diff --git "a/stack-exchange/math_overflow/shard_22.txt" "b/stack-exchange/math_overflow/shard_22.txt" deleted file mode 100644--- "a/stack-exchange/math_overflow/shard_22.txt" +++ /dev/null @@ -1,21650 +0,0 @@ -TITLE: Motivation for relative schemes: why should one work with schemes over a ringed topos? -QUESTION [24 upvotes]: Recently I've been trying to learn more about relative schemes. These were developed in M. Hakim's thesis Topos annelés et schémas relatifs under Grothendieck's guidance and appear in many of later works of the Grothendieck school, such as Berthelot's Cohomologie Cristalline des Schemas de Caracteristique $p>0$ or Illusie's Complexe Cotangent et Déformations I et II. - -Question I. What are some instances in which working in the full generality of a ringed topos gives one more powerful tools than just working with $S$-schemes? - -One example I'm aware of is in Illusie's Complexe Cotangent books. As remarked by Jonathan Wise in this MO question, working with ringed topoi in this setting enables one to study more interesting deformations. -I heard that a modern example might be the Falting topos, which appears in Abbes–Gros–Tsuji's book The p-adic Simpson Correspondence. (I don't really understand this example, however.) - -REPLY [14 votes]: The comments to your question discuss the variation of relative schemes over a topos, vs relative schemes over a site. But it seems your question -stood at the more basic level of the relevance of relative schemes over something else than a scheme. -In Grothendieck's philosophy, a relative scheme $f\colon X\to S$ functions as a family of schemes $X_s$ (for $s\in S$) parameterized by the base $S$, which is a scheme. Both $X$ and $S$ are schemes, and $f$ is a morphism of schemes (possibly with additional properties such as being flat, proper, finite, étale, smooth, etc.) -There are instances where one wants to consider families of schemes which are parameterized by something else, such as a (complex, Berkovich, Huber…) analytic space. For example for formulating GAGA-type theorems: what do complex analytic families of complex varieties in a given projective space look like, when the parameter set is an open disk, say. -In most important cases, those spaces are essentially characterized by rings (eg, local rings, or rings of functions over a compact, affinoid subspace) and sometimes the above study can be reduced to the study of relative schemes over these rings. Such a technique is systematically used in nonarchimedean geometry, for instance. -Nevertheless, it might be interesting to have under disposal a full-fledged theory of relative schemes over bases. -Being over a ringed topos, Monique Hakim's theory can encompass all of the above situations. -In any case, such a theory won't prove the basic (but difficult) results from commutative algebra that are probably needed. In nonarchimedean geometry, some work is needed, for example, to compare the scheme $\mathop{\rm Spec}(A)$ and the affinoid space $\mathscr M(A)$, when $A$ is an affinoid algebra, and similarly for a relative family $X\to \mathop{\rm Spec}(A)$ and its analytification $X^{\mathrm {an}}\to \mathscr M(A)$.<|endoftext|> -TITLE: A question about definable elements in a model of ZFC -QUESTION [6 upvotes]: Let $\langle M,E\rangle$ be a model of $\mathsf{ZFC}$. -Does there exist a $d\in M$ such that, for all $a\in M$, $a\mathrel{E}d$ if and only if $a$ is definable in $\langle M,E\rangle$ without parameters? -A result of J.D. Hamkins, etc. (cf. Pointwise definable models of set theory) shows that, for some models of $\mathsf{ZFC}$, such a $d$ cannot exist. Is there a model in which such a $d$ exists? - -REPLY [2 votes]: For definiteness we say that $a$ is definable in a model $\langle M,E\rangle$ of ZF if there is a formula $\varphi$ with one free variable in the language with one non-logical symbol $\in$ with the property that $\langle M,E\rangle\models\phi(x)$ iff $x=a$. (The Gödel number of such a formula must be standard.) -The condition given by Joel David Hamkins in the last sentence of (v) -of his answer is necessary and sufficient for there to be a set in $M$ consisting of the definable elements of $M$: - -Theorem. For any model $\langle M,E\rangle\models\mathrm{ZF}$, the following are equivalent: - -There exists a $d\in M$ such that for all $a\in M$, $a\mathrel{E}d$ if and only if $a$ is definable in $\langle M,E\rangle$. - -$M$ is an $\omega$-model, and there is $N\in M$ such that $N\prec M$. - - - -Proof: -1 → 2: Suppose $d\in M$, and for all $a\in M$, $a\mathrel{E}d$ if and only if $a$ is definable in $\langle M,E\rangle$. - -Claim. Let $M\models\text{“$\alpha$ is the least ordinal not in $d$”}$. Then $V_\alpha^M\prec M$. - -Proof: Write $V_\alpha$ for $V_\alpha^M$. We prove -$$\forall p_1,\dots,p_n\in V_\alpha\:\bigl(V_\alpha\models\phi(p_1,\dots,p_n)\iff M\models\phi(p_1,\dots,p_n)\bigr)\tag1$$ -by induction on the complexity of $\phi$. For atomic formulas $\phi$, (1) holds. -Suppose that (1) holds for $\phi$ and $\psi$. Then -$$\begin{align*} -V_\alpha\models\phi\land\psi& -\iff(V_\alpha\models\phi\text{ and }V_\alpha\models\psi)\\ -&\iff(M\models\phi\text{ and }M\models\psi) -\iff M\models\phi\land\psi. -\end{align*}$$ -Also, -$$V_\alpha\models\neg\phi\iff ¬V_\alpha\models\phi\iff ¬M\models\phi\iff M\models\neg\phi.$$ -Suppose (1) holds for $\theta(x,\dots)$. If $V_\alpha\models\exists x\,\theta(x,p_1,\dots,p_n)$, then $V_\alpha\models\theta(p,p_1,\dots,p_n)$ for some $p\mathrel{E}V_\alpha$. Therefore $M\models\theta(p,p_1,\dots,p_n)$ and so $M\models\exists x\,\theta(x,p_1,\dots,p_n)$. -Finally, suppose $M\models\exists x\,\theta(x,p_1,\dots,p_n)$ for $p_i$ in $V_\alpha$. Then there is an ordinal $\delta$ in $d$ such that $\{p_1,\dots,p_n\}$ is contained in $V_\delta$. $\beta\mathrel{E}d$ must hold for the least ordinal $\beta$ such that $\delta\mathrel{E}\beta$, $V_\beta$ reflects $\exists x\,\theta$, and $V_\beta$ reflects $\theta$. Therefore $M\models\theta(p,p_1,\dots,p_n)$ for some $p\mathrel{E}V_\beta$, and consequently $V_\alpha\models\theta(p,p_1,\dots,p_n)$, and thus $V_\alpha\models\exists x\,\theta(x,p_1,\dots,p_n)$. -This completes the proof of the Claim. It remains to prove that $M$ is an $\omega$-model. -For each $a\mathrel{E}d$, let $g_a$ be the least Gödel number of a formula which defines $a$ in $V_\alpha$. There is a set $S\in M$ consisting of the $g_a$ for which $a\mathrel{E}d$. Since $S$ is an infinite set of standard natural numbers, its union must be $\omega$. Thus, $\omega\in M$, and $M$ is an $\omega$-model. -2 → 1: Suppose $M$ is an $\omega$-model, $N\in M$ and $N\prec M$. There is a $d\in M$ such that $b\mathrel{E}d$ iff $b\mathrel{E}N$ and -$M\models\text{“$b$ is definable in $N$”}$. Then $b\mathrel{E}d$ if and only if $b$ is definable in $\langle M,E\rangle$.<|endoftext|> -TITLE: Determinantal identities for perfect complexes -QUESTION [17 upvotes]: Let $S$ be a noetherian scheme. Let $V,W$ be vector bundles on $S$. There is a canonical isomorphism of line bundles -$$ -{\rm det}(V\otimes W)\cong{\rm det}(V)^{\otimes{\rm rk}(W)}\otimes{\rm det}(W)^{\otimes{{\rm rk} V}}\,\,\,\,(\ast) -$$ -which is invariant under any base change. See eg -https://math.stackexchange.com/questions/571839/determinant-of-a-tensor-product-of-two-vector-bundles/571906 -for this. -There are similar identities for other tensor operations, eg -$$ -{\rm det}(\Lambda^2(W))\cong{\rm det}(W)^{\otimes({\rm rk}(W)-1)} -$$ -My question is: are there similar identities for perfect complexes in place of $V,W$ ? -Recall that an object in the derived category of ${\cal O}_S$-modules is called perfect if is Zariski locally -isomorphic to a bounded complex of vector bundles. -One can show that if $V^\bullet$ and $W^\bullet$ are bounded complexes of vector bundles (on all of $S$) -then there is an isomorphism -$$ -{\rm det}(V^\bullet\otimes W^\bullet)\cong{\rm det}(V^\bullet)^{\otimes{\rm rk}(W^\bullet)}\otimes{\rm det}(W^\bullet)^{\otimes{{\rm rk}(V^{\bullet})}}\,\,\,\,(\ast\ast) -$$ -where now ${\rm det}(\cdot)$ is the Knudsen-Mumford determinant of perfect complexes. This follows from identity $(\ast)$. However the isomorphism is not canonical. In other words, I don't know how to construct an isomorphism $(\ast\ast)$, which is functorial for isomorphisms in the derived category (or more concretely, for quasi-isomorphisms of bounded complexes of vector bundles on $S$). -In particular, I don't know whether there is an isomorphism $(\ast\ast)$ (even a non canonical one) when $V^\bullet$ and $W^\bullet$ are only assumed to be perfect. -I would be grateful if anyone could share ideas, or direct me to references on this kind of problem. -I am aware of Deligne's work on Picard categories and axiomatic descriptions of determinants but this seems to be of little help. One could try to prove an identity like $(\ast\ast)$ by showing that both sides satisfy the axiomatic properties of determinants (fixing $V^\bullet$ or $W^\bullet$) but such a verification seems difficult and tedious. Another way to proceed might be to write down an isomorphism $(\ast\ast)$ applying $(\ast)$ term by term and to verify functoriality for quasi-isomorphisms directly on the definition of the functoriality of the Knudsen-Mumford determinant but this again is difficult because this functoriality is defined in a very indirect way (see proof of Th. 1 in the paper of Knudsen-Mumford https://www.mscand.dk/article/view/11642 or -How to write down the determinant of a quasi-isomorphism? -). One would expect all the -determinantal identities that are valid for vector bundles to be valid automatically for perfect complexes. There should be a way to show this. - -REPLY [12 votes]: The formula also holds for perfect complexes. This can be deduced from the case of vector bundles, although it requires a lot of structure in that case. Namely, we need to use the fact that the determinant of vector bundles can be promoted to a morphism of $E_\infty$-semirings in algebraic stacks -$$ -\det \colon \mathrm{Vect}\to \mathrm{Pic}^\mathbb{Z}. -$$ -Here, $\mathrm{Vect}$ is the stack of vector bundles, with the ring structure given by $\oplus$ and $\otimes$, and $\mathrm{Pic}^\mathbb{Z}$ is the stack of pairs $(L,n)$ where $L$ is a line bundle and $n$ a locally constant integer. The additive structure on $\mathrm{Pic}^\mathbb{Z}$ comes from the fact that it is the $\infty$-groupoid of invertible objects in the symmetric monoidal stable $\infty$-category $\mathrm{QCoh}(S)$ (or more explicitly, $(L,n)+(L',n') = (L\otimes L', n+n')$ with a sign $(-1)^{nn'}$ in the braiding). The multiplicative structure is more subtle: abstractly it is a square-zero extension of the constant sheaf of rings $\underline{\mathbb Z}$ classified by a certain derivation $\underline{\mathbb Z} \to B\mathrm{Pic}$ induced by the sign map $\pi_1\underline{\mathbb S}=\underline{\mathbb Z}/2 \to \mathbb G_m$. -Once we have this morphism $\det$, it factors through the Zariski sheafification of the group completion of $\mathrm{Vect}$, which coincides with the Zariski sheafification of algebraic K-theory, so we get a morphism of $E_\infty$-rings -$$ -\det \colon K\to \mathrm{Pic}^\mathbb{Z}, -$$ -giving in particular the desired formula for $\det(P\otimes Q)$ for $P,Q\in \mathrm{Perf}(S)$. (Note: this works also for perfect complexes on algebraic stacks, since $\mathrm{Pic}^\mathbb{Z}$ is an fpqc sheaf.) -In principle it is possible to construct the above-mentioned structure by hand, since everything takes place in the 2-category of presheaves of groupoids, but the details are probably quite tedious. Fortunately there is also a very easy way to construct all this structure: the stack $\mathrm{Pic}^\mathbb{Z}$ is simply the $1$-truncation of $K$ as a Zariski sheaf, and $\det$ is the canonical map to the truncation. Indeed, if $R$ is a local ring, the determinant map $\det\colon K(R) \to \mathrm{Pic}^\mathbb{Z}(R)$ exhibits $\mathrm{Pic}^\mathbb{Z}(R)$ as the $1$-truncation of $K(R)$, since it is an isomorphism on $\pi_0$ and $\pi_1$. Since $1$-truncation preserves finite products, it preserves $E_\infty$-ring structures.<|endoftext|> -TITLE: Exact subcategory with trivial Grothendieck group: what are the consequences and examples -QUESTION [5 upvotes]: Let $C$ be (a full) exact subcategory of the category of $R$-modules. We suppose that $C$ is essentially small. If the Grothendieck group $K_{0}(C)=0$, what can be said about the higher groups $K_{n}(C)=0$ ? Is there a non-trivial example of such exact subcategory $C$ ? - -REPLY [6 votes]: There is almost nothing one can deduce about the higher K-groups from just knowing $K_{0}(C)=0$. -As was pointed out in the comments, the restriction to modules over a general ring does not really restrict anything. Any exact category can (as explained) via the Quillen embedding be realized as an extension-closed subcategory of an abelian category and for the latter use Freyd-Mitchell (exactly as explained in the comments). -For producing nearly arbitrary examples: Let D be any idempotent complete exact category, e.g. any abelian category. Take (for example) the Tate category $\underleftrightarrow{\lim}D$ (an alternative notation is $\operatorname{Tate}(C)$) of Sho Saito's paper "On Previdi's delooping conjecture" (https://arxiv.org/abs/1203.0831). This is an exact category. By the previous remarks, it can be realized as a fully exact subcategory of a category of modules over a ring. -As proved in that paper, the nonconnective K-theory just shifts by one degree upwards (Theorem 1.2 of that paper). Since D was idempotent complete, the nonconnective K-theory will agree with usual Algebraic K-theory. -A direct computation will show that $K_0$ of this Tate category vanishes. So whatever K-groups you can find occurring in any idempotent complete exact category, you can make them appear (with a shift by one) in higher K-groups while simultaneously making $K_0$ vanish. -Such ideas were necessary to define nonconnective K-theory for exact categories in the first place, so you can alternatively find out about such constructs in Schlichting's paper "Delooping the K-theory of exact categories". -One does not have to use these Tate category constructions. There are others, e.g. the so-called Calkin category Ind(D)/D, which was used by Drinfeld in his paper on infinite-dimensional vector bundles (https://arxiv.org/pdf/math/0309155.pdf), see Section 3.3.1. The proofs for all these constructions rely on variants of Eilenberg swindles by the way, i.e. a machine to really make all K-groups vanish. One then re-assembles these contractible K-theory spectra in a way such that they become the loop space of an arbitrary input K-theory spectrum. That's the rough idea. In order to not having to do the "re-assembling" by hand (it's tough to exhibit bicartesian squares from scratch) one tries to make localization sequences "become" the squares one needs. -Finally, if your category $D$ happens to have a projective generator, you can also find such for the Tate category above (https://arxiv.org/pdf/1508.07880.pdf, Theorem 1.(2) for n=1). Thus, you can even realize all these examples simply as the projective modules over a certain ring $R$. This produces counter-examples even avoiding the Frey-Mitchell and Quillen embedding steps and one can reasonably explicitly describe these rings. By work of Wagoner, also various infinite matrix rings have these properties.<|endoftext|> -TITLE: Linear algebra underlying quantum entanglement? -QUESTION [12 upvotes]: Hope this question is appropriate. I think I saw certain claims that quantum entanglement is a certain phenomena that can be explained (or modelled) in terms of tensor products in linear algebra. I wonder if this is the case, and if yes, is there some nice mathematical source? If you have your own insight in the question, I would be very happy to learn about it. -I am asking the question because want to mention it in an undergraduate course on representation theory to cheer up students. -PS. Since the proposed suggestions are mainly books (or physics literature), I start to suspect that what I was looking for doesn't exist. I guess, I wanted some short piece of text (say 1-20 pages long), that would be additionally purely mathematical. Basically, some compression of information is needed. How to make a 5 minutes talk out of 10 books? - -REPLY [4 votes]: A concise and a bit more mathy review is that by M Keyl -Fundamentals of quantum information theory which is based on a $C^*$ algebra approach.<|endoftext|> -TITLE: Can every manifold be dominated by a parallelizable one? -QUESTION [19 upvotes]: A closed, oriented $d$-manifold $M$ is said to dominate another such manifold $N$ if there exists a map $M \to N$ of non-zero degree. (This notion should not be confused with the unrelated concept of homotopical domination that Wall's finiteness obstruction relates to.) -This relation turns $d$-manifolds into a poset (not quite; the relation is not antisymmetric even if we identify diffeomorphic manifolds, as there are examples of manifolds that dominate each other and are not homotopy equivalent; but this subtlety is rather irrelevant) with $S^d$ as its smallest element. It has been of interest to geometric topologists at least since the times of Hopf to study this poset. For a survey on this topic, see http://www.map.mpim-bonn.mpg.de/images/c/cf/Brouwer_degree.pdf. -My question is: can every manifold $N$ be dominated by another manifold $M$ that is (stably) parallelizable, i.e., has trivial stable tangent bundle? - -REPLY [10 votes]: I think this follows from the stable Hurewicz map $\pi_n^s(N) \to H_n(N;\mathbb{Z})$ being an isomorphism modulo torsion. If we set $n = \dim(M)$ then some positive multiple of the fundamental class of $M$ is hit. Under the Pontryagin-Thom isomorphism this gives a stably framed $M$ and a continuous $f: M \to N$ such that $f_*([M])$ is a positive multiple of $[N]$.<|endoftext|> -TITLE: How often do you put your research into trash? -QUESTION [9 upvotes]: A soft question. -I am a PhD student, at early stages of my academic career; and have personally experienced the following many times. Sometimes you come up with a result, that you are not quite happy with (for whatever reason, maybe you think is not novel, or is simple, or has no applications, etc.)? How often do you put your research into trash? -Yet another question, I have seen people publishing sometimes papers that they don't feel quite happy about publishing, but every now and then, these papers start getting lots of citations, and being recognized. What do you think about this? Do you think unless the quality is really low you'd go ahead an post, at the very least on arXiv; or do you stick to perfectionism? To some degree, I feel like perfect is the enemy of the good! - -REPLY [14 votes]: I have some strong opinions about this: - -Math is hard and interesting. Don't be snobbish about anyone's math, not least yourself. I have found fruitful research ideas from obscure papers that no-one read, from naive questions by undergraduate students or even my children. Some vague thoughts I had years ago when revisited now become full papers in decent journals since I have more tools and insight to develop them. - -Anything can be turned into a useful experience. That's what I told my kids before any tennis session. If they have to hit with someone weaker, focus on placing the ball perfectly for that person so they can hit it back. That helps them improve their skills and the other kids too. Same for your "weak" results. Try to extend them. Put them in bigger contexts. Write a good introduction. Don't be shy to speculate and ask open questions (but make sure you put some efforts in thinking about them first). These are crucial skills for "bigger" papers too. In this day and age you can always find a non-predatory journal whose quality your note can improve, for instance those published by math societies from countries with little history of math research. Send your note there. And you have done a good thing. - -You can always put them in a file called "Answers to future MO questions"! (-:<|endoftext|> -TITLE: Does every measurable subset of $\mathbb R$ of non zero Lebesgue measure contain arbitrarily long arithmetic progressions? -QUESTION [9 upvotes]: A subset $E$ of $\mathbb R$ is said to contain arbitrarily long arithmetic progressions, if for every natural $n$, there exists $a, d \in R, d$ nonzero, such that $a + kd$ is in $E$ for all natural $k \leq n$. -Does every measurable subset of $\mathbb R$ of non zero Lebesgue measure contain arbitrarily long arithmetic progressions? - -REPLY [10 votes]: Yes. For fixed $n$, we approximate our set $E$ from above by an open set $U=\sqcup \Delta_i$ ($\Delta_i$ are disjoint intervals) with such accuracy that one of intervals $\Delta_i$ satisfies $|E\cap \Delta_i|>(1-\frac1{n+1})|\Delta_i|$, where $|\cdot|$ denotes Lebesgue measure. Now if $\Delta_i=(a,a+(n+1)t)$, we consider $n+1$ sets $E_i:=(E-it)\cap (a,a+t), i=0,1,\ldots,n$. The sum of there measures equals $|E\cap \Delta_i|>nt$, thus there exists a point covered by them all. It corresponds to an arithmetic progression inside $E$ with difference $t$ and $n+1$ terms.<|endoftext|> -TITLE: A question about the Buchsbaum-Eisenbud-Horrocks Conjecture -QUESTION [5 upvotes]: It's known that Mark E. Walker proved the "weaker" version of Buchsbaum-Eisenbud-Horrocks' Conjecture (BEH). Although the claim was stated to hold in arbitrary field $k$, Walker's proof does not seem to include the case when $k$ is of characteristic $2$. - -QUESTIONS. Has weak BEH been proved or disproved when $k$ is of $char=2$? Why does this special characteristic resist, philosophically or technically? - -REPLY [4 votes]: The issue is that the cyclic Adams operation $\psi^2_{cy}$ is not defined when the characteristic of the base field is 2. There are some notes here by Mark Walker himself that do a good job of explaining Adams operations for commutative algebraists.<|endoftext|> -TITLE: Atlas-like websites on specific areas of mathematics -QUESTION [71 upvotes]: In this post, we look for the existing atlas-like websites providing well-presented classifications or database about some specific areas of mathematics. Here are some examples: - -GroupNames: https://people.maths.bris.ac.uk/~matyd/GroupNames - - -Finite groups of order ≤500, group names, extensions, presentations, - properties and character tables. - - -Atlas of Finite Group Representations: http://brauer.maths.qmul.ac.uk/Atlas/v3/ - - -This ATLAS of Group Representations has been prepared by Robert - Wilson, Peter Walsh, Jonathan Tripp, Ibrahim Suleiman, Richard Parker, - Simon Norton, Simon Nickerson, Steve Linton, John Bray, and Rachel - Abbott (in reverse alphabetical order, because I'm fed up with always - being last!). It currently contains information (including 5215 - representations) on about 716 groups [mainly finite simple groups or almost simple]. - - -Atlas of subgroup lattice of finite almost simple groups: http://homepages.ulb.ac.be/~dleemans/atlaslat/ - - -This atlas contains all subgroup lattices of almost simple groups $G$ - such that $S≤G≤Aut(S)$ and $S$ is a simple group of order less than 1 - million appearing in the Atlas of Finite Groups by Conway et al. Some - simple groups and almost simple groups or order larger than 1 million - have also been included, but not in a systematic way. - - -The L-functions and Modular Forms Database: https://www.lmfdb.org/ - - -Welcome to the LMFDB, the database of L-functions, modular forms, and - related objects. These pages are intended to be a modern handbook - including tables, formulas, links, and references for L-functions and - their underlying objects [like field extensions and polynomial Galois groups]. - - -The Inverse Symbolic Calculator: https://isc.carma.newcastle.edu.au/ - - -The Inverse Symbolic Calculator (ISC) uses a combination of lookup - tables and integer relation algorithms in order to associate a closed - form representation with a user-defined, truncated decimal expansion - (written as a floating point expression). The lookup tables include a - substantial data set compiled by S. Plouffe both before and during his - period as an employee at CECM. - -If you know such a website on any area of mathematics, please put it as an answer (with a short description). - -REPLY [2 votes]: http://www.polychora.de/wiki/images/ has renders of cross-sections, vertex figures and their duals, and cellets for many (possibly all) of the 1849 uniform polychora known at the time it was created.<|endoftext|> -TITLE: Reference request: proofs of integrals presented in Erdélyi's *Table of Integral Transforms* -QUESTION [7 upvotes]: I have asked related questions on MSE and received no answer, so: I am looking for proofs of the integrals presented in Erdélyi's Table of Integral Transforms vol. i-ii, specifically proofs of the various gamma integrals presented on page 297-299 of volume 2, for instance: -$$\int_{\mathbb{R}} \frac{dx}{\Gamma(\alpha + x)\Gamma(\beta - x)}= \frac{2^{\alpha + \beta -2}}{\Gamma(\alpha + \beta - 1)},~\Re(\alpha + \beta)>1.$$ -That the proofs for the identities presented in the work were reviewed closely is mentioned in the preface, though I cannot find any separate references containing the proofs. -Link to e-PDFs of the work in question. - -REPLY [10 votes]: Titchmarsh’s Fourier integrals (1937, 7.6.4) has proof and attribution to Ramanujan. - -REPLY [6 votes]: Another approach appears as a comment on your question, so this is just a rip-off trying to make things tidier but surely there are other ways to Titchmarsh and this to prove it -$ -\int_{\mathbb{R}}\frac{dx}{\Gamma(\alpha+x)\Gamma(\beta-x)}\\ -=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{\mathbb{R}}\frac{\Gamma(\alpha+\beta-1)dx}{\Gamma(\alpha+x)\Gamma(\beta-x)}\\ -=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{\mathbb{R}}\frac{(\alpha+\beta-2)!dx}{(\alpha+x-1)!(\beta-x-1)!}\\ -=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{\mathbb{R}}\binom{\alpha+\beta-2}{\alpha+x-1}dx\\ -=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{\mathbb{R}}\frac{1}{2\pi i}\oint_{|z|=1}\frac{(1+z)^{\alpha+\beta-2}}{z^{\alpha+x}}dzdx\\ -=\frac{1}{\Gamma(\alpha+\beta-1)}\oint_{|z|=1}\frac{(1+z)^{\alpha+\beta-2}}{z^{\alpha}}\frac{1}{2\pi i}\int_{\mathbb{R}}z^{-x}dxdz\\ -=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{-\pi}^{\pi}\frac{(1+e^{i\theta})^{\alpha+\beta-2}}{e^{i(\alpha -1) \theta}}\frac{1}{2\pi }\int_{\mathbb{R}}e^{-i\theta x}dxd\theta\\ -=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{-\pi}^{\pi}\frac{(1+e^{i\theta})^{\alpha+\beta-2}}{e^{i(\alpha -1) \theta}}\delta(-\theta)d\theta\\ -=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{-\pi}^{\pi}\frac{(1+e^{i\theta})^{\alpha+\beta-2}}{e^{i(\alpha -1) \theta}}\delta(\theta)d\theta\\ -=\frac{1}{\Gamma(\alpha+\beta-1)}\frac{(1+e^{i0})^{\alpha+\beta-2}}{e^{i(\alpha -1) 0}}\\ -=\frac{2^{\alpha+\beta-2}}{\Gamma(\alpha+\beta-1)} -$<|endoftext|> -TITLE: Paramodular newvectors and twists -QUESTION [6 upvotes]: In the book Local Newforms for GSp(4), Roberts and Schmidt have defined a theory of "new vectors" for smooth representations of $GSp_4$ over a nonarchimedean local field $F$ with trivial central character, using the paramodular congruence subgroups. -If $\pi$ is a generic unramified representation of $GSp_4(F)$ with trivial central char, and $\chi$ is a ramified quadratic character of $F^\times$, then $\pi \otimes \chi$ again has trivial central character; so there must be some vector in $\pi$ whose image in $\pi \otimes \chi$ is the new vector of the twist. - -Can one write down an explicit element of $\mathbf{C}[GSp_4(F)]$ which sends the spherical vector of $\pi$ to the paramodular new vector of $\pi \otimes \chi$? - -For $GL_2$ it's well known that the operator $\sum_{u \in (\mathcal{O}_F / \varpi)^\times} \chi(u)^{-1} \begin{pmatrix} 1 & u/\varpi \\ 0 & 1 \end{pmatrix}$ does the job, so I'm hoping for something like this. There is a paper by Andrianov "Twisting of Siegel modular forms" which seems related, but he doesn't use the paramodular subgroups. - -REPLY [3 votes]: I've stumbled across an answer to this old question of mine so I'm going to answer it myself. The following paper: -Johnson-Leung, Jennifer; Roberts, Brooks, Twisting of paramodular vectors, Int. J. Number Theory 10, No. 4, 1043-1065 (2014). (preprint version, official journal version) -addresses this exact question, giving an explicit (but lengthy) formula for a twisting map from newvectors of $\pi$ to newvectors of $\pi \otimes \chi$ (for any paramodular $\pi$, not necessarily unramified).<|endoftext|> -TITLE: Show that $(\sum_{k=1}^{n}x_{k}\cos{k})^2+(\sum_{k=1}^{n}x_{k}\sin{k})^2\le (2+\frac{n}{4})\sum_{k=1}^{n}x^2_{k}$ -QUESTION [16 upvotes]: Let $x_{1},x_{2},\cdots,x_{n}>0$, show that -$$\left(\sum_{k=1}^{n}x_{k}\cos{k}\right)^2+\left(\sum_{k=1}^{n}x_{k}\sin{k}\right)^2\le \left(2+\dfrac{n}{4}\right)\sum_{k=1}^{n}x^2_{k}$$ - -This question was posted some time ago at MSE. A bounty was placed on it, but no complete solution was received. -The only solution there claims to solve the problem when $n \le 10^9$. -It's easy to see that the inequality can be proved when $(2+n/4)$ is replaced by $n$ (in fact, this follows directly from Cauchy-Schwarz inequality). -Also the LHS is equal to: $\sum_{k=1}^n x_k^2 + 2\sum_{i < j} x_ix_j\cos(i - j)$. -I'm looking for a proof or any reference of this result. -Any help would be appreciated. - -REPLY [2 votes]: It is possible to prove that the inequality holds for sufficiently large $n$ with $1/4$ replaced by $1/4+\epsilon$ for any $\epsilon>0$ . -(Update - see below for a stronger result conditional on the irrationality measure of $\pi$) -In fact we have: -Theorem -For any $\epsilon>0$ and large enough $n$, depending on $\epsilon$ only, the following inequality holds for any real $a_i\geq 0$ -$$|\sum_{k=1}^n a_k e^{ik}|\leq (1/2+\epsilon) \sqrt n (\sum_{k=1}^{n}a_k^2)^{1/2}.$$ -Proof -We know that $$|\sum_{k=1}^n a_k e^{ik}| = e^{ i\theta}\sum_{k=1}^n a_k e^{ik}$$ for some $\theta \in [0,2\pi]$. Hence -$$|\sum_{k=1}^n a_k e^{ik}| = \sum_{k=1}^n a_k e^{i(k+\theta)}=\sum_{k=1}^n a_k\cos(k+\theta).$$ -Define a function $ \cos_{+}:\mathbb{R}\to[0,1]$ by $\cos_{+}(x)=\cos(x)$ if $\cos(x)\geq 0$ and $0$ otherwise. -Then using $a_k\geq0$ and Cauchy's Theorem, -$$\sum_{k=1}^n a_k\cos(k+\theta)\leq\sum_{k=1}^n a_k\cos_+(k+\theta)\leq(\sum_{k=1}^n a_k^2)^{1/2}(\sum_{k=1}^n {\cos_+}^2(k+\theta))^{1/2}.$$ -Now the equidistribution theorem says that $\{\frac{i}{2\pi}\}$ is uniformly distributed in $[0,1]$. Hence by the Riemann integral criterion for equidistribution, -$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n} f\left(s_k\right) = \frac{1}{b-a}\int_a^b f(x)\,dx$$ -with $f(x)={\cos_+}^2(2\pi x+\theta)$, $s_k=\{\frac{k}{2\pi}\}$ and $a=0$, $b=1$. -In other words, -\begin{equation} -\begin{split} -& \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n {\cos_+}^2(k+\theta)=\int_0^1 {\cos_+}^2(2\pi x+\theta)\,dx =\frac{1}{2\pi}\int_{\theta}^{2\pi+\theta} {\cos_+}^2(\phi)\,d\phi \\ & =\frac{1}{2\pi}\int_{-\pi/2}^{+\pi/2} \cos^2(\phi)\,d\phi=1/4. -\end{split} -\end{equation} -Hence -$$\frac{|\sum_{k=1}^n a_k e^{ik}|}{\sqrt n (\sum_{k=1}^n a_k^2)^{1/2} }\leq \left(\frac{1}{n}\sum_{k=1}^n {\cos_+}^2(k+\theta)\right)^{1/2}\rightarrow 1/2 \ \text{ as } \ n\rightarrow \infty$$ -and the result follows. -$\blacksquare$ -In addition I can prove the following estimate, conditional on the irrationality measure of $\pi$ being less than 3: -Theorem -If the irrationality measure for $\pi$, $\mu(\pi)$ is strictly less than 3 then $$\frac{|\sum_{k=1}^n a_k e^{ik}|^2}{ \sum_{k=1}^n a_k^2 }\leq\frac{n}{4}+D.$$ -for some fixed constant $D$. -Proof -We know from the above that -$$|\sum_{k=1}^n a_k e^{ik}|^2\leq (\sum_{k=1}^n a_k^2) \sum_{k=1}^n {\cos_+}^2(k+\theta)$$ for some $\theta \in [0,2\pi]$. -Also, ${\cos_+}^2(x) = (\cos^2 x+\cos x|\cos x|)/2$. -Hence $$\sum_{k=1}^n {\cos_+}^2(k+\theta)=\sum_{k=1}^n (\cos^2 (k+\theta)+\cos (k+\theta)|\cos (k+\theta)|)/2\\=\frac{1}{2}\sum_{k=1}^n \cos^2 (k+\theta)+\frac{1}{2}\sum_{k=1}^n (\cos (k+\theta)|\cos (k+\theta)|)$$. -Clearly $\sum_{k=1}^n \cos^2 (k+\theta)\leq n/2+B$ for some constant $B$ so it remains to bound the other term which is more complicated. -Let $f(x) = |\cos{(x)}| \cos{(x)}$. Here it is noted by Andreas that the expression $|\sin{(x)}| \sin{(x)}$ can be written as a Fourier series, which we modify to provide a series for $f$, -$$ -f(x) = \frac{8}{\pi}\sum_{m=0}^\infty \frac{(-1)^m}{4(2m+1)-(2m+1)^3} \cos((2m+1)x). -$$ -Now we can sum -$$ -S_n = \sum_{k=1}^{n} f{(k + \phi)} = \frac{8}{\pi}\sum_{m=0}^\infty \frac{(-1)^n}{4(2m+1)-(2m+1)^3} \sum_{k=1}^{n}\cos((2m+1)(k + \phi)) -$$ -where -$$ -\sum_{k=1}^{n}\cos((2m+1)(k + \phi)) = -\frac{\sin(n(m + \frac12)) \cdot \cos((1 + n + 2 \phi)(m + \frac12)) }{ \sin(m + \frac12)}. -$$ -Thus -$$ -S_n = \frac{8}{\pi}\sum_{m=0}^\infty \frac{(-1)^n}{4(2m+1)-(2m+1)^3} \frac{\sin(n(m + \frac12)) \cdot \cos((1 + n + 2 \phi)(m + \frac12)) }{ \sin(m + \frac12)}. -$$ -Taking absolute values and using the triangle inequality -$$ -|S_n| \leq \frac{8}{\pi}\sum_{m=0}^\infty \frac{1}{|4(2m+1)-(2m+1)^3|\sin(m + \frac12)|}\\=\frac{8}{\pi}\sum_{m=0}^\infty \frac{|2\cos (m+\frac12)|}{|4(2m+1)-(2m+1)^3||\sin(2 m + 1)|} -\\ \leq \frac{16}{\pi}\sum_{m=0}^\infty \frac{1}{|(2m+1)^3-4(2m+1)||\sin(2 m + 1)|} -\\ =\frac{16}{3 \pi \sin 1}+\frac{16}{15 \pi \sin 3}+\frac{16}{\pi}\sum_{m=2}^\infty \frac{1}{|(2m+1)^3-4(2m+1)||\sin(2 m + 1)|} -\\ \leq \frac{16}{3 \pi \sin 1}+\frac{16}{15 \pi \sin 3}+\frac{16}{\pi}\sum_{m=2}^\infty \frac{1.3}{(2m+1)^3|\sin(2 m + 1)|} -$$ -To complete the estimate we note that Theorem 5 of Max A. Alekseyev's paper "On convergence of the Flint Hills series" implies that if $\mu(\pi)<3$ then $\sum_{n=1}^{\infty}\frac{1}{n^3|\sin n|}$ converges, hence $\sum_{n=1}^{\infty}\frac{1}{(2n+1)^3|\sin (2n+1)|}$ converges also and we have -$$|S_n| \leq C$$ for some fixed constant $C$. Combining the two estimates we have on the assumption that $\mu(\pi)<3$, -$$\sum_{k=1}^n {\cos_+}^2(k+\theta)=\frac{1}{2}\sum_{k=1}^n \cos^2 (k+\theta)+\frac{1}{2}\sum_{k=1}^n (\cos (k+\theta)|\cos (k+\theta)|)\\ \leq n/4+(B+C)/2$$ -and the result is proved. $\blacksquare$ -Unfortunately although most irrational numbers have irrationality measure 2, and this is probably the true value of $\mu(\pi)$, the best upper bound for $\mu(\pi)$ is 7.103205334137 due to Doron Zeilberger and Wadim Zudilin - see here for their paper so we are a long way from being able to prove the inequality this way at least.<|endoftext|> -TITLE: Pros and cons of specializing in an esoteric research area -QUESTION [31 upvotes]: If a mathematician specializes in a popular research area, then there are many job positions available, but at the same time, many competitors who are willing to get such job positions. For an esoteric research area, there are few competitors and job positions. There are very often pros and cons of such research areas. -What are some pros and cons of specializing in esoteric research areas that many people may not know? -Maybe it is a little hard to answer this question in full generality since circumstances vary. Hence, I especially want to listen to examples, personal experiences, and maybe urban legends. -Now, if you are interested, let me tell my personal circumstance to give some context to this question. I am a student from outside of North America who just graduated from my undergraduate institution. Since I decided to study abroad, I applied to several North American universities last December and was admitted to some of them. Now I am wavering between two universities. Denote those universities X and Y. -Among the specific research areas available at X (resp. Y), I am interested in two of them, say, A and B (resp. C and D). -I did some searching in those research areas, and I found out that there are many people who are researching C and D and some of them are in my country. However, there are only a few people interested in A and B and none of them are in my country. Based on my search, I think (with a little exaggeration) there are about 3 universities in the world where a graduate student can specialize in A. B is not as esoteric as A, but still, it seems there are not many people working on B. However, I think C and D are quite major research areas in my field of study. -[In this question, I used 'field of areas' as something in First-level areas of Mathematics Subject Classification and 'specific research areas' as something in Second or Third-level areas of it. ] -At first glance, I prefer X over Y, because I was very interested in A. Also, this is partially (maybe totally) because X is considered more `prestigious’ than Y. However, I’m a little nervous about specializing in esoteric areas such as A and B, because of the number of job positions and this kind of problem. -Anyway, I think it is not a bad idea to ask a question at MO and to listen to the pros and cons of specializing in esoteric fields to make a better decision. Any personal stories or examples will be really helpful. Thanks in advance. - -REPLY [2 votes]: I think it really depends on your goals. The OP says he's thinking about which grad school to attend. If you're goal is to get a higher degree and then a job in industry, you would be well-advised to do research in something considered "useful." That's not a pre-requisite, and I know a ton of abstract homotopy theorists who now work as data scientists, but it's also fair to say that the transition out of academia is easier if you have some skills you can showcase to companies. -If your goal is to be a professor, I think it's important to realize just how hard it is to get a job these days. The AMS publishes data about the number of PhDs and the number of people who get a tenure track academic position every year. Only about 25% of math phds get a tenure track job. Many of those who get a job in a given year are coming from a postdoc rather than directly out of grad school. When we hire, we get more than 400 applicants for a single job, so even if you do everything right, the competition might still prevent you from getting a tenure track job. Really think about this. A lot of people are jaded when they can't get a job. Don't assume it will be easy or even possible. -If you want to go to a research university, I think you'd be well advised to pick a less esoteric field, or to connect your research to new and sexy things (e.g., lots of people are getting jobs now in topological data analysis, whereas almost none are getting jobs in chromatic homotopy theory). -If you are not wedded to the idea of being a professor at a research university, then do not overlook liberal arts colleges. These jobs have a slightly higher teaching load but can still provide a lot of research support, sabbaticals, etc. The essential point for these schools is that teaching is valued more than research, so if you think you might be happy at such a place then it's essential to develop your capabilities as a teacher. Much more important than your area of research is whether or not you can teach statistics or computer science. You get that skill and you will get a job, 100%. You might consider picking up a masters in CS or stats during grad school, as that would help you get a liberal arts job, and would also help if you go to industry. I got a master's degree in CS at the same time that I was getting a PhD in math, and it really did not take a ton of effort. One other point is that liberal arts colleges increasingly want faculty to do research with undergraduate students. Their undergraduate students can be very strong (more like the level of a master's student), but it might be wise to pick a research area that's easier to get into, like graph theory or knot theory, rather than something insanely esoteric. Again, computer science, applied statistics, and data science are very fertile areas for research with undergraduates.<|endoftext|> -TITLE: Is there a compact, connected, totally path-disconnected topological group? -QUESTION [13 upvotes]: There exist homogeneous spaces such as the pseudo-arc, which are compact, connected, and totally path-disconnected. Is there a nontrivial, Hausdorff topological group with the same properties, i.e. that is compact, connected, and totally path-disconnected? What about a metrizable example? - -REPLY [18 votes]: (I'm assuming the groups to be Hausdorff to avoid the discussion degenerate into idle banter.) -The answer is yes: $\{1\}$ is such a group. -The answer to the intended question (which is probably whether there's a nontrivial such group) is no. -Andrew M. Gleason. Arcs in locally compact groups. Proc. Nat. Acad. Sci. U.S.A. 36 (1950), 663-667. Link -1st line of MR review: The author gives an outline of the proof of the following theorem: Every locally compact group which is not totally disconnected contains an arc. - -Edit: the above is for arbitrary locally compact groups, but for compact groups it's significantly easier. Indeed, if $G$ is a compact connected group, it follows from the Peter-Weyl theorem and the basic structure of compact connected Lie groups that there exists a group $H=A\times\prod_{i\in I}S_i$, where $A$ is a compact connected abelian group, and each $S_i$ is a simple, simply connected compact Lie group, and a surjective homomorphism $H\to G$ with totally disconnected kernel (this is for instance in Bourbaki, Lie, Chap 9, appendix). If $G\neq 1$, then $H\neq 1$, and then $\mathrm{Hom}(\mathbf{R},H)\neq\{1\}$ (since either $I$ is non-empty, or $A\neq 1$, and the abelian case is settled by Pontryagin duality as I mentioned in a comment. The composition map $\mathrm{Hom}(\mathbf{R},H)\to \mathrm{Hom}(\mathbf{R},G)$ being injective (because $\mathrm{Ker}(H\to G)$ is totally disconnected), one deduces $\mathrm{Hom}(\mathbf{R},G)\neq\{1\}$. -(Note: Peter-Weyl was established around 1925, and Pontryagin duality in the early 1930's; the basic structure of compact Lie groups was known before these dates; I'm not sure of an early reference for the structural result on compact connected groups but it follows easily so I guess was known to people working on Hilbert's 5th problem in the late 1940's). - -Edit 2: one of the results in Hilbert's 5th problem is that for every connected locally compact group $G$, every neighborhood of $1$ contains a compact normal subgroup $W$ such that $G/W$ is Lie. Also it was proved by Iwasawa around 1950 that every connected Lie group $G$ has a maximal compact subgroup $K$, and that such $K$ is connected. -One this is granted, one reduces from the compact case to the locally compact case as follows: let $G$ be a nontrivial connected locally compact group. Let $W$ be a compact normal subgroup such that $G/W$ is Lie. Let $K/W$ be a maximal compact subgroup of $G/W$. I claim that $K$ is connected. Granting the claim and the compact case, we're done if $K\neq 1$. Otherwise $K=1$ and hence $W=1$, so $G$ is Lie and this case is fine. -If $K$ were not connected, $K$ would have the nontrivial profinite quotient $K/K^\circ$, and hence a nontrivial finite quotient, say with kernel $K'$. Hence there exists a symmetric neighborhood $N$ of $1$ in $G$ such that $NK'\cap K=K$. Let $W'$ be a compact normal subgroup of $G$ contained in $N$, such that $G/W'$ is Lie Since $K$ is maximal compact, we have $W'\subset K$, and $K/W'$ is a maximal compact subgroup of $G/W'$, but it is not connected, contradicting Iwasawa's result.<|endoftext|> -TITLE: Is there a model-independent characterization of the gaunt strict $n$-categories amongst the weak $(\infty,n)$-categories? -QUESTION [6 upvotes]: Recall that a strict $n$-category $C$ is called gaunt if every $k$-morphism in $C$ with a weak inverse is an identity, for all $k$; let $Gaunt_n$ denote the strict 1-category of gaunt $n$-categories. Another way of saying this is that $C \in Gaunt_n$ iff $C$ has the unique right lifting property with respect to the canonical $n$-functor $E_k \to C_{k-1}$, where $E_k$ is the free $k$-equivalence and $C_{k-1}$ is the free $(k-1)$-morphism. Thus we have a characterization of the essential image of the fully faithful functor $Gaunt_n \to Cat_n^{str}$ into the strict 1-category of strict $n$-categories. -Now consider the composite inclusion $Gaunt_n \to Cat_n^{str} \to Cat_{(\infty,n)}$ into the $(\infty,1)$-category of weak $(\infty,n)$-categories. I believe this inclusion is also fully faithful; can we characterize its essential image? We can't repeat the same characterization as before, because the image of the canonical map $E_k \to C_{k-1}$ is already an equivalence in $Cat_{(\infty,n)}$. -Question: What is a (model-independent) characterization of the essential image $Gaunt_n \to Cat_{(\infty,n)}$? -I have some hope that there is a nice answer, from the following model-dependent considerations. In every model of $Cat_{(\infty,n)}$ I've thought about, the objects of $Cat_{(\infty,n)}$ are the fibrant objects of a model structure on some 1-category $\mathcal K$, defined by a (non-unique) right lifting property against the acyclic cofibrations of $\mathcal K$. It seems to me that in all cases, the following facts hold: - -The functor $Cat_n^{str} \to Cat_{(\infty,n)}$ lifts to a canonical functor $Cat_n^{str} \to \mathcal K$; -The composite functor $Gaunt_n \to Cat_n^{str} \to \mathcal K$ is fully faithful; -The essential image of $Gaunt_n \subseteq \mathcal K$ can be characterized as the objects which satisfy the stronger unique right lifting property against the acyclic cofibrations of $\mathcal K$. - -To flesh out (3) a bit, the lifting properties for fibrant objects in these model structures are generally characterized by "horn fillers" and "univalence maps" (a.k.a. "completeness" or "Rezk" or "2-out-of-6" maps). Unique lifting against the horn fillers seems to generally pick out objects which can be thought of as strict $n$-categories presented via a sort of "naive nerve" which doesn't quite handle equivalences appropriately; if in addition an object lifts against the univalence maps, it is forced to be gaunt (and in this case the "naive nerve" coincides with the "genuine nerve"). For example, if the Duskin nerve of a 2-category is univalent, then the 2-category is gaunt. -I find it striking that the model-dependent description of the essential image $Gaunt_n \subseteq \mathcal K$ seems to always take the same form across models $\mathcal K$, and I'm wondering if these parallel characterizations are really model-dependent avatars of something which can be said model-independently. - -REPLY [6 votes]: Alexander Campbell's guess is correct. -Here is a reference. -Lemma 10.2 of this paper - -Clark Barwick, Christopher Schommer-Pries, On the Unicity of the Homotopy Theory of Higher Categories, arXiv:1112.0040 - -shows that $Gaunt_n \simeq \tau_{\leq 0} Cat_{(\infty,n)}$. That is to say they are precisely the $(\infty,n)$-categories $G$ with the property that the space $Map(C,G)$ is homotopically discrete for all $C$. -They can also be described as the localization of $Cat_{(\infty,n)}$ at the single morphism $S^1 \times C_n \to C_n$, where $C_n$ is the free-walking $n$-cell, $S^1$ is the circle, and the map is projection. This description is model independent (for example the $n$-cell can be characterized model independently as in the proof of Lemma 4.8 in the same paper above), however it is easiest to check that this description is correct in a particular model such as Rezk's $\Theta_n$-spaces.<|endoftext|> -TITLE: Simplest diophantine equation with open solvability -QUESTION [18 upvotes]: What is the simplest diophantine equation for which we (collectively) don't know whether it has any solutions? I'm aware of many simple ones where we don't know (whether we know) all the solutions, but all of these that I know have some solution. -Yes, I know that "simplest" is subjective. I'd be satisfied if it could be typeset in one line in a LaTeX document. Also, it would be nice if it could be easily memorized (no seven-digit numbers with nonobvious patterns:) though that's secondary. - -REPLY [12 votes]: It's more complicated than the other answers by MattF and DanielLoughran, but the Erdős–Straus conjecture states that for every integer $n \ge 2$, there exist positive integers $x, y, z$ such that -$$\frac4n = \frac1x + \frac1y + \frac1z$$ -Again, this is a family of Diophantine equations, related to Egyptian fractions, and computer research has verified the conjecture holds for all $n$ up to a rather large number, but whether it holds for all $n$ ($\ge 2$) is an open problem.<|endoftext|> -TITLE: On a statistic for permutations -QUESTION [7 upvotes]: Given a permutation $\pi$ we can write $\pi=s_{i_1} ... s_{i_l}$ as a product of simple transpositions $s_i=(i,i+1)$ in a minimal way. - -Question 1: Is there an "official" name for the permutation statistic given by the cardinality of the set $\{i_1,...,i_l \}$ ? (such as Coxeter length seems to be the official name for $l$) - -See http://www.findstat.org/StatisticsDatabase/St000019 for this statistic. - -Question 2: Is there a reference that the generating function for this permutation statistic restricted to 321-avoiding permutations is given by Catalan's triangle: http://oeis.org/A009766 ? - -REPLY [2 votes]: For Q1, the sequence corresponding to Stat19 is given in A263484 and comes from a 2005 article by Richard Stanley (developing ideas from Comtet) on connectivity sets, connectedness, cut points, etc. -It may not have a standard name because it is the reflection of the decomposition/block number Stat56, i.e., for $\pi \in S_n$, $\text{Stat19}(\pi) = n - \text{Stat56}(\pi)$. See A059438. Another related statistic is global ascents Stat234. -For Q2, the connection between block numbers of 321-avoiding permutations and Catalan's triangle was established by Adin, Bagno, and Roichman, arXiv 1611.06979, later J. Algebraic Combinatorics. The related abstract for Bagno's talk at the Permutation Patterns 2017 conference is more expository.<|endoftext|> -TITLE: Is there an area-preserving diffeomorphism of the disk which is nowhere conformal? -QUESTION [16 upvotes]: This question is a cross-post. Let $D \subseteq \mathbb{R}^2$ be the closed unit disk. - -Does there exist a smooth area-preserving diffeomorphism $f:D \to D$ that does not have conformal points? - -I want the singular values $\sigma_i$ of $df$ to be everywhere distinct, and $\det(df)=1$. - -It is proven here that for any map $f:D \to \mathbb R^2$, the condition of being with distinct singular values is 'generic' in the following sense: There exist $f_n \in C^{\infty}(D, \mathbb{R}^2)$ such that $f_n \to f$ in $W^{1,2}(D, \mathbb{R}^2)$ and $df_n$ has distinct singular values everywhere on $D$. -It seems to me that this approximation procedure, applied to a map $f \in \text{Diff}(D)$, does not guarantee that the $f_n$ will map $D$ into $D$, let alone be diffeomorphisms. (e.g. I think that the convergence $f_n \to f$ cannot be made uniform in general). However, perhaps this genericity phenomena can still be used somehow. - -This answer provides the following example for a one-parameter family of such diffeomorphisms $D\setminus \{0\} \to D \setminus \{0\}$: -$$f_c: (r,\theta)\to (r,\theta+c\log r).$$ This description is given in terms of polar coordinates both in the domain and the range. For each non-zero $c ֿ\in \mathbb R$ we get a diffeomorphism, with fixed distinct singular values whose product is $1$. - -Edit: -Can we answer the infinitesimal version of the question? That is, let $f_t$ be a smooth family of area-preserving diffeomorphisms. Does each $f_t$ has a conformal point? This answer treats the "formally infinitesimal" case. - -REPLY [3 votes]: Let me try to prove an infinitesimal version where an area preserving diffeomorphism is replaced by a Hamiltonian vector field. -Let H be a function in the unit disc, constant on the boundary. Consider the Hamiltonian vector field $(H_y,-H_x)$ and the respective infinitesimal diffeomorphism -$(x,y) \mapsto (x+\epsilon H_y, y-\epsilon H_x).$ -When is it conformal at a point? When the Jacobian is a dilation. That is, if the matrix is -$\begin{pmatrix} -a&b\\ -c&d -\end{pmatrix}$, -then $a=d, b+c=0$. In our case, the Jacobian is -$\begin{pmatrix} -1+\epsilon H_{xy}& \epsilon H_{yy}\\ --\epsilon H_{xx}& 1-\epsilon H_{xy}. -\end{pmatrix}$. -Hence we want to find solutions of -$H_{xx}=H_{yy}, H_{xy}=0.$ -Consider the vector field $V:=(H_{xx}-H_{yy}, 2H_{xy})$. If it has zero on the boundary, we are done. If not, let's calculate it in polar coordinates $(\alpha,r)$ on the boundary circle $r=1$. Chain rule calculations simplify by the fact that, on the boundary, $H_\alpha=0$. My calculation yields: -$V = H_{rr} (\cos 2\alpha, \sin 2\alpha).$ -The index of this field equals 2, so there are two zeroes, multiplicities counted, inside the disc.<|endoftext|> -TITLE: schemes having same reduced underlying space and same cotangent sheaf are isomorphic? -QUESTION [5 upvotes]: Let $X,Y$ be two closed subschemes of $\mathbb{A}^n_{\mathbb{C}}$ for some fixed $n$. Assume that I have an isomorphism of the underlying reduced spaces: -$$f : X_\text{red} \stackrel{\sim}\longrightarrow Y_\text{red}$$ which induces an isomorphism of $\mathcal{O}_{X_\text{red}}$-module: -$$ f^{*} \left(\Omega_{Y/\mathbb{C}} \otimes \mathcal{O}_{Y_\text{red}} \right) \stackrel{\sim}\longrightarrow \Omega_{X/\mathbb{C}} \otimes \mathcal{O}_{X_\text{red}}.$$ -Can I lift $f$ to an actual isomorphism $X \stackrel{\sim}\longrightarrow Y$? Note that in the situation I am interested $Y$ is very singular. If $Y$ were smooth, I guess the theory of infinitesimal liftings would give the answer. -This is related to this question. Indeed, the existence of such a lift would guarantee the isomorphism between the rings introduced by the OP in that other question. - -REPLY [14 votes]: Consider the simplest example: -$$ -X = \mathrm{Spec}(\mathbb{C}[\epsilon]/\epsilon^2), \qquad -Y = \mathrm{Spec}(\mathbb{C}[\epsilon]/\epsilon^3). -$$ -Definitely, $X_{\mathrm{red}} \cong Y_{\mathrm{red}}$. Also, a simple computation shows that -$$ -\Omega_{X/\mathbb{C}} \cong \mathbb{C}, -\qquad -\Omega_{Y/\mathbb{C}} \cong \mathbb{C}[\epsilon]/\epsilon^2, -$$ -hence $\Omega_{X/\mathbb{C}} \otimes \mathcal{O}_{X_{\mathrm{red}}} \cong \Omega_{Y/\mathbb{C}} \otimes \mathcal{O}_{Y_{\mathrm{red}}}$, but of course $X \not\cong Y$.<|endoftext|> -TITLE: Matrix-valued periodic Fibonacci polynomials -QUESTION [5 upvotes]: Consider the Fibonacci polynomials $f_n(x)$, defined by the recursion $f_n(x)=xf_{n-1}(x)-f_{n-2}(x)$ with initial values $f_0(x)=0$ and $f_1(x)=1$. It is well known that the values of these polynomials - are periodic with period $6$ for $x=1$ and $x=-1.$ -There are also some matrices $x$ with integer coefficients for which the sequence $f_n(x)$ is periodic. -For example let $$x=g_k=\left(g(i,j)\right)_{i,j=0}^{k-1}$$ be the matrix with entries $g(i,j)=1$ if $|i-j|=1$ or $i=j=0$ and $g(i,j)=0$ else. -Then the sequence $f_n(x)$ is periodic with period $2(2k+1).$ -Let for example $k=2$. Here we get a sequence with period $10:$ -$\left ( \begin{matrix} 0 & 0 \\ 0 & 0 - \end{matrix} \right )$, $\left ( \begin{matrix} 1 & 0 \\ 0 & 1 - \end{matrix} \right )$, $\left ( \begin{matrix} 1 & 1 \\ 1 & 0 - \end{matrix} \right )$, $\left ( \begin{matrix} 1 & 1 \\ 1 & 0 - \end{matrix} \right )$, $\left ( \begin{matrix} 1 & 0 \\ 0 & 1 - \end{matrix} \right )$, $\left ( \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right )$, $\left ( \begin{matrix} -1 & 0 \\ 0 & -1 - \end{matrix} \right )$, $\left ( \begin{matrix} -1 & -1 \\ -1 & 0 - \end{matrix} \right )$, $\left ( \begin{matrix} -1 & -1 \\ -1 & 0 - \end{matrix} \right )$, $\left ( \begin{matrix} -1 & 0 \\ 0 &-1 - \end{matrix} \right )$, $\left ( \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right )$ -,$\dots$. -There are also some other matrices $x$ with integer coefficients such that the sequence $f_n(x)$ is periodic. I would be interested to find all such matrices with integer coefficients. Is there anything in the literature? - -REPLY [3 votes]: Up to a change of basis, we can assume the matrix is in Jordan form $M = \bigoplus J_i$. It is known that for every polynomial $p$ one has -$$ -p(J_i) = -p(\begin{bmatrix} -\lambda & 1\\ -& \lambda & 1\\ -& & \ddots & \ddots\\ -& & & \lambda & 1\\ -& & & & \lambda\\ -\end{bmatrix}) -$$ -$$ -\begin{bmatrix} -p(\lambda) & p'(\lambda) & \frac{p''(\lambda)}{2} & \dots & \frac{p^{(n-1)}(\lambda)}{n!}\\ -& p(\lambda) & p'(\lambda) & \dots & \frac{p^{(n-2)}(\lambda)}{(n-1)!}\\ -& & \ddots & \ddots &\vdots\\ -& & & p(\lambda) & p'(\lambda)\\ -&&&&p(\lambda) -\end{bmatrix}. -$$ -In particular, periodicity depends on periodicity of the sequence on the eigenvalues of the matrix, and on the sequence of derivatives in case there are nontrivial Jordan blocks. -So if the matrix is diagonalizable, its period is the lcm of all periods of the eigenvalues. If it has Jordan blocks, you have to work out periods of derivatives of the sequence as well. -So you can reduce your problem to a scalar one. Have you solved that completely? Do you know if there are other periodic values other than $\pm 1$ ?<|endoftext|> -TITLE: Constant curvature difference surfaces -QUESTION [5 upvotes]: Surfaces of Constant Mean Curvature CMC and constant Gauss Curvature Product $K$ are well-known in differential geometry of surfaces of revolution. Denote half sum and half difference curvatures as -$$ k_1+ k_2 = 2 H_s; \, k_1- k_2 = 2 \, H_d. $$ -Significance of $H_d$ -Now I assume that $H_d$ may as well be of equal interest with fundamental importance. The assumption is motivated by its representation in Mohr's Circle radius of curvature for stress, shell curvatures, moment of inertia (in Figure) among other such tensors. As is known, material stress failure theories in structural mechanics operate on stress/strain and other tensoral differences as shear entities. It occurs as an important curvature invariant in the equation of Mohr's Circle : -$$\boxed{ (k_n-H_s)^2 +\tau_g^2= H_d^2} $$ - -I made a search in some textbooks that could be accessed. I was not lucky in finding references for the principal curvature difference profiles. Also the surface cannot be viewed as a particular case of CMC surfaces due to its sign change of curvature and so it is distinctly different. The calculation has a different expression and plots differently. -If $\phi$ is slope of tangent to meridian, primes on meridian arc -Const $H=H_s$ Mean curvature Delaunay meridians -$$\phi^{'}+\frac{\cos \phi}{r} =2 H_s$$ -With initial radius $r=r_1$ at $\phi=0$ -$$ \cos \phi = \frac{H_s (r^2-r_1^2)+r_1}{r}$$ -$$(-k_1,k_2)= (H_s(r^2+r_1^2)-r_1,\, H_s(r^2-r_1^2)+r_1)\,$$ -Const $H_d$ Difference curvature meridians -First Order ODE: -$$-\phi^{'}+\frac{\cos \phi}{r} = 2 H_d \tag1$$ -First order ODE has the disadvantage of numerically computing indefinitely below points of singularity on $r=0$ axis so there appear multiple spindle meridian profiles below this symmetry axis. -Second Order ODE: -$$\phi^{''}= 2 H_d \tan\phi\, (\phi^{'} +2 H_d) \tag2 $$ -$$ \cos \phi = \frac{r}{r_1}+2\, H_d\, r\, log \,\frac{r}{r_1}\tag3 $$ -$$(k_1,k_2)=(\frac{1}{r_1}+2H_d(1+log \,\frac{r}{r_1}),\frac{1}{r_1}+2H_d \,log \,\frac{r}{r_1} ) \tag4 $$ -$$ @\,r=0,\phi \rightarrow \pi/2$$ -Second order ODE has the advantage of stopping computation at point of singularity so there appears no profile below symmetry line $r= 0.$ -Shown below are profiles of constant difference $H_d$ of principal curvatures. -Three distinct shapes occur. -Progressive loops $ H_d <-0.5$ above $r=1$ ; -Two types below $r=1$ -Ovaloids between cylinder and sphere $ 0>H_d>-0.5;$ and, -Profiles with Inflection Point for $ H_d >0 $ occurring at $ r=r_1e^{-(1+1/(2 r_1H_d))}. $ -All profiles meet the axis of symmetry normally, however these are not umbilical points. -$ H_d$ Unduloids" /> -Thanks in advance for your comments and for any references available on the topic. - -REPLY [7 votes]: I see a somewhat different description of the surfaces of revolution with $H^2-K$ equal to a positive constant, say $H^2-K=\delta^2$, where $\delta>0$ is constant. Note that scaling a surface by a constant $\lambda\not=0$ produces a new surface with $H^2-K$ divided by $\lambda^2$, so the actual magnitute of $\delta>0$ is not important for the possible shapes of such surfaces. Thus, your pictures strike me as somewhat misleading. -Instead, the key factor determining the shape of the corresponding surface of revolution is another constant, which can be described as follows: First, let $\bigl(f(s),g(s)\bigr)$ be a unit speed curve in the plane, and consider the surface of revolution parametrized by -$$ -X(s,\theta) = \bigl(f(s)\,\cos\theta,\ f(s)\,\sin\theta,\ g(s)\bigr). -$$ -One easily computes that the principal curvatures of this surface are -$$ -\kappa_1 = g'(s)f''(s)-f'(s)g''(s)\quad\text{and}\quad \kappa_2 = -g'(s)/f(s), -$$ -where, of course, $f'(s)^2+g'(s)^2 = 1$. Then the equation we want to study is -$$ -g'f''-f'g'' + g'/f = 2\delta. -$$ -Using the relation $f'f''+g'g'' = 0$, this becomes the relations -$$ -f'' = g'(2\delta-g'/f)\quad\text{and}\quad g'' = -f'(2\delta-g'/f). -$$ -Scaling the surface by $\lambda$ induces the scaling -$$ -(f,g,s,\delta) \mapsto (\lambda f,\lambda g, \lambda s, \delta/\lambda), -$$ -so we can reduce to the case $\delta=\tfrac12$ without loss of generality. Set $h=g'$ -and we have the equation $h' = -f'(1-h/f)$ or $fh'-hf'+ff'=0$, which can be made exact by dividing by $f^2$, yielding the first integral -$$ -\frac{h}{f} + \log|f| = C, -$$ -so $h = f(C-\log|f|)$, and, since $(f')^2 = 1-h^2$, we must have $|f|(C-\log|f|) \le 1$. In particular, we have -$$ -\frac{(\mathrm{d}f)^2}{\bigl(1-f^2(C-\log|f|)^2\bigr)} = (\mathrm{d}s)^2, -$$ -and, since $(\mathrm{d}g)^2 = h^2\,(\mathrm{d}s)^2$, we arrive at the relation -$$ -\bigl(1-f^2(C-\log|f|)^2\bigr)\,(\mathrm{d}g)^2 - f^2(C-\log|f|)^2\,(\mathrm{d}f)^2 = 0 -$$ -It is the constant $C$ that determines the shape of the resulting profile curve and hence the surface. Obviously, the null curves of this quadratic form have to lie inside the region where $f^2(C-\log|f|)^2 \le 1$, which are bounded by lines defined by -$|f|(C-\log|f|) = \pm1$. -For example, when $C>1$, there are three functions $\rho_i$ on $(1,\infty)$ such that -$$ -0 < \rho_1(C) < \mathrm{e}^{C-1} < \rho_2(C) < \mathrm{e}^C < \rho_3(C) -$$ -such that the function $1-f^2(C-\log|f|)^2$ is nonnegative on the three intervals, -$(-\rho_3(C),-\rho_2(C))$, $(-\rho_1(C),\rho_1(C))$, and $(\rho_2(C),\rho_3(C))$. Consequently, there wil be two distinct solutions $(f,g)$: One where $f$ and $g$ are periodic and $|f|$ is bounded by $\rho_1(C)$, and another where $f$ is periodic, oscillating between a minimum of $\rho_2(C)$ and a maximum of $\rho_3(C)$. -Meanwhile, when $C<1$, there is a single function of $C$, say $\rho_4(C)>\mathrm{e}^C$ such that the function $1-f^2(C-\log|f|)^2$ is nonnegative on the interval $(-\rho_4(C),\rho_4(C))$, and there will be a single solution (up to translation and reflection in $g$) for which $f$ lies in this interval. -There is a similar story for $C=1$, except that there is an exact solution, $(f,g)=(1,s)$ (the cylinder of radius $1$) and then two solutions, one with $0 -TITLE: Common basis for permutation matrices -QUESTION [12 upvotes]: How can I check whether there exists a common basis with respect to which two matrices and are permutation matrices? -More explicitly, let $A$ and $B$ be two unitary matrices whose eigenvalues form complete sets of roots of unity (meaning there is a unitary transformation to a permutation matrix for each). How can I check whether their exists a single unitary $U$ such that -$$ -P_A = U A U^\dagger, \quad P_B = U B U^\dagger, -$$ -such that $P_A$ and $P_B$ are permutation matrices? - -REPLY [2 votes]: One partial answer, for which you don't need the knowledge that the spectra are complete sets of roots of unity. -First step : verify that the group $G$ generated by $A$ and $B$ is finite (it should be contained a permutation group of $n$ elements). For this, explore the free group in two letters $a$ and $b$ (together with $a^{-1}$ and $b^{-1}$). Consider the disk $D_k$ of words $w$ of length $\le k$, and let $N_k$ be the number of distinct elements $w(A,B)$ for $w\in D_k$. If $N_k$ comes to exceed $n!$, the basis does not exist. If not, you have reached a $k$ so that $N_{k+1}=N_k$ ; the group is finite and you still have a chance, provided this $N=N_k$ is $\le n!$. -Next step : In the latter case, you must verify that the trace of $w(A,B)$ is a non-negative integer, for every $w\in D_k$. -I suspect that the converse is true, and that it is a result about linear representations of finite groups. Does anyone have a clue ?<|endoftext|> -TITLE: How and when do I learn so much mathematics? -QUESTION [45 upvotes]: I am about to (hopefully!) begin my PhD (in Europe) and I have a question: how did you learn so much mathematics? -Allow me to explain. I am training to be a number theorist and I have only some read Davenport's Multiplicative Number Theory and parts of Vaughan's book on the circle method. I have briefly seen some varieties from Fulton's algebraic curves and I may have read parts of homotopy and homology and differential geometry of smooth manifolds at the level of Hatcher and Lee. Yet, it seems that I am hopelessly ignorant of elliptic curves, modular forms and algebraic number theory. -For example, if I were to try reading Deligne's proof of Weil's conjecture or Tate's thesis, it seems that I would have to do significant amounts of reading. -When I look at some of my professors or other researchers I have interacted with, I notice that they may be publishing in one or two areas, but are extremely knowledgeable in pretty much everything I ask them about. That begs the questions: - -How much reading outside should I be doing outside my "area"? -Is it a good idea to just focus narrowly on my thesis problem at this stage or is it more usual to be working on multiple problems at the same time? -How and how often do you end up learning new areas? - -Sorry if the question is too vague: I just want to have a sense of how to go about being a good mathematician. Also, part of the reason I am asking this question is that when I go to seminars, I understand so little and I see some of my professors ask questions of the speakers even if they don't work in the same area. - -REPLY [7 votes]: My personal experience suggests that reaching reasonable breadth of mathematical scope is achieved through three different mechanisms : -1) Attending talks (seminars, colloquia, workshops) in subjects where you are not an expert. Not only you will eventually grasp some useful bits and pieces of how to think about various mathematical phenomena, you might even find some things useful years later when you encounter something similar in your own research. -2) Trying to browse arXiv regularly. It is getting increasingly difficult, since the amount of daily postings is much higher than 15+ years ago when I was a PhD student. But still, scrolling through titles of papers and occasionally reading the abstract and/or the introduction when something catches your eye can be very useful. A related advice: if you read something useful and/or exciting in a paper, check on MathSciNet what papers cite it; this might bring you to some other interesting things. -3) Try to find a way to interact with your supervisor to broaden your scope. Good supervisors would only limit themselves to a narrow subject of your intended area if you insist on that. Having a mathematical mentor who can share with you his or her broader vision of the field, of interesting things to learn, of interesting events to attend, is perhaps the greatest thing that PhD studies offer. Use it wisely.<|endoftext|> -TITLE: Could Kronecker accept a proof of Goodstein's theorem? -QUESTION [11 upvotes]: A famous result of Goodstein asserts that the Goodstein sequence of integers terminates. -For a precise statement and a short proof, see https://en.wikipedia.org/wiki/Goodstein%27s_theorem. -A well known proof of this theorem uses ordinal number. The idea is to associate with the Goodstein sequence a sequence of ordinal numbers that is decreasing and every decreasing sequence of ordinal numbers terminates after a finite number of steps. -A well-known result of Kirby and Paris asserts that Goodstein's theorem cannot be proved in the Peano arithmetic. My vague understanding is (correct if I am wrong) that any proof of the Goodstein theorem would require a proof of existence of the ordinal number -$$ -\large \omega^{\omega^{\omega^{\omega^{\omega^\ldots}}}} -$$ -and such a large number cannot be constructed in the Peano arithmetic. -I believe this result is of wide interest in the mathematical community. While there have been a lot of discussions related to this theorem on Mathoverflow, the theorem is set deeply in foundations of mathematics and most of the discussions are so hermetic that only people with good understanding of the foundations of mathematics can follow it. I have some working knowledge in the area (as most of us do), but that restricts to a basic understanding of ZFC, practical use of the axiom of choice, and some understanding of basic results in arithmetic of ordinal numbers. With such a limited knowledge I am not able to follow the discussions. -Georg Cantor created foundations of the modern set theory. However, his theory encountered a strong resistance from prominent mathematicians including Kronecker, Poincaré, Weyl and Brouwer. Kronecker described Cantor as a "scientific charlatan", a "renegade" and a "corrupter of youth". -Without the set theory created by Cantor, the proof of Goodstein's theorem based on the arithmetic of ordinal numbers would not exist and hence the known proof would not be accepted by Kronecker, Poincaré, Weyl or Brouwer. -Therefore my question is: - -Question. Is it possible to prove Goodstein's theorem in an "elementary" way so that it would be accepted by Kronecker? - -I have heard that there are "elementary", but rather long proofs using second-order arithmetic (I do not even know what it means). - -Question. Where can I find a proof that does not involve ordinal numbers? - -Since I find these questions of great interest for everyone, I would appreciate a detailed and "elementary" explanation. For example while referring to the second-order arithmetic I would appreciate if someone could explain in details what it really means by showing some examples. -Edit: I replaced Gauss with Kronecker in the question. It seems a better choice. - -REPLY [9 votes]: As a starting point, I recommend that you study the proof of Theorem 1 in my expository article on The Consistency of Arithmetic. (This is not exactly a proof of Goodstein's theorem, but for the purposes of the present discussion I think it's close enough. Start reading on page 25 where it says "Ordinals below $\epsilon_0$.") -I have tried to make this proof as "pedestrian" as possible, avoiding concepts from logic and set theory that might be unfamiliar to the "working mathematician." In particular, my contention is that if you were to present this proof in (say) an undergraduate analysis class, it would seem completely unobjectionable and unremarkable. True, the proof assumes that it is meaningful to talk about infinite sequences, but infinite sequences are routinely bandied about in undergraduate analysis without comment. Also, the induction argument is trickier than most induction arguments, but I don't think that anyone would find anything suspicious about it unless they had been specifically told to look for something suspicious. -Sometimes the terms "PA proof" and "elementary proof" are used interchangeably. I think that this practice can lead to confusion. The class of proofs that are formalizable in PA is of course a very clearly defined class, but it does not correspond very closely to what we intuitively think of as "elementary" proofs. In particular, the $\epsilon_0$ upper bound is, from the point of view of ordinary mathematical practice, a rather artificial bound. For example, we know from reverse mathematics that in a certain precise sense, the Bolzano–Weierstrass theorem goes beyond PA, but in our undergraduate analysis classes we don't set out flares or start blaring emergency sirens when proving Bolzano–Weierstrass. The exact logical strength of various assumptions that are used in ordinary mathematics is something that is not apparent on the surface, and emerges only after a careful logical analysis. -Conversely, it's possible to encode highly sophisticated concepts and proofs in PA—concepts that most mathematicians would instinctively regard as being "not at all elementary." EDIT (in light of the comments): For example, sometimes people ask if Fermat's Last Theorem can be proved in PA. I think that most of them are implicitly trying to ask whether there is an elementary proof of FLT. But as Angus Macintyre has shown (see comments below), it is very likely that all the high-falutin' super-fancy "non-elementary" machinery used in the known proof of FLT can be sufficiently well approximated in PA to push the proof through. If Macintyre is right, which I think he is, then this would show that FLT is provable in PA, while leaving completely unanswered the question of whether there is an "elementary" proof of FLT. -Returning to the question of Gauss or Kronecker, it's a little tough to guess what they would have thought about these things. They might have affirmed that they believed in "potential infinity" and not "actual infinity," but most modern mathematicians regard these two terms as philosophical gobbledygook, making it hard to translate Gauss's beliefs directly into present-day terms. We can probably safely say that they would have balked at ZFC, in particular the axiom of infinity, but the axiom of infinity isn't absolutely necessary for Goodstein's theorem. -Indeed, Gentzen, in his proof of the consistency of PA, made an effort to argue that the reasoning was "finitary." Since "finitary" isn't a completely precise term either, it's hard to be sure that Gauss would have agreed with Gentzen, but it's at least plausible that he would have. Ultimately, like all historical counterfactuals, it's a matter of speculation what Gauss would have thought. But hopefully I've convinced you that there isn't anything more mysterious about Goodstein's theorem than many other routine mathematical theorems.<|endoftext|> -TITLE: Reduced products of complete Boolean algebras -QUESTION [6 upvotes]: I expect that complete Boolean algebras are not closed under reduced products modulo $\kappa$-complete filters, for any regular cardinal $\kappa$. Is it true? And, is a reference for this? - -REPLY [8 votes]: If you write $2$ for the $2$-element Boolean algebra, and $F$ for the filter of cobounded sets in $\kappa$, then $2^\kappa/F$ is just $P(\kappa)/[\kappa]^{<\kappa}$, which is certainly not complete. -Here is an ad hoc proof that it is not complete, but I am sure that deeper reasons could be found. If it were complete, you could find a maximal antichain $ (b_i:i\in \kappa)$. Let $c_i:= \bigvee_{j\ge i} b_j$. The elements $c_i$ are equivalence classes of sets, equivalent mod $F$. Choose representatives $C_i$ in $c_i$. Let $C'_i:= C_i \cap \bigcap_{k -TITLE: Indecomposable vector bundles on elliptic curves -QUESTION [5 upvotes]: Let $X$ be a smooth complex projective curve of genus 3, -$E$ an elliptic curve, and $f: X \to E$ a finite map of -degree 2. Let $L$ be a line bundle on $X$, and $R^0f_*(L)$ -its direct image. Question: is $R^0f_*(L)$ indecomposable? - -REPLY [6 votes]: This depends on the choice of $L$. For instance, if $L = \mathcal{O}_X$ then $f_*L \cong \mathcal{O}_E \oplus L_E$, where $\deg(L_E) = -2$. -On the other hand, if $L$ is a general line bundle of degree 1, then $f_*L$ is indecomposable. -If $L = \mathcal{O}(P)$ for some $P \in X$ the pushforward $f_*L$ is decomposable. Indeed, on the one hand, $f_*L$ is a vector bundle of rank 2 and degree $-1$ (this follows from Riemann-Roch), on the other hand, it has a global section. Therefore, it comes in an exact sequence -$$ -0 \to \mathcal{O}_E \to f_*L \to M \to 0, -$$ -where $M$ is a line bundle of degree $-1$. But -$$ -Ext^1(M,\mathcal{O}_E) = H^1(E,M^\vee) = 0 -$$ -because the degree of $M^\vee$ is 1, hence the sequence splits and -$$ -f_*L \cong \mathcal{O}_E \oplus M. -$$<|endoftext|> -TITLE: How often two iid variables are close? -QUESTION [16 upvotes]: Is there a constant $c>0$ such that for $X,Y$ two iid variables supported by $[0,1]$, -$$ -\liminf_\epsilon \epsilon^{-1}P(|X-Y|<\epsilon)\geqslant c -$$ -I can prove the result if they have a density, of if they have atoms, but not in the general case. - -REPLY [9 votes]: Mateusz Kwaśnicki's guess, that the optimal value of $c$ is 2, is correct! In fact: - -Theorem: Suppose $X,Y$ are i.i.d. random variables on $\mathbb{R}$. Then the limit $\lim_{\varepsilon \rightarrow 0} \varepsilon^{-1} \mathbb{P}[|X - Y| < \varepsilon]$ exists. It is $+\infty$ unless $X$ has a density function $f$ satisfying $||f||_2^2 = \int f^2\ \text{d}x < \infty$, in which case the limit is equal to $2 ||f||_2^2$. - -Restricting to distributions on $[0,1]$, Cauchy-Schwarz says -$$ 1 = \int_0^1 f\ \text{d}x \le \left ( \int_0^1 f^2\ \text{d}x \right )^{1/2} \left ( \int_0^1 1\ \text{d}x \right )^{1/2} = ||f||_2, $$ -with equality iff $f = 1$. That is, the constant $c = 2$ is valid for every distribution, and it is sharp only for the uniform distribution. -Here are a few examples where the limit is infinite: -Example 1: If $X$ has an atom then $\mathbb{P}[|X-Y| < \varepsilon]$ is bounded away from 0. So the limit in question diverges like $\varepsilon^{-1}$. -Example 2: The Cantor ternary function is the CDF of a probability distribution on $[0,1]$ which is nonatomic but not absolutely continuous. A random sample from this distribution is given by $\sum_{k \ge 1} c_k \tfrac{2}{3^k}$ where $\{c_k\}$ are i.i.d. Bernoulli random variables. (Thought of in terms of the "repeatedly take out the middle third" construction of the Cantor set, the $c_k$'s are the choices of left versus right third.) If $\varepsilon = 3^{-n}$, then $|X-Y| < \varepsilon$ iff the first $n$ $c_k$'s agree. Thus -$$ \mathbb{P}[|X-Y| < \varepsilon] = 2^{-n} = \varepsilon^{\log_3 2}. $$ -So the limit in question diverges like $\varepsilon^{-1+\log_3 2} = \varepsilon^{-0.37}$. -Example 3: The power law $f(x) = \tfrac{1}{2\sqrt{x}}$ is an example of a density function which is not in $L^2$. In this case we can evaluate $\mathbb{P}[|X-Y| < \varepsilon]$ exactly (well, if you believe in inverse hyperbolic trig functions). The asymptotic behavior is -$$ \mathbb{P}[|X-Y| < \varepsilon] = \tfrac{1}{2} (-\log \varepsilon) \varepsilon + (\tfrac{1}{2} + \log 2) \varepsilon + O(\varepsilon^2). $$ -So the limit in question diverges logarithmically. -Mateusz's slick argument worked on the block diagonal, a sum of $n$ squares of width $1/n$. This covers about half of the area of the strip $|X-Y| < \tfrac{1}{n}$, which is why the resulting constant is half of optimal. There's probably a hands-on way to extend it, but you start using words like "convolution" and I found it easier to reason in Fourier space. This question is asking for a bound on the density of $X-Y$ at 0. This random variable has nonnegative Fourier transform (characteristic function, if you prefer), and that condition alone is enough to guarantee positive density (shameless self-citation: Lemma 3.1 in this paper). In general, when it makes sense, the density at 0 of a random variable is equal to the integral of its Fourier transform. So our job is to make that concrete and translate it back to a statement about PDFs. -Convention: The Fourier transform (characteristic function) of a random variable $X$ is the function $t \mapsto \mathbb{E}[e^{2 \pi \mathrm{i} X t}]$. The Fourier transform of a function $f$ is $t \mapsto \hat f(t) = \int e^{2 \pi \mathrm{i} x t} f(x)\ \text{d}x$. If $X$ is absolutely continuous (has a density function), then its Fourier transform is equal to the Fourier transform of its density function. With this convention the Plancherel identity reads $\int f \bar g\ \text{d}x = \int \hat f \overline{\hat g}\ \text{d}t$ (no normalization constant). -Lemma: Let $\psi$ denote the Fourier transform of $X$. Then $\psi \in L^2$ iff $X$ is absolutely continuous and its density function $f$ is in $L^2$. Moreover, if this holds, then $||\psi||_2 = ||f||_2$. -Proof of Lemma: The Fourier transform is an isometry of the space $L^2$. The ($\Leftarrow$) direction is immediate: if $X$ has a density function in $L^2$, then certainly its Fourier transform is in $L^2$. For ($\Rightarrow$), if $\psi$ is in $L^2$ then it is the Fourier transform of some function $f \in L^2$. But then $f$ defines the same distribution as $X$, so it is indeed the density function. $\square$ -(This is probably a basic result in some probability textbook?) -Proof of Theorem: Let $\psi$ denote the Fourier transform of $X$. Then the Fourier transform of $X-Y$ is $t \mapsto \psi(t) \psi(-t) = \psi(t) \overline{\psi(t)} = |\psi(t)|^2$. Let $T_\varepsilon$ denote the "triangle filter" $T_\varepsilon(x) = \varepsilon^{-1} (1 - |x|/\varepsilon)$ for $|x| \le \varepsilon$ and $T_\varepsilon(x) = 0$ otherwise. It is a standard calculation that $\hat T_\varepsilon(t) = \operatorname{sinc}^2(\pi t \varepsilon)$. (Here $\operatorname{sinc} x = \tfrac{\sin x}{x}$.) This is integrable, so we can compute the expected value of $T_\varepsilon$ using the Fourier transform. We find -$$ \varepsilon^{-1} \mathbb{P}[|X-Y| < \varepsilon] \ge \mathbb{E}[T_\varepsilon(X-Y)] = \int |\psi(t)|^2 \operatorname{sinc}^2(\pi t \varepsilon)\ \text{d}t. $$ -The integrand is nonnegative and, as $\varepsilon \to 0$, it converges (pointwise) to its (pointwise) supremum $|\psi(t)|^2$. So, by an appropriate incantation of the monotone convergence theorem (the one in this comment applies exactly), the integral converges to $\int |\psi(t)|^2\ \text{d}t = ||\psi||_2^2$. If this is infinite, then also $\lim_{\varepsilon \to 0} \varepsilon^{-1} \mathbb{P}[|X-Y| < \varepsilon] = +\infty$. -Suppose now that $||\psi||_2^2 < \infty$, i.e., $\psi \in L^2$. Consider the "box filter" $B_\varepsilon$ defined by $B_\varepsilon(x) = (2 \varepsilon)^{-1}$ for $|x| < \varepsilon$ and $B_\varepsilon(x) = 0$ otherwise. This has $\hat B_\varepsilon(t) = \operatorname{sinc}(2 \pi t \varepsilon)$. This is bounded and $|\psi|^2$ is integrable. So we can again compute expected values in Fourier space: -$$ (2\varepsilon)^{-1} \mathbb{P}[|X-Y| < \varepsilon] = \mathbb{E}[B_\varepsilon(X-Y)] = \int |\psi(t)|^2 \operatorname{sinc}(2 \pi t \varepsilon)\ \text{d}t. $$ -The integrand converges pointwise to $|\psi(t)|^2$. We don't have nonnegativity this time, but now we know integrability of the bounding function $|\psi|^2$. So we can apply the dominated convergence theorem, concluding that -$$ \lim_{\varepsilon \to 0} (2 \varepsilon)^{-1} \mathbb{P}[|X-Y| < \varepsilon] = \int |\psi(t)|^2\ \text{d}t = ||\psi||_2^2. $$ -The lemma completes the proof. $\square$<|endoftext|> -TITLE: Refined reverse plane partition generating function -QUESTION [6 upvotes]: I have a simple question about the generating function for reverse plane partitions: -$$\sum_{\pi \in RPP(\lambda)} z^{|\pi|}= \prod_{s \in \lambda} \frac{1}{1-z^{h_{\lambda}(s)}}$$ -There's a natural refinement of the right hand side: -$$ -\prod_{s \in \lambda} \frac{1}{1-t z_1^{a_{\lambda}(s)}z_2^{l_{\lambda}(s)}} -$$ -Or perhaps just with $t=z_1,z_2$. -I suspect there should be an equivalent left hand side to this identity - i.e. counting some "refined weight" of the reverse plane partition. Perhaps along diagonals? In a sense there has to be - I'm just not sure what the "statistic" is to count. I wondered if there is a known generating function? -If it helps the right hand side is something like $c_{\lambda}(q,t)$ from Macdonald polynomial theory. -Update: If I write this instead in terms of $w_1 = z_1/z_2$ and $w_2 = z_1z_2$ I actually expect the RHS to be a polynomial in $w_1$ as in something along the lines: -$$ -\sum_{\pi \in RPP(\lambda)} w_2^{|\pi|}P_{|\pi|}(w_1) -$$ -Where $P(w_2)$ is a finite polynomial. -Thanks - -REPLY [6 votes]: Yes, there is a way to introduce certain statistics that lead to this refinement. -First, I'll assume partitions are given as collections of boxes with coordinates $(i,j)\in \mathbb N^2$. The content of the box $(i,j)$ is the quantity $i-j$. A border strip of a partition $\lambda$ is a subset of boxes of $\lambda$ which is a connected skew shape and contains no $2\times 2$ configuration of boxes. Let's call a border strip maximal if its box of largest content $(i_1,j_1)$ satisfies $(i_1+1,j_1)\notin \lambda$, and its box of smallest content $(i_2,j_2)$ satisfies $(i_2,j_2+1)\notin \lambda$. A skew shape $\lambda/\mu$ can be written as a disjoint union of maximal border strips in a unique way. Let $b(\lambda/\mu)$ be the number of border strips that appear in such a decomposition. -The height of a border strip is defined as one less than the number of rows it occupies (a statistic that should be familiar from the Murnaghan Nakayama rule, for example). The height of a skew shape, $\operatorname{ht}(\lambda/\mu)$ is defined as the sum of the heights of all the border strips that appear when writing $\lambda/\mu$ as a union of maximal border strips. Similarly we can define $\operatorname{ht}'(\lambda/\mu)$ by using columns instead of rows. -Now finally, when you have a reverse plane partition $\pi\in RPP(\lambda)$, you can picture it as a 3D stack of boxes. Each horizontal layer is a certain skew shape $\lambda/\mu_i$, for $i=1,2,\dots$. We define $\operatorname{ht}(\pi)=\sum_{i\geq 1} \operatorname{ht}(\lambda/\mu_i)$, $\operatorname{ht}'(\pi)=\sum_{i\geq 1} \operatorname{ht}'(\lambda/\mu_i)$ and $b(\pi)=\sum_{i\geq 1} b(\lambda/\mu_i)$. We can finally state the desired refined formula as -$$\sum_{\pi\in RPP(\lambda)}z_1^{\operatorname{ht}(\pi)}z_2^{\operatorname{ht}'(\pi)}t^{b(\pi)}=\prod_{s \in \lambda} \frac{1}{1-t z_1^{a_{\lambda}(s)}z_2^{l_{\lambda}(s)}}.$$<|endoftext|> -TITLE: How to prove $e^x\left|\int_x^{x+1}\sin(e^t) \,\mathrm d t\right|\le 1.4$? -QUESTION [13 upvotes]: Related question asked by me on Math SE a few days ago: How to prove $e^x\left|\int_x^{x+1}\sin(e^t) \,\mathrm d t\right|\le 1.4$? -A few days ago, somebody asked How to prove $ \mathrm{e}^x\left|\int_x^{x+1}\sin\mathrm e^t \mathrm d t\right|\leqslant 2$? on Math StackExchange. -However, this bound does not appear to be sharp so I was wondering how to find the maxima/minima of $$f(x)=e^x\int_x^{x+1}\sin(e^t) \,\mathrm d t$$ -or at least how to prove $-1.4\le f(x)\le 1.4$. -Some observations, using the substitution $y=e^t$: -$$f(x)=e^x \int_{e^x}^{e^{x+1}} \frac{\sin(y)}y\,\mathrm dy=g(e^x),$$ -where I have defined $$g(z)=z \int_z^{e z} \frac{\sin(y)}y\,\mathrm dy = z (\operatorname{Si}(e z)-\operatorname{Si}(z)).$$ -($\operatorname{Si}$ is the Sine integral.) -So the question reduces to: What are the maxima/minima of $g(z)$ for $z\geq 0$ ? -Using the series of $\mathrm{Si}(z)$, we get -$$g(z)=\sum_{k=1}^\infty (-1)^{k-1} \frac{z^{2k}(e^{2k-1}-1)}{(2k-1)!\cdot(2k-1)}$$ -and here is a plot of $g(z)$: - -Also, notice that $g$ is analytic and $g'(z)=\sin (e z)-\sin (z)+\text{Si}(e z)-\text{Si}(z)$ which might help for the search of critical points (although I don't think that $g'(z)=0$ has closed form solutions). - -REPLY [2 votes]: This can be done with help of Maple in such a way. First, we find the estimated expression explicitly by - a := (exp(x)*int(sin(exp(t)), t = x .. x + 1) assuming x::real; - -$ {{\rm e}^{x}} \left( -{\it Si} \left( {{\rm e}^{x}} \right) +{\it Si} - \left( {{\rm e}^{x+1}} \right) \right) -$ -In fact, the integral is reduced to another integrals. Next, the asymptotics of $a$ is found. Maple is not able to find this asymptotics directly so the change $x=\log y$ should be used: -asympt(simplify(eval(a, x = log(y))), y, 2); - -$-{\frac {\cos \left( y{\rm e} \right) }{{\rm e}}}+\cos \left( y - \right) +O \left( {y}^{-1} \right) - $ -Now we return to $x$ by -eval(%, y = exp(x)); - -$-{\frac {\cos \left( {{\rm e}^{x}}{\rm e} \right) }{{\rm e}}}+\cos - \left( {{\rm e}^{x}} \right) +O \left( \left( {{\rm e}^{x}} \right) -^{-1} \right) $ -The rest is as in the Fedor Petrov's answer.<|endoftext|> -TITLE: How strong is this set theory? -QUESTION [5 upvotes]: In the spirit of this related question, consider a set theory with the following axioms: -Axiom of extension: -$$ \forall x \forall y (\forall z (z \in x \leftrightarrow z \in y) \rightarrow x = y) $$ -Axiom schema of comprehension: -$$ \exists x \forall y (y \in x \leftrightarrow (\phi y \land C y)) $$ -Axiom of construction: -$$ \forall x (\forall y (y \in x \rightarrow C y) \leftrightarrow C x) $$ -where $C$ is a new symbol like $\in$ that intuitively represents a kind of "constructibility". Some properties I've deduced are - -Every hereditarily finite set exists and is constructible (in particular, the empty set $\varnothing$ exists and is constructible, as are all the finite ordinals). -The quasi-universal set $U$ with $\phi y \equiv \top$ exists, is constructible, and contains itself. -The quasi-Russell set $R$ with $\phi y \equiv y \not\in y$ exists, is not constructible (since that yields the Russell contradiction), and does not contain itself. -The quasi-co-Russell set $R^\ast$ with $\phi y \equiv y \in y$ exists and is constructible iff it contains itself, which seems to be independent of the theory. -Let $y^+ \equiv y \cup \{y\}$ and let $\textsf{inductive } x \equiv \varnothing \in x \land \forall y (y \in x \rightarrow y^+ \in x)$ denote that a set is inductive. $C \varnothing$ and $C y \rightarrow C y^+$, so $\textsf{inductive } U$. -Let $\textsf{natural } x \equiv \forall y (\textsf{inductive } y \rightarrow x \in y)$ denote that a set belongs to every inductive set (i.e. is a natural number). Since $\textsf{inductive } U$, $\textsf{natural } x \rightarrow x \in U$. But $x \in U \rightarrow C x$. Thus $\textsf{natural } x \rightarrow C x$. Instantiate the axiom schema of comprehension with $\phi \equiv \textsf{natural}$. Then $\exists x \forall y (y \in x \leftrightarrow \textsf{natural } y)$. That is, the set of natural numbers $\mathbb{N}$ exists, and moreover is constructible. - -My question is this: How strong is this set theory compared to ZFC or other alternative set theories? -Edit: In light of the contradiction pointed out below, one might consider restricting the formulas allowed in the axiom schema of comprehension. For example, we might consider restricting to positive formulas, as in positive set theory. -Edit 2: An interesting alternative system is presented here. - -REPLY [2 votes]: Regarding the latter suggestion to restrict $\phi$ to positive formulas. The resulting theory would be consistent relative to positive set theory, but it would be just a redundant re-exposition of a fragment of it. Simply take $C$ to be the predicate of being "equal to itself" i.e.: -$C(x) \leftrightarrow x=x$, -and comprehension would just be positive comprehension, and axiom of construction would be trivially true! Same would apply if for example we've restricted $\phi$ to stratified formulas, we'd only get Quine's New foundations set theory. -Actually that argument would hold for any kind of a theory that has a naive like comprehension axiom with only syntactical restrictions on the defining formula. -I think the only reasonable adjustments to salvage your approach are those you've had with your original theory, the one you raised this theory in connection with, which I think is as strong as $PA$, if you want to strengthen it, you add an axiom of infinity, or even a more risky approach is to add a universal class of all $C$ objects but with changing the bound on parameters in comprehension (the schema of construction) for all of them to be elements of that universal class. But I don't know if those are consistent, and what would be their consistency strength?<|endoftext|> -TITLE: Integrating the Riemann curvature tensor over a singular 2-disc -QUESTION [6 upvotes]: There's a classic characterization of the Riemann curvature tensor. Say, take a Riemann metric on an open subset $U$ of $\mathbb R^n$. Given a point $p \in U$ and two vectors $v,w \in T_p U$ you compute the holonomy around the boundary of the parallelogram $\{p + t_1 v + t_2 w : 0 \leq t_1 \leq a, 0 \leq t_2 \leq b\}$. The second-order Taylor expansion of this holonomy, as a function of $a$ and $b$ (centred at $a=b=0$) is given by: -$$Id_{T_p U} + ab R_p(v,w)$$ -If you think carefully about this, what it tells you is that if you have any map $D^2 \to N$ where $N$ is a Riemann manifold, you can compute the holonomy around the boundary of the disc by an appropriate "integral" on the interior of the disc, of the Riemann curvature tensor. It's not an integral in the traditional sense, as you are performing a limit of a system of composites of functions. But in spirit, it is reasonably-close to an integral. -Other than being a limit of a composite of a large number of functions, a technical issue is the choice of basepoint and ensuring you are using an appropriate transport of the Riemann tensor back to one tangent space. -I imagine this observation has been written up somewhere in the literature. Who has proven this theorem, and where might it appear? Does this type of groupoid-y integration have a standard name? - -REPLY [2 votes]: There is "How the curvature generates the holonomy of a connection in an arbitrary fibre bundle" by Helmut Reckziegel and Eva Wilhelmus in Results in Mathematics 46 (2006). I'm not sure that is the first place where this is proved properly, but it does do so nicely (in a more general context).<|endoftext|> -TITLE: The Yoneda pairing, hypercohomology, and cup product -QUESTION [5 upvotes]: Let $\mathcal{F}$ and $\mathcal{G}$ be coherent analytic sheaves on $\mathbb{P}^n$. Let $\mathcal{F}_\bullet$ be a locally free resolution of $\mathcal{F}$. In Principles of Algebraic Geometry by Griffiths and Harris, $\text{Ext}^p(\mathcal{F},\mathcal{G})$ is defined as the hypercohomology of the complex $\mathcal{Hom}(\mathcal{F}_\bullet,\mathcal{G})$, i.e., the cohomology of the complex $\bigoplus_{p=k+\ell} C^k(\mathfrak{U},\mathcal{Hom}(\mathcal{F}_\ell,\mathcal{G}))$, see pages 705 and 446. Here $C^\bullet(\mathfrak{U},\mathcal{Hom}(\mathcal{F}_\ell,\mathcal{G}))$ denotes the Čech complex with respect to some affine open cover $\mathfrak{U}$ of $\mathbb{P}^n$. -If I understand correctly, the Yoneda pairing $$\text{Ext}^p(\mathcal{F},\mathcal{G}) \times \text{Ext}^q(\mathcal{G},\mathcal{H}) \rightarrow -\text{Ext}^{p+q}(\mathcal{F},\mathcal{H})$$ should then be induced by the cup product in Čech cohomology. However, I fail to see precisely how this works out. -Edit: To clarify: -What I am primarily interested in is how the Yoneda pairing can be expressed in terms of the concrete representatives when $\text{Ext}$ is realized as hypercohomology. In [HL, Section 10.1.1], there is an explicit cup product on hypercohomology, which then is said to induce a product $Ext^i(F^\bullet,G^\bullet)\otimes Ext^j(E^\bullet,F^\bullet) \to Ext^{i+j}(E^\bullet,G^\bullet)$. It is then also stated that -"If we interpret $Ext^i(E^\bullet,F^\bullet)$ as $Hom_{\mathcal D}(E^\bullet,F^\bullet[i])$, where $\mathcal{D}$ is the derived category of quasi-coherent sheaves, then the cup product for Ext-groups is simply given by composition." To me, it is not clear neither how the cup product induces the product on $\text{Ext}$, nor why this product coincides with composition in the derived category. Any references to where this is discussed in more detail, or hints on how to prove this would be welcome. -[HL] Huybrechts, Lehn: The Geometry of Moduli Spaces of Sheaves - -REPLY [2 votes]: Suppose Č(U)→Hom(Q(F),G)[p] and Č(V)→Hom(Q(G),H)[q] represent -elements in Ext^p(F,G) and Ext^q(G,H). -Here Č(U) denotes the Čech chains for an open cover U and Q(-) is the functor -that computes appropriate replacements (often cofibrant), e.g., locally free resolutions. -Choose a common refinement W of U and V and restrict the above maps -along the morphisms Č(W)→Č(U) and Č(W)→Č(V). -Tensor the two resulting maps together and restrict along the diagonal map, -obtaining a map Č(W)→Hom(Q(F),G)⊗Hom(Q(G),H)[p+q] -The remaining problem is the mismatch between G and Q(G). -The canonical map Q(G)→G goes in the wrong direction, -and so does the induced map Hom(Q(F),Q(G))→Hom(Q(F),G). -This prevents us from composing things in the most obvious way. -If one is not interested in specific models for representatives, -one can observe that Hom(Q(F),Q(G))→Hom(Q(F),G) becomes an isomorphism -in the derived category, so can be tensored with Hom(Q(G),H) -and then inverted and composed with the other map, -yielding a map Č(W)→Hom(Q(F),H)[p+q], as desired. -Otherwise, working in (say) the projective or flat model structure -one can explicitly lift the cofibration 0→Č(U) -against the acyclic fibration -Hom(Q(F),Q(G))[p]→Hom(Q(F),G)[p], -obtaining a new representative Č(U)→Hom(Q(F),Q(G))[p], -and then apply the second paragraph above to get -the desired map Č(W)→Hom(Q(F),H)[p+q].<|endoftext|> -TITLE: Bounding a Fourier coefficient of a non-negative periodic function in terms of its $L^2$-norm -QUESTION [12 upvotes]: This question is motivated by the earlier MO question: Show that $(\sum_{k=1}^{n}x_{k}\cos{k})^2+(\sum_{k=1}^{n}x_{k}\sin{k})^2\le (2+\frac{n}{4})\sum_{k=1}^{n}x^2_{k}$ . -It is a cleaned up asymptotic version of that question. -Let $f$ be a non-negative function, periodic with period $1$, and square integrable on ${\Bbb R}/{\Bbb Z}$. Is it true that -$$ -|{\widehat f}(1)|^2 = \Big| \int_0^1 f(x) e^{-2\pi ix} dx \Big|^2 \le \frac 14 \int_0^1 f(x)^2 dx \ \ ? -$$ -Equality is attained for example when $f(x) = \max(0, \cos(2\pi x))$. -Note that $|\widehat f(1)| =|\widehat f(-1)|$ and, since $f$ is non-negative, $|\widehat f(1)| \le \widehat f(0)$. Therefore -$$ -\int_0^1 f(x)^2 dx = \sum_n |\widehat f(n)|^2 \ge 3 |\widehat f(1)|^2, -$$ -so that the estimate holds with $1/3$ in place of $1/4$. There is a lot of scope to improve this argument, and with a more careful application of Bessel's inequality I could get the constant $1/4+1/4\pi$. But the claimed inequality looks very clean, and I wonder if (i) it is true!, (ii) is known in some context, and (iii) (hopefully) has an elegant proof? - -REPLY [15 votes]: Your conjecture is indeed correct. The proof is based on the following simple yet powerful trick I learned many years ago: $|z| = \sup_{|v| = 1} \Re (zv)$. Therefore it is enough to only bound from above $\Re \left(\int_0^1 vf(x)e^{-2\pi ix}dx\right)$ for all $v\in \mathbb{T}$. And now it is clear by Cauchy-Schwarz that $f$ should be proportional to $\max(0, \Re(ve^{-2\pi ix}))$, and all such functions gives us $\frac{1}{4}$ ($f$ from the OP corresponds to the choice $v = 1$).<|endoftext|> -TITLE: Comprehension factorization for locally small categories -QUESTION [5 upvotes]: It is well-known that there is an orthogonal factorization system on Cat (the category of small categories and functors) where the final functors are left-orthogonal to the discrete fibrations. This factorization is called "comprehensive" in the original paper by Street and Walters. -However, in that paper the authors do not address any size issue. -Can this factorization system be extended to the category of locally small categories? - -REPLY [2 votes]: Discrete fibrations are faithful functors, so any discrete fibration to a locally small category is locally small. -In particular if you take any comprehension factorization: -$$ C \rightarrow D \rightarrow T $$ -If $T$ is locally small, then $D$ is also locally small. so the answer to your question is yes. -Though you have to be aware that the kind of discrete fibrations we are talking about here can have large fibers.<|endoftext|> -TITLE: Are modular representations isomorphic if they're isomorphic after raising to the pth power? -QUESTION [5 upvotes]: Consider algebraic representations of a reductive group $G$ over a field in characteristic $p$. I even want to allow potentially disconnected reductive groups, i.e. $G$ could be a finite group. (However I'm also interested if the behavior in the connected case is different.) -If $V$ and $W$ are two such representations with $V^{\otimes p}$ ismomorphic to $W^{\otimes p}$, are $V$ and $W$ necessarily isomorphic? -In characteristic $\neq p$ this is obviously false because you can tensor with a character of order $p$, but my question is in characteristic $p$. - -REPLY [13 votes]: Here's one way of constructing counterexamples for finite groups. -Suppose $M$ is a periodic $kG$-module with period $p$: i.e., the $p$th syzygy $\Omega^pM$ is isomorphic to $M$, but $\Omega M\not\cong M$. Then -$$(\Omega M)^{\otimes p}\cong \Omega^pM\otimes M^{\otimes (p-1)}\cong M^{\otimes p},$$ -up to projective direct summands. -If $G$ is a $p$-group, so that all projective $kG$-modules are free, then taking the direct sum of $|G|$ copies of $M$ and of $\Omega M$, and adding a free direct summand to whichever is smaller, we can get two modules $X$ and $Y$ of the same dimension, and then $X^{\otimes p}\cong Y^{\otimes p}$ (since they're isomorphic up to free direct summands and have the same dimension). -For example, take $p=2$ and $G$ the quaternion group $Q_8$. The trivial module $k$ has period $4$, so $M=k\oplus\Omega^2k$ has period $2$. $M$ has dimension $10$ and $\Omega M$ has dimension $14$, so in this case we can take the direct sum of two copies of each module (rather than $|G|=8$ copies), to get modules $X=k\oplus k\oplus\Omega^2k\oplus\Omega^2k\oplus kG$ and $Y=\Omega k\oplus\Omega k\oplus \Omega^3k\oplus\Omega^3k$ with $X^{\otimes 2}\cong Y^{\otimes 2}$. -Examples of suitable periodic modules for odd $p$ can be found in -Carlson, Jon F., Periodic modules with large periods, Proc. Am. Math. Soc. 76, 209-215 (1979). ZBL0419.20011.<|endoftext|> -TITLE: How many numbers $\le x$ can be factorized into three numbers which form the sides of a triangle? -QUESTION [8 upvotes]: Note: Posting in MO since it was unanswered in MSE -Definition: We say that a natural number $n$ has triangular divisors if it has at least one triplet of divisors $n = d_1d_2d_3, 1 \le d_1 \le d_2 \le d_3$, such that $d_1,d_2$ and $d_3$ form the sides of a non-degenerate triangle. -Eg: $60$ has triangular divisors because $60 = 3.4.5$ and $3,4,5$ form a triangle. Note that another triplet of divisors of $60 = 1.4.15$ does not form a triangle but because of the triplet $3,4,5$ the number $60$ qualifies a number with triangular divisors. On the other the number $10$ does not have any triplet of triangular divisors. -The first few numbers in this sequence are -$$ -1,4,8,9,12,16,18,24,25,27,32,36,40,45,48,\ldots -$$ -I am interested in the density of these numbers. In the linked questions users commented that the almost all integers are expected this property since because most numbers will have a several small prime factors so the natural density was initially thought to be $1$. -However, quite counter intuitively, in one of the long comment which was posted as an answer in the linked question, it was proved that if $n$ has triangular divisors then the largest prime factor of $n$ is less than $\sqrt n$ which immediately implies that the natural density of numbers with triangular divisors is $ < 1 - \log 2 \approx 0.3069$. Experimentally, the data shows that the natural density approaches $0$. Let $f(x)$ be the number of integers $\le x$ with triangular divisors. We have -$$ -f(46732002) = 3630678 -$$ -The graph of $\frac{f(x)}{x}$ vs $x$ given below shows that the density decreases as $x$ increases. -Experimental data: Let $f(x)$ be the number of integers $\le x$ with this property. The graph of $\frac{f(x)}{x}$ vs. $x$ is shown below. - -A simple curve fitting gives $\frac{a}{\log x}$ as a good fit with $R^2 = 0.9977$ which suggests that $f(x)$ growth rate somewhere close to a constant times the prime counting function $\pi(x)$. This is rather counter intuitive as mentioned in the comments that we expect almost all integers to have this property. -Higher density of even numbers: A curious observation is the there are significantly more even numbers with triangular divisors as compared to odd numbers. Let $f_o(x)$ be the number of odd numbers $\le x$ with triangular divisors. The graph of $\frac{f_o(x)}{f(x)}$ is shown below. - - -Question: How many numbers $\le x$ have triangular divisors and why are even numbers more dense than odd numbers? - -Related question: Reshaping an object into two integer sided cuboids without changing the total volume. - -REPLY [13 votes]: This problem is related to the higher dimensional multiplication table problem (specifically, the three dimensional version). From the work of Tenenbaum, Ford, and Koukoulopoulos this is now well understood in many cases. From their work, one can show that the number of $n$ up to $x$ with triangular divisors is -$$ -\ll \frac{x}{(\log x)^{\alpha}}(\log \log x)^{\frac 12} -$$ -where -$$ -\alpha=1- \frac{2}{\log 3} +\frac{2}{\log 3} \log \frac{2}{\log 3} = 0.27016\ldots. -$$ -The power of $\log \log x$, which is the thrust of the work of Ford and Koukoulopoulos, is probably not correct here. But there should be a corresponding lower bound of size $x/(\log x)^{\alpha +o(1)}$; it should be possible to establish this from the techniques in Koukoulopoulos. -Let me now indicate why an upper bound of this magnitude holds. We want to count $n \le x$ that may be written as $n=abc$ with $a\le b\le c$ and $cx^{\frac 13}/(\log x)$. Note that $b$ (which is $\ge a$ and $\le \sqrt{x/a}$) then lies in the interval $x^{\frac 13}/\log x$ to $x^{\frac 13} (\log x)^{\frac 12}$. Break the possible values for $a$ and $b$ into dyadic intervals $A < a\le 2A$ and $B< b\le 2B$. There are about $(\log \log x)^2$ such dyadic intervals, and both $A$ and $B$ are about $x^{\frac 13}$, and we can assume that $B \le \sqrt{x/A}$. Now we apply Theorem 2.6 from Koukoulopoulos's thesis which gives -$$ -\#\{ n\le 4x: \ ab |n, \ \ A< a\le 2A, \ \ B< b\le 2B\} \asymp \frac{x}{(\log x)^{\alpha} (\log \log x)^{\frac 32}}. -$$ -Since there are $\ll (\log \log x)^2$ such dyadic blocks, we conclude that the number of $n=abc$ with $a>x^{\frac 13}/(\log x)$ is -$$ -\ll \frac{x}{(\log x)^{\alpha}} (\log \log x)^{\frac 12}. -$$ -This gives the claimed upper bound.<|endoftext|> -TITLE: Is Thompson's group definably orderable? -QUESTION [19 upvotes]: Is Thompson's group $F$ definably left-orderable? definably bi-orderable? - -Orderability definitions: Recall that a group $G$ is left-orderable (resp. bi-orderable) if it admits a left-invariant (resp. bi-invariant) total order. If $S$ is a submonoid of $G$ such that $S\cup S^{-1}=G$ and $S\cap S^{-1}=\{1_G\}$ (call this a cone in $G$), then $\le_S$ defined by: $g\le_S h\Leftrightarrow g^{-1}h\in S$ is a left-invariant total order, and $S$ is conjugation-invariant iff $\le_S$ is bi-invariant. Conversely if $\le$ is a left-invariant total order then $S_\le=\{g:g\ge 1\}$ is a cone (which is conjugacy-invariant iff $\le$ is bi-invariant. We have bi-orderable $\Rightarrow$ left-orderable $\Rightarrow$ torsion-free. -Definability: a subset of a group $G$ is definable if it can be described using logical Boolean operators, quantifiers on group elements, group operations, and parameters in the group: for instance for $a,b\in G$, one can consider the definable subset of $G$: $\{x\in G:\forall y\in G,\exists z\in G:[a,z][b,z^3]=[x,y^5] \vee (x^2=y^2=1)\}$ (this example is totally random). Similarly one can define a definable subset of $G^n$ for every $n$. - -A group $G$ is definably (left/bi)-orderable if it admits a (left/bi)-invariant total order that is a definable subset of $G^2$, or equivalently whose positive cone is a definable subset of $G$. - -One motivation is that such a group satisfies a single sentence $\Phi$ such that every group satisfying $\Phi$ is also (left/bi)-orderable (and hence torsion-free). Another is whether one can interpret an infinite total order in Thompson's group. -Examples: -(a) The trivial group is definably bi-orderable (this is the only obvious example). -(b) The cyclic group $\mathbf{Z}$ is not definably left-orderable, and more generally any group with a nontrivial abelian direct factor is not definably left-orderable. (Indeed, for such torsion-free $G$, the theory of $G\times\mathbf{Z}/p\mathbf{Z}$ tends to the theory of $G$ when the prime $p$ tends to infinity, so no $\Phi$ as above can exist). -(c) The Heisenberg group over $\mathbf{Z}$ (or $\mathbf{R}$) is definably bi-orderable. This is not hard but a bit tricky. -(d) The free group $F_n$, $n\ge 2$, which is bi-orderable, is not definably left-orderable (according to an expert I asked, every definable submonoid of a free group is a subgroup). -Concerning the question: actually, the bi-invariant total orders on Thompson's group are classified (by Navas and Rivas, arXiv link; GGD 2010) and this is very explicit: I don't know if any of those is definable. - -REPLY [8 votes]: Yes, Thompson's group $F$ is definably bi-orderable. -Let $a$ be some element of $F$ with the support of $a$ equal to $(0,1/2)$. Let $b$ be some element of $F$ with the support of $b$ equal to $(1/2,1)$. -We will rely upon the following facts - -If $g$ and $h$ are in $F$ then $[a^g,b^h] = 1_F$ if and only if $(1/2)g \leq (1/2)h$. -$F$ acts transitively on the dyadic rationals. - -The set $S_1:= \left\{ f \in F \mid [a,b^f] = 1_F \right\}$ is the set of elements $f$ of $F$ for which $(1/2)f \geq 1/2$. -The set $S_2:= \left\{ f \in F \mid [a^f,b] \neq 1_F \right\}$ is the set of elements $f$ of $F$ for which $(1/2)f > 1/2$. -The set $S_3:= \left\{ f \in F \mid \exists g \in F \text{ with } [a^{gf},b^g] \neq 1_F \right\}$ is the set of elements $f$ of $F$ for which there is some dyadic rational $d \in (0,1)$ with $(d)f>d$ (here $d = (1/2)g$). -The set $S_4:= \left\{ f \in F \mid \exists g \in F \text{ with } [a^{gf},b^g] \neq 1_F \text{ and } \forall h \in F \text{ we have } [a^h,b^{hf}] = 1_F \vee [a^g,b^h] = 1_F\right\}$ is the set of elements $f$ of $F$ for which there is some dyadic rational $d \in (0,1)$ with $(d)f>d$ and for all dyadic rationals $e \in (0,1)$ either $(e)f \geq e$ or $e \geq d$ (here $d = (1/2)g$ and $e = (1/2)h$. -Equivalently $S_4$ is the set of non-identity elements $f$ of $F$ for which the right gradient at the infimum of the support of $f$ is strictly greater than $1$. -The union $\{1_F\} \cup S_4$ forms the cone of a bi-order on $F$. Specifically $\{1_F\} \cup S_4$ is the cone of $\preceq^+_{x^-}$ from the article of Navas and Rivas linked to in the question. -EDIT: Since people seem to be interested in the case of $[F,F]$ and Chehata's group I have added below a slightly stronger argument that applies to them. -Let $G$ satisfy the following - -Every element of $G$ has only finitely many components of support. -For any $0 < u < v < 1$ and $0 < w < x < y < z < 1$ there is $g \in G$ with $w < (u)g < x < y < (v)g < z$. -There is some element $a$ (that we now fix) of $G$ with a single component of support $(p,q)$ bounded away from $0$ and $1$. - -$G$ could be either of $[F,F]$ and Chehata's group. -Fix a non-identity elements $b$ of $G$ with the support of $b$ a proper subset of the support of $a$. -Let $S_5$ be the set of $g$ in $G$ such that $[g,a] = 1_G = [g,b]$. -For $h \in G$ write $\bar{h}$ for the boundary of the support of $h$. If $h$ is in $S_5$ then $\mathrm{supt}(h) \cap \bar{b} = \varnothing = \bar{h} \cap \mathrm{supt}(a)$. It follows that $\mathrm{supt}(h) \cap \mathrm{supt}(a) = \varnothing$. An element of $G$ whose support is disjoint from the support of $a$ is easily in $S_5$ so $S_5$ is the set of elements of $G$ whose supports do not intersect the support of $a$. -Fix a non-identity element $c$ of $G$ with the support of $c$ a subset of $(q,1)$. -Let $S_6$ be the set of $g$ in $G$ such that there exists $h \in G$ with $[a^h,a] = 1_G = [a^h,b]$ and $[a^h,c] \neq 1_G \neq [a^h,a^g]$. -We will now argue that $S_6$ is the set of those $g$ in $G$ with $q < (q)g$. -Let $g$ be in $S_6$. There exists a conjugate $a^h$ of $a$ whose support does not intersect $\mathrm{supt}(a)$ but does intersect $\mathrm{supt}(c)$ and does intersect $\mathrm{supt}(g)$. -Any conjugate of $a$ must have a single component of support so either $\mathrm{supt}(a^h) \subseteq (0,p)$ or $\mathrm{supt}(a^h) \subseteq (q,1)$. Since the support of $a^h$ intersects the support of $c$ we must have $\mathrm{supt}(a^h) \subseteq (q,1)$. Since the support of $a^g$ intersects the support of $a^h$ it follows that the support of $a^g$ intersects $(q,1)$. Now it follows that $q < (q)g$. -If $g$ is in $G$ and $q < (q)g$ then $g$ is in $S_6$ by condition 2. above. -$S_6$ now corresponds to $S_1$ from the original proof and the rest of the construction works similarly.<|endoftext|> -TITLE: Finite concatenation-free languages -QUESTION [7 upvotes]: Suppose, $A$ is a finite alphabet. $L \subset A^*$ is a language. Let's call $L$ concatenation-free iff $\forall u, v \in L$ we have $uv \notin L$. - -Does there exist some function $c: \mathbb{N} \to (0; 1)$, such that for any finite language $L \subset A^*$, there exists a concatenation-free sublanguage $L_0 \subset L$, such that $|L_0| \geq c(|A|)|L|$? - -The only thing I currently know about this problem, is that we can take $c(1) = \frac{1}{3}$. That is a direct consequence of Erdos-Sidon theorem, that states: - -$\forall A \subset \mathbb{Z}$ $\exists$ a sum-free $A_0 \subset A$, such that $|A_0| \geq \frac{|A|}{3}$ - -However, I do not know how to deal with $|A| \geq 2$. -This question on MSE - -REPLY [5 votes]: $c(n)=1/3$ works for every $n$. Let $L$ be a finite language, $A$ be the multiset (say, nondecreasing sequence) of lengths of words in $L$. Then there exists a sum-free submultiser (subsequence) $B$ of $A$ of cardinality $\ge |A|/3$. Take $L_0$ to be the set of all words in $L$ whose lengths are in $B$. $L_0$ is concatenation-free and contains $\ge |L|/3$ words.See the paper "Sum-free subsets" by Alon and Kleitman https://documentcloud.adobe.com/link/track?uri=urn%3Aaaid%3Ascds%3AUS%3A038f5fe3-82b8-4e2a-bd71-53e1612d4dc9, Proposition 1.2.<|endoftext|> -TITLE: Projective module which splits off sequence of submodules, but not the sum -QUESTION [5 upvotes]: Does there exist an example of a module $X$ over some ring $R$ together with submodules $T_i$ such that: - -$X$ is projective, -$X$ splits as an internal direct sum $X\cong T_1\oplus T_2\oplus \ldots \oplus T_n\oplus S_n$ (with some $S_n$) for every $n$, -$X$ does not split off the infinite direct sum $\bigoplus_{i=1}^\infty T_i$, -$R$ is hereditary. - -Remark: I also don't know the answer without the last condition, so this would already be interesting, though for my specific application I definitely need all conditions. - -REPLY [3 votes]: Here is an example if we interpret all direct sums as internal direct sums. -Example. Let $R$ be a discrete valuation ring with uniformiser $\pi$ and fraction field $K$. Let $X = R^{(\mathbf N)}$, and let $T_i$ be the free rank $1$ submodule with basis $\pi e_{i+1}-e_i$. Then the natural map -$$\bigoplus_{i=1}^n T_i \to X$$ -is injective with image $T_{\leq n} = \operatorname{span}(\pi e_2 - e_1, \ldots, \pi e_{n+1} - e_n)$, because the latter clearly has rank $n$. Moreover, -$$S_n = \bigoplus_{i > n} Re_i \subseteq X$$ -is a complement of $T_{\leq n}$: one easily sees that $S_n \cap T_{\leq n} = 0$, and they span $X$ because $e_n = \pi \cdot e_{n+1} - (\pi e_{n+1} - e_n)$, etcetera. But if $T = \bigoplus_{i \in \mathbf N} T_i = \sum_i T_i \subseteq X$, then -\begin{align*} -X/T &\stackrel\sim\to K\\ -e_i &\mapsto \pi^{-i}. -\end{align*} -This surjection does not split because $X$ has no infinitely divisible elements. $\square$ -What's going on is that we wrote $K$ as a filtered colimit of surjections $S_n \twoheadrightarrow S_{n+1}$ of free modules: -$$K = \underset{\substack{\longrightarrow \\ n}}{\operatorname{colim}}\ S_n.$$ -Each $X \twoheadrightarrow S_n$ has a splitting $S_n \hookrightarrow X$, but $X \twoheadrightarrow K$ does not.<|endoftext|> -TITLE: For which G is BLG weak homotopy equivalent to LBG? -QUESTION [8 upvotes]: Let $G$ be a (Edit: path-)connected topological group. Under what additional hypotheses on $G$ is it true that $LBG$ is a classifying space for $LG$? (or, I guess equivalently, when is $LBG \sim BLG$?) Here I'm taking the free loop space, and the compact-open topology on it. I know it is true for strong hypotheses, such as $G$ a compact Lie group. If one can find a model of $BG$ that is locally contractible and paracompact, then by Atiyah and Bott's The Yang-Mills Equations over Riemann Surfaces (doi:10.1098/rsta.1983.0017), Proposition 2.4, I believe it is possible. So, alternatively, for what $G$ is it true that $BG$ can be thus chosen? - -REPLY [16 votes]: [UPDATE: There were some mistakes in the first version. Here is a more careful account.] -I'll work everywhere with CGWH spaces, so I have a Cartesian closed category. -Note that $BLG$ is always path-connected, but $\pi_0(LBG)=\pi_0(G)/\text{conjugacy}$, so we need to assume that $G$ is path-connected. (The question says connected, but this may be a bit stronger; I do not know whether there are connected topological groups that are not path-connected.) -For any space $X$ and any $t\in S^1$ we have an evaluation map $\epsilon_{X,t}\colon LX\to X$. For the special case of the basepoint $1\in S^1$ we write $p_X=\epsilon_{X,1}\colon LX\to X$. This is always a Hurewicz fibration. If $X$ is based we have a fibre sequence $\Omega X\xrightarrow{i_X}LX\xrightarrow{p_X}X$. -Now let $G$ be a topological group. We write $EG$ and $BG$ for the usual simplicial constructions, so $EG$ is contractible and has a free $G$-action with orbit space $BG$. Simplicial methods also give a natural commutative diagram -$\require{AMScd}$ -\begin{CD} -G @>k_{G}>> EG @>r_{G}>> BG\\ -@Vj_GVV @VV l_G V @VV 1 V\\ -\Omega BG @>>> PBG @>>> BG -\end{CD} -Both $EG$ and $PBG$ are contractible, and the bottom row is a Hurewicz fibration. If the top row is also a Hurewicz fibration, we can conclude that $j_G\colon G\to\Omega BG$ is a homotopy equivalence. If the top row is merely a Serre fibration or quasifibration, we can still conclude that $j_G$ is a weak equivalence. I do not know what are the minimal conditions for the top row to be a quasifibration. -Next, given $u\in BLG$ and $t\in S^1$ we have a homomorphism $\epsilon_{G,t}\colon LG\to G$ and thus a map $B\epsilon_{G,t}\colon BLG\to BG$ and thus an element $(B\epsilon_{G,t})(u)\in BG$. We would like to define $f_G\colon BLG\to LBG$ by $(f_G(u))(t)=(B\epsilon_{G,t})(u)$. To justify this we need to check continuity in $t$ and then in $u$. This in turn needs some continuity properties of the functor $B$, which can be proved using some abstract nonsense with CGWH spaces and Cartesian closure. (The initial version of this answer referred to a natural map in the opposite direction, but I think that does not actually exist.) We now want to construct a diagram as follows: -\begin{CD} - B\Omega G @>Bi_G>> BLG @>Bp_G>> BG \\ - @V f'_G VV @V f_G VV @VV 1 V \\ - \Omega BG @>>i_{BG}> LBG @>>p_{BG}> BG -\end{CD} -Constructions that we have already discussed provide all spaces and maps except for $f'_G$. On the top row we note that $(Bp_G)\circ (Bi_G)$ is trivial, and on the bottom row we know that $i_{BG}$ is the fibre of $p_{BG}$, so there is a unique way to fill in $f'_G$. The bottom row is always a Hurewicz fibration. The top row is obtained by applying $B$ to a Hurewicz fibration of topological groups, but it is not clear exactly what we get from that. If the top row is at least a Serre fibration, we see that $f_G$ is a weak equivalence iff $f'_G$ is a weak equivalence. -Finally, define $\tau\colon S^1\wedge S^1\to S^1\wedge S^1$ by $\tau(s\wedge t)=t\wedge s$. From the definitions one can check that the following diagram commutes: -\begin{CD} - \Omega G @>j_{\Omega G}>> \Omega B\Omega G \\ - @V \Omega j_G VV @VV \Omega f'_G V \\ - \Omega^2 BG @>>\tau^*> \Omega^2 BG -\end{CD} -If $j_G$ and $j_{\Omega G}$ are weak equivalences, we conclude that $f'_G$ is also a weak equivalence. -All this assumes that we start with the simplicial definition of $BG$. One could instead consider an axiomatic characterisation of $BG$, which might include the condition that the map $EG\to BG$ is a Serre fibration. The idea should be that $[X,BG]$ should biject with the set of isomorphism classes of principal $G$-bundles over $X$, but one would need to restrict attention to paracompact $X$, or to principal bundles over arbitrary $X$ that admit a numerable trivialising cover. I do not know how the technicalities would work out.<|endoftext|> -TITLE: On complex dynamics in high dimensions -QUESTION [7 upvotes]: I am a fresh Ph.D student and I'm interested in complex dynamics in high dimensions. I have the following questions. - -What research directions are there in several complex dynamics and what problems are in these directions? -In complex dynamics in high dimension, who are famous or active mathematicians should I follow? -If I want to go into this field, what are the necessary background knowledge (e.g complex geometry)? Are there some good survey papers or books to help me? - -Thank you! - -REPLY [15 votes]: Main areas are dynamics of automorphisms (for example, Henon maps), dynamics of endomorphisms, dynamics of foliations, and local dynamics. -Eric Bedford, Tien-Cuong Dinh, John Fornaess, Misha Lyubich, Nessim Sibony, John Smiley, and their students and collaborators. -One-dimensional holomorphic dynamics, functions of several complex variables (especially holomorphic maps and currents), basic complex algebraic geometry, and basics of the general theory of dynamical systems. - -Some books and collections of surveys, -Abate, Marco; Bedford, Eric; Brunella, Marco; Dinh, Tien-Cuong; Schleicher, Dierk; Sibony, Nessim -Holomorphic dynamical systems. -Lectures given at the C.I.M.E. Summer School held in Cetraro, July 7–12, 2008. Edited by Graziano Gentili, Jacques Guenot and Giorgio Patrizio. Lecture Notes in Mathematics, 1998. Springer-Verlag, Berlin; Fondazione C.I.M.E., Florence, 2010. -Fornæss, John Erik; Sibony, Nessim Complex dynamics in higher dimension. Several complex variables (Berkeley, CA, 1995–1996), 273–296, Math. Sci. Res. Inst. Publ., 37, Cambridge Univ. Press, Cambridge, 1999. -Fornæss, John Erik; Sibony, Nessim Complex dynamics in higher dimension. I. Complex analytic methods in dynamical systems (Rio de Janeiro, 1992). Astérisque No. 222 (1994), 5, 201–231.<|endoftext|> -TITLE: $G_{\infty}$ (also known as $E_2$)-operad in terms of trees -QUESTION [7 upvotes]: It's well known that the $A_{\infty}$ and $L_{\infty}$ operads, being resolutions of the associative and Lie operads, admit descriptions as free operads of certain trees. -The description I am referring to for the $A_{\infty}$ operad is described in Voronov's notes: -http://www-users.math.umn.edu/~voronov/8390/lec9.pdf -Similarly the $L_{\infty}$ operad is described in section 2 of the paper by Voronov, Stasheff and Kimura: -https://arxiv.org/pdf/hep-th/9307114.pdf -Does the $G_{\infty}$-operad, the free resolution of the Gerstenhaber (Poisson) operad admit a similar description in terms of trees? How can one define it explicitly? - -REPLY [3 votes]: Yes. All of your examples fall within the techniques of Koszul duality theory for operads, and for this you can consult the standard reference of J.-L. Loday and B. Vallette. But also see G. Ginot's thesis http://www.numdam.org/item/AMBP_2004__11_1_95_0/ for explicit descriptions of $\mathsf{Gers}_\infty$ and $\mathsf{Pois}_\infty$.<|endoftext|> -TITLE: When is a manifold boundary a deformation retract of its open neighborhood? -QUESTION [7 upvotes]: For this question a manifold is a locally upper-Euclidean Hausdorff space; paracompactness or second-countability is not assumed, and boundary may be present. -Let $M$ be a manifold. What are sufficient conditions for there to exist open $U \subset M$ such that $\partial M \subset U$ and $U$ (strongly?) deformation retracts onto $\partial M$? (Could this be true for every manifold?) -Clearly $\partial M$ being collared in $M$ is a fairly general sufficient condition (this includes paracompact manifolds). However, I'd be interested in whether there is a more general sufficient condition than this. - -REPLY [3 votes]: The following theorem and example give (partial) answers. By an $n$-manifold we understand a Hausdorff space whose any point has a neighborhood, homeomorphic to an open subspace of $\mathbb R^n_+=\{(x_1,\dots,x_n)\in\mathbb R^n:x_n\ge 0\}$. -Theorem. For each 1-manifold $M$ the boundary $\partial M$ is a deformation retract of some open set $U\subseteq M$ that contains $\partial M$. -Proof. By the definition of a 1-manifold, for every $x\in\partial M$ there exists a neighborhood $U_x\subseteq M$, homeomorphic to $[0,1)$. Fix a homeomorphism $h_x:U_x\to[0,1)$ such that $h_x(x)=0$. It can be shown that for distinct $x,y\in\partial M$ the sets $U_x,U_y$ are disjoint. Then $U=\bigcup_{x\in \partial M}U_x$ is an open neighborhood of $\partial M$ and the function $H:U\times [0,1]\to U$, $H:(u,t)=h_x^{-1}(t\cdot h_x(u))$ where $u\in U_x$, is a deformation retraction of $U$ onto $\partial M$. -Example. There exists a separable 2-manifold $M$ whose boundary $\partial M$ is homeomorphic to $\mathbb R\times\mathfrak c$. Being non-separable, the boundary $\partial M$ cannot be a retract of a (necessarily separable) neighborhood $U$ of $\partial M$ in $M$. -This example is described by Peter Nyikos as Example 3.6 of his survey paper "The theory of nonmetrizable manifolds" in the Handbook of Set-Theoretic Topology.<|endoftext|> -TITLE: Source of a quote by Ferdinand Rudio -QUESTION [12 upvotes]: I am looking for the source and context of this quote, found e.g. at St Andrews: - -Only with the greatest difficulty is one able to follow the writings of any author preceding Euler, because it was not yet known how to let the formulas speak for themselves. This art Euler was the first to teach. - — F. Rudio - -(My emphasis. Slight variants suggest that it could originally have been in another language.) - -REPLY [21 votes]: The quote is from a speech Rudio gave at the Town Hall in Zürich on the 6th December 1883; The German original is published in Felix Stähelin, Reden und Vorträge (1956, I have not found it online). -An English translation of the full speech is here. The translated quote reads as follows: - -But I cannot move on from reviewing Euler's mathematical work without - having considered an important factor. I have said that mathematics is - a language in which natural phenomena can be described in the simplest - and most comprehensive manner. With this in mind, you will understand - how important it is to express mathematical thoughts themselves as - concisely and clearly as possible. In this respect, Euler's work was - epoch-making. We can be safe to say that the whole form of modern - mathematical thinking has been created by Euler. If you read any - author immediately before Euler, it is very difficult indeed to - understand his terminology, as he has not yet learned how to let the - formulas speak for themselves. This art was not taught until Euler - came along.<|endoftext|> -TITLE: Why is the theory of small categories not algebraic? -QUESTION [9 upvotes]: In "Partial Horn logic and cartesian categories", E. Palmgren and S. J. Vickers state without proof that "The theory of categories is not algebraic." Is there a reference, or an elementary argument, for this fact? In particular, I'm interested in the setting where "algebraic" refers to the multisorted, potentially infinitary setting. -To be precise, this would entail showing that there is no monadic functor from $\mathbf{Cat} \to \mathbf{Set}/S$ for any set $S$. - -REPLY [22 votes]: This follows from two Facts: -1) A category monadic over Set/S is always an exact category. That is it has quotient by equivalence relation that are effective and universal. It is in particular a regular category. This is showed for example here. -2) The category of categories is not a regular category. An explicit example of regularity (and hence exactness) failing in the category of categories (or posets, or topological spaces) can be found in the example section of the nLab page.<|endoftext|> -TITLE: Quotient of a normal quasi-projective variety by a finite group -QUESTION [8 upvotes]: I'm a topologist and not an algebraic geometer, but the following question arose in my work. -Let $X$ be a quasiprojective algebraic variety over $\mathbb{C}$ and let $G$ be a finite group acting on $X$. Since $X$ is quasiprojective, we have the quotient variety $X/G$. -Question: if $X$ is smooth, must $X/G$ be normal? Even better, does this hold if $X$ is assumed not to be smooth, but only normal? -I've googled and searched through various books, but I can't find this anywhere (though as I said, I am not a specialist in algebraic geometry, so it could be the case that something far more general is true and I just don't know the how to translate the general statement into the above rather simple-minded one). - -REPLY [5 votes]: Just an algebraic interpretation of @Simpleton's answer in the case of finite group actions. Let $B$ an integrally closed domain with the field of fractions $L$. Let $G$ be a finite subgroup of ${\rm{Aut}}(L)$. Then the extension $L/L^G$ is Galois and $B^G=B\cap L^G$ is a domain (the superscript denotes the invariant subfield/subring) inside $L^G$. It is easy to see that $L^G$ is the field of fractions of $B^G$. An element of $L^G$ integral over $B^G$ is integral over $B$ and hence lies in $B\cap L^G=B^G$. So $B^G$ is integrally closed. Now recall that normality is a local property which for an affine and irreducible variety $X$ translates to the integral closedness of $\mathcal{O}(X)$. Applying the algebraic result above, if $\mathcal{O}(X)$ is integrally closed, then so is $\mathcal{O}(X/G)=\mathcal{O}(X)^G$.<|endoftext|> -TITLE: Virus community spread mathematical modeling -QUESTION [8 upvotes]: What is the basic math behind the Virus community spread mathematical modeling,and how the time variable;(in these models),interacts with knowledge (data)?. -I am not asking about how the virus is transmitted or how it replicates. - -REPLY [5 votes]: There is actually a very simple model that works reasonably well over time periods with little change. No model can work across all time periods because various communities (or cities) change their behaviour drastically at certain time points. But for each interval during which the community's behaviour is roughly constant, the following model seems to hold quite well for sufficiently large communities: - -$x = a+\frac{n}{2}·\Big(\tanh\big((t-c)·\frac{r}{2}\big)+1\Big)$, where $x$ is the total number of cases and $t$ is time, for some constants $a,n,c,r$. - -What do these constants mean? There are two main kinds of drastic behaviour change that marks the start of the current interval. -The first kind introduces new infected people into the community (imported cases) after it has already stabilized, in which case $a$ is the total number of previous cases and $n$ is the total number of new cases that will appear under current conditions (i.e. if the current interval has no endpoint). The second kind changes the connectivity graph (e.g. lockdown) or the susceptibility of people to the virus (e.g. wearing mask), in which case $a = 0$ and $n$ is the total number of cases that will result under current conditions. -$c$ is simply the inflection point at which the rate of growth is maximum in the current interval. -$r$ is the rate parameter for exponential growth from (roughly) the same starting amount. Specifically, both $n·\exp((t-c)·r)$ and $\frac{n}{2}·\Big(\tanh\big((t-c)·\frac{r}{2}\big)+1\Big)$ are asymptotically the same as $t→-∞$, but diverge from one another as $t→∞$. -Why this relation? Well, consider the following differential equation: - -$\frac{dy}{dt} = s·y·(n-y)$ where $y=x-a$. - -It assumes that the rate of increase in cases is proportional to both the current number $y$ of cases that have been allowed to infect others and the number of cases that would be but have not yet been infected. These are obviously motivated by the fact that the current conditions will lead to some final total and the virus infection rate will be proportional to the number of infected-but-not-isolated cases and the number of to-be-infected cases, and the fact that the infected-but-not-isolated cases is proportional to the allowed-to-infect-others cases. -Solving it gives $\ln(y)-\ln(n-y) = s·t + k$ for some constants $s,k$ and hence we obtain $y = n·\Big(1-\frac1{\exp(s·t+k)+1}\Big) = \frac{n}{2}·\Big(\tanh\big(\frac{1}{2}(s·t+k)\big)+1\Big)$ as desired.<|endoftext|> -TITLE: Dual space of continuous Banach-space-valued functions -QUESTION [8 upvotes]: Let $X$ be a Banach space and $K$ some compact Hausdorff space. I am interested in the dual space of the Banach space -$$C(K; X) = \lbrace f: K \to X, \ f \text{ is continuous}\rbrace, \qquad \lVert f \rVert_\infty := \max_{x \in K} \lVert f(x) \rVert. $$ -In the case, $X = \mathbb C$, it is well known that the dual space is given by the space of all regular Borel measures $\operatorname{rca}(K)$ on the space $K$. -Now I think one can not hope, in the general case that $X$ is a Banach space at least, for this characterization to carry over to the vector-valued case as one needs some kind of lattice structure to define regularity for vector valued measures. Moreover, if one looks at the $L^p$-situation, $1 \leq p <\infty$, one has $L^p(\Omega; X)' = L^{p'}(\Omega; X')$ if and only if the space $X$ has the Radon-Nikodym property. So a natural conjecture for me would be something like the following: -If $X$ is a Banach lattice, maybe with some additional Banach space or lattice properties, e.g., Radon-Nikodym or order continuity of the norm, then one has $C(K; X)' = \operatorname{rca}(K; X')$, where $\operatorname{rca}(K; X)$ denotes the space of $X'$-valued regular Borel measures. -I would expect that people asked and solved this question already. So my question is if a result of this kind of flavour is known? Moreover, are there good references for this kind of results? Thanks in advance! - -REPLY [8 votes]: The natural language to work in here is that of tensor norms. Here I follow Ryan, Introduction to tensor products of Banach spaces. Section 3.2 shows we can identify $C(K;X)$ with the injective Banach space tensor product $C(K) \check\otimes X$. -Thus, we are lead to understand the dual space of $C(K) \check\otimes X$. This (in a lot more generality) can be identified with the integral operators from $C(K)$ to $X^*$, see section 3.5. -Finally, we wish to understand integral operators. Chapter 5 does this. It turns out that every weakly compact operator $T:C(K)\rightarrow X^*$ can be understood as a vector measure $\mu:K\rightarrow X^*$ given by $T(f) = \int_K f(x) \ d\mu(x)$. Of course, we need to know what a vector measure is to fully understand this. Then Proposition 5.28 shows that $T$ is furthermore integral (all integral operators are weakly compact) exactly when $\mu$ has bounded variation. -So, with suitable definitions, the answer is essentially "yes" it all works. You ask about "regularity". For this, see Lemma 5.24. Hahn-Banach, and weak compactness, basically rescues you. -You might also look at the book by Diestel and Uhl, "Vector Measures".<|endoftext|> -TITLE: How to identify cup product with intersection -QUESTION [7 upvotes]: What's the standard generalization and reference for the following statement: -If two oriented submanifolds $L$, $L'$ of an oriented compact manifold $M$ intersect transversally, then the Poincare dual of the fundamental class $[L\cap L']$ equals to the cup product of the Poincare duals of the fundamental classes $[L]$ and $[L']$. -It seems, in algebraic geometry $L$, $L'$ and $L\cap L'$ don't have to be smooth, it is only required that at a generic point of $L\cap L'$ we have that $L$ and $L'$ are smooth and intersect transversally. In the case of non-transversal intersection sometimes more can be said (e.g. multiplicities appear). -I am asking for a topological version of this. -To exclude proofs with differential forms, let's say we want homology/cohomology with integer or arbitrary coefficients. -In fact, I thought of a sketch of a theory along the following lines (in style of Hatcher's book and adopting an approach from Voisin's book): -First define inductively a class of "good subspaces" by saying that a closed subspace $L\subset M$ in a manifold $M$ is a good subspace of dimension $d$ if there is open $U\subset L$ such that $U$ is a submanifold of $M$ and $L\setminus U$ is a good subspace of dimension $d-1$. Let us call manifolds $U\subset L$ for which $L\setminus U$ have smaller dimension generic. -An orientation on a good space is simply an orientation on some generic subset. -Suppose $M$ is oriented and $L$ is an oriented good subspace of $M$ of codimension $n$. We probably can prove that $H^i(M, M-L)=0$ for $i -TITLE: Is every minimal hypersurface in $S^n$ algebraic? -QUESTION [23 upvotes]: Let $S^n$ be the round n-sphere. Wu-yi Hsiang asked in his paper “Remarks on closed minimal submanifolds in the standard riemannian m-sphere” (1967) the follow question - -Is every minimal hypersurface (i.e. codimensional one) in $S^n$ algebraic (i.e. given by intersecting the zero set in $R^{n+1}$ of some homogenous polynomial in n+1 variables with $S^n$) ? - -e.g. the Lawson surface in $S^3$ is given by the imaginary part of $(x_1+ix_2)^m(x_3+ix_4)^l$. -What is the current status of this problem? - -REPLY [18 votes]: The answer to this question is 'no' for most minimal surfaces of revolution in the $3$-sphere. -Consider surfaces in $S^3 = \{\,(z,w)\in\mathbb{C}^2\,|\,|z|^2+|w|^2=1\,\}$ that are invariant under the circle action -$$ -\mathrm{e}^{i\theta}\cdot(z,w) = \bigl(\mathrm{e}^{ip\theta}\,z,\,\mathrm{e}^{iq\theta}\,w\,\bigr) -$$ -where $p\ge q\ge 0$ are relatively prime integers. The ring of polynomials invariant under this circle action is generated by $z\bar z$ and $w\bar w$ (of degree $2$) and the real and imaginary parts of $\zeta = {\bar z}^q\,w^p$ (of degree $p{+}q$). These are subject to a single relation, $\zeta\bar\zeta = (z\bar z)^q(w\bar w)^p$ of degree $2(p{+}q)$. Setting $u = z\bar z-w\bar w$, the mapping $(u,\zeta):S^3\to \mathbb{R}\times\mathbb{C}$ embeds the space of orbits in $S^3$ (where $z\bar z+ w\bar w = 1$) as the algebraic (orbifold) variety $(1{+}u)^q(1{-}u)^p-2^{p+q}\,\zeta\bar\zeta=0$. -A smooth surface in $S^3$ that is invariant under this action and that avoids the circles $z=0$ and $w=0$ can be parametrized in the form -$$ -\bigl(z(\theta,s),w(\theta,s)\bigr) = \left( -\mathrm{e}^{i(p\theta-q^*\phi(s))} -\left(\tfrac12(1+u(s)\right)^{1/2}, -\, -\mathrm{e}^{i(q\theta+p^*\phi(s))} -\left(\tfrac12(1-u(s)\right)^{1/2} - \right) -$$ -for some functions $\phi(s)$, defined modulo $2\pi$, and $u(s)$, satisfying $|u(s)|<1$. In this formula, $q^* = q/(p^2{+}q^2)$ and $p^*=p/(p^2{+}q^2)$. -Calculation shows that such a surface is minimal if and only if the curve $\bigl(u(s),\phi(s)\bigr)$ is, up to reparametrization, a geodesic for the metric -$$ -g = \frac{\bigl((p^2{+}q^2)+(p^2{-}q^2)u\bigr)}{2(1-u^2)}\,\mathrm{d}u^2 + (1-u^2)\,\mathrm{d}\phi^2. -$$ -Moreover, the parametrized surface will be an algebraic surface if and only if $u(s)$ and $v(s) = \tan\phi(s)$ are algebraically related, since, on such a surface, we will have -$$ -\zeta(s) = \mathrm{e}^{i\phi(s)} -\left(\tfrac12(1{+}u(s)\right)^{q/2} -\left(\tfrac12(1{-}u(s)\right)^{p/2} -= \frac{1{+}i\,v(s)}{\sqrt{1{+}v(s)^2}} -\left(\tfrac12(1{+}u(s)\right)^{q/2} -\left(\tfrac12(1{-}u(s)\right)^{p/2} -$$ -When $p\ge q>0$, the metric $g$ is a smooth metric on an orbifold of rotation (the 'poles' of the rotation are the orbifold points, of orders $p$ and $q$ respectively), with Gauss curvature -$$ -K = \frac{2p^2+2q^2+(p^2{-}q^2)\,u}{(\,p^2+q^2+(p^2{-}q^2)\,u\,)^2} > 0. -$$ -(In the special case $(p,q)=(1,0)$, the singularity at the 'pole' $u=-1$ is not an orbifold point, but the metric is, of course, still smooth on the disk $u>-1$.) -Making use of the Clairaut first integral for metrics of revolution, one finds that a geodesic that does not have $\phi$ constant (and hence, does not pass through the orbifold 'poles') and does not have $u$ constant must satisfy a relation of the form -$$ -\frac{m}{1-u^2}\sqrt{\frac{2\bigl((p^2{+}q^2)+(p^2{-}q^2)u\bigr)}{\bigl(1-u^2-4m^2\bigr)}}\,\mathrm{d}u -\frac{\mathrm{d}v}{1+v^2}\, = 0\tag1 -$$ -for some constant $m$ with $0<|m|<\tfrac12$. -Now, it is not hard to show, using Liouville's Theorem, that the first term in $(1)$ is not the differential of any elementary function of $u$ when $p^2{-}q^2>0$. (Obviously, it is the differential of an elementary function when $p^2{-}q^2 = 0$, but, otherwise, it is the differential of a nontrivial linear combination of incomplete elliptic integrals of the third kind in Legendre's terminology.) Hence, $u$ and $v = \tan\phi$ cannot be algebraically related, since expressing $v$, even locally, as an algebraic function of $u$ would express the first term in $(1)$ as the differential of $\tan^{-1}(v)$, which would be an elementary function of $u$. (For those unfamiliar with Liouville's Theorem and the theory of integration in terms of elementary functions, these notes by Brian Conrad give a very useful and clear introduction.) -Meanwhile, the set of values of $m$ satisfying $0< |m| < \tfrac12$ for which the geodesic described by the above relation closes smoothly is dense in the interval $(-\tfrac12,\tfrac12)$, since the integral of the first term in $(1)$ over the interval $u^2<1{-}4m^2$ is not constant in $m$, as it is an odd function of $m$ that is not equal to zero when $m$ is not zero. -Hence there exist such $m$ that yield a closed geodesic on the orbifold of rotation, which lifts to a closed minimal surface in $S^3$ that is not algebraic. -Remark 1: There may be similar arguments that can be made for $S^n$ ($n>3$) using an appropriate symmetry reduction via a group action of cohomogeneity $2$, but not all of the induced metrics on the orbit spaces have a continuous symmetry, and hence the Clairaut integral does not always exist, which was essential in the above argument (so that an appeal could be made to Liouville's Theorem). -Remark 2: There is an interesting example associated to this question: If, instead of the standard metric on $S^3$ of constant sectional curvature $+1$, one considers the 'squashed' (but still smooth and positive definite) metric -$$ -G = \mathrm{d}z\,\mathrm{d}\bar z + \mathrm{d}w\,\mathrm{d}\bar w - (\mathrm{d}f)^2 -$$ -where $f = \tfrac14\sqrt{1+8z\bar z}$ (a smooth function on $S^3$) and sets $(p,q)=(3,1)$, then the metric on the (orbifold) orbit space whose geodesics describe minimal surfaces in this metric that are invariant under the circle action turns out to be the metric on the Tannery pear orbifold (up to a constant multiple). In particular, not only are all the geodesics closed in Tannery's pear, but they are all, in addition, algebraic. Hence, the corresponding minimal tori of revolution in this metric are algebraic as well (using the standard algebraic structure on $S^3$).<|endoftext|> -TITLE: How do you prove that Hochschild cohomology is Morita invariant? -QUESTION [7 upvotes]: I am simply trying to show that $HH^\bullet(A)= HH^\bullet(M_r(A))$ for any matrix ring of $A$. -In Loday's book (Sect 1.5.6) the Morita invariance is explained as follows : it says that if $M$ is an $A$-bimodule, we have -$$HH^\bullet(M_r(A),M_r(M))= HH^\bullet(A,M) $$ -If I put $M=A^*$, we get $HH^\bullet(A)=HH^\bullet(A,A^*)$ (Hochschild cohomology of $A$) -As per the formula, we get the left hand side to be $HH^\bullet(M_r(A),M_r(A^*))$. -We would expect that the left side should give the Hochschild cohomology of $M_r(A)$. Hochschild cohomology of $M_r(A)$ is $HH^\bullet(M_r(A))=HH^\bullet(M_r(A),M_r(A)^*)$. -So it remains to show that $M_r(A^*)=M_r(A)^*$ as $M_r(A)$-bimodules. -While this seems isomorphic as $k$-vector spaces, the isomorphism does not look like it preserves $M_r(A)$-bimodule structure. -Are we getting something wrong? Is the Morita invariance of Hochschild cohomology to be proved in some other way? - -REPLY [5 votes]: It is even a derived invariant. Here is a proof in a special case, but Im not sure whether it works more general (with the same proof?) -Let $A$ and $B$ two noetherian $K$-algebras for a commutative ring $K$ that are projective as $K$-modules. -Then a standard derived equivalence $F=X \otimes_A^L: D^b(A) \rightarrow D^b(B)$ with quasi-inverse $G=Y \otimes_B^L $ induces a derived equivalence between the derived bimodule categories $H=(X \otimes_A^L - ) \otimes_A^L Y : D^b(A^e) \rightarrow D^b(B^e)$ where $A^e=A^{op} \otimes_K A$. -Now the Hochschild cohomology has terms $Ext_{A^e}^l(A,A)$ but $H$ sends $A$ to $B$ and thus preserves Hochschild cohomology. (it also preserves $A^{*}$ in case $A$ is finite dimensional, not sure about the general case)<|endoftext|> -TITLE: Realizing inner automorphisms on Eilenberg-MacLane spaces -QUESTION [12 upvotes]: Let $G$ be a discrete group and let $(X,x_0)$ be a based Eilenberg-MacLane space for $G$, so there is a fixed isomorphism $\pi_1(X,x_0) = G$ and the universal cover $\widetilde{X}$ is contractible. The homology of $X$ is thus the same as the homology of $G$. -Choose some $\gamma \in G$, and let $c_{\gamma} \in \text{Aut}(G)$ be the inner automorphism that conjugates elements of $G$ by $\gamma$. It is standard that $c_{\gamma}$ acts as the identity on $H_k(G)$ for all $k$. -Since $(X,x_0)$ is an Eilenberg-MacLane space for $G$, there is a based map $\psi_{\gamma}\colon (X,x_0) \rightarrow (X,x_0)$ that induces $c_{\gamma}$ on $\pi_1(X,x_0)$, and in fact $\psi_{\gamma}$ is unique up to based homotopy. -Question: Is it possible to write down $\psi_{\gamma}$ in some natural way where in particular it is obvious that it acts as the identity on $H_k(X)$? All the proofs I know that inner automorphisms act trivially on homology are either very algebraic or are based on very specific models of Eilenberg--MacLane spaces that mimic the algebraic constructions, and I'd like to see this fact topologically/geometrically. - -REPLY [11 votes]: Let's assume that the space $X$ is reasonable in the sense that the inclusion $x_0 \hookrightarrow X$ is a cofibration, i.e. has the homotopy extension property. This will hold, for instance, if $X$ is a CW complex and $x_0$ is a vertex. -Represent $\gamma$ by a path $\rho\colon [0,1] \rightarrow X$ with $\rho(0)=\rho(1)=x_0$. We can then use the homotopy extension property to extend $\rho$ to a homotopy $\phi_t\colon X \rightarrow X$ such that $\phi_0 = \text{id}$ and $\phi_t(x_0) = \rho(t)$ for all $t \in [0,1]$. Since $\phi_{1}$ is homotopic to the identity (but through a homotopy where the basepoint moves!), it clearly induces the identity on homology. -So it is enough to prove that $\phi_1\colon (X,x_0) \rightarrow (X,x_0)$ induces $c_\gamma$ on $\pi_1$. Let $(\widetilde{X},\widetilde{x}_0) \rightarrow (X,x_0)$ be the based universal cover of $(X,x_0)$. There is a unique lift $\Phi\colon (\widetilde{X},\widetilde{x}_0) \rightarrow (\widetilde{X},\widetilde{x}_0)$ of $\phi_1$. Identify the orbit of $\widetilde{x}_0$ under the deck group with $G$, so $1 = \widetilde{x}_0$. It is enough to prove that $\Phi(g) = \gamma^{-1} g \gamma$ for all $g \in G$. -The path $\rho$ lifts to a collection of paths that connect $g \in G$ to $g \gamma$ for all $g \in G$. Lifting the homotopy $\phi_t$, we get a homotopy $\widetilde{\phi}_t\colon \widetilde{X} \rightarrow \widetilde{X}$ starting at the identity that slides each $g \in G$ along these paths. You might think that $\widetilde{\phi}_1$ is $\Phi$, but while it is a lift of $\phi_1$, it is not $\Phi$ since it takes the basepoint $1 = \widetilde{x}_0$ to $\gamma$. We have to correct this by multiplying $\widetilde{\phi}_1$ by the element $\gamma^{-1}$ of the deck group, so $\Phi(x) = \gamma^{-1} \cdot \widetilde{\phi}_1(x)$ for all $x \in \widetilde{X}$. In particular, for $g \in G$ we have $\Phi(g) = \gamma^{-1} g \gamma$ as desired.<|endoftext|> -TITLE: Stationary correctness of ultrapowers by low order measures -QUESTION [8 upvotes]: Suppose $U$ is a normal ultrafilter on $\kappa$ of Mitchell order zero, and let $j_U : V \to M$ be the associated embedding. Does there exist a nonstationary $X \subseteq \kappa^+$ such that $X \in M$ and $M \models X$ is stationary? -Note that if $W$ is a normal measure derived from an embedding $i : V \to N$ where $\mathcal P(\kappa^+) \subseteq N$, then $W$ gives a stationary-correct ultrapower, hence the restriction to low Mitchell order. - -REPLY [5 votes]: I once tried and failed to answer this question in the canonical (forgive me) inner models, and your really nice observation about strong cardinals and Mitchell rank helped me finally make some progress. So thanks. Out of gratefulness, I have written the following essay of an answer. I actually only did the version for $L[U]$. -Proposition. Suppose $\kappa$ is a cardinal and $U$ is a normal ultrafilter on $\kappa$. Let $W = U\cap L[U]$. Then in $L[U]$, $\text{Ult}(L[U],W)$ is not stationary correct at $\kappa^+$. -For the proof, we might as well assume $V = L[U]$. Call an ordinal $\alpha < \kappa^+$ relevant if there is a relevance witness for $\alpha$. Relevance witness is a term I made up for a mouse $\mathcal M$ with the following properties: - -The sequence of $\mathcal M$ has a final measure $U^\mathcal M$ which is total on $\kappa$. -$\mathcal M$ projects to $\kappa$ and is sound. -$\mathcal M|\kappa = V_\kappa$. -$\alpha = \kappa^{+\mathcal M}$. - -To the fine-structure averse, I promise that I won't talk much more about the above conditions. They have the key consequence that there is at most one relevance witness for any ordinal $\alpha < \kappa^+$. This is a standard comparison argument which we omit. It barely uses the assumption that $V = L[U]$. It goes through below a strong cardinal (or more generally if $\kappa$ is a cutpoint of the relevant models). It follows, for example, from Lemma 1.11 in Steel's paper "Projectively well-ordered inner models." -Let $S$ be the set of relevant ordinals. Notice that relevance is absolute between $V$ and $\text{Ult}(V,U)$ since $\text{Ult}(V,U)$ is closed under $\kappa$-sequences. Therefore $S\in \text{Ult}(V,U)$. I'm going to show that $\kappa^+\setminus S$, the set of irrelevant ordinals, is stationary in $\text{Ult}(V,U)$ but nonstationary in $V$. Actually, I found it easier to prove the following equivalent statement: $S$ contains a club in $V$, but no club in $\text{Ult}(V,U)$. -Let's start by showing that in $V$, the set of relevant ordinals contains a club. (This is where we really use that $V= L[U]$.) To see this, let $\mathcal P$ be the mouse in our fine-structural hierarchy whose underlying set is $H_{\kappa^{++}}$. Since $V = L[U]$, $\mathcal P$ has a top measure which is total on $\kappa$. For any elementary substructure $H\prec \mathcal P$ containing $\kappa$ and of cardinality $\kappa$, the ordinal $\alpha = H\cap \kappa^+$ is relevant. To see this, let $\mathcal N$ be the mouse given by transtive collapse of $H$. Build the constructible hierarchy over $\mathcal N$ until you reach a mouse $\mathcal M$ that projects to $\kappa$. Then $\mathcal M$ is a relevance witness for $\alpha$. -Finally, let's see that in $\text{Ult}(V,U)$, the set of relevant ordinals does not contain a club. Assume towards a contradiction that it does. In $\text{Ult}(V,U)$, we define $D$ to be the collection of sets $A$ such that there is a club of relevant ordinals $\alpha$ with a relevance witness $\mathcal M$ such that $A\in U^\mathcal M$. -We claim that $D = U$, which contradicts that $D\in \text{Ult}(V,U)$. Clearly $D$ contains $U$, so since $U$ is an ultrafilter, we are finally done if $D$ is a proper filter. This follows immediately from the uniqueness of relevance witnesses, since for any sets $A_0,A_1\in D$, we can intersect clubs to find a single relevant ordinal $\alpha$ and witnesses $\mathcal M_1,\mathcal M_0$ for $\alpha$ such that $A_0\in U^{\mathcal M_0}$ and $A_1\in U^{\mathcal M_1}$. But the uniqueness of relevance witnesses tells us that $\mathcal M_0 = \mathcal M_1$, and so $A_0\cap A_1\neq \emptyset$ since they belong to the proper filter $U^{\mathcal M_0} = U^{\mathcal M_1}$. This concludes the proof. -OK, now that that's over, there are a number of interesting questions. What's consistency strength of stationary correct normal ultrafilters? And at what level of the canonical inner models do they occur? By your observation, it is somewhere below a $2$-strong cardinal. Actually the proof of your observation shows that if $M$ is the first mouse with an extender on its sequence that is not equivalent to a normal ultrafilter, then an initial segment of $M$ thinks there is a proper class of stationary correct normal ultrafilters. So the answer to both the consistency strength question and the inner model theory question lies way below a 2-strong cardinal, somewhere in the hierarchy of Mitchell models with higher order measurables. -I think the above proof generalizes to show that stationary correct ultrafilters do not exist in the minimal inner model for $o(\kappa) = \kappa^++1$ (or in any smaller mouse). You just have to modify Condition 1 in the definition of a relevance witness. Maybe you can make it to $o(\kappa) = \kappa^{++}$. You can also ask whether there's a model of ZFC+ GCH with a stationary correct normal ultrafilter on the least measurable cardinal.<|endoftext|> -TITLE: Linear algebra over non-commutative semirings -QUESTION [5 upvotes]: I'm reading up on linear algebra over semirings, and I'm wondering why people seem to stop short of showing an equivalence between linear transformations between free modules and matrices. -It seems clear to me that over commutative semirings, the usual developments one does for commutative rings go through: a linear transformation $f: R^m\to R^n$ between free left semimodules is such that $f(x+y)=f(x)+f(y)$ and $f(ax)=af(x)$, and then any such $f$ determines a matrix in $R^{n\times m}$ in the usual way; conversely, any such matrix induces a linear transformation. -Now if $R$ is a non-commutative ring, then it seems to be known that the above fails in general, but holds if one works with free bimodules. Similarly, I am quite sure that things go through for free bisemimodules over non-commutative semirings. -Now, two questions: - -Did I say something wrong above? -Why can't I seem to find those things anywhere? Golan's Semirings and affine equations seems like an obvious place to look; he does matrix semirings and linear transformations, but doesn't seem to connect them. Droste et.al.'s Handbook has matrices, but no linear transformations; likewise for all the Bloom/Ésik/Kuich work I know. - -(I'm a bit of an amateur in this area, so forgive me if this is stupid or I missed some literature.) - -REPLY [2 votes]: What I said was mostly right; except that there's no need for the bimodule structure on the $R^n$: -Let $R$ be a semiring and, for $n,m\ge 0$, $R^n$ and $R^m$ the free bimodules over $R$. Say that a function $f:R^m\to R^n$ is left-linear if $f(x+y)=f(x)+f(y)$ and $f(ax)=af(x)$ for all $x,y\in R^m$ and all $a\in R$. -One can easily show that a function $f:R^m\to R^n$ is left-linear iff there is a matrix $M\in R^{m\times n}$ such that $f$ is right-multiplication by $M$, such that $f(x)=xM$ for all $x$. Only the structure of $R^m$ as a right semimodule is needed for this. -Similarly one can define a notion of right linearity, which is then equivalent to left matrix multiplication and uses only the left semimodule structure. -As a last point, if a function is both left and right linear, then its two matrices are transposes of each other and all their elements are in the center of $R$.<|endoftext|> -TITLE: How should one think about the band of a gerbe? -QUESTION [5 upvotes]: Let $X$ be a topological space. Let $\mathcal{F}$ be a fibered category over $X$; seen as an assignment of a category $\mathcal{F}(U)$ for each open $U\subseteq X$. -A fibered catgeory $\mathcal{F}$ over $X$ satisfying some properties, is called a stack over $X$. Further, if $\mathcal{F}(U)$ is a groupoid for each open $U\subseteq X$, we call $\mathcal{F}$ to be a stack of groupoids over the topological space $X$. - -A stack $\mathcal{F}$ over a topological space $X$ is said to be a gerbe - over $X$, if the following conditions are satisfied: - -there exists an open cover $\{U_\alpha\}$ of $X$ such that $Obj(\mathcal{F}(U_\alpha))\neq \emptyset$ for every $\alpha$. -Fix an open set $U\subseteq X$ and objects $a,b$ of $\mathcal{F}(U)$. Then there exists an open cover $\{V_\alpha\}$ of - $U$ such that - $\text{Hom}_{\mathcal{F}(V_\alpha)}\left(a|_{V_\alpha},b|_{V_\alpha}\right)\neq - \emptyset$ for every $\alpha$. - - -Given a gerbe $\mathcal{G}$ over $X$, one can choose an open cover $\{U_\alpha\}$ of $X$ and objects $a_\alpha$ of $\mathcal{G}(U_\alpha)$. This gives sheaves of groups $\underline{\text{Aut}}(a_\alpha)$ for each $\alpha$. Appropriate usage of second condition in the definition of gerbe defines an outer isomorphism of sheaves on $U_{\alpha\beta}$; namely -$$\lambda_{\alpha\beta}:\underline{\text{Aut}}(a_\beta)|_{U_{\alpha\beta}}\rightarrow \underline{\text{Aut}}(a_\alpha)|_{U_{\alpha\beta}}.$$ -Given a gerbe $\mathcal{G}$ on $X$; the collection of sheaves of groupoids $\underline{\text{Aut}}(a_\alpha)$ and outer automorphisms $\{\lambda_{\alpha\beta}\}$ is called the band of the gerbe of $X$. -I understand the definition and some immediate remarks about bands. I could not get the main idea behind the association of a band for a gerbe. Above definition is from the notes Introduction to the language of gerbe and stacks by Ieke Moerdijk. I have had a look at the notes on 1-gerbes and 2-gerbes by Lawrence Breen. -Question : - -How should one think about the band of a gerbe? -It was mentioned in another question that, band of a gerbe was a misguided attempt. Still there are many notes that talks about band of a gerbe. So, is the present day definition different from that of the definition of Giraud? What was the motivation for the change? Is there a better definition or usage or understanding of the notion of a band of a gerbe since then? - -REPLY [6 votes]: if $X$ is a connected topological space without a chosen base point, then what is $\pi_1(X)$? The good answer is of course that it's a groupoid. [Actually, let's also assume that $\pi_n(X)=0$ for $n\ge 2$] -But let's say that that we disallow that good answer, and that we want something more like a group... -Then the next best answer is that it's a group well defined up to isomorphism, where the isomorphism is itself well defined up to inner automorphism. - -Now, sheaffify the above story, and you get what the band of a gerbe is. - -A sheaf of connected spaces with $\pi_n=0$ for $n\ge 2$ is the same thing as a gerbe. -Taking $\pi_1$ (in the "next best answer" way) yields the band of the gerbe. - -REPLY [2 votes]: I will start by explaining the easiest possible case of bundle gerbes, -when the band A (alias structure group) is an abelian Lie group. -A bundle n-gerbe with band A over a smooth manifold M -is a principal bundle over M with its structure ∞-group -being the Lie ∞-group B^n(A). -Here B^n(A) can be described concretely as -a simplicial presheaf S↦UΓ(C^∞(S,A)), -where S ranges over the cartesian site (smooth manifolds diffeomorphic to R^n), -and Γ: Ch→sAb is the Dold–Kan functor and U: sAb→sSet is the forgetful functor. -Bundles over M with structure group B^n(A) can be described as -(derived) maps y(M)→B^{n+1}(A). -Here B^{n+1}(A) was defined above and y(M) denotes the (restricted) Yoneda embedding -of M, i.e., the simplicial presheaf S↦C^∞(S,M), -where the set C^∞(S,M) is turned into a discrete simplicial set. -Such derived maps can be computed by cofibrantly replacing y(M) -and fibrantly replacing B^{n+1}(A) in (say) the local projective model structure -on simplicial presheaves on the cartesian site. -However, B^{n+1}(A) is an objectwise Kan complex and satisfies the homotopy -descent property, so is already fibrant in the local projective model structure. -A cofibrant resolution of y(M) can be written down as the Čech nerve of a good -open cover U of the smooth manifold M. -This is a simplicial presheaf whose presheaf of n-simplices -is the coproduct of representables U_{i_0}∩⋯∩U_{i_n}, -where i ranges over all (n+1)-tuples such that the above intersection is nonempty. -Now computing the derived mapping spaces using the above resolutions -produces the classical Čech complex for bundle gerbes. -For nonabelian gerbes the above formal setup works equally well, -with the proviso that the Lie ∞-group B^n(A) can now be replaced -by any Lie ∞-group G, and instead of B^{n+1}(A) we use B(G), -obtained by using (say) the Dwyer–Kan classifying space functor for simplicial groups -objectwise on the values of the presheaf G. -There is also a nonabelian analog of the functor Γ, -which takes as an input a crossed module or a crossed complex -and produces a simplicial group, which one can use to construct G. -One can then compute the derived mapping space in a similar manner, -the only difference being is that we can no longer convert it to a chain complex -because G need not be abelian. -So the answer to question (1) is that one thinks of the band of an (abelian) -bundle n-gerbe as giving rise to the structure Lie ∞-group B^n(A) -and for question (2) the modern approach dictates that we simply use -simplicial presheaves (or any equivalent formalism, such as quasicategorical ∞-sheaves), -possibly passing from crossed complexes to presheaves of simplicial groups, if desired. - -REPLY [2 votes]: Like for sheaf of groups, you can try to describe the patching data required to glue sheaves of groupoids ("stacks"). A grebe is a sheaf of groupoids, locally equivalent to a sheaf of the form $BG$ for a sheaf of groups $G$. However, since $Hom(BG,BH)$ and $Hom(G,H)$ are not the same, gluing these local $BG$-s together require different descent datum than what you need in order to glue the $G$-s thenselves. In some sense, it is "easier" to glue the classifying groupoids. In fact, $Hom(BG,BH)\cong Hom(G,H)//H$, where here the quotient is the orbit groupoid. It has $\pi_0Hom(BG,BH) = Hom(G,H)/H$ and $\pi_1(Hom(BG,BH),f)=Z_H(Im(f))$, the centralizer of the image of $f$. The gluing data for a grebe takes into account both $\pi_0$ and $\pi_1$. When trying to glue the local $BG$-s in pairs, we need to choose a homotopy class of isomorphisms between their restrictions to the intersection. Namely, for $U_1$ and $U_2$ we need a class in $\pi_0(Hom(BG_1,BG_2))$. But now the compatibility of the choice become an extra structure rather than a property, and the collection of choices of compatibilities of this identifications at triple intersections becomes a torsor for $\pi_1(Hom(BG_1,BG_3))$ where $BG_i$ the the local stack at the open set $U_i$ for $i=1,2,3$. Hence, essentially, the $\pi_1$ controls the degrees of freedom in choosing how the gluing data is compatible. Of course, now this data has to strictly satisfy compatibility on quadraple intersections. -From this description of the descent data for a stack, you immediately see that the band is just what happen when we ignore the $\pi_1$-part. Equivalently, it describes how the local $BG$-s glue when considered as objects of the 1-category of groupoids, where we identify naturally isomorphic functors. While the glued object is not very meaningful, it still gives some partial information about the stack itself, more or less like the components of a top. space gives information about the space itself.<|endoftext|> -TITLE: The homology of the universal covering space, why so difficult to compute -QUESTION [16 upvotes]: Let suppose that we are given a connected CW-complex $X$, such that we know - -All its homology groups. -All its homotopy groups, in particular we know $\pi_{1}(X)$. - -As far as I know there is no spectral sequence converging to the homology of the universal covering $\tilde{X}$ of $X$. Why it is so difficult (I guess there is no a clear method in general) to compute the homology groups of $\tilde{X}$ ? -Edit: I would like to thank all the authors for their answers, I did learn a lot. I had to choose one answer. I am aware that my question was vague enough, but at the and it seems that the answer I was looking for corresponds more to the one given by M. Rivera. - -REPLY [6 votes]: This is not a complete answer to your question, which I am not sure how to answer without having a precise meaning for what you mean by "computing" and what you mean by "knowing" $\pi_1(X)$, as indicated in the comments above, and given that you have not included the data of any action of $\pi_1(X)$. -However, I think you may be interested in taking a look at my paper with Mahmoud Zeinalian "Singular chains and the fundamental group". -In this article, we prove that for any path connected pointed space $(X,b)$, the $E_{\infty}$-coalgebra structure (in fact, the $E_{2}$ part of it) of (a pointed version of) the singular chains $C_*(X,b)$ extending the Alexander-Whitney coproduct determines $\pi_1(X,b)$ functorially and in complete generality. The correct notion of weak equivalence under which this information is preserved is defined via the cobar functor. -In particular, the natural algebraic structure on the singular chains $C_*(X,b)$ determines a chain complex (by purely algebraic means) which calculates the homology of the universal cover of $X$ at $b$. The construction is the following: consider the differential graded connected coassociative coalgebra $C=(C_*(X,b), \partial, \Delta)$ and take its cobar construction $\Omega C$. This is a dg associative algebra such that $H_0(\Omega C)\cong \mathbb{Z}[\pi_1(X,b)]$, the fundamental group ring. There is a natural "twisting cochain" (in the sense of Brown) $\tau: C \to H_0(\Omega C)$ through which we may construct a twisted tensor product $(C \otimes_{\tau} H_0(\Omega C), \partial_{\tau})$. The homology of this chain complex is the homology of the universal cover. Note that the differential $\partial_{\tau}$ uses both the dg coassociative coalgebra structure of $C$ and the algebra structure of $H_0(\Omega C)$ (which in turn, was obtained from the dg coassociative coalgebra structure of $C$). Also, this construction is invariant in the following "Koszul duality" sense: if $f: C\to C'$ is a map of dg coalgebras such that $\Omega f: \Omega C \to \Omega C'$ is a quasi-isomorphism of dg algebras then the induced map $$(C \otimes_{\tau} H_0(\Omega C), \partial_{\tau}) \to (C' \otimes_{\tau} H_0(\Omega C'), \partial_{\tau})$$ is a quasi-isomorphism. This notion is strictly stronger that ordinary quasi-isomorphisms of dg coalgebras. In some sense, this suggests that you need more information than just the homology to determine the action of $\pi_1$ on the universal cover. -This fact can be used to show an extension of classical theorem of Whitehead, namely we can now prove that a continuous map of pointed path connected spaces $f: (X,b) \to (Y,c)$ is a weak homotopy equivalence if and only if the induced map at the level of dg coassociative coalgebras of pointed singular chains $f: C_*(X,b) \to C_*(Y,c)$ becomes a quasi-isomorphism after applying the cobar functor. -Note that I didn't use the homotopy groups of $X$ directly but I used more information than the homology of $X$, namely, the chain level natural algebraic structure of the singular chains. Again, the question of "how computable this is" is a different one, which I am not sure how to formulate precisely (and it may be tricky given that group theory is not decidable).<|endoftext|> -TITLE: Does $\chi(1)^2=|G:Z(G)|$ for irreducible character of a finite group $G$ imply $G$ is solvable? -QUESTION [11 upvotes]: In "Character Theory of Finite Groups" I.M. Isaacs mention the following conjecture: -It is only possible in a solvable group $G$ to have $\chi(1)^2=|G:Z(G)|$ with $\chi \in$ Irr$(G)$. -Is this problem still open? I tried to search for attempts to solve it but didn't find anything. - -REPLY [13 votes]: Such groups are solvable. This has been solved by Howlett and Isaacs himself, in -Howlett, Robert B.; Isaacs, I. Martin, On groups of central type, Math. Z. 179, 555-569 (1982). MR652860 ZBL0511.20002. -The proof uses the classification of finite simple groups.<|endoftext|> -TITLE: Refined Euler characteristic -QUESTION [7 upvotes]: Is there a refinement of Euler characteristic that distinguishes between the torus $S^1 \times S^1$ and the cylinder $S^1 \times [0,1]$? -(The intuition here is that $\chi$ is multiplicative, so that $\chi(S^1 \times S^1) = \chi(S^1) \times \chi(S^1) = 0 \times 0$, which “vanishes twice”.) -Relatedly, is there a setting in which the Euler characteristic emerges as an eigenvalue, or more broadly as a root of an equation? This would provide a sense in which 0 might occur with multiplicity greater than 1. - -REPLY [13 votes]: Look up the Poincare polynomial $p_X(t)$. It is still multiplicative, by the Kunneth formula. The Euler characteristic is $\chi_X=p_X(-1)$. We have $p_{S^1\times S^1}(t)=(t+1)^2$ but $p_{S^1\times [0,1]}(t)=(t+1)$, with the former having a double root at $t=-1$.<|endoftext|> -TITLE: Expansion of $\det(A+B)$ -QUESTION [14 upvotes]: If $A,B\in{\bf M}_n(k)$, then the following formula holds true: -$$\det(A+B)=\sum_{r=0}^n\sum_{|I|=|J|=r}\epsilon(I,I^c)\epsilon(J,J^c)A\binom IJ B\binom{I^c}{J^c}.$$ -In this formula, $I$ and $J$ are ordered (increasingly) $r$-uplets in $[1,n]$, $A\binom IJ$ is the corresponding minor, $I^c$ is the ordered complement of $I$ and $\epsilon(I,I^c)$ is the signature of the permutation thus defined. - -I should like to have a reference for this formula. Thanks in advance. - -REPLY [13 votes]: The formula appears in -Marvin Marcus "determinants of sums". College mathematics journal, March 1990. -https://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/determinants-of-sums -The author notes that it is hard to track it's origins, but mentions previous references in the text, in particular to a book of himself in 1975.<|endoftext|> -TITLE: Bounding an "integral" from below by the Hausdorff measure of the domain -QUESTION [6 upvotes]: Let $(X,d)$ be an arbitrary metric space and $E \subset X$ also arbitrary. Fix $s \in (0,\infty)$. - -Is it true that for any $ \delta > 0 $ and any collection of pairs $\{(A_i,a_i)\}_{i \in \mathbb{N}}$ where $A_i$ are subsets of $X$ and $a_i \in [0,\infty]$, if - $$ -\text{diam} \, A_i \leq \delta \quad for \ \ \ all \quad i\in \mathbb{N}, -$$ - and - $$ -\chi_E(x) \leq \sum_{i} a_i \, \chi_{A_i}(x) \quad for \ \ \ all \quad x \in X \ , -$$ - then - $$ -\mathcal{H}^s_{5\delta}(E) \leq C \sum_i a_i \, (\text{diam} \, A_i)^s \ , -$$ - with a $C$ that does not depend on $\delta$? - -The infimum of all such $ \sum_i a_i \, (\text{diam} \, A_i)^s $ can be viewed as an "integral" of the characteristic function of $E$, or lternatively as the "weighted" Hausdorff measure of $E$. -Notation: -1) $\chi_G$ stands for the characteristic function, -2) $\mathcal{H}^s_{5\delta}$ stands for the approximating Hausdorff measure. The ones that appear in the definition of the Hausdorff measure: $\mathcal{H}^s(G) = \lim_{\delta \to 0} \mathcal{H}^s_{\delta}(G).$ -Remarks: -3) Of course the interesting case is when $ 0 < a_i \leq 1$ because if $a_i > 1$ then the pair $(A_i,a_i)$ can be replaced by $(A_i,1)$. -4) I am interested only in the asymptotic $\delta \to 0$, so feel free to assume $\delta$ is small. -5) I don't care if $5 \, \delta$ becomes $563 \, \delta$. As the joke goes, show it for some $5$ in the natural numbers! -Motivation: A proof of this will be a significant step in my attempt to give a proof of the coarea inequality, also known as the Eilenberg's inequality: Fix arbitrary metric spaces $X$ and $Y$, and pair of non-negative integers $\mu$ and $q$. Then for any lipschitz map $ g: X \to Y$ and any subset $E \subset X$, -$$ -\int^*_Y \mathcal{H}^\mu (g^{-1}(y) \cap E) d\mathcal{H}^q(y) \leq (\text{Lip g)}^q \ \frac{\omega _\mu \omega_q}{\omega_{\mu+q}}\mathcal{H}^{\mu + q}(E) \, . -$$ -Here $\omega_k$ is the volume of unit ball in $\mathbb{R}^k$. - -REPLY [7 votes]: It is true and deserves to be known better (it is morally equivalent to the statement that the Frostman lemma is just an exercise in linear programming duality though there are some pesky details I don't want to go into now). -I'll assume that all $A_j$ are balls $B_j$. It may change 5 to 10 but I don't think you care. -Consider first the case when the cover is finite. Then it is possible to choose the non-negative coefficients $a_j$ so that $\sum_ja_j\chi(B_j)\ge \chi_E$ and $\sum_j a_jd(B_j)^s$ is the minimal possible under this restriction. Now, in this optimal choice, take any ball with $a_j>0$ and consider all other balls $B_i$ that intersect it. -Claim $a_jd(B_j)^s+\sum_i a_id(B_i)^s\ge\frac 12d(B_j)^s$ -Proof If $a_j\ge \frac 12$, there is nothing to do. Otherwise we can replace $a_j$ by $a_j-h$ and all $a_i$ by $a_i(1+2h)$ and still have an admissible choice of coefficients. Since our choice was optimal, we have the inequality -$$ --hd(B_j)^s+2h\sum_i a_id(B_i)^s\ge 0 -$$ -and we are done again. -Now just run the standard Vitali construction throwing out the biggest participating ball $B_j$ with all balls that intersect it and $3B_j$ from $E$ and reoptimizing the remaining picture after each throw (the last step is needed because otherwise the inequality in the claim may involve some balls that are already gone). -The reduction of the infinite case to the finite one is standard. Choose a finite part of the sum $\sum_j a_jd(B_j)^s$ that is at least $0.9$ of the whole sum and consider the set on which the corresponding linear combination of the characteristic functions is at least $1/3$. Then through these pieces away and triple the remaining coefficients. Repeat. The point is that if $\sum_ku_k\ge 1$ then there exists $u_k$ that is at least $3^{-k}$, so no point of $E$ will escape and $\mathcal H^s_{10\delta}$ is countably subadditive.<|endoftext|> -TITLE: How can we find n points on a plane so that as many pairs of points as possible have the same distance? -QUESTION [10 upvotes]: There are $n$ points on the plane, and we need to maximize the number of pairs of points which have the same Euclidean distance. - -REPLY [14 votes]: The number is tabulated at OEIS. It seems that it's only known up to $n=14$ (and some scattered larger values). Links are given there to some papers on the topic. Evidently, no one knows how to do it for general $n$. -Also discussed on math.stackexchange.<|endoftext|> -TITLE: Commutator of finite global dimension algebras -QUESTION [5 upvotes]: Let $A=KQ/I$ be a finite dimensional quiver algebra of finite global dimension. -Is it true that the dimension of $A/[A,A]$ is equal to the number of simples of $A$? -Here $[A,A]$ is the vector space generated by all elements of the form $ab-ba$. -Note that it is known that in general for such $A$ that the dimension of $A/([A,A]+rad(A))$ is equal to the number of simple $A$-modules. Thus the question should be equivalent to asking whether we have $rad(A) \subseteq [A,A]$ in case $A$ has finite global dimension. - -REPLY [6 votes]: Yes. See the result of Section 2.5 of a wonderful paper of Bernhard Keller : -https://webusers.imj-prg.fr/~bernhard.keller/publ/ilc.pdf -(and the references therein).<|endoftext|> -TITLE: Complex cobordism and Chern numbers -QUESTION [11 upvotes]: Let $L$ be the Lazard's universal ring, and $R=\mathbb{Z}[b_1,b_2,\cdots,b_n,\cdots]$, regarded as a graded ring with the degree of $b_i$ equal to $2i$. Let $\theta: L\rightarrow R$ be the homomorphism carrying the universal formal group law $\mu^L$ to the formal group law -$$\mu^R(x_1,x_2)=\exp(\log(x_1)+\log(x_2)),$$ -where the power series -$$\exp(x)=x+\sum_{i\geq 1}b_ix^{i+1},$$ -and $\log(x)$ its inverse, denoted as -$$\log(x)=x+\sum_{i\geq 1}m_ix^{i+1}.$$ -Let $MU$ be the complex cobordism spectrum, and by Quillen's theorem we have the following commutative diagram -$\require{AMScd}$ -\begin{CD} -L @>\theta>> R\\ -@V \cong V V @VV \cong V\\ -\pi_*(MU) @>>h> H_*(MU;\mathbb{Z}) -\end{CD} -where $h$ is the Hurewicz homomorphism. -In Section 9, Part II of -Adams, J. F., Stable homotopy and generalised homology, Chicago Lectures in Mathematics. Chicago - London: The University of Chicago Press. X, 373 p. 3.00 (1974). ZBL0309.55016.** -it is stated that the class $[\mathbb{C} P^n]\in\pi_*(MU)$ is sent to $(n+1)m_n\in H_*(MU;\mathbb{Z})$ by $h$, and it is indicated there that the argument is a Chern number computation, but I am not seeing the argument.** -I would greatly appreciate your help if you could sketch the proof or point out a reference containing a proof. Thank you! - -REPLY [9 votes]: It is a key result that the composite -$$ MU_* \xrightarrow{h} H_*(MU;\mathbb Z) \xrightarrow[\sim]{\Phi^{\vee}}H_*(BU;\mathbb Z),$$ -where $\Phi^{\vee}$ is the dual of the Thom isomorphism $\Phi$, agrees with evaluating on normal Chern numbers. -In other words, $\langle \Phi(c), h([M])\rangle = \bar c(M)$ for all $c \in H^*(BU)$ and for all $[M] \in MU_*$. -(A reference in the real case is the diagram on page 228 of A concise course in algebraic topology by J.P.May. The complex case is identical.) -So the assertion is just a Chern number calculation: -$h([\mathbb CP^n]) = (n+1)m_n$ if and only if, for all $c \in H^{2n}(BU;\mathbb Z)$, $$\langle \Phi(c), (n+1)m_n\rangle = \bar c(\mathbb CP^n).$$<|endoftext|> -TITLE: Minors of low rank skew-symmetric matrix -QUESTION [9 upvotes]: Let $A$ be an $n\times n$ skew-symmetric matrix of rank $r$. -Given subsets $X$ and $Y$ of row and column indices respectively, let $A_{X,Y}$ denote the submatrix of $A$ obtained by only keeping rows with indices in $X$ and columns with indices in $Y$. -Prove that for any subsets $X, Y\subseteq \{1, 2, \ldots, n\}$ each of size $r$, we have -$$\det A_{X,X} \cdot \det A_{Y,Y} = (-1)^r (\det A_{X,Y})^2.$$ -I've heard that this theorem is due to Frobenius, but have not been able to track down a reference that proves this result. - -REPLY [5 votes]: Since no one else has posted a complete answer so far, let me give one. -Note that it is only complete in the sense of answering the OP's question; -several other questions arise that I cannot easily address. -In Section 1, I will prove the main result (Theorem 1), which is more general -than the OP's equality. In Section 2, I will derive the latter from the -former. In Sections 3 and 4, I will generalize the statement and ask further questions. -1. On size-$r$ minors of a rank-$r$ matrix -We fix a field $\mathbb{K}$. (In Section 3, we will generalize this to a -commutative ring.) -Let $\mathbb{N}=\left\{ 0,1,2,\ldots\right\} $. For each $n\in\mathbb{N}$, -let $\left[ n\right] $ denote the set $\left\{ 1,2,\ldots,n\right\} $. -If $A=\left( a_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq m}\in -\mathbb{K}^{n\times m}$ is an $n\times m$-matrix (for some $n,m\in\mathbb{N} -$), and if $I\subseteq\left[ n\right] $ and $J\subseteq\left[ m\right] $ -are arbitrary subsets, then $A_{I,J}$ will denote the submatrix $\left( -a_{i_{x},j_{y}}\right) _{1\leq x\leq p,\ 1\leq y\leq q}$ of $A$, where the -subsets $I$ and $J$ have been written as $I=\left\{ i_{1} -TITLE: Girth and diameter of a graph with minimum degree at least 3 -QUESTION [10 upvotes]: The problem is motivated by generalizing Moore graphs, graphs with maximum possible girth ($2\text{diam}+1$) given the diameter. -Question. Does there exist a graph $G$ with $\text{g}(G)-\text{diam}(G)>8$ and minimum degree at least $3$, where $\text{g}(G)$ and $\text{diam}(G)$ are the girth and diameter of $G$, respectively? -Remark. There's an infinite class of graphs with $\text{g}(G)-\text{diam}(G)=8$: the (point,line) incidence graphs of generalized octagons have $\text{g}(G)=16$ and $\text{diam}(G)=8$. -Bonus points (+200 bounty) for a proof that $\text{g}-\text{diam}$ is unbounded for graphs with minimum degree at least $3$. - -REPLY [7 votes]: The recent preprint "A randomized construction of high girth regular graphs" by Linial and Simkin suggests in its closing remarks that the answer (at least, to your bonus question) is unknown: - -The possible relation between a graph’s girth and its diameter is particularly intriguing. It follows from [12] and Moore’s bound that -$$2 \ge \mathrm{lim\ sup} \frac{\mathrm{girth}(G)}{\mathrm{diam}(G)} \ge 1,$$ -where the lim sup ranges over all graphs where all vertex degrees are $\ge 3$. Nothing better seems to be known at the moment. -Even more remarkably, we do not know whether -$$\mathrm{sup}(\mathrm{girth}(G) − \mathrm{diam}(G))$$ -is finite or not. The sup is over all $G$ in which all vertex degrees are $\ge 3$.