$$ -and -$$ -S_2 = \sum\limits_{k=1}^{\lfloor N/2 \rfloor} \frac{y^k}{(k_0+\lfloor N/2\rfloor)+k}. -$$ -Then, the second sum is multiplied by $y^{\lfloor N/2 \rfloor}$ and is added to the first one. -Since $m$ is roughly $\log_2 n$, summing this up for all $i$, we get $O(n \log^3 n)$. -I still wonder if it is possible to do it faster or simpler...<|endoftext|> -TITLE: Reference request for a certain exponential series -QUESTION [5 upvotes]: I recently encountered the series $$\sum_{d \in \mathbb{Z}} e^{-t^d}t^{kd},$$ for real $0 -TITLE: Is there any hope to prove that $g(x)>-4$ if $f(x)<0$? -QUESTION [6 upvotes]: I have these two functions for $x>0$, $\beta>0$ and $\alpha$ (all reals) -$$ f(x)= \frac{\alpha \; \sin (\beta \; x)}x+4 \cos (\beta\; x) ,\qquad\qquad\qquad\qquad\qquad\qquad\\ -g(x)=\frac{\alpha \cos (\beta \;x)}{\beta \; x^2}+\frac{\alpha \cos (2 \beta \; x)}{\beta \; x^2}-\frac{\alpha\; \cot (\frac{\beta \; x}{2})}x -\frac{4\;\sin (2 \beta \; x)}{\beta\;x} -$$ - -Numerically, I am sure that if $f(x)<0$, then $g(x)>-4$ (a plot of a typical example is attached); I appreciate any hints (if there are any) to prove this analytically. - -REPLY [8 votes]: $\newcommand{\R}{\mathbb R}$Your conjecture is true. -Indeed, -\begin{equation*} -\begin{aligned} - f(x)&=F(a,u):=a\frac{\sin u}u+4\cos u, \\ - g(x)&=G(a,u):=a\frac{\cot(u/2)}{u^2}\,h(u)-4\frac{\sin 2u}u, -\end{aligned} -\end{equation*} -where -\begin{equation*} - a:=\alpha\beta\in\R,\quad u:=\beta x>0, -\end{equation*} -\begin{equation*} - h(u):=\sin2u-\sin u-u. -\end{equation*} -(One of the ways to check the equality $g(x)=G(a,u)$ is by using the identities $\sin u=\frac{2t}{1+t^2}$ and $\cos u=\frac{1-t^2}{1+t^2}$, where $t:=\tan\frac u2$, as well as the identities $\cos2u=2\cos^2u-1$ and $\sin2u=2\sin u\cos u$.) -So, your conjecture can be rewritten as follows: -for $a$ and $u$ as specified above, -\begin{equation*} - F(a,u)<0\implies G(a,u)>-4. \tag{1}\label{1} -\end{equation*} -Note that $F(a,u)=4\not<0$ if $u=2k\pi$ for a natural $k$. So, without loss of generality (wlog) $u\ne 2k\pi$ for any natural $k$, and hence $\cot(u/2)\ne0$ and, moreover, -\begin{equation*} - \cot(u/2)=\frac{\sin u}{2\sin^2(u/2)}\overset{\text{sign}}=\sin u, \tag{2}\label{2} -\end{equation*} -where $\overset{\text{sign}}=$ means the equality in sign. -Also, if $\sin u=0$, then $G(a,u)=0>-4$, so that implication \eqref{1} holds. So, wlog $\sin u\ne0$ and hence $\cos u\ne1$. -Lemma 1: $h<0$. -Lemma 2: $r(u):=\dfrac{u-\sin u}u<1.25$. -These elementary lemmas will be proved later in this answer. -Letting $a_u:=-4u\cot u$, note that - -$F(a,u)<0\iff a0$; - -$F(a,u)<0\iff a>a_u$ $\quad\text{if}\quad$ $\sin u<0$. - - -So, by Lemma 1 and \eqref{2}, -\begin{equation*} - F(a,u)<0\implies G(a,u)>G(a_u,u)=H(u):=R(u) r(u), -\end{equation*} -where -\begin{equation*} - R(u):=\frac{4\cos u}{1-\cos u}. -\end{equation*} -Modulo Lemmas 1 and 2, it remains to show that $H>-4$. If $\cos u>0$, this follows because $r>0$. -If, finally, $\cos u<0$, then $R(u)>-2$ and hence, by Lemma 2, $H(u)=R(u) r(u)>-2\times1.25=-2.5>-4$. $\quad\Box$. - -It remains to prove Lemmas 1 and 2. -Proof of Lemma 1: Recall that $h(u):=\sin2u-\sin u-u$. Note that $\sin2u-\sin u$ is $2\pi$-periodic in $u$, whereas $-u$ is decreasing in $u$. So, it suffices to show that $h<0$ on the interval $(0,2\pi)$. But this follows because -\begin{equation*} - h'(u)=(\cos u-1)(4\cos u+3)\overset{\text{sign}}=-(4\cos u+3), -\end{equation*} -so that the critical points of $h$ are easily found and considered. $\quad\Box$ -Proof of Lemma 2: The inequality $r(u)<1.25$ can be rewritten as $b(u):=u+4\sin u>0$. Note that $\sin u$ is $2\pi$-periodic in $u$, whereas $u$ is increasing in $u$. So, it suffices to show that $b>0$ on the interval $(0,2\pi)$. But this follows because $b'(u)=1+4\cos u$, so that the critical points of $b$ are easily found and considered. $\quad\Box$ - -Remark 1: It follows from this proof that the lower bound $-4$ on $g(x)=G(a,u)$ in \eqref{1} can be replaced by the better lower bound $-2.5$. In fact, the best lower bound on $g(x)=G(a,u)$ in \eqref{1} is $\min_{u>0}H(u)=H(u_*)=-2.1289\dots$, where $u_*=3.5264\dots$. The minimization of $H(u)$ in $u>0$ can be effectively done by, say, the interval method, using the fact that $r$ is increasing on all intervals where $\cos<0$.<|endoftext|> -TITLE: Diagrams for strongly invertible knots with 10 crossings -QUESTION [6 upvotes]: I would like to extend the list of diagrams in the paper 'On strongly invertible knots' by Makoto Sakuma (1986) to knots with 10 crossings and succeeded for all but 8 of them: $10_{49}$, $10_{62}$, $10_{65}$, $10_{112}$, $10_{113}$, $10_{143}$, $10_{152}$, $10_{154}$. -Does anyone know of a method to obtain a symmetric diagram for these knots (or of an already existing list for 10 crossing knots)? -Some more detail: I look for transvergent diagrams for these 8 knots. Rotation about the axis then transforms the diagram into itself but with reversed orientation. If two axes are possible it would be great to have diagrams showing both axes simultaneously but for the moment I am fine with having diagrams with only one of them. -Edit, 24.05.2022: For $10_{112}$ the KLO diagram (see Marc's answer) based on the DT-code taken from Knot Info (and also the Knotscape diagram there) is already in the symmetric form (from the Rolfsen diagram it cannot be easily seen). For $10_{62}$ and $10_{65}$ I tried to modify the diagrams in KLO but was not yet successful. - -Edit, 28.05.2022: Thanks Marc, for finding symmetric diagrams in all cases. For $10_{154}$ I transformed the intravergent diagram into the following transvergent one: - -and the similar case of $10_{152}$ is obtained from this diagram by switching the crossing on the axis. -Therefore my goal to find symmetric diagrams for all strongly invertible knots with 10 crossings is achieved. In this set there are 45 2-bridge knots (these are always strongly invertible). There are 87 strongly invertible prime 3-bridge knots with 10 crossings (if I did not make an error in counting). I used the fact that invertible hyperbolic knots are strongly invertible and the symmetry information in Knot Info ('reversible' and 'fully amphicheiral' knots are invertible; note that there are several conventions for naming symmetries). I would like to present all 132 diagrams in a template fashion similar to the diagrams of symmetric unions in -my paper "The search for nonsymmetric ribbon knots". One aim of this study is a comparison of these two cases ('symmetric' and 'anti-symmetric' diagrams; the first family denoting symmetric unions and the second strongly invertible knots with transvergent diagrams). - -REPLY [6 votes]: Here is a strongly invertible diagram of 10_49. The symmetry axis is almost horizontal. - -This was created by loading the 'standard' diagram of 10_49 to KLO and then using KLO's simplifying method together with some work by hand. (KLO's simplifying method seems to prefer symmetric diagrams this is why I would expect the method to work in most cases.) -We have used similar methods for example -here to create strongly invertible surgery diagrams. I would expect that the same works for the other 7 knots from your list. If you need help with that let me know. -EDIT: For the other knots it works similar. Here are diagrams of them. -10_62 - -10_65 - -10_113 - -10_143 - -10_152 - -10_154 - -(For the last two the symmetry axis is orthogonal to the projection plane.)<|endoftext|> -TITLE: Kneser theorem about the Klein bottle -QUESTION [7 upvotes]: I know that in $1923$ H. Kneser showed that a continuous flow in a Klein bottle without singular points has a periodic trajectory. The original article is this, but does anyone know another old or new proof of this result? (So far I have found the article by Kneser, the article by Nelson G. Markley and some works that use the results of these articles, I am looking for some other idea focused on proving the result mentioned at the beginning) I would really like to read this result, I tried to do it from your original article but the language is too complicated for me. I searched on the internet but found almost nothing about the proof. I asked here but I didn't find any answer even with bounty. I hope to have some help it would help me a lot. - -REPLY [4 votes]: It sounds like you are looking for a textbook. I have read bits of Introduction to the geometry of foliations: Part A by Hector and Hirsch. It is well-written, with pictures! They give Kneser's theorem on pages 62-65, after developing the necessary theory.<|endoftext|> -TITLE: A question regarding isomorphism in cohomology for moduli space of stable bundles over a compact Riemann surface -QUESTION [5 upvotes]: Let $N(n,k)$ denote the moduli space of stable vector bundles of rank $n$ and degree $k$ over a compact Riemann surface $X$, and let $N_0(n,k)$ denote the moduli space where we fix rank $n$ and some fixed determinant bundle of degree $k$. We know that the determinant map $det: N(n,k)\rightarrow Pic^k(X)$ is a proper submersion with fibers isomorphic to $N_0(n,k)$. -In the paper 'The Yang-Mills equations over Riemann surfaces' by M. F. Atiyah and R. Bott, (Phil. Trans. R. Soc. Lond. A 308, 523-615 (19)) the authors prove the following: - -Proposition 9.7. (page 578) For rational cohomology we have $H^*(N(n,k)) \simeq H^*(N_0(n,k))\otimes H^*(Pic^0(X)).$ - -Immediately after this, the authors mention the following : - -"This proposition, which is equivalent to the statement that $\Gamma_n := H^1(X,\mathbb{Z}_n)$ acts trivially on the rational cohomology of $N_0(n,k)$, ..." - -( $\Gamma_n$ actually corresponds to the $n$-torsion line bundles on $X$, and a line bundle $L\in \Gamma_n$ acts on $N_0(n,k)$ by sending $E\mapsto E\otimes L$.) -I want to understand the above equivalence , at least the direction why the statement implies the proposition. I believe that it is somehow related to the monodromy action of $\pi_1(Pic^k(X))\simeq H^1(X,\mathbb{Z})$ on the cohomology of fibers via the $det$ map I mentioned above. If this action is trivial, then one can deduce that the map on cohomologies induced by the inclusion of a fiber is surjective, which would imply the proposition thanks to Leray-Hirsch theorem. But the authors state that instead the action of $\Gamma_n=H^1(X,\mathbb{Z}_n)$ is trivial. Of course, applying UCT we have -$$H^1(X,\mathbb{Z}_n)=H^1(X,\mathbb{Z})\otimes \mathbb{Z}_n \simeq \pi_1(Pic^k(X))\otimes \mathbb{Z}_n,$$ -does this imply that the monodromy action of $\pi_1(Pic^k(X))$ factors through the action of $\Gamma_n$? Is this the reason? -Thanks in advance. - -REPLY [9 votes]: Things are actually simpler. View $\Gamma _n=H^1(X,\mathbb{Z}/n)$ as the group of line bundles $L\in \operatorname{Pic}^{0}(X) $ with $L^{{\tiny \otimes }n}=\mathscr{O}_X$. The -map $N_0(n,k) \times \operatorname{Pic}^{0}(X) \rightarrow N(n,k)\ $ given by $\ (E,L)\mapsto E\otimes L\ $ identifies $N(n,k)$ to the quotient of $N_0(n,k) \times \operatorname{Pic}^{0}(X) $ by $\Gamma _n$. Thus $H^*(N(n,k))$ is the subgroup of $H^*(N_0(n,k))\otimes H^*(\operatorname{Pic}^{0}(X) )$ invariant by $\Gamma _n$. But since $\Gamma _n$ acts by translation on $\operatorname{Pic}^{0}(X) $, it acts trivially on cohomology; -so $H^*(N(n,k))$ is isomorphic to the whole tensor product if and only if $\Gamma _n$ acts trivially on $H^*(N_0(n,k))$.<|endoftext|> -TITLE: Gaps in cardinalities of MAD families -QUESTION [6 upvotes]: Let $[\omega]^\omega$ denote the collection of infinite subsets of $\omega$. We say $a,b\in [\omega]^\omega$ are almost disjoint if $a\cap b$ is finite. A subset $A\subseteq [\omega]^\omega$ is said to be an almost disjoint family if $a, b$ are almost disjoint for all $a \neq b \in A$. A standard application of Zorn's Lemma shows that every almost disjoint family is contained in a maximal almost disjoint family (MAD family) (maximal with respect to $\subseteq$). -A diagonalization argument shows that all infinite MAD families have uncountable cardinality. By ${\frak a}$ we denote the minimum cardinality that a MAD family can have. It is consistent that ${\frak a} < {\frak c} = 2^{\aleph_0}$. -Question. Is it consistent that - -${\frak a} < {\frak c}$, -there is a MAD family $A\subseteq [\omega]^\omega$ with $|A| = {\frak c}$, and -there is a cardinal ${\frak g}$ with ${\frak a} \in {\frak g} \in {\frak c}$ such that there is no MAD family with cardinality ${\frak g}$? - -REPLY [8 votes]: Yes, this is consistent. -Suppose we force to add $\kappa$ mutually generic Cohen reals to a model of $\mathsf{CH}$, where $\kappa$ is some cardinal with uncountable cofinality. In the extension, there are MAD families of cardinality $\aleph_1$ and cardinality $\kappa = \mathfrak{c}$, but there are no MAD families of any intermediate cardinality. -The proof of this is a basic instance of what's known as an "isomorphism of names" argument. I think this result appears as an exercise in Kunen's 1980 book, although I think it's originally due to Arnie Miller. The argument is greatly extended in - -S. Shelah and O. Spinas, "MAD spectra," Journal of Symbolic Logic 80 (2015), pp. 243-262 (link). - -They prove that the set of all cardinals $\kappa$ such that there is a MAD family of size $\kappa$ can be almost completely arbitrary. For example, given any $A \subseteq \omega \setminus \{0\}$, there is a forcing extension in which $A = \{ n < \omega :\, \text{there is a MAD family of size } \aleph_n\}$.<|endoftext|> -TITLE: Are the polyhedral cones the only examples of cones that remains closed when they are added to vector subspaces? -QUESTION [5 upvotes]: Let $C \subset \mathbb{R}^{n}$ be a closed convex cone. If one wants to know whether the linear map $T:\mathbb{R}^{n} \to\mathbb{R}^m$ sends the closed set $C$ to another closed one, $T(C)$, it is needed to prove that $\text{ker } T + C$ is closed. -My concern turns into to know whether there exist good examples of cones, other than polyhedral cones, that are mapped to closed sets by any linear map. Hence, the question is reduced to: - -There exist a closed convex cone $C$, different from any polyhedral cone, such that, for every vector subspace $V$, $V+C$ is closed? - -The article On the closedness of the linear image of a closed convex cone almost answers my question. If one can find an enlightening example of cone $C$ satisfying the condition (SUM-WE) for every subspace, my example is constructed. - -REPLY [5 votes]: The radial cone of $C$ is defined via -$$ -\mathcal R_C(x) := \bigcup_{\lambda > 0} \lambda ( C - x)$$ -for all $x \in C$ -and we can show -$$ -\mathcal R_C(x) = C + \operatorname{span}(x), -$$ -since $C$ is a cone. -By assumption $\mathcal R_C(x)$ is closed for all $x \in C$, since $\operatorname{span}(x)$ is a subspace. But now, -Proposition 2 of Duality of linear conic problems by Shapiro and Nemirovski -implies that $C$ is a polyhedral cone.<|endoftext|> -TITLE: Non-trivial solutions for $-(c^2-d^2)(a^2-b^2)=2(ad-bc)(bd+ac)$? -QUESTION [11 upvotes]: Consider the quartic system in four variables $a,b,c,d\in\mathbb R$: -$$-(c^2-d^2)(a^2-b^2)=2(ad-bc)(bd+ac).$$ - -Does this system admit rational solution with $$abcd(c^2-d^2)(a^2-b^2)(a^2-c^2)(b^2-d^2)\neq0?$$ - - -Is there any easy way to compute for rational solution? - -REPLY [33 votes]: There is no such solution. Let -$$ -Q(a,b,c,d) = 2(ad-bc)(bd+ac) + (a^2-b^2)(c^2-d^2) -$$ -be the difference between the two sides of the equation, -so we seek to solve $Q(a,b,c,d) = 0$. This is a quadratic equation -in each variable, so in a rational solution the discriminant of $Q$ -with respect to each variable is a square. We choose $d$ -(the others work the same way), and find -$$ -{\rm disc}_d(Q) = (2c)^2 (2a^4 - 4a^3b + 4ab^3 + 2b^4). -$$ -Thus either $c=0$ or the second factor is a square. -The former is possible (with $d=0$ as well) but the problem statement -forbids it. The latter gives rise to an elliptic curve, -which turns out to be curve -40.a3 -with rank $0$ over the rational numbers; -its four torsion points correspond to $a=\pm b$ -(each of which gives rise to two points on the curve). -Since the problem statement also forbids $a^2=b^2$ we're done. -How did this problem arise?<|endoftext|> -TITLE: Contracting a set to a ball -QUESTION [5 upvotes]: $\newcommand\R{\mathbb R}\newcommand\S{\mathbb S}$ - -Question 1: Let $S$ be a nonempty measurable subset of $\R^n$. Let $B$ be a closed ball in $\R^n$ such that $m(B)=m(S)$, where $m$ is the Lebesgue measure. Is there a bijective $1$-Lipschitz map from $S$ onto a dense subset of $B$? - - -Question 2: If such a map exists, can we make it measure-preserving? - - -Question 1a: Let $S$ be a nonempty measurable subset of $\S^{n-1}$, the unit sphere in $\R^n$. Let $C\subseteq\S^{n-1}$ be a closed spherical cap such that $m(C)=m(S)$, where $m$ is the Haar measure. Is there a bijective $1$-Lipschitz map from $S$ onto a dense subset of $C$? (The metric on $\S^{n-1}$ with respect to which the $1$-Lipschitz condition is considered is either the geodesic metric on $\S^{n-1}$ or, equivalently, the one induced by the Euclidean metric on $\R^n$.) - - -Question 2a: If a map described in Question 1a exists, can we make it measure-preserving? - -A complete and correct answer to any one of these four questions will be considered a complete and correct answer to the entire post. - -Related, but different, questions and answers can be found on this MathOverflow page of 2018. - -REPLY [7 votes]: Trivial "No" to all: Take the union $S$ of two disjoint balls $B_1$ and $B_2$ of diameters $d$ and $\sqrt{1-d^2}$ respectively. If $f$ maps $S$ to the ball $B$ of diameter $1$, then $f(B_1)$ has diameter $\le d$. If $d$ is small enough, then $B\setminus f(B_1)$ still contains two opposite points on the circumference, so the diameter of it is still $1$ and $f(B_2)$ has no chance to get anywhere close to covering it.