<|endoftext|> -TITLE: Polynomial algebra and its special ideals -QUESTION [8 upvotes]: Consider a polynomial algebra $A=\mathbb{K}[x_1,\ldots,x_n]$ and its ideal $I$, such that $A/I=\mathbb{K}[y_1,\ldots, y_k]$. Is it true that there exist new polynomial generators $z_1,\ldots,z_n$ (in a sense that $A=\mathbb{K}[z_1,\ldots,z_n]$) such that $I=(z_1,\ldots,z_{n-k})$? - -REPLY [9 votes]: This is called the Embedding Problem. I believe that it is false in positive characteristics and unknown in characteristic zero, except a few small cases. See Kraft's review -for the extent of my knowledge of the state of the problem. The counterexample in characteristic $p$ is from this paper: -$$ -K[x,y]/(y^{p^2}-x-x^{2p}) -$$<|endoftext|> -TITLE: NCF, P-points, weak P-points, and cardinalities -QUESTION [5 upvotes]: The post is a bit long, but all the questions are similar or concern the same topic. -Let $\omega^*=\beta\omega\setminus\omega$. A well-known topological definition of a P-point (on $\omega$) is as follows: a point $x\in\omega^*$ is a P-point if every intersection of countably many open neighborhoods of $x$ contains an open neighborhood of $x$. Similarly, a point $x\in\omega^*$ is a weak P-point if $x$ is not in the closure of any countable subset not containing $x$. We have also an equivalent definition of P-points in terms of functions $\omega\to\omega$: an ultrafilter $x\in\omega^*$ is a P-point if for every function $f\colon\omega\to\omega$ there is $A\in x$ such that $f\restriction A$ is either constant or finite-to-one. Since every P-point is a weak P-point, my first question is as follows: -Question 1. Does there exist a similar characterization of weak P-points (i.e. in terms of functions $\omega\to\omega$)? -Kunen proved in ZFC that we always have $2^{\mathfrak{c}}$ weak P-points and at least $\mathfrak{c}$ incomparable in the sense of Rudin-Keisler weak P-points. My next questions concern the total number of (incomparable/incompatible) P-points provided that there is at least one P-point. -Question 2. Assume that a P-point exists. (a) Does there exist another (non-isomorphic or incomparable) P-point? (b) Do there exist $2^{\mathfrak{c}}$ different P-points? (c) Do there exist $\mathfrak{c}$ incomparable P-points? -(It is easy to see that there exist $\mathfrak{c}$ isomorphic P-points.) -Question 3. Assume that there exist $2^{\mathfrak{c}}$ many P-points. (a) Do we have then $2^{\mathfrak{c}}$ (or at least $\mathfrak{c})$ incomparable P-points? (b) Do there exist $2^{\mathfrak{c}}$ (or at least $\mathfrak{c}$, or even $2$) incompatible P-points? -Let us say that two ultrafilters $U,V\in\omega^*$ are near coherent if there exists a finite-to-one function $f\colon\omega\to\omega$ such that $f(U)=f(V)$. The near coherence is an equivalence relation, so we can count the number of the equivalence classes. The Near Coherence of Filters principle (NCF in short) states that there exists only one equivalence class. Blass and Shelah constructed a model of set theory in which the NCF holds (it is now also known to hold in the Miller model). On the other hand, Banakh and Blass proved that either we have finitely many equivalence classes or $2^{\mathfrak{c}}$ (the latter holds e.g. in each model where $\mathfrak{u}\ge\mathfrak{d}$, so e.g. under CH). The next question is in the same spirit as Questions 2 and 3. -Question 4. Assume that we have $2^{\mathfrak{c}}$ many near coherence classes. Does there exist $\mathfrak{c}$ (or $2^{\mathfrak{c}}$) many incompatible in the sense of Rudin-Keisler ordering weak P-points? -It is believed (but not proved so far) that there exists a model with exactly $2$ near coherence classes and for every $n>2$ there is no model with $n$ classes. Also, the NCF implies the existence of a P-point. Thus, my next (and last) question is the following. -Question 5. Assume there are exactly 2 classes of near coherence. Does there exist any P-point? If yes, then are there two that are not compatible? -Thank you very much for the help! - -REPLY [5 votes]: An answer to question 2: Shelah constructed a model with exactly one $P$-point (up to isomorphism). In fact, the one $P$-point is a selective ultrafilter. You can find the construction in section XVIII.4 of Proper and Improper Forcing. -An answer to question 5: If there are exactly $2$ classes of near coherence, then $\mathfrak{u} < \mathfrak{d}$. (You say this in your post, just before Question 4.) In other words, there is a non-principal ultrafilter generated by fewer than $\mathfrak{d}$ sets. Ketonen proved that any ultrafilter generated by fewer than $\mathfrak{d}$ sets is a (non-selective) $P$-point. See - -J. Ketonen, "On the existence of $P$-points in the Stone-Cech compactification of the integers," Fundamenta Mathematicae 92 (1976), pp. 91-94. (link)<|endoftext|> -TITLE: Lax pair of an integrable non-linear PDE -QUESTION [6 upvotes]: The following is a fourth-order non-linear PDE that passes the Painleve integrability test -$$\left(1+x^{2}\right)^{2}u_{xxxx} + 8x\left(1+x^{2}\right)u_{xxx} + 4\left(1+3x^{2}\right)u_{xx}+ t\left(2xuu_{xx} + \left(1+x^{2}\right)\left(uu_{xx}\right)_{x} - 4\left(1+3x^{2}\right)u_{xxt} - 4x\left(1+x^{2}\right)u_{xxxt}\right)=0,$$ -where $u=u(x,t)$. The leading-order behaviour of this PDE is of the form $u=A_{0}\left(x-x_{0}\right)^{-1}$, where $A_{0}=3\left(1+x_{0}^{2}\right)t^{-1}$. I am looking for a Lax pair $[\mathcal{L},\mathcal{M}]$ which satisfies the equation -$$\dot{\mathcal{L}}+[\mathcal{L},\mathcal{M}]=0,$$ -but not able to find the same. I suspect that this may be due to the fact that although few systems may pass the Painleve test, they need not be integrable. Any help/comment(s) about the problem and/or the integrability of the PDE is appreciated. Also, are there any other (stronger) methods using which I can explicitly try and probe for the integrability of the PDE although it passed the Painleve test? - -REPLY [2 votes]: You could try using the Wahlquist-Estabrook prolongation structure technique, per H.D. Wahlquist and F.B. Estabrook, J. Math. Phys 16 (1975) 1-7 (covering the Korteweg-deVries equation), & F.B. Estabrook and H.D. Wahlquist J. Math. Phys. 17 (1976) 1293-1297 (covering the nonlinear Schrödinger equation). -A non-trivial prolongation structure would be a signal that your equation is indeed integrable. To construct a Lax pair, you need to find an explicit representation of the prolongation structure: the paper by R. Dodd and A. Fordy Proc Roy. Soc. Lond. A385 (1983) 389-429 provides a method of doing this. -Since your PDE includes both independent variables in the coefficients, you may also wish to review the note by B.A. Kupershmidt, Nonautonomous form of the theory of Lax equations, Lettere al Nuovo Cimento 33 (1982) 103–107.<|endoftext|> -TITLE: Is the number of commutation classes of reduced words of the longest element of $S_n$ even for $n\geq 3$? -QUESTION [6 upvotes]: Observably, the number of primitive sorting networks on $n$ elements (or the number of commutation classes of reduced words of the longest element of $S_n$) is even for $3\leq n\leq 15$. These are all the values for which it has been computed so far. Is there an abstract way to see that it is always even? -Sequence is at http://oeis.org/A006245 - -REPLY [6 votes]: On the linked OEIS entry I find the following: - -Also the number of mappings X:{{1..n} choose 3}->{+,-} such that for any four indices a < b < c < d, the sequence X(a,b,c), X(a,b,d), X(a,c,d), X(b,c,d) changes its sign at most once (see Felsner-Weil and Balko-Fulek-Kynčl reference). - Manfred Scheucher, Oct 20 2019 - -There is an obvious fixed-point free involution on the set of such maps given by switching "+" and "-", assuming that $n \geq 3$ so that $\binom n 3 > 0$.<|endoftext|> -TITLE: On the global dimension of an endomorphism algebra -QUESTION [6 upvotes]: Let $G_n$ be the elementary abelian 2-group with $2^n$ elements and $R=R_n:=KG$ the group algebra over the field with 2 elements. -Let $M_n$ be the direct sum of all non-projective modules of the form $uR$ for some element $u \in R$ up to isomorphism. Let $B_n:=End_{R_n}(M_n)$. - -Question: For which $n$ does $B_n$ have finite global dimension? - -Interestingly, $B_n$ has Cartan determinant equal to -1 for $n=3$ and has finite global dimension for $n=1$ and $n=2$. So in case $B_3$ has finite global dimension, this would give a counterexample to the Cartan determinant conjecture. Sadly $B_3$ has vector space dimension 813 and thus even with a computer it looks impossible to understand the algebra $B_3$ directly, but maybe there is a trick? - -REPLY [4 votes]: In my answer to does-this-algebra-have-finite-global-dimension-human-vs-computer I recalled how one can go about computing a projective resolution of the simple modules of an endomorphism ring of a module. There is a missing piece of code in QPA to do this in general, but now I have made an attempt to rectify this. The code is not made public yet, but hopefully soon it will. Using this new code it seems that for one simple $B_3$-module, the third and the fourth syzygy of this particular simple $B_3$-module has a common isomorphic direct summand. Hence the projective dimension of this particular simple $B_3$-module is infinite and consequently the global dimension of $B_3$ is also infinite. I am stressing that the calculations only indicating this, I don't have a proof of this. -Addition April 17, 2020: Here is a copy of the GAP-session computing the above (using the newly uploaded additions to QPA as of April 17th, 2020): -gap> G := ElementaryAbelianGroup( 8 ); - -gap> A := GroupRing( GF( 2 ), G ); - -gap> B := AlgebraAsQuiverAlgebra( A )[ 1 ]; -]/]>, (6 generators)>> -gap> elements := Elements( RadicalOfAlgebra( B ) );; -gap> mods := List( elements, m -> RightAlgebraModule( B, \*, RightIdeal( B, [ m ] ) ) );; -gap> Mods := List( mods, m -> RightAlgebraModuleToPathAlgebraMatModule( m ) ); -[ <[ 0 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, - <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, - <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, - <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, - <[ 4 ]>, <[ 4 ]>, <[ 1 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, - <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, - <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, - <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, - <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]> ] -gap> n := Length( Mods ); -128 -gap> -gap> non_isos := [ ]; -[ ] -gap> isos := [ ]; -[ ] -gap> multiplicities := [ ]; -[ ] -gap> for i in [ 2..n - 1 ] do -> if not i in isos then -> Add( non_isos, i ); -> testset := [ i + 1..n ]; -> SubtractSet( testset, isos ); -> num := 0; -> for j in testset do -> if IsomorphicModules( Mods[ i ], Mods[ j ] ) then -> num := num + 1; -> Add( isos, j ); -> fi; -> od; -> Add( multiplicities, num + 1 ); -> fi; -> od; -gap> -gap> basiclist := List( non_isos, i -> Mods[ i ] ); -[ <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, - <[ 2 ]>, <[ 2 ]>, <[ 4 ]>, <[ 2 ]>, <[ 2 ]>, <[ 2 ]>, <[ 1 ]> ] -gap> Length( basiclist ); -22 -gap> test := ProjectiveResolutionOfSimpleModuleOverEndo( basiclist, 1, 4 ); -[ "projdim > 4", [ <[ 3 ]>, <[ 10 ]> ] ] -gap> U := test[ 2 ][ 1 ]; -<[ 3 ]> -gap> V := test[ 2 ][2]; -<[ 10 ]> -gap> decomp := DecomposeModule( V ); -[ <[ 3 ]>, <[ 7 ]> ] -gap> IsomorphicModules( U, decomp[ 1 ] ); -true<|endoftext|> -TITLE: Definition of functions in the induced space from parabolic induction -QUESTION [7 upvotes]: Let $P$ be a parabolic subgroup of a connected, reductive group $G$ over a $p$-adic field. Let $M$ be a Levi subgroup of $P$, and let $N$ be the unipotent radical of $P$. If $(\pi,V)$ is a smooth, irreducible representation of $M$, extend $\pi$ to a representation of $P$ by making it trivial on $N$, and let $\sigma = \operatorname{Ind}_P^G \pi$, the smooth representation of $G$ obtained by parabolic induction. -By definition, a function $f: G \rightarrow V$ lies in the space of $\sigma$ if the following conditions are met: - -$f$ is locally constant. -$f(mng) = \pi(m)f(g)$ for all $m \in M, n \in N, g \in G$. -There exists an open compact subgroup $K$ of $G$, depending on $f$, such that $f(gk) = f(g)$ for all $g \in G$ and $k \in K$. - -Is the third condition redundant in this definition? I know in the general case for smooth induction in totally disconnected groups, it is necessary, but I have thought that since $P \backslash G$ is compact, there should be some way to show the third condition from the first two. I haven't been able to do this. I have seen some authors leave out the third condition in the definition of parabolic induction. - -REPLY [2 votes]: Let $H$ be any subgroup such that $H\backslash G$ is compact. Let $K$ be an open subgroup. Then there are $x_1,...x_n$ such that $G=Hx_1 K \cup...\cup Hx_nK.$ Suppose $f$ satisfies points 1 and 2. Replace $K$ with a smaller $K'$ so that $f(x_i k)=f(x_i)$ for all $i$ and for all $k \in K'.$ Given $a \in G,$ there are $h \in H, i,$ and $k' \in K'$ such that $a=hx_i k'$. Let $k \in K'$. Then $f(ak)=f(h x_i k' k)= \pi(h)f(x_i k' k)=\pi(h)f(x_i k')=f(h x_i k')=f(a).$ -So $f$ satisfies point 3.<|endoftext|> -TITLE: Intersection of primes in a regular local ring -QUESTION [8 upvotes]: Suppose $R$ is a regular local ring of dimension at least 3, and suppose $P_1, P_2$ are dimension 1 primes. Does there necessarily exist a dimension 2 prime $Q$ contained in both? -In other words, is every pair of "little curves" contained in a "little surface"? -If this is false, will it hold if the ring is complete? -I am particularly interested in the mixed characteristic case. - -REPLY [3 votes]: I prove below that the answer is positive in the following special case: $R$ is the localization of a finitely generated polynomial ring over a field $k$. By Cohen's structure theorem this will also cover the case that $R$ is finite dimensional, complete and equicharacteristic. -For the proof it suffices to consider the case that $k$ is algebraically closed and $R$ is the localization of $k[x_1, \ldots, x_n]$ at the maximal ideal generated by $x_1, \ldots, x_n$. let $C_1, C_2$ be two irreducible curves on $k^n$ passing through the origin. It suffices to show that there is an irreducible surface $S$ containing both $C_1$ and $C_2$. -Claim: there is a curve $C$ and non-constant (and therefore dominant) maps $\phi_j: C \to C_j$, $j = 1, 2$. -Proof of the claim: Take any polynomial $f$ which is non-constant on both $C_j$. The fields $k(C_j)$ of rational functions on $C_j$ are finite extensions of $k(f)$, so that there is a common finite extension $L$ of $k(C_1)$ and $k(C_2)$ (see this for a neat trick). Now define $C$ to be the (unique) nonsingular curve such that $k(C) = L$. -It is easy to construct $S$ with $C, \phi_1, \phi_2$ from the claim: take the closure of the image of $\phi: C \times k \to k^n$ given by $\phi(x, t) := t\phi_1(x) + (1-t)\phi_2(x)$.<|endoftext|> -TITLE: Minimal generating set for $S_\omega$ -QUESTION [6 upvotes]: If $G$ is a group and $S\subseteq G$, let $\langle S \rangle$ be the intersection of all subgroups of $G$ containing $S$. -Let $S_\omega$ denote the group of all bijections $f:\omega\to\omega$ with composition. -Is there $M\subseteq S_\omega$ such that $\langle M \rangle =S_\omega$, but for all $m\in M$ we have $\langle M\setminus\{m\} \rangle\neq S_\omega$? - -REPLY [14 votes]: No. -Indeed, F. Galvin proved in 1995 that every countable subset of $S_\omega$ is contained in a finitely generated subgroup (and also $S_\kappa$ for every infinite $\kappa$). By contradiction suppose $M$ exists. Let $I$ be an infinite countable subset of $M$, so $I\subset \langle F\rangle$ for some finite $F$, and hence there exists a finite subset $J$ of $M$ such that $F\subset \langle J\rangle$. Hence, for $g\in I\smallsetminus J$, we have $g\in\langle J\rangle$, and therefore $M\smallsetminus\{g\}$ generates $S_\omega$. -More generally no such $M$ exists in any uncountable group $G$ with uncountable cofinality. Indeed fix $I\subset M$, write finite subsets $I_0\subset I_1\subset I_2...$ with union $I$, and define $M_n=(M\smallsetminus I)\cup I_n$. Then $G =\bigcup\langle M_n\rangle$ (increasing union). By definition of uncountable cofinality, $G=\langle M_n\rangle$ for some $n$: contradiction. Thus this shows that for every generating subset $M$, there exists $M'\subset M$ with $M\smallsetminus M'$ infinite, such that $M'$ generates $G$. -This also shows some stronger consequence: $G$ is not "infinitely independently generated": there is no sequence $(S_n)_{n\in\omega}$ of subsets of $G$ such that $\bigcup_k S_k$ generates $G$, but $\big\langle\bigcup_{k\neq n}S_k\big\rangle\neq G$ for every $n$. The latter condition has the advantage of being purely intrinsic to the poset of subgroups of $G$.<|endoftext|> -TITLE: A variant of Lambert function -QUESTION [7 upvotes]: How to express the solution of $x^{x+1}=a$ using Lambert function? I know that the standard Lambert function can be used to describe the solution of $x^x=a$. I wonder if $x^{x+1}=a$ can be addressed similarly. - -REPLY [12 votes]: Sorry, my answer is wrong. As it was pointed out by "Simply Beautiful Art" $x\ne W(z)$ but $x=e^{W(z)}.$ -It is known that $$W'(z)=\frac{W(z)}{z(1+W(z))}.$$ -Let $z=\log t$. Then $x=W(z)$ is a root of the equation $x^x=t$, in particular $x\log x=\log t=z.$ It means that $$W'(z)=\frac{x}{(x+1)\log t}=\frac{1}{(x+1)\log x}=\frac{1}{\log x^{x+1}},\quad x^{x+1}=e^{\frac{1}{W'(z)}}.$$ -So solution of the equation $x^{x+1}=a$ is $x=W(z)$, where $z$ is defined by $W'(z)=1/\log a.$<|endoftext|> -TITLE: Definition of a system being hyperbolic -QUESTION [5 upvotes]: Consider the $n \times n$ system -$$u_t + A(u)u_x = F(u).$$ -If the eigenvalues of $A$ are all real and distinct the system is called strictly hyperbolic. -What is the relationship between this definition and the existence of a strictly convex entropy? Are they equivalent? - -REPLY [6 votes]: They are not. First of all, the existence of a convex entropy is not meaningful for a system given in this quasi-linear form. The reason is that you might make a change $v=\phi(u)$ of unknown, but the convexity is not preserved by composition by the diffeomorphism $\phi$. -In addition, if $n\ge3$, a generic quasi-linear system does not admit conservation laws $\partial_t\eta(u)+\partial_q(u)=g(u)$, because the compatibility condition $\nabla\eta A=\nabla q$ is over-determined. -Now, if you give yourself a system of balance laws $u_t+f(u)_x=F(u)$, whose principal part is in conservation form, then the notion of convex entropy becomes meaningful, because you authorize only linear change of variables. Once again, a generic system with $n\ge3$ does not admit an entropy. So the system can be hyperbolic without having this convex entropy. -Of course, systems coming from thermodynamics are not generic. They were characterized by Godunov as those for which there are two functions $E(w),M(w)$ with $E$ strictly convex, such that $u=\nabla E(w)$ and $f(u)=\nabla M(w)$. Then the system is symmetric hyperbolic, in the sense that -$$\nabla^2Ew_t+\nabla^2Mw_x=F(u)$$ -where the matrices are symmetric (Hessians), and the first one is positive definite. Such systems do have a convex entropy, namely the Legendre transform $E^*(u)$. Conversely, a system of balance laws equipped with a strictly convex entropy can be written that way. - -REPLY [5 votes]: No, entropy convexity and hyperbolicity are not equivalent conditions. A necessary and sufficient condition for the system of differential equations to possess a strictly convex entropy is that the system is symmetrizable and hence hyperbolic. The symmetrizability condition is stronger than the condition of hyperbolicity, a system may have real eigenvalues and be therefore hyperbolic without being symmetrizable, and therefore without having a strictly convex entropy. -See for example these notes, or theorem 3.2 of this book.<|endoftext|> -TITLE: Containment of $c_0$ in projective tensor products -QUESTION [7 upvotes]: Let $X$ and $Y$ be Banach spaces and denote by $X\hat{\otimes}_\pi Y$ the projective tensor product. -Question: -If $X\hat{\otimes}_\pi Y$ contains an isomorphic copy of $c_0$, must then $X$ or $Y$ contain an isomorphic copy of $c_0$ also? - -REPLY [8 votes]: The answer is no. Bourgain and Pisier have given a counterexample (A construction of $\mathcal{L}_\infty$-spaces and related Banach spaces. Bol. Soc. Bras. Mat. 14, No. 2, 109-123 (1983). See Zbl 0586.46011 https://zbmath.org/?q=an%3A0586.46011 ).<|endoftext|> -TITLE: How to understand the interface of the consistency strength hierarchy, reverse mathematics, and proof-theoretic ordinal analysis? -QUESTION [10 upvotes]: I am aware of three major "hierarchies" of mathematical theories, but I don't know how to relate these hierarchies to one another. Here are the hierarchies I have in mind: - -Consistency strength. My understanding is that here one considers (recursively-enumerable?) theories $T$ of arithmetic (or which interpret the language of first-order arithmetic) of sufficient strength to fix some scheme for the syntax of first-order languages. One (partially) orders these theories by saying that $T > T'$ if $T$ proves the consistency of $T'$ (when $T'$ is coded up according to the aforementioned syntactic scheme). -Reverse mathematics. My understanding is that here one considers theories $T$ of second-order arithmetic (or which interpret the language of second-order arithmetic) and (partially) orders them directly by their implications. -Proof-theoretic ordinal analysis. My understanding is that here one considers theories $T$ of arithmetic (or which interpret the language of first-order arithmetic) and orders them by their proof-theoretic ordinal, i.e. the supremum of all (countable) ordinals $\alpha$ such that there exists a relation $R \subseteq \mathbb N \times \mathbb N$ definable in $T$ such that $T$ proves that $R$ is a well-order (although the sense in which $T$ can even express that $R$ is a well-order if $T$ is first-order is something that I don't quite understand), and $R$ is (externally) isomorphic to $\alpha$. - -Questions: - -Where do the domains of applicability of these hierachies overlap? - -For instance, it seems that the "strongest" theories (like ZFC+ large cardinals) are usually studied in terms of consistency strength as opposed to reverse math or proof theory. I have the impression that proof-theoretic ordinals are most commonly used for relatively weak theories, and that reverse math lies somewhere in the middle. But I'm not even sure where to look for the overlaps in these domains, partly because the sorts of theories considered in each hierarchy are slightly different. - -Where there domains overlap, how do these hierarchies relate? - -In general, I imagine there are no direct implications saying that any of these partial orders refines any of the others (even where their domains of applicability coincide). But I imagine that there are some general tendencies -- a stronger theory in one hierarchy should probably typically be stronger in another as well. - -Should I really be thinking of these 3 hierarchies as "comparable" in the sense that they give some notion of "strength" of a theory? And are there other hierarchies I should have in mind in this regard as well? - -REPLY [11 votes]: Sorry this is a bit disjointed - there's a lot of stuff here. I hope this helps though. - -All of these notions are applicable in all contexts - or at least, all sufficiently rich contexts (we probably want at least to interpret $PRA$). That said, once we get to reasonably strong theories (basically anything above $\Pi^1_2$-$CA_0$) we don't know how to calculate proof-theoretic ordinals, so in practice ordinal analysis doesn't reach (anywhere close to) ZFC-type theories. -Of course, ordinal analysis provides more than just a hierarchy - it assigns a "value" to each theory, independent of what other theories we're considering. The implication and consistency hierarchies don't do this, or at least not directly (see here for some pushback on this claim), so it's not surprising that calculating proof-theoretic ordinals is harder than comparing consistency strengths - although it may be surprising (it was to me) that it's that much harder. - -At this point it's a good idea to get to defining proof-theoretic ordinals properly. There isn't a single definition here, and some definitions (most?) have an element of subjectivity (they require a preexisting notion of "natural ordinal notation"). My favorite definition - which is fully formal - is the following: - -Set $PTO(T)$ to be the smallest $\alpha$ such that there is no (index for a) primitive recursive well-ordering isomorphic to $\alpha$ which $T$ proves is well-ordered. - -This definition makes sense for theories in sufficiently rich languages (e.g. second-order arithmetic and set theory). This isn't too big an issue (e.g. $RCA_0$ is conservative over $I\Sigma_1$ and $ACA_0$ is conservative over $PA$), but we can whip up versions for first-order arithmetic by talking about provable induction schemes (e.g. "$T$ proves $\Sigma_1$-induction along (that notation for) $\alpha$"). Here we have another level of flexibility, namely how much induction along the notation we want; if I recall correctly at this point $\Sigma^0_1$ induction is the standard choice. -But there are many other kinds of proof-theoretic ordinal, and beyond this one notion I really have no relevant competency. -The key point is that we need to talk about only primitive recursive relations. For example, in $\Pi^1_1$-$CA_0$ we can define a canonical relation on $\omega$ of ("true") ordertype $\omega_1^{CK}$, and in $ZFC$ we can go galactically beyond that. The issue is that these aren't really "concrete," and if we're thinking of ordinal analysis as a tool for Gentzen-style consistency proofs we really want to work lower down. - -That said, we could look at very different ways of assigning ordinals to theories - see e.g. here. - - -Implication strength of course behaves quite differently from consistency strength/ordinal analysis. First, differences in language are a bit more significant here, and we have to talk about interpretations/conservative extensions. More importantly, while there are definitely large chunks of well-foundedness (for example: $I\Sigma_n/B\Sigma_n$; the Big Five (+ higher $\Pi^1_k$-$CA_0$s); $KP\omega+\Sigma_n$-Replacement; large chunks of the large cardinal hierarchy) there are important nonimplications amongst the natural theories (plenty in the context of choice fragments over $ZF$, and in reverse mathematics probably the most important - sociologically speaking - being $WKL_0\perp RT^2_2$). -That said, I know of relatively few situations where we have a strict inequality in consistency strength and no corresponding strict implication (or incompatibility: over ZFC, a measurable doesn't imply V=L but it does resolve it). Some occur in the large cardinal hierarchy, and by Montalban/Shore we know that the $n$-$\Pi^0_3$-determinacy hierarchy has lots of incomparabilities with the $\Pi^1_n$-$CA_0$ hierarchy, but it does seem to be pretty rare. -The three contexts mentioned - first-order arithmetic, second-order arithmetic, and set theory - do "glue together" reasonably well in terms of modified implication (= folding in appropriate interpretations to deal with language differences). E.g. $RCA_0$ is conservative over $I\Sigma_1$, and every model of $ATR_0$ is the set of reals of some model of $KP\omega$ (the converse fails though!). The real gulf comes in when we try to get from weak set theories (like $KP\omega$, $Z$, etc. - see here) to ZFC and its ilk. This gap is gigantic, and I know very little about it. - -As a trivial observation, implication strength is also much broader than consistency strength and ordinal analysis. The division between "classifying members/subclasses of a given class" and "comparing different axiom systems" is pretty subjective - there's no reason we couldn't think of either as the other.<|endoftext|> -TITLE: The set of embeddings is open in the strong Whitney topology -QUESTION [6 upvotes]: In Hirsch's book "Differential Topology," he claims in Chapter 2, Theorem 1.4 that the set of $C^1$-embeddings is open in the strong Whitney topology $C^1(M, N)$ where $M$ and $N$ are $C^1$ manifolds. My question is: How do you see that the set $\mathcal{N}_1$ is open? See below for the setup and description of $\mathcal{N}_1$. -$M$ and $N$ are $C^1$ manifolds and we have an embedding $f: M \to N$. We equip $M$ with a locally finite atlas $\{\phi_i, U_i\}$ of $M$ which admits compact sets $K_i \subseteq U_i$ whose interiors cover $M$ too. We find a cover $\{\psi_i, V_i\}$ of $N$ so that $f(U_i) \subseteq V_i$. A lemma says we can find numbers $\epsilon_i > 0$ so that any $C^1$ function $g: M \to N$ that is in the neighborhood -$$\mathcal{N}_0 = \{ g : M \to N \,\vert\, (\forall i) (\forall x \in K_i)(g(K_i) \subseteq V_i)\text{ and }\, \lvert D^k(\psi_i f\phi_i^{-1} (x) - D^k(\psi_i f \phi_i^{-1}(x) \rvert < \epsilon_i \text{ for } k = 0,1 \} -$$ -satisfies $g\vert_{\mathrm{Int}(K_i)}$ is an embedding. -Since $f$ is an embedding, we can find open sets $A_i$, and $B_i$ in $N$ so that $f(K_i)\subseteq A_i$ and $f(M\setminus U_i) \subseteq B_i$ and $A_i\cap B_i = \emptyset$. The claim is that there is an open set $\mathcal{N}_1$ in $C^0(M,N)$ about $f$ such that if $g \in \mathcal{N}_1$ then $g(K_i) \subseteq A_i$ and $g(M\setminus U_i) \subseteq B_i$. -The result is certainly true if $M$ is compact, so I believe it is possible to extend it to the case where $M$ is not compact using the strong Whitney topology. I just can't get it to work. - -REPLY [2 votes]: A proof of this is also contained in Michor: Manifolds of mappings. There Proposition 5.3 establishes that the $C^r$-embeddings are $WO^1$ open in the space $C^r (M,N)$ (no restriction on the manifolds $M,N$). Thus if you set $r=1$ this implies that they are open with respect to the strong Whitney topology, as the $WO^1$-topology in Michor is coarser then the strong $C^1$-Whitney topology. For definitions and a comparison of the topologies see the book, it can be downloaded for free here: https://www.mat.univie.ac.at/~michor/manifolds_of_differentiable_mappings.pdf. -Going a bit in the details concerning the proof (answering to the comment by the OP, you have to decide whether this is an easy way to see that it is open): -Michor defines the set - $$B = \{g \in C^1(X,Y) : d(g(x),f(x)) <\varepsilon (x), d_1 (j_1 f (x),j_1 g (x))< \delta (x), g(K_\alpha) \subseteq A_\alpha , g(X\setminus U_\alpha ) \subseteq B_\alpha \text{for all }x \in X, \text{ for all } \alpha \}.$$ - -For the first two conditions $d(g(x),f(x)) <\varepsilon (x), d_1 (j_1 f (x),j_1 g (x))< \delta (x)$ recall that $d,d_1$ are metrics adapted to the jet-bundles and $\varepsilon, \delta$ are continuous functions. Thus these conditions are open in $WO^1$ by the characterisation of the $WO^1$-topology in Michor 4.4.2. -I claim that the conditions $g(K_\alpha) \subseteq A_\alpha , g(X\setminus U_\alpha ) \subseteq B_\alpha$ are open conditions in the $WO^1$-topology. To see this recall that $K_\alpha$ is compact, the $A_\alpha, B_\alpha$ and $U_\alpha$ are open with $K_\alpha \subseteq U_\alpha$ and the $U_\alpha$ form a locally finite-family. Now $WO^1$ is induced by an embedding from the so called $LO$-topology (recorded as 4.6 in Michor) and for the $LO$-topology, 3.7 Lemma in Michor states that the above conditions are open. - -As pointed out by the OP the second point is still missing details. Since the family of closed sets $X \setminus U_\alpha$ will in general not be locally finite. We can not use for this condition as it is written the cited result. The additional information making this possible is that we have an embedding f to construct a family of neighborhoods which is locally finite and implements a condition similar to the one asked in Michors book. A writeup on how to do this can be found her https://www.math.uni-hamburg.de/home/latschev/lehre/ws17/embeddings.pdf -Thus in conclusion the set B is open in $WO^1$. I admit it is not pretty and easy, but that is the shortest explanation I could come up with in limited time.<|endoftext|> -TITLE: Understanding the definition of stacks -QUESTION [16 upvotes]: First of all I should apologies if this question does not count as a research level one. I asked the same question on MathUnderflow and didn't receive any answer. Let me cross post (copy and paste) it here. -I am trying to understand the notion of sheaves and stacks. Intuitively, sheaves are bit easy to understand as a gluing of compatible families of sets assign to opens sets of a topological space. In other words, it is a contravariant functor $\mathcal{F}:\mathbf{Open}(X)^{\operatorname{op}}\to\mathbf{Set}$ such that -$$\mathcal{F}(U)=\lim\left(\prod_{i\in I}\mathcal{F}(U_i) \rightrightarrows \prod_{j,k\in I^2}\mathcal{F}(U_j\cap U_k)\right)$$ -where $X$ is a topological space, $U$ is an open set of $X$ and $\{U_i\}_{i\in I}$ is any open cover of $U.$ Further, it is easy to imagine a sheaf as a étalé space over $X.$ -Then I started reading about stacks using this notes and nlab as my primary sources. I learned that a stack is a contravariant functor $\mathcal{F}:\mathcal{C}^{\operatorname{op}}\to\mathbf{Grpd}$ satisfying a descent property and, categories fibered in groupoids over $\mathcal{C}$ is an intuitive way to think about stacks, where $\mathcal{C}$ is a site (category equipped with a coverage). Now I have following questions: - -How can I understand the descent property for stack? To be more specific, how can triple fiber products (intersections) appear in the equalizer fork diagram? -What categories fibered in groupoids over $\mathcal{C}$ corresponds to stacks? Is this the correct analogue of étalé space of a stack? - -REPLY [9 votes]: What categories fibered in groupoids over $\mathcal{C}$ corresponds to stacks? - -A category fibered in groupoids over $\mathcal{C}$ is given by a functor $p_{\mathcal{F}}:\mathcal{F}\rightarrow \mathcal{C}$ satisfying certain conditions (I am not writing the definition as I assume you already know what is a category fibered in groupoids); look at Definition 4.2 in the paper Orbifolds as stacks? -Put a Grothendieck topology on the category $\mathcal{C}$; considering it as a site. -Given an object $U$ of the category $\mathcal{C}$, we consider its fiber; a category, denoted by $\mathcal{F}(U)$, defined as -$$\text{Obj}(\mathcal{F}(U))=\{V\in \text{Obj}(\mathcal{F}):\pi_{\mathcal{F}}(V)=U\},$$ -$$\text{Mor}_{\mathcal{F}(U)}(V_1,V_2)=\{(f:V_1\rightarrow V_2)\in \text{Mor}_{\mathcal{F}}(V_1,V_2):\pi_{\mathcal{F}}(f)=1_U\}.$$ -Given a cover $\{U_\alpha\rightarrow U\}$ of the object $U$ (remember that we fixed a Grothendieck topology), we consider its descent category, denoted by $\mathcal{F}(\{U_\alpha\rightarrow U\})$. -An object of the category $\mathcal{F}(\{U_\alpha\rightarrow U\})$ is given by the following data: - -for each index $i\in \Lambda$, an object $a_i$ in the category $\mathcal{F}(U_i)$, -for each pair of indices $i,j\in \Lambda$, an isomorphism $\phi_{ij}:pr_2^*(a_j)\rightarrow pr_1^*(a_i)$ in the category $\mathcal{F}(U_i\times_{U}U_j)$ - -satisfying appropriate cocycle condition. -Now, given a categroy fibered in groupoids $p_{\mathcal{F}}:\mathcal{F}\rightarrow \mathcal{C}$, an object $U$ of $\mathcal{C}$ and a cover $\mathcal{U}(U)=\{U_\alpha\rightarrow U\}$ of $U$ in $\mathcal{C}$, there is an obvious functor $$p_{\mathcal{U}(U)}:\mathcal{F}(U)\rightarrow \mathcal{F}(\{U_\alpha\rightarrow U\})$$ -(which you might have guessed already but let me say that), at the level of objects $$a\mapsto ((a|_{U_\alpha}),(\phi_{ij}))$$ - -A category fibered in groupoids - $p_{\mathcal{F}}:\mathcal{F}\rightarrow \mathcal{C}$ is said to be a - stack if, for each object $U$ of $\mathcal{C}$ and for each cover - $\mathcal{U}(U)=\{U_\alpha\rightarrow U\}$, the functor - $$p_{\mathcal{U}(U)}:\mathcal{F}(U)\rightarrow \mathcal{F}(\{U_\alpha\rightarrow U\})$$ is an equivalence of categories. - -Now, you can ask what does equivalence of categories has anything to do with "sheaf like" properties? For a functor $\mathcal{D}\rightarrow \mathcal{C}$ to be an equivalence of categories, along other things, for each object $d\in \mathcal{D}$ we need an object $c\in \mathcal{C}$ such that, there is an isomorphism $F(c)\rightarrow d$. -Let $((a_\alpha),\{\phi_{\alpha\beta}\})$ be an object of $\mathcal{F}(\{U_\alpha\rightarrow U\})$. For this, by equivalence of categories, gives an element $a\in \mathcal{F}(U)$ that maps to $((a_\alpha),\{\phi_{\alpha\beta}\})$. That is, given an object $U$ of $\mathcal{C}$, an open cover $\{U_\alpha\rightarrow U\}$, for each collection of objects $\{a_\alpha\in \mathcal{F}(U_\alpha)\}$ that are compatible in some sense, there exists an object $a\in \mathcal{F}(U)$, such that, under appropriate restriction of $a$, you get the objects $a_{\alpha}$. This should remind the notion of sheaf on a topological space. This is how a stack is seen as a generalization of sheaf. -References : - -Notes on Grothendieck topologies, fibered categories and descent theory. -Orbifolds as stacks? -How is a Stack the generalisation of a sheaf from a 2-category point of view?<|endoftext|> -TITLE: How to prove the relationship between Stern's diatomic series and Lucas sequence $U_n(x,1)$ over the field GF(2)? -QUESTION [7 upvotes]: I found the bit count of Lucas sequence $U_n(x,1)$ over the field GF(2) is Stern's diatomic series, I want to know the reason? -https://oeis.org/A002487 : Stern's diatomic series -https://oeis.org/A168081 : Lucas sequence $U_n(x,1)$ over the field GF(2) - -REPLY [7 votes]: I always work mod 2. Let $f(n)$ be the bit count (number of nonzero -coefficients) of $U_n(x,1)$. $U_n(x,1)$ satisfies - $$ U_{2n}(x,1) = xU_n(x,1)^2 $$ -$$ U_{2n+1}(x,1) = (U_n(x,1)+U_{n+1}(x,1))^2. $$ -From the first equation $f(2n)=f(n)$, and only odd powers of $x$ -can have nonzero coefficients. From the second equation, only even -powers of $U_{2n+1}$ can have nonzero coefficients. Hence no -cancellation occurs when we add $U_n(x,1)$ and $U_{n+1}(x,1)$, so -$f(2n+1)=f(n)+f(2n+1)$. Thus $f(n)$ satisfies the same recurrence and -same initial conditions $f(0)=0$ and $f(1)=1$ as Stern's diatomic sequence. -Addendum. The two formulas above can be proved by induction on -$n$. Namely, writing just $U_n$ for $U_n(x,1)$, - $$ U_{2n}=xU_{2n-1}+U_{2n-2} = x(U_{n-1}+U_n)^2+xU_{n-1}^2 - =xU_n^2 $$ -and - $$ U_{2n+1} = xU_{2n}+U_{2n-1} = x^2U_n^2+(U_{n-1}+U_n)^2 - =U_n^2+(xU_n+U_{n-1})^2=U_n^2+U_{n+1}^2. $$<|endoftext|> -TITLE: How does the Steenrod algebra act on $\mathrm{H}^\bullet(p^{1+2}_+, \mathbb{F}_p)$? -QUESTION [7 upvotes]: Let $p$ be an odd prime. The $\mathbb F_p$ cohomology of the cyclic group of order $p$ is well-known: $\mathrm{H}^\bullet(C_p, \mathbb F_p) = \mathbb F_p[\xi,x]$ where $\xi$ has degree 1, $x$ has degree 2, and the Koszul signs are imposed (so that in particular $\xi^2 = 0$). As a module over the Steenrod algebra, the only interesting fact is that $x = \beta \xi$, where $\beta$ denotes the Bockstein. The rest of the Steenrod powers can be worked out by hand. -There are two groups of order $p^2$. The $\mathbb F_p$ cohomology of $C_p \times C_p$, including its Steenrod powers, is computable from the Kunneth formula. For the cyclic group $C_{p^2}$, you have to think slightly more, because there is a ring isomorphism $\mathrm{H}^\bullet(C_p, \mathbb F_p) \cong \mathrm{H}^\bullet(C_{p^2}, \mathbb F_p)$, but the Bockstein vanishes on $\mathrm{H}^\bullet(C_{p^2}, \mathbb F_p)$. Still I think the Steenrod algebra action is straightforward to write down. -I want to know about the groups of order $p^3$. The abelian ones are not too hard, I think, and there are two nonabelian groups. The one with exponent $p^2$ is traditionally denoted "$p^{1+2}_-$", and the one with exponent $p$ is traditionally denoted "$p^{1+2}_+$". I care more about the latter one, but I'm happy to hear answers about both. And right now I care most about the prime $p=3$. -The cohomology of these groups was computed in 1968 by Lewis in The Integral Cohomology Rings of Groups of Order $p^3$. Actually, as is clear from the title, Lewis computes the integral cohomology, from which the $\mathbb F_p$-cohomology can be read off using the universal coefficient theorem. For the case I care more about, Lewis finds that $\mathrm{H}^\bullet(p^{1+2}_+, \mathbb Z)$ has the following presentation. (I am quoting from Green, On the cohomology of the sporadic simple group $J_4$, 1993.) The generators are: -$$ \begin{matrix} -\text{name} & \text{degree} & \text{additive order} \\ -\alpha_1, \alpha_2 & 2 & p \\ -\nu_1, \nu_2 & 3 & p \\ -\theta_j, 2 \leq j \leq p-2 & 2j & p \\ -\kappa & 2p-2 & p \\ -\zeta & 2p & p^2 -\end{matrix}$$ -(For the $p=3$ case that I care most about, there are no $\theta$s, since $2 \not\leq 3-2$.) A complete (possibly redundant) list of relations is: -$$ \nu_i^2 = 0, \qquad \theta_i^2 = 0, \qquad \alpha_i \theta_j = \nu_i \theta_j = \theta_k \theta_j = \kappa \theta_j = 0$$ -$$\alpha_1 \nu_2 = \alpha_2 \nu_1, \qquad \alpha_1 \alpha_2^p = \alpha_2 \alpha_1^p, \qquad \nu_1\alpha_2^p = \nu_2 \alpha_1^p,$$ -$$ \alpha_i\kappa = -\alpha_i^p, \qquad \nu_i\kappa = -\alpha_i^{p-1}\nu_i,$$ -$$ \kappa^2 = \alpha_1^{2p-2} - \alpha_1^{p-1}\alpha_2^{p-1} + \alpha_2^{2p-2}, $$ -$$ \nu_1 \nu_2 = \begin{cases} \theta_3, & p > 3, \\ 3\zeta, & p = 3. \end{cases}$$ -From this Green (ibid.), for example, writes down a PBW-type basis. - -Question: What is the action of the Steenrod algebra been on $\mathrm{H}^\bullet(p^{1+2}_+, \mathbb F_p)$? - -I'm not very good at Steenrod algebras. Does the ring structure on the $\mathbb Z$-cohomology suffice to determine the action? For instance, the additive structure of $\mathrm{H}^\bullet(G, \mathbb Z)$ already determines the Bockstein action on $\mathrm{H}^\bullet(G, \mathbb F_p)$. If there is a systematic way to do it, where can I learn to do the computations? - -REPLY [7 votes]: I think the answer you want can be found in the following article: -AUTHOR = {Leary, I. J.}, - TITLE = {The mod-{$p$} cohomology rings of some {$p$}-groups}, -JOURNAL = {Math. Proc. Cambridge Philos. Soc.}, -FJOURNAL = {Mathematical Proceedings of the Cambridge Philosophical - Society}, -VOLUME = {112}, - YEAR = {1992}, -NUMBER = {1}, - PAGES = {63--75}, - ISSN = {0305-0041},<|endoftext|> -TITLE: A generalization of discrete Hilbert's transform (Montgomery's inequality) -QUESTION [7 upvotes]: In the paper "Hilbert's inequality", Montgomery and Vaughan proved that a generalization of the discrete Hilbert transform is bounded in $\ell^2$. The inequality reads as follows -$$ \Big| \sum_{k\neq n}\frac{a_k \overline{b_n}}{\lambda_k-\lambda_n} \Big| \leq \frac{\pi}{\delta} \Big(\sum_{k=1}^{\infty} |a_k |^2 \Big)^{1/2}\Big( \sum_{n=1}^{\infty} |b_n |^2 \Big)^{1/2}, $$ -where $\{a_k\}, \{ b_n \}\in \ell^2 $, $ \lambda_n $ is an increasing sequence of real numbers such that -$$ \delta:= \inf_{k n}| \lambda_k-\lambda_{k+1}|. $$ -Of course $\delta$ is assumed to be strictly positive. Also the constant appearing in the inequality $\pi/\delta$ is optimal. Quite surprisingly all proofs I managed to find use strongly the Hilbert space structure of $\ell^2$. -Therefore I would like to ask if anything is known for the this inequality when considered on $\ell^p, p\neq 2$. Namely, is it true -$$\Big| \sum_{k\neq n}\frac{a_k \overline{b_n}}{\lambda_k-\lambda_n} \Big| \leq C(p,\delta) \Big(\sum_{k=1}^{\infty} |a_k |^p \Big)^{1/p}\Big( \sum_{n=1}^{\infty} |b_n |^q \Big)^{1/q}, $$ -where $10 ?$ -(I wouldn't venture so far as to ask for an optimal constant in this case, given the difficulty of the problem for the classical discrete Hilbert transform.) - -REPLY [8 votes]: One can transfer the continuous $L^p$ theory to this discrete setting without difficulty. -Let's normalise $\sum_k |a_k|^p = \sum_n |b_n|^q = 1$. Consider the two quantities -$$ X_1 := \sum_{k \neq n} \frac{a_k \overline{b_n}}{\lambda_k - \lambda_n}$$ -$$ X_2 := \sum_{k, n} p.v. \int_{{\bf R}^2} \varphi(s) \varphi(t) \frac{a_k \overline{b_n}}{(\lambda_k+s) - (\lambda_n+t)}\ ds dt$$ -where $\varphi$ is a bump function of total mass 1. It is not difficult to show that -$$ p.v. \int_{\bf R} p.v. \int_{\bf R} \varphi(s) \varphi(t) \frac{1}{(\lambda_k+s) - (\lambda_n+t)}\ dt$$ -is equal to $\frac{1}{\lambda_k - \lambda_n} + O_\delta( |k-n|^{-2} )$ when $k \neq n$ and $O_\delta(1)$ when $k=n$, so we have $X_1-X_2 = O_{p,\delta}(1)$ by Schur's test. One can also write $X_2$ as -$$ p.v. \int_{\bf R} \int_{\bf R} \frac{f(x) g(y)}{x-y}\ dx dy$$ -where -$$ f(x) := \sum_k a_k \varphi(x-\lambda_k)$$ -and -$$ g(y) := \sum_n b_n \varphi(x-\lambda_n)$$ -so from the $L^p$ boundedness of the continuous Hilbert transform we have $X_2 = O_{p,\delta}(1)$, and the claim follows.<|endoftext|> -TITLE: Hyperbolic manifolds with infinite cyclic fundamental group -QUESTION [5 upvotes]: It is a well known fact that there is a correspondence between complete hyperbolic $n$-manifolds up to isometry and discrete subgroups of isometries of the hyperbolic space $\mathbb{H}^n$ that act freely on $\mathbb{H}^n$ up to conjugation. -The correspondece is given by $\Gamma < Isom(\mathbb{H}^n)\mapsto \mathbb{H}^n/\Gamma$ and the inverse is given by the map $M\mapsto \pi_1(M)\hookrightarrow Isom(\tilde{M})$ where $\tilde{M}\simeq \mathbb{H}^n$ is the fundamental cover of $M$. -The requirement that $\Gamma$ acts freely on $\mathbb{H}^n$ is equivalent to requiring that there are no elliptic isometries in $\Gamma$ or equivalently if every element in $\Gamma$ has infinite order. -In particular any parabolic or hyperbolic element in $Isom(\mathbb{H}^n)$ generates an infinite cyclic subgroup. This will correspond to manifolds with $\pi_1 M \simeq \mathbb{Z}$. - -What are the complete hyperbolic manifolds with fundamental group $\mathbb{Z}$? Can we say something at least in the case of $3$-manifolds? - -REPLY [7 votes]: This consists in classifying non-elliptic elements of the Lie group $\mathrm{Isom}(\mathbf{H}^n)\simeq\mathrm{PO}(n,1)$ up to conjugacy and inversion. -One can do separately loxodromics and horocyclics ("parabolics"). -Loxodromics: they have two invariants: the translation length (a positive real number), and the transverse isometry, namely an isometry of $\mathbf{H}^{n-1}$ fixing a point (up to conjugation fixing this point), and this is classified by an element of $\mathrm{O}(n-1)$ up to conjugation [and inversion]. -Horocyclics: they are classified by their action on the horosphere (which is a non-geodesic copy of the Euclidean space $\mathbf{R}^{n-1}$, modulo conjugation [and inversion] by the whole group of similarities. Hence, by a non-elliptic isometry of $\mathbf{R}^{n-1}$, modulo conjugation by similarities. In general, the horocyclic is orthogonal direct sum of a nontrivial translation and an element of $\mathrm{O}(n-2)$. Hence horocyclics are classified by conjugacy classes of $\mathrm{O}(n-2)$. -The horocyclic case corresponds to the existence of a cusp in the quotient manifold. If one sticks to orientable manifolds, one should restrict to $\mathrm{SO}(n-1)$ in the loxodromic case and $\mathrm{SO}(n-2)$ in the horocyclic case. -["And inversion" will not play any role since it follows that all isometries are conjugate to their inverse, since this holds in $\mathrm{O}(k)$ for every $k$.] -Let's specify in small dimension: -$n=2$: loxodromics are classified by a positive real number, and a sign (preserving or not the orientation). There's a single horocyclic (orientation-preserving). -$n=3$: loxodromics are classified by a positive real number, and by an element of $\mathrm{O}(2)$ up to conjugation (hence, either a rotation of angle in $[0,\pi]$, or a reflection). Horocyclics: it can be a translation, or a glide reflection. -$n=4$: loxodromics are classified by a positive real number, and by an element of $\mathrm{O}(3)$ up to conjugation (hence a rotation or antirotation of angle in $[0,\pi]$). Horocyclics: classified by some conjugacy class of $\mathrm{O}(2)$. - -Topological classification: -actually, in the orientable case, the quotient manifold is analytically diffeomorphic to $\mathbf{R}^{n-1}\times (\mathbf{R}/\mathbf{Z})$, and in the non-orientable case, it is analytically diffeomorphic to $\mathbf{R}^{n-2}\times (\text{Möbius})$. -Indeed, in both case one sees that the isometry is analytically conjugate to a non-elliptic isometry of the Euclidean space $\mathbf{R}^n$. Such an isometry can be conjugated to have the form $f:(t,y)\mapsto (t+1,Sy)$ with $t\in\mathbf{R}$, $y\in\mathbf{R}^{n-1}$ and $S\in\mathrm{O}(n-1)$. If $S\in\mathrm{SO}(n-1)$, there is a 1-parameter subgroup $(S^t)$ with $S^1=S$, and conjugating $f$ by the analytic self-diffeomorphism $(t,y)\mapsto (t,S^ty)$ yields a translation. If $S\notin\mathrm{SO}(n-1)$, write $f$ as $(t,u,z)\mapsto (t+1,-u,Tz)$ with $z\in\mathbf{R}^{n-2}$ and $T\in\mathrm{SO}(n-2)$. Then conjugating as above only on the last variable conjugates to $(t,u,z)\mapsto t+1,-u,z)$ and this yields the requested description.<|endoftext|> -TITLE: Metrizability of a topological vector space where every sequence can be made to converge to zero -QUESTION [7 upvotes]: This is a follow-up to this answer. -If $E$ is a (real or complex) topological vector space, we say that a sequence $\{x_n\}_{n=1}^\infty$ in $E$ can be made to converge to zero if there exists a sequence $\{\alpha_n\}_{n=1}^\infty$ of strictly positive real scalars such that $\lim_{n\to\infty} \alpha_n x_n = 0$. -In the aforementioned answer, I compared this with two related notions, and I gave a generic example of a space where every sequence can be made to converge to zero: take a metrizable topological vector space and pass to a weaker (i.e. coarser) topology, possibly with a different dual. My question is whether every example is of this form. In other words: -Question. If $E$ is a topological vector space where every sequence can be made to converge to zero, is there a finer linear topology on $E$ that is metrizable? -Maybe it is false in general, but true in certain special cases? (E.g. if $E$ is locally convex/complete/separable, or if $E$ the strong dual of a Fréchet space?) - -REPLY [3 votes]: For the counterexample below we need a little lemma about barrelled spaces (i.e., every barrel = closed, absolutely convex, and absorbing set is a $0$-neighbourhood): -If $(E,\mathcal T)$ is a barrelled locally convex space which has a finer metrizable vector space topology $\mathcal S$ then $(E,\mathcal T)$ is metrizable. -Indeed, let $(V_n)_{n\in\mathbb N}$ be a basis of the $0$-neighbourhoods in $(E,\mathcal S)$ und $U_n =\overline{\Gamma(V_n)}^{\mathcal T}$ the closed absolutely convex hulls of $V_n$. These are barrels in in $(E,\mathcal T)$ and hence $\mathcal T$-neighbourhoods of $0$. On the other hand, every $\mathcal T$-neighbourhood of $0$ contains a closed absolutely convex one which itself contains some $V_n$ (because $\mathcal S$ is finer) and hence some $U_n$. This means that $(U_n)_{n\in\mathbb N}$ is a countable basis of the $0$-neighbourhoods of $(E,\mathcal T)$ which is thus metrizable. -Now the counterexample: Let $I$ be an uncountable index set and -$$E=\{(x_i)_{i\in I}\in \mathbb R^I: \{i\in I: x_i\neq 0\} \text{ is countable}\}$$ -endowed with the relative topology of the product topology on $\mathbb R^I$. According to proposition 4.2.5(ii) in the book Barrelled Locally Convex Spaces of Bonet and Perez-Carreras, $E$ is barrelled. Since every $0$-neighbourhood only gives conditions on finitely many coordinates there is no countable basis of all $0$-neighbourhoods, i.e., $E$ is not metrizable and hence, by the lemma, it does not admit a finer metrizable vector space topology. On the other hand, for every sequence $x^n\in E$ the union $J$ of the supports of the $x^n$ is countable, so that $\{x^n:n\in\mathbb N\}$ is contained in the subspace $F=\mathbb R^J\times \{0\}^{I\setminus J}$ which is metrizable. Therefore, there is a sequence of $t_n>0$ such that $t_nx_n$ converges to $0$ in $F$ and hence also in $E$. - -As you also asked for the particular case of strong duals of Frechet spaces $X$: This is not only covered by the answer of Tomasz Kania but it is rather trivial: If $X$ is not isomorphic to a Banach space there is a fundamental sequence of semi-norms $p_n$ which are pairwise non-equivalent. Then $X'=\bigcup_{n\in\mathbb N} X_n'$ where $X_n'$ is the space of functionals which are continuous with respect to $p_n$. Choosing $\phi_n\in X_n'\setminus X_{n-1}'$ there is no sequence of $t_n>0$ such that $t_n\phi_n\to 0$ because every bounded subset of $X'$ is contained in some $X_n'$. - -EDIT. Another counterexample is the space of absolutely summable families $$\ell^1(I)=\{(x_i)_{i\in I}: \sum_{i\in I} |x_i|<\infty\}.$$ -Every absolutely summable family has only countably many non-zero terms and hence a somehow natural topology (of course, besides the Banach space topology of the $\ell^1$-norm) is that of the inductive limit $\ell^1(I)=\lim\limits_\to \ell^1(J)$ for all countable subsets $J\subseteq I$. This finer topology is barrelled but not metrizable (hence it does not admit a finer metrizable vector space topology) but every sequence can be made bounded.<|endoftext|> -TITLE: The theory of a model of a theory that knows all formulas true in almost all its models -QUESTION [6 upvotes]: We are in first order logic world. -Let $\sigma$ be a finite signature and $T$ a consistent theory of $\sigma$. -Due to Löwenheim–Skolem theorem, we can consider the $\underline{set}$ of all at most countable models of $T$ up to elementary equivalance(does not change what formulas hold). -Let $\mu$ be a nonprincipial ultrafilter of it, denote by $\Omega^T_\mu$ the ultraproduct of -all those models with respect to $\mu$. -Clearly, the theory of $\Omega^T_\mu$ are formulas that are true for almost all with respect to $\mu$ models of $T$. -In particular, this theory, call it $T(\Omega^T_\mu)$, has $T$ as a subset. -My questions are: - -Does the equality $T(\Omega^T_\mu)=T$ hold? -Does it depend on $\mu$? Is the intersection over all $\mu$ equal to $T$? -If they are not equal, what the difference means? Furthemore, we could consider a sequance of theories $T\subset T(\Omega^T_\mu) \subset T(\Omega^{T(\Omega^T_\mu)}_{\mu'})\subset\ldots$. Does it stabilize? -Does $T(\Omega^T_\mu)$ complete $T$ in any nice sense? - -I suspect that all those questions are classical and already answered, but I do not know where -to look for them. Just before pandemia I have finished my first course on mathematical logic and I have been wondering about this object since then. - -REPLY [7 votes]: In this answer, I will basically repeat things that zeb said in their nice answer, but arranged differently (in a way that might or might not be more clear). -Let $\text{Mod}_{\leq \aleph_0}(T)$ be the "set" of all at most countable models of $T$. Your question is a bit ambiguous about what this means - I'll come back to that later. Morally, picking an ultrafilter on $\text{Mod}_{\leq \aleph_0}(T)$ is the same as picking a completion of $T$. -In one direction, if you pick an ultrafilter $\mu$ on $\text{Mod}_{\leq \aleph_0}(T)$, you can take the complete theory of the ultraproduct, $T(\Omega^T_\mu)$ in your notation. This is a complete theory, and it contains $T$, so it's a completion of $T$. -In the other direction, given a completion $T\subseteq T'$, for each sentence $\varphi\in T'$, consider the set of models $M_\varphi = \{M\in \text{Mod}_{\leq \aleph_0}(T)\mid M\models \varphi\}$. Then $F_{T'} = \{M_\varphi\mid \varphi\in T'\}$ has the finite intersection property, so we can extend it to an ultrafilter. And moreover for any ultrafilter $\mu$ extending $F_{T'}$, we have $T(\Omega^T_\mu) = T'$, by Łoś's Theorem. -But now there's a subtlety here: You want $\mu$ to be a non-principal ultrafilter. -If by $\text{Mod}_{\leq \aleph_0}(T)$, you mean a set of isomorphism representatives for the models of $T$, so this set contains just one copy of each model of $T$ up to isomorphism, then if $T'$ has only one (at most countable) model (i.e. if it's the theory of a finite structure, or if it's $\aleph_0$-categorical), then the ultrafilter $\mu$ you get by the above construction will be principal. That is, picking a non-principal ultrafilter on $\text{Mod}_{\leq \aleph_0}(T)$ picks a completion of $T$, but we never get a completion that is $\aleph_0$-categorical or the theory of a finite structure. Worse, if $T$ already has only finitely many (at most countable) models up to isomorphism, then there is no non-principal ultrafilter on $\text{Mod}_{\leq \aleph_0}(T)$ - this corresponds to the fact that $T$ has no completion with more than one (at most countable) model. -These issues go away if $\text{Mod}_{\leq \aleph_0}(T)$ contains infinitely many copies of each model up to isomorphism (e.g. if it's more like the proper class of all at most countable models of $T$). Then when we extend $F_{T'}$ to an ultrafilter $\mu$, we can always make sure $\mu$ is non-principal. -So the answer to your questions are: -(1) $T(\Omega_\mu^T) = T$ if and only if $T$ is already complete. -(2) Yes, it definitely depends on $\mu$ (unless $T$ is complete). If you remove the restriction that $\mu$ be non-principal, the intersection of all theories of the form $T(\Omega^T_\mu)$ is $T$. If you require $\mu$ to be non-principal, this intersection is the set $T_{\text{cof}}$ of all sentences which are in all but finitely many models in $\text{Mod}_{\leq \aleph_0}(T)$. If $\text{Mod}_{\leq \aleph_0}(T)$ contains infinitely many copies of each model up to isomorphism (e.g. if this "set" is actually the proper class of all at most countable models of $T$), then again $T_{\text{cof}} = T$. -(3) The "difference" consists of moving from $T$ to a completion. The chain stabilizes after the first step, i.e. $T(\Omega_\mu^T) = T(\Omega_{\mu'}^{T(\Omega_\mu^T)})$, since $T(\Omega_\mu^T)$ is already complete. -(4) Yes, $T(\Omega_\mu^T)$ is a completion of $T$.<|endoftext|> -TITLE: Sobolev convergence of Fourier series -QUESTION [5 upvotes]: Consider $f\in H^{\sigma}(S^1)=W^{\sigma, 2}$ (the usual Sobolev space on the circle) and let $S_Nf$ be its truncated Fourier series $S_Nf = \sum_{|n|\leq N} \hat{f}(n)e^{2\pi i n x}$. I am looking for the theory for inequalities of the following type: -$$\|f-S_Nf\|_{H^s} \leq CN^{c(s,\sigma)} \|f\|_{H^{\sigma}} \, ,$$ -What values of $s>0$ work for any given $\sigma >0$? What is known of the constant $C$ and of the rate $c(s,\sigma )$? -Motivation: I am familiar with the equivalent theory for orthogonal polynomial expansions (see Canuto and Quarteroni, Math. Comp. 1982), but figured out that the Fourier theory is older, and maybe simpler. - -REPLY [3 votes]: Let us start with pointing out that $f\in H^\sigma$ is equivalent to -$$ -(\langle n\rangle^\sigma\hat f(n))_{n\in \mathbb Z}\in \ell^2(\mathbb Z), -\quad \text{with $\langle n\rangle=\sqrt{1+n^2}$.} -$$ -Then you have for $s<\sigma$ -$$ -\Vert f-S_N(f)\Vert_{H^s}^2=\sum_{\vert n\vert> N}\langle n\rangle^{2s}\vert\hat f(n)\vert^2= -\sum_{\vert n\vert> N}\langle n\rangle^{2\sigma}\vert\hat f(n)\vert^2\langle n\rangle^{2s-2\sigma}\le\langle N\rangle^{2s-2\sigma}\Vert f\Vert^2_{H^\sigma}. -$$<|endoftext|> -TITLE: Suggestions for reducing the transmission rate? -QUESTION [12 upvotes]: What are suggestions for reducing the transmission rate of the current epidemics? -In summary, my best one so far is (once we are down to the stay home rule) to discretize time, i.e., to introduce the following rule for the general populace not directly involved in necessary services: -If members of your household go to public services on a certain day, the whole household should not use any public service for 2 weeks after. That way you still can get infected but cannot infect without knowing it. -Do you have some suggestions? Good models to look at? No predictions please, just advices what to do. -Edit: In more detail: -Model (the simplest version to make things as clear as possible): There are several categories of people $C_i$ that constitute portion $p_i$ of the population and have a certain matrix $A$ of interactions per day. Then, if $x_i(t)$ is the number of ever infected people in category $C_i$ by the time $t$, the driving ODE is -$$ -\dot x(t)=\alpha A[x(t)-x(t-\tau)] -$$ -where $\tau$ is the ("typical") time after which the sick person is removed from the population and $\alpha$ is the transmission probability. In this model the exponential growth is unsustainable if $\alpha\lambda(A)\tau<1$ where $\lambda$ is the largest eigenvalue of $A$. We do not know $\alpha$ (though we can try to make suggestions how to reduce it, most such suggestions are already made by the government). The government can modify $A$ by issuing orders. Some orders merely reduce $a_{ij}$ to $0$, but the government cannot shut essential public services completely this way. -Questions: What is $A$, which entries $a_{ij}$ are most important to reduce, how to issue a sensible order that will modify them, and by how much they will reduce the eigenvalue? -First suggestion for these 4 answers: There are two categories of people: ordinary population that only goes to public services and -public servants that both provide services and go to them. There is only one ("averaged") type of service involving a dangerous client-server interaction and all infection goes there. The portion of public servants in the population is $p$. The server sees $M$ clients a day. Then the current social interaction matrix (say, for the grocery store I've seen yesterday) is $A=\begin{bmatrix}0 & M(1-p)\\ Mp & 2Mp\end{bmatrix}$ (ordinary population does not transmit to ordinary population, servers transmit to clients who can be both servers and clients, clients transmits to servers. The largest eigenvalue is $M(p+\sqrt p)$. The lion's share comes from $\sqrt p$, which is driven by the off-diagonal entries, The order should be issued as above, the effect that ordinary people never come to the service infected, which will remove the left bottom corner and drop the largest eigenvalue to $2Mp$. Assuming $p=1/9$ (not too unrealistic), the drop will be two-fold even if you leave the service organization as it is. -That ends the solution I propose in mathematical language. -In layman terms, the public will completely do this part (you cannot ask for more) and still have some life, and we can concentrate on the models of how servers should be organized. -Edit: -Time to remove the non-relevant part and add some relevant thoughts about what else we can help with plus the response to JCK. -First of all, It is very hard to formulate the orders correctly. The stay home rule really means "avoid all close contacts outside your household except the necessary interactions with public servants providing vital services to you" (and even that version is, probably flawed). It is not about dogs, etc., as the Ohio version reads now. If everybody understood and implemented that meaning, my suggestion could be formulated as I said. However the intended meaning really is -When going to public services, minimize the probability that you can infect others as much as feasible and consider it to be $0$ if in the last two weeks nobody from your household had a contact with a stranger and nobody in the household had any symptoms. -Now it is more to the point, but also more complicated. And if a professional mathematician like myself is so inept, imagine the difficulties of other people. -So, within that model, what would be the best formulation of the order to give? -Second, the set of questions I asked is clearly incomplete. -One has to add for instance "What assumption can be wrong and what effect that will have on the outcome under the condition that the order is given in the currently stated form. I have never seen a book that teaches the influence of the order formulation on the possible model behavior and that may be a crucial thing now. The interaction between the formal logic and differential equations within a given scenario is a non-existing science (or am I just ignorant of something? That reference would really be useful). -Third, if we have a particular question (say, how much to reduce and how organize the public transportation, which is NY and Tel Aviv headache now), what would be a good mathematical model for just that and what would be the corresponding order statement under this model? -The questions like that are endless and if there were ready answers in textbooks, the governments would just implement them already instead of having 7-hour meetings. So I can fairly safely conclude that they are not there. -What I tried with my model example was, in particular, to show that there may be some non-trivial moves in even seemingly optimal situations (strict stay home order and running only the absolutely vital services at the minimal rate that still allows to serve the population) that also make common sense and can be used by everyone right now and right here. Finding such moves can really help now. The main real life question now is "What can I (as government, business, or individual) do to reduce the largest eigenvalue of the social interaction matrix?" Now show me the textbook that teaches that and I'll stop the "ballspitting" and apologize for the wasted time of the people reading all this. - -REPLY [2 votes]: I have one rather practical answer, which does not answer the question in the body, but possibly in the title. -I was very surprised to learn that a (small) hospital in my home town was doing their scheduling ("rostering" might be the correct term, but I don't know) by hand. -They were clever enough to ask for help just in time. Again surprisingly: informally via a mathematician they knew privately. Thus, they now have a professional university team developing a plan that fits their new needs - minimal service, minimizing the probability that a large portion of their staff simultaneously drops out. -I guess that such a strategy might also make sense for other community services, who might not even be aware of this fact. -(Disclaimer: I am not helping with the implementation, but I was lucky to know someone who knew someone who was a professional in this field, who luckily agreed to do it.)<|endoftext|> -TITLE: What is the geometric shape of the Monster sporadic group? -QUESTION [19 upvotes]: Conway made the comment that the Monster group represents the symmetries of a shape in 196,883 dimensions, something like a "star you hang on a Christmas tree." -My question is, What do we know (or conjecture) about the enigmatic shape whose symmetries are captured by the Monster? Is the shape analogous to a star (pointy) or convex like a polygon / polytope, or is it more likely to be something with a different topology, analogous to a high-dimensional torus? -Conversely, what do we know it is not like, what have we ruled out? -P.S. Also, if anyone knows Conway, please wish him well, and maybe they might ask him this question? -Update May 2020 RIP Mr Conway - -REPLY [15 votes]: It is possible that Conway was referring to the generic construction that works for all finite groups equipped with faithful representations, given in the other answers. However, I think it is more likely that Conway was referring to a construction that is specific to the monster, hinted at in Ian Agol's comment. -In section 14 of Conway's 1985 Inventiones paper, "A simple construction for the Fischer-Griess monster group", Conway points out that in the 196883 dimensional representation, there is a collection of distinguished lines fixed by centralizers of elements in conjugacy class 2A (in fact, 196883 decomposes as 1+4371+96255+96256 under the action of such a centralizer). Elements in this conjugacy class are known as Fischer involutions or transpositions, and the lines are called axes of transpositions. The centralizer of a transposition is a double cover of Fischer's Baby Monster sporadic group, so there are $\frac{|\mathbb{M}|}{2|\mathbb{B}|} \sim 9 \times 10^{19}$ of these axes. Choosing a nonzero point on an axis, and taking its orbit yields an arrangement of points (or we may consider the convex hull). I claim that this arrangement has Monster symmetry. -To prove this, we use the fact that for a pair $(x,y)$ of 2A elements, their product lies in 2A if and only if the angle between the axes is a particular value. In fact, the conjugacy class of the product is uniquely determined by the angle, except for classes 3C and 4B, as we can see from the "McKay E8 diagram" in Conway's paper: - -We therefore have a natural construction of a graph from this polytope, whose vertices correspond to axes (or orbit elements on the axes), and whose edges correspond to those pairs whose product lies in class 2A. This graph is the "monster graph" mentioned in Ian Agol's comment, and Griess showed that its automorphism group is precisely the monster. -These axes appear in the theory of vertex algebras, in the following way. The 196883 dimensional representation is naturally embedded as the space of Virasoro primary vectors in the weight 2 subspace of the "Moonshine Module" vertex operator algebra $V^\natural$. On each axis, there is a distinguished "Ising vector" that generates a vertex algebra isomorphic to the $L(1/2,0)$ minimal model. Miyamoto showed that any Ising vector in a vertex algebra yields a "Miyamoto involution", and for $V^\natural$, these are precisely the 2A elements.<|endoftext|> -TITLE: Simple proof that the arithmetic genus is non-negative -QUESTION [6 upvotes]: I take an irreducible and reduced closed curve $C\subseteq \mathbb{P}^n$, defined over an algebraically closed field $k$ and define the arithmetic genus $p_a(C)$ as the integer such that the Hilbert polynomial of $C$ (or of its ideal of polynomials vanishing on it) is -$$h_C(X)=\deg(C) X+1-p_a(C).$$ -I would then like to have a simple proof that $p_a(C)\ge 0$. Of course, one may say that it is because it is the dimension of a vector space obtain using cohomology, but I would like a proof not involving cohomology, as it is for a master work and in my course of algebraic geometry I (unfortunately you will say) did not define the cohomology. It would then take a lot of time for the student to understand the cohomology just to prove this. -In the course, we defined the Hilbert Polynomial of any ideal, computed local intersections, Bézout theorem in $\mathbb{P}^n$ and studied blow-ups of surfaces. We also proved that $p_a(C)=g(C)$ when $C$ is smooth, where $g(C)$ is given by Riemann-Roch (the smallest integer such that $\ell(D)\ge \deg(D)+1-g$ for each divisor $D$ on $C$). I would like the student to use the fact that $p_a(C)\ge 0$ to bound the type of singularities of a plane curve and to show that one can have a resolution by blowing-up the singular points and repeating this process finitely many times. -Thanks for your help. - -REPLY [2 votes]: Mumford (Complex Projective Varieties, section 7) has the following, reasonably simple proof. -Let $d$ be the degree of $\mathrm{C}$, $m$ big enough such that $h_{\mathrm{C}}(m)=\mathrm{dim}_{\mathbf{C}} (\mathbf{C}[\mathrm{T}_0,\ldots,\mathrm{T}_n]/\mathrm{I}(\mathrm{C}))_m$ and $md/2>p_a$. Embed $\mathrm{C}$ into $\mathbf{P}^N$ by the degree $m$ Veronese embedding; let $\mathrm{L}\subset\mathbf{P}^N$ be a linear space containing $\nu_m(\mathrm{C})$ such that $\nu_m(\mathrm{C})$ is nondegenerate in $\mathrm{L}$. Then $\dim(\mathrm{L})=h_\mathrm{C}(m)=md+1-p_a$, in other words $\mathrm{L}\simeq\mathbf{P}^{md-p_a}$, and the degree of $\nu_m(\mathrm{C})$ is $$md=\mathrm{deg}(\nu_m(\mathrm{C}))\geqslant\mathrm{codim}(\nu_m(\mathrm{C}))+1\geqslant md-p_a$$ -since $\nu_m(\mathrm{C})$ is nondegenerate in $\mathrm{L}$.<|endoftext|> -TITLE: Bounds on Fourier coefficients for $GL(3)$ -QUESTION [5 upvotes]: I am referring for instance to this question about coefficients of automorphic forms on $GL(3)$. I know that the Ramanujan on average bound is known and gives -$$\sum_{n^2 m < x} |\lambda(n,m)|^2 \ll x^{1+\varepsilon}.$$ -Is there anything known (in terms of upper bounds, with explicit dependence in the fixed $m$ or $n$) about the partial sums (except trivially bounding by the above) : -$$\sum_{n^2< x} |\lambda(n,m)|^2 \qquad \text{and} \qquad \sum_{m < x} |\lambda(n,m)|^2 ?$$ - -REPLY [2 votes]: On the Ramanujan conjecture $\lambda(m,n) \ll 1$. As there is no cancellation in the second sum, essentially (upto $x^\epsilon$) the best upper bound which one may expect for that is $x^{1+\epsilon}$. -For the first sum for the same reason the best possible bound would be $x^{1/2+\epsilon}$. To prove that note that $\lambda(n,m)=\overline{\lambda(m,n)}$. So -$$\sum_{n^2 -TITLE: Vanishing theorems on a non-compact manifold -QUESTION [5 upvotes]: In complex geometry, various vanishing theorems for cohomology -groups of a hermitian line bundle E over a compact complex manifold X have been found. -My question is -Is there some vanishing theorems over a general noncompact complex manifold exist? (Except shose on Stein manifolds) - -REPLY [5 votes]: A complex manifold of dimension $n$ is non-compact if and only if $H^n(X,{\mathcal F})=0$ for any coherent sheaf ${\mathcal F}$ on $X$. This is the only general vanishing result that I know of on non-compact manifolds. -But other than that, there are some other vanishing theorems on non-compact manifolds. For instance, if $X$ is $q$-complete, then $H^r(X, {\mathcal F})=0,\forall r\geq q$. -Or if $X$ is weakly $1$-complete, and $L$ is a positive line bundle on $X$, then $H^{n,q}(X,L)=0$, $\forall q\geq 1$. -In general, for a non-compact manifold, you know nothing about the cohmology groups, they can even be non-Hausdorff.<|endoftext|> -TITLE: The $S$-unit equation for functions on curves -QUESTION [8 upvotes]: Let $X$ be a smooth projective connected curve over a number field $k$, and let $S \neq \emptyset$ be a finite set of closed points of $X$. The curve $Y = X \setminus S$ is affine, and we denote by $R$ the $k$-algebra of regular functions on $Y$. -The $S$-unit equation for $k(X)$ is the equation $f+g =1$, with $f,g \in R^\times \setminus k^\times$; in other words $f$ and $g$ are two non-constant rational functions on $X$ whose zeros and poles are contained in $S$. -For example, in the case $Y = \mathbb{P}^1 \setminus \{0,1,\infty\}$, the pair of functions $(f,g) = (t,1-t)$ is a solution of the $S$-unit equation. In fact, if $f$ is an homography preserving $\{0,1,\infty\}$ then $1-f$ has the same property, and $(f,1-f)$ is a solution. So there are at least 6 solutions. -Mason proved that there exists an effecitve bound (depending on the cardinality of $S$ and the genus of $X$) on the degrees of the possible solutions $f,g$; see e.g. Zannier, Some remarks on the $S$-unit equation in function fields, Acta Arith. 64 (1993) no. 1, 87--98. -Is it expected that the number of solutions $(f,g)$ is actually finite? Are there methods or algorithms to find these solutions in practice? -I am interested in the following particular cases: - -$X=E$ is an elliptic curve and $S$ is a finite subgroup of $E$; -$X$ is a modular curve and $S$ is the set of cusps of $X$. - -REPLY [5 votes]: The set of solutions to the $S$-unit equation for $k(X)$ is finite. Let me explain why. (You can "theoretically" find all solutions, as the finiteness eventually boils down to the "effective" finiteness result of de Franchis-Severi on maps of curves.) -Let $k$ be a number field, let $X$ be a smooth projective geometrically connected curve over $k$, let $S$ be a finite set of closed points of $X$, and let $Y := X \setminus S$. Let $R = \mathcal{O}(Y)$. -Claim. The set of solutions $(f,g)$ of the $S$-unit equation $f+g =1$ for $X$ (with $f$ and $g$ thus in $R^\times \setminus k^\times $ ) is in bijection with the set of non-constant morphisms $Y\to \mathbb{P}^1_k \setminus \{0,1,\infty\}$. -Proof of Claim. Let $(f,g)$ be a solution of the $S$-unit equation in $k(X)$. Then $f:Y\to \mathbb{G}_{m,k}$ is a non-constant morphism such that $1-f$ also defines a morphism to $\mathbb{G}_{m,k}$. Thus $f(Y) \subset \mathbb{G}_{m,k} \setminus \{1\}$. Conversely, if $f$ is a non-constant morphism from $Y$ to $\mathbb{P}^1_{k}\setminus \{0,1,\infty\}$, then $1-f$ is also such a morphism. This concludes the proof. QED -Let $K$ be an algebraic closure of $k$. Note that $Hom_k(Y,C) \subset Hom_K(Y_K,C_K)$. Thus, -to answer your question, we can work over an algebraically closed field $K$ of characteristic zero. (That is, you can as well let $k$ be any field of characteristic zero.) -The finiteness of the set of solutions will boil down to finiteness results for hyperbolic curves. -Let me recall what a hyperbolic curve is. From now on, let $K$ be an algebraically closed field of characteristic zero. - - -Hyperbolic curves. Let $C$ be a smooth quasi-projective connected curve over $K$. We say that $C$ is hyperbolic if $2g(\overline{C}) - 2 + \#( \overline{C}\setminus C )>0$. Equivalently, $C$ is non-hyperbolic if and only if $C$ is isomorphic to $\mathbb{P}^1_K$, $\mathbb{A}^1_K, \mathbb{A}^1_{K}\setminus \{0\}$, or a smooth proper connected genus one curve over $K$. - - -We will need the following topological lemma on hyperbolic curves. (For your purposes we really just need that $\mathbb{P}^1_k\setminus \{0,1,\infty\}$ has a finite etale cover of genus at least two. This can be proven by considering $\mathbb{P}^1_k\setminus \{0,1,\infty\}$ as an (open) modular curve and taking a modular curve of high enough (even) level. -Topological Lemma. If $C$ is a hyperbolic curve over $K$, then there is a finite etale morphism $D\to C$ with $D$ a smooth quasi-projective connected curve over $D$ such that the genus of $\overline{D}$ is at least two. (This is obvious if $\overline{C}$ itself is of genus at least two. Thus, we reduce to the case that $C = \mathbb{P}^1_K\setminus \{0,1,\infty\}$ or that $C $ is $E\setminus \{0\}$ with $0$ the origin on an $E$ an elliptic curve over $K$. In these two cases, one can explicitly construct $D$. -Hyperbolic curves satisfy many finiteness properties. One of them is the following version of the theorem of De Franchis-Severi. An integral quasi-projective curve is of log-general type if its normalization is of log-general type, i.e., hyperbolic. - - -Theorem. [De Franchis-Severi] Let $C$ be an integral quasi-projective curve over $K$ whose normalization is of log-general type. Then, for every integral quasi-projective curve $Y$ over $K$, the set of non-constant morphisms $Y\to C$ is finite. - - -Proof of Theorem. Note that the normalization $\widetilde{Y}\to Y$ is surjective. Therefore, replacing $Y$ by its normalization if necessary, we may and do assume that $Y$ is smooth. Now, every non-constant morphism $Y\to C$ is dominant and will factor uniquely over the normalization of $C$. Thus, we may and do assume that $C$ is smooth. -Now, we use the Topological Lemma. Thus, let $D\to C$ be a finite etale morphism with $D$ of genus at least two. Let $d:=\deg(D/C)$. If $Y\to C$ is a morphism, then the pull-back $Y':=Y\times_C D$ is finite etale of degree $d$ over $Y$. Since $K$ is algebraically closed of characteristic zero, the set of $Y$-isomorphism classes of finite etale covers $Y'\to Y$ of degree $d$ is finite. Thus, we may and do assume that $C=D$. Now, note that every non-constant morphism $Y\to C$ extends to a non-constant morphism $\overline{Y}\to \overline{C}$. However, there are only finitely many such maps as $\overline{C}$ is of genus at least two. QED -Remark. In the last paragraph of the previous proof we use the finiteness theorem of de Franchis-Severi for compact connected Riemann surfaces of genus at least two. (It just happens to be that this "compact" version implies the analogous "affine" version. This is no longer true in higher dimensions.) The "compact" finiteness result also holds in higher dimensions: if $C$ is a proper variety of general type and $Y$ is a proper variety, then the set of dominant rational maps $Y\dashrightarrow C$ is finite. This was proven by Kobayashi-Ochiai. (You can use this to show that, for every integral quasi-projective variety $Y$ over $K$, the set of non-constant morphisms $Y\to \mathbb{P}^1_K\setminus\{0,1,\infty\}$ is finite.)<|endoftext|> -TITLE: Fully invariant measures for rational functions -QUESTION [5 upvotes]: Let $f(z)$ be a rational function of degree $d \geq 2$, with complex coefficients. I am interested in fully invariant measures for the dynamical system $(\mathbb C_\infty,f)$, where $\mathbb C_\infty$ is the Riemann sphere. By a fully invariant measure, I mean a probability measure $\mu$ such that $f^\ast \mu = d \mu$. (Such a measure also satisfies $f_\ast \mu =\mu$). -We can limit ourselves to ergodic ones (those which are not barycenter of two other fully invariant measures). -Those measures which have finite support are easy to describe: from the classical theory of Montel-Fatou-Julia, there exists a larger finite subset if $E$ such that $f^{-1}(E)=E$ and it has at most two elements; every fully invariant measure with finite support has support in $E$, so there are $0$, $1$ or $2$ finite support fully invariant ergodic measures. -From results of Ljubisch and Freyre-Lopez-Mane, there is also the so-called "natural measure" $\mu$, which is defined as the limit of $(\frac{1}{d}f^\ast)^n \nu$ where $\nu$ is any smooth measure on $\mathbb C_\infty$, or any Dirac measure $\delta_x$ for $x \not \in E$. It is ergodic, has the Julia set for support, and many other nice properties. -My question is - -Are there always other ergodic fully invariant measures? If so, how to prove it and construct one? - -REPLY [5 votes]: The unique measure of maximal entropy $\mu_f$ supported on the Julia set of a rational map $f$ of degree $d \geq 2$ is indeed the unique balanced measure for $f$, i.e., the only probability measure $\mu$ not charging the exceptional set and satisfying $f^*\mu =d \cdot \mu$. As you already noticed in the comments, uniqueness of a measure with this property is explicitly stated in the mentioned paper by Freire, Lopes and Ma\~ne in their Theorem, part (d) (page 46). The proof of this statement is on p. 55 and the argument goes as follows: for any balanced measure $\mu$ it is shown that $\mu$ is absolutely continuous with respect to $\mu_f$ and the ergodicity of $\mu_f$ implies that $\mu=\mu_f$ (existence and ergodicity of $\mu_f$ are proved earlier in the paper). No assumption of non-atomicity, no reference to critical points or classification of Fatou components is employed in the proof of this uniqueness statement. -Another way to prove uniqueness of balanced measure is to use potentials of measures on the Riemann sphere $\mathbb{C}_\infty$ introduced as in -F. Berteloot, V. Mayer, Rudiments de dynamique - holomorphe, Vol. 7 of Cours Sp\'ecialis\'es, - Soci\'et\'e Math\'ematique de France, Paris (2001) -They give a streamlined treatment based on prior results by Fornaess and Sibony, Hubbard and Papadopol, Ueda and others. -Consider the cone $\mathcal{P}$ of functions $U$ plurisubharmonic on $\mathbb{C}^2$ and satisfying $U(tz)=c\log|t|+U(z)$ with a constant $c=c(U) >0$. Each such function defines a positive measure $\mu_U$ on $\mathbb{C}_\infty$ by $\langle \mu_U, \Phi \rangle =\int_{\mathbb{C}_\infty}(U \circ \sigma)\frac{i}{\pi}\partial\bar{\partial}\Phi$ for every smooth test function $\Phi$ with support in the domain of definition of the section $\sigma$ of the natural projection $\Pi: \mathbb{C}^2\setminus \{0\} \to \mathbb{C}_\infty$. Furthermore, every positive measure $\nu$ on $\mathbb{C}_\infty$ is defined by a function $U \in \mathcal{P}$ (unique if required to satisfy $\sup_{\|z\|\leq 1}U(z)=0$), specifically by $U(z)=\int_{\mathbb{C}_\infty}\log\frac{|z_1w_2-z_2w_1|}{\|w\|}d\nu([w])$ (Th\'eor`eme VIII.9 in this reference). This is called the potential of $\nu$. -Now, if a measure $\nu$ is balanced, then its potential $U$ satisfies $F^*U=d\cdot U$ Lemme VIII.12), hence $\frac{1}{d^n}F^{*n}U=U$ for every $n$. Here $F$ denotes a lift of $F$ to $\mathbb{C}^2$. Taking limits in $L^1_{loc}$ as $n \to \infty$ we get $U=G_f$ (Th\'eor`eme VIII.15), the potential of the Lyubich-Freire-Lopes-Ma\~ne measure $\mu_f$. Lifts are not unique, but this does not cause a problem. -If you relax the assumption on a measure supported on Julia set to $f_*\mu = \mu$, then there can be more measures satisfying it, even ergodic ones, besides the measure $\mu_f$. Of course the entropy will be less than $\log d$, sometimes even $0$. For more details on this see -S. P. Lalley, Brownian motion and the equilibrium measure on the Julia - set of a rational mapping, Ann. Probab. 20, 4 (1992), - 1932--1967.<|endoftext|> -TITLE: How can I see the relation between shtukas and the Langlands conjecture? -QUESTION [20 upvotes]: The following bullet points represent the very peak of my understanding of the resolution of the Langlands program for function fields. Disclaimer: I don't know what I'm writing about. - -Drinfeld modules are like the function field analogue of CM elliptic curves. To see this, complexify an elliptic curve $E$ to get a torus $\mathbb{C} / \Lambda$. If $K$ is an imaginary quadratic field, those lattices $\Lambda$ such that $\mathcal{O}_K \Lambda \subseteq \Lambda$ correspond to elliptic curves with CM, meaning that there exists a map $\mathcal{O}_K \to \operatorname{End} E$ whose 'derivative' is the inclusion $\mathcal{O}_K \hookrightarrow \mathbb{C}$. Now pass to function fields. Take $X$ a curve over $\mathbb{F}_q$, with function field $K$, and put $C$ for the algebraic closure of the completion. Then we can define Drinfeld modules as an algebraic structure on a quotient $C / \Lambda$. - -Shtukas are a 'generalisation' of Drinfeld modules. According to Wikipedia, they consist roughly of a vector bundle over a curve, together with some extra structure identifying a "Frobenius twist" of the bundle with a "modification" of it. From Goss' "What is..." article, I gather that some analogy with differential operators is also involved in their conception. - -Shtukas are used to give a correspondence between automorphic forms on $\operatorname{GL}_n(K)$, with $K$ a function field, and certain representations of absolute Galois groups. For each automorphic form, one somehow considers the $\ell$-adic cohomology of the stack of rank-$n$ shtukas with a certain level structure, and I presume this cohomology has an equivariant structure that gives rise to a representation. - - -While this gives me a comfortable overview, one thing I cannot put my finger on is why things work the way they work. I fail to get a grasp on the intuition behind a shtuka, and I especially fail to see why it makes sense to study them with an outlook on the Langlands program. This leads to the following questions. - -Question 1. What is the intuition behind shtukas? What are they, even roughly speaking? Is there a number field analogue that I might be more comfortable with? -Question 2. How can I 'see' that shtukas should be of aid with the Langlands program? What did Drinfeld see when he started out? Why should I want to take the cohomology of the moduli stack? Were there preceding results pointing in the direction of this approach? - -REPLY [11 votes]: I think shtukas are best understood ahistorically. I would start with the modular curves, but specifically with the (geometric) Eichler-Shimura relation. This says that the Hecke operator at $p$, viewed as a correspondence on $X_0(N) \times X_0(N)$, when reduced at characteristic $p$, is just the graph of Frobenius plus the transpose of the graph of Frobenius. The relevance of this fact to the proof of a Langlands correspondence that relates traces of Frobenius acting on cohomology to eigenvalues of Hecke operators acting on cohomology should be unsurprising, even if completing the argument required much brilliant work by many people. -Now for higher-dimensional Shimura varieties, one cannot always generalize this simple geometric Eichler-Shimura relation, and instead must state and prove a suitable cohomological analogue of it. -When we go to the function field world, on the other hand, it pays to be very naive. We want to define a moduli space of some kind of object where Hecke operators act, and Frobenius acts, and these two actions are related. As David Ben-Zvi described in his answer, we understand what kind of object Hecke operators act on, and how - they act on vector bundles, or more generally $G$-bundles, and they act by modifying the bundle at a particular point in a controlled way. Frobenius also acts on $G$-bundles, by pullback. But these actions have nothing to do with each other. -The solution, then, is to force these actions to have something to do with each other in the simplest possible way - demand that the pullback of a $G$-bundle by Frobenius equal its modification at a particular point, in a certain controlled way. In fact, we can freely do this at more than one point, producing a space on which Frobenius acts like any desired composition of Hecke operators at different points. -The actual way Drinfeld came up with the definition was totally different, and did involve differential operators. He first came up with Drinfeld modules by an inspired analogy with the moduli spaces of elliptic curves. He then realized an analogy with work of Krichever, which he realized should lead to an analogous object defined using sheaves, the resulting definition was a shtuka. Precisely, the relation is that a rank $r$ Drinfeld module is the same thing as a $GL_r$-shtuka with two legs (corresponding to the standard representation of $GL_r$ and its dual under the geometric Satake isomorphism, in the modern language) at two points, where one is allowed to vary and the other is fixed at the point "$\infty$", and furthermore where we require that the induced map of vector bundles at the point $\infty$ is nilpotent. -So moduli spaces of Drinfeld modules will be certain subsets of moduli spaces of shtukas. However, the relationship between Drinfeld modules and shtukas is rarely used to study either - the research on both of them is usually quite separate.<|endoftext|> -TITLE: Questions about existence of injections between infinite sets and the sets of all infinite topologies on them -QUESTION [5 upvotes]: 1) If $X$ is an infinite set and $T_X$ the set of all infinite topologies on $X$ is it in general true that there is no injection $f_T:T_X \to X$? -2) What conditions on $X$ assure an injection (if that´s ever possible)? -By infinite topologies, I mean topologies with an infinite number of sets (subsets of $X$) as its elements. - -REPLY [2 votes]: For an infinite set $X$, there are $2^{2^{|X|}}$ topologies with infinitely many elements on $X$. -The upper bound is obvious, the lower bound is given by my comment (and the slight correction in YCor's): for any ultrafilter $U$ on $X$, define $T_U$ to have as opens the empty set and the elements of $U$. It is clear that this forms a topology, and that you can recover $U$ from it. -Moreover, clearly $T_U$ has infinitely many open sets. -So $U\mapsto T_U$ defines an injection $\beta X\to T_X$, and it's well known that $|\beta X| = 2^{2^{|X|}}$. -So there can be no injection $T_X\to X$ (in fact, no injection $T_X\to P(X)$) -YCor's comment below mine elaborates a bit on what one can ask of these topologies and how many we get (Hausdorff, metrizable)<|endoftext|> -TITLE: How did Hilbert prove the Nullstellensatz? -QUESTION [7 upvotes]: All of the many proofs of the Nullstellensatz I have seen use results from long after Hilbert’s time: Zariski’s lemma, Noether normalization, the Rabinowitch trick, model theory, etc. How did Hilbert’s original proof go? - -REPLY [10 votes]: Here is a scan of the relevant pages from the english translation of Hilbert's 1893 paper.<|endoftext|> -TITLE: Sard's theorem and Cantor set -QUESTION [6 upvotes]: Sard's famous theorem asserts that -Theorem. The set of critical values of a smooth function from a manifold to another has Lebesgue measure $0$. -I am asking for the curiosity that is it possible to find such a function whose set of critical values is - -Cantor set or -Any other uncountably infinite set? - -REPLY [2 votes]: By a theorem of Whitney (easy in this 1-dimensional case), any compact subset K in the interval I is the set of zeroes of a -smooth (C infty) nonnegative function f. As Robert said, take a primitive F. Provided -that K has no interior point, the critical values F(K) of F are homeomorphic with K.<|endoftext|> -TITLE: Stationary phase in spherical integral -QUESTION [6 upvotes]: I'm trying to find asymptotics for an oscillatory integral on $\mathbb{S}^{n-1}$, for which my advisor said I should use stationary phase arguments. The particular, he claims that: - -If $\lambda\gg 1$, then - $$I(\lambda,x) = \int_{\mathbb{S}^{n-1}}(x\cdot y)e^{i\lambda(x\cdot y)}\,d\sigma(y),$$ - is $\mathcal{O}(|x|(\lambda |x|)^{-\frac{n-1}{2}})$ when $|x|\geq \lambda^{-1}$. - -When I hear stationary phase, I think of working with operators of the form -$$ -L = \frac{\nabla_y(x\cdot y)}{i\lambda |\nabla_y (x\cdot y)|^2} \cdot \nabla_y, -$$ -since then $L^N[e^{i\lambda (x\cdot y)}] = e^{i\lambda (x\cdot y)}$, for any $N \geq 1$, and I can use integration by parts to move these operators over to the $(x\cdot y)$ term. However, wouldn't that require that the integral be defined over an $n$-dimensional region, rather than an $(n-1)$-dimensional surface? -I wouldn't be struggling so much if the gradient $\nabla_y$ could be taken in Cartesian coordinates. But we have a surface integral in $d\sigma(y)$, meaning that we would need to parametrize our surface with $n-1$ parameters, say $\omega = (\omega_1, \ldots, \omega_{n-1}) \in \Omega \subset \mathbb{R}^{n-1}$, with any derivates now being taken in these new variables. Specifically $\nabla_y$ would generate an $n$-vector, while $\nabla_{\omega}$ would generate an $n-1$-vector. So how should I go about applying the same kind of stationary phase arguments to this new integral? -$$ -I(\lambda, x) = \int_{\Omega} (x \cdot y(\omega)) e^{i\lambda (x\cdot y(\omega))} \,dV(\omega) -$$ -Now that $y$ depends on $\omega$, the gradients $\nabla_\omega (x\cdot y(\omega))$ become much trickier to get a grasp on. I'm particularly struggling trying to argue how we can find regions where $|\nabla_\omega (x\cdot y(\omega))| > 0$, so that our $L$ type operators are properly defined. -Am I going about this all wrong? If I can get the big-oh asymptotics I mentioned above in some other way, it doesn't really matter. I just need to prove these results as a lemma to something bigger. Any help is much appreciated! - -REPLY [3 votes]: You have -$ -I(\lambda, x)=x\cdot\int_{\mathbb S^{n-1}} ye^{i \lambda x\cdot y} d\sigma(y)=x\cdot J(x,\lambda) -$ -and you claim that for $\vert x\vert \lambda \ge 1$, you have -$$ -J(x,\lambda)=O((\vert x\vert \lambda)^{-\frac{n-1}{2}}). -$$ -Indeed, using coordinate charts and a finite partition of unity, you are reduced to the case where -$$ -J(x,\lambda)=\int_{\mathbb R^{n-1}} a(z) e^{i\lambda (x'\cdot z+ x_n\sqrt{1- \vert z\vert^2})}dz, \quad\text{$a\in C^\infty_0(\mathbb R^{n-1})$ supported near $0$, $x=(x', x_n)\in \mathbb R^{n-1}\times \mathbb R$}. -$$ -Let us set $\phi(x,z)=x'\cdot z+ x_n\sqrt{1- \vert z\vert^2}$. We have -$$ -\frac{\partial \phi}{\partial z}= x'-(1- \vert z\vert^2)^{-1/2} z x_n, -$$ -which vanishes at $z=0$ when $x'=0$. Then you calculate the Hessian of $\phi$ at $z=0$ -and get -$$ -\phi''_{zz}=-x_n. -$$ -The stationary phase in $n-1$ dimensions gives the sought estimate with $O((\vert x_n\vert \lambda)^{-\frac{n-1}{2}})$ (note that you know that $\vert x_n\vert \lambda \ge 1$, since you are near $x'=0$).<|endoftext|> -TITLE: Spectrum of "classical" operators -QUESTION [7 upvotes]: Lately, I've been reading a couple of papers from different one-dimensional PDE contexts on which operators like $\mathcal{L}:=-\partial_x^2+c_*+\Phi$ repeatedly appear. Usually, on these contexts $\Phi$ is a smooth exponentially decaying function and $c_*\in\mathbb{R}$ is a positive constant. -I am very surprised that in all of these papers the authors, just by knowing these facts, they immediately conclude that the continuous spectrum of $\mathcal{L}$ is exactly $[c_*,+\infty)$ and the rest of the spectrum consists on a finite number of eigenvalues. My question is, how do they know that the continuous spectrum starts exactly at $c_*$? I've seen this at least five times for different values of $c_*$ and different functions $\Phi$ (all of them smooth with exponential decay). Does anyone have an explanation for this? Or any reference? -A second question (I know that the difficulty of the question can exponentially grow now so I am actually very happy just by understanding the previous part!): What if I consider a non-smooth but still exponentially decaying $\Phi$? For instance $\Phi=e^{-\vert x\vert}$? The previous "result" still holds? - -REPLY [11 votes]: Let $A$ be a self-adjoint operator with domain $D(A)\subset\mathcal H$ ($\mathcal H$ is some Hilbert space). An operator $C$ with $D(A)\subset D(C)$ is called relatively compact with respect to $C$ if $C(A-zI)^{-1}$ is compact for some (hence all) $z\notin\sigma(A)$. Paraphrasing Corollary 2, page 113 Section XIII.4, in [1], we have - -If $C$ is relatively compact with respect to $A$, then $B:=A+C$ is closed with domain $D(B)=D(A)$, and $$\sigma_{ess}(B)=\sigma_{ess}(A).$$ - -In fact, this is elementary once one realises that $C(A-zI)^{-1}:\mathcal H\to \mathcal H$ is compact if and only if $C:D(A)\to\mathcal H$ is compact. -In your case, setting $A = -\partial_x^2 + c_*$ and $C$ the multiplication operator by $\Phi$, is it not hard to prove that $C:H^2(\mathbb R)\to L^2(\mathbb R)$ is compact. We then obtain that -$$ -\sigma_{\mathrm{ess}}(\mathcal{L}) = \sigma_{\mathrm{ess}}(-\partial_x^2 + c_*). -$$ -Since it is well know that $\sigma_{\mathrm{ess}}(-\partial_x^2 + c_*) = [c_*,+\infty)$, the result follows. -For your second question, this answer is yes the result hold for $\Phi(x) = e^{-|x|}$. In fact it will hold in dimension $d\leq 3$ for any $\Phi\in L^2(\mathbb R^d)$. As explained earlier, it is enough to show that multiplication by $\Phi$ is (well-defined and) compact from $H^2(\mathbb R^d)\to L^2(\mathbb R^d)$; let us prove this result. -Let $(f_n)_{n\geq0}$ be a bounded sequence in $H^2(\mathbb R^d)$, $d\leq 3$. The core of the argument is the following fact. - -$(f_n)_{n\geq0}$ is bounded in $L^\infty$ and, up to extraction, converges $L^\infty$-locally to some function $f$. - -Note that $f$ then has to be bounded as well. Once this is established, we see that -$$ \begin{align*} -\limsup |\Phi f_n-\Phi f|_{L^2}^2 -& = \limsup \int|\Phi|^2\cdot|f_n-f|^2 \\\\ -& \leq \limsup |f_n-f|_{L^\infty([-k,k]^d)}^2\cdot\int|\Phi|^2{\mathbf 1}_{[-k,k]^d} \\\\ -& \quad + \limsup (|f_n|_{L^\infty}+|f|_{L^\infty})^2\cdot\int|\Phi|^2{\mathbf 1}_{([-k,k]^d)^\complement} \\\\ -& \leq |\Phi|_{L^2}^2\cdot0 - + 4M^2\cdot\limsup \int |\Phi|^2{\mathbf 1}_{([-k,k]^d)^\complement} -\end{align*} $$ -for $M$ a bound on the norms $|f_n|_{L^\infty}$, and $\limsup$ the limit superior along a convergent subsequence. Because $\Phi$ is in $L^2$, Lebesgue's dominated convergence theorem guarantees that the limit is zero, and $\Phi f_n$ converges to $\Phi f$ as expected. -Let us turn to the proof of the fact. The boundedness of $(f_n)_{n\geq0}$ on compact sets follows from the continuous embeddings from $H^2(\ell+[0,1]^d)$ to $\mathcal C^0(\ell+[0,1]^d)$ for all $\ell\in\mathbb Z^d$ (because $d\leq3$). Since the norm of these injections does not depend on $\ell$, $(f_n)_{n\geq0}$ is in fact uniformly bounded over $\mathbb R^d$ -As for the convergence up to extraction, according to the usual Sobolev embeddings/inequalities, the sequence of restrictions $\big((f_n)_{|[-k,k]^d}\big)_{n\geq0}$ is relatively compact in $\mathcal C^0([-k,k]^d)$ for all $k\geq1$ (we use again that $d\leq3$). We can then extract diagonally a subsequence $(f_{\sigma(m)})_{m\geq0}$ that converges to some continuous function $f$ uniformly over the compact sets, concluding the proof of the fact. - -[1] Reed, M., Simon, B. (1978). Methods of Modern Mathematical Physics. IV Analysis of Operators. New York: Academic Press.<|endoftext|> -TITLE: Where did this presentation of Godel's theorem appear? -QUESTION [11 upvotes]: This question was asked and bountied at MSE, with no response. - -Many years ago I ran into the following proof of Godel's first incompleteness theorem -(here $T$ is an "appropriate" theory of arithmetic.): - -First, we observe that Tarski's undefinability theorem can be slightly tweaked to say that for all $M\models T$, the theory of $M$ can't be the standard part of an $M$-(parameter-freely-)definable set. Next, by the usual representability arguments we have that every computable set is invariantly definable: if $X\subseteq\mathbb{N}$ is computable and $M\models T$ then $X$ is the standard part of a definable set in $M$. Now if $S$ were a computable satisfiable completion of $T$ we would get a contradiction by looking at $M\models S$. If we want, we can then replace "satisfiable" with "consistent" via the completeness theorem. - -My question is: -Where did this argument appear? -I'm not asking whether this is a "genuinely different" argument (although that said, see here). Rather, I'm interested in its presentation in terms of invariant definability. This was a notion first introduced by Kreisel following Godel/Herbrand and subsequently studied by others (see e.g. 1, 2, 3). My vivid recollection is that the argument above appeared as a footnote in the first paper mentioned above, but it turns out it's not there after all. - -REPLY [3 votes]: This argument is essentially similar to the argument of Mel Fitting in his article, "Russell's paradox, Gödel's theorem" Chapter in book: Raymond Smullyan on self reference, 47–66, Outstanding Contributions to Logic, 14, Springer, Cham, 2017. -We had used this article as one of the readings in my Graduate Phil Logic seminar last term. -Fitting argues via Russell's paradox using Tarski's theorem on the non-definability of truth, in a manner that I find very similar to the argument you describe (although he doesn't use your invariant terminology, and perhaps you may regard his argument as not the same as what you intend). Namely, he argues that although provability is representable, truth is not, because of Russell's theorem, and so the incompleteness theorem follows. -Part of his point in that article is that this way of arguing avoids the need for explicit self-reference.<|endoftext|> -TITLE: Movement of repelled particles in a ball -QUESTION [6 upvotes]: EDIT: - -Given a system of $N\geq 3$ charged point particles in $\mathbb{R}^3$ of the same charge which interact according to Coulomb law (thus they repell one from each other). Is it possible that the system remains in a fixed ball all the time? (For $N=2$ this is impossible, and this is what I expect in general.) - -More precisely, denote $m_1,\dots, m_N>0$ the masses of the particles. Assume that the $i$th particle acts on $j$th one with the force -$$\vec F_{ij}=\frac{k_ee_ie_j}{|\vec x_j-\vec x_i|^3}\cdot (\vec x_j-\vec x_i), $$ -where $k_e>0$ is a constant, $e_i$ is a charge of $i$th particle such that $e_ie_j>0$, $\vec x_i$ is the location of the $i$th particle. The equations of motions are -$$m_j\frac{d^2 x_j}{dt^2}=\sum_{i\ne j}\vec F_{ij}, \mbox{ where } j=1,\dots,N.\,\,\,(1)$$ - -The question is whether there is a solution such that for some $R$ one has - $$||\vec x_i(t)||0, \, i=1,\dots, N.$$ - -ADDED: I expect that this is impossible. In fact I expect that not only for Coulomb law, but still in greater generality. Assume that the equations (1) are satisfied when the force $\vec F_{ij}=\vec F_{ij}(x_i,x_j)$ has the same direction as the vector $\vec x_j-\vec x_i$. Assume moreover that if all points are in a fixed ball of the radius $R$ then for some constant $\varepsilon >0$ such that -$$||\vec F_{ij}||>\varepsilon.$$ -Is there a solution of (1) such that all the point are in the ball of radius $R$ for all $t>0$? - -REPLY [7 votes]: If all the particles remained in a bounded domain, the virial theorem would apply. In the case of a radial inverse square power law, it states that twice the asymptotic time average of kinetic energy of the system equals minus the asymptotic time average of its potential energy. However, while the kinetic energy is always nonnegative, the potential energy for repulsive coulomb forces is positive, contradiction.<|endoftext|> -TITLE: Cobordism invariants: topological v.s. geometric -QUESTION [8 upvotes]: Some cobordism invariants are not cohomology classes. Such as the $\mathbb{Z}_{16}$-valued eta invariant $\eta$ of $\Omega_4^{Pin^+}$, the $\mathbb{Z}_8$-valued Arf-Brown-Kervaire invariant ABK of $\Omega_2^{Pin^-}$, and the $\mathbb{Z}_4$-valued quadratic enhancement $q(a)$ of $\Omega_2^{Pin^-}(B\mathbb{Z}_2)$ where $a\in H^1(M,\mathbb{Z}_2)$. The last invariant is defined as follows: Any 2-manifold $M$ always admits a $Pin^-$ structure. $Pin^-$ structures are in one-to-one correspondence with quadratic enhancements -$$q: H^1(M,\mathbb{Z}_2)\to\mathbb{Z}_4$$ -such that -$$q(x+y)-q(x)-q(y)=2\int_M x\cup y\mod4.$$ -In particular, -$$q(x)=\int_M x\cup x\mod2.$$ -There are many more such examples, I only mention these. -We say that a cobordism invariant is topological if it can be defined purely using topological data, for example, if it is a cohomology class. -While we say that a cobordism invariant is geometric if it can be defined purely using geometric data like metric, connection, and curvature. -These two definitions have no confliction, a cobordism invariant can be both topological and geometric. -My question: Determine whether the cobordism invariants mentioned above are topological and geometric. In general, are there any examples of cobordism invariants which are geometric but not topological? Are there any examples of cobordism invariants which are topological but not geometric? -For example, the eta invariant $\eta$ is discussed in this question. -Thank you! - -REPLY [8 votes]: $\newcommand{\ko}{\mathit{ko}} -\newcommand{\MTSpin}{\mathit{MTSpin}} -\newcommand{\Z}{\mathbb Z}$ -As Mike points out in his comment, it's not obvious what it means for a bordism invariant is -“topological” or “geometric.” The bordism invariants you mention can be described -topologically, and some bordism invariants which you might think of as topological also admit geometric -descriptions. -Pontrjagin numbers are oriented bordism invariants which look topological: take a cohomology class on your manifold -and evaluate it on the fundamental class. For example, there's an injective map $\Omega_4^{\mathrm{SO}}\to\mathbb -Z$ sending a closed, oriented $4$-manifold $X$ to $\langle p_1(X), [X]\rangle$; here $p_1$ is the first Pontrjagin -class of $X$. -However, there is an equivalent, “geometric” definition of this invariant: choose a connection on the -vector bundle $TX\to X$, and let $F$ be its curvature. Then one can make sense of $\mathrm{tr}(F\wedge -F)\in\Omega^4(X)$, and Chern-Weil theory proves that -$$ - -\frac{1}{8\pi^2}\int_X \mathrm{tr}(F\wedge F) = \langle p_1(X), [X]\rangle. -$$ -Certainly a connection is geometric data, so this invariant is both “topological” and -“geometric.” - -The pin$^\pm$ bordism invariants you mention admit geometric interpretations (via $\eta$-invariants), but also -topological ones, though the topology is harder to see. First, let's reframe the above bordism invariant in terms -of Thom spectra: the characteristic class $p_1\in H^4(B\mathrm{SO})$ defines via the Thom isomorphism a cohomology -class in $\tilde H^4(M\mathrm{SO})$, hence a map $M\mathrm{SO}\to\Sigma^4 H\mathbb Z$, and upon taking $\pi_4$, we obtain -the map $\Omega_4^{\mathrm{SO}}\to\mathbb Z$ from above. -The point of this reformulation is that by replacing $H\mathbb Z$ (i.e. ordinary cohomology) with other cohomology -theories, we can describe the invariants you've mentioned above. - -As a warm-up, take the Arf invariant of a spin surface, which defines an isomorphism $\Omega_2^{\mathrm{Spin}}\to\mathbb - Z/2$. There are several different ways to define it, but here's one in line with our above description of $p_1$: -we have the Atiyah-Bott-Shapiro map $\mathit{ABS}\colon\MTSpin\to\ko$,1 and upon taking $\pi_2$, this -yields a map $\Omega_2^{\mathrm{Spin}}\to \pi_2\ko\cong\mathbb Z/2$. One can unwind this definition through the -Pontrjagin-Thom isomorphism and obtain a description of the Arf invariant through integration of $\ko$-cohomology -classes. -Next, the Arf-Brown-Kervaire invariant. There is a splitting $\mathit{MTPin}^-\simeq\MTSpin\wedge - \Sigma^{-1}\mathit{MO}_1$, where $\mathit{MO}_1$ is the Thom spectrum of the tautological bundle $\sigma\to - B\mathrm O_1$. Therefore, smashing the Atiyah-Bott-Shapiro map with $\Sigma^{-1}\mathit{MO}_1$, we obtain a map -$$ - \mathit{MTPin}^-\simeq\MTSpin\wedge \Sigma^{-1}\mathit{MO}_1\longrightarrow \ko\wedge\Sigma^{-1}\mathit{MO}_1. - $$ -Taking $\pi_2$, we get a map -$$\Omega_2^{\mathrm{Pin}^-}\to \pi_2(\ko\wedge \Sigma^{-1}\mathit{MO}_1)\cong - \widetilde{\ko}_3(\mathit{MO}_1)\cong\Z/8,$$ -and this is the Arf-Brown-Kervaire invariant.2 This can also be described in terms of a pushforward in -twisted $\ko$-theory. -The same approach works $\Omega_4^{\mathrm{Pin}^+}\to\Z/16$, using this time the splitting -$\mathit{MTPin}^+\simeq\MTSpin\wedge\Sigma\mathit{MTO}_1$; here $\mathit{MTO}_1$ is the Thom spectrum of the -virtual bundle $-\sigma\to B\mathrm O_1$. Smashing the Atiyah-Bott-Shapiro map with $\Sigma\mathit{MTO}_1$ and -taking $\pi_4$ yields a map $\Omega_4^{\mathrm{Pin}^+}\to \widetilde{\ko}_3(\mathit{MTO}_1)\cong\Z/16$. -The same approach works for $\Omega_2^{\mathrm{Pin}^-}(B\Z/2)\to\Z/4$: -$$\Omega_2^{\mathrm{Pin}^-}(B\Z/2)\cong\pi_2(\MTSpin\wedge \Sigma^{-1}\mathit{MO}_1\wedge (B\Z/2)_+),$$ -which maps to $$\pi_2(\ko\wedge \Sigma^{-1}\mathit{MO}_1\wedge (B\Z/2)_+) = \widetilde{\ko}_3(\mathit{MO}_1\wedge - (B\Z/2)_+)\cong\Z/4.$$ - -If you're asking which bordism invariants come from cohomology classes, the answer is that characteristic classes -for $n$-dimensional $G$-bordism live in $H^n(BG;A)$, where $A$ is a coefficient group; a $G$-structure on a -manifold $M$ pulls back this class to $M$, and then we evaluate it on the fundamental class. In general, this won't -capture everything: for example, if two $G$-structures have the same underlying orientation, their values on any -cohomological bordism invariant will agree. Hence, for example, the Arf invariant isn't cohomological, as there are -different spin structures on a torus which induce the same orientation, but have different Arf invariants. This is -why topological descriptions of such bordism invariants use generalized cohomology. - -1: Here, given a $G$-structure $X\to B\mathrm O$, $\mathit{MTG}$ means the Thom spectrum whose homotopy groups are -the bordism groups of manifolds with a $G$-structure on the tangent bundle, rather than on the stable normal -bundle. This distinction is irrelevant for spin bordism, but important for pin$^\pm$ bordism. -2: The Arf-Brown-Kervaire invariant depends on a choice of an 8th root of unity; depending on this choice, we might -need to compose with an automorphism of $\Z/8$ to obtain “the” Arf-Brown-Kervaire invariant. The same -caveat applies to the $\Z/16$ and $\Z/4$ invariants.<|endoftext|> -TITLE: Classification of absolute 2-limits? -QUESTION [8 upvotes]: Let $\mathcal V$ be a good enriching category. Recall that an enriched limit weight $\phi: D \to \mathcal V$ is called absolute if $\phi$-weighted limits are preserved by any $\mathcal V$-enriched functor whatsoever. If this is a familiar concept, please jump to my question at the end. But otherwise, I will provide some context. For instance, - -Idempotent splitting is absolute for $Set$-enrichment and (hence) for any enriching category; -Finite products are absolute for enrichment in abelian groups, commutative monoids, etc; -Limits of Cauchy sequences are absolute for enrichment in $([0,\infty],\geq,+)$; -$\infty$-categorically, finite limits are absolute for enrichment in spectra; -Eilenberg-Moore objects for idempotent monads are absolute for enrichment in $Cat$. - -There are a number of general things to say here: - -(Street) A $\mathcal V$-weight $\phi: D \to \mathcal V$ is absolute if and only if $\phi$ has a left adjoint $\bar \phi: D^{op} \to \mathcal V$ in the bicategory of $\mathcal V$-categories and $\mathcal V$-profunctors. -Following Lawvere and the third example above, a $\mathcal V$-category with all absolute $\mathcal V$-limits is called Cauchy complete. The Cauchy completion of a $\mathcal V$-category $C$ is obtained by freely adjoining absolute $\mathcal V$-limits to $C$, and may be constructed as the $\mathcal V$-category of absolute weights on $C$. This construction exhibits the Cauchy complete $\mathcal V$-categories as a reflective subcategory of $\mathcal V\text{-}Cat$. -Thus a $\mathcal V$-weight $\phi: D \to \mathcal V$ is absolute if and only if it lies in the Cauchy completion of $D$. - -These general facts are useful in working out explicit characterizations of the absolute weights for particular $\mathcal V$: - -An ordinary ($Set$-enriched) category $C$ is Cauchy-complete iff $C$ has all split idempotents; a $Set$-enriched weight $\phi: D \to \mathcal V$ is absolute iff it is a retract of a representable (and in particular an ordinary conical limit is absolute iff the indexing diagram has a cofinal idempotent); -An $Ab$-enriched category $C$ is Cauchy-complete iff $C$ has split idempotents and finite products; -A generalized metric space is Cauchy-complete iff it is Cauchy-complete in the usual sense; -A spectrally-enriched $\infty$-category is Cauchy-complete iff it is stable with split idempotents; - -But I don't know an analogous statement for $Cat$-enrichment. - -Question: When does a 2-category $C$ have all absolute weighted 2-limits? - -Clearly $C$ must have split idempotents and Eilenberg-Moore objects for idempotent monads -- do these suffice? Or are there more absolute 2-limit weights out there which can't be built from these? - -REPLY [4 votes]: A 2-category is Cauchy complete (in the sense you describe) if and only if idempotents splits, which is, if and only if its underlying ordinary category is Cauchy complete. -This holds more generally in this context: the base of enrichement $\mathcal V=(\mathcal V_0,\otimes,I)$ is symmetryc monoidal closed and locally presentable, and the functor $\mathcal V_0(I,-):\mathcal V_0\to\mathbf{Set}$ is (weakly) cocontinuous and (weakly) strong monoidal. (weakly here means that the induced comparison maps need not be bijections, but just surjections.) In this case a $\mathcal V$-category is Cauchy complete if and only if it has splittings of idempotents. This is proven for example in Corollary 3.16 here. -The idea is that, given an absolute $\mathcal V$-weight $\phi\colon\mathcal D\to\mathcal V$, then the category of elements $\mathcal E$ of $\mathcal V_0(I,\phi_0-)\colon \mathcal D_0\to\mathbf{Set}$ is absolute (in the ordinary sense, i.e. $\mathcal E$-colimits are preserved by any functor) and that $\phi$ can be written as a conical colimit -$$ \phi\cong \textit{colim}(\mathcal{E}_{\mathcal V}\to\mathcal D^{op}\to[\mathcal D,\mathcal V] )$$ -indexed on $\mathcal E$. -Examples of such bases are of course $\mathbf{Cat}$ and $\mathbf{Set}$, but also $\mathbf{SSet}$, $\mathbf{Pos}$, and $\mathcal V\textit{-}\mathbf{Cat}$ whenever $\mathcal V$ is locally presentable.<|endoftext|> -TITLE: Has vol. 3A of Cullis's "Matrices and Determinoids" been scanned and vol. 3B been archived? -QUESTION [7 upvotes]: This is a borderline question, but I'm going to risk posing it. -Cuthbert Edmund Cullis (1875?-1955?) was a somewhat obscure British mathematician whose opus magnum was a multi-volume treatise called Matrices and Determinoids. While his notation is somewhat idiosyncratic, it was a fairly systematic exposition of most(?) of the linear algebra known by the 1920s (heavy on determinants and minors, but also including invariant factors and canonical forms, or at least things looking like them), along with lots of new results. Some of his innovations, I believe, would be of interest even to us modern people: in particular, in the first 200 pages of Volume 3A, he seems to algorithmically construct multi-resultants using multisymmetric polynomials (aka MacMahon symmetric polynomials). I have never seen this done anywhere else. -Three volumes of his treatise were published: volume 1, volume 2 and volume 3A. According to his necrologue, he left behind a (plan of?) volume 3B. Thus, I'm wondering: - -Question 1. Has Volume 3A ever been digitized? -Question 2. Is the manuscript of Volume 3B available anywhere? - -The reason for Question 2 is mostly curiosity, but I have a rather practical reason for Question 1: I'm interested in understanding multi-resultants elementarily and also in applying multisymmetric polynomials. The twist is that I have access to Volume 3A in hardcopy for the rest of this week, but before I leave this library I'd like to know if it's worth scanning the doorstopper (700 pages!) or someone has already done that work. - -REPLY [6 votes]: Q1: Volume 3 part 1 (a.k.a. volume 3A) has been digitized and reissued as a paperback by Cambridge UP, see Amazon. The digital version is online in the HathiTrust digital library, but with limited "search only" access because of copyright restrictions. -With some further search I actually found a library in India that has placed the digital version online (a 54 MB download). Quite possibly not copyright compliant, but in the spirit of the National Emergency Library I guess in these difficult times a download can be justified.<|endoftext|> -TITLE: A non nuclear $C^*$ algebra $A$ for which the algebraic tensor product $A\otimes A$ admits a unique $C^*$ norm -QUESTION [9 upvotes]: Is there a non nuclear $C^*$ algebra $A$ for which the minimum and maximum $C^*$ norms on $A\otimes A$ coincide? -A somewhat similar question is discussed here. - -REPLY [12 votes]: Pisier https://arxiv.org/abs/1908.02705 very recently constructed a non-nuclear $C^\ast$-algebra $A$ with the weak expectation property (WEP) and the local lifting property (LLP). By a celebrated result of Kirchberg (see Corollary 13.2.5 in Brown and Ozawa's book) it follows that if $B$ has WEP and $C$ has LLP, then $B\otimes_{\max{}} C = B\otimes_{\min{}} C$. Hence $A\otimes_{\max{}} A = A\otimes_{\min{}} A$.<|endoftext|> -TITLE: Ascending sequences of idempotents in inverse semigroups -QUESTION [5 upvotes]: I've enocuntered the following question in my current research, and I'd appreciate any help you could give me. This is probably well known to experts on the subject. -Let $S = \langle K \rangle$ be a finitely generated inverse semigroup. Recall that the set $E$ of idempotents (i.e. elements $e \in S$ such that $e^2 = e$) is partially ordered via $e \leq f$ when $ef = fe = e$ (idempotents always commute in inverse semigroups). -Question: Can $S$ have an infinite ascending sequence of idempotents? That is, are there $e_n \in E$ such that $e_1 < e_2 < \dots < e_n < \dots$? -By $e < f$ I mean that $e \leq f$ and $e \neq f$. The only instance of the latter behavior I've come accross is the semigroup $S = (\mathbb{N}, \min)$ (and its' relatives), where $n \cdot m := \min\{n, m\}$. In this case we have that $S$ is equal to its' semilattice of idempotents, and $1 < 2 < \dots$, but this semigroup is not finitely generated. - -REPLY [5 votes]: Yes. Indeed, for $X$ a set, let $G_X$ be the group of partial bijections of $X$, that are defined and identity outside a countable subset. I claim that, for $X$ uncountable, every countable subset of $G$ is contained in a (5-generator) finitely generated submonoid (and hence in a finitely generated inverse submonoid). -The claim being granted, and using that the power set of $\omega$ contains a chain isomorphic to $(\mathbf{Q},\le)$, one obtains such a chain of idempotents in a suitable inverse monoid. -Note: the same claim was proved by Sierpinski and Banach in the 1930's for the monoid of all self-maps of every set, and by Galvin (1995) for the group of all permutations of every set. -Now let me prove the claim, inspired by Galvin's proof. Let $(f_n)_{n\in\mathbf{Z}}$ be a sequence in $G_X$. So there exists an infinite countable subset $X_{0,0}$ such that for every $n$, each $f_n$ is defined and identity outside $X_{0,0}$. Choose for all other $(m,n)\in\mathbf{Z}^2$ an infinite countable susbet $X_{m,n}$, pairwise disjoint. Henceforth, all maps are assumed to be defined and identity outside $X'=\bigcup_{m,n}X_{m,n}$. Also fix a bijection $X_{0,0}\to X_{m,n}$ for all $(m,n)\neq (0,0)$, so that we identify $X'$ to $X_{0,0}\times\mathbf{Z}^2$. -Define - -$u$ as the permutation $(x,m,n)\mapsto (x,m+1,n)$; -$r$ as the permutation $(x,0,n)\mapsto (x,0,n+1)$, $(x,m,n)\mapsto (x,m,n)$ for $m\neq 0$; -$f$ as the permutation $(x,m,n)\mapsto (f_m(x),m,n)$ for $n\ge 0$ and $(x,m,n)\mapsto (x,m,n)$ for $n\ge 0$. - -I claim that for every $m$ we have $f_m\in\langle u,u^{-1},r,r^{-1},f\rangle$, where $\langle\cdots\rangle$ means the submonoid generated (actually, it follows that $f_m\in\langle u,r,f\rangle_{\text{inverse-monoid}}$). -Indeed, write $g_m=u^mfu^{-m}$: then $g_m$ is like $f$, but shifted $m$ times to the right. Then one sees that $g_m(r^{-1}g_mr)^{-1}=f_m$, and the claim is proved. -[Note 1: observe that $f_m$ is written as a word of length $\le 2+2(2m+1)=4m+6$ with respect to the given generators: since this only depends on $m$, this shows that $G_X$ is "strongly distorted" (as monoid, and as inverse monoid) and in particular strongly bounded, a.k.a. Bergman's property.] -[Note 2: Probably it's also true for $X$ countable, with some further preliminary lemmas. Also with only two generators.] -[Note 3: from Vagner-Preston, every countable inverse monoid embeds into $G_{\aleph_1}$. As corollary, every countable inverse monoid embeds into a 3-generated one. This is probably well-known?]<|endoftext|> -TITLE: Abelian category equivalent to a non-abelian category -QUESTION [6 upvotes]: I was told that if we have an equivalence of categories $F : \mathcal{A} \rightarrow \mathcal{B}$ with $\mathcal{A}$ abelian, then it is not necessarily true that $\mathcal{B}$ is also abelian. -I would like to know if there are nice examples of an abelian category $\mathcal{A}$ which is equivalent to a non-abelian category $\mathcal{B}$. -Furthermore, are there any conditions over $F$ or $\mathcal{B}$ so that we have "$F : \mathcal{A} \rightarrow \mathcal{B}$ is an equivalence and $\mathcal{A}$ is abelian implies $\mathcal{B}$ is abelian"? - -REPLY [34 votes]: Here is a manifestly invariant definition of an abelian category $\mathcal{C}$. It is a category with finite limits and colimits such that: - -(It is pointed) the map from the initial to the final object is an isomorphism; we denote by 0 any object which is both initial and final. -(It is semiadditive) the map $X \amalg Y \to X\times Y$, given on $X$ by components $(\mathrm{id}_X, X \to 0 \to Y)$ and on $Y$ by components $(Y \to 0\to X, \mathrm{id}_Y)$, is an equivalence. We denote by $X\oplus Y$ the coproduct or product, identified as above. This equivalence produces an abelian monoid structure on all hom sets, where addition arises from $X \to X\oplus X \stackrel{f\times g}{\to} Y\oplus Y \to Y$. -(It is additive) the shearing map $X\oplus X \to X\oplus X$, given by adding the identity map to the projection onto the first component followed by inclusion, is an equivalence. Equivalently, each hom-monoid has the property that it is group-like. -(first isomorphism theorem) if $f: A \to B$ is arbitrary, then the map $A/\mathrm{ker}(f) \to \mathrm{ker}(B \to B/A)$ is an isomorphism. - -Being an abelian category is a property not structure. - -REPLY [26 votes]: What you were told is wrong, for we have the following: -Proposition. If two categories are equivalent and one of them is abelian, then so is the other. -A proof (and some related results) can be found in Satz 16.2.4 in H. Schubert, Kategorien II, Springer, 1970 (likewise in the English version https://www.amazon.com/Categories-Horst-Schubert/dp/3642653669, under the same numbering).<|endoftext|> -TITLE: Invariant subspaces and isotypic decomposition (reference request) -QUESTION [5 upvotes]: If $G$ is a (say) compact group and $V=\bigoplus_{i\in I}V_i$ the isotypic (a.k.a. primary) decomposition of a $G$-module, then - -any $G$-invariant subspace $W\subset V$ writes $W=\bigoplus_{i\in I}(W\cap V_i)$. - -While this isn’t hard to prove (similar to Hoffman-Kunze 1971, §7.5 for a single operator), it seems silly to redo it in a paper. Unfortunately, the only reference I’m familiar with omits the proof (Kirillov 1976, §8.3), and when using it (e.g. Lie groups VIII.3.1) Bourbaki refers to such an abstrusely worded version (Algebra VII.2.2) that unpacking it takes as much work as a direct proof. -Q: What is a good reference to quote for this? Bonus points if the case of non-algebraically closed fields is spelled out. - -REPLY [2 votes]: As suggested by the OP, I am turning my comment into an answer: a better Bourbaki reference is Algèbre VIII (new edition), §4, Proposition 4 d) (unfortunately not yet translated, as far as I know). It works for semi-simple modules over an arbitrary ring.<|endoftext|> -TITLE: Logconcavity of height of Dyck paths -QUESTION [12 upvotes]: A finite sequence $a_i$ is called logconvace in case $a_i^2 \geq a_{i-1} a_{i+1}$. - -Question : For a fixed $n$, is the sequence $a_{n,k}$ giving the number of Dyck paths of semilength $n$ having height $k$ logconcave? (see http://oeis.org/A080936) - -REPLY [2 votes]: A stronger property than log-concavity, is real-rootedness of $\sum_k t^k a_{n,k}$. -However, for $n=4$, this polynomial is $1 + 7 t + 5 t^2 + t^3$ which is not real-rooted.<|endoftext|> -TITLE: On a mysterious reference of Grothendieck -QUESTION [6 upvotes]: These days I found a mysterious page on Google books describing a book entitled On the De Rham cohomology of schemes by Grothendieck, Coates, and Jussila. -At once I thought this was an error and Google books had miss-indexed Exposé IX of Dix Exposés sur la Cohomologie de Schémas. However, the references diverge both in their titles: - -Crystals and the De Rham Cohomology of Schemes -vs. -On the De Rham cohomology of schemes, - -as well as their numbers of pages (53 in Dix Exposés vs. 106 in this reference). -The book is even cited in Cisinski 's Habilitation thesis as - -[Gro66] A. Grothendieck « On the de Rham cohomology of schemes » Publ. Math. IHES 29 (1966), p. 93--103. - -Maybe this is "On the de Rham cohomology of algebraic varieties", but again, the references diverge in page number and title. -So, does this book really exists, and, if so, is it available anywhere? - -REPLY [10 votes]: As an addendum to Donu's answer, I quote from the Table des matieres of Dix Exposés: - -IX GROTHENDIECK (Alexander), Crystals and the De Rham Cohomology of schemes (notes by I. Coates and O. Jussila), IHES Decembre 1966, 54 p. - -So it is certainly Dix Exposés. And "I. Coates" is most likely John Coates. According to Wikipedia Coates was born in '45 (so he would've been 21 at the time), and he moved to Paris to study at the ENS after obtaining a Bachelor's degree from ANU.<|endoftext|> -TITLE: Is a symplectic camel actually prohibited from passing through the eye of a needle? -QUESTION [39 upvotes]: Gromov's symplectic nonsqueezing theorem asserts that in the symplectic space ${\bf R}^{2n}$ with canonical coordinates $p_1,\dots,p_n,q_1,\dots,q_n$, and two radii $0 < r < R$, it is not possible to symplectomorphically map the ball $B(0,R)$ into the cylinder $\{ p_1^2+q_1^2 \leq r^2 \}$. It is often known as the "symplectic camel theorem", with several authors interpreting this theorem as an assertion that a symplectic camel (in the sense of a ball $B(0,R)$ that can only evolve by a continuous one-parameter family of symplectomorphisms cannot pass "through" the eye of a needle of radius $r$ less than $R$. -In the past I have accepted this interpretation uncritically, but on closer inspection it seems that the standard formulation of the nonsqueezing theorem does not actually imply a statement of this form. To make precise what an "eye of a needle" is, one needs a codimension one obstacle to serve as the "needle". I will somewhat arbitrarily choose the hyperplane $\{q_n=0\}$, minus the above-mentioned cylinder, as the needle, in which case one has a precise mathematical question. - -Question. Let $n \geq 2$ and $0 < r < R$, and let $N \subset {\bf R}^{2n}$ denote the "needle" $N := \{ q_n = 0; p_1^2+q_1^2 > r^2 \}$. Does there exist a family $S(t): B(0,R) \to {\bf R}^{2n} \backslash N$, $t \in [0,1]$ of symplectomorphisms varying continuously in $t$ (say in the uniform topology), such that $S(0)$ takes values in the left half-space $\{ q_n<0\}$ and $S(1)$ takes values in the right half-space $\{q_n>0\}$? - -Certainly a counterexample to Gromov's non-squeezing theorem (using a symplectomorphism that is connected to the identity) would allow one to construct a positive answer to this question, by first moving the ball far away from the needle, transforming it into a subset of the cylinder, sliding that subset through the needle and then far on the other side, then undoing the transformation. However it is not clear to me that the non-squeezing theorem in its standard formulation prevents some more exotic way to slide this "camel" through the "needle" (for instance, if it is possible to symplectically deform the ball into an "L-shape" object that resembles the union of two half-cylinders, which could then be maneuvred through the needle). - -REPLY [28 votes]: Eliashberg & Gromov sketched a proof in their paper "Convex symplectic manifolds" (Section 3.4). Written in the 4-dimensional case it says: -For $r>0$ define the subspace $X(r)\subset\mathbb{R}^4$ to be the union of the half-space $\lbrace q_2<0\rbrace$ and the half-space $\lbrace q_2>0\rbrace$ and the 3-ball $\lbrace(p_1,p_2,q_1,0)\;|\;p_1^2+p_2^2+q_1^2r$ there is no 1-parameter family of symplectic embeddings $S_t:(B(0,R),\omega_\text{std})\to (X(r),\omega_\text{std})$ with the image of $S_{t\le0}$ in $\lbrace q_2<0\rbrace$ and the image of $S_{t\ge1}$ in $\lbrace q_2>0\rbrace$. -McDuff & Traynor ("The 4-dimensional symplectic camel and related results") go on to show that $X(r_1)$ and $X(r_2)$ are symplectomorphic if and only if $r_1=r_2$. Oh they also give Eliashberg--Gromov's proof of the camel theorem (Theorem 5.2), reducing it to the monotonicity lemma as in the usual Gromov nonsqueezing theorem.<|endoftext|> -TITLE: The last step in Ahlfors' and Sario's proof of the triangulability of surfaces -QUESTION [7 upvotes]: In their book, Riemann Surfaces, Ahlfors and Sario write, at the bottom of pg. 109 to the top of pg. 110, - -"Consider the sequences $\{V_n\}$ and $\{W_n\}$ introduced by Lemma 46B. We will show that there exist closed Jordan regions $J_n$, - such that $V_n \subset J_n \subset W_n$, whose boundaries $\gamma_n$ have only a finite number of common points. - They will then form a covering of finite character." - -No proof is given for the last sentence. How does one prove it? -The precise content of Lemma 46B and the definition of "covering of finite character" will be given below, -for convenience here. The essence of the question is then extracted. -Establishing the property of the cited last sentence constitutes the last step in the proof of triangulability of surfaces. -The triangulation of a surface is constructed directly from $\{J_n\}$, based on the property in the cited last sentence. See the Remarks below, for further comments on the importance of the property in the cited last sentence. -Regarding the property (in the second cited sentence), "whose boundaries $\gamma_n$ have only a finite number of common points", -the authors mean that for all $m$, $n$: $\gamma_n \cap \gamma_m$ consists of at most a finite number of points. -Here are the given properties of the open Jordan regions, $V_n$, $W_n$, and the desired and established properties of the closed Jordan regions, $J_n$, -all in the connected surface (2-dim second countable manifold), $F$; the enumeration of the properties follows that of the authors. -(An open Jordan region in $F$ is a subset of $F$ whose closure is homeomorphic to a closed disk in the Euclidean plane, in such a manner that the open region corresponds to the open disk. A closed Jordan region is the closure of an open Jordan region.) -Lemma 46B: There exist sequences $V_n$, $W_n$ ($n = 1, 2, ...$) of open Jordan regions in $F$ satisfying: - -(B1) $\overline{V}_n \subset W_n$. -(B2) $\bigcup_n V_n = F$. -(B3) No point of $F$ belongs to infinitely many $\overline{W}_n$. - -Definition of 'covering of finite character': Closed Jordan regions $J_n$ ($n = 1, 2, ...$) in $F$, form a covering of finite character if - -(A0) $\bigcup_n $Int$(J_n) = F$, where Int$(J_n)$ denotes the interior of $J_n$. -(A1) Each $J_n$ meets at most a finite number of others. -(A2) For all $m$, $n$: $\gamma_n \cap \gamma_m$ consists of at most a finite number of points or arcs (possibly both), where $\gamma_n := \partial J_n$. - -The constructed regions ${J_n}$: The closed Jordan regions $J_n$, $n = 1, 2, ...$, in $F$ are constructed recursively in a manner such that for all $n$, -$V_n \subset J_n \subset W_n$ and their boundaries $\gamma_n := \partial J_n$ satisfy: -For all $n$, $\gamma_n$ meets $\gamma_{n-1} \cap \cdots \cap \gamma_1$ in at most a finite number of points. -The essential question: Does (A1) follow from (A0), (B3), and the $\{\gamma_n\}$'s property, -for all $m$, $n$: $\gamma_n \cap \gamma_m$ consists of at most a finite number of points? -Remarks: If the sequences $\{V_n\}$ and $\{W_n\}$ are of finite length, then the whole business is trivial; -so wlog the sequences are of infinite length. The result (A0) follows immediately from (B2) since Int$(J_n) \supset V_n$, for all $n$. -The property (A2) is satisfied by construction. Thus only the verification of (A1) remains. From the purpose of triangulation, it would suffice to show that a subsequence of $\{J_n\}$ satisfies (A1). Thus if $F$ is compact, then we are done. The strength of the approach to triangulation that is adopted by the authors, is that it also handles noncompact surfaces (as well as surfaces with or without boundary). -Once $\{J_n\}$ have been constructed, the sequence $\{V_n\}$ -has no further role to play, since (A0) can play the role of (B2). The same might be said of the sequence $\{W_n\}$ since it follows from (B3) -that no point of $F$ belongs to infinitely many $J_n$. This leads to the following modified question. -Modified question: Does (A1) follow from (A0) if, in addition, no point of $F$ belongs to infinitely many $J_n$? - -REPLY [2 votes]: I received the following observation by email; it constitutes an answer to my question. -The last paragraph of the proof of Lemma 46B shows that the following result, which is stronger than (B3), holds: Each $\overline{W}_n$ meets at most a finite number of the other $\overline{W}_m$'s. Therefore, in particular, the $J_n$'s satisfy (A1) because, by construction, each $J_n \subset W_n$. -Whether or not the weaker condition, (B3), is sufficient for (A1) to hold (as in ``The Essential Question''), remains an open question; but an answer to the latter question is not required for the proof of triangulability of surfaces.<|endoftext|> -TITLE: Equal products of consecutive integers -QUESTION [7 upvotes]: Summary: What are the non-trivial solutions to the question: Find two sequences of consecutive integers whose products are the same. There are four known solutions, all of thich consist of small integers, and there is no clear pattern connecting them. Given the small number of known solutions and the lack of any clear pattern among them, I suspect this question contains more number theory than at first appears. -The "Puzzle Corner" of MIT News for March/April 2020 gives a "speed" problem by Sorab Vatcha: "Find seven consecutive integers whose product equals the product of four consecutive integers." The obvious "speed" solution is: 0, 1, 2, 3, 4, 5, and 6; and 0, 1, 2, and 3. -This leads to the general question of whether there are any "nontrivial" solutions to find two distinct sequences of integers whose products are the same. The obvious trivial solutions involve (1) sequences containing 0, (2) replacing all of the integers in a sequence with their negatives, (3) adding or deleting values 1 or -1, and (4) one sequence has length 1 (and hence contains one integer that is the product of the integers in the other sequence). These criteria reduce the problem to finding two distinct sequences of integers $\ge 2$ of length $\ge 2$ whose products are the same. -There is a less obvious criterion of non-triviality: (5) The two sequences must not overlap. If the sequences overlap, then the overlapping part can be deleted from both of them, yielding a soltion with shorter sequences. This interacts with the prohibition (4) against sequences of length 1: Removing the overlapped part may reduce one sequence to length 1, and the shorter solution may be trivial also. And indeed, there is a large family of solutions constructed this way: If the product of $a \cdots b$ is $P$, then $a \cdots (P-1) = (b+1) \cdots P$. -There are non-trivial solutions. The one with the smallest product is $5 \cdot 6 \cdot 7 = 14 \cdot 15 = 210$. Is there an enumeration of all solutions? -The known non-trivial solutions are: -$5 \cdots 7 = 14 \cdot 15 = 210$, -$2 \cdots 6 = 8 \cdots 10 = 720$, -$19 \cdots 22 = 55 \cdots 57 = 175560$, and -$8 \cdots 14 = 63 \cdots 66 = 17297280$. -I have run a number of computer searches that have not revealed any further solutions: (a) all sequences with products less than $10^{30}$, (b) sequences with numbers less than 10,000,000 and length less than 10, (c) sequences with numbers less than 1,000,000 and length less than 100, (d) sequences with numbers less than 100,000 and length less than 1,000, and (e) sequences with numbers less than 10,000 and length less than 10,000. -See also https://math.stackexchange.com/questions/991728/equal-products-of-consecutive-integers/ and https://math.stackexchange.com/questions/3346618/non-trivial-solutions-to-equal-products-of-consecutive-integers. - -REPLY [3 votes]: The observation is made in many places (see the linked questions and OEIS sequence) that the sequence with larger numbers cannot contain a prime. It should also be noted that sufficiently large solutions cannot contain both three times a prime and two times a different prime. This can be extended using information on maximal gaps to limit how many large primes can appear as factors in a large sequence. An interesting variant is when a product of n consecutive positive integers divides another sequence of n consecutive positive integers. Except for n factorial dividing every such, or a sequence dividing itself, I am not aware of examples beyond the ones listed above and trivial modifications. Listing such pairs of n tuples would go far toward solving this problem. -Gerhard "Unless Erdos Already Did It" Paseman, 2020.03.26.<|endoftext|> -TITLE: Restriction of product of automorphic forms -QUESTION [5 upvotes]: Let $W \subset V$ be quadratic spaces over a number field $F$. -Let $G_n=SO(V)$ and $G_m=SO(W)$ and we consider $G_m$ as a subgoup of $G_n$ via a diagonal embedding. -Let $f$ be an automorphic form of $G_n$ and $g$ an automorphic form of $G_m$. -I am wondering whether the function $h$ on $G_m$ defined by $h(k)=f(k)g(k)$ is automorphic form on $G_m$. -Except for $\mathfrak{z}$-finiteness, I verified that other properties of autumorphic forms does hold. But I am doubtful $\mathfrak{z}$-finiteness hold. -Do you have any idea on this? -Thank you very much! - -REPLY [4 votes]: This is a special case of restricting automorphic forms on a larger group $G$ to a smaller (sub)group $H$, of course. So far as I know, orthogonal groups are not special in this regard, although, yes, there are some obvious natural maps among them. -Certainly if the map $H\to G$ is a $k$-morphism (with groundfield $k$, or, being more careful, over some localization of the ring of integers of $k$), then the restrictions of left $G_k$-invariant functions are left $H_k$-invariant. This is the immediate part. -Under very mild hypotheses and/or normalizations, "right $K$-finiteness" is preserved, as is "moderate growth". -But $\mathfrak z$-finiteness (where $\mathfrak z$ is the center of the universal enveloping algebra) is very rarely preserved. Likewise, and in parallel, if we require automorphic forms to generate irreducibles under right translation, this property will very rarely be preserved under restriction. -The rarity of this is already visible on $\mathbb R$ with the Laplacian: very rarely is the product of two $\Delta$-eigenfunctions $\Delta$-finite... -In the automorphic context, indeed, computing decomposition coefficients of such a restriction (or product) occasionally produces very interesting Euler products. Rankin-Selberg and Langlands-Shahidi et al are instances of this. -EDIT: still, yes, it is true in some rather special situations (see work of Kudla and Rallis on "first occurrence", for example, and "Howe conjectures"), that for suitable $H_1\times H_2\subset G$, for rather degenerate (e.g., "minimal") (automorphic and other) repns on $G$, restriction to $H_1\times H_2$ and projection to certain $\pi_1$-components on $H_1$ produces an irreducible on $H_2$. (Also see Segal-Shale-Weil...) -This is not quite what is happening in the literal "map to Fourier-Jacobi coefficients" story. There we have a two-step-nilpotent abelian radical, and the map is "integrate along the center of that unipotent radical". This is a $G$-hom, so preserves $\mathfrak z_G$ eigenvalues. But it does not immediately promise eigenfunction properties for the Levi component. Yes, holomorphy is preserved, for example. Is this more in the direction of what you're wanting?<|endoftext|> -TITLE: Mapping class group of torus, why is $(ST)^3=S^2$? -QUESTION [6 upvotes]: In the context of topological quantum field theories, I am interested in the mapping class group of a torus. Here I can consider the torus as a square with identified edges and also decorated with directed curves (see below). -The mapping class group is $\mathsf{SL}(2,\Bbb Z)$ and is generated by $S$ and $T$, which correspond to a 90deg rotation and a Dehn twist respectively. I take the Dehn twist to be along the vertical (meridian). These generators should satisfy $S^2=C$, and $(ST)^3=S^2$, where $C$ is conjugation corresponding to flipping both directions of the torus. The first is easy to see - -However, I cannot get the second property, here is my working - -Note in this second case I am not mapping back to the original square using the periodicity to make it clear what transformations I am doing. -$(ST)^3=\Bbb 1$ is true for the modular group $\mathsf{PSL}(2,\Bbb Z)$ but then we should also have $S^2=\Bbb 1$. -Can anyone point out where I am going wrong? - -REPLY [12 votes]: Flip the direction of rotation for $S$, or choose the other meridian for $T$. -We can see this at the level of matrices. Define -$$S_1 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \qquad S_2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$ to be the two choices of our rotation matrix. Similarly define -$$T_1 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}, \qquad T_2 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$ to be the two choices for our Dehn twist. (I suppose there really are four choices - the off-diagonal entries could just as well be negative, depending on the orientation of the meridian you're Dehn twisting around - but I'll ignore that. It's not too hard to deal with that possibility.) -Then you can do the algebra to check that $S_1^2 = S_2^2 = -\mathrm{Id} = C$, and that $(S_iT_j)^3 = -\mathrm{Id}$ if $i = j$ and $+\mathrm{Id}$ otherwise. This tells you that you need to swap one of $S$ or $T$ as described above.<|endoftext|> -TITLE: Free groups are CT-groups -QUESTION [6 upvotes]: A group $G$ is called CT-group if being commutative elements is transitive relation on $G\setminus\{1\}$ i.e. if $ 1 \neq x,y,z\in G $ and $[x,y]=1, [y,z]=1 $ then $[x,z]=1$. -I encountered the fact that free groups are CT-group, but every proof to this is very long and work for much general case (I know its true also for hyperbolic torsion free groups, which is nice but overkill for me). -Any one knows a self contained proof for this? The papers I saw uses arguments from algebraic topology, which are way over my knowledge, I only need the case of free groups. - -REPLY [8 votes]: For a completely self-contained proof see Theorem 1.8.28 in my book https://math.vanderbilt.edu/sapirmv/book/b2.pdf. The idea is that you first prove that if $uv=vu$ in the free semigroup then $u,v$ are powers of some other word $w$ which is trivially proved by induction (Theorem 1.2.9). And then prove a similar fact for free group by looking at possible cancelation in $uv$ or $vu$. Case 1 is when $u$ is not cyclically reduced and case 2 is when $u$ is cyclically reduced.<|endoftext|> -TITLE: About the cohomology of $BG^\delta$. Making a Lie group discrete -QUESTION [9 upvotes]: Let $G$ be a connected Lie group. Recall that the topological group $G^\delta$ is $G$ endowed with the discrete topology. The inclusion $G^\delta \to G$ induces a map between the classifying spaces $\eta: BG^\delta\to BG$. -Question 1 -Let $\eta^*:H^*(BG,\mathbb{Z})\to H^*(BG^\delta,\mathbb{Z})$ be the induced map in integral cohomology. -By Corollary 1 in Milnor, On the homology of Lie groups made discrete, -we get that $\eta^*$ is injective. -On the other hand, by Lemma 10 in the same paper, we learn that the kernel of -$\eta_{\mathbb{Q}}^*:H^*(BG,\mathbb{Q})\to H^*(BG^\delta,\mathbb{Q})$ (notice the rational coefficients here) -is equal to the kernel of $\eta^*_{\mathbb{Q}}:H^*(BG,\mathbb{Q})\to H^*(B\Gamma,\mathbb{Q})$, where $\Gamma -TITLE: Smoothness of the generic fiber implies smoothness over a dense open of the target? -QUESTION [9 upvotes]: I can prove the following result without too much trouble: -Let $f: X \to S$ be a proper flat morphism of finite presentation with $S$ irreducible having generic point $\xi \in S$. Then the following are equivalent, -(1) the generic fiber $X_\xi \to \mathrm{Spec}{\kappa(\xi)}$ is smooth -(2) there exists a smooth fiber $X_s \to \mathrm{Spec}(\kappa(s))$ for some $s \in S$ -(3) there exists a dense open set $U \subset S$ such that $f : X_U \to U$ is smooth. -A possible reference is [EGA-IV-4-12.2.4] then use properness to conclude that the image of the nonsmooth locus is closed. -However, I am interested in the case that $f$ is not proper. It is easy to show that (2) does not imply (3) without properness. E.g. take -$$ X = \mathrm{Spec}{k[x, y, z]/(y(xz - 1))} \to \mathrm{Spec}{k[z]} = \mathbb{A}^1_k $$ -which has only one smooth fiber (over $z = 0$). However, is it true that (1) still implies (3) without properness? Explicitly, is the following true: -Let $f: X \to S$ be a flat morphism of finite presentation with $S$ irreducible having generic point $\xi \in S$. Then if $X_\xi \to \mathrm{Spec}(\kappa(\xi))$ is smooth then there is a dense open set $U \subset S$ such that $f : X_U \to U$ is smooth? -If this is true can somebody point me to a reference? If it is false can somebody provide a counterexample? -Many thanks. - -REPLY [8 votes]: As Will Sawin said in his comment, this is true using Chevalley's theorem. A sketch goes as follows. -Suppose that $X_\xi \to \mathrm{Spec}(\kappa(\xi))$ is smooth. Let $N \subset X$ be the nonsmooth locus. For a morphism of finite presentation, the smooth locus is retrocompact open so $N$ is constructible. Thus by Chevalley's theorem $f(N)$ is constructible so $S \setminus f(N)$ is constructible. However, $\xi \in S \setminus f(N)$ because the generic fiber is smooth so $S \setminus f(N)$ contains a dense open $U \subset S \setminus f(N)$. Then $X_U \to U$ is smooth because $f$ is flat + finite presentation and each fiber above $s \in U$ is smooth since $U$ does not intersect $f(N)$ meaning the fiber over $s \in U$ does not intersect $N$.<|endoftext|> -TITLE: Hopping geodesics -QUESTION [17 upvotes]: Is there a complete metric space $X$ with the following property? - -For any pair of points $p,q\in X$ there is unique minimizing geodesic $[pq]_X$ that connects $p$ to $q$, but the map $(p,q)\mapsto [pq]_X$ is not continous. - -Comments - -For sure such space cannot be compact (or proper). -If we fix one end $p$, then there is a classical example --- the wheel; it is a unit circle with continuum of spikes from to the center $p$ to each point on the circle. -An example of noncomplete space with this property can be constructed by starting with a long circle and applying the following construction countably many times: Given a geodesic space $X$ construct a space $X'$ where to each pair of points $p,q\in X$ such that $|p-q|_X>1$ we add a unit segment with ends in $p$ and $q$. - -REPLY [7 votes]: Yes, such examples do live. The following answer was given in "Metric spaces of non-positive curvature" by Bridson and Haefliger; thanks to GGT and Moishe Kohan.<|endoftext|> -TITLE: Simplicial simple homotopy vs. cellular simple homotopy -QUESTION [10 upvotes]: I recently started reading up on simple homotopy theory. Here is a question I stumbled upon. -In his 1938 Paper Simplicial Spaces, Nuclei and m-Groups Whitehead introduced the notion of elementary expansions and elementary collapses of simplicial complexes. Essentially a simplicial complex $K'$ is obtained from $K$ through an elementary collapse, by removing two simplices $p$ and $q$ from $K$ such that: - -$p$ is a maximal simplex of $K$. -$q < p$ is a (maximal, proper) free face of $p$ (i.e. not contained in any other simplex but itself and $p$). - -An elementary expansion is the obvious inverse operation. There are topological realizations of these operations, with the collapse given by "pushing the free face onto the others". Lets call a map of simplicial complexes a simplicial simple homotopy equivalence, if it is homotopic to the composition of such maps. -Later on he decided that CW-complexes were a more convenient setting to work in (Simple Homotopy Types), and this was where he developed the now famous results on whitehead torsion. This is the setting that most of the following material (such as Cohens Book A course in simple homotopy theory) are presented in and probably most familiar to most topologists. By a cellular simple homotopy equivalence, I mean a map of CW-complexes as in 3. -It seems to me, that it is somewhat folklore knowledge, that the second setting is a generalization of the prior in the following sense: - -Let $K$,$K'$ be abstract simplicial complexes and $f:|K| \to |K'|$ a map between their realizations. Then $f$ is a simplicial simple homotopy equivalence, if and only if it is a cellular simple homotopy equivalence, with respect to the obvious $CW$-structures on $|K|$ and $|K'|$. - -Is this even true? -I imagined this would be a simple consequence of the simplicial approximation theorem, but couldn't figure out an easy proof. I also skimmed most of the papers from the time period I could find on the matter, but didn't really get a satisfying answer. -If yes, I would be really thankful for a reference, or a sketch of a proof. - -REPLY [3 votes]: Turns out I should have read the original material Simplicial Spaces, Nuclei and m-groups a little more carefully. It was in there all along. I'm still supprised nobody ever explicitly stated this though. But I guess it was such common knowledge at the time that nobody bothered. -My question can be rephrased in the following way. For a simplicial complex $K$ (all complexes are taken to be finite) denote by $E_S(L)$ the set of inclusions of simplicial complexes $L \to K$ that are homotopy equivalences, modulo the equivalence relation generated by $L\to K \xrightarrow{s} K' = L\to K' \implies L \to K \sim L \to K'$ for $s$ a composition of (simplicial) elementary expansions (this actually has set size). This is the description Siebenmann chooses in Infinite Simple Homotopy Types. Further denote by $E_C(K)$ the obvious analogon in the CW setting. The latter of course is just the underlying set of the geometric description on the Whiteheadgroup $WH(X)$ as it is constructed in Cohens simple Homotopy Theory for example. As both equivalence relations identify homotopic maps and using the standard mapping cylinder arguments, my question rephrases to: -What is the Kernel of the obvious forgetful map $E_S(L) \to E_C(L) = WH(L) \cong WH( \pi_1(L))$ for $L$ connected. Here I mean kernel in the sense of pointed sets, as the lefthandside will only a posteriori be a group (one could of course bother with proving its a group first..). -The way Whitehead proved in Simple Homotopy Types that the last isomorphism is injective, is effectively by transforming each inclusion $L \to K$ to the form $ L \to K' = L \cup \bigcup e_i^n \bigcup e_i^{n+1}$ through cellular simple expansions and collapses $(n \geq 1)$. The corresponding element in $WH( \pi_1(L))$ is then given by $ \pi_{n+1}( K', K'^{n} \cup L) \to \pi_n(K'^n \cup L, L)$. Explicitly one checks using Hurewicz theorem and universal coverings that this is in fact an isomorphism of free $\mathbb Z (\pi_1(L))$ modules, with basis given by the cells $e_i$. One then shows, that all the elementary matrix operations making the corresponding matrix trivial on the $WH(\pi_1(L))$ side have an analogue on the simple homotopy side, proving injectivity. -What was known to me when I asked the question, was that this works in the cellular category, i.e. with cellular simple equivalences. It turns out, before passing to the cellular setting, Whitehead did the analogous proof in the simplicial world, which turns out to require a lot more technical arguments, but it is completely done in 1 and the statement is implicitly proven in the proof of theorem 20 there. -Very roughly, one first proves that subdivisions are simplicial simple equivalences, so that one can work in the p.l. category instead. Here one has the p.l. mapping cylinder see 1). The attachment of a p.l. cell along a p.l. boundary map, is then to be understood as taking the cylinder along the boundary and then gluing the cell on top of it. (I guess this circumvents problems with pushouts in the p.l. category). He then shows that the same simplicifactions as in the cellular setting are valid in the p.l. setting using such cylinders and simplicial approximation, thus, showing that $E_S(L) \to WH(\pi_1(K))$ and hence $E_S(L) \to E_C(L)$ has trivial kernel. -In particular this gives a positive answer to my original question.<|endoftext|> -TITLE: Random walks on infinite directed regular graphs -QUESTION [6 upvotes]: Let us consider a directed graph $\Gamma=(V,E,s,t)$ ($V$ set of vertices, $E$ set of edges, $s,t: E \rightarrow V$ are the "source" and "target" maps). -Assume that $\Gamma$ is bi-regular, that is there are two integers $d_1 \geq 1$ and $d_2 \geq 1$ such that $|s^{-1}(v)|=d_1$ and $|t^{-1}(v)|=d_2$ for every $v \in D$. Assume that $\Gamma$ weakly connected (there is an undirected path relying any two vertices). -And finally assume that $\Gamma$ has no forward-closed finite subset (i.e. if $S$ is a finite subset of $V$, there is an edge in $E$ with source in $S$ and target not in $S$). In particular $\Gamma$ is infinite. -Consider the standard discrete-time forward random walk on $\Gamma$, starting at $x \in V$: at time $0$ you are at $x$ with probability $1$; for any $n \geq 0$, at time $n+1$, conditional to being at $y$ at time $n$, your odds of being at any of the $d_1$ forward-neighbors of $y$ (i.e. $t(e)$ for $e \in E$, $s(e)=y$) is $1/d_1$. -Let $p^n_{x,y}$ be the probability of being at $y$ at time $n$. - -Is it true that for every $y$, $p_n(x,y) \rightarrow 0$ as $n \rightarrow \infty$? - - -Here are some remarks. This question is related to Fedja's beautiful answer to this question. Fedja proves the result when $\Gamma$ is an undirected regular graph (seen as an undirected graph by replacing each undirected edge by two directed edges going both way). Unfortunately, I have not been able to extend his argument to my directed case. -The hypothesis that $\Gamma$ has no forward-closed finite subset is certainly necessary: If $S$ was such a subset, and $x \in S$, then you would be sure to stay in the finite set $S$ forever, so $\sum_{y \in S} {p^n_{x,y}} = 1$ and one of these $p^n_{x,y}$ at least can not tend to $0$. -The hypothesis that $|s^{-1}(v)|=d_1$ (or at least $s^{-1}(v)$ finite) for all $v$ is necessary to define the random walk, but that $|t^{-1}(v)|=d_2$ (or at least is finite) for all $v$ is also necessary for the theorem to be true. Without it, consider the graph with $V=\mathbb Z$, and for every $a \in \mathbb Z$, there is one directed edge from $a$ to $a+1$ and one directed edge from $a$ to $0$ (called the "speedy return" edge). Thus $|s^{-1}(a)|=2$ for every $a \in \mathbb Z$, but $|t^{-1}(0)|=\infty$. The probability $p^n_{0,0}$ is $\geq 1/2$ since every path that ends up with the "speedy return edge" goes from $0$ to $0$. - -REPLY [5 votes]: Here is a counterexample: -Let $G_1$ be the digraph with vertex set $\mathbb N$, two loops at $0$, an edge from $0$ to $1$, and for every $i \geq 1$ an edge from $i$ to $(i+1)$ and two parallel edges from $i$ to $(i-1)$. Let $G_2$ be any countable digraph in which every vertex has $2$ outgoing edges and $4$ incoming edges, and let $f \colon V(G_2) \to V(G_1)$ be such that $|f^{-1}(0)| = 0$ and $|f^{-1}(i)| = 1$ for every $i \geq 1$. -Let $G$ be the digraph obtained from $G_1 \uplus G_2$ by adding all edges from $v$ to $f(v)$. Then $G$ is bi-regular with $d_1 = 3$ and $d_2 = 4$. Moreover, $G$ is weakly connected and has no finite forward closed sets since every vertex has a forward edge connecting it to the forward ray in $G_1$. -The simple random walk on $G$ almost surely enters $G_1$ after finitely many steps (and remains in $G_1$ thereafter since there are no edges from $G_1$ to $G_2$). But the simple random walk on $G_1$ is just a biased random walk on $\mathbb N$ with bias towards $0$. This random walk is irreducible, aperiodic (because of the loops at $0$), and positive recurrent. Thus $\lim_{n \to \infty} p_n(x, i) = \mu(i)$ independently of the starting point $x$, where $\mu \neq 0$ is the invariant probability measure on $\mathbb N$.<|endoftext|> -TITLE: How to tell, roughly, which PDE's are interesting to analyse? -QUESTION [7 upvotes]: How can one tell which PDE's are, roughly speaking, perhaps more interesting to analyse? -Physical motivation is one reason. For example, the KdV $u_t+u_{xxx} - 6uu_x=0$ for a function $u:\mathbb R\times\mathbb R\to\mathbb R$ is a model for many physical systems, such as the propagation of shallow water waves along water bodies with low height. It is also a completely integrable PDE, which makes its analysis more tempting. But putting aside complete integrability (many studied PDE do not have completely integrable variants) why not study the derivative quasilinear equation $u_t + u_{xxx} - 6uu_{xxx}$ or the another derivative semilinear equation $u_t + u_{xxx} - 6uu_{xx}=0$ (I am only putting the KdV here as an example, and am not really asking this particular question for the KdV)? -Aside from physical motivation, how can one pick PDE to analyse? -Edit: Just to be clear, I am asking this from the perspective of PDE research. When I say "analyse," I mean "do PDE research on." - -REPLY [6 votes]: My impression is that a PDE, or rather a class of PDEs, is interesting to the analysts when it is relevant for some analytical tool. Let me take a few examples. The list is not exhaustive. - -linear constant coefficient PDEs in the whole space ${\mathbb R}^n$ are treated with Fourier analysis. -Elliptic (scalar) PDEs are analyzed with the maximum principle. This is particularly true in the modern treatment. -Linear and semi-linear evolution PDEs are the realm of semi-group theory. -Dispersive linear PDEs obey to Strichartz inequalities. -So-called integrable equations or systems are treated by Lax pairs and spectral theory. -Othere examples in homogenization or in hyperbolic conservation laws are solved by using compensated compactness.<|endoftext|> -TITLE: What is symplectic rigidity? -QUESTION [5 upvotes]: What is an explanation for what the theory of symplectic rigidity is and what kind of questions it can answer? I was led to this after reading about the symplectic non-squeezing theorem of Gromov. - -REPLY [12 votes]: Rigidity, as used throughout mathematics and not just symplectic geometry, indicates that some structure attached to an object captures more data than one would "expect" from the underlying object itself. There are a number of helpful examples already in this Wikipedia entry. (I write "expect" in quotes because often times the "expectation" is far from correct, and might not even match your initial intuition.) -Gromov's non-squeezing theorem is typically the first example most people encounter in symplectic geometry. It states if there exists a symplectic embedding of the standard ball $B^{2n}(R)$ into $\mathbb{R}^{2n}$ such that its image lies completely in the cylinder $Z^{2n}(r) = B^2(r) \times \mathbb{R}^{2n-2}$, then $R \leq r$. However, there certainly exist volume-preserving embeddings $B^{2n}(R) \hookrightarrow Z^{2n}(r)$. Hence, this is an example of symplectic rigidity if you take the setting of volume-preserving geometry as your expectation. In other words, symplectic geometry is more rigid, or stronger, than volume-preserving geometry, since it sees more information. -Volume-preserving geometry is not the only setting from which you can build your "expectation." Depending on the situation, you may take: - -Almost symplectic geometry: $(M^{2n},\omega)$ with $\omega \in \Omega^2(M)$ such that $\omega^n \neq 0$ (we drop the closedness condition). -The underlying smooth geometry - -Obviously, if you have symplectic rigidity when compared with almost symplectic geometry, then you also have symplectic rigidity with respect to volume geometry and smooth geometry, for example. On the other hand, sometimes the problem at hand doesn't have an interpretation in volume-preserving geometry, and so you have to settle for rigidity with respect to smooth geometry. -Here's another example of rigidity: there is no closed exact Lagrangian submanifold embedded in standard $\mathbb{R}^{2n}$. However, in the smooth category, there are many $n$-dimensional closed submanifolds of $\mathbb{R}^{2n}$. So symplectic geometry is more rigid than smooth geometry. (Here's an example where it wouldn't make sense to compare this to volume-preserving or almost symplectic geometry.) -There's a whole industry of rigid invariants nowadays in symplectic geometry; the industrial revolution occurred in 1985 with Gromov's famous paper Pseudo holomorphic curves in symplectic manifolds. In fact, both examples mentioned so far first appeared in that paper. Often, these rigid invariants are tied up intimately with counts of pseudo-holomorphic curves (e.g. Gromov-Witten theory and Quantum Cohomology). Gromov non-squeezing can be seen as a zeroth order approximation to these subtle curve counts: it boils down to the existence of a particular pseudo-holomorphic curve. There's also the Fukaya Category, which is by now a big machine which can be thought of as encoding rigid invariants coming from Lagrangian submanifolds, and is itself intimately tied up with physics via mirror symmetry (this is all a long story in itself). -Finally, I should mention on the flip side that you can ask if symplectic geometry exhibits phenomena in which the weak "expectation" actually matches. This does happen, and is called flexibility. (This is particularly fruitful when combined with contact geometry, in which there are overtwisted discs and loose Legendrians.) As a purely symplectic example of flexibility, for any symplectic manifold $(M^{2n},\omega)$ and any manifold $L$ with $\dim L < n$ there is a forgetful map from the space of isotropic embeddings $\{f \colon L \hookrightarrow M \mid f^*\omega = 0\}$ to the space of formal isotropic embeddings, i.e. embeddings $f \colon L \hookrightarrow M$ and bundle monomorphisms $F \colon TL \rightarrow f^*TM$ (over $L$) with $F^*\omega = 0$. The forgetful map is just given by taking the same underlying embedding and $F = df$. This forgetful map induces a homotopy equivalence (and is hence referred to as an h-principle, where the "h" stands for "homotopy").<|endoftext|> -TITLE: Explicit fundamental domain for the action of $\operatorname{O}(n,1)(\mathbb{Z})$ on $\operatorname{O}(n,1)(\mathbb{R})$ -QUESTION [6 upvotes]: Minkowski computed explicit fundamental domains for the action of $\operatorname{SL}_n(\mathbb{Z})$ on $\operatorname{SL}_n(\mathbb{R})/\operatorname{SO}_n(\mathbb{R})$ for each $n \leq 6$. In the case where $n = 2$, one obtains the familiar fundamental domain for the action of $\operatorname{SL}_2(\mathbb{Z})$ on the complex upper half-plane. The case where $n = 3$ is studied in detail in the paper entitled "Hecke Operators and the Fundamental Domain for $\operatorname{SL}(3, \mathbb{Z})$" by Daniel Gordon et al. -Are there analogous computations in the literature of explicit fundamental domains for the action of the orthogonal group $\operatorname{O}(n,1)(\mathbb{Z})$ on $\operatorname{O}(n,1)(\mathbb{R})$, at least for some small values of $n$? I am particularly interested in the case where $n = 2$. -What I know: I understand that computing such fundamental domains is difficult in general. I'm aware of the construction of Borel and Harish-Chandra via Siegel domains, but I'm not sure whether it's possible to make their construction explicit in the way that Minkowski was able to do. - -REPLY [9 votes]: For information about these groups up through dimension 17, see: -Vinberg, È. B. -The groups of units of certain quadratic forms. (Russian) -Mat. Sb. (N.S.) 87(129) (1972), 18–36. -English translation [Math. USSR-Sb. 87 (1972), 17–35]. -Vinberg shows up through dimension 17 that $O(n,1; \mathbb{Z})$ has a finite index subgroup generated by reflections. He gives an explicit polygon for this reflection group, i.e., a fundamental domain for its action. Then $O(n,1; \mathbb{Z})$ is generated by this reflection group along with the symmetry group of the polygon. -He has a later paper with Kaplinskaja that studies 18 and 19: -Vinberg, È. B.; Kaplinskaja, I. M. -The groups O18,1(Z) and O19,1(Z). (Russian) -Dokl. Akad. Nauk SSSR 238 (1978), no. 6, 1273–1275. -English translation: Soviet Math. Dokl. 19 (1978), no. 1, 194–197. -I believe there are some references for (slightly) higher dimensions as well. You might look at some papers of Allcock, for example. -Edit: For the special case $n=2$, it is reflections in the sides of a $(2,4,\infty)$ triangle, i.e., angles $\pi/2$, $\pi/4$, and one ideal vertex. This can be seen directly from the diagram in Table 5 of Vinberg's paper.<|endoftext|> -TITLE: Contrasting theorems in classical logic and constructivism -QUESTION [13 upvotes]: Is it possible there are examples of where classical logic proves a theorem that provably is false within constructivism? Is so what are some examples? -What are some examples of most contrasting theorems provable in these two logics that does not fall in 1.? - -REPLY [19 votes]: There are several ways one could interpret the word "constructivism" here, and the answer depends on what you meant by it. -Bishop-style constructivism is a generalization of Brouwerian intuitionistim, Russian constructivism, and classical mathematics. It is mathematics done without excluded middle (of course, you can still use excluded middle on those instances that you can prove to hold using other means) and general axiom of choice, but you still have countable choice. Thus, anything you prove in this setting is true in classical mathematics as well. -There are other forms of constructivism which are Bishop-style constructivism extended with additional principles and axioms. These additional principles often contradict classical logic, and so you get consequences that are classically false. Here are some examples: -In the internal language of the effective topos (an older name for this is Russian constructivism) the following are valid statements: - -There are countably many countable subset of $\mathbb{N}$. -There is an increasing sequence in $[0,1]$ that has no accumulation point. -The Cantor space $2^\mathbb{N}$ and the Baire space $\mathbb{N}^\mathbb{N}$ are homeomorphic. -Every map $f : [0,1] \to \mathbb{R}$ is continuous. -There exists a continuous unbounded map $f : [0,1] \to \mathbb{R}$. -There is a covering of $\mathbb{R}$ by intervals $(a_n, b_n)_n$ with rational endpoints such that $\sum_{k = 1}^n |b_n - a_n| < 1$ for all $n \in \mathbb{N}$. -There is a subset $S \subseteq \mathbb{N}$ which is not finite and not infinite. -There exists an infinite binary rooted tree in which every path is finite. -The ordinals form a set, i.e., they are not a proper class. One has to be careful about how ordinals are defined and how to precisely understand the notions of "class" and "set", but these are technical details. - -In the internal language of the realizability topos $\mathsf{RT}(K_2)$ (an older name for this is Brouwerian intuitionisism) the following statements are valid: - -Every map $f : X \to Y$ between complete separable metric spaces is continuous. -Every map $f : [0,1] \to \mathbb{R}$ is uniformly continuous. -Every map $f : \mathbb{R} \to \{0,1\}$ is constant, or equivalently, if $\mathbb{R} = A \cup B$ and $A \cap B = \emptyset$ then $A = \mathbb{R}$ or $B = \mathbb{R}$. - -There are many other examples. I recommend taking the effort to get used to these amazing new worlds of mathematics.<|endoftext|> -TITLE: Is there a connection between the average 'compositeness' of a rational number and $\phi$ (golden ratio)? -QUESTION [6 upvotes]: Let $n\in N$, where $n = p_{1}^{k_{1}}p_{2}^{k_{2}}...p_{m}^{k_{m}}$ for $p_{i}$ prime. -Define the 'density' of $n$ as: -$d(n) = \frac{(p_{1}+1)^{k_{1}}(p_{2}+1)^{k_{2}}...(p_{m}+1)^{k_{m}}}{n}$ -Notice that $d(n)$ gives us a measure of the 'compositeness' of a number - relative to other numbers of a similar size. Notice also that $n_{1} \neq n_{2} \implies d(n_{1}) \neq d(n_{2})$ -Now extend the definition to the rational numbers so that for $r \in Q$, where $r=a/b$, and $a=p_{1}^{k_{1}}p_{2}^{k_{2}}...p_{m}^{k_{m}}$, $b=q_{1}^{l_{1}}q_{2}^{l_{2}}...q_{n}^{l_{n}}$ -Define the density of $r$ as: -$d(r) = \frac{(p_{1}+1)^{k_{1}}(p_{2}+1)^{k_{2}}...(p_{m}+1)^{k_{m}}}{a}.\frac{b}{(q_{1}+1)^{l_{1}}(q_{2}+1)^{l_{2}}...(q_{n}+1)^{l_{n}}}$ -Now order the rational numbers in an $n$ x $n$ grid (the same grid used to prove the countability of the rationals). Denote the average density of the first $n$ rationals in the grid by $D_{n}$ -Does $\lim_{n\to\infty} D_{n}$ exist, and if so what is it? -I have computed this for $n > 500,000$ and it turns out to be 1.61806, which is remarkably close to $\phi$. Is there a relationship between the density definition given above and the golden ratio? - -REPLY [7 votes]: Probably not. I can tell you what the limiting value is when taking averages over $m\times m$ grids themselves, rather than diagonal-counting-sequences; but I suspect the averages are the same. -The average of $d(r)$ over the $m\times m$ grid is simply -$$ -\frac1{m^2} \sum_{a=1}^m \sum_{b=1}^m d\big( \tfrac ab\big) = \bigg( \frac1m \sum_{a=1}^m d(a) \bigg) \bigg( \frac1m \sum_{b=1}^m \frac1{d(b)} \bigg). -$$ -The first sum is a sum over a totally multiplicative function $d$ with the property that $d(p) = 1+\frac1p$. General results about sums of multiplicative functions that are close to $1$ on primes tell us that -$$ -\frac1m \sum_{a=1}^m d(a) \sim \prod_p \bigg( 1 + \frac{d(p)}p + \frac{d(p^2)}{p^2} + \cdots \bigg) \bigg( 1-\frac1p \bigg) = \prod_p \bigg( 1 - \frac{d(p)}p \bigg)^{-1} \bigg( 1-\frac1p \bigg) -$$ -as $m\to\infty$ (where the products are over all primes $p$); in this case, we obtain -$$ -\frac1m \sum_{a=1}^m d(a) \sim \prod_p \bigg( 1 - \frac{1+1/p}p \bigg)^{-1} \bigg( 1-\frac1p \bigg) = \prod_p \bigg( 1 + \frac1{p^2-p-1} \bigg) \approx 2.67411. -$$ -Since $1/d$ is also totally multiplicative, the same procedure gives -\begin{align*} -\frac1m \sum_{b=1}^m \frac1{d(b)} &\sim \prod_p \bigg( 1 - \frac{1/d(p)}p \bigg)^{-1} \bigg( 1-\frac1p \bigg) \\ -&= \prod_p \bigg( 1 - \frac{1/(1+1/p)}p \bigg)^{-1} \bigg( 1-\frac1p \bigg) = \prod_p \bigg( 1 - \frac1{p^2} \bigg) = \frac6{\pi^2}. -\end{align*} -Therefore the average in equation is -$$ -= \frac6{\pi^2} \prod_p \bigg( 1 + \frac1{p^2-p-1} \bigg) \approx \frac6{\pi^2} \cdot 2.67411 \approx 1.62567. -$$<|endoftext|> -TITLE: A problem about real quadratic field -QUESTION [5 upvotes]: I have calculated some real quadratic field 's Hilbert class field with class number $2$,and I found they were satisfied $Gal(H_{K}/Q)\cong Z/2Z\oplus Z/2Z$,here $H_{K}$ is the Hilbert class field of a real quadratic field $K$ whose class number is $2$.Is it true for all quadratic field with class number $2$? How to prove it?(ps:I'm a beginner of algebraic number theory ,and I'm very interested in Hlibert class field .) - -REPLY [11 votes]: This is a consequence of "genus theory". If $K / \mathbf{Q}$ is an abelian extension, then the "genus field" of $K$ is the maximal extension $L / K$ such that $L/K$ is unramified and $L/\mathbf{Q}$ is abelian. You can read more about genus theory here: https://en.wikipedia.org/wiki/Genus_field. -In your case, the fact that $H_K$ is Galois over $\mathbf{Q}$ and $[H_K : \mathbf{Q}] = 4$ implies that $H_K$ must be abelian over $\mathbf{Q}$ (since there are no nonabelian groups of order 4); so $H_K$ coincides with the genus field $L_K$. The genus field of of a quadratic field is always a composite of quadratic fields, so $H_K$ must have Galois group $C_2 \times C_2$. -(More generally, for any quadratic field (real or imaginary), genus theory completely describes the 2-torsion part of the class group.)<|endoftext|> -TITLE: Cofinality of infinitesimals -QUESTION [6 upvotes]: Suppose $\kappa$ is an infinite cardinal and $U$ is a countably incomplete uniform ultrafilter over $\kappa$. Then $\mathbb R^\kappa/U$ is nonstandard. What is the cofinality of the set of infinitesimals of this field? What can we say when $U$ is $\kappa$-regular? -Background information: Recall that $U$ is $\kappa$-regular when there exists a sequence $\langle X_\alpha : \alpha < \kappa \rangle \subseteq U$ such that for any $\beta < \kappa$, $\{ \alpha : \beta \in X_\alpha \}$ is finite. If $U$ is $\kappa$-regular, then I can show that the cofinality of $\mathbb R^\kappa/U$ (rather than infinitesimals) is $>\kappa$. Furthermore, if $\mathbb R^\kappa/U$ is $\delta$-saturated, then the cofinality of the infinitesimals is $\geq\delta$. $\omega_1$-saturation is automatic for ultrapowers by countably incomplete ultrafilters. If the ultrafilter satisfies a property stronger than regularity called goodness, then the ultrapower is $\kappa^+$-saturated. - -REPLY [8 votes]: As pointed out in a comment by James Hanson, the cofinality of the infinitesimals is the same as the coinitiality (i.e., cofinality or the reverse order) $\mu$ of the nonstandard part of $\omega^\kappa/U$. -Even for $\kappa=\omega$, this coinitiality $\mu$ is not decided by the axioms of set theory. Furthermore, even within a single model of set theory, $\mu$ can depend on the particular ultrafilter $U$. -Specifically, if one starts with a model of CH and adds $\lambda$ Cohen reals, the resulting model has nonprincipal ultrafilters $U$ on $\omega$ for which $\mu$ is any regular uncountable cardinal $\leq\lambda$. (The same holds for the cofinality of the whole ultrapower $\omega^\omega/U$, and in fact this cofinality and $\mu$ can be chosen independently.) Similarly, if one adds $\lambda$ random reals to a model of CH, every regular uncountable cardinal $\leq\lambda$ occurs as $\mu$ for some $U$. (But now the cofinality of $\omega^\omega/U$ is $\aleph_1$ because random forcing is $\omega^\omega$-bounding.) -These results were proved by Mike Canjar in his thesis; the MathSciNet data for the published version are: -MR0924678 (89g:03073) Reviewed -Canjar, Michael -Countable ultraproducts without CH. -Ann. Pure Appl. Logic 37 (1988), no. 1, 1–79.<|endoftext|> -TITLE: Finding $Q(\sqrt{-2})$-rational points on $X_0(33)$ -QUESTION [7 upvotes]: Let $K = Q(\sqrt{-2})$. How can I compute the $K$-rational points on the modular curve $X_0(33)$? -Recall that $X_0(33)$ is of genus $3$ and has the following affine model, -$$y^2 +(-x^4-x^2-1)y = 2x^6-2x^5+11x^4-10x^3+20x^2-11x+8.$$ -My attempt at finding $K$-rational points on $X_0(33)$ is as follows: First I find a rational map $f$ from $X_0(33)$ to a quotient curve $E$ of $X_0(33)$ with $E$ an elliptic curve. Second, I determine the preimages of $E(K)$ under $f$. If $E$ is of rank $0$, $E(K)$ is finite. Then I can use a Grobner basis to determine $f^{-1}(x)$ for every $x \in E(K)$. However in my case $E(K)$ is of rank $1$ and as a result it is computationally infeasible to determine a Grobner basis for every $f^{-1}(x)$ with $x \in E(K)$. I am wondering if there is a work-around this issue? -Any help in finding $K$-rational points on $X_0(33)$ would be appreciated. -EDIT: Removed a question after a clarification by Christian Wuthrich. - -REPLY [4 votes]: P. Bruin and F. Najman have determined the exceptional quadratic points on $X_0(33)$. -See Table 8 of https://arxiv.org/pdf/1406.0655.pdf<|endoftext|> -TITLE: Estimate for $\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}\frac{b}{a(ab)_p}$, where $p$ is a large prime -QUESTION [9 upvotes]: Is this estimate true? Can anyone give a proof of it? -$$ -\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}\frac{b}{a(ab)_p}=\frac{1}{2}p\ln^2 p+o(p\ln^2 p)\qquad (p\text{ prime, } p\to\infty) -$$ -where $ -(ab)_p\equiv ab\;(\operatorname{mod}p)$, $0<(ab)_p -TITLE: Convergence in the Caratheodory sense and Hausdorff sense -QUESTION [6 upvotes]: Among Jordan domains, I understand that Caratheodory convergence is weaker than Hausdorff convergence. -But if a sequence of Jordan domains all have rectifiable boundary whose arc length are all $L$, and their Caratheodory limit is also a Jordan domain with the same boundary arc length $L$, does this necessarily imply this sequence converge in the Hausdorff metric? -To my knowledge, examples of sequences converging in the Caratheodory sense but not in the Hausdorff sense do not preserve arc length. ------Edit----- -I realize that Caratheodory convergence is a less commonly known concept, so here I am updating with two equivalent definitions. -A sequence of pointed domains $(\Omega_n,x_n)$ is said to converge to $(\Omega,x)$ in the Caratheodory sense if - -$x_n \to x$, -for all compact $K \subseteq \Omega$, we have $K \subseteq \Omega_n$ for every $n$ sufficiently large, and -for all open connected $U$ containing $x$, if $U \subseteq \Omega_n$ for infinitely many $n$, then $U \subseteq \Omega$. - -The pointed domains $(\Omega_n,x_n)$ converge to $(\Omega,x)$ in the Caratheodory sense if and only if the harmonic measures $\omega(\Omega_n,x_n)$ converge weakly to the harmonic measure $\omega(\Omega,x)$. -The definition of Hausdorff metric can be found at https://en.m.wikipedia.org/wiki/Hausdorff_distance?wprov=sfla1 - -REPLY [2 votes]: The answer is yes. -The boundaries $\partial\Omega_n$ are uniformly 1-Lipschitz on $[0;L]$ under the arc-length parametrization, hence, by Arzela—Ascoli, you can extract a uniformly convergent sub-sequence, say $\partial\Omega_{n_k}\to\gamma$, where $\gamma$ also has a 1-Lipschitz parametrisation on $[0;L]$. Also, since $\Omega_{n_k}\to\Omega$ in Carathéodory sense, we must have $\partial\Omega\subset \gamma$. So, you have a rectifiable Jordan curve $\partial \Omega$ of length $L$ and also another curve $\gamma$ of length $\leq L$ that contains $\partial \Omega$. This certainly implies $\gamma=\partial\Omega$, say, by interpreting the length as the Hausdorff measure of dimension 1. So, any sub-sequence of your domains has a further sub-sequence whose boundaries uniformly converge to $\partial\Omega$, which implies the claim.<|endoftext|> -TITLE: How to explain this property of totient? -QUESTION [7 upvotes]: I am running a program to search for solutions of $$\varphi(pm+1)=\varphi(pm+p+1).$$ -So far, for $m=1,\ldots,327$ solutions have been found (some relatively large). -(in the body of the question, $p$ is a natural number, not necessarily prime) - -I would like to conjecture that for every $m \in \mathbb N$ there exists natural number $p \in \mathbb N$ such that $\varphi(pm+1)=\varphi(pm+p+1)$ - - -Is this known to be true? -If not known to be true, is there some evidence that totient could have (or not have) this interesting property? - -Update: I stopped the program at $m=407$ since it was taking a long long time to find solution for $m=407$ (if there is any), but all $m=1,\ldots,406$ have a solution. -Update 2: Currently, the program is running the code for approx. half an hour just for $m=490$, and yes, the solutions exist for $m=1,\ldots,489$ -Update 3: It seems that it is rare that for some $k \in \mathbb N$ there do not exist some $p$´s such that we do not have $\varphi(pm+k)= \varphi(pm+p+k)$ for all $m$´s, and a research of when that fails to be true seems to be one possible avenue of research. - -REPLY [6 votes]: It might just be a matter of randomness. Seeing no reason for $\varphi(pm+1)$ and $\varphi(p(m+1)+1)$ to be especially related, we might imagine heuristically that -$\varphi(pm+1)$ and $\varphi(p(m+1)+1)$ have probability $\sim \text{constant}/(pm)$ of being equal. Since $\sum_m 1/m = \infty$, it would then be reasonable to expect there to be -infinitely many $m$ for which this is the case. Of course this is not a proof. -This also suggests that if a small $m$ is not found for a particular $p$, you might need to look very far (something like $\exp(\text{constant}/p)$) before finding an $m$ that works.<|endoftext|> -TITLE: Online events during the quarantine -QUESTION [29 upvotes]: With many places on earth subjected to quarantine and large gathering prohibited, there are announcements of online seminars and talks open to people around the world. The talks can be conducted via ZOOM or other platforms. As far as I know, the information about them is spread by the word of mouth, or more precisely personal emails. -Are there some sources/online pages where information about ongoing online events im mathematics is collected? I am interested in probability in particular, but I assume other people would be interested in events related to their research interests. -A somewhat similar question was asked here 10 years ago but the circumstances are different now, and the available technological tools are more advanced. - -REPLY [3 votes]: AMS has started a list, organized by mathematical field, at https://www.ams.org/profession/online-talks.<|endoftext|> -TITLE: Tannakian criterion for reducedness of Tannakian dual group -QUESTION [6 upvotes]: Given an affine group scheme G over a field of positive characteristic. -Question: Is there a simple criterion for G to be reduced in terms of the neutral Tannakian category of its finite dimensional algebraic representations? - -REPLY [4 votes]: Yes, and one could argue as follows, at least if the group scheme is assumed to be of finite type (so it is an algebraic group), and the base field is perfect. Recall that if $k$ is a field of characteristic zero, then any algebraic group is smooth, and that if $k$ is a perfect field of characteristic $p$, then an algebraic group is reduced if and only if it is smooth. -I'll make some statements using dg-categories, because that language simplifies some statements. Let $k$ be a perfect field, and let $A$ be a $k$-algebra. Then $A$ is smooth (in the usual sense) if and only if $A$ is a perfect object of the (dg-)category $\mathrm{BMod}_A(\mathrm{Mod}_k)$ of $A$-$A$-bimodules in $k$-vector spaces. I don't really have a reference for this, but one could view this result as a jazzed up statement of Serre's regularity criterion. (Since $k$ is perfect, $A$ is smooth if and only if it is regular.) This MathOverflow post has some discussion of such criteria: Smooth dg algebras (and perfect dg modules and compact dg modules). -This implies that if $A$ is a $k$-algebra, then $A$ is smooth if and only if $\mathrm{Mod}_A$ is a dualizable $k$-linear dg-category such that the unit $\eta:\mathrm{Mod}_k\to \mathrm{Mod}_A \otimes_k \mathrm{Mod}_A^\vee$ preserves compact objects. Indeed, the dual of $\mathrm{Mod}_A$ as a $k$-linear dg-category is just the category of modules over the opposite algebra (which is $A$ itself if $A$ is commutative), so $\mathrm{Mod}_A \otimes_k \mathrm{Mod}_A^\vee$ is the category of $A$-$A$-bimodules in $k$-vector spaces. The functor $\eta$ sends the unit $k\in \mathrm{Mod}_k$ to $A$ regarded as a bimodule over itself, so $A$ is smooth by the above discussion. -This gives the desired criterion over a perfect field $k$: an algebraic group $G$ is smooth if and only if the dg-category $\mathrm{Rep}(G)$ is a dualizable object in $k$-linear dg-categories, such that the unit $\eta:\mathrm{Mod}_k\to \mathrm{Rep}(G) \otimes_k \mathrm{Rep}(G)^\vee$ preserves compact objects. -If this is too abstract of a criterion, one could also just stop at the point where we appealed to Serre's homological criterion for regularity. This leads to the statement that an algebraic group $G$ over a perfect field $k$ is smooth if and only if $\mathrm{Rep}(G)$ is of finite global dimension, i.e., $\mathrm{Hom}_{\mathrm{Rep}(G)}(V, W[n])$ vanishes for all finite-dimensional representations $V,W\in \mathrm{Rep}(G)$ and all but finitely many $n$.<|endoftext|> -TITLE: Completeness of coefficient functionnals -QUESTION [7 upvotes]: My questions is about Schauder bases and more specifically about coefficient functionals. -Let $(x_n)$ be a Schauder basis of a Banach space $X$. Thus for all $x$ in $X$, $x = \sum f_n(x) x_n$. The $f_n$ are called coefficient functionals. -They are continuous. If $X$ is reflexive, they form a basis of $X^*$ (with Hahn–Banach theorem). -My question is : if $X$ is not reflexive, but if we suppose $X^*$ separable, is $(f_n)$ a basis of $X^*$? - -REPLY [6 votes]: M. Zippin showed that for a Banach space $X$ with a basis, if every basis of $X$ is boundedly complete or if every basis of $X$ is shrinking, then $X$ is reflexive. -The result of Zippin answers you question in the negative (or, rather, the answer is, "not necessarily".) However to complete the picture let us recall the earlier result of R.C. James (alluded to in your question) asserting the following: For a reflexive Banach space $X$ with a basis $(x_n)$, the basis $(x_n)$ is both shrinking and boundedly complete. -Putting the results of Zippin and James together yields the following: -Let $X$ be a Banach space with a basis. The following are equivalent: - -$X$ is reflexive; -every basis of $X$ is shrinking; and -every basis of $X$ is boundedly complete.<|endoftext|> -TITLE: Sliding block code on irreducible sofic shift -QUESTION [6 upvotes]: I was looking at the following exercise from Lind/Marcus book An Introduction to Symbolic Dynamics and Coding that I cannot solve. Can someone give me a hint? -Find an example of a pair of irreducible sofic shifts $X,Y$ and a sliding block code $f:X\rightarrow Y$ such that $f$ is not one-to-one, but the restriction of f to periodic points is one-to-one. -The exercise points to other exercises that I also don't know. -Let $X$ be an irreducible shift of finite type and $f:X\rightarrow Y$ a sliding block code. Prove that $f$ is an embedding if and only if it is one-to-one on the periodic points of $X$. -I know the definitions for sofic, finite type, irreducible and sliding block code; but have no idea how to use them. Why things should be different here for sofic shifts and shifts of finite type? - -REPLY [4 votes]: This is a fun pair of exercises (the first one you mention is 3.2.9 and the second is 2.3.6a)! For 2.3.6a, recode to a $1$-block code $\phi$ on an irreducible edge shift $X$, suppose that $x, x' \in X$ are such that $x \neq x'$ but $\phi(x) = \phi(x')$. Consider two cases: either $x_j \neq x_j'$ for either all $j > i$ or all $j < i$; or there exist $j < i < k$ with $x_j = x_j'$, $x_k = x_k'$. In either case, try to construct a pair of distinct periodic points whose images agree. -Then, for 3.2.9, here's an idea for a construction: try to construct a sofic shift $X$ by labelling an irreducible graph with alphabet $\{ 0, 1, 2 \}$, such that the path presenting $10^m 1$ can be determined by whether $m$ is even or odd, and such that every cycle (other than self-loops) contains a $2$. Then code from $X$ to $Y$ by replacing $2$ by $1$ and leaving the other symbols the same. -EDIT: I've provided an answer because I didn't notice I was on MO rather than Math.SE, but this question should probably be on Math.SE rather than MO. It's an exercise in an introductory textbook, albeit a textbook with some rather tricky exercises.<|endoftext|> -TITLE: Order of product of group elements -QUESTION [5 upvotes]: Let $G$ be a finite non-commutative group of order $N$, and let $x, y \in G$. Let $a$ and $b$ be the orders of $x$ and $y$, respectively. Can we say anything non-trivial about the order of $xy$ in terms of $a$ and $b$? If it helps, you may assume that $a$ and $b$ are coprime. - -REPLY [20 votes]: The following theorem (which does not take the order $N$ of the group $G$ into account) shows that all possible combinations of $a$, $b$ and the order of $xy$ are possible. See Theorem 1.64 from Milne's course notes on group theory. - -For any integers $a,b,c > 1$, there exists a finite group $G$ with elements $x$ and $y$ such that $x$ has order $a$, $y$ has order $b$, and $xy$ has order $c$. - -REPLY [5 votes]: I strongly suspect that the answer is "very little", unless you are looking for bounds on the order of $xy$ in terms of $N$ or something like that. -Example. Let's focus on the example $a = 2$ and $b = 3$. It is well-known that -\begin{align*} -\mathbf Z/2 * \mathbf Z/3 &\stackrel\sim\to \operatorname{PSL}_2(\mathbf Z)\\ -x &\mapsto \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix},\\ -y &\mapsto \begin{pmatrix}0 & 1 \\ -1 & -1\end{pmatrix}, -\end{align*} -where $x$ and $y$ are the generators with $x^2 = 1$ and $y^3 = 1$. The product $xy$ maps to -$$\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}0 & 1 \\ -1 & -1\end{pmatrix} = \begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix},$$ -which has infinite order. For any $n \in \mathbf N$, the surjection -$$\operatorname{PSL}_2(\mathbf Z) \twoheadrightarrow \operatorname{PSL}_2(\mathbf Z/n)$$ -gives a finite group of order roughly $n^3$ in which $xy$ has exact order $n$. -Example. Another example, still with $a = 2$ and $b = 3$, is the group -$$G = \mathbf Z/n \wr S_3 = (\mathbf Z/n)^3 \rtimes S_3,$$ -where multiplication is defined by -$$(a_1,a_2,a_3,\sigma)(b_1,b_2,b_3,\tau) = (a_1+b_{\sigma(1)},a_2+b_{\sigma(2)},a_3+b_{\sigma(3)},\sigma\tau).$$ -Then one easily checks that $x = (1,-1,0,(12))$ has order $2$ and $y = (0,0,0,(123))$ has order $3$, and $xy = (1,-1,0,(23))$ has order $2n$. This time we get any even number as the order of $xy$, inside a group of order $6n^3$.<|endoftext|> -TITLE: What happens if you strip everything but the “between” relation in metric spaces -QUESTION [49 upvotes]: Given a metric space $(X,d)$ and three points $x,y,z$ in $X$, say that $y$ is between $x$ and $z$ if $d(x,z) = d(x,y) + d(y,z)$, and write $[x,z]$ for the set of points between $x$ and $z$. -Obviously, we have - -$x,z\in[x,z]$; -$[x,z]=[z,x]$; -$y \in [x,z]$ implies $[x,y] \subseteq [x,z]$; -$w,y \in [x,z]$ implies: $w\in [x,y]$ iff $y \in [w,z]$. - -My question: -Has the family of objects with such an axiomatic “interval structure” $[\bullet,\bullet]:X \times X \to \mathcal{P}(X)$ satisfying approximately the above conditions been studied somewhere? - -REPLY [9 votes]: This is more of an extended remark than a full answer. -My message is: -The property $[x,x]=\{x\}$ does not necessarily hold, but you can always make it hold by taking a natural quotient. -Your axioms don't imply $[x,x]=\{x\}$, so you have to add it if you want it. -A pseudometric (e.g. a seminorm) will satisfy your axioms, and $[x,x]$ is the set of points distance zero from $x$, and this set can have any size. -In a pseudometric space you can take the quotient by the relation $x\sim y\iff d(x,y)=0$ and you get a metric space. -It turns out the same works with the interval systems. -We can define a relation $\sim$ on $X$ so that $x\sim y\iff x\in[y,y]$. -This is an equivalence relation: - -Reflexivity: -Axiom 1. -Symmetry: -By axioms 1 and 2 $x,y\in[y,x]$. -Thus axiom 4 says $x\in[y,y]\iff y\in[x,x]$. -Transitivity: -Suppose $x\sim y\sim z$. -By axiom 3 $x\sim y$ and $y\sim z$ imply $[y,x]\subset [y,y]$ and $[z,y]\subset [z,z]$. -By axioms 2 and 3 $[y,y]\subset[z,y]$. -Thus $x\in[y,x]\subset[y,y]\subset[z,y]\subset[z,z]$. - -Then we can take the quotient space $X/{\sim}$. -To avoid confusion, let's denote by $(x)$ the quotient set of $x\in X$. -(In fact, $(x)=[x,x]$, but I find it cleaner to keep the separate notation.) -Then declaring $(x)\in[(y),(z)]\iff x\in[y,z]$ defines an interval structure on $X/{\sim}$. -To see this, we need to check two things: - -$[x,y]\subset[x',y]$ when $x\sim x'$. -(The reverse inclusion and varying both $x$ and $y$ follow easily.) -Proof: Since $x\in[x',x']\subset[x',y]$, we have indeed $[x,y]\subset[x',y]$. -$x\in[z,y]\implies[x,x]\subset[z,y]$. -Proof: Since $x\in[z,y]$ and $x\in[x,y]$, applying axiom 3 twice gives $[x,x]\subset[x,y]\subset[z,y]$. - -In the quotient structure we evidently have $[(x),(x)]=\{(x)\}$.<|endoftext|> -TITLE: Newman's proof of the prime number theorem -QUESTION [10 upvotes]: I am teaching a graduate course in Complex Analysis and I am covering Newman's proof of the prime number theorem. I have been using the simplified version in the papers of -Zagier and Korevaar. However, I ran into a problem. -Both papers rely on this theorem: -Theorem -Let $f:[0,\infty)\rightarrow\mathbb{C}$ be bounded and -locally integrable and let -$$ -g(z):=\int_{0}^{\infty}f(t)e^{-tz}dt,\quad\operatorname{Re}z>0. -$$ -Assume that for every $z\in\mathbb{C}$ with $\operatorname{Re}z=0$ there -exists $r_{z}>0$ such that $g$ can be extended holomorphically to $B(z,r_{z} -)$. Then the generalized Riemann integral -\begin{equation} -\int_{0}^{\infty}f(t)\,dt \label{pn1} -\end{equation} -is well-defined and equals $g(0)$. -This theorem is used to prove that the generalized Riemann integral -$$ -\int_{1}^{\infty}\frac{\theta(x)-x}{x^{2}}dx -$$ -converges. Here, $$ -\theta(x):=\sum_{p\text{ prime}\leq x}\log p,\quad x\in\mathbb{R}. -$$ -Everything is fine up to this point. Then the authors use the convergence of this integral to prove that -\begin{equation} -\lim_{x\rightarrow\infty}\frac{\theta(x)}{x}=1. \label{pn limit theta}% -\end{equation} -Their proof is as follows: -Assume by contradiction that -$$ -\limsup_{x\rightarrow\infty}\frac{\theta(x)}{x}>1. -$$ -There there exists an increasing sequence $x_{n}\rightarrow\infty$ such that -$\theta(x_{n})>(1+\varepsilon)x_{n}$ for all $n\in\mathbb{N}$ and for some -$0<\varepsilon<1$. Since $\theta$ is increasing, if $x>x_{n}$, $\theta -(x)\geq\theta(x_{n})>(1+\varepsilon)x_{n}$, and so -\begin{align*} -\int_{x_{n}}^{(1+\varepsilon)x_{n}}\frac{\theta(x)-x}{x^{2}}dx & \geq -\int_{x_{n}}^{(1+\varepsilon)x_{n}}\frac{(1+\varepsilon)x_{n}-x}{x^{2}}dx\\ -& =\int_{1}^{(1+\varepsilon)}\frac{(1+\varepsilon)-s}{s^{2}}ds>0 -\end{align*} -where we made the change of variables $x=x_{n}s$ so $dx=x_{n}ds$. Since -$x_{n}\rightarrow\infty$, by selecting a subsequence we can assume that -$x_{n+1}\geq2x_{n}$ for all $n$. Hence, by summing all the disjoint integrals -on the left-hand side we obtain that -$$ -\int_{\bigcup(x_{n},(1+\varepsilon)x_{n},)}\frac{\theta(x)-x}{x^{2}}% -dx=\sum_{n=1}^{\infty}\int_{x_{n}}^{(1+\varepsilon)x_{n}}\frac{\theta -(x)-x}{x^{2}}dx\\=\sum_{n=1}^{\infty}\int_{1}^{(1+\varepsilon)}\frac -{(1+\varepsilon)-s}{s^{2}}ds=\infty. -$$ -The papers claim that this fact contradicts the fact that the integral converges and proves that -$$ -\limsup_{x\rightarrow\infty}\frac{\theta(x)}{x}\leq1. -$$ -However, this is not the case since all we know is that -$$ -\lim_{T\rightarrow\infty}\int_{1}^{T}\frac{\theta(x)-x}{x^{2}}dx=\ell -\in\mathbb{R}% -$$ -but this does not prevent that -$$ -\int_{1}^{\infty}\frac{(\theta(x)-x)^{+}}{x^{2}}dx=\int_{1}^{\infty}% -\frac{(\theta(x)-x)^{-}}{x^{2}}dx=\infty. -$$ -The typical example is -$$ -\int_{1}^{\infty}\frac{\sin x}{x}dx, -$$ -which exist as an improper Riemann integral but not as Lebesgue integral. -Am I missing something? If not, is there a correct proof? - -REPLY [12 votes]: It doesn't seem to me that either article follows the line of reasoning as you have presented it. Indeed, we do not take the integral over the union of integrals $(x_n,(1+\varepsilon)x_n)$. We do get that the integral over those intervals is infinite, but you correctly note this does not give a contradiction. Instead, the argument goes as follows. -Let me denote -$$F(T)=\int_{1}^{T}\frac{\theta(x)-x}{x^{2}}dx,\quad C=\int_{1}^{(1+\varepsilon)}\frac{(1+\varepsilon)-s}{s^{2}}ds$$ -so that $F(T)$ converges and $C$ is a positive constant. Since $F(T)$ converges, it is in particular Cauchy, so for large enough $x,y$ we have $|F(x)-F(y)| -TITLE: De Rham and Koszul complexes -QUESTION [11 upvotes]: Consider the algebraic de Rham complex of the $n$-dimensional plane: this is merely -$$\ldots\rightarrow Sym(V^*)\otimes\bigwedge^{k}V^*\rightarrow Sym(V^*)\otimes\bigwedge^{k+1}V^*\rightarrow\ldots -$$ -with the standard de Rham differential $fdx_I\rightarrow\sum_{i=1}^n\frac{df}{dx_i}dx_i\wedge dx_I$. -On the other hand, there is a Koszul resolution of the ideal $(x_1,\ldots,x_n)\subset Sym(V^*)$ is defined on a chain complex which is termwise the same as above, but with differential going in the opposite direction: e.g. it sends $dx_{i_1}\wedge\ldots\wedge dx_{i_k}$ to $x_{i_1}dx_{i_2}\wedge\ldots\wedge dx_{i_k}-x_{i_2}dx_{i_1}\wedge dx_{i_3}\wedge\ldots\wedge dx_{i_n}+\ldots+(-1)^{k-1}x_{i_k}dx_{i_1}\wedge\ldots dx_{i_{k-1}}$. -If we call the de Rham differential $d$, and the Koszul differential $e$, then $de+ed$ is scaling by the total degree (exterior and symmetric parts) of the form. -Having some familiarity with different types of cohomology groups that one encounters in algebraic geometry, it seems unusual to me that there would exist two different differentials on the same "complex", and that they would be related in such a way. Thus, I am curious whether this is an instance of some more general phenomenon, for example whether this occurs for certain complexes of sheaves on other spaces. Though I am not an expert in this language, it seems one way to phrase what happens in this example is that there is a single graded algebra with two different DGA structures which are related (by this scaling formula). Are there any/many other examples of this happening? - -REPLY [11 votes]: What you have here is a graded version of an $\mathfrak{sl}_2$-triple. Compare for example the proof of the Lefschetz decomposition and hard Lefschetz theorem in Hodge theory, which is essentially the observation that $L$ and $\Lambda$ form the $e$ and $f$ of an $\mathfrak{sl}_2$-triple, because -$$[L,\Lambda] = (k-n)\operatorname{id}$$ -on degree $k$ forms (so let $h$ act by $k-n$ in degree $k$). -I will first give a definition of a 'graded $\mathfrak{sl}_2$-triple', and then give a conceptual explanation for the example you gave. -Definition. Let $k$ be a field (algebraically closed of characteristic $0$, for simplicity), and let $C$ be the complex -$$\ldots \to 0 \to k \stackrel 0\to k \to 0 \to \ldots$$ -with $k$ in (cohomological) degrees $0$ and $1$. Let $\mathfrak{gl}(C)$ be the graded Lie algebra defined by the graded algebra $\operatorname{Hom}^*(C,C)$ (all endomorphisms of $C = C^0 \oplus C^1$, with its natural grading), and let $\mathfrak{sl}(C)$ be the graded Lie subalgebra obtained by replacing $\operatorname{Hom}^0(C,C)$ by its trace $0$ elements, recalling that -$$\operatorname{tr}\big(f \colon C \to C\big) = \sum_i (-1)^i\operatorname{tr}\big(f^i \colon C^i \to C^i\big).$$ -Explicitly, $\mathfrak{sl}(C)$ has a basis $\{e,f,h\}$ where $h \colon C \to C$ is the identity (which indeed has trace $0$), $e \colon C \to C[1]$ is the map -$$\begin{array}{ccccccccc}\ldots & \to &0 & \to & k & \to & k & \to & \ldots \\ & & \downarrow & & || & & \downarrow & & \\ \ldots & \to & k & \to & k & \to & 0 & \to & \ldots,\! \end{array}$$ -and $f \colon C \to C[-1]$ is the map -$$\begin{array}{ccccccccc}\ldots & \to &k & \to & k & \to & 0 & \to & \ldots \\ & & \downarrow & & || & & \downarrow & & \\ \ldots & \to & 0 & \to & k & \to & k & \to & \ldots.\! \end{array}$$ -With the graded Lie bracket, we get -\begin{align*} -[e,f] &= ef + fe = h,\\ -[h,e] &= 0,\\ -[h,f] &= 0. -\end{align*} -Viewing the complex $C$ with zero differentials as a graded object $C^0 \oplus C^1$, the map $e$ can be represented by the map $(a,b) \mapsto (b,0)$ and $f$ by $(a,b) \mapsto (0,a)$. -Definition. A graded $\mathfrak{sl}_2$-triple on a graded vector space $V$ is a triple $(e,f,h)$ of elements in $\operatorname{Hom}^*(V,V)$ of degrees $1$, $-1$, and $0$ respectively satisfying the identities -\begin{align*} -[e,f] &= ef + fe = h,\\ -[h,e] &= 0,\\ -[h,f] &= 0. -\end{align*} -Example. Let $V$ be a finite dimensional vector space, and let $V \otimes C$ be the two term complex $V \oplus V[-1]$ with zero differential. Then -$$\operatorname{Sym}(V \otimes C) = \operatorname{Sym}(V) \otimes \operatorname{Sym}(V[-1]) = \bigoplus_i \operatorname{Sym}(V) \otimes \left(\bigwedge\nolimits^i V\right)[-i],$$ -since the sign in the graded swap $K \otimes L \stackrel\sim\to L \otimes K$ replaces the symmetriser on $K^{\otimes n}$ by the antisymmetriser in odd degree. This is the complex you're studying, except it has differentials all $0$ for now. -But $\operatorname{Hom}^*(C,C)$ acts on $V \otimes C$, hence so does $\mathfrak{sl}_2(C)$. Then the latter also acts¹ on $T^*(V \otimes C)$ and its quotient $\operatorname{Sym}(V \otimes C)$. - -Lemma. This graded $\mathfrak{sl}_2$-triple is the one described by the OP, with $e$ the Koszul differential, $f$ the de Rham differential, and $h$ multiplication by the total degree. - -Proof. Writing -$$T^n(V \otimes C) = \big(V \oplus V[-1]\big)^{\otimes n}$$ -with elements $(v_1,w_1) \otimes \ldots \otimes (v_n,w_n)$, the action of $\mathfrak{sl}_2(C)$ is given by -\begin{align*} -e \colon T^n(V \otimes C) \to& T^n(V \otimes C)[1]\\ -(v_1,w_1) \otimes \ldots \otimes (v_n,w_n) \mapsto & \sum_{i=1}^n (v_1,-w_1) \otimes \ldots \otimes (v_{i-1},-w_{i-1}) \\ & \otimes (w_i,0) \otimes (v_{i+1},w_{i+1}) \otimes \ldots \otimes (v_n,w_n),\\\\ -f \colon T^n(V \otimes C) \to& T^n(V \otimes C)[-1]\\ -(v_1,w_1) \otimes \ldots \otimes (v_n,w_n) \mapsto & \sum_{i=1}^n (v_1,-w_1) \otimes \ldots \otimes (v_{i-1},-w_{i-1}) \\ & \otimes (0,v_i) \otimes (v_{i+1},w_{i+1}) \otimes \ldots \otimes (v_n,w_n), -\end{align*} -and $h$ is multiplication by $n$. In the above notation, the quotient map -$$T^n(V \otimes C) \to \operatorname{Sym}^n(V \otimes C) = \bigoplus_{i+j=n} \operatorname{Sym}^i V \otimes \left(\bigwedge\nolimits^j V\right)[-j]$$ -is given by -$$(v_1,w_1) \otimes \ldots \otimes (v_n,w_n) \mapsto \sum_{I \amalg J = \{1,\ldots,n\}} v_I \otimes w_J,$$ -where $v_I = \prod_{i \in I} v_i$ and $w_J = w_{j_1} \wedge \ldots \wedge w_{j_s}$ if $\{j_1 < \ldots < j_s\} = J$. Then $f$ descends to the de Rham differential $d$, and $e$ descends to the Koszul differential $e$. Indeed, consider an element -$$(v_1,0) \otimes \ldots \otimes (v_r,0) \otimes (0,w_1) \otimes \ldots \otimes (0,w_s)$$ -lifting $v_I \otimes w_J$ where $I = \{1,\ldots,r\}$. Then $f$ maps this to -$$\sum_{i=1}^r (v_1,0) \otimes \ldots \otimes (v_{i-1},0) \otimes (0,v_i) \otimes (v_{i+1},0) \otimes \ldots \otimes (v_r,0) \otimes (0,w_1) \otimes \ldots \otimes (0,w_s),$$ -which under the quotient $T^n(V \otimes C) \to \operatorname{Sym}^n(V \otimes C)$ maps to -$$\sum_{i=1}^r \left(\prod_{j \neq i} v_j\right) \otimes \big(v_i \wedge w_J\big),$$ -which is the de Rham differential of $v_I \otimes w_J$. On the other hand $e$ takes it to -$$\sum_{j=1}^s (v_1,0) \otimes \ldots \otimes (v_r,0) \otimes (0,-w_1) \otimes \ldots (0,-w_{j-1}) \otimes (w_j,0) \otimes (0,w_{j+1}) \otimes \ldots \otimes (0,w_n),$$ -which under the quotient $T^n(V \otimes C) \to \operatorname{Sym}^n(V \otimes C)$ maps to -$$\sum_{j=1}^s (-1)^{j-1} v_Iw_j \otimes w_{J \setminus\{j\}},$$ -which is the Koszul differential of $v_I \otimes w_J$. Finally, just like on $T^n(V \otimes C)$, on $\operatorname{Sym}^n(V \otimes C)$ the map $h$ is just multiplication by $n$, the total degree of $v_I \otimes w_J$. $\square$ - -¹ The map $\operatorname{Hom}^*(C,C) \to \operatorname{Hom}^*(C \otimes C, C \otimes C)$ by $f \mapsto f \otimes f$ is not linear (nor does it preserve the grading). So we do not get a natural action of $\operatorname{Hom}^*(C,C)$ on $\operatorname{Sym}(V \otimes C)$. However for a graded Lie algebra $L$ acting on complexes $C$ and $D$, there is an action on $C \otimes D$ by -$$\rho_{C \otimes D}(x)(c \otimes d) = \Big(\rho_C(x) \otimes 1 + (-1)^{\deg(x)\deg(c)} \otimes \rho_D(x)\Big)(c \otimes d)$$ -for $x \in L$, where $\rho_C(x)$ and $\rho_D(x)$ are the actions of $x$ on $C$ and $D$ respectively. Similarly one gets actions on $\operatorname{Sym}(C)$, etcetera.<|endoftext|> -TITLE: Some interesting experimental results about the distribution of primes -QUESTION [7 upvotes]: Let's consider the following metric of the gap between consecutive primes -$$m(k)=\frac {p_k^2-p_{k-1}^2} {24}\;\;\;\;\;(k\ge4)$$ -Now, let's define the function -$\delta(k)=m(k)\;\;\;\;$ if $\,m(k)\,$ is prime $\;\;\;\;(1)$ -$\delta(k)=0\;\;\;\;\;\;\;\;\;\;$ otherwise $\,\;\;\;\;\;\;\;\;\;\;\;\;\;(2)$ -If we plot the graph of $\,m(k)\,$ we obtain (up to $\,k=10^4$): - -But, if we plot the graph of $\,\delta(k)$, we obtain something completely different (up to $\,k=10^4$): - - -In the second graph the three different increasing curves seem to have - the following asymptotic behaviors: -$$\sim \frac {p_k} {2}$$ -$$\sim \frac {p_k} {3}$$ -$$\sim \frac {p_k} {6}$$ -Further, the prime numbers are not uniformly distributed among the three curves. The number of primes decreases passing from the highest - to the lowest curve (I'm trying to estimate the number of primes that fall on each curve). -I do not have adequate knowledge of analytic number theory to - explain the phenomenon, therefore I ask for your help. - -Many thanks. - -REPLY [14 votes]: Assume that $k\geq 5$ and $m(k)$ is prime. Observe the factorization -$$\frac{p_k-p_{k-1}}{2}\cdot\frac{p_k+p_{k-1}}{2}=6m(k).$$ -On the left hand side, the second fraction is greater than $6$, hence the first fraction is smaller than $m(k)$. So the first factor is relatively prime to $m(k)$, whence it divides $6$. So $p_k-p_{k-1}$ equals $2$ or $4$ or $6$ or $12$. - -If $p_k-p_{k-1}=2$, then $m(k)=(p_k+p_{k-1})/12\sim p_k/6$. -If $p_k-p_{k-1}=4$, then $m(k)=(p_k+p_{k-1})/6\sim p_k/3$. -If $p_k-p_{k-1}=6$, then $m(k)=(p_k+p_{k-1})/4\sim p_k/2$. -If $p_k-p_{k-1}=12$, then $m(k)=(p_k+p_{k-1})/2$, which is a contradiction, because there is no prime between $p_k$ and $p_{k-1}$. - -This explains the "three increasing curves" $p_k/6$, $p_k/3$, $p_k/2$. The density of these three cases (i.e. the validity of two linear equations in the three primes $p_k$, $p_{k-1}$, $m(k)$) is heuristically known, see e.g. Conjecture 1.2 in Green-Tao: Linear equations in primes.<|endoftext|> -TITLE: Localization of symmetric monoidal categories and geometry -QUESTION [7 upvotes]: I have a series of vague questions, related to localization of symmetric monoidal categories. -Here is the context. Say we are working over a field of characteristic zero. Then the "one category level higher" version of (DG) commutative ring is a (DG) symmetric monoidal category. It is well-known that for $X$ a scheme (or even, IIUC, a Noetherian stack with affine diagonal) $X$ can be recovered from the DG symmetric monoidal category of its quasicoherent sheaves. (Moreover, the functor $\text{Schemes}\to\text{SymMonCat}$ is fully faithful, including in an $\infty$-categorical context. -Now if $R$ is a commutative ring, we say that $S$ is a localization of $R$ if it can be obtained from $R$ by inverting some set of elements. If $R, S$ are both Noetherian, then there is a very nice alternative way of characterizing localizations: - -(*) A map $R\to S$ is a localization if and only if the product map $S\otimes_R S\to S$ (derived tensor product) is an equivalence. - -Now there are (at least) three interesting notions of localization for a DG symmetric monoidal category $\mathcal{C}$ (note all make sense also for just monoidal categories, and that when taking universal objects in the category of categories I'm going to be vague about what I'm requiring from categories: I'm willing to assume compactly generated, idempotent complete, etc.). - - Localization along a morphism $f:X\to Y.$ - Localization along an object $X$ (defined as the universal symmetric monoidal DG category admitting a functor from $\mathcal{C}$ where $X$ is $\otimes$-invertible, maybe satisfying some additional conditions). - "Parallel" localization along a morphism: if $f:X\to Y$ is a morphism, I'm defining this to be the initial category in which $X,Y$ are invertible and there exists a map $f':X^{-1}\to Y^{-1}$ with $f\otimes f' = \text{id}:\mathbb{I}\to \mathbb{I},$ where the equality is understood via an appropriate system of coherences. - -(Of course 2. is a special case of 3.) -Here are some questions. - - Defining $\otimes$ in terms of colimit in the category of symmetric monoidal categories, is there a context where (*) holds for symmetric monoidal categories (i.e., localizations can be characterized by a tensor-idempotence condition)? - Are there "interesting" localizations of this type of the category $\mathcal{C}$ of DG quasicoherent coherent sheaves on a scheme $X$ which do not come from geometric localizations? If no, are there any examples in more general contexts that have been studied or computed in some sense? (For example, what happens if you $\otimes$-localize the category of vector spaces along a two-dimensional vector space?) - By functoriality and universality, localizations of any of the types 1., 2., 3. can be "combined" (in a commutative way), and (by formal nonsense), the result of applying two localizations $\mathcal{C}\to \mathcal{C}_1$ and $\mathcal{C}\to \mathcal{C}_2$ is the colimit $\mathcal{C}_1\otimes_{\mathcal{C}}\mathcal{C}_2$. - -**Definition**. For $\mathcal{C}_1\leftarrow\mathcal{C}\to \mathcal{C}_2$ a pair of localizations of $\mathcal{C}$ as above, set $$\mathcal{C}_{12}: = \mathcal{C}_1\otimes_{\mathcal{C}}\mathcal{C}_2$$ for the "combined" localization. Say that $\mathcal{C}_1$ and $\mathcal{C}_2$ *cover* $\mathcal{C}$ if $\mathcal{C}$ is the limit of the diagram $\mathcal{C}_1\to \mathcal{C}_{12}\leftarrow \mathcal{C}_2.$ My question now is: are there interesting examples of covers of DG symmetric monoidal categories in this sense other than Zariski covers in geometry? - -REPLY [4 votes]: Given a symmetric monoidal functor $\mathcal{C} \rightarrow \mathcal{C}'$, the property that $\mathcal{C}' \otimes_{\mathcal{C}} \mathcal{C}' \rightarrow \mathcal{C}'$ be an isomorphism is equivalent to $\mathcal{C} \rightarrow \mathcal{C}'$ being an epimorphism in the category of symmetric monoidal dg categories. This is a purely formal fact: in any cocartesian symmetric monoidal category $\mathcal{E}$ a map out of the initial object $0 \rightarrow Z$ is an epimorphism if and only if the map $Z \coprod Z \rightarrow Z$ is an isomorphism. Our case follows from this by taking $\mathcal{E}$ to be the category of symmetric monoidal categories over $\mathcal{C}$. -The three notions of localization mentioned in the question are epimorphisms, since a map out of the localization is by definition a map out of $\mathcal{C}$ satisfying a property (namely, that a certain arrow becomes invertible, that a certain object becomes invertible, etc). Hence all three notions satisfy the tensor idempotence condition. -Inverting an arrow $f: X \rightarrow Y$ amounts to passing to the quotient by the ideal generated by the cofiber of $f$. A symmetric monoidal functor $F: \mathcal{C} \rightarrow \mathcal{D}$ maps $\operatorname{cofib}(f)$ to zero if and only if it inverts $1_{\mathcal{C}} \oplus \operatorname{cofib}(f)$. Therefore your first notion of localization is a particular case of the second one. -Your third notion of localization is in fact equivalent to the second one. Given a map $f: X \rightarrow Y$ between invertible objects, any such map $f': X^{-1} \rightarrow Y^{-1}$ is necessarily dual to an inverse to $f$. Hence localizing in your third way along an arbitrary map $f:X \rightarrow Y$ is equivalent to first inverting $X, Y$ and then inverting $f$, which we already observed can be reduced to the second notion. - -There are epimorphisms that do not arise as quotients by ideals: Consider for instance the category $\operatorname{Sh}(M)$ of sheaves of (complexes of) vector spaces on a manifold $M$. Let $x$ be a point in $M$ and $U$ its complement. The star pullback functor $\operatorname{Sh}(M) \rightarrow \operatorname{Sh}(\lbrace x \rbrace)$ exhibits $\operatorname{Sh}(\lbrace x \rbrace)$ as the quotient of $\operatorname{Sh}(M)$ by the ideal of sheaves with vanishing stalk at $x$. Similarly, $\operatorname{Sh}(U)$ is the quotient of $\operatorname{Sh}(M)$ by the ideal of sheaves supported at $x$. The ideal generated by the union of these two ideals is the whole $\operatorname{Sh}(M)$, so we see that $\operatorname{Sh}(\lbrace x \rbrace) \otimes_{\operatorname{Sh}(M)} \operatorname{Sh}(U) = 0$. It follows that the functor $\operatorname{Sh}(M) \rightarrow \operatorname{Sh}(\lbrace x \rbrace) \times \operatorname{Sh}(U)$ satisfies the tensor-idempotence condition, but it doesn't arise as the quotient by an ideal since its right adjoint $\operatorname{Sh}(\lbrace x \rbrace) \times \operatorname{Sh}(U) \rightarrow \operatorname{Sh}(M)$ (given by star-pushforward in each coordinate) is not fully faithful. - -Under tameness conditions all notions of localization agree: I don't know if every epimorphism arises by inverting an object in general, but under certain tameness conditions one can show that this is the case: -Claim: Let $\mathcal{C}$ be a symmetric monoidal dg category compactly generated by its dualizable objects and $\mathcal{C}' -$ be a compactly generated symmetric monoidal dg category equipped with a symmetric monoidal functor $\mathcal{C} \rightarrow \mathcal{C}'$ that preserves compact objects, and such that the map $\mathcal{C}' \otimes_{\mathcal{C}} \mathcal{C}' \rightarrow \mathcal{C}'$ is an isomorphism. Then the functor $\mathcal{C} \rightarrow \mathcal{C}'$ arises by passing to the quotient by an ideal of $\mathcal{C}$. -Sketch of proof: Let $\mathcal{K}$ be the full subcategory of $\mathcal{C}'$ generated under colimits by the image of the functor $\mathcal{C}\rightarrow \mathcal{C}'$. Our conditions guarantee that the right adjoint to the inclusion $\mathcal{C} \rightarrow \mathcal{K}$ is colimit preserving and monadic. Note moreover that $\mathcal{K}$ is a $\mathcal{C}$-module and the functor $\mathcal{C} \rightarrow \mathcal{K}$ is a map of $\mathcal{C}$-modules. Its right adjoint in principle only commutes with the $\mathcal{C}$-action up to natural transformations, but the fact that $\mathcal{C}$ is compactly generated by its dualizable objects guarantees that the natural transformations are isomorphisms, and so the functor $\mathcal{K}\rightarrow \mathcal{C}$ is also a morphism of $\mathcal{C}$-modules. It follows that $\mathcal{K}$ is the category of algebras for a $\mathcal{C}$-linear monad on $\mathcal{C}$, and so we have an identification $\mathcal{K} = A\operatorname{-mod}$ for some algebra $A$ in $\mathcal{C}$. The fact that $\mathcal{C}' \otimes_{\mathcal{C}} \mathcal{C}' = \mathcal{C}'$ implies that the multiplication map $A \otimes A \rightarrow A$ is an isomorphism. This means that $\mathcal{K}$ is in fact the category of algebras for an idempotent $\mathcal{C}$-linear monad, and so it arises as the quotient of $\mathcal{C}$ by an ideal. This whole thing reduces you to understanding the case when $\mathcal{C} \rightarrow \mathcal{C}'$ is fully faithful. Since the canonical map $\mathcal{C}\otimes_{\mathcal{C}}\mathcal{C}' \rightarrow \mathcal{C}'\otimes_{\mathcal{C}} \mathcal{C}'$ is an isomorphism, we have that $\mathcal{C}'/\mathcal{C} \otimes_{\mathcal{C}} \mathcal{C}'$ vanishes. This contains $\mathcal{C}'/\mathcal{C} \otimes_{\mathcal{C}} \mathcal{C} = \mathcal{C}'/\mathcal{C}$ as a full subcategory, so we see that $\mathcal{C}' = \mathcal{C}$ is the trivial localization. - -The algebro-geometric case: The above includes for instance the case of $\mathcal{C} = \operatorname{QCoh}(X)$ for $X$ a separated scheme. Moreover, from the proof we see that the resulting localizations are categories of modules for quasicoherent sheaves of algebras $A$ over $X$ such that the multiplication map $A \otimes A \rightarrow A$ is an isomorphism. In the Noetherian case you can use the result stated in the question to deduce that $A$ is locally a localization of the structure sheaf, so you see that all localizations in this sense are classified by collections of points of $X$ closed under specialization. -If you drop the condition that the functor $\mathcal{C} \rightarrow \mathcal{C}'$ preserves compact objects there are more examples, even in the geometric case. Indeed, any ideal of $\operatorname{QCoh}(X)$ provides an example, and these are classified (in the Noetherian case) by arbitrary collections of (non necessarily closed) points of $X$. This classification goes back to Hopkins, Neeman, and by now there is a whole industry about it - key words being tensor triangular geometry and classification of localizing subcategories. - -Beyond algebraic geometry: If you are looking for interesting covers beyond Zariski covers in algebraic geometry one source could be topology. If you have a manifold $M$ and $U$ is an open set of $M$, the category $\operatorname{Sh}(U)$ is the category of comodules for an idempotent coalgebra in $\operatorname{Sh}(M)$ with underlying sheaf $k_U$, and so it can be obtained as the colimit -$$\operatorname{Sh}(M) \xrightarrow{\otimes k_U}\operatorname{Sh}(M) \xrightarrow{\otimes k_U} \operatorname{Sh}(M)\xrightarrow{\otimes k_U} \ldots .$$ -This is the same diagram you would use to invert $k_U$, and so $\operatorname{Sh}(U)$ in fact results from inverting $k_U$. From this it follows that $\operatorname{Sh}(U)\otimes_{\operatorname{Sh}(M)} \operatorname{Sh}(V) = \operatorname{Sh}(U \cup V)$ for every pair of opens, and so you see that any open cover of $M$ provides a cover of symmetric monoidal categories in your sense. -You can build many more examples by variations of this theme. You could take $M$ to be the union of two manifolds $U, V$ along a closed submanifold (for instance $M$ could be the union of the $x$ and $y$ axes in $\mathbb{R}^2$) and you still have that $\operatorname{Sh}(M)$ is covered by $\operatorname{Sh}(U)$ and $\operatorname{Sh}(V)$. You could even require your sheaves to be constructible with respect to a stratification and $U, V$ to respect the stratification to get covers of categories of modules over quivers.<|endoftext|> -TITLE: Analytic equivalents for primes in arithmetic progressions -QUESTION [8 upvotes]: By way of context: it is known that the prime number theorem $\pi(x) \sim x/\log x$ is (nontrivially) equivalent to the statement that $\zeta(s)$ does not vanish on the line $\Re s=1$. -I would like to have it clarified what are the analogous equivalent statements for primes in arithmetic progressions. In particular, I would love to have specific references to papers or books in which the following equivalences are proved (or at least stated explicitly): - -Dirichlet's theorem, that there are infinitely many primes in any admissible arithmetic progression (mod $q$). is equivalent to the nonvanishing of $L(1,\chi)$ for all Dirichlet characters $\chi\pmod q$. [Edit: as has been pointed out, this nonvanishing is more likely equivalent to the equality of the Dirichlet densities of primes among the reduced residue classes (mod $q$). What analytic statement could be equivalent to the mere infinitude of primes in all such classes?] -The prime number theorem in arithmetic progressions $\pi(x;q,a)\sim x/(\phi(q)\log x)$ is equivalent to the statement that $L(s,\chi)$ does not vanish on the line $\Re s=1$ for all Dirichlet characters $\chi\pmod q$. - -Of course, if these statements themselves are incorrect, I would like to be corrected as well as being pointed to the literature. - -REPLY [3 votes]: Here are two equivalences. -Theorem 1. For each $m \geq 1$, the following are equivalent. -a) For all nontrivial Dirichlet characters $\chi \bmod m$, $L(1,\chi) \not= 0$. -b) For all $a \in (\mathbf Z/m\mathbf Z)^\times$, the set of primes $p \equiv a \bmod m$ has Dirichlet density $1/\varphi(m)$. -Proof of Theorem 1. -We will will compute the Dirichlet density of $\{p \equiv a \bmod m\}$ without assuming (a) and then see why (a) and (b) are equivalent. The trivial Dirichlet character modulo $m$ will be written as $\mathbf 1_m$. -For each Dirichlet character $\chi \bmod m$, $L(s,\chi)$ is analytic for ${\rm Re}(s) > 0$ except for $L(s,{\mathbf 1}_m)$ having a simple pole at $s = 1$. Set -$$ -n(\chi) := {\rm ord}_{s=1}(L(s,\chi)) -$$ -so $n({\mathbf 1}_m) = -1$ and $n(\chi) \geq 0$ for all nontrivial $\chi$. -For ${\rm Re}(s) > 1$ and $(a,m) = 1$, -$$ -\sum_{p \equiv a \bmod m} \frac{1}{p^s} = -\frac{1}{\varphi(m)} \sum_{p}\sum_{\chi} \frac{\chi(p)\overline{\chi}(a)} -{p^s} = -\frac{1}{\varphi(m)} \sum_{\chi}\overline{\chi}(a)\left(\sum_{p} \frac{\chi(p)} -{p^s}\right) -$$ -where the sum on the right run over all primes $p$ and all Dirichlet characters $\chi \bmod m$. -For a Dirichlet character $\chi \bmod m$ and ${\rm Re}(s) > 1$, -$$ -\log L(s,\chi) = \sum_p \frac{\chi(p)}{p^s} + \sum_{p,k\geq 2} \frac{\chi(p^k)}{kp^{ks}} = \sum_p \frac{\chi(p)}{p^s} + O(1), -$$ -where the $O$-constant is $\sum_{p,k \geq 2} 1/(kp^k)$, so -$$ -\sum_{p \equiv a \bmod m} \frac{1}{p^s} = -\frac{1}{\varphi(m)}\sum_{\chi \bmod m} \overline{\chi}(a)\log L(s,\chi) + O(1). -$$ -Now let's bring in the order of vanishing $n(\chi)$ above. For $s$ near $1$, $L(s,\chi) = (s-1)^{n(\chi)}f_\chi(s)$ where $f_\chi(s)$ is an analytic function in a neighborhood of $s = 1$ and $f_\chi(1) \not= 0$. Therefore $f_\chi(s)$ has an analytic logarithm around $s = 1$ (well-defined up to adding an integer multiple of $2\pi i$), so for $s > 1$, -$\log L(s,\chi) = n(\chi)\log(s-1) + \ell_{f_\chi}(s)$, -where $\ell_{f_\chi}(s)$ is a suitable logarithm of $f_\chi(s)$. Thus -$$ -\log L(s,\chi) = n(\chi)\log(s-1) + O_\chi(1) -$$ -for $s$ near $1$ to the right, and plugging this into the above displayed formula, -\begin{align} -\sum_{p \equiv a \bmod m} \frac{1}{p^s} & = \frac{1}{\varphi(m)}\sum_{\chi \bmod m} \overline{\chi}(a)(n(\chi)\log(s-1)+ O_\chi(1)) + O(1) \nonumber \\ -& =\frac{1}{\varphi(m)}\left(\sum_{\chi} \overline{\chi}(a)n(\chi)\right)\log(s-1) + O_m(1). -\end{align} -To compute a Dirichlet density, -we want to divide both sides by $\sum_p 1/p^s$ for -$s$ near $1$ to the right. For such $s$, -$$ -\log \zeta(s) = \sum_p \frac{1}{p^s} + O(1) = -\log(s-1) + O(1). -$$ -Therefore $\sum_p 1/p^s \sim -\log(s-1)$ as $s \to 1^+$, so -dividing through by $\sum_p 1/p^s$ and letting $s \to 1^+$ gives us -\begin{equation} -\lim_{s \to 1^+} -\frac{\sum_{p \equiv a \bmod m} 1/p^s}{\sum_p 1/p^s} = -\frac{1}{\varphi(m)}\left(-\sum_{\chi} \overline{\chi}(a)n(\chi)\right), -\end{equation} -which expresses the Dirichlet density of $\{p \equiv a \bmod m\}$ in terms of the -orders of vanishing $n(\chi)$ as $\chi$ runs over Dirichlet characters mod $m$. -If (a) is true then $n(\chi) = 0$ for all nontrivial $\chi$, so the right side of the above limit calculation -is $(1/\varphi(m))(-n({\mathbf 1}_m)) = 1/\varphi(m)$, which is (b). -Conversely, if (b) is true then -$$ -\sum_{\chi} \overline{\chi}(a)n(\chi) = -1 -$$ -for all $a \in (\mathbf Z/m\mathbf Z)^\times$ by our limit calculation. Why does this imply $n(\chi) = 0$ for nontrivial $\chi$? -Using complex vectors indexed by all the Dirichlet characters mod $m$, let -${\mathbf n}_m = (n(\chi))_\chi$ and -${\mathbf v}_a = (\chi(a))_\chi$ for each $a \in (\mathbf Z/m\mathbf Z)^\times$. The space of all complex vectors $\mathbf z = (z_\chi)_\chi$ has -dimension $\varphi(m)$ and it has the Hermitian inner product -$\langle \mathbf z, \mathbf w\rangle = \frac{1}{\varphi(m)}\sum_{\chi} z_\chi\overline{w_\chi}$ for which the vectors ${\mathbf v}_a$ are an orthonormal basis by the orthogonality relations for Dirichlet characters mod $m$. The above displayed formula -says $\langle {\mathbf n}_m,{\mathbf v}_a\rangle = -1/\varphi(m)$ for all $a$ in $(\mathbf Z/m\mathbf Z)^\times$, so -$$ -{\mathbf n}_m = \sum_{a} \langle {\mathbf n}_m,{\mathbf v}_a\rangle{\mathbf v}_a = -\frac{1}{\varphi(m)}\sum_{a}{\mathbf v}_a. -$$ -For each nontrivial $\chi \bmod m$, the $\chi$-component of -$\sum_{a} {\mathbf v}_a$ is $\sum_a \chi(a)$, which is $0$. So the $\chi$-component of ${\mathbf n}_m$, which is $n(\chi)$, is 0. That is (a). -QED Theorem 1. (I only realized after copying and pasting this that I had already copy and pasted it earlier as an answer to the MO question here.) -Theorem 2. For each $m \geq 1$, the following are equivalent. -a) For all Dirichlet characters $\chi \bmod m$, $L(s,\chi) \not= 0$ when ${\rm Re}(s) = 1$. -b) $\sum_{n \leq x} \chi(n)\Lambda(n) = o(x)$ -for nontrivial Dirichlet characters $\chi \bmod m$ -and $\sum_{n \leq x} \chi_{{\mathbf 1}_m}(n)\Lambda(n) \sim x$, -c) For all $a \in (\mathbf Z/m\mathbf Z)^\times$, $|\{p \leq x : p \equiv a \bmod m\}| \sim (1/\varphi(m))x/\log x$. -Comparing the proof of Theorem 2 below to the sketch in the answer by 2734364041, we will also be using a Tauberian theroem (to prove (b) implies (c)), but we will not need an explicit formula. -Proof of Theorem 2. -We will show (a) is equivalent to (b) and (b) is equivalent to (c). -First we show (a) implies (b). -Set $\psi_\chi(x) = \sum_{n \leq x} \chi(n)\Lambda(n)$ for all $\chi$, so -(b) says $\psi_\chi(x) = o(x)$ for nontrivial $\chi$ and $\psi_{{\mathbf 1}_m}(x) \sim x$. -For $\sigma > 1$, $-L'(s,\chi)/L(s,\chi) = \sum \chi(n)\Lambda(n)/n^s$, for all Dirichlet characters $\chi \bmod m$, -so $\psi_\chi(x)$ is a partial sum of coefficients of $-L'(s,\chi)/L(s,\chi)$. -Since $L(s,{\mathbf 1}_m) \not= 0$ on $\sigma = 1$ by (a), $-L'(s,{\mathbf 1}_m)/L(s,{\mathbf 1}_m)$ is holomorphic on $\sigma \geq 1$ except for a simple pole at $s = 1$ with residue 1 and it has nonnegative Dirichlet series coefficients with $\psi_{{\mathbf 1}_m}(x) = O(x)$. Therefore $\psi_{{\mathbf 1}_m}(x) \sim x$, which is part of (b), by Newman's Tauberian theorem. To get the rest of (b), namely $\psi_\chi(x) = o(x)$ for nontrivial $\chi$, we have $-L'(s,\chi)/L(s,\chi)$ being holomorphic on $\sigma \geq 1$ by (a) and its Dirichlet series coefficients satisfy $|\chi(n)\Lambda(n)| \leq {\mathbf 1}_m(n)\Lambda(n)$ for all $n$, so $\psi_\chi(x) = o(x)$ by a corollary of Newman's Tauberian theorem for $-L'(s,\chi)/L(s,\chi)$ using comparison Dirichlet series $-L'(s,{\mathbf 1}_m)/L(s,{\mathbf 1}_m)$ . -Thus (a) implies (b). -To show (b) implies (a), we will use the following fact. For a function $a(x)$ on $[1,\infty)$ that is bounded and Riemann integrable on $[1,T]$ for all $T \geq 1$, so $f(s) := \int_1^\infty (a(x)/x^s)dx/x$ is absolutely convergent on $\sigma > 1$, if $a(x) \to 0$ as $x \to \infty$ and $f$ extends to a meromorphic function on $\sigma = 1$ then $f$ in fact is holomorphic on $\sigma = 1$. (This is used to show the condition $\psi(x) \sim x$ implies $\zeta(s) \not= 0$ on $\sigma = 1$ by using $a(x) = \psi(x)/x - 1$.) Because of the integral representations -$$ --\frac{L'(s,\chi)}{sL(s,\chi)} = \int_1^\infty \frac{\psi_\chi(x)}{x} \frac{dx}{x^s} -$$ -and -$$ --\frac{L'(s,{\mathbf 1}_m)}{sL(s,{\mathbf 1}_m)} - \frac{1}{s-1} = \int_1^\infty \left(\frac{\psi_{\mathbf 1}(x)}{x} -1\right)\frac{dx}{x^s}, -$$ -for ${\rm Re}(s) > 1$, -where $\chi$ is nontrivial in the first equation. -we can use the above fact when $a(x) = \psi_\chi(x)/x$ for nontrivial $\chi$ and -$a(x) = \psi_{{\mathbf 1}_m}(x)/x - 1$ to conclude that $L'(s,\chi)/L(s,\chi)$ is holomorphic on -$\sigma = 1$ for nontrivial $\chi$ and $L'(s,{\mathbf 1}_m)/L(s,{\mathbf 1}_m)$ is holomorphic on $\sigma = 1$ except for a simple pole at $s = 1$, so -$L(s,\chi)$ is nonvanishing on $\sigma = 1$ and -$L(s,{\mathbf 1}_m)$ is nonvanishing on $\sigma = 1$. -Thus (b) implies (a). -That (b) implies (c) follows from the above integral representation of $-L'(s,\chi)/L(s,\chi)$ for all nontrivial $\chi$ by a standard method to prove (c). -Our last step is showing (c) implies (b). Set $\pi(x;a \bmod m) = |\{p \leq x : p \equiv a \bmod m\}|$ when $(a,m) = 1$ and $\pi_\chi(x) = \sum_{p \leq x} \chi(p)$, -where $\chi$ is a Dirichlet character mod $m$. -Write $\chi$ as a linear combination of -delta-functions on $(\mathbf Z/m\mathbf Z)^\times$: -$\chi = \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\delta_a$. Then -\begin{align*} -\pi_\chi(x) & = \sum_{p \leq x} \chi(p) \\ -& = \sum_{p \leq x} \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\delta_a(p) \\ -& = \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\left(\sum_{p \leq x} \delta_a(p)\right) \\ -& = \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\pi(x; a \bmod m), -\end{align*} -so -$$ -\frac{\pi_\chi(x)}{x/\log x} = \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\frac{\pi(x;a \bmod m)}{x/\log x}. -$$ -By (c), as $x \to \infty$ the right side tends to -$\sum_{a \in ({\mathbf Z}/m{\mathbf Z})^\times} \chi(a)/\varphi(m)$, which is 0 if $\chi$ is nontrivial. Therefore when $\chi$ is nontrivial we have $\pi_\chi(x) = o(x/\log x)$, which implies $\psi_\chi(x) = o(x)$ by -the same argument that $\pi(x) \sim x/\log x$ implies $\psi(x) \sim x$. To show (c) implies -$\psi_{{\mathbf 1}_m}(x) \sim x$, sum the relation in (c) over all $a$ in $(\mathbf Z/m\mathbf Z)^\times$ to -get $\pi(x) \sim x/\log x$, the Prime Number Theorem, which -is equivalent to -$\psi(x) \sim x$, so -$\psi_{{\mathbf 1}_m}(x) \sim x$ since $\psi_{{\mathbf 1}_m}(x) = \psi(x) + O_m(\log x)$. -QED Theorem 2.<|endoftext|> -TITLE: Non-density of continuous functions to interior in set of all continuous functions -QUESTION [5 upvotes]: Let $M$ be an $m$-dimensional manifold and $N$ be an $n$-dimensional manifold. Suppose also that the topology on $N$ can be described by a metric. Thus, the set $C(M,N)$ can be endowed with the topology of [uniform convergence on compacta][2]. -Let $N'\subseteq N$ be a dense subset which is homeomorphic to $\mathbb{R}^n$. In this post's answer's comment it was remarked that $C(D,D-\{0\})$ is not dense in $C(D,D)$ where $D$ is the unit disc. -In general, when is $C(M,N')$ not dense in $C(M,N)$? - -REPLY [5 votes]: 1) About Erz' answer, and Annie's second question. It is a general property that for two differentiable manifolds $M$, $N$, such that $M$ is compact, and given a continuous map $f:M\to N$, any continuous map $g:M\to N$ -close enough to $f$ is homotopic to $f$. This can be proved by many different ways, depending on taste. (One can assume that $N$ is also compact, since -we are only interested in a small neighborhood of the compact $f(M)$ in $N$): -a) Put a Riemannian metric on $N$, let $i>0$ be its injectivity radius: any two points $x,y\in N$ whose distance is less than $i$ are linked by a unique shortest geodesic, and this geodesic depends continuously on the pair $(x,y)$. Then, assuming that $d(f(x),g(x))m$ in $N$, then by Thom's transversality theorem, $C^\infty(M,N')$ is $C^\infty$-dense in $C^\infty(M,N)$, and in particular $C^0$-dense. (As to Erz' second question: since $\dim(M)+\dim(S_i)<\dim(N)$, saying that $f:M\to N$ -is transverse to $S_i$ amounts to say that $f(M)$ does not meet $S_i$).<|endoftext|> -TITLE: Does the expression $x^4 +y^4$ take on all values in $\mathbb{Z}/p\mathbb{Z}$? -QUESTION [9 upvotes]: As the title asks: does there exist $N$ such that, for any prime $p$ larger than $N$, the expression $x^4 +y^4$ takes on all values in $\mathbb{Z}/p\mathbb{Z}$? - -I have been thinking about this problem for days, but I failed to solve it. Does anyone know whether this is true, or does anyone know any partial results about it? -Partial results: -If $p=4k+3$, it easily works -if $p=4k+1$, if $g$ is a primitive root modulo $p$, and $A_i = \left\{ g^k : k \equiv i \pmod{4} \right\}$, then at least three of $A_i (i=0,1,2,3)$ must be expressed. - -REPLY [12 votes]: Expanding on a comment, the curve $X^4+Y^4=aZ^4$ (for $a\ne0$) has genus $3$. So the Hasse-Weil bound says -$$ N_p(a) := \#\bigl\{ [X,Y,Z]\in\mathbb P^2(\mathbb F_p) : X^4+Y^4=aZ^4 \bigr\} $$ -satisfies -$$ \bigl| N_p(a) - p - 1 \bigr| \le 2g\sqrt{p} = 6\sqrt{p}. $$ -Thus -$$ N_p(a) \ge p + 1 - 6\sqrt{p}. $$ -There are at most $4$ points with $Z=0$, so -$$ -\#\bigl\{ (X,Y) \in\mathbb A^2(\mathbb F_p) : X^4+Y^4=a \bigr\} \ge p-3-6\sqrt{p}. -$$ -So you'll always have a solution provided $p\ge3+6\sqrt{p}$, which means that there is always a solution provided $p\ge43$.<|endoftext|> -TITLE: Solving a delay-differential equation related to epidemiology -QUESTION [27 upvotes]: For some inexplicable reason, I have recently been interested in epidemiology. One of the classical and simplistic models in epidemiology is the SIR model given by the following system of first-order autonomous nonlinear ordinary differential equations in the real variables $s,i,r$ satisfying $s,i,r\geq 0$ and $s+i+r=1$: -$$ -\begin{align} -s' &= -\beta i s\\ -i' &= \beta i s - \gamma i\\ -r' &= \gamma i -\end{align} -\tag{$*$} -$$ -(where prime denotes derivative w.r.t. time, $s,i,r$ represent the proportion of “susceptible”, “infected” and “recovered” individuals, $\beta$ is the kinetic constant of infectiousness and $\gamma$ that of recovery; the “replication number” here is $\kappa := \beta/\gamma$). It is easy to see that for every initial value $(s_0,i_0,r_0)$ at $t=0$ the system has a unique $\mathscr{C}^\infty$ (even real-analytic) solution (say $r_0=0$ to simplify, which we can enforce by dividing by $s_0+i_0$); it does not appear to be solvable in exact form in function of time, but since $s'/r' = -\kappa\cdot s$ and $i'/r' = \kappa\cdot s - 1$ (where $\kappa := \beta/\gamma$) it is possible to express $s$ and $i$ as functions of $r$, viz., $s = s_0\,\exp(-\kappa\cdot r)$ and of course $i = 1-s-r$. This makes it possible to express, e.g., the values of $s,i,r$ at peak epidemic (when $i'=0$): namely, $s = 1/\kappa$ and $r = \log(\kappa)/\kappa$); or when $t\to+\infty$: namely, $s = -W(-\kappa\,\exp(-\kappa))/\kappa$ where $W$ is (the appropriate branch of) Lambert's transcendental W function; this is upon taking $i_0$ infinitesimal and $r_0=0$. -Now one of the many ways in which this SIR model is simplistic is that it assumes that recovery follows an exponential process (hence the $-\gamma i$ term in $i'$) with characteristic time $1/\gamma$. A slightly more realistic hypothesis is recovery in constant time $T$. This leads to the following delay-difference equation: -$$ -\begin{align} -s' &= -\beta i s\\ -i' &= \beta i s - \beta (i s)_T\\ -r' &= \beta (i s)_T -\end{align} -\tag{$\dagger$} -$$ -where $(i s)_T = i_T\cdot s_T$ and $f_T(t) = f(t-T)$. When I speak of a “$\mathscr{C}^\infty$ solution” of ($\dagger$) on $[0;+\infty[$ I mean one where the functions $s,i,r$ are $\mathscr{C}^\infty$ and satisfy the equations whenever they make sense (so the second and third are only imposed for $t\geq T$), although one could also look for solutions on $\mathbb{R}$. -I am interested in how ($\dagger$) behaves with respect to ($*$). Specifically, - -Does ($\dagger$) admit a $\mathscr{C}^\infty$ (or better, real-analytic) solution on $[0;+\infty[$ for every initial value $(s_0,i_0,r_0)$ at $t=0$? Or maybe even on $\mathbb{R}$? Is this solution unique? (Note that we could try to specify a solution by giving initial values on $[0;T[$ and working in pieces, but this does not answer my question as it would not, in general, glue at multiples of $T$ to give a $\mathscr{C}^\infty$ solution.) -Can we express this solution in closed form or, at least, express $s$ and $i$ in function of $r$? -Qualitatively, how do the (at least continuous!) solutions of ($\dagger$) differ from those of ($*$), and specifically: - -How do the behaviors compare when for $t\to 0$? -How do the values compare around peak epidemic $i'=0$? -How do the limits when $t\to+\infty$ compare? - -REPLY [13 votes]: After trying many Ansätze in the form of series, I stumbled upon the fact (which I find rather remarkable) that ($\dagger$) admits an exact solution in closed form. To express this, let me first introduce the following notations: - -$\kappa := \beta T$ (reproduction number), which I assume $>1$; -$\Gamma := -W(-\kappa \exp(-\kappa))/\kappa$ the solution in $]0,1[$ to $\Gamma = \exp(-\kappa(1-\Gamma))$, which will be the limit of $s$ when $t\to+\infty$ (both in ($*$) and in ($\dagger$), starting with $i$ and $r$ infinitesimal); -$X := \exp(\beta(1-\Gamma) t)$, a change of variable on the time parameter, in which $s,i,r$ will be expressed. - -The solution is then given by: -$$ -\begin{align} -s &= \frac{(1-\Gamma)^2+\Gamma c X}{(1-\Gamma)^2+c X}\\ -i &= \frac{(1-\Gamma)^4 c X}{((1-\Gamma)^2+c X)((1-\Gamma)^2+\Gamma c X)}\\ -r &= \frac{\Gamma (1-\Gamma) c X}{(1-\Gamma)^2+\Gamma c X} -\end{align} -$$ -where c is an arbitrary positive real parameter which merely serves to translate the solution. -From this we can deduce the following about the behavior at start, peak and end of the epidemic, and how it compares to ($*$) (for which I let $\kappa := \beta/\gamma$ and $\Gamma$ defined by the same formula): - -Initially, $i$ grows like $c\,\exp(\beta(1-\Gamma) t)$ (and $r$ like $\frac{\Gamma}{1-\Gamma}$ times this). This is in contrast to ($*$) where $i$ initially grows like $c\,\exp((\beta-\gamma) t)$ (and $r$ like $\frac{\gamma}{\beta-\gamma}$ times this), i.e., for a given reproduction number $\kappa$, solutions of ($\dagger$) grow faster than those of ($*$). -Peak epidemic happens for $X = \frac{(1-\Gamma)^2}{c\sqrt{\Gamma}}$, at which point we have $s = \sqrt{\Gamma}$ and $i = (1-\sqrt{\Gamma})^2$ and $r = \sqrt{\Gamma}(1-\sqrt{\Gamma})$. This is in contrast to ($*$) where we have $s = \frac{1}{\kappa}$ and $i = \frac{\kappa-\log\kappa-1}{\kappa}$ and $r = \frac{\log\kappa}{\kappa}$; so, for a given reproduction number $\kappa$, solutions of ($\dagger$) have a peak epidemic with fewer uninfected ($s$) than those of ($*$). -The limit when $t\to+\infty$ is the same in ($\dagger$) as in ($*$), namely $s \to \Gamma$.<|endoftext|> -TITLE: Equivalence of antiderivative in L1 sense and in the usual sense -QUESTION [6 upvotes]: We say that$\ f$ is differentiable w.r.t to $L_1$ if there exists a$\ g$ such that: -$$ -\lim_{h\to 0}\left\Vert\frac{f(x+h)-f(x)}{h} - g(x)\right\Vert_1 = 0 -$$ -where $\Vert \cdot \Vert_1$ is the $L_1$ norm. Since $f$ is in $L_1$, the corresponding$\ g$ must be in$\ L_1$ too, and so by Lebesgue, it has an antiderivative $G$ which is differentiable a.e, with $G'(x)=g(x)$. -My question is: does $f=G$ a.e? -Here is my line of thought: if $G$ is in $L_1$, it can be shown that -$$ -\hat{g}{(t)} = 2\pi it\hat{G}{(t)} = 2\pi it\hat{f}{(t)}, -$$ -which then implies that $f=G$ a.e. and so, in order to show that $f=G$ a.e, it is enough to show that$\ G$ is in$\ L_1$, and that's where i got stuck. - -REPLY [5 votes]: Most antiderivative of $g$ are not in $L^1(\mathbb{R})$, in your case only one antiderivative $G_0$ will be in $L^1(\mathbb{R})$, the one actually equal to $f$. All the other antiderivatives $G$ are equal to $G_0 + c$, with $c \neq 0$, which is not in $L^1(\mathbb{R})$. -To proof that there exist one antiderivative $G_0$ in $L^1(\mathbb{R})$. -You start by noticing that your $L^1(\mathbb{R})$ differentiability imply differentiability in the distributional sens so, for $\phi \in \mathcal{C}_{comp}^\infty(\mathbb{R})$, we have -$$ -\langle f',\phi \rangle = \langle g,\phi \rangle. \qquad (1) -$$ -Fix $G$ an antiderivative of $g$, you have $G' = g$ in the distributional sens. -The equation $(1)$ become -$$ -\langle f',\phi \rangle = \langle G',\phi \rangle \implies \langle (f-G)',\phi \rangle = 0 -$$ -and than imply $f-G = c$ a constant. -Choosing $G_0 = G + c$, we have $G_0 = f \in L^1(\mathbb{R})$. -You can further show that there exists a constant $c_0$ such that -$$ -G_0(x) = c_0 + \int_0^x g(y)\, dy. -$$<|endoftext|> -TITLE: Is the spectrum of a "self adjoint" operator real on $\ell^p$? -QUESTION [10 upvotes]: There might be an obvious answer to the question, but it doesn't come to mind. -Suppose we have an infinite matrix $A=(a_{ij})$, which defines a bounded linear operator on $\ell^p$, i.e. for all sequences $(x_i)\in \ell^p, p>1$ -$$ \sum_{i=1}^\infty\big|\sum_{j=1}^\infty a_{ij}x_j\Big|^p \leq C \Vert x \Vert_{p}^p.$$ -For some positive constant $C$. -Furthermore, assume that $a_{ij}=\overline{a_{ji}}$. Is it true that the spectrum of $A$ on $\ell^p $ is real ? - -REPLY [4 votes]: It seems that I have found a counter example myself. -For the Hilbert matrix -$$ H_\lambda:= \big( \frac{1}{1-\lambda+k+n} \big)_{k,n\geq 0}, \lambda < 1 $$ -Rosenblum in "On the Hilbert Matrix I, Proceedings of the AMS" proves that the pointspectrum considered as an operator on $\ell^p, p>2$ contains the set -$$ \{ \pi \sec(\pi u ) : | \Re ( u )| < 1/2-1/p \}. $$ -If one could provide a more elementary counterexample I would be interested in looking at.<|endoftext|> -TITLE: Explicit description of exponentials of étalé spaces -QUESTION [7 upvotes]: It is well known that the category $\mathit{Sh}(X)$ of sheaves of sets on a topological space $ X $ is a topos. -On the other hand, there exists a natural equivalence of categories between $\mathit{Sh}(X)$ and the category $\mathit{Et}(X)$ of étalé spaces over $ X $, so that the category of etale spaces over $ X $ is a topos, too. -The constructions in $\mathit{Sh}(X)$ are translatable to $\mathit{Et}(X)$, and vice versa. -For example, products of sheaves correspond to fiber products of étalé spaces. -The description of the exponentials in $\mathit{Sh}(X)$ is given, e.g., in [Mac Lane and Moerdijk, Sheaves in Geometry and Logic]. The question that arises is - - -What is the explicit construction of exponentials in $Et(X)$? - -REPLY [2 votes]: Let $[E,E’]$ be the internal hom of two étale spaces $\pi : E \to X$ and $\pi’: E’ \to X$. I will write $\mathrm{Hom}_X(A,B)$ for the set of morphisms of étale spaces $A$ and $B$ over $X$, and $A \times_X B$ for the product of $A$ and $B$ as étale spaces over $X$. -To compute the sections of $[E,E’]$, you can use -$$\mathrm{Hom}_X(U,[E,E’]) \simeq \mathrm{Hom}_X(U \times_X E, E’) \simeq \mathrm{Hom}_X(U \times_X E, U \times E’).$$ -The first isomorphism uses the universal property of the exponential, the second isomorphism uses that the image of $U \times E$ is always contained in $(\pi’)^{-1}(U) =U \times E’$, so there is a unique factorization through $U \times E’$. -The elements of the total space correspond to stalks, so the fiber of $[E,E’]$ above $x \in X$ is given by the set -$$[E,E’]_x \simeq \varinjlim_i \mathrm{Hom}_X(U_i \times_X E, U_i \times_X E’)$$ -where $(U_i)_{i \in I}$ is a projective system of open neighborhoods of $x$ such that every open neighborhood of $x$ contains some $U_i$ (for example, if $X$ is a metric space, you can take open balls of shrinking radius). -A basis of open sets on $[E,E’] = \bigsqcup_x [E,E’]_x$ is given by the sets $A_{(U,s)} = \{ s_x : x \in U \}$, for $U$ in some basis of open sets of $X$, $s \in \mathrm{Hom}_X(U \times_X E, U \times_X E’)$, and $s_x$ the element of $[E,E’]_x$ corresponding to $s$. This topology is often not easy to visualize. The spaces $E$ and $E’$ are already typically non-Hausdorff, and even if they are Hausdorff there is no guarantee that $[E,E’]$ will be as well. But maybe it helps to keep in mind that the subspace topology on the fibers is always the discrete topology (as for any étale space). -For example, take $X= \mathbb{R}$, $E = \mathbb{R}-\{0\}$ and $E’ = \mathbb{R}\sqcup\mathbb{R}$ with $\pi$ and $\pi’$ the natural projections. -We first look at $x \neq 0$. For a small open interval $U$ containing $x\neq 0$, there are exactly two maps $E \times_X U \to E’ \times_X U$ (because $E \times_X U \simeq U$ and $E’ \times_X U \simeq U \sqcup U$). -However, if $U$ is a small open interval containing $x=0$, then $E \times_X U = U -\{0\}$ has two components, and because of this there are four maps to $E’ \times_X U \simeq (U-\{0\}) \sqcup (U - \{0\})$. -So in this example, $[E,E’]$ has four points above $x=0$ and two points above each $x \neq 0$. Further, $[E,E’]$ has four global sections corresponding to the four continuous maps $E \to E’$ over $X$. This gives four open sets $A_{(X,s)}$ in $[E,E’]$, with each of these homeomorphic to $\mathbb{R}$. But these four lines are glued together to a single line on $\{ x \in \mathbb{R} : x > 0\}$ iff the corresponding sections agree on $\{ x \in \mathbb{R} : x > 0\}$, and similarly over $\{x \in \mathbb{R} : x < 0 \}$. -Here is a picture: - -The gluing of different lines is supposed to be “instant” (just as in the case of the line with two origins) and as a result the space is non-Hausdorff.<|endoftext|> -TITLE: Is every function $f: \mathbb R \to \mathbb R$ differentiable at at least one point when restricted to some everywhere dense subset of $\mathbb R$? -QUESTION [20 upvotes]: I was doing some fairly simple research a few hours ago and I almost asked a similar question with the word continuous instead of differentiable in the title, but then I found this question asked by Gro-Tsen where there is an affirmative answer to that question. -Apparently, that is the result of Blumberg, that for every $f: \mathbb R \to \mathbb R$ there exists a dense subset $D$ of $\mathbb R$ such that $f|_D$ is continuous. -Blumberg´s paper can be found here and I have slightly did a research of his arguments, however, I am not sure can they be adapted to show that $f$ is differentiable at at least one point when restricted to some everywhere dense subset of $\mathbb R$. -Honestly, I expect that there are some $f$´s which have the property that when restricted to every possible everywhere dense subset of $\mathbb R$ are non-differentiable everywhere on all such sets -However, I am not sure, and that´s why I ask it here, since I think that´s known, because Blumberg´s result is relatively long time ago established (1922). -Here is the question: - -Is it true that for every function $f: \mathbb R \to \mathbb R$ there exists at least one everywhere dense set $D \subseteq \mathbb R$ such that $f|_D$ is differentiable at at least one point? - -REPLY [17 votes]: The answer is no. This is because, if $f: \mathbb R \rightarrow \mathbb R$ is a continuous, nowhere differentiable function, then $f \!\restriction\! Q$ is nowhere differentiable for any dense $Q \subseteq \mathbb R$. -To see this, fix $x \in \mathbb R$ and, aiming for a contradiction, let us suppose $f \!\restriction\! Q$ is differentiable at $x$, say with derivative $c \in \mathbb R$. -Let $\varepsilon > 0$. Because $f \!\restriction\! Q$ is differentiable at $x$, there is some $\delta > 0$ such that for all $y \in Q \setminus \{x\}$ with $|x-y| < \delta$, we have $|\frac{f(y) - f(x)}{y-x} - c| < \varepsilon$. -Because $f$ is not differentiable at $x$, and in particular does not have derivative equal to $c$ at $x$, there is some $z_0 \in \mathbb R \setminus \{x\}$ with $|x-z_0| < \delta$ such that $|\frac{f(z_0) - f(x)}{z_0-x} - c| > 2\varepsilon$. -Because $f$ is continuous on $\mathbb R$, the function $z \mapsto |\frac{f(z) - f(x)}{z-x} - c|$ is continuous on $\mathbb R \setminus \{x\}$. This means that $\lim_{z \rightarrow z_0} |\frac{f(z) - f(x)}{z-x} - c| = |\frac{f(z_0) - f(x)}{z_0-x} - c| > 2\varepsilon$. -Therefore there is some $\eta < \delta-|x-z_0|$ such that if $|z_0-z| < \eta$ then $|\frac{f(z) - f(x)}{z-x} - c| > \varepsilon$. -Let $z \in Q$ with $|z-z_0| < \eta$. Then we have $|x-z| < \delta$ while $|\frac{f(z) - f(x)}{z-x} - c| > \varepsilon$. This contradicts our choice of $\delta$.<|endoftext|> -TITLE: Convex Julia sets -QUESTION [10 upvotes]: Consider the classical Julia set $J_f$ associated with $f(z)=z^2+c$. -Since $J_c$ is completely invariant, -we know that $f^{-1}(J_f) \subseteq J_f$. -Now, let $H_f$ be the convex hull of $J_f$. -Is it true that $f^{-1}(H_f) \subseteq H_f$? -I have done some basic computer experiments, and it seem to hold for $c \in [0,1]^2 \subset \mathbb{C}$. Moreover, I suspect that the natural generalization of the statement above might hold for all polynomial maps. However, I have examples with rational maps where the statement is not true. -As an example, consider $f(z)=z^3-iz + 0.2 + 0.4i$. -The blue points is the Julia set $J_f$ associated with $f$. The shaded region is the convex hull $H_f$ of the Julia set. -Taking a uniform square grid $G$ on $H_f$, and plotting -the points $f^{-1}(G)$ gives the black dots. -As we can see, it is reasonable to guess that $f^{-1}(H_f)\subset H_f$. - -REPLY [10 votes]: Edited: The previously found sufficient condition is indeed necessary, but even better, it is satisfied by all polynomials of degree at least two. Thus the conjecture is true: -Theorem: Let $p$ be a complex polynomial of degree $d \geq 2$ and let $H_p={\rm conv}J_p$, the convex hull of the Julia set $J_p$ of $p$. Then $p^{-1}(H_p) \subset H_p$. -To prove this, I use several lemmata. First, there is the following version of Gauss-Lucas theorem, due to W. P. Thurston, which can be found in the preprint -A. Ch\'eritat, Y. Gao, Y. Ou, L. Tan: \emph{A refinement of the Gauss-Lucas theorem (after W. -P. Thurston)}, 2015, preprint, hal-01157602 -Lemma 1: Let $p$ be any polynomial of degree at least two. Denote by $\mathcal{C}$ the convex hull of the critical points of $p$. Then $p: E \to \mathbb{C}$ is surjective for any closed half-plane $E$ intersecting $\mathcal{C}$. -From this we have the following: -Lemma 2: Let $p$ be any polynomial of degree at least two. Then all zeros of $p'$ belong to $H_p={\rm conv}J_p$, the convex hull of the Julia set $J_p$ of $p$. -Proof: Suppose there is an $x_0 \not \in H_p$ such that $p'(x_0)=0$. By the hyperplane separation theorem , there exists a closed half-plane $E$ such that $x_0 \in E$ and $E \cap H_p = \emptyset$. In particular, $E \cap J_p = \emptyset$. By Lemma 1, $p: E \to \mathbb{C}$ is surjective. Take a $z_0 \in J_p$. Then on one hand $p^{-1}(z_0) \subset J_p$, while on the other hand $p^{-1}(z_0) \cap E \neq \emptyset$, a contradiction. -The next lemma is a modification of Exercise 2.1.15 in - Lars Hörmander: Notions of Convexity -Publisher Springer Science \& Business Media, 2007 (Modern Birkhäuser Classics) -ISBN 0817645853, 9780817645854 -Lemma 3: -Let $p(z)=\sum_{j=0}^d a_jz^j$ be a polynomial in $z \in \mathbb{C}$ of degree $d$. Let $B$ be a closed convex subset of $\mathbb{C}$ containing all zeros of $p'$. Then the set $C_B$ of all $w \in \mathbb{C}$ such that all the zeros of $p(\cdot)-w$ are contained in $B$ is a convex set. -Proof of Lemma: Note that by continuity of roots $C_B$ is closed when $B$ is. Let $w_1,w_2 \in C_B$ and $n_1,n_2 \in \mathbb{N}$ and consider the polynomial (in one complex variable $z$) $P(z):=(p(z)-w_1)^{n_1}(p(z)-w_2)^{n_2}$. Then all zeros of $P$ lie in $B$ (by definition of $C_B$), so the convex hull of zeros of $P$ is contained in $B$. By Gauss-Lucas theorem (standard version), all zeros of $P'$ are contained in $B$. The zeros of $P'$ are respectively all the zeros of $p(z)-w_1$, all the zeros of $p(z)-w_2$ (if $n_1, n_2 >1$), all the zeros of $p'$ and all the zeros of $p(\cdot)-\left (\frac{n_1}{n_1+n_2}w_1 + \frac{n_2}{n_1+n_2}w_2\right)$. By definition of $C_B$, $\frac{n_1}{n_1+n_2}w_1 + \frac{n_2}{n_1+n_2}w_2 \in C_B$. Varying $n_1,n_2$ and using the property that $C_B$ is closed, we get that $tw_1+(1-t)w_2 \in B$ for all $0 \leq t \leq 1$. -Proof of Theorem: Applying the Lemma 3 to $B=H_p={\rm conv}J_p$, we get that the set $C_p=\{w \in \mathbb{C}: \{z : p(z)-w=0\} \subset H_p\}$ is convex. Furthermore, for $w \in J_p$ all solutions of $p(z)-w$ are in $J_p \subset H_p$, so $J_p \subset C_p$. Hence $H_p \subset C_p$, which implies that $p^{-1}(H_p) \subset H_p$. -For the quadratic family $f_c(z)=z^2+c, \ c \in \mathbb{C}$ it is straightforward (without appealing to Lemma 2) to check that the critical point $0$ is the center of symmetry of the Julia set $J_c$, so it is a convex combination of two points in $J_c$.<|endoftext|> -TITLE: A topologically transitive dynamical system without dense orbits -QUESTION [5 upvotes]: By a dynamical system I understand a pair $(K,G)$ consisting a compact Hausdorff space and a subgroup $G$ of the homeomorphism group of $K$. -We say that a dynamical system $(K,G)$ -$\bullet$ is topologically transitive if for every non-empty open set $U\subseteq K$ its orbit $GU=\{g(x):g\in G,\;x\in U\}$ is dense in $K$; -$\bullet$ has dense orbit if for some point $x\in K$ its orbit $Gx$ is dense in $K$. -It is easy to see that a dynamical system is topologically transitive if it has a dense orbit. If the space $K$ is metrizable and nonempty, then the converse also is true. -On the other hand, under $\mathrm{non}(\mathcal M)<\mathfrak c$, there exists a subgroup $G\subset S_\omega$ of cardinality $|G|\le\mathrm{non}(\mathcal M)<\mathfrak c$ that induces a topologically transitive action of the Stone-Cech remainder $\omega^*=\beta\omega\setminus\omega$. The dynamical system $(\omega^*,G)$ does not have dense orbits since the space $\omega^*$ has density $\mathfrak c>|G|$. I am interested if such an example can be constructed in ZFC. - -Problem. Is there topologically transitive dynamical system without dense orbits? - -REPLY [6 votes]: The answer is yes: -there is a topologically transitive dynamical system without dense orbits. -Indeed, -let X be a topological space that is not separable. Let $\ K=X^{\Bbb Z},\ $ and let $\ G\ $ be the group of homeomorphism of $\ K,\ $ induced by shifts $\ s_n\ (n\in\Bbb Z)\ $ of $\ \Bbb Z:\ $ -$$ \forall_{n\in\Bbb Z}\forall_{x\in\Bbb Z}\quad - s_n(x):= x+n $$ -Let $\ p:=(p_n)\in K\ $ be arbitrary, and let -$$ P:=\{p_n:n\in \Bbb Z\} $$ -Then there exists non-empty open $\ U\ $ in $\ X,\ $ disjoint with set $\ P.\ $ Then non-empty open in $\ K\ $ -set $\ W,$ -$$ W\ :=\ \pi_0^{-1}(U)\ $$ -is disjoint with the orbit $\ p.\ $ -On the other hand, let $\ \emptyset\ne G\subseteq K, $ -where $\ G\ $ is open in $\ K.\ $ Then there exists -non-empty $\ H\ $ and integer $ a\ge 0\ $ such that -$\ H\ $ is an open subset of $\ X^{(-a)..a}\ $ (Perl notation "s..t") and -$$ \emptyset\ \ne\ \pi_{(-a)..a}^{-1}(H)\ \subseteq G $$ -Obviously, the orbit of $\ \pi_{(-a)..a}^{-1}(H)\ $, hence of $\ G,\ $ is dense in $\ K.$ -Great!<|endoftext|> -TITLE: Strong chains in $[\omega_2]^{\omega_2}$ mod finite of length $\omega_3$ -QUESTION [5 upvotes]: Probing a bit the difference between $[\omega_1]^{\omega_1}$ and $[\omega_2]^{\omega_2}$ modulo the finite sets: - -Question -Can there exist a family $\langle X_\alpha:\alpha<\omega_3\rangle$ of -sets in $[\omega_2]^{\omega_2}$ such that for $\alpha<\beta<\omega_3$, - -$|X_\beta\setminus X_\alpha|=\omega_2$, and -$ X_\alpha\setminus X_\beta$ is finite? - - -The existence of such a family (a strong chain of length $\omega_3$ in $[\omega_2]^{\omega_2}$ mod finite) implies the existence of a strongly almost disjoint family in $[\omega_2]^{\omega_2}$ of size $\omega_3$: let $A_\alpha:=X_{\alpha+1}\setminus X_\alpha$, and we get a collection of size $\omega_3$ in $[\omega_2]^{\omega_2}$ with pairwise finite intersection. The existence of such a family is consistent, as shown by Baumgartner [1]. -Koszmider [2] showed that you can have long chains (of length $\omega_2$) in $[\omega_1]^{\omega_1}$ modulo the finite ideal, so this question is an attempt to lift his result to $[\omega_2]^{\omega_2}$. (I do not even know the answer for $[\omega_2]^{\omega_2}$ if ``finite'' is replaced by countable, but obviously a consistency result there should be easier to obtain, I just don't know if it's been done.) -There are differences between $\omega_1$ and $\omega_2$ that indicate this question may be interesting. Koszmider [3] showed that one can have a sequence $\langle f_\alpha:\alpha<\omega_2\rangle$ that is strictly increasing mod finite, while Shelah [4] has shown in ZFC that this phenomenon CANNOT happen at $\omega_2$: there is no sequence $\langle f_\alpha:\alpha<\omega_3\rangle$ of functions in $^{\omega_2}\omega_2$ that is strictly increasing mod finite. -[1] Baumgartner, James E., Almost-disjoint sets, the dense set problem and the partition calculus, Ann. Math. Logic 9, 401-439 (1976). ZBL0339.04003. -[2] Koszmider, Piotr, On the existence of strong chains in $(\wp(\omega_1)/ \text{Fin})$, J. Symb. Log. 63, No. 3, 1055-1060 (1998). ZBL0936.03043. -[3] Koszmider, Piotr, On strong chains of uncountable functions, Isr. J. Math. 118, 289-315 (2000). ZBL0961.03039. -[4] Shelah, Saharon, On long increasing chains modulo flat ideals, Math. Log. Q. 56, No. 4, 397-399 (2010). ZBL1200.03031. - -REPLY [3 votes]: No, there can't. The argument is not too different from Shelah's. I emailed you a note.<|endoftext|> -TITLE: Real manifolds and affine schemes -QUESTION [18 upvotes]: I noticed the following strange (to me) fact. If $M$ is a real manifold (smooth or not) and $R = C(X, \mathbb{R})$ is the ring of real functions (smooth functions in the smooth case) then the affine scheme $X = \mathrm{Spec}(R)$ has a natural map $M \to X$ which is a homeomorphism on real points i.e. $M \to X(\mathbb{R})$ is a homeomorphism. Even better, in the smooth case, we can identify $M$ with the ringed space $(X(\mathbb{R}), \mathcal{O}_X)$. Even more better, the functor $M \mapsto \mathrm{Spec}(C(M, \mathbb{R}))$ from the category of smooth manifolds to the category of affine schemes is fully faithful. -Add to this the Serre-Swan theorem which states that there is an equivalence between the category of vector bundles on $M$ and the category of finite projective $R$-modules i.e. vector bundles on $X$. -These facts seem to imply that smooth manifolds may be thought of "as" affine schemes. This observation leads me to ask the following questions: -(1) Do you know of any fruitful consequences or applications of looking at manifolds in this light? -(2) Is there anywhere this identification fundamentally fails? -(3) Is there an algebraic classification for what these rings look like? In particular, if $A$ is an $\mathbb{R}$-algebra then when is $X(\mathbb{R})$ a topological manifold and when can $X(\mathbb{R})$ be given a smooth structure compatible with $\mathcal{O}_{X}$? -(4) What do the "extra" points of $X$ look like? Is there a use for these extra points in manifold theory, the way that generic points have become important in algebraic geometry? -For question (4), I believe that maximal ideals of $R$ should correspond to ultrafilters on $M$ identifying the closed points of $X$ with the Stone-Cech compactification of $M$. What about the other prime ideals? -Many thanks. - -REPLY [17 votes]: (1) This is a highly productive way of looking at smooth manifolds. -It is responsible for synthetic differential geometry and derived smooth manifolds. -Both of these subjects heavily rely on this identification. -Synthetic differential geometry looks at spectra of finite-dimensional -real algebras, and interprets these as geometric spaces of infinitesimal shape. -This allows one to make infinitesimal arguments of the type used by -Élie Cartan and Sophus Lie perfectly rigorous. -(Text)books have been written about this approach: - -Anders Kock: Synthetic differential geometry; -Anders Kock: Synthetic geometry of manifolds; -René Lavendhomme: Basic concepts of synthetic differential geometry; -Ieke Moerdijk, Gonzalo Reyes: Models for smooth infinitesimal analysis. - -Derived smooth manifolds start by enlarging the category of real algebras -of smooth functions to a bigger, cocomplete category -(such as all real algebras, or, better, C^∞-rings, see below). -One then takes simplicial objects in this category of algebras, -i.e., simplicial real algebras. -This category admits a good theory of homotopy colimits, -which are given by derived tensor products and other constructions -from homological algebra. -The opposite category of simplicial real algebras -should be thought of as a category of geometric spaces of some kind. -This category has an excellent theory of homotopy limits, -in particular, one compute correct (i.e., derived) intersections -of nontransversal submanifolds in it, and get expected answers. -For example, this can be used to define canonical representative -of characteristic classes, one could take the Euler class to be -the derived zero locus of the zero section, for example. -Other (potential) applications include the Fukaya ∞-category, -which in presence of nontransversal intersections can be turned -into a category in a more straightforward way. -Dominic Joyce wrote a book about this: - -Dominic Joyce: Algebraic geometry over C^∞-rings. - -Other sources: - -David Spivak: Derived smooth manifolds. -Dennis Borisov, Justin Noel: Simplicial approach to derived differential manifolds. -David Carchedi, Dmitry Roytenberg: On theories of superalgebras of differentiable functions. -David Carchedi, Dmitry Roytenberg: Homological algebra for superalgebras of differentiable functions. -David Carchedi, Pelle Steffens: On the universal property of derived smooth manifolds. - -(2) Yes, for example, Kähler differentials are not smooth differential 1-forms, -even though derivations are smooth vector fields. -This is (one of) the reasons for passing to C^∞-rings instead. -In the category of C^∞-rings, C^∞-Kähler differentials are precisely -smooth differential 1-forms. -(3) Not really, this question was discussed here before, -here is one of the discussions: Algebraic description of compact smooth manifolds?<|endoftext|> -TITLE: A smooth map on a Banach manifold whose pointwise rank is finite but its rank is not globally bounded -QUESTION [6 upvotes]: Is there a connected Banach manifold $M$ and a smooth map $f:M \to M$ such that the rank of $Df_x$ is finite for every $x\in M$ but this rank is not uniformly bounded - -REPLY [5 votes]: Let $\mathbb H$ and $(e_n)_{n\geq0}$ be your favourite second countable Hilbert space and orthonormal Hilbert basis. Mine is the space $\ell^2(\mathbb N)$ of square integrable sequences, with $e_n$ the indicator function of the singleton $\lbrace n\rbrace$. I write, for any $x\in\mathbb H$, $x^n$ for the $n$th coordinate $\langle e_n,x\rangle$ of $x$. -Choose a smooth function $\phi:\mathbb R\to\mathbb R$ such that $\phi$ is identically zero over $(-\infty,1/2)$ and $\phi'(1)\neq0$. Then -$$f:x\mapsto\sum_{n\geq0} \phi(x^n)e_n$$ -is a function as you describe. Indeed, - -if $|x^n|<1/3$ for all $n\geq N$, then in a neighbourhood of $x$ the condition still holds with $1/2$ instead of $1/3$, and the function $f$ has values in the finite dimensional space generated by $(e_0,\cdots,e_{N-1})$ hence its differential has finite rank; -since $x$ can have $|x^n|\geq1/3$ for only a finite number of $n$, $f$ is well-defined and smooth; -the differential of $f$ at $x=e_0+\cdots+e_N$ is -$$Df_x:u\mapsto\sum_{0\leq n\leq N} \phi'(1)u^ne_n,$$ -which has rank $n$. - -Just for fun, here is a stronger topological result. Say that $f$ has finite rank if $Df_x$ has finite rank for all $x$, and bounded rank if the rank of $Df_x$ is bounded uniformly in $x$. We say that $x$ is a nice point for $f$ if there exists a neighbourhood of $x$ over which $f$ has bounded rank, otherwise we say it is bad. - -Theorem -Let $E$ be a subset of $\mathbb H$. Then there exists $f:\mathbb H\to\mathbb H$ of finite rank whose set of nice points is exactly $E$ if and only if $E$ is open dense. - -This makes it possible to have many bad points, for instance there exists an $f$ with finite rank such that for all $\varepsilon>0$, there exists $x$ in the unit ball such that $B(x,\varepsilon)$ contains infinitely many bad points. -The direct implication is a consequence of the Baire category theorem. Indeed, the set $E_n$ of points $x$ such that $Df_x$ has rank at most $n$ is closed, and the set $E=\bigcup_{n\geq0}\operatorname{int}E_n$ of points where the rank is locally bounded is clearly open. To show that $E$ is dense, fix a non-empty open set $U$. Since $f$ has finite rank, $\bigcup_{n\geq0}(E_n\cap U)=U$, hence one of the $E_n\cap U$ must have non-empty interior, which is the same as saying that $\operatorname{int}E_n\cap U$ is non-empty. -In the other direction, this answer shows that in a Hilbert space, any closed set is the vanishing set of some smooth real-valued map with all derivatives uniformly bounded ($\forall k,\exists M,\forall x,\|D^k\rho_x\|\leq M$). Let $d:\mathbb H\to\mathbb R$ be the distance to $E^\complement$; note that it is continuous. Let $E_0$, resp. $E_n$ for $n>0$, be the inverse image by $d$ of $(1,+\infty)$, resp. $(\frac1{n+1},\frac2n)$. Obviously, $E_n$ is open for all $n$ and $E$ is the union of the $E_n$. Moreover, since $E$ is open dense, $x\in E$ if and only if there exists a neighbourhood of $x$ that intersects finitely many of the $\overline{E_n}$. -Let $\rho_n:\mathbb H\to\mathbb R$ be a smooth function with zero set $E_n$ and all derivatives uniformly bounded. Then -$$ x\mapsto \rho_n(x)\sum_{k\leq i\leq \ell}x^ie_i $$ -has rank zero in the complement of $\overline{E_n}$, and at least $k-\ell$ on $E_n$ (because the derivative of $\rho_n$ may only “kill” at most one direction in the image of the projector). Moreover, all its derivatives are uniformly bounded. It means that there exists a sequence $(\varepsilon_n)_{n\geq0}$ of positive numbers such that all the derivatives of the series -$$ f:x\mapsto\sum_{n\geq0}\varepsilon_n\rho_n(x)\left(x^{n^2+1}e_{n^2+1}+\cdots+x^{(n+1)^2}e_{(n+1)^2}\right) $$ -converge uniformly over $\mathbb H$, so that $f$ is well-defined and smooth (see the reference above for more details). -The rank of $Df_x$ is at least $2n$ but finite over $E_n$, and the rank of $Df_x$ is zero (hence finite) at all points $x$ of $E^\complement$. Moreover, $x$ is nice for $f$ if and only if one of its neighbourhoods intersects only finitely many $\overline{E_n}$, if and only if $x\in E$.<|endoftext|> -TITLE: Projective/injective object in functor category -QUESTION [7 upvotes]: Let $\mathcal{C}$ denote the functor category $Fun(\textbf{Man} , \textbf{Ab})$, where $\textbf{Man}$ and $\textbf{Ab}$ denote the category of smooth manifolds and abelian groups respectively. I want to know what are the projective objects or injective objects in this category $\mathcal{C}$? More precisely, I am interested in resolving objects of $\mathcal{C}$ by either projective or injective objects. So for that I need to know what are these objects? Any kind of partial help or comments would be great. -The approach which I thought of is the following: If we can find a category $\mathcal{D}$ with two functors from $ F : \mathcal{C} \rightarrow \mathcal{D}$ and from $ G : \mathcal{D} \rightarrow \mathcal{C}$ such that both are in adjunction. Then under some conditions using the unit or co-unit map, we can resolve objects. - -REPLY [2 votes]: One way to construct injectives on a presheaf category $[\mathscr C^{\operatorname{op}},\mathbf{Ab}]$ is to consider the forgetful functor -$$i^* \colon \big[\mathscr C^{\operatorname{op}},\mathbf{Ab}\big] \to \big[\mathscr C^{\operatorname{disc,op}},\mathbf{Ab}\big]$$ -induced by the inclusion $i \colon \mathscr C^{\operatorname{disc}} \to \mathscr C$ (where $\mathscr C^{\operatorname{disc}}$ is the subcategory with only identity morphisms). If $\mathscr C$ is small, then $i^*$ has left and right adjoints $i_!$ and $i_*$ given by -\begin{align*} -\big(i_! \mathscr F\big)(c) = \bigoplus_{c' \to c} \mathscr F(c'),\\ -\big(i_* \mathscr F\big)(c) = \prod_{c \to c'} \mathscr F(c'). -\end{align*} -In particular, $i^*$ is an exact left adjoint to $i_*$, so $i_*$ takes injectives to injectives [Stacks, Tag 015N]. But in $[\mathscr C^{\operatorname{disc,op}},\mathbf{Ab}]$ injectives are computed pointwise, so this gives a recipe to construct injectives in $[\mathscr C^{\operatorname{op}},\mathbf{Ab}]$. -See for example [Stacks, Tag 01DJ] for a brief discussion, or [SGA IV$_1$, Exp. I, Prop. 5.1] for a more general discussion of adjoints (but without the mention of injectives). -In general the colimit for $i_!$ is taken over the opposite of the comma category $(i \downarrow c)$ (whose objects are $(i(c') \to c)$), which in this case is just a discrete category since $\mathscr C^{\operatorname{disc}}$ is, so we get a direct sum; similarly for $i_*$. - -References. -[SGA IV$_1$] M. Artin, A. Grothendieck, J.-L. Verdier, Séminaire de géométrie algébrique du Bois-Marie 1963–1964. Théorie de topos et cohomologie étale des schémas (SGA 4), 1: Théorie des topos. Lecture Notes in Mathematics 269. Springer-Verlag (1972). ZBL0234.00007. -[Stacks] A.J de Jong et al, The stacks project.<|endoftext|> -TITLE: Support of closed random walk on $\mathbb Z$ -QUESTION [7 upvotes]: I am researching closed random walks on graphs and have the following problem that I haven't been able to find a reference for. -Consider a random walk on $\mathbb Z$ starting at 0 and at each step it moves $-1$ or $+1$ each with probability $1/2$. If the walk has length $2n$ it is well-known that the support (or how many elements of $\mathbb Z$ that are covered by the walk) is $\Theta(\sqrt n)$ with high probability. -Suppose now that our walk is closed, i.e. we condition on that the walk starts and ends at 0. Is it still the case that the walk has support $\Theta(\sqrt n)$ with high probability? -I would be happy if I could just show that the support is at least $\Omega(n^{\varepsilon})$ for some constant $\varepsilon>0$. - -REPLY [4 votes]: One way to do this is as follows. We have to show that -$$P(M_n\ge x|S_n=0)\to1$$ -(as $n\to\infty$) if $x=o(\sqrt n)$, where $S_n$ is the position of the walk at time $n$ and $M_n:=\max_{0\le k\le n}S_k$. By the reflection principle (see e.g. Theorem 0.8) and the de Moivre--Laplace theorem , for natural $x$ such that $x=o(\sqrt n)$, -$$P(M_n\ge x,S_n=0)=P(S_n=2x)\sim P(S_n=0),$$ -whence -$$P(M_n\ge x|S_n=0)=\frac{P(M_n\ge x,S_n=0)}{P(S_n=0)}\to1,$$ -as desired.<|endoftext|> -TITLE: Completeness hypothesis in the positive mass theorem -QUESTION [7 upvotes]: I am trying to understand and further formalize Witten's proof of the positive mass theorem. Dan Lee, in his book "Geometric relativity" did a wonderful job with formalizing and carrying out the details of Parker and Taubes' work, which was already a formalization of Witten's work. -The statement of the theorem in his book is more or less the following: - -Theorem: - Let $(N,g)$ be a complete asymptotically euclidean spin $n$-manifold with nonnegative scalar curvature and $n \geq 3$. - Suppose further that $N$ has a well defined ADM mass. - Then the ADM mass of each end is nonnegative. - Moreover, if the mass of any end is zero, then $(N,g)$ is globally isometric to euclidean space. - -I do not particularly like the completeness hypothesis as in most cases of interest in physics, the manifold is not complete. -Therefore I am wondering why the completeness hypothesis is necessary. -The only place I can find in the proof in his book where the completeness hypothesis is used explicitly is for positive mass rigidity, that is to say to prove that if the mass of any end is zero, then $(N,g)$ is globally isometric to euclidean space. -The completeness hypothesis is almost never stated in other surveys. -Parker and Lee, in their survey on the Yamabe problem state the theorem as follows: - -Theorem: - Let $(N,g)$ be an asymptotically flat Riemannian manifold of dimension $n \geq 3$ such that the ADM mass is well defined, and with nonnegative scalar curvature. - Then its mass $m(g)$ is nonnegative, with $m(g) = 0$ if and only if $(N, g)$ is isometric to $\mathbb{R}^n$ with its Euclidean metric. - -The positive mass rigidity part in this theorem is plainly false, as $\mathbb{R}^n \setminus \{0\}$ satisfies all hypotheses but is not isometric to $\mathbb{R}^n$, so for this part the completeness is necessary. -However, without completeness it is possible to prove that the manifold has to be flat. -Hence I think the following theorem is also true: - -Theorem: - Let $(N,g)$ be an asymptotically euclidean spin $n$-manifold with nonnegative scalar curvature and $n \geq 3$. - Suppose further that $N$ has a well defined ADM mass. - Then the ADM mass of each end is nonnegative. - Moreover, if the mass of any end is zero, then $(N,g)$ is flat. - -Can someone confirm this? - -REPLY [7 votes]: Completeness is necessary. Otherwise you can just take a maximal spatial slice of the negative-mass Schwarzschild solution (i.e. a constant $t$ slice in Boyer-Lindquist coordinates) and it has vanishing scalar curvature with $m < 0$. There are however no complete maximal spatial sections of the negative-mass Schwarzschild solution. - -For Lee and Parker, the implicit assumption of completeness is harmless: for applications to the Yamabe problem your asymptotically flat manifold comes from stereographic projection of a compact (closed) manifold about a point, so is always complete. -In the Witten proof included in the Appendix to Lee and Parker, completeness is used in the integration by parts argument equation (A.5). Here $N_R$ is assumed to be a compact manifold with boundary $S_R$. If your original manifold were not complete, but asymptotically flat, then for sufficiently large $R$ there must be another component of the boundary $N_R$ corresponding to the ends of the terminating geodesics. - -Finally, if you make your definitions carefully, then incomplete manifolds are also ruled out of the Lee and Parker statement. -Recall that their definition of Asymptotically Flat manifold starts with a Riemannian manifold $N$ which is decomposed into some non-compact ends each of which are asymptotically flat in the usual sense, plus a compact portion. -If you start with an incomplete Riemannian manifold (here we are careful in having "manifolds with boundary" being not "manifolds"), and take away the asymptotically flat ends, you will necessarily still be left with something non-compact due to the incompleteness. (Incompleteness will always leave the manifold [which by definition cannot include a boundary] non-compact [Cauchy sequence alone the incomplete geodesic], and finite-length means it cannot be coming from an AF end.) So "complete" can be regarded as implicit.<|endoftext|> -TITLE: Is limit of null-homotopic maps null-homotopic? -QUESTION [10 upvotes]: The question is motivated by my failed comment to this one. -Let $M$ and $N$ be path connected locally compact, locally contractible metric spaces (you may assume that they are manifolds). -Let $\varphi_{n}:M\to N$ be null-homotopic and convergent to $\varphi:M\to N$ in the compact open topology. - -Does it follow that $\varphi$ is null-homotopic? - -Note that homotopy between maps is a path in $C(M,N)$ (for nice $M$, $N$), and so what my question asks is whether the path component of a constant map is closed in $C(M,N)$. -I am waaay out of my depth here, but perhaps there is a continuous or positive lower semi-continuous functional on $C(M,N)$ akin to the topological degree such that null-homotopic maps would be the zero-set of that functional? - -REPLY [2 votes]: Please see the answer to Annie's question. -Non-density of continuous functions to interior in set of all continuous functions<|endoftext|> -TITLE: How to find the associated conservation law from a given symmetry -QUESTION [7 upvotes]: It is a very well-known fact that any conservation law associated with some given PDE has an associated invariance (by Noether's Theorem). However, it is completely mysterious for me how to compute/derive these conservation laws just by knowing the invariances of the equation. For example, the one-dimensional nonlinear wave equation $$ -u_{tt}-u_{xx}+f(u)=0, \qquad (t,x)\in\mathbb{R}\times\mathbb{R}, -$$ -is invariant under space translations. On the other hand, it is "well-known" that associated to this space translation invariance is the momentum conservation of the equation, that is, $$ -P(u,v)(t):=\int_{\mathbb{R}} u_x(t,x)v(t,x)dx=\int_{\mathbb{R}}u_{0,x}(x)v_0(x)dx=P(u,v)(0). -$$ -Nevertheless, I have no idea how to derive this conservation law (generally speaking) just by knowing that the equation is invariant under space translations. What about time-translations for example, what is its associated conservation law? Please, don't misunderstand me, I do know how to explicitly derive the momentum conservation directly from the equation, what I would like to know is how to derive it from the space-translations invariance. Any hint suggested reading or answer is very welcome! - -REPLY [8 votes]: To be blunt, the answer to your question is Noether's theorem (often precised as Noether's first theorem). So, essentially you already knew the answer to your own question. -However, the other answers are missing a degree of pragmatism. The calculation of the conserved current, once you know the Lagrangian and the symmetry is straightforward and mechanical. Namely, suppose you have a Lagrangian density $L[\phi] = L(x,\phi(x),\partial \phi(x), \partial^2\phi(x), \ldots)$, which depends your dynamical field $\phi(x)$. The variational principle will be $S(\phi) = \int L[\phi] \, \mathrm{d}x$, where $\mathrm{d}x$ is the coordinate volume form.1 An infinitesimal local field transformation $\phi^a \mapsto \phi^a + \delta_{\xi}\phi^a$ is allowed to be coordinate and field dependent, $\delta_\xi \phi^a = \xi^a[\phi] = \xi^a(x,\phi(x), \partial \phi(x), \partial^2 \phi(x), \ldots)$, and commutes with coordinate derivatives, namely $\delta_\xi \partial^n \phi^a = \partial^n (\delta_\xi \phi^a) = \partial^n \xi^a[\phi]$ for any $n\ge 0$. The example of time translation $\xi^a[\phi] = \frac{\partial}{\partial t} \phi^a$ is illustrative. -Such a local field transformation is a symmetry of the Lagrangian when its variation vanishes modulo a total divergence, $\delta_\xi L[\phi] = \partial_i J_0^i[\phi]$. The next step is a bit unintuitive, but it makes the calculation of the conserved current mechanical. Consider now the variation $\delta_{\varepsilon \xi}$, where $\varepsilon = \varepsilon(x)$ is an arbitrary function of the coordinates $x^i$. Using integration by parts, we can put the variation of the Lagrangian into the form -$$ \tag{$*$} - \delta_{\varepsilon \xi} L[\phi] - = \varepsilon\partial_i J^i_0[\phi] + (\partial_i\varepsilon) J^i_1[\phi] + \partial_i(-)^i . -$$ -The leading term has to agree with $\delta_\xi L[\phi]$ when we set $\varepsilon \equiv 1$. The desired conserved current corresponding to $\xi$ is -$$ J_\xi^i[\phi] = J_0^i[\phi] - J_1^i[\phi] . $$ -You can get the current in one step if you use integration by parts to directly put the variation of the Lagrangian into the form $\delta_{\varepsilon \xi} L[\phi] = -J_\xi^i[\phi] (\partial_i \varepsilon) + \partial_i(-)^i$, which is a formula that can be found in some physics textbooks on QFT. -The proof of Noether's theorem in this form is also straightforward (and a reshuffling of the standard proof). It only relies on the usual lemma that any density $N[\varepsilon, \ldots]$ that linearly depends on an arbitrary function $\varepsilon = \varepsilon(x)$ (and possibly any other fields) has a unique representative modulo total divergence terms, namely $N[\varepsilon, \ldots] = \varepsilon N_0 + \partial_i(-)^i$, with $N_0$ unique. The Euler-Lagrange equations $E_a[\phi]=0$ are defined by the identity $\delta_\xi = \xi^a E_a[\phi] + \partial_i(-)^i$ for arbitrary $\xi$. So, when $\xi$ is a symmetry, using $(*)$ and one more integration by parts, we find the identity -$$ - \delta_{\varepsilon \xi} L[\phi] - = \varepsilon \xi^a E_a[\phi] + \partial_i(-)^i - = \varepsilon \partial_i J^i_\xi[\phi] + \partial_i(-)^i , -$$ -which implies that $\partial_i J^i_\xi[\phi] = \xi^a E_a[\phi]$, which vanishes when $E_a[\phi] = 0$. In other words, $J^i_\xi[\phi]$ is a conserved current. - -1 If you change the independent coordinates $x^i$, the Lagrangian will change by the appropriate Jacobian. Working with differential forms allows you to keep everything more manifestly invariant.<|endoftext|> -TITLE: How do mathematicians find coauthors? -QUESTION [17 upvotes]: I am totally new to academia so I am really not sure how mathematicians works together, can more experienced mathematicians here shed some light on how you find coauthors? I guess one way to do this is to attend conference. But if one doesn't have the chance to do so, are there other options? - -REPLY [13 votes]: If you want to find a coauthor, go to a pub or a conference dinner, find a nice person, use a pick up line, something like"Do you know that the derived category of an abelian category is also abelian?", After you find common interests, -you propose " Would you be my coauthor?". If the answer is "yes" you are done. If the answer is"no way" or "I already have a coauthor", find some other person or go to another conference dinner.<|endoftext|> -TITLE: A question on the Riemann zeta function -QUESTION [9 upvotes]: Yesterday, a certain very talented and passionate young student from Southern Africa asked me the following question about the Riemann zeta function $\zeta(s)$. He says he "thinks" he knows the answer, but he just wants to hear my views. However, I'm not a number theorist, hence I couldn't answer him. So below is the question: -Consider the Riemann zeta function $\zeta(s)$, and let $\alpha$ be the supremum of the real parts of its zeros. Let $\mu$ denote the Möbius function. Define $S(x)= \sum_{n\leq x} \frac{\mu(n)\log n}{n}$. -Note that -$$\Big(\frac{1}{\zeta(s+1)}\Big)' = -s \int_{1}^{\infty} S(x)x^{-s-1} \mathrm{d}x$$ for $\Re(s)> \alpha-1$, where the prime denotes differentiation. It is known that $S(x)=-1 + o(1)$, thus the above integral converges if and only if $\Re(s)>0$. The student's question is: what does this tell us, if anything, about the value of $\alpha$ ? -PS: Personally, i couldn't verify the above identity, neither could I verify the "known" result that $S(x) = -1 + o(1)$, hence I couldn't answer his question. - -REPLY [7 votes]: The identity -$$\sum_{n=1}^\infty\frac{\mu(n)\log n}{n}=-1$$ -was conjectured by Möbius (1832) and proved by Landau (1899). It is a consequence of the prime number theorem. Not surprisingly, the rate of convergence is determined by the (known) zero-free region of $\zeta(s)$. In particular, -$$S(x)=-1+O_\epsilon(x^{\alpha-1+\epsilon})$$ -holds for any $\epsilon>0$, and $\alpha$ in the exponent cannot be lowered. -Here is a sketch of the proof of the mentioned facts. By Perron's formula, we have (at least for $x\not\in\mathbb{N}$) -$$S(x)=\frac{1}{2\pi i}\int_{1-i\infty}^{1+i\infty}\left(\frac{-1}{\zeta(s+1)}\right)' \frac{x^s}{s}\,ds.$$ -The integration is meant over the vertical line with abscissa $1$. By truncating the integral at some height, and applying the residue theorem appropriately, we can move the line segment of integration to the left with the benefit of $x^s$ being much smaller there. This is the same technique by which the prime number theorem was originally proven. At $s=0$, the derivative inside the integral equals $-1$, while $x^s/s$ has a simple pole with residue $1$. Therefore, as we move the curve of integration to the left of $s=0$, we pick up the main term $-1$. The error term then depends on how far to the left we can move the curve of integration without encountering further poles, i.e. where the zeros of $\zeta(s+1)$ are located. The standard zero-free region already implies my first display. If $\alpha<1$, then we have a much wider zero-free region, and the second display follows. The fact that the exponent is optimal follows by reversing this logic, namely by examining the analytic continuation of the RHS of the OP's formula to the left of $s=0$. -I hope this helps your student, or perhaps this is exactly what she/he had in mind. It is standard material, but a good way to better understand the prime number theorem and its relation to the zeros of $\zeta(s)$.<|endoftext|> -TITLE: Singular Sturm-Liouville problems: criterion for discrete spectrum for zero potential ($q=0$) and Hermite Polynomials -QUESTION [7 upvotes]: There are some known criteria for the Sturm-Liouville Problem -\begin{equation} \tag{1} -\frac {\mathrm {d} }{\mathrm {d} x}\left[p(x){\frac {\mathrm {d} y}{\mathrm {d} x}}\right]+q(x)y=-\lambda w(x)y -\end{equation} -to have a spectrum discrete and bounded below (BD) in the singular case: -If singular endpoints are Limit-Circle and Non-Oscillating (LCNO), results from -(Niessen, H.-D. and Zettl, A. (1992). Singular Sturm-Liouville Problems: The Friedrichs extension and comparison of eigenvalues.) -apply. -There also is the famous criterion by Molchanov and generalizations for $w, p \neq 1$ in (Kwong and Zettl - Discreteness conditions for the spectrum of ordinary differential operators). Various other generalizations exist. -As I understand it, the Molchanov criterion and its generalizations are not applicable to the case $q = 0$ or $q$ constant. Inherently, $q$ must run to infinity at the endpoint in a certain way (details depending on $p$ and $w$). (See related questions For which type of potentials a Schrödinger operator will have discrete spectrum?, Harmonic oscillator discrete spectrum) -My questions: -Is there a theory and criteria for (1) to be BD for the case $q = 0$ if endpoints are in the Limit-Point (LP) case? I specifically appreciate pointers to literature. -Or do I overlook some tweak, allowing to apply Molchanov-style criteria to the case $q = 0$? -What I read and reasoned so far: -As far as I can assess, transformations to normal form are not suitable to work around this particular issue. Also, according to Zettl (Zettl, A. (2010). Sturm-Liouville Theory), it is questionable whether such a transformation preserves the desired properties of the spectrum (discussion in Chapter 10.13). -I looked into various books and papers and googled around, but I almost exclusively found Molchanov-style criteria, most usually setted for the half-line $[a > -\infty, \infty)$. It appears that Schrödinger operators are the main interest in this field and the Molchanov criterion is a good fit for them. I am more interested in self-adjoint operators similar to the one yielding Hermite polynomials: -\begin{equation} \tag{2} -q = 0, \quad w = p = e^{-{\frac {x^{2}}{2}}} -\end{equation} -on the real line with two singular LP endpoints. I used to perceive Hermite Polynomials as an important standard example and it puzzles me that it seems to be so hard to find criteria for their spectrum. I mean criteria that are not totally specific to equation (1) with values (2), but also hold if some other (normal-like?) probability distribution is used in (2), e.g. with adjusted variance, skewness, kurtosis etc. (not expecting the solutions to be (orthogonal) polynomials anymore, but still to have BD spectrum). (Edit: It occurred to me that the case of (2) with adjusted variance is known as the generalized Hermite polynomials.) -E.g. I skimmed through some books by L. L. Littlejohn and A.M. Krall about orthogonal polynomials in context of singular SL theory, e.g. (L. L. Littlejohn and A.M. Krall, Orthogonal polynomials and singular Sturm-Liouville Systems, I) but they do not seem to discuss the discreteness of the spectrum (sorry if I overlooked it, pointers welcome). -Secondary question: -What about the full line vs. the Half line? -It appears that in his original work, Molchanov considered the full line (A.M. Molchanov, Conditions for the discreteness of the spectrum of self-adjoint second-order differential equations.) however, I cannot read Russian and only got this impression from the math terms. Also (Inge Brinck, Self-Adjointness and spectra of Sturm-Liouville operators) assumes the full line, but (Kwong and Zettl - Discreteness conditions for the spectrum of ordinary differential operators) and (D. Hinton, Molchanov’s discrete spectra criterion for a weighted operator) assume the half line. -This gives me the impression that the distinction between full line and half line is not essential, but I did not find it discussed anywhere. A pointer to literature that specifically discusses in what sense results are transferable between the half line and full line case would be welcome. (Niessen, H.-D. and Zettl, A. (1992). Singular Sturm-Liouville Problems: The Friedrichs extension and comparison of eigenvalues.) discuss it a bit, but not in context of criteria for BD in the LP case (or I don't understand how to conclude it). -Thanks in advance, also for hints, comments and improvements! - -REPLY [3 votes]: The case $q=0$ actually has a complete answer, and the reason this works is that we can solve the equation for $\lambda =0$ explicitly then, by $u=1$ and $v=\int_0^x \frac{dt}{p(t)}$. The spectrum is purely discrete if and only if ($w\in L^1$ and) -$$ -\lim_{x\to\infty}\int_0^x wv^2\, dt \int_x^{\infty} w\, dt =0 . \quad\quad\quad\quad (1) -$$ -(This holds in your example, if I managed the asymptotics of the error functions correctly.) -I obtained this by rewriting the SL equation as a canonical system $Jy'=-\lambda Hy$, and the coefficient matrix $H$ we obtain here is given by -$$ -H = w\begin{pmatrix} 1 & v \\ v & v^2 \end{pmatrix} . -$$ -There is a precise criterion for purely discrete spectrum for (general) canonical systems, and this gave me (1). -This will probably all sound quite cryptic if you haven't seen these things before, and there seem to be too many small details to explain them all here. You could take a look at my papers 1, 2 on my homepage for full background information. -Also, if I hadn't had that available, I would have tried to analyze the Prufer equation -$$ -\varphi' = \frac{1}{p}\cos^2\varphi + (\lambda w+q)\sin^2\varphi -$$ -for the phase of the solution vector $(y,py')$. Purely discrete spectrum is equivalent to $\varphi(x)$ staying bounded in $x$ for all $\lambda$. This might also be good enough to get (1).<|endoftext|> -TITLE: Irrationality measure of arctan(1/3) -QUESTION [8 upvotes]: I recently came across the concept of the irrationality measure. It really fascinated me and when I was looking for known values $\mu(x)$ for mathematical constants $x$, I also came across this paper: On the irrationality measure of arctan 1/3 -Unfortunately it could't help me out quite well. -What is the irrationality measure of $\arctan(1/3)$? Are there upper and lower bounds? Which famous constant have known irrationality measures or upper/lower bounds? (Except for the ones, that can easily be found on mathworld e.g. $\pi, \ln(2),\ln(3), \pi^2, \zeta(3)$)? - -REPLY [8 votes]: As mentioned by GH from MO, the irrationality measure -for almost all real numbers is 2. However, -computing it for a particular number -is a notoriously difficult problem. For an irrational algebraic number this measure is indeed 2, but this is a pretty hard theorem by Roth for which he got a Fields medal. -Other then that, to my knowledge the only "famous constant" with the known irrationality measure is the number $e$. It is, again, equal to 2 and this is a very old result essentially known to Euler. (It follows from the expansion of $e$ into a continued fraction.) -For other "famous", or even not that famous, constants only upper bounds are known. -[EDIT] There is actually a pretty good answer here on MO which I missed somehow: -Numbers with known irrationality measures?<|endoftext|> -TITLE: What is the precise relationship between pyknoticity and cohesiveness? -QUESTION [24 upvotes]: Pyknotic and condensed sets have been introduced recently as a convenient framework for working with topological rings/algebras/groups/modules/etc. Recently there has been much (justified) excitement about these ideas and the theories coming from them, such as Scholze's analytic geometry. (Small note: the difference between pyknotic and condensed is essentially set-theoretic, as explained by Peter Scholze here.) -On the other side, cohesion is a notion first introduced by Lawvere many years ago that aims to axiomatise what it means to be a category of "spaces". It has been developed further by Schreiber in the context of synthetic higher differential geometry (and also by Shulman in cohesive HoTT and by Rezk in global homotopy theory, to give a few other names in this direction). -Recently, David Corfield started a very interesting discussion on the relation between these two notions at the $n$-Category Café. The aim of this question is basically to ask what's in the title: - -What is the precise relation between pyknoticity and cohesiveness? - -Along with a few subquestions: - -(On algebraic cohesion) It seems to me that the current notion of cohesion only works for smooth, differential-geometric spaces: we don't really have a good notion of algebraic cohesion (i.e. cohesion for schemes/stacks/etc.) or $p$-adic variants (rigid/Berkovich/adic/etc. spaces). Is this indeed the case? -(On the relevance of cohesion to AG and homotopy theory) Despite its very young age, it's already clear that condensed/pyknotic technology is very useful and is probably going to be fruitfully applied to problems in homotopy theory and algebraic geometry. Can the same be said of cohesion? -(On "condensed cohesion") Cohesion is a relative notion: not only do we have cohesive topoi but also cohesive morphisms of topoi, which recover the former in the special case of cohesive morphisms to the punctual topos. Scholze has suggested in the comments of the linked $n$-CatCafé discussion that we should not only consider cohesion with respect to $\mathrm{Sets}$, but also to condensed sets. What benefits does this approach presents? Is this (or some variant of this idea) a convenient notion of "cohesion" for algebraic geometry? - -REPLY [6 votes]: We have a case of relative cohesion used in an algebraic geometric setting discussed at the nLab. The entry for differential algebraic K-theory interprets - -Ulrich Bunke, Georg Tamme, Regulators and cycle maps in higher-dimensional differential algebraic K-theory (arXiv:1209.6451) - -via cohesion over the base $Sh_\infty\left(Sch_{\mathbb{Z}}\right)$, ∞-stacks over a site of arithmetic schemes. -See also Urs Schreiber's entry, differential cohesion and idelic structure, and arithmetic elements of his research proposal, Higher theta functions and higher CS-WZW holography.<|endoftext|> -TITLE: Map which is null-homotopic on compacts -QUESTION [7 upvotes]: This is the missing ingredient towards answering my previous question. -Let $M$ and $N$ be path connected locally compact, locally contractible metric spaces (you may assume that they are manifolds). It seems the "correct" condition on $N$ is absolute neighborhood retract. Let us also assume that $M$ is $\sigma$-compact, i.e. a union of a sequence of compact sets (and then we can even assume that every compact set in $M$ is contained in an element of that sequence). - -Let $\varphi:M\to N$ be such that for every compact $K\subset M$ the map $\varphi|_{K}$ is null-homotopic. Does it follow that $\varphi$ is in fact null-homotopic? - -The intuition says that if there is a hole in $N$ such that $\varphi$ is wrapped around it, it should be wrapped already on some compact set. -Let me also add a specific case when $\varphi$ is identity map. - -If $N$ is such that the inclusion of every compact $K$ is null-homotopic (meaning $K$ is contractible within $N$), does it follow that $N$ is contractible? - -REPLY [6 votes]: The extent to which the answer to your question is no is analysed by Milnor's exact sequence. You can write $M$ as the colimit of a sequence $M_n \subset M_{n+1}$ of cofibrations with $M_n$ compact (at least if $M$ is a manifold but much more generally). Then there is a "short exact sequence" of pointed sets -$$ -\{1\} \to \textstyle{\lim^1_n} [\Sigma M_n, N]_* \to [M,N]_* \to \lim_n [M_n,N]_* \to \ast -$$ -(in the usual sense that the map of pointed sets on the right is surjective and its fibers are orbits of the action of the group $\lim^1$ which acts on the set in the middle). Brayton Gray used this sequence to construct the example that Mark Grant mentions in the comments above in this paper (since $S^3$ is simply connected there is no difference between pointed and unpointed homotopy classes). -Another reference for the Milnor exact sequence is Bousfield and Kan, Homotopy Limits, Completions and Localizations, Corollary IX.3.3. -Edit Regarding the second question: under the assumptions, $N$ has trivial homotopy groups, i.e. it is weakly contractible. Therefore, if it has the homotopy type of a cell complex (for instance if it is a manifold) then it is contractible.<|endoftext|> -TITLE: Abelian category with enough injectives but not functorially -QUESTION [15 upvotes]: Let $\mathcal{A}$ be an Abelian category with enough injectives. Is it always possible to make the injective embedding functorial? By this I mean that there should exist a functor $I \colon \mathcal{A} \to \mathcal{A}$ and a natural transformation $\operatorname{id} \to I$ such that for all objects $A$, the mapping $A \to I(A)$ is a monomorphism. This should be the same as this definition from the Stacks project. (edit: in a first version, I was requiring the functor to be additive, which is not what I had in mind even in the case of modules, as pointed out by Jeremy Rickard) -The Stacks project distinguishes categories with enough injectives from categories with functorial injective embeddings, so the two notions should be different. But I realized that I cannot think of an example of a category with enough injectives that does not admit functorial injective embeddings. -edit removed a motivation comment that was sparking more discussion than necessary and distracting from the main question. - -REPLY [18 votes]: Since the dual of an abelian category is also an abelian category, the question is equivalent to the same question for projective resolutions. -I will show that the category $\mathbf{Ab}^{\operatorname{f.t.}}$ of finitely generated abelian groups has enough projectives, but no functorial projective cover. The idea is that multiplication by any $n \in \mathbf Z$ is central in $\mathbf{Ab}$, and we do some representation theory to show that $F(\mathbf Z)$ has to have infinite rank by considering the action of multiplication by $n$ on $\mathbf Z/m$ for all $m$. -First some trivial lemmas: -Lemma 1. The category $\mathbf{Ab}^{\operatorname{f.t.}}$ has enough projectives. For $A \in \mathbf{Ab}^{\operatorname{f.t.}}$, the following are equivalent - -$A$ is projective in $\mathbf{Ab}^{\operatorname{f.t.}}$; -$A$ is projective in $\mathbf{Ab}$; -$A$ is finite free. - -Proof. Implications (2) $\Rightarrow$ (1) and (2) $\Leftrightarrow$ (3) are clear. This immediately gives the first statement. For (1) $\Rightarrow$ (3), choose a surjection $F \twoheadrightarrow A$ with $F$ finite free. By assumption (1) it splits, so $A$ is a summand of a finite free module, hence finite free. $\square$ -Lemma 2. Let $F \twoheadrightarrow G$ be an epimorphism of functors $F, G \colon \mathscr C \to \mathscr D$. If $G$ is faithful, then so is $F$. -Proof. Two maps $f, g \colon A \rightrightarrows B$ give a commutative diagram -$$\begin{array}{ccc}F(A) & \twoheadrightarrow & G(A)\\\downdownarrows & & \downdownarrows\\F(B) & \twoheadrightarrow & G(B).\!\end{array}$$ -Since the top map is an epimorphism, we see $F(f) = F(g) \Rightarrow G(f) = G(g)$, which by assumption implies $f = g$. $\square$ -We are now ready for the main result. - -Proposition. Let $F \colon \mathbf{Ab}^{\operatorname{f.t.}} \to \mathbf{Ab}$ be a functor taking every object to a projective object, together with a natural surjection $F \twoheadrightarrow \iota$ onto the inclusion $\iota \colon \mathbf{Ab}^{\operatorname{f.t.}} \to \mathbf{Ab}$. Then $F(\mathbf Z)$ has infinite rank. In particular, there is no such functor landing in $\mathbf{Ab}^{\operatorname{f.t.}}$. - -By Lemma 1, this shows that there is no functorial projective hull on $\mathbf{Ab}^{\operatorname{f.t.}}$. -Proof. First note that Lemma 2 implies that $F$ is faithful, i.e. for all $A, B \in \mathbf{Ab}^{\operatorname{f.t.}}$, the map -\begin{align*} -\operatorname{Hom}(A,B) &\to \operatorname{Hom}(F(A),F(B))\\ -f &\mapsto f_* -\end{align*} -is injective. For any $n > 1$, we can equip every $F(A)_{\mathbf Q} = F(A) \otimes_{\mathbf Z} \mathbf Q$ for $A \in \mathbf{Ab}^{\operatorname{f.t.}}$ with the structure of a $\mathbf Q[x]$-module by letting $x$ act by $n_*$, where $n \colon A \to A$ is multiplication by $n$ (so $x^k$ acts by $(n_*)^k = (n^k)_*$ for $k \geq 0$). For any $m$, the natural surjection $\pi \colon \mathbf Z \to \mathbf Z/m$ gives a commutative diagram -$$\begin{array}{ccc}\mathbf Z & \stackrel{n^k}\to & \mathbf Z \\ \!\!\!\!\!\!{\scriptsize \pi_*}\downarrow & & \downarrow{\scriptsize \pi_*}\!\!\!\!\!\! \\ \mathbf Z/m & \underset{n^k}\to & \mathbf Z/m,\!\end{array}$$ -which by functoriality gives a commutative diagram -$$\begin{array}{ccc}F(\mathbf Z) & \stackrel{n^k_*}\to & F(\mathbf Z) \\ \!\!\!\!\!\!{\scriptsize \pi_*}\downarrow & & \downarrow{\scriptsize \pi_*}\!\!\!\!\!\! \\ F(\mathbf Z/m) & \underset{n^k_*}\to & F(\mathbf Z/m).\!\end{array}$$ -Thus the image of the map $\operatorname{Hom}(\mathbf Z,\mathbf Z/m) \to \operatorname{Hom}_{\mathbf Q}(F(\mathbf Z)_{\mathbf Q},F(\mathbf Z/m)_{\mathbf Q})$ is contained in $\operatorname{Hom}_{\mathbf Q[x]}(F(\mathbf Z)_{\mathbf Q},F(\mathbf Z/m)_{\mathbf Q})$. -If $F(\mathbf Z)$ has finite rank, then $F(\mathbf Z)_\mathbf Q$ has finite length over $\mathbf Q[x]$, hence is supported at finitely many maximal ideals $\mathfrak m \subseteq \mathbf Q[x]$. Since $n$ acts invertibly of order $m$ on $\mathbf Z/(n^m-1)$, the $\mathbf Q[x]$-module $F(\mathbf Z/(n^m-1))_\mathbf Q$ is supported at -$$\mathbf Q[x]\big/\big(x^m-1\big) \cong \prod_{d \mid m} \mathbf Q\big(\zeta_d\big),\tag{1}\label{1}$$ -where $\mathbf Q(\zeta_d)$ is the $d$-th cyclotomic field. Choose $m = p \gg 0$ prime so that $F(\mathbf Z)_\mathbf Q$ is not supported at $\mathbf Q(\zeta_p)$. Then $\operatorname{Hom}_{\mathbf Q[x]}(F(\mathbf Z)_\mathbf Q, F(\mathbf Z/(n^p-1))_\mathbf Q)$ is only supported at $\mathbf Q(\zeta_1) = \mathbf Q[x]/(x-1)$ by (\ref{1}), i.e. the action of $x$ is trivial. But this contradicts faithfulness of $F$: the maps $n^k\pi \colon \mathbf Z \to \mathbf Z/(n^p-1)$ for $k \in \{0,\ldots,p-1\}$ are pairwise distinct, hence the same goes for the $n^k_*\pi_*$. We conclude that $F(\mathbf Z)_\mathbf Q$ cannot have finite length as $\mathbf Q[x]$-module, so $F(\mathbf Z)$ has infinite rank. $\square$<|endoftext|> -TITLE: Can nonstandard fields contain $\mathbb R$ in different ways? -QUESTION [9 upvotes]: Suppose $e : \mathbb R \to F$ is an elementary embedding in the language of ordered fields. Can there exist an elementary embedding $e' : \mathbb R \to F$ such that $e \not= e'$? Note that it would have to be the case that for some undefinable $r \in \mathbb R$, $e(r) \not= e'(r)$, and $e(r) - e'(r)$ is infinitesimal. - -REPLY [2 votes]: In fact $\mathbb{R}$ is elementarily embedded in several ways in any non-archimedean real-closed field which contains it. The proof is more involved than I thought before writing it, but if you don't know the arguments, I think this will make the situation of your question clearer. -Let $K\supset \mathbb{R}$ be a non-archimedean real-closed field. Let $K^{\prec}$ denote the set of infinitesimal elements in $K$, including zero. Those are elements $\varepsilon \in K$ with $\frac{-1}{n+1}<\varepsilon < \frac{1}{n+1}$ for all $n \in \mathbb{N}$. The set of finite elements in $K$, i.e. elements $a$ with $-n -TITLE: Question on Nash's paper on $C^1$ isometric immersions: Why approximating the error tensor $\delta$? -QUESTION [8 upvotes]: I am trying to go through the classical paper by Nash on the existance of $C^1$ isometric immersion of a Riemannian manifold $(M,g)$ (here is the Jstor link: https://www.jstor.org/stable/1969840?seq=1#metadata_info_tab_contents). Let me recall the setup so I can make my question as accurate as possible: -We have a (possibly non-compact) Riemannian manifold $(M,g)$ and a short immersion $z:M\rightarrow \mathbb{R}^k$ wich induces a new metric $h$ on $M$. Shortness means that the tensor $\delta=g-h$ is again possitive definite. -We also have an open cover $N_p$ along with a subordinate partition of unity $\phi_p$ such that every neighborhood $N_p$ meets only finite others. So we now focus on the immersion of $N_p$ into $\mathbb{R}^k$. -Now Nash wants to approximate the tensor $\frac{1}{2}\delta_{ij}$ by another positive tensor $\beta_{ij}$. He does this by first finding a set $M_{\mu,\nu}$ of p.d.s.m (possitive definite symmetric matrices) such that for any p.d.s.m. $A$ we have $A=\sum_{\mu,\nu}C^{*}_{\mu,\nu}M_{\mu,\nu}$ where the coefficients $C^*_{\mu,\nu}$ depend smoothly on $A$. -Here is what is bugging me. Nash writes: -So if I understand correctly, since $\beta_{ij}$ is a smooth choice of a p.d.s.m for any point $x\in N_p$, we can view $\beta_{ij}$ as smooth immersion of $N_p$ into the space of p.d.s.m's , which is an open cone in the vector space of symmetric matrices. My question is: Why can't we choose $\beta_{ij}=\delta_{ij}$? -Again, on top of page $391$ it is stated that we can make $\beta_{ij}$ arbitrarily close to $\delta_{ij}$. But what stops us from having them exactly equal? Any expository notes on this theorem will also be greatly appreciated! - -REPLY [2 votes]: I believe the only issue is that Nash only assumes $g$ to be continuous. So the error tensor is also only continuous, and with the procedure as you suggest it, the functions $a_\nu$ (following Nash's notation) would only be continuous. This would be a problem since $a_\nu$ are one of the building blocks of the "Nash twist" (13), so that the next 'immersion' in the sequence would only be continuous - and hence not even an actual immersion, not something that induces a new metric with a new associated error. The procedure would break down. -Of course, the paper is already extremely interesting if you assume everything to be smooth other than the very final $C^1$ 'isometric immersion' limit of the approximating immersions, so I think this is an ignorable technicality for anyone reading the paper for the first time. -But it is an interesting technicality, since it is roughly of the same nature as the 'loss of derivatives' problem which motivates Nash's famous introduction of smoothing operators into the Newton method, in his subsequent paper on smooth isometric embeddings. -Tangentially, it's worth noting that Gromov generalized this part of Nash's paper in a very appealing way. From page 170 of his "Partial Differential Relations": let $A\subset\mathbb{R}^q$ be a connected embedded $C^\ell$-submanifold, let $V$ be a compact manifold, and let $f_0$ be a $C^r$ map from $V$ into the interior of the convex hull of $A$, with $r\leq\ell$. Then $f_0$ can be written a finite constant convex combination of $C^r$ maps from $V$ into $A$. -In Nash's context, $\mathbb{R}^q$ is something like the vector space of symmetric matrices while $A$ is the submanifold of rank-1 matrices. I find Gromov's statement quite remarkable, since even in simple contexts (e.g. $q=3$, $A$ a sphere) it seems extremely non-intuitive. Strangely, neither Nash's statement (5) nor its proof seem so non-intuitive.<|endoftext|> -TITLE: How to find a finite splitting field $K$ for $G$ such that every indecomposable $KG$-module is absolutely indecomposable -QUESTION [5 upvotes]: Let $G$ be a finite group and let $k$ be a finite field with char$(k)=p$ such that $p\mid |G|$. -If $k$ is a splitting field for $G$, then, no matter which splitting field we take, after extending scalars we always have the same number of isomorphism classes of simple modules. -However, it is not necessarily true that every indecomposable $kG$-module is absolutely indecomposable. -This leads to the following - -Question: Is there a general construction that gives us a finite splitting field $K$ for $G$ (and all subgroups of all factor groups of $G$) such that every indecomposable $KG$-module (and every indecomposable K[H/J]-module where H/J is a factor group of a subgroup of G) is absolutely indecomposable? - -Is there a reference in the literature? -Thanks for the help. - -REPLY [8 votes]: No. -Let $G=C_2\times C_2$, generated by elements $g$ and $h$, and let $K$ be any finite field of characteristic $2$. -Let $V$ be a finite dimensional $K$-vector space of dimension greater than one with an endomorphism $\varphi$ with irreducible characteristic polynomial $p(t)$. -Then $V\oplus V$ can be made into a $KG$-module by letting $g$ act as $\pmatrix{\text{id}_V & \text{id}_V\\0 & \text{id}_V}$ and $h$ act as $\pmatrix{\text{id}_V & \varphi\\0 & \text{id}_V}$, and it is indecomposable as a $KG$-module, but if $\overline{K}$ is an algebraic closure of $K$ then $V\otimes_K\overline{K}$ is a direct sum of two-dimensional $\overline{K}G$-modules corresponding to the roots of $p(t)$ in $\overline{K}$.<|endoftext|> -TITLE: Does the Grothendieck construction produce a 2-category or a category? -QUESTION [8 upvotes]: Let $F : \mathcal{C} \to \mathbf{Cat}$ be a lax 2-functor. Then we can form a category $\int F $ which is the Grothendieck construction on F. -There's a number of resources detailing this construction, but none mentioning if we can get a 2-category, rather than a category this way. It seems like the natural construction would be to generate a 2-category - is this true? I'm looking for a reference. -Namely - it seems like the 2-cells in $\mathcal{C}$ in standard definitions don't really play any role at all in $\int F$, but it seems like they should correspond to 2-cells in $\int F$. -The closest constructions I found were in (https://arxiv.org/abs/2002.06055, Definition 10.7.2.) . They provide a bicategorical Grothendieck construction for a functor $F : \mathcal{C}^{op} \to \mathbf{Bicat^{ps}}$ - but of a rather strange kind. -It does not use 2-cells in $\mathcal{C}$ and it also assumes certain equality of 1-cells in $\mathcal{C}$. -Another potential candidate is https://www2.irb.hr/korisnici/ibakovic/sgc.pdf, but it seems to be on a higher level of abstraction than I'm comfortable with. -It also seems to talk about the Grothendieck construction of functors whose codomain is $\mathbf{2-Cat}$, rather than into $\mathbf{Cat}$. It seems like this is extra structure that is not needed for what I'm asking. -So, in short, if there is a lax functor $F : \mathcal{C} \to \mathbf{Cat}$, is there a way to make the Grothendieck construction into a 2-category? If so - is there reference with an explicit construction, showing in details what all the 2-cells would look like? - -REPLY [9 votes]: The usual Grothendieck construction has for $\mathcal C$ an ordinary category, so it doesn't have any 2-cells (or at least, doesn't have any non-identity 2-cells). Moreover, we actually get not only a category out of it, but an object of the slice (2-)category $\mathcal Cat/\mathcal C$. If $\mathcal C$ weren't an ordinary category, this wouldn't make as much sense since $\mathcal C$ wouldn't be an object of $\mathcal Cat$. (Just like how we can have functors $\mathcal C \to \mathcal Cat$, though, it's still possible to talk about functors from 1-categories to 2-categories). -Nonetheless, you can get a meaningful construction when adding in 2-cells. I have no idea if this construction has a name, but once you get the pattern, it's easy to extend this to any arbitrary level. For this construction, we'll assume we have a (lax) 2-functor $F : \mathcal C \to \mathcal Cat$. -0-cells of $\int F$ are pairs $(c, x)$ where $c$ is an object of $\mathcal C$ and $x$ is an object of $F(c)$. -1-cells $(c, x) \to (c', x')$ are pairs $(f, g)$ where $f : c \to c'$ and $g : x \to_f x'$. $x \to_f x'$ is the set of dependent morphisms from $x$ to $x'$ (terminology adapted from HoTT's dependent paths). Effectively, we use functorality in the types of $x$ and $x'$ to transport from one type to the other. In this case, the type of $x$ is $F(c)$ and the type of $x'$ is $F(c')$ so we can use $F(f)$ to map $x$ into the type of $x'$. -Summing up, $g$ should be a morphism $F(f)(x) \to x'$, i.e. an element of $\hom_{F(c')}(F(f)(x), x')$. -Next, our 2-cells $(f, g) \to (f', g')$ should be pairs $(\alpha, \beta)$ where $\alpha$ is a 2-cell $f \to f'$ in $\mathcal C$. $\beta$ should be a dependent morphism $g \to_\alpha g'$. -Now the types of $g$ and $g'$ are $\hom_{F(c')}(F(f)(x), x')$ and $\hom_{F(c')}(F(f')(x), x')$ respectively. This time, these type are contravariant in the variable we need to transport ($f$ and $f'$), so we'll transport $g'$ to $\hom_{F(c')}(F(f)(x), x')$ via $\hom_{F(c')}(F(\alpha)(x), x')$. -Unpacking this, $g'$ gets sent to $g' \circ F(\alpha)(x)$, so $\beta$ is a morphism $g' \circ F(\alpha)(x) \to g$. This time, though, we're talking about a morphism between morphisms in $F(c')$, which is an ordinary category. So rather than an actual morphism, we'll have an equality $g' \circ F(\alpha)(x) = g$. - -REPLY [3 votes]: It produces a functor between categories. In fact, what is called a fibred category. The construction is detailed in Volume 2 of Borceux's Handbook on Categorical Algebra. -Also Angelo Vistolis Notes on Grothendieck topologies, fibered categories and descent theory is worth looking at.<|endoftext|> -TITLE: Recognizing Lipschitz functions up to change of target metric -QUESTION [12 upvotes]: Let $K$ be a compact subset of $\mathbb{R}^n$ (for simplicity, I am happy to take $K=\overline{B(0,1)}$ for now if it is easier). -Let $f:K \rightarrow \mathbb{R}^m$ be a continuous function. -Is there a new metric $d$ on $\mathbb{R}^m$ (compatible with the topology) such that -$$ f: K \rightarrow (\mathbb{R}^m,d) $$ -is Lipschitz? -(Here $K$ is still equipped with the usual Euclidean metric, and only the metric on the target space $\mathbb{R}^m$ has changed.) -If not, - -What are obstructions? -Are there sufficient conditions? - -REPLY [2 votes]: Here are some suggestions trying to expand erz's and Nik Weaver's comments. I do not have time to work out the details (although stuck at home arrests like many of us), but I'd be glad if nevertheless they turn out to be useful. -Recall that a set $S\subset\mathbb{R}^m$ is $\mathcal{H}^1$- null iff for all $\epsilon>0$ it has a countable partition $\{S_j\}_{j\in\mathbb{N}}$ such that $\sum_{j\in\mathbb{N}}\rm{diam}(S_j)\le\epsilon$. -Say the that a map $f:A\subset\mathbb{R}^m\to\mathbb{R}^n$ is $\mathcal{H}^1$- Absolutely Continuous iff it is continuous and for any $\mathcal{H}^1$-null set $S\subset A$, the image $f(S)$ is $\mathcal{H}^1$-null in $\mathbb{R}^n$ . -[edit 4.24.20]: as user erz pointed out to me, this definition is too weak (the function in Nik Weaver's counterexample satisfies it). A more suitable definition is maybe: iff it is continuous and for any $\epsilon>0$ there exists $\delta>0$ such that for any $S\subset A$, $\mathcal{H}^1(S)<\delta$ implies that $\mathcal{H}^1 (f(S))<\epsilon$ . -The conjecture should be: - -Any $\mathcal{H}^1$-AC map defined on a compact subset $K$ is Lipschitz - up to a choice of an equivalent distance on the target space. - -We may always assume $n=m$ (via inclusion $K\subset \mathbb{R}^n\subset\mathbb{R}^m$, or $\mathbb{R}^m\subset\mathbb{R}^n$ if needed). Then - -It does not seem problematic to extend an $\mathcal{H}^1$-absolutely continuous map $f$ defined on a compact $K$ to a $\mathcal{H}^1$-absolutely continuous map $f:\mathbb{R}^n\to\mathbb{R}^n$ that is also a surjective and proper map (it should sufficient to make it locally lipschitz on the complement of $K$, and equal to the identity outside a large ball. -For $x,y$ in $\mathbb{R}^n$ define - -$$d(x,y):=\inf\{\mathcal{H}^1(S): f(S)\,{ \rm connected},\, \{x,y\}\subset f(S)\}$$ - -Then $d:\mathbb{R}^n\times \mathbb{R}^n\to[0,\infty)$ is clearly symmetric, and satisfies the triangular inequality. -To show $d(x,y)=0$ implies $x=y$, is where the hypotheses enter. The idea should be that $d(x,y)=0$ implies the existence of a $\mathcal{H}$-null subset $S$ such that $f(S)$ is connected and $\{x,y\}\subset f(S)$, which forces $x=y$ since $f(S)$ is also $\mathcal{H}$-null. To this end one should start from a minimizing bounded sequence of compact subsets $S_j$ with $\mathcal{H}^1(S_j)\to0$, and $\{x,y\}\subset f(S_j)$. Compact subsets of a given compact are a compact in the Hausdorff distance, the Hausdorff measure is lower semicontinuous, connected sets are a closed set, so I think it could be done. In fact, it seems OK that the infimum in the definition of this distance be always attained, by the same argument. -By definition of $d$, taking as $S$ the segment $[u,v]$ one has $d(f(u),f(v))\le \mathcal{H}^1([u,v])= \|u-v\|$. -It remains to show that $d$ is topologically equivalent to the Euclidean distance, which seems true, although I see it less clearly at the moment.<|endoftext|> -TITLE: Taylor expansion with remainder on locally convex spaces -QUESTION [6 upvotes]: It is usual to introduce Fréchet and Gâteaux derivatives in Banach spaces. In this context, the familiar Taylor expansion with remainder is also at hand, as you can see on the picture below taken from this reference. Now, I'd like to know if there exists an analogous theorem for locally convex spaces and where should I go to learn it if this is the case. Do we need Fréchet and Gâteaux derivatives as well? Any comments would be helpful. Thanks in advance! -EDIT: Let $E, F$ be locally convex spaces over some field $\mathbb{K}$ and $f: E \to F$. We say that $f$ is Gâteaux differentiable at $x \in E$ if there exists a continuous linear functional $Df[x]$ such that -\begin{eqnarray} -\lim_{\epsilon \to 0}\frac{f(x+\epsilon h)-f(x)}{\epsilon} = Df[x](h) \tag{1}\label{1} -\end{eqnarray} -for every $h \in E$. -The above definition makes sense and it is the definition of Gâteaux differentiability that I know. Now, here the author states a Taylor expansion result (pag. 25, Theorem 1.4.11) but his definition of Gâteaux differentiability is a little different (Definition 1.4.7 on page 24). However, I believe my definition and his definition are equivalent and, if so, this answers my question. Am I missing something here? - -REPLY [5 votes]: First of all the good news: There is an analogous theorem for locally convex spaces. It can be formulated in almost the same way using Gateaux derivatives (more on this notion in a moment). Obviously, the norm condition on the vanishing of the remainder makes no sense in the absence of a norm. However, you can formulate the vanishing of the remainder term of order $n$ via a homogeneity condition on the arguments of the remainder. -Before we come to the bad news, some comments on differential calculus beyond Banach spaces. Frechet derivatives make no sense beyond Banach spaces (again the norm condition breaks ones neck) and there are several (beyond Frechet spaces mutually inequivalent) ways of doing calculus in locally convex spaces. You can choose among dozens of approaches (all summarised in H.H. Keller: Calculus in locally convex spacesa), but most people stick with one of the following approaches: -1. The convenient calculus (popular since Kriegl and Michors book: The convenient setting of global analysis, from 1998 available online here 1 ). To my knowledge this does NOT feature a Taylor like theorem, though you find a section on Jet spaces (which is a more fancy way of talking about Taylor like expansions) -2. Bastiani calculus as used in most writings on infinite-dimensional Lie theory, see e.g. Section 1 of 2 for an overview and more references. In Bastiani calculus I am certain that there is a version of Taylors theorem. I give you a cartoon version since it is technical and long and I do not want to type endlessly: -Let $E,F$ be locally convex spaces, $k\in \mathbb{N}_0$ and $f\colon U \rightarrow F$ a Bastiani $C^k$-map on an open subset of $E$. Then the following holds: -(a) for each $x\in U$, there exists a unique polynomial $P_x^kf \colon E \rightarrow F$ of degree $\leq k$ such that $\delta_0^j(P_x^kf)=\delta_x^jf $ for each $j\leq k$. -(b) $\lim_{t\rightarrow 0} \frac{f(x+ty)-P_x^kf(ty)}{t^k}=0, \forall y \in E$ and the polynomial is unique with this property. -(c) There is a continuous remainder term -$R_k(x,y,t)=\frac{1}{(k-1)!}\int_0^1 (1-r)^{k-1}(\delta_{x+rty}^kf(y)-\delta_x^kf(y))\mathrm{d}r$ satisfying equation (30.14) in the picture you uploaded -On to the bad news: I am certain that it exists because I have a nice pdf in front of me stating the theorem together with all the nitty gritty details and how to prove it. The unfortunate part is that I am not at liberty to share it since this is part of the book manuscript for the forthcoming* book by Glöckner and Neeb on infinite-dimensional Lie theory. This is unfortunately also the only source I know of, where one can find it in this form. You could try to write an email to Glöckner (see 2) and see whether he can either point you towards an older source or wants to share the pdf. -*: Unfortunately, the book was forthcoming from 2005 on. I have seen many revisions but it is anybodys guess when it will finally appear in public.<|endoftext|> -TITLE: Infinity local systems -QUESTION [13 upvotes]: I have seen many references in the (geometric representation theory, symplectic geometry, etc) literature to "infinity local systems". -From what I've been told, given a good cover $\{U_i\}$ of $X$, an infinity local system on a connected space $X$ assigns: --a chain complex $C_i$ to every contractible open set $U_i$ --a chain morphism $F_{ij}: C_i \to C_j$ to every double intersection $U_i \cap U_j$ --a homotopy of chain morphisms $F_{ik} \sim F_{jk} \circ F_{ij}$ for every triple intersection $U_i \cap U_j \cap U_k$ --etc. -EDIT: actually, I think that the above definition is rather an "infinity-sheaf". An infinity local system should be a locally constant "infinity-sheaf". My guess is that the Maurer-Cartan condition mentioned below is precisely encoding the "locally-constant" condition. -Some questions: -(1) What are some good sources to learn about these objects (from the perspective of a second year graduate student with limited exposure to these ideas)? -The only paper I have found is Section 2.1 of "A Riemann–Hilbert correspondence for infinity local systems" (https://arxiv.org/pdf/0908.2843.pdf) In this paper, an infinity local system is defined as a set of maps from a simplicial set $K$ to a differential graded category $C$ which satisfy a certain Maurer-Cartan equation. However, it's not clear to me how to make sense of this definition (e.g. what is the role of Maurer-Cartan? how does this definition match up with the intuitive notion of an infinity local system described above?) -(2) The infinity local systems on a topological space $X$ are supposed to form a dg category. I recently heard a talk in which the speaker claimed (as if it was the most natural thing in the world:) )that $hom(k_X, k_X)=C^*(X;k)$, where $k$ is some field, $k_X$ is the locally constant sheaf with stalk $k$ is degree $0$ and $C^*(X;k)$ is the singular cohomology of $X$ with $k$ coefficients. -Is there a good way to understand why $hom(k_X, k_X)=C^*(X;k)$? More generally, how should one understand the dg category structure on $\operatorname{Loc}_{\infty}(X)$? - -REPLY [3 votes]: Since you tagged this with "symplectic geometry", I'm going to give an answer from a symplectic geometry perspective, which may not be what you're looking for, but (as a symplectic person) I find it is a helpful point of view. This will use the language of $A_\infty$-categories as well as dg-categories. -Given a manifold $X$, take its cotangent bundle $T^*X$. This is a non-compact symplectic manifold. You can consider its Fukaya category of exact embedded Lagrangian submanifolds which agree with the symplectisation of a Legendrian at infinity. The Floer cochain groups (morphisms) between $L_1,L_2$ are free k-vector spaces on intersection points between $L_1$ and $\phi(L_2)$, where $\phi$ is the time 1 map of a suitable Hamiltonian. You have to specify a suitable class of Hamiltonians, and I want to use Hamiltonians which are "quadratic at infinity", in other words they look something like kinetic energy with respect to some Riemannian metric. Since kinetic energy as a Hamiltonian generates geodesic flow, the Floer cochains are something like (k-linear combinations of) geodesics connecting the Lagrangians. -For example, if $L_1$ and $L_2$ are cotangent fibres at $x_1$ and $x_2$ then the Floer complex is something like the free k-vector space on the set of geodesics from $x_1$ to $x_2$. It is then a theorem (of Abbondandolo and Schwartz at the level of homology beefed up to the $A_\infty$ level by Abouzaid) that the Floer complex between two cotangent fibres is quasiisomorphic to chains on the space of paths between $x_1$ and $x_2$ (and to $C_{-*}(\Omega X)$ as an $A_\infty$-algebra when $x_1=x_2$ and concatenation makes sense). -Abouzaid also showed that a cotangent fibres generates this Fukaya category, so you get a fully faithful Yoneda functor from the Fukaya category to the dg-category of $A_\infty$-bimodules over chains on the based loop space. In other words, if you want to compute the Floer complex between two Lagrangians $L_1$ and $L_2$ and you have a cotangent fibre $F$, you can take the $CF(F,F)$-bimodules $CF(F,L_n)\otimes CF(L_n,F)$, $n=1,2$, and take homs between them in the category of $A_\infty$ $CF(F,F)$-bimodules. Since $CF(F,F)\cong C_{-*}(\Omega X)$, this is quasiequivalent to the category of infinity local systems. So this category of infinity local systems is the Fukaya category of $T^*X$. -Now how do you see that $hom(k,k)=C^*(X)$? Well, there is a Lagrangian in $T^*X$ whose Floer complex is $C^*(X)$, namely the zero section. For example, a small Hamiltonian deformation of the zero section is a graph of an exact 1-form $df$, and the intersection points between this and the zero section happen at critical points of $f$; in fact Floer showed that in this case the Floer complex is the Morse complex for a suitable choice of almost complex structures. -What is the Yoneda bimodule corresponding to the zero section? Well the zero section intersects our cotangent fibre at a single point, so $CF$ is just k (our field, considered as a trivial $A_\infty$-module over $C_{-*}(\Omega X)$). Its self homs in the category of $C_{-*}(\Omega X)$-bimodules should therefore compute $C^*(X)$. -The relevant papers of Abouzaid are: -https://arxiv.org/abs/0907.5606 -https://arxiv.org/abs/1003.4449<|endoftext|> -TITLE: Eigenfunctions of the laplacian on $\mathbb{CP}^n$ -QUESTION [14 upvotes]: I want to find explicit formulas for the eigenfunctions of the Laplacian on $\mathbb{CP}^n$ endowed with the Fubini Study metric. -For the first eigenvalue $\lambda_1 = 4(n+1)$, the eigenfunctions are given by the real and imaginary parts of $\phi^{i, j} = \frac{z_i\bar{z_j}}{\sum_{k}|z_k|^2}-\frac{\delta_{i, j}}{n+1}$, and these functions together gives the Veronese isometric embedding of $\mathbb{CP}^n$ into $S^{2(n+1)^2-1}$. -Is there an analogue of this for higher eigenfunctions? Is the $k$-th eigenspace also associated with some embedding of $\mathbb{CP}^n$? Are there explicit formulas listing the higher eigenfunctions like the $\phi$ above? - -REPLY [23 votes]: The $k$-th eigenfunctions are actually easy to describe: In $\mathbb{C}^{n+1}$ with unitary complex coordinates $z_0,z_1,\ldots,z_n$, write $Z = |z_0|^2+\cdots+|z_n|^2$. -Now, for a given $k\ge0$, let $H_k$ be the (real) vector space of real-valued polynomials $p(z,\bar z)$ that are homogeneous of degree $k$ in the $z$-variables and degree $k$ in the $\bar z$-variables and that are harmonic, i.e., they satisfy -$$ -\frac{\partial^2p}{\partial z_0\partial \bar z_0}+\cdots+ -\frac{\partial^2p}{\partial z_n\partial \bar z_n} = 0. -$$ -One easily computes that -$$ -\dim_\mathbb{R}H_k = {{n+k}\choose{k}}^2 --{{n+k-1}\choose{k-1}}^2. -$$ -Then, for $p\in H_k$, the function $f_p:\mathbb{CP}^n\to\mathbb{R}$ given by -$$ -f_p\bigl([z]\bigr) = \frac{p(z,\bar z)}{Z^k} -$$ -is well-defined and is an eigenfunction with eigenvalue $\lambda_k$. Conversely, every eigenfunction with eigenvalue $\lambda_k$ is of this form for a unique $p\in H_k$. -Added remark: I was privately asked how to prove that the description I gave of the eigenvalues and eigenfunctions of $\mathbb{CP}^n$ is correct. I'm sure it's in the literature in various places, but it's easier to just give the argument, using the known description of the eigenvalues and eigenfunctions on the standard sphere. Here's the idea: -Let $S^{2n+1}\subset \mathbb{C}^{n+1}$ be the standard unit $2n{+}1$-sphere with its induced metric and let $\sigma:S^{2n+1}\to\mathbb{CP}^n$ be the quotient mapping $\sigma(z) = [z] = \mathbb{C}z$. Then $\sigma$ is a Riemannian submersion when $\mathbb{CP}^n$ is given the Fubini-Study metric (appropriately scaled; with this choice, the area of a linear $\mathbb{CP}^1\subset\mathbb{CP}^n$ is $\pi$, not $4\pi$). -If $f$ is an eigenfunction on $\mathbb{CP}^n$ with eigenvalue $\lambda_k$, then $\sigma^*(f)$ is an eigenfunction on $S^{2n+1}$ with eigenvalue $\lambda_k$, one that is, moreover, invariant under the circle action $\mathrm{e}^{i\phi}\cdot z = \mathrm{e}^{i\phi}z$ on $S^{2n+1}$. -Now, it is known that, if $\mu_d$ is the $d$-th eigenvalue of $S^{2n+1}$, then any eigenfunction $f$ with this eigenvalue is the restriction to $S^{2n+1}$ of a harmonic polynomial on $\mathbb{R}^{2n+2}$ that is homogeneous of degree $d$. We are looking for such polynomials that are also invariant under the circle action of multiplication by $\mathrm{e}^{i\phi}$. When expressed as a polynomial in the complex coordinates $z_i$ and $\bar z_i$, a polynomial that is invariant under this circle action must have the same number of $z$'s as $\bar z$'s. This can only happen if $d$ is even. Conversely, if $d=2k$, then we get the space $H_k$ as defined above. Thus, $\lambda_k = \mu_{2k}$, and the above description is justified.<|endoftext|> -TITLE: Two Dehn fillings yielding a lens space? -QUESTION [11 upvotes]: Let $M$ be an oriented $3$-manifold with $\partial M$ torus. Suppose that two different Dehn fillings $M(r)$ and $M(r')$ are (oriented) homeomorphic to a lens space $L(p,q).$ Does that imply that $M$ is a solid torus? - -REPLY [13 votes]: This is a case of the oriented knot complement problem in lens spaces, also called the cosmetic surgery problem. If $M(r)\cong M(r')$ preserving orientation, then these form a purely cosmetic (or truly cosmetic) pair. See Problem 1.81 (A) from Kirby's problem list (posed by Steve Bleiler). -You probably want to avoid reducible examples like that given in Mukherjee’s answer (note that this hypothesis is missing also in the statement of Problem 1.81). -Given that assumption, there is no such example which is irreducible and non hyperbolic by a result of Daniel Matignon (he classifies all cosmetic surgeries on such manifolds, and shows that all the examples are chiral cosmetic pairs, i.e. orientation-reversing). -There are examples of cosmetic surgeries on hyperbolic knots in lens spaces that reverse orientation, but I think the orientation preserving case is still open in general. -By the way, maybe the interesting case of this question is for $M$ not a homology circle. Otherwise, the two surgeries must lift to the $p$-fold cyclic cover of $M(r)\cong M(r')\cong L(p,q)$, and hence one would have two surgeries on a knot yielding $S^3$, a contradiction to the knot complement theorem.<|endoftext|> -TITLE: Is there an i.i.d sequence in the unit cube $[-1,1]^d$ with $\mathbb E \left[ \Big \| \sum_{i=1}^N X_N \Big \|_\infty\right] = \sqrt {dN}$? -QUESTION [5 upvotes]: There are loads of concentration results for sums of scalar-valued independent sums $X_1,X_2,\ldots, X_N$ with $\mathbb E[X_n]=0$. For example Hoeffding's Inequality says if all $|X_1|\le 1$ then $\mathbb E\left[ \left|\sum_{i=1}^N X_i\right|\right] = O (\sqrt N)$. -These concentration results can be generalised for example to vector-valued random variables with $\|X_n\| \le 1$ for $\| \cdot \|$ the Euclidean norm. -Suppose instead we have $\|\cdot\|_\infty$ norm bounds. That means $X_1,X_2,\ldots, X_N \in \mathbb R^d$ are independent with $\mathbb E[X_n]=0$ and $\|X_n\|_\infty \le 1$. Since $\|X_n\|_2 \le \sqrt d \|X_n\|_\infty \le \sqrt d$ we can use concentration results for the Euclidean norm to get $\mathbb E\left[ \left\|\sum_{i=1}^N X_i\right\|_\infty\right] = O (\sqrt {dN})$. -Does anyone know an example of when this dependence on $\sqrt d$ actually occurs? -Note: This is a bit informal. What I'm really asking is if the $O(\sqrt{dN})$ bound can be improved. I imagine a negative answer would be a family of examples of i.i.d sequences $X_n^{(d)}$, one for each value of $d$, such that $\mathbb E\left[ \left\|\sum_{i=1}^N X_i^{(d)}\right\|_\infty\right] \ge F(\sqrt {dN})$ for some common $\Omega(\sqrt{dN})$ function $F$. -I suspected such an example might be $X_n = (B_1^1,\ldots, B^1_d)$ for all $B^i_j$ independent and taking values $\pm 1$ with probability $1/2$. However the expectation seems more like $O (\sqrt {\log(d)N})$. To see this apply scalar concentration to each $j$ coordinate to get, up to coefficients: -$$P\left( \Big\|\sum_{i=1}^N X_i \Big\|_\infty t\right)dt \le \int_0^\infty (1-(1-e^{-t^2/N})^d )dt$$ -I don't know if the integral has a closed form. What I do know is the same argument gives -$$\mathbb E \Big\|\sum_{i=1}^N X_i \Big\|_\infty ^2 \le \int_0^\infty (1-(1-e^{-t/N})^d )dt$$ -which I do know how to solve. Substitute $x = e^{-t/N}$ to get $dt = - (N/x)dx$ and the integral becomes -$$ N \int_0^1 \frac{(1-(1-x)^d )}{x}dx$$ -which equals $N$ times the $d$-th harmonic number, which is $O(N \log d)$. So we have $$\mathbb E \Big\|\sum_{i=1}^N X_i \Big\|_\infty ^2 \le O\left(N \log(d) \right)$$ and by the Jensen inequality $$\mathbb E \Big\|\sum_{i=1}^N X_i \Big\|_\infty \le O\left(\sqrt{\log(d) N}\right).$$ - -REPLY [7 votes]: Let $X_i=(X_{i,1},\dots,X_{i,d})$, $S:=(S_1,\dots,S_d)$, $S_j:=\sum_{i=1}^d X_{i,j}/\sqrt n$. Then, by Hoeffding's inequality, for $s\ge0$ -$$P(|S_j|\ge s)\le2e^{-s^2/2},$$ -whence -$$E\|S\|_\infty=\int_0^\infty ds\,P(\|S\|_\infty\ge s) -\le\int_0^\infty ds\,\min(1,2d\,e^{-s^2/2}) -=O(1+\sqrt{\ln d});$$ -here we used the inequality -$$\int_t^\infty ds\,e^{-s^2/2}\le e^{-t^2/2}/2$$ -for $t\ge0$. -So, for $d\ge2$ you get -$$E\Big\|\sum_{i=1}^n X_i\Big\|_\infty=O(\sqrt{n\ln d}),$$ -without the assumption of the independence of the coordinates and without any other additional assumptions.<|endoftext|> -TITLE: Do multiplicative Banach limits exist? -QUESTION [6 upvotes]: Let $(D, \succeq)$ be a directed set, and let $B$ be the space of real-valued bounded functions on $D$. A Banach limit $\ell$ on $D$ is a linear functional that satisfies -$$\sup_{d \in D} \inf_{c \succeq d} f(c) \leq \ell(f) \leq \inf_{d \in D} \sup_{c \succeq d} f(c)$$ -for all $f \in B$. -Banach limits exist by the Hahn-Banach theorem. In fact, if we say instead that our functions on $D$ take values in an abstract Dedekind complete ordered vector space, then the existence of Banach limits is equivalent to the Hahn-Banach theorem. But I am primarily interested in the real-valued case. -Question. Are there Banach limits that satisfy -$$\ell(fg) = \ell(f)\ell(g)$$ -for all $f,g \in B$? If so, does this require anything stronger than the Hahn-Banach theorem? - -REPLY [9 votes]: I do not think that this is the usual definition of Banach limit. (What I know under this name is linear functional on $l_\infty$ which is positive, shift-invariant and extends the usual limit, see the linked Wikipedia article. Of course, the terminology in various sources might differ.) In connection with the question it might be worth mentioning that Banach limit in this sense cannot be multiplicative. -EDIT: The OP mentioned in comments that the definition of Banach limit given in the question (i.e., with directed sets) can be found in Schechter's Handbook of Analysis and Its Foundations, see page 318 and also in Howard, Rubin: Consequences of the Axiom of Choice, see page 63, Form 372. (If we work on arbitrary directed sets, we do not have a natural way to talk about shift-invariance. So this is not a generalization of the notion of Banach limit on $\mathbb N$ mentioned in the first paragraph.) - -Your conditions can be rewritten as1 -$$\liminf_{c\in D} f(c) \le \ell(f) \le \limsup_{c\in D} f(c).$$ -So you want a functional which is between $\liminf$ and $\limsup$ (and therefore extends the usual limit of a net) and is multiplicative. -You can simply take any ultrafilter $\mathcal U$ which contains all tail sets of the directed set $D$. (I.e., for any $d\in D$ you have $d\uparrow=\{c\in D; c\ge D\}\in\mathcal U$.) And then define $\ell$ using limit along this utlrafilter as: -$$\ell(f) = \operatorname{{\mathcal U}-\lim} f(c).$$ -This functional has the properties you want. (Boundedness of $f$ guarantees that the $\mathcal U$-limit exists.2 We get multiplicativity from the fact that $\mathcal U$-limit is multiplicative. And the fact that $\mathcal U$ contains the tail filter helps with the condition about limit inferior and limit superior.) -The same construction is mentioned in the answer to: What is a generalized limit? -In case it helps to find some references for $\mathcal U$-limit (limit along an ultrafilter or, more generally, limit along a filter or a filter base), I will mention my answers to these questions: Where has this common generalization of nets and filters been written down? and Basic facts about ultrafilters and convergence of a sequence along an ultrafilter. -Since you are interested in multiplicative functionals, this might be of interest, too: Every multiplicative linear functional on $\ell^{\infty}$ is the limit along an ultrafilter. - -1For more on limit superior/inferior of a net, see: Limsups of nets and About the notion of limsup and liminf -2Limit along an ultrafilter with values in a compact space always exists. The proof is given, for example, in this answer: Basic facts about ultrafilters and convergence of a sequence along an ultrafilter.<|endoftext|> -TITLE: Limit case of Sobolev space in $1$-D -QUESTION [7 upvotes]: This might look too an elementary question, but I am confined and is not able to find a textbook which answers the following question. - -I have a function $f:{\mathbb R}\rightarrow{\mathbb R}$, such that $f\in L^3({\mathbb R})$ and - $$\int\int\frac{|f(y)-f(x)|^3}{|y-x|^4}dydx<\infty.$$ - May I conclude that $f\in W^{1,3}({\mathbb R})$ ? - -This is a limit case of Sobolev-Slobodeckij space, as $4=1\cdot3+1$. Obviously, the same integral but with exponent $s\cdot3+1$ with $s<1$ is valid, hence $f\in W^{s,3}({\mathbb R})$ - -REPLY [3 votes]: Let summarize the comments there: -In order for the seminorm here to be finite, one needs at least $|f(x)-f(y)| = o(|x-y|)$ when $x-y\to 0$, and this is possible only for constants functions. Since $f∈L^3$, $f=0$ (and so $f∈W^{1,3}$ ...). -If to avoid that one uses the second order difference $f(2y-x)-2f(y)+f(x)$ instead of $f(y)-f(x)$, one could however not conclude that $f∈W^{1,3}$, and this is due to the misleading definition of fractional Sobolev(-Slobodeckij) spaces $W^{s,p}$ since when $s>0$ is not an integer $W^{s,p} = B^s_{p,p} = F^s_{p,p}$ (where the $F^s_{p,q}$ are the Triebel-Lizorkin spaces) while $W^{n,p} = F^n_{p,2}$ when $n$ is an integer. An other fractional extension of Sobolev spaces are the Bessel-Sobolev spaces $H^{s,p}$ where the seminorm is the $L^p$ norm of the fractional Laplacian. They verify $H^{s,p} = F^s_{p,2}$ (For every $s≥0$). -All these spaces are ordered in this way when $p≥ 2$ (with strict inclusion when $p>2$) -$$ -B^s_{p,1} ⊂ B^s_{p,2} ⊂ F^s_{p,2} (=H^{s,p}) ⊂ F^s_{p,p} = B^s_{p,p} ⊂ B^s_{p,\infty}. -$$ -This is well explained for example in the book of Hans Triebel, Theory of Function Spaces II. Springer Basel, 1992.<|endoftext|> -TITLE: Toposes in which countable choice is true but dependent choice isn't -QUESTION [9 upvotes]: I'd like examples of toposes in which Countable Choice is true but Dependent Choice isn't. I'd prefer examples without Excluded Middle. It's hard to find a natural example. - -REPLY [7 votes]: Given any topological group $G$, the topos of sets with a continuous $G$-action is Boolean, and very often violates some choice principle or other. Under the translation between material and structural set theories, such toposes correspond to permutation (or Fraenkel–Mostowski) models of ZFA. -Theorem 8.12 in Jech's Axiom of Choice describes in terms in material sets a model of set theory in which countable choice (i.e. $\mathrm{AC}_{\aleph_0}$) holds, but DC doesn't. [In fact, Jech describes something more general, of which this is the "$<\aleph_1$ case"] -Consider the set $A := \aleph_1^{<\omega} = \bigcup_{n\in \omega} \aleph_1^n$ of finite sequences of countable ordinals. This carries a partial order where $s \leq t$ iff $t$ extends $s$, and is in fact a tree. Consider the automorphism group of this tree, call it $G$. This gets a topology by specifying a filter $F$ generated by an ideal $I\subset P(A)$. A subset $E\subset A$ is in $I$ precisely if it is a countable, bounded-height sub-tree. Then $F$ is the filter generated by the subgroups $\mathrm{Fix}(E) < G$ consisting of automorphisms that fix the subset $E \subset A$ pointwise. -Then the topos of continuous $G$-sets with this topological group is what you wanted.<|endoftext|> -TITLE: Does any derivation of commutative algebra preserve its nil-radical? -QUESTION [14 upvotes]: Given a commutative associative unital algebra over a field of characteristic zero. - -Is it true that any derivation of it preseves its nil-radical? - -More explicitly, let $D$ be a derivation of an algebra $A$. Let $N$ denote the nil-radical of $A$. - -Is it true that $D(N)\subset N?$ - -REPLY [11 votes]: Here is another cute argument (I don't remember where I learned it, I think it is folklore). Let $P\subset A$ be an arbitrary prime ideal. We claim it contains a $D$-stable prime ideal. For this, consider the mod $P$ Taylor map -$$ f\colon A\to (A/P)[[t]] , a \mapsto \sum_{n\geq 0} \frac{D^n(a) \textrm{ mod } P}{n!} t^n.$$ -A quick computation shows that $f$ is a ring map, and that for all $a\in A$ we have $f(D(a)) = \frac{d}{dt}(f(a))$. Therefore, the kernel $Q = \mathrm{ker}(f)$ is a $D$-stable ideal of $A$. Moreover, $Q$ is prime because $(A/P)[[t]]$ is a domain, and we have $Q \subset P$ because the constant term of $f(a)$ is $a \textrm{ mod } P$. -So every prime ideal of $A$ contains a $D$-stable prime ideal. Hence, the intersection of all prime ideals of $A$ equals the intersection of all $D$-stable prime ideals of $A$. But the former is the nilradical, and the latter is clearly $D$-stable.<|endoftext|> -TITLE: Function whose graph is a Borel relation -QUESTION [5 upvotes]: Suppose $f\colon\mathbb{R}^{\omega}\longrightarrow\mathbb{R}$ is a function such that -$$G(f):=\{(x,y)\in\mathbb{R}^{\omega}\times\mathbb{R}\mid f(x)=y\}$$ -is a Borel set. Does it necessarily follow that $f$ is a Borel function? - -REPLY [11 votes]: Yes. This immediately follows from Lusin's separation Theorem. Note that $\mathbb{R}^\omega$ is Polish. -Let $B \subseteq \mathbb{R}$ be a Borel set. The set $f^{-1}(B)=\{x \in \mathbb{R}^\omega \colon \exists y \,\,\, y \in B \land (x,y) \in G(f) \}$ is $\Sigma_1^1$ (analytic). However, $f^{-1}(B)=\{x \in \mathbb{R}^\omega \colon \forall y \,\,\, (x,y) \in G(f) \rightarrow y \in B \}$ is also $\Pi_1^1$ (co-analytic). Therefore, $f^{-1}(B)$ is Borel.<|endoftext|> -TITLE: Image of a polynomial function $x^2+y^2-x+y-axy$ over $\mathbb{F}_p$ -QUESTION [5 upvotes]: Let $p$ be an odd prime and $h(x)=x^2+ax+1$ be an irreducible polynomial over the field $\mathbb{F}_p$. I need to prove that the function -$$\Psi: \mathbb{F}_p^2 \longrightarrow \mathbb{F}_p, \quad (x,y)\mapsto x^2+y^2-x+y-axy$$ -is surjective. I know that is true because the values in the image are in one-to-one correspondence with some conjugacy classes in some groups, but I would like to have an elementary proof of this fact, using the properties of polynomials over finite fields. -I tried to restrict to some suitable subset of the plane, like lines but I can not prove that the values the function takes when restricted to different lines cover all $\mathbb{F}_p$. - -REPLY [11 votes]: It does not matter whether $h$ is reducible or not, it is only important that $a\ne \pm 2$. By affine change of variables (start with $x-\frac a2 y=z$ and use new variables $(z,y)$, then shift them appropriately) we reduce the polynomial $\Psi$ to $x^2-by^2$, where $b=\frac{a^2}4-1$ is non-zero. This is surjective, since for any $\alpha\in \mathbb{Z}_p$ the sets $\{x^2,x\in \mathbb{Z}_p\}$ and $\{by^2+\alpha,y\in \mathbb{Z}_p\}$ have a common element: both contain more than a half of elements of $\mathbb{Z}_p$.<|endoftext|> -TITLE: A boundary Schauder estimate -QUESTION [5 upvotes]: According to Theorem 1.1' in this paper we have the following estimate on classical solutions $u \in C^2(\overline{B_1^+})$ of $-\Delta u = f \text{ in } B_1^+ = B_1 \cap \{x _n \ge 0 \}$ and $u = 0 \text{ on } \partial B_1^+ \cap \{x_n = 0\}$ -$$|D^2u(x) - D^2u(y)| \le C\left(r\lVert u \rVert_{L^\infty(B_{1}^+)} + \int_0^{r} \frac{\omega_f(t)}{t}\,dt + r\int_{r}^{1} \frac{\omega_f(t)}{t^2}\,dt\right) \tag{1}$$ $\forall \, x,y \in B_{1/2}^{+}$ with $r = |x-y|$ where, $\omega_f$ denotes the modulus of continuity of $f$ which we assume is Dini continuous. -The proof is supposed to be similar to proof of Theorem 1.1 which is the interior estimate. Following the lines of this proof we can get all analogous 'boundary' harmonic function estimates in $B_{1}^+$ v.i.a. applying the estimates in Theorem 1.1 to the odd extensions of the harmonic functions to the full ball $B_1$. But in the last step (going along the lines of eqn $(1.13)$ in the paper) it seems we need the estimate $$|D^2u_0(x) - D^2u_0(y)| \le C \lVert u \rVert_{L^{\infty}(B_1^+)}|x-y|, \, \forall \, x,y \in B_{1/2}^+ \tag{2}$$ where, $u_0$ satisfies $-\Delta u_0 = f(0)$ in $B_{1}^+$ and $u_0 = 0$ on $\partial B_1^+ \cap \{x_n = 0\}$ where, $C$ is supposed to be independent of $f$. - -I can't seem to get the estimate $(2)$ for half-ball. - -In case of interior estimate we can consider $v_0 := u_0 - \frac{f(0)}{2n}(1 - |x|^2)$ which is harmonic in $B_1$ and write \begin{align*} |D^2u_0(x) - D^2u_0(y)| = |D^2v_0(x) - D^2v_0(y)| &\le r\lVert D^3v_0\rVert_{L^{\infty}(B_{1/2})} \\ &\le Cr\lVert v_0\rVert_{L^{\infty}(\partial B_{1})} = Cr\lVert u_0\rVert_{L^{\infty}(\partial B_{1})}\end{align*} where, $r = |x-y|$ which proves the interior analogue of $(2)$ using only gradient estimate for harmonic function $v_0$. -But a similar approach doesn't seem to be working for the boundary case (for example, considering $u_0 - \frac{f(0)}{2}x_n^2$ which is harmonic and applying odd extension to this.) - -Is there a different way of approaching the estimate $(2)$? - -Any help is appreciated. Thanks. - -REPLY [6 votes]: One approach is to observe that -$$\|u_0\|_{L^{\infty}(B_1^+)} \geq \frac{1}{16n}|f(0)|.$$ -It suffices by linearity to prove this when $f(0) = 1$. Consider the barrier -$$b(x) = \frac{1}{2n}\left(\left|x - \frac{1}{2}e_n\right|^2 - \frac{1}{8}\right).$$ -Since $b \geq \frac{1}{16n}$ on $\partial B_1^+$, either $u_0 \geq b$ somewhere on $\partial B_1^+$ and we are done, or $u_0 \leq b$ on $\partial B_1^+$ in which case the comparison principle implies that -$$u\left(\frac{e_n}{2}\right) \leq -\frac{1}{16n}$$ -and we are again done. -With this estimate in hand, your suggestion to use reflection and interior estimates for the harmonic function $v_0 = u_0 - \frac{f(0)}{2}x_n^2$ works because -$$\|v_0\|_{L^{\infty}(B_1^+)} \leq (1+8n)\|u_0\|_{L^{\infty}(B_1^+)}.$$ -More precisely, -$$\|D^3u_0\|_{L^{\infty}(B_{1/2}^+)} = \|D^3v_0\|_{L^{\infty}(B_{1/2}^+)} \leq C(n)\|v_0\|_{L^{\infty}(B_1^+)} \leq C(n)(1+8n)\|u_0\|_{L^{\infty}(B_1^+)}.$$<|endoftext|> -TITLE: Poset-troids …? -QUESTION [10 upvotes]: In many respects, - -spanning tree : graph :: linear extension : poset - -For instance, the number of spanning trees/linear extensions is a measure of the "richness" or "complexity" of the graph/poset. Also, the collection of all the spanning trees/linear extensions determines the graph/poset. -(Okay if the graph is not connected we should say "spanning forest" instead of "spanning tree"; but the tree terminology is more common.) -The collection of spanning trees of a graph (thought of as subsets of edges) is a prototypical example of a matroid. Indeed, the notion of matroid can be seen as an abstraction of this kind of collection. -Question: Has anyone ever defined/investigated a "matroid-like" abstraction of the collection of linear extensions of a poset? E.g., collections of permutations of size $n$ satisfying some compatibility condition? -A wildly optimistic guess would be that these could be related to the full flag variety in the way that matroids are related to Grassmannians; however, I take it that "flag matroids" already have an established meaning in this context, which at first glance does not appear to be related to what I am asking (see e.g. Cameron, Dinu, Michałek, and Seynnaeve - Flag matroids: algebra and geometry). - -REPLY [3 votes]: I'm not aware of anything exactly along the lines of your question. It really depends on what properties you're hoping to generalize. -Here's one direction in which one can generalize. Promotion and evacuation are operations on the set of linear extensions of a poset. These operations can be formulated in a more general setting, as Stanley explains in his paper. -Your mention of the relationship between matroids and Grassmannians suggests that you might find the concept of Coxeter matroids interesting, but I don't think Coxeter matroids can plausibly be considered to be generalizations of the set of linear extensions of an arbitrary poset.<|endoftext|> -TITLE: On the conjecture : $|m^2-n^3|>\frac{1}{5}\sqrt[6]{m^2+n^3}$ -QUESTION [8 upvotes]: Let $m,n$ be positive integers, such that $m^2\neq n^3$. I conjecture the inequality -$$|m^2-n^3|>\dfrac{1}{5}\sqrt[6]{m^2+n^3}.$$ -I've tried a lot of numbers, and they all seem to work, but how do I prove it? - -REPLY [18 votes]: As suggested by Joe Silverman, there are counterexamples in my paper - -Rational points near curves and small nonzero $|x^3-y^2|$ via lattice reduction, Lecture Notes in Computer Science 1838 (proceedings of ANTS-4, 2000; W.Bosma, ed.), 33-63. arXiv: math.NT/0005139 (https://arxiv.org/abs/math/0005139) - -The best one there, which I think still holds the record for the largest ratio -$n^{1/2} / |m^2-n^3|$ known, is -$$ -\begin{array}{rcl} -m & \!\! = \!\! & 447884928428402042307918, \cr -n & \!\! = \!\! & 5853886516781223, -\end{array} -$$ -with -$$ -|m^2-n^3| < \frac{1}{52.3}\sqrt[6]{m^2+n^3}. -$$<|endoftext|> -TITLE: Minimum area of the convex hull of the union of a parallelogram and a triangle -QUESTION [7 upvotes]: This question is somewhat dual to my previously stated question about Maximum area of the intersection of a parallelogram and a triangle, where the triangle and parallelogram each is assumed to be of unit area. -Now I ask: - -Question. Suppose a parallelogram is of area 1 (say, the unit square if you prefer) and a triangle is of area $1$ as well, then what is the minimum area of the convex hull of the union of the parallelogram and the triangle? - -It is obvious that the minimum exists and is greater than $1$. The best I can do is $\sqrt{2}$ $-$ is this the minimum? -My example is: the unit square and the right triangle with legs of length $\sqrt2$ each, the triangle's right angle coinciding with the square's corner. (Added by Joe's request.) -In fact, there is a continuous cyclic family of examples, each producing the same area of the convex hull: - -An analogous question can be asked in three dimensions, replacing parallelogram with parallelepiped (or cube, if you prefer) and triangle with simplex or pyramid with a parallelogram base, each of unit volume. - -REPLY [2 votes]: This isn't a full answer, just an algebraic simplification. - -Let a square have vertices $(\pm \frac12, \pm \frac12)$. Suppose a triangle has vertices $(a,b),(c,d),(e,f)$ satisfying: -$$c \le -1/2 \le e \le 1/2 \le a$$ -$$f \le -1/2 \le d \le 1/2 \le b$$ -Then the convex hull of the square and triangle has area -$$\phantom{+}\max(a-c+b-d,\ 2ad-2bc)\,/\,4\\+\max(a-e+b-f,\ 2be-2af)\,/\,4\\+\max(e-c+d-f,\ 2cf-2de)\,/\,4$$ -This surprisingly simple formula is the result of summing the area of the square, the four white triangles bordering it, and any positive areas for the gray triangles. -For instance, the contribution of the triangle on $(a,b),(c,d),(-1/2,1/2)$ is -$$\max(0,\left| -\begin{align} -a\ \ \ \ &\ \ \ \ b &\!1\\ -c\ \ \ \ &\ \ \ \ d &\!1\\ --1/2\ \ &\ \ 1/2 &\!1 -\end{align} -\right|/2)$$ -If the triangle has a positive signed area, then the line between $(a,b)$ and $(c,d)$ is above the corner of the square, and the triangle area should be included in the convex hull. -If the triangle has a negative signed area, then the line is below the corner, and the triangle area should not be included in the convex hull. -I'd welcome an explanation for the overall area formula that doesn't go through a bunch of determinants and algebraic simplification. In any case, the formula is simple enough that we can reasonably ask a computer algebra system to minimize it subject to the inequalities and the area constraint on the triangle. -Furthermore, I think these inequalities hold generically for all minimizing answers to the original problem. If the triangle has vertices $(a,b),(c,d),(e,f)$, then one of them is highest, one of them is lowest, one of them is leftmost and one of them is rightmost. By the pigeonhole principle, two of those designations must coincide. So we can assume without loss of generality that $(a,b)$ is highest and rightmost, $(c,d)$ is leftmost and $(e,f)$ is lowest. Thus at a minimum we have -$$c \le e \le a$$ -$$f \le d \le b$$ -and I think the other inequalities follow too.<|endoftext|> -TITLE: Thom spectrum in the definition of power operations -QUESTION [5 upvotes]: I am reading now Tyler Lawson's $E_n$ ring spectra and Dyer-Lashof operations form the Handbook of Homotopy Theory and I've got a question on the Remark 1.4.19. -We have an operad $\mathcal{O}$ and $\Sigma_k$ acts freely and properly discontinuosly on $\mathcal{O}(k)$. Let $V\subset\mathbb{R}^k$ consisting of elements summing up to $0$ and let $\rho\to B\Sigma_k$ be associated vector bundle of dimension $k-1$ (1). -Now we can form for any $m$ an associated vector bundle $\mathbb{R}^m\otimes\rho$. If we define $P(k)$ as $\mathcal{O}(k)/\Sigma_k$, then there is a virtual bundle $m\rho$ on $P(k)$. Then the Thom spectrum $P(k)^{m\rho}$ is canonically equivalent to the spectrum $\Sigma^{-m}\Sigma^{\infty}_+\mathcal{O}(k)\otimes_{\Sigma_k}(S^m)^k$, where the latter appears in the definition of power operations (2). -So my question(s) are: - -What is the associated vector bundle $\rho$ appearing in (1)? Is it just subbundle of the trivial bundle? -How do we get the equivalence in (2)? - -REPLY [14 votes]: If a group $G$ acts on a vector space $V$, and $X$ is a space where $G$ acts properly discontinuously, then the map -$$ -(V \times X) / G \to X/G -$$ -can be given the structure of a vector bundle: it inherits this structure from the vector space $V$. The pullback of this bundle to $X$ is the trivial bundle $X \times V$, but it is probably not trivial on $X/G$. -(If you prefer to think about vector bundles in terms of classifying spaces, the action of $G$ on $V$ is a group homomorphism $G \to GL(V)$, and there's an induced map $BG \to BGL(V)$. The space $X/G$ has a map to $BG$ classifying the principal $G$-bundle $X \to X/G$, and the composite $X/G \to BG \to BGL(V)$ classifies this associated bundle.) -This recipe for the associated bundle also gives you a recipe for the Thom space: the Thom space is -$$(S^V \wedge X_+)/G$$ -where $S^V$ is the one-point compactification. We sometimes write this as $S^V \wedge_G X_+$. -In the case of this vector bundle $\bar \rho$ that you've written down, there a $\Sigma_k$-equivariant isomorphism between $\Bbb R^k$ and $\Bbb R \oplus V$, where $\Sigma_k$ acts on the factor of $\Bbb R$ trivially. The associated sphere is $S^k$, and this decomposition determines a $\Sigma_k$-equivariant isomorphism of one-point compactifications $S^k \simeq S^1 \wedge S^V$. The Thom space of the bundle $\rho$ associated to $\Bbb R^k$ then satisfies an identity: -$$Th(\rho) \cong S^k \wedge_{\Sigma_k} X_+ \cong S^1 \wedge (S^V \wedge_{\Sigma_k} X_+) \cong S^1 \wedge Th(\bar \rho)$$ -On the level of Thom spectra, we can desuspend both sides and turn this into an identity -$$ -X^{\bar \rho} = \Sigma^\infty Th(\bar \rho) \simeq S^{-1} \wedge \Sigma^\infty Th(\rho) \simeq \Sigma^{-1} X^\rho. -$$ -The version where we multiply the vector bundle by an integer $m$ is roughly the same, except that we get more trivial factors and we have to take care that we get everything with virtual bundles correct when when $m < 0$. -Sorry for the confusion!<|endoftext|> -TITLE: Note rejected from arXiv: what to do next? -QUESTION [81 upvotes]: Short version: A note of mine was rejected by the arXiv moderation (something I didn't even know was possible) on account of being “unrefereeable”. The moderation process provides absolutely no feedback as to why and does not answer questions. I can think of various reasons but I don't know which are actually relevant, and I'm afraid that trying to resubmit the note, even with substantial changes, might get me banned permanently. So I'm looking for advice from people with more experience either in the subject or in dealing with the arXiv, on what to do next (e.g., “forget it, it's crap”, “try to improve it”, “upload it somewhere else”, “do to establish dialogue with the arXiv moderators”, something of the sort), or simply for insight. -[Meta-question here as to whether this question was appropriate for MO.] -The detailed story (this is long, but I thought important to get all the specifics clear; actual questions follow): -A little over a week ago, I asked a question on MO on a delay-differential equation modeling a variant of the classical SIR epidemiological model where individuals recover in constant time instead of an exponential distribution. A little later, I found that I was able to answer my own question by finding an exact closed-form solution to this model: I wrote a short answer here and, since the answer garnered some interest, a longer discussion on the comparison of both models in a blog post (in French). A number of people then encouraged me to try to give this a little more publicity than a blog post. (My main conclusion is that constant-time recovery, which seems a little less unrealistic than exponential-process recovery, gives a faster initial growth, and a sharper and more pronounced epidemiological peak even assuming a given reproduction number, contagiousness and expected recovery time, while still having the same attack rate: in a world where a lot of modeling is done using SIR, I think this is worth pointing out.) -So I wrote a note on the subject, expanding a little more what I could say about the comparison between this constant-time-recovery variant and classical SIR, adding some illustrative graphs and remarks on random oriented graphs. After getting the required endorsements, I submitted this note to the arXiv (on 2020-04-06) in the math.CA (“Classical Analysis and ODEs”) category. The submission simply vanished without a trace, so I inquired and the arXiv help desk told me that the submission had been rejected by the moderators with the following comment: - -Our moderators have determined that your submission is not of sufficient interest for inclusion within arXiv. This decision was reached after examining your submission. The moderators have rejected your submission as "unrefereeable": your article does not contain sufficient original or substantive scholarly research. -As a result, we have removed your submission. -Please note that our moderators are not referees and provide no reviews with such decisions. For in-depth reviews of your work, please seek feedback from another forum. -Please do not resubmit this paper without contacting arXiv moderation and obtaining a positive response. Resubmission of removed papers may result in the loss of your submission privileges. - -I must admit I didn't know there was even such a thing as arXiv moderation (since there is already the endorsement hurdle to cross) or, if there was, I thought it was limited to removing completely off-topic material and obviously off-topic stuff like proofs that Einstein was wrong. And I certainly disagree about the assessment that my note is “unrefereeable” (it would probably not get past the refereeing process in any moderately prestigious journal, but I find it hard to believe that no journal would even consider sending it to a referee). I made an honest effort to try to ascertain whether the exact-form solution I wrote down had been previously known, and could come up with nothing: but of course, this sort of things is very hard to make sure and I can have missed some general theory which would imply it trivially. I also believe the remarks I make near the end of the note concerning the link between extinction probabilities and attack rates of epidemics, extinction probabilities of Galton-Watson processes, and reachable nodes in random oriented graphs, are of interest. -The problem wouldn't be so bad if I could at least have some kind of dialogue with the arXiv moderators, e.g., inquire into how this judgment was made, and what kind of changes would get it reconsidered. But I wrote to moderation at arxiv.org to ask for clarification and got no answer whatsoever. So I'm taking their advice and trying to “seek feedback from another forum”. -Clearly it was a mistake of mine to submit a note with so few references and I should probably have framed the main result as a precise theorem. Hindsight is always 20/20. Now I don't know whether this can be fixed or whether this would be enough: I have now heard chilling stories about how the arXiv is capricious in banning people silently and permanently for trying to upload something which they don't like, which makes me wish to be careful before I try anything like re-uploading. Also, it may have been a mistake of mine to create my arXiv account using a personal rather than institutional email address, I don't know. -It is also possible that the arXiv is currently overwhelmed by papers on the pandemic and have taken a hard line against anything remotely related to epidemiology or Covid-19. -I thought it best not to upload the note to viXra, which would probably classify it in everyone's eyes as crackpotology, so the best I could come up with (beyond self-hosting) was to place it on “HAL Archives Ouvertes”, a web site created by some French institutions which, however, does not have the same goals as the arXiv (it's more about storing than dissemination) and does not seem to provide a way to publish the source files. -Questions: - -Can somebody provide insight as to how the arXiv moderation process works and how they form their decisions, or why my note might have been rejected? -Is there some way to communicate with the arXiv moderators? Is there any way I could ask for a second opinion after improving my note (e.g., to add many references) and without risking a ban? -Is there anything else I might try to do with that note apart from just giving up? (Various people have suggested bioRxiv or even PLOS One, but I wouldn't want to risk being blacklisted by every scientific preprint site in existence in the attempt to make one result public.) - -REPLY [3 votes]: Let me give some observations with respect to your article: - -Your recollections about classical SIR are possibly too detailed. A short summary of the most important findings might suffice (without extensive proofs like that of Proposition 1.5). - -You could cite e.g. a paper like Hethcote's The Mathematics of Infectious Diseases which summarizes many classical results. - -Your Hypothesis 1.6 isn't a hypothesis but an assumption. - -Your emphasis on the behaviour for $t \rightarrow -\infty$ seems a bit too strong to me. Interesting things happen (only?) for $t \rightarrow +\infty$. I missed a discussion of initial conditions, instead. - -With respect to readability I would suggest to use the variable name $s_{\infty}$ instead of $\Gamma$ (which is the classical name of another transcendental function – next to Lambert's $W$) and to stick to the standard symbol $R_0$ instead of $\kappa$. - -Personally I would have been interested why $s_{\infty} = W(r\cdot e^{r})/r$ is the unique solution of $x = e^{r(1-x)}$ with $r = -R_0$. (I had to find out that this is a classical result.) - - -Hope this helps.