<|endoftext|> -TITLE: Determinant of matrix with Stirling numbers as elements -QUESTION [7 upvotes]: After noticing that the determinant of an $n \times n$ matrix $A_n$ with elements $a_{i,j}=i^j$, $1 \le i \le n$, $1 \le j \le n$, is the superfactorial (product of the first $n$ factorials), I wanted to try other cases. -I noticed that with $a_{i,j}=\binom{n+i-1}{j}$: -$$\lvert A_n \rvert = \binom{2n-1}{n}$$ -This one has been proved in this answer using the paper Binomial Determinants, Paths, and Hook Length Formulae by Gessel and Viennot. -After this I tried $a_{i,j}=\left[{n+i-1\atop j}\right]$ (unsigned Stirling numbers of the first kind) and obtained: -$$\lvert A_n \rvert = (n-1)!^{n} \tag{1}\label{1}$$ -and with $a_{i,j}={n+i-1 \brace j}$ (Stirling numbers of the second kind): -$$\lvert A_n \rvert = n!^{n-1} \tag{2}\label{2}$$ -Any hint for proving $(1)$ and $(2)$? -@Peter Taylor found the following generalizations: -With $a_{i,j}=\left[{k+i-1\atop j}\right]$ (unsigned Stirling numbers of the first kind): -$$\lvert A_n \rvert = (k-1)!^{n} \tag{3}\label{3}$$ -and with $a_{i,j}={k+i-1 \brace j}$ (Stirling numbers of the second kind): -$$\lvert A_n \rvert = n!^{k-1} \tag{4}\label{4}$$ - -REPLY [3 votes]: If we take a general sequence of (unsigned) Comtet numbers of the first kind [1, 2] $$c(n, k) = e_{n-k}(\xi_1, \ldots, \xi_n)$$ then the $n \times n$ submatrix with a row offset of $k$ has determinant $$\det_{0 \le i, j < n} \Big( e_{k+i-j}(\xi_1, \ldots, \xi_{k+i}) \Big) = (\xi_1 \cdots \xi_k)^n$$ -Similarly, if we take a general sequence of Comtet numbers of the second kind $$C(n, k) = h_{n-k}(\xi_1, \ldots, \xi_{k+1})$$ then the $n \times n$ submatrix with a row offset of $k$ has determinant $$\det_{0 \le i,j < n} \Big( h_{k+i-j}(\xi_1, \ldots, \xi_{j+1}) \Big) = (\xi_1 \cdots \xi_n)^k$$ -Note in particular that for your question (the Stirling numbers respectively of the first and second kinds, omitting the row $n=0$ and column $k=0$) we can take $\xi_i = i$. -A more natural generalisation of both the binomial and the Stirling examples you give is to take a row offset of $k$ and a column offset of $1$, but the resulting LU factorisations are far uglier. -Proofs -In a comment, Max Muller gave a reference to C. Krattenthaler, Advanced determinant calculus: A complement, Lin. Alg. and its Applications, 411 (2005), 68-166. Theorems 44 and 45 are very much in the same area, certainly generalise the result for Comtet numbers of the first kind. I'm no longer confident that they also cover Comtet numbers of the second kind, but I haven't wrestled with them enough to say that they definitely don't. Regardless, the paper which Krattenthaler references for these results is described as "in press" and, 17 years later, doesn't appear yet to have been published, so I give proofs for the more restricted claims made above. -Theorem: $\det_{0 \le i, j < n} \Big( e_{k+i-j}(\xi_1, \ldots, \xi_{k+i}) \Big) = (\xi_1 \cdots \xi_k)^n$ -Proof: $\Big( e_{k+i-j}(\xi_1, \ldots, \xi_{k+i}) \Big)_{0 \le i, j < n} = L_e U_e$ where $$\begin{eqnarray*}L_e &=& \Big( e_{i-j}(\xi_{k+1}, \ldots, \xi_{k+i}) \Big)_{0 \le i, j < n} \\ -U_e &=& \Big( e_{k+i-j}(\xi_1, \ldots, \xi_k) \Big)_{0 \le i, j < n} -\end{eqnarray*}$$ -To verify the factorisation observe that $$(L_e U_e)_{i,j} = \sum_{a=0}^{n-1} e_{i-a}(\xi_{k+1}, \ldots, \xi_{k+i}) e_{k+a-j}(\xi_1, \ldots, \xi_k) = e_{k+i-j} (\xi_1, \ldots, \xi_{k+i})$$ -Trivially, $\det L_e = 1$ and $\det U_e = e_k(\xi_1, \ldots, \xi_k)^n = (\xi_1 \cdots \xi_k)^n$. -$\blacksquare$ -Theorem: $\det_{0 \le i,j < n} \Big( h_{k+i-j}(\xi_1, \ldots, \xi_{j+1}) \Big) = (\xi_1 \cdots \xi_n)^k$ -Again we use an LU-factorisation as $L_h U_h$ where $$\begin{eqnarray*}L_h &=& \Big( h_{i-j}(\xi_1, \ldots, \xi_{j+1}) \Big)_{0 \le i, j < n} \\ -U_h &=& \Big( h_{k+i-j}(\xi_{i+1}, \ldots \xi_{j+1}) \Big)_{0 \le i, j < n} -\end{eqnarray*}$$ -I don't find that the verification yields to proof by inspection this time because of the overlap in variables between the two halves of each term: -$$(L_h U_h)_{i,j} = \sum_{a=0}^{n-1} h_{i-a}(\xi_1, \ldots, \xi_{a+1}) h_{k+a-j}(\xi_{a+1}, \ldots \xi_{j+1})$$ -So let $P(i,j,k, \vec{\xi}) = \sum_{a=0}^{j} h_{i-a}(\xi_1, \ldots, \xi_{a+1}) h_{k+a-j}(\xi_{a+1}, \ldots \xi_{j+1})$ and we aim to prove by induction on $j$ that $P(i,j,k,\xi) = h_{i+k-j}(\xi_1, \ldots, \xi_{j+1})$. -Base case, $j=0$: $P(i,0,k, \vec{\xi}) = h_{i}(\xi_1) h_{k-j}(\xi_{1}) = h_{i+k-j}(\xi_i)$ -Inductive step, $j > 0$: \begin{eqnarray*} -P(i,j,k, \vec{\xi}) &=& \sum_{a=0}^{j} h_{i-a}(\xi_1, \ldots, \xi_{a+1}) h_{k+a-j}(\xi_{a+1}, \ldots \xi_{j+1}) \\ -&=& h_{i-j}(\xi_1, \ldots, \xi_{j+1}) h_{k}(\xi_{j+1}) + \sum_{a=0}^{j-1} h_{i-a}(\xi_1, \ldots, \xi_{a+1}) h_{k+a-j}(\xi_{a+1}, \ldots \xi_{j+1}) \\ -&=& \xi_{j+1}^k h_{i-j}(\xi_1, \ldots, \xi_{j+1}) + \sum_{a=0}^{j-1} \sum_{b=0}^{k+a-j} \xi_{j+1}^b h_{i-a}(\xi_1, \ldots, \xi_{a+1}) h_{k+a-b-j}(\xi_{a+1}, \ldots \xi_{j}) \\ -&=& \xi_{j+1}^k h_{i-j}(\xi_1, \ldots, \xi_{j+1}) + \sum_{b=0}^{k-1} \xi_{j+1}^b \sum_{a=0}^{j-1} h_{i-a}(\xi_1, \ldots, \xi_{a+1}) h_{k+a-b-j}(\xi_{a+1}, \ldots \xi_{j}) \\ -&=& \xi_{j+1}^k h_{i-j}(\xi_1, \ldots, \xi_{j+1}) + \sum_{b=0}^{k-1} \xi_{j+1}^b P(i,j-1,k-b-1, \vec{\xi}) \\ -&=& \xi_{j+1}^k h_{i-j}(\xi_1, \ldots, \xi_{j+1}) + \sum_{b=0}^{k-1} \xi_{j+1}^b h_{i+k-b-j}(\xi_1, \ldots, \xi_j) \\ -&=& h_{i+k-j}(\xi_1, \ldots, \xi_{j+1}) -\end{eqnarray*} -By design, we have $(L_h U_h)_{i,j} = P(i,j,k, \vec{\xi})$, with the difference in upper limit of the sum being trivial due to the term $h_{k+a-j}(\xi_{a+1}, \ldots \xi_{j+1})$. Then $\det L_h = 1$ and $\det U_h = h_k(\xi_1) \cdots h_k(\xi_n) = (\xi_1 \cdots \xi_n)^k$. -$\blacksquare$ -[1] Louis Comtet, Nombres de Stirling généraux et fonctions symétriques, C. R. Acad. Sc. Paris, t. 275 (1972), Sér. A 747–750. -[2] Hopkins, Speyer, Taylor, Zaimi, Matrices of combinatorial sequences that are inverse in two ways , MO question 418266 and answers<|endoftext|> -TITLE: First Dirichlet eigenvalue on regular polygons -QUESTION [5 upvotes]: Assume $\lambda(P)$ is the first Dirichlet eigenvalue of a regular polygon $P$. Let $u$ be the corresponding eigenfunction, normalized by $\|u\|_{L^2(P)}=1$, and $\partial_{\nu}u$ be its normal derivative on the boundary. Is the following estimate correct: -\begin{eqnarray*} -\lambda(P)\geq \|\partial_{\nu} u\|^2_{L^{\infty}(\partial P)}\vert P\vert\quad? -\end{eqnarray*} -Here $\vert P\vert$ denotes the area of $P$. - -REPLY [6 votes]: There is a more general reason why any such statement will fail: If you consider a $P$ consisting of $N$ separate copies of the same basic region $P_0$, then $|P|=N|P_0|$, while everything else in your inequality is independent of $N$, so the inequality will fail for large $N$. (This problem has a degenerate ground state, but of course you could address this by slightly changing the shapes.) -If you connect the components by thin tubes, then you now have a connected $P$ and are still approximately in the situation described above.<|endoftext|> -TITLE: Has the following problem, resembling the lonely runner conjecture, been studied? -QUESTION [13 upvotes]: Given $n$, what is the smallest value $\delta_n$ satisfying the following: - -For any group of $n$ runners with constant but distinct speeds, -starting from the same point and running clockwise along the unit -circle, there exists a moment, at which the length of each of $n$ -empty circle segments between two consecutive runners does not exceed -$\delta_n$. - -More formally, for $n \in \mathbb{N}$, $M=\{m_1,\dots,m_n\} \subset \mathbb{N}$, and $t \in [0,1]$, we sort the fractional parts of the values $tm_i$ in ascending order: $0 \le \{tm_{i_1}\} \le\dots\le \{tm_{i_n}\} \le 1$. Then we consider the longest empty segment between two consecutive 'runners': -$$ \text{LongestArc}(M,t) := \max\Big\{ \{tm_{i_2}\}-\{tm_{i_1}\}, \dots, \{tm_{i_n}\}-\{tm_{i_{n-1}}\}, 1+\{tm_{i_1}\}-\{tm_{i_n}\} \Big\}.$$ -Finally, we put -$$ \delta_n := \mathop{\smash{\mathrm{sup}}}_{|M|=n} \inf_{t \in [0,1]} \text{LongestArc}(M,t).$$ -Has the problem of finding these values (or, perhaps, an equivalent one) been studied earlier? I have looked through some papers related to the Lonely Runner Conjecture (including this one) but found nothing. -It is clear that $\delta_2=\frac12$. With a little more effort goes $\delta_3=\frac25$, one of the 'extremal' sets here is $M=\{0,1,3\}$. One may check that $\delta_4 \ge \frac13$ by taking $M=\{0,1,2,4\}$. I suspect this bound to be tight. Maybe someone sees a counterexample or a short argument, why is that so? -For large $n$, I believe I can show that $\delta_n \ge \frac{\log n -O(\log \log n)}{n}$. However, I don't have any (decent) upper bound on $\delta_n$ apart from the fact that $\delta_n=o(1)$ as $n \to \infty$. -Finally, I found that the 'dual' problem concerning the values -$$ \epsilon_n := \inf_{|M|=n} \mathop{\smash{\mathrm{sup}}}_{t \in [0,1]} \text{LongestArc}(M,t)$$ -has earlier been considered on MathOverflow (not in the literature, though). One may also state two more 'dual' extremal problems related to the shortest arc (instead of the longest one), but I haven't found anything published about them as well. -I would appreciate any related reference! - -REPLY [11 votes]: OK, since Will thinks it is worth posting, here it goes. This is not an answer, just setting up the triviality level. As usual, the circle will be identified with $\mathbb R/\mathbb Z$. -Let $\psi$ be the piecewise linear $1$ periodic function that is $0$ at $0$, raises to $1$ at $\delta/2$, drops down to $0$ at $\delta$ and then stays $0$ throughout the rest of the period (the well-known "triangle hump", just shifted forward from its usual center). We have the Fourier expansion -$$ -\psi(x)=\frac{\delta}2+\sum_{k\ne 0}c_ke^{2\pi ikx} -$$ -with $\sum_{k\ne 0}|c_k|=1-\frac\delta 2<1$. -We want to find $t$ such that for every runner moving at the speed $v_0\in V$ ($V$ is the set of all speeds), we have -$$ -\Psi_{v_0}(t)=\sum_{v\in V} \psi(vt-v_0t)\ne 0\,. -$$ -This will mean that there is at least one runner in the interval of length $\delta$ ahead of every runner, so the maximal gap is at most $\delta$. -However -$$ -\Psi_{v_0}(t)=\frac{\delta n}2+\sum_{k\ne 0}c_ke^{-2\pi i kv_0t}F(kt) -\\ -\ge -\frac{\delta n}2-\sum_{k\ne 0}|c_k||F(kt)|\ge -\frac{\delta n}2-\sqrt{\sum_{k\ne 0}|c_k||F(kt)|^2} -$$ -where $F(t)=\sum_{v\in V}e^{2\pi i vt}$. -Note now that for each $k\ne 0$, the mean value of $|F(kt)|^2$ (in the sense of almost-periodic functions, i.e., $\lim_{T\to\infty}\int_{-T}^T\dots\,dt$) is $n$, so the mean value of the sum under the square root is $0$. Consider the functions $\varphi_v(x,t)$ that are 1-periodic in $x$ and given by $\chi_{[0,\delta]}(vt-x)-\delta$ on the period and $\Phi(x,t)=\sum_v\varphi_v(x,t)$. If for some $t$ we have a $2\delta$-gap, then $$ -\int_0^1 |\Phi(x,t)|^2\,dx\ge \delta(-n\delta)^2=n^2\delta^3\,. -$$ -Thus, if we have that gap for all $t$, we must have -$$ -\operatorname{Mean}_t\int_0^1 |\Phi(x,t)|^2\,dx\ge n^2\delta^3 -$$ -but we have the joint in $x,t$ ortogonality of $\varphi_v$, so the left hand side is just $n\delta(1-\delta)$. -The issue is that when $\delta -TITLE: The difference between consecutive primes in arithmetic progressions -QUESTION [9 upvotes]: Let $\pi(x)=\sum_{p\leq x}$ denote the prime counting function. A well known result of Baker, Harman, and Pintz on prime gaps states that for $x\geq y\geq x^{0.525}$ we have that -$$\pi(x+y)-\pi(x)\gg \frac{y}{\log x}.$$ -Let $\pi(x;q,a)$ denote the number of primes $p\leq x$ such that $p\equiv a\pmod{q}$. Is there an analogous version of this theorem for $\pi(x;q,a)$? - -Question: What is the smallest growing function $f(x,q)$ such that for any $x\geq y\geq f(x,q)$ we have that $$\pi(x+y;q,a)-\pi(x;q,a)\gg \frac{y}{\log x}?$$ - -Corollary 13.8 of Montgomery and Vaughn's Multiplicative Number Theory states that under Generalized Riemann Hypothesis $$\psi(x;q,a)=\sum_{\begin{array}{c} -n\leq x\\ -n\equiv a\ (q) -\end{array}}\Lambda(n)=\frac{x}{\phi(q)}+O\left(x^{1/2}\log^{2}x\right).$$ -From this Corollary, it follows that under GRH the desired lower bound holds for $$f(x,q) = \phi(q)x^\frac{1}{2}\log^{2}(x).$$ -I'm aware of a 1933 result of Heilbronn (MR1545353), which I believe (my German isn't great) proves that for fixed $q,a$, there exists $\delta>0$ such that for $x\geq y\geq x^{1-\delta}$ $$\pi(x+y;q,a)-\pi(x;q,a)\sim \frac{y}{\phi(q)\log x}.$$ -Are there any more recent results on this topic? - -REPLY [4 votes]: Unconditionally, one can take $f(x,q)=x^{0.525}$, provided that $q$ is in the Siegel-Walfisz range $q\leq (\log x)^A$. There exists a (suitably small) constant $\delta>0$ such that one can take $f(x,q)=x^{0.525}$ for all except a density zero subset of $q\leq x^{\delta}$. -This follows from a certain Bombieri-Vinogradov-type result for a carefully constructed lower bound for the indicator function of the primes (due to Kumchev, but with $0.525$ replaced by $0.53$) combined with the second moment bound for $L(s,\chi)^2$ times a Dirichlet polynomial (due to Harman, Watt, and Wong). Full details on how to combine these two results can be found in the work of Alweiss and Luo (see Theorem 2.3 therein). One then takes $Q = (\log x)^A$.<|endoftext|> -TITLE: Set theory / Formal logic of Baba is You -QUESTION [7 upvotes]: ''Baba is You'' is a recent puzzle game in which the player builds a set of rules by pushing squares with words written on them. If we leave aside the combinatorial difficulty of how to move the blocks around without getting stuck, the game seems to be unique insofar as it is determined entirely by rules which the player builds. The total set of rules can be enlarged or made smaller (for example, if we have Key Is Defeat and push Key away from the rest of the rule, the number of rules is reduced by one). -A rule is made from at least three words (where the second connecting word is usually ‘is’) and a word can be a noun, an adjective, a conjunction, or a verb. For example, in Key is Open, the word ‘Open’ is a verb, implying that the key will remove a door if it is manipulated onto the same square as the door. In order for a level to have a winning strategy, each level must have a Win condition and a You condition. -I am trying to think if there is something mathematically interesting about the way that one changes the logical rules to solve a puzzle. For me, the game is a bit paradoxical because it definitely appeals to mathematicians but I do not think there is anything actually mathematical to be said about it in terms of set theory and formal logic (besides the trivial combinatorics of most effectively finding a solution). -The mathematical aspect seems to be that often one needs to make creative changes to the underlying rules in order to arrive at a scenario which can be solved (perhaps something like writing a proof by finding the correct language with the logical truths which you need). This seems more philosophical than mathematical to me though. - -REPLY [2 votes]: The developer has a GDC talk where he talks about the mechanics which you might find interesting. My impression is it's a lot of random hacks, which may fit with your description.<|endoftext|> -TITLE: How would you work out this integral as a series? -QUESTION [6 upvotes]: The integral is: -$$f(a) = \int\limits_{-\infty}^\infty \frac{x e^{-a^2 x^2}}{\tanh(x)}dx$$ -which seems to converge for all $a>0$. But I don't know how to get a sense of the function $f(a)$ such as writing it as a convergent series. The usual Taylor series has infinity for each term. Any ideas? -Edit: -I believe (using answers below) that when $a$ is close to zero we have (doing a substitution): -$$f(a) = \frac{1}{a^2}\int\limits_{-\infty}^\infty \frac{x e^{-x^2}}{\tanh (x/a)}dx$$ -But using $\tanh(x/a)\rightarrow\operatorname{sign}(x)$ as $a\rightarrow 0^+$. So the above should become: -$$f(a) \approx \frac{2}{a^2}\int\limits_{0}^\infty x e^{-x^2}dx = \frac{1}{a^2}$$ -So that gives the behaviour of $f(a)$ when $a$ is small. But I don't know how to give extra terms. (Also not sure if this is mathematically correct). From numerical calculations I find that near $0$, -$$ -f(a)\approx \frac{1}{a^2} + \frac{\pi^6}{6} - \frac{\pi^4 }{60}a^2+\frac{\pi^6 }{252}a^4+... = \frac{1}{a^2}\sum\limits_{n=0}^\infty \frac{B_{2n} a^{2n}\pi^{2n}}{n!} -$$ -although apparently this doesn't converge? -Comment -The answers below are two asymptotic series depending on whether $a$ is small or large. These give good approximations if we truncate the summation before begins to diverge. In the mid-range, when $a^2=1/\pi$, these two sums become term-by-term equal and the closest the truncated sum get to the true answer of $f(1/\sqrt{\pi})$ is to about 1% error. Using both these sums, we can know any value to within about 1%-2% error, and if $a$ is small or large then much more accurately. - -REPLY [11 votes]: To obtain the series in $a$, separate off the leading term proportional to $1/a^2 $ and expand the Gaussian instead of the hyperbolic cotangent: -\begin{eqnarray} -f(a)&=& 2\int_{0}^{\infty } dx\, x \ e^{-a^2 x^2 } \left[ \coth x -1 +1 \right] \\ -&=& 2\int_{0}^{\infty } dx\, x \ e^{-a^2 x^2 }+ -2\sum_{n=0}^{\infty } \frac{(-a^2)^{n}}{n!} \int_{0}^{\infty} dx\, x^{2n+1} [\coth x -1] \\ -&=& \frac{1}{a^2 } + 2\sum_{n=0}^{\infty } \frac{(-a^2)^{n}}{n!} \frac{ (-1)^n \pi^{2n+2} B_{2n+2} }{2n+2} \\ -&=& \frac{1}{a^2} + \frac{1}{a^2} \sum_{n=1}^{\infty } \frac{(\pi a)^{2n} B_{2n} }{n!} \\ -&=& \frac{1}{a^2} \sum_{n=0}^{\infty } \frac{(\pi a)^{2n} B_{2n} }{n!} -\end{eqnarray} -Also this is only an asymptotic expansion - the Bernoulli numbers diverge more rapidly than $n!$ at large $n$.<|endoftext|> -TITLE: Does the base-10 representation of $2^n$ contain all 10 digits for all sufficiently large integers $n$? -QUESTION [12 upvotes]: Does the base-10 representation of $2^n$ contain all 10 digits for all sufficiently large integer $n$? - -In general, let $x_{k}$ denote the base-$k$ representation of the positive integer $x$. We say $x$ is $k$-powerful if $x^n_{k}$ contains all of the $k$ digits for all sufficiently large integers $n$. -For example, it's easy to check that 2 is 2-powerful, but 3 is not 3-powerful. The title question asks whether 2 is 10-powerful. -For any given $x$ and $k$, can we decide if $x$ is $k$-powerful? - -REPLY [7 votes]: Heuristically, one would expect the answer to be yes. There's an existing partially explicit version of this mentioned in Richard Guy's "Unsolved Problem in Number Theory" entry F24, which is that for $n> 86$, $2^n$ always contains a zero in its base 10 expansion. There are some related questions also in that entry and some other entries in the book as well. For example, there's a conjecture that every sufficiently large power of 2 contains 0s, 1s and 2s in its base 3 expansion. The reasonable generalization (which I have not seen stated explicitly but seems to be implicit in all of these), is that if $a$ and $b$ are relatively prime integers both greater than 1, then for all sufficiently large $n$, $a^n$ has in its base $b$ expansion all of $0, 1, \dotsc b-1$. It is also plausible that the sufficiently large is a not very fast growing function, something like being true for all $n \geq a^2b^2$. Edit: See Emil's comment below: as written this doesn't include the case $2$ and $10$; one wants that there's a prime $p$ such that $p$ divides exactly one of $a$ and $b$. That includes the relatively prime case and also cases like $2$ and $10$. -However, the set of $a$ and $b$ where we can prove anything like this is small and mostly relegated to when $b=2$. In that case, some of these results are implicitly very old, dating back to actually the middle ages. For example, any power of $3$ greater than $3$ must contain both $1$ and $0$ in its base $2$ expansion. This follows, since if not, $3^n$ would have to be of the form $2^k-1$, and Gersonide's theorem that $8$ and $9$ are the largest power of 2 next to a power of $3$ then applies. For larger bases, Mihailescu's proof of Catalan's conjecture shows that the same is essentially true if one has $b=2$ and $a$ is any number greater than $2$.<|endoftext|> -TITLE: Sum of product of characters of $S_{n+m}$ over $S_n$ -QUESTION [6 upvotes]: I asked this in MSE first, here, but it is not getting any attention. -Let $S_n$ be the group of permutations acting on the set $\{1,...,n\}$. -Given $R_1,R_2$ two irreps of $S_n$ with characters $\chi_{R_i}$, I know that -$$ \sum_{a\in S_n}\chi_{R_1}(a)\chi_{R_2}(ab)=n!\frac{\chi_{R_1}(b)}{\chi_{R_1}(1)}\delta_{R_1R_2},$$ -when $b\in S_n$. -What I want to know is what is the result of -$$ \sum_{a\in S_n}\chi_{R_1}(a)\chi_{R_2}(ab),$$ -where now $R_1$ is an irrep of $S_n$ but $R_2$ is an irrep of $S_{n+m}$, with $b\in S_{n+m}$ and $a$ having $\{n+1,...,n+m\}$ as fixed points. Can this be expressed in terms of the representation induced from $R_1$, for example? - -REPLY [3 votes]: I would rather write $\chi_{R_1}(a)=\chi_{R_1^*}(a^{-1})$. This allows us to see the sum -$$\frac{1}{n!}\sum_{a\in S_n}\chi_{R_1^*}(a^{-1})\chi_{R_2}(ba)$$ as the trace on $\mathrm{Hom}_{\mathbb{C}}(R_1,R_2)$ of the map $f\mapsto \frac{1}{n!}\sum_{a\in S_n}bafa^{-1}=b\frac{1}{n!}\sum_{a\in S_n}afa^{-1}$. If $b=1$, this is projection to the space of invariants, so its trace is the dimension of this space $\delta_{R_1,R_2}$. If $b\neq 1$, then it's projection to invariants, followed by composition with $b$. If you choose a basis consisting of $b$ (which is sent to $\chi_{R_1}(b)/\chi_{R_1}(1)$ times itself), and then a basis of the other isotypic components in $\mathrm{Hom}_{\mathbb{C}}(R_1,R_2)$ (which are all killed), you can see this gives the factor of $\chi_{R_1}(b)/\chi_{R_1}(1)$ (incidentally, this seems to be a restatement of the proof that this ratio is an algebraic integer). -For the generalization, we still have that $$\frac{1}{n!}\sum_{a\in S_n}\chi_{R_1^*}(a^{-1})\chi_{R_2}(ba)$$ as the trace on $\mathrm{Hom}_{\mathbb{C}}(R_1,R_2)$ of the map $f\mapsto \frac{1}{n!}\sum_{a\in S_n}bafa^{-1}=b\frac{1}{n!}\sum_{a\in S_n}afa^{-1}$. However, this projection might have a much larger image, since there might be many homomorphisms in $\mathrm{Hom}_{S_n}(R_1,R_2)$, since $R_2$ isn't irreducible as an $S_n$-module. My impulse is that there's no easy expression for this trace, but I can't say I'm sure.<|endoftext|> -TITLE: How to read an article and make it actually useful? -QUESTION [34 upvotes]: I've been wondering for a while: how should mathematicians read an article in order to "take most" from it? -For example, when I did my Master's thesis I based it on an article (I'm into analysis) and of course I analyzed every and each part of it, extending some results in there and filling some gaps that were "left to the reader" I guess, or were thought to be sufficiently trivial by the authors. -Now I'm doing my phd and I have to choose a specific topic (more or less I have an idea but I haven't exactly set up my mind yet). The fact is I've had to look up some articles to get some ideas and possible topics of research, and by now and I quickly understood that reading and trying to understand the whole thing is impossible. I mean they often have like 60+ articles/books in the bibliography, so I mainly read the introduction and the results, skipping proofs entirely (or almost entirely). Basically what I try to do is to get an idea of the general path taken by the authors, skipping all the technicalities are ok and noting on the side the techniques/results mentioned that I don't know about. Then I quickly look up on the internet what the idea of these techniques is, and that's it. Obviously, a couple of days pass and most of it is gone, except maybe the very very general idea/result it obtained (but only if it isn't too technical). -I mean it seems a bit shallow, but I can't come up with with better ways to read them, they are so stuffed I really can't keep up. So how would a professional mathematician read an article about a topic he's interested in without going crazy, trying to learn most from it? Maybe in n years from now if one continues this path one can expect to be so well-versed in a very specific topic so that research articles about it become way easier to read? - -REPLY [40 votes]: Before you decide how to read a particular paper, you have to decide why you're reading that particular paper. -You said you're doing your Ph.D. You might want to take a look at a couple of other MO questions, such as When is one 'ready' to make original contributions to mathematics? and On starting graduate school and common pitfalls... and How to escape the inclination to be a universalist or: How to learn to stop worrying and do some research. If the reason you're reading these papers is that you think you have to absorb an infinite amount of mathematical knowledge before you can start doing research, then the first thing you need to do is: Stop thinking that. -One reason for reading a paper is that you have a specific research problem that you are trying to solve, and you believe that the paper contains a technique that will help you solve your problem. In such a situation, you should be constantly thinking, "Does this technique in fact help me solve my problem, and if so, how does the proof work?" If you determine that the paper does not help you solve your problem, I'd recommend abandoning ship, especially at this stage of your career. Your time is precious, and although mastering the techniques of the paper might come in handy one day, you have bigger fish to fry at the moment. On the other hand, if the paper does seem to solve your problem, then don't be afraid to sink a lot of time into it to understand it thoroughly. If some argument is logically needed for the proof of your own result, then you need to understand it completely, even if you end up just citing the other paper without reproducing the argument in your own paper. -Another reason for reading a paper is to increase your general store of mathematical knowledge. This is a laudable goal, but again, your time is precious. You first need to triage to decide how much effort to invest. - -Option 0 is not to read it at all. This is what you will decide to do with most papers, because there are too many papers out there to even read the abstracts of all of them. If a paper is too far from your own area of work and looks like it will require significant effort to absorb, then drop it and move on. - -Option 1 is to read just enough to get a general sense of what it is about, so that when you later run into a situation where you might need it, a bell will ring in your head ("Someone has done some relevant work on this") and you can return to the paper later. The important thing here is that even though you're not spending a lot of time on the paper, you should make some effort to put at least a "stub" of the paper in your long-term memory, so that when the occasion arises, the memory of this paper will be triggered. How you do this is up to you; maybe you can keep a log of some sort in which you write short notes to yourself about various papers you read, and re-read the log periodically to refresh your memory. - -Option 2 is to read the paper more seriously, to understand in some detail what the results are, but stopping short of completely mastering the proofs. This is the sort of reading one tends to do when you're looking around for a new topic to work on, and think that this paper might be something you'll jump into and do research on yourself. You'll need to keep asking yourself, "Am I really going to work on this topic?" If you decide that the answer is no, then you should go to Option 1. If you decide that the answer is yes, then you may wish to proceed to Option 3 below. - -Option 3 is to study the paper in full detail and master it. As I mentioned above, you should generally do this only when you're sure that understanding the proofs is needed to do your own research. - - -In short, the key is to triage, triage, triage. Decide as soon as you can why you are reading the paper and whether it is meeting your needs. Don't be afraid to drop the paper if it isn't meeting your needs, and don't be afraid of investing a lot of time into the paper if it is meeting your needs.<|endoftext|> -TITLE: Can we interpret arithmetic in set theory, with exactly PA as the ZFC provable consequences? -QUESTION [25 upvotes]: There are many interpretations of arithmetic in set theory. The -Zermelo interpretation, for example, begins with the empty set and applies the singleton operator as successor: -$$0=\{\ \}$$ -$$1=\{0\}$$ -$$2=\{1\}$$ -$$3=\{2\}$$ -and so on... The von Neumann interpretation, in contrast, is guided by the idea that every number is equal to the set of smaller numbers. -$$0=\{\ \}$$ -$$1=\{0\}$$ -$$2=\{0,1\}$$ -$$3=\{0,1,2\}$$ -In each case, one equips the numbers with their arithmetic structure, addition and multiplication, and in this way one arrives at the standard model of arithmetic -$$\langle\mathbb{N},+,\cdot,0,1,<\rangle$$ -In those two cases, there isn't much at stake because ZFC proves that they are isomorphic and hence satisfy exactly the same arithmetic assertions. In each case, the interpretations provably satisfy the axioms PA of Peano arithmetic. -But furthermore, each of these interpretations satisfies many additional arithmetic properties strictly exceeding PA, such as Con(PA) and Con(PA+Con(PA)), which are all ZFC theorems. -My main point, however, is that these further properties are not provable in PA itself, if consistent, and so my question is whether there is an interpretation of arithmetic in set theory realizing exactly PA. -Question. Is there an interpretation of arithmetic in set theory, such that the ZFC provable consequences of the interpretation are exactly the PA theorems? -What I want to know is whether we can interpret arithmetic in ZFC in such a way that doesn't carry extra arithmetic consequences from ZFC? - -REPLY [24 votes]: The standard terminology is that an interpretation $I$ of a theory $U$ in a theory $T$ is faithful if for all sentences $\phi$ in the language of $U$, -$$T\vdash\phi^I\iff U\vdash\phi.$$ -(Here and below, I will assume all theories to be recursively axiomatized.) A classical result of Feferman, Kreisel, and Orey [1] states: - -Theorem 1 (Feferman, Kreisel, Orey). Let $T$ be a reflexive theory which includes a modicum of arithmetic and which is $\Sigma_1$-sound. Then all theories interpretable in $T$ are faithfully interpretable in $T$. - -They also show that conversely, if a $\Sigma_1$-sound theory $U$ is faithfully interpretable in an essentially reflexive theory $T$, then $T$ is itself $\Sigma_1$-sound. - -Corollary: PA is faithfully interpretable in ZFC if and only if ZFC is $\Sigma_1$-sound. - -Further results on faithful interpretability in reflexive theories were obtained by Lindström [2]; for example, he proved the following characterization: - -Theorem 2 (Lindström). Let $T$ and $U$ be essentially reflexive theories. Then $U$ is faithfully interpretable in $T$ if and only if $U$ is $\Pi_1$-conservative over $T$ (which is, by itself, equivalent to the interpretability of $U$ in $T$), and $T$ is $\Sigma_1$-conservative over $U$. - -Generalizing Theorem 1 in a different direction, a theory $T$ is called trustworthy if all theories that are interpretable in $T$ are faithfully interpretable in $T$. Thus, Theorem 1 states that $\Sigma_1$-sound reflexive theories are trustworthy. -A complete characterization of trustworthiness was given by Visser [3], who in particular proves that the assumption of reflexivity is unnecessary. (Note that the criterion below does not rely on a fixed interpretation of arithmetic, unlike notions such as $\Sigma_1$-soundness or reflexivity.) - -Theorem 3 (Visser). A theory $T$ is trustworthy if and only if it has an extension $T'$ (in the same language) such that Robinson’s arithmetic $Q$ has a $\Sigma_1$-sound interpretation in $T'$. - -He also derives from this an earlier result attributed to (but apparently unpublished by) H. Friedman: - -Corollary. A consistent finitely axiomatized sequential theory is trustworthy. - -References -[1] Solomon Feferman, Georg Kreisel, and Steven Orey: 1-Consistency and faithful interpretations, Archiv für mathematische Logik und Grundlagenforschung 6 (1962), pp. 52–63, doi 10.1007/BF02025806. -[2] Per Lindström: On faithful interpretability, in: Computation and Proof Theory (Börger, Oberschelp, Richter, Schinzel, Thomas, eds.), Lecture Notes in Mathematics vol. 1104, Springer, 1984, doi 10.1007/BFb0099490. -[3] Albert Visser: Faith & falsity, Annals of Pure and Applied Logic -131 (2005), pp. 103–131, doi 10.1016/j.apal.2004.04.008.<|endoftext|> -TITLE: How to count the total zeros of a complex polynomial outside a closed curve? -QUESTION [8 upvotes]: Set up -Suppose $\gamma$ a simple closed curve, oriented in a counterclockwise direction. $f(z)$ is a complex polynomial -$$ -f(z)=a_nz^{n}+a_{n-1}z^{n-1}+\cdots+a_0. -$$ -We already know that the integral -$$ -N=\frac{1}{2\pi i}\oint_{\gamma}{\frac{f'(z)}{f(z)}dz} -$$ -which we called the winding number, gives the total zeros $N$ of $f(z)$ inside the closed curve $\gamma$. Now I want to know the total zeros $M$ outside $\gamma$ and this can be done exactly by the fundamental theorem of algebra, which leads to -$$ -M=\mathrm{total\ zeros\ of}\ f(z)\ \mathrm{in\ whole\ plane}\ -N. -$$ -My question is: is there an "integral way" instead of the "algebraic way" to count the number of zeros $M$ outside $\gamma$ like what we did for $N$? - -REPLY [5 votes]: This is just an extended comment to answer your concerns. Maybe it's helpful to treat $f$ as a black-box function, where all we know is that it's meromorphic on $\mathbb CP^1$ (of course, this means it's actually rational, but let's suppress that for now). In particular, we happen to know there's a pole at $\infty$. We can calculate the order by taking a sufficiently small circle around it (ie, a sufficiently large circle) that doesn't contain any other zeroes or poles of $f$. Then we can calculate the mentioned integral, call it $-n$. By symmetry, $n$ will equal the number of zeroes $-$ poles on the other side of the circle, ie all the others besides $\infty$. If we know $f$ is entire on $\mathbb C$ this tells us that the total number of zeroes of $f$ is equal to the order of the pole at $\infty$, and we can calculate this quantity directly with an integral on a sufficiently large circle, which is what you wanted.<|endoftext|> -TITLE: Do there exist elliptic curves over $H_K$ having everywhere good reduction and CM by $\mathcal{O}_K$? -QUESTION [5 upvotes]: For $K$ a number field, denote by $\mathcal{O}_K$ its ring of integers and by $H_K$ its Hilbert class field. -For which imaginary quadratic field $K$ does there exist an elliptic curve $E$, defined over $H$, with complex multiplication by $\mathcal{O}_K$ having everywhere good reduction (on $H$)? - -By Fontaine's Il n'y a pas de variété abélienne sur Z, corollary of Théorème B, there do not exist such curves for $K=\mathbb{Q}(\sqrt{-1})$ or $\mathbb{Q}(\sqrt{-3})$. - -REPLY [10 votes]: Here is an example. -Let $K = \mathbf{Q}(\sqrt{-21})$. Then the class group of $K$ is $C_2 \times C_2$ and its Hilbert class field is $H_K = \mathbf{Q}(\sqrt{-1}, \sqrt{3}, \sqrt{7})$. In particular, $H_K$ is a CM-field and its maximal totally real subfield $H_K^+$ is $\mathbf{Q}(\sqrt{3}, \sqrt{7})$. -The LMFDB database has an elliptic curve 4.4.7056.1-1.1.a1 over $H_K^+$ which has everywhere good reduction and has CM by $\mathcal{O}_K$. Base-extending this from $H_K^+$ to $H_K$ gives the example you seek. -Similar examples should exist whenever $H_K$ has class number 1 and all units of $H_K$ are in the kernel of the norm map to $K$. (The class number condition may well not be needed, but the condition on the units certainly is). I have no idea if there are infinitely many such fields $K$, but there definitely some! This happens for $\mathbf{Q}(\sqrt{-d})$ for $d = 21, 33, 42, 57, 66, 77, 93$ (and no others for $d < 100$).