<|endoftext|> -TITLE: Fibre preserving maps of Borel constructions -QUESTION [6 upvotes]: Let $G$ be a discrete group with universal principal bundle $EG\to BG$, and let $X$ and $Y$ be left $G$-spaces. An equivariant map $\overline{f}:X\to Y$ induces a fibre-preserving map $f:EG\times_G X\to EG\times_G Y$ between Borel constructions, as in the following diagram. -$\require{AMScd}$ -\begin{CD} - X @>\overline{f}>> Y\\ - @V V V @VV V\\ - EG\times_G X @>>f> EG\times_G Y \\ - @V V V @VV V\\ - BG @>=>> BG -\end{CD} -I believe some sort of converse to be true. That is, if $f:EG\times_G X\to EG\times_G Y$ is a fibre-preserving map between Borel constructions, it induces a $G$-equivariant map $\overline{f}:X\to Y$. Here I am identifying $G=\pi_1(BG,\ast)$, which acts (up to homotopy) on the fibres. - -Does anyone know of a reference to a precise statement along these lines? - -Edit (in response to user51223's comment): I'm mainly interested in the existence question: Is it true that there exists an equivariant map $\overline{f}:X\to Y$ if and only if there exists a map $f:EG\times_G X\to EG\times_G Y$ over $BG$? (So $\overline{f}$ doesn't have to be the induced map on fibres.) - -REPLY [2 votes]: Similar to Granja's answer: let $X$ be a point, and $Y = EG$, and let $f$ be induced by the diagonal on $EG$.<|endoftext|> -TITLE: Cartesian product of Banach spaces: all norms such that the inclusion is an isometry are equivalent? -QUESTION [6 upvotes]: Let $\mathcal{A}$ be an arbitrary (typically infinite-dimensional) Banach space with norm $\|\cdot\|_{\mathcal{A}}$ and let $\mathcal{A}^{n}$ be its Cartesian product. I came across the following statement and wonder whether it is true or not: -All norms $\|\cdot\|_{\mathcal{A}^{n}}$ on $\mathcal{A}^{n}$ such that $\|(0,...,0,\cdot,0,...,0)\|_{\mathcal{A}^{n}}=\|\cdot\|_{\mathcal{A}}$ for all $i=1,...,n$ (i.e. the inclusion is an isometry) are equivalent. -It is clear that for every $x=(x_{1},...,x_{n})\in\mathcal{A}^{n}$ it holds: -\begin{equation} \|x\|_{\mathcal{A}^{n}}\leq\sum_{i=1}^{n}\|x_{i}\|_{\mathcal{A}}=:\|x\|_{1} -\end{equation} -Therefore, $\|\cdot\|_{1}$ is stronger than $\|\cdot\|_{\mathcal{A}^{n}}$. -What what about the other direction? Is it true? -Thanks for your help. - -REPLY [16 votes]: It is actually not true in general that such a norm $\Vert \cdot \Vert_{\mathcal{A}^n}$ must be complete, despite the fact that the contrary is presented as fact in reputable sources in the literature (see, e.g., Section B.4.11 of Albrecht Pietsch's book Operator Ideals [the version published in 1980 by North-Holland] for the case $n=2$]). -A reference for the fact that $\Vert \cdot \Vert_{\mathcal{A}^n}$ need not be complete is the paper of Eve Oja and Peeter Oja, On the completeness of Cartesian products of Banach spaces (in Russian). Acta et commentationes Universitatis Tartuensis, 661 (1984), 33−35. The English summary of the Oja-Oja paper can be read here. -I think the Oja-Oja paper is hard to come by, so when I learnt of its existence many years ago I believe I just derived my own example to satisfy myself. The Oja-Oja paper also notes that equivalence (hence also completeness) is assured when an additional condition is assumed. In particular it follows easily from the triangle inequality that if there exists a constant $c>0$ such that $$ \Vert (-x_1,\ldots,-x_{i-1},x_i,-x_{i+1},\ldots,-x_n)\Vert_{\mathcal{A}^n}\leq c \Vert (x_1,\ldots,x_{i-1},x_i,x_{i+1},\ldots,x_n)\Vert_{\mathcal{A}^n}$$ for all $x_1,\ldots, x_n\in\mathcal{A}$ then $\Vert\cdot\Vert_{\mathcal{A}^n}$ is $(1+c)$-equivalent to the norm $\Vert\cdot\Vert_1$ as defined in your question. -(If I have time I might come back to this later and give some details of how to construct a counterexample, but I recall that it's not particularly difficult). -Edit: Bill Johnson's comment below notes a nice way of obtaining a counterexample.<|endoftext|> -TITLE: The stabilizers of the canonical boundary action of hyperbolic groups -QUESTION [5 upvotes]: My question is that -Is every stabilizer of the canonical boundary action of a hyperbolic group on its Gromov boundary a finitely generated group? -I guess every stabilizer is a (finitely generated) virtually cyclic group, but I do not have a proof nor a reference. -More generally, let G be a countable group that is relatively hyperbolic to subgroups $P_1,\ldots, P_n$. Under which conditions we can conclude that -every stabilizer of the canonical boundary action on its boundary is a (finitely generated) virtually cyclic group or finitely generated virtually abelian group? -Please see section 2 in https://arxiv.org/pdf/1502.04834.pdf for relative hyperbolic groups and their boundaries. -Thank you very much! - -REPLY [4 votes]: This is following my comment, which was getting too long. -Using the notations of the paper you are citing, there are two kinds of points in the boundary : elements of the Gromov boundary $\partial \Gamma$ of the fine graph $\Gamma$ on which $G$ acts and elements in $V_\infty$, which are vertices of $\Gamma$ of infinite valence. The former are called conical limit points and the later are called parabolic limit points. -This boundary equivariantly agrees with the Gromov boundary $\partial X$ of any proper Gromov hyperbolic space on which $G$ acts via a geometrically finite and minimal action (if you want to remove the word minimal, you need to take the limit set $\Lambda G$ of $G$ instead of $\partial X$). This is Proposition 9.1 combined with Theorem 9.4 in Bowditch's paper relatively hyperbolic groups. In particular, you can choose any fine graph $\Gamma$ on which $G$ acts which satisfies Bowdtich's Definition 2 (which is the definition in the paper you're referring to). -To simplify the following, choose $\Gamma$ to be the coned-off graph with respect to the parabolic subgroups $P_1,...,P_n$, or if you prefer Osin's formulation, the Cayley graph $\mathrm{Cay}(G,S\cup P_1\cup ... \cup P_n)$, where $S$ is any finite generating set, which is quasi-isometric to the coned-off graph. Then, the action of $G$ on this graph $\Gamma$ is acylindrical (this is Proposition 5.2 in Osin's paper that AGenevois indicated in their comment). -Now take any point $\xi$ in the Gromov boundary of $\Gamma$ and let $H$ be its stabilizer. Then, the action of $H$ on $\Gamma$ also is acylindrical and $H$ cannot contain infinitely many independent loxodromic elements, so it is virtually cyclic, by Theorem 1.1 of Osin's paper. This settles conical limit points. On the other hand, let $\xi$ be a parabolic limit point. By point (3) of Bowditch's Definition 2, the stabilizer of $\xi$ is exactly one of the peripheral subgroups, that is, with our notations, is one of the conjugates of the $P_k$.<|endoftext|> -TITLE: Are "étalé spaces" a thing for probability spaces? -QUESTION [9 upvotes]: Let $PX$ be a $\sigma$-algebra on the set $X$, and let $j : PX \to {\sf Set}_{/X}$ be the tautological functor that sends an event $E\subseteq X$ to itself, regarded as a function with codomain $X$. Now, the category ${\sf Set}_{/X}$ is cocomplete, thus $j$ has a unique cocontinuous extension to a pair of functors -$$ -J : [PX^\text{op},{\sf Set}] \leftrightarrows {\sf Set}_{/X} : N -$$ here, $J$ is the left Kan extension of $j$ along the Yoneda embedding and $N$ is its right adjoint. -Most of you will have noticed that I am copying the exact procedure that yields the equivalence between sheaves on a topological space $X$ (so, a subcategory of $[OX^\text{op},{\sf Set}]$: open subsets instead of events, but the idea is the same) and étalé spaces (those $(E,p : E\to X) \in {\sf Top}_{/X}$ that are local homeomorphisms). - -I am interested in the properties of the adjunction $(J,N)$. - -I have no exact request apart a little bit of help wrapping my head around this construction, with particular attention to what is different from the topological case. -For the moment, let me just add something about the functor $J = \text{Lan}_yj$: the Kan extension can be written acting on the presheaf $F$ as -$$ -\int^{E\in PX} FE \otimes jE -$$ -(here $\otimes$ is a tensor in ${\sf Set}_{/X}$, so –I think– a coproduct of as many copies of $E \hookrightarrow X$ as there are elements in $FE$; now you perform a suitable quotient on the coproduct of all these $FE \otimes jE$; the colimit is done in Set and then the universal property there yields a unique structure of object in Set/X for the colimit, because –if I remember well that a category is connected if and only if its twisted arrow category is– the colimit defining the coend is over a connected category). -I know this might appear a naive question, but I have always found this construction very specific to sheaf theory (to the point that the very name "sheaf" comes from a pictorial representation of how the functor $[OX^\text{op},{\sf Set}] \to {\sf Top}_{/X}$ acts: all in all, the colimit breaks into the coproduct $\coprod_{x\in X} \text{colim}_{U\ni x} FU$ of all the "stalks" of $F$, that are fibers "stemming" from the "root" $x$... The stalks are then tied together by a certain topology on the disjoint union). - -Is there a similar visual intuition for how $J$ acts on a... well, how would you call it? a pre—? - -REPLY [8 votes]: Etale Spaces can be used to analyze Giry-algebras ($\mathcal{G}$-algebras), and hence (for a fixed object $X$) probability spaces on $X$ as follows. First note that your functor $j$ above should read $j: \Sigma_X \rightarrow \mathbf{Meas}/X$, which is analogous to the topological case (requiring continuous functions rather than just set functions). Here, $\mathbf{Meas}$ is the category of separated measurable spaces - meaning $(2, Discrete)$ is a coseparator of the elements of $X$. Now suppose $\pi_X:\mathcal{G}(X) \rightarrow X$ is a Giry-algebra. (The reason we require separated measurable spaces is that because if $X$ is not separated, then there are no $\mathcal{G}$-algebras on $X$.) -Now the slice category $\mathbf{Meas}/X$ is cocomplete, and one has the same construction as you have noted above, which is just Thm. 2, pp41-42 of Sheaves in Geom. & Logic (SGL), so we have the cited adjunction between the left-Kan extension, $J$, and the functor $N$. Now fix the object $\pi_X$ in $\mathbf{Meas}/X$, and using the adjunct pair $J \dashv N$, look at the universal arrow from $J$ to the object $\pi_X$, i.e., the counit of the adjunction at $\pi_X$. $N(\pi_X)$ is the ''sections functor'', i.e., $N(\pi_X)(U) = \{s: U \rightarrow \mathcal{G}(X) \, | \, \pi_X \circ s = id_U\}$, and $J(N(\pi_X)) = \pi_X$. (Provided I haven't done something foolish, this is just applying the argument in equation 8-10 of the text SGL, p 42, taking $E=\pi_X$ and $P$=sections functor.) -OK. This is all ''standard fare'', and I haven't said anything that answers your question regarding how you interpret presheaves, etc. - and I've yet to work it out. Sheaf spaces are constructed with a horizontal and vertical slice interpretation. Toward that same end note the following two points: -(1) every $\mathcal{G}$-algebra, such as $\pi_X$, specifies a super convex space structure on the underlying set of $X$, via $\sum_{i=1}^{\infty} \alpha_i x := \pi_X( \sum_{i=1}^{\infty} \alpha_i \delta_{x_i})$. More specifically, there is a functor, $\mathbf{Meas}^{\mathcal{G}} \rightarrow \operatorname{\mathbb{R}_{\infty}-\mathbf{SCvx}}$, from $\mathcal{G}$-algebras to $\mathbb{R}_{\infty}$-coseparated super convex spaces. ($\mathbb{R}_{\infty}$ is the one-point extension of the real line $\mathbb{R}$ by a point ''$\infty$'', and that set has the obvious super convex space structure, i.e., $(1-r) u + r \infty = \infty$ for all $r \in (0,1]$.) (The object $\mathbb{R}_{\infty}$ ''arises'' as follows. Every convex space is either a geometric convex space (meaning it embeds into a real vector space), a discrete convex space, or a mixture of the two (which is most common). A geometric space is coseparated by the unit interval $[0,1]$. A discrete space is coseparated by $\mathbf{2}$. In $\mathbf{SCvx}$ there is a map $\mathbf{2} \rightarrow \mathbb{R}_{\infty}$, taking $0 \mapsto \infty$ and $1 \mapsto 0$. The space $\mathbb{R}_{\infty}$ can therefore coseparate any super convex space. (Borger & Kemp showed $\mathbb{R}_{\infty}$ is a coseparator for $\mathbf{Cvx}$, and by restricting to $\operatorname{\mathbb{R}_{\infty}-\mathbf{SCvx}}$ it is a coseparator for that category also.)) -(2) The object $\pi_X: \mathcal{G}(X) \rightarrow X$ is a (weak) terminal object in $\mathbf{Meas}/X$ because if $f: Y \rightarrow X$ is an object in $\mathbf{Meas}/X$ then the composite $\eta_X \circ f: Y \rightarrow \mathcal{G}(X)$ is an arrow to $\pi_X$. We know how the $\sigma$-algebra structure of $\mathcal{G}(X)$ is constructed - via the evaluation maps $ev_U: \mathcal{G}(X) \rightarrow \mathbb{R}$. -Now to the main point. The idea is that the fibers over $x \in X$, which are the ''vertical slices'', specify a (coseparated) super convex space, while the horizontal slices specify the measurable structure. Taking $X = \mathbb{R}_{\infty}$, the $\mathcal{G}$-algebra is the expectation operator, -$\mathbb{E}: \mathcal{G}(\mathbb{R}_{\infty}) \rightarrow \mathbb{R}_{\infty}$, sending $P \mapsto \int_{x \in \mathbb{R}_{\infty}} x dP$. (Taking $P$ to be the half-Cauchy distribution, it is clear why we need $\infty$.) -Now suppose $X$ is an arbitrary (separated) measurable space with the $\mathcal{G}$-algebra $\pi_X$. Then a commutative square, corresponding to a $\mathcal{G}$-algebra morphism $\hat{f}: \pi_X \rightarrow \mathbb{E}$ is specified by a measurable function $f: X \rightarrow \mathbb{R}_{\infty}$, which under the induced super convex space structures on $X$ and $\mathbb{R}_{\infty}$ is also a countably affine map (which is easy enough to verify directly). -This gives the basic idea of how you interpret (part of) the construction you are referring to. -Let me add some context. Your coend formulation is correct - but you can view it from a slightly different point of view. (The coend formulation is Prob. 5 on Page 223, CWM, MacLane.) Let me use MacLanes's notation. Let S be any presheaf, $S: \Sigma_X^{op} \rightarrow \mathbf{Set}$, and take $T: \Sigma_X \rightarrow \operatorname{\mathbb{R}_{\infty}-\mathbf{SCvx}}/\mathcal{G}(X)$ to be given by $U \mapsto (\mathcal{G}(U) \hookrightarrow \mathcal{G}(X))$. Since $\operatorname{\mathbb{R}_{\infty}-\mathbf{SCvx}}$ is cocomplete, so is the slice over $\mathcal{G}(X)$, and an element of that slice category is any ''kernel map'' $k: A \rightarrow \mathcal{G}(X)$. Then the tensor product of $S$ and $T$, which is the coend, is valued in $\operatorname{ \mathbb{R}_{\infty}-\mathbf{SCvx}} / \mathcal{G}(X)$, i.e., the tensor product is a kernel map. (Note that the functor $T$ is just the composite of $j$ and the functor $\hat{\mathcal{P}}: \mathbf{Meas}/X \rightarrow \operatorname{ \mathbb{R}_{\infty}-\mathbf{SCvx}} / \mathcal{G}(X)$ which is induced by the functor $\mathcal{P}: \mathbf{Meas} \rightarrow \operatorname{ \mathbb{R}_{\infty}-\mathbf{SCvx}}$ which is just the Giry monad viewed as a functor into the category of super convex spaces.<|endoftext|> -TITLE: Conway's lesser-known results -QUESTION [230 upvotes]: John Horton Conway is known for many achievements: -Life, the three sporadic groups in the "Conway constellation," surreal numbers, his "Look-and-Say" sequence analysis, the Conway-Schneeberger $15$-theorem, the Free-Will theorem—the list goes on and on. -But he was so prolific that I bet he established many less-celebrated -results not so widely known. Here is one: -a surprising closed billiard-ball trajectory in a regular tetrahedron: - -          - - -          - -Image from Izidor Hafner. - - - -Q. What are other of Conway's lesser-known results? - - -Edit: Professor Conway passed away April 11, 2020 from complications of covid-19: -https://www.princeton.edu/news/2020/04/14/mathematician-john-horton-conway-magical-genius-known-inventing-game-life-dies-age - -REPLY [3 votes]: Conway polynomials are irreducible polynomials that provide a basis for finite fields of order $p^n$. -While there is a unique finite field of order $p^n$, there are many ways of representing its elements as polynomials, all resulting in the same field (up to isomorphism). -Conway polynomials provide a standardize choice of basis. -They are nice because they satisfy a certain compatibility condition with respect to subfields (i.e., fields of order $p^m$ with $m$ dividing $n$). -Formally, the Conway polynomial for $\mathbb{F}_{p^n}$ is defined as the lexicographically minimal monic primitive polynomial of degree $n$ over $\mathbb{F}_p$ that is compatible with the Conway polynomials of all its subfields (see Wikipedia for more details). -Of course, imposing the lexicographical ordering is a convention and is necessary to make them unique. -Conway polynomials are very useful when performing computations using computer algebra systems. -They also provide portability among different systems. -Computing them in general is hard, however for many small cases they have been tabulated -(e.g., see the extensive tables computed by Frank Lübeck). -These tables are available, for example, in Sage.<|endoftext|> -TITLE: Is a gluing of homeomorphic Mazur manifolds diffeomorphic to $S^4$? -QUESTION [10 upvotes]: A recent paper proves the existence of homeomorphic but not diffeomorphic Mazur manifolds (see also examples of exotic pairs of contractible Stein manifolds). -Let's call them $M_1$ and $M_2$. If we glue $W= M_1 \cup_{\partial M_1=\partial M_2} M_2$, then we get a manifold homeomorphic to $S^4$ by Freedman's theorem. Doubling $M_1$ or $M_2$ yields the standard smooth $S^4$ by a theorem of Mazur. But I'm wondering if $W$ is diffeomorphic to $S^4$? -Actually, I don't know if twisting the double of an Akbulut cork can yield an exotic $S^4$? Any exotic pair of manifolds are related by twisting along a cork, so I guess I'm asking whether anything more is known when the complementary contractible manifolds are homeomorphic? - -REPLY [10 votes]: It is diffeomorphic to $S^4$. I have drawn the Kirby picture, please let me know if it is not clear. -(Also I'm sorry that I don't know how to draw Kirby picture in computer so I do old fashioned drawing on notebook) -The Key ideas of the proof are, -1)When we upside down a compact 4 manifold $M$ with 0,1 and 2 handle, then 0 handle becomes 4 handle and 1 handle become 3 handle. Now when we upside down the 2 handle, the co-core become the new attaching circle, which is an unknotted meridian to the orginial attaching cirle of the 2 handle. And the new attaching framing is trivial,i.e, 0. [This is what we use to draw the kirby picture of $M_1\cup M_2$, using the fact that they have identical boundary] -2) 0 framed meridian of a knot always helps to resolve all the crossings of a knot (handle slides).[ This is what we use to simplify the Kirby picture] -3) If a 1 handle and a 2 handle (in the picture a dotted 1 handle and the attaching circle of the 2 handle) geometrically intersect at one point, then they cancel each other. Similarly a 0 framed unknotted 2 handle get canceled by a 3 handle. [this is what we used in last two steps to cancel all but only one 0 handle and one 4 handle and thus we get $S^4$].<|endoftext|> -TITLE: Continuous dependence on initial parameters of an ODE for non-Lipschitz functions? -QUESTION [5 upvotes]: For ODEs, the standard theorem of continuous dependence of initial parameters deals only with functions that are Lipschitz. Do there exist more general results holding for non-Lipschitz functions? If so, could someone direct me to some resources on the topic? - -REPLY [4 votes]: The continuous dependence on initial conditions and parameters (and even on the right-hand side in the compact-open topology) is a consequence of the uniqueness. See Theorem 3.2 of Chapter II in Hartman's "Ordinary differential equations" (I call this statement the Kamke lemma). Note that without the uniqueness, the question of continuous dependence (in the classical sense) is incorrect. So when you have this correctness, you automatically have the continuous dependence.<|endoftext|> -TITLE: History of well founded relations -QUESTION [14 upvotes]: I have changed the title in the hope of attracting the attention of someone who knows about the history of set theory as well as intuitionistic logic: -Who was the first to state the definition of well-foundedness intuitionistically as the induction scheme? -$$ \forall \phi.\quad\frac{\forall x.(\forall y. y\prec x\Rightarrow \phi y)\Rightarrow \phi x}{\forall x.\phi x} $$ -While I'm here, are the following historically correct? - -Euclid's Elements Book VII Proposition 31 (the Euclidean algorithm for prime factorisation) says that an infinite descending sequence is impossible amongst the natural numbers. -What other forms of induction and recursion were stated before the 17th century? -Fermat, Pascal and Wallace in the 1650s stated induction in the form of the base case and induction step. -Cantor 1897 (earlier?) proved that, for any two well ordered sets, one is uniquely equivalent to an initial segment of the other. -Mirimanoff 1917 was the first to recognise the importance of the absence of infinite descending sequences in the consistency of set theory. -von Neumann 1925 proposed the first version of the axiom of foundation, that the system of set theory is the minimal one. -Zermelo 1930 first asserted the axiom of foundation as the lack of infinite descending sequence of elements. -Zermelo 1935 was the first to study well-foundedness in the abstract, as a tool for proof theory. -von Neumann 1928 was the first to prove the recursion theorem for ordinals. - -I am currently working on the categorical reformulation of von Neumann's recursion theorem, for well founded coalgebras: -http://www.paultaylor.eu/ordinals -This includes the full bibliographical details of the above references. -I can read French and Italian reasonably fluently, but unfortunately not German. -Even some guesses would be appreciated! - -REPLY [2 votes]: Emmy Noether must fit in there somewhere. Computer Scientists always mention her when talking about the foundations for making sure that iterative and recursive algorithms terminate. Unfortunately, I don't know of any translations of her work from the German. Bibliography at https://enacademic.com/dic.nsf/enwiki/9878553 where the only relevant work that I see has the note "By applying ascending and descending chain conditions to finite extensions of a ring, Noether shows that the algebraic invariants of a finite group are finitely generated even in positive characteristic.".<|endoftext|> -TITLE: Classification of finitely generated modules over non-commutative rings -QUESTION [5 upvotes]: Let $\Lambda$ be a commutative integral ring with an automorphism $\sigma$ (I have in mind $\mathbb Z_p[[t]]$ and $\sigma(t) = (1+t)^\alpha - 1$ with $\alpha \in \Lambda^\times$) and $R = \Lambda\{F\}$ with $F\lambda = \sigma(\lambda)F$ for $\lambda \in \Lambda$. -Is there a classification of finitely generated modules over $R$ that are free and finite as modules over over $\Lambda$? I allow faithfully flat base changes of $\Lambda$ so that we can assume it's fraction field is algebraically closed (among other things). -Ultimately, I am only interested in the eigenvalues of F, if that makes sense. -When $\Lambda$ is a field, there is a classification similar to the standard one over PID's in chapter three of "The theory of rings" by Nathan Jacobson. -What about the general case or at least my specific example? Or even when $\Lambda$ is a PID? Ideally, I would want any finitely generated module to be isomorphic to a direct sum of modules generated by one element, perhaps up to finite kernel and cokernel. - -REPLY [4 votes]: I haven't thought about base changes, but the original problem for $\alpha=1$ (so $\sigma$ is the identity map and $R=\mathbb{Z}_p[[t]][F]$ is just a polynomial ring over $\mathbb{Z}_p[[t]]$, and classifying $R$-modules that are finitely generated and free over $\mathbb{Z}_p[[t]]$ up to isomorphism is equivalent to classifying square matrices over $\mathbb{Z}_p[[t]]$ up to conjugacy) is a "wild" problem (i.e., if you could classify these then you could classify pairs of matrices over some field up to simultaneous conjugacy), and so is probably intractable. -In fact, Theorem 2 of -Gudivok, P. M.; Oros, V. M.; Rojter, A. V., Representations of finite $p$-groups over the ring of formal power series with integral $p$-adic coefficients, Ukr. Math. J. 44, No. 6, 678-689 (1992); translation from Ukr. Mat. Zh. 44, No. 6, 753-765 (1992). ZBL0787.20006. -shows that the classification of representations of the cyclic group $C_{p^2}$ over $\mathbb{Z}_p[[t]]$ is a wild problem, and this is the subproblem of classifying those matrices whose $p^2$th power is the identity. -For $\alpha\neq1$, I think it should still be a wild problem, as the problem of classifying $R$-modules that are finitely generated and free over $\mathbb{Z}_p[[t]]$ should be at least as hard as classifying representations of finite cyclic groups over $\mathbb{Z}_p$, and this is a wild problem for $G=C_{p^3}$ ($p$ odd) and $C_{16}$ ($p=2$) (see the main theorem of -Dieterich, Ernst, Group rings of wild representation type, Math. Ann. 266, 1-22 (1983). ZBL0506.16021.)<|endoftext|> -TITLE: Do there exist general conditions for cyclicity of unit groups of quotient rings (generalizations of the primitive root theorem)? -QUESTION [5 upvotes]: Let $R$ by a commutative ring with $1$, and $I \subset R$ a non-zero integral ideal in $R$. When $R$ has finite quotients, and $I = P$ is prime in $R$, the group of units $(R/P)^{\times}$ of the finite ring $R/P$ is cyclic as $R/P$ is a finite field. Do there exist known sufficient and necessary conditions on $R$ and $I$ in general or for certain classes of unital rings for cyclicity of $(R/I)^{\times}$ ? In particular, do there exist more general analogues of the primitive root theorem, which answers this question for $R = \mathbb{Z}$ in terms of number-theoretic criteria on the positive generators of the principal ideals $I = (n)$? - -REPLY [4 votes]: $\newcommand{\F}{\mathbb{F}} \newcommand{\Z}{\mathbb{Z}}$ -[Throughout this answer all rings will be commutative (and unital!).] -It seems that van Dobben de Bruyn has essentially rediscovered a theorem of Gilmer: - -Gilmer, Robert W., Jr. Finite rings having a cyclic multiplicative group of units. Amer. J. Math. 85 (1963), 447-452. -A couple of preliminary comments: (i) In van Dobben de Bruyn's answer, we may as well take $I = (0)$: that is, he is giving necessary and sufficient conditions on a finite commutative ring to have cyclic unit group. (ii) A finite ring $R$ is indeed Artinian, hence a finite product $\prod_{i=1}^r R_i$ of local rings $R_i$, each of which must have prime power order. As seen in his answer, we quickly find that $R^{\times}$ -is cyclic iff each $R_i^{\times}$ is cyclic and $\# R_1^{\times},\ldots, \# R_r^{\times}$ -are pairwise coprime. Thus the critical case is the classification of finite local rings with cyclic unit group. Here is Gilmer's result: - -Theorem Let $R$ be a finite local ring. Then $R^{\times}$ is cyclic iff $R$ is isomorphic to one of the following rings: - (A) A finite field $\F$. - (B) $\Z/p^a \Z$ for an odd prime number $p$ and $a \in \Z^+$. - (C) $\Z/4\Z$. - (D) $\Z/p\Z[t]/(t^2)$ for a prime number $p$. - (E) $\Z/2\Z[t]/(t^3)$. - (F) $\Z[t]/\langle 2t,t^2-2 \rangle$, a $\Z/4\Z$-algebra of order $8$. - -To compare Gilmer's classification to van Dobben de Bruyn's it is helpful to observe that the local rings of order $p^2$ are $\F_{p^2}$, $\Z/p^2\Z$ and $\Z/p\Z[t]/(t^2)$ and to know the six local rings of order $8$. -By the way, Gilmer's Theorem appears as Theorem 5.14 in this expository note of mine, where it used to derive Theorem 5.15, a 2013 result of Hirano-Matsuoka that explicitly determines the product over all elements of the unit group of a finite ring. (Thus it is a generalization of Wilson's Theorem that $(p-1)! \equiv -1 \pmod{p}$. It seems weird that it is so recent.) I wanted to include the proof of Gilmer's Theorem in the note, but it is rather long and computational. The proof of van Dobben de Bruyn looks a bit shorter! -A final comment leading to a question: It turns out that all the rings in Gilmer's classification are principal, i.e., every ideal is principal. (This is obvious except for (F), in which case you can see my paper if you don't want to do the calculation yourself.) In other words, for a finite ring $R$ the property that the unit group be cyclic forces every $R$-submodule of $R$ to be cyclic. Is that just a coincidence, or can it be proved directly?<|endoftext|> -TITLE: Is the connected centralizer of a semisimple element in a connected reductive group also a centralizer? -QUESTION [5 upvotes]: Let $G$ be a connected reductive algebraic group defined over an algebraically closed field and let $g\in G$ be semisimple. Write $C=\mathrm{C}_G(g)$ and $C^\circ=\mathrm{C}_G(g)^\circ$ for the centralizer of $g$ and for its identity component, respectively. - -Question Does there exist $h\in G$ such that $\mathrm{C}_G(h)=C^\circ$? - -The statement is true in the case where $C^\circ$ is the Levi factor of a parabolic subgroup of $G$. The proof I know is by computing dimensions. -Fix $T$ to be a maximal torus containing $g$, and let $\Sigma$ (resp $\Phi$) denote the root system of $C^\circ$ (resp of $G$) with respect to $T$. Then, for $h\in T$, it holds that $\mathrm{C}_G(h)=C^\circ$ if and only if $\{\alpha -\in \Phi:\alpha(h)=1\}=\Sigma$ and $\mathrm{Stab}_W(h)$ (the stabilizer of $h$ in the Weyl group of $G$) is generated by the reflections $s_\alpha$ for $\alpha\in \Sigma$. -Given a root systems $\Sigma\subseteq \Psi\subseteq\Phi$ and subgroups $\langle{s_\alpha:\alpha\in \Sigma}\rangle\subseteq S\subseteq W$, put -$$T_\Psi^S=\{h\in T: \{\alpha\in \Phi:\alpha(h)=1\}=\Psi\text{ and }\mathrm{Stab}_W(h)\supseteq S\}.$$ -Then, using the assumption that $\Sigma$ is the root system of a Levi subgroup in several key steps, one can show that $\dim T_\Psi^S$ attains a maximum if and only if $\Psi=\Sigma$ and $S=\langle{s_\alpha:\alpha\in \Sigma}\rangle$. Therefore $T_\Sigma^{\langle s_\alpha:\alpha\in \Sigma\rangle}\setminus (\bigcup T_\Psi^S)$ is non-empty, and contains the $h$ we are seeking. -However, this proof falls apart completely if we take $\Sigma$ to be an arbitrary (closed) subsystem of $\Phi$. -I would very much appreciate if anyone could either suggest a different proof for this statement, which hopefully extends to general centralizers, or otherwise provide a counterexample for my question. Thank you. - -REPLY [3 votes]: So after thinking about this for a few extra days, I think I found a counterexample. I'm recording it here, as community wiki, in case it would be of interest for anyone in the future. - -Let $k$ be an algebraically closed field of characteristic not $2$ and $G=\mathrm{PGSp}_{2n}(k)=\mathrm{GSp}_{2n}(k)/k^\times$, where $\mathrm{GSp}_{2n}(k)$ is the group of similitudes of the standard symplectic form, i.e. -$$\mathrm{GSp}_{2n}(k)=\{x\in\mathrm{GL}_{2n}(k):x^tJx=\lambda J\text{ for some }\lambda\in k^\times\}\text{ where }J=\begin{pmatrix}0&I_n\\-I_n&0\end{pmatrix},$$ -(I don't know if this notation is common). -$G$ is simple of adjoint type with maximal torus $T=\lbrace d(t_1,t_2):=\left[\begin{smallmatrix}t_1\\&t_2\\&&t_1^{-1}\\&&&t_2^{-1}\end{smallmatrix}\right]:t_1,t_2\in k^\times\rbrace$, (here $[\cdot]$ denotes the class mod $k^\times$ of a matrix; note that $d(\lambda t_1,\lambda t_2)=\lambda d(t_1,t_2)$ implies $\lambda=\pm 1$) and root system with simple roots: -$$\alpha(d(t_1,t_2))=t_1/t_2\text{ and } \beta(d(t_1,t_2))=t_2^2$$ -(the other positive roots are $\alpha+\beta$ and $2\alpha+\beta$). -Consider the subsystem $\Sigma=\lbrace\pm \beta,\pm(2\alpha+\beta)\rbrace$ (viz. the long roots). Then $\Sigma$ is the root system of a pseudo-Levi subgroup of $G$ which is isomorphic to $(\mathrm{GL}_n(k)\times\mathrm{GL}_n(k))/k^\times$, and one can easily verify that -$$(\star)\quad \beta(d(t_1,t_2))=(2\alpha+\beta)(d(t_1,t_2))=1\:\iff\: d(t_1,t_2)\in\lbrace\left[\begin{smallmatrix}1\\&1\\&&1\\&&&1\end{smallmatrix}\right],\left[\begin{smallmatrix}1\\&-1\\&&1\\&&&-1\end{smallmatrix}\right]\rbrace.$$ -Let $g$ be the non-central element in this set. Then, a standard computation, taking into account that $g=-g$ in $G$, shows that $C_G(g)$ is disconnected (the non-identity connected component is generated by the coset of the Weyl group element permuting $t_1$ and $t_2$). On the other hand, $Z(C_G(g))$ consists of precisely the two elements on the RHS of $(\star)$, so there exists no $g\ne h\in Z(C_G(g))$ such that $C_G(h)\subseteq C_G(g)$, and, in particular, the question above has a negative answer in this case.<|endoftext|> -TITLE: Paths through convergent sequences in $\Delta$-generated spaces -QUESTION [7 upvotes]: So-called $\Delta$-generated spaces are topological spaces in which paths "determine" the topology of the space. In particular, $X$ is $\Delta$-generated if a set $U\subseteq X$ is open (resp. closed) if and only if $\alpha^{-1}(U)$ is open (resp. closed) in $[0,1]$ for all paths, i.e. continuous functions, $\alpha:[0,1]\to X$. The $\Delta$-generated spaces form a coreflective convenient category of topological spaces that relate to some familiar properties. For instance, it is true in general that: - -first countable and LPC $\Rightarrow$ $\Delta$-generated $\Rightarrow$ sequential and LPC - -where LPC abbreviates "locally path connected." -Definition: Let's say that that a topological space $X$ is path-sequential if for every convergent sequence $\{x_n\}\to x$ in $X$, there is a path $\alpha:[0,1]\to X$ such that $\alpha(0)=x$ and $\alpha(1/n)=x_n$ for all $n\in\mathbb{N}$. -With some basic arguments, it becomes clear that we have: - -first countable and LPC $\Rightarrow$ sequential and path-sequential $\Rightarrow$ $\Delta$-generated $\Rightarrow$ sequential and LPC - -The first and third implications are definitely not reversible. -Question: Must every $\Delta$-generated space be path-sequential? -I am really just interested in the case where $X$ is Hausdorff or is at least a US-space, i.e. a space where convergent sequences have unique limits. -Note: it's easy to see that if $X$ is $\Delta$-generated and a US-space, then for every convergent sequence $\{x_n\}\to x$ there exists a path $\alpha:[0,1]\to X$ such that $\alpha(0)=x$ and $\alpha(1/k)=x_{n_k}$ for some subsequence $\{x_{n_k}\}$. - -REPLY [3 votes]: The answer to the question is negative. To construct a counterexample, choose a maximal almost disjoint infinite family $\mathcal A$ of infinite subsets of $\omega$. -Endow $\mathcal A$ with the discrete topology and consider the product $[0,1]\times \mathcal A$. For every subset $A\subseteq \omega$, let $$2^{-A}=\{0\}\cup\{2^{-n}:n\in A\}.$$ -Let $X$ be the topological sum $2^{-\omega}\cup([0,1]\times\mathcal A)$, and $\sim$ be the smallest equivalence relation on the space $X$ such that $0\sim (0,A)$ and $2^{-n}\sim(2^{-n},A)$ for every $A\in\mathcal A$ and $n\in A$. It can be shown that the quotient space $Y=X/_\sim$ is a required counterexample: $Y$ is $\Delta$-generated but not path-sequential (the latter follows from the fact that $S$ is not contained in a path-connected compact subspace of $Y$). -To be sure that everything works, let us write down the proof of the following -Fact. The space $Y$ is $\Delta$-generated. -Proof. The space $Y$ can be identified with the union $$2^{-\omega}\cup\bigcup_{A\in\mathcal A}([0,1]\setminus 2^{-A})\times\{A\},$$ endowed with a suitable topology. Let $q:X\to Y$ be the quotient map. -Take any non-closed set $C\subset Y$. If there exists some $y\in(\bar C \setminus C)\setminus 2^{-\omega}$, then there exists a unique set $A\in\mathcal A$ such that $y\in ([0,1]\setminus 2^{-A})\times\{A\}$. In this case for the map $\gamma_A:[0,1]\to Y$, $\gamma_A(t)\mapsto q(t,A)$, has the desired property: $\gamma_A^{-1}(C)$ is not closed in $[0,1]$. -So, we assume that $\bar C\setminus C\subseteq 2^{-\omega}$. First assume that $2^{-n}\in\bar C\setminus C$ for some $n\in\omega$. Choose two real numbers $a,b$ such that $2^{-n-1} -TITLE: Could the 4-color theorem be proven by contracting snarks? -QUESTION [7 upvotes]: Suppose someone came up with an algorithm that could take any snark and perform edge contraction to result in the Peterson graph. If an inspection of the algorithm reveals that it works as claimed, would the algorithm be sufficient to prove the 4CT? - -REPLY [15 votes]: Yes, the 4-colour theorem is true if and only if every snark is non-planar (this is due to Tait). -Showing that a snark has a Petersen minor would be enough to show that it is non-planar.<|endoftext|> -TITLE: An inequality involving square roots and sums -QUESTION [9 upvotes]: I've been trying to prove (maybe even disprove) the following inequality: -$$ -\sum_{n=1}^{N} \frac{a_n}{\sqrt{\sum_{i=1}^{n}a_i}} \leq C \sqrt{\sum_{n=1}^{N}a_n} -$$ -Where $ a_1,...,a_N\geq 0 $ are some non-negative numbers, and $C$ is an absolute constant. -Help will be much appreciated. - -REPLY [18 votes]: For every $n\in\{1,\dotsc,N\}$, we have -$$2\sqrt{\sum_{i\leq n} a_i}-2\sqrt{\sum_{i\leq n-1} a_i}=\frac{2a_n}{\sqrt{\sum_{i\leq n} a_i}+\sqrt{\sum_{i\leq n-1} a_i}}>\frac{a_n}{\sqrt{\sum_{i\leq n} a_i}}.$$ -Summing these up, we obtain the inequality with $C=2$. It is also straightforward to see that for $C<2$ the inequality fails, hence $C=2$ is the optimal constant. - -REPLY [14 votes]: Rewrite your inequality as -$$lhs:=\sum_{n=1}^N \frac{s_n-s_{n-1}}{\sqrt{s_n}}\le C\sqrt{s_N},$$ -where $s_n:=\sum_{i=1}^n a_i$. Note that -$$\sum_{n=1}^N \frac{s_n-s_{n-1}}{\sqrt{s_n}}$$ -is a lower Riemann sum for the integral -$$\int_0^{s_N}\frac{ds}{\sqrt s}=2\sqrt{s_N}.$$ -So, -$$lhs\le2\sqrt{s_N},$$ -as desired. - -In the above proof, it was tacitly assumed that $a_i>0$ for all $i$. This can be obviously extended to the case when we only know that $a_i\ge0$ for all $i$ -- assuming that, by continuity, -$\frac{a_n}{\sqrt{\sum_{i=1}^{n}a_i}}:=0$ whenever $a_n=0$.<|endoftext|> -TITLE: Results that are widely accepted but no proof has appeared -QUESTION [75 upvotes]: The background of this question is the talk given by Kevin Buzzard. -I could not find the slides of that talk. The slides of another talk given by Kevin Buzzard along the same theme are available here. -One of the points in the talk is that, people accept some results but whose proofs are not publicly available. (He says this leads to wrong conclusions, but, I am not interested in wrong conclusions as of now. All I am interested is are results which are accepted as true but without a detailed proof, or with only a partial proof.) - -What are results that are widely accepted to be true with no detailed proof, or only a partial proof? - -I am looking for situations where $A$ has asserted in print that he/she has a proof of $X$, but hasn't published a proof of $X$, and then $B$ publishes a proof of $Y$, where the proof depends on the validity of $X$. For example as in page 20,21,22 of the slides mentioned above. -Edit: Please give reference for the following: - -Where the result is announced? -Where the result is used? - -Edit (made after Per Alexandersson's answer) : I am not looking for "readily available but not formally published". As mentioned by Timothy Chow, "there are many more examples if "readily available but not formally published" counts.". - -REPLY [6 votes]: A long time ago, M. Ajtai, J. Komlós, M. Simonovits, and E. Szemerédi announced a proof (for large $k$) of the Erdős–Sós conjecture that every graph with average degree -more than $k-1$ contains all trees with $k$ edges as subgraphs, but the proof has not yet appeared as of this writing (2022). -What do I mean by "a long time ago"? Reinhard Diestel, in the notes to Chapter 7 of Graph Theory (5th edition), gives a date of 2009. But Václav Rozhoň, in A local approach to the Erdős-Sós conjecture, says that the result was announced in the early 1990's. - -EDIT: I found another reference, Local and mean Ramsey numbers for trees, by B. Bollobás, A. Kostochka, and R. H. Schelp (J. Combin. Theory Ser. B 79 (2000), 100–103), which says, "It was -announced recently that M. Ajtai, J. Komlós, and E. Szemerédi confirmed -the Erdős–Sós conjecture for sufficiently large trees."<|endoftext|> -TITLE: Is $Tor_A(k,k)$ a bicommutative Hopf algebra? -QUESTION [7 upvotes]: Let $A$ be a commutative (or graded commutative) algebra over a field $k.$ In some sources, such as Mcleary's book on spectral sequences, Corollary 7.12, pg. 248, it is claimed that $\text{Tor}_A(k,k)$ is a bicommutative Hopf algebra. Is this a typo? Obviously, $\text{Tor}$ has a commutative algebra structure, but is it true that the coaddition is cocommutative? - -REPLY [9 votes]: This is not true. Consider the algebra $A=T(V)/V^{\otimes 2}$, it is a commutative algebra whose augmentation ideal has zero multiplication. We have $\mathrm{Tor}_A(k,k)\cong T(V[1])$ with the shuffle product and deconcatenation coproduct, so the coproduct is very much noncommutative (it is the coproduct of the cofree conilpotent coalgebra co-generated by $V[1]$).<|endoftext|> -TITLE: Equivalence of σ-convex hull and closed convex hull -QUESTION [10 upvotes]: Let $X$ be a locally convex topological space, and let $K \subset X$ be a compact set. Recalling that the standard convex hull is defined as -$$\text{co}(K) = \Big\{ \sum_{i=1}^n a_i x_i : a_i \geq 0,\, \sum_{i=1}^n a_i = 1,\, x_i \in K \Big\},$$ - define the $\sigma$-convex hull as -$$\sigma\text{-}\mathrm{co}(K) = \Big\{ \sum_{i=1}^\infty a_i x_i : a_i \geq 0,\, \sum_{i=1}^\infty a_i = 1,\, x_i \in K \Big\},$$ -where the summation is to be understood as convergence of the sequence in the topology of $X$. -I would like to understand conditions under which $\sigma\text{-}\mathrm{co}(K)$ is exactly the closure of $\mathrm{co}(K)$. In particular, does this property hold for any separable normed space $X$, or are further constraints on $X$ (and $K$?) required? -The motivation for this question is Choquet's theorem, which allows one to write -$$\overline{\mathrm{co}}(K) = \Big\{ \int x d\mu(x) : \mu \in M(K) \Big\}$$ -with $M(K)$ standing for probability measures on $K$ for any compact subset $K$ in a normed space. I would like to understand the "countable" version of this theorem as presented above, but I could not find any references nor do I have an idea about how one could prove it. - -REPLY [5 votes]: Wlod AA gave a good counterexample for the case when $K$ is not required to be compact, here I give a counterexample $K$ compact, first in a locally convex space, and then for a(n infinite-dimensional) separable normed space, and (after an edit) for all infinite-dimensional Banach spaces. -There is a standard counterexample if $X$ is only required to be locally convex, which is to take $X = C([0,1])^*$ with the weak-* topology, and to take $K$ to be the set of unital ring homomorphisms $C([0,1]) \rightarrow \mathbb{R}$. Making free use of the Riesz representation theorem to consider elements of $C([0,1])^*$ as measures on $[0,1]$, the elements of $K$ are the Dirac $\delta$-measures. Now, for each element $\mu$ of $\sigma\mbox{-}\mathrm{co}(K)$, there exists a countable set $S \subseteq [0,1]$ such that $\mu([0,1]\setminus S) = 0$. However, $\overline{\mathrm{co}}(K)$ consists of $P([0,1])$, the set of all positive unital linear functionals on $C([0,1])$, i.e. all probability measures on $[0,1]$, and so Lebesgue measure is an element of $\overline{\mathrm{co}}(K) \setminus \sigma\mbox{-}\mathrm{co}(K)$. -To get this to happen in a normed space, we will use $\ell^2$, and embed $P([0,1])$ affinely and continuously into it. First, observe that we can affinely embed $P([0,1])$ into $[0,1]^{\mathbb{N}}$, getting each coordinate by evaluating at $x^n$ (including $n = 0$). This is injective because polynomials are norm dense in $C([0,1])$, and continuous by the definition of the weak-* topology. We can then embed $[0,1]^{\mathbb{N}}$ into $\ell^2$ by the mapping: -$$ -f(a)_n = \frac{1}{n+1}a_n -$$ -this is affine and continuous from the product topology on $[0,1]^\mathbb{N}$ to the norm topology on $\ell^2$ (in fact, it defines a continuous linear map from the bounded weak-* topology on $\ell^\infty$ to the norm topology on $\ell^2$). We use $e$ for the composition of these two embeddings, and it is affine and continuous on $P([0,1])$. -A continuous injective map from a compact Hausdorff space to a Hausdorff space is a homeomorphism onto its image, and as we also preserved convex combinations by making the embedding affine, we have that $\overline{\mathrm{co}}(e(K)) = e(\overline{\mathrm{co}}(K)) = e(P([0,1]))$, while, taking $\lambda$ to be the element of $P([0,1])$ defined by Lebesgue measure, $e(\lambda) \in e(P([0,1]))$, but $e(\lambda) \not\in e(\sigma\mbox{-}\mathrm{co}(K)) = \sigma\mbox{-}\mathrm{co}(e(K))$. - -Added in edit: -As Bill Johnson points out, there is an injective bounded map from $\ell^2$ into any infinite-dimensional Banach space $E$. By the same argument used to transfer the example to $\ell^2$, this allows us to transfer the example to $E$. -In the other direction, the convex hull of a compact subset $K$ of a finite-dimensional space is compact (using Carathéodory's theorem we can express the convex hull of $K$ as the continuous image of the compact set $K^{d+1} \times P(d+1)$, where $d$ is the dimension. Therefore the $\sigma$-convex hull and closed convex hull of $K$ coincide. -All together, this means: - -If $E$ is a Banach space, the statement "for all compact sets $K \subseteq E$, the closed convex hull equals the $\sigma$-convex hull" is equivalent to "$E$ is finite-dimensional". - -There are, however, complete locally convex spaces in which every bounded set, and therefore every compact set, is contained in a finite-dimensional subspace, and for which, therefore, the $\sigma$-convex and closed convex hulls of compact sets coincide. One example is the space $\phi$ of finitely supported functions $\mathbb{N} \rightarrow \mathbb{R}$, topologized as an $\mathbb{N}$-fold locally convex coproduct of $\mathbb{R}$ with itself, or equivalently as the strong dual space of $\mathbb{R}^{\mathbb{N}}$.<|endoftext|> -TITLE: The Angel and Devil problem with a random angel -QUESTION [16 upvotes]: In the classic version of Conway's Angel and the Devil problem, an angel starts off at the origin of a 2-D lattice and is able to move up to distance $r$ to another lattice point. The devil is able to eat a lattice point, preventing the angel from ever moving to that point. The angel and devil take turns, and the devil wins if the angel is at some point no longer able to move. The question then is for what values of $r$ does the angel win and for what values does the devil win? This problem was essentially solved completely by distinct proofs by Kloster and Máthé that the 2-angel can escape. (It is easy to see that a 1-angel can be trapped.) One can generalize this problem to higher dimensions; note that if for a given choice of $r$ the angel can escape in $d$ dimensions, then the angel will escape for any higher $d$. -What I'm interested in is the situation where the angel moves randomly (uniformly distributed among all possible legal moves) but the devil has a pre-determined strategy (allowed to depend on $r$ and $d$ but not allowed to depend on any choices the angel has made). For what $r$ and $d$ can the devil beat an angel with probability 1? -It isn't too hard to see that if $d=2$ the devil can win with probability 1. Here's the basic strategy the devil uses: Pick a very fast growing sequence of positive integers, $a_1$, $a_2$, $a_3 \cdots$. The devil works in stages: At each stage, the devil eats a square of side length $a_n$ centered about the origin, and with thick walls of thickness $r$. Each such square requires about $4ra_n+r^2 \sim 4ra_n$ moves by the devil. But by the standard result that a random walk is with probability 1 never much more than the square root of the number of steps away from the origin, in the time the devil has taken to eat the $a_n$ square, the angel with probability 1 will have only moved about $\sqrt{4r}\sqrt{a_n}$ steps from the origin. So, the devil just creates larger and larger squares of this sort, and eventually the angel will be trapped. (This by itself will put the angel in a finite region, but trapping in a finite region is essentially the same as being unable to move since the devil can go back and fill in these squares ever so slowly, say eating a single lattice point near the origin before moving on to start making each new large square. -This construction fails for 3 dimensions. To make a cube of that size takes about $6a_n^2$ steps, so the angel has a high probability of being near the boundary. -Question 1: can this strategy or a similar one be modified to work for $d=3$? My guess is yes for $d=3$, but I don't have a proof. I also don't have any intuition for higher dimension. -One standard observation which simplifies the analysis of the original problem is that one may without loss of generality assume that the angel never returns to the same lattice point. If it did, it would have used a suboptimal strategy, since it is back where it was earlier but with the devil having eating out a few lattice point. So, we can define another variant of the problem with an angel which chooses randomly, but only out of lattice points it has not yet reached. -Question 2: Given this non-repeating angel, is there a strategy for the devil to win with probability 1? -I suspect that the answer for $d=2$ is that the same basic strategy should still work; my suspicion here is that with probability 1, the angel's distance at $k$ steps should be bounded by $k^{(\frac{1}{2}+\epsilon)}$ in which case the same proof would go through. But I'm much less certain about what happens here if $d=3$. - -REPLY [3 votes]: Doesn't the non-repeating angel trap itself with probability 1 even without the devil's help? I mean there is always going to be some finite set of moves the angel can make that leave it nowhere to go. Since it has infinitely many tries, it should make this sequence of moves eventually.<|endoftext|> -TITLE: Approximating a projection by a sum of elementary tensors with a certain property -QUESTION [5 upvotes]: Let $A$ and $B$ be two C$^{*}$-algebras and suppose we have a non-zero projection $p\in A\otimes B$. (We can assume $A$ is nuclear, so that there is only one possible tensor product.) -Does there exist a choice of elements $a_{1},\ldots,a_{n}\in A$ and $b_{1},\ldots,b_{n}\in B$ such that: - -$\left\|\left(\sum_{i=1}^{n}a_{i}\otimes - b_{i}\right)-p\right\|<\frac{1}{2}$; -If $\pi$ is a non-zero irreducible representation of $A$ and $(\pi\otimes\operatorname{id})(p)=0$, then $\pi\left(\sum_{i=1}^{n}|a_{i}|\right)=0$? - -All I can deduce is that for a given choice of $a_{i}$'s and $b_{i}$'s, and and irreducible representation $\pi$ of $A$, we have -$$ -\left\|\sum_{i=1}^{n}\pi(a_{i})\otimes b_{i}\right\|=\left\|(\pi\otimes \operatorname{id})\left(\left(\sum_{i=1}^{n}a_{i}\otimes b_{i}\right)-p\right)\right\|\leq \left\|\left(\sum_{i=1}^{n}a_{i}\otimes b_{i}\right)-p\right\|<\frac{1}{2}, -$$ -provided that $(\pi\otimes\operatorname{id})(p)=0$. - -REPLY [6 votes]: Yes, this is possible, assuming that $A$ (or $B$) is nuclear. The same argument below (using that exact $C^\ast$-algebras are locally reflexive) also works if $A$ or $B$ is exact and the tensor product is the spatial (aka minimal) tensor product. -The role of nuclearity is that any two-sided closed ideal $J \subseteq A\otimes B$ is the closed linear span of all tensor products $I_A \otimes I_B$ of two-sided closed contained in $J$, see Corollary 9.4.6 in Brown and Ozawa's book. Now, let $J_A \subseteq A$ be the intersection of all two-sided, closed ideals $J \subseteq A$ such that $p \in J \otimes B$ and let $I:= \bigcap J \otimes B$ where the intersection is indexed by such $J$. Clearly $J_A \otimes B \subseteq I$. For the converse (which is not as trivial as it a priori looks), take $I_A \otimes I_B \subseteq I$ where $I_A$ and $I_B$ are two-sided closed ideals in $A$ and $B$ respectively. Then $I_A \subseteq J_A$ and $I_B \subseteq B$, so $I_A \otimes I_B \subseteq J_A \otimes B$. By nuclearity of $A$, $I$ is the closed linear span of all such $I_A\otimes I_B$, so $I = J_A \otimes B$. In particular, $p\in J_A \otimes B$. -Now, let $a_1,\dots, a_n \in J_A$ and $b_1,\dots, b_n \in B$ such that -\begin{equation} -\| \sum_{i=1}^n a_i \otimes b_i - p \| < 1/2. -\end{equation} -Let $\pi \colon A \to \mathcal B(\mathcal H)$ be an irreducible representation such that $(\pi \otimes \mathrm{id}_B)(p) = 0$. The image of $(\pi \otimes \mathrm{id}_B)$ is canonically $\pi(A) \otimes B$, so by nuclearity of $A$ (or $B$) and exactness of maximal tensor products, it follows that -\begin{equation} -0 \to (\ker \pi)\otimes B \to A \otimes B \to \pi(A) \otimes B \to 0 -\end{equation} -is exact, so the kernel of $(\pi \otimes \mathrm{id}_B)$ is $(\ker \pi )\otimes B$. Hence $p \in (\ker \pi)\otimes B$ so $a_1,\dots, a_n \in J_A \subseteq \ker \pi$ by construction of $J_A$. -If $A$ or $B$ is exact, we get $\ker (\pi \otimes \mathrm{id}_B)$ either by using that $A$ is locally reflexive or $B$ is exact to get that -\begin{equation} -0 \to (\ker \pi) \otimes_{\min{}} B \to A \otimes_{\min{}} B \to \pi(A) \otimes_{\min{}} B \to 0 -\end{equation} -is exact.<|endoftext|> -TITLE: How to explain this prime gap bias around last digits? -QUESTION [6 upvotes]: My question is related to this article by Oliver and Soundararajan (article about a bias in the distribution of the last digits of consecutive prime numbers). -After trying some python experimental researches on the prime gaps, I notice a bias in the gaps around each last digits {1,3,7,9}. -Could this bias be intrinsically linked to the bias found in the distribution of the last digits of consecutive prime numbers? -Experimental results -Here some experimental plot results up to the 3,500,000th prime number. -Two set-ups are calculated: -1. Cumulative and average gaps -2. Cumulative and occurence of particular gap size - -1. Cumulative and average gaps - -1.1 Cumulative/average prime gaps after last digits: - - + Cumulative/average prime gaps before last digits - -1.2 Cumulative/average prime gaps before + after last digits - - - -2. Cumulative and occurence of a particular gap size - -2.1 Cumulative/occurence of gaps size of 10 after last digits - - - - Animated cumulative/occurence per incremented gap sizes after last digits - -3. Misc - -3.1 Additional research on a waveform script of the sieve of Ératosthène, -colored by last digits: - + plot 2 - - -The script is on colab.research and is ready to fork - -REPLY [10 votes]: Yes, it can be analyzed in the same way. Since the effects you're measuring are similar to each other, I'm only going to address the first one, cumulative prime gaps after primes with a fixed last digit. In Lemke Oliver and Soundararajan's paper, they count pairs of consecutive primes $p$ and $p_{\text{next}}$ up to $X$ satisfying $p \equiv a$ and $p_{\text{next}} \equiv b \pmod{10}$, or a sum of the form -$$\sum_{\substack{h > 0 \\ h \equiv b-a}} \#\{ p \le x : p \equiv a \pmod{10}, p_{\text{next}} - p = h\}.$$ -This setting can be described by two variations to the above sum. First, by summing the prime gaps, each term is being weighted by $h$. Second, since it's all prime gaps after primes with a fixed last digit, the expression is also summed over possible choices of $b$. -Armed with these modifications their analysis should follow through very similarly. For example, the proof of Prop. 2.1 in their paper can be carried out, but instead of considering -$$F_{q,\chi}(s) = \sum_{h \ge 1} \frac{\chi(h)}{h^s} \mathfrak S_q(\{0,h\}),$$ -weighing each term by a factor of $h$ means that instead we're considering $F_{q,\chi}(s-1)$. The poles of this function have shifted, so the corresponding sum for $S_0(q,0;H)$ in the statement of Prop. 2.1 would have a leading term of linear size in $H$, rather than log size. -As Lucia commented, the bias seems to be on a smaller scale than the bias from the paper. This reduction in bias should be a result of summing over all values of $b$, which has some averaging effect. As a test of this, we can consider the Main Conjecture on page 2 and sum the $\log \log x /\log x$ term over all values of $b$ with fixed $a$; the result is that it disappears entirely, leaving the $1/\log x$ term. -Another crude experiment can be done by taking the data on page 2 of their paper, assuming that the gap is larger than zero but otherwise as small as possible, and computing what the corresponding sum would be in this case. For $a = 1$, this would be the sum $$10 \cdot \pi(x_0;10,(1,1)) + 2 \cdot \pi(x_0;10,(1,3)) + 6 \cdot \pi(x_0;10,(1,7)) + 8 \cdot \pi(x_0;10,(1,9)).$$ To compare to your data, one can take this a step further by treating the total of these sums as a ``total prime gap'' and taking the percentage of the total for each. The result of doing this with the values from page 2 is: -\begin{array}{c|cc} -a & \text{Sum} & \text{Percent of total} \\ -\hline -1 & 149655728 & 25.96 \\ -3 & 165705632 & 28.75 \\ -7 & 125284384 & 21.73 \\ -9 & 135805698 & 23.56 \\ -\end{array} -A huge limitation here is that these sums are heavily influenced by the chance that numbers coming immediately after $a$ have of being prime (in particular, they are in increasing order beginning at 5 mod 10). Lemke Oliver and Soundararajan explicitly address this effect, showing that it doesn't account for the full bias. And, in fact, this doesn't quite match the data that you have. Nevertheless, some discrepancy is visible even in this crude model.<|endoftext|> -TITLE: Fourier transform on Minkowski space -QUESTION [9 upvotes]: Physicists Some people like to define the "Fourier transform" on Minkowski space as $\hat f(\xi) = \int e^{i \eta(x,\xi)} f(x) dx$, where $\eta(x,\xi)$ is the Minkowski form. I'm used to thinking of the Fourier transform as a canonical isomorphism $L^2(K) \to L^2(\hat K)$ where $K$ is a locally compact abelian group and $\hat K$ is its Pontryagin dual. But this "Minkowski-Fourier transform" doesn't seem to arise in this way. -Questions: - -Is there an abstract framework in which to understand this "Minkowski-Fourier transform"? For instance, is there a general theory of "Fourier transforms" on spaces equipped with a nondegenerate symmetric bilinear form? Is there a relationship between this "Minkowski-Fourier transform" and the representations of a suitable "Heisenberg group"? -Which properties of the usual Fourier transform on Euclidean space are shared by the "Minkowski-Fourier transform"? For instance, what is a precise statement of the Fourier inversion formula in this context? -Is there a good reference for the mathematical properties of the "Minkowski-Fourier transform"? - -Perhaps it's worth adding that physicists seem to be a bit blase about using this "Minkowski-Fourier transform", and treat it as though it were an ordinary Fourier transform. - -REPLY [10 votes]: Well you seem to have worked it out but I wrote most of this before your comment happened: I claim there isn't any material difference between your "Minkowski space Fourier transform" and the usual Fourier transform on ${\mathbb R}^n$: in fact write $$ \hat f(\xi)\equiv \int e^{i\eta(x,\xi)} f(x) dx $$ for any non-degenerate bilinear form $\eta$. Then there exists another such form $\eta^{-1}$ so $\eta(x,\eta^{-1}\zeta)= \langle x,\zeta\rangle$ where $\zeta \in ({\mathbb R}^n)^\star$. Clearly $$ \hat f(\eta^{-1} \zeta)=({\mathcal F}f)(\zeta)\,,$$ where $\mathcal F$ is the usual --- ``Euclidean'' --- Fourier transform. -In physics texts this $\zeta$ variable is the down-index momentum ($k_\mu$ in e.g. Peskin and Schroeder) while $\xi$ is the up-index momentum ($k^\mu$ in e.g. Peskin and Schroeder). Derivatives play perfectly nice with up/down index notation, which allows one to be blase about whether the Fourier transform involves the up- or down-index $k$. -Mathematically speaking you're taking the Pontryagin dual of Minkowski space seen as a group of translations, which is exactly the same as that of the corresponding Euclidean space. More abstractly the conserved charge in the sense of Noether's theorem corresponding to a translation is the down-index momentum rather than the up-index one.<|endoftext|> -TITLE: Abelian category from the category of Hopf algebras -QUESTION [7 upvotes]: The kernel of a Hopf algebra map $\phi:H_1 \to H_2$ is in general not a Hopf -sub-algebra of $H_1$. Is there some replacement or alteration of the notion -of a kernel in the Hopf algebra setting. Same question for cokernels. So can we construct an abelian category from the category of Hopf algebras (over a fixed field $k$). - -REPLY [2 votes]: This is a delicate matter, and there are in principle many possible answers depending on exactly what you wish to apply things to. -Frequently, one of two variations of the Hopf subalgebra mentioned by Konstantinos is used: Given $\pi\colon H\to K$ a morphism of Hopf algebras, we have the left coinvariants and right coinvariants (of $\pi$) defined respectively by: -$$ {}^{\text{co}\,\pi}H = \{ h\in H \ | \ (\pi\otimes\operatorname{id})\Delta(h) = 1\otimes h\},\\ -H^{\text{co}\,\pi} = \{ h \in H \ | \ (\operatorname{id}\otimes\pi)(\Delta(h))= h\otimes 1\}.$$ -These need not be the same subobjects of $H$, however. -One could then define a short exact sequence of Hopf algebras (over $k$) -$$ k\longrightarrow K \overset{i}{\longrightarrow} H \overset{\pi}{\longrightarrow} L\longrightarrow k$$ -to be a sequence of morphisms of Hopf algebras such that - -$i$ is injective and $\pi$ is surjective; -$\ker(\pi) = H\, i(K)^+$; -$i(K) = {}^{\text{co}\,\pi}H$. - -Note that the more classical notion of kernel is still involved: it will generally not be enough to conduct any of the usual arguments you'd like to do with a "short exact sequence" to assume only the first and third (or first and second) conditions. The first and third are enough when $H$ is faithfully coflat, and the first and second are enough when $H$ is faithfully flat. -This is a good definition for short exact sequences and results that normally rely on them, such as looking for Jordan-Holder analogues. As such, coinvariants-as-kernels will be prominent when working in a category of Hopf algebras. But as I said it's not the only answer, and in other categorical contexts (generalizing/lifting to functors between representation categories, namely) it is the wrong answer. The categorical kernels and cokernels of $\pi\colon H\to K$ are -$$ \text{Hker}(\pi) = \{ h\in H \ | \ h_{(1)}\otimes \pi(h_{(2)})\otimes h_{(3)} = h_{(1)}\otimes 1 \otimes h_{(2)}\},\\ -\text{Hcoker}(\pi) = K/(K \pi(H^+) K).$$ -And this kernel (and cokernel) is not necessarily the same thing as the left or right coinvariants (and associated cokernel). By applying counits to the left/right of the defining relation, we see that $$\text{Hker}(\pi)\subseteq {}^{\text{co}\,\pi}H\cap H^{\text{co}\,\pi},$$ at least. Though equality (of all three subobjects) can happen, and assuring it does is usually the key to making sure attempts at defining exact sequences of tensor categories works out well. -Sonia Natale recently posted a review of these notions, and the related issues with tensor categories, that should be helpful.<|endoftext|> -TITLE: CW structure on infinite-dimensional manifolds -QUESTION [6 upvotes]: It is well-known (due to this work of Palais, I believe) that Banach manifolds are dominated by countable CW complexes. It then follows (due Whitehead, as indicated by Milnor in this work) that they have the homotopy type of CW complexes. - -Are there conditions ensuring that a Banach or Fréchet manifold admits a genuine CW complex structure? -What are the main examples of that? - -Actually, I'm working with Banach fiber bundles, which are Serre fibrations. I would like to regard them as Hurewicz fibrations to use HLP (homotopy lifting property). Due to this answer of Peter May (and the references cited by him), I know that it is enough to get a CW structure. -Any help is welcome. - -REPLY [6 votes]: Let $M$ be a Hilbert manifold. Then there exists a Riemannian metric $g$ on $M$ such that the induced metric $d$ is complete (See https://arxiv.org/pdf/1610.01527.pdf by Biliotti and Mercuri). Thus $M$ has the topology of a complete metric space. By the Baire category theorem $(M,d)$ is a Baire space: It cannot be written as a countable union of closed sets with empty interior. -If $M$ had the structure of a CW complex, it does. It must clearly be an infinite dimensional $CW$-complex, but such a thing is the countable union of its $n$-skeleta, which have empty interior. -With this argument you can treat also other infinite dimensional manifolds: using the fact that seperable Frechet spaces are homeomorphic to seperable Hilbert space, one can show that a Frechet manifold is homeomorphic to a Hilbert manifold.<|endoftext|> -TITLE: Bourbaki's definition of the number 1 -QUESTION [61 upvotes]: According to a polemical article by Adrian Mathias, Robert Solovay showed that Bourbaki's definition of the number 1, written out using the formalism in the 1970 edition of Théorie des Ensembles, requires -2,409,875,496,393,137,472,149,767,527,877,436,912,979,508,338,752,092,897 $\approx$ 2.4 $\cdot$ 1054 -symbols and -871,880,233,733,949,069,946,182,804,910,912,227,472,430,953,034,182,177 $\approx$ 8.7 $\cdot$ 1053 -connective links used in their treatment of bound variables. Mathias notes that at 80 symbols per line, 50 lines per page, 1,000 pages per book, this definition would fill up 6 $\cdot$ 1047 books. (If each book weighed a kilogram, these books would be about 200,000 times the mass of the Milky Way.) -My question: can anyone verify Solovay's calculation? -Solovay originally did this calculation using a program in Lisp. I asked him if he still had it, but it seems he does not. He has asked Mathias, and if it turns up I'll let people know. -(I conjecture that Bourbaki's proof of 1+1=2, written on paper, would not fit inside the observable Universe.) - -REPLY [15 votes]: Mathias, Grimm, and some people who've contributed to this Q&A seem to take these numbers as evidence of the impracticality of working formally in Bourbaki's theory. YCor explicitly called it unpractical, and said that Bourbaki shared that belief, which may be true. In light of that, I think it's worth writing a bit about the premise that seems to underlie the question. -If you define $F_1=F_2=1,\; F_{n+2}=F_n+F_{n+1}$, and eliminate $F$ from the expression $F_{250}$ by substitution, you get a tree of about $1.6\times 10^{52}$ nodes, or a string of that many symbols in Bourbaki/Polish notation. -I don't think it follows that the definition is a bad one. It's hard to see how you could avoid the blowup without introducing complications that would interfere with the definition's intended use. -You could object that this isn't a comparable situation because $250>1$, but that scarcely matters. Bourbaki happened to put a bunch of structure underneath $1$, but even if $1$ is primitive in your proof system, as soon as you do anything remotely interesting with it, you will have the same problem. The mere statement of your upper bound in Ramsey theory, never mind the proof, will have over Graham's-number symbols in it. -There is a simple way to get a better measure of the complexity of $F_{250}$ without changing the definition: merge common subtrees. The result is a dag of 249 nodes, of which 248 are additions: one for $F_3$, one for $F_4$, $\ldots$ -Here's what happens when you merge common subtrees of the expressions from Grimm page 514. (The one called M is Solovay's, and the one called SS is mentioned in Timothy Chow's answer.) - - - - -expression -tree size -tree links -dag size -dag links - - - - -SS -57330670440×1050 -21634097377×1050 -13153 -876 - - -SM -315628276×1050 -114233082×1050 -13015 -876 - - -S -171713×1050 -64721×1050 -8061 -544 - - -M -24098×1050 -8718×1050 -7971 -544 - - - - -I calculated these with a Python program that constructs the complete dags in memory and then collects statistics with and without duplication counts. The construction consumes a whopping several megabytes of RAM due to CPython's inefficient object representation, and takes a noticeable fraction of a second due to CPython being slow. If my code were published in a series of books, and each book weighed 1 kilogram, the total mass of all the books would be a few grams. -The dags are still much larger than the code that made them. The remaining bloat can be blamed on the definition of $\exists$, which uses its body twice in a way that precludes sharing, and on complicated constructions that can't be shared because they have children, particularly ordered pairs. These problems can be solved by introducing parametrized subtrees. One approach is to express the dag as a tree without duplication by introducing a "structural let" that assigns names to subtrees, and then permit the named subtrees to have holes whose values are specified at each use point. (Or you can keep the dag and add $\phi$ nodes – see Appel.) With this extra flexibility you can write any of these expressions in just a few hundred nodes. -I want to stress that these concise expressions are fully formally equivalent to the original strings. Not a single double negative has been eliminated. Given the concise formula and an index, you can compute the symbol at that index in the original string. -In this framework – which is just a small subset of what's available in a proof system like Coq – it hardly matters what's primitive. If Bourbaki's pairing operator $\supset$ isn't primitive, you can define it once at the top of the expression; the space cost is constant regardless of how many times it's used. If the expression is a proposition whose proof will be formally checked, the axioms of $\supset$ need to be proven and checked as theorems only once each, adding a constant startup time regardless of how many times they're used. These costs aren't fundamentally different from the costs of implementing $\supset$ as primitive. Only if you convert to the normal form will you see a difference, and there is no reason ever to do that. It's an abstraction violation. -To summarize: - -These gigantic normal forms show up in all formal systems, not just Bourbaki's. -They aren't a good measure of practicality or anything else.<|endoftext|> -TITLE: Existence of a quasi-isometric residually finite group? -QUESTION [17 upvotes]: It's, by now, more or less well known that residual finiteness is not a quasi-isometry invariant for finitely generated groups (see here for an example). Thus the following question makes sense: -Question: given a finitely generated group $G$, is there a finitely generated group $H$ such that $H$ is residually finite and $G$ and $H$ are quasi-isometric? That is, does the class of groups quasi-isometric to $G$ always contain residually finite groups? -Acknowledging this question is fairly open-ended, I would be glad with any reference in this direction. Thanks in advance. The following tries to motivate the above via some C*-algebraic questions. - -Motivation (and guess): Should the above be known true, then one would be able to manually build so-called quasi-diagonalizing projections for the reduced group C*-algebra of $G$, that is, finite rank projections $p_n \in \mathcal{B}(\ell^2 G)$, with $p_n \rightarrow 1$ in the strong operator topology and -$$ ||p_n \lambda_g - \lambda_g p_n|| \rightarrow 0 $$ -for every $g \in G$, where $\lambda$ is the left regular representation of $G$. Indeed, note that if $H$ is QI to $G$ and res. finite and amenable, then, by work of Orfanos there are projections $q_n$ for $H$ as above, and those can be pullback-ed to projections $p_n$ for $G$. Thus, since this C*-question is still open, my guess is that the original question is open as well. - -REPLY [19 votes]: Take any finitely-presented group $G$ with undecidable word problem. Then $G$ is not quasi-isometric to any finitely generated group with decidable word problem, in particular, to any residually-finite group. (Note that finite presentability and decidability of the WP are quasi-isometry invariant. The latter is because quasi-isometries preserve the equivalence class of the Dehn function and WP is for a finitely-presented group decidable iff the Dehn function is recursive.)<|endoftext|> -TITLE: If $\zeta(s)=0$ with $\Re(s)=\frac{1}{2}$, is then $|\hat{\zeta}(s,3)|^2=\frac{1}{2}$? -QUESTION [15 upvotes]: Helmut Hasse has proved that for $s \in \mathbb{C}-\{1\}$ the Riemann zeta function can be written as: -$$\zeta(s)=\frac{1}{1-2^{1-s}}\sum_{n=0}^\infty\frac{1}{2^{n+1}}\sum_{k=0}^n(-1)^k\ {n \choose k}\ \frac{1}{(k+1)^{s}}$$ -Let $K(a,b) = \frac{\gcd(a,b)}{a+b}$. Then we might define: -$$\hat{\zeta}(s,a)=\frac{1}{1-2^{1-s}}\sum_{n=0}^\infty\frac{1}{2^{n+1}}\sum_{k=0}^n(-1)^k\ {n \choose k} K(k,a)^{s}$$ -Then $\hat{\zeta}(s,1) = \zeta(s)$. -I looked at some non-trivial zeros of $\zeta$ and computed numerically in SAGEMATH: -If $\zeta(s)=0$ with $\Re(s)=\frac{1}{2}$, is then $|\hat{\zeta}(s,3)|^2=\frac{1}{2}$ or is this just some bug / numerical curiosity in my code? -Here is the SAGE Code: -def K(a,b): - return gcd(a,b)/(a+b) - -def zzeta(s,a=1,N=400): - return 1/(1-2**(1-s))*sum([ 1/2**(n+1)*sum([ (-1)**k*binomial(n,k)*K(k,a)**s for k in range(n+1)]) for n in range(0,N+1)]) - - -zz = zeta_zeros() - -for z in zz[0:100]: - print( abs(zzeta(1/2+z*I,3))**2) - -And here is some of the ouptut: -0.5000000001016516 -0.49999999980871757 -0.49999999990048116 -0.4999999999150287 -0.5000000001425052 -0.5000000003259728 -0.4999999999688171 -0.5000000000737418 -0.5000000004006991 -0.5000000003452242 -0.49999999982210547 -0.5000000000784613 -0.49999999939079287 -0.5000000012082603 -0.5000000000591772 -0.5000000002545129 -0.4999999996870637 - -Some data for $a=2$: -0.5027599264329976 -0.4120370789607451 -0.3869873346989051 -0.2811880366841188 -0.2594539123560252 -0.7742279955909737 -0.18310290214474273 -0.4006464279725155 -0.2723910071065707 -0.16780149194046892 - -REPLY [5 votes]: The intent of this post is to share some results as requested in the comments above and assumes the following definitions from the original question above. - -(1) $\quad\zeta(s)=\underset{N\to\infty}{\text{lim}}\left(\frac{1}{1-2^{1-s}}\sum_\limits{n=0}^N\frac{1}{2^{n+1}}\sum_\limits{k=0}^n(-1)^k\left( -\begin{array}{c} - n \\ - k \\ -\end{array} -\right)\frac{1}{(k+1)^s}\right)$ -(2) $\quad\hat{\zeta}(s,a)=\underset{N\to\infty}{\text{lim}}\left(\frac{1}{1-2^{1-s}}\sum_\limits{n=0}^N\frac{1}{2^{n+1}}\sum_\limits{k=0}^n(-1)^k\left( -\begin{array}{c} - n \\ - k \\ -\end{array} -\right)K(k,a)^{s}\right),\qquad K(a,b)=\frac{\gcd(a,b)}{a+b}$ - -The following figure illustrates formula (2) above for $|\hat{\zeta}(s,3)|$ plotted along the critical line $s=\frac{1}{2}+i\ t$. The red discrete portion of the plot represents the evaluation of formula (2) for $|\hat{\zeta}(s,3)|$ at the first 10 non-trivial zeta zeros in the upper half plane. In the plot in the figure below formula (2) was evaluated using an upper limit of $N=400$. - - -Figure (1): Illustration of formula (2) for $|\hat{\zeta}(s,3)|$ evaluated along the critical line - -The following conjectured formulas based on my Questions related to globally convergent formulas for the Dirichlet eta function $\eta(s)$ seem to evaluate considerably faster than formulas (1) and (2) above and perhaps provide some additional insight. - -(3) $\quad\zeta(s)=\underset{N\to\infty}{\text{lim}}\left(\frac{1}{\left(1-2^{1-s}\right)\ 2^{N+1}}\sum\limits_{n=0}^N (-1)^n\ \frac{1}{(n+1)^s}\sum\limits_{k=0}^{N-n} \left( -\begin{array}{c} - N+1 \\ - N-n-k \\ -\end{array} -\right)\right)$ -(4) $\quad\hat{\zeta}(s,a)=\underset{N\to\infty}{\text{lim}}\left(\frac{1}{\left(1-2^{1-s}\right)\ 2^{N+1}}\sum\limits_{n=0}^N (-1)^n\ K(n,a)^{s}\sum\limits_{k=0}^{N-n} \left( -\begin{array}{c} - N+1 \\ - N-n-k \\ -\end{array} -\right)\right)$ - -The following figure illustrates formula (4) above for $|\hat{\zeta}(s,3)|$ plotted along the critical line $s=\frac{1}{2}+i\ t$. The red discrete portion of the plot represents the evaluation of formula (4) for $|\hat{\zeta}(s,3)|$ at the first 10 non-trivial zeta zeros in the upper half plane. In the plot in the figure below, formula (4) was evaluated using an upper limit of $N=400$. - - -Figure (2): Illustration of formula (4) for $|\hat{\zeta}(s,3)|$ evaluated along the critical line - -Formula (5) below is from the accepted answer to this question posted by Lucia. -(5) $\quad\hat{\zeta}(s,3)=\frac{2^{-s}-1}{1-2^{1-s}}+\zeta(s)+\frac{3^s-1}{3^s} \zeta(s)$ - -The following figure illustrates formula (5) above for $|\hat{\zeta}(s,3)|$ and $\left|\frac{2^{-s}-1}{1-2^{1-s}}\right|$ in blue and orange respectively where both functions are evaluated along the critical line $s=\frac{1}{2}+i\ t$. The red discrete portion of the plot represents the evaluation of formula (5) for $|\hat{\zeta}(s,3)|$ at the first 10 non-trivial zeta zeros in the upper half plane. - -Figure (3): Illustration of formula (5) for $|\hat{\zeta}(s,3)|$ evaluated along the critical line<|endoftext|> -TITLE: On tilting and cotilting modules -QUESTION [5 upvotes]: Let A be an Artin algebra and assume all modules are basic, then a classical result says that tilting modules $T$ are in bijection with complete cotorsion pairs $(T^{\perp}, \check{ add(T)})$ (with some additional assumptions on the cotorsion pairs) , where $T^{\perp}= \{ X | Ext_A^i(T,X)=0 $ for all $i >0 \}$ and $\check{ add(T)}$ denotes the full subcategory of modules $Y$ with a finite $add(T)$ coresolution. -Dually we have the same for cotilting modules $U$ with the dual properties and cotilting modules are in bijection with complete cotorsion pairs $( ^{\perp} U, \hat{add(U)})$ (with some additional assumptions on the cotorsion pairs). -Assume we have a tilting module $T$ and a cotilting module $U$ such that the following two conditions are satisfied: - -$T^{\perp} \cap \hat{add(U)} = \hat{add(U)}$ -$^{\perp}U \cap \check{add(T)} = \check{add(T)}$. - -Is it then true that $A$ is Gorenstein if and only if $U=T$? (or maybe this is true up to adding another condition?) -Note that being Gorenstein is equivalent to all tilting modules being cotilting and also that every module of finite projective dimension has finite injective dimension and vice versa. -So the direction $U=T$ implies Gorenstein is easy. -For the other direction we can use that $A$ being Gorenstein gives us that $T$ is some cotilting module and 1. also gives us that $Ext^i(T,U)=0$ for all $i>0$. Now it would be enough to show that also $Ext^i(U,T)=0$ for all $i>0$ to conclude that $T=U$. But I do not see a good approach to prove that. - -REPLY [3 votes]: We have $^\perp(T^\perp) = \check{\operatorname{add}}\, T$, and $(\check{\operatorname{add}}\, T)^\perp = T^\perp$, and $(^\perp U)^\perp = \hat{\operatorname{add}}\, U$, and $^\perp(\hat{\operatorname{add}}\, U) = {^\perp U}$. Condition 1. and 2. are equivalent to - -$\hat{\operatorname{add}}\, U \subseteq T^\perp$ -$\check{\operatorname{add}}\, T \subseteq {^\perp U}$. - -Taking $^\perp$ on the correct sides and using the above identities, it follows that 1. and 2. are equivalent. So it is enough to have one of the conditions, say 2. Take your favorite Gorenstein algebra which is not selfinjective. Then choose $U = D(A)$ and $T = A$, which implies that $\check{\operatorname{add}}\, T = \mathcal{P}(A)$ and $^\perp U = \operatorname{mod}\, A$, that is, 2. is satisfied. But $T \neq U$. So additional conditions have to be added in order for this to be true.<|endoftext|> -TITLE: Backward uniqueness for a heat equation with a drift -QUESTION [5 upvotes]: Consider heat equation with a drift (=reaction-diffusion equation) -$$ -\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}+f(t,u(t,x)), \quad t\ge0,\, x\in [0,1] -$$ -with periodic or Dirichlet boundary conditions. Here $f$ is globally bounded and Lipschitz in the second argument. Is it true that if $u$,$v$ are two solutions of this PDE with $u(1)=v(1)$, then $u(0)=v(0)$? How can we prove this? -The case $f=0$ is classical; it can be found, e.g., in Evans book. There, the proof goes by showing that the second derivative of log energy is nonnegative. However, it is not clear to me at all how this technique can be adapted here, since $f$ might have no time derivatives. -An equivalent formulation of the above problem is the following: suppose we are given a bounded function $w\colon [0,1]\times \mathbb{R}_+\to\mathbb{R}$ with Dirichlet or periodic boundary conditions. Suppose that $w(0)\equiv 0$ and -$$ -\Bigl|\frac{\partial w}{\partial t}+\frac{\partial^2 w}{\partial x^2}\Bigr|\le C|w| -$$ -everywhere. Is it true that $w(t)\equiv0$ for all $t\ge0$? -UPD: I was told that one can use an appropriate Carleman bound. Do you know whether there are any Carleman estimates suitable for this problem? - -REPLY [2 votes]: The result follows by an extension of the method of logarithmic convexity which is well-known for the heat backward problem. -Let $H$ be a Hilbert space. Consider the following inequality -\begin{equation} -\|\partial_t u + Au\| \leq \alpha\|u\|, \qquad \text{ on } (0,T), \qquad (1) -\end{equation} -with $\alpha=\mathrm{const}>0$. -We assume that $$A=A_+ + A_-, \qquad (2)$$ -where $A_+$ is a linear symmetric operator on $H$ with domain $D(A)$ and $A_-$ is skew-symmetric such that: -\begin{align} -\|A_- u\|^2 &\leq c(\|A_+ u\| \|u\|+ \|u\|^2), \qquad \qquad \qquad \qquad (3)\\ -\partial_t \langle A_+ u, u\rangle &\leq 2\langle A_+ u, \partial_t u\rangle + c(\|A_+ u\|\|u\|+\|u\|^2) \,\qquad (4). -\end{align} - -Theorem. Let $u(t)\in D(A)$, $u\in C^1([0,T];H)$ be a solution of the differential inequality $(1)$, where $A$ satisfies the conditions $(2)$-$(4)$. Then - \begin{equation} -\|u(t)\| \leq C_1 \|u_0\|^{1- \lambda(t)} \|u(T)\|^{\lambda(t)}, \qquad 0\leq t\leq T, \qquad (*) -\end{equation} - for a constant $C_1=C_1(\alpha,T)>0$ and $\lambda(t)=\dfrac{1-e^{-Ct}}{1-e^{-CT}}$ with $C$ depending on $\alpha$. - -In your case, $A=A_+=\Delta$,$\; A_-=0$ and it is easy to check that conditions $(1)$-$(4)$ hold. From $(*)$ it is clear that if $w(0)=0$ then $w(t) \equiv 0$ for all $t\ge 0$, which answers your question (second formulation). -This is Theorem 3.1.3 pp 51 in - -V. Isakov, Inverse Problems for Partial Differential Equations. Springer, (2017). - -It also applies to more general elliptic operators, (see Example 3.1.6 in the same reference).<|endoftext|> -TITLE: Minimal good cover of the torus -QUESTION [14 upvotes]: Recall that an open cover $\mathfrak{U} = \{ U_\alpha \}$ of a manifold $M$ is called a good cover if all possible finite intersections $U_{\alpha_1} \cap ... \cap U_{\alpha_n}$ are contractible. -Question: What is the minimum number of open sets required for a good cover of the 2-dimensional torus? -The picture below provides a good cover of the torus (i.e. opposite sides of the parallelogram identified as usual) using 7 open sets (i.e. take sufficiently small open neighbourhoods of the hexagons). Can one do any better than 7? If not, how does one prove that 7 is optimal? - -REPLY [22 votes]: You can't do any better than $7$. This follows from -Karoubi, Max; Weibel, Charles A., On the covering type of a space, Enseign. Math. (2) 62, No. 3-4, 457-474 (2016). ZBL1378.55002. -in particular Theorem 5.3 in the arXiv version. -The strict covering type of a space $X$ is the minimal cardinality of a good cover, denoted $\operatorname{sct}(X)$. This is not a homotopy invariant, so Karoubi and Weibel introduce the covering type, defined by -$$ -\operatorname{ct}(X)=\min\{\operatorname{sct}(X')\mid X'\simeq X\} -$$ -Obviously $\operatorname{ct}(X)\leq \operatorname{sct}(X)$. In Theorem 5.3 they use cohomological arguments to show that $\operatorname{ct}(T^2)=7$.<|endoftext|> -TITLE: $GL_1(\mathcal{E}'(\mathbb{R}))$ open in $\mathcal{E}'(\mathbb{R})$? -QUESTION [7 upvotes]: Let $\mathcal{E}'(\mathbb{R})$ be algebra of all compactly supported distributions on $\mathbb{R}$, equipped with the strong dual topology $\beta(\mathcal{E}',\mathcal{E})$, and with the usual operations of addition and convolution. -Is the set ${{\textrm{GL}}}_1(\mathcal{E}'(\mathbb{R}))$ of invertible elements open in $\mathcal{E}'(\mathbb{R})$? -(As an example of an element in ${{\textrm{GL}}}_1(\mathcal{E}'(\mathbb{R}))$ which does not have support $\{0\}$, we have that $\delta_n$ belongs to ${{\textrm{GL}}}_1(\mathcal{E}'(\mathbb{R}))$ -because $\delta_n \ast \delta_{-n}=\delta_0$.) - -REPLY [3 votes]: Perhaps Jochen Wengenroth's comments already give the answer, but here's a direct argument. -By Paley-Wiener for distributions, the Fourier transform $\tilde{\Delta}(k)$ of a distribution of compact support $\Delta(x) \in \mathcal{E}'(\mathbb{R})$ is entire, of exponential type, with at most asymptotic polynomial growth on the real axis. Since convolutions become products under the Fourier transform, $\Delta(x) \in \mathrm{GL}_1(\mathcal{E}'(\mathbb{R}))$ only if $\tilde{\Delta}(k)$ is pointwise invertible and $\tilde{\Delta}(k)^{-1}$ has the same analytic properties. But by the Hadamard factorization theorem this can only be if $\tilde{\Delta}(k) = e^{i k c_1 + c_0}$, with $c_1 \in \mathbb{R}$, $c_0\in \mathbb{C}$. So indeed, $\mathrm{GL}_1(\mathcal{E}'(\mathbb{R}))$ consists only of scaled translates of $\delta_0$. -I'm no big expert on the topology of $\mathcal{E}'(\mathbb{R})$, but I suspect this set is too small to be open.<|endoftext|> -TITLE: Are etale morphisms "strongly formally etale"? -QUESTION [5 upvotes]: EDIT: Thanks to Harry Gindi and Anonymous below for insightful comments, I've refined the definitions here. Recall that a formally etale morphism of schemes $Y \to X$ is a morphism which has the unique right lifting property with respect to all nilpotent thickenings $Z \to W$. One salient feature of nilpotent thickenings is that they are universal homeomorphisms. -Definition: Say that a morphism of schemes $Y \to X$ is strongly formally etale if it has the unique right lifting property with respect to all universal homeomorphisms $Z \to W$. That is, for every commutative square as below, there exists a unique diagonal filler $W \to Y$, as indicated, making the two triangles commute. -$$\require{AMScd} \begin{CD} Z @>>> Y \\ @VVV \nearrow @VVV\\ W @>>> X \end{CD}$$ -By definition, then, if $Y \to X$ is strongly formally etale, then $Y \to X$ is formally etale. The converse presumably does not hold. However, etale morphisms have an additional finiteness condition (etale = formally etale + locally of finite presentation) which makes me hope for an affirmative answer to the first question below: -Questions: - -Let $Y \to X$ be an etale morphism. Then is $Y \to X$ strongly formally etale? -Does there exist standard terminology for "strongly formally etale"? -Is there a characterization of the class of morphisms which have the unique left lifting property with respect to all etale morphisms? How about (strongly) formally etale? - -REPLY [6 votes]: This follows from the well-known "topological invariance of the étale site"; see e.g. Tag 04DZ. -Indeed, write $S' \to S$ for $Z \to W$, and consider the pullbacks of the étale map $Y \to X$ to étale maps $T \to S$ and $T' \to S$. We need to show that a section to $T' \to S'$ uniquely arises by pullback from a section of $T \to S$. This is a special case of Tag 0BTY. -Presumably the same should hold for weakly étale affine morphisms, since these are very close to ind-étale; see Tag 097Y. The only "topological invariance of the pro-étale site" I have seen is Lemma 5.4.2 is Bhatt–Scholze, which is only on the level of sheaves (rather than an actual equivalence of sites).<|endoftext|> -TITLE: Geometric Langlands: From D-mod to Fukaya -QUESTION [11 upvotes]: This post is rather wordy and speculative, but I promise there is a concrete question embedded within. For experts, I'll open with a question: -Question: Given a compact Riemann surface $X$, why does one prefer the category of D-modules on the space/stack $Bun_{X}(G)$ instead of the (Fukaya) category of Langrangians on the space/stack of $G$-local systems $\mathcal{L}_{X}(G).$ -Much of what follows is explaining what I mean by this question. -Let $G$ be a connected reductive complex Lie group and $G^{\vee}$ the Langlands dual group. Let $X$ be a compact Riemann surface (smooth projective curve over $\mathbb{C}$). The central players in the geometric Langlands conjecture are the stacks of $G$-bundles on $X,$ denoted $Bun_{X}(G),$ and the stack of $G^{\vee}$-local systems on $X,$ which I'll denote by $\mathcal{L}_{X}(G^{\vee}).$ -The best hope (using the words of Drinfeld) is that there is a (derived) equivalence between the category of D-modules on $Bun_{X}(G)$ and the category of quasi-coherent sheaves on $\mathcal{L}_{X}(G^{\vee}).$ This best hope is by now entirely dashed, and, to name one main player, Gaitsgory has expended a tremendous amount of effort to both indicate its failure, and conjecture a solution, along the way writing some very technical and interesting papers to give evidence. -Meanwhile, Witten, Kapustin, and others have made efforts to indicate how to restore some symmetry to this nebulous blob of conjectures, by arguing that they have natural interpretations as dimensional reductions of certain four dimensional super-symmetric gauge theories. -In the course of reading about various aspects of this story, I was struck by a point that Witten has made many times in talks and in print that I want to ask about here. I'm going to speak rather prosaically from here on forward, and welcome an explanation about why my simplifications don't make any sense. -Starting from Hitchin's work in the late 1980's, it became clear that the space of $G$-local systems $\mathcal{L}_{X}(G)$ is basically the cotangent bundle of $Bun_{X}(G).$ If you take stacky language seriously enough, this can be made reasonably precise. Taking this as true, there are many concrete, though not immediately applicable, theorems which indicate that the category of D-modules on $Bun_{X}(G)$ is equivalent to the Fukaya category of Langrangians in its co-tangent bundle: the latter of which is $\mathcal{L}_{X}(G).$ -Using this quasi-logic, the best hope geometric Langlands conjecture posits an equivalence of (derived) categories between the Fukaya category of $\mathcal{L}_{X}(G)$ and the category of quasi-coherent sheaves on $\mathcal{L}_{X}(G^{\vee}).$ -In other words, geometric Langlands becomes the statement that $\mathcal{L}_{X}(G)$ and $\mathcal{L}_{X}(G^{\vee})$ are mirror partners in the sense of homological mirror symmetry. -Question: Has this interpretation been taken seriously somewhere in the mathematical literature, and if not, is there a good reason it hasn't been? -As one final comment. There is (at least) one major advantage to putting the conjecture into this language. Both the Fukaya category of $\mathcal{L}_{X}(G)$ and the category of quasi-coherent sheaves on $\mathcal{L}_{X}(G^{\vee})$ can be defined without recourse to the complex/algebraic structure on $X.$ This is because both of these spaces/stacks are naturally complex symplectic, and this structure is independent of the complex/algebraic structure on $X.$ -With this in mind, it makes more sense to refer to $\mathcal{L}_{\Sigma}(G)$ and $\mathcal{L}_{\Sigma}(G^{\vee})$ where $\Sigma$ is a connected, oriented, smooth surface. An advantage of this is that now topological symmetries (diffeomorphisms) of $\Sigma$ act naturally on these spaces/stacks and the subsequent categories. The study of these symmetries is what people in my field call the study of the mapping class group of $\Sigma,$ and there are many deep open questions about the mapping class group, that might find a natural home in the aforementioned discussion. - -REPLY [4 votes]: More a comment than an answer: -I think the lagrangian-to-sheaf dictionary is more “immediately applicable” than is commonly supposed. In particular, it’s possible to immediately apply the dictionary to construct some (plausibly Hecke eigen) sheaves from smooth Hitchin fibers. See the following note: -https://arxiv.org/abs/2108.13571<|endoftext|> -TITLE: A sum of two binomial random variables -QUESTION [11 upvotes]: Let $p\in(0,1)$, $n$ a positive even integer, $k,l\in\{0,\dots,n\}$, and $X_k\sim \text{Binomial}(k,p)$, $Y_{n-k}\sim \text{Binomial}(n-k,1-p)$ independent random variables. I would like to prove that -$$ -\Pr(X_k+Y_{n-k}=l)\leq\Pr(X_{n/2}+Y_{n/2}=n/2). -$$ -This question can be stated analytically. Setting $c=(1-p)/p$, define: -$$ -f_{n,c}(k,l)=c^{l+k}\sum_{i=\max(0,k+l-n)}^{\min(k,l)}\binom{k}{i}\binom{n-k}{l-i}c^{-2i}. -$$ -Prove that $f_{n,c}$ attains its maximum at $k=l=n/2$, for any even $n$ and $c>0$. - -REPLY [7 votes]: Here is a (surprising) proof using Cauchy-Schwarz and "rearrangement". -The following lemma will be the key. -Lemma -: Let $X,Y$ be independent integer-valued rvs, then \begin{align*} -(a)\; &\mbox{ for any } z: \;\mathbb{P}(X+Y=z)^2\leq \big(\sum_x\mathbb{P}(X=x)^2\big)\,\big(\sum_y \mathbb{P}(Y=y)^2\big)\\ -(b)\; &\sum_z\mathbb{P}(X-Y=z)^2=\sum_z\mathbb{P}(X+Y=z)^2\end{align*} -Proof: (a) apply Cauchy-Schwarz to -$\mathbb{P}(X+Y=z)=\sum_x \mathbb{P}(X=x)\mathbb{P}(Y=z-x)$ -(b) let $(X^\prime,Y^\prime)$ be distributed as $(X,Y)$, and independent of $(X,Y)$. Then -\begin{align*} - \sum_z\mathbb{P}(X-Y=z)^2=\mathbb{P}(X-Y=X^\prime-Y^\prime)=\mathbb{P}(X-X^\prime=Y-Y^\prime)\\ - \sum_z\mathbb{P}(X+Y=z)^2=\mathbb{P}(X+Y=X^\prime+Y^\prime)=\mathbb{P}(X-X^\prime=Y^\prime-Y) -\end{align*} -Since $Y-Y^\prime$ and $Y^\prime-Y$ are identically distributed, and independent of $(X,X^\prime)$ the right hand sides above are equal. End Proof -Now to your question above. Let $n=2m$. We have to show that -\begin{align*} \mathbb{P}(X_k+Y_{2m-k}=\ell)\leq \mathbb{P}(X_m+Y_m=m)\end{align*}For the right hand side above we have (using 1. below) -\begin{align*} \mathbb{P}(X_m+Y_m=m)=\mathbb{P}(X_m=X_m^\prime)=\sum_l \mathbb{P}(X_m=l)^2\end{align*} -To transform the left hand side, observe that well known properties of the binomial distribution give: -\begin{align} 1.\; &Y_k \mbox{ is distributed as}\; k-X^\prime_k, \mbox{where $X_k^\prime$ is distributed as $X_k$,}\\&\mbox{ and independent of $X_k$}\\ -2.\; &\mbox{ $X_{m+j}$ resp. $Y_{m+j}$ are distributed as $X_m+X_j$ resp. $Y_m+Y_j$, where}\\ -&\mbox{ the summands are independent} -\end{align} -Using 1. gives that $\sum_l \mathbb{P}(X_m=l)^2=\sum_l \mathbb{P}(Y_m=l)^2$ and further that -$$ \mathbb{P}(X_k+Y_{2m-k}=\ell)=\mathbb{P}(X_{2m-k}+Y_k=2m-\ell)\;,$$ so we may w.l.o.g. assume that $k\leq m$. -Using 1. and 2. we have -$$\mathbb{P}(X_k+Y_{2m-k}=\ell)=\mathbb{P}(X_k-X_{m-k}+Y_m=\ell+k-m)$$ -where $X_k,X_{m-k}$ and $Y_m$ on the right hand side are independent. -Using part (a) of the lemma (with $X=X_{k}-X_{m-k}$ and $Y=Y_m$) -gives $$\mathbb{P}(X_k+Y_{2m-k}=z)^2 \leq \big(\sum_x \mathbb{P}(X_k-X_{m-k}=x)^2\big)\big(\sum_y\mathbb{P}(Y_m=y)^2\big)$$ -Finally using part (b) of the lemma on the first factor gives -$$\sum_{x}\mathbb{P}(X_k-X_{m-k}=x)^2=\sum_x\mathbb{P}(X_k+X_{m-k}=x)^2=\sum_x\mathbb{P}(X_m=x)^2$$ -so that ultimately -$$\mathbb{P}(X_k+Y_{n-k}=z)^2\leq \big(\sum_x \mathbb{P}(X_m=x)^2\big)\big(\sum_y\mathbb{P}(Y_m=y)^2\big)=\big(\sum_x \mathbb{P}(X_m=x)^2\big)^2\;,$$ -as desired.<|endoftext|> -TITLE: idea and intuition behind triangulated category -QUESTION [6 upvotes]: I have some trouble in understanding the significance of some axiom of triangulated category. -If someone could explain to me each axiom with some intuition, and explain to me the intuition behind the translation functor -I would be -very grateful !! -Thank you in advance - -REPLY [8 votes]: TR3 is redundant: it is implied by the other axioms. TR4 is confused by the initial octahedral shape: it starts with a square in which one side is the identity. If one instead starts with a triangle given by a pair of composable arrows and their composition and writes it as a braid, one readily sees that it is just describing the behavior of triangles under composition. The description becomes obvious in the topological framework that was Verdier's starting point (he credited work of Puppe). For the redundancy claim, the braid description, and the behavior of distinguished triangles under products, see https://www.sciencedirect.com/science/article/pii/S0001870801919954 or https://www.math.uchicago.edu/~may/PAPERS/AddJan01.pdf.<|endoftext|> -TITLE: Descent of vector bundle along branched cover of curve -QUESTION [5 upvotes]: Suppose $\pi:C'\to C$ is a branched cover of compact Riemann surfaces such that the associated extension of function fields is Galois with group $G$ -- so that $\pi$ presents $C$ as the quotient $C'$ by the action of $G = \text{Aut}(C'/C)$. Now, let $\rho:G\to GL(W)$ be a finite dimensional complex representation of $G$. Below, we identify locally free coherent sheaves with holomorphic vector bundles. -Let $\underline W$ be the trivial vector bundle on $C'$ with fibre $W$. Define the vector bundle $W^\rho$ on $C$ to be the subsheaf of $\pi_*\underline W$ whose sections over $U\subset C$ are the $G$-equivariant holomorphic functions $U' = \pi^{-1}(U)\to W$. We want to compute $c_1(W^\rho)$ as follows. We have a natural map $\varphi:\pi^*W^\rho\to\underline W$ on $C'$ (coming from the adjunction counit $\pi^*\pi_*\underline W\to\underline W$) which is an injective map of coherent sheaves. Now, by looking at the zeros of the determinant of $\varphi$ (which occur exactly at the critical points of $\pi$), we can figure out the value of $c_1(W^\rho) = \frac1{|G|}c_1(\pi^*W^\rho) = -\frac1{|G|}\cdot\dim H^0(C',\text{coker }\varphi)$. -Carrying out this computation explicitly, we seem to get the following answer. Given any branch point $p\in C$ of $\pi$, pick a preimage $p'\in C$. Let $G_{p'}\subset G$ be the (necessarily cyclic) stabilizer group of $p'$, of order $n_p$. For $0\le i -TITLE: Regular limit points of possible cofinalities -QUESTION [6 upvotes]: Let $A$ be a non-empty set of regular cardinals such that $\vert A\vert <\text{min}\ A$, and $\{\nu_i\mid i -TITLE: Abelianization of general linear group of a polynomial ring -QUESTION [13 upvotes]: For $K$ a field, is it known what the abelianization of $GL_2(K[X])$ is? - -REPLY [6 votes]: As an alternative to YCor's beautiful answer, one can use the following theorem of P. M Cohn [1, Theorem 9.5]. - -Theorem. Let $R$ be a ring which is quasi-free for $\text{GE}_2$ and denote by $N$ the ideal generated by all $\alpha - 1$ with $\alpha \in U(R)$. Then there is a split exact sequence - $$0 \rightarrow R/N \rightarrow \text{GE}_2(R)^\text{ab} \rightarrow U(R)^\text{ab} \rightarrow 0.$$ - and the mapping $\alpha \mapsto \begin{pmatrix} \alpha^\text{ab} & 0 \\ 0 & 1 \end{pmatrix}$ induces a splitting. - -In the above statement - -$U(R)$ denotes the unit group of $R$, -$G^\text{ab}$ denotes the abelianization of a group $G$ and $g \mapsto g^\text{ab}$ the abelianization homomorphism (in particular in the case of $U(R)\to U(R)^\text{ab}$), -$\text{GE}_2(R)$ is the subgroup of $\text{GL}_2(R)$ generated by the elementary matrices $\begin{pmatrix} 1 & r \\ 0 & 1 \end{pmatrix}$, $\begin{pmatrix} 1 & 0 \\ s & 1 \end{pmatrix}$ with $r,s \in R$ and the diagonal matrices $\begin{pmatrix} \alpha & 0 \\ 0 & \beta \end{pmatrix}$ with $\alpha, \beta \in U(R)$. - -I will not expand the definition of a quasi-free ring for $\text{GE}_2$ but just mention that a discretely normed ring is quasi-free for $\text{GE}_2$. In particular $K[X]$ is quasi-free for $\text{GE}_2$ if $K$ is a field. -Note that if $\text{SL}_2(R)$ is generated by the elementary matrices, e.g. $R$ is Euclidean, then $\text{GE}_2(R) = \text{GL}_2(R)$. This holds in particular for $R = K[X]$ with $K$ a field since $K[X]$ is Euclidean (only) in this case. -If $K$ is the field with two elements, then the above theorem yields an isomorphism of the additive group of $K[X]$ with $\text{GL}_2(K[X])^\text{ab}$. If $K$ is a field with more than two elements, the same theorem yields an isomorphism of $\text{GL}_2(K[X])^\text{ab}$ with $U(K)$. - -[1] P. M. Cohn, "On the structure of the $\mathrm{GL}_2$ of a ring", 1966 (MSN).<|endoftext|> -TITLE: Why is $-\int_{-\infty}^\infty \log\left[1+2f'(x)(1-\cos\phi)\right]\,dx$ equal to $\phi^2$? -QUESTION [10 upvotes]: I came across this integral involving the derivative $f'(x)$ of the Fermi function $f(x)=(1+e^x)^{-1}$: -$$I(\phi)=-\int_{-\infty}^\infty \log\left[1+2f'(x)(1-\cos\phi)\right]\,dx.$$ -I'm pretty certain that $I(\phi)=\phi^2$ for $|\phi|<\pi$, periodically repeated, as in the plot. It means that only the ${\cal O}(\phi^2)$ term in a Taylor expansion of the integrand has a nonzero contribution, but I am unable to prove this. (Mathematica returns a polylog function, and will not simplify it further.) Any help would be much appreciated, I'm hoping such a simple answer will have a simple derivation --- perhaps without needing special functions? - -REPLY [9 votes]: It is clear that $I(0)=0$, hence by Leibniz's integration rule, it suffices to show that -$$\int_{-\infty}^\infty\frac{-2f'(x)\sin\phi}{1+2f'(x)(1-\cos\phi)}\,dx=2\phi,\qquad |\phi|<\pi.$$ -We can assume, without loss of generality, that $0<\phi<\pi$. By a bit of algebra, we can rewrite the last equation as -$$\int_{-\infty}^\infty\frac{e^x\sin\phi}{e^{2x}+2(\cos\phi)e^x+1}\,dx=\phi.$$ -By the change of variables $u=(e^x+\cos\phi)/\sin\phi$, this becomes -$$\int_{\frac{\cos\phi}{\sin\phi}}^\infty\frac{du}{u^2+1}=\phi,$$ -which in turn can be verified by calculus: -$$\int_{\frac{\cos\phi}{\sin\phi}}^\infty\frac{du}{u^2+1}=\arctan(\infty)-\arctan\left(\frac{\cos\phi}{\sin\phi}\right)=\frac{\pi}{2}-\left(\frac{\pi}{2}-\phi\right)=\phi.$$ -The proof is complete. - -REPLY [4 votes]: We have -$$I(t)=-\int_{-\infty}^\infty l_x(t)\,dx,$$ -where -$$l_x(t):=l(t):=\ln(1+2f'(x)(1-\cos t)).$$ -Next, -$$l_x''(t)=l''(t)=-\frac{2 e^x \left(c \left(e^{2 x}+1\right)+2 e^x\right)}{\left(2 c e^x+e^{2 x}+1\right)^2},$$ -where $c:=\cos t\in(-1,1]$ for $|t|<\pi$. -So, by substitution $e^x=u$, for $c\in(-1,1]$, -$$I''(t)=-\int_{-\infty}^\infty l''_x(t)\,dx=2,$$ -which implies $I(t)=t^2$ for $|t|<\pi$ (since $l_x(0)=0=l'_x(0)$ and hence $I(0)=0=I'(0)$).<|endoftext|> -TITLE: Dense graphs where every maximal independent set is large -QUESTION [5 upvotes]: A maximal independent set of a graph $G$ is a subset of vertices $S$ such that each vertex of $G$ is either in $S$ or adjacent to some vertex in $S$, and no two vertices in $S$ are adjacent. Consider graphs of $n$ nodes that are dense, i.e., there are $m$ edges, where $m \ge n^{1+\epsilon}$, for some constant $\epsilon>0$. -Update: -As pointed out in the comments, one can take $n/2$ isolated vertices and then any dense graph of high $\ge 5$ on the remaining $n/2$ vertices. However, I'm more interested in regular graphs where every node has the same degree. I've updated my question is as follows: - -Does there exist a family of regular dense graphs of girth $\ge 5$ where every maximal independent set has a size of at least $\Omega(n)$? - -Note that the girth $\ge 5$ condition rules out obvious candidates such as the complete bipartite graph which has girth $4$. - -REPLY [3 votes]: No, the object you’re looking for does not exist. -A result of Harutyunyan, Horn, and Verstraete (see Theorem 4.15 of this survey) states the following - -Theorem: There is a constant $c > 0$ such that every $d$-regular graph $G$ on $n$ vertices of girth at least 5 satisfies $i(G) \leq \dfrac{n(\log(d)+c)}{d}$. - -Here, $i(G)$ is the independence domination number, which is the size of the smallest maximal independent set. -Setting $m=n^{1+\varepsilon}$, this bound gives $i(G) \leq O(n^{1-\varepsilon} \log(n)) = o(n)$.<|endoftext|> -TITLE: Yoneda extensions in derived categories -QUESTION [9 upvotes]: If given an abelian category $\mathcal{A}$, we can consider the bounded derived category $D^b(\mathcal{A})$. For two objects $A,B \in \mathcal{A}$, we know that there is a natural identification between -$$\text{Ext}_{\mathcal{A}}^i(A,B)$$ -and -$$\text{Hom}_{D^b(\mathcal{A})}(A,B[i])$$ -using Yoneda extensions. This is proven in Verdier's thesis (see also group of Yoneda extensions and the EXT groups defined via derived category). -I wondered whether one in general can identify the morphisms in the derived category with Yoneda extensions in the following way: -Given two complexes $E^\bullet, F^\bullet \in D^b(\mathcal{A})$, can we identify the set of morphisms $$\text{Hom}_{D^b(\mathcal{A})}(E^\bullet, F^\bullet[i])$$ -with sequences of morphisms in the derived category $D^b(\mathcal{A})$ -$$F^\bullet \to Z_{i-1}^\bullet \to \dots \to Z_0^\bullet \to E^\bullet $$ -such that the above sequence breaks into distinguished triangles, i.e. there exist distinguished triangles -$$F^\bullet \to Z_{i-1}^\bullet \to G_{i-1}^\bullet, \quad G_{i-1}^\bullet \to Z_{i-2}^\bullet \to G_{i-2}^\bullet, \dots$$ -In the case that $E^\bullet,F^\bullet$ are complexes concentrated in degree 0 (i.e. objects in $\mathcal{A}$), this is precisely the Yoneda extension with the $Z_j^\bullet, G_j^\bullet$ also complexes concentrated in degree 0. -For $i=1$ and $E^\bullet, F^\bullet$ arbitrary, this also holds true since a morphism $E^\bullet \to F^\bullet[1]$ can be completed to a distinguished triangle $$G^\bullet \to E^\bullet \to F^\bullet[1]$$ which we can rotate to obtain $$F^\bullet \to G^\bullet \to E^\bullet.$$ I am curious about the general case. - -REPLY [4 votes]: There is such a sequence, but it's not very interesting. -Given an element of $\text{Hom}_{D^b(\mathcal{A})}(E,F[i])$, then in the same way you describe, this gives a distinguished triangle -$$F\to Z_{i-1}\to E[-i+1].$$ -Then you can just take the distinguished triangles $E[-i+1]\to0\to E[-i+2]$, $E[-i+2]\to0\to E[-i+3]$, $\dots$, up to $E[-1]\to0\to E$, which can be spliced to give a sequence -$$F\to Z_{i-1}\to Z_{i-2}\to\dots\to Z_0\to E,$$ -where $Z_{i-2}=\dots= Z_0=0$. -This might seem like cheating, but I don't think it is. In the case of (a nonzero element of) $\text{Ext}_{\mathcal{A}}^i(A,B)$, the only reason you can't have zero objects in your sequence -$$B \to Z_{i-1} \to Z_{i-2}\to\dots \to Z_0 \to A$$ -is that you insist that the $Z_j$ are not just objects of $D^b(\mathcal{A})$, but of $\mathcal{A}$. If you drop this requirement, then you just have too much freedom.<|endoftext|> -TITLE: How to understand the combinatorial Laplacian $\Delta$ which is defined on the graph? -QUESTION [6 upvotes]: I have a question about the combinatorial Laplacian $\Delta$ which is defined by -$$\Delta(u,v)=c(u)1_{u=v}-c(u,v)$$ -where $u, v$ are some vertices in the graph $G=(V, E)$, and $c(u,v)$ is a conductance function defined on the edge $uv$ (i.e. weighted functions). - -If I define a function $F: V\to \mathbb{R}$, we can define the gradient $\nabla F(e)$ by - $$\nabla F(uv):=c(u,v)(F(v)-F(u))$$. But how to understand the $\Delta F(uv)$ by the combinatorial Laplacian $\Delta$? Actually, textbook claims that - $$\nabla \cdot \nabla F= -\Delta F$$ - -I have no idea to prove that. -The divergence $\nabla\cdot f$ is defined by -$$\nabla\cdot f(v)=\sum_{e} f(e).$$ - So $\nabla\cdot \nabla F(v)=\sum_{xy} c(x, y)(F(y)-F(x))$. - -REPLY [10 votes]: Just to add an (in my opinion important) piece of information. Say $F$ is a function on the vertices of a graph, so $F:V \to \mathbb{R}$. Then $\nabla F$ is a function from the edges to $\mathbb{R}$ (here I see an edge as a pair of vertices $(x,y)$, so edges are oriented): -$$\nabla F (x,y) := F(y) - F(x)$$ -Now this definition is very natural in many ways. For example, you would expect that the integral of the gradient of a function along a path is just the difference of the values of the function at the end of this path. And this holds here: if $\vec{p}$ is an oriented path (say from $a$ to $b$) then $\sum_{\vec{e} \in \vec{p}} \nabla F(\vec{e}) = F(b) - F(a)$. You can add a weight to the edges, but this is (in my opinion) not the important point for intuition. -Here is the important piece of information: if your graph has bounded degree$^*$, $\nabla$ defines an operator from $\ell^2V$ to $\ell^2E$. (The pairing on $\ell^2V$ is just $\langle f \mid g \rangle_V = \sum_{v \in V} f(v)g(v)$. Same pairing on $\ell^2E$ just that the sum is over the edges) -So you may ask, what is the adjoint of this operator? Well the defining property can be tested on Dirac masses (which are a nice basis of our spaces): -$$ -\langle \nabla^* \delta_{\vec{e}} \mid \delta_x \rangle = -\langle \delta_{\vec{e}} \mid \nabla\delta_x \rangle -$$ -So this is $+1$ if $\vec{e}$ has $x$ as target, $-1$ if $x$ is the source and 0 otherwise. Extended by linearity this gives: (here $G(x,y)$ is a function on the edges) -$$ -\nabla^* G(x) = \sum_{y \in N(x)} G(x,y) - \sum_{y \in N(x)} G(y,x) -$$ -where $y \in N(x)$ means $y$ is a neighbour of $x$. -(If your edges are not oriented, it is natural to consider only alternating functions on the edges, that is $G(x,y) = -G(y,x)$; the above expression simplifies then a bit) -The rest is just a computation: -$$ -\begin{array}{rl} -\nabla^* \nabla F(x) -&= \displaystyle \sum_{y \in N(x)} \nabla F(x,y) - \sum_{y \in N(x)} \nabla F(y,x) \\ -&= \displaystyle \bigg( \sum_{y \in N(x)} [F(y) - F(x)] \bigg) - \bigg( \sum_{y \in N(x)} [F(x) - F(y)] \bigg) \\ -&= \displaystyle 2 \bigg( \sum_{y \in N(x)} [F(y) - F(x)] \bigg) \\ -&= \displaystyle 2 \bigg( \big[ \sum_{y \in N(x)} F(y) \big] - \deg(x) F(x) \bigg) \\ -\end{array} -$$ -And that's the formula for the Laplacian (when the conductance is 1). -Note that I got a difference of a factor of 2 (because my definition of divergence is a bit different). But having a divergence which is the adjoint of the gradient, is a very important point, in my opinion. -If you add a weight to the edges, the computation are slightly more complicated, but it's just [possibly painful] bookkeeping. -$^*$ if you have weighted edges you could have an infinite number of edges as long as their weight is bounded -EDIT: a small addendum, for the case where the edge have a weight, as I realised there are many ways to add a weight in the above setup: - -you can add it to the definition of the gradient (but then the property that the integral along a curve is the difference of the values at the ends fail) - -you can add it to the definition of the divergence - -you can add it to the norm on $\ell^2E$ - - -I would recommend doing using the third one (which is the most natural: since the edge have a weight, incorporate it the norm in $\ell^2E$). This means the inner product on $\ell^2E$ is -$$\langle f \mid g \rangle = \sum_{\vec{e} \in E} c(\vec{e}) f(\vec{e}) g(\vec{e}) $$ -Because edges can be written as pair of vertices $(x,y)$ this reads -$$\langle f \mid g \rangle = \sum_{(x,y) \in E} c(x,y) f(x,y) g(x,y) $$ -[In your context, you probably want $c(x,y) = c(y,x)$.] -Now if you look at -$$ -\langle \nabla^* \delta_{\vec{e}} \mid \delta_x \rangle = -\langle \delta_{\vec{e}} \mid \nabla\delta_x \rangle -$$ -then this is $c(y,x)$ if $\vec{e}$ has $x$ as target, $-c(x,y)$ if $x$ is the source and 0 otherwise. Extended by linearity this gives: (here $G(x,y)$ is a function on the edges) -$$ -\nabla^* G(x) = \sum_{y \in N(x)} c(x,y) G(x,y) - \sum_{y \in N(x)} c(y,x) G(y,x) -$$ -If you assume $c(x,y) = c(y,x)$ and $G(x,y) = -G(y,x)$ (as you should in the unoriented case), you get: -$$ -\nabla^* G(x) = 2 \sum_{y \in N(x)} c(x,y) G(x,y) -$$ -Then, direct computation yields -$$ -\begin{array}{rl} -\nabla^* \nabla F(x) -&= \displaystyle 2 \sum_{y \in N(x)} c(x,y) \nabla F(x,y) \\ -&= \displaystyle 2 \bigg( \sum_{y \in N(x)} c(x,y) [F(y) - F(x)] \bigg) \\ -&= \displaystyle 2 \bigg( \sum_{y \in N(x)} [ c(x,y) F(y) - c(x,y) F(x)] \bigg) \\ -&= \displaystyle 2 \bigg( \big[ \sum_{y \in N(x)} c(x,y) F(y) \big] - \big[ \sum_{y \in N(x)} c(x,y) \big] F(x) \bigg) \\ -&= \displaystyle 2 \bigg( \big[ \sum_{y \in N(x)} c(x,y) F(y) \big] - c(x) F(x) \bigg) \\ -\end{array} -$$ -where $c(x)$ is a short-hand for $\sum_{y \in N(x)} c(x,y)$. -This is the Laplacian (up to a sign). The fact that you put a "$-$" sign or not depends entirely on your taste: if you want a Laplacian with negative spectrum, you should put a "$-$", otherwise don't (it's a standard trick to see that $A^*A$ has positive spectrum).<|endoftext|> -TITLE: Reference request: Recent progress on the conjugacy problem for torsion-free one-relator groups? -QUESTION [9 upvotes]: I am aware that the Spelling Theorem of B. B. Newman implies that one-relator groups with torsion are hyperbolic, and thus have a solvable conjugacy problem. My understanding is that for one-relator groups without torsion, the conjugacy problem is still open, though the most recent reference I have for this is over 20 years old. -Has there been any recent developments in this area? Any references for work in this area would be appreciated. -Thanks. - -REPLY [9 votes]: Yuhasz approach was to prove that every one-relator group satisfies certain version of small cancelation condition and hence has solvable conjugacy problem. Of course if the relator satisfies $C'(1/6)$ the group is hyperbolic and we are done. I think from what he proved the CAT(0) - conditions $C(3)-T(6)$ and $C(4)-T(4)$ are Ok too but I do not think all cases are covered in his paper. It was long since I looked at his text, so I may forget something. -Victor Guba knows the text well.<|endoftext|> -TITLE: Sullivan minimal model in the case of $H^1(V)\neq 0$ -QUESTION [5 upvotes]: Is there a simple construction of a Sullivan minimal model $\Lambda U \rightarrow V$ in the case that $H^1(V)\neq 0$? Do you have a reference? I envisage a degree-wise construction as in the case of $H^1(V)=0$ explaining how to deal with the new phenomenon. The existence is proven very generally in Felix, Halperin, Thomas, Rational Homotopy Theory but I find the proof very complicated and not transparent enough in this simple case (they prove that relative Sullivan models exist and that any relative Sullivan algebra is isomorphic to a product of a minimal relative Sullivan algebra and an acyclic algebra) - -REPLY [3 votes]: Gelfand and Manin explain it very nicely in their book "Methods of Homological Algebra", last chapter.<|endoftext|> -TITLE: Abstract mathematical concepts/tools appeared in machine learning research -QUESTION [5 upvotes]: I am interested in knowing about abstract mathematical concepts, tools or methods that have come up in theoretical machine learning. By "abstract" I mean something that is not immediately related to that realm. For instance, a concept from mathematical optimization does not qualify since optimization is directly related to the training of deep networks. In contrast, to me Topological Data Analysis is a non-trivial example of applying algebraic topology to data analysis. -Here are few examples that I have encountered in the literature (all in the context of deep learning). - -Betti numbers have been utilized to introduce a complexity measure -which could be used for comparing deep and shallow architectures: -https://www.elen.ucl.ac.be/Proceedings/esann/esannpdf/es2014-44.pdf -A connection between Sharkovsky's Theorem and the expressive power of deep networks: -https://arxiv.org/pdf/1912.04378.pdf -An application of Riemannian geometry: -https://arxiv.org/pdf/1606.05340.pdf -Algebraic geometry naturally comes up in studying neural networks with polynomial activation functions. This paper discusses functional varieties associated with such networks: -https://arxiv.org/abs/1905.12207 - -I find it useful to compile a list of such research works on ML that draw on pure math. - -REPLY [5 votes]: Probably, one the most striking is the "UMAP" (Uniform manifold approximation and projection) - a method of dimensional reduction in machine learning. The authors of the method use CATEGORY THEORY for its discovery. Well, there are certain discussions to what extent category theory is really required, see John Baez blog and references there in, still it is the author's original viewpoint how the method has been discovered. (The algorithm/implementation can be understood without category theory). -The method become quite quickly very popular - gaining 748 citations in two years according to google scholar. And found applications in many fields including bioinformatics (UMAP Nature), as well as capable to produce beautiful images MO355631. -It is similar to previously widely used method - tSNE (T-distributed stochastic neighbor embedding), however, quite often produces better results with less computational efforts, thus beating the predecessor both in quality and speed. -The documentations can be found here: UMAP docs.<|endoftext|> -TITLE: Is $J_1$ a subquotient of the monster group? -QUESTION [8 upvotes]: Edit: I was able to make a 3D diagram of the happy family if anyone is interested! -https://www.youtube.com/watch?v=_4IjnIcECoQ -I'm working on a twitter thread about the monster group, because I saw an interview with John Conway and he was very interested in the monster group. -Here is a link to the wikipedia article on the monster: -https://en.wikipedia.org/wiki/Monster_group -I'm very interested in the diagram of the sporadic groups. (Also if anyone would like to provide some basic understanding of the sporadic groups that would be helpful.) I'm trying to make a 3D version of the diagram with wire... -When I look at the diagram -I can understand that the sporadic groups that are maximal are the ones that are circled. And $M$ is the monster, with $B$ the baby monster. And it looks like $J_1$ is a subquotient of the maximal group $O'N$. I'm not very familiar with subquotients, so I am wondering if $J_1$ is a subquotient of $M$? That is, for my purposes, do I need to include $O'N$ and $J_1$ in the diagram of the monster? - -REPLY [15 votes]: No $J_1$ is not involved in (i.e. is not a subquotient of) the Monster. -The six sporadic simple groups listed here as "Pariahs" are precisely those that are not involved in the Monster, namely $J_1$, O'N, $J_3$, Ru, $J_4$, Ly. -There is a lot of information about the Monster and its subgroups in Section 5.8 of Robert Wilson's book, "The Finite Simple Groups''. -Note that, apart from a few uncertainties about a few possible groups ${\rm PSL}_2(q)$ for some small values of $q$, the maximal subgroups of the Monster are now all known, so locating which other simple groups are involved in it is moderately straightforward, using the ATLAS of Finite Simple Groups or the weblink above. -In fact Robert Wilson proved in a 1986 paper "Is $J_1$ a subgroup of the Monster" in Bull. London Math. Soc. 18, 349-150 that $J_1$ is not a subgroup, which was much more difficult at the time, because much less was known about the maximal subgroups of the Monster.<|endoftext|> -TITLE: Intercept the missile -QUESTION [20 upvotes]: A stealth missile $M$ is launched from space station. You, at another space station far away, are trusted with the mission of intercepting $M$ using a single cruise missile $C$ at your disposal . -You know the target missile is traveling in straight line at constant speed $v_m$. You also know the precise location and time at which it was launched. $M$, built by state-of-the-art stealth technology however, is invisible (to you or your $C$). So you have no idea in which direction it is going. Your $C$ has a maximum speed $v_c>v_m$. -Can you control trajectory of $C$ so that it is guaranteed to intercept $M$ in finite time? Is this possible? - -I can think of 3 apparent possibilities: - -Precise interception is possible. (It is possible in two dimensions, by calibrating your missile's trajectory to the parameters of certain logarithm spiral). -Precise interception is impossible, but for any $\epsilon\gt 0$, paths can be designed so that $C$ can get as close to $M$ as $\epsilon$ in finite time. -There's no hope, and your chance of intercepting or getting close to $M$ diminishes as time goes by. - - -This question is inspired by a similar problem in two dimensions by Louis A. Graham in his book Ingenious Mathematical Problems and Methods. - -REPLY [8 votes]: There is no hope. Timothy Budd already explained why staying on the sphere $S_t$ of possible locations of $M$ will not work; so this is to explain why leaving it will not help. What is enough to show is the following claim: at large times, if $t_1 -TITLE: Complexity of a combinatorial constraint -QUESTION [6 upvotes]: For two $k$-partitions $X,Y\in k^\omega$ of $\omega$ -(seen as functions $\omega\rightarrow k$), -we say $X,Y$ are almost disjoint -iff $X^{-1}(i)\cap Y^{-1}(i)$ is finite -for all $i -TITLE: Is a cosparse action on a CAT(0) cube complex an essential action? -QUESTION [5 upvotes]: Let $X$ be a CAT(0) cube complex. -(From Sageev and Wise's Cores for Quasiconvex actions) -A group $G$ acts cosparsely on a CAT(0)-cube complex $X$ if there exists a compact space $K$ and finitely many quasiflats $F_{1} , \ldots , F_{r}$ each quasi-isometric to $\mathbb{E}^{m}$ for some $m$ such that - -$X = GK \cup_{i} GF_{i}$ -Each hyperplane in $X$ crosses -$GK$. -$hF_{i} \cap kF_{j} \subset GK$ unless $i=j$ and $k^{-1}h -\in $ Stabiliser$(F_{i})$ -Quasiflats are $D$-isolated in the -sense that $hF_{i} \cap kF_{j}$ has diameter $< D$ unless $hF_{i} = -kF_{j}$. - -(From Caprace and Sageev's Rank Rigidity of CAT(0)-cube complexes) -Let $\Gamma $ be a group acting on $X$. -A hyperplane $\hat{\mathfrak{h}}$ of $X$ is called $\Gamma$-essential if both halfspaces contain $\Gamma$-orbit points of any vertex $v$ arbitrarily far away from $\hat{\mathfrak{h}}$. -An action on a CAT(0)-cube complex is called essential if every hyperplane is $\Gamma$-essential. -Are cosparse actions essential actions? -Sageev and Wise's Cores for quasiconvex actions (Prop 7.4) states that when a group $G$ is hyperbolic relative to virtually-free abelian groups, a proper and cosparse action on a CAT(0)-cube complex $X$ can be reduced to $G$ acting properly and cocompactly on a convex subspace of $X$ (which may not be a subcomplex and with convexity with respect to the CAT(0)-metric). -Lemma 3.1 of Caprace and Sageev's Rank Rigidity for CAT(0)-cube complexes implies that if $X$ strictly contains a $\Gamma$-invariant convex subcomplex, then the action of $\Gamma$ is not essential. - -REPLY [3 votes]: Being a cosparse action is not a rigid property. If $G \curvearrowright X$ is any cosparse action on a CAT(0) cube complex, then $G\curvearrowright X \times [0,1]$ is again cosparse but it is not essential. For instance, any action of a finite group on a CAT(0) cube complex which is not a single vertex is cosparse but not essential. -If you are interested in less trivial example, consider the action of $\mathbb{Z}$ on $\mathbb{R}^2 \times [0,1]$ through the translation by $(1,1,0)$. The action is already cosparse but not essential (nor cocompact). But you can make the example more complicated. The quotient $Q:=(\mathbb{R}^2 \times [0,1]) / \mathbb{Z}$ is a product of $[0,1]$ with a non-positively curved square complex which is topologically a cylinder and which has two hyperplanes. Consider the pointed sum $Q \vee \mathbb{S}^1$. Its fundamental group (a free group $\mathbb{F}_2$) acts on its universal cover (a CAT(0) cube complex which is a tree of quasiflats $\mathbb{R}^2 \times [0,1]$), and this action is cosparse but not essential.<|endoftext|> -TITLE: Finding $q(x)$ such that $p(q(x))$ is reducible over $\mathbb{Q}[x]$ -QUESTION [10 upvotes]: Let $p(x) \in \mathbb{Z}[x]$, such that $\deg (p) \ge 3$. - Can we always find $q(x) \in \mathbb{Z}[x]$, such that $\deg (q) < \deg(p)$ and $p(q(x))$ is reducible over $\mathbb{Q}[x]$? - -Is there any algorithm to find $q(x)$? -Note that the degree of $q$ is less than degree of $p$. -I'm looking for a proof or any reference of this result. -Any help would be appreciated. - -REPLY [13 votes]: Note: in this answer, I have inadvertently disregarded your requirement for $q$ to have integral coefficients. I do however prove that a $q$ with rational coefficients does exist, so I will just let this answer be on here for the moment. - -To answer your main question, yes this is possible (and simple to prove). We may assume $p$ irreducible, otherwise there is nothing to prove. Let $\alpha$ be a zero of $p$ and consider $K = \mathbb{Q}(\alpha)$. Now the subset $S$ of $K$ consisting of elements $\beta$ such that $\mathbb{Q}(\beta)=K$ is equal to the complement in $K$ of a finite number of lower-dimensional $\mathbb{Q}$-vector spaces; in particular it is non-empty. Likewise, the subset $T$ of $K$ consisting of all elements that are linearly independent over $\mathbb{Q}$ with $\{1, \alpha\}$ is also a complement of a finite number (in this case just one) of lower-dimensional vector spaces (here we need the degree of $p$ to be at least $3$). -Now take any $\beta$ in $S \cap T$ (which is non-empty by what I wrote above): since $\mathbb{Q}(\beta)=K$, we have $\alpha=q(\beta)$ for some $q$ in $\mathbb{Q}[X]$ whose degree we can choose to be $< \deg (p)$ by subtracting the appropriate multiple of the minimal polynomial of $\beta$. Since $\beta$ is in $T$, we have that the degree of $q$ is $>1$. Then $\beta$ is a root of the polynomial $p \circ q$, so the latter must have a factor of degree $\deg (p)$, whereas its degree is $\deg(p)\deg(q)>\deg(p)$. Hence it is reducible.<|endoftext|> -TITLE: Skew differential graded algebra -QUESTION [7 upvotes]: A sigma, or skew, derivation is a natural generalisation of the -notion of derivation depending on an algebra automorphism $\sigma$ which -when equal to $id = \sigma$ reduces to the usual notion of a -derivation. For a precise definition see here -https://planetmath.org/SigmaDerivation -Does there exist a notion of skew differential graded algebra -in the literature? If so where do these objects arise? -EDIT: To confirm I am asking if there exists a graded analogue of skew derivation algebra. So an $\mathbb{N}_0$-graded algebra $A = \bigoplus_{k \in \mathbb{N}_)0} A_k$, together with a degree $1$ map $d$ satisfying $d^2 = 0$, and a skew analogue of the -graded Leibniz rule: -$$ -d(a \wedge b) = da \wedge \sigma(b) + (-1)^k \sigma(a)db, ~ a \in A_k -$$ - -REPLY [2 votes]: This edited version of the "skew Leibniz rule" has appeared in geometry: if $\varphi: N\to M$ is a map of (super) manifolds, a section $X$ of the pullback bundle $\varphi^\star TM$ is a linear map $C^\infty(M)\to C^\infty(N)$ satisfying precisely your skew-leibniz rule: $$ X(fg)=X(f)\varphi^\star g +(-1)^{\deg X\deg f} (\varphi^\star f) X(g)$$ -(See J. Nestruev, Smooth manifolds and observables, paragraph 9.47.) -So if you put $\mathbb N_0$-gradings on your structure sheaves (what is $\mathbb N_0$? Positive integers?) and pick an $X$ of degree $+1$ which squares to zero you seem to arrive at your setup. (Identifying $\varphi^\star$ with $\sigma$ and $C^\infty(M)$ with $A$.) -On the algebraic side there is the (comparatively more obscure) notion of $({\bf l},{\bf r})$-coderivation of Berglund (Definition 3.2) which should be dual to your proposal in the case ${\bf l}={\bf r}=\sigma$.<|endoftext|> -TITLE: Every homotopy class contains at least a harmonic representative -QUESTION [5 upvotes]: Let $(M^3,g)$ be a closed, connected and oriented Riemannian $3$-manifold. A circle-valued map $v : M \to S^1$ is harmonic iff the gradient $1$-form $\omega_v = v^* d\theta \in \Omega_1(M)$ is harmonic in the Hodge sense: $d \omega_v = 0$ and $\delta \omega_v = 0$. It can be seen that this happens precisely when $v$ minimizes the Dirichlet energy in its homotopy class $[v] \in [M:S^1]$. Thus, by Hodge theory, each homotopy class of a circle-valued map contains a harmonic representative. -My question is whether every homotopy class of $S^2$-valued maps contains a harmonic representative. More precisely: given $u : M \to S^2$ a smooth map, does there exist a harmonic map $u_0 : M \to S^2$ such that $u$ is smooth and homotopic to $u$? -A parallel question: if $u_0 : M \to S^2$ is any harmonic map, can we say that $u_0^* \sigma \in \Omega_2(M)$ is a harmonic $2$-form, where $\sigma$ is the area form of (the round) $S^2$? - -REPLY [4 votes]: As Andy says, the answer is 'no': It is known that there is no harmonic map of degree $1$ from the torus to the $2$-sphere. I forget who first observed this. (Amended after Andy's comment: It's originally due to J. C. Wood in the early 1970s, see Andy's comment for the exact reference.) -If I have time, I can put in the argument, but the essential outline of the argument is this: -There are two kinds of harmonic maps from the torus to the $2$-sphere. Those that are conformal and those that are not. -If it is conformal, then, up to reversing the orientation on the torus, it is a holomorphic map, and it is well-known that a non-constant holomorphic map from the torus to the $2$-sphere has degree at least 2. (In fact, there is such a holomorphic map of any degree $d\ge 2$.) -If it is not conformal, then a simple calculation shows that the degree of the mapping is zero. (Essentially, one produces an explicit $1$-form on the torus whose differential is the pullback of the area form on the $2$-sphere.) -Thus, there is no harmonic map of degree 1 from the torus to the $2$-sphere.<|endoftext|> -TITLE: Descent for $K(1)$-local spectra -QUESTION [6 upvotes]: For odd primes, we have an equalizer diagram for the $K(1)$- local sphere given by -$$L_{K(1)}S \rightarrow K{{ \xrightarrow{\Psi^g}}\atop{\xrightarrow[i_K ] {}}} K$$ -where $g$ is a topological generator of $\mathbb{Z}_p^{\times}$ and $\Psi^g$ denotes the Adams operation. Now we can apply the functor $Mod(-): \textrm{Spectra} \to \textrm{Symmetric monoidal infinity categories} $ and then apply the functor $L_{K(1)}$. This gives us a diagram of the form -$$L_{K(1)}Sp \rightarrow Mod_{K(1)}(K){{ \xrightarrow{\Psi^g}}\atop{\xrightarrow[i_K ] {}}} Mod_{K(1)}(K)$$ where $L_{K(1)}Sp$ are the $K(1)$-local spectra and $Mod_{K(1)}(K)$ are the $K(1)$-local $K$ modules. -Now is this an equalizer diagram? -I have been told that this is most likely true, so I was just wondering if there was a reference for this statement. Thanks in advance. - -REPLY [7 votes]: It's not quite true: need to require a $p$-adic continuity condition for the $\Psi^g$-semilinear automorphism of the $K(1)$-local $K$-module. You can see https://arxiv.org/pdf/2001.11622.pdf Proposition 3.10 for a slight variant which also works at the prime 2.<|endoftext|> -TITLE: Riemannian manifold as a metric space -QUESTION [15 upvotes]: I am looking for a reference to the following simple statement; it must be classical. (It is easy to proof, but I want to have a reference.) - -A metric space $X$ that corresponds to a Riemannian manifold $(M,g)$ completely determines the underlying smooth manifold $M$ and the metric tensor $g$. - -REPLY [14 votes]: It was proven by Dick Palais. - -MR0088000 (19,451a) Reviewed Palais, Richard S. On the - differentiability of isometries. Proc. Amer. Math. Soc. 8 (1957), - 805–807. - 53.2X - -MathSciNet - -@article {MR88000, - AUTHOR = {Palais, Richard S.}, - TITLE = {On the differentiability of isometries}, JOURNAL = {Proc. Amer. Math. Soc.}, FJOURNAL = {Proceedings of the American - Mathematical Society}, - VOLUME = {8}, - YEAR = {1957}, - PAGES = {805--807}, - ISSN = {0002-9939}, MRCLASS = {53.2X}, MRNUMBER = {88000}, MRREVIEWER = {K. Krickeberg}, - DOI = {10.2307/2033302}, - URL = {https://doi-org.ucc.idm.oclc.org/10.2307/2033302}, } - -According to Palais, if I read his paper correctly, Myers and Steenrod proved the differentiability of isometries but Palais obtained an explicit description of smooth functions on the manifold from the metric geometry.<|endoftext|> -TITLE: Who introduced direct limits? -QUESTION [28 upvotes]: The general notion of a direct limit of a commuting system of embeddings, indexed by pairs in a directed set, has seen heavy use in set theory. It is the same notion as in category theory. I was surprised to find that the general definition does not appear in the book on model theory by Chang and Keisler (MR1059055). Who was the originator of this idea? - -REPLY [27 votes]: The definition of a direct limit of groups was given by Pontrjagin in his 1931 paper Über den algebraischen Inhalt topologischer Dualitätssätze<|endoftext|> -TITLE: Renormalization group strategies -QUESTION [5 upvotes]: Before introducing block spin transformations in chapter four of Random Walks, Critical Phenomena and Triviality in Quantum Field Theory, the authors state the following: - -"In this chapter we sketch a specific method for constructing scaling - (continuous) limits $G^{*}(x_{1},...,x_{n})$ of reescaled correlations - $G_{\theta}(x_{1},...,x_{n})$ as $\theta \to \infty$, namely the - Kadanoff block spin transformations. They serve as a typical example - of "renormalization group transformations". Of course there are many - other incarnations of the renormalization group strategy [...]." - -Well, as the name suggests, the aim of the book is to address QFT but some of these techniques are also applicable and useful to rigorous statistical mechanics. For instance, I think more useful to statistical mechanics is the infinite volume limit than the continuum limit. -What are the most important "incarnations" of the renormalization group strategy in statistical mechanics? To what types of problems they fit and what are their limitations? Are they all related? - -REPLY [3 votes]: In statistical mechanics one is mostly interested in some fixed probability measure for some spin configurations on the infinite volume lattice $\mathbb{Z}^d$. The two main problems related to such a measure are P1) the construction of the infinite volume limit and P2) the study of the long distance behavior of correlations for this measure. -In QFT one looks at continuum limits. There are two types of continuum limits L1) limits $\theta\rightarrow\infty$ where the unit lattice measure is fixed and L2) limits where the unit lattice measure varies with the UV cutoff/inverse lattice spacing $\theta$. Massive QFTs are typically obtained via L2). CFTs are obtained via L1) or what probabilists mean by a "scaling limit" in a rather strict sense. Some non massive continuum limits also involve L1), e.g., the RG trajectory going from a Gaussian fixed point to a nontrivial infrared fixed point/CFT. For an example of rigorous result on the latter, see my article "A Complete Renormalization Group Trajectory Between Two Fixed Points". -RG methods are useful for all of the above problems P1), P2), L1) and L2). From a statistical mechanics point of view, L2) is not that interesting. However P2) essentially is the same as L1).<|endoftext|> -TITLE: Infra-Pták space that is not Pták -QUESTION [6 upvotes]: From reading the literature of the 1970s heyday of locally convex spaces, it seems that it was an important open question whether there is an infra-Pták (i.e. $B_r$-complete) space that is not Pták (i.e. $B$-complete). Is this still open? - -REPLY [4 votes]: Valdivia constructed counterexamples in his paper Br-Complete Spaces which are not B-Complete.<|endoftext|> -TITLE: Lower bound on exponential sums -QUESTION [8 upvotes]: Let $k\geq 2$. Consider the following norm of exponenetial sum: -$$ -I(N,p,k)=\int_0^1\int_0^1 \left|\sum_{n=0}^N e^{2\pi i (n x+n^k y)}\right|^p dxdy. -$$ -Bourgain mentioned on Page 118 of -https://math.mit.edu/classes/18.158/bourgain-restriction.pdf -that $I(N,6,2)\gtrsim N^3\log N$, where he referenced the following article: -https://www.researchgate.net/publication/259308546_The_method_of_trigonometric_sums_in_number_theory. -But I did not find an explicit result in the above article that leads directly to the lower bound above. -So my questions are: - -What is the idea to prove the above lower bound? The famous Vinogradov's mean value theorem deals with upper bounds of $I(N,p,2)$, but not lower bounds. -What is a reasonably sharp lower bound for $I(N,p,3)$, or particularly, $I(N,6,3)$? Note that this may not be in the direct form of Vinogradov's mean value theorem, as the $n^2$ term is missing here. - -REPLY [7 votes]: The result for $I(N,6,2)$ was proved by Rogovskaya N. N. in the article An asymptotic formula for the number of solutions of a certain system of equations. The proof is elementary. Main idea is to replace the system $$x_ 1+x_ 2+x_ 3=y_ 1+y_ 2+y_ 3,\quad x^ 2_ 1+x^ 2_ 2+x^ 2_ 3=y^ 2_ 1+y^ 2_ 2+y^ 2_ 3 -$$ -by $$a_1+a_2+a_3=0,\quad a_1b_1+a_2b_2+a_3b_3=0,$$ -where $a_i=x_i-y_i$ and $b_i=x_i+y_i$. Then you can count solutions of the last equation and sum the result over $a_i.$ The answer is nice -$${\mathcal N}(P)=18\pi^{-2}P^ 3\log P+{\mathcal O}(P^ 3).$$ -Probably this is the only case when trigonometric integral was calculated explicitely.<|endoftext|> -TITLE: DW, state sum models, and fully extended TQFTs -QUESTION [8 upvotes]: I am interested in state sum models and their relations with some other of TQFTs, especially the fully extended TQFTs and the Dijkgraaf-Witten TQFTs (generalized, in the sense that finite-group-bundles are replaced by higher bundles over higher algebraic structures). Forgive about my naiveness, but I immaturely suspect maybe three of them are the same. I hope I will get an answer one day, and this post is my first step. -I don't have much access to the experts of this field, and therefore am not sure how much this has been developed. Perhaps the answers are in written papers. In any case, I think a complete answer is too much to hope. If you have any relevant paper in mind, please point them out with short comments. Thank you so much in advance! -1. Crane-Yetter and Dijkgraaf-Witten -Crane-Yetter theory is a well-known $4$-D state sum model. According to Manuel Bärenz's edit on nLab, it can be realized as a generalized DW theory, based on quantum groups instead of finite groups. -Q1.1 Can you give a formal reference for that statement? -Q1.2 Can (m)any other state sum models be interpreted as a generalized DW theory? We have to be flexible here: fields can be higher bundles. -Q1.3 In contrast, can generalized Dijkgraaf-Witten theories be realized as state sum models? -2. Dijkgraaf-Witten and Fully Extended TQFTs -By Domenico Fiorenza's edit on nLab (sec. 2), Dijkgraaf-Witten models are fully extended. -Q2.1 Is there a formal reference for this statement? -Q2.2 Can fully extended TQFTs be realized as generalized Dijkgraaf-Witten models? -3. Fully Extended TQFTs and State Sum Models -Q3.1 I have been trying to find evidence why CY is fully extended and what the point is associated to, but in vain. The best answer I have heard is that physicists believe state sum models should automatically be fully extended. If this is true, I really want to know how and why it should work. -Q3.2 On the other hand, by cobordism hypothesis (proved by Lurie 2009), any fully extended TQFT is determined by the assignment at the point. I have a feeling that if the target category is "finite" enough, then this might be interpreted as a state sum model. Would you share your understanding? (EDIT: an answer by Kevin Walker suggested that some standard techniques bring you a state sum model from fully extended ones. Another possibly related post can be found here). - -REPLY [8 votes]: Let me try to answer your questions at least in part. My apologies for references I've missed. For an overview of the ideas without references, you might enjoy Pavel Safranov's talks at the intro conference at MSRI. -1.What exactly do you mean by Dijkgraaf-Witten? If you asked me what Dijkgraaf-Witten was (and this is more a statement about myself than about mathematical definitions) I would have said that it was Crane-Yetter theory for a pointed braided tensor category Vec(A,q). Without this I can't answer your questions. -2.1 This is certainly "known" but I'm having a little trouble tracking down the exact reference. I think Freed-Hopkins-Lurie-Teleman for finite groups is doing what you want, based on earlier work of Freed. -2.2 I really don't understand what you're asking here. If Dijkgraaf-Witten is a special case of fully extended TFTs, then by the definition of generalization fully extended TFTs are generalized Dijkgraaf-Witten... -3.1 Crane-Yetter attached to a modular tensor category C is a fully local 4-dimensional TFT which assigns to the point C inside the Haugseng-Johnson-Freyd-Scheimbauer Morita 4-category of E2 algebras in the 2-category of LFP categories. This is "known" to experts, but not fully in print anywhere. There is an incomplete construction given in unfinished notes by Walker, this uses somewhat non-standard definitions for higher categories and TFTs so even when its completed one can argue about whether it translates into other definitions. There's also some unfinished work of Freed-Teleman in this direction. The 012-dimensional part of Crane-Yetter is constructed and computed in work of Ben-Zvi-Brochier-Jordan building on Ayala-Francis's factorization homology. I have work with Brochier-Jordan and also with us and Safranov showing that non-degenerate braided fusion categories are fully dualizable and actually invertible, and thus by the cobordism hypothesis give framed fully local TFTs which we think of as a framed version of Crane-Yetter, but we don't have serious calculations of what this theory yields in high dimensions. Furthermore, as Arun points out in comments, in order to turn this into the usual Crane-Yetter you need to understand how the ribbon structure on a modular tensor category gives an SO(4)-fixed point structure. You can see some results in this direction in my MSRI talk on joint work in progress with Douglas-Schommer-Pries, but we're still a ways off from giving a full answer. -3.2 A good reference for how to get a state-sum out of a fully extended theory is Orit Davidovich's PhD thesis. In principal there's no problem doing this, but in practice there's plenty of interesting questions along these lines to which we don't yet know the answer. For example, you should be able to use my work with Douglas-Schommer-Pries to get a framed Turaev-Viro state sum model attached to a (not necessarily spherical) fusion category. But we don't even have a guess for what exactly that should look like. Or, after you've analyzed the SO(3)-fixed points enough, you should be able to show that the TFTs coming from spherical fusion categories via the oriented version of the cobordism hypothesis and our work agrees with Turaev-Viro thereby showing that Turaev-Viro theories are fully extended. I expect within 5-10 years we will have all of this understood, but we don't yet, and there may be other approaches that are more direct. -Two other helpful references (with extensive bibliographies) which I didn't mention specifically are recent papers by Schommer-Pries about invertible TFTs and Reutter's paper on semisimple theories.<|endoftext|> -TITLE: How to prove some identities about infinite product? -QUESTION [5 upvotes]: Recently, I read one paper titled Modular equations and approximations to π by Ramanujan, in which there are some formulas for $q=\pi i \tau$( where $\tau=x+yi, y>0$, hence $|q|<1)$ : -$$\prod_{n=1}^\infty\left(1+q^{2n-1}\right)=2^{\frac{1}{6}} q^{\frac{1}{24}}(kk')^{-\frac{1}{12}} ~~~ (1)$$ -and -$$ \prod_{n=1}^\infty\left(1-q^{2n-1}\right)= 2^{\frac{1}{6}} q^{\frac{1}{24}}k^{-\frac{1}{12}}k'^{\frac{1}{6}} ~~~~(2)$$ -where $k=k(\tau)$ is the Jacobi modulus， $k^2(\tau)=\lambda(\tau)$, the elliptic modular function, and $k'=\sqrt{1-k^2}.$ -The following result can be calculated by Mathematica: -$$\left(1+e^{-\pi }\right)\left(1+e^{-3 \pi }\right)\left(1+e^{-5 \pi }\right) \cdots=2^{\frac{1}{4}} e^{-\pi / 24}.$$ -But I do not know how to prove these formulas (1) and (2). I would appreciate if someone could give some suggestions. - -REPLY [3 votes]: First， by theta function we have $$ k=\frac{\theta_2}{\theta_3},k'=\frac{\theta_4}{\theta_3},$$ where -$$ \theta_2=2q^{\frac{1}{4}} G \prod (1+q^{2n})^2; ~(1)$$ -$$ \theta_3= G \prod (1+q^{2n-1})^2;(2)$$ -$$ \theta_4= G \prod (1-q^{2n-1})^2;(3)$$ -and -$$ G= \prod (1-q^{2n})^2.$$ -So we have $RHS=2^{\frac{1}{6}}q^{\frac{1}{24}}(\frac{\theta_2 \theta_4}{\theta^2_3})^{-\frac{1}{6}}=(\frac{2\theta_3^2}{\theta_2\theta_4})^{\frac{1}{6}}q^{\frac{1}{24}}.$ -It is enough to prove the following : -$$ \prod(1+q^{2n-1})^2=(\frac{2\theta_3^2}{\theta_2\theta_4})^{\frac{1}{3}}q^{\frac{1}{12}}$$ -Put (1),(2),(3) into the above identity, Jacobi triple product Identity is obtained. Hence the result is established.<|endoftext|> -TITLE: Does the "three-set-lemma" imply the Axiom of Choice? -QUESTION [21 upvotes]: Consider the following curious statement: - -$(S)$ $\;$ Let $X$ be a non-empty set and let $f:X \to X$ be fixpoint-free (that is $f(x) \neq x$ for all $x\in X$). Then there are subsets $X_1, X_2, X_3 \subseteq X$ with $X_1\cup X_2\cup X_3 = X$ and $$X_i \cap f(X_i) = \emptyset$$ for $i \in \{1,2,3\}$. - -There are easy examples showing that one cannot get by using $2$ subsets only. Statement $(S)$ can be proved using the axiom of choice. -Question. Does $(S)$ imply (AC)? - -REPLY [22 votes]: To complement godelian’s answer, the three-set lemma is not provable in ZF alone, as it implies the axiom of choice for families of pairs. This holds even if we allow any finite (or even just well ordered) number of sets instead of three. -If $\{P_i:i\in I\}$ is a family of disjoint two-element sets, put $X=\bigcup_{i\in I}P_i$, and let $f\colon X\to X$ be defined such that it maps each element to the other element in the same pair. Then if $\alpha$ is an ordinal and $X=\bigcup_{\beta<\alpha}X_\beta$ where $f[X_\beta]\cap X_\beta=\varnothing$ for each $\beta$, we have $|X_\beta\cap P_i|\le1$ for all $\beta<\alpha$ and $i\in I$, thus the following is a selector: $s(i)=$ the unique element of $X_\beta\cap P_i$, where $\beta<\alpha$ is the least such that the intersection is nonempty. - -EDIT: Here is an exact characterization. (Again, the argument also applies if we allow any well-ordered number of sets instead of three.) - -Theorem. Over ZF, the three-set lemma is equivalent to the axiom of choice for families of finite sets. - -As mentioned in godelian’s answer, the right-to-left implication can be found in - -K. Wiśniewski: On functions without fixed points, Commentationes Mathematicae 17 (1973), no. 1, pp. 227–228. DML-PL - -See also Andreas Blass’s comment above. -For the left-to-right implication, let $F$ be a family of nonempty finite sets. We may assume $F$ is closed under (nonempty) subsets. Put -$$C=\{\langle A,h\rangle:A\in F,|A|\ge2,h\text{ is a cyclic ordering of }A\},$$ -where a cyclic ordering of $A$ is a permutation $h\colon A\to A$ which forms a single cycle of length $|A|$. Let $X$ be the disjoint union $\sum_{\langle A,h\rangle\in C}A$, and define $f\colon X\to X$ as the corresponding union $\sum_{\langle A,h\rangle\in C}h$. Let us fix $X_1,X_2,X_3\subseteq X$ as given by the three-set lemma. -By induction on $n$, we will construct a selector $s_n$ on $F_n=\{A\in F:|A|=n\}$. Then $\bigcup_ns_n$ will be the desired selector on $F$. -The base case $n=1$ is trivial. Assume that $n\ge2$, and we have already constructed $\{s_m:m -TITLE: "Conjugacy classes–irreducibles" bijection, but for permutation representations -QUESTION [7 upvotes]: The linear representation theory (over say $\mathbb{C}$ for concreteness) of a finite group $G$ is "the same as" its character theory. Characters are naturally functions on conjugacy classes of elements ("class functions"), and the number of irreducible representations is the same as the number of conjugacy classes of elements. But in general there is no canonical bijection between conjugacy classes of elements and irreducible representations. -Now let's consider permutation representations, i.e., $G$-sets. The analog of characters for permutation representations are "marks". Marks are naturally functions on conjugacy classes of subgroups of $G$, and the number of "irreducible" $G$-sets is the same as the number of conjugacy classes of subgroups of $G$. But now we do have a canonical bijection between conjugacy classes of subgroups and irreducible $G$-sets: each conjugacy class $H$ naturally determines an orbit $G/H$. -Question: Is there a high-level explanation for this apparent "difference" between the linear representation theory and the permutation representation theory of finite groups? - -REPLY [2 votes]: The difference is apparent, but not perhaps real. We are not saying that there isn't a 'canonical' basis for the set of class functions, just that it isn't the irreducible characters. -Here is an arguable definition of a canonical basis for the set of class functions. Let $G$ be finite, and $x\in G$ of order $n$. Define $\phi_x$ to be the character of $\langle x\rangle$ that maps $x$ to $\mathrm{e}^{2\pi \mathrm{i}/n}$, and let $\chi_x$ be the induction of $\phi_x$ to $G$. Notice that if $x$ and $y$ are conjugate in $G$ then $\chi_x=\chi_y$, and indeed the $\chi_x$ form a basis for the set of class functions. -That's pretty natural a basis to choose, it is just far from the irreducible characters in general. I don't know if this answers your question.<|endoftext|> -TITLE: Polynomial inequality of sixth degree -QUESTION [11 upvotes]: There is the following problem. - -Let $a$, $b$ and $c$ be real numbers such that $\prod\limits_{cyc}(a+b)\neq0$ and $k\geq2$ such that $\sum\limits_{cyc}(a^2+kab)\geq0.$ - Prove that: - $$\sum_{cyc}\frac{2a^2+bc}{(b+c)^2}\geq\frac{9}{4}.$$ - -I have a proof of this inequality for any $k\geq2.6$. -I think, for $k<2.6$ it's wrong, but my software does not give me a counterexample -and I don't know, how to prove it for some $k<2.6$. -It's interesting that without condition $\sum\limits_{cyc}(a^2+kab)\geq0$ the equality occurs also for $(a,b,c)=(1,1,-1)$. -My question is: What is a minimal value of $k$, for which this inequality is true? -Thank you! - -REPLY [2 votes]: We want to show that your inequality does not hold for $k\in[2,13/5)$. In view of the identity in your answer, it is enough to show that for each $k\in[2,13/5)$ there is a triple $(a,b,c)\in\mathbb R^3$ with the following properties: $a=-1>b$, -\begin{align}s_4&:=\sum_{cyc}(2a^3-a^2b-a^2c) \\ -&=a^2 (2 a-b-c)+b^2 (-a+2 b-c)+c^2 (-a-b+2 c)=0, -\end{align} -$$s_3:=a b + b c + c a<0,$$ -and -$$s_2+k s_3=0,$$ -where -$$s_2:=a^2 + b^2 + c^2.$$ -Indeed, then the right-hand side of your identity will be -$$\frac{20}{3}\sum_{cyc}(a^4-a^2b^2)(13/5-k)s_3<0,$$ -so that your identity will yield -$$\sum_{cyc}\frac{2a^2+bc}{(b+c)^2}<9/4.$$ -For each $k\in(2,13/5)$, the triple $(a,b,c)$ will have all the mentioned properties if $a=-1$, $b$ is the smallest (say) of the 6 real roots $x$ of the polynomial -$$P_k(x):=-18 - 15 k + 4 k^2 + - 4 k^3 + (36 k + 6 k^2 - 12 k^3) x + (-27 - 9 k - 21 k^2 + - 6 k^3) x^2 + (18 + 60 k - 10 k^2 + 8 k^3) x^3 + (-27 - 9 k - - 21 k^2 + 6 k^3) x^4 + (36 k + 6 k^2 - 12 k^3) x^5 + (-18 - 15 k + - 4 k^2 + 4 k^3) x^6, $$ -and -$$c=\tfrac12\, (k - b k) - \tfrac12\, \sqrt{-4 - 4 b^2 + 4 b k + k^2 - 2 b k^2 + b^2 k^2}.$$ -For $k=2$, $(a,b,c)=(-1,0,1)$ will be such a triple. -So, we are done. - -This result was obtained with Mathematica, as follows (which took Mathematica about 0.05 sec):<|endoftext|> -TITLE: For what sets $X$ do there exist a pair of functions from $X$ to $X$ with the identity being the only function that commutes with both? -QUESTION [13 upvotes]: It is not too difficult to show that if $X$ is an infinite set, then there exists a two-element subset of the group $\operatorname{Sym}(X)$ with trivial centralizer iff $\lvert X\rvert \leq \lvert\mathbb{R}\rvert$. -My question is if this is true if we replace $\operatorname{Sym}(X)$ with $\operatorname{End}(X)$. -I.e., for what infinite sets $X$ do there exist functions $f,g: X \rightarrow X$, such that if $h:X \rightarrow X $ satisfies $fh = hf$ and $gh = hg$, then $h = I$? The same argument from the $\operatorname{Sym}(X)$ case shows that it is true when $|X| \leq \mathbb{R}$ (and was given as a problem in the 6th Romanian Masters of Mathematics competition). But is it false for $|X| > |\mathbb{R}|$? - -REPLY [11 votes]: The answer is no: for every set $X$ there exists a pair in the monoid $X^X$ of self-maps of $X$, with centralizer reduced to $\{\mathrm{id}\}$. -(I first left my original "groupwise" answer because it's easier and because it has other follow-up questions. It's now deleted and copied as an answer to another question). -For $X$ empty take $(\mathrm{id},\mathrm{id})$. For $X$ finite nonempty, take a constant, and a cycle. So henceforth I assume that $X$ is infinite. -(a) First I use Sierpiński-Banach theorem [cf. here and here] that every countable subset (here just finite is fine) of $X^X$ is contained in the subsemigroup generated by a 2-element subset. This reduces to proving that there is a finite (actually 6-element) subset $\Sigma\subset X^X$ with trivial centralizer. -(b) Next I split $X$ as union of two subsets $Y,Z$ of the same cardinal. Let $f,g\in X^X$ have image equal to $Y$ and $Z$ respectively. If $u$ commutes to $f$, then $u$ stabilizes $\mathrm{Im}(Y)$, and similarly with $g$, $Z$. I'll therefore assume $f,g\in\Sigma$, and hence every $u$ in the centralizer of $\Sigma$ stabilizes both $Y$ and $Z$. -(c) It was proved in [VPH] that there exists a "strongly rigid" binary relation on $Y$: a subset $R\subset Y^2$ (actually, $R$ being subset of a well-ordering) such that the only endomorphism $u$ of $(Y,R)$ is the identity. (Here endomorphism means that $u\times u:Y^2\to Y^2$ maps $R$ into itself.) Clearly the cardinal of $R$ is that of $|Y|=|X|$. -Choose a partition $Z=Z'\sqcup Z''$ of $Z$ in subsets of the same cardinal. -Choose a bijection $i$ from $R$ to $Z'$. Define self-maps $p,q$ of $X$ as follows. On $Y$, $p$ and $q$ are chosen as injective maps into $Z''$. Also $p$ and $q$ are defined on $Z'$ by: for $(y,y')\in Y^2$ and $z=i(y,y')$, $q(z)=p(y)$ and $p(z)=q(y')$. Finally, extend $p,q$ arbitrarily choosing maps $Z''\to Y$. -Then, for $(y,y')\in Y^2$, we have $(y,y')\in R$ if and only if there exists $z_1,z,z_2\in Z$ such that $p(y)=z_1$, $q(z)=z_1$, $p(z)=z_2$, $q(y')=z_2$. [Intuition: this is a "$\stackrel{p}\to\stackrel{q}\leftarrow\stackrel{p}\to\stackrel{q}\leftarrow$ path" from $y$ to $y'$] -Indeed $\Rightarrow$ works by construction with $z_1=p(y)$, $z=i(y,y')$, $z_2=q(y')$. Conversely, suppose that such elements exist; write $(Y,Y')=i^{-1}(z)$, so $(Y,Y')\in R$. By definition $p(z)=q(Y')$ and $q(z)=p(Y)$. So $q(Y')=q(y')$ and $p(Y)=p(y)$. By injectivity of $p$ and $q$ on $Y$, we have $(y,y')=(Y,Y')\in R$. -As a consequence, if $u$ stabilizes $Z$ and $Y$ and commutes with $p$ and $q$, then $u$ preserves $R$ on $Y$. -Next we define similarly $p',q'$ from a strongly rigid binary relation on $Z$. -Then the above proves that the centralizer of $\{f,g,p,q,p',q'\}$ in $X^X$ is reduced to $\{\mathrm{id}\}$. -[VPH] Vopěnka, P.; Pultr, A.; Hedrlín, Z. -A rigid relation exists on any set. -Comment. Math. Univ. Carolinae 6 (1965), 149–155. - -Informal outline: the hard step is the above reference (existence of a strongly rigid binary relation). Then, the 0th step is Sierpinski-Banach (which is not hard) to pass from 6 to 2. The second is quite trivial: there exists a pair such that centralizing this pair implies preserving each component of a partition into two moieties. The third step is to encode a binary relation into a pair of self-maps using such a $\stackrel{p}\to\stackrel{q}\leftarrow\stackrel{p}\to\stackrel{q}\leftarrow$ path and the "coloring" by the 2-component partition.<|endoftext|> -TITLE: Can a torsion-free group be quasi-isometric to a torsion group? -QUESTION [15 upvotes]: I have looked around in the literature on group theory and geometric group theory and this looks to be an open question as far as I can tell (by torsion group, I mean as usual a group in which every element has finite order). -I was wondering if anyone has recently made any progress on this question or if there is some review article which looks at the possibilities? - -REPLY [2 votes]: This is one of many open questions in geometric group theory related to quasi-isometries. Proving things about invariance under quasi-isometries is generically quite tricky, as quasi-isometries do not even need to be continuous. Some other open questions: - -Is the Haagerup property invariant under quasi-isometries? (see comment by YCor for recent work on this one) -Is the rapid decay property invariant under quasi-isometries? -Is the property of having uniform exponential growth invariant under quasi-isometries? -Are random finitely presented groups quasi-isometry rigid? -How can fundamental groups of compact $3$-manifolds be classified up to quasi-isometry?<|endoftext|> -TITLE: Books to develop a unified view of statistics and information theory? -QUESTION [7 upvotes]: I hope to understand the connection between statistics and information theory in a deep philosophical sense. -I suppose the best place to start would be David MacKay's Information Theory, Inference, and Learning Algorithms, but I was curious what else there may be. Are there any other good books out there like this? What are your recommendations? - -REPLY [3 votes]: The booklength tutorial by Shannon award winner Imre Csiszár and Paul Shields is freely available online here: -Information Theory and Statistics: A Tutorial -I. Csiszár, Rényi Institute of Mathematics, Hungarian Academy of Sciences -There is also an article in the International Encyclopeadia of Statistical Science by E. Haroutunian which seems to be behind a paywall.<|endoftext|> -TITLE: What kind of object are the solutions of the Knizhnik-Zamolodchikov Equations -QUESTION [7 upvotes]: I am reading about the KZ equations in Kassel's Quantum groups. In definition XIX.3.1 (page 455) he defines the differential system $(KZ_n)$ as -$$ -dw = \frac{h}{2\pi\sqrt{-1}} \sum_{1 \leq i -TITLE: Subgroup of $\mathrm{GL}_n$ stabilizing linear subspace skew-symmetric matrices -QUESTION [9 upvotes]: I am currently reading "Schiffer variations and the generic Torelli theorem for hypersurfaces" by Voisin, where it is claimed that the subgroup of $\mathrm{SL}_{2m}$ ($m \geq 3$) which preserves a generic subspace of $\bigwedge^2 \mathbb{C}^{2m}$ of dimension bigger than $3$ must be finite. There are no references given in the paper, where it is claimed that this fact is easily checked. Unfortunately (for me), I am not able to prove it. -Here "preserves" means that $g\cdot w\cdot{\vphantom g^tg} \in W$, for any $w \in W$. Is this fact indeed well-known? I am looking for a reference of this fact, or a quick proof. - -REPLY [5 votes]: In the following, a linear group is a (closed) subgroup $G$ of some $GL(V)$. Now the stabilizer of a generic $d$-space in $V$ is the same as the stabilizer in $G\times GL(d)$ of a generic vector of $V\otimes\mathbb C^d$. Thus, in our case we have to investigate the generic stabilizer of $GL(n)\times GL(d)$ acting on $\wedge^2\mathbb C^n\otimes\mathbb C^ d$ with $1\le d\le n(n-1)/4$ and determine when it is infinite. -Determining the generic isotropy group $H$ for a linear action of a reductive group $G$ is a classical problem of invariant theory. It was pretty much settled by the Vinberg school approx. 50 years ago. More specifically: In -Andreev, E. M.; Vinberg, È. B.; Èlašvili, A. G.: Orbits of highest dimension of semisimple linear Lie groups. (Russian) Funkcional. Anal. i Priložen. 1 1967 no. 4, 3–7 -the authors derive a numerical criterion for $H$ to be infinite (the first theorem of that paper). The argument is really ingeneous and I recommend reading the paper. From their criterion they derive the classification of irreducible simple linear groups with infinite $H$. The table became later notorious because the very same table showed up in various different classification projects (see, e.g., the appendix to Mumford-Fogarty(-Kirwan)). -Probably, the numerical criterion alone is already enough to settle the finiteness of $H$ in the case at hand. But that is not necessary. In the follow-up paper -Èlašvili, A. G. Stationary subalgebras of points of general position for irreducible linear Lie groups. (Russian) Funkcional. Anal. i Priložen. 6 (1972), no. 2, 65–78 -Èlašvili (almost, see below) settles the case of arbitrary irreducible linear groups. The answer for non-simple groups is given in Tables 5 and 6. The cases $\wedge^2\mathbb C^n\otimes\mathbb C^ d$ with infinite $H$ are $(n,d)=(n,1), (n,2), (4,3), (5,3)$, and $(6,3)$. So $d\ge4$ does not occur, as claimed. -There is more to say. Sifting through the tables one will notece that Robert Bryant's case $(n,d)=(6,3)$ is missing in Table 5. This case along with three more missing cases was pointed out in -Popov, A. M. Irreducible semisimple linear Lie groups with finite stationary subgroups of general position. (Russian) Funktsional. Anal. i Prilozhen. 12 (1978), no. 2, 91–92. -See the second paragraph. As for reliability of these results, the whole classification was recovered in -Knop, Friedrich; Littelmann, Peter Der Grad erzeugender Funktionen von Invariantenringen. (German) [The degree of generating functions of rings of invariants] Math. Z. 196 (1987), no. 2, 211–229 -There we dealt with a slightly broader classification problem where we nevertheless had to do all calculations from scratch. So we were not using Èlašvili's tables directly. That's why I am pretty positive that the tables are complete.<|endoftext|> -TITLE: What aspects of math olympiads do you find still useful in your math research? -QUESTION [24 upvotes]: I was rereading the book Littlewood's Miscellany and this passage struck me: - -It used to be said that the discipline in 'manipulative skill' bore - later fruit in original work. I should deny this almost absolutely - such - skill is very short-winded. My actual experience has been that after a - few years nothing remained to show for it all except the knack, which has - lasted, of throwing off a set of (modern) Tripos questions both suitable - and with the silly little touch of distinction we still feel is called for; - this never bothers me as it does my juniors. (I said 'almost' absolutely; - there could be rare exceptions. If Herman had been put on to some - of the more elusive elementary inequalities at the right moment I can - imagine his anticipating some of the latest and slickest proofs, perhaps - even making new discoveries.) - -I would like to ask a question to former math olympiad students who now are actively involved in math research. Do you find the training for olympiads useful in later research career as a mathematician? - -REPLY [8 votes]: I'd point out a trivial thing: often, you just need to sit down and do a calculation, or a case-by-case analysis, etc. I mean something that can't be directly fed to Mathematica, requiring higher-level reasoning, still technical enough so that you need to crunch it with pen and paper. -Olympiad kids are trained to concentrate on such tasks and do them quickly and cleanly. Or course, any research mathematician, knowing the problem boils down to a computation, will be eventually able to do it. But they might spend more time, get distracted, make a mistake, spend even more time because of that, etc. Even more importantly, you often don't know in advance whether the outcome of your computation will solve the problem at hand. The quicker and more confident you are at such things, the more you can try.<|endoftext|> -TITLE: Is there a finite extension with a non-trivial class group of any PID? -QUESTION [6 upvotes]: Let $R$ be a PID with infinitely many prime ideals. Does there always exist a finite extension $R\subset R'$ with $R'$ being a Dedekind domain with a non-trivial class group? - -REPLY [7 votes]: Counterexample. Let $S$ be an infinite set of primes (of $\mathbf Z$) of density $0$. Let $R$ be the localisation of $\mathbf Z$ away from $S$, i.e. the elements $\tfrac{a}{b}$ with $p \nmid b$ for all $p \in S$. Then -$$\operatorname{Spec} R = S \cup \{\eta\}$$ -where $\eta$ is the generic point. -Let $R \subseteq R'$ be a finite extension of Dedekind domains, let $K = \operatorname{Frac} R'$, and write $\mathcal O_K$ for the (usual) ring of integers in $K$. Then $R'$ is the localisation of $\mathcal O_K$ away from $S$, the natural map $\operatorname{Cl}(\mathcal O_K) \to \operatorname{Cl}(R')$ is surjective, and the primes of $\mathcal O_K$ lying above $S$ have density $0$. -Let $H$ be the Hilbert class field of $K$. The isomorphism $\operatorname{Cl}(\mathcal O_K) \stackrel\sim\to \operatorname{Gal}(H/K)$ takes prime ideal classes $[\mathfrak p]$ to the Frobenius $\operatorname{Fr}_\mathfrak p$ at $\mathfrak p$, and the Chebotarev density theorem implies that every ideal class $[I] \in \operatorname{Cl}(\mathcal O_K)$ can be represented by a positive density set of primes. -In particular, $[I]$ contains a prime $\mathfrak p$ not above $S$, hence maps to $0$ under $\operatorname{Cl}(\mathcal O_K) \to \operatorname{Cl}(R')$ since $\mathfrak pR' = R'$. Since $[I]$ is arbitrary and $\operatorname{Cl}(\mathcal O_K) \to \operatorname{Cl}(R')$ surjective, we conclude that $\operatorname{Cl}(R') = 0$. $\square$ -Remark. It might be possible to make a more elementary argument if you choose the $S$ to be sufficiently sparse. For example you could try to take $S = \{p_1,p_2,\ldots,\}$ with $p_{i+1} > 2^{p_i}$ or some other bound. (You have to produce a lot of relations in $\operatorname{Cl}(\mathcal O_K)$ involving relatively small primes. I tried this but couldn't quite work it out by hand.)<|endoftext|> -TITLE: Classification of symplectic resolutions -QUESTION [7 upvotes]: A. Okounkov said, "symplectic resolutions are Lie algebras of the 21st century." Is there a conjecture on the classification of symplectic resolutions? Do Braverman-Finkelberg-Nakajima Coulomb branches give most known examples of symplectic singularities (and do BFN Coulomb branches have explicit descriptions)? Where can one find a list of all known examples of symplectic resolutions? What are the consequences of the classification of symplectic resolutions in representation theory etc.? Is classification of symplectic resolutions a very hard problem (or, if it is intractable, is there a nice class of symplectic resolutions analogous to semisimple Lie algebras that can be classified)? What are some directions in this problem that can be approachable (cf. results of Bellamy-Schedler)? Also, is there an object "Lie group of the 21st century" which fits into an analogy [Lie group of the 21st century] : [symplectic resolution (Lie algebra of the 21st century)] = Lie group : Lie algebra (I suppose quantizations of symplectic resolutions loosely correspond to universal enveloping algebras in this analogy)? - -REPLY [4 votes]: Here is an answer by Gwyn Bellamy, which he let me post here: -1) Is there a conjecture on the classification of symplectic resolutions? No, not that I am aware of. I think this is the wrong question anyway. Rather, one should first try to classify all conic symplectic singularities. There is an amazing result of Namikawa that says that if you bound the degrees of your algebra of functions on the singularity then there are only countably many isomorphism classes. So it is not inconceivable that a classification is possible. I believe that Namkiawa is trying to develop such a classification program. See in particular the papers of his PhD student T. Nagaoka. I think if we had such a classification then it would be relatively straightforward to decide when they admit symplectic resolutions. -2) Do Braverman-Finkelberg-Nakajima Coulomb branches give most known examples of symplectic singularities? Maybe. First, it is not known how many of these are actually conic (to fit into (1)). If we consider first the Higgs branch rather than the Coulomb branch then I think it is a reasonable question to ask if most conic symplectic singularities can be realised as Hamiltonian reductions of a symplectic vector space with respect to a (possibly disconnected) reductive group. One gets all nilpotent orbit closures of classical type this way for instance (I don’t know if this is still true for more general Slodowy slices). Now if this is the case and we believe symplectic duality then one should also realise most conic symplectic singularities as coulomb branches. I think there’s a slight issue here though. The definition as given by BFN does not work so well for disconnected groups. For instance if we take the gauge group to finite then the coulomb branch is just a point. Another way to see that one probably can’t get many quotient singularities (V/G for G \subset Sp(V) finite) is that the coulomb branch is always rational (has same field of fractions as affine space). I don’t think V/G is always rational even for type E Kleinian singularities, so can’t be realised via BFN construction. Maybe there is a way to modify their construction. -3) Do BFN Coulomb branches have explicit descriptions? No (though I am not an expert) outside the quivers gauge theories of finite type (or affine type A) there is no geometric or moduli description. -4) The case of quotient singularities is the one I am most familiar with (work with Travis). Here the classification of symplectic resolutions is almost complete, except for a finite number of exceptional groups. I believe that a PhD student of U. Thiel is looking at these. We also know precisely when quiver varieties admit symplectic resolutions, and I believe there is a classification due to Fu/Namikawa for (normalizations of) nilpotent orbit closures. -5) Also, is there an object "Lie group of the 21st century" which fits into an analogy [Lie group of the 21st century] : [symplectic resolution (Lie algebra of the 21st century)] = Lie group : Lie algebra? Yes, I would say this picture is very well understood. See the Asterique article by Braden-Licata-Proudfoot-Webster and subsequent work by Losev.<|endoftext|> -TITLE: The lion and the zebras -QUESTION [27 upvotes]: The lion plays a deadly game against a group of $N$ zebras that takes place in the steppe (= an infinite plane). The lion starts in the origin with coordinates $(0,0)$, while the $N$ zebras may arbitrarily pick their starting positions. The lion and the group of zebras move alternately: - -In a lion move, the lion moves from its current position to a position at most 1 unit away. -In the zebras move, one of the zebras moves from its current position to a position at most 1 unit away. - -The lion wins if for any $\varepsilon\gt 0$, it can get within $\varepsilon$ of a zebra in finite number of moves. Otherwise the zebras win. -There're only 2 possibilities: - -Zebras win for all $N\geq 1 $. -$\exists M$, such that lion wins for all $N\geq M$. - -Which possibility is true? (Heuristics are welcome too) - -Source: I found this lovely little game from here, where the case for $N=100$ is discussed but remains inconclusive. You may also want to check this, where the zebras have been shown to have a winning strategy if the $\varepsilon$ requirement is dropped (i.e. the lion needs to actually catch a zebra to win instead of just getting within $\varepsilon$ to it). - -Edited the question following Ycor's advice in the comment. - -REPLY [9 votes]: Zebras win for all $N$. -I didn't realize Lawrence's answer in the source is actually sound (or so I think, when I really took some time to read it through this morning). Below I basically adopt Lawrence's strategy for $N$, with schematic drawings to make the argument easier to follow. -The following is a winning starting position for the zebras. - -where $a$ is the distance, to be determined, at which the zebras are able to keep the lion away. Each lane is of width $4+2a$ with zebras horizontally centered. -Strategy for the zebras: -Each zebra mentally draws a square with itself at the center. We specify zebras' strategy to win in each possible situation below: - - -If the lion is at the boarder or outside of the square, stay put. -As soon as the lion is inside the square but outside the $2a$ strip marked by the pair of dotted lines (boarder included), zebra move 1 unit vertically away from it. -As soon as the lion is inside the square and the strip of dotted lines, zebra move 1 unit vertically away from it. - -The lion never wins by staying in Situation $1$ and $2$. -Strategy under Situation $3$ -Situation 3 merits more analysis, because there the zebra can't keep going indefinitely without the lion closing in eventually. How far can it keep going before falling into the $a$ radius of the lion? The answer is that it can go at least as far as $L$, as shown below: - -By Pythagorus, we have $L=1+\frac{1}{2a}$. Notice $L$ can be made as large as we want by adjusting $a$ accordingly. Of course $L$ has to be an integer by zebras' strategy stated above. Let's give a wide margin and say it can go at least as far as -$$L^{*}=L/2=\frac{1}{2}+\frac{1}{4a} \;\;\;\;\; (1)$$ -Now the idea for a strategy in situation 3 is this: the zebra choose some point $s$ along its vertical escape path (of length $L^{*}$), at which it flees horizontally away from the lion. The point $s$ should be chosen so that all the other zebras are far away enough from this horizontal escape path. In that case, if the lion changes target during its horizontal pursuit, the escaper would be able to escape to the center of an unoccupied vertical lane before the lion reaches the new target's square, thereby forcing the game back to situation $1$. -How can this be achieved? Notice to escape to the center of the nearest unoccupied lane, a zebra will have to cross a distance at most $N(4+2a)$. Let's take -$$L^{*}=2N(N(4+2a)+2+2a) \;\;\;\;\;(2)$$ -Then by the pigeonhole principle, there exists $s$ along the vertical escape path whose nearest vertical distance to another zebra is at least $\frac{L^{*}}{2N}=N(4+2a)+2+2a$. If the zebra turns and flees horizontally at this $s$, the lion will be at least $N(4+2a)$ away vertically from any other zebra's square, as shown below. - -And we're done! If the lion keeps its horizontal chasing, the zebra just keeps running. The horizontal distance between the pair will always be greater than $1/2$ (by $(1)$). If the lion switches target during this chase, it can't reach its new target's square before the old target reaches the center of an unoccupied vertical lane, as shown above. -Solving $(1)$ and $(2)$ gives -$$a= \frac{\sqrt{256N^4 + 256N^3 + 48N^2 +1} + 1 - 16N^2 - 8N}{16(N^2 + N)}$$ -The lion will not be able to get within this radius of any zebra. - -If my calculation below is correct, by widening the lane (and enlarging the squares accordingly), the zebras can keep the lion at arbitrarily large distance. Let's take the size of the lane and squares to be $2k+2a$, equations $(1)$ and $(2)$ becomes -$$L^{*}=\frac{(a + k - 1)^2 - a^2}{4a}\;\;\;\;\;\;\;\;\;\;\;\;\; (1)'$$ -$$L^{*}=2N(N(2k+2a)+k+2a)\;\;\;(2)'$$ -Solving $(1)'$ and $(2)'$ for $a$ we have -$$a=\frac{\sqrt{q} + k - 8kN^2 -4kN - 1}{16N(N + 1)}$$ -where -$$q= 64k^2N^4 + 64k^2N^3 + 16k^2N^2 + 8k^2N + k^2 - 16kN^2 - 24kN - 2k + 16N^2 + 16N + 1$$ -Clearly, $\displaystyle{\lim_{k \to \infty} a(N,k) = \infty}$. -So it seems this game is really skewed to the zebras' side.<|endoftext|> -TITLE: Identity involving zonal polynomials and $\operatorname O(N)$ irrep dimensions -QUESTION [14 upvotes]: $\DeclareMathOperator\U{U}\DeclareMathOperator\O{O}$Schur functions $s_\lambda(x)$ with $\lambda\vdash n$ are simultaneously the irreducible characters of the unitary group $\U(N)$ and proportional to Jack polynomials $J_\lambda^{\alpha}(x)$ with parameter $\alpha=1$, $J_\lambda^1(x)=\frac{n!}{\chi_\lambda(1)}s_\lambda(x)$, where $\chi_\lambda$ are the irreducible characters of the permutation group $S_n$. As a function of $N$, $p_\lambda^{\U}(N)=J_\lambda^1(1^N)$ is a monic polynomial. -Two important relatives of the Schur functions are the irreducible characters of the orthogonal group $\O(N)$, call them $o_\lambda(x)$, and the zonal polynomials $Z_\lambda(x)$, which are Jack polynomials with parameter $\alpha=2$. Both $p_\lambda^{\O}(N)=\frac{n!}{\chi_\lambda(1)}o_\lambda(1^N)$ and $p_\lambda^Z(N)=Z_\lambda(1^N)$ are monic polynomials in $N$. -I have been led to conjecture the following relation between the reciprocals of these polynomials: -$$ \sum_{\lambda \vdash n}\frac{\chi_{2\lambda}(1)G_{\lambda\gamma}}{p_\lambda^Z(N)}=\frac{(2n)!}{2^nn!}\frac{\chi_\gamma(1)}{p_\gamma^{\O}(N)},$$ -where $2\lambda=(2\lambda_1,2\lambda_2,\dotsc)$ and -$$ G_{\lambda\gamma}=\sum_{\mu\vdash n}C_\mu \omega_\lambda(\mu)\chi_\gamma(\mu).$$ -Here $C_\mu$ is the size of the conjugacy class in $S_n$ of elements with cycle type $\mu$, and $\omega_\lambda(\mu)$ are zonal spherical functions of $S_{2n}$ with respect to the hyperoctahedral group. -This conjectured relation appeared in connection with immanants of random elements from $\O(N)$ (Oliveira and Novaes - On the immanants of blocks from random matrices in some unitary ensembles) and also with commutators of random elements from $\O(N)$ (Palheta, Barbosa, and Novaes - Commutators of random matrices from the unitary and orthogonal groups). -It is easy to prove that this relation holds for large $N$, but it seems to be true for every finite $N$. -Is this relation known? If not, any idea how to prove it? (Something similar holds for the symplectic analogues, but I omit it for simplicity.) -EDIT: -I don't know if this sheds any light into the conjecture, but, since $s_\gamma=\frac{1}{n}\sum_\mu C_\mu\chi_\gamma(\mu)p_\mu$ and $p_\mu=\frac{2^nn!}{(2n)!}\sum_\lambda \chi_{2\lambda}(1)\omega_\lambda(\mu)Z_\lambda$, it follows that the quantities $\chi_{2\lambda}(1)G_{\lambda\gamma}$ are in fact the coefficients in the expansion of Schur functions into Zonal polynomials, $s_\gamma=\frac{2^n}{(2n)!}\sum_\lambda \chi_{2\lambda}(1)G_{\lambda\gamma}Z_\lambda$. - -REPLY [3 votes]: This conjecture has now been proved by Valentin Bonzom, Guillaume Chapuy and Maciej Dołęga in their paper $b$-monotone Hurwitz numbers: Virasoro constraints, BKP hierarchy, and O(N)-BGW integral.