<|endoftext|> -TITLE: A differential equation governing compositional inversion -QUESTION [7 upvotes]: Looking for references for the following theorem. -Given the formal Taylor series/exponential generating function -$$T(z) = \sum_{n \ge 1} a_n \; \frac{z^n}{n!},$$ -for which the indeterminates $a_n$ and the independent variable $z$ may be complex, its formal compositional inverse -$$T^{(-1)}(z) = \sum_{n \ge 1} b_n \; \frac{z^n}{n!}=\sum_{n \ge 1} Prt_n(a_1,a_2,...,a_n) \; \frac{z^n}{n!} $$ -satisfies the differential equation -$$-\frac{\partial}{\partial a_{n}}\;T^{(-1)}(z)= \frac{\partial}{\partial z}\;\frac{(T^{(-1)}(z))^{n+1}}{(n+1)!}.$$ -Edit: Don't quite follow Pietro's derivation (I already had two other derivations of this theorem), but Pietro's inspires the following third derivation; -With $S(\omega,a_n) = T^{(-1)}(\omega,a_n) = z$ and $\omega = T(z,a_n)$, then $z = S(T(z,a_n),a_n)$, -so, suppressing the $a_n$ in the arguments, -$$\partial_{a_n} z = 0 = \partial_T \;S(T(z))\; \partial_{a_n} \;T(z) + \partial_{a_n} \;S(T(z))$$ -$$= \partial_T \;S(T(z))\; \frac{z^n}{n!} + \partial_{a_n} \;S(T(z))$$ -$$=\partial_T \;S(T(z))\; \frac{(S(T(z)))^n}{n!} + \partial_{a_n} \;S(T(z))$$ -$$=\partial_T \; \frac{(S(T(z)))^{n+1}}{(n+1)!} + \partial_{a_n} \;S(T(z)),$$ -implying -$$\partial_{\omega} \; \frac{(S(\omega))^{n+1}}{(n+1)!} + \partial_{a_n} \;S(\omega)=0.$$ - -Instances I had already found of this integrable hierarchy of diff ids / conservation laws are in the section "5.4 Deformation of flow equations" on p. 57 and in Theorem 5.4 on pp. 66 of "On Emergent Geometry of the Gromov-Witten Theory of Quintic Calabi-Yau Threefold" by Jian Zhou. -One reason I ask for more references is that this diff id implies some interesting relations w.r.t. free probability theory, associahedra, and the inviscid Burgers-Hopf differential equation, so I'm hoping to see some perspectives from others on these and will give due credit on such in any of my postings in the future. - -Edit (5/28/22): The derivation is more accurately expressed as -$$\partial_{a_n} \; z=0 =\partial_{a_n} \; S(T(z,a_n),a_n) =\partial_{a_n} \; S(\beta_1,\beta_2)$$ -$$= (\partial_{\beta_1} \; S(\beta_1,\beta_2)) \;\partial_{a_n}\beta_1 +(\partial_{\beta_2} \; S(\beta_1,\beta_2)) \; \partial_{a_n}\beta_2$$ -$$ = (\partial_{\beta_1} \; S(\beta_1,\beta_2)) \;\frac{z^n}{n!} +\partial_{\beta_2} \; S(\beta_1,\beta_2) $$ -$$ = (\partial_{\beta_1} \; S(\beta_1,\beta_2)) \;\frac{( S(\beta_1,\beta_2))^n}{n!} +\partial_{\beta_2} \; S(\beta_1,\beta_2) $$ -$$ = \partial_{\beta_1} \; \;\frac{(S(\beta_1,\beta_2))^{n+1}}{(n+1)!} +\partial_{\beta_2} \; S(\beta_1,\beta_2) $$ -which may be expressed as -$$0 = \partial_{u} \; \;\frac{(S(u,a_n))^{n+1}}{(n+1)!} +\partial_{a_n} \; S(u,a_n).$$ - -REPLY [5 votes]: I do not have a reference, but I’d say it is a plain instance of the Implicit Function Theorem for formal power series. I add the computation below, in case you had a different one. -We choose our index $n\ge1$ and see $T$ as a power series in $z$ and $a:=a_{n}$, that is, as an element of $\mathbb{C}[[z, a]]$. The corresponding compositional inverse of $T(z, a)$, is the formal series $S(z, a)$ well defined as implicit function by -$$z=T(S(z, a), a) $$ -(with $a_1\neq0$). -Differentiating w.r.to $a $ and multiplying by $\partial_{1 }S(z,a) $ we get -$$0=\Big[\partial_1 T\big(S(z, a ), a \big)\partial_2S(z,a)+\partial_2 T\big(S(z,a),a\big)\big] \partial_1S(z,a)=$$ -$$=\partial_1 T\big(S(z, a ), a \big)\partial_1S(z,a) \partial_2S(z,a) +\partial_2T\big(S(z,a),a\big) \partial_1S(z,a)=$$ -$$= \partial_1\big[T (S(z, a ), a)\big]\partial_2S(z,a) + \frac{S(z,a)^{n}}{n!} \, \partial_1S(z,a)=$$ -$$=\partial_2S(z,a) + \partial_1\bigg(\frac{S(z,a)^{n+1}}{(n+1)!}\bigg).$$<|endoftext|> -TITLE: Boundary of a $4$-manifold and the fundamental group -QUESTION [7 upvotes]: I am trying to learn $4$-manifolds with boundaries and I don't know much about this topic so these questions may be silly. Given a $4$-manifold $M$ with a boundary say $N$, - -Assume $\pi_1(N)$ is known, is there a way to compute $\pi_1(M)$ or is it possible to say any features of $\pi_1(M)$? To be more precise how the boundary is influencing the fundamental group of the manifold? - -Can I put some restriction on $N$ so that $\pi_1(M)$ becomes finite? - - -I apologize for asking too many questions in a single post. Any techniques, references, or suggestions will be helpful. -Thanks in advance! - -REPLY [4 votes]: Let me expand the comment of @RyanBudney by giving concrete examples. -Given coprime positive integers $p,q$ and $r$, Brieskorn spheres are important classes of $3$-manifolds, they can be defined as the links of singularities at the origin: $$\Sigma(p,q,r) = \{ x^p +y^q +z^r =0 \} \cap S^5 \subset \mathbb C^3.$$ -The fundamental groups of Brieskorn spheres are well-known, see for instance the paper of Milnor. -However, the following question is a very central problem in low-dimensional topology, see Kirby's problem list, Problem 4.20: - -Which homology 3-spheres (in particular Brieskon spheres) bound smooth contractible 4-manifolds? - -In some cases, Brieskorn spheres may bound Mazur type contractible manifolds which can be built by a single $0$-, $1$-, and $2$-handle. -Once you know the Kirby diagram of a Mazur manifold, you can both compute the fundamental groups of the $4$-manifold and its boundary $3$-manifold, see the dicussion in this question.<|endoftext|> -TITLE: Is there a strongly noncommutative fusion category? -QUESTION [6 upvotes]: A fusion category is called noncommutative if its Grothendieck ring is noncommutative. Let us call a fusion category strongly noncommutative if every fusion category Morita equivalent to it (i.e. same Drinfeld center up to equiv.) is noncommutative. -Question: Is there a strongly noncommutative fusion category (say over $\mathbb{C}$)? -If so, what are the known examples? -Note that if $G$ is a finite group then $Vec(G)$ is not strongly noncommutative (even if $G$ is noncommutative) because it is Morita equivalent to $Rep(G)$ which has a commutative Grothendieck ring. Moreover, the Extended Haagerup fusion categories are also not strongly noncommutative because they form a Morita equivalent class and one of them is commutative. -This post is in the same spirit than this one about strongly simple fusion categories. The main difference is that I know examples of strongly simple fusion categories whereas I do not know a single strongly noncommutative fusion category. -The next step would be about fusion categories which are both strongly simple and strongly noncommutative. - -REPLY [9 votes]: Consider the symmetric group group $G = S_3$ of order $6$. Then $\mathrm{H}^3_{\mathrm{gp}}(G;\mathrm{U}(1)) \cong \mathbb Z/6\mathbb Z$. Choose a generator $\omega \in \mathrm{H}^3_{\mathrm{gp}}(G;\mathrm{U}(1))$. Then $\omega$ restricts nontrivially to every nontrivial subgroup of $G$. -It follows that $\mathbf{Vec}^\omega[G]$ is not Morita-equivalent to any other fusion category. (Recall that, for any $G,\omega$, fusion categories equivalent to $\mathbf{Vec}^\omega[G]$ are indexed by pairs consisting of a subgroup $H \subset G$ together with a 2-cochain $\psi$ on $H$ solving $\mathrm{d}\psi = \omega|_H$.) -But $G$ is noncommutative.<|endoftext|> -TITLE: What is an example of a meager space X such that X is concentrated on countable dense set? -QUESTION [6 upvotes]: A topological space $X$ is concentrated on a set $D$ iff for any open set $G$ if $D\subseteq G$, then $X\setminus G$ is countable. -What is an example of a separable metrizable (uncountable) meager (meaning a countable union of nowhere dense subsets, also known as a set of first category) space $X$ such that $X$ is concentrated on countable dense set? - -REPLY [10 votes]: ADDED LATER -The answer to your question is that there is such a space $X$ if and only if $\mathfrak{b} = \aleph_1$. -If $\mathfrak{b} = \aleph_1$, then there is such a space. -To see this, first note that $\mathfrak{b} = \aleph_1$ if and only if there is an uncountable subset of the irrationals concentrated on $\mathbb Q$. This is proved by van Douwen in section 10 of his article in the Handbook of Set Theoretic Topology. By modifying his argument slightly, we can show that if $\mathfrak{b} = \aleph_1$ then there is an uncountable subset of the Cantor space concentrated on a countable subset of the Cantor space. -[The proof of this goes as follows. If $\mathfrak{b} = \aleph_1$, then one can construct via transfinite recursion a length-$\omega_1$ sequence $\langle f_\alpha :\, \alpha < \omega_1 \rangle$ of functions that is unbounded with respect to $\leq^*$, but also with the property that $\alpha < \beta$ implies $f_\alpha \leq^* f_\beta$. This means that a subset of this sequence is $\leq^*$-bounded if and only if it is countable. It's a well-known fact that the space $\omega^\omega$, endowed with the usual product topology, is homeomorphic to the space of irrational numbers. It is also fairly well known that the irrationals are homeomorphic to $C \setminus D$, where $C$ is the Cantor space and $D$ is some countable relatively dense subset of $C$. Let $Y$ be the image of your sequence of functions under some homeomorphism $\omega^\omega \rightarrow C \setminus D$. If $U$ is an open set containing $D$, then $C \setminus U$ is a compact subset of $C$. It's not too difficult to show that every compact subset of $\omega^\omega$ is $\leq^*$-bounded above by some function (or even more -- it is $\leq$-bounded), and this means $C \setminus U$ contains only countably many points of $Y$.] -Let $Y$ be, as in the previous paragraph, an $\aleph_1$-sized subset of the Cantor space $C$ that concentrates on a countable $D \subseteq C$. Let $X = Y \cup \mathbb Q$. As a subspace of the reals, this set $X$ meets all your requirements. -A proof that if there is a space $X$ as described in your post then $\mathfrak{b} = \aleph_1$. -Suppose there is an uncountable separable metrizable space $X$ that is concentrated on a countable set $D \subseteq X$. Let $\{ d_0, d_1, d_2, \dots \}$ be an enumeration of $D$. For each $x \in X \setminus D$, define a function $f_x : \omega \rightarrow \omega$ by setting $f_x(n) = \min\{ k :\, \mathrm{dist}(x,d_n) > \frac{1}{k} \}$. -Let $Y$ be a subset of $X \setminus D$ with $|Y| = \aleph_1$. The set of functions $\{ f_x :\, x \in Y \}$ is an $\aleph_1$-sized subset of $\omega^\omega$, and I claim it is unbounded with respect to $\leq^*$. In other words, I claim this set of functions witnesses $\mathfrak{b} = \aleph_1$. -To see this, suppose instead there is some $g \in \omega^\omega$ such that $f_x \leq^* g$ for all $x \in Y$. By a pigeonhole argument, there is some function $h \in \omega^\omega$, differing from $g$ in only finitely many places, such that $f_x \leq h$ for uncountably many $x \in Y$. Let $U = \bigcup_{n \in \omega} B_{1/h(n)}(d_n)$. This is an open set containing $D$, and our definition of the $f_x$'s ensures that $x \notin U$ if and only if $f_x \leq h$. Thus there are uncountable many $x$'s with $x \notin U$, contradicting the fact that $Y$ concentrates on $D$. -Let me point out that this argument is essentially contained in an earlier MO post by Taras Banakh found here. (A version of the argument seems to be in van Douwen's article too, and I don't know whether it really originated there either, but anyway I first learned it from Taras' post.) -$$$$ -ORIGINAL POST -Here is a consistent example of a subspace of $\mathbb R$ with these properties. -First, there is an uncountable subset $X$ of $\mathbb R$ concentrating on $\mathbb Q$ if and only if $\mathfrak{b} = \aleph_1$. This is proved in chapter 10 of van Douwen's article in the Handbook of Set-Theoretic Topology. Note that $X$ concentrates on $\mathbb Q$ if and only if every subset of $X$ does, so if $\mathfrak{b} = \aleph_1$ then there is an $\aleph_1$-sized set of reals concentrating on $\mathbb Q$. -Now suppose $\aleph_1 = \mathfrak{b} < \mathrm{non}(\mathcal M)$ (the least size of a non-meager subset of $\mathbb R$). This situation is consistent -- it happens in the random real model, for example. In such a model, let $X$ be an $\aleph_1$-sized set concentrating on $\mathbb Q$. Adding countably many points to $X$ if necessary, we may (and do) assume $X$ is dense in $\mathbb R$. Then (viewed as a subspace of $\mathbb R$) it meets all your requirements. -$$$$ -A sketch of an argument as to why $\mathfrak{b} = \aleph_1$ implies there is an $\aleph_1$-sized set concentrating on $\mathbb Q$: -Since you may not have access to van Douwen's article, here is a sketch of the idea I quoted above. If $\mathfrak{b} = \aleph_1$, then one can construct via transfinite recursion a length-$\omega_1$ sequence $\langle f_\alpha :\, \alpha < \omega_1 \rangle$ of functions that is unbounded with respect to $\leq^*$, but also has the property that $\alpha < \beta$ implies $f_\alpha \leq^* f_\beta$. It's a well-known fact that the space $\omega^\omega$, endowed with the usual product topology, is homeomorphic to the space of irrational numbers. Let $X$ be the image of your sequence of functions under any such homeomorphism. If $U$ is an open set containing $\mathbb Q$, then $\mathbb R \setminus U$ is a $\sigma$-compact subset of the irrationals. It's not too difficult to show that every $\sigma$-compact subset of $\omega^\omega$ is $\leq^*$-bounded above by some function, and this means $\mathbb R \setminus U$ can only contain countably many points of $X$.<|endoftext|> -TITLE: Does there exist a study of entire functions which satisfy $|F(x+iy)| \leq a e^{-bx^2}e^{cy^2}$? -QUESTION [5 upvotes]: I recently successfully extended a certain result where I use analytic functions which satisfy the following property: $F: \mathbb C \to \mathbb C$ is entire and there exist constants $a,b,c>0$ such that -$$ -|F(x+iy)| \leq a e^{-bx^2}e^{cy^2} -$$ -for every $x,y \in \mathbb R$. -I was wondering if functions with this property are studied somewhere in the literature. In particular do these functions have maybe a special name? Certainly they are of order less or equal than 2. - -REPLY [9 votes]: The inequality in the post can be rewritten as -$$\left|F(z)e^{\frac{b+c}{2}z^2}\right|\leq ae^{\frac{c-b}{2}|z|^2},\qquad z\in\mathbb{C}.$$ -For $cb$, we see that studying the functions in the post is essentially the same as studying the entire functions $G:\mathbb{C}\to\mathbb{C}$ satisfying $|G(z)|\leq e^{|z|^2}$ for all $z\in\mathbb{C}$.<|endoftext|> -TITLE: Can Fock spaces be replaced by arbitrary Hilbert spaces under some hypothesis to justify path integrals? -QUESTION [5 upvotes]: I was reading this post from PSE and it reminded me an old question of mine, in which the use of creation and annihilation operators were discussed. Both questions got answers which agreed on the fact that in QFT one is not always assuming a Fock space as the underlying Hilbert space of the theory. That made me think about the role of Fock spaces to justify path integrals in some contexts. Let me elaborate. -To keep the ideas simple and keep the notations from my previous linked post, I will deal with fermionic models. Suppose our Hilbert space $\mathscr{H}$ has dimension $n < +\infty$. In many-body quantum theory, one considers its associated Fock space $\mathcal{F}^{-}(\mathscr{H})$ which has dimension $2^{n}$. There is a vast literature about rigorous treatments of many-body systems. This Fock space identifies, in a natural way, with a Grassmann algebra with finitely many generators $\psi_{1},...,\psi_{2^{n}}$ by simply identifying each creation operators on the Fock space with a generator of the Grassmann algebra and the vacuum vector $\Omega$ with $1$. In this setting, one can give mathematical meaning to the path integral representation by using: -(i) The fact that the trace of an operator $A$ on $\mathcal{F}^{-}(\mathscr{H})$ has a natural "translation" in terms of Grassmann integrals and -(ii) The Lie-Trotter formula for bounded operators. -At least from my perspective, every reference I know that discusses path integrals for fermions does it in the above setting: finite-dimensional Hilbert spaces of one-particle system, finite dimensional Fock spaces and so on. However, as stressed before, in a more general setting, we couldn't assume a Fock space as the underlying Hilbert space. The fact that (at least to my knowledge) no reference discusses path integral formulations in more abstract setting makes me think this approach either only works for many-body systems, in which Fock spaces can be used, or it is just easier/more intuitive to do this way. -My point is: I don't see why Fock spaces are so much necessary apart from being more intuitive. So, can't we justify the path integral method in a more abstract way, under some additional hypothesis? Suppose $\mathcal{F}^{-}(\mathscr{H})$ is replaced by an arbitrary finite-dimensional complex Hilbert space $\mathscr{H}$, with creation and annihilation operators $\varphi^{\dagger}(x)$ and $\varphi(x)$ and some Hamiltonian $H$ on $\mathscr{H}$. If we assume: -(1) These creation and annihilation operators satisfy CAR relations -(2) The existence of a unique vacuum vector $\Omega$, which is the ground state for $H$ -(3) $\mathscr{H}$ is spanned by successive applications of creation operators to $\Omega$ -Then: -Q1: Doesn't the above hypothesis fully define a fermionic system? -Q2: Aren't (1) to (3) enough to establish a isomorphism between $\mathscr{H}$ and a finitely generated Grassmann algebra? -Q3: Doesn't the path integral formulation, using the Lie-Trotter formula hold in the very same way as before? -In summary: Are the hypothesis (1) to (3) enough to replace the Fock space by an arbitrary Hilbert space and still get the same analysis? -The hypothesis (1) to (3) seems exactly the kinds of hypothesis physicists are assuming when studying their fermionic models, but the lack of discussions about path integrals on these more abstract settings (i.e. outside Fock spaces and many-body quantum mechanics) makes me question whether there is something wrong with the above. - -REPLY [5 votes]: As soon as you assume the structure of a CAR algebra, then you are automatically dealing with a Fock space. To define a CAR algebra structure, it must be generated by something, and that something is the single-particle Hilbert space $\mathscr{H}$. -Let $\mathscr{H}$ be a complex Hilbert space, and let $\mathscr{A}$ be the CAR algebra it generates. There is a unique state $\Omega$ that is a $0$-eigenstate of all annihilation operators. Then the GNS representation based on $\Omega$ is unitarily equivalent to the Fock space $\mathcal{F}^-(\mathscr{H})$. Now, this is all quite independent of the details of the Hamiltonian $H$. A free field Hamiltonian is compatible with and preserves this structure. We could, if we wanted, consider a more general Hamiltonian, and it would generate an automorphism of the algebra $\mathscr{A}$, but it would not preserve the CAR structure, in the sense that it would not preserve $\mathscr{H}$ or the mapping from $\mathscr{H}$ to creation and annihilation operators. -In infinite-dimensional systems, the situation is even worse. Nothing prevents us from considering the CAR algebra of the field on a particular time-slice. But the interacting Hamiltonians we would want to write down are symmetric but not self-adjoint on the Fock space, so they do not generate a unitary time evolution. No one knows (or at least there is no general agreement) how to write down relations, such as the canonical anticommutation relations, that would define the algebra of interacting fields. So we are forced to approximate with free fields and do perturbation theory.<|endoftext|> -TITLE: Do polynomial values rarely have large multiple prime factors? -QUESTION [7 upvotes]: I am interested in the following set-up: -Let $F \in \mathbb{Z}[x_1,\dots,x_n]$ be a fixed irreducible homogeneous polynomial of degree $d$ and consider the quantity -$$N_{\delta}(B)=\#\{(x_1,\dots,x_n) \in \mathbb{Z}^n: \vert x_1\vert ,\dots,\vert x_n\vert \le B, \exists p \ge B^{\delta}: p^2 \mid F(x_1,\dots,x_n)\}$$ -where $\delta$ is some (small) positive constant. Of course the trivial bound is $N_{\delta}(B) \ll B^n$ and my question is whether we can get a power saving $$N_{\delta}(B) \ll B^{n-c(\delta)}$$ -as $B \to \infty$, for some positive constant $c(\delta)$, possibly depending on $F,n,d$ as well. -Heuristically, the probability that a number is divisible by $p^2$ is $\frac{1}{p^2}$, so we would expect $\frac{B^n}{p^2}$ values divisible by $p^2$ so that we could hope for a bound of the type -$$N_{\delta}(B) \ll \sum_{p \ge B^{\delta}} \frac{B^n}{p^2} \ll B^{n-\delta}.$$ -But of course this is very far from being a rigorous argument... -(It definitely seems like a question that has to have appeared before, but unfortunately I could not locate any reference.) -Addendum (31.05.): If this seems to hard to answer, does anyone at least have an intuition -a) whether or not this ought to be true (or whether there is something obvious I am missing) and -b) if true, whether or not this ought to be in reach of existing technology? - -REPLY [4 votes]: The question of showing that a multivariable polynomial takes on many squarefree values, or more generally controlling the size of the largest square factor, is referred to in the literature as the "squarefree sieve". -As you have already observed, standard sieve methods only allow you to filter out prime factors up to a certain size -- so one needs an additional ingredient to get control over the case of large prime factors. Generally this additional ingredient has come from diophantine geometry. There's various work out there, starting with that of Granville in the one-variable case, obtaining an exact asymptotic for the squarefree sieve conditional on the abc conjecture. Without the abc conjecture, things get harder, and unconditional results are mainly known in low degree cases or where the polynomial has additional symmetries that can be exploited.<|endoftext|> -TITLE: Set theory without the empty set -QUESTION [12 upvotes]: Has there ever been a set theory without an empty set? Is this possible? - -I ask because we usually take the empty set to exist axiomatically or obtain it through separation and a nonempty set together with the standard parameter-free predicate $X\neq X$, but it seems possible to have a 'set theory' without an axiom asserting the existence of an empty set or an axiom of separation. -I put 'set theory' in quotations because such a nonstandard axiomatization might not really deserve to be called a set theory per-se (it wouldn't prove the existence of intersections of disjoint sets), but more formally I mean - -Has a theory in the language of set theory whose axioms do not prove the existence of an empty set ever been explored? - -REPLY [3 votes]: I’ve explored a “set theory whose axioms do not prove the existence of an empty set,” the Incomprehensive Set Theory in my “Naive View of the Russell Paradox” (https://arxiv.org/abs/2103.00090), but for axiomatic parsimony, not with the motives I suspect you’re looking for. I note that “Any assumptions about set existence, even that any sets at all exist, will be noted explicitly. What follows will be trivial unless a lower or an upper exists, and the main interest is when an empty set and a universal set exist.” I imagine Incomprehensive Set Theory plus the existence of the Universal Set is consistent and does not prove the existence of the empty set, but, as noted, that is not necessarily of much interest.<|endoftext|> -TITLE: A "negative" standard system -QUESTION [5 upvotes]: For $\mathcal{M}$ a (countable) nonstandard model of $\mathsf{PA}$, let $\mathsf{SS}(\mathcal{M})$ be the set of sets of natural numbers coded by elements of $\mathcal{M}$. There are various ways to define this, for example $$\{X\subseteq\mathbb{N}:\exists a\in\mathcal{M}\;\forall k\in\mathbb{N}\;(k\in X\iff p_k\vert a)\}$$ (where $p_k$ is the $k$th prime and we conflate $\mathbb{N}$ with its canonical image in $\mathcal{M}$). -I'm curious about how this could differ from the following analogue: let $$\mathsf{SS}^-(\mathcal{M})=\{X\subseteq\mathbb{N}:\exists a\in\mathcal{M}\;\forall k\in\mathbb{N}\;[k\in X\iff \exists n\in\mathbb{N}\;(p_k^n\not\vert a)]\}.$$ -Intuitively, elements of $\mathbb{N}$ are prevented from entering $X$ by corresponding primes dividing $a$ "too much." (This version has come up in a separate problem I'm playing with, and I'd like to understand it better.) -It's easy to show that $\mathsf{SS}(\mathcal{M})$ is the set of elements of $\mathsf{SS}^-(\mathcal{M})$ whose complements are also in $\mathsf{SS}^-(\mathcal{M})$, and it's not hard to show that every $\mathcal{M}$ has an elementary extension $\mathcal{N}$ such that $\mathsf{SS}^-(\mathcal{N})=\mathsf{SS}(\mathcal{N})$. However, this still leaves a lot open. In particular: - -Is there an $\mathcal{M}$ with $\mathsf{SS}(\mathcal{M})\not=\mathsf{SS}^-(\mathcal{M})$? - -Equivalently per the above, is there an $\mathcal{M}$ whose $\mathsf{SS}^-$ is not closed under complementation? - -REPLY [3 votes]: If $\mathcal{M}$ is a nonstandard model of PA then for any set $X \in \mathsf{SS}(\mathcal{M})$, $X' \in \mathsf{SS}^-(\mathcal{M})$. Thus if $\mathsf{SS}(\mathcal{M}) = \mathsf{SS}^{-}(\mathcal{M})$ then $\mathsf{SS}(\mathcal{M})$ is closed under the Turing jump. Since not every Scott set is closed under the jump, $\mathsf{SS}(\mathcal{M})$ is not always equal to $\mathsf{SS}^-(\mathcal{M})$. -Claim. If $X \in \mathsf{SS}(\mathcal{M})$ then $X' \in \mathsf{SS}^-(\mathcal{M})$. -Proof. Suppose $X$ is in $\mathsf{SS}(\mathcal{M})$. Thus there is some $a$ such that -$$ -\forall k \in \mathbb{N}\, (k \in X \iff p_k \mid a). -$$ -Now let $b$ be a fixed nonstandard number in $\mathcal{M}$ and let $c$ be a nonstandard number such that -$$ -\forall n, k < b\, (p_k^n \mid c \iff \varphi^a_k(k) \text{ does not converge in $\leq n$ steps}). -$$ -Here, using $a$ as an oracle for $\varphi_k(k)$, means that when $\varphi_k(k)$ asks a question about $m$ to the oracle, we check if $p_m \mid a$ to determine the answer. Note that the existence of such a $c$ can be proved in PA. -Now let $Y = \{k \in \mathbb{N} \mid \exists n \in \mathbb{N} \, (p_k^n \nmid c)\}$. I claim that $Y = X'$. To show this it is enough to show that for all $n, k \in \mathbb{N}$, $p_k^n \mid c$ if and only if $\varphi^X_k(k)$ does not converge in $\leq n$ steps. -Note that for $n$ and $k$ standard natural numbers, the convergence or nonconvergence of $\varphi^X_k(k)$ within $n$ steps is witnessed by a standard natural number (encoding the transcript of the computation). And since running $\varphi^X_k(k)$ for $n$ steps never requires asking questions of the oracle at nonstandard numbers, there is no difference between using $X$ and $a$. Thus $\mathcal{M}$ can check that the witness to convergence or divergence of $\varphi^X_k(k)$ within $n$ steps is also a witness to the convergence or divergence of $\varphi^a_k(k)$ within $n$ steps and therefore by definition of $c$, $p_k^n \mid c$ if and only if $\varphi^X_k(k)$ does not converge in $\leq n$ steps.<|endoftext|> -TITLE: Is the "equidistant curve" to an algebraic curve algebraic? -QUESTION [16 upvotes]: Let $ L \subseteq \mathbb{R}^2 $ be a smooth real algebraic curve. Let's fix some parameter $ \delta \in \mathbb{R} $ and for every point $ (x,y) \in L $ define $$ L_{\delta}(x,y) = (x,y) + \delta n(x,y), $$ where $ n(x,y) $ is a normal vector to $L$ at $(x,y)$. The equidistant curve is then $$ L_\delta = \{L_\delta (x,y) : (x,y) \in L\}. $$ -The idea is very simple: we take a normal vector with length $ \delta $ and "roll" it along the curve. Note that this is not the same as the set of points at the same distance $\delta$ from $L$. -I am trying to find the equation of $L$. But I do not even understand if it is algebraic. For example, let's consider the parabola $ L$ given by $y - x^2 = 0 $. It is easy to find a parametrization of $ L_\delta $: -$$ x = t - \delta \frac{2t}{\sqrt{4t^2+1}}, $$ -$$ y = t^2 + \delta \frac{1}{\sqrt{4t^2+1}}. $$ -But how do I find the equation oh $x,y$ from this one? I looked up some literature about computational algebraic geometry. As I understood, there are only theorems for the case when the parametrization is rational in the variable $t$, which is not the case here. $ \mathbb{R} $ is not algebraicaly closed, maybe this is also an issue. -For me the question is interesting when $L$ is an algebraic surface in $ \mathbb{R}^n $, but the case of curves on the plane looks so simple that I am sure somebody has already solved it. Any help would be appreciated! - -REPLY [25 votes]: Yes, $L_\delta$ is algebraic. You can find its equations by elimination theory as follows: Let $L$ be defined by the polynomial equation $F(x,y) = 0$. Now consider the polynomial equations -$$ -F(x,y)=(u{-}x)-aF_x(x,y)= (v{-}y)-aF_y(x,y)= a^2(F_x(x,y)^2+F_y(x,y)^2)-\delta^2=0. -$$ -for (x,y,a,u,v). This is 4 equations for 5 unknowns. You can now use elimination theory to find a polynomial equation $G(u,v)=0$ in the ideal of the above system of equations that does not involve $x$, $y$, or $a$. This will be the algebraic equation of the parallel curve you want. -In your example of $y-x^2$, you find, for example, that the curve has degree $6$ and has the equation -$$ -\begin{aligned} -0=&16\,{u}^{6}+16\,{u}^{4}{v}^{2}-40\,{u}^{4}v + \left( -48\,{b}^{2}+1 - \right) {u}^{4}\\ -&\quad-32\,{u}^{2}{v}^{3}+ \left( -32\,{b}^{2}+32 \right) {u -}^{2}{v}^{2}+ \left( 8\,{b}^{2}-2 \right) {u}^{2}v+ \left( 48\,{b}^{4} --20\,{b}^{2} \right) {u}^{2}\\ -&\quad+16\,{v}^{4}+ \left( -32\,{b}^{2}-8 - \right) {v}^{3}+ \left( 16\,{b}^{4}-8\,{b}^{2}+1 \right) {v}^{2}+ - \left( 32\,{b}^{4}+8\,{b}^{2} \right) v\\ -&\quad-16\,{b}^{6}-8\,{b}^{4}-{b}^{2 -} -\end{aligned} -$$ -where, to save typing, I have writteen $b$ instead of $\delta$. -Remark: If you apply this method to Willie Wong's 'counterexample', you get the equation -$$ -0 = \bigl((u-v)^2-2b^2\bigr)\bigl((u+v)^2-2b^2\bigr), -$$ -i.e., the equation of the 4 lines $u\pm v = \pm b\sqrt2$, as expected.<|endoftext|> -TITLE: On a proof involving Young symmetrizers acting on tensor spaces -QUESTION [5 upvotes]: I hope this is not too elementary for this site, but I already asked something similar on MSE which has not received any attention whatsoever. I am extremely unfamiliar with the algebraic/representation theoretic area where Young tableaux are frequently used, so I am posting it here slightly rewritten as compared to the MSE question hoping someone can help me understand this without me having to learn an entire area from skratch just for this. - -My motivation is that I want to understand a proof given by Krupka in a paper (page 77 at formula (8) ) regarding total divergence equations in the calculus of variations. -If I remove the algebraic problem from its context, there is a finite $m$ dimensional real vector space $V$ and a tensor $T\in \left(S^r V\right)^{\otimes m}\otimes V\subseteq V^{\otimes mr+1}$ given in index notation as $$ T^{I_1...I_mj}, $$with each multiindex $I_k=(i_{1,(k)},\dots,i_{r,(k)})$ standing for a block of $r$ indices in which $T$ is symmetric. -This tensor observes the additional symmetry $$ T^{I_1...(I_k|...I_m|j)}=0\qquad(\sharp), $$i.e. when symmetrized over $I_kj$ for any $k=1,2,\dots,m$, it gives zero. -Krupka now proves that $$ T^{I_1...I_mj}=0\qquad(\ast). $$ -As the proof technique Krupka gives, verbatim - -We prove $(\ast)$ by showing that all Young diagrams, defining the Young decomposition of the expression on the left of $(\ast)$, vanish. - -I already have some trouble interpreting this sentence, but I'll ask about it later. -Then the proof proceeds as follows: - -Since $T$ is symmetric in each of the blocks $I_k$, only those Young tableaux contribute in which every element of each block $I_k$ lies on one row. -If $j$ is in the same row as one of the $I_k$, then by $(\sharp)$, the tableau gives zero. Thus $j$ goes to the last box of the tableau. -If any row contains at least two multiindices $I_k$ then by taking column permutations, it is possible to bring $j$ to be in one row with an intact multiindex $I_k$, and we again get zero by 2. -Thus the only possible nonzero tableau is $$\begin{matrix} I_1 \\ I_2 \\ \vdots \\ I_m \\ j\end{matrix}, $$ but the first column of this tableau has $m+1$ boxes to antisymmetrize on and as $\dim V=m$, hence this also gives zero. Therefore $T=0$. - - -Due to my unfamiliarity with Young symmetrizers, Schur modules etc. although I understand the general thought behind this proof, I am having trouble getting a precise and complete understanding. - -Just the general method of attack. From that I know the relevant tensor spaces can be decomposed as $$ \left(S^rV\right)^{\otimes m}\otimes V=\bigoplus_\lambda S^\lambda(V), $$where the $\lambda$ are numbered Young tableaux with $mr+1$ boxes and the $S^\lambda(V)$ are the corresponding Schur modules. The sum is over an appropriate set of Young tableaux. $$\ $$As far as I am aware all tableaux of the same shape give isomorphic Schur modules, and this sum has multiplicities. But tableaux with the same shape but different numbering give subspaces of $V^{\otimes mr+1}$ that are not literally the same as point sets. $$\ $$If $c_\lambda:V^{\otimes mr+1}\rightarrow S^\lambda(V)$ is the corresponding Young symmetrizer (symmetrizes the rows first, then antisymmetrizes the columns), then a scalar multiple of $c_\lambda$ is a projection, hence we should have $$ T=\sum_\lambda n_\lambda c_\lambda T. $$ So I suppose the idea behind the proof is to show that all Young symmetrizers $c_\lambda$ give zero when applied to $T$ right? -The first step 1. in the list above is somewhat problematic for me. At first I assumed that if there is a tensor symmetric in a number of indices, then any Young symmetrizer is zero on it whenever the symmetric indices are not in the same row of the tableau. But this is manifestly false, for example if the tensor $\phi^{ijk}$ is symmetric in $ij$ but the symmetrizer corresponding to the tableau $$\begin{matrix} i & k \\ j\end{matrix}$$ gives (up to a scalar multiple) $$ (c_\lambda\phi)^{ijk}=\phi^{kji}-\phi^{kij}, $$ which is nonzero in general. So why can we disregard every tableau in which a multiindex $I_k$ is spread over multiple rows? - - -If my concerns above are answered then I understand the rest of the proof. So my questions are as above. However basically any answer that helps me understand Krupka's proof is helpful to me. -One thing I'd like to remark is that with the first point 1. it would be helpful to use Young symmetrizers with the alternative convention, i.e. antisymmetrize first and then symmetrize last. However then Step 2. would be ruined. Also Krupka later on (page 79 at formula (11) ) deduces that an object is antisymmetric in a set of indices from a similar argument except where Step 4. does not give a zero symmetrizer. Hence, Krupka certainly uses the "antisymmetrize last" convention. - -Edit: As mentioned in the comments, I'd prefer to learn how this proof works as opposed to providing some alternative proof for the same result, since I wish to explore some of its implications and also generalize it. -One thing I thought about is that there is a rule $$ S_\lambda V\otimes S^rV=\bigoplus_\mu S_\mu V $$where the $\mu$ are tableaux obtained from $\lambda$ by adding $r$ boxes such that no two added boxes are in the same column. -Hence we can exclude those tableaux in which different elements of $I_k$ are in the same column. I then thought we can make column permutations to only consider tableaux in which each $I_k$ is on one row. -But this is also false, at least without some further arguments, because