<|endoftext|> -TITLE: Serre duality in families -QUESTION [6 upvotes]: In Ravi Vakil's lecture notes ("Foundations of Algebraic Geometry", Classes 53 and 54) one can find a relative version of Serre duality (Exercise 6.1), namely: -"Suppose $\pi: X\rightarrow Y$ is a flat projective morphism of -locally Noetherian schemes, of relative dimension $n$. Assume all of the geometric fibers -are Cohen-Macaulay. Then there exists a coherent sheaf $\omega_{X/Y}$ on $X$, along with a trace -map $R^n\pi_\ast\omega_{X/Y}\rightarrow\mathcal{O} _Y$ such that, for every finite rank locally free sheaves $\mathcal{F}$ on $X$, each of -whose higher pushforwards are locally free on $Y$, -$$R^i\pi_\ast\mathcal{F}\times R^{n-i}\pi_\ast(\mathcal{F^\vee\otimes\omega}_X)\rightarrow R^n\pi_\ast\mathcal{\omega}_X\rightarrow\mathcal{O}_Y$$ -is a perfect pairing." -For citing purposes, I'd like to have a more canonical reference (i.e. paper or textbook) of this result, but was yet unable to find any. Moreover, I'd actually like to have that result for a flat proper morphism instead of a flat projective morphism. Is it also true in this case? -I'm sorry if this question is trivial, I'm not really familiar with algebraic geometry. -Thank you! - -REPLY [10 votes]: In this case, Serre duality in families = Grothendieck duality. At the level of generality you are asking, I suggest the paper: -Kleiman, Steven L.: Relative duality for quasicoherent sheaves. Compositio Math. 41 (1980), no. 1, 39–60. -http://www.numdam.org/item/?id=CM_1980__41_1_39_0 -But if you are seriously interested in the topic you should get into the derived category formulation. A modern exposition is in -Lipman, Joseph; Hashimoto, Mitsuyasu: Foundations of Grothendieck duality for diagrams of schemes. Lecture Notes in Mathematics, 1960. Springer-Verlag, Berlin, 2009. -Lipman's contribution is available also at: -https://www.math.purdue.edu/~lipman/Duality.pdf<|endoftext|> -TITLE: Commutative ring $R$ with no nontrivial idempotents, with a localization $R_r$ with infinitely many idempotents -QUESTION [8 upvotes]: I am looking for a commutative ring $R$ with $1$ such that $R$ has no idempotents and there exists $r\in R$ such that the localization ring $R_r$ has infinitely many idempotents. - -REPLY [17 votes]: Let $k$ be a field and take $$R=\{(a_i)\in\prod\limits_{i\in\mathbb{N}}k[t]\mid a_i(0)=a_j(0)\text{ for all }i,j\}$$ -An idempotent in this ring has to be sent to $0$ or $1$ under the map $R\xrightarrow{(a_i)\mapsto a_i(0)}k$ hence has to be equal to $(0,0,...)$ or $(1,1,..)$. However, if we invert $r=(t,t,t,..)$ then each of the elements $(0,\dots,0,t,0,\dots)r^{-1}$ gives an idempotent in the localization. -Geometrically, this is analogous (but is not exactly equivalent as taking spectrum does not take infinite products to disjoint unions) to gluing infinitely many affine lines by their origins to get something connected that splits into infinitely many connected components after removing the origin.<|endoftext|> -TITLE: Is there a connection of prime numbers and extreme value theory? -QUESTION [9 upvotes]: As most others are, so am I fascinated by primes. -By the theorem of Euclid and the sieve of Eratosthenes the $ k \ge 2$ - th prime is given by: -$$ p_k = \min_{x>1,\gcd(x,p_1 \cdots p_{k-1})=1} x $$ -which might be seen as one definition of the $k$-th prime. -Thank God this process is deterministic, so that each time we apply it, with $p_1=2$ we get the same prime numbers. -Suppose now that we add some randomness to this process: -$$y = \min_{x_i > 1, \gcd(n, x_i)=1} (x_1, \ldots, x_k)$$ -where we draw each $x_i$ with replacement and independent of each other from $1,\cdots,n$. -I read that the extreme value theory is concerned with $\min, \max$ of iid random variables as $k$ goes to infinity. -Another hint might be the Gumbel distribution which shows values $\zeta(2), \zeta(3)$ in variance and mean where $\zeta$ is the Riemann zeta function. -Yet another hint comes from empirical connection of primes to extreme value theory: -https://arxiv.org/abs/1301.2242 -My question is, if one can make this heuristic more precise, if this is not asked too much. (So maybe there is a way to think of a random process where one can apply the extreme value theory? ) -Thanks for your help! -Reference: -https://arxiv.org/abs/1301.2242 -https://en.wikipedia.org/wiki/Gumbel_distribution -https://stats.stackexchange.com/questions/220/how-is-the-minimum-of-a-set-of-random-variables-distributed -https://en.wikipedia.org/wiki/Extreme_value_theory -Edit: -I did some computations based on the following model: -Given $N$, choose with replacement $y_1,\cdots,y_m$ from the set $\{ x | \gcd(x,N)=1, N \ge x>1\}$ which has $\phi(N)-1$ elements. -The probability that $Y$ is the least prime $p$ not dividing $N$ is: -$$P(Y=p) = \frac{1}{\phi(N)-1}$$ -The probability $P(Y_{\min}=y)$ is: -$$P(Y_{\min}=y)=(1-F(y-1))^m - (1-F(y))^m$$ -where -$$F(y) = \sum_{a \le y} P(Y=a) = \frac{1}{\phi(N)-1} \cdot \chi(N,y)$$ -and -$$\chi(N,y) = |\{a | a \le y, a > 1, \gcd(a,N)=1 \}|$$. -Hence the expected value of $Y^{(N,m)}_{\min}$ is given by: -$$E(Y^{(N,m)}_{\min}) = \sum_{k=1,k>1,\gcd(k,N)=1}^N k \cdot P(Y_{\min}=k)$$ -For $m \rightarrow \infty$ we "should" have: -$$\lim_{m \rightarrow \infty} E(Y^{(P_k,m)}_{\min})=p_{k+1}$$ -where $P_k$ is the $k$-th primorial. -It would be nice, if someone with more experience in probability theory and number theory can have a look at this computation. -I also did some computations with SAGEMATH to this theoretical consideration: -def PK(k): - return prod(primes(nth_prime(k))) - -def FF(N,m,k): - return 1/(euler_phi(N)-1)*len([ a for a in range(1,k+1) if gcd(a,N)==1 and a > 1]) - -def EE(N,m): - EN = euler_phi(N) - return sum([k*((1-FF(N,m,k-1))**m-(1-FF(N,m,k))**m) for k in range(1,N+1) if gcd(k,N)==1 and k>1]) - - -for n in range(3,6+1): - print nth_prime(n),EE(PK(n),PK(n)).n() - -5 5.00000000000000 -7 7.03931627655029 -11 11.0222933198474 -13 13.0321521439774 - -Thanks for your help. - -REPLY [2 votes]: This answer is maybe two years too late, but better too late then not answered: -Let $N \ge 3$ be a natural number. Consider the set $\Omega_N:= \{y \in \mathbb{N}| 1 < y \le N, \gcd(y,N) =1 \}$. -For a subset $A \in \Omega_N$ we define the probability as: -$$P(A) = \frac{|A|}{|\Omega_N|} = \frac{|A|}{\phi(N)-1}$$, -where $\phi$ is the Euler totient function, defined as $\phi(N) = \{ k | 1 \le k \le N , \gcd(k,N)=1 \}$. From the definition it follows that for $y \in \Omega_N$ we have: -$$P(y) = P(\{y\}) = \frac{1}{\phi(N)-1}$$ -and we also have: -$$P(Y \le y ) = \frac{|\{k \in \Omega_N| k \le y\}|}{\phi(N)-1} = \frac{\chi(N,y)}{\phi(N)-1}$$ -where $\chi(N,y) = |\{a| 1 < a \le y, \gcd(a,N)=1\}|$ -We draw with replacement and independent of each other with equal probability $\frac{1}{\phi(N)-1}$ some $m$ numbers $y_1,\cdots,y_m$ from $\Omega_N$ and we let now -$$Y_{\min} := \min_{y_i \in \Omega_N} \{ y_1,\cdots,y_m\}$$ -It follows then, that: -$$P(Y_{\min} \le y) = 1-(1-P(Y \le y))^m$$ -and so: -$$P(Y_{\min} = y )= P(Y_{\min} \le y) - P(Y_{\min} \le y-1) = (1-P(Y \le y-1))^m-(1-P(Y \le y))^m$$ -Let $r := \min( \Omega_N ) $. We notice that $r$ is the smallest prime, which does not divide $N$. -We also notice that $\chi(N,r) = 1$ since the set $\{a | 1 < a \le r, \gcd(a,N)=1 \}$ is equal to the set $\{r\}$, by definition of $r$. -From this las observation, we also observe that $\chi(N,r-1) = 0$. -Hence we get: -$$P(Y_{\min}=r) = (1-P(Y \le y-1))^m-(1-P(Y \le y))^m = (1-\frac{\chi(N,r-1)}{\phi(N)-1})^m-(1-\frac{\chi(N,r)}{\phi(N)-1})^m $$ -$$= (1-0)^m-(1-\frac{1}{\phi(N)-1})^m = 1-(1-\frac{1}{\phi(N)-1})^m$$ -We further then observe that: -$$\lim_{m \rightarrow \infty} P(Y_{\min} = r) = \lim_{m \rightarrow \infty} 1-(1-\frac{1}{\phi(N)-1})^m = 1$$ -The expected value $E(Y_{\min})$ is given by: -$$E(Y_{\min}) = \sum_{ y \in \Omega_N} y \cdot P(Y_{\min} = y)$$ -It follows that: -$$\lim_{m \rightarrow \infty} E(Y_{\min}) = \sum_{ y \in \Omega_N} y \cdot \lim_{m \rightarrow \infty} P(Y_{\min} = y)$$ -$$ = r \cdot 1 + 0 \cdot \sum_{ y \in \Omega_N, y \neq r} y = r = \min(\Omega_N)$$ -For $N = P_k =$ the $k$-th primorial, it follows that (with $k \ge 2$): -$$\lim_{m \rightarrow \infty} E(Y_{\min}) = \min(\Omega_{P_k}) = p_{k+1},$$ -since the prime $p_{k+1}$ is the smallest prime $q$ which does not divide the primorial $P_k$ and for which $1 < q \le P_k$ holds.<|endoftext|> -TITLE: Automorphisms of a modular tensor category -QUESTION [5 upvotes]: I would like to ask for references on automorphisms of a modular tensor category, that do not change the objects. Some special cases, such as automorphisms of a quantum double, are also helpful. - -REPLY [4 votes]: For most quantum group categories all braided autoequivalences are classified by Cain Edie-Michell in this paper. The kind you're interested in, which is called "gauge auto-equivalences" there, almost never exist. One example where it does happen is Cor. 3.2 for the adjoint subcategory of sl_3 at level 3, but I'm not sure if that one is braided. -Many of the arguments there build on an idea from my paper with Grossman Thm 5.5 which is that if you have a planar algebraic description of your tensor category then gauge autoequivalences in particular give you an automorphism of the planar algebra, and you can often see that no such automorphism exists.<|endoftext|> -TITLE: Freyd-Mitchell for $k$-linear categories -QUESTION [9 upvotes]: I don't know much about the proof of the Freyd–Mitchell embedding theorem and I could not find an answer to my question looking naïvely online, but at the same time I feel like this is the kind of question to which someone who knows some of the details of the proof might be able to answer immediately, so it's probably worth trying. Here it is: - -Can the Freyd-Mitchell embedding theorem be made stronger for $k$-linear abelian categories (where $k$ is a field), saying that not only, if $\mathcal{A}$ is a small abelian $k$-linear category, there exists a ring $R$ and a full, faithful, exact functor $F: \mathcal{A} → \text{$R$-$\mathrm{Mod}$}$, but that, moreover, $R$ can be assumed to be a $k$-algebra and $F$ to be $k$-linear? -More in general (also for non-$k$-linear categories): can one say anything about $R$? Is there even a unique "minimal" $R$ (up to Morita equivalence)? - -REPLY [7 votes]: Well, if $\mathcal{A}$ is a small $k$-linear abelian category, then the embedding is given by the following: -First we put $\mathcal{A}$ inside $\mathcal{L}(\mathcal{A},\operatorname{Ab})$, the category of left exact additive functors from $\mathcal{A}$ to the category of abelian groups $\operatorname{Ab}$, by considering the contravariant Yoneda embedding $\mathcal{Y} : \mathcal{A} \longrightarrow \mathcal{L}(\mathcal{A},\operatorname{Ab})$ which sends $A$ to $\operatorname{Hom}_{\mathcal{A}}(A,{-})$. Since $\mathcal{A}$ is $k$-linear, we may show that $\mathcal{L}(\mathcal{A},\operatorname{Ab})$ is also $k$-linear and that $\mathcal{Y}$ is a $k$-linear functor. ($\mathcal{Y}$ is also exact.) -Now, $\mathcal{L}(\mathcal{A},\operatorname{Ab})$ is a complete abelian $k$-linear category possessing an injective cogenerator. Then we apply the duality functor $D$ in $\mathcal{L}(\mathcal{A},\operatorname{Ab})$ and we obtain a covariant (exact) $k$-linear embedding $D \mathcal{Y} :\mathcal{A} \longrightarrow \mathcal{L}(\mathcal{A},\operatorname{Ab})^{op}$. -Finally, we know that $\mathcal{L}(\mathcal{A},\operatorname{Ab})^{op}$ is a cocomplete abelian category possesing a projective generator $P$, and we take a certain coproduct of copies of $P$, obtaining an object $Q$. Then we take the ring $R = \operatorname{End}(Q)$, which is a $k$-algebra and we consider the exact embedding $T : \mathcal{L}(\mathcal{A},\operatorname{Ab})^{op} \longrightarrow {\operatorname{Mod}}R$ defined by $T(X) = \operatorname{Hom}(Q,X)$, which is also $k$-linear. -Therefore, the embedding of $\mathcal{A}$ into ${\operatorname{Mod}}R$ is given by $TD \mathcal{Y} : \mathcal{A} \longrightarrow {\operatorname{Mod}}R$ and it is a $k$-linear functor. -Remarks: I took Mitchell's book "Theory of Categories" (MSN) as a reference for this answer.<|endoftext|> -TITLE: Orientations of triples of points in the plane -QUESTION [5 upvotes]: Given a finite indexing-set $I$ and a collection $P = \{P_i: \ i \in I\}$ of points in the plane no three of which are collinear, let $I_{(3)}$ denote the set of ordered triples of distinct elements of $I$, and let $f_P$ be the function from $I_{(3)}$ to $\{1,-1\}$ such that $f_P(i,j,k)$ is 1 (resp. $-1$) if the points $p_i,p_j,p_k$ lie in counterclockwise (resp. clockwise) order on the circle going through the three points. Call an $f$ that is of the form $f_P$ for some $P$ “achievable”. Is achievability a local condition, in the sense that there exists a fixed $k$ with the property that a function $f: I_{(3)} \rightarrow \{1,-1\}$ is achievable iff its restriction to $I’_{(3)}$ is achievable for all $k$-element subsets $I’ \subseteq I$? -The smallest unachievable $f$, with $|I|=4$, has $f(1,2,3)=f(1,4,2)=f(2,4,3)=f(3,4,1)$ (associated with the faces of a tetrahedron). To see why it can’t be achieved, note that the three lines through $P_1$, $P_2$, and $P_3$ divide the plane into seven regions; the specified $f$ would correspond to points in the eighth, nonexistent region. -This question is a sharpened version of my earlier question -Axiomatizing orientation in the complex plane -somewhat in the spirit of the question Arrangements of points in the plane . - -REPLY [4 votes]: If I understand your function correctly, this is the so called „order type“ of a point set, introduced by Goodman and Pollack, see e.g. this survey, which also contains the references to everything that I mention in the following. The question is now whether there is a number k s.t. if for order type of size n every partial order type of size k is realizable, then the whole order type is realizable. -My short answer: probably not -The slightly longer version: -I believe I have seen a construction of a non-realizable order type of size n where every partial order type of size n-1 is realizable. However, I dis not find this construction anymore, so I might be confusing it with a different setting. If I find it later, I will update my answer. -There are also other reasons for my „probably not“-answer. The first is that deciding whether an order type is realizable is NP-hard as shown by Shor (in fact, Mnëv has shown that it is ETR-hard, that is, the question whether a system of polynomial equations and inequalities has a solution in the reals is reducible in polynomial time to the question whether an irder type is realizable). If the above number k would exist, it would imply a polynomial time algorithm for order type realizability, proving P=NP=ETR. -There is also the related setting of allowable sequences, which is the setting in yor second related question Arrangements of points in the plane. In this setting, there is an example of a configuration that is not realizable, but every subconfiguration is, see Theorem 2.1 and Figure 2.3 in the survey by Goodman and Pollack. If you allow for collinearities (taking the value 0 in your function), this construction can be adapted to order types by placing additional points at the intersections of the „diagonals“.<|endoftext|> -TITLE: The Serre duality theorem intuition -QUESTION [5 upvotes]: It is a well known fact that proper scheme $X$ over $k$ has a up to isomorphism unique dualizing sheaf (EGA I, Hartshorne). -This dualizing sheaf $\omega_X$ comes with two striking properties: -(i) There is a homomorphism $t : H^n(X, \omega_X ) \to k$ (also called the trace) such that for every coherent -$\mathcal{O}_X$-Module $\mathcal{F}$ the following holds: There exists a canonical bilinear map -$$ \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) \times H^n(X, \mathcal{F}) \to H^n(X, \omega_X) $$ -which gives an isomorphism: -$$ \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) \cong H^n(X, \mathcal{F})^*$$ -by composing with t. Here, * means the dual vector space over $k$. -(ii) In addition for every integer $i ≥ 0$ and coherent $\mathcal{F}$ , there exists a canonical isomorphism: $\operatorname{Ext}^i(\mathcal{F}, \omega_X) \cong H^{n-i}(X, \mathcal{F})^*$ if and only if $X$ is Cohen– -Macaulay. -In other words (i) means $H^n(X, \mathcal{F})^*$ is representable. Now $H^n(X, \mathcal{F})^*$ carries structure of a $k$-vector space. If we assume that $\dim_k H^n(X, \mathcal{F}) < \infty$, then $H^n(X, \mathcal{F}) \cong H^n(X, \mathcal{F}) ^*$ and in following we will not differ between $H^n(X, \mathcal{F})$ and it's dual. -A one dimensional $k$- subspace $V_1 \subset H^n(X, \mathcal{F})$ correspond in high tec language to an orbit of a non zero vector $v \in H^n(X, \mathcal{F})$ by action of $k$ on $H^n(X, \mathcal{F})$ via multiplication, ie $V_1= k \cdot v$. -Now my first question is what are the special morphisms in $\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) $ which correspond to $V_1$ aka to the orbit of $k$-action on $v$. How are they related to each other as objects in $\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) $? -In other words if $\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) $ inherits the $k$-action from $H^n(X, \mathcal{F})$, are the elements from the same orbit related to each other in certain "deep" way? Any intuition how one can think about these orbits (except of the boring answer "lines in $H^n(X, \mathcal{F})$")? -The second question is if we take $\mathcal{F}= \omega_X$, then $id_{\omega_X} \in \operatorname{Hom}_(\omega_X,\omega_X)$. Is it's image in $H^n(X, \mathcal{F})$ "special" in certain way? What can we say about this element considered as vector? - -REPLY [12 votes]: First of all, dualizing sheaves are unfortunately not treated in EGA. The treatment in Hartshorne has some limitations. Perhaps some of them are related to your questions. -For pointers to more recent and complete treatments of duality, I suggest you to look at the MO question "Serre duality in families". -Let me start by your second question. The defining property of $\omega_X$, namely -$$ - H^n(X, \mathcal{F})^* \cong -\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) -$$ -expresses the fact that the functor $H^n(X, -)^*$ is representable. This means that there is a representing pair $(\omega_X, \int_X)$ with $\int_X \colon H^n(X, \omega_X) \to k$ a canonical isomorphism that, as you explain, indices the previously displayed isomorphism. This is what you denote "$t$" in your post. Notice that $\int_{X}$ is what corresponds to $\operatorname{id_{\omega_X}}$ in the isomorphism -$$ -H^n(X, \omega_X)^* \cong -\operatorname{Hom}_{\mathcal{O}_X}(\omega_X,\omega_X) -$$ -The fact that the representing object of a functor is unique up to unique isomorphism means that there is no choice for it, once you have a concrete description of $\omega_X$ it forces a unique description of $\int_X$. -How to get such a description? On the projective space $\mathbb{P}^n_k$ one gets a characterization of $\Omega_{\mathbb{P}^n_k|k}$ as $\mathcal{O}_{\mathbb{P}^n_k}(-n-1)$ and from this characterization a canonical isomorphism: -$$ -\int_{\mathbb{P}^n_k} \colon H^n(\mathbb{P}^n_k, \Omega^n_{\mathbb{P}^n_k|k}) \longrightarrow k -$$ -Therefore $\omega_{\mathbb{P}^n_k|k} = \Omega^n_{\mathbb{P}^n_k|k}$. Once you get this description you extend it to other projective varieties and with a little more work to proper varieties over $k$. This is explained under the assumption that $k$ is perfect in J. Lipman's blue book, Dualizing sheaves, differentials and residues on algebraic varieties, Astérisque No. 117 (1984). -Now for your fist question. The pairing -$$ -\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) \times H^n(X, \mathcal{F}) \to k -$$ -that assigns to a linear map $\varphi \colon \mathcal{F} \to \omega_X$ and a cohomology class $\alpha \in H^n(X, \mathcal{F})$ the element $\int_{X} (\alpha \circ \varphi)$ where "$\circ$" denotes Yoneda composition. So the fact that $H^n(X, \omega_X)$ is 1-dimensional means, essentially, that the integral is unique up to "rescaling", so any time you take a multiple of $\alpha$ you are basically recalling it. In a perhaps more abstract point of view one may interpret $\alpha \colon \mathcal{O}_X \to \mathcal{F}[n]$ (in the derived category), therefore $\alpha \circ \varphi \colon \mathcal{O}_X \to \omega_X[n]$ is just a scalar multiple of the "volume form": the element in $H^n(X, \omega_X)$ whose image by $\int_{X}$ is $1 \in K$. -The fact that you are dealing with canonical maps suggests that one should avoid identifying a space with its dual, unless there is canonical choice of the isomorphism. This is crucial in this theory. -What is fascinating to me is that this story makes sense in any characteristic, and that there is an interesting counterpoint between the abstract aspects (dualizing sheaves, representable functors) and the more concrete ones in the sense of computations with cohomology classes, traces and differentials.<|endoftext|> -TITLE: How to prove the determinant of a Hilbert-like matrix with parameter is non-zero -QUESTION [16 upvotes]: Consider some positive non-integer $\beta$ and a non-negative integer $p$. Does anyone have any idea how to show that the determinant of the following matrix is non-zero? -$$ -\begin{pmatrix} -\frac{1}{\beta + 1} & \frac{1}{2} & \frac{1}{3} & \dots & \frac{1}{p+1}\\ -\frac{1}{\beta + 2} & \frac{1}{3} & \frac{1}{4} & \dots & \frac{1}{p+2}\\ -\frac{1}{\beta + 3} & \frac{1}{4} & \frac{1}{5} & \dots & \frac{1}{p+3}\\ -\vdots & \vdots & \vdots & \ddots & \vdots \\ -\frac{1}{\beta + p + 1} & \frac{1}{p+2} & \frac{1}{p+3} & \dots & \frac{1}{2p+1} -\end{pmatrix}. -$$ - -REPLY [10 votes]: Rows linearly dependent means for some $c_1$, $\ldots$, $c_{p+1}$ the non-zero rational function -$\sum_{k=1}^{p+1} \frac{c_k}{x+k}$ has $p+1$ roots $\beta$, $1$, $2$, $\ldots$, $p$, not possible, since its numerator has degree at most $p$.<|endoftext|> -TITLE: Considering each half of factorization of weak equivalence separately -QUESTION [5 upvotes]: I have been working through the proof of Theorem 3.4.1 in the article https://arxiv.org/pdf/1211.2851.pdf (pages 35-6), but there is one technical detail still unclear to me. -Specifically, we have constructed a commutative diagram - -in $\mathbf{sSet}$ such that - -$A \to B$ is a cofibration, -$w$ is a weak equivalence, -$E_i\to A$ is a Kan fibration, and -$\overline{E}_2 \to B$ is a Kan fibration. - -We now want to prove the claim (c) that $\overline{E}_1$ is a fibration over $B$ and $\overline{w}$ is a weak equivalence. The authors state that by factoring the weak equivalence $w$ as a trivial cofibration followed by a trivial fibration, we may prove (c) assuming that $w$ is either a trivial cofibration or a trivial fibration. I'm sure this is obvious to most, but I can't see why this suffices. I would be grateful for an explanation. - -REPLY [7 votes]: Note that the construction $w \mapsto \overline{w}$ is functorial. Thus, if $w = v u$ is a factorization as a trivial cofibration followed by a trivial fibration, we have $\overline{w} = \overline{v} \,\overline{u}$. -Now if (c) holds under the two stated assumptions, the statement of (c) can be applied to $u$ and $v$. Applied to $v$, we see that $\overline{v}$ is a weak equivalence and its domain is a fibration over $B$. The latter conclusion now means that we can apply (c) to $u$, concluding that $\overline{u}$ is a weak equivalence and its domain is a fibration over $B$. But the domain of $\overline{u}$ is $\overline{E_1}$, and $\overline{w} = \overline{v} \,\overline{u}$ so it is also a weak equivalence.<|endoftext|> -TITLE: Is the category $\operatorname{sVect}$ an "algebraic closure" of $\operatorname{Vect}$? -QUESTION [10 upvotes]: $\DeclareMathOperator\sVect{sVect}\DeclareMathOperator\Vect{Vect}$The category $\sVect_k$ of (let's say finite-dimensional) super vector spaces can be obtained from the category $\Vect_k$ of (finite-dimensional) vector spaces by formally adjoining an "odd line square root" $\Pi k$ to the unit object $k \in \Vect_k$ -- see Prop 2.6 in Rezk - The congruence criterion for power operations in Morava E-theory. Here "square root" means that $\Pi k \otimes \Pi k \cong k$, and "odd line" means that the braiding $\Pi k \otimes \Pi k \to \Pi k \otimes \Pi k$ is given by the scalar $(-1)$. -It's not hard to see that $\sVect_k$ has odd line square roots for all even line objects (where "even line" means that the braiding is the identity)—the only even line object being $k$ itself again. So $\sVect_k$ can be characterized as the closure of $\Vect_k$ under the operation of adding odd line square roots for even line objects. This is analogous to $\mathbb C$ being the closure of $\mathbb R$ under the operation of adding square roots for all elements. -But in the case of $\mathbb C$ and $\mathbb R$, much more can be said—$\mathbb C$ is in fact algebraically closed, i.e., closed under the operation of adding roots for all polynomials. Can something analogous be said for the case of $\sVect_k$? -Question 1: Is there a reasonable sense in which the symmetric monoidal $k$-linear category $\sVect_k$ is "algebraically closed"? -I'm primarily interested in the case $k = \mathbb C$. -Here is an attempt to make the question more precise. One way of saying that $\mathbb C$ is algebraically closed is that for every injective map of finitely-generated commutative $\mathbb R$-algebras $A \to B$ and every map $A \to \mathbb C$, there is an extension $B \to \mathbb C$. This motivates the following somewhat more precise question: -Question 2: Is there a reasonably large class of symmetric monoidal $k$-linear functors $A \to B$ between $k$-linear symmetric monoidal categories with the property that any symmetric monoidal $k$-linear functor $A \to \sVect_k$ extends to $B \to \sVect_k$? -Finally, here's a guess at a class of maps $A \to B$ which might possibly do the trick: -Question 3: In particular, let $A \to B$ be a conservative strong symmetric monoidal $k$-linear functor where $A$, $B$ are symmetric monoidal $k$-linear categories with duals for all objects. Then does any strong symmetric monoidal $k$-linear functor $A \to \sVect_k$ admit a lift $B \to \sVect_k$? -This question bears some similarities to Is super-vector spaces a "universal central extension" of vector spaces?, and the "algebraic closure" idea even appears there in a comment of André Henriques, attributed to Alexandru Chirvasitu. -Remark: It might be better to assume that the $k$-linear categories under consideration are also abelian (with bicocontinuous $\otimes$) and that the functors under consideration are exact. Or perhaps some other variation of this flavor. -Edit: I'm mostly interested in characteristic zero, but my intuition is that in characteristic $p$, it would be reasonable to replace "algebraically closed" above with "separably closed", though I don't really know what that would mean in this categorified context. - -REPLY [11 votes]: $\newcommand\sVec{\mathrm{sVec}}\newcommand\Vec{\mathrm{Vec}}$Yes. Over an algebraically closed field of characteristic $0$, $\sVec$ is the algebraic closure of $\Vec$. By "algebraic closure" of $K$ I mean a weakly-terminal object of the category of not-too-large non-zero commutative $K$-algebras. (An object is weakly terminal if it receives maps from all other objects, and terminal if that map is unique.) With this definition, the statement "$\sVec$ is the algebraic closure of $\Vec$" is a summary of Deligne's theorem on the existence of super fibre functors. This interpretation of Deligne's theorem is due to my paper Spin, statistics, orientations, unitarity. (I had the opportunity to ask Deligne last fall if he had been aware of this interpretation of his theorem. He said no, he had been focused on the question "what distinguishes categories of representations of groups?", but that he liked my interpretation.) -Actually, I'm not sure that the weak terminality condition that I use deserves the name "algebraic closure". The issue is that $\sVec$ is not weakly terminal among finitely generated symmetric monoidal categories: you need to include some growth conditions on powers of a generating object. In my paper, I only look at "finite dimensional" extensions of $\Vec$, which is good enough for the usual theory of algebraic closures of fields, but doesn't use the full strength of Deligne's theorem. -In positive characteristic $p\geq 5$, $\sVec$ is not weakly terminal among finite-dimensional extensions of $\Vec$, as observed by Ostrik in On symmetric fusion categories in positive characteristic. But Ostrik does show that $\sVec$ is weakly terminal among separable extensions of $\Vec$, and so is the "separable closure" but not the "algebraic closure". So the category of vector spaces over an algebraically closed field of positive characteristic is not "perfect". -In unpublished work joint with Mike Hopkins, I have also established the 2-categorical version of the statement. Namely, the symmetric monoidal 2-category "$2{\sVec}$" of supercategories and superfunctors is the "separable closure" of the 2-category "$2{\Vec}$" of (linear) categories and functors. The 3-categorical version of the statement is false: we know a separable symmetric monoidal 3-category which does not emit a symmetric monoidal functor to the 3-category of super-2-categories. -Actually, there is one important piece of the story that I haven't worked out. In my paper cited above, I gave a quick-and-dirty definition to the word "field": I said a symmetric monoidal category is a "field" if all the symmetric monoidal functors that it emits are faithful and essentially injective. Under this definition, $\Vec$ and $\sVec$ are fields, so I felt it was good enough. But if you are not working over an algebraically closed base, then $2{\Vec}$ is not a field for this definition, which I do not like. I am still in the process of working out a good higher-categorical version of the word "field". -In the meantime, I would say that yes, $\sVec$ is "algebraically closed", but I would not say that it is "the" algebraic closure, since without a definition of "field", the weak-terminality definition does not characterise a unique object. - -Added in response to comments: -Deligne proves the following stronger result than mere existence. Suppose that $C$ is a reasonable (i.e. linear over your algebraically-closed characteristic-zero ground field, some size constraints, rigid, etc.) symmetric monoidal category category. Then the category of all symmetric monoidal functors $C \to \sVec$ is a groupoid (this requires that $C$ is rigid), and $\pi_0$ of this groupoid is $\operatorname{Spec}(\operatorname{End}(1_{C}))$, where $1_{C}$ is the unit object in $C$. I will write $\operatorname{Spec}(C)$ for the whole groupoid. (A better name would be $\operatorname{Spec}(C)(\sVec)$.) -In particular, if $A \to B$ is a functor of reasonably small symmetric monoidal categories, then you get a map $\operatorname{Spec}(B) \to \operatorname{Spec}(A)$ of groupoids. The question Tim asks above is whether a point in $\operatorname{Spec}(A)$ can be lifted against this map to $\operatorname{Spec}(B)$. This is a question that can be asked just in terms of $\pi_0$ of these groupoid. -Said another way, a functor $F : A \to \sVec$ extends a long a functor $A \to B$ if and only if the induced map $F(1) : \operatorname{End}(1_A) \to \mathbb{C}$ extends along $\operatorname{End}(1_A) \to \operatorname{End}(1_B)$. The answer is "not always": the point $F(1) \in \operatorname{Spec}(\operatorname{End}(1_A))$ might not be in the image of $\operatorname{Spec}(\operatorname{End}(1_B))$. But this is the only obstruction.<|endoftext|> -TITLE: Why are orthogonal matrices so often denoted $Q$? -QUESTION [6 upvotes]: I apologize for the stupid question in the title. Of course, we can baptize a particular given matrix as we want but, for example, the QR-decomposition has a fixed meaning. -My humble guess is that somebody had the idea to use $O$ for orthogonal Matrices and, since this clashes with $0$, then changed to $Q$. -To state an explicit Question: What is the first appearance in the literature of the term QR decomposition? - -REPLY [4 votes]: The explanation that "The letter Q is a substitute for the letter O from orthogonal " may or may not be what John Francis had in mind when he introduced The QR Transformation: A Unitary Analogue to the LR Transformation (1961), but the use of these two letters to indicate the decomposition of a matrix goes back further. Here is a reference by A.H. Clifford from 1942: - -The fact that Q and R are subsequent letters in the alphabet makes this a natural combination, perhaps more natural than the O $\mapsto$ Q switch. - Golub and Uhlig suggest that Francis’s QR algorithm may have been influenced by Rutishauser’s qd algorithm, where "q" stands for "quotient". Since this lower case q refers to a vector rather than a matrix, it does not seem a likely explanation for the choice of the letter Q.<|endoftext|> -TITLE: General Tarski-Seidenberg Theorem -QUESTION [6 upvotes]: The Tarski-Seidenberg Theorem states that the polynomial image of a semi-algebraic set is semi-algebraic. A semi-algebraic subset of a Euclidean space $\Bbb{R}^n$ is by definition a finite union of subsets of the form -$$ -\{P_1=\dots=P_k=0, Q_1>0,\dots,Q_l>0\} -$$ -where $P_i$'s and $Q_j$'s belong to $\Bbb{R}[x_1,\dots,x_n]$. I wonder is there a general coordinate-free version of this theorem for morphisms of real varieties? (By a real variety I mean the set of $\Bbb{R}$-points of a variety over $\Bbb{R}$.) - -REPLY [11 votes]: The most abstract version of the Tarski-Seidenberg theorem I know of is the following - -Let $f:A\to B$ be a morphism of finite presentation of commutative rings. Then the induced map -$$f^*:\operatorname{Sper}B\to \operatorname{Sper}A$$ -sends constructible sets to constructible sets. - -Here $\operatorname{Sper}$ is the real spectrum of the ring, i.e. the set of all pairs $(p,<)$ where $p$ is a prime ideal and $<$ is an order on the residue field at $p$. -It is well known (e.g. theorem 7.2.3 in Bochnack-Coste-Roy Real Algebraic Geometry) that if $A$ is an algebra of finite presentation over $\mathbb{R}$, the boolean algebra of constructible subsets of $\operatorname{Sper}A$ is in natural bijection with the semialgebraic subsets of the real points of the variety $\operatorname{Spec}A$.<|endoftext|> -TITLE: Homomorphisms from higher rank lattices with infinite center to $\mathbb{Z}$ -QUESTION [7 upvotes]: Suppose that $\Gamma$ is an irreducible lattice in a semi-simple real Lie group $G$ of higher rank (with infinite center!), is every homomorphism $\Gamma \to \mathbb{Z}$ trivial? -The case where $G$ has finite center follows easily from Margulis Normal subgroup Theorem. The simplest example I can think of where this question is relevant is the lift of $SL_2(\mathbb{Z}(\sqrt{2}))$ to the universal covering of $SL_2(\mathbb{R})\times SL_2(\mathbb{R})$. -Also, any reference where a discussion about lattices in semi-simple real Lie group of higher rank with infinite center would be appreciated. I only know of Ch.9 Sec.6 in Margulis Book, where I couldn't find an answer to this question. -Thank you! - -REPLY [2 votes]: This is a follow-up of Uri's answer. My goal is just to confirm that (for any $p$) the $L^p$-integrability of lattices in a connected semisimple Lie group $G$ follows from the $L^p$-integrability of lattices in $G/Z(G)$. The non-trivial ingredient that is needed is that the central extension $G\to G/Z(G)$ is represented by a bounded $2$-cocycle. The argument (which I think I learned from Nicolas Monod) is at least in Proposition 7.1 of my paper with Tim de Laat https://arxiv.org/abs/1401.3611 -The fact that this central extension is represented by a bounded $2$-cocycle follows, for simple Lie groups, from the well-known classical work of Guichardet-Wigner, see also the paper Shtern, A. I. Bounded continuous real 2-cocycles on simply connected simple Lie groups and their applications. Russ. J. Math. Phys. 8 (2001), no. 1, 122–133. -The case of semisimple Lie groups follows by decomposing into simple parts.<|endoftext|> -TITLE: How can we know the well-foundedness of $\epsilon_0$? -QUESTION [5 upvotes]: I think the question can be quite philosophical, but I see that $WF(\epsilon_0)$ is widely accepted as one of the attributes of the natural numbers. - -Gentzen proved $Con(PA)$ with $PRA+WF(\epsilon_0)$. -The proofs of some theorems of arithmetic, such as Goodstein's theorem or the termination of Hydra Game, essentially rely on $WF(\epsilon_0)$. - -However, I'm curious if there ever is an justification about this. I'm aware of that ZFC provides such justification, but also I couldn't convict myself whether the set $\omega$ in ZFC (one of the interpretation of it) really gives us the natural number $\mathbb{N}$. -(Just to be clear: the statement of $WF(\epsilon_0)$ itself doesn't require any set theory - it can be coded into arithmetic statement.) -On the other hand, highly unlikely, but if ever $WF(\epsilon_0)$ turns out to be equivalent with $Con(PA)$ or $Con(PA+Con(PA))$, all of which have $\mathbb{N}$ as a model, we know that it is true. -If I understand formalism correctly, even the strictest formalists wouldn't deny these consistency statements because they can't make any deduction without having actual natural numbers or strings, which is equivalent to having PA. -I am relatively new on metamathematics field, and I learned logic from a formalist. ZFC seems just another random formal theory to me, except that I can't do second-order logic without some decent set theory. -So my question is this: is there any non set-theoretical justification for $WF(\epsilon_0)$, which involves the natural number $\mathbb{N}$? - -REPLY [12 votes]: Let me use the notation $\omega_n$ for an exponential tower of $\omega$'s of height $n$, so $\omega_{n+1}=\omega^{\omega_n}$. Then $\epsilon_0$ is the supremum of $\{\omega_n:n\in\omega\}$. PA proves well-foundedness of $\omega_n$ for each individual $n$, but it needs a separate proof for each $n$. If you believe PA (which you apparently do) and you believe in the natural numbers (which, unlike your formalist teacher, you also apparently do), then you should accept that "for all $n$, $\omega_n$ is well-founded." The well-foundedness of $\epsilon_0$ then follows, because, if there were an infinite decreasing sequence in $\epsilon_0$, its first term and therefore all its terms would be below $\omega_n$ for some $n$.<|endoftext|> -TITLE: Kneser subgraph with high chromatic number -QUESTION [10 upvotes]: For positive integers $n\geq 2k$, it is known that the chromatic number of the Kneser graph $K_{n,k}$ is $n-2k+2$. Moreover, the Schrijver graph $S_{n,k}$ (definition in the same link), which is a subgraph of $K_{n,k}$, also has chromatic number $n-2k+2$. The number of vertices of $S_{n,k}$ is $\binom{n-k+1}{k}$. -Is it known whether there is a subgraph of $K_{n,k}$ with chromatic number $n-2k+2$ whose number of vertices is polynomial in $n$ and $k$? - -REPLY [3 votes]: The general answer is no, at least when $n$ is close to $2k$. -Theorem 3 in this paper shows that a graph with odd girth $\geq 2d+1$ and chromatic number $>m$ contains more than -$$ - \frac{(m+d)(m+d+1)\dots(m+2d-1)}{2^{d-1}d^d} -$$ -vertices. The parameters for the Kneser graph $K_{n,k}$ are $m=n-2k+1$ and $d=\bigl\lceil\frac k{n-2k}\bigr\rceil$, so, say, for $n\approx 2k+\sqrt k$ the bound is already exponential in $\sqrt k$.<|endoftext|> -TITLE: An orientable surface that cannot be embedded into $\Bbb R^3$? -QUESTION [7 upvotes]: I previously asked this question on MSE, without success. - -By Whitney's embedding theorem, every 2-dimensional manifold (aka. a surface) can be embedded into $\Bbb R^4$. -Now, Wikipedia states in this paragraph that we can even embedd into $\Bbb R^3$ if the surface is - -compact and orientable, or -compact and with non-empty boundary. - -In the second bullet point, it is clear that I cannot drop either of the conditions. -It is not clear to me why I can't drop "compact" in the first bullet point. - -Question: Is there an orientable but non-compact surface that does not embedd into $\Bbb R^3$? - -REPLY [3 votes]: It seems to me that every orientable surface is indeed embeddable in $\mathbb{R}^3$. By Ian Richards' classification theorem (https://www.ams.org/journals/tran/1963-106-02/S0002-9947-1963-0143186-0/S0002-9947-1963-0143186-0.pdf), we know that a non-compact orientable surface is determine by the pair $Y\subset X$ where $X$ is its space of ends (homeomorphic to a compact subset of the Cantor space) and $Y$ is the closed subset of ends with genus. -When genus is finite, i.e. $Y=\varnothing$, the embedding is easily realised as a genus-$g$ surface with a copy of $X$ removed. When genus is infinite, one simply starts with a sphere with a copy of $X$ removed, and adds smaller and smaller handles accumulating to all points of $Y$ (but not to points of $X\setminus Y$). -To ensure this can be done, observe that $Y$ has a countable dense subset. By taking a countable basis of neighborhoods of each of them, we get a countable family $(U_n)_{n\in\mathbb{N}}$ of open subsets of $\mathbb{S}^2\setminus X$ and we want to put one handle in each of them, but no family of handles should approach any point outside $Y$. We put a handle in $U_1$ (i.e. we remove two discs from $U_1$ and glue a handle to them), then inductively add one in the smallest $U_n$ not containing both gluing circle of any previous handle.<|endoftext|> -TITLE: When are all signatures represented by units? -QUESTION [6 upvotes]: What conditions on a number field imply that every signature is represented by a unit? Are there conditions that imply that it is not the case that every signature is represented by a unit? -For example, if K is a cubic cyclic number field with odd class number then all signatures are represented by units by Theorem V in [Armitage, J., & Fröhlich, A. (1967). Classnumbers and unit signatures. Mathematika, 14(1), 94-98. doi:10.1112/S0025579300008044]. -What are some references for articles on this topic? - -REPLY [4 votes]: The structure of the group of units modulo totally positive units is studied here; see also the references in this article. Since then, relevant articles by Dummit and Dummit & Voight have appeared.<|endoftext|> -TITLE: Minimal number of "words" that contain all possible pairs of letters in all position pairs -QUESTION [7 upvotes]: Defining a word as sequence of ordered letters ($1$..$q$ letters) of length L, -what is the minimal number of words such that among the entire list of words, -at every pair of positions, I can find any two letters? -for example, for $q=3$ and $L=2$ -here is the minimal list: -$$1 1, -2 2, -3 3, -1 2, -2 3, -3 1, -1 3, -2 1, -3 2,$$ -total $q^2$ words are needed. -but for $L=3$ the minimal number is still $q^2$, obtained by: -$$1 1 1, -2 2 1, -3 3 1, -1 2 2, -2 3 2, -3 1 2, -1 3 3, -2 1 3, -3 2 3,$$ -for $L=4$ the number is different... -what is the minimal number of words for $(q,L)$ and specifically, what is the asymptotic value for $L\gg 1$? -Thanks for you answers! - -REPLY [2 votes]: First, let me elaborate the upper bound from @kodlu’s answer. If $q$ is a power of a prime, then $q^2$ words suffice for $L=q+1$ (for $L=q$ it is almost trivial, improving by $1$ needs just a bit more). Then, doubling $L$ increases the number of words by $q(q-1)$, so for those values of $L$ it suffices to take $q(q-1)\log_2\frac L{q+1}+q^2$ words. -Let me show a somewhat close lower bound. Let $w$ be the number of words; set $k=w-q(q-1)+1$. Take any $k$ words. Let $v_i$ be a vector composed from all $i$th entries of the $k$ words. If two of those vectors, say $v_i$ and $v_j$, coincide, this means that at most $w-k=q(q-1)-1$ words differ in positions $i$ and $j$, so not all pairs are covered. -Thus, we have $L$ distinct vectors in $[q]^k$, so $L\leq q^k$, or $w\geq q(q-1)+1+\log_qL$. -Therefore, the growth rate is indeed logarithmic (but the constant at the logarithm is yet unclear). -Addendum. Let me present an example for $L=q+1$, when $q$ is a power of a prime. -Consider an affine plane $\mathbb F_q^2$. All lines in it are partitioned into $q+1$ classes $C_1,\dots,C_{q+1}$ of mutually parallel lines (one class consists of all lines with equations of the form $ax+by=c$ with fixed $(a:b)$). Enumerate the lines in each class by numbers from 1 to $q$. -For each point $p\in \mathbb F_q^2$, take a word $w_1\dots w_{q+1}$ where $w_i$ is the number of the line in $C_i$ passing through $p$. Then, for any two classes $C_i$ and $C_j$ and for any two lines in them, the lines meet at a unique point, which means exactly what we need to get in the $i$th and $j$th positions.<|endoftext|> -TITLE: Is Axiom of Choice equivalent to its version for families of sets, indexed by ordinals? -QUESTION [5 upvotes]: Is Axiom of Choice equivalent to the following statement? - -Axiom of Ordinal Choice: For any ordinal $\lambda$ and any indexed family of sets $(X_\alpha)_{\alpha\in\lambda}$ there exists a function $f:\lambda\to\bigcup_{\alpha\in\lambda}X_\alpha$ such that $f(\alpha)\in X_\alpha$ for all $\alpha\in\lambda$? - -REPLY [9 votes]: It is strictly weaker than choice. This is explained in Asaf Karagila's answer at MSE: the $L(\mathbb{R})$ of $L$ + $\aleph_1$-many Cohen generics witnesses this. -(There the principle is phrased for well-ordered index sets, but you can always pass from a well-ordered index set to an ordinal index set.) - -As Taras Banakh comments below, there is a superficially-similar fact: namely that over NBG without global choice, global choice is equivalent to $Ord$-indexed choice. The proof is simple: for $\alpha\in Ord$ let $C_\alpha$ be the set of well-orderings of $V_\alpha$. From a choice sequence $(\triangleleft_\alpha)_{\alpha\in Ord}$ we get a well-ordering of $V$ as follows: set $x\prec y$ iff $rk(x) -TITLE: Equivalence between Ramanujan and Selberg conjectures -QUESTION [6 upvotes]: At first the Ramanujan conjecture for automorphic forms and the Selberg conjecture appear to be understood as totally independent. However, they are now known to be tighyly connected once viewed in the light of global adelic automorphic representations. I would like to understand how far this is true. I recall the different versions, say in $GL(2)$. -Ramanujan-Petersson conjecture. Let $f$ be a modular form (we could state it for a Maass form too) of weight $k$ and level $1$, if the $a_n$ denotes its Fourier coefficients then -$$a_n \ll n^{(k-1)/2}.$$ -Selberg conjecture. Let $f$ be a Maass form and $\lambda_1$ its smallest nonzero eigenvalue for the Laplacian. Then -$$\lambda > \frac 14.$$ -Now, there is a "more modern" version in terms of automorphic representations. -"Automorphic" Ramanujan conjecture. Let $\pi$ be an automorphic cuspidal representation of $GL(2, \mathbb{A})$. It decomposes by Flath's theorem as $\pi \simeq \bigotimes_v \pi_v$. Then $\pi_v$ is tempered for all place $v$. -Are these three formulation equivalent, or only in a mild sense? More precisely, - -does the automorphic version imply the two others? -do the two "local" versions imply the automorphic one? -are the two "local" version equivalent? (or: does one of them imply the automorphic version, by a kind of rigidity of the automorphic representations) - -(I haven't found proper answer to these questions and would be glad to know if I missed papers or notes about it.) - -REPLY [9 votes]: Let $f$ be an automorphic form corresponding to an automorphic representation $\pi =\otimes_v \pi_v$ of $GL_2(\mathbb A_{\mathbb Q})$. -For an unramified prime $p$, the following are equivalent (for $f$ holomorphic of weight $k$): - -$|a_p| \leq 2p ^{(k-1)/2 }$. -For all $n$, $|a_{p^n} |\leq (n+1) p^{n (k-1)/2} $. -For all $n$, $|a_{p^n} |\ll p^{n ((k-1)/2 +\epsilon) } $. -$\pi_p$ is tempered. - -There is an analogous local equivalence for Maass forms, where for the standard normalization you should take $k=1$. There is also, a more complicated, statement about ramified primes. -For the place $\infty$, the following are equivalent (for $f$ Mass) - -$\lambda \geq \frac{1}{4}$ -$\pi_{\infty}$ is tempered. - -For holomorphic forms, $\pi_{\infty}$ is automatically tempered. -Putting this together, we see that $\pi$ is tempered at all places if and only if $a_n \ll n^{ (k-1)/2+ \epsilon}$ for all $n$, plus some additional conditions at the ramified primes, and (if $f$ is Maass) $\lambda_1 \geq 1/4$. -Because all cuspidal automorphic representations of $GL_2(\mathbb A_{\mathbb Q})$ correspond to cuspidal holomorphic or Maass forms, the Ramanujan conjecture for $GL_2(\mathbb A_{\mathbb Q})$ is equivalent to these statements for all holomorphic or Maass forms simultaneously. -In summary, the automorphic one implies the two local ones, the two local ones imply the automorphic (at least away from ramified primes), but the local ones are not equivalent in any meaningful sense.<|endoftext|> -TITLE: Element of Weyl chamber contracting $\mathbb{A}^n_k$ to a point -QUESTION [5 upvotes]: Let $G$ be a connected reductive group over an algebraically closed field $k$ of characteristic 0. Fix a Borel subgroup $B$ and a maximal torus $T \subset B$. Let $P \subset G$ be a parabolic subgroup containing $B$, and let $L \subset P$ be a Levi subgroup containing $T$. -Suppose given an action of $L$ on $\mathbb{A}^n_k$ such that (1) the commutator subgroup $[L,L]$ fixes $\mathbb{A}^n_k$, and (2) $\mathbb{A}^n_k$ is a toric variety for some quotient $T'$ of the torus $L/[L,L]$. In other words, $T' \cong \mathbb{G}_m^n$, and under this isomorphism, $T'$ acts on $\mathbb{A}^n_k$ in the natural way. The origin $0 \in \mathbb{A}^n_k$ is fixed by $T'$, and there are "many" one-parameter subgroups $\lambda': \mathbb{G}_m \to T'$ such that -$\lim_{t \to 0} \lambda(t) \cdot z = 0$ for all $z \in \mathbb{A}^n_k$. On the other hand, the composition $T \to L \to T'$ of the inclusion map and the quotient map allows us to associate to any one-parameter subgroup $\lambda: \mathbb{G}_m \to T$ a one-parameter subgroup $\lambda': \mathbb{G}_m \to T'$ (given by the composition of $\lambda$ with the map $T \to T'$). -My question is this: does there exist some choice of $\lambda: \mathbb{G}_m \to T$ such that (1) -the corresponding one-parameter subgroup $\lambda': \mathbb{G}_m \to T'$ contracts all of $\mathbb{A}^n_k$ to $0$, and (2) $\lambda$ lies in the Weyl chamber of our fixed Borel subgroup $B$ (equivalently, $\langle \lambda, \alpha \rangle > 0$ for all positive roots $\alpha$ of $G$ with respect to $T$ and the base given by $B$). -In the case where $G = P = \mathrm{SL}_n$, $B = L$ is the subgroup of upper triangular matrices, and $T$ is the subgroup of diagonal matrices, everything works out very nicely. In this case, $T = T'$ and $\lambda(t) = \mathrm{diag}(t^{m_1},\dots,t^{m_n})$ for some $m_i \in \mathbb{Z}$, and the condition that $\lim_{t \to 0} \lambda(t) z = 0$ for all $z \in \mathbb{A}^n_k$ is the statement that $m_i > 0$ for all $i$. The positive roots of $G$ with respect to $T$ are the maps $T \to \mathbb{G}_m$ sending $\mathrm{diag}(t_1,\dots,t_n) \mapsto t_i/t_j$ for $i < j$, so we may pick for $\lambda$ any choice of $m_i > 0$ such that $m_i > m_j$ for $i < j$. -In general, I think if we embed $G$ in $\mathrm{SL}_n$ and pick a basis where $T$ and $T'$ are diagonalized, we might be able to do something similar. I'm not convinced that this works, though. Essentially, the condition that $\lambda$ contracts all of $\mathbb{A}^n_k$ to a point says that -$\langle \lambda, e_i \rangle > 0$ for all $1 \leq i \leq r$, where the characters $e_i: T \to \mathbb{G}_m$ factor through $T'$ and form a certain basis for the character group $\mathcal{X}(T')$. It's not clear to me that these hyperplanes will in general intersect all the hyperplanes -$\langle \lambda,\alpha \rangle > 0$ for $\alpha$ a positive root. I'm not particularly comfortable with arguments about roots of reductive groups and one-parameter subgroups, so I would really appreciate any suggestions you may be able to give me! - -REPLY [2 votes]: The answer to your question, as stated, is No. Take $G={\rm SL}_2$, $P=B$, $T'=T$. -Note that you do not specify the isomorphism $T'\to {\Bbb G}_m^n$. Let us take the following isomorphism: -$$T'\to {\Bbb G}_m\colon\,{\rm diag}(s,s^{-1})\mapsto s^{-1}\text{ for }s\in k^\times.$$ -Then our torus $T=T'$ acts on $\Bbb A^1$ by -$${\rm diag}(s,s^{-1})\colon\, x\mapsto s^{-1}x.$$ - Now if you take $s=\lambda(t)=t^m$, then your condition (1) that $\lambda$ contracts all of $\mathbb{A}^1_k$ to $0$ means that $m<0$, while your condition (2) that $\lambda$ lies in the Weyl chamber of our fixed Borel subgroup $B$ means that $m>0$. Contradiction....<|endoftext|> -TITLE: Computational version of inverse sumset question -QUESTION [5 upvotes]: Let $p$ be prime and $\mathbb{F}_p$ the finite field with $p$ elements. Suppose we have a set $B\subseteq \mathbb{F}_p$ satisfying $|B| -TITLE: On Glaeser's Theorem for non-smooth functions -QUESTION [8 upvotes]: Glaeser's Theorem says that a $C^\infty$ function $F$ on $\mathbb R^n$ which is invariant under permutation of the variables is a smooth function of the symmetric polynomials of $(x_1, \dots, x_n)$. -Question 1: What remains (if anything) of this statement if $F$ is $C^k$ ? -Question 2: In the statement above, is it clear that you can replace $\mathbb R^n$ by a symmetric open subset of $\mathbb R^n$? - -REPLY [9 votes]: A more direct reference (quoted in Rumberger's paper) is: G. Barbançon, Le théorème de Newton pour les fonctions de classe $C^r$. Ann. Sci. École Norm. Sup. 5 (1972), 435–458. He proves that a symmetric function of class $C^{nr}$ on $\Bbb{R}^n$ is a function of class $C^{r}$ of the symmetric polynomials.<|endoftext|> -TITLE: On the moduli stack of abelian varieties without polarization -QUESTION [6 upvotes]: (I am especially interested in abelian surfaces and characteristic 0). - -How bad is the moduli stack of abelian varieties (with no polarization or level structure)? Is it an Artin stack? DM (Deligne-Mumford) stack? -How bad is the is the stack of abelian varieties with full 2 level structure (so with a basis for $A[2]$)? -Consider the maps from either 2) above or stack of principally polarized abelian varieties to 1) above. Are these maps smooth, are the geometric fibers finite (ie, are there only finitely many principal polarizations on an abelian variety)? - -Neither moduli space is a stack because every point has the automorphism $-1$ but the same is true for the moduli stack of elliptic curves and that is still a DM stack and not too bad. -Even in characteristic 0, the CM locus is higher dimensional so the "especially stacky" locus has high dimension but I don't know how serious the problem is. -For the second question, while $-1$ fixes the two level structure, I suppose a generic CM automorphism doesn't fix it so perhaps the second stack is very nice, or atleast almost as nice as that of elliptic curves? - -REPLY [14 votes]: First, when defining the stack you will have the issue that there are formal deformations of abelian varieties which do not extend to families of abelian varieties over any reduced scheme. These are the deformations that do not respect any polarization. (In the complex analytic world these correspond to deformations of complex tori) So unless you have some very strange definition of the functor, the local structure of this stack will be at least as bad as the formal limit $\lim_{n\to \infty} \operatorname{Spec} k[x]/x^n$. I think this rules out ever having a smooth morphism from a scheme, and thus rules out being an Artin stack. -For $E$ a non-CM elliptic curve, the automorphism group of $E^n$ is $GL_n(\mathbb Z)$. This shows that for $n>1$, points of this moduli stack can have infinitely many automorphisms. In particular, the diagonal is not quasicompact. -Level $2$ structure doesn't help with this at all, you just get the group of $n\times n$ matrices congruent to $1$ mod $2$. -The map from the stack of principally polarized abelian varieties to this stack is not smooth because deformations can kill a principal polarization, and the fibers are not finite, again because of examples like $E^n$, whose principal polarizations are in bijection with $n \times n$ symmetric positive definite integer matrices with determinant $1$. -In summary: There's a reason you haven't heard of this stack before. -P.S. You shouldn't worry so much about automorphisms of the generic point, which almost never cause problems in practice. It's everything else that you should worry about!<|endoftext|> -TITLE: Examples of noetherian local rings $R$ such that $K'_0(R)$ is not isomorphic to $\mathbb Z$ -QUESTION [5 upvotes]: Does there exist a simple example of a commutative noetherian local ring $R$ such that $K'_0(R) = K_0(\mbox{Mod-}R)$ (by $\mbox{Mod-}R$ I mean the abelian category of finitely generated $R$-modules) is not isomorphic to $\mathbb Z$? - -REPLY [7 votes]: This is just to flesh out @Tom Goodwillie example. -For any reasonable scheme $X$ and an open set $U$, one has a natural exact sequence, -$$K_0(X-U)\to K_0(X)\to K_0(U)\to 0.$$ -Taking $X=\operatorname{Spec} (\mathbb{Q}[x,y]/xy)_{(x,y)}$ (or a number of similar examples) and $U$ the punctured spectrum, we note that the punctured spectrum is two points and thus $K_0(U)=\mathbb{Z}^2$. The kernel is generated by the closed point, but going mod $x+y, x+y^2$, one can easily see that 2 and 3 times the closed point is zero in $K_0(X)$. So, we get $K_0(X)=\mathbb{Z}^2$.<|endoftext|> -TITLE: Mostow Rigidity Theorem and reconstruction from fundamental group -QUESTION [8 upvotes]: The Mostow Rigidity Theorem is phrased in terms of a relationship between isometries and isomorphisms of fundamental groups, which raises an obvious question. Given the fundamental group of a complete finite-volume hyperbolic manifold of dimension $> 2$, is it possible to reconstruct the hyperbolic manifold? - -REPLY [11 votes]: How is your fundamental group $\Gamma$ given to you? As a presentation in terms of generators and relations? Here is an answer for hyperbolic $3$-space that can probably be generalized to $\mathbb{H}^n$, $n \ge 4$, with some effort. -In short, you compute the $\mathrm{SL}_2(\mathbb{C})$ character variety. The relevant ideas and facts I use below are in Culler and Shalen's famous paper [CS]. -Here is an algorithm. - -Compute the algebraic set $\mathrm{Hom}(\Gamma, \mathrm{SL}_2(\mathbb{C}))$ from your presentation. This is naturally an affine algebraic set in $\mathbb{C}^{4 d}$, where $\Gamma$ has $d$ generators. -Then, there is an explicit set of elements $\sigma_1, \dots, \sigma_n \in \Gamma$ for which the trace function -$$ -t(\rho) = \left(\mathrm{tr}(\rho(\sigma_1)), \dots, \mathrm{tr}(\rho(\sigma_n))\right) \in \mathbb{C}^n -$$ -has image the $\mathrm{SL}_2(\mathbb{C})$ character variety, $X(\Gamma)$. See Proposition 1.4.1 in [CS]. -Compute this algebraic set in $\mathbb{C}^n$ and find its irreducible components (maybe in Macaulay?). -Mostow rigidity tells you that a discrete and faithful representation $\rho : \Gamma \to \mathrm{SL}_2(\mathbb{C})$ determines an isolated point of $X(\Gamma)$. You might get several isolated points in $X(\Gamma)$ from different lifts from $\mathrm{PSL}_2$ to $\mathrm{SL}_2$ and from different Galois conjugates of the discrete and faithful representation, not to mention possibly other random isolated points. (Small nitpick: In the finite volume, noncompact case you need to also cut out the equations that traces of peripheral elements are all $2$ in order to get isolated points. You need to cut away deformations related to Dehn filling.) -There is a standard way described in [CS] to algorithmically lift a point on $X(\Gamma)$ back to a representation $\rho$. Basically you choose a lift so that your first generator is upper-triangular with the right trace and $1$ in the upper right entry. You make the second generator lower-triangular of the right trace, where now the lower left entry is going to depend on the trace of the product of the two generators, etc. Compute such a $\rho$ for each isolated point of $X(\Gamma)$. -Now, compute a Dirichlet or Ford domain in $\mathbb{H}^3$ for each $\rho(\Gamma)$ from your list of isolated points. One of these will terminate to give you a fundamental domain for the complete structure. - -Now you have the representation associated with the complete structure and a fundamental domain. So if that is your desired notion of "reconstructing" the manifold, there you go. -A last remark on the finite volume case. Your presentation oracle also needs to enumerate the conjugacy classes of $\mathbb{Z}\oplus \mathbb{Z}$ subgroups - these are precisely the peripheral subgroups. (The manifold is homeomorphic to the interior of a compact manifold with boundary a union of essential tori.) As described in [CS], Thurston proved that the dimension of the character variety at the complete structure equals the number of cusps, and the complete structure is cut out by making each peripheral subgroup unipotent (in fact, setting the traces of these elements all equal to $2$ cuts out the desired point). You can probably make finding these algorithmic in just the presentation. The rank of the abelianization of $\Gamma$ is an upper bound for the number of peripheral subgroups (by Poincaré duality), so one knows when enough subgroups have been found. This point is irrelevant in higher dimensions. -[CS] Culler, Marc; Shalen, Peter B. Varieties of group representations and splittings of 3-manifolds. Ann. of Math. (2) 117 (1983), no. 1, 109–146.<|endoftext|> -TITLE: Nonnegativity implies $\langle Lf,f\rangle\geq \int f^2-(\int fg)^2$ for $g\geq 0$ -QUESTION [5 upvotes]: I have been a lot of time trying to understand a key step on a paper about spectral analysis but I have no clue how to prove it (and the authors only said "by standard analysis"). Let me state the question (this is how I interpret actually, I tried to clean it since I prefer to avoid all the details and definitions of the paper). Consider the following 1D-Schrödinger operator $$ -L:=-\partial_x^2+c_1-c_2\Phi -$$ -where $c_1,c_2>0$ and $\Phi$ is a positive Schwartz function. -Now, I can prove that due to the specific structure of the operator and $\Phi$, $L$ is a nonnegative operator (zero is its first eigenvalue, which is simple). Moreover, the eigenfunction associated to its first eigenvalue (zero), let's say $\zeta$, is a positive Schwartz function (it's some power of $\Phi$ actually). Here is where everything become a bit dark for me. Is it true that if I consider a nonnegative function $g\in L^2\setminus\{0\}$, then, due to the spectral information of $L$ above, there exist $\lambda>0$ (depending on my election of $g$) such that for all $f\in H^1(\mathbb{R})$ it holds: $$ -\langle Lf,f\rangle\geq \lambda\int f^2-\dfrac{1}{\lambda}\left(\int fg\right)^2? -$$ -I am quite surprised that it seems that I can "choose" this $g$ (as soon as restricting myself to nonnegative functions not identically zero). Does anyone has any idea on how to prove it? Or maybe some recommended references? -PS: Notice that since the eigenfunction associated to the zero eigenvalue is positive, and we are also - assuming $g$ positive, then $g$ cannot be orthogonal to $\zeta$. Thus, you cannot choose $g\perp \zeta$ and then try to choose $f=\zeta$ to make the left-hand side equals zero (while the right-hand side would remain strictly positive). - -REPLY [4 votes]: Under your assumptions the essential spectrum is $[c_1, \infty[$, hence the part of the spectrum in $[0,c_1[$ is discrete. Let $\mu>0$ be the second eigenvalue (the first is 0), if it exists, or $c_1$. Then $(Lh,h) \ge \mu (h,h)$ if $h$ is orthogonal to $\zeta$. Next assume by contradiction that $(Lf_n, f_n) +(f_n,g)^2 \le n^{-1}\|f_n\|^2$ and $\|f_n\|=1$ and split $f_n=c_n \zeta+h_n$ with $(\zeta, h_n)=0$. Then $(Lf_n,f_n)=(Lh_n,h_n) \to 0$, hence$\|h_n\| \to 0$ and $|c_n| \to 1$. Next $(f_n,g)=(c_n \zeta+h_n,g) \to 0$, which is impossible since $|c_n| \to 1$ and $(\zeta,g)>0$.<|endoftext|> -TITLE: Is there a known closed form expression for this integral? -QUESTION [6 upvotes]: I am interested in the following integral: -$$f(x,y) = \int_{\mathbb{S}^d} \max(0,x^Tw)\cdot\max(0,y^Tw) \, dw, \qquad x,y\in\mathbb{S}^d,$$ -where $\mathbb{S}^d\subset\mathbb{R}^{d+1}$ is the $d$-dimensional unit sphere. Here, $x^Ty$ denotes the dot-product/inner-product between $x$ and $y$. -Is there a known closed-form expression for this integral? It is clear that $f(x,y)$ is isotropic, i.e., if $Q$ is an orthogonal transformation then $f(x,y) = f(Qx,Qy)$. Therefore, I believe that $f(x,y) = \phi(x^Ty)$ for some function $\phi : [-1,1]\rightarrow \mathbb{R}$. Ideally, I would know $\phi$. - -REPLY [7 votes]: As you've pointed out, the only parameter that matters here is the angle $\theta$ between $x$ and $y$. To see how, consider instead the Gaussian integral: -$$ -I(x,y)=\frac{1}{(2\pi)^{(d+1)/2}}\int_{u\in\mathbb{R}^{d+1}}\max(0,x^Tu)\cdot\max(0,y^Tu)\exp(-\frac{1}{2}u\cdot u)du -$$ -The integral you are interested in is obtained by changing to $(d+1)$ dimensional spherical coordinates, and easily integrating over the radial coordinate. As such, it will be enough to just focus on the Gaussian integral. -I like computing these types of integrals by forming an orthonormal basis of $\mathbb{R}^{d+1}$ like so: -$$ -b_1=(x+y)/\|x+y\|\\ -b_2=(x-y)/\|x-y\| -$$ -...and where the remaining basis elements are chosen to be orthogonal to $b_1$ and $b_2$. With respect to this basis we have: -$$ -x=(a,b,0,...,0)\\ -y=(a,-b,0,...,0) -$$ -where $a=|\cos(\theta/2)|$ and $b=|\sin(\theta/2)|$. Of course the components of $u$ change as does the Gaussian measure, but the Gaussian measure is invariant under rotations so I'll suppress relabeling the components of $u$ with respect to the new basis. After easily integrating over the last $(d-1)$ coordinates of $u$ we write the integral $I(x,y)$ as: -$$ -\frac{1}{2\pi}\int_{(u_1,u_2)\in R}(a\cdot u_1+b\cdot u_2)\cdot (a\cdot u_1-b\cdot u_2)\exp(-\frac{1}{2}(u_1^2+u_2^2))du_1du_2, -$$ -where $R=\{(u_1,u_2)|(a\cdot u_1+b\cdot u_2)>0\textrm{ and }(a\cdot u_1-b\cdot u_2)>0\}$ is the region where the two maximums are nonzero. After a change of coordinates $v_1=au_1$ and $v_2=bu_2$ we obtain: -$$ -\int_{(v_1,v_2)\in R'}\frac{\left(v_1^2-v_2^2\right) e^{\frac{1}{2} - \left(-\frac{v_1^2}{a^2}-\frac{v_2^2}{b^2}\right)}}{2 \pi a b}dv_1dv_2 -$$ -where $R'=\{(v_1,v_2)|(v_1+v_2)>0\textrm{ and }(v_1-v_2)>0\}$. From here, change to polar coordinates $(r,t)$ and integrate over the radial coordinate to obtain: -$$ -\int_{-\pi/4}^{\pi/4}\frac{a^3 b^3 \cos (2 t)}{\pi \left(a^2 \sin ^2(t)+b^2 \cos ^2(t)\right)^2}dt -$$ -Using Mathematica for example, we obtain: -$$ -\int_{-\pi/4}^{\pi/4}\frac{a^3 b^3 \cos (2 t)}{\pi \left(a^2 \sin ^2(t)+b^2 \cos ^2(t)\right)^2}dt = \frac{a b+(a-b) (a+b) \tan ^{-1}\left(\frac{a}{b}\right)}{\pi } -$$ -From here, you can then express everything in terms of the inner product of $x$ and $y$ along with the norms of $x$ and $y$. -Of course, hopefully the result can be checked! May I ask where this integral came up? I bet there are others who can obtain something similar using some pretty slick symmetry arguments. -Cheers!<|endoftext|> -TITLE: Bialynicki-Birula decompositions and fixed points -QUESTION [10 upvotes]: I was reading Luna's paper Toute variété magnifique est sphérique and stumbled on a few facts about Bialynicki-Birula decompositions and fixed points that I don't understand. -Here is the setup. Let $G$ be a connected reductive group over an algebraically closed field $k$ (of characteristic 0, though I'm not certain if that matters), and fix a Borel subgroup $B \subset G$ and a maximal torus $T \subset B$. Denote by $B^-$ the opposite Borel group to $B$ containing $T$. Let $X$ be an irreducible, normal, complete $G$-variety, and suppose that $X$ has finitely many $G$-orbits. Given any one-parameter subgroup $\lambda: \mathbb{G}_m \to T$ and any $y \in X^T$, we write -$$X(\lambda,y) = \{x \in X\ |\ \lim_{t \to 0} \lambda(t)x = y\}.$$ -Here are the claims Luna makes that I don't understand: -(1) The fixed point set $X^T$ is finite. -(2) We are mainly interested in the case where $\lambda$ is in the Weyl chamber of $B$ (i.e.\ where $\langle \lambda, \alpha \rangle > 0$ for all positive roots $\alpha$), so that $X(\lambda,y)$ is $B$-stable. Luna states that for a "sufficiently general" such $\lambda$, we will have $X^{\mathbb{G}_m} = X^T$, where $\mathbb{G}_m$ acts on $X$ via its image under $\lambda$. He also states that in this case, the $X(\lambda,y)$ for various $y \in X^T$ form the Bialynicki-Birula decomposition of $X$. -(3) With $\lambda$ satisfying the conditions in (2), if $X(\lambda,y)$ is open, then $y$ is fixed by the opposite Borel subgroup $B^-$. (Luna doesn't say anything about this, but I'm also curious: is it true that if $y$ is fixed by $B^-$, then $X(\lambda,y)$ is open?) -All of these statements seem pretty reasonable to me, and I've worked them out in the case where $X = \mathbb{P}(V)$, $G = \mathrm{SL}(V)$, and $B$ (resp. $T$) is the subgroup of upper triangular (resp. diagonal) matrices. In this case, everything is clear using projective coordinates, but I don't know how to make these types of arguments without appealing to coordinates like that. Any proofs (or references to proofs) would be much appreciated! - -REPLY [2 votes]: I think all of these should be easy enough to resolve. First note that (1) is a triviality from your assumption that $G$ has finitely many orbits on $X$, because a maximal torus of $G$ can only have finitely many fixed points on $G/H$. -Now recall that by a beautiful result of Sumihiro (in the case $G$ is an connected linear algebraic group) given a normal $G$-variety $X$ we and an orbit $Y \subset X$ we can find a $G$-stable opens $Y \subset U \subset X$ such that $U$ is isomorphic to a $G$-stable locally closed subset of $\mathbb{P}(\rho)$ for $\rho$ a finite dimensional representation of $G$, a well-known corollary allows us to let $U$ be affine when $T$ is a split $k$-torus. Since all of your questions are about the local structure of orbits in $X$ unless I am misreading you, I will just assume $X \subset \mathbb{P}(\rho)$ is locally closed for $(V, \rho)$ some fixed representation from now on. -For (2), $T$ acts on $V$ as such $V \cong \oplus_i V_{\chi_i}$ for $\chi_i$ various distinct characters of $T$, then let $\lambda$ be sufficiently general in the sense that $\langle \lambda, \chi_i - \chi_j \rangle \neq 0$ for $i \neq j$. Then the eigenvalues $\langle \lambda, \chi_i \rangle$ are distinct so $\lambda$ and $T$ induce the same eigendecomposition of $V$ and thus a point $x \in \mathbb{P}(V)$ is fixed by one iff it is fixed by the other. To me the definition you gave is the definition of the B-B decomposition, so you will have to elaborate on what definition you are working with if you want me to show that this induces the B-B decomposition. -For (3) lets take $y \in X^T$ such that $X(\lambda, y) = U$ is the big cell, and take $\lambda$ as above to be a regular cocharacter of $T$ wrt $B$. We know that for any $b \in B^-$ we have that $\text{lim}_{t \to 0} \lambda(t)^{-1}b\lambda(t) \in T$ by explicit computation of the BB decomposition for $G$ (alternatively if you use the dynamic approach to parabolics this is by definition). -Further considering $A: B^- \to X$ the action map $A(b) = b\cdot y$ we know that there is an open subscheme $V$ of $B^-$ such that $V \cdot y \subset U$. But for $b \in B^-$ $\text{lim}_{t \to 0} \lambda(t)^{-1} b \cdot y = \text{lim}_{t \to 0} b^{\lambda(t)^{-1}} \lambda(t)^{-1} \cdot y = y$, since $y$ is torus-stable. But then for $b \in V$ we have that $b \cdot y$ has both limits $\text{lim}_{t \to 0} \lambda(t) \cdot b \cdot y$ and $\text{lim}_{t \to \infty} \lambda(t) \cdot b \cdot y$ defined, which defines a map $\mathbb{P}^1 \to X$, the image of this map lies inside of $B^- \cdot y$, which is affine because $B^-$ is solvable, therefore it is constant. Thus $b\cdot y$ is a fixed point for $\lambda$, thus by $(2)$ it is a fixed point for $T$, for any $b \in V$. Because the stabilizer of $y$ is closed and $V$ is dense in $B^-$ we are done, $B^- \cdot y = y$. -Hope this helps, let me know if anything is unclear.<|endoftext|> -TITLE: Generators and relations for the 2-dimensional unoriented cobordism category -QUESTION [9 upvotes]: It is very well known in the field of TQFT that the 2-dimensional oriented cobordism category is generated by the disk and the pair of pants (each going in both directions), subject to a finite set of relations. Those generators and relations are equivalent to the morphisms and axioms of Frobenius algebras. -What is the analogue statement in the unoriented case? It is easy to see that it suffices to add the Moebius strip to the generators, but is there a provably sufficient set of relations for them? -I feel like this must be worked out somewhere, but I'm having trouble to find anything. If someone could give a reference where generators and relations are listed, this would be helpful! - -REPLY [12 votes]: My initial answer was wrong, here's the correct version plus a reference: Turaev-Turner -New generating morphisms: The Mobius strip $\emptyset \rightarrow S^1$ and the "orientation reversing" diffemorphism of the circle $S^1 \rightarrow S^1$. -New relations: Orientation reversal is an involution. Orientation reversal plays well with all the other generators. Orientation reversal composed with a twice punctured unoriented surface is itself. The punctured Klein bottle is both the product of two Mobius strips, and a composition of copairing, orientation reversal, and pants. -Algebraically, you have a Frobenius algebra involution $\phi: A \rightarrow A$ and an element $\theta \in A$ such that: - -$\phi(\theta v) = \theta v$ -$(m \circ (\phi \otimes id) \circ \Delta)(1) = \theta^2$. - -It actually follows from these definitions, see Lemma 2.8, that $\theta^3 = (m \circ \Delta)(\theta)$. That is the simplest relation you can state without making reference to the orientation reversing map, but you need the orientation reversing map in order for $\mathrm{Hom}(S^1,S^1)$ to be correct, which is why my initial guess was wrong.<|endoftext|> -TITLE: Has the exponentiation of ordinals a nice geometric model? -QUESTION [16 upvotes]: It is well known that the sum $\alpha+\beta$ of two ordinals $\alpha,\beta$ can be defined "geometrically" as the order type of the sum $(\{0\}\times \alpha)\cup(\{1\}\times\beta)$ endowed with the lexicographic order. -Also the product $\alpha\cdot\beta$ of ordinals $\alpha,\beta$ is the order type of the Cartesian product $\beta\times\alpha$ endowed with the lexicographic order. - -What about the exponentiation of ordinals? -Does $\alpha^\beta$ have some nice "geometric'' or combinatorial model? -Maybe as some set of (partial) functions endowed with a suitable well-order? - -REPLY [21 votes]: Ordinal exponentiation is a special case of linear order exponentiation. For any linear order $L$, element $a\in L$, and ordinal $\beta$ we can define the $\beta$th power of $L$ at $a$, which I'll call "$L_a^\beta$," as the set of functions $f:\beta\rightarrow L$ such that all but finitely many $\alpha\in\beta$ have $f(\alpha)=a$. The ordering on this set is given by looking at the last point of difference: $$f\trianglelefteq g\iff f=g\mbox{ or } f(\max\{x:f(x)\not=g(x)\}) -TITLE: The lattice of analogues of Robinson's $Q$ -QUESTION [8 upvotes]: This question was asked and bountied at MSE without response. - -Call a sentence $\varphi$ in the language of arithmetic $Q$-like iff $\mathbb{N}\models\varphi$ and $\{\varphi\}$ is essentially incomplete. The standard example is of course the conjunction of the finitely many axioms of Robinson's $Q$, but this is of course not unique - and indeed the partial order $\mathfrak{Q}$ of equivalence classes of $Q$-like sentences under entailment as in the Lindenbaum algebra ($\varphi\le\psi\iff\vdash\varphi\rightarrow\psi$) is not linear. On the positive side, $\mathfrak{Q}$ is clearly a distributive lattice, and every countable partial order embeds into each element of $\mathfrak{Q}$'s lower cone (see my comment below). -My question is: - -What exactly is $\mathfrak{Q}$, up to isomorphism? - -There's an obvious candidate, based on the idea that everything that can happen does in this sort of situation: the (countable) random distributive lattice (that is, the Fraisse limit of the set of finite distributive lattices). However, I'm having trouble proving this. Even showing that $\mathfrak{Q}$ has no greatest element isn't trivial, as far as I can see. (EDIT: by "isn't trivial" I now mean "I can't.") -(As a quick remark, note that essentially undecidable theories need not come from elements of $\mathfrak{Q}$: Robinson's $R$ is essentially undecidable but each of its finitely axiomatizable subtheories has a computable completion.) - -REPLY [5 votes]: $\mathfrak{Q}$ is the countable random distributive lattice. -Emil Jeřábek has already pointed in his comments that there are only two possibilities for $\mathfrak{Q}$. Either there are no greatest element in $\mathfrak{Q}$ and it is the countable random distributive lattice. Or there is the greatest element in $\mathfrak{Q}$ and $\mathfrak{Q}$ is the countable random distributive lattice with appended greatest element. So I'll only need to show that there exist no sentence $\varphi_0$ such that $\mathbb{N}\models\varphi_0$ and for any $\varphi$, if $\mathbb{N}\models \varphi$, then -$$\varphi\text{ is essentially undecidable }\iff \vdash \varphi\to \varphi_0.$$ -Indeed assume for a contradiction that $\varphi_0$ exist. -To simplify things as much as possible here I'll consider $\mathbb{N}$ to have the signature consisting of the constant $0$ and predicates $\mathsf{Succ}(x,y)$, $\mathsf{Add}(x,y,z)$, $\mathsf{Mul}(x,y,z)$, and $x\le y$; it is possible to modify the argument so that it will work with the standard signature $0,S,+,\times$, but it would add additional complications. Let us consider the class $\Pi_1^{-}$ of all formulas of the form $\forall x\;\theta(x)$, where all quatifiers in $\theta$ are $x$-bounded. Note that the set of all true $\Pi_1^{-}$ sentences is $\Pi_1$-complete. -For any $\Pi_1^{-}$ arithmetical sentence $\psi$ of the form $\forall x \;\theta(x)$ let us consider the sentence $\psi^\star$: -$$\mathsf{Q}^{-}\land \forall x\;(\theta(x)\to \exists y\;(\mathsf{Succ}(x,y)).$$ -Here $\mathsf{Q}^{-}$ should be a version of $\mathsf{Q}-\text{"totality of $S,+,\times$"}$ in our signature. The key properties of $\psi^\star$ that we will need are the following: - -if $\psi$ is false, then $\psi^\star$ has a finite model; -if $\psi$ is true, then any model of $\psi^\star$ contains $\mathbb{N}$ as an initial segment; -$\mathbb{N}\models \psi^\star$, regardless of whether $\psi$ were true or not. - -Notice that any sentence $\varphi$ (in our finite signature) with a finite model isn't essentially undecidable. And that by the standard argument (that uses a pair of recursively inseparable sets) we see that if any model of a sentence $\varphi$ contain $\mathbb{N}$ as an initial segment, then $\varphi$ is essentially undecidable. To conclude, $\psi^{\star}$ is always true and is essentially undecidable iff $\psi$ is true. -Under the assumption that $\varphi_0$ exists we see that $$\{\psi\in \Pi_1^{-}\mid\mathbb{N}\models \psi\}=\{\psi\in \Pi_1^{-}\mid \psi^{\star}\text{ is essentially undecidable}\}=\{\psi\in \Pi_1^{-}\mid \vdash \psi^{\star}\to \varphi_0\}$$ is $\Sigma_1$. But on the other hand it should be $\Pi_1$-complete, contradiction. -For the sake of completeness let me sketch my reconstruction of Emil's argument. Observe that by Gödel's first incompleteness theorem $\mathfrak{Q}$ has no least element. By Rosser's theorem, for any pair $a<_{\mathfrak{Q}}b$ the interval $[a,b]$ is a countable atomless Boolean algebra. By a standard back and forth argument it is easy to show that for a countable distributive lattice $K$, if all non-trivial intervals in $K$ are countable atomless Boolean algebra, then there are only 4 possibilities for $K$: - -$K$ is the random distributive lattice; -$K$ is the random distributive lattice with appended $0$; -$K$ is the random distributive lattice with appended $1$; -$K$ is the random distributive lattice with appended $0$ and $1$.<|endoftext|> -TITLE: How long for Brownian motion to "fill-out" a torus in d-dimensions? -QUESTION [13 upvotes]: I've been taken by the concise result1 -that (roughly!), on a $2$-dimensional torus $\mathbb{T}^2$, the time it takes -to visit nearly every point (within $\epsilon$, as $\epsilon \to 0$) is: $\frac{2}{\pi}$. -My question is: - -Q. Is the same situation known for the $d$-dimensional - torus, $\mathbb{T}^d$? What is the time it takes for a Brownian-motion particle to visit within $\epsilon \to 0$ of - every point of $\mathbb{T}^d$? - -This is probably known—or known to be unknown—so this is just -a reference request. - -1 -Dembo, Amir, Yuval Peres, Jay Rosen, and Ofer Zeitouni. "Cover times for Brownian motion and random walks in two dimensions." Annals of Mathematics (2004): 433-464. -Annals link. - -REPLY [8 votes]: A very general answer, in dimension $d\geq 3$, - is in the following paper of Dembo, Peres and Rosen -https://projecteuclid.org/euclid.ejp/1464037588: -for compact $d$-dimensional manifolds, -$$C_\epsilon(M) /\epsilon^{2-d}\to V(M)$$ -where $V(M)$ is the volume. -The $d$ dimensional case for $d\geq 3$ is much simpler than $d=2$ because in the latter, you have -long range (=logarithmic) correlations arising from the recurrence of BM. -In passing, let me mention that Belius's paper mentioned in Josiah Park's answer gives a more refined answer, namely a limit law for a centered version of -$\sqrt{C_\epsilon}$ (with no scaling), -for the random walk case. Similar results can be written -for the manifold case in $d\geq 3$. For the -critical case of $d=2$, higher precision than the law of large numbers is available but as of this writing, the limit law for $\sqrt{C_\epsilon}-E\sqrt{C_\epsilon}$ has not been derived.<|endoftext|> -TITLE: Can one-sided derivatives always exist, but never match? -QUESTION [12 upvotes]: Is there a continuous function $f : \mathbb{R} \to \mathbb{R}$ which has left and right derivatives everywhere, but where those derivatives are unequal at every point? - -REPLY [14 votes]: No, that cannot happen. -Let's use a Baire category argument. More precisesly: a pointwise limit of a sequence of continuous functions $\mathbb R \to \mathbb R$ is continuous everywhere except for a meager set [= set of first category]. ref. -Let $f : \mathbb R \to \mathbb R$ be continous. Assume the left-hand derivative $f^-(x)$ and the right-hand derivative $f^+(x)$ exist everywhere. Let -$$ -f_n(x) = \frac{f(x+1/n)-f(x)}{1/n} -$$ -Then each $f_n$ is continous and $f_n(x) \to f^+(x)$ everywhere. Therefore, $f^+$ is continuous everywhere except for a meager set. Similarly, $f^-$ is continuous everywhere except for a meager set. So there is a point $a$ such that $f^+$ and $f^-$ are both continuous at $a$. -By assumption, $f^-(a) \ne f^+(a)$. Replacing $f$ by $-f$, if necessary, we may assume WLOG that $f^-(a) > f^+(a)$. Adding a linear function to $f$, if necessary, we may assume WLOG that $f^-(a) > 0 > f^+(a)$. Because $f^+, f^-$ are continuous at $a$, there is $\delta > 0$ so that -$$ -\forall u \in [a-\delta,a+\delta] \quad f^-(u) > 0\text{ and } - f^+(u) < 0 . -$$ -Now, consider a point $u \in [a-\delta,a+\delta]$. Because $f^-(u) > 0$, there is $\alpha_u < u$ so that -$$ -\forall x\in(\alpha_u,u),\quad \frac{f(u)-f(x)}{u-x} > 0, \text{ so } f(x) < f(u). -$$ -Because $f^+(u) < 0$, there is $\beta_u > u$ so that -$$ -\forall x\in(u,\beta_u),\quad \frac{f(x)-f(u)}{x-u} < 0, \text{ so } f(x) < f(u) . -$$ -Thus, there is a neighborhood $(\alpha_u,\beta_u)$ of $u$ such that $f(x) < f(u)$ for all $x \in (\alpha_u,\beta_u)\setminus\{u\}$. So $f$ has a strict local maximum at $u$. But $f$ is continuous on $[a-\delta,a+\delta]$, and therefore achieves its minimum value at some point $u \in [a-\delta,a+\delta]$. A contradiction.<|endoftext|> -TITLE: Can conservativity depend on the universe? -QUESTION [5 upvotes]: Question 1: Let $F: C \to D$ be a conservative, $\kappa$-cocontinuous functor between small, $\kappa$-cocomplete categories. Is the induced functor $Ind_\kappa(F): Ind_\kappa(C) \to Ind_\kappa(D)$ also conservative? -Terminology: I think this is pretty self-explanatory, but to be clear: - -$\kappa$ is a regular cardinal. Things are perhaps most familiar when $\kappa = \aleph_0$. -A $\kappa$-cocomplete category is a category with $\kappa$-small colimits, i.e. colimits indexed by categories with fewer than $\kappa$-many morphisms. -A $\kappa$-cocontinuous functor is a functor preserving $\kappa$-small colimits. -$Ind_\kappa(C)$ is obtained from $C$ be freely adjoining $\kappa$-filtered colimits, or by the formula $Ind_\kappa(C) = Fun_\kappa(C^{op},Set)$, where $Fun_\kappa(A,B)$ is the category of $\kappa$-continuous functors from $A$ to $B$. -The induced functor $Ind_\kappa(F)$ is defined by left Kan extension along the Yoneda embedding. - -I'm a little worried that Question 1 is too much to ask for, so here's an even milder variant. - -Let $\kappa$ be inaccessible. -Let $Pr^L(\kappa)$ be the category (really a $(2,1)$-category) of categories which are locally presentable with respect to the universe $V_\kappa$. That is, $Pr^L(\kappa)$ consists of categories of the form $Ind_\lambda^\kappa(C)$ where $C$ is a $\kappa$-small and $\lambda$-cocomplete category with $\lambda < \kappa$; here $Ind_\lambda^\kappa$ is the free cocompletion under $\kappa$-small, $\lambda$-filtered colimits. The morphisms are left adjoint functors. -Similarly, let $Pr^L$ be the $(2,1)$-category of locally presentable categories and left adjoint functors. -Then we have a functor $Ind_\kappa: Pr^L(\kappa) \to Pr^L$. - -Question 2: Does the functor $Ind_\kappa: Pr^L(\kappa) \to Pr^L$ preserve conservative functors? -Question 2 is asking whether the property of a left adjoint between locally presentable categories being conservative depends on which universe we work in. -In the setting of either question, it's clear that if $Ind_\kappa(F)$ is conservative, then $F$ is conservative. So in either case, if the answer to the question is affirmative, then we have "$F$ conservative $\Leftrightarrow$ $Ind_\kappa(F)$ conservative", which would be reassuring. - -REPLY [4 votes]: Probably not the best one can do, and what follows might be a bit 'overkill', but it answer the question about dependency on universe, and it is a nice argument. -Also if you know how the proof of the left transfer results I will use below works, it might give some idea on how to prove more general case of the result. -Theorem: Let $\kappa$ be an uncountable regular cardinal. Let $F: \mathcal{C} \to \mathcal{D}$ be a strongly $\kappa$-accessible left adjoint functor between locally $\kappa$-presentable categories. Then $F$ is conservative if and only if the restriction of $F$ to the full subcategory of $\kappa$-presentable objects is conservative. -(Here strongly $\kappa$-accessible = sends $\kappa$-presentable objects of $\mathcal{C}$ to $\kappa$-presentable objects of $\mathcal{D}$) -Proof: The weak factorization system (isomorphisms, all maps) on $\mathcal{D}$ is ($\kappa$-)combinatorial: it is generated by the empty set of maps. -Hence the left transfered weak factorization system on $\mathcal{C}$ along $F$ exists and is also $\kappa$-combinatorial (I mean by this that is is cofibrantly generated by maps between $\kappa$-presentable objects). -By definition, the left class of this weak factorization system is the set of maps such that $F(i)$ is an isomorphism. But if the restriction of $F$ to $\kappa$-presentable objects is conservative it means that all generators are isomorphisms hence, the left class only contains isomorphisms, so $F$ is conservative. -Note: the finer version of the left transfer theorem that I'm using which specify the presentability rank can be found as Theorem B.8.(4) in this paper of mine, but at the end of day it mostly follows from an analysis of the proof of the existence of left transfer available in the litterature (for e.g. in Makkai and Rosicky Cellular categories.)<|endoftext|> -TITLE: For $x$ irrational, is $a_{n} =\sum_{k=1}^{n}(-1)^{⌊kx⌋}$ unbounded? -QUESTION [26 upvotes]: For $x$ irrational, define $a_{n} :=\sum_{k=1}^{n}(-1)^{⌊kx⌋}$. Can you prove that $\left\{a_n\right\}$ is unbounded? -I feel that it is not easy to treat every irrational $x$. -I have asked in S.E. and it seems that it is an advanced question. Thus I go for help here. - -REPLY [4 votes]: As Ville Salo already wrote in his answer, your question can be phrased in terms of the difference between the number of elements of the sequence $y, 2y, \dots, ky$ which are contained in $[0,1/2]$, and $k$ times the length of $[0,1/2]$. Here $y = x/2$. In the language of discrepancy theory, you are asking whether the interval $[0,1/2]$ is a so-called bounded remainder set of the sequence $(n y~\text{mod}~1)_{n \geq 1}$. However, it is known that it is not: the only intervals with bounded remainder are those whose length is in $\mathbb{Z} + y \mathbb{Z}$, and since $y$ is irrational in your question the length of $[0,1/2]$ is not of such type. Bounded remainder sets were classified in this paper: H. Kesten, On a conjecture of Erdös and Szüsz related to uniform distribution mod 1, Acta Arith. 12(1966), 193–212.<|endoftext|> -TITLE: Distribution of degree in graphs: when is the friendship paradox the paradox it wants to be? -QUESTION [12 upvotes]: $\DeclareMathOperator\deg{deg}\DeclareMathOperator\ndeg{ndeg}\newcommand\abs[1]{\lvert#1\rvert}$The friendship paradox goes most people have fewer friends than their friends have on average. The original paper Feld - Why your friends have more friends than you do has a simple counter example: (Figure 5, p. 1474). -My question is how exceptional is this example? -Even in some random graph models where some answer is tractable, the analysis of the distribution of a neighbour's degree deals with expectation. That is, in some models it is explicitly shown that the average difference between the mean number of friends of friends and the number of friends is positive, which is much less likely to get press attention I think. -Given any graph $G=(V,E)$ and $v\in V$, let $N(v) = \{w\in V:\{v,w\} \in E\}$ and $\deg(v) = \abs{N(v)}$. This is the number of friends of $v$. Now consider $\ndeg(v)= \sum_{w \in N(v)}\deg(w)$, the total friends of friends of $v$ and the distribution of -$$ -f(v) = \frac{\ndeg(v)}{\deg(v)} - \deg(v). -$$ -For a friendship paradox, -$$ -g = \abs{\{v:f(v)>0\}}-\abs{\{v:f(v)<0\}}>0. -$$ -For the graph pictured above, -$$ -\{f(v)\}_{v\in \{A,B,C,D,E,F\}}=\left\{ 1,1,-\frac13-\frac13,-\frac13,-\frac13 \right\}. -$$ -Clearly $g = -2$, and no paradox. But, the mean of $f$ is $1/9>0$. -The mean of $f$ is shown to be positive in the popular configuration model. And this is usually said to be evidence/proof of the friendship paradox. (Tangential question: is $g>0$ in this model?) -Now the question(s): - -Is the mean of $f$ positive for any graph? -What are some equivalent conditions stated in terms of graph properties to $g >0$? - -REPLY [10 votes]: I can answer question 1, at least. The mean of $f$ is always positive (unless every connected component of the graph $G$ is regular, in which case the mean of $f$ is zero). To see this, we rewrite the sum of the values of $f$ as -$$\sum_{v \in V} f(v) = \sum_{v \in V} \sum_{w\text{ s.t. }\{v,w\} \in E} \left(\frac{\deg(w)}{\deg(v)} - 1\right) = \sum_{\{v,w\} \in E} \left(\frac{\deg(w)}{\deg(v)} + \frac{\deg(v)}{\deg(w)} - 2\right),$$ -and each summand is positive by AM-GM. More explicitly, we have -$$\sum_{v \in V} f(v) = \sum_{\{v,w\} \in E} \frac{(\deg(v)-\deg(w))^2}{\deg(v)\deg(w)} \ge 0.$$<|endoftext|> -TITLE: Detecting weak equivalence on free loop space homology -QUESTION [13 upvotes]: Given $f:X \to Y$ a continuous map between two spaces (unpointed CW-complexes) such that $f$ induces an isomorphism in homology with integer coefficient, and $f$ induces an isomorphism on homology of the free loop spaces: $H_*(X^{S^1}) \to H_*(Y^{S^1})$ (also with integer coefficient). Does $f$ has to be a weak homotopy equivalence ? -If I further assume that $f$ induces an equivalence $H_*(X^{K}) \to H_*(Y^{K})$ for each finite CW-complex $K$, is $f$ a weak homotopy equivalence ? -Note that working in the unpointed settings makes a big difference here: It was showed by Arlin and Christensen that, contrary to the case of pointed connected space, for any small set of spaces $E$, there are maps that induces an isomorphism $\pi_0(X^K) \to \pi_0(Y^K)$ for all $K \in E$ without being weak equivalences. It is unclear to me if adding higher homology groups makes a big difference or not. -Edit: The case of groupoids $X=BG$ and $Y=BH$ is already an interesting example. $X^{S^1}$ is the (classyfing space of) the groupoid corresponding to the action of $G$ on itself by conjugation. Hence given $f:G \to H$ a morphism of group. Saying that it is a bijection on the $H_0$ of the loop space gives that $f$ induces a bijection between the set of conjugation classes of $G$ and $H$, and the fact that it is a bijection on the $H_1$ of the loop spaces gives that for all $g \in G$, $f$ induces an isomorphisms between the ablianization of the centralizer of $g$ and the abelianization of the centralizer of $h$. -That does not seem to be quite enough to conclude that $f$ is an isomorphisms, but this is already pretty restrictive. I havn't been able to formulate the fact that $f$ induce bijections on the higher $H_i$ in simple terms. -Now, if we start looking a $H_0(X^K)$ and $H_0(Y^K)$ for $K$ a finite complex, it seems that in this case all the interesting information is in the case where $K$ is a wedge of circle. Being a bijection on $H_0(X^K)$ and $H_0(Y^K)$ means that: -1) $f$ induces a bijection between $G^n/G$ and $H^n/H$ for all $n \in \mathbb{N}$, where $G$ and $H$ acts on $G^n$ and $H^n$ by the diagonal conjugation action. -2) For every finite family of elements $g_1,\dots,g_n$ in $G$, $f$ induce a bijection between the abelianization of the centralizer of $\{g_1,\dots,g_n\}$ and the abelianization of the centralizer of $\{f(g_1),\dots,f(g_n) \}$. - -REPLY [3 votes]: I believe these questions were studied and answered in the this 1984 paper: -HIROSHIMA MATH. J. 14 (1984), 359-369 -On the set of free homotopy classes and Brown's construction -Takao MATUMOTO, Norihiko MINAMI and Masahiro SUGAWARA -They also have counterexamples that seem to be the same as in the recent preprint of Arlin and Christensen that you mention. -Take a look and see if they do what you want.<|endoftext|> -TITLE: Residually finite group with dense finite index subgroups -QUESTION [8 upvotes]: Let $G$ be a locally compact Hausdorff topological group whose underlying abstract group is residually finite. Let $H\subset G$ denote the intersection of all finite-index, closed subgroups. Is there an example of such a $G$ where $H$ is not trivial? Is there an example where $H=G$ (i.e., every finite-index subgroup is dense)? -My motivation for asking this question comes from the study of automorphism groups of connected, locally finite graphs. Therefore, the closer the example is to being of this type the better. E.g., an example which is an automorphism group of some structure on $\mathbb{N}$, say a hypergraph structure, would be extremely interesting; an example where $G$ is separable would be more useful than an example with cardinality larger than the continuum, and so on. - -REPLY [2 votes]: Yes it exists (at least the weaker version). Namely, here is a way to produce second-countable abelian, totally disconnected locally compact groups whose underlying abstract group is residually finite, but in which the intersection of finite index open subgroups is not trivial. -Fix $p$ prime and an infinite countable set $I$. Let $\mathbf{Z}_p$ be the $p$-adic group. Let $A$ be the divisible closure of $K=\mathbf{Z}_p^I$ in $\mathbf{Q}_p^I$ (i.e., those sequence with bounded denominator) and endow $A$ with the group topology making $K$ a compact open subgroup (with its usual product topology). The example will be a suitable subgroup of $A$ containing $K$. -Lemma: let $F$ be a finite subset of $A$ and $\bar{F}$ its image in $A/K$. Let $U$ be a nonempty open subset of $K$, disjoint from $\langle F\rangle+pK$, and $n\ge 1$. Then there exists $x\in A$ whose image $\bar{x}$ in $A/K$ has order $p^n$, such that $\langle \bar{x}\rangle\cap \bar{F}=\{0\}$, and such that $p^nx\in U$. -Proof: choose $z\in U-pK$ (this is possible: as $I$ is infinite $pK$ has empty interior in $K$) and set $x=p^{-n}z$. Then $\bar{x}$ has order $p^n$. -If $\langle \bar{x}\rangle\cap \bar{F}=\{0\}$ fails, then $p^\ell\bar{x}\in\bar{F}$ for some $\ell$ with $1\le\ell -TITLE: On which regions can Green's theorem not be applied? -QUESTION [50 upvotes]: In elementary calculus texts, Green's theorem is proved for regions enclosed by piecewise smooth, simple closed curves (and by extension, finite unions of such regions), including regions that are not simply connected. - -Can Green's theorem be further generalized? In particular, are there regions on which Green's theorem definitely does not hold? - -REPLY [24 votes]: There is a fun reverse definition that is used for so called "currents" in geometric measure theory, objects for which then in Green's theorem always ends up trivially being true. But then using the resulting theory, one can then show that Green's theorem is always true in a more proper sense, whenever the two integrals in it are defined, even if only in a very weak measure theoretic way. -Let $\Omega \subset \mathbb{R}^2$ be a $\mathcal{H}^2$-measureable set ($\mathcal{H}^2$ is the 2-dimensional Hausdorff measure).¹ Then we can define the corresponding linear operator -$$ -\begin{array}{rccl} -[\Omega] : & C_c^\infty(\mathbb{R}^2) &\to &\mathbb{R}\\ -& f & \mapsto & \displaystyle\int_\Omega f dx -\end{array} -$$ -This is what is called a $2$-current, i.e. an element of the topological dual $\mathcal{D}_2 := C_c^\infty(\mathbb{R}^2)'$. The theory of currents is quite similar to those of distributions, but a bit more geometrical. In particular, for any $T \in \mathcal{D}_2$ we define its boundary by -$$ -\begin{array}{rcl} -\partial T: & C_c^\infty(\mathbb{R}^2;\mathbb{R}^2) & \to & \mathbb{R}\\ -& F &\mapsto & T(\operatorname{curl}F) -\end{array} -$$ -The resulting object is what is called a $1$-current.² As one would expect, those correspond to integral along suitably generalized curves. Using this definition, Green's theorem is always automatically true, since $\partial[\Omega](F) = [\Omega](\operatorname{curl} F)$ by definition. -One can of course then reformulate the question to make it interesting again. Let $\Gamma$ be a $\mathcal{H}^1$-measurable set of locally finite measure and $\tau: \Gamma \to \mathbb{R}^2$ some $\mathcal{H}^1$-measurable "unit tangent" orienting that set. Then we can define similarly -$$ -\begin{array}{rccl} -[\Gamma] : & C_c^\infty(\mathbb{R}^2;\mathbb{R}^2) &\to &\mathbb{R}\\ -& F &\mapsto& \displaystyle\int_\Gamma F \cdot \tau d \mathcal{H}^1 -\end{array} -$$ -which is likely the weakest way to give some sense to the other integral in Green's theorem. Now the question is, for which $\Omega$ does this commute, i.e. $\partial [\Omega] = [\partial \Omega]$? Here the topological boundary $\partial \Omega$ is easy enough to define, but it turns out that the key-question here is what is the tangential vector $\tau$? The resulting notion is that of rectifiability. Roughly the condition for $\tau$ to be "the" tangential vector of $\Gamma$ at $x$ is that for any double-cone in direction $\tau$, most of $B_\epsilon(x)\cap \Gamma$ lies in that cone (the details are technical). If such a $\tau$ exists $\mathcal{H}^1$-almost everywhere, then $\Gamma$ is called rectifiable. -Now there are some further minor details involving the orientation of $\tau$ and possible multiplicity, but fundamentally it turns out that whenever the topological boundary $\partial \Omega$ is rectifiable, then there is a matching $\tau$ such that $\partial$ and $[\,]$ commute, i.e. Green's theorem holds. -The proper citation for all of this is Federer's "Geometric measure theory", but as it is one of those books, I'd recommend picking up Morgan's "Geometric measure theory: A beginner's guide" first. -¹Asking if Green's theorem holds for non-measurable sets should probably only be done by Zen Buddhists. -²As you can see, the number refers to the "dimensionality" of the object. To be more precise one should actually use differential forms of the dimension 2 in place of $f$ (resp. dimension 1 in place of $F$). This, and using the exterior derivative instead of $\operatorname{curl}$, also give the right generalisation to higher dimensions.<|endoftext|> -TITLE: A naive question about the prime number theorem -QUESTION [13 upvotes]: Let $\psi(x)=\sum_{n\leq x} \Lambda(n)$, where $\Lambda(n)$ is the von Mangoldt function. -Then as Chebyshev showed, the following equality holds -$$\sum_{n\leq x} \psi(x/n)=x\log(x)-x+O(\log(x)).$$ -My question is, how far can one go towards proving the prime number theorem by only -using the above estimate and the fact that $\psi$ is increasing? Alternatively, is there a well-known example of an increasing function $\psi(x)$ for which the above equality holds, but $\lim_{x\to \infty}\frac{\psi(x)}{x}\neq 1?$ Thank you very much. - -REPLY [14 votes]: Theorem 1 in the following paper of Ingham shows that the stated estimate, together with $\psi$ being positive and nondecreasing, is 'enough' to deduce that $\psi(x)/x \to 1$: -A. E. Ingham: Some Tauberian theorems connected with the prime number theorem. J. London -Math. Soc. 20, 171–180 (1945). Full text (paywalled) -In fact, this remains true even if one only has the weaker error term $o(x)$ in place of $O(\log{x})$. Of course, one has to be careful about what `enough' means here; Ingham's proof still uses the nonvanishing of $\zeta(s)$ on $\Re(s)=1$, which is the key input in the usual proofs of the prime number theorem.<|endoftext|> -TITLE: Acting with a finite number of rotations on a set of positive measure can you fill almost the whole circle? -QUESTION [18 upvotes]: Let $E\subset S^1$ have positive Lebesgue measure. Do there exist finitely many rotations -$r_1, r_2, \dots ,r_n$ such that $r_1E\cup r_2E\cup \dots\cup r_nE$ has measure $2\pi$? Or is there a counterexample? - -REPLY [17 votes]: The answer is no. Take a compact set $E$ with positive measure buth empty interior and assume that $K=r_1 E\cup r_2 E \cdots \cup r_n E$ has measure $2\pi$. Then $K$ would be dense in $S^1$, hence equal to $S^1$, since it is closed. But this is impossible, since $K$ has empty interior, too.<|endoftext|> -TITLE: Anomaly in QFT physics v.s. determinant line bundle -QUESTION [18 upvotes]: In a quantum field theory (QFT) lecture, a math-physics professor explains the anomaly in physics, say the non-invariance of the partition function of an anomalous theory under background field transformation, can be regarded as a section of a complex line bundle over the space of background field. -In math terminology, he explained that the so-called anomaly in physics is the determinant line bundle in math. - -How precise is the analogy: - -anomaly in QFT physics v.s. determinant line bundle - -can you provide a few examples between the twos? - -REPLY [16 votes]: The partition function should assign to each possible field configuration $\Phi$ (or field history) in your quantum field theory a number $Z(\Phi)$. That is, it should be a function on the collection of fields configurations, and from that function you can derive lots of quantities in the field theory. -It can happen, however, that in order to come up with, or write down, the number $Z(\Phi)$, you need to make some auxiliary choices. Often these choices work only for certain fields, rather than for all fields at the same time. For instance, in a gauge theory with fermions you might need to choose a real number $\lambda$ which is not in the spectrum $\sigma(D_A)$ of the Dirac operator coupled to the gauge field $A$. In general, such a choice cannot be made for all gauge potentials $A$ simultaneously, and hence exists "only locally" on the collection of fields. Different local choices of auxiliary information will lead to different values of what you compute as $Z(\Phi)$, and it usually turns out that the transformation law between the values of $Z(\Phi)$ for different auxiliary choices is that of a section of a line bundle on the collection of fields. -What people normally want in gauge theory is for the partition function to be well-defined on the space $\mathcal{A}/\mathcal{G}$ of gauge potentials modulo gauge transformations. While any line bundle on $\mathcal{A}$ is trivialisable (since $\mathcal{A}$ is an affine space), this is not true on the quotient $\mathcal{A}/\mathcal{G}$. Let us say that our partition function can be understood as a section $Z$ of a line bundle $L \to \mathcal{A}/\mathcal{G}$. Then, any trivialisation of $L$ allows us to translate $Z$ into a function on $\mathcal{A}/\mathcal{G}$, and hence into an actual partition function. The QFT anomaly can hence be described as the obstruction to the existence of a trivialisation of $L$ -- this is a class in $H^2(\mathcal{A}/\mathcal{G};\mathbb{Z})$. Often this class can be computed, like in the case of the Dirac anomaly. -Some nice mathematical references, in my opinion, are https://arxiv.org/abs/hep-th/9907189, https://arxiv.org/abs/math-ph/0603031v1, and for a more conceptual perspective, https://arxiv.org/pdf/1212.1692.pdf.<|endoftext|> -TITLE: Graph metric approximating Euclidean metric -QUESTION [9 upvotes]: I've been reading Wolfram's recent articles about graph/mesh/grid structures as an analogy for physical space, and it seems to me that there will be a problem getting the notion of distance to work out, since the natural metric on eg a square grid or triangle grid is L1, even in the limit of a very dense grid becoming a patch of the plane. -Is there any simple rule for generating a mesh-like graph that, with the length of all edges considered to the be the same, in the limit of very fine scale, closely approximates Euclidean distance in the plane? -Looking around, I was able to find this paper (https://projecteuclid.org/download/pdf_1/euclid.cmp/1104286245) which shows that if you allow the length of each edge in the graph to match its actual length in the plane, there is a solution with a simple production rule, but is it possible with all graph edges considered to be the same length? - -REPLY [4 votes]: Take $M=\mathbb{R}^2\times[0,h]^k$ for fixed $k\ge0$ and $h\gtrsim 1$. Spatter $M$ with dots according to a uniform probability distribution with unit density. Form a graph by connecting each dot with its $2(2+k)$ nearest neighbors. There will be topologically disconnected pieces of the graph, which should be finite in size. Cut these out, leaving some defects like Swiss cheese holes. If $k$ and $h$ are large, then the size and frequency of these holes is small. We can make a model with $k=0$, but larger $k$ should greatly reduce the prevalence of these holes. If we like, we can restrict our attention to some large defect-free disk in $\mathbb{R}^2$, or we can just live with the holes and make an infinite model. -Let $\ell$ be the number of steps between points on the graph, and $r$ the Euclidean distance in $\mathbb{R}^2$, ignoring the extra $k$ dimensions. Then I conjecture that there is some constant $\alpha$ such that at large distances, the relative expected error $E(\alpha\ell-r)/r$ approaches zero. Let $d=\alpha\ell$. -This model has the advantage that $E(d-r)$ is manifestly invariant under rotation and translation in $\mathbb{R}^2$. The model also degrades gracefully at small scales. At $r\lesssim h$, it begins to act like $\mathbb{R}^{2+k}$, balls have a reasonable topology, and the triangle inequality is still some kind of reasonable approximation. This reasonable behavior at small scales is a good property to have because the OP says the motivation for the question was Wolfram's ideas, in which the laws of physics operate on a spacetime grid, as in Conway's game of life. In that kind of system, the laws of physics operate between a point and its neighbors. Therefore it would be undesirable to have too pathological a geometry at short scales.<|endoftext|> -TITLE: How much time does a function spend above or below its average value around a point? -QUESTION [5 upvotes]: Given a locally integrable function $f: \mathbb R \to \mathbb R$, define $ -K: \mathbb R \times \mathbb R+ \to \mathbb R$ by -$$ -K(x, r) := -\begin{cases} -1, & \text{if }f(x) > \dfrac{1}{2r}\displaystyle\int\limits_{B_{r}(x)}f\,\mathrm{d}x\\ --1, & \text{if }f(x) < \dfrac{1}{2r}\displaystyle\int\limits_{B_{r}(x)}f\,\mathrm{d}x\\ -0,& \text{if }f(x) = \dfrac{1}{2r}\displaystyle\int\limits_{B_{r}(x)}f\,\mathrm{d}x -\end{cases} -$$ -Define $U: \mathbb R \to [-1, 1]$ and $L: \mathbb R \to [-1, 1]$ by -$$ -\begin{split} -U(x) &:= \limsup_{r \to 0} \frac{1}{2r}\int\limits_{B_r(0)} K(x, t)\,\mathrm{d}t\\ -L(x) &:= \liminf_{r \to 0} \frac{1}{2r}\int\limits_{B_r(0)} K(x, t)\,\mathrm{d}t -\end{split} -$$ -Intuitively, $U$ and $L$ represent the weighted proportion of time a function spends above or below its average value in an infinitesimal neighbourhood of a point. -i) Is it true that for any locally integrable function, $U = L$ a.e? -ii) Is it true that $U = 1, 0, or -1$ a.e.? - -REPLY [2 votes]: For i) the Brownian motion gives a counter example. -We choose $f(t)=W_t$ with $W_t$ a Brownian on $\mathbb{R}$. Because one can calculte $U(0)$ and $L(0)$ from the Brownian motion restricted to any neighbourhood of $0$, we can apply Blumenthal’s 0-1 law: There exists $c_1,c_2\in \mathbb{R}$ such that $\mathbb{P}(U(0)=c_1)=1$ and $\mathbb{P}(L(0)=c_2)=1$. By symetry we have $c_1 =-c_2$. Moreover for any $00$ -$$\mathbb{P}\Big(\frac{1}{2r}\int_{B_r(0)}K(0,t)dt \geq a\Big)>\epsilon$$ Therefore $c_1\geq a$ and we conclude that $c_1=1$ and $c_2=-1$. -Finally by translation invariance of the Brownian motion we have for all $t\in \mathbb{R}$ $$\mathbb{P}(U(t)=1)=\mathbb{P}(L(t)=-1)=1$$<|endoftext|> -TITLE: Ramsey theory in infinite-dimensional projective spaces -QUESTION [6 upvotes]: Let $\mathbb{F}_q$ be a finite field and $k$ be a positive integer. If we colour each point of the infinite-dimensional projective space $\mathbb{F}_q \mathbb{P}^{\infty}$ with one of $k$ colours, can we necessarily find an infinite-dimensional monochromatic projective subspace? - -There are a couple of observations that hint that the answer is 'yes': -Observation 1: The statement for $\mathbb{F}_2$ is true, and is equivalent to Hindman's theorem. The reason for asking this question is that it would provide an interesting generalisation of Hindman's theorem. -Observation 2: If we weaken the problem to finding an $n$-dimensional monochromatic projective subspace (where $n$ is finite), the statement is also true (and follows, for example, from the 'Vector Space Ramsey Theorem' in the paper Ramsey's Theorem for Spaces by Joel H. Spencer). - -REPLY [3 votes]: After further investigation, it appears that disappointingly the answer is 'no', and a proof appears in Lemma 2.4 of Partition Theorems for Subspaces of Vector Spaces (Cates and Hindman, 1975).<|endoftext|> -TITLE: Is $A_5$ the only finite simple group with only 4 distinct sizes of orbits under the action of the automorphism group? -QUESTION [20 upvotes]: Given a finite group $G$, let $\eta(G)$ denote the number of distinct -sizes of orbits on $G$ under the action of ${\rm Aut}(G)$. -It happens that there are infinitely many non-abelian finite simple groups -$G$ such that $\eta(G) = 5$. For example, this holds for the groups -${\rm PSL}(2,p)$ for prime numbers $p \geq 7$. -In contrast, is it true that $G = {\rm A}_5$ is the only non-abelian finite -simple group such that $\eta(G) = 4$? -- -GAP computations suggest this may be true. -Remark: I know that among the alternating groups, the sporadic groups and -the simple groups of Lie type except for those of type $^2{\rm A}$, ${\rm A}$, ${\rm B}$, -${\rm C}$, ${\rm D}$, $^2{\rm D}$ and $^3{\rm D}_4$, there are no further -groups $G$ with $\eta(G) \leq 4$. -Update (April 30, 2020): Meanwhile this question has 3 deleted answers. - -REPLY [3 votes]: Let me see if I can write down a proof that the answer is YES for most of the $A_n$-groups. The method should work for the rest of the $A_n$-groups and, indeed, all of the other groups you mention. Specifically, I'll prove -Proposition: If $G={\rm PSL}_n(q)$ with $n\geq 5$ and $q\geq 2$, then $\mu(G)>4$. -Proof: In what follows, I'll write $q=p^f$ where $p$ is a prime, $f$ a positive integer. Then $|{\rm Aut}(G)|/ |{\rm PGL}_n(q)|=2f$. Recall that a primitive prime divisor of $p^{df}-1$ is a prime that divides $p^{df}-1$ but not $p^k-1$ for any integer $kdf$. It's written down as Lemma 2.7 of a paper of mine with Azad and Britnell. This is important because it means that primitive prime divisors of $p^{df}-1$ do not divide $2f$ provided $df>1$. -So now our job is to find five elements, $g_1,\dots, g_5\in G$, with different orbit sizes under ${\rm Aut}(G)$. Write $o(g_i)$ for the the orbit size of $g_i$. In what follows we note down what makes each $o(g_i)$ definitely different to the others. - -Let $g_1=1$. Then $o(g_1)=1$. -Let $g_2$ be central in a Sylow $p$-subgroup of $G$. Then $o(g_i)$ is not divisible by $p$. -Let $g_3$ be an element whose centralizer is a maximal torus in ${\rm PGL}_n(q)$ of size $\frac{q^n-1}{q-1}$. Then $o(g_3)$ is not divisible by a ppd of $q^n-1$. -Let $g_4$ be an element whose centralizer is a maximal torus in ${\rm PGL}_n(q)$ of size $q^{n-1}-1$. Then $o(g_4)$ is not divisible by a ppd of $q^{n-1}-1$. -Let $g_5$ be an element whose centralizer is a maximal torus in ${\rm PGL}_n(q)$ isomorphic to $(q^{n-2}-1)\times(q-1)$. Then $o(g_5)$ is not divisible by a ppd of $q^{n-2}-1$. - -Some remarks: - -the existence of tori of this size can be seen directly, but is written down explicitly in the paper of Buturlakin and Grechkoseeva. -The fact that a ppd of $q^{n-2}-1$ does not divide $q^n-1$ follows from the fact that $n\geq 5$ and the fact that $gcd(q^{n-2}-1, q^n-1)$ divides $q^2-1$. Similarly for the other pairs of ppds. -We are implicitly using the lower bound on ppd's used above -- so that the action of $|{\rm Out}(G)|$ doesn't mess things up. -In theory one should check what happens when $(q,n)\in\{(2,6), (2,7), (2,8)\}$ -- here Zsigmondy's theorem fails for one of the tori mentioned above and a ppd does not exist. But I've excluded $q=2$ so this doesn't arise (see next comment). -Finally the fact that there really are elements that are centralized by these maximal tori is a fairly straightforward eigenvalue argument. The only problem arises when $q=2$ and we have the torus $(q^{n-2}-1)\times (q-1)$, hence I've excluded $q=2$ from the statement of the theorem. QED - -Extra remarks - -Dealing with $q=2$ should be easy. For instance if $n$ is odd, then you can substitute that torus of size $(q^{n-2}-1)\times (q-1)$ with one of size $(q^{n-2}-1)\times (q+1)$ (maybe this will work for $n$ even too but I haven't checked). -Likewise $n=2,3,4$ can be done by hand (I guess). -The whole argument carries over to the unitary groups just by changing a few signs in the size of the tori. The other classical groups will require that you choose tori of different sizes.<|endoftext|> -TITLE: Does a finite morphism to $\mathbb{P}^n$ necessarily split at some height one point in the etale locus? -QUESTION [6 upvotes]: Does there exist a finite morphism $\pi\colon X\to \mathbb{P}_{\mathbb{C}}^n$, that does not admit a rational section along any prime divisor $D\subset\mathbb{P}^n$ in the locus where $\pi$ is etale? - -REPLY [3 votes]: Let $f:X\to Y$ be a finite surjective morphism of smooth (projective) varieties over complex numbers. Let $B\subset Y$ be the branch locus and let $q\in Y-B$. Let $f^{-1}(q)=\{p=p_1,\ldots, p_n\}$, where $n=\deg f$. Take $D\subset X$, a smooth divisor passing through $p$, but not passing through $p_i, i>1$. Let $E=f(D)$. Then, one can check that $E$ is smooth at $q$ and $f:D\to E$ is birational. This gives on an open set of $E$ the section you desire (or not desire, as the question is framed).<|endoftext|> -TITLE: Analysis of solutions to a nonlinear ODE -QUESTION [6 upvotes]: Consider the following ODEs: -$\phi^2=\phi''\sqrt{1-\phi'^2}$, or $\phi^2=-\phi''\sqrt{1-\phi'^2}$. -Is there any theory (e.g. comparison theorems) which analyzes solutions of the above ODEs? I am only interested in nonnegative solutions, i.e. $\phi\geq 0$. Actually one can write down a solution $\phi(t)=\cos t,\ -\frac{\pi}{2}\leq t\leq \frac{\pi}{2}$ to the second equation. I want to know whether one can find explicit solutions to the first equations, as well as other solutions to the second one (of course excluding those coming from time translations of $\cos t$). If one can not find explicit formula for the solution, I am also interested in asymptotic properties/existence time of the solution. In particular, I want to know if there exist solutions defined on the entire real line. - -REPLY [15 votes]: Edited on May 2, 2020: The OP pointed out that I had not addressed a special case (namely $C=1$ below), so I am amending my answer to address this and reorganizing so that the $C=1$ case gets addressed naturally when it comes up. — RLB -We can assume that $\phi$ is not constant, since the only constant solution is $\phi\equiv0$. -Thus, assume that the solution has $\phi'\not=0$ on some interval $I$. Multiply the equation $\phi^2 = \pm\phi''\sqrt{1-\phi'^2}$ by $\phi'$, then integrate both sides to get that there is a constant $C$ such that -$$ -\phi^3 = C^3 \mp (1-\phi'^2)^{3/2}. -$$ -The case $C=0$ corresponds to $\phi^2 = 1-\phi'^2$, which, since $\phi'$ is assumed nonzero on $I$ implies that $\phi = \cos(t-t_0)$ for some constant $t_0$. We can thus set $C=0$ aside and assume that $C\not=0$. -We have $C^3{-}1\le \phi^3\le C^3{+}1$, so $C\ge -1$, otherwise there cannot be any nonnegative solutions. If $C = -1$, then $\phi\le0$ and, since we are only interested in non-negative solutions, the only solution in this case is $\phi\equiv0$, so we can assume henceforth that $C>-1$. -Both equations (with either sign) can be studied as special cases of the polynomial differential equation -$$ -(\phi^3-C^3)^2 - (1 - \phi'^2)^3 = 0, -$$ -so that is what we will do. Set $\phi^3-C^3 = u^3$ where $|u|\le 1$ -and note that this implies $(\phi')^2 = 1-u^2$. We then have -$$ -\pm 1 = \frac{u^2u'}{(1-u^2)^{1/2}(C^3+u^3)^{2/3}}, -$$ -so -$$ -t_1-t_0 = \pm\int_{u(t_0)}^{u(t_1)} -\frac{u^2\,du}{(1-u^2)^{1/2}(C^3+u^3)^{2/3}}. -$$ -For $C\not=0,1$ (remember that $C>-1$), the integral -$$ -\int_{-1}^1\frac{u^2\,du}{(1-u^2)^{1/2}(C^3+u^3)^{2/3}} -= \int_{-1}^1\frac{u^2\,du}{(1-u^2)^{1/2}(C+u)^{2/3}(C^2-Cu+u^2)^{2/3}} -$$ -converges. However, when $C=1$, the denominator contains a factor of $(1+u)^{1/2+2/3} = (1+u)^{7/6}$, so the integral diverges at $u=-1$. -In the case $C=1$, we can write down a parametrization of the graph $\bigl(t,\phi(t)\bigr)$ in the form $\bigl(t(v),\phi(t(v))\bigr)$ where -$|v|<\sqrt2$ with -$$ -\phi(t(v)) = \bigl(1+(1-v^2)^3\bigr)^{1/3} -\quad\text{and}\quad -t(v) = \int_0^v \frac{2(1-s^2)^2\,ds}{(2-s^2)^{7/6}(1-s^2+s^4)^{2/3}}. -$$ -Note that, while $t$ is a strictly increasing function of $v$ and $|t|\to\infty$ as $|v|\to\sqrt2$, $t'(\pm1) = 0$, and $\phi$ is not -a smooth function of $t$ where $v = \pm 1$. (It is, however, continuously once differentiable there, see below.) -Meanwhile, when $C\not=1$, a solution $\phi$ exists for all time and is periodic, as $\phi$ oscillates between $(C^3{-}1)^{1/3}$ and $(C^3{+}1)^{1/3}$. The period of $\phi$ is -$$ -p(C) = 2\int_{(C^3{-}1)^{1/3}}^{(C^3{+}1)^{1/3}}\frac{d\xi}{\sqrt{1-(\xi^3-C^3)^{2/3}}} -= \int_{-1}^1\frac{2u^2\,du}{(1-u^2)^{1/2}(C^3+u^3)^{2/3}}. -$$ -Of course, $p(0) = 2\pi$ and $p(C)$ has a series expansion $2\pi\,C^{-2}+\tfrac{175}{576}\pi\,C^{-8}+\cdots$ when $C>1$. -Near a minimum at $t=t_0$, the solution has a series expansion of the form -$$ -\phi(t) = (C^3{-}1)^{1/3}\left(1+\tfrac12(C^3{-}1)^{1/3}(t-t_0)^2+\tfrac1{24}(C^3+1)(C^3{-}1)^{2/3}(t-t_0)^4 + \cdots\right) -$$ -Near a maximum at $t=t_1$ the solution has a series expansion of the form -$$ -\phi(t) = (C^3{+}1)^{1/3}\left(1-\tfrac12(C^3{+}1)^{1/3}(t-t_1)^2-\tfrac1{24}(C^3-1)(C^3{+}1)^{2/3}(t-t_1)^4 + \cdots\right) -$$ -Note that, when $C\not=0$, $\phi$ is not $C^2$ when it attains the intermediate value $C$. In fact, if $\phi(t_2) = C$ -and $\phi'(t_2) = 1$, then $\phi$ has a series expansion in powers of $(t{-}t_2)^{1/3}$: -$$ -\phi(t) = C +(t{-}t_2) - \frac{C^{4/3}}{10}\,\bigl(3(t{-}t_2)\bigr)^{5/3} -- \frac{C^{8/3}}{280}\,\bigl(3(t{-}t_2)\bigr)^{7/3} + \cdots. -$$ -Meanwhile, if $\phi(t_3) = C$ -and $\phi'(t_3) = -1$, then $\phi$ has a series expansion in powers of $(t{-}t_3)^{1/3}$: -$$ -\phi(t) = C -(t{-}t_2) + \frac{C^{4/3}}{10}\,\bigl(3(t{-}t_2)\bigr)^{5/3} -+ \frac{C^{8/3}}{280}\,\bigl(3(t{-}t_2)\bigr)^{7/3} + \cdots. -$$ -Finally, note that $\phi''<0$ when $C<\phi<(C^3+1)^{1/3}$ while $\phi''>0$ when $(C^3-1)^{1/3}<\phi -TITLE: A second order nonlinear ODE -QUESTION [7 upvotes]: In my research (in differential geometry) I recently came across the following nonlinear second order ode: -$$\frac{f''(x)}{f'(x)}-\frac{2}{x}+\frac{f'(x)+1}{2f(x)-x-1}+\frac{f'(x)-1}{2f(x)+x}=0$$ -It actually arose from the symmetry reduction of some pde. I know from an analysis of the equation that there exists a 1-parameter family of solutions. Moreover I also know two explicit solutions; -$$f(x)=x+\frac{1}{2}$$ -$$f(x)=\frac{1}{4}+\frac{1}{4}(1+3x)\sqrt{(1+2x)}$$ -The existence of these 2 solutions, expressible in terms of elementary functions, makes me wonder if one can in fact find more (if not all) explicit solutions to this ode. Note that both these solutions are well defined at $x=0$, although the ode itself is singular at that point! It is not too hard to show that any solution well-defined at $x=0$ requires $f(0)=\frac{1}{2}$ and $f'(0)=1$. -As far as I am aware there are no standard tricks for these type of fully nonlinear odes. I have been trying to simplify the ode by various substitutions but without any success. -I was hoping that someone might be able to spot a clever transformation, or even argue that it is impossible to find any other explicit solutions. I would also be interested to know of any references where such a class of ode that might have been studied. -This also leads me to ask if there is any general theory known about when can a solution to an ode (say of second order) be expressed in terms of elementary functions, or is it just a case-by-case study? Thanks! - -REPLY [6 votes]: This ODE has some very interesting properties. If one clears fractions and writes it out as -$$ -x(x+2y)(x-2y+1)\,y'' = (4x^2-8y^2+3x+4y)\,y' + x(4y-1)\,(y')^2, -\tag1 -$$ -one recognizes this as the equation for geodesics of a projective connection in the complement $D$ (which has $7$ components) of the three lines $x = 0$, $x+2y=0$, and $x-2y+1=0$ in the $xy$-plane. Moreover, because the right hand side of (1) has no terms of degree $0$ or $3$ in $y'$, it follows that the lines $x=x_0$ and $y=y_0$ are geodesics of this projective connection in $D$ and should be regarded as 'solutions' of the equation. This is probably most evident if one writes the equation in parametric form for a curve $\bigl(x(t),y(t)\bigr)$, in which case, the equation becomes -$$ -x(x{+}2y)(x{-}2y{+}1)\,\bigl(\dot y\,\ddot x-\dot x\,\ddot y\bigr) + x(4y{-}1)\,\dot x\,\dot y^2 + (4x^2{-}8y^2{+}3x{+}4y)\,\dot x^2\,\dot y =0 -\tag2 -$$ -Contrary to what the OP claims, there is a $2$-parameter family of solutions that are regular at $x=0$. If one looks for a formal power series solution in the form -$$ -y(x) = a_0 + a_1\,x + a_2\,x^2 + a_3\, x^3 + \cdots,\tag3 -$$ -then one finds, by examining the three lowest terms in the equation, that one must have either -$$ -(i)\ \ a_1 = a_2 = 0,\qquad (ii)\ \ a_0 = 0,\ a_1 = -1, -\quad\text{or}\quad (iii)\ \ a_0=\tfrac12,\ a_1 = 1. -$$ -Since $y(x)$ is a solution if and only if $\tfrac12 - y(x)$ is a solution, the second and third cases are essentially the same, so I will treat only the first two cases henceforth. -In the first case, one finds that there is a formal power series solution of the form -$$ -y(x) = \tfrac14(1{+}a) + \frac{(a^2{-}1)b}{12} x^3 -\frac{b}{4}\,x^4 - \frac{b}{5}\,x^5 - \frac{a(a^2{-}1)b^2}{72}\,x^6 + \cdots + p_k(a,b)\,x^k + \cdots,\tag4 -$$ -where $p_k(a,b)$ is a (unique) polynomial in constants $a$ and $b$. Moreover, this series has a positive radius of convergence for each $(a,b)$. [Proofs that these and similar series listed below have positive radii of convergence can be based on techniques in the book Singular Nonlinear Partial Differential Equations by R. Gérard and H. Tahara.] Note that the symmetry $y(x)\mapsto \tfrac12 - y(x)$ corresponds to the symmetry $(a,b)\mapsto (-a,-b)$. -In the second case (and, similarly, via the symmetry $y\mapsto \tfrac12 - y$, the third case), one finds that there is a formal power series solution -$$ -y(x) = - x + \frac{b}{2}\,x^2 - \frac{b}{5}\,x^3 + \frac{b(b{+}3)}{10}\,x^4 - - \frac{4b(13b{+}25)}{175}\,x^5 + \cdots + q_k(b)\,x^k + \cdots, -\tag5 -$$ -where $q_k(b)$ is a (unique) polynomial in $b$ of degree at most $\tfrac12 k$. This series has a positive radius of convergence for every $b$. The value $b=0$ gives the solution $y(x) = -x$ and the value $b=-5/4$ gives the solution $y(x) = \tfrac14 - \tfrac14(1+3x)(1+2x)^{1/2}$. Note that (4) with $a=-1$ and (5) give two distinct $1$-parameter families of solutions passing through the point $(x,y)=(0,0)$, where the two singular lines $x=0$ and $x+2y=0$ meet. -As for analytic solutions meeting the singular line $x+2y=0$, there exist two distinct $2$-parameter families of series solutions: The first is given in parametric form by -a formal power series -$$ -\begin{aligned} -x(t) &= a + a(2a{+}1)\,t\,,\\ -y(t) &= -\frac{a}{2} + a(2a{+}1)b\,t^2\left(1 + \frac{2(5a{-}4b{+}2)}{3}\,t +\cdots + p_k(a,b)\,t^k + \cdots \right), -\end{aligned} -\tag6 -$$ -where $p_k(a,b)$ is a (unique) polynomial in $a$ and $b$ and where the $y$-series in $t$ has a positive radius of convergence for every $(a,b)$. The second $2$-parameter family can be written in the form -$$ -\begin{aligned} -x(t) &= a + a(2a{+}1)\,t^3\,,\\ -y(t) &= -\frac{a}{2} + (2a{+}1)\,t^2\left(b + a\,t + \frac{2b^2}{5}\,t^2 +\cdots + q_k(a,b)\,t^k + \cdots \right), -\end{aligned} -\tag7 -$$ -where $q_k(a,b)$ is a (unique) polynomial in $a$ and $b$ and where the $y$-series in $t$ has a positive radius of convergence for every $(a,b)$. Interestingly, these solutions with $b\not=0$ have a cusp singularity at $t=0$, while, when $b=0$, only the terms involving $t^{3k}$ remain, so that $x(t)$ and $y(t)$ are analytic functions of $t^3$. -Note, however, that these two series solutions degenerate at the special values $a=0$ and $a=-\tfrac12$. The value $a = 0$ corresponds to the point $(x,y)=(0,0)$, where the singular lines $x=0$ and $x+2y=0$ cross, while the value $a=-\tfrac12$ corresponds to the point $(x,y) = (-\tfrac12,\tfrac14)$, where the singular lines $x+2y=0$ and $x-2y+1=0$ cross. -Finally, through the singular point $(x,y) = (-\tfrac12,\tfrac14)$, where the singular lines $x+2y=0$ and $x-2y+1=0$ cross, there are two convergent series solutions with one parameter: The first is -$$ -\begin{aligned} -x(t) &= -\frac{1}{2} + t\,,\\ -y(t) &= +\frac{1}{4} + b\,t^3 -3b\,t^4 + \cdots + f_k(b)\,t^k + \cdots , -\end{aligned} -\tag8 -$$ -where $f_k(b) = -f_k(-b)$ is a polynomial in $b$. The second series is -$$ -\begin{aligned} -x(t) &= -\frac{1}{2} + b\,t^2 + \frac{b^2(5b{+}32)}{16}\,t^4 + \cdots + g_k(b)\,t^{2k} + \cdots\,,\\ -y(t) &= +\frac{1}{4} + t\, , -\end{aligned} -\tag9 -$$ -where $g_k(b)$ is a polynomial in $b$. Note that the value $b=2$ in this latter series corresponds to the known solution(s) represented by solving $(y-\tfrac14)^2-(1+3x)^2(1+2x) = 0$ for $y$ as a function of $x$. -It is very interesting that through every point of the $xy$-plane, there passes at least one $1$-parameter family of solution curves, and, along two of the singular lines, there can be two distinct $1$-parameter families of solution curves. -One more interesting feature should be noted: Because the equation defines a projective structure on $D$, each solution curve in $D$ comes equipped with a canonical projective structure, i.e., a 'developing map' to $\mathbb{RP}^1$ that is unique up to linear fractional transformation and provides a local parametrization of the curve. This developing map extends analytically across the points where such a curve crosses one of the three singular lines, but the developing map is no longer a local diffeomorphism at such places; its differential vanishes to second or third order at such places.<|endoftext|> -TITLE: Can a locally presentable category have a proper class of accessible localizations? -QUESTION [6 upvotes]: Question: What is an example of a locally presentable category $\mathcal C$ such that there exists a proper class of accessible localizations $(\mathcal C \to \mathcal D_i)_{i < ORD}$? -In other words, $(D_i)_{i < ORD}$ should be a proper class of full reflective subcategories of $\mathcal C$ which are accessibly embedded in $\mathcal C$. -I'm also interested in the infinity-categorical setting, though I suspect it doesn't make much difference. - -REPLY [8 votes]: A limit closure of a set of objects of a locally presentable category $\mathcal K$ is reflective. In this way one gets an increasing chain of reflective full subcategories of $\mathcal K$. If this chain stops $\mathcal K$ has a cogenerator. Since a category of groups does not have a cogenerator, it has a proper class of reflective full subcategories. These subcategories are accessibly embedded under Vopěnka's principle. I do not expect that Vopěnka's principle is needed but, at this moment, I do not see how to avoid it.<|endoftext|> -TITLE: Preserve unbounded sets between different cofinality -QUESTION [5 upvotes]: Working in ZFC, let $\kappa,λ$ be cardinals with $\kappa>λ$, and assume that $\kappa$ is regular. -We say that a function $F:\kappa^n→λ$, for some finite $n$, is preserving unbound, if for all $a⊆\kappa$ unbounded, we have that the closure of $F[a^n]$ under $F$ is unbounded in $λ$. -Is there always preserving unbound function? -For $\mbox{cof}(λ)=ω$ this is easy: take $n=1$ and map every element from $\kappa\setminusλ$ into some some element of $λ$, and map an element from $λ$ to somewhere greater using some fixed cofinal sequence of $λ$, i.e. fix some cofinal $(λ_i\mid i\in\omega)$, and for $k$ be the minimal $k$ such that $x∈λ_k$, map $x$ to some element of $\lambda_{k+1}\setminusλ_k$. -The problem arise when $\mbox{cof}(λ)>ω$. In this case we cannot "climb" a cofinal. -In this case for $n=1$ there are clearly no preserving unbound function. Indeed if $F:\kappa→λ$ be any function, then there exists some $x∈λ$ such that $F^{-1}(x)$ is unbounded in $\kappa$. -A special case is that $λ$ is regular, in this case preserving unbound becomes: - -A function $F:\kappa^n→λ$, for some finite $n$, such that if for all $a⊆\kappa$ unbounded, we have $F[a^n]$ is unbounded in $λ$. - -Indeed if $F[a^n]$ is bound, then $|F[a^n]|<λ$, then $|\bigcup\{F[a^n],F[F[a^n]],...\}|<λ$, so the closure is also bounded. -The I am most interested in the special cases where $λ$ is indeed regular and that $\kappa=λ^+$, in particular $\kappa=ω_{k+1},λ=ω_k$ for finite $k$ - -REPLY [4 votes]: In the case that $\kappa=\lambda^+$, I claim that $n=2$ suffices to produce such a function, as follows. Fix, for each ordinal $\beta<\kappa$ a one-to-one function $f_\beta:\beta\to\lambda$. Then define $F:\kappa^2\to\lambda$ by setting $F(\alpha,\beta)=f_\beta(\alpha)$ if $\alpha<\beta$. (It doesn't matter how you define $F(\alpha,\beta)$ for $\alpha\geq\beta$.) Now consider any unbounded $A\subseteq\kappa$; I want to show that $F(A^2)$ is unbounded in $\lambda$, for which it suffices to show that $F(A^2)$ has cardinality $\lambda$. Since $A$ is unbounded in $\kappa$, it has a $\lambda$-th element $\beta$. As $\alpha$ ranges over the $\lambda$ members of $A$ that are $<\beta$, $F(\alpha,\beta)=f_\beta(\alpha)$ takes $\lambda$ distinct values, because $f_\beta$ is one-to-one. And all of these $\lambda$ values are in $F(A^2)$ because $\beta$ and all the $\alpha$'s under consideration are in $A$. -I'm reasonably sure a similar argument (iterating this idea) will show that $n=q+1$ suffices when $\kappa$ is the $q$-fold successor of $\lambda$. Unfortunately, I don't have time right now to write down the argument.<|endoftext|> -TITLE: Torsion group with finitely many elements of order 2 but infinitely many elements of order 4 -QUESTION [7 upvotes]: Does there exists a group $G$ satisfying all the following conditions? - -$G$ is finitely generated, -$G$ is of bounded torsion (has finite exponent), -$G$ has finitely many elements of order $2$, -$G$ has infinitely many elements of order $2^l$ for some $l$ (for example $l=2$). - -By 1. and 2. $G$ is a quotient of a free Burnside group. -Near-examples of such groups violating only Condition 2. are generalized dicyclic group $Dic(A,y)$, where $A$ is a finitely generated abelian group. -References on somehow related problems are welcome. - -REPLY [6 votes]: Here is an example with infinitely many $4$-torsion elements and only one $2$-torsion element. (The smallest number of generators I can do is $4$, and the exponent is pretty big and not a power of $2$.) -Example. Pick $n$ odd and $d \geq 2$ such that $B(d,n)$ is infinite. Let $p$ be an odd prime. Set -$$A = \mathbf Z/2 \oplus \bigoplus_{g \in B(d,n)} (\mathbf Z/p)e_g,$$ -with distinguished element $y = (1,0,\ldots)$ or order $2$. Then the generalised dicyclic group $\operatorname{Dic}(A,y) = A \amalg Ax$ satisfies all criteria except 1: - -$\operatorname{Dic}(A,y)$ is not finitely generated because it has an index $2$ subgroup $A$ that is not finitely generated. -Every element of $\operatorname{Dic}(A,y)$ is killed by $4p$. -The only element of order $2$ is $y \in A$. -Every element in $Ax$ has order $4$, and $Ax$ is infinite by assumption. - -Finally, the group $B(d,n)$ acts on $A$ fixing $y$ by $(g,e_h) \mapsto e_{gh}$. Thus this extends to an action on $\operatorname{Dic}(A,y)$, and we take $G$ to be the semidirect product -$$G = \operatorname{Dic}(A,y) \rtimes B(d,n).$$ -Then all criteria are satisfied: - -If $x_1,\ldots,x_d$ are the standard generators of $B(d,n)$, then $x, e_1, x_1,\ldots,x_d$ generate $G$. Indeed, they generate the quotient group $B(d,n)$, hence we get all elements $ge_1g^{-1} = e_g$, hence we get everything. -Since $B(d,n)$ has exponent $n$ and $\operatorname{Dic}(A,y)$ has exponent $4p$, we conclude that $G$ has exponent dividing $4pn$. -and 4. Since $B(d,n)$ has odd exponent, all $2$-power torsion happens in $\operatorname{Dic}(A,y)$. - -So $G$ is an example. $\square$ -Remark. I have no idea if there are also examples of $2$-power exponent. - -REPLY [2 votes]: Here's an elaboration on R. van Dobben de Bruyn's answer. -Let $A$ be an abelian group. Define the group $\mathrm{Di}(A)$ as follows: (a) consider the direct product $A'=A\times C_2$, denoting $y$ the nontrivial element of $C_2$. (b) perform the semidirect product $A''=C_4\ltimes A$, with $\pm$-action. Denote by $z$ the element of order 2 of the acting $C_4$: it remains central in $A''$, and so does $y$. (c) Obtain $\mathrm{Di}(A)$ (generalized dicyclic group on $A$) by modding out $A''$ by the central subgroup of order 2 $\langle z^{-1}y\rangle$. Still denote by $y$ the image of $y$ in $\mathrm{Di}(A)$, it it central of order $2$. -Note that $\mathrm{Di}(A)/\langle y\rangle$ is the dihedral product $C_2\ltimes_\pm A$. -In $A''$, denoting by $t$ a generator of $C_4=\{1,t,z,t^{-1}\}$ and writing $A$ additively, we have $(t^{\pm},a)^2=(z,0)$, $(z,a)^2=(1,a)^2=(1,2a)$ for $a\in A'$. In particular, $\eta=(z,y)$ (which is killed in $\mathrm{Di}(A)$ is not a square. Hence the elements of order $\le 2$ in $\mathrm{Di}(A)$ are the images of elements of order $2$ in $A''$, which are the elements of the form $(1,a),(z,2a)$ with $2a=0$. In particular, if $A$ has no element of order $2$, these elements are $(1,0),(1,y),(z,0),(z,y)$, which in $\mathrm{Di}(A)$ is reduced to $\{1,y\}$. That is, if $A$ has no element of order $2$ then the only element of order $2$ in $\mathrm{Di}(A)$ is $y$. -Also this shows that all elements $(t^\pm,a)$ map to elements of order $4$ in $\mathrm{Di}(A)$, which therefore has infinitely many elements of order $4$ if $A$ is an arbitrary infinite abelian group. -Now the construction $A\mapsto\mathrm{Di}(A)$ is clearly functorial under group isomorphisms. Hence, if $\Gamma$ acts on $A$ by automorphisms, then it naturally acts on $\mathrm{Di}(A)$ by automorphisms: in steps: (a) extend in the trivial way the action to $A'=A\times C_2$, then (b) extend in the trivial way to $C_4\ltimes A'$ (acting trivially on $C_4$): this works because the $C_4$-action, by $\pm$, commutes with the $\Gamma$-action on $A'$; finally this action fixes $\eta=(z,y)$ and hence passes to the quotient to an action on $\mathrm{Di}(A)$, defining a semidirect product $\mathrm{Di}\rtimes\Gamma$. -From what's preceding, we immediately get: if $\Gamma$ has no element of order $2$, then the only element of order $2$ in $\mathrm{Di}(A)\rtimes\Gamma$ is $y$; if $\Gamma$ is infinite then it contains $\mathrm{Di}(A)$ hence has infinitely many elements of order $4$. -Finally we can choose $A=C_n^{(\Gamma)}=\bigoplus_{\gamma\in\Gamma}C_n$ for some odd $n>1$, and choose $\Gamma$ of finite odd exponent $q>1$. Here $n$ and $q$ are unrelated, can be chosen equal or not (they could be chosen even for the construction but then this will produce infinitely elements of order $2$). Then the resulting group $$G=\mathrm{Di}(C_n^{(\Gamma)})\rtimes \Gamma$$ works: it has a single element of order $2$, infinitely of order $4$, and has exponent dividing $nq$. Actually modding out by $\langle y\rangle$, the resulting group admits, as subgroup of order $2$, the standard wreath product $C_n\wr \Gamma$ (in particular, all elements of order $4$ in $G$ lie in the nontrivial coset of the unique subgroup of index $2$). -(In general— arbitrary abelian group $A$, arbitrary $\Gamma$-action on $A$, we always get this subquotient killing the center $\langle y\rangle$ and passing to a subgroup of index $2$, which yields the semidirect product $A\rtimes\Gamma$. For instance, if we want $G$ to have Kazhdan's Property T, the permutational action is hopeless, but possibly some choice of $\Gamma$-module works, namely we need $A\rtimes\Gamma$ to have Kazhdan's Property T.)<|endoftext|> -TITLE: Is the acyclic chromatic number bounded in terms of the book thickness? -QUESTION [5 upvotes]: ISGCI says that the chromatic number of a graph is upper bounded in terms of the book thickness. -https://www.graphclasses.org/classes/par_32.html -This can be improved by saying that the book thickness bounds the degeneracy. -A further improvement would be that the book thickness bounds the acyclic chromatic number. -Is it true? Is there a reference? - -REPLY [5 votes]: I believe that "book thickness bounds the acyclic chromatic number" was -established in this paper: - -Dujmovic, Vida, Attila Pór, and David R. Wood. "Track layouts of graphs." Discrete Mathematics and Theoretical Computer Science 6, no. 2 (2004). - arXiv abs. - -In the Abstract they say, - -"As corollaries we - prove that acyclic chromatic number is bounded by both queue-number and stack-number." - -And then later (Section 5), they say, - -"Note that stack-number is also called page-number and book-thickness." - -Not sure if this is relevant to your quesiton, but -the acyclic chromatic number is not bounded by geometric thickness -$\overline{\theta}(G)$.<|endoftext|> -TITLE: Doubly periodic 4 color theorem? -QUESTION [23 upvotes]: Let $G$ be a graph embedded (without crossings) on a torus $T$. It's fairly well known that this implies the chromatic number of $G$ is at most 7. If I lift $G$ to the universal cover of $T$, we get a doubly periodic planar graph $\tilde{G}$ and of course the four color theorem tells us there is a four coloring of $\tilde{G}$. -With a little work I can improve this slightly to say that for any such $G$ there is a finite cover $\widehat{T}$ such that the corresponding cover $\widehat{G}$ is four colorable. My question is: Can this be done uniformly in $G$? If so, how small can we take the cover? -Concretely: Does there exist a covering map $T' \to T$ such the pull back to $T'$ of any graph embedded on $T$ can be properly four colored? Which covers work and what is the minimal degree of such a cover? -I was especially interested in the case where $T = \mathbb{R}^2/\mathbb{Z}^2$ and $T'$ was the 4-fold cover $\mathbb{R}^2/(2\mathbb{Z})^2$ but would be interested in hearing about any case. -EDIT: Since I thought this was a fun question I thought about it more and did some more searching through literature. Here are my current best partial results: -1) For a surface $\Sigma$ of genus $g$ there exists a degree $36^g$ cover such that any graph embedded on $\Sigma$ becomes $6$-colorable when pulled back to the cover. -2) For genus 1, any graph embedded on a torus becomes $5$-colorable when pulled back to the $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ cover described above. - -REPLY [10 votes]: As Michael Klug points out in the comments, I've thought about related questions before. I'll make a few comments on the question. -Firstly, the usual reduction allows one to consider triangulations on a surface: if a graph $G$ does not induce a triangulation of $\Sigma$, then we can complete it to a triangulation $G'$ so that if $G'$ (or a cover $\hat{G'}$ induced by a cover $\hat{\Sigma}$) is 4-colorable then so is $G$ (or $\hat{G}$). -So let's assume that $G$ induces a triangulation of $\Sigma$. Then the dual graph $G^*$ (with respect to the embedding in $\Sigma$) is a cubic graph. If $G^*$ is 3 edge-colorable (i.e. has a Tait coloring), then one can see that a $\mathbb{Z}/2\times \mathbb{Z}/2$-cover $\hat{\Sigma}\to \Sigma$ will give a lift of $G$ which is 4-colorable. To prove this, identify the three colors with the non-zero elements of the Klein 4-group $V=\mathbb{Z}/2\times \mathbb{Z}/2$. Then coloring the vertices of $G$ corresponds to coloring the faces of $G^* \subset \Sigma$. If we color one face of $G^*$ by $0\in V$, then each time we cross an edge of $G$, we change the color by adding the element of $V$ corresponding to the edge coloring. This is locally well-defined near a vertex, but globally might have holonomy in $V$. So passing to a 4-fold cover $\hat{\Sigma}\to \Sigma$ induced by this holonomy, we get a pulled-back graph $\hat{G}$ which is 4-colorable. (In the planar case, there is no holonomy, and hence Tait's observation that Tait colorings suffice). -Thus it suffices to consider 3-edge colorings of cubic graphs in $\Sigma$. The Snark theorem implies that if the graph $G^*$ is not 3-edge colorable, then there is a Petersen minor (that is, a copy of the Petersen graph embedded topologically in $G^*$). The Petersen graph is non-planar, so must be embedded in an essential way in $\Sigma$ (not isotopic into a disk). Hence any Petersen subgraph of $G^*$ will not lift to some 2-fold cover of $\Sigma$. However, passing to a cover to which no Petersen subgraph lifts, there may be new Petersen subgraphs of $\hat{G^*}$ created. Nevertheless, one can ask if there is a finite cover $\hat{\Sigma}\to \Sigma$ such that the preimage of any embedded cubic graph in $\Sigma$ is not a Snark? Seems implausible, but it is a natural question to ask when thinking about virtual Tait coloring. -One can weaken the condition of Tait coloring, allowing passage to a finite-sheeted cover. If a cubic graph $G^*$ has a perfect matching (also called a 1-factor, a degree 1 regular subgraph spanning the vertices), then the complementary subgraph is a 2-factor, i.e. a regular subgraph of degree 2 containing every vertex, homeomorphic to a union of circles, each component a cycle graph . If the 2-factor is also bipartite (2-colorable, r every component has an even number of edges), then we may 2-color the 2-factor and use a third color for the 1-factor to get a Tait coloring of $G^*$. Then we can look for a 2-factor $C\subset G^* \subset \Sigma$ such that every non-bipartite component of $C$ is a non-trivial curve on $\Sigma$. In this case, we can pass to a $2^{2g}$-fold cover in which ever non-separating curve has each component of the preimage an even-index cover, and every separating essential curve has preimage components non-separating, and repeat, to get a finite cover for which the preimage of every essential curve is an even index cover on each component. Then the preimage of a 2-factor with the above properties will be a bipartite 2-factor, and hence the preimage graph will be 3-colorable (and a further 4-fold cover will give a 4-colorable dual triangulation). -One knows that every bridgeless cubic graph has a perfect matching (or 1-factor, and hence a 2-factor), known as Petersen's theorem. One could try to modify the proof to try to show that a graph $G^*\subset \Sigma$ has a 2-factor with odd cycles all essential. But I didn't see how to do this. In any case, it seems possibly easier to find a controlled cover of $\Sigma$ where the preimage of every cubic graph has a 2-factor with essential odd cycles. -Another special case is triangulations of even degree. Then we can try to 3-color the vertices. Once one 3-colors the vertices of a triangle, there is a unique way to continue the coloring, well-definied locally around a vertex because of the even degree hypothesis. This may have non-trivial holonomy, but passing to an $S_3$-cover (of index 6), we get a preimage which is a 3-colorable graph. This works e.g. for $K_7\subset T^2$. -Ultimately, this problem ought to be as hard as the 4-color theorem itself. Given a large graph embedded in a disk, one ought to be able to insert it into a disk on a surface $\Sigma$ of genus $>0$ as a subgraph. Coloring the graph larger graph in a finite-sheeted cover will induce a coloring of the planar graph. So I think one will likely have to use the 4-color theorem or parts of its proof as an essential ingredient in resolving this question. -One reduction I've contemplated is to make the surace the boundary of a handlebody, and pass to the universal cover of the handlebody. The preimage of the boundary is a planar surface, so the preimage of the graph $\tilde{G}$ is 4-colorable. The -space of 4-colorings of $\tilde{G}$ is a closed subset of the Cantor set $4^\tilde{V}$, where $\tilde{V}$ is the vertex set of $\tilde{G}$. The covering translations form a rank $g$ free group. If there is a probability measure on the space of colorings which is invariant under the free group action, then I can show that there is a finite-sheeted cover (induced by a cover of the handlebody) which is 4-colorable, using a theorem of Lewis Bowen. However, I haven't been able to show the existence of such a probability measure (again, this may require non-trivial input from the proof of the 4-color theorem). One could do a similar thing with 2-factors of cubic graphs, where every contractible cycle is bipartite, and ask for an invariant probability measure on these. This approach, if it worked, would likely not give a uniform finite-sheeted cover.<|endoftext|> -TITLE: Are Conway's combinatorial games the "monster model" of any familiar theory? -QUESTION [25 upvotes]: This is related to this question about a "mother of all" groups, and so seemed like it'd fit in better at MO than MSE. -If I understand the answer to that question correctly, the surreal numbers have a nice characterization as being the "monster model" of the theory of ordered fields (and I think also the real-closed fields), which means that every ordered field embeds into the surreal numbers. In the answer to the question above, Joel David Hamkins gave an interesting example of what the monster model of the theory of groups would look like, which has the property that every possible group is a subgroup of this group (which caused it to be dubbed the "Hamkins' All-Encompassing Group-Like Thing," or I suppose HAEGLT, in the comments). -This question, then, is about Conway's formalization of combinatorial games, of which the surreal numbers are embedded. Conway's games are much more general than the surreal numbers, and have (among other things) the following structure: - -There is a commutative sum of two games (which agrees with the sum on surreal numbers) -For any game, there is an additive inverse (so we have an abelian group) -There is a partial order on the games -There are nilpotent games, such as the star $\{*|*\}$ game of order 2, as seen in Conway's analysis of Nim - -My question is, are the Conway games the monster model of the theory of... well, anything familiar related to the above? Abelian groups? Partially ordered abelian groups? Something else? -To my precise, I am sure there is probably some way to devise some artificial theory that the games are technically a monster model of. What I am wondering is if they are a monster model of some familiar algebraic theory that people use all the time, or perhaps some such theory with just a bit of added structure. Since they generalize the surreals in a fairly "natural" way, it seems intuitive that they might be a monster model of some equally "natural" theory that is more general than that of ordered fields. -EDIT: I previously wrote that the surreal multiplication can also be extended to a commutative product on the entire theory of games, as shown on (page 412 of this book). However, this is apparently not entirely true, as written in the comment below, as there is some subtlety with the equality relation. - -REPLY [28 votes]: In On a conjecture of Conway (Illinois J. Math. 46 (2002), no. 2, 497–506), Jacob Lurie -proved Conway's conjecture that the class $G$ of games together with Conway's addition defined thereon is (up to isomorphism) the unique "universally embedding" partially ordered abelian group, i.e. -for each such subgroup $A$ of $G$ whose universe is a set and any such extension $B$ of $A$, there is an isomorphism $f:B\rightarrow G$ that is an extension of the identity on $A$. The terminology "universally embedding", which is due to Conway, is unfortunate since it is sometimes confused with "universal". For partially ordered abelian groups "universally embedding" implies "universal", but I haven't checked if they are equivalent (though I suspect they're not). For ordered fields the notions are not equivalent; whereas $\mathbf{No}$ is up to isomorphism the unique "universally embedding" ordered field, it is not up to isomorphism the unique universal ordered field (though it is of course universal). I point this out in my paper Number systems with simplicity hierarchies: a generalization of Conway's theory of surreal numbers (J. Symbolic Logic 66 (2001), no. 3, 1231–1258). In that paper I further suggest the terminology universally extending in place of universally embedding. As an example of the potential for confusion using Conway's terminology, I point out (p. 1240) that Conway's terminology led Dales and Wooden (Super-real ordered fields, Clarendon Press, Oxford. 1996, p. 58) to mistakenly claim that $\mathbf{No}$ is up to isomorphism the unique universal ordered field. -Edit (4/30/20): For the sake of completeness it is perhaps worth adding that while Lurie's result is the deeper of the two, David Moews proved that Conway's class of games with addition (but without the order relation) is (up to isomorphism) the unique universally embedding Abelian group. See Moews's The abstract structure of the group of games in Richard J Nowakowski (ed.) More games of no chance, MSRI Publications no. 42, Cambridge University Press, Cambridge, 2002, pp. 49-57.<|endoftext|> -TITLE: Comparing $X+Y$ and $X-Y$ for independent random variables with values in an abelian locally compact group -QUESTION [6 upvotes]: Let $G$ be an abelian locally (separable?) compact group with Haar measure $\mu$. Inspired by the interesting proof of A sum of two binomial random variables : -Let $X$ and $Y$ be $G$-valued independent random variables with $\mu$-density $f_X$ resp. $f_Y$. Let $f_{X \pm Y}$ be the $\mu$-density of $X \pm Y$. Is it always true that -$$\int f_{X-Y}^2 d\mu = \int f_{X+Y}^2 d\mu \ ?$$ -This holds true for $G = \mathbb{Z}$ (here both integrals are finite) -. - -REPLY [6 votes]: The answer is yes. Indeed, let $dx:=\mu(dx)$ for brevity. We have -\begin{align} -I_{X,Y}&:=\iiint f_X(x'+y'-y)f_Y(y)f_Y(y')f_X(x')dx'\,dy\,dy' \\ -&=\iint f_{X+Y}(x'+y')f_Y(y')f_X(x')dx'\,dy' \\ -&=\iint f_{X+Y}(t)f_Y(t-x')f_X(x')dt\,dx' \\ -&=\int f_{X+Y}(t)^2dt. -\end{align} -Also, -\begin{align} -I_{X,Y}&=\iiint f_X(x'-u'+u)f_Y(-u)f_Y(-u')f_X(x')dx'\,du\,du' \\ -&=\iiint f_X(x'+u-u')f_{-Y}(u)f_{-Y}(u')f_X(x')dx'\,du\,du' \\ -&=\iiint f_X(x'+u'-u)f_{-Y}(u)f_{-Y}(u')f_X(x')dx'\,du\,du' \\ -&=I_{X,-Y} =\int f_{X-Y}(t)^2dt, -\end{align} -by what was shown in the previous display. -Thus, -$$\int f_{X+Y}(t)^2dt=\int f_{X-Y}(t)^2dt,$$ -as desired.<|endoftext|> -TITLE: History of the notion of irreducible representation -QUESTION [15 upvotes]: I am looking for the earliest references where the study of irreducible representations appears. There has been many articles and books on the history of representation theory. A fundamental feature of this theory, is that in good situations where one is dealing with a semisimple category, one can decompose objects into simple ones, or here, irreducible representations. My understanding is that the introduction of this circle of ideas is usually credited to Frobenius around the end of the 19th century. -However, the decomposition of tensor products of irreducible representations of $SL_2$ can be found in the article by Paul Gordan "Beweis, dass jede Covariante und Invariante einer binären Form eine ganze Function mit numerischen Coefficienten einer endlichen Anzahl solcher Formen ist." in J. reine angew. Math. 323 (1868), 323-354. -It is written in old fashioned language that can be difficult to decipher, but the Clebsch-Gordan decomposition for $SL_2$ is basically there in Section 2 of that article. One could also ask: when was it realized that it was important and very useful to decompose general representations in terms of irreducibles? Reading the proof in that article, one can only conclude that Gordan was very well aware of that. -Also note that for the irreducible representations for $SL_n$, one can find a description already in the article by Alfred Clebsch "Ueber eine Fundamenthalaufgabe der Invariantentheorie", in -Abhandlungen der Königlichen Gesellschaft der Wissenschaften in Göttingen 17 (1872), 3-62, and its shorter follow-up "Ueber eine Fundamentalaufgabe der Invariantentheorie", Math. Ann. 5 (1872), 427-434. -Are there earlier references about irreducible representations? - -REPLY [6 votes]: I convert my comments to an answer per Abdelmalek’s request: - -Dieudonné attributes the classification of irreducible $sl_2$-modules to Cayley (1856). -Also the theory of spherical and cylindrical harmonics should qualify as prehistory — told in e.g. Heine (1878, pp. 1–10), Burkhardt (1902–1903, Chap. V). -The words “irreducible” and “degree” hint at another root: if $G$ is finite, decomposing its regular representation $L$ on $\mathbf C[G]$ (elements $x=\smash{\sum x_g\delta^g}$, product $\smash{\delta^g\cdot\delta^h}=\smash{\delta^{gh}}$, $L(x)y=x\cdot y$) amounts to factoring the “group determinant” $\det(L(x))$ into irreducible polynomials in the $x_g$. - -(“When was it realized that it was important and very useful to decompose general representations in terms of irreducibles?” inadvertently evokes a whole other question involving the origin of Fourier analysis, going back to at least D. Bernoulli, not to mention celestial epicycles, Pythagorean ideas on musical harmony; or Lang’s Algebra’s casting of Jordan form as “Representation Theory of One Endomorphism” (or the monoid algebra $k[\mathrm X]$ of $(\mathbf N,+)$, with irreducibles etc.) — but I understand the intent was to keep it non-commutative.)<|endoftext|> -TITLE: closed ideals in L(L_1) -QUESTION [7 upvotes]: Denote $L_1=L_1[0,1]$ The lattice of closed ideals in $\mathcal{L}(L_1)$ includes the chain -$$ -\{0\}\subsetneq\mathcal{K}(L_1)\subsetneq\mathcal{FS}(L_1) -\subsetneq\mathcal{J}_{\ell_1}(L_1)\subsetneq\mathcal{S}_{L_1}(L_1) -\subsetneq\mathcal{L}(L_1). -$$ -I believe it's still an open question whether the lattice contains any more distinct elements than the above listed. -One strategy for producing many closed ideals very quickly is to use Schreier-Rosenthal operator ideals, and this has worked, for instance, in the cases $\mathcal{L}(\ell_1\oplus\ell_\infty)$. In particular, given a countable ordinal $\xi$, we could seek some special operator $A\in\mathcal{R}_\zeta(L_1)\setminus\mathcal{R}_\xi(L_1)$ for some $\zeta>\xi$. -Here's how it worked for $\mathcal{L}(\ell_1\oplus\ell_\infty)$. Let $T_\xi$ denote the $\xi$-Tsirelson space, and let $I_{1,\xi}:\ell_1\to T_\xi$ be the formal identity. Of course we can find an embedding $J_{\xi,\infty}:T_\xi\to\ell_\infty$. Then $J_{\xi,\infty}\circ I_{1,\xi}$ is class $\mathcal{R}$ but not class $\mathcal{R}_\xi$. By a result of Beanland/Freeman we can find $\zeta>\xi$ such that it is class $\mathcal{R}_\zeta$ as well. Then use transfinite induction. -Adapting this approach for $\mathcal{L}(L_1)$ presents serious difficulties. Of course $L_1$ contains a complemented copy of $\ell_1$, so we can still use a formal identity map from $\ell_1$ onto any space with an unconditional basis. But we can't use $T_\xi$ because it doesn't embed into $L_1$. (We know this because every infinite-dimensional subspace of $L_1$ contains a copy of $\ell_p$ for some $1\leq p\leq 2$.) -But perhaps there is a way around this. For any scalar sequence $(a_n)$ we can define -$$\|(a_n)\|_\xi=\|(a_n)\|_2\vee\sup_{F\in\mathcal{S}_\xi}\|(a_n)_{n\in F}\|_1$$ -This generates a 1-unconditional basic sequence whose closed linear span we denote $X_\xi$. Is it possible that $X_\xi$ (or something like it) embeds into $L_1$? We would also need to make sure that $\ell_1$ doesn't embed into $X_\xi$, but that seems fairly likely given the latter's construction. -In general, to make this work we need to find a space $X_\xi$ with a basis which is $\xi$-equivalent to $\ell_1$ and embeds into $L_1$, but doesn't contain a copy of $\ell_1$. Frankly, I believe this is asking for too much. Nevertheless, I thought it would be worth seeing what y'all thought before I give up on the idea. And if there is no such space, perhaps that would be worth proving anyway. -Thanks guys! - -REPLY [6 votes]: Your post is awfully technical for MO, IMO. -The lattice of closed ideals in $L(L_1)$ contains at least a continuum of elements: -https://arxiv.org/abs/1811.06571<|endoftext|> -TITLE: Exponential towers of $i$'s -QUESTION [24 upvotes]: It's well known that the expression $i^i$ takes on an infinite set of values if we understand $w^z$ to mean any number of the form $\exp (z (\ln w + 2 \pi i n))$ where $\ln$ is a branch of the natural logarithm function. -Since all values of $i^i$ are real, all values of $i^{i^i}$ (by which I mean $i^{\left( i^i \right)}$) are on the unit circle, and in fact they form a countable dense subset of the unit circle. -I can't figure out what's going on with $i^{i^{i^i}}$, though. I've posted a pdf version of Mathematica notebook at https://jamespropp.org/iiii.pdf containing images starting on page 2. Each image shows the points in the set of values of $i^{i^{i^i}}$ lying in an annulus whose inner and outer radii differ by a factor of 10. For instance, here's what we see in the annulus whose inner and outer radii are 100 and 1000 respectively. - -Can anyone see what's going on? -Also, what happens for taller exponential towers of $i$'s? Does the set of values of $$i^{i^{i^{{\ \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}\ }^i}}}$$ become dense in the complex plane once the tower is tall enough? The following picture makes me think that for a tower of height five, the set of values is everywhere dense. - -REPLY [11 votes]: Let $f : \mathcal{P}(\mathbb{C}) \rightarrow \mathcal{P}(\mathbb{C})$ be the function on the powerset of the complex numbers defined by: -$$ f(A) = \left\{ \exp\left( \frac{\pi i}{2}(4n + 1)z \right) : z \in A, n \in \mathbb{Z} \right\} $$ -Then your question is asking about $f^k(\{1\})$ and its (topological) closure. -One thing to note is that, for any set $A$, we have $\overline{f(A)} = \overline{f\left(\overline{A}\right)}$, where the overline denotes closure. This follows from the fact that the expression $\exp\left( \frac{\pi i}{2}(4n + 1)z \right)$ is continuous as a function from $\mathbb{C} \times \mathbb{Z}$ to $\mathbb{C}$. So, if we want to study the closure of $f^n(\{1\})$, we are allowed to take the closure after each application of $f$. -The OP remarked that $\overline{f^3(\{1\})}$ is the unit circle. Then the set: -$$ B :=\left\{ \frac{\pi i}{2}(4n + 1)z : z \in \overline{f^3(\{1\})}, n \in \mathbb{Z} \right\} $$ -is precisely the union of a circle of radius $\frac{\pi i}{2} k$ for each odd integer $k$. Next, we need to take the image of this set $B$ under $\exp$. As the function $\exp$ is many-to-one, this is best understood by first reducing $B$ modulo $2 \pi i$ so that it lies in the strip with imaginary part $[-\pi, \pi)$. -Here's a picture of the first six circles in $B \mod 2 \pi i$: - -The intersection of $B \mod 2 \pi i$ with a line of fixed real part is a set of points which contains two limit points (the intersection with the light grey lines in the above plot). The closure of $B \mod 2 \pi i$ therefore contains all of these circles together with the two light grey lines (and nothing else). -$\overline{f^4(\{1\})}$ is therefore just the image of this set under $\exp$ (viewed as a bijective function from the strip to the punctured complex plane) together with the origin. -Now let's consider $\overline{f^5(\{1\})}$. Recall that the first step is to take the union of lots of homothetic copies of $\overline{f^4(\{1\})}$: -$$ B' :=\left\{ \frac{\pi i}{2}(4n + 1)z : z \in \overline{f^4(\{1\})}, n \in \mathbb{Z} \right\} $$ -If we pull this back through the inverse of the $\exp$ map (so we're working on the strip instead of the complex plane), this corresponds to taking the union of lots of translates of $B \mod 2 \pi i$. Specifically, we are interested in the set of points: -$$ \log(B') :=\left\{ w + \log(i) + \log\left(\frac{\pi}{2}(4n + 1)\right) : w \in (B \mod 2 \pi i), n \in \mathbb{Z} \right\} $$ -Now, the set $\{ \log\left(\frac{\pi}{2}(4n + 1)\right) : n \in \mathbb{N} \}$ has the very desirable property that the set is unbounded to the right and the gaps between the points become arbitrarily small. Also, the intersection of every horizontal line with the original set $B \mod 2 \pi i$ (the one in the picture above) is unbounded to the left. As such, it follows that their convolution is dense in every horizontal line, and therefore dense in the whole strip. -Consequently, $\overline{f^5(\{1\})}$ is the entire complex plane as you suspected.<|endoftext|> -TITLE: Heat flow, decay of the Fisher information, and $\lambda$-displacement convexity -QUESTION [8 upvotes]: In the whole post I will work in the flat torus $\mathbb T^d=\mathbb R^d/\mathbb Z^d$ and $\rho$ will stand for any probability measure $\mathcal P(\mathbb T^d)$. This question is strongly related to two of my previous posts, universal-decay-rate-of-the-fisher-information-along-the-heat-flow and improved-regularization-for-lambda-convex-gradient-flows. - -Fact 0: the quadratic Wasserstein distance $W_2$ induces a (formal) Riemannian structure on the space of probability measures, which gives a meaning to Wasserstein gradients $\operatorname{grad}_{W_2}F(\rho)$ of a functional $F:\mathcal P(\mathbb T^d)\to\mathbb R$ at a point $\rho$ -Fact 1: the heat flow $\partial_t\rho_t=\Delta\rho_t$ -is the Wasserstein gradient flow -$$ -\dot\rho_t=-\operatorname{grad}_{W_2}H(\rho_t) -$$ -of the Boltzmann entropy -$$ -H(\rho)=\int_{\mathbb T^d}\rho\log\rho -$$ -Fact 2: the Boltzmann entropy is $\lambda$-(displacement) convex for some $\lambda$. -Its dissipation functional is the Fisher information, -$$ -F(\rho):=\|\operatorname{grad}_{W_2} H(\rho)\|^2_{\rho}=\int _{\mathbb T^d}|\nabla\log\rho|^2 \rho -$$ -Fact 3: for abstract metric gradient flows (in the sense of [AGS]) and $\lambda$-convex functionals $\Phi:X\to\mathbb R\cup\{\infty\}$ one expects a smoothing effect for gradient flows $\dot x_t=-\operatorname{grad}\Phi(x_t)$ in the form -\begin{equation} -|\nabla\Phi(x_t)|^2\leq \frac{C_\lambda}{t} \Big[\Phi(x_0)-\inf_X\Phi\Big] -\tag{R} -\end{equation} -at least for small times, where $C_\lambda$ depends only on $\lambda$ but not on $x_0$ see e.g. [AG, Proposition 3.22 (iii)]. -Fact 3': with the same notation as in Fact 3, an alternative regularization can be stated as -\begin{equation} -|\nabla\Phi(x_t)|^2 \leq \frac{1}{2e^{\lambda t}-1}|\nabla\Phi(y)|^2 +\frac{1}{(\int_0^te^{\lambda s}ds)^2} dist^2(x_0,y), -\,\, -\forall y\in X -\tag{R'} -\end{equation} -Fact 4: in the Torus the Fisher information decays at a universal rate, i-e there is $C=C_d$ depending on the dimension only such that, for all $\rho_0\in \mathcal P(\mathbb T^d)$ and $t>0$, the solution $\rho_t$ of the heat flow emanating from $\rho_0$ satisfies -\begin{equation} -F(\rho_t)\leq \frac{C}{t} -\tag{*} -\end{equation} -This follows from the Li-Yau inequality [LY], see this post of mine and F. Baudoin's answer. - - -Question: is there more to ($*$) than just the convexity of the Boltzmann functional? If the driving functional were upper-bounded $\Phi(x_0)\leq C$ (for all $x_0\in X$) in the regularization estimate (R) then we would immediately get the universal decay $|\nabla \Phi(x_t)|^2\leq \frac{C}{t}$. - However, in the specific context of Facts 0-2 it is clearly not true that the Boltzmann entropy is upper-bounded. In fact there are many probability measures with infinite entropy, take e.g. any Dirac mass. Since (R) is optimal, I guess that one cannot simply deduce (*) from general $\lambda$-convexity arguments, and there is more than meets the eyes. But is there any connection? Note that both the Li-Yau inequality and the displacement convexity of the Boltzmann entropy strongly rely on the nonnegative Ricci curvature of the underlying torus. - -I tried desperately to use any modified regularization estimate (e.g. R' and variants thereof instead of R), but to no avail so far. I am starting to believe that there is no direct implication, and that the work of Li-Yau is really profoundly ad-hoc (don't get me wrong, I just mean that their results cannot be generalized for abstract gradient-flows, and that their result/proof really leverages the specific structure and setting of the heat flow in Riemannian manifolds, not just any gradient flow). I would immensely appreciate any input or insight! - -[AG] Ambrosio, L., & Gigli, N. (2013). A user’s guide to optimal transport. In Modelling and optimisation of flows on networks (pp. 1-155). Springer, Berlin, Heidelberg. -[AGS] Ambrosio, L., Gigli, N., & Savaré, G. (2008). Gradient flows: in metric spaces and in the space of probability measures. Springer Science & Business Media. -[LY] Li, P., & Yau, S. T. (1986). On the parabolic kernel of the Schrödinger operator. Acta Mathematica, 156, 153-201. - -REPLY [3 votes]: I won't say that it is impossible, but I don't see how to obtain $(\ast)$ using only general theory. There might be a different strategy that works, but I can tell you why I don't think the Li-Yau estimate can be proven using general properties of convexity. In particular, Li-Yau relies on some careful estimates involving the Laplace-Beltrami operator (and some hard analysis), which I don't think general theory can "see". -For a detailed write up of the Li-Yau estimate, I recommend Lectures on Differential Geometry by Schoen and Yau, which was very helpful for me.  From a high level overview, the idea is to let $u$ be a non-negative solution to the heat equation, consider $\log (u + \epsilon)$ and try to bound its derivative. To do this, you consider the point which maximizes $ | \nabla \log (u + \epsilon) |^2$ and use the Bochner formula. Bochner's formula has a correction term due to the curvature, but when the manifold is Ricci positive, this has a favorable sign and we can ignore it (or use something like a barrier function to sharpen the estimate). The key insight is actually a clever use of the Cauchy-Schwarz inequality to eke out a little bit extra from the second derivative terms. It's elementary, but also a stroke of genius, and allows everything else to work. -If you read proofs of Li-Yau, the logarithm tends to appear near the end. However, it was helpful for my intuition to realize that this is not ad hoc; there was always going to be a logarithm because we are using the maximum principle applied to the function $\dfrac{|\nabla u|^2}{(u+\epsilon)^2} = | \nabla \log(u+\epsilon)|^2$. -The fact that $\nabla u$ and $u$ are raised to the same power here is crucial. When the power of $\nabla u$ is less than $u$, integrating out the resulting inequality gives a bounded function (which is significantly less useful). There's this really delicate balancing act in order for everything to work, and logs play an essential role. As a brief aside, I suspect you get different powers of $\nabla u$ and $u$ if you try the Li-Yau strategy with the porous media equation (I'm not entirely sure of this though). -So back to your question about whether this can be done using general properties of gradient flows. It might be a lack of imagination on my part, but it's hard for me to see how this would work. There's several essential steps that rely on hard analysis. For instance, you really need the Cauchy-Schwarz step to work and the resulting function that you get from integrating out should be unbounded. Furthermore, while it's possible to sharpen the estimate, the original version is already fairly sharp, in that there is not a whole lot of wiggle room. As such, while it's possible to adapt the argument to elliptic operators or to include lower order terms, it does seem like there is genuinely more here than the general theory.<|endoftext|> -TITLE: Metrics on torus without closed contractible geodesics -QUESTION [7 upvotes]: It is easy to see that any closed geodesic on a flat 2-torus is noncontractible. -Further the same holds true for a torus of revolution. -Indeed either a closed geodesic is a meridian and therefore noncontractible, or it intersects all the meridians transversally and noncontractible as well. -The same argument shows that if a torus admits a geodesic foliation, then it has no contractible closed geodesic. -Are there other examples? - -Are there examples of Riemannian metrics on $\mathbb{T}^2$ without closed null-homotopic geodesics and without a geodesic foliation? - -A commnet of Dima Burago: It is impossible to find an example by taking a small perturbation of a flat metric. Indeed, by KAM theory any such perturbation has a geodesic foliation. - -REPLY [6 votes]: This is a comment, not an answer, but it's too long to fit into a comment window. -I don't know of a 'generic' condition on metrics on the torus that would guarantee that there are no null-homotopic closed geodesics, but then I'm not sure what 'generic' is supposed to mean. -For example, would it suffice to find some sort of 'open' condition (for example, that a metric be 'sufficiently close' in some appropriate sense to being flat) that would imply that there are no null-homotopic closed geodesics, or are we supposed to be trying to decide whether, given any metric on the torus, there is another metric arbitrarily 'close' to it that has no null-homotopic closed geodesics? (This second meaning is what I take to be the literal meaning of 'generic', which I think is unlikely to hold.) -Meanwhile here is an interesting, 'nongeneric' example that has no nontrivial symmetries (which is, itself, a 'generic' condition in a sense) and has no null-homotopic closed geodesics: -Let $a$ and $b$ be 'generic' positive smooth functions on $\mathbb{R}$, periodic of period $2\pi$, and consider the metric -$$ -g = \bigl(a(x)+b(y)\bigr)(\mathrm{d}x^2 + \mathrm{d}y^2). -$$ -which is doubly periodic and hence defines a metric on the torus $\mathbb{T} = \mathbb{R}/(2\pi\mathbb{Z}^2)$. -I will show that $g$ has no closed geodesics in $\mathbb{R}^2$. -To do this, note that this is a metric of Liouville type and hence all the unit speed geodesics satisfy the equations (i.e., 'conservation laws' or 'first integrals') -$$ -\bigl(a(x)+b(y)\bigr)(\dot x^2+\dot y^2) = 1 -\quad\text{and}\quad -\bigl(a(x)+b(y)\bigr)\bigl(b(y)\,\dot x^2 - a(x)\,\dot y^2) = c -$$ -for some constant $c$. Given such a geodesic, let $\theta$ be the function on the geodesic that satisfies $\cos\theta = \dot x\sqrt{a(x)+b(y)}$ and $\sin\theta = \dot y\sqrt{a(x)+b(y)}$. This $\theta$ is well-defined up to an integer multiple of $2\pi$ and gives the 'slope' of the geodesic in the standard $xy$-coordinates. Then we have -$$ -b(y)\,\cos^2\theta - a(x)\,\sin^2\theta = c. -$$ -Notice that, because $a$ and $b$ are strictly positive, this equation implies that, for any given $c$, there is an open set of directions $\phi$ such that $\theta$ cannot attain either value $\phi$ or $\phi+\pi$, an impossibility for a closed curve in the $xy$-plane, since a closed curve has to have a tangent perpendicular to any given direction. -Thus, $g$ has no closed geodesics in $\mathbb{R}^2$ and hence the induced metric on $\mathbb{T}$ has no closed null-homotopic geodesics. -Of course, you can object that Liouville metrics are not 'generic'. But it might be that there is some 'open' condition on a lattice-periodic metric on the plane such that no geodesic satisfying this condition can ever 'turn all the way around', as in this Liouville case. If one can find such a condition, then this would, I take it, be an answer to Anton's question. -Moreover, there are `higher' Liouville metrics, i.e., for which the geodesic flow has higher degree polynomial conservation laws, and, as far as I know, there is no limit on how high the degree of such a conservation law might be. It's possible that one could find examples of arbitrary complexity, so that these metrics are as 'generic' as you could possibly want.<|endoftext|> -TITLE: B-model and Hochschild cohomology -QUESTION [9 upvotes]: In "On the Classification of Topological Field Theories" in Example 1.4.1, Lurie introduces the B-model with target an (even dimensional) Calabi-Yau variety $X$: The Hochschild cohomology $\operatorname{HH}^*(X)$, together with the "canonical trace map" $\operatorname{HH}^*(X) \rightarrow k$, form a graded commutative Frobenius algebra; and by a classical theorem that can, as he states, analogously applied to the graded setting, this defines a 2-dimensional Topological Quantum Field Theory $Z$ with values in complexes such that $Z(S^1)=\operatorname{HH}^*(X)$. -While I can roughly follow his reasoning (although it would be helpful if someone could provide a short explanation on how this trace map occurs), as far as I knew the states of the B-model on $X$ were related to the (bounded) derived category $\operatorname{D}^b(\operatorname{Coh(X)})$, and I don't really see why Lurie's definition should describe the B-model as I know it, or why Hochschild cohomology occurs here. As Lurie only uses this statement as a motivation for introducing extended TQFTs with values in chain complexes, he doesn't give any further explanation or references, so I wondered if anyone here could. Thanks in advance and greetings, -Markus Zetto - -REPLY [3 votes]: For correct attribution, one should at least mention the paper which the preprint of Moore and Segal itself quotes as the source for the particular case of the algebraic description of open-closed TFTs which they give without proof. The following paper was published about 5 years before the preprint of Moore and Segal: -C. I. Lazaroiu, On the structure of open-closed topological field theory in two dimensions, Nucl.Phys.B603 (2001) 497-530, https://arxiv.org/abs/hep-th/0010269 -Notice that the spaces of boundary states in the B-model with CY manifold target X are Z-graded, since the physically correct category of B-type topological D-branes is not D^b(X) but rather its shift completion. The Z/2Z grading in Lazaroiu's axioms is the mod 2 reduction of that natural Z-grading -- and this Z/2Z grading is ignored by Moore and Segal. Shift completions were explained in: -C. I. Lazaroiu, Graded D-branes and skew-categories, JHEP 0708 (2007) 088, https://arxiv.org/abs/hep-th/0612041 -(which also explains the twist-completion that occurs, for example, in B-type orbifold LG models). -One should also note that the argument using HH(D^b(X)) and the holomorphic HKR isomorphism is not due to Lurie but to Kontsevich. Furthermore this argument (while very nice and natural) is not needed for the B-model, since the closed string state space of the B-model is known independently from localization of the path integral in the bulk sector, which produces the space of polyvector fields endowed with its (Dolbeault) differential and Serre trace. This goes back to Witten and to the paper of BCOV in the early 1990's. The advantage of the polyvector approach is that it provides an off-shell model which also appears in the BCOV theory and has a dual description using a BV bracket.<|endoftext|> -TITLE: Criterion for generic polynomials -QUESTION [5 upvotes]: Generic polynomials, which are recalled below, play an important role in the constructive aspects of the inverse Galois problem. -Definition. Let $P(\mathbf{t},X)$ be a monic polynomial in $\mathbb{Q}(\textbf{t})[X]$ with $\textbf{t} = (t_1,\dots, t_n)$ and $X$ being indeterminates, and let $\mathbb{L}$ be the splitting field of $P(\textbf{t},X)$ over $\mathbb{Q}(\textbf{t})$. Suppose that: - -$\mathbb{L}/\mathbb{Q}(\textbf{t})$ is Galois with Galois group $G$, and that -every $L/\mathbb{Q}$ with Galois group $G$ is the splitting field of a polynomial -$P(\mathbf{a},X)$ for some $\textbf{a} = (a_1,\dots, a_n) \in \mathbb{Q}^n$. - -We say that $P(\textbf{t},X)$ parametrizes $G$-extensions of $\mathbb{Q}$ and $P(\textbf{t},X)$ is a parametric polynomial. The parametric polynomial $P(\textbf{t},X)$ is generic if $P(\textbf{t},X)$ is parametric for $G$-extensions over any field containing $\mathbb{Q}$. -Much of the literature on such polynomials concerns their existence and their constructions. -Question. Is there is a useful criterion for determining whether a monic polynomial $P(\mathbf{t},X)$ in $\mathbb{Q}(\textbf{t})[X]$ is generic or not? In particular, I have encountered the following polynomial in my research -$$ -g_t(x) := x^3 + 147(t^2 + 13t + 49)x^2 + 147(t^2 + 13t + 49)(33t^2 + 637t + 2401)x + 49(t^2 + 13t + 49)(881t^4 + 38122t^3 + 525819t^2 + 3058874t + 5764801), -$$ -and I would like to know whether this is a generic polynomial for $\mathbb{Z}/3\mathbb{Z}$-extensions. (For my purposes, it would be enough to know that this is a parametric polynomial for $\mathbb{Z}/3\mathbb{Z}$-extensions.) It is easy to verify condition (1) in the definition using Magma, but I do not know of a way to test condition (2). -Any references and/or suggestions are greatly appreciated! - -REPLY [7 votes]: I believe there is no good way to determine in general if a polynomial $P(\mathbf{t},X)$ is generic. In fact, given a number field $K$ and a univariate polynomial $P(t,x) \in \mathbb{Q}[t,x]$, the problem of determining whether this is some $t \in \mathbb{Q}$ for which $P(t,x)$ has a root in $K$ is quite hard (and in some cases boils down to determining the $\mathbb{Q}$-points of a curve of genus greater than $1$). -The polynomial that you give is a generic $\mathbb{Z}/3\mathbb{Z}$-extension of $\mathbb{Q}(t)$ -however. If you let $f(t,x) = x^{3} - tx^{2} + (t-3)x + 1$, be the example of a generic $\mathbb{Z}/3\mathbb{Z}$ extension given in the comments by Daniel Loughran, one can verify (using Magma for example) that your polynomial and $f(\frac{49}{t} + 8, x)$ define the same degree $3$ extension of $\mathbb{Q}(t)$. -This proves the claim, expect possible for the cyclic cubic extension obtained from $f(0,x) = x^{3} - 3x + 1$. But this cubic extension is also obtained from $f(-20,x)$. -EDIT: The OP asked for some info about how I found this. I did a fair amount of stumbling around to come up with $f(49/t + 8,x)$. First, I noticed that the discriminant of the maximal order of $\mathbb{Q}(t)[x]/(g_{t}(x))$ was $(t^{2} + 13t + 49)^{2}$, while the discriminant of the maximal order of $\mathbb{Q}(t)[x]/(f(t,x))$ was $(t^{2} - 3t + 9)^{2}$, and replacing $t$ with $t+8$ gives two polynomials with discriminants that are in the same square class. However, these do not define isomorphic extensions - specializing for lots of values of $t$ reveals that in most cases the discriminant of the ring of integers for $g_{t}(x)$ and that for $f(t+8,x)$ differ by a factor of $49$. This suggests that for a fixed $t$, if $K_{1}$ is the number field defined by $f(t+8,x)$ and $K_{2}$ is that defined by $g_{t}(x)$, then $K_{2} \subseteq K_{1}(e^{2 \pi i / 7} + e^{-2 \pi i /7})$ [note that $\mathbb{Q}(e^{2 \pi i /7} + e^{-2 \pi i / 7})$ is the unique cyclic extension with discriminant $49$]. One can verify that this is true. I suspected that there was a linear fractional transformation in $t$ that would take $f(t+8,x)$ to a polynomial defining one of the other cubic subfields of $K_{1}(e^{2 \pi i / 7} + e^{-2 \pi i /7})$. Looking at several specific cubic fields and searching for the values of $t$ (there are many) that result in $f(t+8,x)$ and $g_{t}(x)$ defining them leads to the suggestion that $t \to 49/t$ would work.