one cannot make column permutations in a tableaux freely without changing the Young symmetrizer since $c_{\pi\cdot\lambda}=\pi\cdot c_\lambda\cdot\pi^{-1}$, and if $\pi$ is a column permutation then $\pi\cdot c_\lambda=\pm c_\lambda$ but the additional factor of $\pi^{-1}$ on the right still remains. -I thought the answer was something simple I missed, but I strongly think now Krupka's proof is flawed or at least incomplete. The results he derives in the linked paper are known to be correct however (there are alternative proofs), so I suspect there is a way to make this proof sound. I am still quite interested in that. - -REPLY [3 votes]: I don't know what the paper is doing -- its use of Young diagrams definitely -looks vague to me, but maybe I would understand it better after reading (e.g.) -Towber. But here is an approach that avoids any combinatorics. -I generalize the problem: First, $\mathbb{R}$ becomes any field $K$ (not sure -if fieldness is really necessary); second, I replace the symmetric power -$S^{r}V$ by the whole symmetric algebra $SV$, of which the symmetric power -$S^{r}V$ is just a direct addend. Thus, the situation is as follows: -We are given a field $K$ and an $m$-dimensional $K$-vector space $V$. Let $SV$ -denote the symmetric algebra of the vector space $V$. -We let $\left[ m\right] :=\left\{ 1,2,\ldots,m\right\} $. For each -$i\in\left[ m\right] $, we let $\mu_{i}:\left( SV\right) ^{\otimes -m}\otimes V\rightarrow\left( SV\right) ^{\otimes m}$ be the $K$-linear map -that sends each pure tensor $a_{1}\otimes a_{2}\otimes\cdots\otimes a_{m}\otimes -b\in\left( SV\right) ^{\otimes m}\otimes V$ to -\begin{align*} -a_{1}\otimes a_{2}\otimes\cdots\otimes a_{i-1}\otimes a_{i}b\otimes -a_{i+1}\otimes a_{i+2}\otimes\cdots\otimes b -\end{align*} -(where the product $a_{i}b$ is taken in the symmetric algebra $SV$). This map -$\mu_{i}$ can be viewed as a "partial symmetrization": It multiplies the $b$ -onto the $a_{i}$ in $SV$ (which, if $\operatorname*{char}K=0$ and if you view -$SV$ as a subspace of the tensor algebra $TV$, corresponds to symmetrizing the -tensor $a_{i}\otimes b$). -Now, you claim: - -Theorem 1. We have $\bigcap_{i\in\left[ m\right] }\operatorname*{Ker} -\mu_{i}=0$. - -Proof. It is well-known that we can identify the symmetric algebra $SV$ with -a polynomial ring $K\left[ u_{1},u_{2},\ldots,u_{m}\right] $ in $m$ -variables. To do so, we fix a basis $\left( e_{1},e_{2},\ldots,e_{m}\right) -$ of $V$; then, there is a unique $K$-algebra isomorphism -\begin{align*} -f:K\left[ u_{1},u_{2},\ldots,u_{m}\right] & \rightarrow SV,\\ -u_{i} & \mapsto e_{i}. -\end{align*} -Consider this $f$. Let $K\left[ u_{1},u_{2},\ldots,u_{m}\right] -_{\operatorname*{lin}}$ denote the degree-$1$ homogeneous component of -$K\left[ u_{1},u_{2},\ldots,u_{m}\right] $. Then, $f\left( K\left[ -u_{1},u_{2},\ldots,u_{m}\right] _{\operatorname*{lin}}\right) =S^{1}V$ (the -$1$-st symmetric power of $V$). We identify $S^{1}V$ with $V$. -Thus, via the isomorphism $f^{\otimes m}$, we can identify the tensor power -$\left( SV\right) ^{\otimes m}$ with -\begin{align*} -\left( K\left[ u_{1},u_{2},\ldots,u_{m}\right] \right) ^{\otimes m} & -\cong\bigotimes_{i=1}^{m}\left( K\left[ u_{i,1},u_{i,2},\ldots -,u_{i,m}\right] \right) \\ -& \cong K\left[ u_{1,1},u_{1,2},\ldots,u_{m,m}\right] -\end{align*} -(a polynomial ring in $m^{2}$ indeterminates, named $u_{i,j}$ for all pairs -$\left( i,j\right) \in\left\{ 1,2,\ldots,m\right\} ^{2}$). Likewise, via -the isomorphism $f^{\otimes\left( m+1\right) }$, we can identify our tensor -product $\left( SV\right) ^{\otimes m}\otimes V$ with -\begin{align*} -& \left( K\left[ u_{1},u_{2},\ldots,u_{m}\right] \right) ^{\otimes -m}\otimes K\left[ u_{1},u_{2},\ldots,u_{m}\right] _{\operatorname*{lin}}\\ -& \cong\bigotimes_{i=1}^{m}\left( K\left[ u_{i,1},u_{i,2},\ldots -,u_{i,m}\right] \right) \otimes K\left[ x_{1},x_{2},\ldots,x_{m}\right] -_{\operatorname*{lin}}\\ -& \cong\left\{ \text{all polynomials in the }m^{2}+m\text{ indeterminates} -\right. \\ -& \ \ \ \ \ \ \ \ \ \ \left. u_{1,1},u_{1,2},\ldots,u_{m,m},x_{1} -,x_{2},\ldots,x_{m}\text{ that are}\right. \\ -& \ \ \ \ \ \ \ \ \ \ \left. \text{homogeneous of degree }1\text{ in the last -}m\right. \\ -& \ \ \ \ \ \ \ \ \ \ \left. \text{indeterminates }x_{1},x_{2},\ldots -,x_{m}\right\} . -\end{align*} -For each $i\in\left[ m\right] $, the map $\mu_{i}:\left( SV\right) -^{\otimes m}\otimes V\rightarrow\left( SV\right) ^{\otimes m}$ corresponds -(under these two identifications) to the map that takes a polynomial $P$ in -the $m^{2}+m$ indeterminates $u_{1,1},u_{1,2},\ldots,u_{m,m},x_{1} -,x_{2},\ldots,x_{m}$ and replaces the last $m$ indeterminates $x_{1} -,x_{2},\ldots,x_{m}$ by $u_{i,1},u_{i,2},\ldots,u_{i,m}$. (Check this -- it's -particularly easy here since $\mu_{i}$ is a $K$-algebra homomorphism when -extended to $\left( SV\right) ^{\otimes\left( m+1\right) }$ in the obvious -way.) We denote the image of $P$ under the latter map by $P\left( -\mathbf{x}\mapsto\mathbf{u}_{i}\right) $. -Thus, $\bigcap_{i\in\left[ m\right] }\operatorname*{Ker}\mu_{i}$ corresponds -to the set of all polynomials $P$ in the $m^{2}+m$ indeterminates -$u_{1,1},u_{1,2},\ldots,u_{m,m},x_{1},x_{2},\ldots,x_{m}$ that are homogeneous -of degree $1$ in the last $m$ indeterminates $x_{1},x_{2},\ldots,x_{m}$ and -satisfy -\begin{align*} -P\left( \mathbf{x}\mapsto\mathbf{u}_{1}\right) =P\left( \mathbf{x} -\mapsto\mathbf{u}_{2}\right) =\cdots=P\left( \mathbf{x}\mapsto\mathbf{u} -_{m}\right) =0. -\end{align*} -Our goal is thus to show that the only such polynomial $P$ is $0$. -This is easiest to achieve by another change of viewpoint: We let $S$ be the -commutative ring $K\left[ u_{1,1},u_{1,2},\ldots,u_{m,m}\right] $, and we -let $Q$ be the field of fractions of $S$. Any polynomial $P$ in the $m^{2}+m$ -indeterminates $u_{1,1},u_{1,2},\ldots,u_{m,m},x_{1},x_{2},\ldots,x_{m}$ over -our original field $K$ can then be viewed as a polynomial in the $m$ -indeterminates $x_{1},x_{2},\ldots,x_{m}$ over $S$, thus also as a polynomial -in the $m$ indeterminates $x_{1},x_{2},\ldots,x_{m}$ over $Q$ (since $S$ is a -subring of $Q$). Moreover, if this polynomial $P$ is homogeneous of degree $1$ -in the last $m$ indeterminates $x_{1},x_{2},\ldots,x_{m}$, then viewing it as -polynomial in the $m$ indeterminates $x_{1},x_{2},\ldots,x_{m}$ over $Q$ -yields a degree-$1$ homogeneous polynomial in these $m$ indeterminates, i.e., -a linear form on the $m$-dimensional $Q$-vector space $Q^{m}$. Finally, the -condition -\begin{align*} -P\left( \mathbf{x}\mapsto\mathbf{u}_{1}\right) =P\left( \mathbf{x} -\mapsto\mathbf{u}_{2}\right) =\cdots=P\left( \mathbf{x}\mapsto\mathbf{u} -_{m}\right) =0 -\end{align*} -tells us that the values of this linear form on the $m$ vectors $\left( -\begin{array} -[c]{c} -u_{1,1}\\ -u_{1,2}\\ -\vdots\\ -u_{1,m} -\end{array} -\right) $, $\left( -\begin{array} -[c]{c} -u_{2,1}\\ -u_{2,2}\\ -\vdots\\ -u_{2,m} -\end{array} -\right) $, $\ldots$, $\left( -\begin{array} -[c]{c} -u_{m,1}\\ -u_{m,2}\\ -\vdots\\ -u_{m,m} -\end{array} -\right) $ are all $0$. Since these $m$ vectors are $Q$-linearly independent -(because the matrix they form has determinant $\det\left( \left( -u_{i,j}\right) _{i,j\in\left[ m\right] }\right) \neq0$), this entails that -the linear form must be $0$. Hence, $P$ must be $0$. This proves Theorem 1. -$\blacksquare$<|endoftext|> -TITLE: Is every second-countable Hausdorff space symmetrizable? -QUESTION [8 upvotes]: Let us recall that a symmetric on a set $X$ is any function $d:X\times X\to[0,\infty)$ such that -for every $x,y\in X$ the following two conditions are satisfied: -$\bullet$ $d(x,y)=0$ if and only if $x=y$; -$\bullet$ $d(x,y)=d(y,x)$. -A topological space $X$ is called symmetrizable if there exists a symmetric $d$ on $X$ such that a subset $U\subseteq X$ is open if and only if for every $x\in X$ there exists $\varepsilon>0$ such that $B_d(x,y)\subseteq U$, where $B_d(x,\varepsilon)=\{y\in X:d(x,y)<\varepsilon\}$. -By the Urysohn Metrization Theorem, every regular second-countable space is metrizable and hence symmetrizable. -Question 1. Is each Hausdorff second-countable space symmetrizable? -A weaker version of Question 1 is also of interest: -Question 2. Is each countable first-countable Hausdorff space symmetrizable? - -Added in Edit. I have just realized that Question 2 has a simple affirmative answer (which I present below), so only Question 1 remains open. -Added in Edit 31.05.2022. Ups: Question 1 has a negative answer! - -REPLY [13 votes]: I have just realized that the first question has a simple affirmative answer. -Theorem 1. Every countable first-countable $T_1$ space $X$ is symmetrizable. -Proof. For every $x\in X$, fix a neighborhood base $(U_n(x))_{n\in\omega}$ such that $U_{n+1}(x)\subseteq U_n(x)$ for all $n\in\omega$. Since $X$ is countable, there exists a linear order $\le$ on $X$ such that for every $x\in X$ the initial interval $\{y\in X:y\le x\}$ is finite. -It is easy to see that the topology of $X$ is generated by the symmetric -$$d(x,y)=\inf\big\{2^{-n}:\max\{x,y\}\in U_n(\min\{x,y\})\big\}.\quad\square$$ - -Important Remark. After looking at a reference Dave L. Renfro suggested in a comment as possibly being relevant, I discovered that affirmative answers to both my questions follow from -Theorem 2.9 (Arhangelski, 1966): Every first-countable $T_1$ space with a $\sigma$-discrete (closed) network is symmetrizable. -But reading the proof of this theorem I discovered that it works only for regular spaces. Arhangelski writes that one loses no generality assuming that the $\sigma$-discrete network is closed, i.e., consists of closed sets. But after taking the closure of elements of a network in a $T_1$-space (even in a Hausdorff space), the network property can be destroyed. So, Question 2 seems to stay open and not answered even for Hausdorff (not mention $T_1$) spaces. - -Added in Edit 31.05.2022. To my big surprise (and contrary to what was claimed by Arhangelski in his Theorem 2.9), I have just discovered that Question 1 has negative answer! -First let us prove that the symmetrizability of second-countable spaces is equivalent to the perfectness. Recall that a topological space $X$ is perfect if every closed subset of $X$ is of type $G_\delta$. -Theorem 2. A second-countable (Hausdorff) $T_1$-space $X$ is symmetrizable if (and only if) it is perfect. -Proof. If $X$ is symmetrizable and Hausdorff, then the first-countability of $X$ implies that $X$ is semi-metrizable and perfect, see the paragraph before Theorem 9.8 in Gruenhage's "Generalized metric spaces". -Conversely, if a second-countable $T_1$-space $X$ is perfect, then each open subset of $X$ is an $F_\sigma$ in $X$, which implies that $X$ has a countable closed network. Now we can apply Arhangelski's Theorem 2.9 to conclude that $X$ is symmetrizable. $\square$ -Let $\mathrm{non}(\mathcal M)$ denote the smallest cardinality of a nonmeager set in the real line. -Example 1. There exists a second-countable Hausdorff space of cardinality $\mathrm{non}(\mathcal M)$ which is not perfect and hence not symmetrizable. -Proof. Take any nonmeager linear subspace $L$ of cardinality $\mathrm{non}(\mathcal M)$ in $\mathbb R^\omega$ such that for every $n\in\omega$ the intersection $L_n=L\cap(\{0\}^n\times\mathbb R^{\omega\setminus n})$ is dense in $\{0\}^n\times\mathbb R^{\omega\setminus n}$. Consider the quotient space $X=L_\circ/_\sim$ of $L_\circ=L\setminus\{0\}$ by the equivalence relation $\sim$ defined by $x\sim y$ iff $\mathbb R x=\mathbb Ry$. Since the space $L_\circ$ is Baire and the quotient map $q:L_\circ\to X$ is open, the space $X$ is second-countable and Baire. It is easy to check that the closure of every nonempty set in $X$ contains the set $q[L_n\setminus\{0\}]$ for some $n\in\omega$. This implies that the space $X$ is superconnected in the sense that for every nonempty open sets $U_1,\dots,U_n$ in $X$ the intersection of their closures $\overline U_1\cap\dots\cap\overline U_n$ is not empty. -Now take any disjoint nonempty open sets $U,V$ in $X$. Assuming that $V$ is of type $F_\sigma$, we can apply the Baire Theorem and find a nonempty open set $W\subseteq V$ whose closure in $X$ is contained in $V$. Then $\overline{U}\cap\overline{W}=\emptyset$, which contradicts the superconnectedness of $X$. $\square$ -The cardinality $\mathrm{non}(\mathcal M)$ is the above example can be lowered to $\mathfrak q_0$, where $\mathfrak q_0$ is the smallest cardinality of a second-countable metrizable space which is not a $Q$-space (= contains a subset which is not of type $G_\delta$). -A topological space is submetrizable it it admits a continuous metric. -Each submetrizable space is functionally Hausdorff in the sense that for any distinct elements $x,y\in X$ there exists a continuous function $f:X\to\mathbb R$ such that $f(x)\ne f(y)$. -Example 2. There exists a submetrizable second-countable space $X$ of cardinality $\mathfrak q_0$, which is not symmetrizable. -Proof. By the definition of the cardinal $\mathfrak q_0$, there exists a second-countable metrizable space $Y$, which is not a $Q$-space and hence contains a subset $A$ which is not of type $G_\delta$ in $X$. Let $\tau'$ be the topology on $X$, generated by the subbase $\tau\cup\{X\setminus A\}$ where $\tau$ is the topology of the metrizable space $Y$. It is clear that $X=(Y,\tau')$ is a second-countable space containing $A$ as a closed subset. Since $\tau\subseteq\tau'$, the space $X$ is submetrizable. Assuming that $X$ is symmetrizable and applying Theorem 2, we conclude that $X$ is perfect and hence the closed set $A$ is equal to the intersection $\bigcap_{n\in\omega}W_n$ of some open sets $W_n\in\tau'$. By the choice of the topology $\tau'$, for every $n\in\omega$ there exists open sets $U_n,V_n\in \tau$ such that $W_n=U_n\cup(V_n\setminus A)$. It follows from $A\subseteq W_n=U_n\cup(V_n\setminus A)$ that $A=A\cap W_n=A\cap U_n\subseteq U_n$. -$$A=\bigcap_{n\in\omega}W_n=A\cap\bigcap_{n\in\omega}W_n=\bigcap_{n\in\omega}(A\cap W_n)=\bigcap_{n\in\omega}(A\cap U_n)\subseteq \bigcap_{n\in\omega}U_n\subseteq \bigcap_{n\in\omega}W_n=A$$ -and hence $A=\bigcap_{n\in\omega}U_n$ is a $G_\delta$-set in $X$, which contradicts the choice of $A$. This contradiction shows that the submetrizable second-countable space $X$ is not symmetrizable. $\square$ -On the other hand we have the following partial affirmative answer to Question 1. -Theorem 3. Martin's Axiom implies that every second-countable $T_1$ space of cardinality $<\mathfrak c$ is perfect and hence symmetrizable. -Proof. It is known that Martin's Axiom implies that every second-countable $T_1$-space $X$ of cardinality $\mathfrak c$ is a $Q$-space, which means that every subset of $X$ is of type $G_\delta$. In particular, $X$ is perfect and by Theorem 2 is symmetrizable. $\square$<|endoftext|> -TITLE: Compact complex non-Kähler manifolds with nef canonical bundle -QUESTION [5 upvotes]: Are there examples of compact complex manifolds $X$ with $K_X$ nef, but $X$ is not Kähler? Perhaps even non-Moishezon examples? -Here, nef can be defined as follows: For any $\varepsilon>0$ there is a Hermitian metric $h_{\varepsilon}$ on $K_X$ with curvature $\Theta_{h_{\varepsilon}} \geq - \varepsilon \omega$, where $\omega$ is a positive-definite real $(1,1)$--form on $X$. - -REPLY [8 votes]: Let $X$ and $Y$ be compact complex manifolds. Note that $K_{X\times Y} \cong \pi_1^*K_X\otimes \pi_2^*K_Y$. If $Y$ has trivial canonical bundle, then $K_{X\times Y} \cong \pi_1^*K_X$. Now the pullback of a nef line bundle is again nef, see Proposition 1.8 (i) of Compact Complex Manifolds with Numerically Effective Tangent Bundles by Demailly, Peternell, and Schneider. So one can construct many examples by choosing $X$ with $K_X$ nef and $Y$ non-Kähler with $K_Y$ trivial. -Example: Let $X$ be a curve of genus $g > 1$ and $Y$ be a primary Kodaira surface. Then $X\times Y$ is a non-Kähler threefold with $K_{X\times Y}$ nef. Note that $X\times Y$ is also not Moishezon as it contains $Y$ as a complex submanifold and $Y$ is not Moishezon. -For more examples of non-Kähler manifolds with $K_Y$ trivial, see the introduction of Non-Kähler Calabi-Yau Manifolds by Tosatti. As is pointed out in Proposition 1.1, if $K_Y$ is trivial (or even just torsion), then it admits a metric $h$ with curvature $\Theta_h = 0$ and hence $K_Y$ is nef, so we don't even need to take a product with $X$ in the above construction.<|endoftext|> -TITLE: Is there a proof of strong normalisation that uses ordinal numbers? -QUESTION [10 upvotes]: I am currently trying to find a proof for strong normalisation of an extension of $\lambda$-calculus. -I've tried several approaches and one would be to assign an ordinal number $\operatorname{cs}(t)$ to each term $t$ in the calculus, and then show that this assigned ordinal number at least does not increase under any reduction and is reduced in certain cases. Then one could conclude that these certain cases can only occur a finite number of times in each reduction chain. -Is there a proof of strong normalisation for any calculus which uses ordinal numbers in this way? - -REPLY [7 votes]: As far as I know, such a strong normalization proof is not known even for the simply-typed $\lambda$-calculus. It is mentioned as Problem 26 in the TLCA List of Open Problems. -(It must be noted though that the list does not seem to have been updated since 2014, so the problem might have been solved in the meantime. I personally doubt it, but who knows!).<|endoftext|> -TITLE: On entire functions with polynomial Schwarzian derivative -QUESTION [9 upvotes]: The Schwarzian derivative of an entire holomorphic function $f$ is defined as -$$Sf:=\left(\frac{f^{''}}{f'}\right)'-\frac{1}{2}\left(\frac{f^{''}}{f'}\right)^2.