<|endoftext|> -TITLE: Homogeneous vector bundles on Abelian varieties -QUESTION [6 upvotes]: I recently encountered a result about vector bundles on Abelian varieties, which I found interesting. It characterizes homogeneous (translation invariant) vector bundles on Abelian varieties. More precisely any such vector bundle is in the form of $\bigoplus_L L\otimes U_L $, where $U_L$ is a unipotent vector bundle i.e. constructed by successive extensions of trivial line bundle and $L$ is an algebraically trivial line bundle. I wasn't able to find a stand-alone proof of this fact. All proves eventually refer to "M. Miyanishi, Some remarks on algebraic homogeneous vector bundles, in: Number theory, algebraic geometry and commutative algebra, 71–93, Kinokuniya, Tokyo, -1973.". It seems that this reference doesn't exist any more! (at least online). I'd appreciate if anyone knows where to find a proof of this fact. -What I was also interested was understanding the homogeneous sub-bundles of a homogeneous vector bundle. I was wondering whether the proof implies existence of any characterization of such sub-bundles or not? (like does it necessarily imply a homogeneous sub-bundle of $\bigoplus_{L\in A} L\otimes U_L $, is something like $\bigoplus_{L\in B \subseteq A} L\otimes U'_L $, where $U'_L$ is a unipotent sub-bundle of $U_L$ with unipotent quotient. ) - -REPLY [4 votes]: I don't have access to Miyanishi's article either (at least during lockdown), but as Ulrich suggested, one can look at Mukai's paper "Duality between D(X) and $D(\hat X)$...", where he introduces the Fourier-Mukai transform. I'll denote this transform by $\mathcal{F}$. On page 159, Mukai gives a proof of the characterization of homogeneous bundles you mentioned. Before that in ex 2.9 and 3.2, he proves that - -Theorem. $\mathcal{F}$ induces an equivalence between the category of homogenous bundles on an abelian variety $X$, and the category of coherent sheaves on the dual $\hat X$ with finite support. The unipotent bundles correspond to sheaves on $\hat X$ supported at the origin. - -Given a homogenous bundle $V$, the points of the support of $\mathcal{F}(V)$ are the line bundles $L$ in your decomposition. The $U_L$ can also be recovered by taking inverse transform of the translate of $\mathcal{F}(V)_L$ back to the origin. Given all of this, it seems clear that homogenous subbundles are as you describe.<|endoftext|> -TITLE: When is the Radon-Nikodym derivative locally essentially bounded -QUESTION [5 upvotes]: Let $\mu\lll\nu$ be $\sigma$-finite Borel measures, which are not finite, on a topological space $X$. Under what conditions is $0<\operatorname{ess-supp}(\frac{d\mu}{d\nu}I_K)<\infty$ for every compact subset $\emptyset\subset K\subseteq X$. -In other words when is $\frac{d\mu}{d\nu} \in L^{\infty}_{\nu,\mathrm{loc}}(X)$? - -REPLY [4 votes]: Let $$f:=\frac{d\mu}{d\nu}.$$ Then -$$f\in L^{\infty}_{\nu,loc}(X)\iff\text{$\forall$ compact $K\subseteq X$ $\exists$ $c_K\in(0,\infty)$ $\forall$ Borel $A$ we have $\mu(A\cap K)\le c_K\nu(A\cap K)$.}$$ -Indeed, for the $\Rightarrow$ implication, take any compact $K\subseteq X$. Then $\exists$ $c_K\in(0,\infty)$ such that $f\le c_K$ $\nu$-a.e. on $K$. So, for any -Borel $A$ we have -$$\mu(A\cap K)=\int_{A\cap K}f\,d\nu\le c_K\nu(A\cap K),$$ -as desired. -Vice versa, for the $\Leftarrow$ implication, take any compact $K\subseteq X$ and suppose that $\mu(A\cap K)\le c_K\nu(A\cap K)$ for some $c_K\in(0,\infty)$ and all Borel $A$. Let now $A:=f^{-1}((c_K,\infty))$, so that $f>c_K$ on $A$. Then -$$\mu(A\cap K)=\int_{A\cap K}f\,d\nu\ge c_K\nu(A\cap K),$$ -and the latter inequality is strict (and hence contradicts condition $\mu(A\cap K)\le c_K\nu(A\cap K)$) if $\nu(A\cap K)>0$. So, $\nu(A\cap K)=0$, that is, $f\le c_K$ $\nu$-a.e. on $K$, as desired. - -Similarly, for any compact $K\subseteq X$, -$$\operatorname{esssup}_Kf>0\iff \text{ $\exists$ $b_K\in(0,\infty)$ $\exists$ Borel $A$ such that $\mu(A\cap K)\ge b_K\nu(A\cap K)$.}$$<|endoftext|> -TITLE: the fractional integration method of the proof of Stein-Tomas theorem? -QUESTION [6 upvotes]: In Schalg's Classical multilinear and Harmonic analysis, he presented two methods of the proof of Stein-Tomas theorem, one of which is called the fractional integration method. As a matter of fact, in order to prove -\begin{equation} -\lVert f * \hat\mu \rVert_{L^{p'}(\mathbb{R}^d)}\le C \lVert f \rVert_{L^p(\mathbb{R}^d)}, \quad \text{for } p=\frac{2d+2}{d+3}, \quad d\ge 3, -\end{equation} -where $\hat{\mu}\triangleq K$ is the Foureir transform of Lebesgue measure of surface with nonvanishing Gaussian curvature (and we may assume it is just the Fourier tranform of Lebesgue measure of unit sphere $\mathbb{S}^{d-1}$), -he tore the coordinates into two pieces $x=(x',t)$, where $x'=(x_1,...,x_{d-1})$, then -\begin{equation} -f*\hat{\mu} (x) =\int_{\mathbb{R}}\int_{\mathbb{R}^{d-1}} K(x'-y',t-s) f(y',s) dy'ds. -\end{equation} -Thus we may restrict our attention on the behavior of $K(x',t)$ with respect to $x'$. More precisely, if we assume that $(Ug)(x')= \int_{\mathbb{R}^{d-1}} K(x'-y',t)dt$, then Schlag claimed that $U(t)$ satifies -\begin{equation} -\lVert U(t) \rVert_{L^1(\mathbb{R}^{d-1}) \to L^\infty(\mathbb{R}^{d-1})} \le C |t|^{d-1}, \quad \lVert U(t) \rVert_{L^2(\mathbb{R}^{d-1}) \to L^2(\mathbb{R}^{d-1})} \le C <\infty, -\end{equation} -where $C>0$ is independent of $t\in \mathbb{R}$, and then we can use Riesz-Thorin interpolation theorem and then use Hardy-Littlewood-Sobolev inequality to get the desired estimates. -And my question is how to check the second estimate (i.e. the uniform bound of $L^2 \to L^2$), Schlag said it suffices to check that $K(\hat{\cdot},t) \in L_{\xi'}^\infty L_t^\infty ( \mathbb{R}^{d-1} \times \mathbb{R})$, where $K(\hat{\cdot},t)$ means the Fourier transform of $K(x',t)$ w.r.t. $x'$. For example, in $d=3$, then the Fourier transform of unit sphere can be represented by $\hat{\sigma}(x)=\frac{\sin{|x|}}{|x|}$ explicitly, but how can I check that -\begin{equation} -K(\xi',t)= \int_{\mathbb{R}^2} e^{-2\pi i x' \cdot \xi'}\frac{\sin{|(x',t)|}}{|(x',t)|} dx' \in L_{\xi'}^\infty L_t^\infty ( \mathbb{R}^{2} \times \mathbb{R}) \quad? -\end{equation} - -REPLY [2 votes]: Let's first clarify the definitions (also, there are some typos in your post, perhaps you should consider correcting them). -For $\xi\in\mathbb{R}^d$ we shall write $\xi=(\xi',\xi_d)$ with $\xi'\in\mathbb{R}^{d-1}$. -For a tempered distribution $T$ we shall denote its distributional Fourier transform by $\widehat{T}$. We will use the same symbol for distributions on $\mathbb{S}(\mathbb{R}^d)$ and $\mathbb{S}(\mathbb{R}^{d-1})$, it will be clear from the context. -We work with a surface of the form -$$ -M=\{(x', \psi(x')): x'\in U\} -$$ -for some bounded open set $U\subset\mathbb{R}^{d-1}$ (can think $M=\mathbb{S}^{d-1}$). The surface measure on $M$ is given for $f\in\mathbb{S}(\mathbb{R}^d)$ by -$$ -\int_{\mathbb{R}^d}f(x)d\mu(x)=\int_{U}f(x', \psi(x'))\sqrt{1+|\nabla \psi(x')|^2}dx'. -$$ -Note that $\sqrt{1+|\nabla \psi(x')|^2}\simeq 1$, which means that this factor is harmless. -We define -$$ -K(\xi)=\widehat{\mu}(\xi),\qquad \xi\in\mathbb{R}^d. -$$ -Next for a fixed $t\in\mathbb{R}$ we consider a locally integrable function $K_t$ on $\mathbb{R}^{d-1}$ given by -$$ -K_t(\xi'):=K(\xi',t),\qquad \xi'\in\mathbb{R}^{d-1}. -$$ -We $\textbf{shall show that}$ the distributional Fourier transform of $K_t$ coincides with an $L^\infty$ function on $\mathbb{R}^{d-1}$ which is bounded uniformly in $t\in\mathbb{R}$. -$\textbf{Solution:}$ -Using the definition of a Fourier transform of a distribution and then applying Fubini's theorem, we get for $\varphi\in\mathbb{S}(\mathbb{R}^{d-1})$ -\begin{align*} -\langle \widehat{K_t}, \varphi\rangle&=\langle K_t, \widehat{\varphi}\rangle=\int \widehat{\mu}(\xi',t)\widehat{\varphi}(\xi')d\xi'=\int_{\mathbb{R}^{d-1}}\int_{\mathbb{R}^{d}}e^{-2\pi i(x'\xi'+x_d t)}d\mu(x',x_d) \widehat{\varphi}(\xi')d\xi'\\ -&=\int_{\mathbb{R}^{d}}e^{-2\pi i x_d t}\left(\int_{\mathbb{R}^{d-1}}e^{-2\pi i x'\xi'}\widehat{\varphi}(\xi')d\xi'\right)d\mu(x',x_d)\\ -& =\int_{\mathbb{R}^{d}}e^{-2\pi i x_d t}\varphi(x')d\mu(x',x_d)=\int_U e^{-2\pi i \psi(x') t}\varphi(x')\sqrt{1+|\nabla \psi(x')|^2}dx'\\ -&=: -\langle F_t, \varphi\rangle, -\end{align*} -where $F_t(x')=\chi_U(x')\sqrt{1+|\nabla \psi(x')|^2}e^{-2\pi i\psi(x') t}$. Clearly $F_t(x')\in L_{x'}^\infty L_t^\infty ( \mathbb{R}^{d-1} \times \mathbb{R})$, so the claim is proved.<|endoftext|> -TITLE: What graph's minimum vertex cover equals twice the maximum matching? -QUESTION [9 upvotes]: Matching: https://en.wikipedia.org/wiki/Matching_(graph_theory) -Vertex Cover: https://en.wikipedia.org/wiki/Vertex_cover -It is easy to see that -$$\texttt{minimum vertex cover} \leq 2 \texttt{ maximum matching}$$ -I want to know that for what kind of graphs the equality is hold in the above inequality. -As an instance, $C_3$ is an example. - -REPLY [5 votes]: Answer. Such a graph $G$ is a disjoint union of odd complete graphs. -Obviously such graphs satisfy the equality $$\texttt{minimum vertex cover} = 2 \texttt{ maximum matching}.\quad (\star)$$ -Assume that $G=(V,E)$ satisfies $(\star)$. Denote by $k$ the size of maximal independent set in $G$, then $$k=|V|-\texttt{minimum vertex cover}=|V|-2\cdot\texttt{maximum matching}=\\ -\texttt{ minimum number of vertices not covered by a matching}.$$ -On the other hand, by Tutte — Berge formula, if $k$ is the minimum number of vertices not covered by a matching, then there exists a subset $U\subset V$ such that $G-U$ has $|U|+k$ odd connected components. If $|U|>0$, then taking a vertex from each component we get an independent set with more than $k$ vertices. Therefore $U=\emptyset$, $G$ has $k$ odd components and if $G$ has also an even component, we again may take an independent set with more than $k$ vertices. Also if one of these connected components $C$ is not a complete graph, we may take to not-connected vertices in $C$ and a vertex from each other component, again having too large independent set. That is.<|endoftext|> -TITLE: What is the asymptotic of the irregular blue curve? Is it $(8x)^{1/2}$ or is it something else? -QUESTION [5 upvotes]: From Terry Tao's post here there is the statement: -"Conversely, if one can somehow establish a bound of the form -$$\displaystyle \sum_{n \leq x} \Lambda(n) = x + O( x^{1/2+\epsilon} ) \tag{1}$$ -for any fixed ${\epsilon}$, then the explicit formula can be used to..." -I don't know about the word "fixed", but the irregular behaviour of the blue curve below gives plenty of room for an ${\epsilon}$, if it is true that the asymptotic is $(8x)^{1/2}$, and if it is also true that it bounds the partial sums of the Möbius transform of the Harmonic numbers minus $x$. But we don't know and can't conclude any such bounds from this question. I am only asking about the asymptotics of a certain sum that is connected to / a truncated absolute value version of the numerators of the expansion of the primes. -Let: -$$\varphi^{-1}(n) = \sum_{d \mid n} \mu(d)d \tag{2}$$ -Then for $n>1$: -$$\Lambda(n) = \sum\limits_{k=1}^{\infty}\frac{\varphi^{-1}(\gcd(n,k))}{k} \tag{3}$$ -Form the table: -$$A(n,k)=\sum_{\substack{i=k\\\ n \geq k}}^n \varphi^{-1}(\gcd (i,k)) \tag{4}$$ -From numerical evidence it appears that: -$$\sum _{k=1}^{x} \text{sgn}\left(\left(\text{sgn}\left(x+\sum _{j=2}^k -|A(x,j)|\right)+1\right)\right)+1 \sim (8x)^{1/2} \tag{5}$$ -Is it true or is the asymptotic something else? -Question: - -The complicated sign formula in $(5)$ comes from what we are really -doing which is to ask: What is the asymptotic of the least $k$ for which -the function $F(x)$: -$$F(x)=x+\sum _{j=2}^k -|A(x,j)| \tag{6}$$ -is negative? For $k=1..x$. - -Plot of the numerical evidence where the irregular blue curve is that least $k$ for which the function $F(x)$ is negative and thereby also the LHS of (5) while the smooth red curve is the conjectured asymptotic $(8x)^{1/2}$: - -Efficient Mathematica program to generate the plot. Setting nn=10000 gives the plot above: -(*start*) -(*Mathematica*) -Clear[a, f, p]; -nn = 1000; -p = 0; -f[n_] := n*Log[n]^p; -(*Clear[f];*) -(*f[n_] := n*Log[n]^4/(Pi*8)^2/8;*) -a[n_] := DivisorSum[n, MoebiusMu[#] # &]; -Monitor[TableForm[ - A = Accumulate[ - Table[Table[If[n >= k, a[GCD[n, k]], 0], {k, 1, nn}], {n, 1, - nn}]]];, n] -TableForm[B = -Abs[A]]; -Clear[A]; -B[[All, 1]] = N[Table[f[n], {n, 1, nn}]]; -TableForm[B]; -TableForm[B1 = Sign[Transpose[Accumulate[Transpose[B]]]]]; -Clear[B]; -Quiet[Show[ - ListLinePlot[ - v = ReplaceAll[ - Flatten[Table[First[Position[B1[[n]], -1]], {n, 1, nn}]], - First[{}] -> 1], PlotStyle -> Blue], - Plot[Sqrt[8*f[n]], {n, 1, nn}, PlotStyle -> {Red, Thick}], - ImageSize -> Large]] -ListLinePlot[v/Table[Sqrt[8*f[n]], {n, 1, nn}]] -(*end*) - -Variant of the Mathematica program above: https://pastebin.com/GJ81MQez -Inefficient Mathematica program to generate the LHS in (5): -Clear[varphi]; -nn = 20; -constant = 2*Sqrt[2]; -varphi[n_] := Total[Divisors[n]*MoebiusMu[Divisors[n]]]; -Monitor[TableForm[ - A = Table[ - Table[Sum[If[n >= k, varphi[GCD[i, k]], 0], {i, k, n}], {k, 1, - nn}], {n, 1, nn}]];, n] -Table[1 + - Sum[Sign[(1 + Sign[x + Sum[-Abs[A[[x, j]]], {j, 2, k}]])], {k, 1, - x}], {x, 1, nn}] - -which starts: -{2, 3, 4, 5, 6, 5, 7, 7, 10, 7, 11, 10, 11, 10, 11, 11, 14, 13, 14, 13} -For my own memory to remember where to start editing tomorrow I write this Mathematica program: -Clear[varphi]; -nn = 40; -varphi[n_] := Total[Divisors[n]*MoebiusMu[Divisors[n]]]; -Table[1 + - Sum[Sign[(1 + - Sign[x + - Sum[-Abs[ - Sum[If[x >= j, varphi[GCD[i, j]], 0], {i, j, x}]], {j, 2, - k}]])], {k, 1, x}], {x, 1, nn}] - -There are previous efforts related to this question. Here is one of them. -A construction: -$$\sqrt{x} \log ^2(x)=\sqrt{x} \left(x-\left(\sqrt{x}-\log (x)\right) \left(\sqrt{x}+\log (x)\right)\right)$$ - -REPLY [2 votes]: If one rewrites $(4)$ as: -$$A(n,k)=\sum_{i=1}^n \varphi^{-1}(\gcd (i,k))$$ -Then one finds empirically that the mean of the $k$-th column in $A(n,k)$ is: -$$\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n A(n,k) = -\frac{\varphi^{-1}(k)}{2} \tag{*}$$ -and that the period length of column $k$ is $k$. -In https://oeis.org/A173557 on Jun 18 2020, Vaclav Kotesovec says that the Dirichlet generating function for: $$a(k)=|\varphi^{-1}(k)|$$ -is: -$$\frac{\zeta(s)\zeta(s-1)}{\zeta(2s-2)} \prod_{p \text{ prime}} \left(1 - \frac{2}{(p + p^s)}\right)$$ -and per email he explained to me that by choosing the residue at $s=2$ he finds that: -$$\sum_{k=1}^{n} a(k) \sim c \frac{n^2}{2}$$ -or equivalently: -$$\sum_{k=1}^{n} |\varphi^{-1}(k)| \sim c \frac{n^2}{2} \tag{**}$$ -where: $$c = A307868 = \prod_{p \text{ prime}} \left(1-\frac{2}{(p+p^2)}\right) = 0.471680613612997868...$$ -In the question above we ask what is the least $k$ for which the function $F(n)$: -$$F(n)=n-\sum _{j=2}^k|A(n,j)|$$ -becomes negative. Since the columns are periodic and the average of the periods in the $k$-th column is as said above: $-\frac{\varphi^{-1}(k)}{2}$, this is then approximately equal to asking what is the least $k$ for which the function $G(n)$: -$$G(n)=n-\sum _{j=2}^k\frac{|\varphi^{-1}(j)|}{2} \tag{***}$$ -becomes negative. -Combining $(*)$ and $(**)$: -$$\sum _{j=1}^k\frac{|\varphi^{-1}(j)|}{2} \sim c\frac{k^2}{4}$$ -Setting $j=2$ in the lower summation index: -$$\sum _{j=2}^k\frac{|\varphi^{-1}(j)|}{2} \sim c\frac{k^2}{4}-1$$ -Inserting into $(***)$: -$$G(n)=n-\left(c\frac{k^2}{4}-1\right)$$ -The least $k$ for which $G(n)$ becomes negative or changes sign is when $G(n)=0$. Solving: -$$c\frac{k^2}{4} - 1 = n$$ for $k$ gives the answer: - -$$k(n) = \frac{2 \sqrt{n+1}}{\sqrt{c}} = \sqrt{8.4803146 (n+1)}$$ - -where as the conjectured asymptotic for the least $k$ when $F(n)$ becomes negative was: -$$k(n) \sim \frac{2 \sqrt{n}}{\sqrt{\frac{1}{2}}}=\sqrt{8n}$$ -for comparison.<|endoftext|> -TITLE: Why do we say the Fitting subgroup/generalized Fitting subgroup control the structure of a group? -QUESTION [6 upvotes]: I’m learning the Fitting subgroup these days. I’m interested in this topic and particularly in the role that it plays in the structure of groups. Many people on MSE mentioned that the Fitting subgroup/generalized Fitting subgroup controls the structure of a group. Here are some quotes. -@Stephan mentioned in the comment that: - -For a given Fitting subgroup $F$ there are just a finite number of subgroups $U$ which contain the center $Z(F)$, and also $\mbox{Aut}(F)$ is finite, and if $V \le \mbox{Aut}(F)$ also there are just a finite number of homomorphisms $\varphi : U \to \mbox{Aut}(V)$, so there are just a finite number of solvable groups that could be constructed as a semidirect product $G = V \ltimes_{\varphi} U$, in particular so that $U = C_G(F)$, in this way we have a bound on the number of groups that are possible. And that might mean "the Fitting subgroup controls the structure". - -I can understand this. But as a beginner, I want to make sure this is a right way to think. So my first question is: Is his understanding correct? -@Geoff mentioned in the nice answer that ($E(G)$ below refers to the layer of a group): - -The automorphism group of $E(G)$ has a normal subgroup $K$ consisting of the automorphisms which fix every component, and ${\rm Aut}(E(G))/K$ is a permutation group of degree $n,$ where $G$ has $n$ components. - Also, $K/{\rm Inn}(G)$ is isomorphic to a subgroup of a direct product of outer automorphism groups of finite simple groups. Thus the structure of $F^{*}(G)$ controls the structure of $G$ to a large extent. - -I got stuck here. He gave a good answer but I still got a few questions. He said $K$ is a subgroup of ${\rm Aut}(E(G))$, but he then used the notation “$K/E(G)$”. I wonder if $K/E(G)$ is well-defined since $K$ was not defined to contain $E(G)$. If not, it must be a typo, so here comes my second question: What did he intend to refer to by “$K/E(G)$”? (EDIT: I now know that it actually should be “$K/{\rm Inn}((E(G))$”) And could you explain why ${\rm Aut}(E(G))/K$ is permutation group of degree $n$ and why $K/{\rm Inn}(G)$ is isomorphic to a subgroup of a direct product of outer automorphism groups of finite simple groups? I also wonder how did the “THUS” come, namely how was it concluded that $\mathbf{F^*(G)}$ controls the structure of $G$ by giving some properties of $\mathbf{E(G)}$. I know it’s my problem and I know $F^*(G)=F(G)E(G)$. -My third question: Many people mentioned the outer automorphism group when talking about the structure of a group, but it seems to be quite a hard stuff to understand. What do I need to know about the outer automorphisms in terms of constructing a group? -Here are three questions and you can help me with commenting on or answering anyone of them. Or anything that you think can help me understand the importance of the Fitting subgroup in controlling the structure of a group is welcome. Any help is sincerely appreciated! Thanks! - -REPLY [5 votes]: I'll try and give my point of view on the question in the title. -The key facts that answer your question are the following: - -If $G$ is a finite solvable group, then $C_G(F(G))=Z(F(G))$. -If $G$ is a finite group, then $C_G(F^*(G))=Z(F^*(G))$. - -The first fact is classical -- I don't know who should be credited. The second is more modern although I'm still unsure who to credit. In Finite Group Theory, Aschbacher suggests that Bender, Gorenstein and Walter, and Wielandt all did important work related to this. Bender was the first to define $F^*(G)$, and I think of that second fact as "Bender's theorem" although that might be my mistake. -So, why does this answer your question? Because given a group $G$, the quotient $G/Z(F^*(G))$ is isomorphic to a subgroup of ${\rm Aut}(F^*(G))$. Which means that you've chopped the group $G$ up into two pieces, both of which are "controlled" by $F^*(G)$ -- one bit is the center of $F^*(G)$, the other is a subgroup of the automorphism group of $F^*(G)$. -Of course you still have work to do to understand the group $G$: in general if you have two finite groups $H_1$ and $H_2$, then there might be many groups $G$ that have a normal subgroup $N$ such that $N\cong H_1$ and $G/N\cong H_2$. But, despite this, knowing how a group breaks into "pieces" (normal subgroup & quotient) is usually a good start to studying it. -Let's take a couple of examples. - -Suppose that your group $G$ has $F^*(G)\cong C_p^n$, i.e. $F^*(G)$ is an elementary-abelian group of order $p^n$. Then $Z(F^*(G))=F^*(G))$ and ${\rm Aut}(F^*(G))\cong {\rm GL}_n(p)$. If $G/Z(F^*(G))$ is as big as possible -- i.e. it's isomorphic to ${\rm GL}_n(p)$ -- then in almost all cases your group $G={\rm AGL}_n(p)$, an affine group, and in particular the quotient $G/F^*(G)$ is split, and $G$ is determined up to isomorphism. However for certain values of $n$ and $p$, some weird things can happen. -At the other extreme suppose that your group $G$ has $F^*(G)\cong {\rm SL}_2(9)$, a quasisimple group. In this case $Z(F^*(G))=C_2$ and ${\rm Aut}(F^*(G))\cong \mathrm{P\Gamma L}_2(9)$. There are lots of funny things that can happen here. Studying $\mathrm{P\Gamma L}_2(9)$, one can see that $G/F^*(G)$ can, in principle, be isomorphic to one of the following groups $A_6$, $S_6$, ${\rm PGL}_2(9)$, $M_{10}$ or $\mathrm{P\Gamma L}_2(9)$. Let's look at two possibilities: (a) if $G/F^*(G)=S_6$, then the group $G$ is a "double cover of $S_6$", and it is well-known that there are two of these up to isomorphism, although the two different groups behave very similarly from many points of view; (b) on the other hand if $G/F^*(G)=M_{10}$, then,... you've made a mistake. There aren't any groups with $F^*(G)={\rm SL}_2(9)$ and $G/F^*(G)=M_{10}$ even if, in theory, it looks like there could be. This example, along with a lot of other interesting stuff is discussed in the Isoclinism chapter of Conway et al's ATLAS. - -I chose these examples because the nature of $F^*(G)$ is as far from each other as possible -- one elementary-abelian, one "very non-abelian" -- but hopefully you can see that in both cases knowledge of $F^*(G)$ allows you to have very strong information about the full group $G$. -Notice that the discussion above does not apply to an arbitrary normal subgroup of a group $G$: there are an infinite number of groups, up to isomorphism, which have a normal subgroup isomorphic to ${\rm SL}_2(9)$. The difference here is that, thanks to the two facts I stated at the top, knowing the structure of $F^*(G)$, gives you information about a normal subgroup $N=Z(F^*(G))$ and the quotient $G/N$. -It seems reasonable to sum this up by saying that the generalized Fitting subgroup controls the structure of the group...<|endoftext|> -TITLE: Seeking to understand meaning of "von Neumann spectrum" in a paper of Bader–Furman–Shaker -QUESTION [5 upvotes]: In attempting to understand the paper "Superrigidity, Weyl groups, and actions on the circle" of Uri Bader, Alex Furman and Ali Shaker (linked at Furman's page) -I find that towards the end of the proof of Lemma 2.2 they have a situation where $G$ is a locally compact second countable group, $X$ is a Lebesgue space on which $G$ has a measure-class-preserving action, and $Y$ is a probability space on which $G$ has a measure-preserving action, and they speak of an isomorphim (a Banach space isomorphism, presumably) between $(L^{\infty}(Y))^{G}$ and $(L^{\infty}(X \times Y))^{G}$, and then say that this induces a Lebesgue space isomorphism between the corresponding von Neumann spectra, and I'm just wondering if anyone can clarify what "von Neumann spectrum" means in this instance. Presumably not just a subset of $\mathbb{C}$ associated with some Banach space operator. It seems as though it's a little difficult to resolve this question just by Googling. - -REPLY [4 votes]: Let me explain what we mean by the term "von Neumann spectrum". -Before doing so, let me recall the better known Gelfand duality: -the functor $X\mapsto C(X)$, from -the category of compact Hausdorff topological spaces to the category of unital commutative C*-algebras, establishes an equivalence of categories. The functor in the other direction is called the Gelfand spectrum. In particular, the compact space associated with a unital commutative C*-algebra $A$ under this functor is called the Gelfand spectrum of $A$. -Consider now the category of Lebesgue spaces. -Here a Lebesgue space is a standard Borel space endowed with a measure class -and a morphism of such is a class of a.e defined measure class preserving Borel maps, where two such are considered equivalent if they agree a.e. -Recall also that a von Neumann algebra is a unital C*-algebra which is a dual space as Banach space. Let us say that a von Neumann algebra is separable if its predual (which is uniquely defined) is separable wrt the Banach space topology (note that an infinite dimensional von Neumann algebra is never separable as a Banach space, so this terminology should cause no confusion). -There is an obvious functor from Lebesgue spaces to separable commutative von Nuemann algebras, $X\mapsto L^\infty(X)$. By the term von Neumann duality (which is not standard, but it should be) I refer to the fact that this functor establishes an equivalence of categories. The functor in the other direction is called the von Neumann spectrum. In particular, the Lebesgue space associated with a separable commutative von Nuemann algebra $A$ under this functor is called the von Neumann spectrum of $A$. -Again, let me stress that the term "von Neumann spectrum" is not completely standard, but it should be. I currently do not have a handy reference for the above discussion, but any book dealing with the foundations of the theory of von Neumann algebras should cover it. In particular, the above should be discussed when one decomposes a general von Neumann algebra as a direct integral of factors over its center. Note that the center is a commutative von Neumann algebra. The measured space carrying the direct integral is its spectrum. -Before concluding this explanation, I want to briefly expand on something that was not explicitly asked, but is very much related. It is a crucial fact that the variety of objects in the category of Lebesgue spaces is quite dull: up to isomorphisms there is a unique atomless Lebesgue space. But this doesn't mean the category itself is dull, only that its reachness in its morphisms. By analogy, think of the category of separable infinite dimensional Hilbert spaces, where again you have a unique class of objects but a reach variety of morphisms. So really, the miracle is in the group of automorphisms $\text{Aut}(X)$ and things get interesting when you consider representations of groups into this target. Studying these is (one aspect of) Ergodic Theory. It turns out that when one studies such representations of locally compact second countable groups, the von Neumann duality extends equivariantly. This fact is known by the slogan "Mackey's point realization theorem". But I guess I went far enough and end this discussion now. - -Finally, let me make some comments about the paper under consideration and the relevant framework. -First thing, you should know that this paper was not published, for some unfortunate circumstances. In particular, it was never refereed and it might have some rough edges. You should handle it with some care. Overall it is solid, but there might be some glitches in the presentation. -Second, the specific framework under which we considered lemma 2.2 was changed in time. We found that it is convenient to replace the "ergodic with unitary coefficient" assumption with the "metrically ergodic" assumption, -defined in section 2 here. -The latter is formally stronger, but we find it easier to handle (and generalize) in our later work. -In particular, an analogue of lemma 2.2 is given in lemma 3.7 here.<|endoftext|> -TITLE: What is known about the functional square root of the Riemann zeta function? -QUESTION [6 upvotes]: Let us consider the Riemann zeta function $\zeta(s)$, where $s$ can take on values on the domain $\mathbb{R}_{>1}$: -$$\zeta(s) := \sum_{n=1}^{\infty} \frac{1}{n^{s}} .$$ -I wonder what is known about the functional square root(s) of the Riemann zeta function defined on the aforementioned domain (*). In other words, I'm curious about the properties of the function(s) $f$ such that $$f(f(s)) = \zeta(s). \qquad \qquad (1)$$ -Questions - -Has a closed-form solution been found for $f$ in equation $(1)$ ? -If not (which I expect), have partial results been found for such a function? Properties like existence, (non)uniqueness, continuity, or results about the functional square root of the partial sums? $$f(f(s)) = \sum_{n=1}^{k} \frac{1}{n^{s}} \qquad \qquad k \in \mathbb{Z}_{>0}$$ -If so, I would be grateful if you have some pointers to relevant articles or other sources. - -(Cross-post from MSE.) -(*) Edit as per Gerald Edgar's answer, this condition should be changed. We must define $f$ on $(0, \infty) \cup X$ for some subset $X \subset \mathbb{R} \setminus (0,\infty) $. Then $f$ must map $(1,\infty)$ bijectively onto $X$, and $X$ iself onto $(0,\infty)$. Under these conditions, there is still a possibility that $f$ is both continuous and real-valued. I am interested in the properties of such an $f$. - -REPLY [6 votes]: Note that $\zeta$ maps $(1,\infty)$ bijectively onto itself, $\zeta$ is continuous, $\zeta$ is decreasing on $(1,\infty)$. -Suppose $\zeta = f \circ f$ where $f$ also maps $(1,\infty)$ onto itself. A continuous injective function $f$ from an interval to an interval is either increasing everywhere or decreasing everywhere. (This is from the intermediate value theorem.) But in either case, $f\circ f$ is increasing, so $f \circ f \ne \zeta$. -Now of course we can define $f$ on some larger set, say $(1,\infty) \cup X$, where $X$ has the power of the continuum. Let $f$ map $(1,\infty)$ bijectively onto $X$ and $X$ bijectively onto $(1,\infty)$. Easy.<|endoftext|> -TITLE: A question about comparison of positive self-adjoint operators -QUESTION [8 upvotes]: I have the following question but have no idea on its proof (one direction is trivial): - -Let $A$ and $B$ be (bounded) positive self-adjoint operators on a complex Hilbert space $H$. Prove that - $$\limsup_{n \to \infty} \|A^n x\|^{1/n} \le -\limsup_{n \to \infty} \|B^n x\|^{1/n}$$ - holds for every $x \in H$ if and only if $A^n \le B^n$ for each positive integer $n$. - -Any suggestion? -Edit: I suspect the result maybe wrong, for example, if the two limits are equal, then it implies that $A=B$, too strong to be true; anyway, I don't know if the limit (in)equality is so strong. Maybe at most we can say $A^n \le B^n$ for large enough integer $n$. -And for the hard part, it suffices to show $A \le B$ by replacing $A$ with $A^n$ etc. and a similar limit inequality holds. A friend of mine using some trick arguments shows this holds when $H=\mathbb{C}^3$, a good evidence. - -REPLY [5 votes]: The condition $A^n \leq B^n$ for all $n$ defines the spectral order on the positive part of $B(H)$, usually written $A \preceq B$. It makes the positive part of any von Neumann algebra a complete lattice. It's equivalent to saying that $P_{[0,t]}(B) \leq P_{[0,t]}(A)$ for all $t > 0$, where -$P_S(A)$ is the spectral projection of $A$ for $S$. -Suppose $\limsup \|A^nx\|^{1/n} \leq \limsup \|B^nx\|^{1/n}$ for all $x$. The set of $x$ for which the left limsup is $\leq t$ is precisely the range of $P_{[0,t]}(A)$; this is easy to see if you take $A$ to be a multiplication operator. Thus the inequality implies $P_{[0,t]}(B) \leq P_{[0,t]}(A)$ for all $t$, i.e., $A \preceq B$. -(You can see that $P_{[0,t]}(B) \subseteq P_{[0,t]}(A)$ for all $t > 0$ implies $A \leq B$ by noting that $\langle f(A)x,x\rangle \leq \langle f(B)x,x\rangle$ for any simple function $f$ of the form $\sum a_i\chi_{[t_i, t_i + 1)}$. Taking a limit as $f$ approaches the function $t \mapsto t$ yields $\langle Ax,x\rangle \leq \langle Bx,x\rangle$. Also, $P_{[0,t]}(B) \subseteq P_{[0,t]}(A)$ for all $t$ implies the same for $B^n$ and $A^n$, so we actually get $A^n \leq B^n$ for all $n$.)<|endoftext|> -TITLE: Finitely presented non-residually amenable groups without free subgroups -QUESTION [5 upvotes]: Does there exist a finitely presented group that does not contain a nonabelian free group and is not residually amenable? - -REPLY [4 votes]: Let $G$ be the group of piecewise homographic self-homeomorphisms of the real projective line $\mathbb{P}=\mathbb{P}^1(\mathbf{R})$, and $G_\infty$ the stabilizer of $\infty$. So $G_\infty$ can be viewed as the group of self-homeomorphisms of $\mathbf{R}$, piecewise homographic (i.e. piecewise of the form $x\mapsto\frac{ax+b}{cx+d}$) with finitely many breakpoints. -The group $G$, or rather its $C^1$-subgroup was studied a long time ago as automorphism group of a Moulton plane (a fake projective plane); it was also considered by Greenberg. -It is not hard to show that $G_\infty$ has no nonabelian free subgroup (it's an immediate adaptation of the analogous result in the piecewise affine case, e.g., in Thompson's group $F$; see [M, Theorem 14] for a short sketch including all ideas). -Monod discovered that $G_\infty$ is non-amenable. This is quite involved but intuitively, the idea is as follows. Let $A$ be a countable dense unital subring of $\mathbf{R}$, and $G(A)$ the elements of $G$ with breakpoints in $A$ and acting on pieces as $\mathrm{PGL}_2(A)$. Then $G(A)$ induces on $\mathbb{P}$ the same equivalence relation as its subgroup $\mathrm{PGL}_2(A)$. Carrière and Ghys proved that this equivalence relation is non-amenable (in a suitable sense, pertaining to topological dynamics of groups). Then $G_\infty(A)$ also induces the same equivalence, the only difference being $\{\infty\}$ being apart; this tiny difference doesn't affect non-amenability of the action. Now a non-amenable equivalence relation can't be induced by an amenable group. So $G_\infty(A)$ is non-amenable. In this way Monod produced finitely generated subgroups of $G_\infty(A)$ that are not amenable (while without non-abelian free subgroup). -Next Lodha and Moore produced such subgroup that are in addition finitely presented, using explicit generators. I claim their group $\Gamma$ has its third derived subgroup $\Gamma'''$ containing every nontrivial normal subgroup. -Let $\Gamma_{0}$ be the subgroup of compactly supported elements in $\Gamma$ (the support being viewed in $\mathbf{R}$): this is the kernel of the homomorphism consisting in taking germs at $\pm\infty$ (which is valued in a metabelian group). Hence $\Gamma''\subset\Gamma_{0}\subset$ and in particular $\Gamma_0$ is non-trivial. -Their group $\Gamma$ contains a fixed-point-free self-homeomorphism $\rho$, so for every compact subset $K$ there exists $n$ such that $\rho^n(K)\cap K$ is empty. This implies, by a simple commutator trick (see for instance Lemma 3.3.4 in Burillo's book) that every normal subgroup of $\Gamma$ contains $[\Gamma_0,\Gamma_0]$. In particular, $\Gamma'''=[\Gamma_0,\Gamma_0]$. -Thus the intersection of nontrivial normal subgroups of $\Gamma$ is nontrivial. So if $\Gamma$ is not P (P any property passing to subgroups), then it's not residually P. So $\Gamma$ is not residually amenable. -Probably a closer look shows that $[\Gamma_0,\Gamma_0]=\Gamma'''$ is simple, but it would need some more details: the above is enough (it implies that $[\Gamma_0,\Gamma_0]$ has no nontrivial proper conjugacy $\Gamma$-invariant subgroup and in particular is characteristically simple). Maybe $\Gamma''$ itself is simple but this might be significantly more technical. -[CG] Y Carrière, E. Ghys. Relations d'équivalence moyennables sur les groupes de Lie. Comptes Rendus Acad. Sci. t.300 Sér.I no.19, 1985, 677–680. (French) link at Ghys' webpage -[LM] Y. Lodha, J.T. Moore. A nonamenable finitely presented group of piecewise projective homeomorphisms. -Groups Geom. Dyn. 10 (2016), no. 1, 177–200. arXiv link<|endoftext|> -TITLE: Class of lattices that excludes $M_3$? -QUESTION [10 upvotes]: It is well known that a lattice is distributive iff it excludes as a sublattice $N_5$ (the pentagon) and $M_3$ (three unordered elements with a top and bottom). Further, a lattice that only excludes $N_5$ is modular, and consequently a lattice excluding $N_5$ but including $M_3$ is modular but not distributive. -What, if anything, is known about the class of lattices that excludes $M_3$? Does it relate to classes of structures of any particular interest? (Note this is not the same as asking about "nonmodular varieties", since those are varieties which include $N_5$). - -REPLY [10 votes]: As bof says, the class $\mathcal K$ of lattices omitting $\mathbf M_3$ as a sublattice is a quasivariety that is not a variety. As bof also says, this implies that $\mathcal K$ is closed under the formation of sublattices, products, and reduced products, but the class is not closed under the formation of homomorphic images. -But $\mathcal K$ is closed under some types of homomorphic images. -A. (Projection onto a direct factor) -If $L\times L'\in \mathcal K$, then $L, L' \in \mathcal K$. Every quasivariety of lattices has this property, but there exist quasivarieties of algebraic structures that do not have this property. -B. (Quotients of finite members) If $L\in \mathcal K$ is finite, then any homomorphic image of $L$ is also in $\mathcal K$. This property typically fails for quasivarieties, even quasivarieties of lattices. It holds for $\mathcal K$ because $\mathbf M_3$ is projective in the class of finite lattices.<|endoftext|> -TITLE: Why are faithful actions called faithful and who first called them faithful? -QUESTION [9 upvotes]: Sorry for this question. I asked this on MSE and HSM but no one answered and I decided to post it here that is full of experts. - -I want to know why are faithful actions called faithful and who first called them faithful? -Definition: An action $G$ on $X$ is faithful when ${g_1 \neq g_2 \Rightarrow g_1 x \neq g_2 x}$ for some ${x \in X}$ (different elements of $G$ act differently at some point). - -REPLY [16 votes]: The German word is treu, and I would look to papers by Hermann Weyl for its introduction. E.g. Quantenmechanik und Gruppentheorie (1927, p. 16): - -Da das Gruppenschema aus der Darstellung abstrahiert wurde, ist die Darstellung getreu, d.h. verschiedenen Elementen entsprechen verschiedene Abbildungen $U$, oder, was dasselbe besagt, $U(s)$ ist $= \mathbf1$ nur für $s = \mathsf1$. - -or The theory of groups and quantum mechanics (1931, p. 114 — note the scare quotes): - -The realization is said to be faithful -when to distinct elements of the group correspond distinct -transformations: -$$ -T(a)\ne T(b)\text{ when } a\ne b. -$$ -In accordance with the fundamental equation (2.1) the necessary -and sufficient condition for “faithfulness” is that $T(a)$ shall be -the identity only if $a$ is the unit element. - -The same definition occurs in Wigner (1931, p. 79), van der Waerden (1931, p. 178; 1932, p. 32), and in French, Bauer (1933, p. 75).<|endoftext|> -TITLE: What is special to dimension 8? -QUESTION [25 upvotes]: Dimension $8$ seems special, as the partial list below might indicate. -Is there any overarching reason that dim-$8$ is "more special" than, say, dim-$9$? -Surely it isn't it, in the end, simply because $8=2^3$, but $9=3^2$? Or that $\phi(8)=4$ but $\phi(9)=6$? - -The sphere packing problem in dimension -$8$: arXiv. -Annals Journal. -Quanta article. -Octonians (Wikipedia). -John Baez: The Octonions. -E8. See also Garrett Lisi's $E_8$ Theory (Wikipedia). -Wolchover, N. "The Peculiar Math That Could Underlie the Laws of Nature." Quanta magazine (2018). -De Giorgi's conjecture: -Abstract: "A counterexample for $N\ge9$ has long been believed to exist. ...we prove a counterexample [...] for $N\ge9$." -Del Pino, Manuel, Michal Kowalczyk, and Juncheng Wei. "Annals of Mathematics (2011): 1485-1569. -DP,MKM,JW. "On De Giorgi’s conjecture and beyond." PNAS 109, no. 18 (2012): 6845-6850. -Both the Snake-in-a-Box and the Coil-in-a-box problems have been solved -for $d \le 8$: arXiv abs. -For $d>8$, only lower bounds are known. -Bott Periodicity: -"period-$8$ phenomena" -(as per @Meow's comment). -The Simons minimal cone, -a $7$-dimensional cone in $\mathbb{R}^8$ -(as per @DeanYang's comment). - -REPLY [4 votes]: Some special properties of dimension 8, in addition to the ones you identify: - -Bernstein's problem holds up to dimension $n=8$. The only function of $\mathbb{R}^{n-1}$ whose graph in $\mathbb{R}^n$ is minimal is a linear function. This fails in dimension $n=9$, with failure due to the existence of the Simons cone in dimension 8, so it's related to your last bullet point. -There are 4 infinite families of Euclidean reflection groups, with exceptional ones only up to dimension 8. This is related to the existence of the exceptional simplex reflection groups and exceptional Lie algebras. - - - -There are 4 infinite families of holonomy groups of Riemannian manifolds, with two exceptional cases of $G_2$ and $Spin(7)$, the latter being in dimension 8. -As pointed out by @YCor, triality holds for $Spin(8)$. $Spin(8)$ has three 8-dimensional irreducible representations which are permuted by the $S_3$ action associated with the symmetries of the $D_4$ Dynkin diagram. -Cohn and Kumar found various tight simplices including a maximal 15 point tight simplex in $\mathbb{HP}^2$ which is 8 dimensional. A simplex in this case refers to a collection of equidistant points. - -There are several other examples in the comments of phenomena where 8 dimensions is the first dimension in which the phenomenon appears (or is known to appear), but I've listed examples that seem to be special to dimension 8 (and most seem to be connected to the phenomena that you've already identified).<|endoftext|> -TITLE: Recreational mathematical papers -QUESTION [8 upvotes]: Sometimes it is nice to get a less technical paper on mathematics to read and learn something different for a change. These papers often make us discover some new curiosity, to think about the process of learning mathematics or even about mathematics itself and this could be a really nice way to quickly 'scape' our daily problems to amuse ourselves with what I will call here recreational mathematics. As an example, I could mention this paper discussing our unability of seeing beyond three dimensions, which is something really curious. What are other interesting recreational mathematical papers to read during our break time? -Note: Maybe the term 'recreational' fits better when applied to puzzles or something like this but I didn't know what else to call it. - -REPLY [7 votes]: Two of my favorite recreational papers are: -J-F Mestre, R Schoof, L Washington and D Zagier, -Quotients homophones des groupes libres. Homophonic quotients of free groups -Experiment. Math. 2 (1993), no. 3, 153–155. -Norman Wildberger, Real fish, real numbers, real jobs, The Mathematical Intelligencer 21(2), (June 1999), 4-7. Here is a youtube recording of the author reading this paper. -In fact the Intelligencer and the American math monthly have many papers of this type. Also a mainstream mathematical journal St Peterburg Mathematical Journal has a permanent department "Easy reading for professionals".<|endoftext|> -TITLE: Is a 8-dimensional quadratic form recognized by its Lie algebra, modulo equivalence and scalar multiplication? -QUESTION [7 upvotes]: Question. Let $K$ be a field of characteristic zero (large characteristic should be fine too). Let $q,q'$ be two non-degenerate quadratic forms on $K^n$ with $n=8$. Suppose that the Lie algebras $\mathfrak{so}(q,K)$ and $\mathfrak{so}(q',K)$ are isomorphic (these are simple of type $D_4$, 28-dimensional). Does it follow that $q$ is equivalent to some nonzero scalar multiple of $q'$? -A restatement of the question is whether $\mathrm{SO}(q)$ and $\mathrm{SO}(q')$ being isogeneous over $K$ implies the same conclusion. -This is asking the converse of an obvious fact (since $\mathfrak{so}(q,K)$ and $\mathfrak{so}(tq,K)$ are equal for every nonzero scalar $t$. By an elementary argument (see this MO answer), the converse holds for $n\ge 3$ with the possible exception $n=8$ (while it fails for $n=2$ as soon as $K$ has a non-square). The difficulty comes from the existence of triality, namely automorphisms $\mathfrak{so}(q,K)$ not induced by $\mathrm{O}(q,K)$. -The argument can be used to give a positive answer if "the" absolute Galois group of $K$ does not admit as quotient a group of order 3 or 6. This applies to the reals, in which case we can also argue using the signature of the Killing form. An ad-hoc argument can also probably be done for $p$-adic fields. -(In the comments to the linked answer, some hints were given towards a positive answer for $n=8$. I don't know if they're enough to conclude but obviously if so they should be promoted to a full answer.) - -REPLY [9 votes]: Yes: Proposition C.3.14 in Brian Conrad's article Reductive group schemes is that $SO(q)$ determines $q$ up to similarity for all $q$ of dimension $> 2$. (This was pointed out by @user74230 in a comment somewhere.) -This could be viewed as a special case of a more general phenomenon where there is a simple algebraic group $G$ acting on a vector space $V$ and a $G$-invariant homogeneous polynomial $f$ on $V$ so that the twisted forms of $G$ are in bijection with twisted forms of $f$ up to similarity, see Bermudez and Ruozzi Classifying forms of simple groups via their invariant polynomials, where the fact about quadratic forms is stated as Proposition 7.2.<|endoftext|> -TITLE: Kan fibrant replacement for a sphere -QUESTION [5 upvotes]: To compute the simplicial homotopy group of a space $X$, we find a Kan fibrant replacement $X \to Y$ and calculate for that for $Y$, which can be implemented in a computer program. -Computing homotopy groups for spheres are fundamentally hard, and I believe the problem lies in the difficulty of finding their Kan fibrant replacement. -Could you please show why it's hard, for the easiest possible nontrivial case? - -REPLY [6 votes]: Computing homotopy groups for spheres are fundamentally hard, and I believe the problem lies in the difficulty of finding their Kan fibrant replacement. - -Computing the fibrant replacement for simplicial sets is quite -easy: it is given by the Kan fibrant replacement functor Ex^∞. Explicitly, n-simplices in the fibrant replacement of a simplicial set X are maps Sd^k Δ^n → X, for some k≥0. -Here Sd^k denotes the k-fold barycentric subdivision of a simplicial set. -We allow to increase k by further subdividing, this does not change the simplex. -This description is very simple and can be easily programmed into a computer. -The problem is, however, is that the number of simplices grows exponentially -with k, and we also do not have an efficient way to get an a priori upper -bound for k. So some problems are bound to be computationally undecidable, -such as the problem of computing whether π_1 of a simplicial set is trivial or not.<|endoftext|> -TITLE: Question on a subcategory being extension-closed -QUESTION [6 upvotes]: In the article "Homological theory of noetherian rings" by Idun Reiten from 1996, it was stated that it seems to be not known whether the subcategory $\operatorname{Tr}(\Omega^i(\mathrm{mod}\text{-}A))$ (which includes by defintion all projective modules) is extension-closed for $i>1$, where $\operatorname{Tr}$ denotes the Auslander-Bridger transpose and $A$ an Artin algebra. - -Question: Is there some more background on this question? Are there partial solutions for special classes of algebras for this question or is it even answered somewhere in the literature? - -REPLY [3 votes]: Using experiments with QPA, it seems a counterexample has been found. It is quite complicated and surprisingly no simple counterexample seems to work. Maybe someone is interested to check it (I hope there is no mistake). There might be a simplificiation or even a simpler counterexample. -Let $A=KQ/I$ be the finite dimensional quiver algebra over a field $K$ where -\begin{tikzcd} - 2 \arrow[r, "z", shift left=0.75ex] & 1 \arrow[loop, distance=3em, out=35, in=-35, "x"] \arrow[l, "y", shift left=0.75ex] -\end{tikzcd} -(how to use tikz here?) -$Q$ is the quiver with 2 vertices 1 and 2 and there is a loop x from 1 to 1, an arrow y from 1 to 2 and an arrow z from 2 to 1. -The relations are $I=\langle xy, yz, zx, x^3 \rangle$. The algebra is representation-finite and special biserial with 13 indecomposable modules. -Let $S_i$ denote the simple $A$-modules. -$A$ has vector space dimension 7 over the field $K$. Let $e_i$ denote the primitive idempotents of the quiver algebra $A$ corresponding to the vertices $i$ for $i=1,2$. -The indecomposable projective $A$-module $P_1=e_1 A= \langle e_1, x ,y , x^2 \rangle$ has dimension vector $[3,1]$ and the indecomposable projective $A$-modules $P_2=e_2 A= \langle e_2, z, zy \rangle $ has dimension vector $[1,2]$. We remark that $P_2$ is also injective. -For finite dimensional algebras we have $D Tr=\tau$, the Auslander-Reiten translate and since $D$ is a duality, the subcategory $Tr (\Omega^2(\mod-A))$ will be extension closed if and only if $\tau ( \Omega^2(\mod-A))$ is extension closed. We will thus look at the subcategory $\tau ( \Omega^2(\mod-A))$ in the following. -We use the notation $\tau_i:=\tau(\Omega^{i-1})$ in the following for the higher Auslander-Reiten translates. -Let $I=(x+z)A$ denote the right ideal generated by $x+z$, which has vector basis basis $\langle x+z, x^2, zy\rangle$. -$M_1:=A/I$ is an indecomposable $A$-module with dimension vector $[2,2]$. -Then it is easy to see that $\tau_3(M_1) \cong M_1$ and thus $M_1 \in \tau ( \Omega^2(\mod-A))$. -Now let $M_2:=P_2/S_2= e_2 A/zy A$, which is an indecomposable $A$-module with dimension vector $[1,1]$. -Again it is easy to see that $\tau_3(M_2) \cong M_2$ and thus $M_2 \in \tau ( \Omega^2(\mod-A))$. -Now $\dim(Ext_A^1(M_2,M_1))=1$ and there is (up to isomorphism) a unique short exact sequence that is not split: -$$0 \rightarrow M_1 \rightarrow W \rightarrow M_2 \rightarrow 0.$$ -If we show that $W$ is not in $\tau ( \Omega^2(\mod-A))$ then we have shown that the subcategory $\tau ( \Omega^2(\mod-A))$ is not extension closed. -Now $W \cong P_2 \oplus U$, where $U=A/((x+y+z)A)$. -Here the ideal $(x+y+z)A$ has vector space basis $\langle x+y+z,y,x^2,zy \rangle$ and thus $U$ has dimension vector $[2,1]$. -Now let $f_1: P_2 \rightarrow M_2$ be the projective cover of $M_2$ and let -$f_2: U \rightarrow M_2$ be the canonical non-zero map $A/(xA+yA+zA) \rightarrow e_2 A/zyA=M_2$, which is given by the canonical surjections $g: A \rightarrow e_2 A \rightarrow e_2 A/zyA$ and noting that here $(xA+yA+zA)$ is in the kernel of $g$ which induces the map $f_2$. -Let $f:=f_1 \oplus f_2$, then it is easy to see that the kernel of $f$ is isomorphic to $M_1$ and since $W$ is not isomorphic to $M_1 \oplus M_2$ the short exact sequence is not split. -Now we show that the direct summand $U$ of $W$ is not in $\tau ( \Omega^2(\mod-A))$ or equivalently that $K:=\tau^{-1}(U)$ is not in $\Omega^2(\mod-A)$. -Now a module $T$ is a direct summand of a module of the form $X \oplus P$ for $X$ an $n$-th syzygy module and a projective module $P$ if and only if $T$ is a direct summand of a module of the form $P' \oplus \Omega^n(\Omega^{-n}(T))$. -Now for $T=K$, the module $\Omega^2(\Omega^{-2}(T))$ is isomorphic to $S_2 \oplus V$, where $V$ is an indecomposable $A$-module with dimension vector $[2,0]$. -Thus $K$ is not a direct summand of $P' \oplus \Omega^2(\Omega^{-2}(K))$ for any projective module $P'$ and thus $K$ is not a 2-th syzygy module. -Thus the subcategory $\tau ( \Omega^2(\mod-A))$ is not extension-closed.<|endoftext|> -TITLE: Complex plane minus Cantor set admits non-constant bounded harmonic function -QUESTION [9 upvotes]: Let $K\subset [0,1]$ denote the usual 1/3 Cantor set. I know that $\mathbb{C}\backslash K$ has no non-constant bounded analytic function, since the singularity $K$ can be removed. However, a statement I am reading says that $\mathbb{C}\backslash K$ admits a non-constant bounded harmonic function. Why is this true? Any help would be appreciated. - -REPLY [9 votes]: Since the Cantor set $K$ has Hausdorff dimension $\log2/\log 3<1$, it is a removable set for bounded analytic functions, and so, as you say, there is no bounded analytic function outside of $K$. But it does not mean that there are no bounded harmonic functions outside of $K$. Actually, any point of $K$ is regular for the Dirichlet problem. So, choosing any non-constant continuous function $u$ on $K$, there exists a unique harmonic function $f$ in $\overline{\mathbb C}\setminus K$ such that -$$ -\forall\zeta\in K,~\lim_{z\to\zeta}f(z)=u(\zeta). -$$ -Then $f$ is a non-constant bounded harmonic function in $\overline{\mathbb C}\setminus K$. Note that $K$ is of positive capacity so there is no reason that $f$ can be extended to $\mathbb C$ (and actually it cannot be extended). -A proof that any point of $K$ is regular for the Dirichlet problem, that uses Wiener criterion, is given p.100 of -J.B. Garnett, D.E. Marshall, Harmonic measure. New Mathematical Monographs 2. Cambridge University Press, Cambridge, 2005.<|endoftext|> -TITLE: A tree with prime vertices -QUESTION [14 upvotes]: Let us construct a simple (undirected) graph $T$ as follows: -$\quad$ Let the set of all primes be the vertex set of $T$. For each prime $p$, take the least prime $q>p$ such that $2(p+1)-q$ is prime (such a prime q should exist in view of Goldbach's conjecture), and then set an edge connecting $p$ and $q$. -Clearly the graph $T$ contains no circle. If it is connected then it is a tree. -QUESTION. Is the above graph $T$ a tree? -In Feb. 2013, I constructed the graph $T$ and conjectured that $T$ is indeed a tree. For example, the path connecting $2$ and $191$ is -\begin{align*}2&\to 3\to 5\to 7\to 11\to 13\to 17\to 19\to 23\to 29\to 31\to 41, -\\ &\to43\to 47\to 53\to 61\to 71\to 73\to 89\to 97\to 107\to 109 -\\&\to 113\to 127\to 149\to 151\to 167\to 173\to 181\to 191. -\end{align*} -Any ideas towards the solution of the question? Your comments are welcome! - -REPLY [3 votes]: This does not give a complete answer. This provides a strategy for conditional approach. -Given that the nature of the problem is asking for $q$ and $2(p+1)-q$ being simultaneously primes, I think that the question can be approached conditionally (similar to zz7948). -A slight modification is shifting the focus to finding $p0$ and a prime $q_0$ such that the number $N(q)$ of primes $pq_0$ is connected to some smaller prime. -If we can show (computation) that all primes $q\leq q_0$ are connected, then all primes will be connected. -Once we obtain that all primes are connected, we now start to remove the edges to include only the smallest $q$ with $q>p$ and $2p+2-q$ are both primes. -Then, the issue is, whether or not Prim's algorithm of finding spanning tree of a connected graph gives the desired graph.<|endoftext|> -TITLE: Which curves and surfaces are realizable by linkages? references? -QUESTION [11 upvotes]: Ok, so I try to formulate rigorously the question in the title, for which I am asking for references. My definitions may be flawed, so feel free to adjust/correct them! I care about dimensions 2 and 3 below, but feel free to mention general d as well. -Let $d\ge 1$ be an integer, $G=(V,E)$ be a graph, $w:E\to [0, \infty)$ be a weight function. A realization in $\mathbb R^d$ with graph $G$ and weight $w$ is a map $P:V\to\mathbb R^d$ with the further property that $|P(v)-P(v')|=w(v,v')$ whenever $\{v,v'\}\in E$. I will identify such $P$ with its image, I hope it's not a problem. -Edit (May 11, 2020): -As pointed out by Misha, this below definition is not correct. The action of isometries makes the set of all realizations of a linkage always cover all $\mathbb R^d$. He indicates a paper in which a more inclusive definition is formulated in $d=2$. - -(Previous "wrong" definition: I say that a set $A\subset \mathbb R^d$ is realizable by linkages if there exists $G,w$ as above and an cover of $A$ by open sets of $\mathbb R^d$ such that for every $U\subset\mathbb R^d$ in the cover there exist $G,w$ such that that the union of all (images of) realizations of $G,w$ in $\mathbb R^d$ intersected with $U$ coincides with $A\cap U$.) - -To "fix the problem", following the we will allow a subset of vertices of $G$ to be kept fixed in $\mathbb R^d$. In dimension $2$ this apparently generalizes the definition in the above paper, but I think that the result of the paper still allows to reply positively to the $d=2$ case of the question, with little extra work. -Revised definition: We say that $A\subset \mathbb R^d$ is realizable by linkages if there exists a cover of $A$ by open sets of $\mathbb R^d$ such that for every $U\subset\mathbb R^d$ in the cover there exist $G=(V,E)$ and $w$ as above, a subset $F\subset V$, and a map $\phi: F\to\mathbb R^d$, such that the union of all (images of) those realizations of $G,w$ which restricted to $F$ equal $\phi$, intersected with $U$, coincides with $A\cap U$. -Question: Say $d=2$ or $d=3$. Is it true that all algebraic sets $A\subset\mathbb R^d$ are realizable by linkages? What are references for this? -(Note: as of May 11 2020, it appears to me that the case $d=2$ is nicely treated in the answers given, while the case $d=3$ is not yet treated, possibly due to the previously bad definition.) -I found some mention of this, without references on Branko Grünbaum's "Lectures on lost mathematics", dated around 1975, and he says there that $d=2$ case is known, but does not give references, and $d=3$ case is a question by Hilbert which is open (but again no references there). - -REPLY [10 votes]: Erik Demaine and I also included a proof for $d=2$ in Geometric Folding Algorithms: -Linkages, Origami, Polyhedra, Chapter 3. -There we asked if there is a planar (non-crossing) linkage that -"signs your name" (traces any semi-algebraic region), a question -posed by Don Shimamoto in 2004. - -          - - -This was recently settled positively by Zachary Abel in his Ph.D. thesis: -any polynomial curve $f(x,y) = 0$ can be traced by a non-crossing linkage. - -Abel, Zachary Ryan. "On folding and unfolding with linkages and origami." PhD diss., Massachusetts Institute of Technology, 2016. - MIT link.<|endoftext|> -TITLE: What is a module over a Boolean ring? -QUESTION [15 upvotes]: Recall that a (unital) Boolean ring is a (unital) commutative ring $A$ where every element is idempotent; it follows that $A$ is of characteristic 2. There is an equivalence of categories between Boolean rings and Boolean algebras; the Boolean algebra corresponding to a Boolean ring $A$ (which I'll continue to call "$A$") has the same elements as $A$, and multiplication corresponds to "AND", while addition corresponds to "XOR". -Recall also that Stone duality gives an equivalence between the opposite of the category of Boolean algebras and totally disconnected compact Hausdorff spaces. Under this equivalence, a Boolean algebra $A$ is sent to the space $Spec A$ of ultrafilters on $A$, and $A$ is recovered as the algebra of clopen subspaces of $Spec A$. -Question: Let $A$ be a Boolean ring. Let $M$ be an $A$-module. How can the data of $M$ be described in terms of the Boolean algebra $A$, or better yet in terms of the topological space $Spec A$? -One thing to say is that $M$ is naturally an $\mathbb F_2$-vector space, and the $A$-module structure on $M$ corresponds to a representation of $A$ as a sublattice of the lattice of $\mathbb F_2$-subspaces of $M$. This is nice as far as it goes, but I'd really like a description which doesn't mention vector spaces at all, just like the usual definitions of Boolean algebras or totally disconnected compact Hausdorff spaces don't mention rings at all. For instance, it would be nice if this could be described as some kind of representation of the Boolean algebra $A$ on the powerset lattice of a set or something like that. -One possible direction: If $M$ is an $A$-module, then there is a natural preorder on $M$ where $m \leq m'$ iff there is $a \in A$ such that $m = am'$. The set $Spec M$ of ultrafilters on this preorder carries a natural topology with subbasis given by the sets $\hat m = \{p \in Spec M \mid m \in p\}$, for $m \in M$. There is a natural continuous map $Spec M \to Spec A$ given by $p \mapsto \{a \in A \mid \exists m \in p (am = m)\}$. This yields a faithful functor $Spec : Mod_A^{op} \to Top_{/Spec A}$. I wonder if there some additional structure / properties on $Spec M$ which can turn this functor in an equivalence? - -REPLY [16 votes]: Theorem: Given $A$ a boolean ring/boolean algebra then there is an equivalence of categories between the category of $A$-modules and the category of sheaves of $\mathbb{F}_2$-vector spaces on Spec $A$. The equivalence sends every sheaf $\mathcal{M}$ of $\mathbb{F}_2$-vector space to its space of section, $\Gamma(\mathcal{M})$ which is a module over $\Gamma(\mathbb{F}_2) = A$. -Proof: Spec $A$ has a basis of clopen given by the elements of $A$. Using Grothendieck comparison lemma, this allows to give a more algebraic description of sheaves on Spec $A$ as: -For each $a \in A$ a set $F(a)$, with (functorial) restriction maps) $F(a) \to F(b)$ when $b \leqslant a$ such that the natural maps: -$$ F(a \cup b) \to F(a) \times F(b) $$ -is an isomorphism when $a \cap b = 0$. -This allows to exhibit the inverse construction: Given a $A$-module $M$, we define -$$ \mathcal{M}(a) = aM $$ -with the restriction map $aM \to bM$ being given by multiplication by $b$. One easily check that if $a \cap b = 0$ then $aM \times bM \simeq (a \cup b) M$ hence $\mathcal{M}$ is a sheaf of $\mathbb{F}_2$-vector space such that $\Gamma(\mathcal{M}) = \mathcal{M}(1) = M$. -Conversely, starting from any sheaf of $\mathbb{F}_2$-vector spaces $\mathcal{M}$ we have $\mathcal{M}(1) = \Gamma(\mathcal{M}) = \mathcal{M}(a) \times \mathcal{M}(\neg a)$, and the action of $a \in A$ on $\Gamma(\mathcal{M})$ is the identity on $\mathcal{M}(a)$ and zero on $\mathcal{M}(\neg a)$. It follows that $a\Gamma(\mathcal{M}) = \mathcal{M}(a)$, and from there we easily check that the two constructions are inverse of each others. -Note that the exact same argument proves more generally that: -Theorem: If $X$ is a stone space, and $\mathcal{A}$ is a sheaf of rings on $X$, then there is an equivalence of categories between sheaves of $\mathcal{A}$-modules and $\Gamma(\mathcal{A})$-modules. -In particular sheaves of abelian groups on $X$ corresponds to module over the ring of locally constant integer valued functions on $X$. -Even more generally (but the proof is more involved) the same conclusion holds if $X$ is an arbitrary locally compact space, $\mathcal{A}$ is a "c-soft" sheaf of rings and $\Gamma$ is replaced by the "compactly supported section" functor. I prove this as proposition 5.1 of this paper, which is about generalizing this sort of theorem when $X$ is not a space but a topos satisfying apropriate local finiteness assumption, but I'm convince this had been observed before, I just do not know a reference for it.<|endoftext|> -TITLE: What is the name of the real form corresponding to the quaternionic symmetric space? -QUESTION [5 upvotes]: Let $G$ be a compact simple Lie group. Choose a system of positive roots, and let $\mathrm{SU}(2) \subset G$ correspond to the highest root, and $\mathbb{Z}/2 \subset \mathrm{SU}(2)$ the centre. The centralizer of this $\mathbb{Z}/2$ inside of $G$ is a subgroup $H \subset G$ of shape $\mathrm{SU}(2) \circ K = (\mathrm{SU}(2) \times K) / (\mathbb{Z}/2)$. The Dynkin diagram for $H$ can be found by drawing the affine dynkin diagram for $G$, and deleting the node(s) adjacent to the affine root. The now-isolated affine root is the copy of $\mathrm{SU}(2)$, and the rest of the Dynkin diagram for $K$. In the type-A case, $H$ is reductive but not simple, picking up a $\mathrm{U}(1)$ factor; this is because in that case the affine root had two neighbours, not just one. The list of $H$'s is available in the Wikipedia article Quaternion-Kähler symmetric space, because the quotient spaces $G/H$ are precisely the quaternionic symmetric spaces. -Standard arguments then say that $G$ has a real form with maximal compact $H$. It is not the compact form (except for $G = H = \mathrm{SU}(2)$), and it is usually not the split real form. Rather, it is a third canonical real form for any group. For the classical series, it is $\mathrm{SU}(2,n-2)$, $\mathrm{SO}(4,n-4)$, and $\mathrm{Sp}(1,n-1)$. If I am reading Wikipedia correctly, then, together with $\mathrm{SO}(3,n-3)$, these are the real forms that admit quaternionic discrete series representations. - -Does this canonical real form have a standard name in the literature? - -REPLY [2 votes]: In the Crelle paper by Gross and Wallach “On quaternionic discrete series representations, and their continuations” (J. Reine Andgew. Math. 481 (1996) 73–123, available here), the form you mention is described in §3 and called “the quaternionic real form”. -There are other references to “quaternionic real form” found by Google, and all seem to refer to the same thing. I have seen it regularly at least to denote the form of $E_8$ with Cartan index $-24$. So I think we can say that “quaternionic real form” is a reasonably standard term. -(There is still at least some possibility of confusion as this question uses the term, albeit with quotes, to denote a different real form of the $D_n$ series.)<|endoftext|> -TITLE: Natural $\Pi_1$ sentence independent of PA -QUESTION [7 upvotes]: Order invariant graphs and finite incompleteness by Harvey Friedman gives an example of a combinatorial/non-metamathematical $\Pi_1$ sentence that is independent of ZFC. Is there a simpler example of a combinatorial/non-metamathematical $\Pi_1$ sentence that is independent of PA? - -REPLY [3 votes]: You may look at Shelah's paper - ``On logical sentences in PA''. -For a modern exposition of Shelah's work and an alternative example see ``Independence in Arithmetic: The Method of (L, n)-Models''<|endoftext|> -TITLE: Prokhorov theorem on non Polish spaces -QUESTION [6 upvotes]: It is well known that if $X$ is a Polish space and $\mathcal{F} \subset \mathcal{M}_+(X)$ (the set of finite positive Radon measures on $X$) is uniformly tight and bounded in mass, it is relatively compact w.r.t. to the weak topology, i.e. the coarsest topology on $\mathcal{M}_+(X)$ w.r.t. the maps $\mu \mapsto \int_X \varphi \text{ d} \mu$ are continuous for every $\varphi \in C_b(X)$, the continuous bounded real functions on $X$. -I am intersted in a similar statement but in the general case of a Hausdorff topological space $X$. -A possible way to introduce a topology on $\mathcal{M}_+(X)$, when $X$ is a Hausdorff topological space, is to consider the coarsest topology on $\mathcal{M}_+(X)$ w.r.t. the maps $\mu \mapsto \int_X \varphi \text{ d} \mu$ are lower semi continuous for every $\varphi \in LSC_b(X)$, the lower semi continuous and bounded real functions on $X$. -In the book of Schwarz "Radon measures on arbitrary topological spcaes" it is proven that, in this topology, uniform tightness and boundedness in mass together again imply relative compactness. -I am wondering, how much can I enrich the topology on $\mathcal{M}_+(X)$ and still have that Prokhorov theorem holds? -For example, if $X$ is a Hausdorff topological space and I endow $\mathcal{M}_+(X)$ with the coarsest topology w.r.t the maps $\mu \mapsto \int_X\varphi \text{ d}\mu$ are continuous for every $\varphi \in LSC_b(X) \cup USC_b(X)$, what happens? -Here $LSC_b(X)$ (resp. $USC_b(X)$) is the set of the lower (resp. upper) semi continuous bounded real functions on $X$. - -REPLY [2 votes]: This seems to be a pure general topology problem. Enriching the topology will necessarily destroy the Prohorov property whenever the enrichment matters. If you have two nested Hausdorff topologies and a set is relatively compact in the finer topology, then the closure is the same under both topologies and the trace topologies coincide on the closure. -To see this, let $\tau$ and $\tau'$ be Hausdorff topologies on $X$ such that $\tau\subseteq\tau'$ and let $R\subseteq X$ be relatively compact in the topology $\tau'$. Trivially, $R$ is also relatively compact in the coarser topology $\tau$ and the $\tau'$-closure of $R$ is a subset of the $\tau$-closure of $R$. Let $x$ be a point in the $\tau$-closure of $R$ and $\langle x_\alpha\rangle$ be a net in $R$ converging to $x$ under $\tau$. Since $R$ is relatively compact under $\tau'$, a subnet will $\tau'$ converge to a point $x'$. But this subnet will also converge to $x'$ under $\tau$. Since $\tau$ is Hausdorff , $x=x'$ and the closure of $R$ is the same under both topologies. Let $C$ be this compact closure. The identity is a continuous function from $(C,\tau')$ to $(C,\tau)$ and will therefore map compact subsets of $C$ to compact subsets of $C$. But these are exactly the closed subsets of $C$ under both topologies, so both topologies must coincide on $C$.<|endoftext|> -TITLE: Is the tensor product of chain complexes a Day convolution? -QUESTION [9 upvotes]: Recently, Jade Master asked whether the tensor product of chain complexes could be viewed as a special case of Day convolution. Noting that chain complexes may be viewed as $\mathsf{Ab}$-functors from a certain $\mathsf{Ab}$-category $\mathsf{C}$, Yuri Sulyma suggested¹ that maybe we could obtain the tensor product of two chain complexes as a Day convolution by endowing $\mathsf{C}$ with the monoidal structure given by $[n]\otimes_\mathsf{C}[m]\overset{\mathrm{def}}{=}[n+m-1]$. -Questions: Is this affirmation true? More precisely: - -Given two chain complexes $X_\bullet$ and $Y_\bullet$ on an abelian category $\mathcal{A}$, is their Day convolution as $\mathsf{Ab}$-functors from $(\mathsf{C},\otimes_\mathsf{C})$ the usual tensor product of chain complexes $\otimes_{\mathsf{Ch}(\mathcal{A})}$? -If not, is there some other monoidal structure on $\mathsf{C}$ for which Day convolution gives $\otimes_{\mathsf{Ch}(\mathcal{A})}$? -If this fails too, is there perhaps another way to view $\otimes_{\mathsf{Ch}(\mathcal{A})}$ as a special case of some general construction in enriched category theory? - - -¹Note that his account is protected and hence his reply is not public. - -REPLY [17 votes]: The answer to the question posed in the title of your post is yes, the tensor product of chain complexes is a Day convolution product. The important thing to note is that, to define a Day convolution monoidal structure on the $\mathcal{V}$-enriched functor category $[\mathcal{C},\mathcal{V}]$ (where $\mathcal{V}$ is a complete and cocomplete symmetric monoidal closed category, e.g. $\mathbf{Ab}$), we needn't demand $\mathcal{C}$ to be a monoidal $\mathcal{V}$-category: it suffices for $\mathcal{C}$ to be a promonoidal $\mathcal{V}$-category. This is the generality at which Day convolution was originally defined in Day's thesis, which may be found here (see also his earlier paper in the Reports of the Midwest Category Seminar IV, where the word "premonoidal" was used). -A promonoidal structure on a small $\mathcal{V}$-category $\mathcal{C}$ consists of tensor product and unit "profunctors", i.e. $\mathcal{V}$-functors $P \colon \mathcal{C}^\mathrm{op}\times\mathcal{C}^\mathrm{op} \times \mathcal{C} \to \mathcal{V}$ and $J \colon \mathcal{C} \to \mathcal{V}$, together with associativity and unit constraints subject to the usual two -"pentagon" and "triangle" axioms. Given a promonoidal structure on $\mathcal{C}$, we may construct the Day convolution monoidal structure on $[\mathcal{C},\mathcal{V}]$, whose tensor product is given at a pair of $\mathcal{V}$-functors $F,G \in [\mathcal{C},\mathcal{V}]$ by the coend -$$F\ast G = \int^{A,B \in \mathcal{C}} P(A,B;-) \otimes FA \otimes GB$$ in $\mathcal{V}$, -and whose unit object is the $\mathcal{V}$-functor $J \in [\mathcal{C},\mathcal{V}]$, and so on. This monoidal structure on $[\mathcal{C},\mathcal{V}]$ is biclosed (i.e. the tensor product $\mathcal{V}$-functor has a right $\mathcal{V}$-adjoint -- equivalently, preserves (weighted) colimits -- in each variable). In fact, every biclosed monoidal structure on $[\mathcal{C},\mathcal{V}]$ arises in this way from some promonoidal structure on $\mathcal{C}$. (For instance, one recovers the $\mathcal{V}$-functor $P$ from the tensor product $\ast$ by $P(A,B;C) = (\mathcal{C}(A,-) \ast \mathcal{C}(B,-))C$.) -So, since the $\mathbf{Ab}$-category $\mathbf{Ch}$ of chain complexes is (equivalent to) an $\mathbf{Ab}$-enriched functor category $[\mathcal{C},\mathbf{Ab}]$ (for the $\mathbf{Ab}$-category $\mathcal{C}$ described in the question to which you linked), and since the standard monoidal structure on $\mathbf{Ch}$ is $\mathbf{Ab}$-enriched and biclosed, this monoidal structure must be the Day convolution monoidal structure for some promonoidal structure on $\mathcal{C}$. And it isn't too hard to describe that promonoidal structure. For instance, (presuming I haven't bungled the calculation) the functor $P$ is defined on objects by $$P(i,j;k) = \begin{cases} \mathbb{Z} & \mathrm{if\,\,} i+j=k, \\ -\mathbb{Z} \oplus \mathbb{Z} & \mathrm{if\,\,} i+j=k+1, \\ -\mathbb{Z} & \mathrm{if\,\,} i+j=k+2, \\ -0 & \mathrm{else}. -\end{cases}$$<|endoftext|> -TITLE: Has anyone seen this generalization of the snake lemma? Is it useful? -QUESTION [15 upvotes]: I originally posted this question on MSE (link), but was suggested to post here instead. -While learning about spectral sequences a friend of mine found a proof of the snake lemma using spectral sequences. We noticed that the proof works equally well for larger bicomplexes. Particularly if you have an exact (anti)-commutative diagram - -you get an -exact sequence. - -We also have a little write-up of the proof here. We looked around, but couldn't find any reference to this anywhere, and no one else we talked to had really thought about it before. While toying with this we realized that the hypothesis is quite strong. That is, it is pretty difficult to find any interesting exact bicomplexes of the right size. I'm starting to suspect that there might not really be any interesting examples of this, and that that is the reason we haven't found anything about it anywhere. -So we're wondering, has anyone seen this before? Can anyone think of any non-trivial examples or applications of this? - -REPLY [8 votes]: Spectral sequences are obtained by applying the snake lemma ad nauseam, essentially. You are doing that (finitely many times) in your spectral sequence argument. -Let me illustrate. Consider the case you have a diagram that has three exact rows and four exact columns, and start with the top right corner space $C_{1,4}$, and pick $x$ there. -If $d_v(x)=0$, pull this back to some $x'\in C_{1,3}$ and apply $d_v(x')\in C_{2,3}$. Then $d_h$ of this is zero, so there is some $x''$ in $C_{2,2}$ such that $d_h(x'') = d_v(x')$. Apply $d_v$ again to fall into $C_{3,2}$. Then take the class of this, and this is your map. -So, in general, you start with $x_{1,n} \in C_{1,n}$ in the kernel of $d_v$, -pull it back to $x_{1,n-1}$, apply $d_v$, pull it back, apply $d_v$, keep going, until you reach $C_{n+1,1}$. You can do this because your rows are exact, and this (I think) is all that you need, you don't need your columns to be exact (as in the snake lemma). -You can do this in general for any $n$, with no spectral sequences, its just a longer version of the snake lemma. The spectral sequence formalism is packaging neatly, but this -is what is going on.<|endoftext|> -TITLE: Is the preimage of a nonreduced subscheme via a proper map nonreduced? -QUESTION [11 upvotes]: Let $f\colon X \to Y$ a surjective proper map between smooth varieties over an algebraically closed field $k$ of characteristic zero. Let $Z\subset Y$ be a closed non-reduced subscheme. Is the preimage $f^{-1}(Z)$ nonreduced? -(The situation I am interested in is the resolution of an ideal sheaf, but I do not know if this hypothesis helps.) -Notes: -1- this is a refining of this question, asked by Shende and answered by Sawin. In the answer, $Z$ was supported on the singular locus of $Y$. In my question, I am adding smoothness and properness hypotheses. -2- The following I think is an important example. Let $f\colon X\to Y$ be the blow-up of a smooth point $p$ on a surface. Let $Z$ be isomorphic to $Spec(k[x]/x^2)$, supported at $p$, and $x$ goes to a tangent direction $v$. Then $f^{-1}(Z)$ is the exceptional divisor (with reduced scheme structure) union an embedded point at $v$, so it is indeed nonreduced. However, if you remove $v$, then the map is still surjective but $f^{-1}(Z)$ is reduced. So properness is important. - -REPLY [3 votes]: If $Z$ is not generically reduced, then its pullback is not reduced. (The argument can show that the pullback is not generically reduced, but not the way I wrote it.) -Lemma: To show that the pullback of $Z$ is not reduced, it suffices to check that there is a smooth curve $C$ mapping to $Y$ such that the pullback of $\mathcal I_Z$ and $\sqrt{\mathcal I_Z}$ to $C$ differ, where $\mathcal I_Z$ is the ideal sheaf of $Z$ and $\sqrt{\mathcal I_Z}$ is its radical. -Proof: By the valuative criterion, you can map some ramified cover of $C$ to $X$, lifting the map to $Y$, and then pulling back to this ramified cover shows that $f^* \sqrt{\mathcal I_Z}\neq f^* \mathcal I_Z$. Now because $f^* \mathcal I_Z$ contains a power of $f^* \sqrt{\mathcal I_Z}$, it follows that $f^* \mathcal I_Z$ is not radical and thus $f^* Z$ is not reduced. -Now let's check that if $Z$ is generically non-reduced, there exists such a $C$. To do this, work locally near the generic point of $Z$, so that the radical just becomes the maximal ideal at this point. Since $Z$ is not reduced, the map from the ideal of $Z$ to the Zariski cotangent space at the generic point (i.e. generators of this maximal ideal) is not surjective, so there exists a nonzero vector in the Zariski tangent space which is perpendicular to the image of the ideal of $Z$. -Pick a smooth curve $C$ whose tangent vector is that nonzero vector. The pullback of the maximal ideal to $C$ will have multiplicity $1$ while the pullback of $I_Z$ will have multiplicity $>1$, so they will be distinct. - -However, this approach (showing equality between $\mathcal I_Z$ and $\sqrt{\mathcal I_Z}$) will not work in general. Let $Z$ be the vanishing locus of $$x_1 x_3, x_2 x_4, (x_1 x_2 - x_2 x_3), (x_2 x_3 - x_3 x_4), (x_3 x_4 - x_1 x_4) $$ -as well as $x_i x_j x_k$ for all triples $i,j,k$, not all equal. The point is that the induced reduced subscheme of $Z$ is the vanishing locus of $$x_1x_2,x_1x_3,x_1x_4, x_2x_3,x_2x_4, x_3,x_4$$ and $Z$ differs from this by an embedded point. -I claim the pullback of $Z$ and its induced reduced subscheme to the blowup of $\mathbb A^4$ at $0$ are equal (but neither is reduced since they each contain a double neighborhood of the exception divisor). -In a typical affine chart $x_1 =a_1 z, x_2 =a_2 z, x_3 =a_3 z, x_4 = z$, the pullback of the induced reduced is the vanishing locus of $z^2 (a_1,a_2,a_3)$ and the pullback of $Z$ is the vanishing locus of $$z^2 (a_1a_3, a_2, (a_1a_2-a_2a_3), (a_2a_3 - a_3), (a_3 - a_1)) $$ -which equals $z^2(a_1,a_2,a_3)$ since we can cancel $a_2a_3$ with $a_2$ and get $a_3$ then cancel $a_3$ and get $a_1$. -By the symmetry rotating the four variables, all the affine charts look like this, so the pullbacks are equal as ideal sheaves.<|endoftext|> -TITLE: Surfaces with non-constant negative curvature -QUESTION [11 upvotes]: Are there any nice models of surfaces with non-constant negative curvature, analogous to the Poincare disk for constant negative curvature. I have found lots of general results and theory but no nice clean models. - -REPLY [13 votes]: If you just want examples for which it's not hard to figure out how the geodesics behave, here's a class of examples with negative and non-constant curvauture in the plane where the geodesics are relatively easy to understand: -Let $a$ and $b$ be smooth functions on $\mathbb{R}$ such that $a(x)+b(y)>0$ for all $x,y\in\mathbb{R}$ and consider the metric -$$ -g = \bigl(a(x) + b(y)\bigr)(\mathrm{d}x^2 + \mathrm{d}y^2) -$$ -on $\mathbb{R}^2$. The curvature of this metric is -$$ -K = \frac{a'(x)^2+b'(y)^2-\bigl(a''(x)+b''(y)\bigr)\bigl(a(x)+b(y)\bigr)} -{2\,\bigl(a(x)+b(y)\bigr)^3} -$$ -It's easy to choose $a$ and $b$ so that $K<0$. For example, $a(x) = x^2+1$ and $b(y) = y^2+1$ gives a complete metric on $\mathbb{R}^2$ that has non-constant negative curvature $K = -4/(x^2{+}y^2{+}2)^{3}<0$. -Note that, taking $a$ (respectively, $b$) to be a constant gives a metric $g$ that has a Killing vector field, namely $\partial/\partial x$ (respectively, $\partial/\partial y$), but, for generic choices of $a$ and $b$, the metric $g$ will have no Killing vector field. -As for geodesics, the good thing about these metrics (called Liouville metrics in the literature) is that their geodesic flows are integrable: Any unit speed geodesic $(x(t),y(t))$ satisfies -$$ -\bigl(a(x)+b(y)\bigr)\bigl(\dot x^2+\dot y^2\bigr) = 1 -\quad\text{and}\quad -\bigl(a(x)+b(y)\bigr)\bigl(b(y)\,\dot x^2- a(x)\,\dot y^2\bigr) = c -$$ -for some constant $c$. (Note that, when either $a$ or $b$ is constant, this second 'first integral' of the geodesic equations specializes to the well-known 'Clairaut integral' for surfaces of revolution.) -In particular, -$$ -\bigl(b(y)-c)\bigr)\,\dot x^2 - \bigl(a(x)+c)\bigr)\,\dot y^2 = 0, -$$ -and, assuming that you are in a region when $a(x){+}c$ and $b(y){-}c$ are both positive, -$$ -\frac{\mathrm{d}x}{\sqrt{a(x)+c}} \pm \frac{\mathrm{d}y}{\sqrt{b(y)-c}}=0, -$$ -which gives two foliations of this region by geodesics, which can be found by quadrature. -In any case, you will have good qualitative control over these geodesics and can draw some nice pictures. -Added remark (12 May 2020): As an example of what one can do with this more explicit information, you might be interested in this answer of mine to an old question about Riemannian surfaces for which one can compute an explicit distance function.<|endoftext|> -TITLE: Can learning Riemann surfaces be more beneficial than numerical analysis for an analyst? -QUESTION [10 upvotes]: I am in master program of mathematics, specialized in PDE and numerical analysis. Now I am trying to decide which classes to take for next semester. Of course I want to become an expert in my field, but I am also interested in Geometry. I learned some differential geometry in my bachelor class, and I want to take the next tier, which is Riemann surfaces in my university program. But I am not very sure this will be the right choice, since if I don't take this, I will have some time to take more numerical analysis courses which directly connect to my area. But in same time I feel like I can self-study some numerical methods in the future. So now, the question is: - -Can taking a Riemann surface course be more beneficial (which is a very vague concept) over taking more numerical analysis course? -If I want to do research in my future career (i.e. proceed to PhD) will there be any topics in analysis related to Riemann surfaces? - -REPLY [10 votes]: Painleve - It came to appear that, between two truths of the real domain, the easiest path quite often passes through the complex domain. -Hadamard - It has been written that the shortest and best way between two truths of the real domain often passes through the imaginary one." -Read first "The Magic Wand Theorem of A. Eskin and M. Mirzakhani" by Anton Zorich for motivation on Riemann sufaces. -And, "A Singular Mathematical Promenade" by Etienne Ghys (particularly p. 87-93 and glance at the Wikipedia article on monodromy). -More prosaically: -To understand integral transform solutions (Mellin, Laplace, Fourier) to pdes, you need to understand poles and branch cuts of complex functions. -Solutions to Laplace's equation in two dimensions are called harmonic functions that are the real and complex components of a complex function and give mutually orthogonal contour lines on the complex plane and Riemann sphere. -From "THE THEOREM OF RIEMANN-ROCH AND ABEL’s THEOREM" by Siu: -The theorem of Riemann-Roch and Abel’s theorem could be interpreted as answering the question: for which configuration of charges, dipoles, or multipoles on a compact Riemann surface of genus ≥ 1 would the flux functions (whose level curves are the flux lines and which are the harmonic conjugates of the electrostatic potential functions) in the case of the theorem of Riemann-Roch, or their exponentiation after multiplication by 2πi in the case of Abel’s theorem, be single-valued on the Riemann surface so that the flux lines are closed curves? -At a more basic level for numerical analysis, in understanding convergence of real power series and, therefore, series solns. to pdes, you need to understand singularities (poles and branch cuts in the complex domain) and these involve Riemann surfaces. -Same for Newton (finite difference) and sinc function (Nyquist-Shannon) interpolations of sequences of real/complex numbers and their numerical analytic continuations and for asymptotic series a la Poincare. (Norlund, Poincare, and Berry wrote well on these topics.) -Helps in understanding convolutions, Dirac Delta functions and their derivatives, and, therefore, fractional calculus and operational calculus. -Necessary in understanding basic string theories. -The list is endless. Without such knowledge, you live (perhaps blissfully) in Abbott's Flatland. -The examples really suggest that you may be imposing a gratuitous, restrictive dichotomy--there is plenty of synergy between the study of numerical analysis and Riemann surfaces and both provide paths to other intriguing areas of the grand, evolving tapestry of mathematics, engineering, and science. (Of course, if you are looking where the money is in America, well I suggest a medical degree or starting a munitions factory.)<|endoftext|> -TITLE: Game on a square grid -QUESTION [5 upvotes]: Not research level, comments are welcome. -Consider the following game: - -The board is the vertices of an $n$ by $n$ square grid. -Two players take moves in turns. -A move is picking two vertices and drawing a straight line between them. -If the line intersects another line or passes through a third vertex, -the game ends and the player who made the move loses the game. Two or more lines are allowed to end at the same vertex. - -Is there winning strategy depending on $n$? -Partial result: -We believe if we take the board to be the vertices of regular polygon, -the first player always wins, even if they don't have any skills -except finding a non-losing move if it exists. - -REPLY [11 votes]: Doesn't a symmetry argument make this question rather simple? -If $n$ is odd, the second player (blue) can always mirror the first player's (red) moves (reflected through, or rotated by $\pi$ radians around the origin of the grid, i.e. its center vertex), so the first player loses. - -If $n$ is even, the first player can draw one of the diagonals of the central square (orange; this move can't be mirrored) and can then mirror all of the second player's moves, so the second player loses.<|endoftext|> -TITLE: Relationship between fans and root data -QUESTION [6 upvotes]: A (split) reductive linear algebraic group is equivalently described by combinatorial information called a root datum. -A toric variety is described by combinatorial information called a fan. -Both correspondences use the character lattice. -The reference: -http://u.cs.biu.ac.il/~margolis/Linear%20Algebraic%20Monoids/Renner-%20Lin.%20Alg.%20Monoids.pdf -says that spherical varieties are a nice class of objects that include -all my favorite spaces (e.g. symmetric spaces, toric varieties) . And, moreover, that a spherical variety is equivalent to combinatorial information called a colored fan. Is there any way of recovering a root datum from a colored fan? Or is a reductive group actually given as part of the data of a colored fan? -Are fans/ Toric varieties and root data/ reductive groups both special cases of a larger pattern (for example, colored fans/ spherical varieties)? - -REPLY [8 votes]: (1) A (connected) reductive group $G$ over an algebraically closed field $k$ is described by a combinatorial object called the based root datum ${\rm BRD}(G)$. -(2) A spherical homogeneous space $Y=G/H$ is a homogeneous space on which a Borel subgroup $B$ of $G$ acts with an open Zariski-dense orbit. It is described (uniquely at least in characteristic 0) by its homogeneous combinatorial invariants. These combinatorial invariants constitute an additional structure on ${\rm BRD}(G)$. -(3) A spherical embedding $G/H\hookrightarrow Y^e$ is a normal $G$-variety $Y^e$ containing a spherical homogeneous space $G/H$ as an open dense $G$-orbit. It is described by its colored fan, which is an additional structure on the homogeneous combinatorial invariants. -By spherical varieties one means spherical homogeneous spaces and spherical embeddings. -Therefore, I think that the based root datum of $G$ should be regarded as a part of data describing the $G$-variety $Y^e$. -In the case when $G=T$ is a torus, we take $H=1$, and then the spherical embeddings of $G/H=T$ are the same as the toric varieties for $T$, and the corresponding colored fans are just fans. -Reference: Nicolas Perrin, On the geometry of spherical varieties. - -REPLY [4 votes]: Not an answer, but: you can construct a fan from a root system. Let $R$ be a root system in an Euclidean space, and let $\Lambda_R$ be the root lattice with dual lattice $\Lambda_R^\vee$. The fan $\Sigma$ in $\Lambda_R^\vee$ associated to $R$ consists of the Weyl chambers of $R$ and all their faces. For instance, if $R=A_1$, then the associated toric variety is $\mathbf{P}^1$. I don't know how to determine when a fan comes from a root system, but I'm guessing someone here does.<|endoftext|> -TITLE: what is the number of paths returning to 0 on the hexagonal lattice -QUESTION [7 upvotes]: I am looking for an estimation of the number of paths of length $n$ going from 0 to 0 on the hexagonal (or honeycomb) lattice. -I can find plenty on references on self avoiding paths, but I am looking into every paths. Is it considerably more difficult? Has anyone a reference? - -REPLY [11 votes]: This is answered by Ian Agol here, with the reference "All Roads Lead to Rome-Even in the Honeycomb World", Brani Vidakovic, Amer. Statist. 48 (1994) no. 3, 234-236. -An exact formula is -$$ p(n) = \sum_{k=0}^m \binom{2k}{k} \binom{m}{k}^2$$ -if $n= 2m$ is even, and $0$ otherwise. -This is sequence A002893 on OEIS. -According to OEIS, the number of paths is asymptotic to -$$ p(n) \sim \frac{1}{2\pi n} 3^{n + 3/2}$$ -when $n$ is even, -which agrees with the estimate given by shurtados. -In the above reference, Vidakovic proves that $p(n) \geq C \cdot {3^n}/{n}$ for some constant $C$.<|endoftext|> -TITLE: Less fundamental applications of Zeta regularization: -QUESTION [6 upvotes]: As we all know, zeta regularization is used in Quantum field theory and calculations regarding the Casimir effect. - -Are there less fundamental applications of zeta function regularization? By "less fundamental" I mean - it 'naturally' pops up in more of an artificially / purely mathematically ideal constructed scenario. - -Thanks! - -REPLY [5 votes]: Zeta function regularization computes the asymptotics of smoothed sums. -https://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/ -Also, the regularized determinant of the Laplacian is related to the Ray-Singer analytic torsion, which is equal to the Reidemeister torsion: -https://en.wikipedia.org/wiki/Analytic_torsion<|endoftext|> -TITLE: Natural number solutions for equations of the form $\frac{a^2}{a^2-1} \cdot \frac{b^2}{b^2-1} = \frac{c^2}{c^2-1}$ -QUESTION [5 upvotes]: Consider the equation $$\frac{a^2}{a^2-1} \cdot \frac{b^2}{b^2-1} = \frac{c^2}{c^2-1}.$$ -Of course, there are solutions to this like $(a,b,c) = (9,8,6)$. -Is there any known approximation for the number of solutions $(a,b,c)$, when $2 \leq a,b,c \leq k$ for some $k \geq 2.$ -More generally, consider the equation $$\frac{a_1^2}{a_1^2-1} \cdot \frac{a_2^2}{a_2^2-1} \cdot \ldots \cdot \frac{a_n^2}{a_n^2-1} = \frac{b_1^2}{b_1^2-1} \cdot \frac{b_2^2}{b_2^2-1}\cdot \ldots \cdot \frac{b_m^2}{b_m^2-1}$$ -for some natural numbers $n,m \geq 1$. Similarly to the above question, I ask myself if there is any known approximation to the number of solutions $(a_1,\ldots,a_n,b_1,\ldots,b_m)$, with natural numbers $2 \leq a_1, \ldots, a_n, b_1, \ldots, b_m \leq k$ for some $k \geq 2$. Of course, for $n = m$, all $2n$-tuples are solutions, where $(a_1,\ldots,a_n)$ is just a permutation of $(b_1,\ldots,b_n)$. - -REPLY [2 votes]: Above equation shown below, has solution: -$\frac{a^2}{a^2-1} \cdot \frac{b^2}{b^2-1} = \frac{c^2}{c^2-1}$ -$a=9w(2p-1)(18p-7)$ -$b=4w(72p^2-63p+14)$ -$c=3w(72p^2-63p+14)$ -Where, w=[1/(36p^2-7)] -For, $p=0$ we get: -$(a,b,c)=(9,8,6)$<|endoftext|> -TITLE: Jack function in power symmetric basis -QUESTION [5 upvotes]: In Macdonald's book, the Jack symmetric function $J_{\lambda}(x_1,\ldots, x_n)$ for a partition $\lambda$ -is defined by three properties (orthogonality, triangularity, and normalization). In the following paper (http://www-math.mit.edu/~rstan/pubs/pubfiles/73.pdf) its existence and uniqueness appear as Theorem 1.1. -The Jack symmetric functions can be seen as eigenfunctions for the operators -$$D(\alpha)= \alpha/2 \sum_{i=1}^{n}x_i^2\frac{\partial}{\partial_i^2}+\sum_{i\neq j}\frac{x_i^2}{x_i -x_j}\frac{\partial}{\partial x_i}$$ -with the eigenvalues as given in Theorem 3.1 of the above paper. -Now a recent paper of Chapuy and Dolega -(https://arxiv.org/pdf/2004.07824.pdf) - defines -the following operator, which is given in the power symmetric basis, -$$D_{\alpha}^{'}= \alpha/2 \sum_{i,j\geq 1}ij p_{i+j}\frac{\partial^2}{\partial p_i \partial p_j} -+ 1/2 \sum_{i,j\geq 1}(i+j) p_{i}p_{j}\frac{\partial^2}{\partial p_{i+j}}+(\alpha -1)\sum_{i\geq 1}\frac{i(i -1)}{2}p_i\frac{\partial}{\partial p_i}$$ -and it defines the Jack symmetric functions to be those function which are eigenfunctions to these operators; they give the eigenvalues in terms of $\alpha$, which takes a nice form. In the paper it appears in Proposition 5.1. -My question is how to derive this operator from Stanley's paper and to apply it on the Jack symmetric function. Do we need to express the Jack symmetric function in the power symmetric basis? I cannot do it in the case of a general partition. Also, the operator in Stanley's paper involves finitely many variables, but in the Chaupy and Dolenga paper it involves infinitely many. I hope someone can give me more details. - -REPLY [3 votes]: This explanation can be found in Macdonald's book "Symmetric Functions and Hall Polynomials" by looking at Ex.VI.4.3. -Note that Stanley's Laplace-Beltrami operator $D(\alpha)$ depends on $n$ and acts on the algebra $\mathbb{Q}(\alpha)\otimes \Lambda^n$, where $\Lambda^n$ denotes the algebra of symmetric polynomials in $n$ variables $x_1,\dots,x_n$. Because of that I prefer to modify your notation and denote Stanley's operator by $D_n(\alpha)$. Let me introduce a modified version of this operator $D'_n(\alpha)$, which also acts on $\mathbb{Q}(\alpha)\otimes \Lambda^n$ and is defined by setting -$$ D'_n(\alpha)f := \big(D_n(\alpha)-(n-1)\deg(f)\big)f$$ -for homogenous $f$ and extended by linearity. -Recall that the algebra $\Lambda$ of symmetric functions is defined as the projective limit of $\Lambda_n$ with respect to the morphism $\rho_n : \Lambda_{n+1}\to \Lambda_n$ which kills the last variable: -$$\rho_n(f)(x_1,\dots,x_n) := f(x_1,\dots,x_n,0).$$ -It is easy to check that -$$\rho_n D'_{n+1}(\alpha) = D'_{n}(\alpha)\rho_{n},$$ -so you can define an operator $D'_\alpha := \lim D'_{n}(\alpha)$ which acts on $\mathbb{Q}(\alpha)\otimes \Lambda$. Stanley's computations used in the proof of his Theorem 3.1 give you immediately an expression for $D'_\alpha$ as a differential operator in power-sums.<|endoftext|> -TITLE: Are the AHSS and Adams spectral sequence the same when computing connective Morava K-theory of a space? -QUESTION [18 upvotes]: Let $k(n)$ be the $n$th connective Morava K-theory, with $k(n)_* = \mathbb F_p[v]$ where $|v| = 2p^n-2$. If $X$ is a space or a spectrum (assumed bounded below), one can compute $k(n)_*(X)$ using either the classical Adams spectral sequence or the even more classical Atiyah-Hirzebruch spectral sequence. -Both spectral sequences are spectral sequences of modules over $k(n)_*$. (In the ASS the bidegree of $v$ is $(1,2p^n-1)$.) Both spectral sequences start with $k(n)_* \otimes H_*(X;\mathbb F_p)$, at $E_1$ for the ASS, and $E_2$ for the AHSS. Both have first possible nontrivial differential given by the formula $ d(x) = vQ_n(x)$, -where $Q_n$ is the $n$th Milnor primitive in the Steenrod algebra (acting on homology by going down in degree by $2p^n-1$). -So it seems that these must really be the same spectral sequence. Is this true? Is this a fact in the literature? (I am a tad bothered by the fact that the AHSS arises from an increasing filtration of $k(n) \wedge X$ while the ASS arises from a decreasing filtration of $k(n) \wedge X$.) - -REPLY [12 votes]: Tyler's comment answers my question. A bit more detail: the Postnikov tower of $k(n)$ is an Adams resolution, because the `bottom class' map $k(n) \rightarrow H\mathbb F_p$ is onto in mod $p$ cohomology; indeed $H^*(k(n);\mathbb F_p) = A_p//E(Q_n)$. -The appendix by Greenlees and May has the details that two spectral sequences converging to $\pi_*(Y \wedge X)$, one coming from filtering $Y$ by its Postnikov tower and the other by filtering $X$ by its skeleta, agree.<|endoftext|> -TITLE: Maximal distance of $2d+1$ points on a sphere -QUESTION [8 upvotes]: If one arranges $2d$ points on the sphere $\mathbf S^{d-1}\subset\Bbb R^d$ as the vertices of the regular octahedron, then one can achieve a minimal spherical distance of $\pi/2$ between any two points, and this is best possible. -What if I want to arrange $2d+1$ points on $\mathbf S^{d-1}$ as far apart as possible from each other? What are the best known upper and lower bounds on the minimal distance between any two points in such an arrangement, and is there anything known about how these arrangements look like? - -REPLY [4 votes]: Turns out kodlu's idea works in all dimensions, regardless of the existence of any Hadamard matrices. -Consider all coordinate permutations of -$$(1,...,1,-d)\in\Bbb R^{d+1}\quad\text{and}\quad (-1,...,-1,d)\in\Bbb R^{d+1}.$$ -This gives $2d+2$ points in $\Bbb R^{d+1}$, but all these lie in the $d$-dimensional subspace orthogonal to $(1,...,1)$. -The smallest angle between two of these is attained e.g. between $a=(1,...,1,-d)$ and $b=(-1,...,-1,d,-1)$, whose cosine turns out to be -$$\cos\measuredangle(a,b)=\frac{\langle a,b\rangle}{\|a\|\|b\|} = \frac{\overbrace{(-1)+\cdots+(-1)}^{d-1}+d+d}{\underbrace{1+\cdots+1}_d+d^2} = \frac{d+1}{d(d+1)}=\frac1d.$$<|endoftext|> -TITLE: Moduli of smooth curves in $|\mathcal{O}_{\mathbb{P}^1\times\mathbb{P}^1}(2,2)| $ and their invariants -QUESTION [12 upvotes]: It is well known that any smooth curve -$C\in |\mathcal{O}_{\mathbf{P}^1\times\mathbf{P}^1}(2,2)| $ has geometric genus equal to 1, so its isomorphism class is determined by its $j$-invariant. Nevertheless we have that $\dim(\mathbf P(H^0(\mathcal{O}_{\mathbf{P}^1\times\mathbf{P}^1}(2,2))))=8$ and $\dim(\operatorname{Aut}(\mathbf P^1\times\mathbf P^1))=6$. So the dimension of the GIT quotient -$$ -\mathbf P(H^0(\mathcal{O}_{\mathbf{P}^1\times\mathbf{P}^1}(2,2)))//\operatorname{Aut}(\mathbf P^1\times\mathbf P^1) -$$ -is 2. If we assume that the base field has characteristic 0, then by Castelnuovo theorem the quotient above is a rational surface. So, at least for a dense subset, there should be two numbers $I_1,I_2$ parametrizing the orbits of the action of $\operatorname{Aut}(\mathbf P^1\times\mathbf P^1)$. -My questions are: -1) Are these numbers known (in terms of the coefficients of the defining polynomial of $C$)? -2) How are they related to the $j$-invariant? -Thanks in advance. - -REPLY [10 votes]: [EDITED to exhibit $j$ as a rational function of $J_2,J_3,J_4$, -and to fix various local errors etc.] -The action of ${\rm SL_2} \times {\rm SL_2}$ on the $9$-dimensional space of -$(2,2)$ forms has a polynomial ring of invariants, with generators in degrees -$2,3,4$. If we write a general $(2,2)$ form $P(x_1,x_2;y_1,y_2)$ as -$(x_1^2, x_1 x_2, x_2^2) M_3 (y_1^2, y_1 y_2, y_2^2)^{\sf T}$ where -$M_3$ is the $3 \times 3$ matrix -$$ -M_3 = \left( \begin{array}{ccc} - a_{00} & a_{01} & a_{02} \cr - a_{10} & a_{11} & a_{12} \cr - a_{20} & a_{21} & a_{22} -\end{array} \right) \; , -$$ -then $J_k$ ($k=2,3,4$) can be taken to be the $x^{4-k}$ coefficient of the -characteristic polynomial $\chi^{\phantom.}_{M_4}$ of the $4 \times 4$ matrix -$$ -M_4 = \left( \begin{array}{cccc} - \frac12 a_{11} & -a_{10} & -a_{01} & 2 a_{00} \cr - a_{12} & -\frac12 a_{11} & -2 a_{02} & a_{01} \cr - a_{21} & -2 a_{20} & -\frac12 a_{11} & a_{10} \cr - 2 a_{22} & -a_{21} & -a_{12} & \frac12 a_{11} -\end{array} \right) . -$$ -This matrix is characterized by the identity -$$ -P(x_1,x_2;y_1,y_2) = -(z_{11},z_{12},z_{21},z_{22}) M_4 (z_{22},-z_{21},-z_{12},z_{11})^{\sf T} -$$ -where each $z_{ij} = x_i y_j$, together with the requirement that -$M_4$ has trace zero and becomes symmetric when its columns are listed -in reverse order and columns $2,3$ are multiplied by $-1$. -The invariants of degree $2$ and $3$ can also be written as -$$ -J_2 = -\frac12 a_{11}^2 + 2(a_{01} a_{21} + a_{10} a_{12}) -- 4 (a_{00} a_{22} + a_{20} a_{02}), -\quad -J_3 = -4 \det M_3; -$$ -of course $J_4 = \det M_4$. -The (Jacobian of the) genus-$1$ curve $P=0$ is isomorphic with the -elliptic curve -$$ -y^2 = x (x-J_2)^2 - 4 J_4 x + J_3^2. -$$ -In particular this lets us compute the $j$-invariant of this curve -as a rational function of $J_2,J_3,J_4$: -$$ -j = \frac{256 (J_2^2 + 12 J_4)^3}{16 J_2^4 J_4 - 4 J_2^3 J_3^2 - - 128 J_2^2 J_4^2 + 144 J_2 J_3^2 J_4 + 256 J_4^3 - 27 J_3^4} \, . -$$ -One way to obtain these results is as follows. First compute the -Hilbert series of the invariant ring. We find that it is -$1 / \bigl( (1-t^2) (1-t^3) (1-t^4) \bigr)$; this suggests -a polynomial ring of invariants with generators of degrees $2,3,4$, -and shows that if we find independent invariants $J_2,J_3,J_4$ -of those degrees then ${\bf C}[J_2,J_3,J_4]$ is the full invariant ring. -Now use the basis $\{z_{ij}\}$ of the four-dimensional space, -call it $Z$, of sections of ${\cal O}(1,1)$; it is well-known that -$\{z_{ij}\}$ embeds ${\bf P}^1 \times {\bf P}^1$ into ${\bf P}^3$ -as the quadric $z_{11} z_{22} = z_{12} z_{21}$, -identifying ${\rm SL_2} \times {\rm SL_2}$ with -the special orthogonal group ${\rm SO}(Q)$ where -$Q$ is the quadratic form $z_{11} z_{22} - z_{12} z_{21}$. -This identifies $P$ with some other quadratic form $\tilde P$ -in the $z_{ij}$, determined uniquely modulo ${\bf C} Q$. -Now $Q$ is nondegenerate, so it identifies $Z$ with its dual $Z^*$, -and thus identifies quadratic forms on $Z$ with self-adjoint maps -$T: Z \to Z$, with $Q$ itself mapping to the identity map. -It is known that generically ${\rm SO}(q)$ orbits of such $T$ -are determined by their spectrum, and thus by the characteristic polynomial -$\chi^{\phantom.}_T$. There is a unique translate $\tilde P + cQ$ of trace zero, -represented by the above matrix $M_4$. Hence the coefficients -$J_2,J_3,J_4$ of $\chi^{\phantom.}_{M_4}$ are invariant and independent, -as claimed. -To identify the elliptic curve, write $C$ as a double cover of -one of the ${\bf P}^1$'s by taking the discriminant of $P$ with respect to -the other ${\bf P}^1$, and then use classical formulas for the -Jacobian of a genus-$1$ curve $y^2 = {\rm quartic}$. -The formulas, though not pretty to look at, are short enough -to let us identify the coefficients with polynomials in $J_2,J_3,J_4$. -The resulting curve has a rational point whose $x$-coordinate is -a multiple of $J_2$; translating $x$ to put this point at $x=0$ -yields the model $y^2 = x (x-J_2)^2 - 4 J_4 x + J_3^2$ exhibited above. -The visible rational point $(x,y) = (0,J_3)$ -ought to correspond to the difference between the divisors -${\mathcal O(0,1)}_C$ and ${\mathcal O(1,0)}_C$, -but I haven't checked this.<|endoftext|> -TITLE: Mirzakhani's hyperbolic method generalized to moduli space of stable maps -QUESTION [6 upvotes]: I've been learning about Mirzakhani's use of hyperbolic geometry to compute Weil-Petersson volumes of moduli space of curves, and the application to proving Virasoro constraints for a point. Why have these methods not been directly extended to higher dimensional target spaces? I don't know much about the structure of the moduli space of stable maps into a projective variety. Is there expected to exist a reasonable definition of volume for these spaces? I'm further confused by the fact that "topological recursion" introduced by Eyndard-Orantin, and inspired by Mirzakhani's recursion relations, was used to compute Gromov-Witten invariants for toric Calabi-Yau 3-folds. I suppose my question is: to what extent might one expect that intersection theory on the moduli space of stable maps into a projective variety is related to the hyperbolic geometry of the domain Riemann surface? - -REPLY [3 votes]: Mirzakhani's computation of volumes of deformation spaces is very heavily based on the work of Greg McShane (McShane's identity). McShane's theory has been extended to other deformation spaces, see, for example -Labourie, François; McShane, Gregory, Cross ratios and identities for higher Teichmüller-Thurston theory, Duke Math. J. 149, No. 2, 279-345 (2009). ZBL1182.30075. -(a search of Google Scholar for "McShane Labourie" reveals a fair bit of other work). So, for the last question, the answer is "not very", but feel free to read the papers...<|endoftext|> -TITLE: Is Minkowski sum of boundary convex again? -QUESTION [7 upvotes]: Consider a closed, bounded and convex set $C \subset \mathbb{R}^{2}$ and denote its boundary with $\partial C$. It is very well-known that the Minkowski sum of two convex sets is convex again. What about the Minkowski sum of its boundary? -Is the Minkowski sum $\partial C + \partial C$ again a convex set and how can one prove that? -Does this property hold in other dimensions? - -REPLY [11 votes]: Yes, $\partial C + \partial C$ is convex since it equals $2C$. Equivalently, every point in $z \in C$ is a midpoint of two boundary points. This is obvious if $z \in \partial C$. Otherwise, let $f :S^{n-1} \to \mathbf{R}$ be the continuous function which sends $u$ to the length of the segment going from $z$ to $\partial C$ in direction $u$. Since $n > 1$, this function takes equal values at a pair of antipodal points (a very simple corollary to Borsuk-Ulam, if you want), which gives the desired property.<|endoftext|> -TITLE: Prime-like numbers that avoid Green-Tao? -QUESTION [7 upvotes]: I would like to understand the conditions that support -the Green-Tao Theorem, which established that -the primes contain arbitrarily long arithmetic progressions. -I am wondering: - -Q. Is it difficult to define an infinite - set $S$ of natural numbers that mimics the - $\sim n / \log n$ distribution of - the primes (and perhaps other prime properties), but yet fails to contain arbitrarily long arithmetic progressions? - -For example, $S$ might be the -lucky numbers -(OEIS A000959), -or -Hawkins' "random primes." -Do either of these avoid arbitrarily long arithmetic progressions? -My intuition is that it is easier to avoid such progressions -than it is to establish their existence. - -MSE question unanswered: -Do lucky numbers contain arbitrarily long arithmetic progressions? - -REPLY [11 votes]: This is likely impossible. -Indeed the largest sets known to be free of arbitrarily long arithmetic progressions asymptotically satisfy $| [ A \cap [1, n] | \lesssim_{k} n / \log^{k} n$ for all $k>1$, and it is widely believed that these examples are near maximal. -The Erdos-Turan conjecture mentioned in the comments is almost equivalent to the claim that any set of relative density greater than $n/ \log n$ contains arbitrary long arithmetic progressions. As remarked above, this is likely true for even sparser sets.<|endoftext|> -TITLE: Why 'excedances' of permutations? -QUESTION [20 upvotes]: For a permutation $\pi=\pi_1\pi_2\cdots\pi_n$ written in one-line notation, an index $i$ for which $\pi_i > i$ is usually called an 'excedance.' To me, this seems like a mispelling of what should be 'exceedance': many dictionaries list 'exceedance' as a valid word, but none I can find consider 'excedance' a correct spelling of any word. Also, as far as I know, 'excedance' is pronounced like 'exceedance' (that is, ex-ceed-ance). But, while both 'excedance' and 'exceedance' are to some extent used for this permutation concept, it seems that 'excedance' is much more common. -Question: Does anyone know the origin of the spelling 'excedance' for this permutation concept? Is it an error which has become standard? - -REPLY [45 votes]: Mea culpa. Comtet used the term excédence. When writing EC1 I needed an English term for this concept. For some reason I didn't like the word exceedance. I thought it looked better without the double e, analogous to proceed and procedure. Thus I made up the word excedance.<|endoftext|> -TITLE: Surjectivity of the Abel-Prym map -QUESTION [9 upvotes]: It is well known that the Abel-Jacobi map restricted to $\text{Eff}_g(C)$ surjects onto the Jacobian $\text{Jac}(C)$, since every divisor of degree $g$ is effective. -Is there an analogous statement for Prym varieties? That is, given an unramified double cover $\widetilde C\to C$ with involution $\tau$, consider the map $f:\text{Eff}_d(\widetilde{C})\to\text{Prym}(\widetilde{C}/C)$ given by $f(D)=D-\tau(D)$. Is $f$ surjective if, for instance, $d=g-1$? - -REPLY [9 votes]: First of all, note that your definition is not correct: when $d$ is odd, the image of your map does not land in the Prym variety -- you have to add a constant term. When this is done, the answer is yes, for the following reason. Let $X$ be the image of $\tilde{C} $ in $P:=\operatorname{Prym}(\tilde{C}/C ) $. Let me put $h:=g-1=\dim P$. What you want to prove is that the addition map $X^{h}\rightarrow P$ is surjective, that is, of degree $>0$. Now this degree is computed by the Pontryagin product $[X]^{*h}$, where $[X]$ is the class of $X$ in $H^{2h-2}(P,\mathbb{Z})$. We know that this class is $2\dfrac{\theta ^{h-1}}{(h-1)!} $, where $\theta $ is the class of the principal polarization. -So we just have to prove that $\theta ^{*h}\in H^{2h}(P,\mathbb{Z})$ is nonzero. This is true for any principally polarized abelian variety $(P,\theta )$ of dimension $h$: it suffices to prove it for a Jacobian $J\Gamma $, and this amounts to say that the Abel-Jacobi map $\Gamma ^h\rightarrow J\Gamma $ is surjective, as you recall in your post.<|endoftext|> -TITLE: The origin(s) of the word "elliptic" -QUESTION [34 upvotes]: The word elliptic appears quite often in mathematics; I will list a few occurrences below. For some of these, it is clear to me how they are related; for instance, elliptic functions (named after ellipses, see here) are the functions on elliptic curves over $\mathbb C$. For others, I do not know if there is a relationship at all. - -Ellipses -Elliptic integrals -Elliptic functions -Elliptic curves -Elliptic genera (in the sense of Hirzebruch) -Elliptic (as opposed to parabolic or hyperbolic) isometries of the hyperbolic plane -Elliptic partial differential operators, elliptic PDEs -Elliptic cohomology - -I am interested in the etymology of this word, in particular, the origins of the different usages listed above. More precisely, I was wondering whether there is, in a way, a single "strain" for all uses of elliptic in mathematics, going all the way back to ellipses in Euclidean geometry. - -REPLY [5 votes]: The origins of all of the companion terms parabola, hyperbola, and ellipse were coined by Apollonius of Perga, in his classic text "On Conic Sections." (He was born about 262 BC, approximately 25 years after the birth of Archimedes.) -The terms we use are direct descendants of the Greek words. Now, the reason they have the names they is that in the construction of conic sections produced by passing a plane through a cone, a parabola (something literally "thrown beside") differs from a hyperbola (something thrown over) and an ellipse (left out). These are in reference to a "parameter" (Apollonius's term) something "measuring alongside," a parameter exceeding, and a parameter being deficient by (leaving out), an amount. -We use those same Greek words in a variety of unrelated contexts, of course. Hyperbolae refers to statements that are over the top, excessive, and by extension unbelievable. Parabolic seems to be generally in reference to the conic section, but ellipse means something that has been left out, in the same way that Apollonius used it in reference to the construction of his conic section. Generally we think of an ellipse as a figure that can be described as an oval, with two foci. (A circle is a degenerate ellipse, because there is only one focus.) However the reason Apollonius called an ellipse "deficient" is because the same parameter was lacking, in the same way that it was excessive in the hyperbola. -The classic text "On Conic Sections" is part of the Great Books of the Western World Collection, which comprise the core of the Great Books programs taught at St. John's College (Santa Fe, NM, and Annapolis, MD), the University of Chicago, and a few other offshoots -- St. Mary's Moraga, CA for example. -Marklan<|endoftext|> -TITLE: A property of varieties between unirational and retract rational -QUESTION [10 upvotes]: EDIT: The vague question Q1 below is partially answered, while the concrete question Q2 seems to be still open. -Let $V$ be a geometrically integral variety over a field $K$. -I consider the following properties: -(1) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ for $n={\rm dim}(V)$. -(2) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ for some $n$. ($V$ is unirational) -(3) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ with geometrically -integral generic fiber, for some $n$. -(4) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ with a right inverse $V\dashrightarrow\mathbb{P}_K^n$, for some $n$. ($V$ is retract rational) -(5) There exists a birational map $\mathbb{P}_K^n\dashrightarrow V\times\mathbb{P}_K^m$ for some $m,n$. ($V$ is stably rational) -(6) There exists a birational map $\mathbb{P}_K^n\dashrightarrow V$ for some $n$. ($V$ is rational) -We have that $(6)\Rightarrow(5)\Rightarrow(4)\Rightarrow(3)\Rightarrow(2)\Leftrightarrow(1)$. For curves all of these properties are equivalent, but they diverge in higher dimension. From browsing the literature I gather that it is known that $(2)\not\Rightarrow(4)$ and $(5)\not\Rightarrow(6)$, and it is expected that $(4)\not\Rightarrow(5)$. -However, I am interested in property (3), which I could not find anywhere in the literature. -Q1: Does property (3) occur in the literature? Does it have a name? Is it equivalent to either (2) or (4)? -EDIT: As the answer of Daniel Loughran shows, the Châtelet surfaces described below are examples for $(3)\not\Rightarrow(5)$ over $K=\mathbb{R}$. -Phrased more down to earth, here is a very concrete question I am interested in: -Q2: Is every intermediate field $F$ of $\mathbb{R}(X,Y,Z)/\mathbb{R}$ which is algebraically closed in $\mathbb{R}(X,Y,Z)$ purely transcendental over $\mathbb{R}$? -Of course this is clear for $F$ of transcendence degree $0$, $1$ or $3$ over $\mathbb{R}$, so it is really just a question about surfaces. The equivalent question over $\mathbb{C}$ has a positive answer, as every unirational complex surface is known to be rational. The closest to a counterexample I found in the literature is the surface over $\mathbb{R}$ given by $x^2+y^2=f(z)$ with $f$ of degree $3$ with three real roots, which I think satisfies (2) but not (5), but I don't know if it satisfies (3). -EDIT: As explained in the answer of Daniel Loughran, such a surface $V$ satisfies (3). But it seems not clear whether the $n$ in $\mathbb{P}_K^n\dashrightarrow V$ with geometrically integral generic fiber can be chosen to be 3, which would be needed to answer Q2 negatively. - -REPLY [4 votes]: Let $k$ be a field of characteristic $0$, $a \in k$ and $f$ a separable polynomial of degree $3$. -The projective surface $X$, given as the minimal smooth compactification of the affine surface -$$X: \quad x^2 - ay^2 = f(z)$$ -you have written down is an example of a Châtelet surface. (Note that $X(k) \neq \emptyset$ always as there is a rational point at infinity). These have been studied in great detail by Colliot-Thélène and his collaborators. The key paper relevant to your question is: -Arnaud Beauville, Jean-Louis Colliot-Thélène, Jean-Jacques Sansuc and Peter Swinnerton-Dyer - Variétés Stablement Rationnelles Non Rationnelles, Annals of Mathematics. -Such surfaces are non-rational provided $a$ is not a square in any of the residue fields of the irreducible factors of $f$. Moreover, in the above paper, it is shown that they are stably rational provided certain assumption hold (e.g. $f$ is irreducible with Galois group $S_3$). But as remarked in the comments, there are examples which are also not stably rational. -They prove this using universal torsors -$$T \to X.$$ -For a good overview of the theory of universal torsors, I would recommend the book -Skorogobatov - Torsors and rational points -I will just remark that these are torsors under the Néron-Severi torus, in particular the generic fibre is geometrically integral. -A sufficient criterion for the existence of a universal torsor is $X(k) \neq \emptyset$; but as already explained we have this property so universal torsors exist. There may be many universal torsors in general; but the twists of a given torsor gives a parametristation of the rational points of $X$. So there is always some torsor with a rational point. But it turns out that such torsors $T$ are birational to a complete intersection of two quadrics in projective space, which is shown to be a rational variety (details in the above paper). So this shows that (3) holds. -Altogether, this shows that there are Châtelet surfaces which satisfy (3), (5), but not (6), and also those which satisfy (3) but not (5) nor (6). This seems to give a complete answer to your question. -Further details on these results and constructions can be found in the seminar Bourbaki report: -Laurent Moret-Bailly - Variétés stablement rationnelles non rationnelles<|endoftext|> -TITLE: Free category with product and coproduct -QUESTION [11 upvotes]: Is there a known description of the free category with both product and coproduct? -That is, given a small category $C$, I want to consider a category $U C$ which has product and coproduct, a functor $C \to U C$ and such that $UC$ is universal for functor preserving both product and coproduct. The case $C = \emptyset$ is already interesting. -I'm also happy to focus on finite product and finite coproduct, especially if it avoids some size problems, though I don't think this is essential. -My guess is that this category should be a category of two-player games (player and opponent) with morphisms being simulation and where outcome of the game are marked by objects of $C$ (if $C = \emptyset$ we should just have a win/lose outcome): -The coproduct of a family of games is the game where the player first chooses which game he wants to play in the family, while their product is the game where the opponent chooses which game he wants to play. The initial object is the game where player loses at the start, and the final object is the one where opponent loses at the start. -But the details of this, and especially the proper definition of the morphisms are a bit involved, so I'm curious whether this has been worked out somewhere. -Of course, as soon as we assume compatibility between product and coproduct (for e.g. distributivity) there are simple description, but here I'm interested in the completely unconstrained situation. - -REPLY [12 votes]: The general problem of giving a categorical construction of the free category with finite coproducts and products (or "free sum–product category") seems to still be open, though there are several works on special cases of the problem. -Cockett–Santocanale's On the word problem for ΣΠ-categories, and the properties of two-way communication gives a good introduction to the problem. They state: - -There have been, directly or indirectly, a number of contributions towards -our goal in this paper. [...] These related results, however, work only for the fragment without units – or, more precisely, for the fragment with a common initial and final object. As far we know, there is no representation theorem for the full fragment with distinct units. - -Special cases of interest include: - -A syntactic construction of the free category with finite coproducts and products (and various characterisations): Finite sum–product logic, Cockett and Seely. -A construction of the free category with coproducts, products and a zero object: Coherence Completions of Categories and Their Enriched Softness, Hu and Joyal. -A construction of the free category with nonempty finite coproducts and products: A canonical graphical syntax for non-empty finite products and sums, Hughes. - -Joyal has two related papers (at least for general colimits and limits), but unfortunately without explicit constructions or proofs: - -Free bicomplete categories. -Free bicompletion of enriched categories.<|endoftext|> -TITLE: Origin of the term 'index of a subgroup' -QUESTION [6 upvotes]: The index of a subgroup $H$ in a group $G$ is the number of distinct cosets of $H$ in $G$. -Why did someone decide to call this an 'index'? What's the rationale for this? - -REPLY [13 votes]: The short answer is Cauchy, with only justification: “for short”. Burnside’s Theory of groups of finite order (1897) has a useful glossary stating, p. 382 (my bold): - -The ratio of the order of a sub-group $H$, to the order of the group - $G$ containing it, is called by French writers the “indice,” by German - the “Index,” of $H$ in $G$. The phrase is most commonly used of a - substitution group in relation to the symmetric group of the same - degree. - -The French writers in question are those mentioned in Burnside’s preface: Serret’s Cours d’Algèbre Supérieure (1866) (“the first connected exposition of the theory”) has, p. 287: - -432. $\vphantom{\frac{\mathrm N}m}$ To abbreviate the discourse, I will call index of a system of conjugate substitutions formed with $n$ letters, the quotient obtained by dividing the product $\mathrm N=1.2.3\dots n$ by the number which expresses the system’s order. If $m$ denotes the index of a conjugate system of order $\mu$, one will have $m=\frac{\mathrm N}\mu$. - -And already Cauchy’s memoir Sur le Nombre de Valeurs qu’une Fonction peut acquérir, lorsqu’on y permute de toutes les manières possibles les quantités qu’elle renferme, J. École Polytechnique 10 (1815) 1–28 (from which “the theory of groups of finite order may be said to date”) has, p. 6: - -if one represents by $R$ the total number of essentially different values of the function $K$, $M$ being the number of values equivalent to $K_\alpha$ (...) one will have $RM=N$ (...) Hence $R$ (...) can only be a factor of $N$, that is, of the product $1.2.3\dots n$ (...) For short, I will henceforth call index of the function $K$, the number $R$ which indicates how many essentially different values this function can assume. - -Note that the index is to “indicate” (but this could be said of any named count, especially ratio...), and that the paper’s title itself restates this very definition — it could be shortened “Sur l'Indice”.<|endoftext|> -TITLE: Is $\mathcal{S}(\mathbb{R}^n)$ a tame Fréchet space? -QUESTION [8 upvotes]: Hamilton's paper "The Inverse Function theorem of Nash and Moser" (1982, Bull. Amer. Math. Soc, vol. 7, n. 1, page $137$) proves that $C^{\infty}(M)$ is a tame Fréchet space when $M$ is a compact manifold. It was asked here on MO if this space is tame in the non-compact case, and in the second answer, there was an argument that for a space to be tame, every sufficiently large taming semi-norm would be a norm (which is false in $C^{\infty}(M)$ for the non-compact case), but this is true for $\mathcal{S}(\mathbb{R}^n)$ (the space of functions that are rapidly decreasing, along with its derivatives). So to disprove tameness, a different argument would be necessary. -So the question is, is $\mathcal{S}(\mathbb{R}^n)$ a tame Fréchet space (with the usual Fréchet topology)? Is there a reference? - -REPLY [10 votes]: Using Fourier series, $C^\infty(S^1)$ is ismorphic to the sequence space -$$s=\{(x_k)_{k\in \mathbb N}\in \mathbb C^{\mathbb N}: \sum_{k=1}^\infty k^{2n} |x_k|^2 <\infty \text{ for all $n\in\mathbb N$}\}$$ -and this space is isomorphic to $\mathscr S(\mathbb R^d)$. -This is very classical, a modern treatment (with much information about structural properties of Fréchet spaces) is in the book Introduction to Functional Analysis of Meise and Vogt.<|endoftext|> -TITLE: Postnikov invariants of the Brauer 3-group -QUESTION [15 upvotes]: Given a commutative ring $k$ there is a bicategory with - -algebras over $k$ as objects, -bimodules as morphisms, -bimodule homomorphisms as 2-morphisms. - -This is a monoidal bicategory, since we can take the tensor product of algebras, and everything else gets along nicely with that. -Given any monoidal bicategory we can take its core: that is, the sub-monoidal bicategory where we only keep invertible objects (invertible up to equivalence), invertible morphisms (invertible up to 2-isomorphism), and invertible 2-morphisms. -The core is a monoidal bicategory where everything is invertible in a suitably weakened sense so it's called a 3-group. -The particular 3-group we get from a commutative ring $k$ could be called its Brauer 3-group and denoted $\mathbf{Br}(k)$. It's discussed on the $n$Lab: there it's called the Picard 3-group of $k$ but denoted as $\mathbf{Br}(k)$. -Like any 3-group, $\mathbf{Br}(k)$ has homotopy groups which I will call $\pi_1, \pi_2, \pi_3$ (though there are choices of where we start numbering). These are well-known things: - -$\pi_1$ is the Brauer group of $k$. -$\pi_2$ is the Picard group of $k$. -$\pi_3$ is the group of units of $k$. - -My question is whether people have studied, or computed, the Postnikov invariants involving these things. The simplest is the map -$$ a : \pi_1^3 \to \pi_2$$ -coming from the associator in the monoidal category of $k$-algebras (with isomorphism classes of bimodules as morphisms). Since the associator obeys the pentagon identity this is a 3-cocycle on $\pi_1$ with values in its module $\pi_2$, so it gives an element of $ H^3(\pi_1, \pi_2)$. -Is this element trivial? If not, what is it? -But in fact $\mathbf{Br}(k)$ is not just a 3-group but also a symmetric monoidal bicategory. So, it's what I call a symmetric 3-group, though some others call it a Picard 2-category. These have a number of other Postnikov invariants: - -Nick Gurski, Niles Johnson, Angélica M. Osorno and Marc Stephan, Stable Postnikov data of Picard 2-categories. - -Has anyone figured out any of these for $\mathbf{Br}(k)$? - -REPLY [4 votes]: Let me see if I understand what Jacob says in the comments. I think his argument can be summarized as: the Brauer 3-group is étale-locally an Eilenberg-MacLane spectrum, hence étale-locally an $\mathbb{Z}$-module spectrum, hence an $\mathbb{Z}$-module spectrum, hence the Postnikov tower splits. Do I have that right? -If so, I want to point out that the situation changes with one tweak, which is to allow superalgebras, super bimodules, etc. When $k = \mathbb{R}$ the super Brauer 3-group has a nontrivial homotopy operation $\pi_1 \to \pi_3$ given by taking the super dimension of the zeroth Hochschild homology of a superalgebra in the super Brauer group, and I computed an example of it taking a nontrivial value here. This implies that the Postnikov tower can't split but I don't know what the $k$-invariants are. I suppose following Jacob we could try to work things out as homotopy fixed points of the $\text{Gal}(\mathbb{C}/\mathbb{R})$-action on the super Brauer 3-group over $\mathbb{C}$ but this is beyond me.<|endoftext|> -TITLE: Residually amenable groups -QUESTION [6 upvotes]: I have two questions about residually amenable groups: - -Is every finitely presented amenable group residually elementary amenable? -Given $n$, is the free Burnside group of exponent $n$ on two generators residually amenable? - -Regarding 1., I know that Grigorchuk constructed an example of a finitely presented amenable group that is not elementary amenable, but I was unsure if it could be residually elementary amenable. Regarding 2., I believe it is an open question whether or not the free Burnside group on two generators is sofic, so if the answer to 2. is known, I'm guessing that it would be in the negative. - -REPLY [3 votes]: (From my initial comments) - -No, there's a finitely presented ascending HNN extension of Grigorchuk's group, and it's an isolated group (it has a unique minimal nontrivial normal subgroup— more precisely every proper quotient is metabelian [Sapir-Wise]), so it's not residually elementary amenable. See details and references §5.7 in my 2007 J. Algebra paper with Guyot and Pitsch on isolated groups ArXiv link. - -As confirmed by Mark Sapir for each given exponent and number $\ge 2$ of generators the question of residual amenability of the given Burnside group is open, except in the few cases where it's known to be locally finite (exponent $\le 4$ and $6$).<|endoftext|> -TITLE: Is there a fusion category not Grothendieck equivalent to a unitary one? -QUESTION [8 upvotes]: We refer to the book Tensor categories by Etingof-Gelaki-Nikshych-Ostrik (MR3242743) for the notion of (unitary) fusion category. Two fusion categories are Grothendieck equivalent if they have the same fusion ring. -Question: Is there a fusion category not Grothendieck equivalent to a unitary one? -The following citations (coming from above book) are almost on-topic, but not exactly, because they are about the existence of a certain structure on certain fusion categories, whereas above question relaxes up to Grothendieck equivalence, which is much weaker. -On page 284: - -We note that we do not know an example of a fusion category over - $\mathbb{C}$ which does not admit a Hermitian structure, or a - pseudo-unitary fusion category which does not admit a unitary - structure. - -On page 76: - -Does every semisimple tensor category admit a pivotal structure? A - spherical structure? This is the case for all known examples. The - general answer is unknown to us at the moment of writing (even for - fusion categories over ground fields of characteristic zero). - -REPLY [6 votes]: Yes, according to Andrew Schopieray. He just provided a categorifiable fusion ring, of rank 6 and multiplicity 2, without pseudounitary categorification (so without unitary categorification), in the following preprint called Non-pseudounitary fusion. -https://arxiv.org/abs/2010.02958<|endoftext|> -TITLE: Maximum size of $k$-Sidon set over $\mathbb{F}_2^n.$ -QUESTION [6 upvotes]: Fix $k \in \mathbb{N}$, $k \ge 2.$ -Does there exist a subset $A \subset \mathbb{F}_2^n$ such that $|A| \ge c 2^{n/k}$ with some absolutely positive constant $c,$ and satisfying - $$ a_1 + a_2 + \dots + a_k \neq b_1 + b_2 + \dots + b_k $$ -for every pair of distinct $k$-element subsets $\{a_1,...,a_k\} \neq \{b_1,...,b_k\}$ of $A$ ? - -REPLY [4 votes]: Yes, such $A$ exist for all $k$, -and one can even take $c=1/2$ independent of $k$. -It is enough to prove that if $n=km$ for some integer $m$ then -there exists such a subset $A$ of size $2^m$, because $A$ will then work for -each $n \in [km, k(m+1))$, and $2^{n/k} < 2^{m+1}$ for all such $n$. -Identify ${\bf F}_2^n$ with the vector space $F^k$ -where $F$ is a finite field of $2^m$ elements. -(Alas I cannot use the usual $k$ for such a field . . .) -Let $A$ consist of all vectors $(a,a^3,a^5,\ldots,a^{2k-1})$ with $a \in F$. -The desired result will then follow once we prove: -Proposition. Let $A = \{a_1,\ldots,a_k\}$ and $B = \{b_1,\ldots,b_k\}$ be -any $k$-element subsets of a field $F$ of characteristic $2$. -If $\sum_{j=1}^k a_j^r = \sum_{j=1}^k b_j^r$ for each $r=1,3,5,\ldots,2k-1$ -then $A=B$. -Proof: For any finite subset $S$ of $F$ and any integer $r \leq 0$ define -$p_r(S) = \sum_{s \in S} s^k$. We thus assume that $p_r(A)=p_r(B)$ for each -$r=1,3,5,\ldots,2k-1$. Since $x \mapsto x^2$ is a field homomorphism, -we have $p_{2r}(S) = p_r(S)^2$, so our hypothesis implies that in fact -$p_r(A)=p_r(B)$ for all positive integers $r \leq 2k$. -Now let $\alpha,\beta \in F[t]$ be the polynomials -$\alpha = \prod_{j=1}^k (1 + a_j t)$, $\beta = \prod_{j=1}^k (1 + b_j t)$. -Then $\alpha'/\alpha$ has Taylor expansion -$$ -\sum_{j=1}^k \frac{a_j}{1 + a_j t} -= \sum_{j=1}^k (a_j + a_j^2 t + a_j^3 t^2 + a_j^4 t^3 + \cdots) -= \sum_{r=1}^\infty p_r(A) \, t^{r-1}, -$$ -and likewise $\beta'/\beta = \sum_{r=1}^\infty p_r(B) \, t^{r-1}$. -These Taylor expansions agree through the $t^{2k-1}$ term, so -$\alpha'/\alpha - \beta'/\beta = O(t^{2k})$; since -$\deg(\alpha' \! \beta - \alpha \beta') < 2k$, this implies that -$\alpha'/\alpha = \beta'/\beta$. Therefore $(\alpha/\beta)' = 0$, -so $\alpha / \beta \in F(t^2)$. Since $A$ and $B$ may not have -repeated elements it follows that $\alpha / \beta$ is a constant, -whence $A=B$ as claimed. QED<|endoftext|> -TITLE: Retracting off a compact set -QUESTION [7 upvotes]: Let $K$ be a compact set in $\mathbb{R}^n$ and let $U$ be a bounded open set that contains $K$. You may assume both are connected. - -Can we always find an open $V$ such that $K\subset V\subset\overline{V}\subset U$ such that $U\backslash K$ retracts on $U\backslash V$? - -For example, if we somehow find $V$ such that $\overline{V}$ is homeomorphic to the closed ball, then choosing any point of $x\in K$ to be its center, we can radially repel all the points of $V\backslash\{x\}\supset V\backslash K$ onto $\partial V$ and leave every other point fixed. -I tried to cover $K$ with cubic neighborhoods and run an induction over the number of cubes, but it is getting too messy. Perhaps there is a more clever way. - -REPLY [5 votes]: Let us show how to find such a retraction for $n=2$ (I do not know if this method generalizes to higher dimensions). -Given a compact set $C\subset\mathbb R^2$ and an open neighborhood $U\subseteq\mathbb R^2$ of $C$, choose a triangulation on $\mathbb R^2$ so fine that no triangle of the triangulation meets $C$ and $\mathbb R^2\setminus U$ simultaneously. -Replacing the triangulation by a finer triangulation, we can assume that for each triangle $T$ with $T\not\subseteq C$, one vertex of $T$ does not belong to $C$. -How to find such a triangulations? Assuming that $T\not\subseteq C$, we can find an interior point $v$ of $T$ that does not belong to $C$ and replace the triangle $T$ by 3 subtriangles having $v$ as a vertex. -Also we can assume that either $T\subseteq\mathbb R^2\setminus C$ or $T$ has a vertex in $C$. Assuming that $T$ has no vertices in $C$ but $T\cap C\ne\emptyset$, we can choose a point $c\in T\cap C$ and replace the triange $T$ by two or three triangles having $c$ as a vartex. -Therefore, we lose no generality assuming that each triangle $T$ of the triangulation has one of the following properties: -1) $T\subseteq C$; -2) $T\cap C=\emptyset$; -3) $T$ has one vertex in $C$ and one vertex outside of $C$; -4) If two vertices $u,v$ of $T$ do not belong to $C$, then the side $[u,v]$ does not intersect $C$. -A triangle $T$ will be called difficult if it has one vertex say $u$ outside $C$, two vertices $v,w$ in $C$ and the side $[v,w]$ is not a subset of $C$. -In this case choose any point $c[v,w]\in [v,w]\setminus C$. The points $c[v,w]$ can be chosen so that for two difficult triangle sharing the common side $[v,w]$ the point $c[v,w]$ is the same. -Now for every triangle $T$ of the triangulation we define a function $r_T\setminus C:T\to T\setminus C$ such that $r_T\circ r_T=r_T$ as follows. In case (1), let $r_T$ be the empty map and in case (2) $r_T$ be the idenity map of $T$. -In the remaining cases, the triangle $T$ has one vertex in $C$ and one vertex outside of $C$. If the triangle $T$ is not difficult, then it has two vertices $u,v$ such that the side $[u,v]$ either is contained in $C$ or is disjoint with $C$. If $[u,v]$ is contained in $C$, then let $r_T:T\setminus C\to\{w\}$ be the constant map into the unique vertex $w\notin C$ of $T$. -If $[u,v]\cap C=\emptyset$, then the third vertex $w$ of $T$ belongs to $C$ and we can apply the Urysohn lemma to find a function $r_T:T\setminus C\to[u,v]$ such that $r_T[[w,u]\setminus C]=\{u\}$, $r_T[[w,v]\setminus C]=\{v\}$, and $r_T(x)=x$ for every $x\in [u,v]$. -It remains to consider the case of a difficult triangle $T$. Such a triangle has one vertex $u$ outside of $C$, two vertices $v,w$ in $C$ and the point $c[v,w]\in [v,w]\setminus C$. Two cases are possible. -1) There exists a path $\gamma:[0,1]\to T\setminus C$ such that $\gamma(0)=u$ and $\gamma(1)=c[v,w]$. We can assume that $\gamma$ is injective and hence its image $A_T=\gamma[0,1]$ is an arc with endpoints $u$ and $c[v,w]$. Using the Urysohn Lemma, we can find a continuous function $r_T:T\setminus C\to A_T$ such that $r_T[([u,v]\cup[u,w])\setminus C]\subseteq \{u\}$, $r_T[[v,w]\setminus C]\subseteq\{c[v,w]\}$ and $r_T(a)=a$ for every $a\in A_T$. -2) No such a path $\gamma$ exists. Then the points $u$ and $c[v,w]$ belong to distinct connected components of $T\setminus C$. In this case we can choose a continuous map $r_T:T\setminus C\to\{u,c[v,w]\}$ such that $r_T[([u,v]\cup[u,w])\setminus C]\subseteq\{u\}$ and $r_T[[v,w]\setminus C]\subset\{c[v,w]\}$. -The definitions of the maps $r_T$ ensure that they agree on the intersections of their domains. Consequently, the union $r=\bigcup_T r_T$ of these maps is a continuous function $r:\mathbb R^2\setminus C\to\mathbb R^2\setminus C$ such that $r\circ r=r$. So, $r$ is a retraction onto the closed subset $F$ which can be written as the union of the triangles of the triangulation that do not intersect $C$, some vertices of the triangles that intersect $C$ and the arcs $A_T$ of difficult triangles (of the first type). -The choice of the triangulation $T$ (as sufficiently fine) implies that $V=\mathbb R^2\setminus F$ is a neighborhood of $C$ with $\bar V\subset U$. Then $r{\restriction}U\setminus C$ is the required retraction of $U\setminus C$ onto $U\setminus V$.<|endoftext|> -TITLE: How to cite a MathSciNet review -QUESTION [11 upvotes]: I want to cite a MathSciNet review. Is there a standard format for that? (I couldn't find anything) - -REPLY [7 votes]: I checked with the head of copy-editing at Mathematical Reviews and the Mathematical Reviews librarian. Their joint answer is (using the paper mentioned by Carlo Beenakker): -If you want to add something to a reference list, the form would be: -Udrişte, Constantin, review of ``Optimal approximations by piecewise smooth functions and associated variational problems,'' (in Comm. Pure Appl. Math. 42 (1989), no. 5, 577–685, by David Mumford and Jayant Shah), Mathematical Reviews/MathSciNet MR0997568 https://mathscinet.ams.org/mathscinet-getitem?mr=997568. -If you want to mention something in text, -Constantin Udrişte (see his review published in Mathematical Reviews/MathSciNet [MR0997568]) related the Mumford-Shah results to Griffiths' law of cracks in solid mechanics. -I hope this helps. - -REPLY [5 votes]: Mathematical Reviews can be cited like any other journal, with a review number instead of volume and page numbers: -J. Smith. Review of the article “Regular doodads are widgits” by J. Doe. Mathematical Reviews 123456 (2009). -@article {review, - AUTHOR = {Smith, J.}, - TITLE = {Review of the article ``{R}egular doodads are widgits'' by {J}. {D}oe}, - JOURNAL = {Mathematical Reviews}, - VOLUME = {123456}, - YEAR = {2009}, - ISSN = {0025-5629}, - URL = {https://mathscinet.ams.org/mathscinet-getitem?mr=123456}, -}<|endoftext|> -TITLE: Fourier series of $\log(a +b\cos(x))$? -QUESTION [10 upvotes]: By numerical computation it seems like, if $a_0 < a_1$: -$$ -\begin{multline} -\log({a_0}^2 + {a_1}^2 + 2 a_0 a_1 \cos(\omega t)) = \log({a_0}^2 + {a_1}^2) \\ -+ \frac{a_0}{a_1}\cos(\omega t) -- \frac{1}{2}\frac{{a_0}^2}{{a_1}^2}\cos(2\omega t) -+ \frac{1}{3}\frac{{a_0}^3}{{a_1}^3}\cos(3\omega t) -- \frac{1}{4}\frac{{a_0}^4}{{a_1}^4}\cos(4\omega t) \ldots -\end{multline} -$$ -If $a_0 > a_1$, the two term must be exchanged. -I'm quite confident in this solution but I cannot find a way to prove it mathematically... -Would be nice to know the reason of this result! - -REPLY [7 votes]: To apply the solution proposed by @ChristianRemling to the more general case stated in the title you need to do the following: -The formula implies $a > 0$ and $|b| < a$. To solve the problem, we reformulate the formula as: -$$ -\begin{align} -r|1 + qe^{ix}|^2 -&= r(1 + qe^{ix})(1 + qe^{-ix}) \\ -&= r(1 + 2q\cos x + q^2) \\ -& = a + b\cos x \\ -\end{align} -$$ -This requires to solve the following system: -$$ -\begin{align} -a &= r(1 + q^2) \\ -b &= 2rq -\end{align} -$$ -From the two possible solutions we take the one where $|q|<1$ which is mandatory for the Taylor expansion. -$$ -\begin{align} -q = \frac{a - \sqrt{a^{2} - b^{2}}}{b} \\ -r = \frac{a + \sqrt{a^{2} - b^{2}}}{2} -\end{align} -$$ -Finally, Taylor series expansion solve the problem: -$$ -\begin{align} -\log (a + b\cos x) -&= \log r -+ 2\log | 1 + qe^{ix} | \\ -&= \log r -+ 2 \sum {(-1)}^{n-1} \frac{q^n}{n} e^{inx} \\ -&= \log r -+ 2 \sum {(-1)}^{n-1} \frac{q^n}{n} \cos nx -\end{align} -$$<|endoftext|> -TITLE: Differential geometry of Donaldson-Thomas invariants -QUESTION [7 upvotes]: The Donaldson-Thomas invariants are defined by Thomas in the paper A holomorphic Casson invariant for Calabi-Yau 3-folds, and bundles on K3 fibrations, following the proposal in Gauge theory in higher dimensions by Donaldson and Thomas. Although the initial proposal was motivated by differential geometry, the rigorous definition uses tools from algebraic geometry, including moduli spaces of (semi-)stable sheaves and perfect obstruction theory. -The most successful application of DT theory so far seems to be the enumeration of curves, i.e. considering the DT invariants of ideal sheaves, see e.g. 13/2 ways of counting curves. So actually the question is two-fold: - -Is there a way of counting solutions to some version of (perturbed) Hermitian-Yang-Mills equations which could presumably recover DT invariants of ideal sheaves? -As Gromov-Witten theory also makes sense for symplectic manifolds, is there a symplectic (actually, almost complex) counterpart of DT theory? - -REPLY [2 votes]: A symplectic theory should exist but is still some way off, even for curve counting invariants. See arXiv/1712.08383 by Doan-Walpuski for some progress for stable pairs.<|endoftext|> -TITLE: Any exact faithful functor is represented by a unique projective generator -QUESTION [7 upvotes]: In the book 'Tensor Categories' by Pavel Etingof, Shlomo Gelaki, Dmitri Nikshych and Victor Ostrik on page 10 it says: -'Conversely, it is well known (and easy to show) that any exact faithful functor $F : \mathcal{C} \rightarrow \text{Vec}$ is represented by a unique (up to a unique isomorphism) projective generator $P$.' -But I could not find any proof of that fact. Can someone tell me how to prove it or where I can find a proof? -Note: Here $\mathcal{C}$ is a finite k-linear abelian category for some field k. - -REPLY [5 votes]: Let $\mathcal C$ be the category of finite dimensional left modules over a finite dimensional ring $R$. Let $G: \mathcal C \to \mathrm{Vec}$ be an exact and faithful functor to finite dimensional vector spaces. We use $V^*$ to denote the dual vector space. For motivation, notice that if we had a representing object $M$, we would have $$G(R^*) = Hom_R(M,R^*) = Hom_R(R, M^*) = M^*.$$ -Now $G(R^*)$ is a right $R$ module via the action of $R$ by left multiplication. So we define $P:= G(R^*)^*$ to be the dual left module and consider the functor $Hom_R(P,-)$. -This functor is tautologically left exact, and it takes the injective left module $R^*$ to $Hom_R(P,R^*) = Hom_R(R, G(R^*)) = G(R^*)$. Any other finite left module $M$ admits an injective presentation $$0 \to M \to R^{* \oplus a} \to R^{* \oplus b} \to $$ -dual to the presentation of $M^*$ as a right $R$ module. So by left exactness, we see that there is a natural isomorphism $Hom_R(P,-) \simeq G(-)$. -Thus $G$ is represented by $P$. Since $G$ is right exact $P$ is projective, and since it is faithful $P$ is a generator.<|endoftext|> -TITLE: Categories of modules generated under coproducts by a small set? -QUESTION [9 upvotes]: Question 1: For which rings $R$ does there exist a small set $S \subseteq Mod_R$ such that every module $M \in Mod_R$ is a direct sum of modules in $S$? -Equivalenty, for which rings $R$ does there exist a cardinal $\kappa$ such that every module $M \in Mod_R$ is a direct sum of modules generated by $\leq \kappa$-many elements? -Faith [1] says that "$Mod_R$ has a basis" if it answers to Question 1. -Background: - -Kothe showed that if $R$ is an artinian principal ideal ring, then every $R$-module is a direct sum of cyclic modules (i.e. $R$ answers to my question with $\kappa = 1$). - -The general question with $\kappa=1$ seems to be generally known as "Kothe's problem". - -Cohen and Kaplansky showed that the converse holds if $R$ is commutative. Warfield extended this to show that if $R$ is commutative, then $R$ answers to my question if and only if $R$ is an artinian principal ideal ring. - -So the question is only interesting when $R$ is noncommutative. - -Nakayama constructed a (noncommutative) ring $R$ such that every module is a direct sum of cyclic modules, and yet $R$ is not an artinian principal ideal ring. - -So Warfield's theorem apparently doesn't extend in the most straightforward way to the noncommutative setting. - -Faith and Walker showed that there exists a cardinal $\kappa$ such that every injective right $R$-module is a direct sum of $\leq \kappa$-generated modules if and only if $R$ is right Noetherian. Faith ([1] cf. also Griffiths Thm 2.2) then showed that if $R$ answers to Question 1, then $R$ is right Artinian. - -So any ring answering to Question 1 is right Artinian, though not necessarily a principal ideal ring by Nakayama's result. -I happen to be interested in a special case: -Question 2: Same as Question 1, but assuming that $R$ is hereditary. -[1] Faith, "Big Decompositions of Modules" AMS Notices 18, Feb 1971 p. 400 - -REPLY [8 votes]: The rings satisfying your condition (for right modules) are the right pure semisimple rings. There are many equivalent conditions. You can find a lot of information in Section 4.5 of the book -Prest, Mike, Purity, spectra and localisation., Encyclopedia of Mathematics and its Applications 121. Cambridge: Cambridge University Press (ISBN 978-0-521-87308-6/hbk). xxviii, 769 p. (2009). ZBL1205.16002. -or you might find it easier to access the older paper -Prest, Mike, Rings of finite representation type and modules of finite Morley rank, J. Algebra 88, 502-533 (1984). ZBL0538.16025. -As you say, such a ring must be right artinian. -It is known that a ring is both left and right pure semisimple if and only if it has finite representation type (i.e., every module is a direct sum of indecomposable modules, and there are finitely many isomorphism types of indecomposable module), which is a left/right symmetric condition. And there is a longstanding conjecture about a strengthening of this. -Pure Semisimplicity Conjecture: A right pure semisimple ring has finite representation type. -Or equivalently this says that pure semisimplicity should be a left/right symmetric condition. There are many positive results for particular classes of rings. -In Question $2$ you say that you are particularly interested in the hereditary case. This doesn't make the conjecture easier, as Herzog proved that if there is a counterexample then there is a hereditary counterexample. Combining this with a result of Simson, it turns out that to prove the conjecture it would be enough to prove that a right pure semisimple hereditary ring is left artinian. -There is a lot of work on rings of finite representation type, especially hereditary ones. The fundamental result is Gabriel's theorem classifying the finite dimensional hereditary algebras over an algebraically closed field with finite representation type as those Morita equivalent to path algebras of quivers whose underlying graph is a disjoint union of simply laced Dynkin diagrams. There are many generalizations; one for general hereditary artinian rings is -Dowbor, P.; Ringel, Claus Michael; Simson, D., Hereditary Artinian rings of finite representation type, Representation theory II, Proc. 2nd int. Conf., Ottawa 1979, Lect. Notes Math. 832, 232-241 (1980). ZBL0455.16013.<|endoftext|> -TITLE: Resources for divergent / asymptotic series -QUESTION [10 upvotes]: This series is divergent; therefore, we may be able to do something with it. -- Oliver Heaviside -[Edit (1/14/21) from the answer by Count Iblis to a recent MO-Q on math vids: An enthusiastic intro is that to the set of lectures by Carl Bender "Perturbation and Asymptotic Series." ] -Other than the usual references given in Wikipedia and Mathworld, which resources have you found helpful as intros to the topic and for advanced exploration? -I'll prime the pump with - -"Divergent series:taming the tails" by M. V. Berry and C. J. Howls (cf. also refs in this MO-Q) - -Sporadic examples in Heaviside's publications, see Heaviside's Operational Calculus, a post by Ron Doerfler. - -A Singular Mathematical Promenade by Etienne Guys - -Sum Divergent Series by the user mnoonan, a series of posts at The Everything Seminar - -"Euler's constant: -Euler's work and modern developments" by Jeffrey Lagarias - -"Uniform asymptotic methods for integrals" by Nico Temme - -"On the Specialness of Special Functions (The Nonrandom Effusions of the Divine Mathematician)" by -Robert W. Batterman - - -For one example of the importance of such series, see the relation between the Harer-Zagier formula and the asymptotic expansion of the digamma function in Chapter 5 "The Euler characteristic of the moduli space of curves" of the course notes "Mathematical ideas and notions of quantum field theory" by Etingof. - -REPLY [2 votes]: As far as on-line-available things go, I've attempted to modernize some arguments and give examples of asymptotics of integrals (both Watson's Lemma and easy Laplace/saddle-point examples), as well as asymptotics for ordinary differential equations, both regular and certain irregular singular points. On-line, as well as a chapter in my Cambridge Univ Press book of 2018 (http://www.math.umn.edu/~garrett/m/v/current_version.pdf). For earlier, separate treatments, see http://www.math.umn.edu/~garrett/m/mfms/notes_2019-20/05e_asymptotics_of_integrals.pdf, http://www.math.umn.edu/~garrett/m/mfms/notes_2013-14/11b_reg_sing_pt.pdf, and http://www.math.umn.edu/~garrett/m/mfms/notes_2013-14/11c_irreg_sing_pt.pdf. Those notes (and the book, on-line or not) have substantial bibliographic/historical references.<|endoftext|> -TITLE: Is the square root of a monotonic function whose all derivatives vanish smooth? -QUESTION [12 upvotes]: Let $g:[0,\infty] \to [0,\infty]$ be a smooth strictly increasing function satisfying $g(0)=0$ and $g^{(k)}(0)=0$ for every natural $k$. - -Is $\sqrt g$ is infinitely (right) differentiable at $x=0$? - -I know that $\sqrt g \in C^1$ at zero*, and that in complete generality, one cannot expect for $\sqrt g$ to be even $C^2$. However, in the counter-example given in the linked question, $g$ was not monotonic. - -Does this additional assumption of (strict) monotonicity save us? I tried to look at the literature, but did not find a treatment of this particular case. - -*The proof that $\sqrt g \in C^1$ goes by rewriting $g(x)=x^2h(x)$ where $h \ge 0$ is smooth (this is possible since $g(0)=g'(0)=0$). -Edit: -As pointed out by Igor Rivin, it seems that theorem 2.2 (on page 639) here (pdf) does the job. It states that - any square root of $f$ "precised up to order $m$" is of class $C^m$. (The definition of a "square root precised up to order $m$" is Definition 1.1 on page 636). -This certainly settles the issue. -However, I think it would still be nice to find a simpler approach, since here we assume much more-the strict monotonicity is a much stronger assumption than those assumed in the paper. - -Comment: -If we assume that $g''>0$ in a neighbourhood of zero (which implies that $g'>0$), then $\sqrt g \in C^2$. (details below). - -I think that there is a chance for smoothness under the additional assumption that $g^{(k)}>0$ in a neighbourhood of zero for every $k$, but I am not sure. The calculations become quite messy even when trying to establish $\sqrt g \in C^3$. - - -A proof $\sqrt g \in C^2$ when $g',g''>0$ near zero: (We use these assumptions when applying L'Hôpital's rule). -$$\sqrt{g}'' = \frac{g''}{2\sqrt{g}} - \frac{(g')^2}{4g^{3/2}}.$$ -Thus it is enough to prove that $(g'')^2/g\to 0$ and $(g')^4/g^3\to 0$. -$$ -\lim_{x\to 0^+} \frac{(g'')^2}{g} = \lim_{x\to 0^+} 2\frac{g''g^{(3)}}{g'} = \lim_{x\to 0^+} 2\frac{g''g^{(4)}+(g^{(3)})^2}{g''} = 0, -$$ -where in the last equality we applied $\frac{(h')^2}{h}\to 0$ above for $h=g''$. -$$ -\lim_{x\to 0^+} \frac{(g')^4}{g^3} = \lim_{x\to 0^+} \frac{4(g')^2g''}{3g^2} = \lim_{x\to 0^+} \frac{8(g'')^2 + 4g' g^{(3)}}{6g} = \lim_{x\to 0^+} \frac{2g' g^{(3)}}{3g} = \lim_{x\to 0^+} \left(\frac{2g^{(4)}}{3} + \frac{2g''g^{(3)}}{3g'}\right)=\lim_{x\to 0^+} \frac{2g''g^{(3)}}{3g'} = \lim_{x\to 0^+} \frac{2g^{(4)}}{3}+\frac{2(g^{(3)})^2}{3g''} = 0,$$ -where in the first row we used the first calculation, and in the second we again applied $\frac{(h')^2}{h}\to 0$ to $h=g''$. - -REPLY [4 votes]: The answer is yes, by the results of -Bony, Jean-Michel; Colombini, Ferruccio; Pernazza, Ludovico, On square roots of class (C^m) of nonnegative functions of one variable, Ann. Sc. Norm. Super. Pisa, Cl. Sci. (5) 9, No. 3, 635-644 (2010). ZBL1207.26004. -Here is the math review: - -Clearly the condition is fulfilled in the OP (for any $m$).<|endoftext|> -TITLE: A recommendation for a book on perverse sheaves -QUESTION [8 upvotes]: I would like to learn about perverse sheaves. -I will be grateful if someone could recommend me a book with the following structure. - -Introduction to basic homotopy theory (derived category and t-structure) -Introduction to sheaves -Introduction to perverse sheaves - -REPLY [8 votes]: Pramod Achar is working on a book on perverse sheaves and applications in representation theory. It's a great book! -EDIT (2021): The book has now been published by the AMS: -Perverse Sheaves and Applications to Representation Theory.<|endoftext|> -TITLE: High sum of fractional parts -QUESTION [6 upvotes]: Let $n\geq 2$ and $x_1,\ldots,x_n > 0$ be such that $x_1+\cdots+x_n =1$. Is it true that there must exist a positive integer $k$ such that $$\{x_1k\}+\cdots+\{x_nk\} = n-1?$$ -This looks closely related to the density of the fractional part. -Note that the quantity $\{x_1k\}+\cdots+\{x_nk\}$ is always an integer, since it equals $k-\lfloor x_1k\rfloor - \dots - \lfloor x_nk\rfloor$. Also, as each term is strictly less than one, $n-1$ is the highest value the sum can take. - -REPLY [10 votes]: If the $x_i$ are rational, take the lcm of the denominators and decrease it by $1$. -If they are not necessarily rational, act similarly: using, e.g., Kronecker's theorem, take a $k$ such that all the $kx_i$ are sufficiently close to integers, and decrease that $k$ by $1$.<|endoftext|> -TITLE: Existence of injective compact operators -QUESTION [5 upvotes]: We know that if $X$ is a separable Banach space, then for every infinite dimensional Banach space $Y$, there exists an injective compact operator from $X$ to $Y$. -My query is for every Banach space $X$ (need not be separable ) do there exist a Banach space $Y$ and an injective compact operator $T:X\to Y$? - -REPLY [12 votes]: No, for cardinality reasons. The range of a compact operator is norm-separable hence has cardinality continuum (if non-zero). It is then enough to take $X$ to have bigger cardinality, for example, $X = \ell_\infty^*$. Then you have no chance of building such operators. -Another possibility for counterexamples comes with non-separable reflexive spaces (or, more generally, WCG spaces), which contain non-separable weakly compact sets. Injective bounded linear operators are then homeomorphic embeddings of such sets with respect to the weak topology, so cannot be compact.<|endoftext|> -TITLE: Does every cofinal branch through Kleene's O compute true arithmetic? -QUESTION [9 upvotes]: My question concerns cofinal branches through Kleene's $O$, which is a set of natural numbers and a computably enumerable relation $<_O$ on this set that provides -ordinal denotations for any desired computable ordinal. For every number $n\in O$, the $<_O$-predecessors of $n$ in $O$ are a computably enumerable set of natural numbers that is well-ordered by $<_O$, -representing a computable ordinal, and every computable ordinal is -represented in this way. Meanwhile, the set $O$ itself is not computable nor even hyperarithmetic, for -it is $\Pi^1_1$-complete. -I am interested specifically in the complexity of cofinal branches through -Kleene's $O$. Let us say that $z$ is a cofinal branch through $O$ -if $z\subset O$, the members of $z$ are linearly ordered by $<_O$, -and $z$ contains one index of every computable ordinal rank. -My intuition is that such branches must have high Turing degree, -but I haven't been able to prove this. For example, because of the -close connection between $O$ and computable ordinals, it would seem -reasonable to suppose that every cofinal branch can compute WO, the -set of Turing machine programs computing a well-ordered relation on $\mathbb{N}$. -Question 1. Does every cofinal branch through Kleene's $O$ -compute a $\Pi^1_1$-complete set of natural numbers? -An affirmative answer would imply, in particular, that every cofinal branch $z$ could compute $O$ itself. -Failing this, perhaps every branch can at least compute the set TA -of true arithmetic assertions. -Question 2. Does every cofinal branch through Kleene's $O$ -compute true arithmetic? -In other words, if I have a cofinal branch $z$ through Kleene's -$O$, and I use $z$ as an oracle, can I compute whether a given -arithmetic sentence is true in the standard model? -This question arose recently in the seminar I am running with Wesley Wrigley, in connection with some of his work, which concerns the Feferman-Spector theorem, asserting that there are some cofinal branches through $O$ for which the theory that arises by iteratively adding consistency statements is not complete, even for $\Pi^0_1$ arithmetic truth. Notice that the issue of whether this theory is incomplete, however, is not the same as the question whether the path itself, when used as an oracle, can compute true arithmetic. - -REPLY [10 votes]: Goncharov, Harizanov, Knight and Shore investigated the Turing degrees of $\Pi^1_1$ cofinal branches (which they call "paths through $\mathcal{O}$"). They showed there is a $\Pi^1_1$ cofinal branch which does not compute $\emptyset'$, so certainly doesn't compute true arithmetic. On the other hand, H. Friedman showed there is a $\Pi^1_1$ cofinal branch which computes $\mathcal{O}$ (reference can be found in the GHKS paper).<|endoftext|> -TITLE: Are twin primes the only solution to this equation? -QUESTION [5 upvotes]: Let $p,q_i, i \ge 1$ be primes, $m$ a positive integer. -The equation -$$ -p.\prod_{i=1}^m(q_i-1)-(p-1).\prod_{i=1}^mq_i=2 -$$ -for $m=1$ has all twin primes $p,q_1=p+2$ as solution. -Are there solutions for all $m\ge2$ ? - -REPLY [4 votes]: Given an odd prime $p$ and integer $m \geq 2$, we can set $q_1=p+2$ and then, for each $i$ with $2 \leq i \leq m$, -$$ -q_i = \frac{ 4+ (p-1) \prod_{j=1}^{i-1} q_j}{2}. -$$ -One can check that -$$ -p \cdot \prod_{i=1}^m (q_i-1) - (p-1) \cdot \prod_{i=1}^m q_i =2, -$$ -and each $q_i$ is a polynomial in $p$, of degree $2^{i-1}$ (with coefficients in $\mathbb{Z}[1/2^{i-1}]$). -It is plausible that this construction should produce infinitely many values of $p$ for which $p$ and each $q_i$ are simultaneously prime. That being said, I may be missing some local obstruction. While this very likely leads to infinitely many examples for $m \in \{ 2, 3, 4 \}$, the smallest with $m=4$ being -$$ -(p,q_1,q_2,q_3,q_4)=(3,5,7,37,1297), -$$ -examples with $m=5$ are rather thinner on the ground : the smallest is with $p=512351711$.<|endoftext|> -TITLE: Computation of fraction field of formal series over the integers -QUESTION [21 upvotes]: What is the fraction field $K$ of the domain $\mathbb Z[[X]]$? -It is strictly smaller than the field of Laurent series $L=\operatorname {Frac}\mathbb Q[[X]]$, since $\sum_{i\geq 0}\frac {X^i}{i!}\in L\setminus K$. -Indeed a Laurent series $f(X)=\sum_{i\in \mathbb Z} q_iX^i \in L \; (q_i=0 \operatorname {for} i\lt\lt 0)$ must have as its coefficients rational numbers whose denominators involve only finitely many primes (which depend only on $f(0)$) in order to belong to $K$. -But is that necessary condition sufficient? -Edit -No, that necessary condition is not sufficient. -For example if $p$ is a prime the series $\sum_{i\geq 0}\frac {X^i}{p^{i^2}}$ does not belong to $K=\operatorname {Frac}\mathbb Z[[X]]$, even though the denominators in its coefficients only involve the single prime $p$. -This follows from Elad Paran's Example 2.3(c) quoted by Arno in his very pertinent answer below. - -REPLY [2 votes]: This is actually a remark on Arno Fehm's answer, but too long for a comment (the remark is due to P. Samuel). Let $p$ be a prime, and let us look at power series $u(X)=\sum_{n\geq 1}a_nX^n$ in $\mathbb{Q}((X))$ satisfying $u^2-pu+X=0$. This is equivalent to $a_1=p^{-1}$, $a_n=a_1a_{n-1}+\ldots +a_{n-1}a_1$. So the $a_i$ are uniquely determined, and of the form $\dfrac{b_i}{p^i} $ with $b_i\in\mathbb{Z}$. Still $u$ does not belong to $K$, because it is integral over $\mathbb{Z}[[X]]$ which is integrally closed. If it were in $K$, it would belong to $\mathbb{Z}[[X]]$, and this is clearly not the case.<|endoftext|> -TITLE: Positive set theory and the "co-Russell" set -QUESTION [16 upvotes]: This is a more focused version of a question which was asked at MSE a couple years ago, but is still unanswered there. That question asks about a broad range of theories, whereas this version focuses on a single one. -Let $S=\{x: x\in x\}$ be the "dual" to Russell's paradoxical set $R$, and consider the question "Is $S\in S$?" If we try to work purely "naively" there does not seem to be an immediate argument one way or the other (dually to the situation for $R$). However, in light of Lob's theorem it's a bit premature to leap to the conclusion that there actually are no such arguments. -So it remains plausible that there is a "reasonable" set theory in which the question "Is $S\in S$?" is decided in a nontrivial way. Note that the nontriviality requirement rules out both $\mathsf{ZF}$- and $\mathsf{NF}$-style set theories: the former prove $S=\emptyset$, while the latter prove that $S$ is not a set in the first place. -The most appealing candidate to me appears to be $\mathsf{GPK}_\infty^+$. This theory proves that $S$ is a set and is nonempty. It also (in my opinion) has a very attractive intuition behind it, as well as an interesting model theory. So my question is: - -Does $\mathsf{GPK}_\infty^+$ decide whether $S\in S$? - -REPLY [3 votes]: I will show that $\mathsf{GPK}_\infty^+$ is consistent with $S \in S$, under the assumption of a weakly compact cardinal. Specifically, I'll show that the 'standard models' of $\mathsf{GPK}_\infty^+$ (which require weakly compact cardinals) satisfy $S \in S$. I've made very little progress with proving anything from $\mathsf{GPK}_\infty^+$ directly, although this is probably my failing. -The main idea of the argument is very simple: Show that these models satisfy that for every set $X$, there exists a set $Y$ satisfying $Y = X \cup \{Y\}$. You then note that $S$ must be a fixed point of this operation, so it must be the case that $S \in S$. -For the sake of completeness, I will give the construction of a model of $\mathsf{GPK}_\infty^+$ in a way that should feel familiar to classical set theorists. I'm also doing this because I couldn't find a presentation of this construction that I liked online, although I did use the Stanford Encyclopedia of Philosophy article Alternative Axiomatic Set Theories to get the general idea. - -Construction -Recall that a forest is a partially ordered set in which the set of predecessors of any element is well-ordered. A tree is a forest with a unique minimal element. I'll use the word path to refer to a downwards-closed, linearly ordered set. (I can't remember if there's a standard term for this.) The height of a path is its order type as an ordinal. The height of a node $x \in T$ is the height of the path $\{ y \in T : y < x\}$. To make the presentation more uniform between successor and limit stages, we'll refer to paths as having children, rather than just nodes. A node $x$ is the child of a path $A$ if $\{ y \in T : y < x\} = A$. A path $A$ is the parent of a node $x$ if $x$ is the child of $A$. Finally, a branch is a path of maximal height. (The forest we will construct will not have any dead paths that are shorter than the forest itself.) -We will build a forest (although really it's just a tree and an isolated branch corresponding to the empty set) whose nodes of height $\alpha$ are labeled with sets of paths of height $\alpha$. (Note that this isn't circular, because the paths of height $\alpha$ do not contain nodes of height $\alpha$.) Every set of paths of height $\alpha$ will be attached to precisely one node of height $\alpha$. We will use the convention that lowercase letters refer to sets of paths and that uppercase letters refer to paths of sets. -At stage $0$, there is a unique path of length $0$, the empty path $\Lambda$, and there are two sets of such paths, the empty set and the singleton $\{\Lambda\}$, so these are the roots of the two trees in our forest. -At non-zero stage $\alpha$, given the forest built up to height $\alpha$, we add each set $x$ of paths of length $\alpha$ as a node. The parent of the node $x$ is the unique path $A$ with the property that for any node $y \in A$, - -for every $B \in x$, there exists a $C \in y$ such that $B$ extends $C$ and -for every $C \in y$, there exists a $B \in x$ such that $B$ extends $C$. - -It is not hard to show by induction that this is well defined. -Now we build this forest up to a height $\kappa$, where $\kappa$ is a weakly compact cardinal. The elements of the model are the branches of the forest, and the element of relation is defined by $A \in B$ iff for every $\alpha < \kappa$, $A \upharpoonright \alpha \in B(\alpha)$. This structure is typically referred to as the $\kappa$-hyperuniverse, but I'll just call it $M$. - -Argument -I claim that $M$ satisfies the following: - -For every $X \in M$, there exists $Y \in M$ such that $Z \in Y$ if and only if either $Z \in X$ or $Z = Y$. - -It's easy to see how this resolves the status of $S \in S$. Let $S' = S \cup \{S'\}$. Clearly by construction, $S' \supseteq S$ and $S' \in S'$. Therefore we have $S' \in S$, but this implies that $S \supseteq S'$, so $S = S'$ and $S \in S$. I think the most reasonable approach to resolving your question would be to prove this claim directly from $\mathsf{GPK}_\infty^+$, but I can't even show that $\mathsf{GPK}_\infty^+$ entails the existence of a Quine atom. (Or that any two Quine atoms are equal, for that matter.) Someone probably knows how to do this, though, and I'd be interested to see it. -Proof of claim. To prove this, consider the path corresponding to $X$. Let $Y$ be the $\kappa$-indexed sequence of nodes on the forest defined inductively by $Y(\alpha) = X(\alpha) \cup \{ Y \upharpoonright \alpha\}$ for each $\alpha < \kappa$. We need to show that this is well defined. - -For $\alpha = 0$, $Y \upharpoonright 0 = \Lambda$, so $Y(0) = X(0) \cup \{\Lambda\}$. - -For a non-zero $\alpha$, assuming we've shown that $Y \upharpoonright \alpha$ is actually a path on the forest, we immediately get that $Y(\alpha) = X(\alpha) \cup \{ Y \upharpoonright \alpha\}$ is a node of height $\alpha$ in the forest. We just need to show that $Y(\alpha)$'s parent is $Y\upharpoonright \alpha$. Fix a node $Y(\beta) \in Y \upharpoonright \alpha$ (for some $\beta < \alpha$). For any $A \in Y(\alpha)$, either $A \in X(\alpha)$ or $A = Y\upharpoonright \alpha$. In the first case, by the fact that $X\upharpoonright \alpha$ is a path, there must exist a $B \in X(\alpha)$ such that $A$ extends $B$. In the second case, $Y \upharpoonright \beta \in Y(\beta)$ and is extended by $Y\upharpoonright \alpha$ (by the induction hypothesis). Essentially the same argument gives that for any $B \in Y(\beta)$, there is an $A \in Y(\alpha)$ extending $B$. - - -Therefore $Y$ is a branch of the forest and corresponds to an element of $M$. $Y$ is clearly a superset of $X$ and clearly contains $Y$ as an element. We need to show that $Y = X \cup \{Y\}$. -Suppose that $Z \in Y$. For each $\alpha < \kappa$, we get that either $Z \upharpoonright \alpha = Y \upharpoonright \alpha$ or $Z \upharpoonright \alpha \in X(\alpha)$. Because of the way forests/trees work, if there is an $\alpha$ such that $Z \upharpoonright \alpha \neq Y \upharpoonright \alpha$, then this must also be true for any $\beta \in (\alpha,\kappa)$. Therefore, if $Z \neq Y$, then for some $\alpha < \kappa$, we have that $Z \upharpoonright \beta \in X(\beta)$ for all $\beta \in (\alpha,\kappa)$. It's not hard to show that the same must be true for any $\beta <\kappa$, so we have that $Z \in X$, proving the claim.<|endoftext|> -TITLE: Number of real roots of irreducible polynomials that are solvable by radicals -QUESTION [19 upvotes]: Let $n \geq 3$ be a natural number. Define the set $X_n$ as the set of natural numbers that appear as the number of real roots an irreducible polynomial of degree $n$ over $\mathbb{Q}$ which is solvable by radicals can have. -Example: In case $n=p$ is a prime, we have $X_p= \{1,p \}$. -Is $X_n$ or the cardinality $|X_n|$ also known for other values? -It would be interesting to see the beginning of the sequence $a_n=|X_n|$ for small values of $n$, maybe it appears in the oeis. - -REPLY [6 votes]: Jensen On the number of real roots of a solvable polynomial - includes a proof of: -Loewy’s theorem. Let $K$ be a real number field and $f(X)$ an irreducible polynomial in $K[X]$ of odd degree $n$. If $p$ is the smallest prime divisor of $n$ and the Galois group of $f(X)$ over $K$ is solvable, then $r(f) = 1$ or $n$ or satisfies the inequalities $p ≤ r(f)≤ n−p +1.$<|endoftext|> -TITLE: Basis vs Schauder basis in normed spaces -QUESTION [8 upvotes]: Following the conventions from Heil: "A Basis Theory Primer" and Albiac, Kalton: "Topics in Banach Space Theory", we might define a basis of an (infinite-dimensional) normed space $V$ as a sequence $(e_n)$ in $V$, such that for any $x \in V$ there is a unique sequence of scalars $(\lambda_n)$, such that $x = \sum_n \lambda_n e_n$ (converging in norm), whereas for a Schauder basis we demand that these coefficients are produced by linear continuous functionals $(e^*_n)$, such that $e^*_m(e_n) = \delta_{mn}$ and $\lambda_n = e^*_n(x)$. -Now, if we work in a separable Banach space, these two notions coincide (theorem 4.13 in Heil and theorem 1.1.3 in Albiac, Kalton), but what if $V$ is a separable normed space which is not complete? Is there a simple, instructive example in which these linear functionals $(e^*_n)$ exist but fail to be continuous? - -REPLY [9 votes]: You ask for an instructive example, so I'll be long winded. -Suppose $(x_n)$ is basis for a normed space $(X,\|\cdot\|)$. The partial sum projections $S_n$ are well defined but might be discontinuous. Define a new, larger, norm on $X$ by $|x| := \sup_n \|S_n x\|$. Then $(x_n)$ is a Schauder basis for $(X,|\cdot|)$. Note that this is the first step in the proof that every basis for a Banach space is a Schauder basis. The harder part of the proof is to show that $(X,|\cdot|)$ is complete when $(X,\|\cdot\|$ is complete. When $(X,\|\cdot\|)$ is not complete, $(X,|\cdot|)$ may not be complete, but if you are building a basis for $(X,\|\cdot\|)$ that is not a Schauder basis, it would be nice to have one such that the basis is a Schauder basis for some natural complete norm on $X$. Also, you would like your non Schauder basis to be in a some natural normed space; let's say an inner product space that is a non closed subspace of $\ell_2$ under the usual $\ell_2$ norm, $\|\cdot\|_2$. Well, $\ell_2$ has lots of non closed subspaces that are complete under a natural norm, the most used being $\ell_1$ under its norm $\|\cdot\|_1$. It should not be too hard to find a Schauder basis for $\ell_1$ such that at least one of the biorthogonal functionals is not in $\ell_2$. Perhaps the simplest such example is $x_1 = e_1$ and $x_n = e_1+e_n$ when $n>1$. (Here $(e_n)$ is the standard unit vector basis.) Notice that the biorthogonal functionals in $\ell_\infty = \ell_1^*$ are given by $x_1^* =(1,-1,-1,\dots)$ and, for $n>1$, $x_n^* = (0,0,\dots,1,0,\dots)$, where the $1$ is in the nth coordinate. Only $x_1^*$ is not continuous in the $\ell_2$ norm--this makes it easy to verify that $(x_n)$ is a basis for $(\ell_1,\|\cdot\|_2$).<|endoftext|> -TITLE: In the two-person Killing the Hydra game, what is the winning strategy? -QUESTION [52 upvotes]: My question is which player has a winning strategy in the -two-player version of the Killing the Hydra game? -In their amazing paper, - -Kirby, Laurie; Paris, Jeff, Accessible independence results for Peano arithmetic, Bull. Lond. Math. Soc. 14, 285-293 (1982), ZBL0501.03017, - -Laurie Kirby and Jeff Paris introduced the Killing the Hydra game, -in which one attempts to kill the Hydra by cutting off its heads. -At stage $n$, when you make a cut, just below a head, the Hydra grows $n$ copies of -itself, copies of the position from one lower node (if any) up to the node preceding the neck that had been cut, and whatever is above that node. To illustrate, here are some initial moves in the Hydra -game: - -The Hydra game involves some fascinating issues in mathematical -logic, because of its connection with Goodstein's -theorem. -Specifically, what Kirby and Paris proved is -Theorem. - -Every strategy in the Killing the Hydra game will eventually succeed in -killing the Hydra; and -This fact is not provable in Peano Arithmetic (PA). - -My question here, however, is concerned with the natural -two-player version of the game. Specifically, given a finite -Hydra tree, we play a two-player version of the Killing the Hydra -game, where each player makes a cut on their turn, and the Hydra -grows new heads according the original Hydra rules. The first -player without a move loses---you want to cut the very last head. -Question. Which player has a winning strategy? What is the -winning strategy? -Since every play of the game will lead eventually to a win for one -of the players, it follows by the fundamental theorem of finite -games that one of the players will have a winning strategy. Which -player is it that has the winning strategy? And what are the -winning moves? -The Kirby-Paris theorem is quite robust with respect to the game rules, for it works even when the Hydra grows many more than $n$ copies at stage $n$, or fewer; but I expect that the two-player version might be sensitive to such changes in the rules. Please provide an answer for any reasonable version of the game to which the Kirby-Paris theorem still applies. - -REPLY [25 votes]: We can think of this as a game of "omega-nim;" to more precise since the game you are describing is impartial, operating under the normal play convention, and finite we have that the Sprague-Grundy Theorem applies. In other words, to every "hydra-ordinal" there is an "omega-nimber." -Already this suggests thinking of the plus signs in the hydra as being something roughly analogous to the heaps in Nim. Let me make this precise. - -Definition The set of hydras (or hydrae?), $\mathcal{H}$, is defined recursively as - -$0 \in \mathcal{H}$ -$ n \in \mathbb{N} \implies \omega^n \in \mathcal{H}$ -$ \kappa_0,...,\kappa_{n-1} \in \mathcal{H} \implies \sum_{i}\omega^{\kappa_i} \in \mathcal{H}$ - - -We define the "winner function" as: - -Definition We denote by $w(\kappa, n, i )$ the "winner of the game $\kappa$ at the $n^\text{th}$ step if it is currently $i^\text{th}$ player's ($i \in \{0,1\}$) turn to move," i.e. $n $ is the number of hydra heads that will grow out if player $i$ cuts a head off of $\kappa$ and $w(\kappa, n, i )$ is the winner under optimal play. Since $w(\kappa, n, i )= 1- w(\kappa, n, 1- i )$ we will sometimes just consider the case $w(\kappa, n) \overset{\text{def}}{=}w(\kappa, n, 0) $ for simplicity. - -The answer to your question is to compute $w(\kappa,1,i)$; but of course, in order to answer it we are going to need to define the following - -Definition A strategy is a $\sigma: (\mathcal{H} \setminus \{0\}) \times \mathbb{N} \longrightarrow \mathcal{H} $ such that $\sigma(\kappa,n)$ is a legal position at step $n+1$ that succeeds a legal position at $n$. Equivalently we could have allowed for a "virtual position/move" $-1$ and defined $\sigma': \mathcal{H} \times \mathbb{N} \longrightarrow \mathcal{H} \cup \{-1\}$ as $\sigma'(0,n) = -1$ and $\sigma'(\kappa,n) = \sigma(\kappa,n)$ otherwise. We let $\mathcal{S}$ be the set of all such strategies. - -Let us try to define the winning $\sigma$ by taking cases on the different possible "heaps." -The heap has size 0 or 1 -Since $w(0,n,i) = 1-i$ the strategy $\sigma(0,n)$ is meaningless (that's we defined $\sigma $ on $(\mathcal{H} \setminus \{0\})$); likewise we see that $w(1,n,i) = i$ so that $\sigma(1,n)$ is forced to be $0$. This easily generalizes to the following case -The hydra is a natural number -It is straightforward to prove by induction that $w(k,n,i) = 1- ((k+i) \% 2)$ where $\%$ is remainder after division since by the rules of the game no new hydras grow after cutting a head off a hydra of the form $\omega^0 + \omega^0 +... + \omega^0 = \omega \cdot k $. Likewise $\sigma(1,n)$ -The hydra is of the form $\kappa + 1$ -If $\kappa ' = \left( \sum_{i}\omega^{\kappa_i}\right) + 1 = \kappa + 1$ then $w(\kappa' ,n,i) = 1 - w(\kappa' - 1 ,n,i) = 1-w(\kappa ,n,i) $. This can proven by a "sum of games" style proof. The idea is the following: if it is $i$'s turn then either: - -making a cut on $\kappa$ wins the game, -making a cut on $\omega^0 = 1$ wins the game, -or none of the above - -but these are respectively true if and only if - -The cut $\sigma(\kappa,n)$ loses the game (for the game $\kappa$ not $\kappa+1$) for all $\sigma$ - -proof by induction using the following two observations: 1) $\omega^0=1$ (or "parity") is a "loop-invariant" of the game and 2) $\sigma(\kappa,n) < \kappa$ (see the proof of thm 2 of [Kirby; Paris] for the proof of the inequality) 3) eventually we will hit $\kappa \in \omega $ for any strategy $\sigma$ ( once again see [Kirby; Paris]) which is the previous case - -the position $\kappa$ is a losing position, but this is true iff the first case is true -or none of the above, but this is true iff the first case is false - -Therefore the game is lost or won regardless of what move is made; it only depends on the "parity." -The hydra is of the form $\kappa + \lambda $ -By a proof by induction and using the fact that $\sigma(\kappa,n) < \kappa$ and $\sigma(\lambda,n) < \lambda $ (see thm 2 of [Kirby; Paris]) we have that -\begin{equation} -w(\kappa + \lambda , n ) = w(\kappa , n ) \oplus w( \lambda , n ) \oplus 1 . -\end{equation} -Since we have already proven it for $\lambda =1$ and we can assume $\kappa > 2$ we can take the following cases in the induction - -Player 1 cuts $\kappa$ then Player 2 cuts $\kappa$ -Player 1 cuts $\kappa$ then Player 2 cuts $\lambda$ -Player 1 cuts $\lambda$ then Player 2 cuts $\kappa$ -Player 1 cuts $\lambda$ then Player 2 cuts $\lambda$ - -Which all lead to smaller cases to which we can apply the induction hypothesis. -Lets verify for simple examples: since $\sigma(\omega,n) = n $ we have that $w(\omega,n) = n \% 2 $ and also $w(0,n) =1$ which agrees with $w(\omega+ 0 ,n) = n \% 2 = w(\omega , n ) \oplus w( 0 , n ) \oplus 1$. Similarly $w(\omega+ 1 ,n) = n \oplus 1 = w(\omega , n ) \oplus w( 1 , n ) \oplus 1$ and more generally we have that $w(\omega+ k ,n) = n \oplus k = w(\omega , n ) \oplus w( k , n ) \oplus 1$. Likewise by a second induction we have that -\begin{equation} -w\left(\sum_i \kappa_i , n \right) = 1 \oplus \bigoplus_i w( \kappa_i , n ) . -\end{equation} -The hydra is of the form $\omega ^ \kappa $ -The point here is that is $\omega ^ \kappa \neq \omega ^ {\lambda +1}$ for all $\lambda \in \mathcal{H}$ then all of the cuts are made inside of the $\kappa$. By induction, we also see that if $\omega ^ \kappa = \omega ^ {\lambda +1}$ then $w(\omega^\kappa,n)$ only depends on the parity of $n$ since if $\sigma$ is the "subtract 1" cut then $\sigma(\omega^\kappa,n) = \omega^\lambda \cdot n$ and $w(\omega^\lambda \cdot n, n+1, 1 ) = \bigoplus_i w( \omega^\lambda , n+1 ) $ by the previous section, so that $w(\omega^\kappa, n ) = \min\{1 \oplus \bigoplus_i w( \omega^\lambda , n + 1) ,1 \oplus w(\kappa, n )\}$ by a similar proof to the last section. The second term corresponds to the following: if player 0 can lose/win the game $\kappa$ then $\kappa$ corresponds to an odd/even game, but when the time finally comes to split $\omega^1$ player 1 will be left with an even/odd game.