$$ -In the following, we only consider entire holomorphic functions. -If $Sf=0$, then it is well known that $f$ must be a linear function. -If $Sf$ is a constant, then it is also well known that $f$ takes the form $f(z)=a+be^{cz}$, where $a,b,c$ are complex numbers. -My question is the following: Let $f$ be a locally injective entire function (the injectiveness is to guarantee that $Sf$ is also entire), if $Sf$ is a polynomial of degree bigger than or equal to $1$, what does $f$ look like? Can anyone give an example? - -REPLY [15 votes]: The answer is this: -$$f(z)=\int_{z_0}^z e^{Q(\zeta)}d\zeta,$$ -where $Q$ is a polynomial, and this is the general form of -an entire function whose Schwarzian is a polynomial. The crucial fact that $f$ has finite order. -Then $f'$, a function of finite order without zeros must be of the form $e^Q$. -This is a special case of the theorem of R. Nevanlinna which characterizes -solutions of the Schwarz equation -$Sf=P$, where $P$ is a polynomial. The Schwarz differential equation is equivalent to the linear differential equation -$$w''+(P/2)w=0,$$ -namely, the general solution $f=w_1/w_2$, where $f_1,f_2$ are two -linearly independent solutions of the linear differential equation. -And the fact that all solution of the linear differential equation with -polynomial coefficient are entire functions of finite order, is classical and well-known. It follows from asymptotic expansions of these solutions for large values of the independent variable, or in a simpler way, from the Wiman-Valiron theory. -Ref. R. Nevanlinna, Analytic functions, Chap.XI,3. -Remark. For generic $P$, the general solution $f$ of $Sf=P$ is -meromorphic but not entire. The condition that it has an entre solution is somewhat complicated: -$$Q''-(1/2){Q'}^2=P,$$ -so this is a condition for a Riccati equation to have a polynomial -solution $Q'$.<|endoftext|> -TITLE: Uniqueness of the set of decomposing spheres in prime decomposition of a 3-manifold -QUESTION [6 upvotes]: At the end of Section 1.1 of 3-manifold groups it is written that "the decomposing spheres are not unique up to isotopy, but two different sets of decomposing spheres are related by ‘slide homeomorphisms’" and I am trying to understand what the author meant. -The given references do not treat the topic (maybe Theorem 3 of The homotopy type of homeomorphisms of 3-manifolds, but if it does, then it is written in a quite mysterious way for my knowledge). -I guess that "two different sets of decomposing spheres are related by ‘slide homeomorphisms’" means that for every two sets of decomposing spheres there exists a slide homemomorphism sending one set of spheres in the other one. -However, it is clear that two sets of decomposing spheres containing a different number of spheres cannot be related by slide homeomorphism. -Maybe the author meant that the two sets of decomposing spheres have to be minimal (but in this case, notice that the decomposing system of spheres treated in the mysterious article that I cited before are not minimal). -Can someone cite some article stating the exact result? -(In the case the result I am looking for is Theorem 3 in the mysterious article, then can someone explain me what this theorem is saying exactly?) - -REPLY [7 votes]: Here is a simple example. -Suppose that $A$, $B$, and $C$ are closed, connected, oriented, prime three-manifolds, so that no two are homeomorphic and none are the three-sphere. (For example, lens spaces with fundamental groups of distinct sizes.) We define a manifold $M = A \ \# \ B \ \#\ C$. Implicit in this a pair of spheres: say $S$ separating $A$ from $B \ \#\ C$ and $S'$ separating $A \ \# \ B$ from $C$. -"Clearly" $M$ is homeomorphic to $B \ \# \ C \ \#\ A$. However, here we find a different pair of spheres: say $T$ separating $B$ from $C \ \#\ A$ and $T'$ separating $A$ from $B \ \# \ C$. "Clearly" there is no homeomorphism of $M$ sending $S \cup S'$ to $T \cup T'$ because the two systems have different separation properties. - -So reading "slide homeomorphisms" as "homeomorphism of the ambient space" is wrong. I believe that the offending sentence would be clearer if written as follows. - -The decomposing spheres are not unique up to isotopy, but two -different sets of decomposing spheres are related by ‘sphere -slides’. - -The proof is somewhat complicated, and relies on finding the correct "sphere surgery sequence". My suggested references are Hatcher's Notes on basic three-manifold topology and the second would be Casson's Three-dimensional topology.<|endoftext|> -TITLE: Expected value of min of variables - what informations do I need? -QUESTION [7 upvotes]: I encountered a problem where I need to compute: -$$\mathbb{E}(U) = \mathbb{E}(\min(X_1, .. , X_6))$$ -The problem is that I have little information on the $X_i$. Basically I know $\mathbb{E}(X_i)$ and $\operatorname{var}(X_i)$ which are identical for all $i$; say $\mu$ and $\nu$ respectively. One may assume for simplicity that the $X_i$ are independant. -I found this question that answers the problem as long as we know the distributions; the thing is that I don't know these. -Question: Can we figure out an upper bound on $\mathbb{E}(U)$ which is not $\mathbb{E}(U) \le \mu$ with little information? -Edit from comments: $X_i$ are non-negative, finite, integer. - -REPLY [6 votes]: One has an upper bound of $\lfloor \mu \rfloor + (\mu - \lfloor \mu \rfloor)^6 $, and this is best possible - i.e. one can obtain $\mathbb E(U)$ arbitrarily close to this. -To see this, let's first consider the case where we know the $X_i$ are integer-valued with mean $\mu$ but the variance is unrestricted. Then we can obtain $\mathbb E(U) =\lfloor \mu \rfloor + (\mu - \lfloor \mu \rfloor)^6 $ by a distribution with probability $1- (\mu - \lfloor \mu \rfloor)$ on $\lfloor \mu \rfloor $ and $(\mu - \lfloor \mu \rfloor)$ on $\lfloor \mu \rfloor +1$. -This is optimal since we are trying to maximize $\sum_{n=1}^\infty \mathbb P(X \geq n)^6$ given $\sum_{n=1}^\infty \mathbb P(X \geq n)=\mu$, for which increasing the larger values of $\mathbb P(X \geq n)$ and decreasing the smaller values always gives an improvement, so an optimum is obtained when the larger values are all $1$ and can't be increased and the smaller values are all $0$ and can't be decreased, i.e. when the probability distribution is supported on at most two values. -Let's now see that we can achieve $\mathbb E(U)$ arbitrarily close to $\lfloor \mu \rfloor + (\mu - \lfloor \mu \rfloor)^6 $ with a given variance $\nu$. Our previous construction works for $\nu= (\mu - \lfloor \mu \rfloor) (1- (\mu - \lfloor \mu \rfloor))$, and it is not possible to have variance smaller than this, so it suffices to handle the case when $\nu$ is larger. -If we shift $\epsilon$ probability mass from $\lfloor \mu \rfloor$ to $\lfloor \mu \rfloor+1$ and $\epsilon/m$ mass from $\lfloor \mu \rfloor$ to $\lfloor \mu \rfloor+m$, we have not changed the mean but have raised the variance by $\epsilon (m+1)$. Taking $$\epsilon = \frac{\nu- (\mu - \lfloor \mu \rfloor) (1- (\mu - \lfloor \mu \rfloor)) }{ m+1} $$ -we see that for $m$ sufficiently large, the distribution is still well-defined after the mass shift, and taking $m$ sufficiently large we may take $\mathbb E(U)$ arbitrarily close to its initial value.<|endoftext|> -TITLE: Product of the entries of a matrix -QUESTION [5 upvotes]: Given a $n \times n$ matrix $A = (a_{ij})$, I was wondering if there was any theory or research interest relevant to the term -$$ \prod_{i,j} a_{ij}$$ -the product of all the entries of the matrix. - -REPLY [10 votes]: Here is another instance of this quantity arising, which is in a similar vein to that of Abdelmalek Abdesselam's answer: -If $A$ is an $n \times n$ real orthogonal matrix, then $\big|\prod_{i,j} a_{i,j}\big| \leq n^{-n^2/2}$. Conversely, equality holds if and only if $A$ is a multiple of a Hadamard matrix (so it is conjectured that equality is attained for some real orthogonal matrix whenever $n$ is a multiple of $4$, but this is of course a long-standing open problem). - -REPLY [9 votes]: By using the arithmetic-geometric mean inequality, if each entry $a_{i,j}$ in $A$ is positive, we can bound several quantities related to $A$ below by the product $\prod_{i,j}a_{i,j}$. The geometric-arithmetic mean inequality states that $(x_1\dots x_n)^{1/n}\leq\frac{1}{n}(x_1+\dots+x_n)$ whenever $x_1,\dots,x_n$ are non-negative. Therefore, $n(x_1\dots x_n)^{1/n}\leq x_1+\dots+x_n$ whenever $x_1,\dots,x_n$ are non-negative. -Suppose $A$ is a matrix with non-negative entries. Then we obtain the following bound for the permanent of $A$: -$$\text{per}(A)=\sum_{\sigma\in S_{n}}\prod_{k=1}^{n}a_{k,\sigma(k)}\geq n!\cdot(\prod_{\sigma\in S_{n}}\prod_{k=1}^{n}a_{k,\sigma(k)})^{1/n!}=n!\cdot\big((\prod_{i,j}a_{i,j})^{(n-1)!}\big)^{1/n!}$$ -$$=n!\cdot(\prod_{i,j}a_{i,j})^{1/n}.$$ -Here, equality is reached if and only if every entry in $A$ is the same. While this inequality is easy to prove, the Van der Waerden's conjecture is a result that was proven in 1980 that strengthens this inequality whenever $A$ is doubly stochastic. -If $A$ is doubly stochastic, then by again applying the geometric-arithmetric mean inequality, we obtain -$\prod_{i,j}a_{i,j}\leq n^{-n^2}.$ -Van der Waerden's conjecture states that -$$n!\cdot(\prod_{i,j}a_{i,j})^{1/n}\leq\frac{n!}{n^n}\leq \text{per}(A)$$ -whenever $A$ is doubly stochastic. -For stochastic matrices, the product of all entries can be interpreted in terms of Markov chains. -Observation: Suppose that $(X_r)_r$ is an irreducible aperiodic Markov chain with underlying set $\{1,\dots,n\}$ and with transition matrix $A$. Furthermore, suppose that every entry in $B$ is $1/n$. -For almost all tuples $(y_r)_r\in\{1,\dots,r\}^{\omega}$, we have -$$\lim_{N\rightarrow\infty}P(X_0=y_0,\dots,X_N=y_N)^{1/N}=\prod_{i,j}a_{i,j}^{n^{-2}}\leq 1/n.$$ -If each entry in $A$ is positive, then the spectral radius $\rho(A)$ of $A$ is an eigenvalue of $A$. -The $i,j$-th entry in $A^{N}$ is the sum of all products of the form -$a_{i,i_{1}}\dots a_{i_{N-1},j}$. However, the geometric mean value of -$a_{i,i_{1}},\dots,a_{i_{N-1},j}$ is about $(\prod_{i,j}a_{i,j})^{1/n^2}$, so the geometric mean value of the product $a_{i,i_{1}}\dots a_{i_{N-1},j}$ is about $(\prod_{i,j}a_{i,j})^{N/n^2}$. And since the $i,j$-th entry is the sum of $n^{N-1}$ many factors, we estimate that the $i,j$-th entry in $A^N$ is about $n^{N-1}(\prod_{i,j}a_{i,j})^{N/n^2}$ which is about -$[n\cdot(\prod_{i,j}a_{i,j})^{1/n^2}]^{N}$. Therefore, we have -$$n\cdot(\prod_{i,j}a_{i,j})^{1/n^2}\leq\rho(A).$$<|endoftext|> -TITLE: Mapping Out(F_n) to the mapping class group -QUESTION [13 upvotes]: Let $\mathrm{Out}(F_g)$ denote the automorphism group of a free group, and $\mathrm{Mod}_g$ the mapping class group of a closed oriented genus $g$ surface. Is there a map, as indicated with the dashed arrow below, making the following diagram commute? -$$ -\begin{array}{ccc} - \mathrm{Out}(F_g) & \dashrightarrow & \mathrm{Mod}_g \\ - \downarrow & & \downarrow \\ - \mathrm{GL}(g,\mathbf Z) & \to & \mathrm{Sp}(2g,\mathbf Z) \\ -\end{array} -$$ -The lower horizontal map is induced by the functor $V \mapsto V \oplus V^\ast$ taking a vector space of dimension $g$ to a $2g$-dimensional symplectic vector space. -I do have a reason for asking but it's too long of a story to write here. - -REPLY [4 votes]: This works for $g=2$, but it’s a very special case. $Out(F_2)\cong GL_2( -\mathbb{Z}) \cong Mod_1 \cong Mod_{1,1}$, the mapping class group of a pointed torus. This is realized by the linear action of $GL_2(\mathbb{Z})$ on $T^2=\mathbb{R}^2/\mathbb{Z}^2$ fixing the origin. Taking the oriented blowup of the action at the origin (blowup by rays), one obtains an action of $GL_2(\mathbb{Z})$ on the surface $\Sigma_{1,1}$, a genus 1 surface with one boundary component. Then $GL_2(\mathbb{Z})$ acts on $\Sigma_{1,1}\times [-1,1]$, which is homeomorphic to a genus 2 handlebody and boundary homeomorphic to the double of $\Sigma_{1,1}$ along its boundary, ie $\Sigma_2$ the closed connected orientable surface of genus 2. The first homology splits as a direct sum into $H_1(\Sigma_{1,1} \times \{1\})$ and $H_1(\Sigma_{1,1}\times \{-1\})$, in such a way that the action of $GL_2(\mathbb{Z})$ acts by the dual action on the second factor since the identification by the product with $[-1,1]$ reverses orientation. Hence this gives the sort of homomorphism you seek in this case.<|endoftext|> -TITLE: On homeomorphic subsets of $\mathbb{R}^3$ with non-homeomorphic complements -QUESTION [8 upvotes]: Let $A,B$ be two homeomorphic topological subspaces of $\mathbb{R}^3$ such that their complements $\mathbb{R}^3 - A, \mathbb{R}^3 - B$ are not homeomorphic to each other. Must $A \cong B$ contain a homeomorphic image of the Cantor set? -(It is known that there are homeomorphic images $A,B$ of the Cantor set such that $\mathbb{R}^3 - A, \mathbb{R}^3 - B$ are not homeomorphic, see e.g. https://www.sciencedirect.com/science/article/pii/016686418690060X) - -UPDATE 1: -The answer is no, as Wojowu's answer shows. This leads to -Question 2: Must $A \cong B$ contain a homeomorphic image of $\mathbb{Q}$? - -UPDATE 2: -After Wojowu's answer, the interesting question remaining is -Question 3: Let $A,B$ be two closed, countable, topological subspaces of $\mathbb{R}^3$ homeomorphic to each other. Must their complements $\mathbb{R}^3 - A, \mathbb{R}^3 - B$ be homeomorphic to each other? - -REPLY [15 votes]: Let $A=\mathbb Q^3$ and $B=\{0\}\times\mathbb Q^2$. It is a classical result that they are homeomorphic (both homeomorphic to $\mathbb Q$), and their complements are not homeomorphic as $\mathbb R^3-B$ contains a subset homeomorphic to an open ball, while $\mathbb R^3-A$ doesn't (e.g. by the invariance of domain theorem). -However, since $A,B$ are countable, they don't contain a copy of the Cantor set. - -The answer to the new question is also negative. Indeed let $A=\{(0,0,n)\mid n\in\mathbb N\}$ and $B=\{(0,0,1/n)\mid n\in\mathbb N\}$. Both of these are discrete and countable, so are homeomorphic. On the other hand, the complement of $A$ is a manifold, while $(0,0,0)$ in $\mathbb R^3-B$ has no Euclidean neighbourhood.<|endoftext|> -TITLE: How hard is it to compute the Davenport constant? -QUESTION [8 upvotes]: The Davenport constant $D(G)$ of a finite abelian group $(G,+)$ is the least positive integer $k$ such that every sequence in $G$ of length $k$ has a zero-sum (nonempty) subsequence. -It seems that the Davenport constant is explicitly known only in a handful of cases (such as when $G$ is cyclic, or is a $p$-group, or has rank $2$). Is anything known about how hard it is to compute the Davenport constant? - -REPLY [8 votes]: The Davenport constants for all groups of order less than thirty-two is computed in arXiv.1702.02997, using GAP. The algorithm is presented and the complexity is discussed in section 6. The run-time varies by orders of magnitude from one group to another, dependent on the number of equivalence classes that need to be considered.<|endoftext|> -TITLE: Is this function rational? -QUESTION [5 upvotes]: Let -$$ -F=\sum_{i\ge0}\frac1{(T+2)^i}\left(\frac T{T+1}\right)^{3^i}\in\mathbb F_3\left(\!\!\left(\frac1T\right)\!\!\right).$$ -Does $F$ belong to $\mathbb F_3(T)$? -Here, truncations of the series do not give a good approximation of $F$. - -REPLY [11 votes]: It is not rational. -First of all we denote $1/T=x$, then $$F=\sum_i \left(\frac{x}{1+2x}\right)^i(1+x)^{-3^i}\in \mathbb{F}_3((x)).$$ -Now denote $x/(1+2x)=y$. Note that rationality in $x$ is equivalent to rationality in $y$ (and also that $\mathbb{F}_3((x))=\mathbb{F}_3((y))$). We have $x=y/(1-2y)$; $1+x=(1-y)/(1-2y)$; $$(1+x)^{-1}=\frac{1-2y}{1-y}=\frac{1+y}{1-y}=1-(y+y^2+y^3+\ldots).$$ -Thus $$F=\sum_i y^i(1-y^{3^i}-y^{2\cdot 3^i}-\ldots)=\frac1{1-y}-\sum_{i\geqslant 0, k\geqslant 1}y^{i+k\cdot 3^i}.$$We should prove that the sequence of coefficients $a_n$ of the power series $$\sum_{i\geqslant 0, k\geqslant 1}y^{i+k\cdot 3^i}=\sum a_ny^n$$ -is not eventually periodic modulo 3 (since rationality of a power series with coefficients in a finite field is equivalent to eventual periodicity of its coefficients). -We have $a_n=|A(n)|$, where $A(n)$ is the set of all non-negative integers $i1$ and $A(k+3^{n-1})=A(k)\sqcup \{k\}$ for all $n>k>1$. This yields $a_{k+3^{k-1}}=a_k$ and $a_{k+3^{n-1}}=a_k+1$ for all $n>k>1$. -Now assume that, on the contrary, $a_{n+T}\equiv a_{n} \pmod 3$ for certain integer $T>0$ and all $n\geqslant n_0$. Consider the powers of 3 of the form $3^{m-1}$, $m>n_0$. By pigeonhole principle, there exist two of them which are congruent modulo $T$, say, $T$ divides $3^{n-1}-3^{k-1}$ for certain $n>k>n_0$. Then $T$ divides $(k+3^{n-1})-(k+3^{k-1})$ but from above we have $a_{k+3^{n-1}}=a_k+1=a_{k+3^{k-1}} +1$. This contradicts to our periodicity assumptions.<|endoftext|> -TITLE: Convex solutions of the Poisson equation -QUESTION [6 upvotes]: Let $D$ be a planar, bounded, convex open domain. Given a positive function $f:D\to(0,+\infty)$, let us consider the Poisson equation -$$\Delta u=f\quad\hbox{in }D.$$ -Not specifying any boundary condition, it admits a bunch of solutions $u=u_0+v$, where $u_0$ is a particular one (for instance the solution of the Dirichlet BVP with $u_0=0$ on $\partial D$), and $v$ is an arbitrary harmonic function. - -Does there exist a convex solution (actually strongly convex in the sense that ${\rm D}^2u(x)$ is positive definite for every $x\in D$) ? - -Let me refine the question as follows. For this, I denote $\nu$ and $\tau$ the unit normal and tangential vector fields along $\partial D$. - -Does there exist a solution of the Laplace equation, satisfying the 2nd-order boundary condition ${\rm D}^2u(\nu,\tau)=0$ along $\partial D$ ? If so, is it a convex function ? - -REPLY [6 votes]: I apologize for having posted this question too early. I realize that the answer to the first question is negative. -Actually suppose that $D=D(0;R)$ is a disk and $f=f(r)$ is a radial function. If a convex solution $u$ existed, then certainly $u_\theta(x):=u(R_\theta x)$ would be another one, where $R_\theta$ is a rotation. Then -$$U:=\frac1{2\pi}\int_0^{2\pi}u_\theta\,d\theta$$ -is again a convex solution, now a radial one $U=v(r)$. It satisfies $v''+\frac1r v'=f$, and the spectrum of ${\rm D}^2U$ is $\{v'',\frac1rv'\}$. Thus the existence of a convex solution for every smooth positive radial $f$ amounts to saying that $rv''+v'\ge0$ over $(0,R)$ for $v'(0)=0$, implies $v',v''\ge0$. This is obviously false.