<|endoftext|> -TITLE: Analogy between Stiefel-Whitney and Chern classes -QUESTION [16 upvotes]: There is a clear similarity between Stiefel-Whitney and Chern classes, if one replaces base field $\mathbb R$ with $\mathbb C$, coefficient ring $\mathbb Z/2$ with $\mathbb Z$ and scales the grading by a factor of $2$. For instance, both can be defined by the same axioms (functoriality, dimension, Whitney sum, value on the tautological bundle over $\mathbb P^1$). -Is there a deep reason behind this correspondence? The best explanation I have so far is the structure of classifying spaces. Being Grassmanians, they admit Schubert cell decomposition (which is essentially an algebraic fact). For cohomology of complex Grassmanians, differentials vanish for dimensionality reasons, and for real ones they vanish when reduced mod $2$. -There is a number of similar phenomena, for instance, $BO(1,\mathbb R) = K(\mathbb Z/2, 1)$ and $BU(1) = K(\mathbb Z, 2)$ which says that in both cases, topological line bundles are completely determined by their first characteristic class. -Also, I have been told that Thom polynomials for Thom-Boardman singularities of maps between real or complex manifolds have the same coefficients when expressed in $w_i$ and $c_i$. Can these facts be explained in a similar way? - -REPLY [7 votes]: Any rank $n$ real vector bundle $E\to X$, $X$ compact $CW$-complex is $\newcommand{\bZ}{\mathbb{Z}}$ is $\bZ/2$-oriented and, as such it has a $\bZ/2$-Thom class $\tau_E\in H^n_{cpt}(E,\bZ/2)$. Then -$$w_n(E)=\zeta^*\tau_E\in H^n(X,\bZ/2),$$ where $\zeta:X\to E$ is the zero-section -Any complex vector bundle $E\to X$, $X$ compact $CW$-complex of complex rank $n$ is $\bZ$-oriented and, as such it has a $\bZ$-Thom class $\tau_E\in H^{2n}_{cpt}(E,\bZ)$. (Note that $2n$ is the real rank of $E$.) Then -$$c_n(E)=\zeta^*\tau_E\in H^{2n}(X,\bZ).$$ -Thus in both cases the top Stieffel-Whitney class and the top Chern class are Euler classes, with different choices of coefficients. $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\bP}{\mathbb{P}}$ -To get the rest of the Stieffel-Whitney/Chern classes on then needs to rely on some basic facts -$$ H^\bullet(\bR\bP^n,\bZ/2)\cong\bZ/2[w]/(w^n+1),\;\; H^\bullet(\bC\bP^n,\bZ)\cong\bZ[c]/(c^n+1),\tag{1} -$$ -where $w\in H^1(\bR\bP^n,\bZ/2)$, $c\in H^2(\bC\bP^n,\bZ)$ are the Euler classes of the (duals) of the tautological line bundles. -These results suffice to construct the Stieffel-Whitney/Chern classes. This is the approach pioneered by Gronthendieck for the construction of Chern classes. For details see Chapter 5 of these notes.. As explained in Example 4.3.5. of these notes duality, under the guise of Thom isomorphism is also responsible for the isomorphisms (1). -One could claim that duality or Thom isomorphism is what makes things work. What is behind Thom isomorphism? As described in Bott-Tu, this follows from two basic facts about cohomology. The first is the Poincare lemma with compact supports -$$H^k_{cpt}(\bR^n, G)=\begin{cases} 0, & k\neq n,\\ -G, &k=n. -\end{cases} -$$ -and the second is the Mayer-Vietoris principle which, roughly specking says that one can recover the cohomology of an union from the cohomology of its parts. View this as a local-to-global principle, a way a patching local data to obtain global information. The orientability condition is the one that allows the local-to-global transition. - -REPLY [4 votes]: Let me try a high-brow answer using equivariant stable homotopy theory. -By the stable Thom isomorphism, the integral (co)homology of $BU$ agrees with that of $MU$; likewise the $\mathbb{Z}/2$-(co)homology of $BO$ agrees with that of $MO$. -Let $H\underline{\mathbb{Z}}$ be the $C_2$-equivariant Eilenberg--Mac Lane spectrum for the constant Mackey functor and let $M\mathbb{R}$ be the Real cobordism spectrum. Using the theory of Hu and Kriz, we see that $H\underline{\mathbb{Z}}$ is Real oriented and thus the $RO(C_2)$-graded groups $H\underline{\mathbb{Z}}^{\bigstar}M\mathbb{R}$ agree with $H\underline{\mathbb{Z}}^{\bigstar}[[ \overline{c}_1, \overline{c}_2, \dots ]]$, where $\overline{c}_i \in H\underline{\mathbb{Z}}^{i+i\sigma}M\mathbb{R}$. The $\overline{c}_i$ define thus maps $M\mathbb{R} \to \Sigma^{i+i\sigma}H\underline{\mathbb{Z}}$, which forget to the usual Chern classes $c_i\colon MU \to H\mathbb{Z}$. -The geometric fixed points of $M\mathbb{R}$ are $MO$, while those of $H\underline{\mathbb{Z}}$ are $\prod_{k\geq 0}\Sigma^{2k}H\mathbb{Z}/2$ and thus come with a projection $p\colon \Phi^{C_2}H\underline{\mathbb{Z}} \to H\mathbb{Z}/2$. Thus combining $\Phi^{C_2}$ with $p$, the $\overline{c}_i$ define maps $MO \to \Sigma^iH\mathbb{Z}/2$, which we claim to be the $w_i$. -I have not thought through this, but I guess one way to show a thing like this is to use homology instead. We have $\pi_{\bigstar}H\underline{\mathbb{Z}} \otimes M\mathbb{R} \cong \pi_{\bigstar}H\underline{\mathbb{Z}}[\overline{b}_1, \dots]$. Write $\pi_*\Phi^{C_2}H\underline{\mathbb{Z}} = \mathbb{Z}/2[u]$. Then -$$H_*(MO; \mathbb{Z}/2)[u]\cong (\Phi^{C_2}H\mathbb{Z}/2)_*(MO) \cong \pi_* \Phi^{C_2}(H\underline{\mathbb{Z}} \otimes M\mathbb{R}) \cong \mathbb{Z}/2[u][\Phi^{C_2}\overline{b}_1, \dots] $$ -Here, we use that we can get the geometric fixed points by inverting one element $a_{\sigma}$ and that $M\mathbb{R}$ is of finite type. Killing the $u$ on both sides, gives that we obtain indeed the whole homology of $MO$ by this geometric fixed points construction. Thus, I expect that by playing around with dual (co)homology classes, one should get indeed that one obtains the $w_i$ from the $\overline{c}_i$. -(Why was it natural to use the projection $p$? Indeed, when we apply geometric fixed points to the $\overline{c}_i$ all other components are $0$. This we can see by lifting the $\overline{c}_i$ to $BP\mathbb{R}^{\bigstar}M\mathbb{R}$ and observing that the geometric fixed points of $BP\mathbb{R}$ are just $H\mathbb{Z}/2$.)<|endoftext|> -TITLE: Sendov's conjecture -QUESTION [10 upvotes]: It has been more than fifty years for famous Sendov's conjecture which states that if $p(z)$ is a polynomial of degree $n$ having all its zeros in the unit disc $|z|\leq 1$ then each of the n roots is at a distance no more than 1 from at least one critical point. It has also been more than twenty years for its proof for polynomials of degree $n=8$. I believe, after that no progress on greater values of $n.$ Some claims are there on arxiv platform, which are not yet validated by the experts I feel. There is a review article http://www.indianmathsociety.org.in/mathstudent-part-1-2019.pdf#page=101 -It is very sad and unfortunate to hear that Sendov passed away on January 19, 2020 and a nice tribute article is published in Journal of Approximation Theory, see https://www.sciencedirect.com/science/article/pii/S0021904520300423?via%3Dihub -There is another paper on the conjecture for high degree of polynomials, see https://www.ams.org/journals/proc/2014-142-04/S0002-9939-2014-11888-0/home.html -A new approach of solving this problem using Grace Theorem by constructing apolar polynomials is given in the article 'A remark on Sendov Conjecture' published in Comptes rendus de l’Acade'mie bulgare des Sciences, Vol 71 (6), 2018, pp.731-734. -Let me know if anything more significant latest contributions? - -REPLY [8 votes]: A 2010 status report is given by D. Khavinson et al. in Borcea's variance conjectures on the critical points of polynomials. A 2019 update is in A note on a recent attempt to prove Sendov's conjecture, by N.A. Rather and Suhail Gulzar.<|endoftext|> -TITLE: The $\infty$-category of natural transformations as an end -QUESTION [7 upvotes]: Let $\mathcal{C}$ be an $\infty$-category viewed as a fibrant scaled simplicial set with all 2-simplices thin and let $\mathfrak{C}\!at_{\infty}$ be the $\infty$-bicategory of $\infty$-categories. A model for $\mathfrak{C}\!at_{\infty}$ is given by applying the marked version of the homotopy coherent nerve to $\operatorname{Set}_{\Delta}^{+}$. -We define the $\infty$-bicategory of functors $\operatorname{Fun}(\mathcal{C},\mathfrak{C}\!at_{\infty})$ as in the $(\infty,1)$-case. Thin 2-simplices are given by maps of simplicial sets $\mathcal{C}\times \Delta^2_{\sharp} \to \mathfrak{C}\!at_{\infty}$ where $\Delta^2_{\sharp} $ denotes the maximally scaled 2-simplex. -Given two functors $F,G: \mathcal{C} \to \mathfrak{C}\!at_{\infty}$ we can use Proposition 2.4 in here to compute a model for $\operatorname{Nat}(F,G)$ in terms of a suitably lax slice construction. -I would like to show that $\operatorname{Nat}(F,G)\cong \lim_\limits{\operatorname{Tw}(C)}\operatorname{Fun}(F(-),G(-))$, which should obviously be true but I am having trouble constructing the universal cone for this functor. - -REPLY [3 votes]: Let $\mathcal{C}$ be an $\infty$-category and $\mathbb{D}$ be an $\infty$-bicategory. In joint work with W. Stern (Enhaced twisted arrow categories ) we show that there is a natural equivalence of $\infty$-categories -$$\operatorname{Nat}(F,G) \simeq \lim_{\operatorname{Tw}(\mathcal{C})^{\operatorname{op}}} \operatorname{Map}_{\mathbb{D}}(F(-),G(-))$$ -The "op" comes from the fact that we are using right fibration $\operatorname{Tw}(\mathcal{C}) \to \mathcal{C} \times \mathcal{C}^{\operatorname{op}}$.<|endoftext|> -TITLE: Slicing up $\mathbb{N}\setminus\{1\}$ -QUESTION [5 upvotes]: Let $\mathbb{N}$ denote the set of positive integers. For any prime $p\in\mathbb{N}$ let $p\mathbb{N} = \{np: n\in \mathbb{N}\}$. Is there a partition ${\cal P}$ of $\mathbb{N}\setminus\{1\}$ such that for all $B \in {\cal P}$ and every prime $p\in\mathbb{N}$ we have $|B \cap p\mathbb{N}|=1$? - -REPLY [5 votes]: Place each $n$ that is not a prime power into its own $B=B(n)$. Then fill the rest of $B(n)$ with powers of primes that do not divide $n$.<|endoftext|> -TITLE: Which convex bodies can be captured in a knot? -QUESTION [13 upvotes]: Which convex bodies can be captured in a knot? - -This question is based on the discussion in "Is it possible to capture a sphere in a knot?". -We assume that the knot is made from unstretchable, infinitely thin rope. -Comments: - -By the construction Anton Geraschenko, the question is equivalent to existence of a graph embedded in the surface of the convex body that locally minimize the total length. Such embedding exists on some convex bodies, for example on an equilateral triangle shown on the diagram (and on anything sufficiently close). - - - -According to the original question a ball cannot be captured (in fact it cannot be captured in a link with 3 components). Moreover a circular disk cannot be captured see my answer (thanks to Wlodek Kuperberg for asking). Likely the same idea works for all convex bodies of revolution. Maybe all convex bodies of general position can be captured. - -REPLY [2 votes]: A circular disc cannot be captured. -Likely the same idea can be used to show that any convex body of revolution cannot be captured. -It can be proved using a small variation of the idea as in the original answer. -It is sufficient to show that infinitesimal Möbius tranform $m$ of the disc can shorten the wrapping length while preserving the crossing pattern. -Once it is proved, moving in this direction will eventually allow the disc to escape. -Choose a Möbius tranform $m$ of the unit disc that is close to the identity map. -Denote by $u$ its conformal factor. -Denote by $U(r)$ the average value of $u(x)$ for $|x|=r$. -It is sufficient to show that $U(r)\le 1$ and the inequality is strict for $r<1$. -In this case by rotating the knot and applying the Möbius tranform will shorten the length. -Consider the circle $C_r$ of radius $r$ centered at the origin. -Denote by $r'$ the radius of $m(C_r)$. -Observe that $U(r)=r'/r$. -Evidently $r'\le r$ if $r\le 1$ and $r'< r$ if $r< 1$, whence the result.<|endoftext|> -TITLE: Is regularization of infinite sums by analytic continuation unique? -QUESTION [6 upvotes]: There are ill-posed summations that we can assign values to, take for concreteness, -$$ S = \sum_{k=0}^\infty k $$ -to which we can assign $-1/12$ by several methods. Is there a fundamental and rigorous reason why these methods have to agree on the same value? -Perhaps the most powerful means we have available is the zeta function regularization, where we attach $\zeta(-1)$ (Riemann zeta function), which formally represents $S$, to the summation. If we have another analytic function that formally represents $S$ and analytically continues to assign a value to $S$, must the value always be the same, $-1/12$? -More concretely, suppose I have a function $\xi(s) = \sum_{k=0}^\infty f_s(k)$ such that $f_1(k) = k$ and $f_s(k)$ is a "nice" function (for the case of zeta function, $f_s(k)=k^{-s}$ is the exponential function in $s$). Suppose that $\xi(s)$ can be analytically continued to a meromorphic function with a value at $s=-1$. Must it be the case that $\xi(-1)=-1/12$? -More generally, what can be said about other divergent sums? If there is an analytic continuation that defines the value of such sums, must it be unique? Where can I look up known uniqueness results and techniques? - -REPLY [4 votes]: As the other answer has pointed out, $-1/12$ is not the only value that can obtained with analytic continuation. However, it is the unique constant term of the asymptotic expansion of the smoothed partial sums, which perhaps explains why it is the most "natural" value. -Let $\eta$ be any Schwartz function such that $\eta(0) = 1$. Then -\begin{align} -\sum_{n=1}^\infty n^s \eta(n \varepsilon) -&= \zeta(-s) + O(\varepsilon) + \frac{1}{\varepsilon^{s+1}} \int_0^\infty x^s \eta(x) dx -\end{align} -Therefore, by choosing for any given $s$ an $\eta$ that makes the last integral zero, we get -\begin{align} -\sum_{n=1}^\infty n^s -&= \sum_{n=1}^\infty n^s \lim_{\varepsilon \rightarrow 0^+} \eta(n \varepsilon) \\ -&\overset{!}{=} \lim_{\varepsilon \rightarrow 0^+} \sum_{n=1}^\infty n^s \eta(n \varepsilon) \\ -&= \lim_{\varepsilon \rightarrow 0^+} \left( \zeta(-s) + O(\varepsilon) \right) \\ -&= \zeta(-s) -\end{align}<|endoftext|> -TITLE: Solutions of PDE under changing topology -QUESTION [6 upvotes]: Let suppose we have a PDE on a manifold. I'm interested in the following question. How does the space of solutions of this PDE change when the topology of the manifold changes? For example in 2D we consider how the kernel of some partial operator changes under handle attachments. Do people study this type of problems? -Thank you! - -REPLY [10 votes]: Yes, this has been studied intensively for quite a while. In particular, people who work in gauge theory (as applied to 4-manifolds) have studied the effect of surgeries on the solutions to the ASD Yang-Mills equations and Seiberg-Witten equations using essentially analytic techniques. An early result of this kind by Donaldson shows that taking connected sum with ${\mathbb C}P^2$ (with its standard complex orientation) can change the number of solutions (counted with signs) from something non-zero to $0$. In other words, the moduli space (as an oriented $0$-manifold) can change under a simple topological operation on the manifold. -There is a vast literature on the subject, where one can detect much more subtle changes in the smooth topology of $4$-manifolds. You might look up the Fintushel-Stern knot surgery formula. There are also many results in other contexts such as moduli spaces of holomorphic curves. -Of course the most basic example is perhaps what you were alluding to in 2D. The dimension of the linear equations for harmonic forms (aka homology groups, by Hodge theory) depend on the topology of the underlying manifold.<|endoftext|> -TITLE: Cutting up the Bring surface into six pairs of pants -QUESTION [12 upvotes]: The Bring sextic, with 120 automorphisms, is the numerically most symmetric compact Riemann surface of genus 4. To cut it up into six pairs of pants, we need to cut along nine disjoint geodesic loops. How short can those loops be, and how symmetric can we make the decomposition? -I am studying the Bring sextic, by the way, because one can elegantly map all of the possible shapes of equilateral pentagons in the Euclidean plane to the points of the Bring sextic. (Calling it "the Bring sextic", by analogy with "the Klein quartic", obviates the clumsy need to choose between "the Bring surface" over $\mathbb{R}$ and "the Bring curve" over $\mathbb{C}$.) -For comparison, the Bolza quintic is the numerically most symmetric compact Riemann surface back in genus 2, with 48 automorphisms. The systole of the Bolza quintic is $2\operatorname{arccosh}(1+\sqrt{2})\approx 3.057$, and there are $12$ systolic loops (that is, loops of that length). There are triples of systolic loops that are disjoint, and such a triple cuts up the Bolza quintic into two isometric pairs of pants. In that decomposition, the three cuffs of one pair of pants are sewn to the three cuffs of the other, and all three twists, as a fraction of the systole, are $\operatorname{arccosh}((5+4\sqrt{2})/7)/(2\operatorname{arccosh}(1+\sqrt{2}))\approx 0.3213$. -Genus 3 is even prettier. The famous Klein quartic is the symmetry champion, with 168 automorphisms. It has $21$ systolic loops, each of length $8\operatorname{arccosh}(\frac{1}{2}+\cos(2\pi/7))\approx 3.936$. Some sextuples of them are disjoint, and such a sextuple cuts up the Klein quartic into four isometric pairs of pants. The 3-regular graph giving the sewing of the cuffs is $K_4$, the edge-graph of a tetrahedron, and all six twists are $1/8$. -With those cases as context, what happens in genus 4? The Bring sextic has $20$ systolic loops, each of length $2\operatorname{arccosh}((9+5\sqrt{5})/4)\approx 4.603$. Since our loops must be disjoint, however, we can take at most six of those $20$. Cutting along those six breaks the Bring sextic into three pieces, each with four loops of boundary. We need to cut each of those three pieces along another geodesic loop, to split it into two pairs of pants. -The Bring sextic has $30$ loops of length $2\operatorname{arccosh}((11+5\sqrt{5})/4)\approx 4.796$ (only slightly longer than systolic). For each of our three current pieces, there is one of those $30$ loops that splits it into two pairs of pants, leaving us with six isometric pairs of pants overall. The 3-regular graph giving the sewing of the cuffs in the resulting decomposition is $K_{3,3}$, a graph famous for its nonplanarity. The twists of the six systolic loops are $1/6$, while the twists of the three longer loops are $1/4$. -Is this the most symmetric way to cut up the Bring sextic? Or are there other decompositions that can compete with it for simplicity and symmetry? - -REPLY [2 votes]: There is an alternative strategy for cutting up the Bring sextic into pairs of pants that might compete with the decomposition above in some ways. -Of the $30$ loops of the second-shortest length $2\operatorname{arccosh}((11+5\sqrt{5})/4)\approx 4.796$, there are sets of six that are disjoint. Cutting along such a set of six breaks the Bring sextic up into three pieces, each of which has four loops of boundary --- as happened also in the decomposition above. -The Bring sextic has $10$ loops of the third-shortest length, which is $2\operatorname{arccosh}((26+10\sqrt{5})/4)\approx 6.368$. For each of our current three pieces, there are two of those $10$ third-shortest loops, either of which can cut up that piece into two pairs of pants. With three binary choices, we get eight decompositions into isometric pairs of pants. The 3-regular graph that gives the sewing of the cuffs is $K_{3,3}$ in four of those eight decompositions, but is the edge graph of a triangular prism in the other four. The twists are somewhat simpler in all eight of those decompositions, however --- which might make them more attractive than the decomposition above for some purposes: The twists along the six shorter loops are $1/4$, as for the three loops of that same length in the decomposition above, but the twists along the three longer loops are $0$.<|endoftext|> -TITLE: What is the geometric realization of the the nerve of a fundamental groupoid of a space? -QUESTION [6 upvotes]: It can be easily seen that there exists a functor $F:Top \rightarrow Grpd$ from the category of topological spaces to the category of groupoids defined as follows: -Obj: $X \mapsto \pi_{\leq 1}(X)$, where $\pi_{\leq 1}(X)$ is the fundamental groupoid of $X$. -Mor: ($f:X \rightarrow Y) \mapsto F(f):\pi_{\leq 1}(X) \rightarrow \pi_{\leq 1}(Y)$ where the functor $F(f)$ is defined as follows: -Obj: $x \mapsto f(x)$ -Mor: $([\gamma]:x \rightarrow y) \mapsto [f(\gamma)]:f(x) \rightarrow f(y) $ where $[\gamma]$ is the homotopy class of path $\gamma$ in $X$ and $[f(\gamma)]$ is the homotopy class of path $f (\gamma)$ in $Y$. -Also it is not difficult to see that $F$ is well behaved with homotopy (for example in the chapter 6 of http://www.groupoids.org.uk/pdffiles/topgrpds-e.pdf)) that is in the sense that if $f,g: X \rightarrow Y$ are homotopic then the induced functors $F(f)$ and $F(g)$ are naturally isomorphic. -Also using this functor $F$ one can construct a 2-funntor $\tilde{F}: 1Type \mapsto Gpd$ where $1Type$ is the 2-category consisting of homotopy 1-types, maps and homotopy class of homotopies between maps and $Gpd$ is the 2-category consist of Groupoids, functors and natural transformations. Now according to Homotopy hypothesis of dimension 1 as mentioned in http://math.ucr.edu/home/baez/homotopy/homotopy.pdf this $\tilde{F}$ is an equivalence of 2-categories. -So from the above mentioned observations I felt that the functor $F$ is an interesting object of study. -Now if we consider the following sequence of functors: -$$ -X \stackrel{F}{\mapsto} \pi_{\le 1}(X) \stackrel{N}{\mapsto} N(\pi_{\le 1}(X)) \stackrel{r}{\mapsto} r(N(\pi_{\le 1}(X))) -$$ -where $N$ is the nerve functor and $r$ is the geometric realization functor. -My question is the following: -How the topological spaces $X$ and $r(N(\pi_{\leq 1}(X)))$ are related? It may be possible that my question does not make much sense when $X$ is a general topological space but then, does there exist any specific class of topological spaces $X$ which has a "good relation" with $r(N(\pi_{\leq 1}(X)))$? -I would be also very grateful if someone can refer some literature in this direction. -Thank you! - -REPLY [4 votes]: The inclusion of groupoids into simplicial sets is fully faithful. Its left adjoint, $\Pi_1$ is given by left Kan extension of the functor $\Delta\to \mathcal{Gpd}$ sending the n-simplex to the contractible groupoid with objects $\{0,...,n\}$. -The entirety of the data of the homotopy type of the space $X$ is contained in its singular simplicial set, which is canonically a Kan complex. In particular, the fundamental groupoid functor you've written above is canonically isomorphic to the composite $\Pi_1 \circ \operatorname{Sing}$. Then we have a universal natural transformation $$\operatorname{Sing} \to N\circ \Pi_1\circ \operatorname{Sing}$$ given by the unit of the adjunction $Π_1\dashv N$. Taking geometric realizations, we obtain a span -$$\lvert \Pi_1 \rvert \cong \lvert N\circ \Pi_1\circ \operatorname{Sing}\rvert \leftarrow \lvert \operatorname{Sing}\rvert \xrightarrow{\simeq} \operatorname{id_{\mathbf{Top}}},$$ -where the righthand map is the counit of the adjunction between simplicial sets and topological spaces $\lvert \bullet \rvert \dashv \operatorname{Sing}$, and is a natural weak homotopy equivalence by a theorem of Quillen. -So the lefthand map exhibits the nerve of the fundamental groupoid as stage 1 of the Postnikov System as mentioned by Denis in the first comment. - -REPLY [2 votes]: A very rough argument that can be (easily) formalized is as follows: -We have a notion of $\infty$-groupoids. These are like groupoids, but they have homotopies between morphisms, homotopies between homotopies, and so on. Every topological space presents an infinity groupoid by taking the objects to be points, morphisms to be paths, morphisms between morphisms to be homotopies of paths, etc. -If one takes the connected components of this we get the path components of our space. If one takes the connected components of the automorphism group of a point, we get the fundamental group. If one takes the connected components of the morphisms from a constant path to itself we get the second homotopy group, and so on. -Hence this $\infty$-groupoid can be seen as presenting all the homotopical information of our space. Now the fundamental groupoid is given by taking this $\infty$-groupoid and taking connected components of the morphisms between points to get an actual groupoid. Now we just remarked that the higher homotopy information (homotopy groups after the first) are all contained in the connected components of the morphism sets. By discretizing these sets we are removing all higher homotopical information. -So what should we expect when we realize it? Well we should expect that $\pi_0 , \pi_1$ are that of are spaces, but $\pi_n$ for $n>1$ is trivial. This is exactly what happens, we get the first Postnikov space for $X$, i.e. $K(\pi_1(X),1)$ (or really a disjoint union of these for each path component).<|endoftext|> -TITLE: May-McClure "A reduction of Segal conjecture" -QUESTION [5 upvotes]: I am looking for a digitalized version of paper by J.P. May and J. McClure A reduction of Segal conjecture, as I need it to understand some lemma from Kuhn's Tate Cohomology and Periodic Localization of Polynomial Functors. The paper was published in Current Trends in Algebraic Topology, Canadian Mathematical Society Conference Proceedings, 1982. -I hope this is not breach of any rules - given the current situation, I cannot go to University's library and check the hard copy, and I cannot find any version of this article on the internet. Any help would be highly appreciated! -(If nobody has an access to a digitalized version, could sbdy quote what Corollary 4 is saying?) - -REPLY [4 votes]: I found it on professor May's web site at http://math.uchicago.edu/~may/PAPERS/42.pdf - Since this link might disappear, it is also archived at the Wayback Machine.<|endoftext|> -TITLE: Analog of the Birkhoff's ergodic theorem for the sequence of squares -QUESTION [5 upvotes]: Consider a dynamical system $(X, \mathcal{B}(X), \mu, T)$ where $(X, \mathcal{B}(X), \mu)$ is a measure space and $T$ is a measure-preserving, invertible transformation. -Then by the classical Birkhoff's ergodic theorem if $p\ge 1$, then for any $f\in L^p(X, \mu)$ the sequence -$$ -\mathcal{M}_N f(x):=\frac{1}{N}\sum_{n=0}^N f(T^n x) -$$ -converges for almost every $x\in X$. -$\textbf{Question:}$ Is it true that the sequence -$$ -\mathcal{A}_N f(x):=\frac{1}{N}\sum_{n=0}^N f(T^{n^2} x) -$$ -is convergent a.e. for $f\in L^p(X,\mu)$ and $p\ge 1$? -I'll be more than happy to see the answer to my naive question for a particular case when: $X=\mathbb{Z}$, $\mu$ is a counting measure and $T$ being a regular shift operator $Tf(x)=f(x+1)$. In this case -$$ -\mathcal{A}_N f(x):=\frac{1}{N}\sum_{n=0}^N f(x+n^2), \qquad x\in\mathbb{Z} -$$ -for $f\in \ell^p(\mathbb{Z})$. - It seems that the problem reduces to the study of the boundedness of the maximal function: -$$ -f\mapsto\sup_N \mathcal{A}_N |f|. -$$ -Is there a smart way of getting this boundedness from the corresponding result in the continuous setting? I tried to apply some known transference principles, but it seems to me, that the fact that there are large gaps between squares, namely $(n+1)^2-n^2\simeq n$, causes some trouble. Please excuse me if I'm overlooking something obvious here. - -REPLY [3 votes]: If $X=\mathbb{Z}$, $\mu$ is the counting measure, and $T$ is the shift operator given by $Tf(x)=f(x+1)$, then for all real $p\ge1$, $f\in\ell^p(\mathbb{Z})$, and $x\in\mathbb{Z}$, by Hölder's inequality, -$$ -|\mathcal{A}_N f(x)|\le \frac1N\,\sum_{n=0}^N|f(x+n^2)| -\le\frac1N\,\|f\|_p\,(N+1)^{1-1/p}\to0 -$$ -and hence $\mathcal{A}_N f(x)\to0$ as $N\to\infty$.<|endoftext|> -TITLE: Integrals of products of fractional parts -QUESTION [13 upvotes]: Let $((x)):=x-\lfloor x \rfloor -1/2$, where $\lfloor x \rfloor $ denotes the greatest integer $\le x$. -Let $a,b,c,...$ denote arbitrary natural numbers. It is clear that -$$ \int_0^a ((x/a)) dx =0.$$ -A little exercise shows that -$$ \int_0^{ab} ((x/a))((x/b)) dx =\frac{\gcd(a,b)^2}{12}.$$ -It appears that -$$ \int_0^{abc} ((x/a))((x/b))((x/c)) dx =0,$$ -but I have not been able to determine a general formula for -$$ \int_0^{abcd} ((x/a))((x/b))((x/c))((x/d))dx.$$ -I would be interested in a general formula for the last integral, and for similar integrals with a larger number of factors. It seems that these integrals, or the corresponding sums, such as -$$ \sum_{k=0}^{ab-1} ((k/a))((k/b)) = \frac{\gcd(a,b)^2-1}{12},$$ -should be in the literature, but I have not been able to find a reference. - -REPLY [6 votes]: These quantities are studied under the name Franel Integrals in the literature. If we change notation so that $\psi(x) = x−⌊x⌋−1/2 $ is the saw tooth function, we may define the "fourth order" Franel integral as -$I(a,b,c,d)=\int_{0}^{1} \psi(ax) \psi(bx) \psi(cx) \psi(dx) dx.$ -Expanding $\psi(x)$ in a Fourier series we see that: -$I(a,b,c,d) = \frac{1}{16 \pi^4} \sum_{\substack{s,t,u,v \in Z - \{0\}\\ sa+tb+uc+ue=0}} \frac{1}{stuv} =: \frac{1}{16 \pi^4} L(a,b,c,d)$ -There is no explicit general closed form evaluation of this quantity but there are many special cases known, including -$L(1,1,1,1) = \frac{\pi^4}{5},$ -$L(a,1,1,1) = \left(\frac{1}{3a} - \frac{2}{15a^3} \right) \pi^4,$ -$L(a,a,b,b) = \frac{\pi^4}{9} + \frac{8 (a,b)^4 \zeta(4)}{a^2b^2}.$<|endoftext|> -TITLE: When does QCoh have 'enough perfect complexes'? -QUESTION [15 upvotes]: Let $X$ be a derived fpqc stack on the $\infty$-category of connective spectral affine schemes $\mathbf{Aff}^{\mathrm{cn}}=(\mathbf{Ring}^{\mathrm{cn}}_{E_\infty})^{\mathrm{op}}$, that is to say, a functor $X:(\mathbf{Aff}^{\mathrm{cn}})^{\mathrm{op}}\to \mathcal{S}$ satisfying fpqc descent. Then we can define its $\infty$-category of quasicoherent sheaves formally by a Kan extension. -Say that a symmetric monoidal stable $\infty$-category $\mathcal{C}$ 'has enough perfect objects' if its full subcategory of dualizable objects is dense (that is to say the induced functor $\mathcal{C}\to \operatorname{Ind}(\operatorname{Perf}(\mathcal{C}))$ is fully faithful). -Are there examples of fpqc stacks $X$ as above for which $\operatorname{QCoh}(X)$ does not have enough perfect objects? -What about if we restrict our question to ((Quasi)-Geometric Stacks, Artin Stacks, Deligne-Mumford Stacks, Algebraic Spaces, Schemes)? -For certain, this is true for quasicompact quasiseparated schemes and algebraic spaces, as well as quasi-Geometric stacks $X$ such that $\operatorname{QCoh}(X)$ is compactly generated and the structure sheaf is a compact object (proven in Lurie, Spectral Algebraic Geometry). - -REPLY [10 votes]: Robert Thomason was the first person to draw attention to this question, before derived schemes and infinity categories. I believe that he proved that for a quasi-compact and quasi-separated scheme that $D_{qc}=\textrm{Ind}(\textrm{Perf})$. For example, see Thomason-Trobaugh section 2.3, though at first glance it appears that only proves the weaker statement that it has enough perfect complexes. -Somewhere he gives two examples to show the necessity of the two hypotheses. Consider an affine scheme with a point of infinite codimension, say, $X=\textrm{Spec}\;k[x_1,x_2,…]$. Let $U$ be the complement of the point. It is not quasi-compact. Let $Y=X\cup_U X$ be $X$ with the point doubled. It is not quasi-separated. A perfect complex is built from finitely many operations, so its support has finite codimension, so they do not notice the points of infinite codimension, so the three schemes all have the same perfect complexes. But they have different quasi-coherent complexes, such as the skyscrapers on the origins. In particular, $Y$ has two such sheaves, but they cannot be distinguished by perfect complexes. Whereas $U$ has too few sheaves, so it fails the strong hypothesis of equivalence of categories, but it still has enough perfects: $\textrm{QCoh}(U)\subset \textrm{QCoh}(X)=\textrm{Ind}(\textrm{Perf(X)})=\textrm{Ind}(\textrm{Perf(U)})$ - -Is there a finite dimensional example? For example, consider this non-quasi-compact scheme built from varieties. Let $Z_0=\mathbb A^2$ and $x_0=0\in Z_0$. Let $Z_{n+1}$ be the blow up of $Z_n$ at $x_n$ and let $x_{n+1}$ be a point in the exceptional fiber. Let $U_n=Z_n-\{x_n\}$. Then $U_n$ is open in $U_{n+1}$ and let $U'=\bigcup U_n$. Does it satisfy $D_{qc}=\textrm{Ind}(\textrm{Perf})$? I believe that it can be compactified by adding a 2-dimensional valuation ring. If so, we could double that point to get a non-quasi-separated scheme. Would it fail to have enough perfects?<|endoftext|> -TITLE: Cartesian dissimilarity of a function $\ f:A^3\to A^3\ $ and its inverse -QUESTION [6 upvotes]: Let $\ A\ $ be an arbitrary set. Let $\ |A|>1\ $ (to avoid triviality). Let each of the functions $\ f_k:A^{\{1\ 2\ 3\}}\to A\ $ depend on all three arguments for $\ k=1\ 2\ 3,\ $ while each of the functions $\ g_k:A^{\{1\ 2\ 3\}}\to A\ $ does not depend on $k$-th variable, for each $\ k=1\ 2\ 3.$ -I'll explain "dependent/independent" at the end of this note. It'll be preceded by "diagonal product $\ \triangle\ $ of functions (here, of three of them). -Assume also that -$$ f:=f_1\triangle f_2\triangle f_3\ \ \text{and} - \ \ g:=f_1\triangle f_2\triangle f_3\, :\,A^3\to A^3 $$ -are inverse one to another. There are (historic) examples when $\ A\ $ is countable or of cardinality continuum, and for all infinite cardinals except, possibly, for the weird ones. There are examples (by analogy) whenever $\ |A|\ $ is finite and odd; there should be some examples when $|A|$ is not a power of $2$, but -- I conjecture -- never when it is: -.............................................................. -CONJECTURE:  cardinality $\ |A|, $ when it's finite, is not a (non-trivial) power of $\ 2\ $ (is different from $\ 2^k\ $ for any natural $k=1\ 2\ \ldots$). ------------------------------------------------------------ -EXAMPLES: -Let $\ A = \Bbb Q\ $ or $\ \Bbb R\ $ or $\ \Bbb Z_n\ $ for arbitrary odd $\ n>1.\ $ Define: - -$\ f_1(a\ b\ c)\ :=\ b+c-a $ -$\ f_2(a\ b\ c)\ :=\ a+c-b $ -$\ f_2(a\ b\ c)\ :=\ a+b-c $ - -and - -$\ g_1(a\ b\ c)\ :=\ \frac{b+c}2 $ -$\ g_2(a\ b\ c)\ :=\ \frac{a+c}2 $ -$\ g_2(a\ b\ c)\ :=\ \frac{a+b}2 $ - -Then, let $\ f\ $ and $\ g\ $ be defined as above. This establishes the odd cardinality case. -Remark:  Every finite abelian group $\ X\ $ of odd order -($\ |X|\, $ -- odd) will do in place of $\ Z_n\,$ (with odd $n$). -The finite even cardinalities different from $\ 2^n\ (n\in\Bbb N),\ $ present a mixed story in an analogy to other combinatorial themes which admit exotic examples. This would be the complementary conjecture -- the mixed picture. ------------------------------------------------------------ -Diagonal product of functions (or morphisms) -Consider set $\ X\ $ and sets $\ Y_q\ $ and functions -$\ f_q:X\to Y_q\ (q\in Q).\ $ Then, the diagonal product -$\ f:=\triangle_{q\in Q} f_q :X\to\prod_{q\in Q} Y_q\ $ -is given by: -$$ \forall_{q\in Q}\quad \pi_q\circ f\ := f_q $$ -i.e. -$$ \forall_{q\in Q}\,\forall_{x\in X}\quad (f(x))(q)\ := f_q(x) $$ ------------------------------------------------------------ -Dependent / independent -Let $X\ Y\ T\ $ be arbitrary sets, and $\ s\in T.\ $ Elements $\ x\in X^T\ $ -are functions $\ x: T\to X$. -A function $\ f:X^T\to Y\ $ does not depend on (is independent of) variable -$\ s\ \Leftarrow:\Rightarrow$ -$$ \exists_{f_s\in X^{T\setminus\{s\}}}\,\forall_{x\in X^T}\quad - f(x)=f_s(x|T\setminus\{s\}) $$ -Otherwise, $\ f\ $ depends on variable $\ s,\ $ i.e. -$$ \exists_{w\ x\,\in\,X^T}\ \ (\, w|T\!\setminus\!\{s\}\,=\, - x|T\!\setminus\!\{s\}\quad\text{and}\quad f(w)\ne f(x) \,) $$ -==================================== -PS. In the style of Q&A, I've provided only the special case of the general question about the number of independent variables of a function and its inverse, $\ f\ $ and $\ g.\ $ Actually, we want to know the whole structure of sets of independent variables for $\ f\ $ and $\ g.\ $ This question is as fundamental as it goes, hence it belongs to the theory of the Foundations of Mathematics. (Please, someone reattach the related tag to my question). - -REPLY [4 votes]: $\newcommand{\F}{\mathbb{F}}$ -Your conjecture is false, I will construct a counterexample for all powers of two $2^n$ with $n \ge 2$. -Let us identify $A$ with $\F_{2^n}$. An example from OP corresponding to the odd numbers, is a linear mapping. Our functions also would be linear. Let $G$ be a matrix constructed from the functions $g_1, g_2, g_3$, as rows and $F$ be its inverse, constructed of the functions $f_1, f_2, f_3$. Then we want the following things: matrix $G$ has zeroes on the diagonal and matrix $F$ does not have zero entries. If we start with the matrix $G$ with the zeroes on the diagonal then the elements of $F$ are product of some two elements of $G$, divided by the determinant. So as long as all non-diagonal elements of $G$ are non-zero and the determinant is non-zero we won. This can be achieved for all $n \ge 2$ by, for example, the following construction: -Let $a \in \F_{2^n}$, $a\ne 0, 1$ and consider functions -$$g_3(x, y, z) = x + y,$$ -$$g_2(x, y, z) = x + z,$$ -$$g_1(x, y, z) = y + az.$$ -then the functions $f$ are -$$f_1(x, y, z) = \frac{1}{a+1}(ay-x+z),$$ -$$f_2(x, y, z) = \frac{1}{a+1}(-ay+az+x),$$ -$$f_3(x, y, z) = \frac{1}{a+1}(y+x-z).$$ -Since $a\ne 1$ we have $a+1\ne 0$ (we are in the field of characteristic two) and so it is a working example. -For the case $N = 2^kM$, $M$ odd, $M > 1$ we can just copy $2^k$ times the construction for $M$ from the OP. It leaves only the cases $N = 1, 2$. For $N = 1$ there are obviously no solutions and for $N = 2$ one can simply bruteforce all the potential cases (as was done by Ycor in the comments while I was writing my answer). -EDIT here is a simpler construction which doesn't need finite fields but only modular arithmetics and work for all $m \ge 3$. We will work in $\mathbb{Z}/m\mathbb{Z}$. Consider $$g_3(x, y, z) = x +y,$$ $$g_2(x, y, z) = x + z,$$ $$g_1(x, y, z) = y-2z.$$ Then the functions $f$ are -$$f_1(x, y, z) = -z+2y+x,$$ -$$f_2(x, y, z) = 2z-2y-x,$$ -$$f_3(x, y, z) = z-y-x.$$<|endoftext|> -TITLE: Representation of central extension -QUESTION [7 upvotes]: Let $G$ be a finite abelian group of rank $n$ and $H\rightarrow G$ a central extension with cyclic finite kernel. -Is it true that we can find a faithful representation $H\rightarrow {\rm GL}_{k(n)}(\mathbb{C})$ where $k(n)$ only depends on $n$? -I feel something like this must be true from the fact that $G$ should admit an irreducible projective representation into ${\rm PGL}_{k(n)}(\mathbb{C})$ where $k(n)$ only depends on $n$. - -REPLY [2 votes]: Here is an elementary proof of a related general fact. Let $H$ be any finite nilpotent group with $H^{\prime} = Z(H)$ cyclic of order $m$. Let $z$ be a generator of $Z(H)$. Then $\langle z \rangle $ has a (faithful) linear character $\lambda$ such that $\lambda(z)$ is a complex primitive $m$-th root of unity. -Note then that all irreducible constituents of ${\rm Ind}_{Z(H)}^{H}(\lambda)$ are faithful. For let $\chi$ be one such. Then by Frobenius recipirocity (and Clifford' Theorem), ${\rm Res}^{H}_{Z(H)}(\chi) = \chi(1)\lambda$, so that ${\rm Res}^{H}_{Z(H)}(\chi)$ is certainly faithful. On the other hand, if $\chi$ were not faithful, then ${\rm ker} \chi$ contains a minimal normal subgroup $M$ of $H$. Since $H$ is nilpotent, $M \leq Z(H)$ ( for $M \cap Z(H) \neq 1$ and $M$ is minimal), contrary to the fact that $M \leq {\rm ker} \chi$ and ${\rm Res}^{H}_{Z(H)}(\chi)$ is faithful. -Furthermore, ( as is well-known, and may be found in the character theory text of I.M. Isaacs for example), since $H$ is nilpotent, it follows that if $\theta$ is any faithful irreducible character of $H$, then $\theta$ vanishes identically outside $Z(H)$. For choose $a \in H \backslash Z(H)$ and choose $b \in H \backslash C_{H}(a)$ we have -$[a,b] = w$ for some $1 \neq w \in Z(H)$. Then $b^{-1}ab = wa$. -Hence $\theta(a) = \theta(b^{-1}ab) = \theta(wa)$. But by Schur's lemma, $w$ is represented by a scalar matrix in any representation affording character $\theta$, and the scalar, $\alpha$ say, is not $1$ as $\theta$ is faithful . Hence $\theta(a) = \theta(wa) = \alpha \theta(a) $, so that $\theta(a) = 0$. -Hence for our previous faithful irreducible character $\chi$ of $H$, the orthogonality relations yield $|H| = \sum_{ h \in H}|\chi(h)|^{2} = |Z(H)|\chi(1)^{2}$, so that $\chi(1) = \sqrt{[H:Z(H)]}.$<|endoftext|> -TITLE: Flatness of the Hitchin system? -QUESTION [10 upvotes]: The Hitchin fibration is a central topic of study in modern geometry. It seems to be folklore knowledge that the morphism from the coarse moduli space of semi-stable Higgs bundles to the Hitchin base (the direct sum of spaces of global sections of powers of the canonical bundle) is flat. Where is this proven? - -REPLY [7 votes]: If you are okay working with the Hitchin map defined on the moduli stack of Higgs bundles rather than the coarse moduli space, then you can find a proof of this statement in the paper The global nilpotent variety is Lagrangian by V. Ginzburg, at least in the case that the genus of the base curve is at least $2$. The precise reference is Corollary 9. -The idea is to show that the global nilpotent cone (the most singular fiber of the Hitchin map) has the same dimension as $Bun_G$. This implies that the Hitchin map is equidimensional and that the stack of Higgs bundles $T^*Bun_G$ is lci. Since the Hitchin base is non-singular, this implies flatness by miracle flatness (Stacks Project Lemma 00R4). Now the stack of semi-stable Higgs bundles is an open substack so the restriction of the Hitchin map to this locus is also flat. -Edit: I think the result also holds for the coarse moduli space of semi-stable Higgs bundles but it seems to be a bit subtle. I think equidimensionality of the coarse Hitchin map follows from that on the stack. Then by miracle flatness the map is flat if and only if the coarse space is Cohen-Macaulay. Locally the coarse space of semi-stable Higgs bundles looks like a quotient $V//H$ where $H$ is reductive and $V$ is affine chart for the stack (in the smooth topology). Thus the question becomes when is $V//H$ Cohen-Macaulay? -For $V$ non-singular this is the Hochster-Roberts theorem. However, it can fail in general even when $V$ itself is Cohen-Macaulay (in fact even when $V$ is a complete intersection). See for example the last paragraph of example $I$ here. In this case we are saved by the fact that moduli space of semi-stable Higgs bundles has symplectic singularities which are in particular Cohen-Macaulay. See for example this paper. -It seems to me then that being in the symplectic setting is used not just for the dimension bounds but also to ensure that the moduli space is Cohen-Macaulay so I'm not sure what to expect for Higgs bundles valued in an arbitrary line bundle $L$.<|endoftext|> -TITLE: Decomposing a polynomial ring into Specht Modules -QUESTION [9 upvotes]: Let $S_{\pi}$ where $\pi$ is an integer partition of $n$, denote the Specht module corresponding to $\pi$. -I am trying to decompose the set of all homogeneous polynomials in $x_1,x_2,...,x_n$ generated linearly (over any field of characteristic zero) by the monomials of the form $x_i^2x_jx_k$ ($i,j,k$ are distinct), into Specht modules. I managed to do it for the polynomials generated by each of the following classes of monomials with $i,j,k,l$ distinct: $x_i^3x_j,x_i^2x_j^2,x_i^4,x_ix_jx_kx_l$. -Once it is achieved for ${x_i}^2x_jx_k$ a decomposition is successfully found for the space of degree 4 homogeneous polynomials in $n$ variables where $n$ is large enough, say $n\ge20$. This is the aim. - -First $x_i^3x_j$ $(i\ne j)$: We know that $x_i^3x_j=\displaystyle \frac{x_i^3x_j+x_ix_j^3}2+\frac{x_i^3x_j-x_ix_j^3}2$ -(a) The terms of the form $\displaystyle \frac{x_i^3x_j+x_ix_j^3}2$ generate linearly a space isomorphic as $S_n$-modules (the module action is by permuting indices) to the homogeneous square-free degree 2 polynomials. This is isomorphic to $S_{(n-2,2)}\oplus S_{(n-1,1)}\oplus S_{(n)}$. -(b) The terms of the form $\displaystyle\frac{x_i^3x_j-x_ix_j^3}2$ generate linearly a space isomorphic as $S_n$-modules to the second exterior power of a vector space generated by $\{x_1,x_2,...,x_n\}$ via $x_i\wedge x_j\mapsto\displaystyle\frac{x_i^3x_j-x_ix_j^3}2$. Thus this is isomorphic to $S_{(n-2,1,1)}\oplus S_{(n-1,1)}$. -So the decomposition is -$\displaystyle S_{(n-2,2)}\oplus S_{(n-2,1,1)}\oplus 2S_{(n-1,1)}\oplus S_{(n)}$ -where "$2$" indicates that we have two copies of $S_{(n-1,1)}$. -$x_i^4$: This is simply a vector space generated by $x_i^4$, and is a direct sum of the standard and the trivial representations of $S_n$ that is $S_{(n-1)}$ and $S_{(n)}$. Thus the decomposition is $S_{(n-1,1)}\oplus S_{(n)}$. -$x_ix_jx_kx_l$: These generate the module isomorphic to module $M_\lambda$ as in Bruce Sagan's book "The Symmetric Group" where $\lambda=(n-4,4)$ which one figures is just $S_{(n-4,4)}\oplus S_{(n-3,3)}\oplus S_{(n-2,2)}\oplus S_{(n-1,1)}\oplus S_{(n)}$. -$x_i^2x_j^2$: These generate the module isomorphic to module $M_\lambda$ where $\lambda=(n-2,2)$ which one figures is just $S_{(n-2,2)}\oplus S_{(n-1,1)}\oplus S_{(n)}$. - -The reason behind showing above is that these are alarmingly simple deductions though I can't seem to find one nearly as slick for the class $x_i^2x_jx_k$. I found that there is a submodule isomorphic to $M_{(n-3,3)}$ inside this class of polynomials. But that is the closeset I could get. -I tried dimension count as well, because the homogeneous degree-4 polynomials are of dimension ${n+3\choose 4}$. -The terms in the decompositions including the $M_{(n-3,3)}$ above sum up to -$\displaystyle S_{(n-4,4)}\oplus 2S_{(n-3,3)}\oplus 4S_{(n-2,2)}\oplus S_{(n-2,1,1)}\oplus 6S_{(n-1,1)}+5S_{(n)}$. -This has a total dimension $n^4-2n^3+35n^2+38n$ which, subtracted from ${n+3\choose4}$ is $\frac13(n^3-3n^2-4n)$ which should be the sum of the dimensions of the remaining irreducible components (that is, copies of Specht modules). -But this counting technique leads to many possible decompositions and I am kind of out of ideas on this. Could anyone help? - -REPLY [4 votes]: In general, these decompositions can be computed using the Pieri rule. (This is essentially the same as Mark's answer) -Fix an integer partition of $d$ as $\lambda = 1^{e_1} \dots r^{e_r}$. Then consider the $S_n$-set of ordered partitions into $r+1$ blocks $[n] = b_0 \sqcup \dots \sqcup b_r$ such that the $i$th block has size $e_i$ for $i > 0$ and the $0$th block has size $n - \sum_{i} e_i$. This set has a canonical bijection with a set of monomials, given by $$b_0 \sqcup \dots \sqcup b_r \mapsto (\prod_k (\prod_{i \in b_k} x_i)^k.$$ -The case you are asking about is $\lambda = 2^1 1^2.$ -Let us write $V_\lambda$ for the vector space with basis the $S_n$ set associated to $\lambda$. Then we can write $V_\lambda$ as an induced representation as follows $$V_\lambda = {\rm Ind}^{S_n}_{S_{n - \sum_{i} e_i} \times \prod_{i = 1}^r S_{e_i}} \left(\bigotimes_{i= 0}^r {\rm triv} \right).$$ -The effect of this induction product can be computed via the Pieri rule, starting with the integer partition $(n- \sum_i e_i)$ corresponding to the trivial representation of $S_{n - \sum_i e_i}$ and adding horizontal strips of length $e_1, \dots, e_r$. This should translate into a combinatorial rule for multiplicities involving skew tableaux with content specified by the $e_i$, but I have not thought this through carefully. -In your case, we begin with the partition $(n-3)$. Multiplying by the horizontal strip $(e_1) = (2)$ we get $(n-1) + (n-2,1) + (n-3,2)$. Multiplying this by $(e_2) = (1)$ we get $$((n-1) + (n-2,1) + (n-3,2))*(1)$$ $$ = (n) + (n-1,1) + (n-2,2) + (n-1,1) + (n-2,2) + (n-2,1,1) + (n-2,2) + (n-3,3) + (n-3,2,1)$$<|endoftext|> -TITLE: A characterization of constant functions -QUESTION [23 upvotes]: In How to recognize constant functions. Connections with Sobolev spaces (Russian Math Surveys 57 (2002); MSN), H. Brezis recalls the following fact: - -Let $\Omega\subset{\mathbb R}^N$ be connected and $f:\Omega\rightarrow{\mathbb R}$ be measurable, such that - $$\int\int_{\Omega\times\Omega}\frac{|f(y)-f(x)|}{|y-x|^{N+1}}\,dx\,dy<\infty.$$ - Then $f$ is constant. - -He adds - -The conclusion is easy to state, but I do not know a direct, elementary, proof. Our proof is not very complicated but requires an “excursion” via the Sobolev spaces. - -My question is whether there is such an elementary proof in the special case of one space dimension ($N=1$, $\Omega$ an interval). - -REPLY [9 votes]: Not quite an answer, but too long for a comment. -Let me make my life easier a bit and take $\Omega=\mathbb{R}^N$ while increasing the exponent slightly. Namely, I will assume that -$$ -I:=\ \int_{\mathbb{R}^{2N}}\ \frac{|f(x)-f(y)|}{|x-y|^{\alpha}}\ d^Nx d^Ny -$$ -is finite, where $\alpha>N+1$. -Then we have, just by the triangle inequality involving the midpoint, $I\le J$ where -$$ -J:=\ \int_{\mathbb{R}^{2N}}\ -\frac{\left|f(x)-f\left(\frac{x+y}{2}\right)\right| -+\left|f\left(\frac{x+y}{2}\right)-f(y)\right|}{|x-y|^{\alpha}}\ d^Nx d^Ny -$$ -$$ -=\ 2^{N+1-\alpha}\ I\ , -$$ -by a trivial change of variable. Since $I\in [0,\infty)$ and $I\le 2^{N+1-\alpha}\ I$ with $N+1-\alpha<0$, we immediately get $I=0$. -The OP's case is clearly a borderline/endpoint one where the above argument just happens to break down. Perhaps one can get a logarithmic improvement, by using smarter estimates. -The above simple idea is just a "Sobolev-ish"-flavored (as opposed to "Hölder-ish") adaptation of the classic proof of Hölder with exponent greater than one in 1D implies constant, by subdividing the interval $[x,y]$ into $k$ pieces and taking $k\rightarrow\infty$.<|endoftext|> -TITLE: In theory, how would Oneiric numbers be defined? -QUESTION [15 upvotes]: Background I am not a professional mathematician. I am researching Surreal numbers & games for fun (I think they are truly beautiful). If this question is not appropriate here, I beg forgiveness & ask that it please be migrated it to MSE. -I read More Infinite Games by John H Conway recently & have not been able to stop thinking about the following line (page 4): - -Note that $\uparrow$ is not a number: it is the value of a game, which - is a more subtle concept. Also note that $\frac{1}{\uparrow}$ is not defined since it would be - bigger than all surreal numbers and there are no such numbers. (In fact, it does - exist but is one of the Oneiric numbers.) - -Unfortnuately, the only thing on the internet about the Oneiric numbers is this paper. If my current understanding is correct, a part of the problem is that the reciprocal of games is not (yet) defined. As previously stated, I'm not an expert & have no clue if working out the reciprocals of games is an impossible (or just very difficult) task, or even if it has already been done. -It has been suggested by a couple people that I try to find someone who knew Conway & might have some idea of what he was thinking. I would be elated to do so. Alternatively, perhaps someone who is knowledgable about the Surreals might venture to take a crack at a definition. -Thanks for taking the time to read this question, I hope it isn't a waste of anyone's time. - -REPLY [9 votes]: I emailed John Conway about this very thing over ten years ago. His response (paraphrasing) was along the lines of RP’s comment; if you treat 1/up as a formal entity nothing breaks, but there wasn’t more to the theory than that. I do not have an account, and I don’t have the email still, so I neither want nor expect the bounty. One other detail I remember is that he represented 1/up with the Roman numeral I.<|endoftext|> -TITLE: How strong a set theory is necessary for practical purposes in sheaf theory? -QUESTION [13 upvotes]: Is it known how much of ZFC is actually necessary for the basic, familiar constructions and theorems in sheaf theory, along the lines of section II.1 (and its exercises) in Hartshorne's "Algebraic Geometry" textbook? -I apologize for this strange question. Here is the motivation behind it: I am a mathematician who works automatically and implicitly in ZFC, like most users of this website. Last year some members of the philosophy department at my university showed me some old questions in metaphysics which seemed to me as though they ought to be approachable axiomatically. Indeed they were, and some of the structures studied by the philosophers turn out to be (non-obviously) equivalent to sheaves of sets on a certain topological space. I was able to make some progress on those problems by using some classical, elementary ideas in sheaf theory, particularly the equivalence of sheaves of sets and local homeomorphisms. My colleagues in philosophy are supportive of this, but they suggest that unrestrained use of the ZFC axioms to resolve questions in philosophy may open the door for argument from philosophers who have various kinds of skepticism about set theory, and that it is best to offer the argument in a way that uses only as weak of a fragment of ZFC as possible. -I spent a while looking over the proofs of the theorems from sheaf theory that I am using, and it appears to me that the proofs all work in ZAC (Zermelo set theory with choice), but perhaps one can go weaker still. The outcome will be better if this kind of work is done by someone that knows more set theory than I do, so I would be happiest if someone else has already figured out how strong a set theory is necessary for elementary sheaf theory. Is this already out there, somewhere, in the literature? I am not even aware of a journal where investigations of this kind would appear, but perhaps people working in some areas of foundations of mathematics have someplace where they figure out such things. -I apologize for my ignorance of foundations of mathematics, and I also apologize again for this very strange question. - -REPLY [13 votes]: Colin McLarty has looked into this - -The large structures of Grothendieck founded on finite order arithmetic, Review of Symbolic Logic 13 issue 2 (2020) pp. 296--325, doi:10.1017/S1755020319000340, arXiv:1102.1773. - -with abstract (emphasis added): - -The large-structure tools of cohomology including toposes and derived categories stay close to arithmetic in practice, yet published foundations for them go beyond ZFC in logical strength. We reduce the gap by founding all the theorems of Grothendieck’s SGA, plus derived categories, at the level of Finite-Order Arithmetic, far below ZFC. This is the weakest possible foundation for the large-structure tools because one elementary topos of sets with infinity is already this strong. - -In the arXiv version the abstract claimed all of EGA's theorems as well, but I haven't investigated why this was removed. Certainly it is generally considered that the axiom of Replacement is not needed for 'generic' (i.e. non-logical/set-theory) mathematics, hence for algebraic geometry generally. In particular, it's generally accepted (though one might have to check specific statements that seem very strong) that ETCS is sufficient as a foundation, roughly equivalent to, but slightly weaker than, what you call ZAC.<|endoftext|> -TITLE: Characteristic polynomial of checker matrix -QUESTION [6 upvotes]: For every integer $n > 0$, let $C_n$ be the $4n \times 4n$ matrix having $1$'s in all positions $(i, j)$ such that $i - j$ is even, $3$'s in the two diagonals determined by $|i - j| = 2n + 1$, and $0$'s everywhere else. For example, we have -$$C_2 = \begin{bmatrix} -1 & 0 & 1 & 0 & 1 & 3 & 1 & 0 \\ -0 & 1 & 0 & 1 & 0 & 1 & 3 & 1 \\ -1 & 0 & 1 & 0 & 1 & 0 & 1 & 3 \\ -0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ -1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\ -3 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ -1 & 3 & 1 & 0 & 1 & 0 & 1 & 0 \\ -0 & 1 & 3 & 1 & 0 & 1 & 0 & 1\end{bmatrix} .$$ -I'd like to prove a formula for the characteristic polynomial of $C_n$. From some numerical experiments, I believe that it is -$$(\lambda - 3)^{2n - 2} (\lambda + 3)^{2n - 2} (\lambda^2 - (2n-3)\lambda - 3) (\lambda^2 - (2n+3)\lambda + 3),$$ -but I failed to prove that. -Any suggestion is welcome. Thanks. -Note 1. What makes things difficult are the $3$'s. If instead of them there were $0$'s, then we would have a circulant matrix, and using the theory of circulant matrices the characterist polynomial would be easily proved to be $\lambda^{4n - 2}(\lambda - 2n)^2$. -Note 2. Following Pat Devlin's suggestion, I checked the eigenspace of $\lambda = 3$ and it seems to be spanned by the row vectors of the following matrix $(2n-2)\times 4n$ matrix -$$\begin{bmatrix}\begin{matrix}-1 \\ -1 \\ \vdots \\ -1\end{matrix} & I_{2n-2} & \begin{matrix}0 & 0 & -1\\ 0 & 0 & -1 \\ \vdots \\ 0 & 0 & -1\end{matrix} & I_{2n-2}\end{bmatrix} .$$ -This shouldn't be difficult to prove, and similarly for the eigenspace of $\lambda = -3$. But I have no idea how to deal with the eigenvalues related to the factor $(\lambda^2 - (2n-3)\lambda - 3) (\lambda^2 - (2n+3)\lambda + 3)$. - -REPLY [6 votes]: Ok! Your conjecture is true. -Let $W$ be the space spanned by the eigenvectors for $\lambda \in \{-3, 3\}$ as described in my comments. Let $V$ be the subspace of $\mathbb{R}^{4n}$ consisting of vectors of the form -$$V = \{(a,b,a,b,a,b, \ldots, a, x, y, b, a, b, \ldots, a,b)\},$$ -where the entries corresponding to $x,y$ are in positions $2n$ and $2n+1$ of the vector. (So $V$ is the orthogonal complement of $W$.) -Let $T : V \to \mathbb{R}^4$ by $T(\vec{v}) = (a,b,x,y)$ in the obvious way (so $T$ is an isomorphism). -We can check that $V$ is invariant under the action of $C_{n}$. And moreover, that $$T \circ C_{n} \circ T^{-1} \begin{pmatrix}a\\b\\x\\y \end{pmatrix} = \begin{pmatrix}(2n-1)a +y + 3b\\(2n-1)b+x+3a\\(2n-1)b+x\\(2n-1)a+y \end{pmatrix}.$$ -Thus, $C_{n}$ restricted to $V$ is isomorphic to the above linear map on $\mathbb{R}^4$, namely -$$\begin{pmatrix}a\\b\\x\\y\end{pmatrix} \mapsto \begin{pmatrix}2n-1 & 3 & 0 &1\\ -3 & 2n-1 & 1 &0\\ -0 & 2n-1 & 1 &0\\ -2n-1 & 0 & 0 &1\end{pmatrix} \begin{pmatrix}a\\b\\x\\y\end{pmatrix},$$ -and this map has the desired remaining four eigenvalues as in your conjecture.<|endoftext|> -TITLE: Unknown work of Nöbeling on topological/Hausdorff dimension -QUESTION [9 upvotes]: Let $\mathcal{H}^n$ denote the Hausdorff measure, $\dim_H X$ the Hausdorff dimension, and $\dim X$ the topological dimension of $X$. -A well known result of -Szpilrajn (He changed his name to Marczewski while hiding from Nazi persecution) proved in [S] asserts that -if $\mathcal{H}^{n+1}(X)=0$, then the topological dimension of $X$ is at most $n$. -Szpilrajn's proof is reproduced in [Theorem 7.3, HW] and [Theorem 8.15, H]. -Szpilrajn however, mentions in [S] that his argument is based on Nöbeling's proof of a weaker result that topological dimension is bounded from above by the Hausdorff dimension of a metric space. However, he did not provide any reference to Nöbeling's work. -There is also no reference to Nöbeling's work in the book by Hurewicz and Wallman. - -Question. Does anybody know the reference to the original work of Nöbeling? - -[H] J. Heinonen, -Lectures on analysis on metric spaces. Universitext. Springer-Verlag, New York, 2001. -[HW] W. Hurewicz, H. Wallman, Dimension Theory. Princeton Mathematical Series, v. 4. Princeton University Press, Princeton, N. J., 1941. -[S] E. Szpilrajn, La dimension et la mesure, Fund. Math. 28 (1937), 81--89. - -REPLY [12 votes]: So, the sought for paper is: -Nöbeling, G., Hausdorffsche und mengentheoretische Dimension, Ergebnisse math. Kolloquium Wien 3, 24-25 (1931). -And here is a ``translation" (to English and to modern math exposition standards, if such a thing exists.) What is called the set-theoretical dimension is defined inductively: dim of the empty set is $-1$, and we set $\dim M = k$ if $k$ is the least integer with the property that every point of $M$ has arbitrarily small (open) neighborhoods whose dim is $k-1$. -Hausdorff and set-theoretical dimension. By George Nöbeling. -Let $M$ be a subset of the Euclidean $\mathbb{R}^n$. One covers $M$ by finitely or countably (infinitely) many balls $K_j$, with diameters $d_j < \rho$, for $p \leq n$ (any non-negative number) form the sum $ \sum d_j^p $. Let $ L_p(\rho, M)$ be the infimum of such sums for all such coverings. Put -$$ -L_p (M) = \lim_{\rho \to 0} L_p (\rho, M) . -$$ -Obviously there is exactly one number $p= p (M)$ such that for every $q > p$, $L_q (M) = 0$ and for every number $q q-1 $, it follows from the induction hypothesis that -$$ -\forall x \in (0,r_0], \; \lim_{i \to \infty} s_i(x) = \infty \, . -$$ -Thus, -$$ - \sum_j \int_0^{r_0} f_{ij}(x) \, dx = \int_0^{r_0} s_i(x) \, dx \xrightarrow{i \to \infty} \infty \, . -$$ -Now observe that $f_{ij} (x) = d_{ij}^{q-1}$ for $x$ from an interval whose length is at most $d_{ij}$ -- the diameter of $K_{ij}$ -- and otherwise $f_{ij} (x) = 0.$ Therefore, -$$ -\sum_j d_{ij}^{q} = \sum_j \int_0^{d_{ij}} d_{ij}^{q-1} \, dx \geq \sum_j \int_0^{r_0} f_{ij}(x) \, dx \, . -$$ -and hence, -$$ -\sum_j d_{ij}^{q} \xrightarrow{i \to \infty} \infty. -$$ -Since this is true for any coverings $K_{ij}$, we conclude -$$ -L_q(M) = \infty \implies p(M) \geq q \, . -$$ -Since $q -TITLE: Reverse mathematics of Cousin's lemma -QUESTION [17 upvotes]: This paper by Normann and Sanders apparently caused a stir in the reverse mathematics community when it came out a couple years ago. It says that Cousin's lemma, which is an extension of the Heine-Borel theorem, requires the full strength of second-order arithmetic (SOA) to prove. It also says this lemma is helpful in mathematically justifying the Feynman path integral. So this is an apparent counterexample to both - -a) The reverse mathematics precept that theorems of classical analysis can (usually?) be proved using one of the "Big Five" subsystems of SOA, with the strength of subsystem required forming a useful classification of such theorems; and -b) Solomon Feferman's argument that scientifically useful mathematics can generally be handled by relatively weak axioms, generally not stronger than PA / ACA0. - -I don't exactly have a mathematical question about the Normann-Sanders paper, but would like to know if it has impacted the reverse mathematics program, and what its significance is seen as. Could path integrals really require such powerful axioms? -Also, Cousins' lemma is traditionally fairly easily proved using the completeness property of the real numbers. The issue is that the completeness property is a second-order property of the reals (i.e. it uses a set quantifier), and SOA is a first-order theory of the reals, that doesn't have sets of reals. In classical analysis though, the completeness axiom really does refer to sets of reals, and this result shows that converting a completeness-based proof to a first-order proof isn't so easy (I haven't read the paper closely and have no idea right now how to prove Cousin's lemma in SOA). Is that significant? -I can understand that the (second order) induction axiom from the Peano axioms translates naturally to the induction schema in first order PA, making induction proofs work about the same way as before. I'd be interested to know why analysis is identified with SOA instead of something that allows sets of reals (needed for functions anyway), since there's not such a straightforward translation of the completeness axiom. Analysis=SOA goes back a long way, since the Hilbert program aimed to prove CON(SOA) once it was done with the consistency of arithmetic. Reverse mathematics came much later. -Thanks! - -REPLY [3 votes]: Sam's answer is, of course, the definitive one. For curiosity's sake, I'll mention that Rod Downey, Noam Greenberg and I have recently been looking at Cousin's lemma for restricted (i.e. cardinality $\mathfrak{c}$) classes of functions. -We have a proof of Cousin's lemma in $\mathsf{ATR}_0 + \Delta^1_2$-induction that works for "any function", to the extent that "any function" can be formalised in this system. This effectively gives an upper bound for Cousin's lemma for functions appropriately definable in second-order arithmetic. This doesn't contradict what Sam said above, since you need third-order arithmetic to talk about all functions, and then $\mathsf{ATR}_0 + \Delta^1_2$-induction is not enough. -We've specifically focused on continuous functions, Baire functions and Borel functions. So far, we've proved: - -Cousin's lemma for continuous functions is equivalent to $\mathsf{WKL}_0$; -Cousin's lemma for Baire 1 functions is equivalent to $\mathsf{ACA}_0$; -Cousin's lemma for Baire $2$ functions (and hence for Borel functions) implies $\mathsf{ATR}_0$. - -These results are in my thesis as well as a joint paper we've written. -To answer the OP's questions, Dag Normann and Sam Sanders' results do contradict (a). But to me, this is not such a surprise. As Sam notes, arbitrary discontinuous functions can't be formalised in second-order arithmetic. So, to the extent that classical analysis deals with discontinuous functions, we can't even formalise it in second-order arithmetic, let alone prove its theorems in, say, $\Pi^1_1$-$\mathsf{CA}_0$. This just shows that SOA is sufficient for most, but not all, of mathematics (it can't really handle topology, set theory, etc). -As for (b), it also contradicts this, if you assume that it is really necessary to consider Cousin's lemma for arbitrary gauges. As Russell A. Gordon shows in his book, "The Integrals of Lebesgue, Denjoy, Perron, and Henstock", it is enough to consider measurable gauges, and maybe this can be further restricted to Borel gauges (I don't know). In that case, maybe $\mathsf{ATR}_0 + \Delta^1_2$-induction is enough for the physicists. -Updated to add new results 2021/05/10<|endoftext|> -TITLE: Does the category of local rings with residue field $F$ have an initial object? -QUESTION [6 upvotes]: Let $F$ be a field. Does the category $C_F$ of local rings $R$ equipped with a surjective morphism $R\longrightarrow F$ have an initial object? -This is, for instance, true if $F=\mathbb{F}_{p}$ for some prime $p$: If $R$ is a local ring with residue field $\mathbb{F}_{p}$, then any $x\in\mathbb{Z}\setminus(p)$ must map to something invertible under the morphism $\mathbb{Z}\longrightarrow R$. Hence that morphism factors as $\mathbb{Z}\longrightarrow\mathbb{Z}_{(p)}\longrightarrow R$; thus $\mathbb{Z}_{(p)}$ is the initial object. -But what happens in the more general case? I guess it should be true at least if $F$ is of finite type over either $\mathbb{Q}$ or $\mathbb{F}_{p}$ (where $p$ is a prime), but I have no idea how to prove it. - -REPLY [6 votes]: A typical approach is to ask for your universal property among complete DVRs, not among all local rings, because there you have a very nice positive result. Given a perfect field $k$ of characteristic $p$, the Witt ring $W(k)$ is initial among complete DVRs of characteristic $0$ equipped with an isomorphism of residue field with $k$. This is Theorem II.5.4 of Serre's book "Local Fields."<|endoftext|> -TITLE: Induction step in Bóna and Ehrenborg's proof that the generating function of the alternating runs has -1 as a root of a certain multiplicity -QUESTION [10 upvotes]: This is a crosspost of a question I asked on Mathematics SE four months ago. Periodically bumping it and placing a bounty on it to attract more attention were to no avail. There are some comments underneath it by @darijgrinberg that I found useful, but the only conclusion in that discussion was that the proof really is confusing. - -I am having trouble with Lemma 2.3 in the paper "A combinatorial proof of the log-concavity of the numbers of permutations with $k$ runs" by Bóna and Ehrenborg, and I was hoping to get some help in clarifying how the proof works. -Let $n > 1$. We say that the permutation $p \in \mathfrak{S}_n$ has $k$ alternating runs (or just $k$ runs) if $p$ changes direction at $k-1$ points, that is, if there are $k$ values of $i \in \left\{2,3,\ldots,n-1\right\}$ such that either $p_{i-1} < p_i > p_{i+1}$ or $p_{i-1} > p_i < p_{i+1}$. Let $r(p)$ be the number of runs of $p$, and $$R_n(x) := \sum_{p \in \mathfrak{S}_n} x^{r(p)}.$$ The authors give a combinatorial proof of the fact that $-1$ is a root of $R_n(x)$ with multiplicity $m = \lfloor (n - 2) / 2\rfloor$ for all $n \geq 4$. -Let $1 \leq j \leq m+1$ and $p \in \mathfrak{S}_n$. We say that $p$ is $j$-half-ascending if the last $j$ disjoint consecutive pairs of elements in the sequence $p_1,p_2,\dotsc,p_n$ are ascending, that is, if $p_{n+1-2i} < p_{n+2-2i}$ for all $1 \leq i \leq j$. For a $(j+1)$-half-ascending permutation $p \in \mathfrak{S}_n$, define $r_j(p)$ to be the number of runs of the subsequence $p_1,p_2,\dotsc,p_{n-2j}$, and $s_j(p)$ to be the number of descents of the subsequence $p_{n-2j},p_{n-2j+1},\dotsc,p_n$. Define $t_j(p) := r_j(p) + s_j(p)$, and let $$R_{n,j}(x) := \sum_{p} x^{t_j(p)},$$ where the sum is taken over all the $(j+1)$-half-ascending permutations $p$ in $\mathfrak{S}_n$. (In the paper, the sum is apparently taken over all $p \in \mathfrak{S}_n$, but that is surely a typo because $t_j(p)$ does not make sense for all $p \in \mathfrak{S}_n$.) - -Lemma 2.3. For all $n \geq 4$ and $1 \leq j \leq m$, we have $$\frac{R_n(x)}{2(x+1)^j} = R_{n,j}(x).$$ - -The proof goes by induction. For the base case, we first note that $p$ and $p^c$ (the complement of $p$) have the same number of runs, so that -$$ -\frac{R_n(x)}{2} = \sum_{p : p_{n-3} < p_{n-2}} x^{r(p)}. -$$ -Let $I$ be the involution on $\mathfrak{S}_n$ that swaps $p_{n-1}$ and $p_n$. One can check that for every $p \in S_n$ such that $p_{n-3} < p_{n-2}$, $r(p)$ and $r(I(p))$ differ by $1$ (it suffices to check this only for the permutations in $\mathfrak{S}_4$, since it is only the last $4$ terms in the sequence $p_1,\dotsc,p_n$ that are really relevant here). Let $q \equiv q(p)$ be that permutation in the set $\{ p, I(p) \}$ with smaller number of alternating runs. Then, -$$ -\sum_{p : p_{n-3} < p_{n-2}} x^{r(p)} = \sum_{q(p)} \bigl(x^{r(q)} + x^{r(q)+1}\bigr) = (1+x) \sum_{q(p)} x^{r(q)}. -$$ -Let $q' \equiv q'(p)$ be that permutation in the set $\{ p, I(p) \}$ which is $2$-half-ascending. Then, it turns out that $r(q) = t_1(q')$ (again, it suffices to only check this for the permutations in $\mathfrak{S}_4$). So, -$$ -\sum_{q(p)} x^{r(q)} = \sum_{q'(p)} x^{t_1(q')} = R_{n,1}(x). -$$ -Hence, $R_n(x)/(2(1+x)) = R_{n,1}(x)$, as required. -Now, this is what the authors say regarding the induction step: - -Now suppose we know the statement for $j-1$ and prove it for $j$. As above, apply $I$ to the two rightmost entries of our permutations to get pairs as in the initial case, and apply the induction hypothesis to the leftmost $n-2$ elements. By the induction hypothesis, the string of the leftmost $n-2$ elements can be replaced by a $j$-half-ascending $n-2$-permutation, and the number of runs can be replaced by the $t_{j-1}$-parameter. In particular, $p_{n-3} < p_{n-2}$ will hold, and therefore we can verify that our statement holds in both cases ($p_{n-2} < p_{n-1}$ or $p_{n-2} > p_{n-1}$) exactly as we did in the proof of the initial case. . . - -This is quite confusing to me; I cannot interpret this paragraph in a way to actually make the proof work. I would be happy to even just see how the inductive step works in the case $n = 6$. - -References - -Bóna, Miklós; Ehrenborg, Richard, A combinatorial proof of the log-concavity of the numbers of permutations with $k$ runs, J. Comb. Theory, Ser. A 90, No. 2, 293-303 (2000). ZBL0951.05002. - -Bóna, Miklós, Combinatorics of permutations, Discrete Mathematics and its Applications. Boca Raton, FL: CRC Press (ISBN 978-1-4398-5051-0/hbk; 978-1-4398-5052-7/ebook). 458 p. (2012). ZBL1255.05001. - -REPLY [10 votes]: I will try to write up a more transparent proof. -Update: My new, direct proof is at https://people.clas.ufl.edu/bona/files/altruns.pdf.<|endoftext|> -TITLE: Universal model category as a $\text{sSet}$-enriched co-completion -QUESTION [6 upvotes]: It's a standard fact that given a small category $\mathcal{C},$ the category of pre-sheaves $\text{Psh}(\mathcal{C})$ is the free co-completion of it. -I'm sure this can be done not only for $\text{Set}$-enriched categories but for general $\mathcal{V}$-enriched categories, with the appropriate notions of $\mathcal{V}$-enriched colimit, and functor preserving the enrichment, and I just found it in section 4.4. of Kelly's Basic Concepts of Enriched Category Theory. -So now the question is: can one prove proposition 2.3 in this paper about simplicial pre-sheaves on $\mathcal{C}$ being the universal model category on $\mathcal{C}$ by just doing the $\mathcal{V}$-enriched co-completion with $\mathcal{V}=\text{sSet}$? - -REPLY [3 votes]: Given $C$ a small category (eventually, a small simplicial category) I denote by $UC$ the projective model structure on the category of simplicial presheaves on $C$ as in the paper. Using the kind of argument you have in mind we obtain the following theorem: -Theorem: If $M$ is a simplicial model category, then there is an equivalence of categories between: - -(Simplicial) Functors $C \to M$ taking values in the full subcategory of cofibrant objects. -Simplicial left Quillen functor $UC \to M$. - -In one direction, the equivalence is simply given by restricting to the Yoneda embedding $ C \to UC$ as representable are cofibrant in the projective model structure, this forces the composite functor $C \to UC \to M$ to take values in cofibrant objects. In the converse direction, one takes the unique simplicial left adjoint functor $UC \to M$ and check, using the axiom of simplicial model category for $M$ that this is a left Quillen functor. -However, this is not what the paper you mention proves. -There, they start from a model category $M$ that is not assumed to be a simplicial model category, and a functor $C \to M$ not assumed to takes values in cofibrant objects. And construct a left Quillen functor $UC \to M$ by considering (and choosing) a cofibrant simplicial resolution of the functor $C \to M$ they started from. In particular, the "uniqueness" of the left Quillen functor obtained this way, is only up to homotopy (to be more precise, up to a contractible space of choices). - -One abstract way to understand the relation between the two is as follows: -Given $M$ a combinatorial left proper model category, there is a Quillen equivalent simplicial model structure on the category $sM$ on the category of simplicial objects of $M$, (this is explained in the paper "Replacing model categories with simplicial one" by Dugger) -The evaluation at $[0]$ gives a left Quillen equivalence $sM \to M$ -One way to understand the non-simplicial theorem is that if you start from $C \to M$, you can see it as a functor $C \to sM$ taking values in constant simplicial objects, then take a levelwise cofibrant replacement to obtain a functor $C \to sM$ taking value in cofibrant object, apply the "simplicial theorem" to get a Quillen functor $UC \to sM$ and finally, post compose with Quillen functor $sM \to M$ that evaluate at $[0]$. -Now for the model structure on $sM$ to exist we need $M$ to be combinatorial and left proper, if you are willing to work with a left semi-model structure instead it is enough to assume that $M$ is an accessible model category (no properness assumption). -But in some sense the central observation of the paper you quote, is that, even if the model structure on $sM$ cannot be constructed, the overall construction make sense with no assumption $M$ (other than being a model category, I guess they also need functorial factorization, I do not remember).<|endoftext|> -TITLE: Are projective modules over a certain localised Laurent polynomial ring free? -QUESTION [11 upvotes]: Let $R=\mathbb{Z}[t^{\pm 1}]$ be the ring of Laurent polynomials, and let $S \subset R$ be the multiplicative subset generated by the polynomial $t-1$. I am interested in the ring $S^{-1}R=\mathbb{Z}[t^{\pm 1},(t-1)^{-1}]$ obtained by inverting $t-1$. More specifically, I know that finitely generated projective $R$-modules are free (e.g. by the Quillen-Suslin theorem) and I would like to know whether finitely generated projective $S^{-1}R$-modules are free? - -REPLY [10 votes]: The answer is yes. Given any projective module $P$ over $S^{-1}A$, where $A=\mathbb{Z}[t]$ (and works for many other rings too), it is the localization $S^{-1}M$ of a projective module over $A$. The reason is, you can always find such a finitely generated module $M$ with $S^{-1}M=P$, but you may replace $M$ with its double dual without affecting the localization, but any reflexive module over $A$ is projective (and thus free, by Seshadri's theorem, which precedes Quillen-Suslin by many years). -To answer your questions in the comments below, double dual of any finitely generated module over $A$ is reflexive. Since $P$ is projective (and hence reflexive), it follows that if $S^{-1}M=P$,then so is $S^{-1}(M^{**})$. For your last question, for any Noetherian ring $A$ and $S\subset A$ a multiplicatively closed set, given any finitely generated module $P$ over $S^{-1}A$, there exists a finitely generated module $M$ over $A$ such that $S^{-1}M=P$. Further, if $P$ reflexive, then you may replace $M$ by $M^{**}$ and thus assume it is reflexive.<|endoftext|> -TITLE: In the category of sigma algebras, are all epimorphisms surjective? -QUESTION [40 upvotes]: Consider the category of abstract $\sigma$-algebras ${\mathcal B} = (0, 1, \vee, \wedge, \bigvee_{n=1}^\infty, \bigwedge_{n=1}^\infty, \overline{\cdot})$ (Boolean algebras in which all countable joins and meets exist), with the morphisms being the $\sigma$-complete Boolean homomorphisms (homomorphisms of Boolean algebras which preserve countable joins and meets). If a morphism $\phi: {\mathcal A} \to {\mathcal B}$ between two $\sigma$-algebras is surjective, then it is certainly an epimorphism: if $\psi_1, \psi_2: {\mathcal B} \to {\mathcal C}$ are such that $\psi_1 \circ \phi = \psi_2 \circ \phi$, then $\psi_1 = \psi_2$. But is the converse true: is every epimorphism $\phi: {\mathcal A} \to {\mathcal B}$ surjective? -Setting ${\mathcal B}_0 := \phi({\mathcal A})$, the question can be phrased as following non-unique extension problem. If ${\mathcal B}_0$ is a proper sub-$\sigma$-algebra of ${\mathcal B}$, do there exist two $\sigma$-algebra homomorphisms $\psi_1, \psi_2: {\mathcal B} \to {\mathcal C}$ into another $\sigma$-algebra ${\mathcal C}$ that agree on ${\mathcal B}_0$ but are not identically equal on ${\mathcal B}$? -In the case that ${\mathcal B}$ is generated from ${\mathcal B}_0$ and one additional element $E \in {\mathcal B} \backslash {\mathcal B}_0$, then all elements of ${\mathcal B}$ are of the form $(A \wedge E) \vee (B \wedge \overline{E})$ for $A, B \in {\mathcal B}_0$, and I can construct such homomorphisms by hand, by setting ${\mathcal C} := {\mathcal B}_0/{\mathcal I}$ where ${\mathcal I}$ is the proper ideal -$$ {\mathcal I} := \{ A \in {\mathcal B}_0: A \wedge E, A \wedge\overline{E} \in {\mathcal B}_0 \}$$ -and $\psi_1, \psi_2: {\mathcal B} \to {\mathcal C}$ are defined by setting -$$ \psi_1( (A \wedge E) \vee (B \wedge \overline{E}) ) := [A]$$ -and -$$ \psi_2( (A \wedge E) \vee (B \wedge \overline{E}) ) := [B]$$ -for $A,B \in {\mathcal B}_0$, where $[A]$ denotes the equivalence class of $A$ in ${\mathcal C}$, noting that $\psi_1(E) = 1 \neq 0 = \psi_2(E)$. However I was not able to then obtain the general case; the usual Zorn's lemma type arguments that one normally invokes to give Hahn-Banach type extension theorems don't seem to be available in the $\sigma$-algebra setting. I also played around with using the Loomis-Sikorski theorem but was not able to get enough control on the various null ideals to settle the question (some subtle issues reminiscent of "measurable selection theorems" seem to arise). However, Stone duality seems to settle the corresponding question for Boolean algebras. - -REPLY [24 votes]: $\require{AMScd}$ -A 1974 paper of R. Lagrange, Amalgamation and epimorphisms in $\mathfrak{m}$-complete Boolean algebras (Algebra Universalis 4 (1974), 277–279, DOI link), settled this affirmatively. In the cited paper, Lagrange shows that for any infinite cardinal $\mathfrak{m}$, the category of $\mathfrak{m}$-complete Boolean algebras with $\mathfrak{m}$-complete morphisms has the strong amalgamation property, which implies that epimorphisms are surjective. He remarks that the proof works just as well for complete Boolean algebras, and I'd also add that it can be adapted for plain old Boolean algebras. If I understand your meaning of an abstract $\sigma$-algebra correctly, this is the result you are after. -Let $\mathcal{C}$ be a concrete category so that we can meaningfully talk about epimorphisms being surjective. -$\mathcal{C}$ is said to have the strong amalgamation property if for every span $C \xleftarrow{f} A \xrightarrow{g} B$ of monomorphisms (aka amalgam), there exists an object $D \in \mathcal{C}$, and a commutative diagram of monomorphisms -$$ -\begin{CD} -A @> g>> B\\ -@VfVV @Vg'VV \\ -C @>f'>> D, -\end{CD} -$$ -such that $g'(B) \cap f'(C) = g'g(A) = f'f(A)$ -Further restrict attention to a variety of algebraic structures, so that being a monomorphism is equivalent to mapping underlying sets injectively and every morphism canonically factors as a surjection followed by a monomorphism. Then the strong amalgamation property immediately implies that epis are surjective (see the corollary in Lagrange's paper). -I think this is a good — or at least thought-provoking — approach to addressing your question in light of your bounty comment that you are "looking for a canonical answer." For varieties, we see that the solution to the strong amalgamation problem always supplies a canonical non-unique extension: given a proper monomorphism $A \rightarrow B$, we get the strong amalgamation $D$ of $B \leftarrow A \rightarrow B$ together with distinct monomorphisms(!) $B \rightarrow D$ that agree on $A$. -Moreover, the solution to the strong amalgamation problem might be considered canonical in and of itself. Basically, Lagrange's method is a three-part construction: (1) Embed in the best coproduct available, which is in the ambient category of Boolean algebras (2) quotient that coproduct in order to force the desired intersection property of strong amalgamation (which has the awesome effect of restoring morphisms to our actual category) (3) Complete this quotiented coproduct, so that the whole embedding is now in-category. In other words, do the best you can with the coproduct you have... and then error-correct in the only sensible way. I guess that feels canonical. -On this last point, it might be interesting to compare this construction with solutions to the strong amalgamation problem in other varieties, in particular (finite) groups and Lie Algebras over fields.<|endoftext|> -TITLE: Riesz–Markov–Kakutani representation theorem for compact non-Hausdorff spaces -QUESTION [11 upvotes]: Let $X$ be a compact Hausdorff topological space, and $\mathcal C^0 (X) = \{f:X\to\mathbb{R}; \ f \text{ is continuous }\}$. It is well known that for any bounded linear functional $\phi: \mathcal C^0(X)\to\mathbb{R},$ such that $\phi(f)\geq 0$ if $f\geq 0$ ($\phi$ is called a positive linear functional), then there exists a unique regular Borel measure $\mu$, such that -$$\phi(g) = \int g\ \mathrm d\mu, \ \forall \ g\in \mathcal C^0(X). $$ -This result follows from a direct application of Riesz–Markov–Kakutani representation theorem. -If we drop the Hausdorff hypothesis (only assuming $X$ as compact topological space). Then we can lose the uniqueness of the measure that represents the linear functional. A famous example is the compact topological space "$[0,1]$ with to origins". In this case the functional $\phi: \mathcal C^0(X)\to\mathbb{R}$, $\phi(f) = f(0)$ can be written as $\int f\ \mathrm{d}\delta_0$ or $\int f\ \mathrm{d}\delta_{0'}.$ -I would like to know if we still have the existence of a measure that represents the functional. In other words, I would like to know if the following theorem is true - -Possible Theorem: Let $(X,\tau)$ be a compact non-Hausdorff space, and $\Lambda : \mathcal C^0(X)\to\mathbb{R}$ a positive bounded linear functional, then there exists a measure $\mu: \mathcal B(\tau)\to \mathbb{R}$ (where $\mathcal B(\tau)$ is the smallest $\sigma$-algebra such that $\tau\subset \mathcal B(\tau))$, such that - $$\Lambda(f) = \int f\ \mathrm{d}\mu, \ \forall \ f\in \mathcal C^0(X).$$ - -Can anyone help me? -I have searched online but I was not able to find a result in the non-Hausdorff case. - -REPLY [9 votes]: The answer is yes. -First, it follows from the following result and the Riesz–Markov–Kakutani representation theorem that we can always find a suitable Baire measure representing a positive linear functional. - -Theorem: Let $X$ be any topological space. Then there exists a completely regular Hausdorff space $Y$ and a continuous surjection - $\tau:X\to Y$ such that the function $g\mapsto g\circ\tau$ is an - isomorphism from $C_B(Y)$ onto $C_B(X)$. - -This is Theorem 3.9 of "Rings of continuous functions" (1960) by Gillman and Jerison. -So the problem reduces to the question whether a Baire measure on a compact topological space can be extended to a Borel measure. We can do this using the following result, which specializes the very abstract Theorem 2.6.1 of "Convex Cones" (1981) by Fuchssteiner and Lusky. - -Theorem: Let $X$ be a non-empty compact topological space and $L:\mathcal{C}^0_+(X)\to\mathbb{R}$ be an additive function on the - cone of nonnegative continuous functions on $X$ such that - $L(g)\leq\max g$ for all $g$. Then there exists a Borel probability - measure $\nu$ on $X$ such that $$L(g)\leq\int g~\mathrm d\nu$$ for - all $g\in \mathcal{C}^0_+(X)$. - -For nonzero $\Lambda$, let $L=1/\Lambda(1)\cdot \Lambda$. Then the measure $\mu=\Lambda(1)\cdot\nu$ does the trick. -It should be noted that the resulting Borel measure need not be regular. For non-Hausdorff $X$, there is no point in going beyond Baire measures.<|endoftext|> -TITLE: Can we recover an inner model of CH after forgetting some generic information? -QUESTION [6 upvotes]: Suppose $\kappa$ is an inaccessible cardinal. Let $G \times H$ be $\mathrm{Col}(\omega_1,{<}\kappa) \times \mathrm{Add}(\omega,\kappa)$-generic over $V$. Let $X \subseteq \kappa$ be $\mathrm{Add}(\kappa,1)$-generic over $V[G][H]$. Since $X$ codes every bounded subset of $\kappa$ as an interval-subsequence, $V[X] \models \kappa = \omega_2 = 2^\omega$. Does there exist an inner model of $V[X]$ with the same cardinals satisfying CH? -Note: By arguments similar to those of Section 2.1 here, $G \notin V[H][X]$. - -REPLY [3 votes]: The following answers the question as posed, but is a bit unsatisfactory since we will find a choiceless inner model. -In $V[X]$, let $F = \{ x \subseteq \omega_1 : \forall \alpha < \omega_1(x \cap \alpha \in V) \}$. Clearly $\mathcal P(\omega_1)^{V[G]} \subseteq F$. We claim that $\mathcal P(\omega_1)^{V[G]} = F$ using: -Lemma (Mitchell): For all $\lambda$, $\mathrm{Add}(\omega,\lambda)$ has the $\omega_1$-approximation property. -This means that any $x \subseteq \omega_1$ which is in $V[G][H] \setminus V[G]$ must have some initial segment not in $V[G]$, and thus not in $V$. -We consider the model $V(F) \subseteq V[G] \cap V[X]$. Since $\mathbb R^{V[G]} = \mathbb R^V$, $V(F)$ satisfies CH. Since it has the same subsets of $\omega_1$ as $V[G]$, it satisfies $\kappa = \omega_2$. By standard homogeneity arguments, $V(F)$ does not have a well-ordering of $F$. -At least we can say that $V[X]$ satisfies weak square $\square^*_{\omega_1}$. (The motivation for the question had to do with the tree property.)<|endoftext|> -TITLE: Topos extensions -QUESTION [17 upvotes]: In set theory, starting from a model $V$ of $ZFC$, a forcing notion $\mathbb{P}$, and a generic filter $G \subset \mathbb{P}$ over $V$, we can find a generic extension which is a model of $ZFC$ and is the smallest model, having the same ordinals as $V$ such that $V[G] \supseteq V$ and $G \in V[G].$ - -My question is: what is the corresponding construction in terms of toposes, if we start with an arbitrary topos $T$. - -Giving references is appreciated. - -REPLY [10 votes]: The informal analogue, is simply the notion of topos of sheaves. -If I work in a "ground" topos (whose object I call set), then a "forcing extention" would be just a Grothendieck topos, that is a topos of sheaves on a small site. -If you want to adopt an external point of view and start from an elementary topos $\mathcal{E}$ , then a forcing extension of $\mathcal{E}$ is a topos $\mathcal{F}$ that can be obtained as the category of $\mathcal{E}$-valued sheaves on an internal site in $\mathcal{E}$, where internal site means " a category object in $\mathcal{E}$ endowed with a "topology". -The simplest way to define the word "topology" here is to say that it is a Lawvere-Tierney operator in the topos of $\mathcal{E}$-valued presheaves on the category object. But one can also define it in a more Grothendieckian style using collection of subobjects of power objects satisfying the internal version of the axioms of a topology. -It is a well known theorem of topos theory that the topos that can be obtained from $\mathcal{E}$ this way are exactly the topos endowed with a bounded geometric morphism $\mathcal{F} \to \mathcal{E}$. (see section B3.3 of P.T.Johnstone Sketches of an elephant). -The best way to get a feeling of why this is a good analogy is to look at the topos theoretic proof of the independence of the continuum hypothesis in MacLane and Moerdijk "Sheaves in geometry and logic" (section VI.2). -However, it is not a perfect analogy: First as, pointed out by Andreas Blass in comment, the standard set theoretic forcing corresponds only the case of a double negation topology on a poset. Though it can be shown that any sheaves topos admit a cover by one of this form, so this is not a strong restriction. -But there is a more subtle difference: Informally, in set theory people construct a a model that contains a "generic filter", in topos theory we construct the model that contains "the universal (generic) filter" (in the sense of classifying toposes). The point here is that the toposes obtained this way are not well-pointed in general, so they can not directly corresponds to a model of ZFC. -If you want a more precise analogy you need to combine the construction of the topos of sheaves with a construction that reproduce a model of ZFC out of a topos. For this I recommend to look at Mike Shulman's paper that give a very good exposition to the topic.<|endoftext|> -TITLE: Category theory and set theory: just a different language, or different foundation of mathematics? -QUESTION [70 upvotes]: This is a question to research mathematicians, as well as to those concerned with the history and philosophy of mathematics. -I am asking for a reference. In order to make the reference request as precise as possible, I am outlining the background and nature of my questions here: -I did my Ph.D. in probability & statistics in 1994, and my formal mathematics education was completely based on set theory. Recently, I got interested in algebraic topology, and have started to read introductory texts like Allen Hatcher, or Laures & Szymik, and others. -I am struck by the broad usage of category theory and started to wonder: -(1) Is category theory the new language of mathematics, or recently the more preferred language? -(2) Recognizing that set theory can be articulated or founded through category theory (the text from Rosebrugh and Lawvere), is category theory now seen as the foundation of mathematics? -(3) Is the choice between category theory language and set theory language maybe depending on the field of mathematics, i.e. some fields tend to prefer set theory, others category theory? -Edit: On (3), if such a preference actually exists, what is the underlying reason for that? -Would someone be able to give me a good reference for questions like this? I would be very grateful for that. -Later Edit: Just adding the link to a great, related discussion on MO: Could groups be used instead of sets as a foundation of mathematics? It discusses the question whether every mathematical statement could be encoded as a statement about groups, a fascinating thought. -Could groups be used instead of sets as a foundation of mathematics? - -REPLY [2 votes]: The underlying reason of the utility of category is structural. This arose in the abstract algebra approach of Noether where the notion of a structure preserving map was isolated - a morphism. -This idea is so basic to mathematics now, that it is baked into category theory whereas in set theory, the notion of a function is a derived concept, never mind that of structure preserving map. -Just as Marx-Engels is said to have turned Hegelian idealism upside-down (not quite true), Eilenberg-Maclane can be said to have turned set theory upside down to construct category theory. -I would also say neither set theory nor category theory is "the language of mathematics". The only language of mathematics is language itself. To see this merely remove all language from any paper and see how easy it is to understand. It isn't usually - it turns into symbol sludge. -One other aspect of category theory has an avatar in physics. This is the notion of covariance which Einstein used so effectively in thinking through what GR entails. One can say that category theory is also a general theory of covariance. -As for a reference, check out the SEP article on Category Theory.<|endoftext|> -TITLE: Finite group ${\rm Sp}_4({\Bbb F}_3)$: involutions coming from a 4-dimensional complex representation -QUESTION [5 upvotes]: I am interested in the finite unitary reflection group $G= G_{32}$, the group No. 32 in Table VII on page 301 of the paper: -Shephard, G. C.; Todd, J., A. Finite unitary reflection groups. Canad. J. Math. 6 (1954), 274–304. -This is a group of order $2^7 3^5 5 = 155520$. Its commutator subgroup $H=(G,G)$ is of index 3, and a computer calculation shows that $H$ is isomorphic to ${\rm Sp}(4,3):={\rm Sp}_4({\Bbb F}_3)$, the symplectic group of of $4\times 4$ matrices over the finite field ${\Bbb F}_3$. -This group $G$ is given with a faithful 4-dimensional complex representation - $$\rho: G\to {\rm GL}(4, {\Bbb C}).$$ -Moreover, it is is stable under the standard complex conjugation in ${\Bbb C}^4$, and so we obtain an involutive automorphism (an automorphism of order 2) $\ \sigma\colon H\to H$. -I am trying to guess this involution $\sigma$ and to compute the first nonabelian cohomology set $H^1(\langle\sigma\rangle, H)$. A computer calculation shows that the $H^1$ is trivial, and I would like to understand this without computer. - -Question 1. What are the nontrivial 4-dimensional complex representations - of the finite group ${\rm Sp}(4,3)$? -Question 2. What are the involutive automorphisms of ${\rm Sp}(4,3)$ ? In particular, is it true that all nontrivial involutive automorphisms of ${\rm Sp}(4,3)$ - come from elements of order 2 in the projective symplectic group ${\rm PSp}_4({\Bbb F}_3)$ ? -Question 3. Which of those involutive automorphisms of $H={\rm Sp}(4,3)$ can come from the complex conjugation in a 4-dimensional complex representation of $H$? - -Feel free to migrate this elementary question to Mathematics StackExchange.com. - -REPLY [6 votes]: We can take $H={\rm Sp}(4,3)$ to be the group $\{ A \in {\rm GL}(4,3) \mid AFA^{\mathsf T} = F\}$, where $$F=\left(\begin{array}{rrrr}0&0&0&1\\0&0&1&0\\0&-1&0&0\\-1&0&0&0\end{array}\right)$$ is the matrix of the preserved symplectic form. -The the matrix $$C =\left(\begin{array}{rrrr}1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1\end{array}\right),$$ satisfies $CFC^{\mathsf T}= -F$, it normalizes and induces an involutory outer automorphism of $H$, and $\langle H,C \rangle$ is the conformal symplectic group, which I prefer to denote by ${\rm CSp}(4,3)$ (although it is sometimes written as ${\rm GSp}(4,3)$). -There are two dual 4-dimensional complex representations of $H$, which are interchanged by the outer automorphism induced by $C$, so this appears to be the automorphism that you are looking for. -From your description, I think the only possible structure of the group $G$ is the direct product $H \times C_3$. -To answer you specific questions, I am not sure what you are looking for in Question 1. -For Question 2, the full automorphism group group of $H$ is the image of ${\rm CSp}(4,3)$ mod scalars, which we can denote by ${\rm PCSp}(4,3)$: it has order $2|{\rm PSp}(4,3)| = 51840$. The involutory automorphism in question is an outer automorphism, and is not induced by an element of ${\rm PSp}(4,3)$. -For Question 3, I am not completely sure. There are actually two conjugacy classes of involutory outer automorphisms of $H$, one of which is induced by the matrix $C$ above, and I am not sure whether both can be induced by complex conjugation or only one of them. -An example of an element of ${\rm CSp}(4,3)$ that induces an involutory automorhism from the other class is $$C' =\left(\begin{array}{rrrr}0&0&1&0\\1&0&0&-1\\-1&0&0&0\\0&1&1&0\end{array}\right).$$ This has order 4 in ${\rm CSp}(4,3)$, but its square is $-I$, so it induces an involutory automorphism. It is interesting that its centralizer in $H$ has order $720$, whereas the centralizer of $C$ has order $48$. That might be useful in deciding which automorphism is induced by complex conjugation.<|endoftext|> -TITLE: Detecting affine varieties in the Grothendieck ring -QUESTION [6 upvotes]: Is there anything special about the classes of affine varieties in the Grothendieck ring of varieties (over $\mathbb{C})$?. Is there some specialisation that allows us to distinguish classes of affine varieties from general classes? -After R. van Dobben de Bruyn's answer, it might be more interesting to consider the modified version: is there anything special about classes that can be represented by an irreducible smooth affine variety? - -REPLY [16 votes]: Being affine is not invariant under scissors relations. In other words, it is possible that $[X] = [Y]$ in the Grothendieck ring, where $X$ is affine and $Y$ is not. -For example, the diagonal $\Delta \subseteq \mathbf P^1 \times \mathbf P^1$ is ample, so $X = \mathbf P^1 \times \mathbf P^1 \setminus \Delta$ is affine. But -$$[X] = [\mathbf P^1 \times \mathbf P^1] - [\mathbf P^1] = (\mathbf L + 1)^2 - (\mathbf L + 1) = \mathbf L^2 + \mathbf L = [\mathbf P^2 - p],$$ -and $\mathbf P^2 - p$ is not affine for any point $p \in \mathbf P^2$. -If you allow $k$-schemes with multiple components, every effective class has an affine representative. For example, a class of the form $[X]$ can be made affine by cutting $X$ into locally closed pieces that are affine (e.g. using Noetherian induction). -It could still be an interesting question which classes have a representative that is irreducible and affine (or irreducible smooth affine, or ...).<|endoftext|> -TITLE: For an abelian scheme, $R^pf_* \Omega^q$ is locally free and its formation is compatible with any base change -QUESTION [6 upvotes]: Let $k$ be a field, $\bar{R} \to R$ a local homomorphism of artinian local rings with the residue fields $k$, $I$ its kernel, $A/R$ an abelian scheme, and $\mathscr{T}$ its tangent sheaf. -Let $A_0 = A \times_R k$. -Assume that $\mathfrak{m}_\bar{R} I = 0$. -Then $H^2(A, \mathscr{T}_{A/R} \otimes_R I) \cong H^2(A_0, \mathscr{T}_{A_0/k}) \otimes_k I$? -This is a part of the proof of (2.2.4.1) of Kai-Wen Lan's "Arithmetic Compactifications of PEL-Type Shimura Varieties". -To show it, I need the following proposition: - -Let $S$ be a scheme, $f : A \to S$ an abelian scheme of relative dimension $g$. - Then the sheaf $R^pf_* \Omega^q$ is locally free. - And this formation commutes with any base change. - -Are there "elementary" proof of this? -I know this is (2.5.2) of Berthelot, Breen, Messing's Théorie de Dieudonné Cristalline. -But its proof is too hard for me, since it heavily relies on the theory which I don't know. -And I know that this post shows it elementary. -But it uses the formally smoothness and the pro-representability of the deformation of "abelian schemes + polarization", which is what I want to show using this highlighted statement. -So it is a circular reasoning for me. - -REPLY [3 votes]: Let $A/R$ be an abelian scheme of relative dimension $g$ over an Artinian local ring $(R, \mathfrak m, \kappa)$. I am going to give you a proof that works if the characteristic of $\kappa$ is not $2$. -Denote $f : A \to \text{Spec}(R)$ the structure morphism. By the usual trick (see for example here) we have $\Omega_{A/R} \cong \mathcal{O}_A^{\oplus g}$. Thus $\Omega_{A/R}^q$ is isomorphic to the free $\mathcal{O}_A$-module of rank ${g \choose q}$. Hence it suffices to prove that $H^i(A, \mathcal{O}_A)$ is a free $R$-module of rank ${g \choose i}$. Namely, we already know that formation of $K = Rf_*\mathcal{O}_A$ in the derived category $D(R)$ commutes with base change (by very general cohomology and base change results, see for example the exposition in Mumford's book on Abelian varieties) and freeness of its cohomology will imply it is the direct sum of its cohomology sheaves. -Denote $[2] : K \to K$ the pullback by multiplication by $2$ on $A$. By cohomology and base change (see above) we know that $K \otimes_R^\mathbf{L} \kappa$ is isomorphic to $\wedge^*(\kappa^{\oplus g})$. It follows that $K$ can be represented in $D(R)$ by a complex of the shape -$$ -K^\bullet : -R \to R^{\oplus g} \to \ldots \to R^{\oplus g} \to R -$$ -See for example here. Moreover, the map $[2] : K \to K$ in $D(R)$ can be represented by a map of complexes $t^\bullet : K^\bullet \to K^\bullet$ by usual homological algebra. Calculating on the special fibre we see that $t^i \bmod \mathfrak m$ is multiplication by $2^i$ on $\wedge^i(\kappa)$. A bit of elementary algebra then shows that the differentials of $K^\bullet$ have to be zero (look at what happens to the ``leading terms''). -PS: In char 2 you may be able to use the trick with the shearing map, but I didn't try.<|endoftext|> -TITLE: Sum over 0-1 matrices -QUESTION [30 upvotes]: I stumbled across the following formula when working on a research problem in theoretical computer science. I am looking for a simple proof of it, or any idea which might prove useful. -I checked its correctness up to $N=5$ with a computer. Brendan McKay (see comment) was able to check its correctness up to $N=8$. -This question was first asked on Maths StackExchange two weeks ago. - -Basic version -Let $\mathcal M_N$ be the set of all 0-1 square matrices without any row/column of zeros (we want all the denominators to be non-zero in the formula below). One could also define $\mathcal M_N$ to be A227414. -$$ -\sum_{M \in \mathcal M_N} -\frac{\det(M)^2 \cdot (-1)^{\|M\|_0 - N}} -{\prod_{i=1}^N\Big(\sum_{j=1}^N M_{i,j}\Big)\prod_{j=1}^N\Big(\sum_{i=1}^N M_{i,j}\Big)} = 1 -$$ -where $\|M\|_0 = \sum_{i,j} M_{i,j}$ is the number of non-zero entry of $M$. - -Weighted generalization -Note that the formula is also true when "positive weights" are associated to every coefficient. Let $P$ and $Q$ be two matrices with positive coefficients. Alternatively one can think of $P$ and $Q$'s coefficients to be indeterminates ($P_{i,j} = x_{i,j}$ and $Q_{i,j} = y_{i,j}$ for all $i,j$). -Let $M \circ P$ (resp. $M \circ Q$) be the elementwise product of $M$ and $P$ (resp. $Q$). -$$ -\sum_{M \in \mathcal M_N} -(-1)^{\|M\|_0 - N} \cdot -\frac{\det(P \circ M)} -{\prod_{j=1}^N\sum_{i=1}^N [P \circ M]_{i,j}} -\cdot -\frac{\det(Q \circ M)} -{\prod_{i=1}^N\sum_{j=1}^N [Q \circ M]_{i,j}} = 1 -$$ -This version might help to understand how the sum does simplify. When $P$ and $Q$'s coefficients are indeterminates, the sum is a rational function which is identically equal to 1. -Here is some python code to check (empirically) my claim (slow when $N > 4$). -from sympy import Matrix, Symbol -from itertools import product -import random - -N = 2 -P = Matrix([random.randint(1,100) for _ in range(N*N)]).reshape(N,N) -Q = Matrix([random.randint(1,100) for _ in range(N*N)]).reshape(N,N) -print(P) -print(Q) - -prettyprint = "= (-1)^%d * (%d / %d) * (%d / %d)" - -result = 0 -for p in product([0,1], repeat=N**2): - MP = Matrix(p).reshape(N, N).multiply_elementwise(P) - MQ = Matrix(p).reshape(N, N).multiply_elementwise(Q) - dP, dQ = MP.det(), MQ.det() - if dP * dQ != 0: - vP, vQ = 1, 1 - for i in range(N): - vP *= sum(MP[:,i]) - vQ *= sum(MQ[i,:]) - val = (-1) ** (sum(p)-N) * dP * dQ / (vP * vQ) - print(p, val, prettyprint%(sum(p)-N, dP, vP, dQ, vQ)) - result += val -print(result) - -And for those of you who don't want to run this program, here is one output. -Matrix([[19, 33], [49, 7]]) -Matrix([[11, 53], [7, 86]]) -(0, 1, 1, 0) 1 = (-1)^0 * (-1617 / 1617) * (-371 / 371) -(0, 1, 1, 1) -77/1240 = (-1)^1 * (-1617 / 1960) * (-371 / 4929) -(1, 0, 0, 1) 1 = (-1)^0 * (133 / 133) * (946 / 946) -(1, 0, 1, 1) -817/3162 = (-1)^1 * (133 / 476) * (946 / 1023) -(1, 1, 0, 1) -77/2560 = (-1)^1 * (133 / 760) * (946 / 5504) -(1, 1, 1, 0) -2597/4352 = (-1)^1 * (-1617 / 2244) * (-371 / 448) -(1, 1, 1, 1) -42665/809472 = (-1)^2 * (-1484 / 2720) * (575 / 5952) -1 - - -An easier/intermediate formula? -The two formula above are not entirely satisfying, because of the no-zero-row/column constraint in the sum. -Let $\mathcal H_N$ be the set of all $N$ by $N$ matrices, such that coefficient $(i,j)$ is either $a_{i,j}$ or $b_{i,j}$. -For all $M \in \mathcal H_N$, define $A(M)$ to be the number of $a_{i,j}$ coefficients in $M$. -Intuitively, $\mathcal H_N$ is an hypercube and $(-1)^{A(M)}$ tells you if you are on an even or an odd "level". -Let $P$ and $Q$ be two square matrices of size $N$, where $P_{i,j} = x_{i,j}$ and $Q_{i,j} = y_{i,j}$. -$$ -\sum_{M \in \mathcal H_N} -(-1)^{A(M)} \cdot -\frac{\det(P \circ M)} -{\prod_{j=1}^N\sum_{i=1}^N [P \circ M]_{i,j}} -\cdot -\frac{\det(Q \circ M)} -{\prod_{i=1}^N\sum_{j=1}^N [Q \circ M]_{i,j}} = 0 -$$ -Which means that the sum on the odd and even "levels" of the hypercube are equal. -I believe this formula might be easier to prove, because of additionnal symmetries. Sam Hopkins' idea to use a sign reversing involution (see comment) might be helpful. -And perhaps it is a first step towards one of the formula above (where we need to subtract the terms with a row/column of $a$'s). -Here is some python code to check (empirically) my claim (slow when $N > 4$). -from sympy import Matrix -from itertools import product -import random - -prettyprint = "= (%d / %d) * (%d / %d)" -def getVal(v): - global P, Q, prettyprint - MP = Matrix(v).reshape(N, N).multiply_elementwise(P) - MQ = Matrix(v).reshape(N, N).multiply_elementwise(Q) - dP, dQ = MP.det(), MQ.det() - if dP * dQ == 0: return 0 - vP, vQ = 1, 1 - for i in range(N): - vP *= sum(MP[:,i]) - vQ *= sum(MQ[i,:]) - val = dP * dQ / (vP * vQ) - print(val, prettyprint%(dP, vP, dQ, vQ)) - return val - -N = 2 -H = [[random.randint(1,100) for _ in range(2)] for i in range(N*N)] -P = Matrix([random.randint(1,100) for _ in range(N*N)]).reshape(N,N) -Q = Matrix([random.randint(1,100) for _ in range(N*N)]).reshape(N,N) -print(H) -print(P) -print(Q) - -result = 0 -for p in product([0,1], repeat=N**2): - print(p, end=" ") - v = [ H[i][x] for i,x in enumerate(p)] - result += getVal(v) * (-1) ** int(sum(p)) -print(result) - -And for those of you who don't want to run this program, here is one output. -[[9, 52], [11, 59], [14, 41], [34, 93]] -Matrix([[26, 19], [46, 29]]) -Matrix([[83, 21], [36, 24]]) -(0, 0, 0, 0) 164595168/4703083825 = (96128 / 1049210) * (493128 / 1290960) -(0, 0, 0, 1) 445610545/3950948198 = (496502 / 2551468) * (1550880 / 2675808) -(0, 0, 1, 0) -24390009/3154893688 = (-163450 / 2533400) * (268596 / 2241576) -(0, 0, 1, 1) 727415911/51716164040 = (236924 / 6160720) * (1326348 / 3626424) -(0, 1, 0, 0) 5083920/3367826693 = (-491200 / 1849946) * (-14904 / 2621520) -(0, 1, 0, 1) -54813491/10541005478 = (-90826 / 3352204) * (1042848 / 5433696) -(0, 1, 1, 0) 601772499/5328265880 = (-1883482 / 4466840) * (-1219212 / 4551912) -(0, 1, 1, 1) 4820101/1199800856 = (-1483108 / 8094160) * (-161460 / 7364088) -(1, 0, 0, 0) 42513838767/149126935925 = (1198476 / 2385220) * (3405432 / 6002040) -(1, 0, 0, 1) 116044834723/250555941884 = (3511748 / 5800376) * (9516888 / 12440592) -(1, 0, 1, 0) 24887838735/336049358857 = (938898 / 3869410) * (3180900 / 10421724) -(1, 0, 1, 1) 419726686285/2203457293574 = (3252170 / 9409628) * (9292356 / 16860276) -(1, 1, 0, 0) 67073493/1168097623 = (611148 / 4205572) * (2897400 / 7332600) -(1, 1, 0, 1) 18295610183/80432973676 = (2924420 / 7620728) * (9008856 / 15198480) -(1, 1, 1, 0) -545598897/35835002665 = (-781134 / 6822466) * (1693092 / 12732060) -(1, 1, 1, 1) 166078396717/3536747545430 = (1532138 / 12362684) * (7804548 / 20597940) -0 - - -[Edit 05/20] I realized that the formula is true with two different weights (before we had $P = Q$). The description has been updated accordingly. -[Edit 05/24] I included Timothy Chow's remark (we can choose $P$ and $Q$'s coefficients to be indeterminates, and get a rational function identically equal to 1). -[Edit 05/24] I updated the description of the basic version to adress Brendan McKay's comment. Before the set $\mathcal M_N$ was (awkwardly) defined as the set of invertible 0-1 matrices. -[Edit 05/25] I included a new formula, which might be an easier/intermediate step. - -REPLY [10 votes]: This is a result of joint efforts with Fedor Petrov. -First, we show that the L.H.S. of the general version does not depend on $P$ and $Q$, and then we compute that constant for some properly chosen $P$ and $Q$. The elements of $\mathcal M$ are called admissible matrices. -Part 1. We show that the L.H.S. does not depend on $P_{11}$ and $Q_{11}$; the rest is similar. -Perform the following transform. In each matrix $P\circ M$, add to the first row all other rows (the determinant does not change) --- denote the resulting matrix by $(P\circ M)^r$. Then expand the determinant of $(P\circ M)^r$ by the first row; for this purpose, denote by $(P\circ M)^r_{[ij]}$ the cofactor of $(P\circ M)^r_{ij}$. The profit is that in each summand, one of the factors in the denominator cancels out. Notice here that $(P\circ M)^r_{[1j]}=(P\circ M)_{[1j]}$. -Perform the same with the columns of $Q\circ M$, denoting by $(Q\circ M)^c$ the matrix obtained by adding all columns to the first one. -We get -$$ - \sum_{M \in \mathcal M_N} - (-1)^{\|M\|_0 - N} \cdot - \frac{\det(P \circ M)} - {\prod_{j=1}^N\sum_{i=1}^N (P \circ M)_{ij}} - \cdot - \frac{\det(Q \circ M)} - {\prod_{i=1}^N\sum_{j=1}^N (Q \circ M)_{ij}}\\ - =\sum_{M \in \mathcal M_N} - (-1)^{\|M\|_0 - N} \cdot - \frac{\det(P \circ M)^r} - {\prod_{j=1}^N(P\circ M)^r_{1j}} - \cdot - \frac{\det(Q \circ M)^c} - {\prod_{i=1}^N(Q\circ M)^c_{i1}}\\ - =\sum_{M \in \mathcal M_N} - (-1)^{\|M\|_0 - N} \cdot - \sum_{s=1}^N - \frac{(P\circ M)^r_{1s}(P\circ M)^r_{[1s]}} - {\prod_j(P\circ M)^r_{1j}} - \cdot - \sum_{t=1}^N - \frac{(Q\circ M)^c_{t1}(Q\circ M)^c_{[t1]}} - {\prod_i(Q\circ M)^c_{i1}}\\ - =\sum_{s=1}^N\sum_{t=1}^N \Sigma_{st}, -$$ -where -$$ - \Sigma_{st}=\sum_{M \in \mathcal M_N} - (-1)^{\|M\|_0 - N} \cdot - \frac{(P\circ M)_{[1s]}} - {\prod_{j\neq s}(P\circ M)^r_{1j}} - \cdot - \frac{(Q\circ M)_{[t1]}} - {\prod_{i\neq t}(Q\circ M)^c_{i1}}. - \qquad\qquad(*) -$$ -In fact, we show that none of the $\Sigma_{st}$ depends on $P_{11}$ or $Q_{11}$. -If $s=t=1$, this is clear: in this case no term in $(*)$ depends on those entries. -Assume now that $(s,t)\neq (1,1)$. The only part in a summand in~$(*)$ which depends on $m_{ts}$ is its sign. So we may pair up the matrices differing in the $(t,s)$th entries: the sum of corresponding terms is $0$. There is an exception, when $m_{ts}$ is the unique non-zero element in the $t$th row or in the $s$th column of $M$: in this situation the pair is not admissible. We consider the first case; the second is similar. -If $t>1$ (and $m_{ts}$ is the unique non-zero is the $t$th row), then $(P\circ M)_{[1s]}=0$, so the term vanishes. -Assume that $t=1$ (and hence $s>1$). Then $(P\circ M)^r_{11}$ does not depend on $P_{11}$, as $m_{11}=0$. Hence the term does not depend on $P_{11}$. Also, it clearly does not depend on $Q_{11}$. This finishes part 1. -$\\$ -$\let\eps\varepsilon$ -Part 2. It remains to compute the value of the L.H.S. for some pair of matrices $P$ and $Q$. We set $P_{ij}=Q_{ij}=\eps^{i+j}$ and check the limit of the L.H.S. as $\eps\to+0$. -In this case, the only term in the expansion of $\det(P\circ M)$ that counts is the product of the topmost nonzero elements in all columns (if this term exists in that expansion). Indeed, this term, when divided by $\prod_{j=1}^N(P\circ M)^r_{1j}$, tends to $\pm1$, while all other terms tend to $0$. -Hence, we are interested only in those matrices $M\in\mathcal M$ in which the topmost $1$s of the columns stand in different rows, and, similarly, the leftmost $1$s of the rows stand in different columns. Call these matrices good. -Take any good matrix. In contains the unique $1$ in the first row (say, $m_{1s}=1$) and the unique $1$ in the first column (say, $m_{t1}=1$). If $s,t>1$, then -$$ - \lim_{\eps\to+0}\frac{\det(P \circ M)} {\prod_{j=1}^N(P\circ M)^r_{1j}} - \cdot - \frac{\det(Q \circ M)}{\prod_{i=1}^N(Q\circ M)^c_{i1}} -$$ -does not depend on $m_{ts}$, so we may again pair up such (good!) matrices differing in the $(t,s)$th entry; the sum of the corresponding two terms is $0$. -Otherwise, $s=t=1$, and we know the first row $[1,0,\dots,0]$ and the first column $[1,0,\dots,0]^T$ of $M$. Consider now the unique ones in the second row/column, and proceed in the same way further. At the end, the only unpaired good matrix will be $M=I$, for which the limit is $1$. Hence, the sought value is $1$ as well.<|endoftext|> -TITLE: Picard group of symplectic group modulo orthogonal group -QUESTION [5 upvotes]: Let $Sp(2n)$ be the group of complex symplectic $2n\times 2n$ matrices, and $O(2n)$ the group of complex orthogonal $2n\times 2n$ matrices. -Consider $Sp(2n)\cap O(2n)\subset Sp(2n)$ and the quotient $X=Sp(2n)/(Sp(2n)\cap O(2n))$. How could one compute the Picard group of $X$? -EDIT. Consider the action of $Sp(2n)$ on the projective space $\mathbb{P}^N$ of $2n\times 2n$ matrices modulo scalar given by $Sp(2n)\times\mathbb{P}^N\rightarrow\mathbb{P}^N$, $(A,Z)\mapsto AZA^t$. The stabilizer $H$ of the identity is then given by those matrices in $Sp(2n)$ such that $AA^t = \lambda I$ for some $\lambda\in\mathbb{C}^{*}$. -Let $X = Sp(2n)/H$ be the orbit of the identity in $\mathbb{P}^N$. -How could one compute the Picard group of $X$? -Consider for instance the case $n = 1$. Since any $2\times 2$ symmetric matrix with non-zero determinant has a multiple that is symplectic the orbit $X$ is $\mathbb{P}^2\setminus C$ where $C\subset\mathbb{P}^2$ is the conic parametrizing matrices with zero determinant. So, in this case, $Pic(X) \cong \mathbb{Z}/2\mathbb{Z}$. - -REPLY [5 votes]: Answer: ${\rm Pic\,} X={\Bbb Z}/2{\Bbb Z}$; see Corollary 4 below. -Theorem 1. Let $G$ be a simply connected semisimple group over a field $k$ of characteristic 0. - Let $H\subset G$ be an algebraic subgroup defined over $k$, not necessarily connected. Set $X=G/H$. - Then there is a canonical isomorphism ${\rm Pic\,} X={\widehat H}(k)$, where ${\widehat H}(k) ={\rm Hom}_k(H,{\Bbb G}_{m})$ - is the character group of $H$. - -Proof. First assume that $H$ is connected. We deduce the theorem from results of the paper -J.-J. Sansuc, Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres. -J. Reine Angew. Math. 327 (1981), 12–80. -By Proposition 6.10 of this paper, there is a natural exact sequence of abelian groups -$${\widehat G}(k)\to {\widehat H}(k)\to{\rm Pic\,} X\to {\rm Pic\,} G.$$ -Clearly we have ${\widehat G}(k)=0$. By Sansuc's Lemma 6.9(iv), we have ${\rm Pic\,} G=0$ (here Sansuc refers to a paper by Fossum and Iversen). -We obtain an isomorphism ${\widehat H}(k)= {\rm Pic\,} X$, as required. -Now we do not assume that $H$ is connected. We deduce Theorem 1 from a general result of -M. Borovoi and J. van Hamel, Extended equivariant Picard complexes and homogeneous spaces. Transform. Groups 17 (2012), 51-86. -Since ${\rm Pic\,} G_{\bar k}=0$ and $X$ has $k$-points, by Theorem 2 (Theorem 7.1) of this paper there is a canonical isomorphism -$$ {\rm Pic\,} X=H^1(k,[{\widehat G}({\bar k})\to {\widehat H}({\bar k})\rangle).$$ -Here ${\bar k}$ is an algebraic closure of $k$, ${\widehat H}({\bar k})={\rm Hom}_{\bar k}(H,{\Bbb G}_{m})$, and similarly for ${\widehat G}({\bar k})$. -Further, $[{\widehat G}({\bar k})\to {\widehat H}({\bar k})\rangle$ denotes the complex of ${\rm Gal}({\bar k}/k)$-modules -$$\dots \to 0\to {\widehat G}({\bar k})\to {\widehat H}({\bar k})\to 0\to \dots$$ -with ${\widehat H}({\bar k})$ in degree 1, and $H^1(k,[{\widehat G}({\bar k})\to {\widehat H}({\bar k})\rangle)$ denotes the first Galois hypercohomology of this complex. -In our case ${\widehat G}({\bar k})=0$, and therefore, -$$ {\rm Pic\,} X=H^1(k,[0\to {\widehat H}({\bar k})\rangle)=H^0(k,{\widehat H}({\bar k}))={\widehat H}(k),$$ -as required. -This looks like killing a fly with a bazooka, and there should be an elementary proof of Theorem 1. -Construction 2. -The class in ${\rm Pic\,} X$ corresponding to a character -$$\chi\colon H\to{\Bbb G}_m$$ -is described as follows. We consider the direct product $G\times {\Bbb G}_m$ and the injective homomorphism -$$\iota_\chi\colon H\to G\times {\Bbb G}_m,\quad h\mapsto (h,\chi(h)).$$ -Further, we consider the quotient $Y_\chi:=(G\times {\Bbb G}_m)/\iota_\chi(H)$ and the projection map -\begin{gather*}\pi\colon\, Y_\chi=(G\times {\Bbb G}_m)/\iota_\chi(H)\,\longrightarrow\, G/H=X,\quad \\ -[g,c]\,\mapsto\, [g]\quad \text{for }g\in G,\ c\in{\Bbb C}^\times.\end{gather*} -The group ${\Bbb G}_m$ acts on the fibers of $\pi$ by $c'\cdot [g,c]=[g,c'c]$ for $c'\in{\Bbb C}^\times$. -We see that $\pi\colon Y_\chi\to X$ is a principal ${\Bbb G}_m$-bundle over $X$. -To $\chi$ we associate the class of $Y_\chi$ in ${\rm Pic\,} X$. -We compute the character group $\widehat H$ of the stabilizer $H={\rm Sp}(2n)\cap{\rm GO}(2n)$, where -$$ {\rm GO}(2n)=\{A\in{\rm GL}(2n,{\Bbb C})\mid A^t A=\lambda_A I,\ \lambda_A\in{\Bbb C}^\times\}.$$ - -Proposition 3. For $H={\rm Sp}(2n)\cap{\rm GO}(2n)$ - we have ${\widehat H}={\Bbb Z}/2{\Bbb Z}$. - -Proof. -We compute the group $H$. -We write the equations for $A\in H$: -$$ -A^t A =\lambda_A I,\qquad A^t J A=J, \qquad\text{where } J= -\begin{pmatrix} 0 & I_n\\ -I_n &0 \end{pmatrix}. -$$ -We obtain -$$\lambda_A A^{-1} J A=J, \quad\text{whence } \lambda_A J A=AJ.$$ -Let $x$ be an eigenvector of $J$ with eigenvalue $\mu$. -Then -$$ Jx=\mu x,$$ -whence -$$AJx=\mu Ax,\qquad \lambda_A JAx=\mu Ax,\qquad Jy=\lambda_A^{-1} \mu y, \text{ where }y=Ax.$$ -We see that $y$ is an eigenvector of the matrix $J$ with eigenvalue $\lambda_A^{-1}$. -Thus $\lambda_A^{-1}\mu$ is an eigenvalue of $J$ as well. -Since our matrix $J$ has only two eigenvalues $i$ and $-i$, we conclude that $\lambda_A$ can take values only $1$ and $-1$. Thus we obtain a homomorphism -$$\lambda\colon H\to \mu_2,\quad A\mapsto \lambda_A.$$ -Consider the matrix -$$ S=i\begin{pmatrix} 0 & I_n \\ I_n & 0\end{pmatrix}. $$ -An easy calsulation shows that -$$ S^t S=S^2=-I,\qquad S^t J S=SJS=J.$$ -Thus $S\in H$, $\lambda_S=-1$. -We obtain a short exact sequence -$$ 1\to H_1\to H\to \mu_2\to 1,$$ -where $H_1={\rm Sp}(2n)\cap{\rm SO}(2,n)$ and where -the homomorphism $\lambda\colon H\to\mu_2$ is surjective because $\lambda_S=-1$. -We have $H=H_1\cup S\cdot H_1$. -The group $H_1$ was computed by Sasha in his answer: -it is isomorphic to ${\rm GL}(n,{\Bbb C})$ acting on $V=L_1\oplus L_2$ by -$B\mapsto (B,B^{-1})$. The linear operator $S$ permutes the subspaces $L_1$ and $L_2$, and it acts on the normal subgroup $H_1$ of $H$ as follows: -$$ S\cdot (B,B^{-1}) \cdot S^{-1}=(B^{-1},B).$$ -Hence -$$ S\cdot (B,B^{-1}) \cdot S^{-1}\cdot (B,B^{-1})^{-1}=(B^{-2},B^2).$$ -It follows that the commutator subgroup $(H,H)$ of $H$ is $H_1$. -Thus -$${\widehat H}=\widehat{H/H_1}=\widehat{\mu_2}={\Bbb Z}/2{\Bbb Z},$$ -as required. The nontrivial element of the character group ${\widehat H}$ is the character -$$\lambda\colon H\to \mu_2\hookrightarrow{\Bbb G}_m,\quad -A\mapsto \lambda_A\in {\Bbb C}^\times.$$ - -Corollary 4. For $X={\rm Sp}(2n)/({\rm Sp}(2n)\cap {\rm GO}(2n))$ we have ${\rm Pic\,} X={\Bbb Z}/2{\Bbb Z}$.<|endoftext|> -TITLE: The relation between t-structures and derived category -QUESTION [10 upvotes]: Let $\mathcal{D}$ be a triangulated category and a $t$-structure $(\mathcal{D}^{\leq 0},\mathcal{D}^{\geq 0})$ on $\mathcal{D}$. The heart of the $t$-structure, $\mathcal{A}=\mathcal{D}^{\leq 0} \cap \mathcal{D}^{\geq 0}$, is an abelian category. -I know that in general there is not a natural functor from the derived category of the heart and the triangulated category . -But it's seem to be very linked ... -But if you consider stable infinite category or Grothendieck derivator is there such functor ? - -REPLY [5 votes]: I had reason to think about this a few years ago. When $\mathcal D$ arises as the derived category of an abelian category (with a possibly exotic $t$-structure), a construction of a realization functor $D^b(A) \to \mathcal D$ can be found already in Beilinson-Bernstein-Deligne-Gabber. (For them $\mathcal D$ is the derived category of constructible sheaves, equipped with the perverse $t$-structure, so $A$ is the category of perverse sheaves.) -Their construction of the realization functor can in fact be imitated in the $\infty$-categorical setting, too, and in this case it works more generally for an arbitrary stable $\infty$-category $\mathcal D$. The way B-B-D-G construct the functor is they use the filtered derived category $\mathcal{D}F$. If $\mathcal D$ is the derived category of an abelian category $B$, then $\mathcal D F$ is the category of complexes in $B$ with a bounded filtration, localized at filtered quasi-isomorphisms. There is an induced $t$-structure on $\mathcal D F$ from a $t$-structure on $\mathcal D$, such that if the $t$-structure on $\mathcal D$ has heart $A$ then the heart of the $t$-structure on $\mathcal D F$ is isomorphic to the abelian category $\mathrm{Ch}^b(A)$ of bounded chain complexes in $A$. Thus we can consider the composition $\mathrm{Ch}^b(A) \to \mathcal D F \to \mathcal D$ where the first is the inclusion of the heart and the second forgets the filtration. A spectral sequence argument shows that this composition takes quasi-isomorphisms in $\mathrm{Ch}^b(A)$ to equivalences in $\mathcal D$, so there is an induced functor $D^b(A) \to \mathcal D$ which is the one we want. -The point of the above is that for a general triangulated category $\mathcal T$ there is no sensible triangulated category $\mathcal T F$ of filtered objects in $\mathcal T$, but if $\mathcal T$ happens to be the derived category of an abelian category we can write down the filtered derived category by hand. This is not a problem in the world of stable $\infty$-categories. -I believe that an analogous argument works also for complexes that are not necessarily bounded (using instead filtrations that are unbounded to the left or right or both). But if we want there to exist a functor $\mathcal DF \to \mathcal D$ that forgets the filtration then we need $\mathcal D$ to have sequential limits or colimits, and one should be careful with the spectral sequence argument.<|endoftext|> -TITLE: Commutation relations between covariant and Lie derivatives -QUESTION [8 upvotes]: I am currently working on extrinsic riemannian geometry and I am looking for a sort of commutation relation between the covariant and Lie derivatives. -To be more precise : considering an hypersurface $H \subset M$ of a riemannian manifold, $\nu$ a vector field normal to $H$ and $S$ its shape operator (or Wiengarten operator) defined by $SX = \nabla_X \nu$, you can consider normal geodesics emanating from $H$ as geodesics veryfing $\gamma(0) \in H$, $\dot\gamma(0) = \nu$. Writing the parameters of these geodesics $r$, you get a vector field $\partial_r = \dot\gamma$. If $(x^1,\ldots,x^n)$ are local coordinates on $H$, then you have Fermi coordinates $(r,x^1,\ldots,x^n)$ on $M$. -We have the Ricatti equation, where $R_{\partial_r} = R(\partial_r,\cdot)\partial_r$ : -\begin{align*} -\mathcal{L}_{\partial_r}S=\partial_r S = -S^2 - R_{\partial_r} -\end{align*} -(in fact, the equation is still true while replacing $\mathcal{L}_{\partial_r}$ by $\nabla_{\partial_r}$, it's a property of the shape operator). -I want to find a differential equation for $\nabla_{\partial_j}S$ where $\partial_j = \frac{\partial}{\partial x^j}$. My idea is to differentiate the Ricatti equation with respect to $\nabla_{\partial_j}$ and use a sort of commutation relation to get a differential equation involving $S$, $\nabla_{\partial_j}S$, $R_{\partial_r}$, etc. with variable $r$. -So, my question is : do we have a nice relation between $\nabla_{\partial_j} \mathcal{L}_{\partial_r} S$ and $\mathcal{L}_{\partial_r}\nabla_{\partial_j}S$ ? -Thank you for reading me. -Edit -I recently tried something : expanding the lie derivative to the connexion itself. That is : -\begin{align} -\mathcal{L}_{\partial_r} \left( \nabla_j S) \right) &= \left(\mathcal{L}_{\partial_r}\nabla_j\right) S + \nabla_j \left( \mathcal{L}_{\partial_r}S\right) -\end{align} -In Einstein Manifolds, Besse, there is a formula for the derivative of the connection with respect to the metrics, in the direction of a symmetric tensor, that is : -\begin{align} -g\left((\nabla'(g)\cdot h)(X,Y),Z\right) &= \dfrac{1}{2}\left(\nabla_Xh (Y,Z) + \nabla_Yh(X,Z) - \nabla_Zh (X,Y) \right) -\end{align} -With that in mind, and recalling that $\mathcal{L}_{\partial_r}g = 2g\left(S\cdot,\cdot\right)$, something is appearing. I would post somthing if this answers the original question. - -REPLY [6 votes]: I recently answered my question by finding a formula I wasn't aware of. -Let $\nabla$ be a connexion and $X$ a vector field. Then $\mathcal{L}_X\nabla$ is a tensor and -\begin{align} -\mathcal{L}_X\nabla &= -i_X\circ R^{\nabla} + \nabla^2X -\end{align} -where $R^{\nabla}(U,V) = \nabla_{[U,V]} - [\nabla_U,\nabla_V]$ is the curvature tensor of $\nabla$, and $\nabla_{U,V}^2X = \nabla_U\nabla_VX - \nabla_{\nabla_UV}X$. Applying this to $\nabla_{\partial_j}S$ we get -\begin{align} -\mathcal{L}_{\partial_r}\left(\nabla_{\partial_j}S\right) &= \mathcal{L}_{\partial_r}(\nabla)(\partial_j,S) + \nabla_{[\partial_r,\partial_j]}S + \nabla_{\partial_j}(\mathcal{L}_{\partial_r}S) -\end{align} -and using the above formula and the Riccati equation for $S$ leads to the wanted linear differential equation.<|endoftext|> -TITLE: Electromagnetic energy in Lovelock gravities -QUESTION [7 upvotes]: To fix ideas, let us recall that General Relativity describes gravitational phenomena on a 4-dimensional pseudo-Riemannian manifold $(X,g_{ab})$ with field equations that relate the energy-momentum tensor $T_{ab}\,$ of the matter distribution to the geometry of spacetime via the so called Einstein tensor: -$$ \mathrm{Ric}_{ab} - \frac{\mathrm{scal}}{2} g_{ab} \, = \, 8 \pi\, T_{ab} \ . $$ -In this setting, the presence of an electromagnetic field is mathematically encoded with a closed 2-form $F_{ab}$. -The field equations for a spacetime with no matter and an electromagnetic field $F_{ab}$ read as follows: -$$ \mathrm{Ric}_{ab} - \frac{\mathrm{scal}}{2} g_{ab} \, = \, 8 \pi \left( F_{a \alpha}F^{\alpha}_{\ \, b} - \frac{1}{4} F^{\alpha_1 \alpha_2} F_{\alpha_1\alpha_2} g_{ab} \right) \ . $$ -In other words, the energy-momentum contribution of the electromagnetic field is measured by this tensor (sometimes called the Maxwell tensor of $F_{ab}$): -$$ \mathsf{M}_{ab} := \, F_{a\alpha}F^{\alpha}_{\ \, b} - \frac{1}{4} F^{\alpha_1\alpha_2} F_{\alpha_1\alpha_2} g_{ab} \ . $$ -My question is: - -Is there an analogue of this Maxwell tensor $\mathsf{M}_{ab}\,$ on Lovelock gravities? - - -To be more precise, Lovelock gravities are higher dimensional analogues of General Relativity, where the vacuum field equations of these theories are now defined to be: -$$ \mathrm{Ric}^{(2q)}_{ab} - \frac{\mathrm{scal}^{(2q)}}{2} g_{ab}\, = \, 0 \ , $$ where -$$ \mathrm{Ric}^{(2q)}_{ab} := \, \delta_{a \beta_2 \dots \beta_{2q}}^{\alpha_1 \alpha_2 \dots \alpha_{2q}} R_{\alpha_1 \alpha_2 b}^{\beta_2} R_{\alpha_3 \alpha_4}^{\beta_3 \beta_4} \dots R_{\alpha_{2q-1} \alpha_{2q}}^{\beta_{2q-1 2q}} \ , $$ -$$ \mathrm{scal}^{(2q)} := \, g^{\alpha \beta} \mathrm{Ric}^{(2q)}_{\alpha \beta} \qquad , \qquad \delta^{\alpha_1 \dots \alpha_{2q}}_{\beta_1 \dots \beta_{2q}} = \mathrm{det} (\delta^{\alpha_i}_{\beta_j}) \ , $$ and -$q$ may run from 0 to the integer part of $(\dim X - 1) /2$ (the case $q=0$ is trivial, and the case $q=1$ recovers Einstein's equation). -My question is then: - -Are there tensors $\widetilde{\mathsf{M}}^{(2q)}_{ab}\,$ that can be coupled into Lovelock equations, so that they define a reasonable theory of electromagnetism? - -Of course, these tensors $\widetilde{\mathsf{M}}^{(2q)}_{ab}\,$ should be defined using $g_{ab}$ and $F_{ab}$, and the vacuum field equations -$$ \mathrm{Ric}^{(2q)}_{ab} - \frac{\mathrm{scal}^{(2q)}}{2} g_{ab}\, = \, \widetilde{\mathsf{M}}_{ab}^{(2q)} \ $$ should impose restrictions on their divergence, etc. - -REPLY [5 votes]: The coupling of electromagnetism (including Born-Infeld nonlinearities) to Lovelock gravity has been studied in Magnetic Branes in Third Order Lovelock-Born-Infeld Gravity. The nonlinearities in the Maxwell Lagrangian are introduced to obtain a finite value for the self-energy of a pointlike charge. Earlier works (cited in that reference) have worked out the linear limit.<|endoftext|> -TITLE: How not to use J-holomorphic curves -QUESTION [13 upvotes]: The field of symplectic topology is filled with subtle traps for the unwary, particularly when it comes to the analysis of $J$-holomorphic curves. So that the next generation of symplectic topologists can avoid committing the sins of their elders, it would be desirable to make a collection of such traps: how not to use holomorphic curves. For instance, the failure of somewhere-injectivity to hold for $J$-holomorphic curves with boundary; the failure of the reparametrization action to be differentiable on Sobolev spaces; inconsistency of gluing with different models of strip-like ends etc. -Of course, there is no need to name names. -If it wasn't obvious, I am looking for: superficially convincing arguments or statements concerning $J$-holomorphic curves that have been used (or almost used) in papers in symplectic topology, and which, on closer inspection, fail to hold for somewhat subtle reasons. I think the answer by Jonny Evans is a great example. - -REPLY [9 votes]: One classic thing to do is to take a sequence of holomorphic curves with a tangency condition and assume that the tangency condition still holds in the limit: if the limit is a multiple cover with a branch point where you had your tangency condition then this can fail. As a PhD student, I once used this to prove that all complex curves in $CP^2$ are diffeomorphic to the sphere (fortunately I realised in time that this remarkable result was not correct so it never made it off my blackboard).<|endoftext|> -TITLE: Is the derived category of local systems equivalent to the derived category of sheaves of vector spaces with local system cohomology? -QUESTION [5 upvotes]: Let $k$ be a field and $X$ a topological space. -Write $\mathrm{Sh}(X)$ for the category of sheaves of vector spaces on $X$, and $\mathrm{Loc}(X)$ for the subcategory of local systems of finite dimensional $k$-vector spaces. -The category not local systems is an abelian category, so we can form the derived category $D(Loc(X))$. This is the category of complexes of local systems on $X$ with quasi-isomorphisms inverted. -We can also consider the subcategory $D_{\mathrm{Loc}}(X)$ of $D(X):=D(\mathrm{Sh}(X))$ consisting of complexes whose cohomology sheaves are local systems on $X$. -I have two questions: - -Is $D_{\mathrm{Loc}}(X)$ a triangulated subcategory of $D(X)$? More precisely is it closed under taking mapping cones? -Under what hypotheses (if any) are $D_{\mathrm{Loc}}(X)$ and $D(\mathrm{Loc}(X))$ equivalent? - -REPLY [2 votes]: UPD: I didn't notice that you're asking about finite dimensional local systems, so this answer doesn't really answer your question. The easy way to fix that is to consider the category $D^b_f(Loc(X))$ of complexes of representations of $\pi_1$ with finite dimensional cohomology; in order to treat the case of an honest category of finite dimensional representations, it seems, the use of (derived) algebraic completion is unavoidable. - -I think that both statements are true if $X$ is sufficiently nice and $K(\pi, 1)$, and if you consider bounded derived categories. Sufficiently nice here means that Exts between two local systems in the category of local systems and in the category of sheaves coinside with each other. I think it is enough to assume that $X$ is locally contractible; from this assumption follows the existence of the universal cover $\tilde X$. -The cone of two bounded complexes with locally constant cohomology has locally constant cohomology, because on a locally contractible space $X$ the category of locally constant sheaves is a so called Serre subcategory of the category of sheaves --- abelian subcategory closed under extensions. Then the long exac sequence of cohomology objects shows you that a cone of two complexes with locally constant cohomology has locally constant cohomology as well. -There are at least two ways to show that Exts between two local systems in two categories are the same. One can show that there are enough local systems, that are projective (resp., injective) as local systems, which are adapted to the functor $\mathrm{Ext}^*(-, V)$ (resp. $\mathrm{Ext}^*(V, -)$), where $V$ is also a local system. For projective local systems we can just take the regular representation of $\pi_1(X)$, let's call it $P$. For an injective, one can take the representation $\prod_{g \in G} \mathbb{Q}g$, where $\mathbb{Q}$ is some injective hull of your base ring. The $G$-action here is by multiplying to $g^{-1}$ from the right. Let's denote this module by $Q$. I think, by using the rule $v \mapsto \prod_{g \in G} g\otimes gv$ one can embed any $G$-module $V$ into $Q \hat\otimes V$, that is, infinite product of $Q$ ($V$ in the tensor product is considered with trivial $G$-action). -Now, both $P$ and $Q$ come from $\tilde X$, that is, if $\pi: \tilde X \longrightarrow X$ is the universal cover, then $P = \pi_! \underline{\mathbb{Z}}_{\tilde X}$ (where $\mathbb{Z}$ is the base ring) and $Q = \pi_*\underline{\mathbb{Q}}_{\tilde X}$. Since $\pi$ is a covering, there are no higher derived functors of $\pi_*$ and $\pi_!$. -Now one can use the adjointness $\mathrm{Ext}_{X}(F, Q) = \mathrm{Ext}_{\tilde X}(\pi^*F, \underline{\mathbb{Q}}_{\tilde X})$. Since $\mathbb{Q}$ is injective, $\tilde X$ is contractible and $\pi^*F$ is a locally constant (hence constant) sheaf, this vanishes in higher degrees. Or, you can show that $P$ is adapted by using the fact that for a covering map $\pi_!$ is left adjoint to $\pi^*$ (a fact which I think I know how to prove, but I was unable to find a reference for it in the case of an infinite covering. You also need to assume $X$ to be locally compact to use this, i suppose). -Now, using the induction on the length of a complex, you can show that the inclusion functor from $D^b(Loc(X))$ into $D^b(X)$ is a fully faithful embedding. -And since local systems generate $D^b_{Loc}(X)$ as a triangulated subcategory of $D^b(X)$, it is precisely the essential image of this functor. This is more or less standart; i think that an appendix to this Positselski's paper has a good treatment of that stuff. -Sadly, I don't know what will happen if one wants to consider unbounded categories.<|endoftext|> -TITLE: Examples of improved notation that impacted research? -QUESTION [26 upvotes]: The intention of this question is to find practical examples of improved mathematical notation that enabled actual progress in someone's research work. -I am aware that there is a related post Suggestions for good notation. The difference is that I would be interested especially in the practical impact of the improved notation, i.e. examples that have actually created a better understanding of a given topic, or have advanced actual research work on a given topic, or communication about results. -I would be interested in three aspects in particular -(1) Clarity and Insights: Improved and simplified notation that made structures and properties more clearly visible, and enabled insights for the researcher. -(2) Efficiency and Focus: Notation that created efficiencies (e.g., using less space and needed less time, dropped unnecessary or redundant details). -(3) Communication and Exposition: Improved notation that supported communicating and sharing new definitions and results. And notation that evolved and improved in the process of communication. Would you have any practical examples of this evolving process, including dead-ends and breakthroughs? -Edit: Have received great examples in the answers that illustrate what I am interested in. Very grateful for that! - -REPLY [3 votes]: I'm a bit surprised that no one mentioned the Einstein summation convention<|endoftext|> -TITLE: Characterizing discrete quantum groups -QUESTION [7 upvotes]: Let $M$ be a von Neumann algebra, and let $\Delta$ be a unital normal $*$-homomorphism $M \rightarrow M \mathbin{\bar\otimes} M$ that satisfies the coassociativity condition $(\Delta \mathbin{\bar\otimes} \mathrm{id}) \circ \Delta = (\mathrm{id} \mathbin{\bar\otimes} \Delta ) \circ \Delta$. Assume that $M$ is an $\ell^\infty$-direct sum of finite type I factors. Are the following conditions equivalent? - -The pair $(M, \Delta)$ is a von Neumann algebraic quantum group, in the sense of Kustermans and Vaes [2, definition 1.1]. -There exists a unital normal $*$-homomorphism $\varepsilon\colon M \rightarrow \mathbb{C}$, with support projection $e \in M$, such that: - - -(a) $(\varepsilon \mathbin{\bar\otimes} \mathrm{id}) \circ \Delta = \mathrm{id}$ -(b) $(\mathrm{id} \mathbin{\bar\otimes} \varepsilon) \circ \Delta = \mathrm{id}$ -(c) For every projection $p \in M$, if $p \otimes 1 \geq \Delta(e)$, then $p = 1$. -(d) For every projection $p \in M$, if $1 \otimes p \geq \Delta(e)$, then $p = 1$. - - -The question is motivated by quantum predicate logic [3, section 2.6]. I am extending my preprint [1] to include a few more examples, but the example of discrete quantum groups is the one it really needs. -[1] A. Kornell, Quantum predicate logic with equality. arXiv:2004.04377 -[2] J. Kustermans & S. Vaes, Locally compact quantum groups in the von Neumann algebraic setting, Math. Scand. 92 (2003), no. 1. -[3] N. Weaver, Mathematical Quantization, Studies in Advanced Mathematics, Chapman & Hall/CRC, 2001. - -REPLY [8 votes]: Also the implication (2) $\Rightarrow$ (1) holds and can be proven as follows. -Denote by $\mathcal{C}$ the category of all finite dimensional, nondegenerate $*$-representations of $M$. The morphisms are the intertwining linear maps. Turn $\mathcal{C}$ in a $C^*$-tensor category by defining $\pi_1 \otimes \pi_2$ to be equal to $(\pi_1 \otimes \pi_2) \circ \Delta$. Then $\varepsilon$ is a unit object for $\mathcal{C}$. We identify the set $I$ of minimal central projections of $M$ with the representatives for irreducible objects of $\mathcal{C}$. For $p \in I$, we pick a finite dimensional Hilbert space $H_p$ such that $Mp = B(H_p)$. -The main point is to prove that $\mathcal{C}$ is rigid, i.e. that every irreducible object has a conjugate with a solution for the conjugate equations. Let $p \in I$. Hypothesis (c) is saying that the join of all the left support projections of elements of the form $p((\text{id} \otimes \omega)\Delta(e))$ equals $p$. Similarly, the join of all the right support projections of elements of the form $((\omega \otimes \text{id})\Delta(e))p$ equals $p$. It follows that we can pick $q,r \in I$ such that -$$(\Delta(e)(r \otimes p) \otimes 1) (1 \otimes (p \otimes q)\Delta(e)) \neq 0 \; .$$ -So we can pick morphisms $V : \mathbb{C} \rightarrow H_p \otimes H_q$ and $W : \mathbb{C} \rightarrow H_r \otimes H_p$ such that -$$(W^* \otimes 1) (1 \otimes V) \neq 0 \; .$$ -Since this expression defines a morphism between the irreducible objects $r$ and $q$, we conclude that $r=q$ and that $V$ and $W$ may be chosen such that -$$(W^* \otimes 1) (1 \otimes V) = 1_q \; .$$ -Then -$$(1 \otimes (V^* \otimes 1)(1 \otimes W)) W = (1 \otimes V^* \otimes 1)(W \otimes W) = W \; .$$ -It follows that $(V^* \otimes 1)(1 \otimes W)$ is nonzero, thus a multiple of $1_p$ and hence, equal to $1_p$. We have proven that $\mathcal{C}$ is rigid. -One could already conclude at this point that $(M,\Delta)$ is a discrete quantum group by making a detour via Woronowicz' Tannaka-Krein theorem. -One can also repeat the proof of the first basic properties of rigid $C^*$-tensor categories and then directly verify Van Daele's axioms cited above. One first proves Frobenius reciprocity. By definition, $p$ is contained in $q \otimes r$ if and only if $\Delta(p) (q \otimes r) \neq 0$. Using Frobenius reciprocity and denoting by $A \subset M$ the dense $*$-subalgebra spanned by the $Mp$, $p \in I$, one gets that the linear span of $\Delta(A) (1 \otimes A)$ is contained in the algebraic tensor product $A \otimes_{\text{alg}} A$. One can thus define Van Daele's map -$$T : A \otimes_{\text{alg}} A \rightarrow A \otimes_{\text{alg}} A : T(a \otimes b) = \Delta(a) (1 \otimes b) \; .$$ -Given $a \in A$ and $p \in I$, take $V$ as above. Define $b \in A \otimes Mp$ such that -$$\Delta(a)_{13} (1 \otimes V) = (b \otimes 1)(1 \otimes V) \; .$$ -Then -$$(T(b) \otimes 1)(1 \otimes V) = (\Delta \otimes \text{id})\Delta(a) (1 \otimes V) = a \otimes V \; .$$ -Therefore, $T(b) = a \otimes p$. It follows that $T$ is surjective. One can reason similarly for $\Delta(A)(A \otimes 1)$ and conclude that Van Daele's definition of discrete quantum groups is satisfied.<|endoftext|> -TITLE: Non-perturbative Renormalization in the sense of Polchinski's equation. Do we have a mathematical formulation? -QUESTION [8 upvotes]: My question is about mathematical treatment of exact renormalization group in the sense of Polchinski's flow equation. In a heuristic form, Polchinski's equation looks like: $\partial_t S[\phi] = \frac{\delta}{\delta \phi} \cdot \Delta \cdot \frac{\delta}{\delta \phi} S[\phi] - \frac{\delta}{\delta \phi}S[\phi]\cdot \Delta \cdot \frac{\delta}{\delta \phi}S[\phi]$, where $S[\phi]$ is the action, $\phi$ is the field, $\Delta$ contains a cut-off version of parametrix associated with the classical equation of motion of the field (for example Klein Gordon equation for scalar bosonic fields). In addition, t is the renormalization "time" which is the log of energy scale, and $\Delta$ depends on t since it involves high-energy cut-off. -Note that Polchinski's equation is meant to be a non-perturbative field-theoretic formulation of the Wilsonian renormalization. -By now, I think there are several expositions of mathematical aspects of renormalization. For example, Costello's formulation of perturbative renormalization in BV formalism. We also have Kreimer-Connes formulation of BPHZ renormalization emphasizing on Hopf algebra structure and non-commutative aspects. However neither seems to work with non-perturbative aspects of renormalization as in the sense of Polchinski. -Therefore, my question is whether there is an attempt to study Polchinski equation in a rigorous mathematical setting. If we write $S[\phi]$ in terms of formal power series in $\phi$, and study the corresponding equations term by term, an important point is that this set of infinitely many equations hold "within path integral" which means they are to be understood as equations of operators. Therefore preferably as a first step, I want to ask if there are formulations of differential equations in operator algebras. -To be more concrete, as a toy model, if we consider finite dimensional self-adjoint matrices $A$ and perturb it by self-adjoint matrices $A_s = A + sK$, where $K$ is self adjoint, and we consider $f \in B^1_{\infty,1}(\mathbb{R})$ (Besov space), it is known that we have a formulation of differentiation in terms of double operator integral. Then what can we say about solution to the ODE: $\frac{d}{ds}f(A_s) = B$ where B is a fixed self adjoint matrix. -Furthermore, suppose we keep the above setting and take $f\in B^2_{\infty,1}(\mathbb{R})$, we can now consider second order derivative in terms of a multi-operator integral. Then following the usual Laplacian on $\mathbb{R}^n$ and let $\{ E_{ij} \}$ be a basis of n-dim Hermitian matrices, consider the analog of Laplacian $\Sigma_{ij}\frac{d^2}{ds^2}f(A^{ij}_s)$ where $A^{ij}_s = A + sE_{ij}$. Can we formulate and prove a maximum principle for such an analog of Laplacian? -Anyhow, this is a super long question. Any thoughts are helpful. The comment on operator differential equation is just my very pre-mature thoughts. Any direction regarding the Polchinski's equation (not necessarily related to operator algebra) is greatly appreciated. Thanks a lot for the help. - -REPLY [7 votes]: Good question! Before going further in your investigations on rigorous nonperturbative implementations of the renormalization group (RG) philosophy used for the construction of QFTs in the continuum, you should at least read my previous answer -https://physics.stackexchange.com/questions/372306/what-is-the-wilsonian-definition-of-renormalizability/375571#375571 -If you have time, also look at -What mathematical treatment is there on the renormalization group flow in a space of Lagrangians? -There is nothing sacrosanct about Polchinski's approach. It is just one among many ways of implementing the RG method. It is characterized by the use of a continuous flow, or ODE on a very infinite-dimensional space of effective actions/potentials. It looks nonperturbative but it isn't really because no one was able to find norms on space of functionals (functions of functions) where one can prove a local in time well posedness result for this ODE. As far as I know, there is only one rigorous nonperturbative result with a continuous flow RG: the article "Continuous Constructive Fermionic Renormalization" by Disertori and Rivasseau in Annales Henri Poincaré 2000. -They didn't use Polchinski's equation but the older Callan-Symanzik equation, and this only works for Fermions which are easier than Bosons because perturbation series converge in the case of Fermion (with cutoffs). -As to what has been done rigorously with Polchinski's equations, it concerns rigorous proofs of perturbative renormalizability, i.e., in the sense of formal power series. A good introduction to this is the book "Renormalization: An Introduction" by Manfred Salmhofer. For recent results with this approach see the works of Christoph Kopper and Stefan Hollands. -For nonperturbative rigorous results for Bosons, at least for what is known so far, one has to abandon the idea of continuous flow as in Polchinski's equation and work with a discrete RG transformation. Again there are several approaches. If you want something that is closest to the Polchinski philosophy, then you might want to look at the approach by David Brydges and collaborators developed over many years and culminating with a series of five articles in J. Stat. Phys., with Gordon Slade and Roland Bauerschmidt. A pedagogical introduction is now available with the book "Introduction to a Renormalisation Group Method" by these three authors.<|endoftext|> -TITLE: Graph with path of length $\geq n$ along grid diagonals - a known result in graph theory? -QUESTION [11 upvotes]: Is the following lemma a well known result in graph theory? -I am studying a basic existence result that appears to be simple yet powerful. I have not seen it stated as an important result in graph theory. I have consulted Reinhard Diestel's "Graph Theory" (5th edition, 2017), but could not find it there. So I wanted to ask this question on MO: -Definition: Given an $n\times n$ grid with $n^2$ unit squares. If you randomly place exactly 1 diagonal in each unit square, these diagonals (together with the vertices of the grid) form a graph $G$. -Existence Lemma: $G$ always contains a path of length $\geq n$. -Above you can see a small example on a $6\times 6$ grid. There is a great graphical example for large $n$ by Joseph O’Rourke https://mathoverflow.net/a/112090/156936 -I would be grateful if you could let me know whether this is a well known result, specifically in graph theory. -Is there maybe some more general result from graph theory that implies this particular case? I would be very interested in that. - -REPLY [2 votes]: I think Timothy Chow's comment is right that there is no result about planar graphs with your lemma as an explicit corollary. -I believe the following 2007 research paper by Guido Helden might be of use to you: http://publications.rwth-aachen.de/record/62349/ It is about hamiltonicity of maximal planar graphs and planar triangulations, and starts with a very good exposition.<|endoftext|> -TITLE: Why is this "the first elliptic curve in nature"? -QUESTION [32 upvotes]: The LMFDB describes the elliptic curve 11a3 (or 11.a3) as "The first elliptic curve in nature". It has minimal Weierstraß equation -$$ -y^2 + y = x^3 - x^2. -$$ -My guess is that there is some problem in Diophantus' Arithmetica, or perhaps some other ancient geometry problem, that is equivalent to finding a rational point on this curve. What might it be? - -Edit: Here's some extra info that I dug up and only mentioned in the comments. Alexandre Eremenko also mentions this in an answer below. The earliest-known example of an elliptic curve is one implicitly considered by Diophantus, in book IV of Arithmetica, problem 24 (Heath's translation): "To divide a given number into two numbers such that their product is cube minus its side". Actually this is a family of curves over the affine line, namely $y(a-y)= x^3-x$, though Diophantus, in his usual way, only provides a single rational point for the single curve corresponding to $a=6$. This curve is 8732.b1 in the L-functions and modular forms database (the Cremona label is 8732a1). So presumably the comment about 11a3 is not meant to mean "historically first". - -REPLY [6 votes]: I asked Kevin Buzzard to ask John Coates directly, and it's basically as people have surmised: the moniker is due to the fact the curve appears first in Cremona's book as it has the smallest possible conductor, and it has the smallest coefficients. It is not due to historical priority, as Coates knows of 8th/9th century Arabic manuscripts that discuss $y^2 = x^3 - x$, whereas the first occurrence of the "first curve in nature" is apparently a book of Fricke on elliptic functions (I think from 1922, but I'm not sure).<|endoftext|> -TITLE: Current status of axiomatic quantum field theory research -QUESTION [19 upvotes]: Axiomatic quantum field theory (e.g. the wightman formalism and constructive quantum field theory) is an important subject. When I look into textbooks and papers, I mostly find that the basic constructions involve functional analysis and operator algebra. Now, modern developments in the field of quantum field theory involve subjects such as algebraic geometry, topology, and knots. Mostly modern developments deal with supersymmetric quantum field theory. -My question is: What is the status of current research on the mathematical foundations of the dynamics of "non topological" quantum field theory? Are the older approaches to the subjects, such as the ones in Glimm and Jaffe's and Wightman's books, obsolete? - -REPLY [10 votes]: The Axiomatic Quantum Field Theory from the 1950's has been rebranded as Algebraic QFT (keeping the same abbreviation), and the emphasis has shifted somewhat from quantum fields to local observables. The Wightman axioms for fields are then replaced by the Haag–Kastler axioms for the algebra of observables. For two relatively recent overviews see Current trends in axiomatic quantum field theory (1998) and Algebraic quantum field theory (2006). A concise summary is at nLab. An alternative approach to AQFT is FQFT (Functorial QFT), which focuses on states rather than observables.<|endoftext|> -TITLE: How is class of composition of two quadratic fields is related class numbers of quadratic field? -QUESTION [5 upvotes]: Let $K_1=\Bbb Q(\sqrt{d_1})$ , $K_2=\Bbb Q(\sqrt{d_2})$ and $K=\Bbb Q(\sqrt{d_1},\sqrt{d_2})$.Suppose $h_1,h_2,h$ be class number of $K_1,K_2,K$ respectively. -(i) Can we express $h$ in terms of $h_1,h_2$? -(ii) Knowing the divisibility properties of $h_1,h_2$, I want help with concluding about the divisibility of $h_1,h_2.$ - -REPLY [10 votes]: I assume you mean $K_1=\mathbb Q(\sqrt{d_1})$. -1) If by $K$ you mean $\mathbb Q(\sqrt{d_1d_2})$, then there is no simple relation -between $h$ and $h_1$ and $h_2$. -2) If by $K$ you mean the quartic biquadratic field $\mathbb Q(\sqrt{d_1},\sqrt{d_2})$, a theorem of Herglotz says that $h=h_1h_2h_3/2^j$, where -$h_3$ is the class number of $\mathbb Q(\sqrt{d_1d_2})$ and $j=0,1,2$ which can -be computed in terms of the units of $K$.<|endoftext|> -TITLE: Degrees of syzygies of points in $\mathbb P^2$ -QUESTION [8 upvotes]: Let $X$ be a collection of points in $\mathbb P^2$ over the complex numbers. Let $I_X$ be the defining ideal. I am interested in knowing when: - -The syzygies of $I_X$ contains no linear forms. Since we are in $\mathbb P^2$, this just says that the Hilbert-Burch matrix contains no (non-zero) linear entries. $(*)$ - -One obvious case when this happens is when we take two general curves $F,G$ of degrees $a,b\geq 2$ and let $X=V(F)\cap V(G)$. Then the syzygies is just the Kozsul relation and has degrees $a,b$. I don't know further examples and would like to know if there are interesting geometric conditions that would imply $(*)$. -One obvious necessary condition is that the generators of $I_X$ have degree at least $2n-2$, where $n$ is the number of generators. -Also, perhaps if you fix the degree $d$ of $X$, then $(*)$ defines a closed subscheme (? I am not sure about this) of the Hilbert scheme $\mathbb P^{2[d]}$. If so, then knowing it's dimension would be nice. - -REPLY [3 votes]: I asked David Eisenbud and was told about Exercises 12 and 13 of Chapter 3 in his book "Geometry of Syzygies". Putting together, they show that for a generic set of $n$ points $X$ in $\mathbb P^2$, the syzygies of $I_X$ has no linear forms if and only if $n= \binom{2s+1}{2}+s$, for some positive integer $s$. It is unclear if a complete characterization can be found.<|endoftext|> -TITLE: Digitalized version of "Cours de topologie algébrique professé en captivité" -QUESTION [19 upvotes]: It is historically known that Jean Leray gave a course on algebraic topology while captive in the Officer's detention camp XVI in Edelbach, Austria during WW2. (References to this topic include an article by Sigmund, Michor and Sigmund in the mathematical Intelligencer, 27/2 (2005), 41-50.) -Is there a digitalized copy of the course material somewhere? - -REPLY [25 votes]: The course has been published in the Journal de Mathématiques Pures et Appliquées, volume 24 (1945), and can be found here: first part and second part and third part. -A quote from Leray's obituary: - -The prisoners in the camp were mostly educated men, career or reserve - officers, many of them still students. As in several other camps, a - "university" was created and Leray became its rector. Classes were - taught, exams were given, and degrees granted, with some degree of - recognition by French authorities of the time. As for research, to - fight the feeling that he might be losing the best productive years of - his life, Leray wanted to resume his work. But he was confronted with - a dilemma. If he continued working in fluid mechanics, he might be - forced to collaborate with the German war effort. Instead, he decided - to pursue some ideas in algebraic topology that he had foreseen during - his collaboration with Schauder. - - -pages 95-167 - -pages 169-199 - -pages 201-248<|endoftext|> -TITLE: The first part of the Hilbert sixteenth problem for elliptic polynomials -QUESTION [5 upvotes]: A polynomial $P(x,y)\in \mathbb{R}[x,y]$ is called an elliptic polynomial if its highest homogeneous part does not vanish on $\mathbb{R}^2\setminus\{0\}$. -Inspired by the first part of the Hilbert 16th problem we ask that: - -Is there an elliptic polynomial $P(x,y)\in \mathbb{R}[x,y]$ of degree $n$ which has a level set $P^{-1}(c)$ with more than $n$ connected components? - Can elliptic polynomials produce $M$-curves? - -REPLY [2 votes]: You can get all the way up to $\binom{n-1}{2}+1 = g+1$, where $g$ is the genus. This is the maximal number of connected components a real curve of genus $g$ can have, so this is optimal. -To do this, I will use Viro's patchworking method. Itenberg and Viro already give an example of how to use patchworking to build a plane curve with $g+1$ connected components, so I will just show how to tweak it to use an elliptic polynomial. -I can't write a better explanation of patchworking than the one I just linked, so I'll assume you read it. -Take the triangulation from Figure 7. . -Along each edge on the outside of the figure, there are $2n$ small triangles. Pair them into $n$ pairs, and merge each pair with its common neighbor to make $n$ trapezoids. Leave the rest of the figure as before. It is easy to check that this subdivision is still coherent. The resulting plane curve has the same topology as the original figure, but the big loop which crossed the line at infinity $2n$ times before is now disjoint from it. -Poking around Viro's website, I came across slides from a talk on Hilbert's 16th problem. On page 48, he says that Hilbert, in 1891, found a construction of a curve with $g+1$ real components by perturbing a union of two conics. . -The slides don't give a detailed explanation of Hilbert's construction, but it looks like it would produce an elliptic polynomial.<|endoftext|> -TITLE: Notions of Lie 2-groupoids -QUESTION [6 upvotes]: The term Lie $2$-groupoid is used in the literature in more than one context. Some examples are given below: - -Ginot and Stiénon's paper $G$-gerbes, principal $2$-group bundles and characteristic classes defines a Lie $2$-groupoid to be a double Lie groupoid in the sense of Brown - Determination of a double Lie groupoid by its core diagram satisfying certain conditions. -Rajan Amit Mehta and Xiang Tang's paper From double Lie groupoids to local Lie $2$-groupoids says that "The simplicial approach to $n$-groupoids goes back to Duskin [Higher-dimensional torsors and the cohomology of topoi: the abelian theory] in the discrete case, and the smooth analogue appeared in [Henriques - Integrating L-infinity algebras]." They define Lie $2$-groupoid to be a simplicial manifold $X=(X_k)$ with some condition on the horn maps. -Matias del Hoyo and Davide Stefani's paper The general linear $2$-groupoid defines a Lie $2$-groupoid to be a Lie $2$-category and a $2$-category satisfying certain conditions. -The n-lab page on Lie $2$-groupoids says "a Lie 2-groupoid is a 2-truncated $\infty$-Lie groupoid". No further discussion is given there. Clicking the link $\infty$-Lie groupoid takes you to the page Lie $n$-groupoid, which does not contain many details. - -Questions : - -Are these notions genuinely different notions introduced for different purposes or are they all the same candidate wearing different outfits? -Are there other notions of Lie $2$-groupoids in the literature? - -REPLY [5 votes]: For your first question: -They are essentially all the same thing: some globular, some simplicial (taking the nerve goes from the former to the latter). The only subtlety is perhaps in the requirement on the maps $d_{2,0}, d_{2,2}$ to be surjective submersions in the del Hoyo–Stefani paper. This is not unusual for the simplicial approach to n-groupoids in non-finitely complete sites: the Kan condition (which is what these maps are about) shouldn't be just encoded by a certain map being an epimorphism, but being a cover of some sort. This is also the content of Definition 1.2 in Henriques' paper (note that you have linked to the arXiv version 2 of the paper, and that's what I'm referring to. Version 1 has slightly different material and you should also check it out). I don't recall offhand if this is automatic for the globular definition as in Ginot–Stiénon's paper. -So I would say with some confidence that 2.–4. are the same, apart from dealing with a 2-groupoid vs the nerve of a 2-groupoid, and 1. might be very slightly more general, though only in a small technical hypothesis that a certain small class of surjective maps are not required to be submersions a priori. I didn't dig into the techicalities of Brown–Mackenzie (cited by Ginot–Stiénon) to see if one gets submersions automatically. Ideally the nLab page on Lie 2-groupoids would get updated by some friendly soul who wants to put in the work to explain the more elementary viewpoint. -On a historical note, apart from the technical hypothesis, these definitions essentially go back to Charles Ehresmann, who defined (strict) 2-categories, double categories, internal categories (including Lie categories) and so on. Focusing specifically on the case 2-groupoids/double groupoids/Lie groupoids arose slowly, partly driven by Brown, Mackenzie, Pradines, Haefliger, .... and so on. -As far as different notion of 2-groupoids go, one can consider the composition functor on hom-groupoids to be a map of their associated stacks, hence an anafunctor as opposed to an internal functor. This viewpoint is implicit in - -Christian Blohmann, Stacky Lie groups, Int. Mat. Res. Not. (2008) Vol. 2008: article ID rnn082, 51 pages, doi:10.1093/imrn/rnn082, arXiv:math/0702399 - -where the one-object case is considered (the general stacky notion is discussed by Breen in Bitorseurs et cohomologie non abélienne. In The Grothendieck Festschrift, Vol. I, Progr. Math., vol. 86, pp. 401–476 (1990)). More explicitly this notion is considered in - -Chenchang Zhu, Lie $n$-groupoids and stacky Lie groupoids - arXiv:math/0609420 - -(with a more general theory in her followup paper 0801.2057) using the language of stacks. In general, what this means is that if you take the viewpoint on internal 2-groupoids as in Definition 2.1 of - -A bigroupoid's topology (or, Topologising the homotopy bigroupoid of a space), Journal of Homotopy and Related Structures 11 Issue 4 (2016) pp 923–942, doi:10.1007/s40062-016-0160-0, arXiv:1302.7019. - -(there given for topological bigroupoids, but one could repeat the definition mutatis mutandis for Lie 2-groupoids, taking the natural transformations $a,r,l,e,i$ to identities), then the hom-groupoid is a Lie groupoid over the square of the manifold of objects, with some properties, and then composition is an anafunctor between Lie groupoids over this manifold. And so on. This provides and extra layer of weakness to the structure. This viewpoint (in the special case of Lie 2-groups) was also used by Chris Schommer-Pries. One should view this as some kind of internal enrichment of a Lie groupoid in the (cartesian monoidal) category of differentiable stacks, much like ordinary 2-groupoids are groupoids enriched in the category of groupoids.<|endoftext|> -TITLE: Can the place of publication be harmful to one's reputation? -QUESTION [21 upvotes]: What can be said about publishing mathematical papers on e.g. viXra if the motivation is its low barriers and lack of experience with publishing papers and the idea behind it is to build up a reputation, provided the content of the publication suffices that purpose. -Can that way of getting a foot into the door of publishing be recommended or would it be better to resort to polishing door knobs at arXiv to get an endorsement? -Personal experience or that of someone you know would of course also be interesting to me. - -REPLY [29 votes]: Yes, the place of publication can absolutely hurt your reputation. Specifically, I can tell you from having served on many hiring committees (and from conversations with professors at other universities about their hiring committees and tenure processes), that publications in predatory journals can hurt you. I'm talking specifically about journals whose model is to get the author to pay them, and whose peer review standards are a joke. Publications in journals like that can be interpreted as an author trying to side-step the normal process, or unethically inflate their numbers. -It may be hard to break into the absolute top journals with your first few papers (unless you have a famous advisor/coauthor or went to a prestigious school). But there are plenty of good journals around and after a track record of publishing in good journals you will have less difficulty publishing in top journals (of course, it'll always be extremely hard to publish in the Annals and other super elite journals). For people starting out, I recommend at least checking Beall's list of predatory publishers to be sure you don't end up publishing somewhere that might be frowned upon later in your career. Also, don't let fear paralyze you from trying. Lots of editors and referees will go gently on new PhDs. I wish this was even more common, rather than pushing young people out of academia.<|endoftext|> -TITLE: Proofs where higher dimension or cardinality actually enabled much simpler proof? -QUESTION [71 upvotes]: I am very interested in proofs that become shorter and simpler by going to higher dimension in $\mathbb R^n$, or higher cardinality. By "higher" I mean that the proof is using higher dimension or cardinality than the actual theorem. -Specific examples for that: - -The proof of the 2-dimensional Brouwer Fixed Point Theorem given by by Aigner and Ziegler in "Proofs from the BOOK" (based on the Lemma of Sperner). The striking feature is that the main proof argument is set up and run in $\mathbb R^3$, and this 3-dimensional set-up turns the proof particularly short and simple. -The proof about natural number Goodstein sequences that uses ordinal numbers to bound from above. -The proof of the Finite Ramsey Theorem using the Infinite Ramsey Theorem. - -In fact, I would also be interested in an example where the theorem is e.g. about curves, lattice grids, or planar graphs $-$ and where the proof becomes strikingly simple when the object is embedded e.g. in a torus, sphere, or any other manifold. -Are you aware of proofs that use such techniques? - -REPLY [3 votes]: Let $P$ be a convex polytope in $\mathbb{R}^d$ with vertices $v_1,\dots,v_n\in \mathbb{Z}^d$. A nice trick that helps to visualize, understand and prove that the number of lattice points in dilations $tP$ $(t\in\mathbb{N})$ is a polynomial in $t$, called the Ehrhart polynomial of $P$, is to add one dimension a consider the cone over $P$: -$$\mathrm{cone}(P)=\{r_1(v_1,1)+\cdots+r_n(v_n,1)\mid r_1,\cdots,r_n\ge0\}\subset\mathbb{R}^{d+1}.$$ -Then the dilated polytope $tP\subset\mathbb{R}^d$ corresponds to the intersection of $\mathrm{cone}(P)$ with the hyperplane $\{(x_1,\dots,x_{d+1})\in\mathbb{R}^{d+1}\mid x_{d+1}=t\}$. This allows to work with some generating functions associated to polyhedra which simplify for cones.<|endoftext|> -TITLE: Equality in spectral inclusion theorem -QUESTION [5 upvotes]: I asked this question on Math SE but didn't receive any response. -Let $(T_t)$ be a $C_0$-semigroup on a Banach space $X$ with generator $A.$ If $\lambda_0\in \mathbb{C}$ is such that $e^{\lambda_0 t}$ is a pole of $R(\cdot,T(t)),$ then $\lambda_0$ is a pole of $R(\cdot,A)$ and $$k(e^{\lambda_0 t},T(t))\geq k(\lambda_0,A)\tag{1}$$ where $k(\cdot,\cdot)$ denotes the order of the corresponding pole. This is proved for example in Theorem IV.3.6 in the book by Engel and Nagel. -My question: Suppose we a priori know that $\lambda_0$ is a pole of $R(\cdot,A)$ and $e^{\lambda_0 t}$ is a pole of $R(\cdot,T(t))$ for all $t\geq 0.$ Are there any known conditions which guarantee an equality in $(1)$ for at least some $t>0?$ - -REPLY [4 votes]: Here is a positive solution if the semigroup is eventually compact: -Consider, say, the open right halfplane -$$ - H := \{\lambda \in \mathbb{C}: \, \operatorname{Re}\lambda > \operatorname{Re}\lambda_0 - 1\}. -$$ -Then $H$ contains only finitely many spectral values of $A$, and all these spectral values are poles of the resolvent with finite-rank spectral projections. So if we denote the sum of these finitely many spectral projections by $P$, then the range $PX$ is a finite dimensional space, and the restriction of the semigroup to the complementary subspace $\ker P$ only contains spectral values with real part $\le \operatorname{Re}\lambda_0 - 1$. -All these properties follow from Corollary V.3.2 in the semigroup book of Engel and Nagel (2000). -The restriction of the semigroup to $\ker P$ is also eventually compact, so the spectral mapping theorem holds for it [op. cit., Corollary IV.3.12(i)]. Hence, $e^{t\lambda_0}$ is not in the spectrum of $T_t|_{\ker P}$ for any $t$. -So we only need to deal with the restriction of the semigroup to the finite dimensional space $PX$. Hence, we may asume from now on that $X$ itself is finite dimensional and that $A$ is a matrix. After a coordinate transformation, $A$ is in Jordan normal form. The matrix $A$ has distinct eigenvalues $\lambda_0, \dots, \lambda_m$, and for each $k \in \{0, \dots, m\}$ there are Jordan blocks $J_{k,1}, \dots, J_{k,\ell_k}$. -Since $A$ is the direct sum of all these Jordan blocks, we can compute $e^{tA}$ by computing $e^{tJ_{k,i}}$ for all Jordan blocks separately. -Now, if $t \in [0,\infty)$ is such that the numbers $e^{t\lambda_0}, \dots, e^{t\lambda_m}$ are all distinct, then out of all direct summands -$$ - e^{tJ_{0,1}} , \dots, e^{tJ_{0,\ell_0}} , \quad \dots \quad , e^{tJ_{m,1}}, \dots, e^{tJ_{m,\ell_m}} -$$ -of $e^{tA}$, only the matrices $e^{tJ_{0,1}} , \dots, e^{tJ_{0,\ell_0}}$ have the eigenvalue $e^{t\lambda_0}$. Hence, the dimension of largest Jordan block in $e^{tA}$ that belongs to the eigenvalue $e^{t\lambda_0}$ (i.e., the order of the pole $e^{t\lambda_0}$ of $R(\cdot,e^{tA})$) cannot be larger than the largest dimension of the matrices $e^{tJ_{0,1}} , \dots, e^{tJ_{0,\ell_0}}$ - which is the same as the largest dimension of the matrices $J_{0,1}, \dots, J_{0,\ell_0}$ (and this is, in turn, the order the pole $\lambda_0$ of $R(\cdot,A)$). -This solves the question for all those times $t$ for which the numbers $e^{t\lambda_0}, \dots, e^{t\lambda_m}$ are distinct.<|endoftext|> -TITLE: How do you check that your mathematical research topic is original? -QUESTION [35 upvotes]: Sorry if this question is not well-suited here, but I thought research in mathematics can be identified from other science field, so I wanted to ask to mathematicians. -I am just starting graduate study in mathematics (and my bachelor was in other field) so I have no research experience in mathematics. Recently I came up with a problem by myself, and thought it was interesting so devoted some time to draw the results. Now I have some results, but I am not sure this is already studied somewhere by someone. I tried to lookup some possible keywords in Google Scholar, but have nothing. -I am sure for more mature mathematicians they already know their fields of study and recent trends of the research, so it won't be a difficult problem to know this is original or not. But if you come up with some idea that are not seems to belong to any field, how do you know your result is original or not trivial result? -Thank you in advance! - -REPLY [2 votes]: It's not that hard nowadays to do a fairly exhaustive literature search to make sure something hasn't been done before. This is a skill which can take a bit of time to develop but will help and saves a lot of time in the long run, especially if you do not have someone you can show the result to. This will also help you to be independent, something which is starting to become increasingly important in academia (especially in mathematics). -Now with Google, Arxiv and many other tools which people in the past did not have access to it is much easier to check if something has already been done, although these checks are usually not foolproof. After a while you will probably also develop intuition for what is already a 'reasonably obvious' idea. -For example, a few weeks ago it occurred me that it would be nice to construct a quantum mechanical system whose Berry phase was a BPS t'Hooft-Polyakov monopole. It wasn't too hard to do some digging around and find that this had already been done: I encourage you to learn how to do this effectively if you do not already know how. -Edit: See this paper by Tao et al. where they investigate in great detail whether or not a result they have found in linear algebra is new or not and if not, exactly what is the history of the discovery. In Figure 1, they provide a very detailed tree of all the citations and previous mentions/studies of the result. Tao does not work directly in the field of linear algebra and his 3 co-authors are neutrino physicists, so how could they possibly have found this out except by doing some digging and deep searching around? OK, the graph might have been crowdsourced, but I doubt everyone contributing to it was exactly an expert.<|endoftext|> -TITLE: Lattice structure in the root poset -QUESTION [12 upvotes]: Let $W$ be a Coxeter group with simple generators $s_1$, $s_2$, ..., $s_r$. Let $\Phi^+$ be the corresponding positive root system, with $\alpha_i$ the positive root corresponding to $s_i$. Bjorner and Brenti, Combinatorics of Coxeter Groups, Chapter Four define the root poset to be a partial order on $\Phi$ as follows: -If $\beta \in \Phi$ and $s_i \beta - \beta \in \mathbb{R}_{>0} \alpha_i$, then $\beta < s_i \beta$. The root poset is then the transitive closure of this relation. -Bjorner and Brenti, Exercise 4.15 asks: - -Is the positive root poset $(\Phi^+, \leq)$, with a bottom element appended, a meet-semilattice? - -I can't find the answer to this exercise. Can someone help? -What I would actually like to know is: - -Are intervals $[\beta, \gamma]$ in the root poset lattices? - -Below, some bibliographic notes: - -I cheated slightly above: Bjorner and Brenti, as well as the sources below, actually only order $\Phi^+$, not $\Phi$. But I see no reason not to extend the order to the negative roots. -This poset was introduced in Henrik Eriksson's PhD thesis and, independently, by Brink and Howlett: - -Brink, Brigitte; Howlett, Robert B., A finiteness property and an automatic structure for Coxeter groups, Math. Ann. 296, No. 1, 179-190 (1993). ZBL0793.20036. - -This poset is not the same as defining $\beta \leq \gamma$ if $\gamma - \beta$ is in the positive span of the $\alpha_i$; a condition which is also sometimes called the root poset. -Let $\beta$ be a positive root and $t$ the corresponding reflection. Then $s_i \beta - \beta \in \mathbb{R}_{>0} \alpha_i$ if and only if $s_i$ is an inversion of $s_i t s_i$. Thus, we can define this relation in a purely Coxeter theoretic way, without mentioning root systems. - -REPLY [3 votes]: John Stembridge recently pointed me towards his very nice paper Quasi-Minuscule Quotients and Reduced Words for Reflections, which gives a lot of insight into the root poset. -Here is what I understand from his paper. -Let $t$ be a reflection and let $\beta$ be the corresponding root. -Then $\phi: w \mapsto - w \beta$ is a surjective poset map from the weak order interval $[e,t]$ to the interval $[-\beta, \beta]$ in the root poset. -Of course, $[e,t]$ is a lattice, so, if $\phi$ is a poset isomorphism, then $[-\beta,\beta]$ is a lattice as well. -It turns out that $\phi$ is an isomorphism if and only if it is a bijection. Theorem 2.6 in Stembridge gives a necessary and sufficient condition for $\phi$ to be a bijection, but one important thing to note is that this always occurs if the Dynkin diagram is a forest (in particular, in all finite types and all affine types other than $\tilde{A}$). -I'm not sure if anyone but me cares about this old question, but I worked through the first example of a non-lattice in $\tilde{A}_2$ from this perspective. Let $t = (s_1 s_2 s_3)^2 s_1 (s_1 s_2 s_3)^{-2}$. I have drawn the interval $[e,t]$ in the image below: $[e,t]$ are the labels on the triangles inside the diamond region. In the bottom of the diamond, I have labeled each triangle with a reduced word; in the top, the words got too long so I just put black dots. The edges of the Hasse diagram are dual to the triangular tiling; moving down the page is going down in the lattice. - -The root corresponding to $t$ is $\beta:=4 \alpha_1 + 3 \alpha_2 + 3 \alpha_3$. The map from $[e,t]$ to $[-\beta, \beta]$ identifies $(s_1 s_2 s_3)^2$ with $(s_1 s_3 s_2)^2$ (both indicated with light shading) and identifies $(s_1 s_2 s_3)^2 s_1$ with $(s_1 s_3 s_2)^2 s_1$ (dark shading). If we had chosen a larger reflection, we would have gotten a larger diamond $[e,t]$, and then $[-\beta, \beta]$ would be obtained by quotienting this diamond by a one dimensional group of translations so that the poset again had width $3$. -Now, look at the elements $x$ and $y$ (in red). The meet $x \wedge y$ in $\tilde{A}_2$ is $s_1 s_2 s_3 s_1 s_2 s_1$ (marked in black). However, $x$ dominates one of the dark shaded regions and $y$ dominates the other one so, in the quotient $[-\beta, \beta]$, the dark shaded element is a second, incomparable, lower bound for the images of $x$ and $y$.