<|endoftext|> -TITLE: Holonomy bounded in terms of area and the curvature -QUESTION [5 upvotes]: I suppose the following result follows -from Ambrose-Singer theorem, but I cannot -find a reference, and the arguments I found -in the literature are usually weaker. The idea -is that holonomy over a null-homotopic loop is bounded -by the supremum of the curvature times the area of the -2-dimensional surface segment bounded by the loop. -THEOREM-CONJECTURE -Let $D$ be a unit disk, and $(B, \nabla)$ a trivial -vector bundle on $D$ with connection (not necessarily -orthogonal). Assume that the curvature -$R$ of $\nabla$ is uniformly bounded, -that is, $R(x, y)$ belongs to a compact subset $K$ -in $End(B_m)$ for all $x, y \in T_m D$ of length 1. -Then the holonomy of $\nabla$ around the boundary -of $D$ is bounded by a uniform constant which -depends on $K$ only. -I think I can prove this, but there are -some segments of the proof which are tricky -and take too much effort. -Can someone please point me to a reference, -or to some relevant papers. Many thanks in advance. - -REPLY [6 votes]: There are in fact more precise versions, expressing the parallel translation around a loop as the identity map plus a curvature integral over a homotopy. References: -Section 3.1 of Werner Ballmann's lecture notes on vector bundles: -http://people.mpim-bonn.mpg.de/hwbllmnn/archiv/conncurv1999.pdf -Deane Yang's notes "Holonomy equals curvature" on his web page -https://cims.nyu.edu/~yangd/papers/holonomy.pdf -Buser-Karcher, Gromov's almost flat manifolds, page 92.<|endoftext|> -TITLE: Asymptotics for $\int\exp( x t / \log t)dt$ -QUESTION [5 upvotes]: What is the asymptotic growth rate of $$f(x) = \int_e^\infty e^{ - x t / \log t} dt$$ as $x \to 0$? -As an example of what is meant by "growth rate" consider $$g(x) = \int_e^\infty e^{-x t} dt = \frac {e^{-ex}} x \approx x^{-1}.$$ -By comparing to $g$, it is easy to see that $f(x) \approx x^{-1 - o(1) }$, but we would like to know what the lower order correction is. -We are guessing that $f(x) \approx \log^p(1/x)/x$ for some $p >0$, but unsure how to show this or find $p$. - -REPLY [4 votes]: According to Maple, -$$ f(x) = -\frac{\gamma+\log x}{x} -+\tfrac14 e^2 x - \tfrac19 e^3x^2 -+ \tfrac{1}{32}e^4x^3 - \tfrac{1}{150}e^5x^4 -+ O(x^5).$$ -A convergent series is available. The exact value of the integral is $\frac{1}{x}(e^{-ex} + \mathrm{Ei}(1,ex))$ where Ei is the exponential integral. Using the known Taylor series of the exponential integral, we get -$$f(x) = -\frac{\gamma+\log x}{x} + - \sum_{j=2}^\infty \frac{(-1)^j(j-1)}{j\,j!} e^j x^{j-1}.$$<|endoftext|> -TITLE: What is the name for the construction of this poset related to coherence of degeneracies of the simplex category? -QUESTION [5 upvotes]: I present you a family of posets here. I don't say the posets themselves have a conventional name. However I'm sure the general construction of this kind has received some terminology, related to Grothendieck construction, comma categories, category of elements, etc. I can't exactly nail it, so it would be very helpful if you could do it for me. -As usual, write $\Delta$ for the simplex category: the category of inhabited finite linearly ordered set and order-preserving sets. Let $\Delta_-$ and $\Delta_+$ denote the wide subcategory of degeneracy maps and face maps, respectively, of $\Delta$. -Let $\sigma$ be a simplex of the simplicial nerve $N(\Delta_-)$: -$$ -\sigma\colon [n_0]\twoheadrightarrow [n_1] - \twoheadrightarrow \dotsb \twoheadrightarrow [n_k], -$$ -where each $\twoheadrightarrow$ lies in $\Delta_-$. -I would like to define a poset $P(\sigma)$ as: -$$ -P(\sigma) := \left\{ \tau \overset{d}{\rightarrowtail} \tau' \overset{u}{\subset} \sigma \right\}. -$$ -Before we define an order on this set, we need to clearify the meaning of the symbols here. Firstly $\tau' \overset{u}{\subset} \sigma$ denotes a face $\tau'$ of $\sigma$ in the nerve $N(\Delta_-)$. It is determined by a face map $u\colon [l] \rightarrowtail [k]$ in $\Delta_+$, and we have -$$ -\tau' = u^*(\sigma)\colon [n_{u(0)}]\twoheadrightarrow [n_{u(1)}] - \twoheadrightarrow \dotsb \twoheadrightarrow [n_{u(l)}]. -$$ -Secondly, $\tau \overset{d}{\rightarrowtail} \tau'$ denotes the vertex-wise family of face maps, i.e. -$$ -\tau\colon [m_0]\twoheadrightarrow [m_1] - \twoheadrightarrow \dotsb \twoheadrightarrow [m_l] -$$ -is a diagram in $\Delta_-$ and $d_i\colon [m_i] \rightarrowtail [n_{u(i)}]$, for $i=0,1,\dotsc, l$, are face maps in $\Delta_+$ which, in $\Delta$, commutes all the squares. -We need to define an order on $P(\sigma)$. Given $\tau \overset{d}{\rightarrowtail} \tau' \overset{u}{\subset} \sigma$ and $\omega \overset{e}{\rightarrowtail} \omega' \overset{v}{\subset} \sigma$, -we say -$$ -\left(\omega \overset{e}{\rightarrowtail} \omega' \overset{v}{\subset} \sigma\right) \le \left(\tau \overset{d}{\rightarrowtail} \tau' \overset{u}{\subset} \sigma\right), -$$ -iff we have $\omega \overset{\exists f}{\rightarrowtail} \exists\omega'' \overset{\exists w}{\subset} \tau$ in the commutative way, i.e. $v=u\circ w$ in $\Delta_-$ and -$e_i = d_{w(i)}\circ f_i$. -I feel a strong déjà-vu looking at this, but I can't write it down into a conventional categorical construction. This clearly looks like a slice category, so if we can name the category with its objects simplices of the simplicial nerve $N(\Delta_-)$ and its morphisms $\bullet \rightarrowtail \bullet \subset \bullet$, we are done. However I can't go beyond that point, so your help would be very helpful. - -REPLY [2 votes]: Here’s one way to see it, if I’m not misunderstanding your definition. - -For a small category $\newcommand{\C}{\mathbf{C}}\C$, take its categorical nerve $\newcommand{\N}{\mathbf{N}}\N\C$ to be the functor $\newcommand{\op}{\mathrm{op}} \Delta^{\op} \to \mathbf{Cat}$ defined by $(\N\C)_k = \C^{[k]}$; and take its semi-nerve $\N_{+}\C$ to be the restriction of this to to $\Delta_{+}$. - -The Grothendieck construction $\int_{\Delta_{+}} \N_{+}\C$ is a split fibration over $\Delta_{+}$. Its objects are strings $\sigma_0 \to \cdots \to \sigma_n$ in $\C$; its morphisms are $(\tau \overset{g}{\to} \tau' \overset{u}{\subseteq} \sigma)$, where $u$ is a face map and $\tau'$ is the restriction of $\sigma$ along $u$, and $g$ is a ladder in $\C$ from $\tau$ to $\tau'$. - - -Taking $\C := \Delta$, this is very nearly the category you describe in your last para (and so its slices are nearly the category you want overall), but it’s a bit more general: its objects are arbitrary strings in $\Delta$ (not just degeneracies) and its vertical maps are arbitrary ladders, not necessarily of face maps. -So we cut down to a subcategory. This can be done already at the level of the functor $\N_{+}\C : \Delta_{+}^{\op} \to \mathbf{Cat}$, before taking the Grothendieck construction. Suppose $\C$ has two distinguished wide subcategories of maps; call them $a$, $b$. Then write $\N^{a,b}\C : \Delta^\op \to \mathbf{Cat}$ (and $\N_{+}^{a,b}$ similarly) for the functor where $(\N^{a,b}\C)_k$ is the subcategory of $\C^[k]$ whose objects are strings of $a$-maps, and whose arrows are ladders of $b$-maps. -Then $\int_{\Delta_{+}} \N_{+}^{-,+}\Delta$ is the category you describe in the last paragraph; and its slices are the posets you want overall. -In particular, $\Delta$ is playing two different roles here, which can be generalised separately: - -the base of the categorical semi-nerve, $\Delta_{+}$ along with its inclusion into $\mathbf{Cat}$; - -the target of the categorical semi-nerve, $\Delta$ with its two distinguished wide subcategories. - - -(This categorical (semi-)nerve is an instance of a well-established construction, the generalised nerve/realisation; and the Grothendieck construction is of course very standard. Cutting down to a subcategory of the nerve in this particular way is not something I’ve seen before.)<|endoftext|> -TITLE: What is the difference (if any) between "fourier transform" and "SO(3) fourier transform"? -QUESTION [5 upvotes]: What is the difference (if any) between "fourier transform" and "SO(3) fourier transform"? -I searched on Google but couldn't find a satisfiable answer. -Thanks in advance :) - -REPLY [3 votes]: In addition to what Carlo Beenakker stated, you can find a detailed introduction to Fourier Transforms, Spherical Harmonics and Wigner-D matrices in Chapter 3 of a 2017 PhD thesis at Bielefeld University linked here. -It focuses on applications for visual robot navigation (and therefore especially on real-valued implementations), but a brief summary of the general idea is given. For a purely math based focus, you can check the references.<|endoftext|> -TITLE: Publishing papers that became classics before they were submitted -QUESTION [20 upvotes]: Sometimes the following happens: a result is proven, but the author never submits a paper for publication. In some cases, a preprint appears. In some cases, the proof is so short that it can be presented at a conference or lecture series, and the community is convinced. Maybe it later becomes incorporated in the work of others, book or lecture notes expositions appear, other work generalizes or improves upon it etc. The result is widely accepted, and the community gives due credit. -Now assume that some years later, the author decides to submit the paper to a prestigious journal. How should the editor treat it, depending on the circumstances listed above, importance of the result, etc.? The two obvious extreme points of view are "reject, since the result is not new" and "treat the paper as proving a new result with a somewhat delayed submission", but there's a whole spectrum in between, e.g., "what was good for Annals 20 years ago is not anymore, but OK for a lesser journal". -I would be especially interested in precedents that are publicly known or can be shared publicly. - -REPLY [16 votes]: This happened in my career. In 1984 I published two preprints, joint with M. Lyubich, in Russian and two short announcements based on these preprints (also in Russian). These had a substantial following. In the early 1990s we wrote a paper and submitted it to a journal. The paper was rejected on the grounds that "this is a survey of the well-known results". -(Which was essentially true: by that time the results were already well-known since many people read and used our preprints). So we submitted the paper to another journal, and in 1992 the paper was published. It is my most cited paper. -The situation is quite common in the subject of holomorphic dynamics. -In 1982 Douady and Hubbard published a large preprint (>200 pages) -"Etude dynamique des polynomes complexes" and announced the main results -in a CR note. They wanted to write a book, but this plan never materialized. This preprint established the foundation of the study of Mandelbrot set. - -REPLY [11 votes]: There are countless examples, especially if one includes the publication of the Nachlass of a deceased mathematician. Let me mention just one example that I have been looking forward to. -The book Integration in Finite Terms: Fundamental Sources (probably not the best hyperlink, but I don't think it has a DOI yet) is slated to be published soon. When I requested more information from one of the editors, Michael Singer, he provided me with a sneak peek at an early draft of the Preface. After explaining that the book contains four items, the first two of which are reprints (of a paper by Rosenlicht and a book by Ritt, both entitled Integration in Finite Terms), the Preface goes on to say: - -The third is a revised version of Robert Risch’s unpublished On the integration of elementary functions which are built up using algebraic -operations. This latter paper reduced the problem of finding elementary -antiderivatives of elementary functions to a problem in arithmetic algebraic geometry: the problem of determining if a point on the jacobian of a curve is of finite -order. Risch presented a solution of this latter problem in [Ris70] in the context -of deciding the elementary integrability of an algebraic function in [Ris70]. The -present paper includes the complete proof. The commentary of Clemens G. Raab -discusses the impact of this paper and further developments. The final paper is the -unpublished thesis of Barry M. Trager Integration of Algebraic Functions. -This latter paper provided practical algorithms for performing the integration in finite terms of algebraic functions after the theoretical results of Risch. It is followed -by the commentary of Barry Trager, who gives further insight into the methods of -this thesis as well as subsequent developments. - -The OP also asks what an editor should do in such situations. Although MO usually frowns upon opinion-based questions, I will offer an opinion anyway. The mathematical community should do more than it currently does to encourage the publication of, or at least the wide distribution of, such material. As a community, we often encourage people to read the masters, and we also take some pride in the "democratic" nature of our subject, meaning that mathematics is open to all. Well, it's not open to all if some of the writings of the masters are available only to a privileged inner circle. Publication of important and highly influential unpublished material does not solve all such problems but it is a step in the right direction. Currently I think we place too much emphasis on novelty. I like what Harold Edwards said in his book Riemann's Zeta Function about Siegel's monumental effort to study Riemann's Nachlass and present his findings to the mathematical public. - -One wonders whether anyone else would ever have unearthed this treasure if Siegel had not. It is indeed fortunate that Siegel's concept of scholarship derived from the older tradition of respect for the past rather than the contemporary style of novelty. - -REPLY [11 votes]: G.A.Margulis announced his super -rigidity and arithmeticity in an ICM talk (which he could not attend). At the next ICM where he received the Fields medal, this was one of the major cited works. He later gave a different proof in an appendix to a Russian translation of M.S Raghunathan's book on discrete subgroups of Lie groups. Later an English translation of this appendix was published in the Inventiones. -Here is a link to the review https://mathscinet.ams.org/mathscinet-getitem?mr=739627<|endoftext|> -TITLE: Is it true that $\operatorname{2-colim}_U \textsf{QCoh}(U) = \textsf{Vect}(K_X)$, as $U$ shrinks to the generic point? -QUESTION [5 upvotes]: Let $X$ be an integral scheme with function field $K$. If $U\subset X$ is an open subscheme, we may consider the restriction functor -$$\textsf{QCoh}(X) \to \textsf{QCoh}(U).$$ -I don't know much about 2-categories (I'm probably thinking about (2,1)-categories, to be more precise), but I wonder if we may consider a 2-colimit of this and obtain an equivalence of categories -$$\operatorname{2-colim}_U\textsf{QCoh}(U) \xrightarrow{\sim} \textsf{Vect}(K).$$ -If this is true, I wonder moreover if the natural functor -$$\textsf{QCoh}(X) \to \operatorname{2-colim}_U\textsf{QCoh}(U)\cong \textsf{Vect}(K)$$ -is the one which sends a quasi-coherent module over $X$ to its stalk on the generic point. - -REPLY [6 votes]: Let $x$ be a point in a scheme $X$. There are two posets, namely the poset of affine opens containing $x$, $A(x)$, and the poset of opens containing $x$, $O(x)$. -The inclusion $A(x)^{op} \to O(x)^{op}$ is (homotopy) cofinal - by Quillen's theorem A, it suffices to show that for any open $U$, $A(x)_{/U}$ is weakly contractible. But it is a co-directed poset : for any $V,W \to U$, $V\cap W$ is an open and therefore contains some affine open $x\in O\subset V\cap W\subset U$; therefore it is weakly contractible. -It follows that, in the $(2,1)$-category of presentable $1$-categories, $colim_{U\in O(x)^{op}}QCoh(U) \simeq colim_{U\in A(x)^{op}} QCoh(U)$. -Now note that the latter diagram in fact lives in the category of $E_0$ presentable categories, that is, presentable categories with a chosen basepoint (namely $O_U$) that are equivalent to something of the form $(Mod_R,R)$ for some (commutative) ring $R$. -By a $(2,1)$-categorical analogue of theorem 4.8.5.11. in Lurie's Higher algebra (which says that $R\mapsto (Mod_R,R)$, as a functor from rings to $E_0$ presentable categories, is fully faithful and colimit preserving) (it's actually not an analogue, it follows strictly from this theorem by specializing to $Set$-modules in presentable $(\infty,1)$-categories), and using the fact that $A(x)$ is weakly contractible, so colimits over it in $E_0$-objects are computed underlying, it follows that $colim_{U\in A(x)^{op}} QCoh(U)\simeq colim_{U\in A(x)^{op}} Mod_{O_U(U)}$ is equivalent to modules over $colim_{U \in A(x)^{op}} O_U(U)= colim_{U\in A(x)^{op}} O_X(U)= O_{X,x}$. -Note that here the colimit is taken in ordinary associative rings, but it's a filtered colimit, so it can also be taken in commutative rings. -In conclusion, for any point $x$, $colim_{U\in O(x)^{op}} QCoh(U)$ is equivalent to modules over $O_{X,x}$. -If $x$ is a generic point, then $O(x)$ is the category of all opens, and so you get the corresponding statement for $colim_U QCoh(U)$. -Note that here I've used the following version of "$2-\mathrm{colim}$": homotopy colimits in the $(2,1)$-category (in the sense of "$(\infty,1)$-categories with $1$-truncated mapping spaces") of presentable $1$-categories. I'm assuming that this shouldn't matter too much \ No newline at end of file