<|endoftext|> -TITLE: Minimal generation of simple groups and Ore's conjecture -QUESTION [6 upvotes]: The well known Ore's conjecture (now established) states that every element of a finite non-abelian simple group $G$ is a commutator of a pair of elements. Also we know that $G$ is $2$-generated. -I am trying to find out what is known about: given any $1 \neq x \in G$, can it be a commutator of two generating elements, i.e., $x = [a,b]$ so that $G$ is generated by $a, b$ as well. -If the answer is negative, are there known restrictions on the conjugacy class of $x$ for which this happens? -The question is motivated from the action of $G$ on Riemann surfaces that yield orbit genus $1$ corresponding to minimal signatures for the group. - -REPLY [2 votes]: Here is a character-theoretic argument which shows that for each $n > 1$, whenever an involution $t \in {\rm SL}(2,2^{n})= G$ has $t = [a,b]$, then $a,b \in B = N_{G}(S)$, where $S$ is the unique Sylow $2$-subgroup of $G$ containing $t$. -In general, when $G$ is a finite group and $x \in G$, then the number of ordered pairs $(a,b) \in G \times G$ with $x = [a,b]$ is expressible as $\sum_{ \chi \in {\rm Irr}(G)} \frac{|G|\chi(x)}{\chi(1)},$ where ${\rm Irr}(G)$ is the set of complex irreducible characters of $G$. This formula was probably known to W. Burnside (in fact, the fact that $x$ is a commutator if and only if the sum is positive appears in Burnside's book, and was later important in the proof of the Ore conjecture). -Letting $T,S, B, G$ be as above, we note that $B$ is a Frobenius group of order $2^{n}(2^{n}-1)$ , and has $2^{n}-1$ irreducible characters of degree $1$ (each with $t$ in their kernel), and one irreducible character of degree $2^{n}-1$ taking value $-1$ at $t$. Hence the number of order pairs $(a,b) \in B \times B$ with $t = [a,b] $ is -$(2^{n}(2^{n}-1) [ (2^{n}-1) - \frac{1}{2^{n}-1}] = 2^{3n}-2^{2n+1}$. -On the other hand, $G$ has one irreducible character of degree $1$, the trivial character, one irreducible character of degree $2^{n}$ (which vanishes at $t$), $2^{n-1}$ irreducible characters of degree $2^{n}-1$, all taking value $-1$ at $t$, and $2^{n-1}-1$ irreducible characters of degree $2^{n}+1$, all taking value $1$ at $t$. -Hence the number of ordered pairs $(a,b) \in G \times G$ with $t = [a,b]$ is -$2^{n}(2^{2n}-1) [ 1 - \frac{2^{n-1}}{(2^{n}-1)} + \frac{2^{n-1}-1}{2^{n}+1}],$ which also turns out to be $2^{3n} - 2^{2n+1}$. -Hence all ordered pairs $(a,b) \in G \times G$ with $t = [a,b]$ actually lie in $B \times B$, so it is not possible to express $t = [a,b]$ where $\langle a,b \rangle = G.$<|endoftext|> -TITLE: Is $[X, \_]$ a homology theory? -QUESTION [7 upvotes]: Let $X$ be a CW-spectrum. It is well-known that $[\_ ,X]$ is a generalized cohomology theory and, by Brown's representability theorem, every generalized theory is $H$ represented by a spectrum (namely, $H$ has this form). -What about $[X, \_]$? Is it a homology theory? (I do not claim every homology is corepresented by a spectrum.) - -REPLY [21 votes]: This holds only for compact objects (i.e. finite CW spectra), since it is easy to see that additivity fails otherwise (the other axioms of homology theories are satisfied). The usual way to obtain a homology theory from a spectrum $X$ is to consider $\pi_*(X\otimes -)$, note that for compact $X$ your $[X,-]$ is of this form as well, since -$$ -[X,-] = \pi_*(DX\otimes -) -$$ -by Spanier-Whitehead duality.<|endoftext|> -TITLE: q-series identity related to Jackson-Slater, proof required -QUESTION [15 upvotes]: The question: -I have been trying to prove the following $q$-series identity for quite some time now: -$$ -\sum_{k \geq 0} \frac{q^{2k^2}}{(q)_{2k}} = \sum_{m,k \geq 0} \frac{q^{m^2 + 3k m + 4k^2}}{(q)_k(q)_m} \left( 1 - q^{2m + 4k +1} \right) \tag{*} -$$ -Where for a non-negative number $k$ we let $(q)_k = (1-q)...(1-q^k)$. I have verified this conjecture with SageMath to order $q^{2000}$. -I am no expert on the topic of $q$-series and partition identities so I have no intuition to decide if this should be an easy consequence of some well known $q$-hypergeometric summation formula or not. I am hoping that if that is the case, some expert here will point me in the right direction or share a proof. -I can prove that for every $n \geq 0$ the coefficient of $q^n$ on the RHS of (*) is bigger or equal to the same coefficient on the LHS, so only the other inequality is required. -Some background: -For a history of this identity please refer to the article we just posted with this conjecture [7]. I will collect below many things I know about this identity. -Here is a reason (besides pride and the list of techniques below that I have tried) that led me to believe that it may not be a trivial identity: -Sums like either of the two terms of the RHS of (*) are known as Nahm sums. In this particular case this is a Nahm sum associated to the matrix $(\begin{smallmatrix}8 & 3 \\ 3 & 2\end{smallmatrix})$. There is a famous conjecture [4] regarding modularity of these sums. This particular matrix is known to have the right asymptotic at $q \rightarrow 1$ for it to be modular. The LHS of (*) is the character of the Ising model vertex algebra which is known to be modular. Zagier studied this particular type of Nahm sums in [5] and did not find a modular sum with only one term and the matrix as in the RHS of (*). What this identity is saying is that by allowing linear combinations of Nahm sums of a fixed matrix that satisfies the right assymptotics, one may find modular sums. -Here are some techniques that I have tried so far: - -The theory of Bailey pairs and Bailey chains. None of the Bailey pairs with parameters found in Sills recent article [1] works. -Polynomial approximations. Most of the RR type identities admit proofs along the lines of finding a polynomial identity that in certain limit gives the wanted identity. There are well known polynomial approximations to the LHS of this identity. For example one can look at the Santos Polynomials [2]. The RHS admits several polynomial approximations, for example one could take -$$ -P_L(q) = \sum_{m=0}^L q^{m^2} \binom{L}{m}_q \sum_{k = 0}^m q^{2k^2 +km} \binom{m}{k}_q -$$ -where the $q$-binomials in this setting is defined as $\binom{m}{n}_q = \frac{(m)_q}{(n)_q(m-n)_q}$. -Two variable generalizations. Most of the identities in Slater's list [3] admit two variable generalizations. In Sills list [1] mentioned above for example, there is a two variable generalization of Slater's identity (39) (which goes by the name in the title of this post), which has the LHS of (*) as one of the members. The RHS of (*) also comes with a natural two-variable generalization (see please [7] for this generating function). However I could not find a natural way of matching this two-variable generalization to any of the known expressions equal to the LHS of (*). -Functional equations: the generating series (or rather its two variable generalization alluded above) satisfies a functional equation of order $11$ so in principle if one were to guess the two variable version of the LHS a computer could decide if it satisfies this functional equation. This equation can be written as a system of $5$ (or even 2) variables and of order $2$ as written in our article. -Combinatorial description: The LHS of (*) counts naturally partitions satisfying some difference conditions due to a theorem of Hirschorn. There are many more families of partitions that are counted by the same generating function, a list below to the equivalent forms of this side should give you an idea. The RHS of (*) also is the generating function of a family of partitions. It arises by studying a Gröbner basis for an ideal sheaf in an affine arc space. As is typically the case in these Gröbner bases computations, the partitions involved are just horrible. It is somewhat of a miracle that we could find the generating series. - -Here are some things that I have not tried: - -The character of the Ising model satisfies an explicit modular differential equation. I do not know how to check that the RHS of (*) satisfies this equation or not. I do not even know how to compute $q \frac{d}{dq}$ of it. -Relations with other similar conjectures: if you look into our paper you will find similar conjectural expressions for the characters of other modules for the Ising vertex algebra. Since these are known to satisfy certain algebraic relations, perhaps these can be exploited to prove all of these identities at once. - -Here are some equivalent forms for both sides of the identity: -The LHS of (*) is very well known and admits the following well known forms: -$$ -\begin{aligned} -% -\sum_{k \geq 0} \frac{q^{2k^2}}{(q)_{2k}} &= \prod_{n=1}^\infty \frac{1}{1-q^n} \sum_{m \in \mathbb{Z}} \left( q^{12m^2+m} - q^{12m^2+7m+1} \right) \\ -% -&= \frac{1}{2} \left( \prod_{m=1}^\infty \left( 1+q^{m-1/2} \right) + \prod_{m=1}^\infty \left( 1-q^{m-1/2} \right) \right) \\ -% -&= \prod_{k=1}^\infty \frac{(1+q^{8k - 5})(1+q^{8k-3})(1-q^{8k})}{1-q^{2k}} \\ -&= \sum_{k = (k_1, k_2, \ldots, k_8) \in \mathbb{Z}^8_{\geq 0}} \frac{q^{k^T C_{E_8}^{-1} k}}{(q)_{k_1} \dots (q)_{k_8}} -\end{aligned} -$$ -This last form is striking, here $C_{E_8}$ is the Cartan matrix for the simple Lie algebra $E_8$. -The RHS can be easily written as -$$ -\sum_{m \geq 0} \frac{q^{m^2}}{(q)_m} \sum_{k =0}^m q^{2k^2 + km} \binom{m}{k}_q (1 - q^k + q^m) -$$ -Perhaps a better approach to try to use the trick of Section 5 of [6], the RHS (*) is also equivalent to -$$ -\sum_{m \geq 0} \frac{q^{m^2}}{(q)_m} \left( 1 - q^{2m +1} \right) \sum_{0 \leq 2k \leq m} \frac{(q^{-m}, q^{1-m}; q^2)_k}{(q;q)_k} q^{(m+1)k} -$$ -Finally, to show a not-so-straightforward equivalent form, by using Ismail-Garret-Statton generalization of the Rogers-Ramanujan identity one can sum first $m$ in (*) to obtain the equivalent form: -$$ -\sum_{k \geq 0} (-1)^k \frac{q^{\frac{k(k+1)}{2}}}{(q)_k} \left( \frac{q^{2k}a_{3k} - a_{3k+2}}{(q,q^4;q^5)_{\infty}} - \frac{q^{2k} b_{3k} -b_{3k+2}}{(q^2,q^3;q^5)_{\infty}}\right) -$$ -where $a_k$, $b_k$ are the Schur polynomials defined by the recursion $x_{k+2} = x_{k+1} + q^k x_k$ and the initial conditions -$$ -a_0 = b_1 = 1, \qquad a_1 = b_0 = 0. -$$ -References: -[1] A. Sills Identities of the Rogers-Ramanujan-Slater Type. International Journal of Number Theory. Vol. 03, No. 02, pp. 293-323 (2007) ] -[2] G. E. Andrews and J. P. Santos. Rogers-Ramanujan type identities for partitions with attached odd parts. The Ramanujan Journal, 1:91–99, 1997. -[3] L. J. Slater. Further identities of the rogers-ramanujan type. Proc. London Math. Soc., 54(2):147–167, 1952. -[4] W. Nahm. Conformal field theory and torsion elements of the Bloch group. In Frontiers in number theory, physics, and geometry. II, pages 67–132. Springer, Berlin, 2007. -[5] D. Zagier. The dilogarithm function. In Frontiers in number theory, physics, and geometry. II, pages 3–65. Springer, Berlin, 2007. -[6] G. Andrews, K. Bringmann, K. Mahlburg Double series representations for Schur's partition function and related identities Journal of Combinatorial Theory, Series A Volume 132, May 2015, Pages 102-119 -[7] J. van Ekeren, R. Heluani The singular support of the Ising model - -REPLY [6 votes]: The following proof is due to George E. Andrews (who is now a coauthor on the mentioned paper). -Consider the following families of polynomials indexed by $n$ -$$ -S_n = \sum_{k \geq 0} q^{2k^2} \binom{n-k}{2k}_q -$$ -$$ -T_n = \sum_{m,k\geq 0} q^{m^2+3k m+4k^2} \left( \binom{n-3k-m}{k}_q -\binom{n-4k-m}{m}_q - q^k \binom{n-3k-m-1}{k}_q\binom{n-4k-m-1}{m-1}_q \right). -$$ -The limit as $n \rightarrow \infty$ of $S_n$ is the LHS of the desired identity while the limit as $n \rightarrow \infty$ of $T_n$ is the RHS of the desired identity. -For every $n \geq 0$ one can check that $S_n = T_n$. We used the Mathematica package qMultiSum which was provided by RISC to check that both sides satisfy the order 8 recursion -$$ -q^{4n+15}S_n + q^{2n+11}(1+q)(S_{n+3} - S_{n+4}) - q^3 S_{n+5} + -(1+q+q^2)(qS_{n+6} - S_{n+7}) + S_{n+8} = 0. -$$ -In fact the same technique works for the three conjectures of our paper so now we have a PBW basis for the Ising model and three new quasi-particle sum expressions for the three characters of the minimal model at central charge $c=1/2$ -I'll leave this question open since it still uses a computer to check the recursion. Perhaps someone can find a proof by pen and paper.<|endoftext|> -TITLE: Whitney sum formula for topological Pontryagin classes -QUESTION [5 upvotes]: Is there a Whitney sum formula for topological rational Pontryagin classes? I thought the answer is yes, but now I cannot find a reference. Is it even true? The PL case would also be of interest. - -REPLY [8 votes]: Yes. A simple argument is that $BO \to BTOP$ is a rational equivalence and an H-space map (in fact even an infinite loop map), so it follows from the Whitney sum formula for vector bundles. -Edit: The argument for this is as follows. Let $\mu : BTOP \times BTOP \to BTOP$ be the map corresponding to Whitney sum of (stable, topological) bundles. The question is whether the identity -$$\mu^* p_n = \sum_{ a + b = n} p_a \otimes p_b$$ -holds. As the map $BO \to BTOP$ is a rational equivalence and an H-space map, it is equivalent to verify this equation in the cohomology of $BO$ instead, where it indeed holds.<|endoftext|> -TITLE: Rings $R$ such that every [regular] square matrix with entries in $R$ is equivalent to an upper triangular matrix -QUESTION [6 upvotes]: Let $\text{M}_n(R)$ be the ring of $n$-by-$n$ matrices with entries in a commutative unital ring $R$. Theorem III in - -C.R. Yohe, Triangular and Diagonal Forms for Matrices over Commutative Noetherian Rings, J. Algebra 6 (1967), 335-368 - -provides a characterization of the Noetherian rings $R$ with the property that every matrix in $\text{M}_n(R)$ is equivalent to a diagonal matrix: It turns out that this is the case if and only if $R$ is a direct sum of PIDs and completely primary PIRs, where ``completely primary'' means a local ring with a nilpotent maximal ideal. With this in mind, here are my questions: - -(1) Is there any analogous characterization for the rings $R$ with the property that every matrix in $\text{M}_n(R)$ is equivalent to an upper triangular matrix? (2) Does the property hold for any choice of $R$? (3) And if the answer to the previous question is no, what if we restrict attention to the regular matrices in $\text{M}_n(R)$? - -[EDIT] The answer to questions (2) and (3) is in the negative, as shown by Mohan in their answer, and that's good to know. On the other hand, I'm really hoping for someone to come up with a reference where it's actually proved that, if $R$ is taken from a class of commutative rings that is sufficiently interesting and sufficiently larger than direct sums of PIDs and completely primary PIRs, then every regular square matrix with entries in is equivalent to an upper triangular matrix. This would not be a characterization in the vein of Yohe's theorem, but still... [END OF EDIT] -Every matrix $A \in \text{M}_n(R)$ can be brought to upper triangular form by elementary row transformations; that is, there exist elementary matrices $E_1, \ldots, E_k \in \text{M}_n(R)$ such that $E_1 \cdots E_k A$ is an upper triangular matrix. But the elementary matrix corresponding to a row-multiplying transformation need not be invertible in $\text{M}_n(R)$; although it is definitely invertible in $\text{M}_n(\mathcal Q(R))$ when $A$ is regular, with $\mathcal Q(R)$ being the total ring of fractions of $R$. Unfortunatley, I don't see how this helps answering any of my questions (I'm especially interested in the last one). -Glossary. By a ``regular matrix'', I mean a regular element in the multiplicative monoid of $\text{M}_n(R)$; or equivalently, a matrix $A \in \text{M}_n(R)$ whose determinant is a regular element of $R$. An element $a$ in a (multiplicatively written) monoid $H$ is regular (or cancellable) if the functions $H \to H: x \mapsto ax$ and $H \to H: x \mapsto xa$ are both injective. - -REPLY [3 votes]: As Luc Guyot mentioned, check out Kaplansky's paper Elementary Divisors and Modules from 1949. -Kaplansky calls a ring Hermite when every $1 \times 2$ matrix is equivalent to a diagonal matrix, and shows that equivalently a ring is Hermite iff for every matrix $M$ there exists an invertible matrix $U$ such that $MU$ is upper triangular. -To get a finite product decomposition result like Yohe's, we obviously need to assume that $R$ has finitely many minimal primes. -On the other hand, we have the following - -Theorem If $R$ is a Bézout ring with finitely many minimal primes then there is a finite set of idempotents $e_i$ such that the ring $e_iR$ is a Hermite ring with a unique minimal prime and $R \cong \prod e_iR$. Hence $R$ is Hermite. - -This is Theorem 2.2 in Elementary Divisor Rings and Finitely Presented Modules by Larsen, Lewis, and Shores. -Maybe we can say more about the structure of these summands. For example, it is easy to show that in any Bézout ring with a unique minimal prime $P$, the ideal $P$ is essential unless $R$ is a domain. Indeed, if $I \cap P = 0$, then $\operatorname{Ann}(a) = P$ for any $a \in I$ and clearly any $a \in I$ is nonzero in every localization of $R$. Since $R$ locally has totally ordered ideals, this implies that $P$ is locally $0$, hence $0$, i.e. $R$ is a domain. -So to sum up - -Conclusion Let $R$ be a ring with finitely many minimal prime ideals. Then the following are equivalent: -$\ \ (1)$ $R$ is a finite direct product of Bézout rings each of which is either a domain or has a unique minimal prime ideal which is essential. -$\ \ (2)$ Matrices over $R$ are equivalent to triangular matrices. - -I'm not sure if you can say very much in general about the structure of Bézout rings with essential unique minimal prime ideal. But, here's one observation: A Bézout ring with a unique minimal prime ideal $P$ has the property that every non-nilpotent element divides every nilpotent element. -Indeed, let $b$ be nilpotent and let $a$ be not nilpotent, i.e. $b \in P$ and $a \notin P$. By Bézoutness, pick $c,d,u,v,r$ such that $ac + bd = r$, $ru = a, rv = b$. Deduce that $r \notin P, v \in P$. It follows that $cv + du -1 \in P$, and we deduce that $du$ is a unit. Therefore $a \mid b$. -Note that if $R$ is additionally Noetherian, we can easily deduce that $P$ is the unique prime ideal of $R$, which exactly recovers Yohe's result. To see this, note first that it suffices to assume $R$ is local with maximal ideal $M$, in which case it has totally ordered ideals and $\bigcap_n M^n = 0$, which implies $M^n \subseteq P$ for some $n$ and thus $M \subseteq P$. Without the Noetherian hypothesis, $R$ can have infinite Krull dimension. -As for your 3rd question, I don't have much to say off the top of my head except that it seems really hard. From your own observations, any ring which is its own total ring of fractions (i.e. regular elements are units) will have this property, and that is quite a broad class of rings, members of which sometimes have ostensibly very little in common.<|endoftext|> -TITLE: Mean curvature flow and knot theory -QUESTION [8 upvotes]: I am wondering if the mean curvature flow of one-dimensional submanifolds of $\mathbb{R}^3$ is understood well enough to give some perspective on (and hopefully a proof of) something like the Fary-Milnor theorem. -For reference, the Fenchel theorem (1929) says that if $c:S^1\to\mathbb{R}^3$ is a smooth embedding, then the total curvature is at least $2\pi$. The Fary-Milnor theorem (1949/50) says that if $c$ forms a nontrivial knot, then the total curvature is at least $4\pi$. -Steven Altschuler ("Singularities of the curve shrinking flow for space curves", JDG 1991) showed that if $c_t:S^1\to\mathbb{R}^3$ is a one-parameter family of smooth embeddings satisfying the mean curvature flow, then -$$\frac{d}{dt}\int_{S^1}|\kappa_t|\leq -\int_{S^1}\tau_t^2|\kappa_t|$$ -where $\tau_t$ is the torsion of $c_t$. So if $\int_{S^1}|\kappa_0|<4\pi$ then certainly $\int_{S^1}|\kappa_t|<4\pi$ for all positive $t$. So it seems like one could hope for a proof of the Fary-Milnor theorem which is more or less directly analogous to Hamilton-Perelman's proof of the Poincaré conjecture or of the topological classification of closed 3-manifolds with nonnegative and positive scalar curvature. -The problem would rest on understanding the singularities of the mean curvature flow. Altschuler seems to have shown that a singularity of the mean curvature flow in this setting is characterized by blowup of the curvature (just like Ricci flow), and that a nontrivial tangent flow is given either by an Abresch-Langer solution or the grim reaper solution. This seems to be directly parallel to the Perelman-Brendle theorem which says that the analogous blowup limit of a finite-time singularity of a Ricci flow on a compact 3-manifold is either a quotient of shrinking round cylinders or the Bryant soliton, which was (in a weaker version) Perelman's breakthrough result. -So it seems like the key ingredients are there. Can they be put together? It seems like the basic problem is that I don't know what the "surgery" analogue might look like or how it might be relevant. -So, more generally, forgetting about Fary-Milnor theorem in particular, could one hope to use mean curvature flow for any sort of application in knot theory? Perhaps the proper analogue of the Hamilton-Perelman approach would decompose a given knot as a connected sum of prime knots, and would give some canonical representation thereof? This seems to be comparable to the geometrization conjecture, although the existence and uniqueness of such a decomposition seems to be already known in knot theory. -In order that I have a reasonably concrete question: - -Ricci flow with surgery on 3D compact manifolds is closely analogous to mean curvature flow of mean-convex surfaces in $\mathbb{R}^3$ (Brendle-Huisken). Is there an analogy, or a conjectural analogy, to mean curvature flow of curves in $\mathbb{R}^3$, or in 3-manifolds? Are there (conjectural) applications in knot theory? - -I couldn't find any references on the web. I am aware of Perelman's use of mean curvature of curves in a Ricci flow background to show finite extinction time of Ricci flow on simply-connected 3-manifolds, but the only detailed version which I know to exist of this, Morgan-Tian's book, seems to have some basic errors (cf. Bahri "Five gaps in mathematics" and some followups on the arxiv) - -REPLY [4 votes]: This is really a comment that's too long to fit in 500 characters, but let me try to explain the obstacles I mentioned in the comments. Sorry for how large the pictures are. I'll edit this if I figure out how to resize them. -Curve shortening flow three dimensions can self- intersect, which is the main roadblock. However, there are a few other issues that seem challenging, even if we can overcome this first problem. I'll try to give some pictures (all of which are from Wikipedia) to demonstrate the challenge of understand the limits of spatial CSF from initial data. -Suppose we start with a knot and we somehow know a priori that it does not self intersect under CSF. The question is whether the limiting curve can be used as some sort of model for that knot. The issue we immediately run into is the result of Altschuler you mentioned; the limiting curves are planar, so cannot be knotted. To get around this, one might imagine that the solution is to consider the limiting curve along with some sort of crossing diagram. In some cases, this might work, but I suspect that most of the time it does not.  -For instance, if we start with a trefoil knot, it might be the case that the limit is what happens on the left side of the following picture by Au [1], where the circle is covered twice. This is a trefoil with minimal total curvature, but it might not be what we hoped for. It's not the Abresch-Langer solution that genuinely looks like a flattened trefoil, but this is pretty much the best case scenario otherwise. Even here, I don't think this is what happens generically. - -For a less ideal example, suppose we start with a figure-8 knot with rotational symmetry, as shown below. - - This also curvature at least curvature at least $2 \pi$. However, here the "minimal diagram" looks instead like this (rotated by 90 degrees). - -This is not going to be the limit of CSF, which tells us that for this initial condition, either CSF has a local singularity or else becomes unknotted. It seems to be the latter, but a proof would take some careful analysis. -Somewhat related to all of this is that we can't easily rule out Type 2 singularities, which are essentially local kinks. In two dimensions, surgery might be useful for local singularities because they only happen when the curve crosses itself. If you cut around an intersection to make two curves, you expect to be able to continue the flow. For spatial curves, it's much less clear when Type 2 singularities emerge. Perhaps there is a version of Grayson's theorem that gives sufficient conditions for an initially embedded curve to develop a Type 1 singularity, but I'm not aware of it. There is a version if the curve is contained on the surface of a sphere, but that's not directly relevant here.  -Au, Thomas Kwok-Keung, On the saddle point property of Abresch-Langer curves under the curve shortening flow, Commun. Anal. Geom. 18, No. 1, 1-21 (2010). ZBL1217.53067.<|endoftext|> -TITLE: completeness of $\mathcal M(\Omega)$ without any topological assumptions? -QUESTION [5 upvotes]: Let $(\Omega,\Sigma)$ be a measurable space (no reference measure is chosen!), and $V$ a finite-dimensional normed vector space. -Note carefully that I am not choosing any topology on $\Omega$, so the $\sigma$-algebra $\Sigma$ is a priori not induced by any Borel structure whatsoever. -The total variation $|\mu|$ of a $V$-valued measure is defined as -$$ -|\mu|(E)=\sup\limits_{\pi}\sum\limits_{E_i\in E}|\mu(E_i)|, -$$ -where the supremum is taken among all possible disjoint partitions $\pi=\cup E_i$ of the measurable set $E\in\Sigma$. (the set-function $|\mu|$ is always a positive measure, see [Rudin, real and complex analysis]). For an arbitrary measure we denote -$$ -\|\mu\|:=|\mu|(\Omega). -$$ -We denote $\mathcal M(\Omega)$ the set of all measures with finite total variation $\|\mu\|<\infty$, and $\mathcal M(\Omega)$ is therefore a normed vector space with the above total variation norm. - -Question: is $(\mathcal M(\Omega), \|\cdot\|)$ automatically a Banach space? - -When $\Sigma$ is the Borel algebra then this is true of course, because we can identify $\mathcal M(\Omega)$ with the topological dual $C_b(\Omega;V^\ast)^\ast $ and the dual of a Complete vector space is automatically complete (and in fact $\|\mu\|=\sup\limits_\phi \int \phi(x)\cdot d\mu(x)$ with $\cdot$ denoting the finite-dimensional $V,V^\ast$ pairing). However, I've never seen the statement written in this full generality, so I'm wondering whether this is actually true or not? - -REPLY [6 votes]: Indeed $(\cal{M}(\Omega),\|\cdot\|)$ is a Banach space. For $V = \mathbb{R}$ or $V = \mathbb{C}$ you can find this result in Dunford/Schwartz (1957), Linear Operators I, ch. III.7.4, in particular p. 161. For arbitrary Banach space $V$ this also holds true, but with a sligthly different norm (see p. 160). For finite dimensional $V$ this norm is equivalent to your norm. Proof as in Lemma III.1.5 of Dunford-Schwartz.<|endoftext|> -TITLE: An extension of symplectomorphism group -QUESTION [6 upvotes]: $\DeclareMathOperator\GL{GL}\DeclareMathOperator\Sp{Sp}$Let $\omega=\sum dx_i\wedge dy_i$ be the standard symplectic structure of $\mathbb{R}^{2n}=\mathbb{R}^{n}\times \mathbb{R}^n$. -We consider the following two -extensions of $\Sp(2n,\mathbb{R})$, the group of linear isomorphisms of $\mathbb{R}^{2n}$ preserving $\omega$: -1) Let $G$ be the group of all $A\in \GL(2n,\mathbb{R})$ which map all isotropic subspaces to isotropic subspaces. (A closed subgroup of the general linear group containing the symplectic group $\mathrm{Sp}(2n,\mathbb{R})$.) -2) Let $H$ be the group of all elements in $\GL(2n,\mathbb{R})$ which map all symplectic subspaces to symplectic subspaces. -A non-linear construction as above can be introduced on a symplectic manifold $M$: $G(M)$ is the group of all diffeomorphisms of $M$ whose derivatives map isotropic subspaces to isotropic subspace. And similarly $H(M)$ is the group of all diffeomorphisms whose linear parts map symplectic subspaces to symplectic subspaces. - -Is there a terminology for Lie groups $G$, $H$, $\bar{H}$ and their Lie algebras and also the corresponding structures on symplectic manifolds? Is there any relation between $G$ and $\bar H$? What can be said about their first fundamental groups, as it is customary to compute the first fundamental groups of classical Lie groups? -Inspired by the concept of symplectic vector fields as Lie algebra of symplectomorphism group, what can be said about the Lie algebra of $G(M)$ and some dynamical interpretations on a symplectic manifold? - -REPLY [4 votes]: Short answer: $G = H$ is the group of conformal symplectic linear maps. What follows is a proof of this (which I've simplified slightly from what I originally wrote): - -1) $G$ is the group of conformal symplectic linear maps -First, some notation. We write $\{e_i\} \cup \{f_j\}$ for the standard basis for $\mathbb{R}^{2n}$, i.e. with $\omega(e_i,e_j) = 0$, $\omega(f_i,f_j) = 0$, and $\omega(e_i,f_j) = \delta_{i,j}$. -Suppose $\phi \in G$. We begin by singling out the isotropic subspaces $A = \mathbb{R}^n \times \{0\}$ and $B = \{0\} \times \mathbb{R}^n$. The following lemma is one way you might think about the group $\mathrm{Sp}(2n,\mathbb{R})$ in the first place: as being defined by its action on an isotropic subspace and a choice of isotropic complement. -Lemma: Given $\phi$, there is a unique element $\psi \in \mathrm{Sp}(2n,\mathbb{R})$ such that the following three properties hold: - -$A = (\psi \circ \phi)(A)$, with $(\psi \circ \phi){\big|}_{A} = \mathrm{id} \colon A \rightarrow A$ -$B = (\psi \circ \phi)(B)$ - -Given this lemma, we have that $\psi \circ \phi = \begin{pmatrix}\mathrm{id} & 0 \\ 0 & T\end{pmatrix}$, and it suffices to find conditions on $T \in \mathrm{GL}(n,\mathbb{R})$ so that isotropic subspaces are preserved by $\psi \circ \phi$. It suffices to consider just two cases: - -For $i \neq j$, the span of $e_i$ and $f_j$ is isotropic, so its image under $(\psi \circ \phi)$, the span of $e_i$ and $Tf_j$, must also be isotropic. This proves that $T$ is diagonal. -Also for $i \neq j$, we can use the subspace spanned by $e_i + f_j$ and $e_j + f_i$, which will give us the condition that the diagonal entries of $T$ are all equal. Hence $T = c \cdot \mathrm{id}$ for some constant $c \neq 0$. - -Finally, we see that with $T = c \cdot \mathrm{id}$, $(\psi \circ \phi)^*\omega = c \cdot \omega$, which clearly preserves isotropic subspaces. So this property completely characterizes $G$. -In terms of nomenclature, $G$ might be called the group of conformally symplectic linear operators, and the nonlinear theory would fall under the umbrella of (locally) conformal symplectic geometry, which is slightly more general than what you ask (i.e. $\omega$ itself does not need to be symplectic, only (locally) conformally symplectic, which is the more natural setting). I know only a little about the theory, and am far from an expert on the history, so I'll say nothing more that there exists literature on the subject. -As for computations of classical invariants, we see that $G \cong \mathbb{R}^* \times \mathrm{Sp}(2n,\mathbb{R})$, so there's nothing interesting to say about computing its standard invariants that can't already be said for $\mathrm{Sp}(2n,\mathbb{R})$. - -2) $G=H$ -Suppose $\omega(v,w) \neq 0$ and $\omega(v,x) = 0$. Then $\omega(v,w+tx) \neq 0$ for all $t \in \mathbb{R}$, so $v$ and $w+tx$ always span a symplectic subspace. Hence, if $\phi \in H$, since $\phi$ preserves symplectic subspaces, we require $$0 \neq \omega(\phi(v),\phi(w+tx)) = \omega(\phi(v),\phi(w)) + t\omega(\phi(v),\phi(x))$$ for all $t \in \mathbb{R}$. Hence, we must have $\omega(\phi(v),\phi(x)) = 0$ whenever $\omega(v,x) = 0$. In particular, $\omega$ preserves isotropic subspaces, so $H \subseteq G$. Meanwhile, every conformal symplectic linear map certainly preserves symplectic subspaces, so $G \subseteq H$. Hence $G=H$. -Remark: Even though the limit of symplectic subspaces may not be symplectic, as in the comments, this is not enough to show $H$ is closed, since if a sequence of elements of $H$ realizes this collapsing of a symplectic subspace onto a non-symplectic one, it can (and must by what I've just written) limit to a singular matrix.<|endoftext|> -TITLE: Reference for homotopy colimit = total complex -QUESTION [6 upvotes]: I'm looking for a reference for the following fact: -take a simplicial chain complex $ X:\Delta^{op}\to Ch_{\geq 0}(\mathcal A)$ for $\mathcal A$ a nice abelian category (say, cocomplete with enough projectives, although I'm willing to add more hypotheses, because the $\mathcal A$ I want to use it for is the category of connective modules over some connective dga ); then a model for the homotopy colimit of $X$ is the total complex of the bicomplex associated to $X$. -I know a proof (I haven't checked the details so I'm not sure it works for an arbitrary dga - it works at least for discrete rings), so that's not what I'm looking for (except if you have an especially short and elegant one, then it wouldn't hurt to see it); I'm mostly looking for a reference. -I know the result is mentioned in Dugger's A primer on homotopy colimits (proposition 19.9) but there seems to be no proof in there - so I'll add the criterion that the reference should contain a proof. -This may be related to this question, which relates the total complex and the diagonal - since there is an answer with a reference there, it would also suffice to provide a reference for the fact that the diagonal is a model for the homotopy colimit (actually, this would be enough for other reasons : one can use the diagonal model for simplicial objects that land in $\mathcal A$, and then use homotopy cofinality of $\Delta^{op}\to \Delta^{op}\times \Delta^{op}$ to get the result for an arbitrary simplicial chain complex). -For the latter, I know references for simplicial sets, but not for simplicial $\mathcal A$-objects (and in the case of a discrete ring, one may use this as well via the usual adjunction). -The answers given here seem to be unsatisfactory given the comments below. -Here, the question itself provides a sketch of proof for $\mathbb Z$ which I think can be adapted to the general case, but the adjunction that is mentioned does not seem crystal clear to me (if you could explain it, that would also be great) and it's not a reference. - -REPLY [2 votes]: See Problem 4.23 and Problem 4.24 (with proofs) of Ulrich Bunke's Differential cohomology. -The underlying abstract machinery for computing homotopy (co)limits -via homotopy (co)ends is presented by -Sergey Arkhipov and Sebastian Ørsted -in Homotopy (co)limits via homotopy (co)ends in general combinatorial model categories.<|endoftext|> -TITLE: Pairs of integers whose product is one more or less than a prime -QUESTION [5 upvotes]: Given a positive integer N it is often possible to pair each of the integers 1, 2, 3, ..., N with a different integer between N + 1 and 2 N so that the product of each pair is one less or more than a prime. -For example, if N = 10, such a pairing is (1,12), (2,11), (3,14), (4,13), (5,16), (6,15), (7,18) (8,17), (9, 20), and (10, 19). -Is this possible for infinitely many N? For all, but a finite number of N? - -REPLY [2 votes]: Note that any such pairing gives products greater than $N$ and at most $2N^2$, and thus all the products must be even or the number 3. So with one exception, odd numbers at most N are paired with even numbers greater than N, and for N greater than 2 odd numbers greater than N are paired with even numbers less than N. So what pairings do not lead to products of the desired form? -Let r be even with r not adjacent to a prime. The smallest r start with 26, then 34,50,56,64,76,86,92,94,116,118,120,122,124,134. A paring is a factorization of one of these special even numbers (p,q) with the product being r and one of the factors being odd. Skipping the case that the odd number is 1, we have in most cases these numbers have one odd factor, with the exceptions in the previous list being 50,64, and 120. In this small sample, we have on average little more than one forbidden edge for each candidate. This supports the notion that for small N one can always find a mapping. -Also, not every disallowed edge needs to be considered. If 210 were a disallowed product (it isn't) then (2,105) would be a problem edge only for N in (52,105), and (14,15) only for N=14, so not all seven possibilities (excluding (1,210) are in play for any given N. So the number of bad edges blocking a matching appears to be small. Again ignoring 1, these edges are (5,10),(2,13),(2,17),(2,25),(7,8),(4,19),(4,23)(2,47), on up. This means one problem edge for N=5, 3 for N=7, 2 for N=8, 3 for N=9, and so on. Given that when N=10 there are fifty edges with five coming from each vertex, having three problem edges seems like nothing. (If we include 1, there are more problem edges, but there are also many edges coming from 1 on the order of 2N/ln(2N) many.) -It seems reasonable to estimate (and challenging to prove, which I won't do now) that for every N, that the product graph (which associates to each k less than n those j in (N,2N] for which kj is adjacent to a prime) for N described above has each node with a degree about N/(2ln N). This does not prove there is a matching, but I think it makes it very likely, especially since the (set of) vertices connected to a j less than N will often be quite different to (set of) those connected to a k less than N. Unfortunately, we need something like Linnik's theorem to show no isolated vertices in this graph. -Gerhard "Not Strengthening Linnik's Theorem Yet" Paseman, 2020.05.24.<|endoftext|> -TITLE: Exponential decay of voltage potential difference -QUESTION [11 upvotes]: Consider the following adjacency matrix of a complete graph: -$$A=(e^{-|i-j|})_{1\leq i\neq j\leq n}$$ -with 0 on the diagonal. Let $D=diag\{d_1,...,d_n\}$ be the degree matrix where $d_i=\sum_{j\neq i}e^{-|i-j|}$. Then $L=D-A$ is the Laplacian. Let $L^\dagger$ be the Moore-Penrose inverse of the Laplacian. I'm interested in the following quantity -$$a_{ij}=|(e_1-e_2)^TL^\dagger(e_i-e_j)|$$ -where $e_i=(0,0...,0,1,0,...0)$ with 1 on the ith coordinate. I conjecture that $a_{ij}$ will decay exponentially when both $i$ and $j$ moves away from 1 and 2. Something like $a_{ij}\leq C _1e^{-C_2\min\{i,j\}}$ where $C_1,C_2$ are some constants. From the physics point of view, $a_{ij}$ is the voltage potential difference between $i$ and $j$. It is intuitive that when they are far away from the source, 1 and 2, they should be very small given the structure of the graph. -In fact, my simulation shows that as long as $i,j\neq1,2$, $a_{ij}$ suddenly becomes extremely close to 0. There seems to be no decay, but an acute cut. This phenomenon holds for slight perturbation of $A$, keeping the decaying property. -Is this conjecture true? How can we prove it? What is the rate of decay? -Another quantity that is also interesting is -$$\sum_{i\neq j}e^{-|i-j|}a_{ij}$$ -which is the weighted average of potential differences. How can we bound this? For this quantity, I conjecture it is bounded by some constant instead of growing with $n$. The physical meaning of this quantity is the sum of all currents in each edge. -(Update) -Enlightened by the discussion with @Abdelmalek Abdesselam below. We have the Neuman series representation: -$$a_{ij}=|(e_1-e_2)^TD^{-1/2}\sum_{k\geq0}\left(D^{-1/2}AD^{-1/2}-\alpha D^{1/2}JD^{1/2}\right)^kD^{-1/2}(e_i-e_j)|$$ -where $J$ is the matrix of all 1s and $\alpha$ is some constant to be chosen. We want to choose $\alpha$ such that the power of the matrix decays fast. How can we achieve this and bound the entries of $D^{-1/2}AD^{-1/2}-\alpha D^{1/2}JD^{1/2}$? A possible choice is $\alpha=1/tr(D)$. - -REPLY [3 votes]: Edit: This turns out to be quite simple. Observe that $a_{1i} / a_{2i} = q$ does not depend on $i \in \{3, 4, \ldots, n\}$. Thus, if $x_1 = 1$, $x_2 = -q$ and $x_i = 0$ for $i \in \{3, 4, \ldots, n\}$, then we clearly have $L x = c e_1 - c e_2$, where $c = \sum_{i=3}^n a_{1i}$. It follows that $$L^\dagger (e_1 - e_2) = c^{-1} x + \operatorname{const}.$$ -I leave the previous version of this answer below, as it provides a way to evaluate $L^\dagger$ explicitly. - - -I cannot say I understand what is really going on here, but at least I have a proof that $a_{ij} = 0$ when $i, j \ge 3$. (I leave my previous comment/answer, as it contains some related stuff that is not included here.) - -Notation: Every sum is a sum over $\{1, 2, \ldots, n\}$. We write $q = e^{-1}$ (and actually any $q \in (0, 1)$ will work). Given a vector $x = (x_i)$ we write -$$ \Delta x_i = x_{i+1} + x_{i-1} - 2 x_i $$ -if $1 < i < n$. - -Given a vector $(x_i)$, we have -$$ Lx_i = \sum_j q^{|i-j|} (x_i - x_j) = b_i x_i - \sum_j q^{|i-j|} x_j ,$$ -where -$$ b_i = \sum_j q^{|i-j|} = \frac{1 + q - q^i - q^{n+1-i}}{1 - q} $$ -Therefore, when $1 < i < n$, we have -$$ \begin{aligned} - \Delta Lx_i & = \Delta (b x)_i - \sum_j (q^{|i-j+1|}+q^{|i-j-1|}-2q^{|i-j|}) x_j \\ - & = \Delta (b x)_i - (q + q^{-1} - 2) \sum_j q^{|i-j|} x_j + (q^{-1} - q) x_i \\ - & = \Delta (b x)_i + (q + q^{-1} - 2) L x_i - (q + q^{-1} - 2) b_i x_i + (q^{-1} - q) x_i \\ - & = (q + q^{-1} - 2) L x_i + b_{i+1} x_{i+1} + b_{i-1} x_{i-1} - ((q + q^{-1}) b_i - (q^{-1} - q)) x_i . -\end{aligned} $$ -A short calculation shows that -$$ b_{i+1} + b_{i-1} = ((q + q^{-1}) b_i - (q^{-1} - q)) $$ -(which looks somewhat miraculous, but there must be some insightful explanation for that). Thus, -$$ \Delta Lx_i = (q + q^{-1} - 2) L x_i + b_{i+1} (x_{i+1} - x_i) + b_{i-1} (x_{i-1} - x_i) . $$ -Suppose that $x_i = L^\dagger y_i$ for some vector $(y_i)$ such that $\sum_i y_i = 0$. Then $L x_i = L L^\dagger y_i = y_i$. Write $c = q + q^{-1} - 2$. We then have -$$ \Delta y_i - c y_i = b_{i+1} (x_{i+1} - x_i) + b_{i-1} (x_{i-1} - x_i) . $$ -In particular, the following claim follows. - -Proposition 1: If $1 < i < n$, $y_{i-1} = y_i = y_{i+1} = 0$ and $x_i = x_{i+1}$, then $x_{i-1} = x_i$. - -The above result will serve as an induction step. To initiate the induction, we need to study the $i = n$, which is slightly different. In that case: -$$ \begin{aligned} - Lx_{n-1} - Lx_n & = b_{n-1} x_{n-1} - b_n x_n - \sum_j (q^{|n-j-1|}-q^{|n-j|}) x_j \\ - & = b_{n-1} x_{n-1} - b_n x_n - (q^{-1} - 1) \sum_j q^{|n-j|} x_j + (q^{-1} - q) x_n \\ - & = b_{n-1} x_{n-1} - b_n x_n + (q^{-1} - 1) L x_n - (q^{-1} - 1) b_n x_n + (q^{-1} - q) x_n \\ - & = (q^{-1} - 1) L x_n + b_{n-1} x_{n-1} - (q^{-1} b_n - (q^{-1} - q)) x_n . -\end{aligned} $$ -This time we have -$$ q^{-1} b_n - (q^{-1} - q) = b_{n-1} , $$ -and hence -$$ Lx_{n-1} - Lx_n = (q^{-1} - 1) L x_n + b_{n-1} (x_{n-1} - x_n) . $$ -Again we consider $x_i = L^\dagger y_i$ for some vector $(y_i)$ such that $\sum_i y_i = 0$, and we write $d = q^{-1} - 1$. We then have -$$ (y_{n-1} - y_n) - d y_n = b_{n-1} (x_{n-1} - x_n) . $$ -As a consequence, we have the following result. - -Proposition 2: If $y_{n-1} = y_n = 0$, then $x_{n-1} = x_n$. - -For $y = e_1 - e_2 = (1, -1, 0, 0, \ldots)$, we immediately obtain the desired result. - -Corollary: If $y = e_1 - e_2$ and $x = L^\dagger y$, then $x_3 = x_4 = x_5 = \ldots = x_n$. Consequently, $a_{ij} = x_i - x_j = 0$ whenever $i, j \ge 3$. - - -Another consequence of the above result is that if $L^\dagger = (u_{ij})$, then $$u_{i+1,j+1}-u_{i,j+1}-u_{i+1,j}+u_{i,j} = 0$$ whenever $i + 1 < j$ or $j + 1 < i$. Moreover, it should be relatively easy to use Propositions 1 and 2 to evaluate $u_{ij}$ explicitly, and in particular to prove that -$$ u_{ij} = v_{\max\{i,j\}} + v_{\max\{n+1-i,n+1-j\}} + \tfrac{1}{4} |i - j| $$ -when $i \ne j$, where $(v_i)$ is an explicitly given vector (in terms of products/ratios of $b_i$, I guess). - -Final remark: there is a corresponding result in continuous variable: the Green function for the operator $$L f(x) = \int_0^1 e^{-q |x - y|} (f(x) - f(y)) dy$$ has zero mixed second-order derivative. The proof follows exactly the same line, and is in fact somewhat less technical.<|endoftext|> -TITLE: Books on foliations -QUESTION [6 upvotes]: I am looking for resources (books, notes, lecture video, etc. anything will do although printed material in English is preferable) on foliations which satisfy some or all of the following constraints. - -Prerequisites: I am familiar with algebraic topology (in the geometric style, as in Hatcher), differential topology (as in Guillemin-Pollack), and Riemannian geometry (as in do Carmo) along with all other standard undergraduate topics (by which I mean content covered in standard textbooks in real and complex analysis, linear and basic algebra, commutative algebra, classical algebraic geometry, point set topology, curves and surfaces). I am also familiar with characteristic classes (Morita's differential forms book and Madsen-Tornehave), basic symplectic geometry (da Silva), and basic topological/measure theoretic dynamics (earlier chapters of Brin-Stuck). - -I am also familiar with physics (general relativity using Caroll, classical mechanics using Goldstein, etc.) if it helps. -Ideally, I am looking for two types of resources. (Notice that the two are not mutually exclusive.) - -An exposition of the theory which has a strong geometric taste (much like Hatcher's books) ideally with a lot of pictures and concrete examples. Ideally, the book connects new ideas introduced in the book with older ideas (described in "prerequisites" above). -Collection of problems which allows one to practice applying the theory. I prefer exercises which are not just filling in technical details which the author did not have time for. Instead, I prefer something which allows one to a.) learn key heuristics, and ideally b.) get a sense on why the theory will be important later in one's studies. - -So far, I have the following books: - -Tamura, Topology of Foliations: An Introduction -Calegari, Foliations and the Geometry of 3–Manifolds - -REPLY [13 votes]: Geometric Theory of Foliations by César Camacho and Alcides Lins Neto in Portuguese, or in English thanks to Sue Goodman's fantastic translation. -I think it does almost everything you're asking for in terms of pictures, examples, and lovely exercises beyond just letting readers fill in details, though some theorems do leave proofs to the reader. -As is pointed out in the comments the two books by Candel and Conlon and your inclusion in your post of Calegari's book are amazing, but for rather different reasons. -Candel and Conlon could be said to provide the opposite of a geometric experience, in the sense that it provides an opportunity to learn the theory by manipulating the symbols of differential geometry's formalism. The degree of precision and generality is probably unrivaled. -Calegari paints a quite different picture, giving a sophisticated and contextualized presentation. His choice of material to collect in that book seems to have been prescient, in that the book was completed 5 years prior to the publication of, what is now known as, the L-space conjecture in ~2012. The content in Calegari's book is a fantastic preparation for understanding the state of the art techniques in low dimensional foliation theory to attack the L-space conjecture. Worth mentioning is that a proof of the L-space conjecture would provide us with a new, Ricci free, proof of the Poincaré conjecture.<|endoftext|> -TITLE: Conjectured primality test for specific class of $N=k \cdot 6^n+1$ -QUESTION [5 upvotes]: Can you provide a proof or a counterexample for the claim given below? -Inspired by Theorem 5 in this paper I have formulated the following claim: - -Let $N=k \cdot 6^n+1$ , $k<6^n$ and $\operatorname{gcd}(k,6)=1$. Assume that $a \in \mathbb{Z}$ is a 6-th power non-residue . Let $\Phi_n(x)$ be the n-th cyclotomic polynomial, then: -$$N \text{ is a prime iff } \Phi_2\left(a^{\frac{N-1}{2}}\right)\cdot \Phi_3\left(a^{\frac{N-1}{3}}\right) \equiv 0 \pmod{N} $$ - -You can run this test here. I have tested this claim for many random values of $k$ and $n$ and there were no counterexamples . -Test implementation in PARI/GP without directly computing cyclotomic polynomials. -EDIT -More generally we can formulate the following claim: - -Let $N=k \cdot (p \cdot q)^n+1$ , where $p$ and $q$ are distinct prime numbers, $k<(p \cdot q)^n$ and $\operatorname{gcd}(k,p\cdot q)=1$. Assume that $a \in \mathbb{Z}$ is a $p \cdot q$-th power non-residue . Let $\Phi_n(x)$ be the n-th cyclotomic polynomial, then: -$$N \text{ is a prime iff } \Phi_p\left(a^{\frac{N-1}{p}}\right)\cdot \Phi_q\left(a^{\frac{N-1}{q}}\right) \equiv 0 \pmod{N} $$ - -You can run this test here. -Test implementation in PARI/GP without directly computing cyclotomic polynomials. -EDIT 2 -It seems that this claim can be generalized even further: - -Let $N=k \cdot b^n+1$ , $k -TITLE: Roadmap for Quillen 1 -QUESTION [8 upvotes]: Question -Suppose you grasped and enjoyed reading Quillen's "Higher Algebraic K-theory I". Now, if you could go back in time to when you started studying algebraic topology and create a reading list / roadmap with the above paper as a goal, what would this plan look like? -Here's another version of the question: suppose you're the PhD advisor of a student that's recently finished their undergrad degree and in terms of books background has only read and completed most of the exercises in: - -algebraic geometry: all of Q. Liu's book -algebraic topology: first two chapters of W. Massey's intro book -differential geometry: first half of J. Lee's smooth manifolds book - -Which books and papers (and in what order) should this student master in order to understand (or at least appreciate) most of Quillen 1? -Outlook -The above questions are likely quite vague wrt to "how do I learn modern algebraic k theory?", but hopefully they're somewhat concrete by stating i) the goal [Quillen 1] and ii) the starting maths background. -If it helps, assume a secondary goal is to eventually focus on studying/appreciating arithmetic problems like Parshin's conjecture. - -REPLY [3 votes]: For someone who is comfortable with Algebraic Geometry (on the level of Liu's book) but less comfortable with Topology, I'd recommend simply using Srinivas book on Algebraic K-theory. -It's a textbook and it first introduces K-theory "axiomatically", then presents a number of applications in algebraic geometry, and only then starts with the proofs which really need some background in topology. Also, the book pretty much covers a great deal in Quillen I, so if the student works through that book, it could act as a self-contained replacement of (most of) Quillen I. -So, if the student reads Srinivas and Quillen I simultaneously, that should make it waaaay easier. -As an aside: I don't think any Differential Geometry/Differential Toplogy is needed for any of this whatsoever. I think when people use Serre-Swan stuff or Bott periodicity to motivate why one might find Algebraic K-theory cool, they usually do this to address an audience that likes these things already. So for somebody who doesn't know these things in the first place, it's not needed to teach them that first. Rather, try to get them curious by relying on their background. For an Algebraic Geometry student, this might be a question like: How does the localization sequence for Chow groups (or Pic, for somebody on the level of Liu) continue to the left? (leading ultimately to Bloch's higher Chow groups and all that stuff)<|endoftext|> -TITLE: Action of fundamental group on homotopy fiber -QUESTION [6 upvotes]: For a Serre fibration of pointed topological spaces $f:X \to B$, there is an action of $\pi_1\left(B,b_0\right)$ on the fiber $F$. The construction of this action I'm familiar with uses a lift $F\times I \to X$ of the map $F \times I \xrightarrow{\pi} I \xrightarrow{\gamma} B$ for any $\left[\gamma \right] \in \pi_1 \left(B,b_0 \right)$. -Now for a general map between two $\infty$-groupoids $f:X \to B$, we can use some version of the Grothendieck construction to construct a map $\phi_f : B \to \operatorname{Grp}_\infty$, and then for an element $\left[\gamma \right] \in \pi_1 \left(B,b_0 \right)$, $\phi_f \left(\gamma \right)$ is an automorphism of $\phi_f \left(b_0 \right)$, which, I guess, generalizes the previous definition (please correct me if that is already false). Is there an explicit description of this automorphism for a specific $\gamma$? (i.e. in terms of pullbacks / pushouts / sections of the maps $f,\gamma$ etc.) I'm particularly interested in writing down obstructions for triviality of this action. - -REPLY [6 votes]: (This answer is written in a model-independent fashion -- translate to your favourite formalism). -For every path $\gamma:[0,1]\to B$ you get an isomorphism in the homotopy category $X_{\gamma0}\xrightarrow{\sim} X_{\gamma1}$ (where with $X_b$ I denote the homotopy fiber over $b\in B$). Probably the easiest and most geometric way of constructing it is to consider the space of lifts. -Let $\operatorname{Sec}_\gamma(f)$ be the space whose objects are sections up to homotopy over $\gamma$ -$$\operatorname{Sec}_\gamma(f)=\{(\tilde\gamma,H)\mid \tilde\gamma:[0,1]\to X,\ H:f\tilde\gamma\sim \gamma\}$$ -(this is nothing else but the homotopy fiber over $\gamma$ of the map $X^{[0,1]}\to B^{[0,1]}$). Then you have a zig zag of maps -$$ X_{\gamma0}\xleftarrow{ev_0} \operatorname{Sec}_\gamma(f)\xrightarrow{ev_1} X_{\gamma1}$$ -where the two maps are evaluation at 0 and 1 respectively. Both maps are homotopy equivalences (this requires some proof, but it's not terribily hard: they're both trivial fibrations when $f$ is a fibration), and so you can define the map in the homotopy category as $ev_1 \circ ev_0^{-1}$. -In order to prove that the action of a loop is trivial, you'll have to prove that $ev_0$ and $ev_1$ are homotopic. I'm not aware of a general way of attacking this problem, but of course studying the behaviour of the two maps on various algebraic invariants can often provide obstructions. - -REPLY [4 votes]: Let $f:E\to B$ be a map of based spaces, and let $F$ be the homotopy fiber. Here is another way of constructing the action of $\Omega B$ on $F$. By definition, there is a homotopy pullback square -$$\require{AMScd} -\begin{CD} -F @>>> \ast\\ -@VVV @VVV \\ -E @>>> B.\\ -\end{CD}$$ -Taking homotopy pullbacks along the inclusion $\ast\to B$ produces a map to the above homotopy pullback square from the following one: -$$\require{AMScd} -\begin{CD} -\Omega B\times F @>>> \Omega B\\ -@VVV @VVV \\ -F @>>> \ast.\\ -\end{CD}$$ -The two morphisms in this square are the projections. The action of $\Omega B$ on $F$ is just the map between the top left corners of these squares; let's call this map $\mu$. This action is not just the projection: this construction shows that there is a homotopy pullback square -$$\require{AMScd} -\begin{CD} -\Omega B\times F @>{\mathrm{pr}}>> F\\ -@V{\mu}VV @VVV \\ -F @>>> E;\\ -\end{CD}$$ -if $\mu$ was just projection onto $F$, then the space $E$ in the bottom right corner would have to be replaced by $F\times B$. -(Note that this diagram shows that the composite $\Omega B \times F\to F\to E$ is trivial on $\Omega B$. You can also see this by the explicit model of this map in spaces: this composite just sends a pair $(\gamma, [e\in E, p:\ast\to f(e)])$ to $e$.) I don't know of general methods to show that the action is trivial. (Remark: one potential advantage of phrasing the construction in this way is that it works in any ($\infty$-)category with finite homotopy limits.)<|endoftext|> -TITLE: Does a compact Lie group have finitely many conjugacy classes of maximal Abelian Lie subgroups? -QUESTION [7 upvotes]: Let $G$ be a compact Lie group. An Abelian Lie subgroup $A \leq G$ is a maximal Abelian Lie subgroup if, for any Abelian Lie subgroup $A'$ such that $A \leq A' \leq G$, then $A' = A$. -Of course any maximal torus of $G$ (there is only one, up to conjugacy classes) is a maximal Abelian Lie subgroup, but there are other ones too, for example the Klein 4-group in $\mathrm{SO}(3)$. -What I'm wondering is if the number of conjugacy classes of maximal Abelian Lie subgroups of any compact Lie group $G$ is always finite? - -REPLY [7 votes]: Yes it's true. It essentially follows from a lemma quoted in the linked answer by user Qayum Khan, which I quote verbatim: - -Neighboring subgroups theorem [1942] - Any compact subgroup $H$ of an arbitrary Lie group $G$ admits a neighborhood $O$ in $G$ such that any subgroup contained in $O$ is $G$-conjugate to a subgroup of $H$. - -Now let by contradiction $(A_n)$ be a sequence of pairwise non-conjugate closed maximal abelian subgroups in a compact Lie group. Let $A$ be a limit point in the Hausdorff topology; this is a compact abelian subgroup. By the above result, for $n$ large enough $A_n$ is conjugate to a subgroup of $A$. By maximality, this means that for $n$ large enough, $A_n$ is conjugate to $A$. This contradicts the non-conjugation.<|endoftext|> -TITLE: Which books should I read in order to be prepared to study information geometry? -QUESTION [10 upvotes]: At the moment, I am preparing my master's thesis (in statistics) and I intend to keep studying in order to pursue a doctoral degree. To be precise, I am mainly interested in studying Information Geometry. -Having said that, I would like to be advised as to which books should I study in order to prepare myself to get acquainted to this subject. -More precisely, could someone tell me which books of differential geometry, probability and statistics should I read in order to introduce myself to it? -A progressive list of readings would be appreciated. - -REPLY [8 votes]: The lecture notes by Frank Nielsen are succinct and fairly self-contained and maybe good to get a first idea. The books [1,2] by Amari can serve for a more in-depth study and contain a fair bit of differential geometry background. In order to get a flavour for some of the applications, I would suggest [3]. -EDIT: Some additional remarks on the differential geometry background -It is quite easy to get bogged down here. I would suggest first reading ref. 1 which has a nice overview of the most important concepts in chapter 1, and then reading additional texts as the need arises (see the comment by @Gabe below). In my view, it is important to have a good grasp of the fundamentals of smooth manifolds and have the geometric intuition for connections and metrics as additional structures on the manifold, i.e. what purpose they serve. It is good to have some experience in Riemannian geometry, enough to understand where and how dually flat connections are different from the Levi-Civita connection. Less important, at least for a first contact with information geometry, are all the global aspects, fibre bundles etc. Of course, if you are interested in those topics for their own sake, there is a lot of interesting stuff. Personally, I learned a lot from [4] but any book which covers the same range of topics would be fine, there are many texts which cater to different styles/preferences. For more guidance on geometry textbooks, there are a bunch of related questions here on MO. - -Amari, S., & Nagaoka, H. (2007). Methods of Information Geometry. American Mathematical Society. -Amari, S. (2016). Information Geometry and Its Applications (Vol. 194). Tokyo: Springer Japan. -Nielsen, Frank, Frank Critchley, and Christopher TJ Dodson. Computational Information Geometry. Springer: Berlin, Germany, 2017. -Lee, J. M. (2009). Manifolds and Differential Geometry (Vol. 107). American Mathematical Society. - -Please let me know if you need further or more specific suggestions.<|endoftext|> -TITLE: Smoothening pseudo-Anosov flows -QUESTION [5 upvotes]: A topological Anosov flow on a closed 3-manifold can be replaced by a smooth Anosov flow using an argument of Fried: use Markov partitions to find a surface of section, put in other terms, one can blow up some closed orbits so that the flow is a suspension of a pseudo-Anosov map on a surface with boundary. Then take a smooth representative of this pseudo-Anosov map within its isotopy class, and blow down the orbits to get back the original manifold. -Is this statement (or some analogue of it) still true for pseudo-Anosov flows? The above argument doesn't work anymore since the singular orbits cannot lie in the interior of a Markov rectangle, so one cannot find a surface of section near these singular orbits in the same way. Is an alternate way of thinking about all of this that generalizes more easily to pseudo-Anosov flows? Any help is appreciated! - -REPLY [4 votes]: In fact, the argument by Fried that you describe is incomplete. It is unclear that such a scheme can be made to work and there is evidence that it may not work (why should the blow down be smooth, and if it were, why would it be a smooth Anosov flow rather than a smooth topological Anosov flow). -However, the result has been clarified and rigourously proved recently in Mario Shannon's thesis http://www.theses.fr/2020UBFCK035# -It is likely that his techniques also respond to your problem and are the technique you seem to be looking for.<|endoftext|> -TITLE: Boundedness of total current in electrical network -QUESTION [7 upvotes]: Consider the following symmetric matrix (adjacency matrix): -$$A=(a_{ij})_{1\leq i,j\leq n}$$ -such that $a_{ij}=a_{ji}, a_{ii}=0$ and $a_{ij}=0$ for $|i-j|\geq k$ where $k\geq3$. We also have $1\leq a_{ij}\leq2, 0<|i-j|0$ independent of $n$ such that -$$\sum_{i,j}a_{ij}|x_i-x_j|\leq C$$ -The physical meaning of the conjecture is that, if we flow 1 unit of current from node 1 to node 2, the sum of current of each edge in the given electrical network is bounded. -Simulation result indicates that this is indeed the case. However, I somehow can only prove bound involving $n$. I think this may be related to the bandwidth of a graph? -This is also related to my previous question: -Exponential decay of voltage potential difference -I have an intuitive idea of why this is the case. Since the total current flows out of node 1 is 1, then this 1 unit of current gets split into at least $k$ parts to the neighbors of 1. For each neighbor of 1, when the current flows out, it again gets split into at least $k$ parts and this somehow forms a geometric series that is summable. -Another idea is to use induction. The physical intuition is to examine how will the total current change when we gradually add new nodes and connections to the existing network. Adding new connections will decrease the effective resistance between 1 and 2 but new current will flow in new edges, they can somehow be balanced making the total current bounded. - -REPLY [5 votes]: Edit: By request, I have added some explanations at the end. The first bullet may be helpful (it introduces a little notation). I also misread the question, and used a constant $k=3$ (instead of $k\ge 3$). This is now fixed, but $k$ must be fixed; for now the resulting bound depends on it... -Edit 2: I have added an idea on how to eliminate this issue, and (in comments) how to sharpen another bound. But the result here still depends on $k$. -Let's think of the problem as occuring on a multigraph (some edges are doubled) and interpret the Laplacian using hitting probabilities of a random walk. -The graph is more or less a long line segment, with edges between some nearby pairs of points. The vertex set is $\{1,\ldots,n\}$. Your solution $x$ is, up to a multiplicative constant $\alpha$ in $[0,1]$, the unique function $h:\{1,\ldots,n\}\rightarrow \mathbb{R}_+$ which is harmonic except at $\{1,2\}$ (by which I mean $(Lh)(k)=0$ for $k\notin \{1,2\}$) and satisfies $h(1)=0,h(2)=1$. This $h$ is nonnegative, taking values in the unit interval. -In fact, $h(i)$ is the probability that a random walk on the graph starting from $i$ will reach $2$ before $1$. You wish to prove $|h(i)-h(j)|$ decays at least exponentially in $\min(i,j)$ at a rate independent of $n$. This can be seen as follows: the probability that a random walk starting from $i$ and one starting from $j$ pass through a common vertex tends exponentially to $1$ in $\min(i,j)$, and if they pass through a common vertex with probability $\ge p$ then $|h(i)-h(j)| \le 2-2p$. -To see the last point, let's denote the random walk starting at $i$ and at $j$ by $W_i$ and $W_j$ respectively. Note that: $$|h(i) - \sum_{u=3}^n P(W_i \text{ intersects $W_j$ at vertex u and at no vertex $\ell > u$})\cdot h(k)| < P(W_i,W_j\text{ do not meet})$$ since the events "intersecting at vertex $k$ and no higher-index vertex" are mutually disjoint and each value of $h$ is in $[0,1]$ (then apply the law of total probability and the fact $h$ is a hitting probability). The same holds for $W_j$ and $h(j)$. -To see that the probability of intersecting does in fact approach $1$ exponentially, suppose $j>i$. Consider the random walk starting from $j$, and look at the first time it reaches a vertex $\ell\le i$. Then $i-\ell -TITLE: Are nuclear operators closed under extensions? -QUESTION [6 upvotes]: Given $X_i, Y_i$ Banach spaces, $f_j, g_j, T_i$ bounded linear operators for $i=1,2,3$ and $j=1,2$. We have the following diagram -$\require{AMScd}$ -\begin{CD} -0 @>>> X_1 @>f_1>> X_2 @>f_2>> X_3 @>>> 0\\ -@V VV @V T_1 VV @V T_2 VV @V T_3 VV @V VV \\ -0 @>>> Y_1 @>>g_1> Y_2 @>>g_2> Y_3 @>>> 0 -\end{CD} -with two horizontal topologically short exact sequences. If $T_1$ and $T_3$ are nuclear operators, does it imply that $T_2$ is nuclear as well? A reference to problems of this general form, would be most welcome. - -REPLY [7 votes]: The answer is no: you can even have $T_1=T_3=0$ and $T_2$ equal to the identity $id$ on an infinite dimensional Banach space. -Indeed, consider the following commutative diagram with exact rows: -$$\begin{CD} -0@>>> 0 @>0>> X @>id>> X @>>> 0\\ -&&@V0VV @VV{id}V @VV0V\\ -0@>>>X @>>id> X @>>0> 0 @>>> 0 -\end{CD} -$$ -See this paper for related results. - -REPLY [5 votes]: Of course, Yemon was faster than me. But I want to emphasize that -the point is very elementary linear algebra: The simple commutative diagram -$\begin{CD} -0 @>>> \mathbb R @>f_1>> \mathbb R^2 @>f_2>> \mathbb R @>>> 0\\ -@V VV @V T_1 VV @V T_2 VV @V T_3 VV @V VV \\ -0 @>>> \mathbb R @>>f_1> \mathbb R^2 @>>f_2> \mathbb R @>>> 0 -\end{CD}$ -with the natural inclusion $f_1(x)=(x,0)$ and projection $f_2(x,y)=y$ in the rows shows shows that $T_2$ is not at all determined by $T_1$ and $T_3$. If $T_2$ is given by a matrix $\begin{bmatrix} a&b\\ 0&c\end{bmatrix}$ you get nothing for $b$.<|endoftext|> -TITLE: An inequality for the Hessian of eigenfunctions of the Laplacian on compact manifolds -QUESTION [8 upvotes]: Let $(M,g)$ be a compact Riemannian manifold, and let $\Delta$ be the Laplace-Beltrami operator. Let $\lambda_1 >0$ be the first positive eigenvalue. That is, there exists a non-trivial function $\phi$ satisfying $\Delta \phi + \lambda_1\phi = 0$, and that $\lambda_1$ is the smallest such real number. Then one can always normalize the eigenfunction such that $\max_M\phi = 1$ and $\min_M = -k\geq -1$. I have two questions: - -Can one further choose an eigenfunction such that the maximum is attained at a unique (or at worse an isolated) point? For example, for a sphere $\mathbb{S}^2$ with the round metric, the eigenfunctions are given by $\cos d$, where $d$ is the distance to a fixed point (say the north pole N). The north pole in this case is the unique maximum. On the other hand if we take the product $\mathbb{S}^2\times\mathbb{S}^2$ with the product of the round metrics, and if we let $d_1$ and $d_2$ be the distance functions (from the respective north poles) on each factor, then each $u_i = \cos d_i$ is an eigenfunction, but the point where they attain maximums are clearly is unique (for instance $u_1$ takes maximum value on $\{N\}\times\mathbb{S}^2$). But in this case one can take $u = (\cos d_1 + \cos d_2)/2$ to be a noramlized eigenfunction with a unique maxima. So my question is whether such a normalized eigenfunction exist on all compact Riemannian manifolds $(M,g)$. -If further the Ricci curvature is bounded below, say $\mathrm{Ric}_g\geq n-1$, where $n$ is the dimension, it is well known that $\lambda_1\geq n$. then does there exist an eigenfunction $u$ such that $$\mathrm{Hess}(u) \geq -\frac{\lambda_1}{n}ug?$$ If the answer to this is affirmative, then such an eigenfunction will clearly have an isolated maximum (since the Hessian will be negative definite at the maximum), hence answering the first question also in affirmative. It is of course well known (Obata) that $\lambda_1 = n$ if and only if $(M,g)$ is isometric to the round sphere. Moreover, in this case we necessarily have that for any eigenfunction $u$, both sides of the above inequality must in fact be equal. This follows from a simple integration by parts argument. On the other hand, suppose $\lambda_1>n$, then once again taking the example of $\mathbb{S}^2\times\mathbb{S}^2$ it is clear that the above inequality of the Hessian will not hold for all eigenfunctions. My guess is that the Hessian inequality is too strong, and is likely wrong. But a counter example would be nice. - -REPLY [3 votes]: The answer to question 1 is no. On a two-dimensional torus $\mathbb{S}^1 \times \mathbb{S}^1$, the first nontrivial eigenspace consists of $\mathrm{Span}\{\cos(x),\sin(x),\cos(y),\sin(y)\}$. Each eigenfunction in this eigenspace is separable, i.e. it is the product of a lower-dimensional function and a constant function.<|endoftext|> -TITLE: Are all classes Stiefel-Whitney classes? -QUESTION [10 upvotes]: When I thought of this question, I was sure it must have been asked before on this site, but I could't find anything. Maybe my search skills are lacking, or maybe the question is obvious and it's my math skills that are lacking. Anyway, here it goes. -For a $CW$-complex $X$ let $sw^*X$ be the subring of $H^*(X,\mathbb{F}_2)$ generated by all classes which are Stiefel-Whitney classes of some vector bundle over $X$. It is not hard to see that $sw$ is a proper subfunctor of mod 2 homology. For example (and this might be overkill) if you take the right dimensional sphere $S^n$, then by Bott periodicity, $KO(S^n)=0$, so $sw^*S^n=0$. -Now let $SW^*X$ be the subring generated by all classes which are either Stiefel-Whitney classes of some vector bundle over $X$, or suspensions or desuspensions of such classes. -$\textbf{Edit}$: Perhaps it wasn't clear from context, but I want $SW^*$ to be a functor, so I force it to be closed under pullbacks. For that reason I am puzzled by Nicholas Kuhn's suggested answer below. Also, we know in retrospect that $H\mathbb{F}_2^*X$ is a summand in $MO^*X$, and that thing is sort of tautologically built out of characteristic classes... -Is $SW^*X=H^*(X,\mathbb{F}_2)$? -I suppose the question is equivalent to something like: does the identity map of $K(\mathbb{F}_2,n)$ factor, stably, through some $BO(m)$? - -REPLY [9 votes]: A 1968 paper in Topology by Anderson and Hodgkin shows that -$KO^*(K(\mathbb F_2, n)) = 0$ if $n \geq 2$. This implies that if $n \geq 2$, then no nonzero classes in $H^*(K(\mathbb F_2,n);\mathbb F_2)$ are SW classes. (And of course, $BO(1) = K(\mathbb F_2,1)$.)<|endoftext|> -TITLE: Are compact objects in presheaf categories finite colimits of representables? -QUESTION [13 upvotes]: An object $x$ in a category $\mathsf{C}$ is called compact or finitely presentable if $$\mathrm{hom}(x,-) : \mathsf{C} \to \mathsf{Set}$$ preserves filtered colimits. This concept behaves best when $\mathsf{C}$ has all filtered colimits, e.g. when it is the category of presheaves on some small category $\mathsf{X}$: -$$ \mathsf{C} = \mathsf{Set}^{\mathsf{X}^{\mathrm{op}}} $$ -Every representable presheaf is compact. In general, any finite colimit of compact objects is compact. Thus, any finite colimit of representables is compact. -My question is about the converse: in the category of presheaves on a small category, is every compact object a finite colimit of representables? - -REPLY [4 votes]: Here is another perspective on the problem, using some big guns (Gabriel-Ulmer duality). -Given a small category $C$, let $K$ be its free finite cocompletion. This means $K^{op}$ is the free finite completion of $C^{op}$, which means in turn that for any functor $F: C^{op} \to \mathbf{Set}$, there is a finitely continuous (or left exact) functor $\tilde{F}: K^{op} \to \mathbf{Set}$ that extends $F$ along the canonical inclusion $i: C^{op} \to K^{op}$, and this extension is unique up to unique isomorphism. Put differently, restriction along $i$ induces an equivalence -$$\mathrm{Lex}(K^{op}, \mathbf{Set}) \to \mathrm{Cat}(C^{op}, \mathbf{Set}).$$ -In particular, the presheaf category $\mathrm{Cat}(C^{op}, \mathbf{Set})$ is locally finitely presentable. By the way, it's well known that the free finite cocompletion $K$ of a small category $C$ is simply the category of finite colimits of representables: see section 5.9 of Kelly's Basic Concepts of Enriched Category Theory. -On the other hand, Gabriel-Ulmer duality assures us that given a locally finitely presentable category $A$, there is up to equivalence only one finitely complete category $L$ for which $A \simeq \mathrm{Lex}(L, \mathbf{Set})$. Even better, Gabriel-Ulmer duality gives a recipe for obtaining $L$: it is the dual of the category of compact objects in $A$, meaning objects $a$ such that $A(a, -): A \to \mathbf{Set}$ preserves filtered colimits. -Putting all this together, this shows that the category of compact objects in the category of presheaves over $C$ is equivalent to the free finite cocompletion of $C$, or to the category of finite colimits of representable presheaves.<|endoftext|> -TITLE: Path integral derivation of extended TQFT -QUESTION [9 upvotes]: I know this isn't exactly a math question, but I am asking it here anyway. We define an extended TQFT to be a functor (preserving tensor products) from the $\left(\infty,n\right)$-category of cobordisms to a suitable $\left(\infty,n\right)$-category of vector spaces. -The original Atiyah-Witten definition was a functor for the category of $n$ dimensional cobordisms to $\mathrm{Vect}_{\mathbb C}$. This definition was justified from the path integral in physics. -Can we similarly get an physicsy intuition of extended TQFT from a path integral-like formulation? Any references to a general construction from physics that gives rise to such a functor, starting from the path integral? -Note: I don't want specific example relating to Chern-Simons theory or any other TQFT, but a general construction deriving the extended TQFT axioms from the path integral, or something similar. - -REPLY [5 votes]: The original motivation for extended TQFTs (as introduced by Freed, Lawrence, Baez-Dolan) is indeed giving a finer form of locality, as explained by Dmitri Pavlov. However I think there are two quicker, and arguably more physical, ways to see n-categorical structure in n-dimensional QFTs. -The first is not really about the states of a QFT (as axiomatized in the Atiyah-Segal formalism) but their algebras of observables (extending the distinction between geometric and deformation quantization in the context of quantum mechanics). Namely the theory of factorization algebras as developed in the book of Costello-Gwilliam extracts from the same data as the path integral an n-dimensional factorization algebra of observables. In the topological context such a factorization algebra is the same as an $E_n$ algebra, which is the same as a very connected $(\infty,n)$-category (one with one object, one 1-morphism, ...all the way down). -The second comes from thinking of what ELSE there is in a QFT beyond the path integral -- the most important being the structure of defects of various dimensions. Of these the richest is the notion of a boundary theory (or "boundary condition") for a QFT, which is very loosely "things we can put on the boundary and couple to our theory" -- something like a QFT of one dimension lower that lives on the boundary of manifolds where the bulk carries our given QFT. -In any case, boundary theories in an $n$-dimensional QFT naturally form something like an $(n-1)$-category, which in the cobordism hypothesis formalism for extended TQFT is closely related to what you'd attach to a point. Namely, as morphisms between any two boundary theories you can consider codimension 2 defects that are interfaces between the two theories (think of dividing the boundary of a half-space in $R^3$ into upper and lower halves with a 1-dim interface on the intersection). As 2-morphisms you can consider interfaces between interfaces, and so on and so forth. -To me this is the most compelling way to see that higher categorical structure is physically natural/meaningful. A (somewhat criminal) paraphrase of the cobordism hypothesis says that a fully extended TQFT is determined by its collection of boundary conditions. [Really boundary theories are morphisms between the unit and the object attached by the TQFT to a point, which in general needn't determine this object, but it's a decent ansatz.]<|endoftext|> -TITLE: Levi-Civita connection from idempotents -QUESTION [5 upvotes]: Let $(M,g)$ be a closed Riemannian manifold. Let $V$ be a smooth complex vector bundle over $M$. We can write $V$ as the range of an idempotent $E$ in a matrix algebra $M_n(C^\infty(M))$ acting on a trivial bundle $M\times\mathbb{C}^n$. Here $C^\infty(M)$ denotes the complex-valued functions on $M$. -Let $D$ be the trivial connection on $M\times\mathbb{C}^n$ defined by applying the de Rham differential to each entry separately. Then the expression $EDE$ defines a connection on $V$. -Some books call the connection $EDE$ the "Levi-Civita connection" on $V$, "by analogy to the classical situation", and I am trying to understand how this analogy works. In particular, when $V$ is the tangent bundle $TM$ (ignoring the fact that this is a real vector bundle), the Levi-Civita connection depends on the metric $g$, while the expression $EDE$ does not seem to depend on $g$. -Question: Is there a precise sense in which $EDE$ produces the "classical" Levi-Civita connection on $TM$? - -REPLY [3 votes]: $\newcommand{\R}{\mathbb{R}}$ -Yes, it uniquely defines a connection in the case where $V$ is the tangent bundle(the connection won't depend on the choice of orthogonal idempotent) and its the Levi Civita connection. But in general different idempotents from the same metric will define different connections(I think, I'm not sure on this one though). -Let $ V^l \to M^k $ be the vector bundle of rank $l$ over a $k$ manifold and take an isometric embedding of the vector bundle into $\R^{2n}$ with the standard metric. Let $s$ be a section of the vector bundle. Let $e: M \to Mat_{n\times n}$ be the orthogonal projection onto the tangent bundle of $M$. -If we differentiate $s$ in a given direction using the ambient $\R^{2n}$, (we can just take all the directions into account by taking the exterior derivative $ds$), then the result will not be in the vector bundle any more. But $eds$ will be. Hence we define the derivative of $s$ in the $v$ direction to be $\nabla_v s =e\frac{ds}{dv}$ -Let us consider the case where $E$ is the tangent bundle of the manifold and $M^k \subset \R^n$. It is possible to write down the expression $e\frac{ds}{dv}$ just using the metric on the manifold. -To show this we show that $eds$ satisfies the axioms of the Levi Civita connection: - -If $X,Y,Z$ are sections of our vector bundle, then -\[ X \langle Y,Z \rangle = \langle \nabla_X Y, Z \rangle + \langle -Y, \nabla_X Z \rangle \] - -Proof: -\[ \nabla_X Y-\nabla_Y X=[X,Y] \] -If we expand the rhs, \[ \langle \nabla_X Y, Z \rangle + \langle Y, \nabla_X Z \rangle = \langle e \frac{dY}{dX}, Z \rangle + \langle Y , e \frac{dZ}{dX} \rangle = \] -\[\langle \frac{dY}{dX}, e^T Z \rangle + \langle e^TY , \frac{dZ}{dX} \rangle \] -Now $e^T=e$ because $e$ is an orthogonal projection. So this is -\[=\langle \frac{dY}{dX}, eZ \rangle + \langle eY , \frac{dZ}{dX} \rangle=\langle \frac{dY}{dX}, Z \rangle + \langle Y , \frac{dZ}{dX} \rangle=X \langle Y,Z \rangle \] -\[ \nabla_X Y-\nabla_Y X= e \frac{dY}{dX} -e \frac{dX}{dY}=e[X,Y]=[X,Y] \] -Exercise: -Using the above identities, and cyclically permuting $X,Y,Z$ write down an expression for $\nabla_X Y$ in terms of the inner product. -Using the above expression, write down a matrix $A$ of 1-forms in terms of the inner product such that -\[\nabla s=ds-As \] -If you do this exercise, the matrix $A$ you get will be the matrix of christoffel symbols. -Another good exercise is to write out what $A$ is in terms of the idempotent. (You should get $(1-e)de$) -Note that you can't cyclically permute the indices $X,Y,Z$ for an arbitrary vector bundle. I don't think that you get a canonical connection on a vector bundle just given the metric(I think this was one of your questions), and at the very least, the trick above doesn't work. -Aside: I came up with this answer after reading through the comment above about it being in Dirac's General relativity book. I've been trying for the past few months and I can't quite tell what he is doing. I did see though that his argument factors through proving that the axioms of the Levi Civita connection are satisfied without saying that that is what he's doing.<|endoftext|> -TITLE: Topological proof that a Vitali set is not Borel -QUESTION [16 upvotes]: This question is purely out of curiosity, and well outside my field — apologies if there is a trivial answer. Recall that a Vitali set is a subset $V$ of $[0,1]$ such that the restriction to $V$ of the quotient map $\mathbb{R}\rightarrow \mathbb{R}/\mathbb{Q}$ is bijective. It follows easily from the definition that $V$ is not Lebesgue measurable. Now suppose you don't know the Lebesgue measure (which, after all, is not that easy to construct). Is there a topological proof that $V$ is not a Borel subset of $\mathbb{R}$? - -REPLY [18 votes]: Sometimes a convenient substitute for Lebesgue measurability is the property of Baire. Just like Lebesgue measurability, the class of sets with this property is a $\sigma$-algebra containing the open subsets - indeed, open sets clearly have property of Baire, this class is closed under countable unions since meager sets are closed under countable unions, and it's closed under complements essentially since the boundary of an open set is nowhere dense. -It of course follows that every Borel set has the property of Baire. It remains to show the Vitali set doesn't have a property of Baire. Indeed, assume it was, then either $V$ is meager, or it is comeager in some open interval. The former cannot hold, since $[0,1]$ is contained in countable union of translates of $V$, so would be meager itself. In the other hand, if it were comeager in some open interval $I$, then for any rational $q$ the intersection $I\cap(V+q)$, since it's contained in $I\setminus V$, is meager, from which we would conclude $V$ is meager. -This might be a slightly overly detailed answer, but I did try to make it reasonably self-contained - it indeed is quite easier than if we had to build up the theory of Lebesgue measure from scratch!<|endoftext|> -TITLE: hyperbolic quotient of hyperbolic group -QUESTION [10 upvotes]: I have a memory of hearing about a result (or perhaps a conjecture), possibly due to Gromov, that, if $G$ is a hyperbolic group and $g \in G$ has infinite order, then the quotient group $G/\langle (g^n)^G \rangle$ is hyperbolic for all sufficiently large $n > 0$. -I have been searching for references, but without success. Can anyone help?. -$\mathbf{Edit}$: After looking at the references in the answer by Mikael de la Salle, I see that I did not state this result correctly. Rather than the statement being for all sufficiently large $n>0$, it should be that there exists and $N>0$ such that $G/\langle (g^{nN})^G \rangle$ is hyperbolic for all $n > 0$. The result stated applies only to non-elementary hyperbolic groups, but for an elementary hyperbolic group this quotient is finite, and so it remains correct. - -REPLY [10 votes]: This is contained in at least Delzant's paper Sous-groupes distingués et quotients des groupes hyperboliques. [Distinguished subgroups and quotients of hyperbolic groups] Duke Mathematical Journal, vol. 83 (1996), no. 3, pp. 661–682, and also in Ol'shanskii's paper SQ-universality of hyperbolic groups, Mat. Sb. 186 (1995), no. 8, 119–132. -I am not an expert, but the first lines of the papers seem to indicate that this result was announced by Gromov, but that the proofs were not all convicing.<|endoftext|> -TITLE: Union of Schubert cells being affine -QUESTION [8 upvotes]: Let $k$ be a field of characteristic zero, $G$ be a reductive group with a Borel $B$ and $\mathcal{F}:=G/B$ the associated flag variety. Let $W$ be the Weyl-group of G. -Then let $S \subset W$ and $Z=\bigcup_{w \in S} C(w) \subset \mathcal{F}$ where $C(w)=BwB/B$ is the Schubert cell associated to $w$. -I'm interested to know when $Z$ is an affine scheme. This is for example the case if all $w \in S$ have the same length. Is this the only case? - -REPLY [3 votes]: This is essentially an extension of my comment, just to answer the actual "is this the only case?" question. It is not, $Z$ will be affine whenever $S$ is an antichain in the Bruhat order. Indeed, this condition means that no $C(w)$ with $w\in S$ intersects the closure $\overline{C(w')}$ for any other $w'\in S$ which shows that $C(w)$ is open in $Z$. Hence the $C(w)$ are the irreducible components of $Z$ and are also affine, this renders $Z$ affine itself (Hartshorne, Exercise 3.3.2). -Of course, the more interesting underlying question is whether this condition is necessary, I might update this answer if I come up with a proof. (Any algebraic geometers here? Is it at all possible for an affine space to be embedded into an affine variety as a proper open subset? If not, this would give us the answer.)<|endoftext|> -TITLE: Matrix inequality : trace of exponential of Hermitian matrix -QUESTION [6 upvotes]: I want to know whether the following inequality holds or not. -\begin{align} -(\mathrm{Tr}\exp[(A+B)/2])^2\leq(\mathrm{Tr}\exp A)(\mathrm{Tr}\exp B)\tag{1} -\end{align} -where $A, B$ are Hermitian matrices of the dimension $D$. -Note that if $A$ and $B$ commute, we can see (1) holds using the simultaneously diagonalizing basis and Cauchy-Schwarz inequality. The problem is the case where $A$ and $B$ do not commute. - -REPLY [10 votes]: You can prove it using the Golden-Thompson inequality $Tr (e^{A+B}) \leqslant Tr(e^{A} e^{B})$ and then applying the Cauchy-Schwarz inequality.<|endoftext|> -TITLE: Conceptual explanations of the class numbers for the first few $\mathbb{Q}(\sqrt{p})$ with odd conductor -QUESTION [12 upvotes]: It's known that the class number of $\mathbb{Q}(\sqrt{p})$ is $1$ for all primes $p<229$. -Question: What would it be like for conceptual explanations of $h(\mathbb{Q}(\sqrt{p}))=1$ for the first few primes of form $4k+1$ (equivalently, $\mathbb{Q}(\sqrt{p})$ having odd conductor)? -To clarify the question: - -A conceptual explanation should treat the first few primes simutaneously, instead of a case-by-case analysis where a case is a single prime. (A case-by-case analysis with a finite number of cases that a priori covers the whole range of primes is allowed, e.g. the cases being p=1, 5 or 9 mod 12.) -For "the first few primes of form $4k+1$", I mean such continuous primes up to a bound, e.g. $5,13,17$ but not $5,17,29$. The argument should be able to cover primes in such a way. -To avoid trivialities, the conceptual explanations should cover at least $5, 13, 17$ and $29$. - -An example of conceptual explanation would be like: - -By Example 2.9 of Masley's paper Class numbers of real cyclic number fields with small conductor, the class number of such fields are odd. -The Minkowski bound gives $h(\mathbb{Q}(\sqrt{p}))<3$ for $p<36$. Thus we have established $h(\mathbb{Q}(\sqrt{p}))=1$ for the first few primes of form $4k+1$: $5,13,17$ and $29$. -This explanation also works for cyclic cubic fields of conductor $7$ and $13$. - -Bonus for explanations that are not specialized on real quadratic fields, e.g. the explanation presented above. - -REPLY [2 votes]: Some complementary heuristics, too long for a comment. -Again start from the fact that the class number $h$ of $K = {\bf Q}(\sqrt{p})$ -is odd if $p$ is a prime of the form $4k+1$. This time we compare with Dirichlet's class number formula, which here gives -$$ -L(1,\chi_p) = \frac{2\log \epsilon}{\sqrt p} h, -$$ -where the character $\chi_p$ is the Legendre symbol $\chi_p(n) = (n/p)$, -and $\epsilon$ is the fundamental unit of $K$. -We expect that $L(1,\chi_p) \approx 1$, so large $h$ go with small $\epsilon$. -A unit $\epsilon > 1$ in a real quadratic field of discriminant $D$ -must be at least as large as $\frac12(m + \sqrt{m^2 \pm 4})$ -for some odd integer $m$, with $D = m^2 \pm 4$. -If $D$ is prime then we must use the plus sign -(unless $m=3$, but then the fundamental unit is $(1+\sqrt5)/2$). -$\epsilon > \sqrt{p} - O(1/\sqrt{p})$ and -So, $2\log \epsilon / \sqrt{p} > \log p - O(1/p)$. -Setting $L(1,\chi_p) \approx 1$ and $h=3$ in the class number formula gives -$\sqrt{p} \approx 3 \log p$; the solution $p \approx 289$ is of about -the right size for the minimal example of $h>1$. -We're actually closer here than we deserved to be: -the solution of $L(1,\chi_p) \sqrt{p} = 3 \log p$ is quite sensitive to -the size of $L(1,\chi_p)$, and $L(1,\chi_{229}^{\phantom.}) = 1.075+$ -is unusually close to $1$; for example, $L(1,\chi_p) > 2$ for -$p = 193, 241, 313, 337$, while $L(1,\chi_p) < 0.4$ for -$p = 173, 293, 677, 773$. -Most of the early examples of $h > 1$ have small $\epsilon$, -either with $p=m^2+4$ as above or the next-smallest possibility, -$\epsilon = m + \sqrt{p}$ with $p=m^2+1$. Indeed -this -LMFDB list of fields ${\bf Q}(\sqrt p)$ with $p<2000$ and $h>1$ -begins with -$$ -229 = 15^2 + 4,\ -257 = 16^2 + 1,\ -401 = 20^2 + 1,\ -577 = 24^2 + 1, -$$ -$733 = 27^2 + 4$, -and then two exceptions $p=761$ and $p=1009$ and nine further $p$ -of which all but $1429, 1489, 1901$ are not of the form $m^2+4$ or $m^2+1$. -Moreover $229$ is the only second-smallest prime of the form $p = m^2 + 4$ -that satisfies our analytic bound $p > 63$ --- and the smallest is -$p = 173$, which was our example of an unusually small $L(1,\chi_p)$. -Likewise the next two examples were $293 = 17^2 + 4$ and $677 = 26^2 + 1$, -which are conjectured to be the largest discriminants $p = m^2+4$ and $p = m^2+1$ -for which ${\bf Q}(\sqrt p)$ has class number $1$.<|endoftext|> -TITLE: Is there a known condition for partial sums of a decreasing positive sequence to take all values up to the total sum? -QUESTION [13 upvotes]: Let $a_0>a_1>\cdots>0$ have the property that, for each positive $a<\sum_{n\in\Bbb N}a_n$ (admitting $\infty$ for the sum), there is $A\subset\Bbb N$ such that $a=\sum_{n\in A}a_n$ . Are there known necessary and sufficient conditions on the $a_n$ (not involving arbitrary partial sums) for this property? To illustrate, $a_n=1/(n+1)$ and $a_n=1/2^n$ possess the required property, but $a_n=1/(2+\varepsilon)^n$ does not for any $\varepsilon>0$. -(This question is adapted from one asked on Mathematics Stack Exchange two months ago which received no answer.) - -REPLY [17 votes]: It is necessary and sufficient that - -$\lim_{n \rightarrow \infty}a_n = 0$, and -$a_n \leq \sum_{m > n}a_m$ for all $n$. - -In other words: the terms go to zero, and no term is bigger than the sum of all the following terms. -Necessity: First, it is necessary that $\lim_{n \rightarrow \infty}a_n = 0$ because you cannot form any sum smaller than $\lim_{n \rightarrow \infty}a_n$. -Suppose $a_n > \sum_{m > n}a_m$ for some $n$. Let $\varepsilon$ be some number with $0 < \varepsilon < a_n - \sum_{m > n}a_m$. Then I claim there is no $A$ such that $\sum_{m \in A}a_m = a_1+a_2+\dots+a_{n-1}+a_n - \varepsilon$. To see this, just consider two cases: (1) if $\{a_1,\dots,a_n\} \subseteq A$, then $\sum_{m \in A}a_m$ is too big, because $\varepsilon > 0$, and (2) if $a_i \notin A$ for some $i \leq n$, then $\sum_{m \in A}a_m$ is too small, because -$\sum_{m \in A}a_m \,\leq\, (a_1+a_2+\dots+a_n) - a_i +\sum_{m > n}a_m \,\leq\, (a_1+a_2+\dots+a_n) -a_n + \sum_{m > n}a_m < (a_1+a_2+\dots+a_n) - \varepsilon.$ -Sufficiency: Suppose $a_n \leq \sum_{m > n}a_m$ for all $n$, and let $c$ be any number with $0 \leq c \leq \sum_{m \in \mathbb N}a_m$. Then we can construct the desired $A \subseteq \mathbb N$ recursively, as follows. If it has already been decided for all $m < n$ whether $m \in A$ or not, then put $n \in A$ if and only if $a_n + \sum_{m \in A \cap \{1,2,\dots,n-1\}} a_m \leq c$. (In other words, put $n \in A$ if and only if putting $n \in A$ does not make the sum too big.) Once we have built $A$ according to this rule, it is clear that $\sum_{m \in A}a_m \leq c$, because none of the finite partial sums exceeds $c$. -Now suppose, aiming for a contradiction, that $\sum_{m \in A}a_m < c$, and let $\varepsilon = c - \sum_{m \in A}a_m$. There is some $N$ such that $a_n < \varepsilon$ for all $n \geq N$. For each such $n$, we have $\sum_{m \in A \cap \{1,2,\dots,n-1\}} a_m \leq \sum_{m \in A} a_m = c - \varepsilon$, and hence $a_n + \sum_{m \in A \cap \{1,2,\dots,n-1\}} a_m < c$. By our rule for constructing $A$, this means $n \in A$ for all $n \geq N$. In other words, $A$ is a co-finite subset of $\mathbb N$. -Let $n$ denote the largest member of $\mathbb N \setminus A$. (Note that $\mathbb N \setminus A \neq \emptyset$, because $\sum_{m \in A}a_m < c \leq \sum_{m \in \mathbb N}a_m$.) Then $\left( \sum_{m \in A \cap \{1,2,\dots,n-1\}} a_m \right)+a_n \leq \left( \sum_{m \in A \cap \{1,2,\dots,n-1\}} a_m \right)+ \sum_{m > n}a_m = \sum_{m \in A}a_m < c$. This is the contradiction we were after, because this tells us that we should have had $n \in A$, although $n$ was supposed to be the largest number not in $A$. - -REPLY [6 votes]: In order for all $a\in(0,a_0+a_1+\cdots)$ to be representable as partial sums of of the $a_i$'s, it is necessary that -\begin{equation*} - a_\infty:=\lim_n a_n=0; \tag{1} -\end{equation*} -otherwise, no $a\in(0,a_\infty)$ is a partial sum of the $a_i$'s. So, assume (1). -A sufficient condition is that for all natural $n$ -\begin{equation*} - a_{n-1}\le a_n+a_{n+1}+\cdots \tag{2} -\end{equation*} -for all $n$. Indeed, assume (2) holds. Take any $a\in(0,a_0+a_1+\cdots)$. Successively define -\begin{equation*} - S_0:=\{k\ge0\colon a_k\le a\},\quad k_1:=\min S_0 -\end{equation*} -and, for $j\ge2$, -\begin{equation*} - S_{j-1}:=\{k>k_{j-1}\colon a_k\le a-s_{j-1}\},\quad k_j:=\min S_{j-1}, -\end{equation*} -where -\begin{equation*} - s_j:=a_{k_1}+\cdots+a_{k_j}. -\end{equation*} -If $S_{j-1}=\emptyset$ for some $j=1,2,\dots$, then, by (1), ($j\ge2$ and) $s_{j-1}=a$, so that we are done. -It remains to consider the case when $S_{j-1}\ne\emptyset$ for all $j=1,2,\dots$, so that we have $0\le k_10$ and all $j$. By the construction, for each $j$ either -(i) $k_j=k_{j-1}+1$ or -(ii) $a_{k_j-1}>a-s_{j-1}$. -In case (ii), $a_{k_j-1}>a-s_{j-1}=a-s_j+a_{k_j}\ge h+a_{k_j}$. So, if case (ii) holds for infinitely may $j$'s, then, letting $j\to\infty$ and recalling (1), we get $0\ge h+0$, a contradiction. -So, case (i) holds eventually, for all large enough $j$. Then for some natural $m$ and $n$ we have $s_{m-1}+a_n+a_{n+1}+\cdots -TITLE: Curvature of nonsymmetric metric tensors? -QUESTION [6 upvotes]: Consider a smooth manifold $M$ of arbitrary dimension. We have notions of psuedo-Riemannian or Riemannian metrics on a manifold, and they differ in the slightest way of being positive-definite or not. However, what happens if we drop positive-definite AND symmetry? For example, if we had a nondegenerate bilinear form $g_p: T_p M \times T_p M \to \mathbb{R}$ that varied smoothly between points. Has this been explored in depth? It appears to me at surface level that one could still concoct connections, curvature, and possibly a notion of parallel transport in this flavor of 'smooth geometry'. A motivation for me to ask is as follows. Suppose $R$ is an $S$-algebra where $\Omega_{R/S}$ is reflexive and the canonical isomorphism $\phi: \Omega_{R/S} \to \Theta_{R/S}$ is an isomorphism of $R$-modules (i.e. nonsingular varieties). There exists a canonical map $\Omega_{R/S} \times \Theta_{R/S} \to R$, which is $R$-bilinear and nondegenerate, and is given by $\langle \omega, V \rangle = l(\omega)$ where $l:\Omega_{R/S} \to k$ such that $l \circ d_{R/S} = V$. This induces a morphism $$\Theta_{R/S} \times \Theta_{R/S} \xrightarrow{\phi^{-1} \times 1}\Omega_{R/S} \times \Theta_{R/S} \to R.$$ Natural questions that arise are is this composition $R$-bilinear nondegenerate, and when is it symmetric? Which symmetric $R$-bilinear forms factor through $\phi^{-1} \times 1$? When we work with a manifold and have a metric tensor that is bilinear and nondegenerate, just how interesting is this flavor of curvature (whatever it is supposed to mean)? - -REPLY [7 votes]: A few remarks: -First, in a sense, (special cases of) this (are) is very commonly studied. Because a bilinear differential form $g$ as you have defined it can naturally be written as a sum $g = \sigma + \alpha$ where $\sigma$ is symmetric and $\alpha$ is skew-symmetric, you are, equivalently, asking about the geometry of the pair $(\sigma,\alpha)$. The most famous example is that of Kähler geometry, i.e., where $\sigma$ is positive definite and $\alpha$ is nondegenerate and parallel with respect to the Levi-Civita connection of $\sigma$, but there are many variations on this (pseudo-Kähler, Hermitian, unitary...) that essentially have the same nature. In general, when $\sigma$ is non-degenerate, you know that there is the canonical Levi-Civita connection of $\sigma$, but, of course, depending on the assumptions you make about $\alpha$, there could be other `canonical' connections. -Second, it is not clear how you ought to define 'non-degenerate' for a general $g$, because there are different notions, and it might well be that what you want to call 'non-degenerate' depends on the applications you have in mind. For example, if $M=\mathbb{R}^2$, and $g = \mathrm{d}x\otimes\mathrm{d}y$, then $g$ is 'degenerate' in the naïve sense (since it has tensor rank $1$ instead of $2$), but both $\sigma = \tfrac12(\mathrm{d}x\otimes\mathrm{d}y+\mathrm{d}y\otimes\mathrm{d}x)$ and $\alpha=\tfrac12(\mathrm{d}x\otimes\mathrm{d}y-\mathrm{d}y\otimes\mathrm{d}x)$ are non-degenerate in the usual senses for symmetric and anti-symmetric quadratic differential forms. In particular, since $\sigma$ has a 'functorial' connection (being non-degenerate), $g$ does also. You'd want to count such a $g$ as 'non-degenerate', no? -Finally, as Dmitry has already remarked, you do need some hypotheses beyond simple algebraic 'non-degeneracy' in order to have a 'functorial' (i.e., invariant under all diffeomorphisms)construction of a connection. Such considerations will enter into any discussion of the right hypotheses, as Vit's discussion above makes clear. I'll just add that sometimes it's important to take higher order derivatives of $g$ into consideration in order to define a class of structures for which a 'functorial' connection exists. For example, as Dmitry pointed out, a closed non-degenerate $2$-form on $M$ does not determine a 'functorial' affine connection on $M$. However, if you drop the assumption 'closed' and replace it with an appropriate higher order non-degeneracy condition, there sometimes is an associated functorial connection. For example, there is an open set $\mathcal{F}$ of germs of $2$-forms on $4$-manifolds that is perserved by (local) diffeomorphisms that does have the property that any $2$-form $\alpha$ on $M^4$ whose germ at every point belongs to $\mathcal{F}$ possesses a functorial torsion-free connection $\nabla^\alpha$, but the definition of $\mathcal{F}$ (and of $\nabla^\alpha$) depends on higher derivatives of $\alpha$ than just first derivatives. (In particular, $\mathrm{d}\alpha$ will be nowhere vanishing if $\alpha$ has its germs belonging to $\mathcal{F}$ at every point.) -Remark: I was asked about how the above $\mathcal{F}$ is defined and how it works. Here's a quick sketch of the result of applying Cartan's method of equivalence to this question: -Start with a non-degenerate $2$-form $\alpha$ on a $4$-manifold $M$. First, there exists a unique $1$-form $\beta$ on $M$ such that $\mathrm{d}\alpha = \beta\wedge\alpha$. Let $\gamma = \mathrm{d}\beta$. Then $0 = \mathrm{d}(\beta\wedge\alpha) = \gamma\wedge\alpha$. Second, there exists a uniqe function $F$ on $M$ such that $\gamma^2 = F\,\alpha^2$, and the first condition defining $\mathcal{F}$ is that $F$ should be nowhere vanishing. For simplicity, I'm going to continue the analysis under the assumption that $F<0$ (there is a similar branch when $F>0$, but I'll leave that to the interested reader). Set $F = -f^2$ where $f>0$. Then, using $\gamma\wedge\alpha= 0$ and $\gamma^2 = -f^2\,\alpha^2$, we see that $(\gamma\pm f\alpha)^2 = 0$, but $(\gamma+f\alpha)\wedge(\gamma-f\alpha) = -2f^2\,\alpha^2\not=0$, so, setting $\gamma\pm f\,\alpha = \pm 2f\,\alpha_{\pm}$, we have -$$ -\alpha = \alpha_+ + \alpha_-\quad\text{and}\quad -\gamma = f\,(\alpha_+ - \alpha_-), -$$ -where $\alpha_\pm$ are a pair of (nonvanishing) decomposable $2$-forms whose wedge product is nonvanishing. -Finally, there are unique decompositions -$$ -\beta = \beta_+ + \beta_- \quad\text{and}\quad \mathrm{d}f = \phi_+ + \phi_- -$$ -where $\beta_\pm \wedge\alpha_{\pm} = \phi_\pm \wedge\alpha_{\pm} = 0$, and so there will be unique functions $g_\pm$ such that -$$ -\beta_\pm\wedge\phi_\pm = g_\pm\,\alpha_\pm\,. -$$ -The final 'open' conditions on $\alpha$ needed to define $\mathcal{F}$ are that $g_+$ and $g_-$ be nonvanishing. -In this case, the $1$-forms $\beta_+$, $\beta_-$, $\phi_+$, and $\phi_-$ define a coframing on $M$ that is functorially associated to $\alpha$. Once one has such a 'canonical' coframing, it is easy to define a connection, in fact, a large family of connections, on $M^4$ (for example, one of these connections (with torsion) will make the given coframing parallel) including some that are torsion-free.<|endoftext|> -TITLE: What is known about the duals of cyclic polytopes? -QUESTION [5 upvotes]: What is known about the duals of cyclic polytopes, in particular, their facets (or equivalently, the vertex-figures of cyclic polytopes)? - -In even dimensions, all facets of the dual are combinatorially equivalent. Are these facets themselves duals of cyclic polytopes? -I think this cannot be true in odd dimensions $\ge 5$. For example, a 5-dimensional cyclic polytope is 2-neighborly, so its vertex figures are only 1-neighborly (is this true?), but 4-dimensional cyclic polytopes are 2-neighborly as well. So when does it happen that the facets are again duals of cyclic polytopes, and when are they all combinatorially equivalent? -In general, is there some kind of classification of the combinatorial types of these facets? - -REPLY [3 votes]: Many properties of the vertex figures of cyclic polytopes can be obtained from Gale evenness condition. -Let $P=C(n,d)$ be a cyclic $d$-polytope, and let $v_1<\cdots -TITLE: Size issues (small/large categories) when defining stacks in the Algebraic/differentiable/topological setting -QUESTION [11 upvotes]: Angelo Vistoli in the notes Notes on Grothendieck topologies, fibered categories and descent theory starts the section of category theory with the following note: - -We will not distinguish between small and large categories. More - generally, we will ignore any set-theoretic difficulties. These can be - overcome with standard arguments using universes. - -Question : Which of the notions introduced in Angelo Vistoli's notes assumes that the category is small? In particular their application to Algebraic/differentiable/topological stacks? -For example, Behrang Noohi puts the following extra condition in his notes on topological stacks: - -Throughout the paper, all topological spaces are assumed to be compactly generated. - -This could be because, the category $\text{Top}$ of all topological spaces is not a small category. -Are there any places one has to be careful to not allow large categories? -Some references to support this question : - -nlab says "In technical terms, a site is a small category equipped with a coverage or Grothendieck topology". It also says (Remark $2.3$ at same page) "Often a site is required to be a small category. But also large sites play a role." -David Metzler in Topological and Smooth Stacks defines (page $2$) a site as a small category equipped with Grothendieck topology. It further says "We will want to discuss, for example, “the category of stacks on the category of all topological spaces,” but strictly speaking this does not exist, since the category of topological spaces does not have a set of objects, but rather a proper class. To avoid this problem we will consider throughout some fixed category $\mathbb{T}$ of topological spaces which has a set of objects, or at least, is equivalent to such a category". - -So, it "looks like", even though one can define a site over a large category, and then a stack over a site (which was defined on a large category), one often restricts (for computational purposes or personal interests) to a small categories and stacks on them. Is this what it is or am I misunderstanding something here? - -REPLY [7 votes]: Are there any places one has to be careful to not allow large categories? - -No. For the purposes of forming the 2-category of algebraic/topological/differentiable stacks, or more generally, some kind of presentable stacks over a large category there are no size issues. Naively, the 2-category of stacks on $S$ is carved out from the presheaf category $[S^{op},\mathbf{Cat}]$ (or $[S^{op},\mathbf{Gpd}]$), which does present size issues for $S$ not essentially small. However, the 2-category of presentable stacks (of groupoids, say, which is the case you are looking at) is equivalent to the bicategory of internal groupoids and anafunctors (and transformations). This can be defined elementarily from the 2-category of internal groupoids, functors and natural transformations. Given a quite weak size condition on the site structure—that is, the size of generating sets of covering families—this bicategory is even locally essentially small. The only case 'in the wild' that I know of that fails this weak condition is the fpqc topology on categories of schemes, and algebraic geometers are a bit wary of that: see tag 0BBK. They are happy to say a single presheaf (of sets, modules, groupoids) is a stack for the fpqc topology, but generally talk about sheaves/stacks for the fppf topology at the finest: see the definition in tag 026O. -Added For a large site not satisfying the condition WISC, the sheafification or stackification functors might not exist. This problem, however, does not impact considering presentable stacks, only when one is wanting to think about arbitrary stacks. For an example of how bad this can get, Waterhouse's paper - -Basically bounded functors and flat sheaves, Pacific Journal of Mathematics 57 (1975), no. 2, 597–610 (Project Euclid) - -gives an example of a presheaf on the fpqc site that does not admit any sheafification. The following quote from the Stacks Project is relevant: - -The fpqc topology cannot be treated in the same way as the fppf topology. Namely, suppose that R is a nonzero ring. We will see in Lemma 34.9.14 that there does not exist a set $A$ of fpqc-coverings of $Spec(R)$ such that every fpqc-covering can be refined by an element of $A$. If $R=k$ is a field, then the reason for this unboundedness is that there does not exist a field extension of $k$ such that every field extension of $k$ is contained in it. -If you ignore set theoretic difficulties, then you run into presheaves which do not have a sheafification, see [Theorem 5.5, Waterhouse-fpqc-sheafification]. A mildly interesting option is to consider only those faithfully flat ring extensions $R\to R'$ where the cardinality of $R'$ is suitably bounded. (And if you consider all schemes in a fixed universe as in SGA4 then you are bounding the cardinality by a strongly inaccessible cardinal.) However, it is not so clear what happens if you change the cardinal to a bigger one. (Tag 022A)<|endoftext|> -TITLE: Can a covering space of the $p$-adic disc split over the circle? -QUESTION [12 upvotes]: Let $D = {\rm Sp}\, \mathbb{C}_p\langle x\rangle$ be the affinoid unit disc over $\mathbb{C}_p$. -Is there an example of a connected finite etale cover of $D$ whose restriction to the "unit circle" ${\rm Sp}\, \mathbb{C}_p\langle x, x^{-1}\rangle \subseteq D$ is disconnected? - -REPLY [5 votes]: $\newcommand{\Sp}{\mathrm{Sp}\,}\newcommand{\cO}{\mathcal{O}}\newcommand{\fm}{\mathrm{m}}$There seem to exist such examples and this does not contradict de Jong's argument because his proof only shows that the homomorphism $G:=\pi_1(\Sp C\langle x^{\pm 1}\rangle)\to H:=\pi_1(\Sp C\langle x\rangle)$ satisfies the following property: if an action of $H$ on a finite set becomes trivial when restricted to $G$ then it is trivial. Equivalently, the normal subgroup generated by the image of $G$ is equal to the whole of $H$. -Take $$A= C\langle x,y\rangle/(y^{p+1}-xy-p^{1/2})$$ The discriminant of this polynomial in $y$ is $-(p+1)^{p+1}p^{p/2}-p^px^{p+1}$ up to a unit and is invertible in $C\langle x\rangle$. Therefore, $C\langle x\rangle \to A$ is a finite etale extension. -To check that $\Sp A$ is connected it is enough to show that the polynomial $y^{p+1}-xy+p^{1/2}$ is irreducible in $C\langle x\rangle[y]$. If $y^{p+1}-xy-p^{1/2}=f_1(y)\cdot f_2(y)$ is a factorization into a product of monic polynomials then both $f_1,f_2$ have to lie in $\cO_C\langle x\rangle[y]$ so their reductions modulo the maximal ideal $\fm_C\subset\cO_C$ provide a factorization of $(y^{p}-x)y$. Hence, we may and will assume that $\deg f_1=1$ and $y^{p+1}-xy-p^{1/2}$ has a root $f(x)\in \cO_C\langle x\rangle$. A root has to have $f(0)^{p+1}=p^{1/2}$ and, taking the derivative of the equation $f(x)^{p+1}-xf(x)+p^{1/2}=0$ we also get $(p+1)f'(x)f(x)^p-f(x)-xf'(x)=0$ so $f'(0)=\frac{1}{p+1}f(0)^{1-p}\not\in \cO_C$ which is a contradiction. Therefore, $\Sp A\to D$ is a connected finite etale cover. -On the other hand, this cover splits over $C\langle x,x^{-1}\rangle $: we can find a root of $y^{p+1}-xy-p^{1/2}=0$ by the Hensel's lemma arguing by induction on $n$: suppose that $y_n\in \cO_C/p^{n/2}[x^{\pm 1}]$ is a root of this equation modulo $p^{n/2}$ that reduces to $0$ mod $p^{1/2}$. To lift it over $p^{(n+1)/2}$ it is enough to be able to supply, for any given $a\in\cO_C/p^{1/2}[x^{\pm 1}]$, an element $z$ such that $(p+1)y_n^p\cdot z-x\cdot z=a$ in $\cO_C/{p^{1/2}}[x^{\pm 1}]$. This is possible because $y_n^p-x$ is a sum of a nilpotent and invertible element, hence is invertible. - -This example is motivated by the behavior of the $p$-torsion in a family of elliptic curves: given such family $\mathcal{E}\to S$ over a $p$-adic formal scheme $S$ over $\cO_K$ let $S_0(p)\to S_K$ be the etale covering of the generic fiber parametrizing $1$-dimensional $\mathbb{F}_p$ subspaces in $\mathcal{E}_K[p]$. When restricted to the generic fiber of the ordinary locus $S_{ord}$ this covering admits a section by the theory of canonical subgroup but it need not have one over the whole of $S_K$. In other words the restriction of the representation $\pi_1(S_K)\to GL_2(\mathbb{F}_p)$ to $\pi_1(S_{ord})$ lands inside a Borel subgroup. -Using Katz-Mazur-Drinfeld integral model the etale cover $S_0(p)\to S_K$ extends to a flat cover of $S$ itself and the special fiber splits into two irreducible components exactly as in the example above.<|endoftext|> -TITLE: Cobordism monopole Floer homology -QUESTION [6 upvotes]: From the famous book: Monopole and three manifold, Kronheimer and Mrowka(https://www.maths.ed.ac.uk/~v1ranick/papers/kronmrowka.pdf). It is known that: -Let $Y$ be a closed oriented $3$ manifold, choosing a spinc structure $\mathfrak s$ and metric $g$ and a generic perturbation $p$, one can construct the monopole Floer homology groups: -$$\check{HM}_*(Y,\mathfrak s, g,p),~\hat{HM}_*(Y,\mathfrak s, g,p),~\overline{HM}_*(Y,\mathfrak s, g,p).$$ -The groups are graded over a set $\mathbb J_s$ admitting a $\mathbb Z$ action.( details are given in Section 20-22). We define the negative completions(Definition 23.1.3 of the book) by -$$\check{HM}_\bullet(Y,\mathfrak s, g,p),~\hat{HM}_\bullet(Y,\mathfrak s, g,p),~\overline{HM}_\bullet(Y,\mathfrak s, g,p).$$ -If we want to consider all spinc -structures at the same time, we need to consider the completed monopole Floer homology -$$\check{HM}_\bullet(M,F;\mathbb F)=\bigoplus_\mathfrak s\check{HM}_\bullet(M,F,\mathfrak;\mathbb F).$$ -To show that these homology groups are independent of the metric and the perturbation, the authors gave a property: a cobordism between 3-manifolds gives rise to homomorphisms between -their monopole Floer homologies(see Section 23-26). They construct a homomorphism from $\check{HM}_\bullet(Y,g_1,p_1)$ to $\check{HM}_\bullet(Y',g_2,p_2)$, where there is a cobordism from $Y$ to $Y'$. -Q I do not understand the two points below: - -Why the authors use the negative completion, where we need it? -If we just want to show that the monopole Floer homology $\check{HM}_*(Y,\mathfrak s)$ is independent of the metric and perturbation, can we just using the trivial cobordism $[0,1]\times Y$ to show a homomorphism $\check{HM}_*(Y,\mathfrak s,g_1,p_1) \to \check{HM}_*(Y,\mathfrak s, g_2,p_2)$? The homomorphism is given by counting the number of solutions of the zero-dim moduli space $M([a_1],W^*,[b_2])$, where $W^*=(-\infty,0]\times Y\cup I\times Y\cup[1,\infty)\times Y$, and $[a_1]$ and $[b_2]$ are the critical points of $(Y,\mathfrak s,g_1,p_1)$ and $(Y,\mathfrak s,g_2,p_2)$ respectively. I think the arguments of Section23-25 also work before taking the negative completion . - -PS Let $G_*$ be an abelian group graded by the set $\mathbb J$ equipped with a $\mathbb Z$-action. Let $O_a(a\in A)$ be the set of free $\mathbb Z$-orbits in $\mathbb J$ and fix an element $j_a\in O_a$ for each $a$. Consider the subgroups -$$G_*[n]=\bigoplus_a\bigoplus_{m\geq n} G_{j_a-m},$$ - which form a decreasing filtration of $G_*$. We define the negative completion of $G_*$ as the topological group $G_\bullet\supset G_*$ obtained by completing with respect to this filtration. - -REPLY [3 votes]: The first bullet is definitely explained in the book! Surely around where it was introduced, it has to do with summing over all spin-c structures. We need to pass to the completion because the 4-manifold can have infinitely many spin-c structures that would need to be used. -The second bullet, yes. In general we should not expect results concerning completions of a graded group to also hold for the uncompleted group. But here we consider the trivial cobordism, and spin-c structures on $[0,1]\times Y$ are the same as spin-c structure on $Y$, so no completion is needed in this situation.<|endoftext|> -TITLE: Partial sums of $\sum_0^\infty z^n$ -QUESTION [15 upvotes]: Let $z$ be a complex number with $|z|<1$. For every subset $A\subset\mathbb N$, the series $\sum_{m\in A}z^m$ is convergent. Denote $S(A)\in\mathbb{C}$ its sum and $\Sigma_z$ the set of all numbers $S(A)$. Remark that the cardinal of $\Sigma_z$ is (likely) that of ${\cal P}({\mathbb N})$, the continuum. - -Is it possible that $\Sigma_z$ be a neighbourhood of the origin ? - -REPLY [12 votes]: The number $z = i/\sqrt2$ seems to work! -Given $x \in [-2/3,4/3]$ we can find a "negabinary" expansion -$$x = \sum_{k=0}^\infty (-1)^k\frac{b_k}{2^k},$$ where each $b_k \in \{0,1\}$. -Similarly, given $y \in [-2/3\sqrt2,4/3\sqrt2]$ we can find -$$y = \frac{1}{\sqrt2}\sum_{j=0}^\infty (-1)^j\frac{c_j}{2^j},$$ -where each $c_j \in \{0,1\}$. -Therefore, -$x + iy = \sum_{n \in A} z^n$ where -$$A = \{2k : b_k = 1\} \cup \{2j+1 : c_j = 1 \}.$$ - -As explained in comment contributions by Bart Michels, Jochen Wegenroth and myself, $|z| \geq 1/\sqrt2$ is necessary. By definition, $$\Sigma_z = z\Sigma_z\cup(1+z\Sigma_z).$$ -If $\mu$ denotes Lebesgue measure, then $\mu(z\Sigma_z) = |z|^2\mu(\Sigma_z)$ thus $\mu(\Sigma_z) \leq 2|z|^2\mu(\Sigma_z)$. Since $\Sigma_z$ is compact, it has finite measure and thus if $\mu(\Sigma_z)>0$ then we must have $|z|^2 \geq 1/2$. -It remains open whether $|z|\geq1/\sqrt2$ and $z \notin \mathbb{R}$ is sufficient for $\Sigma_z$ to contain $0$ in its interior.<|endoftext|> -TITLE: What do we learn from the Wronskian in the theory of linear ODEs? -QUESTION [32 upvotes]: For a real interval $I$ and a continuous function $A: I \to \mathbb{R}^{d\times d}$, let $(x_1, \dots, x_d)$ denote a basis of the solution space of the non-autonomous ODE -$$ - \dot x(t) = A(t) x(t) \quad \text{for} \quad t \in I. -$$ -The mapping -$$ - \varphi: I \ni t \mapsto \det(x_1(t), \dots, x_d(t)) \in \mathbb{R} -$$ -is usually called the Wronskian of the basis $(x_1,\dots,x_d)$, and it seems to be an obligatory topic in every ODE course or book that I've seen. -So in an ODE course that I am currently teaching, I'm facing the following problem: -(1) Despite its prevalence in courses and textbooks, I've rarely (not to say never) encountered any situation where the Wronskian of an ODE is used in a way that sheds non-trivial insight onto the problem at hand - in particular not in any of the books where I've read about it. (Of course, I have also searched on the internet for it, but without any success.) -(2) I feel quite uneasy to teach a concept which I am unable to motivate properly. -(3) I'd feel even more uneasy to just omit it from the course, since chances are that my not knowing of an application of the Wronskian is just due to my ignorance. -Well, what I did is to merely mention the Wronskian in a remark - but of course (and fortunately) I did not get away with it, because quite soon a student asked what the Wronskian is good for. -So this is the -Question: What is the Wronskian (in the context of linear ODEs) good for? -Remarks. - -One can show that $\varphi$ satisfies the differential equation -$$ - \dot \varphi(t) = \operatorname{tr}(A(t)) \varphi(t), -$$ -and since this a one-dimensional equation we have the solution formula -$$ - (*) \qquad \dot \varphi(t) = e^{\int_{t_0}^t \operatorname{tr}(A(s)) \, ds} \varphi(t_0) -$$ -for it (for any fixed time $t_0$ and all $t \in I$). This is nice - but still I can't see how to explain to my students that it is useful. -I've often seen discussions to the end that $(*)$ implies that "the Wronskian is non-zero at a time $t_0$ if and only if it is non-zero at every time $t$" - but I find this somewhat straw man-ish: the fact that $(x_1(t), \dots, x_d(t))$ is linearly independent at one time $t_0$ if and only if it is linearly independent at every time $t$ is an immediate consequence of the uniqueness theorem for ODEs, without any reference to the Wronskian. -One can give a geometrical interpretation of $(*)$: For instance, if all the matrices $A(t)$ have trace $0$, and it follows that the (non-autonomous) flow associated with our differential equation is volume preserving. However, I'm not convinced that this serves as a sufficient motivation to give the mapping $t \mapsto \det(x_1(t), \dots, x_d(t))$ its own name and to discuss it in some detail. -Maybe a word on the notion "good for" that occurs in the question: I'm pretty comfortable with studying and teaching mathematical objects just in order to better understand them, or for the sake of their intrinsic beauty. However, whenever we do so, this usually happens within a certain theoretical context - i.e., we build a theory, introduce terminology, and this terminology somehow contributes to the development (or to our understanding) of the theory. -Some my question could be rephrased as: -"I'm looking either (i) for applications of the Wronskian of ODEs to concrete problems (within or without mathematics) or (ii) for ways in which the concept 'Wronskian' facilitates our understanding of the theory of ODEs (or of any other theory)." -The term 'Wronskian' also seems to be used with a more general meaning (see for instance this Wikipedia entry). However, I am specifically interested in the Wronskian for the solutions of a linear ODE. - -REPLY [7 votes]: This is in the same spirit as Piyush Grover's comment. The determinant $\det(x_1(t),\ldots,x_n(t))$ definitely deserves a name (not only in the context of linear ODE's). In such a lecture students could (and ,in my opinion, should) learn the meaning of the divergence of a vector field $F$. Having learned Picard-Lindelöf they are ready to understand the flow $\phi(t,x)$ as he solution of the initial value Problem $\phi'(t,x)=F(\phi(t,x))$, $\phi'(0,x)=x$, and for a small cube -$x+[0,r]^n$ you can throw the edges into the flow to get after a short time almost the parallelepiped with edges $\phi(t,x+re_j)-\phi(t,x)$ whose (oriented) volume compared to the volume of the cube is $$v(t,r)=\det[\phi(t,x+re_1)-\phi(t,x),\ldots,\phi(t,r+e_n)-\phi(t,x)]/r^n$$ If you take the derivative $\partial_t$ at $0$ and the limit $r\to 0$ you get the divergence of the vector field (no problem to take time dependent vector fields). -The desire to make this more precise also motivates the theorems about differentiability of the solutions of initial value problems with respect to the initial values. Then you can throw quite arbitrary small sets into the flow and compare the evolved (oriented) volume with the original one by calculating them with the $n$-dimensional substitution rule.<|endoftext|> -TITLE: Size of a family of sets of $k$-separated functions over $\{0,1,\ldots,n-1\}$ -QUESTION [5 upvotes]: For $n\ge 1$ we write $[n]$ to denote the set $\{0,1,\ldots,n-1\}$. Let $2^{[n]}$ be the set of all functions from $[n]$ to $\{0,1\}$. Let $\mathcal{F}$ and $\mathcal{G}$ be two nonempty subsets of $2^{[n]}$. Fix $1\le k\le n$. Take $0\le j \le n-k$. We say that $f,g\in 2^{[n]}$ disagree over $[j,j+k)$ if the restrictions of $f$ and $g$ to $\{j,j+1,\ldots,j+k-1\}$ are different, that is, $f|_{\{j,j+1,\ldots,j+k-1\}}\neq g|_{\{j,j+1,\ldots,j+k-1\}} $. We say that $f\in 2^{[n]}$ disagree with a nonempty $\mathcal{G}\subset 2^{[n]}$ over $[j,j+k)$ if for every $g\in\mathcal{G}$ we have that $f$ and $g$ disagree over $[j,j+k)$. That is, for some fixed $j$ the function $f$ disagress over $[j,j+k)$ with all functions in $\mathcal{G}$. -We say that $\mathcal{F}$ and $\mathcal{G}$ are $k$-separated if there exists $0\le j \le n-k$ such that -either there exists $f\in\mathcal{F}$ which disagrees with $\mathcal{G}\subset 2^{[n]}$ over $[j,j+k)$ or there exists $g\in\mathcal{G}$ which disagrees with $\mathcal{F}\subset 2^{[n]}$ over $[j,j+k)$. -Let $\mathscr{T}_{n,k}$ be the largest possible family of pairwise $k$-separated nonempty subsets of $2^{[n]}$, that is, $\mathscr{T}_{n,k}$ is a family of maximal cardinality among all families $\mathscr{F}$ such that if $\mathcal{F},\mathcal{G}\in \mathscr{F}$ and $\mathcal{F}\neq \mathcal{G}$ then $\mathcal{F}$ and $\mathcal{G}$ are $k$-separated. -Keeping $k$ fixed I am looking for an asymptotic behaviour of the cardinality, $|\mathscr{T}_{n,k}|$ of $\mathscr{T}_{n,k}$ as $n\to\infty$. In particular, I hope that -$$ -\frac{|\mathscr{T}_{n,k}|}{2^{2^n}}\to 1\text{ as }n\to\infty. -$$ - -REPLY [3 votes]: 1: Hopes. Let me begin by taking away your hope - that is, disproving the conjectured asymptotic, $|{\mathscr{T}_{n,k}}|/2^{2^n} \to 1$. -I will say that a family $\mathcal{F}$ is $k$-rich if for each $j \in [n]$ and $w \in \{0,1\}^k$, there is $f \in \mathcal{F}$ with $f|_{[j,j+k)} = w$. Accordingly, I will say that a pair $(j,w) \in [n] \times \{0,1\}^k$ "bad" for $\mathcal{F}$ if $f|_{[j,j+k)} \neq w$ for all $f \in \mathcal{F}$, so that $\mathcal{F}$ is $k$-rich if it has no "bad" pair $(j,w)$. -Claim. No two $k$-rich families are $k$-separated. -Proof. For any sequence $g \in \{0,1\}^n$, any $j \in [n]$ and any $k$-rich family $\mathcal{F}$, there exists $f \in \mathcal{F}$ such that $g|_{[j,j+k)} = f_{[j,k+j)}$. Hence, $g$ does not disagree with $\mathcal{F}$ over $[j,j+k)$. The rest follows by unwinding definitions. -Claim. The number of families that are not $k$-rich is $o(2^{2^n})$. -Proof. Recall taht for each family $\mathcal{F}$ that is not $k$-rich, there exist a "bad" pair $j \in [n]$, $w \in \{0,1\}^k$. Given $j \in [n]$, $w \in \{0,1\}^k$, the number of $f \in \{0,1\}^{[n]}$ with $f|_{[j,j+k)} \neq w$ is $(1-2^{-k})2^n$. Hence, the number of families $\mathcal{F}$ for which the pair $(j,w)$ is bad is $2^{(1-2^{-k})2^n}$. By the union bound, the number of families that are not $k$-richis at most $2^{2^n} \times {2^k n}/{2^{2^{n-k}}}$. If $k$ is kept constant and $n \to \infty$, then ${2^k n}/{2^{2^{n-k}}} \to 0$. -2: Asymptotics. We will show that $\lim_{n \to \infty} \log |\mathscr{T}_{n,k}|/n = 2^{k(1+o(1))}$, where the $o(1)$ term tends to $0$ as $k \to \infty$. -For a given family $\mathcal{F}$, let $B(\mathcal{F})$ denote the set of ``bad'' pairs $j,w$: -$$B(\mathcal{F}) = \{ (j,w) \ : \ f_{[j,j+k)} \neq w \quad \forall f \in \mathcal{F}\} \subset [n] \times \{0,1\}^k.$$ -Claim: Two families $\mathcal{F}, \mathcal{G}$ are $k$-separated if and only if $B(\mathcal{F}) \neq B(\mathcal{G})$. -Proof: Suppose that $f \in \mathcal{F}$ and $j$ is such that $f$ disagrees with $\mathcal{G}$ on $[j,j+k)$. Then $(j,f|_{[j,j+k)}) \in B(\mathcal{G}) \setminus B(\mathcal{F})$. Conversely, if $(j,w) \in B(\mathcal{G}) \setminus B(\mathcal{F})$ then there exists $f \in \mathcal{F}$ such that $w=f|_{[j,j+k)}$, and since $(j,w) \in B(\mathcal{G})$, $f$ disagrees with $\mathcal{G}$ on $[j,j+k)$. -Since each two sets in $\mathscr{T}_{n,k}$ are $k$-separated, each of them corresponds to a different subset of $[n] \times \{0,1\}^k$, and so $$|\mathscr{T}_{n,k}| \leq 2^{n2^k}.$$ -In the opposite direction, let $\mathscr{B}$ denote the family of all sets $B \subset [n] \times \{0,1\}^k$ such that $k \mid j$ and $\sum_{i} w_i \equiv 1 \bmod{2}$ for all $(j,w) \in B$. -Claim: For each $B \in \mathscr{B}$ there exists a family $\mathcal{F}_B$ such that $B(\mathcal{F}_B) = B$. -Proof: Let $\mathcal{F}$ consis of all sequences $f \in \{0,1\}^n$ such that $f|_{[j,j+k)} \neq w$ for each $(j,w) \in B$. Clearly, $B \subset B(\mathcal{F})$ so it remains to show that $B(\mathcal{F})$ contains no other pair. Let $(i,u) \in [n] \times \{0,1\}^k \setminus B$ be any such tentative pair. We need to construct $f \in \mathcal{F}$ with $f_{[i,i+k)} = u$. On each interval $[j,j+k)$ with $k \mid j$ and $[i,i+k) \cap [j,j+k) = \emptyset$, put $f|_{[j,j+k)} = 0^k$. If $i < j < i+k$, $k \mid j$, then set $f|_{[i,j)} = u_{[0,j-i)}$, $f_j = \sum_{t= -TITLE: What is the state-of-the-art for solving polynomials systems over fields that are not algebraically closed? -QUESTION [8 upvotes]: I am not working in the field of algorithmic algebraic geometry - yet, for my current work, I need some results from it. -More specifically, what is the state-of-the-art when it comes to solving (whatever "solving" means in this case) system of polynomials of fields that are not algebraically closed, whose ideal has dimension $>0$? -Could you recommend a survey paper that summarizes what has been achieved so far? -For the case of $0$-dimensional ideals, there seems to exist many heavily cited papers, like "Solving Zero-dimensional Algebraic Systems" by D. Lazard, which seem mostly to be concerned with finding ways of to display the system of polynomials in a nice way (e.g. triangularly). Are these articles already superseded, or does it make sense to read them? -Edit: In particular, I'm interested in the field $\mathbb{R}$, since most of my example will come from here (but $\mathbb{Q}$ might be also useful; and perhaps even the ring $\mathbb{Z}$; I don't yet know where the results I will get for $\mathbb{R}$ will take me). -Also worth making more precise: In the case of positive dimension of the ideal, I'm interested in methods that tell me, if I project to whole, infinite solution space down to a single variable and I'm interested in, in what set this variable lies. More formally, if $V(f_1,\ldots,f_s)\subseteq F^n$ is my solution variety, with $f_i \in F[x_1,\ldots,x_n]$, and I'm interested in some specific variable, say $n_0$, what methods are there that describe $\mathop{\rm proj}_{n_0}(V(f_1,\ldots,f_s))$? - -REPLY [3 votes]: For the reals, I particularly like the book by Sturmfels mentioned by Alexandre Eremenko. For the rational numbers, you can hardly do better than Bjorn Poonen's book Rational Points on Varieties, which is available for browsing via his homepage. -For dimension $1$ specifically, Poonen also has a set of lecture notes on rational points on curves, although I always have trouble finding it. Moreover he has several expository articles (listed as such on his page) dealing with rational points on curves. -Restricted to the case of the field of rational numbers and dimension $1$ alone, this is a huge question. Restricting only to the field of rational numbers makes it even huger. Dropping any restrictions on the field entirely makes it well-nigh impossible to answer in full geberality...<|endoftext|> -TITLE: Semisimplicity for tensor products of representations of finite groups -QUESTION [20 upvotes]: Let $G$ be a group and $k$ a field of characteristic $p>0$. Let $$\rho_i: G\to GL(V_i),~ i=1,2$$ be two finite-dimensional semisimple $k$-representations of $G$, with $\dim(V_1)+\dim(V_2) -TITLE: Is there a prefix-continuous bijection between finite words and eventually zero words? -QUESTION [6 upvotes]: Let -$$ X = \{x \in \{0,1\}^{\omega} \;|\; \exists m: \forall i \geq m: x_i = 0\} $$ -(one-way infinite eventually zero words). Let $\{0,1\}^*$ denote the finite (not necessarily nonempty) words over $\{0,1\}$, and write $\{0,1\}^{\leq k} = \{w \in \{0,1\}^* \;|\; |w| \leq k\}$ where $|w|$ denotes length. -Is there a bijection $\phi : X \to \{0,1\}^*$ such that -$$ \exists n \in \mathbb{N}: \forall a \in \{0,1\}: \forall x \in \{0,1\}^{\mathbb{N}}: \exists b, c \in \{0,1\}^{\leq n}: \exists y \in \{0,1\}^*: \phi(x) = b \cdot y \wedge \phi(a \cdot x) = c \cdot y $$ -holds, where $\cdot$ is concatenation? -This is a kind of coarse uniformity / bornologousness assumption: $\phi$ needs to be bornologous between the two sets, seen as metric spaces with the path metric of the graph structure where $x$ and $y$ are adjacent if $y = ax$ or $x = ay$ for some $a \in \{0,1\}$. This seems vaguely familiar to me but I don't know from where, and I'm not seeing how to construct $\phi$. The straightforward idea of cutting out the zero tail doesn't work because it's not surjective, and I run into trouble trying to fix that. But I also didn't manage to prove impossibility because there's a lot of freedom. -The question arises in some (leisurely) research, so asking here instead of math.SE even if it might be safer to start there with this one. Geometric group theory tag because this is related to Thompson's $V$, even if I didn't elaborate and I doubt it's useful (every countable group acts freely on $\{0,1\}^*$). - -REPLY [6 votes]: I believe the following works, but I might be missing something. -If $x$ has at least two 1s, then $\phi(x)$ is the sequence cut just before the last 1: -$$\phi(0011101000\cdots) = 001110 $$ -If $x$ has at most one 1, then $\phi(x)$ is the sequence cut before the one, with an additional zero: -$$ \phi(000\cdots) = \varnothing,\qquad \phi(001000\cdots) = 000 $$ -The inverse bijection is $\psi(\varnothing)=000\cdots$, $\psi(y)=y\cdot1000\cdots$ when $y$ has at least one 1, and $\psi(y\cdot0)=y\cdot 1$ for $y$ consisting only of zeros (possibly $y=\varnothing$). -Then $\phi(a\cdot x)=a\cdot\phi(x)$ if $x$ has at least two 1s, and the case of $000\cdots$ is unimportant, up to taking a large $n$ (although $n=1$ works so far). Now if $x$ consists only of zeros, -$$\phi(0\cdot y\cdot 1000\cdots)=0 \cdot y\cdot 0=00\cdot y\qquad\text{ and }\qquad\phi(1\cdot y\cdot 1000\cdots)=1\cdot y$$ -while $\phi(y\cdot1000\cdots)=y\cdot0=0\cdot y$. -So the hypotheses are satisfied with $n=1$.<|endoftext|> -TITLE: a square root inequality for symmetric matrices? -QUESTION [7 upvotes]: In this post all my matrices will be $\mathbb R^{N\times N}$ symmetric positive semi-definite (psd), but I am also interested in the Hermitian case. In particular the square root $A^{\frac 12}$ of a psd matrix $A$ is defined unambigusouly via the spectral theorem. -Also, I use the conventional Frobenius scalar product and norm -$$ -:=Tr(A^tB), -\qquad -|A|^2:= -$$ - -Question: is the folowing inequality true - $$ -|A^{\frac 12}-B^{\frac 12}|^2\leq C_N |A-B|\quad ??? -$$ - for all psd matrices $A,B$ and a positive constant $C_N$ depending on the dimension only. - -For non-negative scalar number (i-e $N=1$) this amounts to asking whether $|\sqrt a-\sqrt b|^2\leq C|a-b|$, which of course is true due to $|\sqrt a-\sqrt b|^2=|\sqrt a-\sqrt b|\times |\sqrt a-\sqrt b|\leq |\sqrt a-\sqrt b| \times |\sqrt a+\sqrt b|=|a-b|$. -If $A$ and $B$ commute then by simultaneous diagonalisation we can assume that $A=diag(a_i)$ and $B=diag(b_i)$, hence from the scalar case -$$ -|A^\frac 12-B^\frac 12|^2 -=\sum\limits_{i=1}^N |\sqrt a_i-\sqrt b_i|^2 -\leq \sum\limits_{i=1}^N |a_i-b_i| -\leq \sqrt N \left(\sum\limits_{i=1}^N |a_i-b_i|^2\right)^\frac 12=\sqrt N |A-B| -$$ -Some hidden convexity seems to be involved, but in the general (non diagonal) case I am embarrasingly not even sure that the statement holds true and I cannot even get started. Since I am pretty sure that this is either blatantly false, or otherwise well-known and referenced, I would like to avoid wasting more time reinventing the wheel than I already have. - -This post and that post seem to be related but do not quite get me where I want (unless I missed something?) - -Context: this question arises for technical purposes in a problem I'm currently working on, related to the Bures distance between psd matrices, defined as -$$ -d(A,B)=\min\limits_U |A^\frac 12-B^\frac 12U| -$$ -(the infimum runs over unitary matrices $UU^t=Id$) - -REPLY [10 votes]: The classical operator generalization of the scalar inequality $|\sqrt{a}-\sqrt{b}|^2 \leq |a-b|$ is the Powers-Størmer inequality, which involves two different norms : the trace norm $\|X\|_1 = \operatorname{Tr}|X|$ and the Froebenius norm $\|X\|_2 = (\operatorname{Tr}(X^* X))^{\frac 1 2}$, where $|X| = (X^* X)^{\frac 1 2}$ is the usual absolute value of matrices. It says that for all positive matrices $A,B$ (or operators on a Hilbert space), -$$ \|\sqrt{A} - \sqrt{B}\|_2^2 \leq \|A-B\|_1.$$ -It implies a positive answer to your question with $C_N = \sqrt{N}$, because by Hoelder's inequality, $\|A-B\|_1 \leq \sqrt{N} \|A-B\|_2$. The constant is optimal (take $A=\operatorname{Id},B=0$). -The Powers-Størmer surprisingly does not have a wikipedia page yet, but it probably appears in most textbooks on operator algebras or matrix analysis. The original reference is R. T. Powers, E. Størmer, Free states of canonical anticommutation relations, Commun. Math. Phys. 16, 1-33 (1970).<|endoftext|> -TITLE: The correct homotopically relevant notion of ideals of dg-algebras (or $\mathbb E_1$-rings) -QUESTION [6 upvotes]: I'm trying to figure out what an ideal of a, say, dg-algebra (or, if you prefer, $\mathbb E_1$-ring) $R$ is in a homotopically relevant fashion, but I can't actually figure it out. I can assume that $R$ is concentrated in cohomologically nonpositive degrees (or homologically nonnegative degrees). I have stumbled upon a few possibilities: - -There is a notion of monomorphism in an $\infty$-category; hence, I would consider the derived category $\mathsf D(R)$ of $R$-dg-modules as an $\infty$-category and say that $I \to R$ is an "ideal" if it is a monomorphism according to that notion. This notion is also used in Spectral algebraic geometry (Remark C.2.3.4. page 1965) in the framework of Grothendieck prestable $\infty$-categories. -On the other hand, I find other sources such as this and this. From the first one, I quote: - - -There are other things that are weird for commutative ring spectra. Quite often, we end up working with ideals in the graded commutative ring of homotopy groups, but as we saw above, this is not a suitable notion of ideal.There is a notion of an ideal in the context of (commutative) ring spectra [53] due to Jeff Smith, but still several algebraic constructions do not have an analogue in spectra. - - -Given that, I'm pretty confused. Perhaps the notion of monomorphism (1) is fine, but in the case of commutative ring spectra it does not work really well, hence the issues I found (2)? I've tried to skim through some literature on derived algebraic geometry, but still I couldn't find any satisfying answer... - -REPLY [3 votes]: At least in commutative situations, I would argue that a good notion is simply an ideal in $H^0(R)$. -For example, the theory of local cohomology works just as well as it does for commutative rings, as long as you do it with respect to ideals in $H^0(R)$. -Similarly, you can take "derived quotients" with respect to a finite sequence of elements in $H^0(R)$, by taking the Koszul complex with respect to such a sequence. -See for example my very recent paper from last week: -"Koszul complexes over Cohen-Macaulay rings" -https://arxiv.org/abs/2005.10764<|endoftext|> -TITLE: Generalized figures of constant width -QUESTION [9 upvotes]: Is it known which plane figures $Q$ can rotate touching three given circles $A$, $B$, and $C$? - - -This question was asked by Lazar Lyusternik in 1946, there is only one reference to this paper that solves the problem in one limit case. - -Do you know of any other research on this question? - -(I learned this question from Sergei Tabachnikov.) - -REPLY [10 votes]: I have looked at Goldberg's paper referenced by J. J. Castro in his excellent answer. It turns out that there is a simpler (and more general way) to generate Goldberg's non-circular solutions, so I thought that I would just mention that. -The idea is to consider the 'rotor' (Goldberg's term) as fixed and as the envelope of a circle in periodic motion. It's a bit easier if you use complex notation, i.e., think of $\mathbb{R}^2$ as $\mathbb{C}$. Consider a circle of radius $r>0$ and center $z_0=1$, parametrized by $z(s) = 1 + r\mathrm{e}^{is}$, and move it by a circle of rigid motions: $R_t(z) = \mathrm{e}^{it}z + a(t)$, where $a$ is periodic of period $2\pi/n$ for some integer $n\ge 3$. Now take the two envelopes of this $1$-parameter family of circles $R_t\bigl(z(s)\bigr)$. Because of the periodicity of $a$, these will also be envelopes of the circles of radius $r$ centered at the other $n$-th roots of unity, and hence these envelopes will have the property that, when they are moved by the inverse of $R_t$, they will remain tangent to the $n$ circles of radius $r$ centered on the $n$-th roots of unity. -A simple computation shows that the two envelopes of this family are given by -$$ -E_\pm(t) = \mathrm{e}^{it}+a(t) \pm \frac{r\bigl(\mathrm{e}^{it}-ia'(t)\bigr)}{\bigl|\mathrm{e}^{it}-ia'(t)\bigr|}. -$$ -(Note that this is well-defined as long as $\mathrm{e}^{it}-ia'(t)$ never vanishes, which can be ensured, for example, by choosing $a$ so that $|a'(t)|<1$.) -Setting $a=0$ (or a constant) gives the trivial circle solution. Below, I give an example of $E_-$ (the interior envelope) with $n=3$, $r=1/2$, and $a(t) = \tfrac17\sin(3t)$, or, more precisely, $R_{-s}(E_-)$ as $s$ varies between $0$ and $2\pi$. (Replacing this by $a(t) = \tfrac16\sin(3t)$ yields a nonconvex solution. In general, when $a$ is allowed to be large, the envelopes will have cusps.)<|endoftext|> -TITLE: If $X$ is separable space then $X^∗$ is separable in all topologies $\tau$ such that $(X^∗,\tau)^∗ =X$? -QUESTION [6 upvotes]: Let $(X,\|.\|_{X})$ be a separable Banach space and the associated dual space is denoted -by $X^*$. By $w^*$ we shall indicate the weak$-*$ topology on $X^*$. -Let $B_{X^∗}= \{x^∗ \in X^∗ : \|x^∗\|_{X^∗}\leq 1\}$. Since $X$ is separable, the set $B_{X^∗}$ furnished with the relative $w^∗-$topology is compact (by the Alaoglu theorem) and metrizable (see Theorem I.5.85). Note that -$$ -X^*=\bigcup_{n}{nB_{X^*}} -$$ -hence $X^∗_{w^∗}$ (the space $X^∗$ furnished with the $w^∗-$topology) is separable. -Can we say that : $X^∗$ is separable in all topologies $\tau$ such that $(X^∗,\tau)^∗ =X$? - -REPLY [5 votes]: Yes. Separability of a locally convex space is equivalent to the existence of a dense countable-dimensional subspace. All compatible topologies for a dual pair (i.e. those that give the same continuous linear functionals) agree on the closures of convex sets.<|endoftext|> -TITLE: Scalar product of random unit vectors -QUESTION [5 upvotes]: Let $X,X'$ be two random vectors on the sphere $S^{d-1}$. What is the distribution of their dot product $X\cdot X'$ in the following cases: - -$X,X'$ independent with uniform distribution on the sphere $S^{d-1}$ -$X\in S^{d-1}$ deterministic, $X'$ uniformly distributed on $S^{d-1}$ -? - -REPLY [4 votes]: As noted in the comments, by the spherical symmetry, the distribution of the dot product in both parts of your question is the same that of $X\cdot(1,0,\dots,0)$. Moreover, the distribution of $X$ is the same as that of the random vector -$$\frac{Z}{\sqrt{Z_1^2+\dots+Z_d^2}},$$ -where $Z=(Z_1,\dots,Z_d)$ is a standard normal random vector. -So, the distribution of the dot product in question is the same that of -$$R:=\frac{Z_1}{\sqrt{Z_1^2+\dots+Z_d^2}}.$$ -The distribution of $R$ is obviously symmetric, and the distribution of $R^2$ is the beta distribution with parameters $\frac12,\frac{d-1}2$. It follows that the probability density function (pdf) $f_R$ of $R$ is given by -$$f_R(r)=\frac{\Gamma \left(\frac{d}{2}\right)}{\sqrt{\pi }\, \Gamma - \left(\frac{d-1}{2}\right)}\,\left(1-r^2\right)^{\frac{d-3}{2}}\, 1\{|r|<1\},$$ -and the dot product in question has the same pdf.<|endoftext|> -TITLE: SOS polynomials with integer coefficients -QUESTION [27 upvotes]: A well known theorem of Polya and Szego says that every non-negative univariate polynomial $p(x)$ can be expressed as the sum of exactly two squares: $p(x) = (f(x))^2 + (g(x))^2$ for some $f, g$. Suppose $p$ has integer coefficients. In general, its is too much to hope that $f, g$ also have integer coefficients; consider, for example, $p(x) = x^2 + 5x + 10$. Are there simple conditions we can impose on $p$ that guarantee that $f, g$ have integer coefficients? - -REPLY [48 votes]: There is the following result of Davenport, Lewis, and Schinzel [DLS64, Cor to Thm 2]: - -Theorem. Let $p \in \mathbf Z[x]$. Then the following are equivalent: - -$p$ is a sum of two squares in $\mathbf Z[x]$; -$p(n)$ is a sum of two squares in $\mathbf Z$ for all $n \in \mathbf Z$; -Every arithmetic progression contains an $n$ such that $p(n)$ is a sum of two squares in $\mathbf Z$. - - -Criterion 3 is really weak! For example, it shows that in 2, we may replace $\mathbf Z$ by $\mathbf N$. Because it's short but takes some time to extract from [DLS64], here is their proof, simplified to this special case. -Proof. Implications 1 $\Rightarrow$ 2 $\Rightarrow$ 3 are obvious. For 3 $\Rightarrow$ 1, factor $p$ as -$$p = c \cdot p_1^{e_1} \cdots p_r^{e_r}$$ -with $p_j \in \mathbf Z[x]$ pairwise coprime primitive irreducible and $c \in \mathbf Q$. We only need to treat the odd $e_j$ (and the constant $c$). Let $P = p_1 \cdots p_r$ be the radical of $p/c$, and choose $d \in \mathbf N$ such that $P$ is separable modulo every prime $q \not\mid d$. Suppose $P$ has a root modulo $q > 2d\operatorname{height}(c)$; say -$$P(n) \equiv 0 \pmod q$$ -for some $n$. Then $P'(n) \not\equiv 0 \pmod q$, hence $P(n+q) \not\equiv P(n) \pmod{q^2}$. Replacing $n$ by $n+q$ if necessary, we see that $v_q(P(n)) = 1$; i.e. there is a $j$ such that -$$v_q\big(p_i(n)\big) = \begin{cases}1, & i = j, \\ 0, & i \neq j.\end{cases}.$$ -If $e_j$ is odd, then so is $v_q(p(n))$, which equals $v_q(p(n'))$ for all $n' \equiv n \pmod{q^2}$. By assumption 3 we can choose $n' \equiv n \pmod{q^2}$ such that $p(n')$ is a sum of squares, so we conclude that $q \equiv 1 \pmod 4$. If $L = \mathbf Q[x]/(p_j)$, then we conclude that all primes $q > 2d\operatorname{height}(c)$ that have a factor $\mathfrak q \subseteq \mathcal O_L$ with $e(\mathfrak q) = f(\mathfrak q) = 1$ (i.e. $p_j$ has a root modulo $q$) are $1$ mod $4$. By Bauer's theorem (see e.g. [Neu99, Prop. VII.13.9]), this forces $\mathbf Q(i) \subseteq L$. -Thus we can write $i = f(\theta_j)$ for some $f \in \mathbf Q[x]$, where $\theta_j$ is a root of $p_j$. Then $p_j$ divides -$$N_{\mathbf Q(i)[x]/\mathbf Q[x]}\big(f(x)-i\big) = \big(f(x)-i\big)\big(f(x)+i\big),$$ -since $p_j$ is irreducible and $\theta_j$ is a zero of both. Since $f(x)-i$ and $f(x)+i$ are coprime and $p_j$ is irreducible, there is a factor $g \in \mathbf Q(i)[x]$ of $f(x)+i$ such that -$$p_j = u \cdot N_{\mathbf Q(i)[x]/\mathbf Q[x]}(g) = u \cdot g \cdot \bar g$$ -for some $u \in \mathbf Q[x]^\times = \mathbf Q^\times$. Applying this to all $p_j$ for which $e_j$ is odd, we get -$$p = a \cdot N_{\mathbf Q(i)[x]/\mathbf Q[x]}(h)$$ -for some $h \in \mathbf Q(i)[x]$ and some $a \in \mathbf Q^\times$. By assumption 3, this forces $a$ to be a norm as well, so we may assume $a = 1$. Write $h = \alpha H$ for $\alpha \in \mathbf Q(i)$ and $H \in \mathbf Z[i][x]$ primitive. Then -$$p(x) = |\alpha|^2 H \bar H,$$ -so Gauss's lemma gives $|\alpha|^2 \in \mathbf Z$. Since $|\alpha|^2$ is a sum of rational squares, it is a sum of integer squares; say $|\alpha|^2 = |\beta|^2$ for somce $\beta \in \mathbf Z[i]$. Finally, setting -$$F + iG = \beta H,$$ -we get $p = F^2 + G^2$ with $F, G \in \mathbf Z[x]$. $\square$ - -Footnote: I am certainly surprised by this, given that the version for four squares is clearly false. Indeed, the condition just reads $p(n) \geq 0$ for all $n \in \mathbf Z$. But the OP's example cannot be written as any finite sum of squares in $\mathbf Z[x]$, because exactly one of the terms can have positive degree. (However, it might be different in $\mathbf Q[x]$.) - -References. -[DLS64] H. Davenport, D. J. Lewis, and A. Schinzel, Polynomials of certain special types. Acta Arith. 9 (1964). ZBL0126.27801. -[Neu99] J. Neukirch, Algebraic number theory. Grundlehren der Mathematischen Wissenschaften 322 (1999). ZBL0956.11021.<|endoftext|> -TITLE: faithful modules over a finite dimensional commutative algebra -QUESTION [5 upvotes]: Let $A$ be a commutative algebra over a field $k$ which is finite dimensional as a vector space over $k$. Let $M$ be a faithful $A$-module. Does it follow that $dim_k(M)\geq dim_k(A)$? - -REPLY [9 votes]: $A$ is a product of local Artinian rings, so the question is local. The best result I know is in this paper of Gulliksen, who proved that if the socle dimension is at most $3$, then the length of any faithful module is at least the length of the ring. So the answer is yes if $A$ is the product of local Artin rings of socle dimension at most $3$. He also gave counter example when the socle dimension is bigger than $3$.<|endoftext|> -TITLE: Classification of the behaviours of the logistic map -QUESTION [5 upvotes]: On this this wikipedia page, it is claimed that the iterative sequence $x_{n+1}=rx_n(1-x_n)$ (the logistic map) starting at a point $[0,1]$ and where $r$ ranges in $[0,4]$ behaves differently according to $r$. For example: - -For $r\in[0,1)$, $x_n \to 0$, for all $x_0$; -For $r\in [1,2)$, $x_n \to \frac{1-r}{r}$, for all $x_0$; -For $F -TITLE: 1d TQFT minus connection =? -QUESTION [7 upvotes]: Correct me if I am wrong but I believe at least conceptually (maybe even rigorously) data of a 1-dimensional TQFT and of a vector bundle with connection are equivalent. -Going into more detail (and consequently making more and more mistakes), a vector bundle with connection allows to assign to a point the fibre over that point, and to a path the monodromy (or holonomy? Yes I am that ignorant) along this path. -Three questions, but closely related: -Does there exist equally layman-ish description of going back from a TQFT to a vector bundle? -Clearly this requires the base to be at least a smooth manifold. Is there known any TQFT-like object (and maybe also connection-like object) that would work in the non-smooth context? Say, for topological manifolds, or even arbitrary finite CW-complexes? -What if one leaves manifolds alone but removes the connection? Is there known a version of TQFT that would work for vector bundles with arbitrarily severe restrictions (say, very-very nice algebraic vector bundles on very-very good algebraic varieties) but without any additional structure? -Two remarks: -Sort of minimal version of the question is whether a vector bundle $p:E\to[0,1]$ comes with any kind of map (relation? correspondence?) between $p^{-1}(0)$ and $p^{-1}(1)$. Vague association with motives comes to mind but that's all my mind offers. -Obviously I am tempted to ask the same about 2D-TQFT. But I am (almost) successfully resisting this temptation. - -REPLY [4 votes]: Going into more detail (and consequently making more and more mistakes), a vector bundle with connection allows to assign to a point the fibre over that point, and to a path the monodromy (or holonomy? Yes I am that ignorant) along this path. - -Yes, this is pretty much correct. -You can indeed assign holonomy to paths. -Furthermore, there is an equivalence between 1-dimensional smooth TFTs -over X and vector bundles with connections over X. -Precise definitions with proofs can be found in the paper -https://arxiv.org/abs/1501.00967. - -Does there exist equally layman-ish description of going back from a TQFT to a vector bundle? - -Yes. -The underlying vector bundle without a connection can be recovered -by evaluating the smooth TFTs at the smooth family of points given by the manifold itself. -The connection is recovered by differentiating the parallel transport map. - -Clearly this requires the base to be at least a smooth manifold. Is there known any TQFT-like object (and maybe also connection-like object) that would work in the non-smooth context? Say, for topological manifolds, or even arbitrary finite CW-complexes? - -Yes. Replace the site of smooth manifolds with the site of topological manifolds -or the site of finite CW-complexes. - -What if one leaves manifolds alone but removes the connection? Is there known a version of TQFT that would work for vector bundles with arbitrarily severe restrictions (say, very-very nice algebraic vector bundles on very-very good algebraic varieties) but without any additional structure? - -Yes, this is the (∞,1)-version of 1-dimensional TFTs. -(“Holonomy” is now not a strict functor, but an (∞,1)-functor, -which no longer produces a connection.) -See, for example, the survey by Lurie -and the more recent work by Chris Schommer-Pries.<|endoftext|> -TITLE: A generalization of integral Poincaré duality -QUESTION [5 upvotes]: In this paper, Felix, Halperin and Thomas define the notion of a Gorenstein space over a field $\mathbb{k}$: - -An augmented differential graded algebra $R$ over $\mathbb{k}$ is Gorenstein if $\text{Ext}_R(\mathbb{k},R)$ is concentrated in a single degree and has $\mathbb{k}$-dimension one. -$X$ is Gorenstein over $\mathbb{k}$ if the cochain algebra -$C^*(X,\mathbb{k})$ is Gorenstein. - -This definition is motivated by their subsequent results on this being a generalization of the notion of a Poincaré duality space. -Does there exist a parallel notion of a Gorenstein space over $\mathbb{Z}$ which similarly generalizes Poincaré duality over $\mathbb{Z}$? -EDIT: Alternatively, is it thought that no such generalization is available, so that one needs to use the machinery of symmetric spectra to get such a generalization over $\mathbb{Z}$? - -REPLY [3 votes]: Prior to Dwyer-Greenlees-Iyengar, Dwyer and myself (independently) defined Gorenstein conditions for group rings over the sphere $S[G]$, i.e., the suspension spectrum of a topological group. -The definition easily extends to the case of a morphism $R\to k$. -In the orientable case the definition goes like this: -$R\to k$ is said to be Gorenstein of dimension $d$ if: -1) $k$ is finitely dominated as an $R$-module, i.e., $k$ is a retract up to homotopy of a finite $R$-module (this finiteness condition shouldn't be ignored!), and -2) the derived mapping spectrum $\hom_R(k,R)$ is weakly equivalent as an $R$-module to $k[-d] := \Sigma^{-d}k$. -If one wishes to have an unoriented Gorenstein condition, then it seems to me one needs to replace $k[-d]$ by a twisted version of it. In the special case when $R$ is an augmented $k$-algebra spectrum, we can simply require that the $k$-module $\hom_R(k,R)$ is equivalent to $k[-d]$ as a $k$-module (but not necessarily as an $R$-module). -If $R$ is isn't a $k$-algebra, we can fix another $R$-module structure on $k$, call it $k^\xi$, and require that $\hom_R(k,R)$ is -equivalent to $k^\xi[-d]$ as an $R$-module. -Returning the the group ring case $S[G]$, we have: -Theorem. Assume in addition $\pi_0(G)$ is finitely presented. Then -following are equivalent: -1) $S[G] \to S$ is Gorenstein in dimension $d$. -2) $BG$ is a (finitely dominated) Poincaré duality space in dimenson $d$. -And yes, there is a parallel story over $\Bbb Z$, but it is enough to work with dgas instead of ring spectra (for example, the derived $\hom_{R}(\Bbb Z,R)$ is a differential graded module for a discrete ring homomorphism $R\to \Bbb Z$).<|endoftext|> -TITLE: Some basic questions on quotient of group schemes -QUESTION [7 upvotes]: Let $S$ be a fixed base scheme and $G, H$ be group schemes over $S$. Since I am mainly interested in commutative group schemes over fields, we may assume that $G,H$ are commutative and $S$ is a field if this helps. -(1) Let $f:G\to H$ be a morphism of group schemes. To define the cokernel of this map, we need to choose which topology to work with. Some people use the fppf topology (as in van der Geer & Moonen's book) and other people use the fpqc topology (as in Cornell-Silverman). My question is: what is the difference of those two topologies in terms of group schemes? Is fppf quotient and fpqc quotient of group schemes different? Which topology do people prefer when they are working with group schemes? -(2) Let $H$ be a (normal) closed subgroup scheme of $G$. I think there are at least three plausible definitions of the quotient $G/H$: - -Categorical quotient: Since $H$ naturally acts on $G$, we can think categorical quotient $G/H$ of the action $H\times G\to G$. - -Fppf/fpqc quotient: $G/H$ represents the quotient of $H\to G$ in the category of fppf/fpqc sheaves. - -Naive quotient: A group scheme $G/H$ with a surjective (wrt fppf/fpqc topology) map $p:G\to G/H$ such that kernel of $p$ is the inclusion $H\to G$ - - -Are they equivalent in some good situations? In van der Geer & Moonen's book, it is proved that a fppf quotient is also a categorical quotient. But I cannot find proof nor prove other directions. -context of the question (2): Let $f:A\to B$ be an isogeny of abelian varieties with kernel $\ker f$. Then we have the dual exact sequence $0\to \widehat{B}\to \widehat{A}\to \widehat{\ker f}\to 0$. In Milne's book on abelian variety, to prove the dual exact sequence, consider $0\to \ker f\to A\to B\to 0$ as an exact sequence in the category of commutative group schemes over a field and use a long exact sequence with $\text{Hom}(-, \mathbb{G}_m)$. To use the long exact sequence, we need to prove $B$ is $A/\ker f$ as a fppf/fpqc quotient (In fact I don't know which topology to work with. This is why I ask the question (1)...). However, I only know that $B$ is the `naive quotient (3)' $A/\ker f$. -(3) Is the category of commutative group schemes over a field an abelian category? This statement is in Milne's book on abelian variety, but I cannot find proof. The main point is existence of cokernel, i.e. representability of fppf/fpqc quotient. However, I only know the following theorem in Cornell & Silverman, -Theorem. Let $G$ be a finite type $S$-group scheme and let $H$ be a closed subgroup scheme of $G$. If $H$ is proper and flat over $S$ and if $G$ is quasi-projective over $S$, then the quotient sheaf $G/H$ is representable. -and this is too weak to prove our statement. -Also one more quick question: do you know any good reference dealing with sufficiently general group schemes? I know Shatz's paper in Cornell-Silverman, Tate's paper in Cornell-Silvermann-Stevens, and Stix's lecture note, but they focus on finite flat group schemes. Also, I know some other articles & books which mainly focus on affine algebraic groups. Are there some more general references? -Thank you for reading my stupid questions. - -REPLY [2 votes]: So I think for this sort of question (quotients of flat finitely presented group schemes) the best is to use the theory of algebraic stacks and spaces. I am not an expert, so if someone could double check this that would be great. -Let $G$ be an fppf group scheme over a scheme $S$, and $H$ an fppf subgroup scheme of $G$. -Let $\mathcal{X}=[G/H]$ be the stack quotient. Since $G \times H \to G \times G$ is an fppf groupoid, it is algebraic and $G \to \mathcal{X}$ is an fppf presentation of $\mathcal{X}$. Since the inertia is $H$, it is fppf, so $\mathcal{X}$ is a gerbe over the fppf sheaf quotient $G/H$ (which is an algebraic space), and so $\mathcal{X} \to G/H$ is smooth. So $G \to G/H$ is fppf, where $G/H$ is the quotient in algebraic spaces (or in fppf sheafs). -Now if $G/H$ is a nice space, for instance qs (this is always the case in practice, for instance it is if $H \to G$ is qc), then it contains an open subscheme. If the base $S$ is a field, then since $G$ acts transitively in $G/H$ by acting on this subscheme we get that $G/H$ is a scheme (this is the same trick as for proving that a group algebraic space over a field is a group scheme. In fact we also have that an abelian algebraic space over a base $S$ is always an abelian scheme but this is harder to prove). -Remark: if $H \to G$ is proper, then $[G/H]$ is separated.<|endoftext|> -TITLE: Invertibility of discrete Laplacian -QUESTION [6 upvotes]: In QFT and Statistical Mechanics the discrete Laplacian usually plays a key role when we want to discretize the theory. However, few books (at least to my knowledge) really work the properties of this operator in details, so I'm trying to figure out some of these properties myself. -Let $\Lambda := \epsilon Z^{d}/L\mathbb{Z}^{d}$ be a finite lattice where $\epsilon> 0$ and $L > 1$ are integers such that $L/\epsilon \in \mathbb{N}$ is even. An scalar field over the lattice $\Lambda$ is simply a function $\phi : \Lambda \to \mathbb{C}$, so that the space of all fields is $\mathbb{C}^{\Lambda}$. Because the lattice is a quotient space, we're dealing with periodic boundary conditions. Thus, we can introduce the discrete Laplacian as the linear operator $-\Delta: \mathbb{C}^{\Lambda} \to \mathbb{C}^{\Lambda}$ defined by: -$$(-\Delta \phi)(x) := \frac{1}{\epsilon^{2}}\sum_{k=1}^{d}[2\phi(x)-\phi(x+\epsilon e_{k})-\phi(x-\epsilon e_{k})]$$ -with $\{e_{1},...,e_{d}\}$ being the canonical basis for $\mathbb{R}^{d}$. -Now, let $\langle \phi, \varphi \rangle_{\Lambda} := \epsilon^{d}\sum_{x\in \Lambda}\overline{\phi(x)}\varphi(x)$ be an inner product on $\mathbb{C}^{\Lambda}$. If I'm not mistaken, the follwing identity holds: -\begin{eqnarray} -\langle \phi, -\Delta \phi\rangle_{\Lambda} = \sum_{x\in \Lambda}\sum_{y\sim x}|\phi(x)-\phi(y)|^{2} = \sum_{x\in \Lambda}\sum_{y\sim x}(\overline{\phi(x)}-\overline{\phi(y)})(\phi(x)-\phi(y)) \tag{1}\label{1} -\end{eqnarray} -where $y\sim x$ denotes that $|x-y| = 1$, where $|\cdot|$ is the maximum 'norm' on $\mathbb{Z}^{d}$. -My point is the following. We could have assumed $\phi = 0$ outise $\Lambda$ as a boundary condition, instead of our periodic one. In this case, I know that the discrete Laplacian is positive in the sense that: -$$\langle \phi, -\Delta \phi \rangle_{\Lambda} > 0 \quad \mbox{if} \quad \langle \phi, \phi \rangle_{\Lambda} > 0$$ -and I'd expect the same property with periodic bondary conditions. However, because of the first equality in relation (\ref{1}), it seems that if we take $\phi$ to be constant everywhere, say $\phi(x) = 1$ for every $x \in \Lambda$, it'd follow that $\langle \phi, -\Delta \phi \rangle_{\Lambda} = 0$ even with $\langle \phi, \phi\rangle_{\Lambda} > 0$. This would lead to a non-invertibility of this operator. I don't know if this is a known fact that I just didn't know yet or if my reasoning is not correct, but I'd appreciate any help here. - -REPLY [5 votes]: Your reasoning is correct in that the discrete Laplacian for periodic boundary conditions has a zero mode. On the space of fields satisfying $\sum_x\phi(x)=0$, its spectrum is, however, strictly positive, and it can be inverted on that space. This is all well known.<|endoftext|> -TITLE: Why do elementary topoi have pullbacks? -QUESTION [11 upvotes]: In the book of Szabo "Algebra of Proofs", Definition 13.1.9 introduces an elementary topos as a cartesian closed category with a subobject classifier. On the other hand, many other sources including Johnstone add to this definition that the category should contain limits of finite diagrams. For the proof that the requirement on limits of finite diagrams can be removed, Szabo refers the reader to the paper "Colimits in topoi" by Robert Pare who writes in the second paragraph of the section "Preliminaries on topoi" that the existence of finite limits follows from the existence of equalizers which can be derived from appropriate application of the subject classifier. But for finding a monomorphism from the subobject classifier we should have the corresponding pullbacks in the category. Why such pulbacks exist? The definition of a subobject classifier works only in one direction: given a monomorphism it yields the characteristic morphism. But for the opposite direction (from a characteristic morphism to a monomorphism) the definition does not say anything on the existence of the corresponding pullbacks. - -The Question. Is it true that a cartesian closed category with a subobject classifier indeed has pullbacks? - -If yes, could you provide a (desirably simple) proof? Thank you. - -REPLY [19 votes]: I'll give a counter-example to the claim that having a subobject classifier and being cartesian closed implies the existence of all finite limits. However, this is based on the definition of sub-object classifier given on wikipedia (linked in the comment above) that I would consider as incorrect: -The wikipedia definition (at the time this is written) only asks that for every monomorphism $U \hookrightarrow X$ there is a unique map $X \to \Omega$ such that $U$ is the pullback of the universal subobject $1 \hookrightarrow \Omega$, but it does not ask that every map $X \to \Omega$ be the classifier of some subobject (i.e. that all pullbacks of the universal subobject exist). -If you add the requirement that every map to $\Omega$ classify something, i.e. that pullback of the map $1 \to \Omega$ exists, then it follows that pullbacks of all monomorphisms exist. Moreover pullbacks of monomorphisms, and the existence of finite products imply (in a $1$-category) the existence of all finite limits: A fiber product $B \times_A C$ can be recovered as the pullback of the monomorphism $A \to A \times A$ along $B \times C \to A \times A$. -Consider the category $C$ of finite sets that are not (isomorphic to) the three element sets, with all functions between them. (feel free to replace three by any odd prime). - -$C$ has products: if $|A \times B| = 3$ then $|A|=3$ or $|B|=3$, so $C$ is stable under product in the category of sets. As it is a full subcategory it follows that these are products in $C$ as well. -$C$ has a subobject classifier in the sense of Wikipedia's definition, given by the usual $1=\{\top\} \to \Omega = \{ \bot, \top \}$. Indeed given any mono $A \subset B$ in $C$, its classifying map $B \to \Omega$ in set is also a classyfing map in $C$. -$C$ do not have a subobject classifier in the sense of what I would consider the correct definition: the map $4 \to \Omega$ classying $3 \subset 4$ does not have a pullback, indeed if the pullback $P$ existed there should be exactly three maps $1 \to P$, which is the case for no objects of $C$. -In particular, this is an example of a pullback in $C$ that does not exists. -$C$ is cartesian closed. If $X,Y \in C$ then their exponential $X^Y$ in Set is also in $C$ as $|X^Y|=|X|^{|Y|}=3$ has a unique solution given by $|X|=3$ and $|Y|=1$ hence never happen for $X \in C$. Again as $C$ is a full subcategory stable under product this implies that these are exponential objects in $C$.<|endoftext|> -TITLE: Finitary monads on $\operatorname{Set}$ are substitution monoids. Finitary monads on $\operatorname{Set}_*$ are...? -QUESTION [7 upvotes]: $\DeclareMathOperator\Fin{Fin}\DeclareMathOperator\Lan{Lan}\DeclareMathOperator\Set{Set}$ - -The present question is intimately related to another question. - -It is well known that the category of functors $\Fin \to \Set$ is equivalent to the category of finitary endofunctors $\Set \to \Set$; in this equivalence, finitary monads correspond to what are called substitution monoids on $[\Fin,\Set]$, i.e., to monoids with respect to the monoidal structure -$$ -F \diamond G = m\mapsto \int^n Fn \times G^{*n}m \tag{$\star$} -$$ where $G^{*n}$ is the functor -$$ -m \mapsto \int^{p_1,\dotsc, p_n} Gp_1 \times \dotsb \times Gp_n \times \Fin(\sum p_i, m). -$$ -More precisely, the equivalence $[\Set,\Set]_{\omega} \cong [\Fin,\Set]$ can be promoted to a monoidal equivalence, and composition of endofunctors corresponds to substitution of presheaves in the following sense: let $J : \Fin \to \Set$ be the inclusion functor, then -$$ -\Lan_J(F\diamond G) \cong \Lan_JF \circ \Lan_JG \tag{$\heartsuit$} -$$ -and -$$ -(S\circ T) J \cong SJ \diamond TJ\tag{$\clubsuit$} -$$ for two finitary endofunctors $S,T : \Set \to \Set$. (Kan extending along $J$ and precomposing an endofunctor of $\Set$ with $J$ is what defines the equivalence.) - -I would like to prove the same exact theorem, replacing everywhere the cartesian category of sets with the monoidal category of pointed sets and smash product, but I keep failing. - -The equivalence of categories -$$ -[\Fin_*,\Set_*]\cong [\Set_*,\Set_*] -$$ remains true; and this equivalence must induce an equivalence between the category of finitary monads on pointed sets, and the category of suitable "pointed substitution" monoids, that are obtained from the iterated convolution on $[\Fin_*,\Set_*]$ as -$$ -F\diamond' G = m \mapsto \int^n Fn \land G^{*n}m -$$ where $\land$ is the smash product, and $G^{*n}$ iterates the convolution on $[\Fin_*, \Set_*]$ induced by coproduct on domain, and smash on codomain: -$$ -G^{*n}m = \int^{p_1,\dotsc,p_n} Gp_1 \land \dotsb\land Gp_n \land \Fin_*(\bigvee p_i, m) -$$ where $\bigvee p_i$ is the coproduct of pointed sets, joining all sets along their basepoint. -This would be the perfect equivalent of $(\star)$. -However, trying to prove the isomorphisms $(\heartsuit)$, $(\clubsuit)$, I find that it is not true that $\Lan_J(F\diamond G) \cong \Lan_JF \circ \Lan_JG$. I am starting to suspect that the generalisation is false as I have stated it, or that it is true in a more fine-tuned sense. -So, I kindly ask for your help: - -To what kind of monoids on $[\Fin_*,\Set_*]$ do finitary monads on pointed sets correspond? - -Edit: I am led to believe this construction is not a particular instance of an enriched Lawvere theory, because in that framework a theory isn't what it is in mine: - -for Power, if $\mathcal V$ is a locally finitely presentable base of enrichment, a theory is an identity on objects functor $\mathcal V_\omega ^{op}\to \mathcal L$ from the subcategory of finitely presentable objects to $\mathcal L$ strictly preserving cotensors; if $\mathcal V = \Set_*$ with smash product, cotensors in $\mathcal V^{op}$ are tensors in $\mathcal V$, thus smash products. -Instead, for me, a theory is a bijective-on-objects functor $\Fin_* \to \mathcal L$ that sends coproduct into smash product (or even a more general monoidal structure on $\mathcal L$). - -Or at least, this is what I was led to believe trying to back-engineer the equivalence between finitary monads and Lawvere theories in the case of pointed sets. - -REPLY [3 votes]: Rather than checking to see if that is a substitution monoid, I think you might have an easier time using Rory Lucyshyn-Wright’s notion of a eleutheric system of arities, seen here. It’s a relatively straightforward condition to check, with several equivalent statements. -Edit: To give a (very brief) description of how this works out: a monoidal subcategory of $\mathcal{V}$ is a symmetric monoidal subcategory $j:\mathcal{J} \to \mathcal{V}$, and a $\mathcal{J}$-ary Lawvere theory is a $\mathcal{V}$-category with a bijective-on-objects, power-preserving functor $\mathcal{J} \to \mathcal{T}$. A system of arities is eleutheric if for every $T: \mathcal{J} \to \mathcal{V}$, the left Kan extension $Lan_jT$ exists and is preserved by $\mathcal{V}(J,-), J \in \mathcal{J}$. This ends up being just enough to ensure that every $\mathcal{J}$-ary theory induces a free $\mathcal{T}$-algebra monad on $\mathcal{V}$.<|endoftext|> -TITLE: Explaining the "free left fibration" functor for infinity categories -QUESTION [5 upvotes]: This is a cross-post from here -I am reading A. Mazel-Gee's paper "All about the Grothendieck construction". In that paper he explains that the left adjoint ${\mathrm{Cat}_{\infty}}_{/\mathcal{C}}\to \mathrm{coCFib}(\mathcal{C})$ (from $\infty$-categories over $\mathcal{C}$ to cocartesian fibrations over $\mathcal{C}$) to the forgetful functor is the functor that sends $F:\mathcal{D}\to \mathcal{C}$ to the "free cocartesian fibration on F" -$$\mathrm{Fun}([1],\mathcal{C})\times_{\mathcal{C}}\mathcal{D}\to\mathcal{C}$$ -I am now wondering if there is a similar explicit description for the left adjoint ${\mathrm{Cat}_{\infty}}_{/\mathcal{C}}\to \mathrm{LFib}(\mathcal{C})$. This would be the composite of the previous functor with the reflexive localization $L:\mathrm{coCFib}(\mathcal{C})\to \mathrm{LFib}(\mathcal{C})$. Now by the results in the paper we have -a commutative diagram of large $\infty$-categories $$\require{AMScd}\begin{CD}\mathrm{Fun}(\mathcal{C},\mathrm{Cat}_\infty) @>{(=)^{gpd}\circ -}>> \mathrm{Fun}(\mathcal{C},\mathcal{S})\\ -@V{Gr}V{\simeq}V @V{Gr}V{\simeq}V \\ -\mathrm{coCFib}(\mathcal{C}) @>{L}>> \mathrm{LFib}(\mathcal{C}),\end{CD}$$ where $\mathcal{S}$ is the $\infty$-category of spaces, $Gr$ denotes the Grothendieck construction and $(=)^{gpd}$ is the groupoidification functor. -This implies by the naturality of the Grothendieck construction that the fibers of $L(\mathcal{D}\to\mathcal{C})$ over $x$ identify with $(\mathcal{D}_x)^{gpd}$. But it is not straight-up groupoidification as that would take us to $\mathcal{S}_{/\mathcal{C}^{gpd}}$. If I understand correctly the description of the Grothendieck construction as a lax colimit then the functor L should be some kind of "free groupoidification of the fibers". But this is not as explicit as I would like : can we describe this process without referring to the functor by which the coCartesian fibration is classified ? -On the level of model categories, this is presented by the Quillen adjunction $${\mathrm{Set}_{\Delta}^+}_{/\mathcal{C}^\sharp} \leftrightarrows {\mathrm{Set}_\Delta}_{/\mathcal{C}}$$ between the functor forgetting the marked edges and the functor marking all edges ; the model structures are the marked one and the covariant one, respectively. Therefore the functor $L$ is given by a fibrant replacement of a coCartesian fibration $\mathcal{D} \to \mathcal{C}$ in ${\mathrm{Set}_\Delta}_{/\mathcal{C}}$. Do we have explicit such replacements ? - -REPLY [2 votes]: As suggested by David White, I emailed A. Mazel-Gee. Let me paraphrase his answer : -We claim that given a cocartesian fibration $F:\mathcal{D}\to\mathcal{C}$, the free left fibration $LF:\mathcal{E}\to\mathcal{C}$ on $F$ is just given by inverting in $\mathcal{D}$ the morphisms sent to equivalences in $\mathcal{C}$. We will use Corollary 3.11 in this paper by Ayala & Francis. The natural map $\mathcal{D}\to \mathcal{E}$ is a map of coCartesian fibrations so we just have to check that the induced maps on fibers $\mathcal{D}_x\to\mathcal{E}_x$ for $x\in\mathcal{C}$ are localizations. But as I said in my original post, we have $\mathcal{E}_x=(\mathcal{D}_x)^{gpd}$ ; thus $\mathcal{D}\to \mathcal{E}$ is a localization. Now left fibrations reflect equivalences so a morphism in $\mathcal{D}$ get inverted in $\mathcal{E}$ if and only if it gets inverted in $\mathcal{C}$. - -REPLY [2 votes]: It's actually something you already know: It is the fibrewise groupoidification of the free cartesian fibration. The free cartssian fibration functor sends a functor $$p:A\to B\mapsto p': A\downarrow B\to B.$$ This is totally classical and due originally iirc to Ross Street. The thing to look for is the "slice 2-monad".<|endoftext|> -TITLE: Langlands dual group in math vs. Goddard-Nyuts-Olive dual group in physics -QUESTION [9 upvotes]: Given a group $G$, there is a so-called Langlands dual group $G^{∨}$. -Given a group $G$, there is also a so-called Goddard-Nyuts-Olive dual group $G^{'}$ that relates to the magnetic charge. - -Are the two $G^{∨}$ and $G^{'}$ defined in different settings? Or are their definitions related? What are the intuitions and motivations behind their definitions? -Are the two $G^{∨}$ and $G^{'}$ exactly the same? -What are the constraints on $G$ to give such groups: $G^{∨}$ and $G^{'}$? (eg, compact or not, simple, semi-simple, connected, simply-connected, Lie group or not, topological group, etc.) - - -A Reference on Goddard-Nyuts-Olive dual group: -P. Goddard(Cambridge U.), J. Nuyts(UMH, Mons), David I. Olive(CERN and Bohr Inst.) -Dec 1, 1976, Published in: Nucl.Phys.B 125 (1977) 1-28 -DOI: 10.1016/0550-3213(77)90221-8 -https://www.sciencedirect.com/science/article/pii/0550321377902218?via%3Dihub - -REPLY [8 votes]: The groups $G^{\vee}$ and $G'$ are the same. It is clear from the description in the paper you refer to. -Their definitions are the same as well, just swapping roots and coroots. You can do it in compact Lie groups or reductive algebraic groups. -There is further definition of $G^{\vee}$ from geometric Satake. Its physical significance has been looked into by Kapustin and Witten. They also revisit the Goddard-Nuyts-Olive work, you refer to, from the hills of Geometric Langlands.<|endoftext|> -TITLE: Is $L(\mathbb{Z}*\mathbb{Z}_{2})$ a free group factor? -QUESTION [8 upvotes]: This is a reference request for something that is likely to be well-known to operator algebraists. I will not, therefore, include the technical definition of free product of finite von Neumann algebras, but instead refer the reader to Ching - Free products of von Neumann algebras for the definition. -Theorem 3.5 of Dykema - Interpolated free group factors (letting $A=L(\mathbb{Z})$ and $B=\mathbb{C}$) gives that $M_{2}(L(\mathbb{\mathbb{Z}}))*L(\mathbb{Z}_{2})\cong M_{2}(L(\mathbb{F}_{3}))$. Is it known whether or not $L(\mathbb{Z}*\mathbb{Z}_{2})\cong L(\mathbb{Z})*L(\mathbb{Z}_{2})$ is a free group factor, or interpolated free group factor? -I am, of course, interested in related results like the one quoted above, as well, if the original question is still unknown. Please feel free to provide references as answers. - -REPLY [9 votes]: This should just be a comment- but for some reason I couldn't add a comment. -It seems to me that using Corollary 5.3 of this paper by Dykema, we indeed get a positive answer to your question. -Corollary 5.3 states that $L(G \ast H) \cong L(F(2-|G|^{-1}-|H|^{-1}))$, if $G$ and $H$ are nontrivial amenable groups, with $|G|+|H| \geq 5$.( $\infty ^{-1}=0$). -So $L(\mathbb Z \ast \mathbb Z_2)= L(F(1.5))$ according to the above formula (provided I subtracted correctly). -EDIT: I also found Theorem 1.1 in this paper to be very interesting. It relates to reduced $C^{\ast}$-algebras of free product groups.<|endoftext|> -TITLE: Baire category theorem for uncountable unions -QUESTION [9 upvotes]: Any compact Hausdorff space $X$ is a Baire space: -if the set $X$ is a meager set (meaning a countable union of nowhere dense subsets, -also known as a set of first category), -then $X$ is empty. -I am interested in analogues of this theorem for uncountable unions. -Specifically, suppose a compact Hausdorff space $X$ -is partitioned into a disjoint family $\{Y_i\}_{i∈I}$ of nowhere dense subsets. -To exclude trivial counterexamples such as partitions into singleton subsets, -assume that for any subset $J⊂I$ the union $⋃_{i∈J}U_i$ -is a set with the Baire property (meaning it is the symmetric difference -of an open set and a meager set). -If $I$ is countable, then the condition involving the Baire property is trivially satisfied. -Furthermore, any countable collection of nowhere dense subsets -can be easily adjusted to a countable disjoint collection of nowhere dense subsets with the same union -by replacing $Y_i$ with $Y_i∖⋃_{j -TITLE: Examples of Stokes data -QUESTION [13 upvotes]: I'm trying to learn Stokes data but can't find an example to get my teeth into it. -Background. It's well-known that on a complex manifold $X$, there is the Riemann Hilbert equivalence -$$\text{regular holonomic D modules}\ \stackrel{\sim}{\longrightarrow} \ \text{perverse sheaves}$$ -which for instance sends the regular linear ODE $Pf=0$ to its sheaf of solutions, which forms a local system. As I understand it the point of Stokes data is to give something like -$$\text{holonomic D modules}\ \stackrel{\sim?}{\longrightarrow} \ \text{perverse sheaves + Stokes data}$$ -and it should send the linear ODE $Pf=0$ to its sheaf of solutions (plus extra data). -For instance, take $X=\mathbf{P}^1$. Then the above equivalence should send (ignoring shifts) -$$\mathscr{D}_X1 \ \longrightarrow \ \mathbf{C}$$ -$$\mathscr{D}_Xe^{1/x} \ \longrightarrow \ \mathbf{C}.$$ -These D modules are given by the ODEs $y'=0$ and $y'+y/x^2=0$. So the fact that they are sent to the same local system is not a counterexample of RH, since the second is irregular. I gather that under the $?$ map, $\mathscr{D}_Xe^{1/x}$ is sent to $\mathbf{C}$ along with some extra data at the irregular point $x=0$. -Question. What explicitly is the Stokes data of $\mathscr{D}_Xe^{1/x}$ (and in similar cases)? Is there an obvious relation to Stokes lines of the associated ODE? - -REPLY [3 votes]: To complement Loïc Teyssier's excellent answer, this is the algebro-geometric interpretation of Stokes data, first in the case of $e^{1/x}$. -0. A zeroeth approximation: Stokes data is the information that as $x\to 0$, -$$e^{1/x}\ \longrightarrow \ \begin{cases} -0& \text{if }\text{arg}x\in (-\pi/2,\pi/2)\\ -\infty & \text{if }\text{arg}x\in (\pi/2,3\pi/2) -\end{cases}.$$ -Here $x\to 0$ along rays (lines to the origin of constant argument). So Stokes data remembers how the limiting behavoir of the solution approaching the singular point depends on the argument. -Let's turn this into sheaf language. Take ODE on the disk $X=\Delta$ with singular point $0$, and local system of solutions $\mathscr{L}$ on $\Delta\setminus 0$. To be able to talk about the limiting behavior of solutions as $x\to 0 $ along rays, take the real oriented blowup at $0$ -$$\pi \ :\ \widetilde{X}\ \longrightarrow \ X,$$ -then $\pi^{-1}\mathscr{L}$ is a local system containing this information. Identify the fibre above $0$ with $S^1$. Write $\mathscr{V}$ for the restriction of $\mathscr{L}$ to $S^1$; this is where that information is stored. -1. A first approximation: Stokes data is a subsheaf -$$\mathscr{V}^0\ \subseteq \ \mathscr{V}$$ -given by the solutions with at worst a finite order pole in the given direction. Thus a germ $f$ lies in $\mathscr{V}^0_\theta$ if the size of $f(re^{i\theta})$ is bounded by $r^{-n}$ for some $n$ (this is not quite true, this needs to hold for a sector containing $\theta\in S^1$). In the $e^{1/x}$ example, this is -$$\mathbf{C}_{(-\pi/2,\pi/2)}e^{1/x} \ \subseteq \ \mathbf{C}_{S^1}e^{1/x}.$$ -The actual definition asks for (a little) more information about the limiting behavior. -2. A second approximation: Stokes data is a collection of subsheaves -$$\mathscr{V}^\alpha\ \subseteq\ \mathscr{V}$$ -for every $\alpha\in \Omega^1_\Delta(\star 0)$ a meromorphic one form on $\Delta$ with poles only at $0$. A germ $f$ lies in $\mathscr{V}^\alpha_\theta$ iff -$$f(re^{i\theta}) e^{-\int \alpha}$$ -is bounded by $r^{-n}$ in a small sector containing $\theta$. -These subsheaves fit together to form a filtration, in that -$$\mathscr{V}^\alpha_\theta\ \subseteq \ \mathscr{V}^\beta_\theta$$ -whenever $e^{\int\alpha}e^{-\int \beta}$ has aforementioned boundedness property on a sector containing $\theta$. This gives a partial order on $\Omega^1(\star 0)_\theta$, for which the above is a filtration (a lie: you need to replace $\Omega^1(\star 0)$ by its quotient by the forms with at worst simple poles). Moreover, there's a grading on $\mathscr{V}_\theta$ for which this is the associated filtration. -$\infty$. Stokes data as in $2$ a filtration of $\mathscr{V}$ by a partially ordered sheaf, but using a slightly different indexing poset: you replace the Zariski fibre $\Omega^1(\star 0)_\theta$ with the etale fibre. In practice this means that you consider $\alpha=\sum_{n\ge n_0} a_n x^{n/k}dx$ for all $k\in\mathbf{N}$ and instead of just $k=1$. -So e.g. it contains information that -$$e^{1/x}e^{\int \frac{dx}{\sqrt{x}^5}}\ =\ e^{1/x-2/3\sqrt{x}^3} \ \longrightarrow\ \begin{cases} -0&\text{if }\theta\in \pm(\pi,2\pi/3)\\ -\infty&\text{if }\theta\in (-2\pi/3,2\pi/3) -\end{cases}$$ -where $\sqrt{x}$ is the positive square root defined off the negative reals. - -In this language, Stokes lines are just the phenomenon that $f e^{-\int\alpha}$ flips between satisfying and not satisfying the boundedness condition for only finitely many angles $\theta$, so you can see the Stokes lines directly in the sheaves $\mathscr{V}^\alpha$. -e.g. in the $e^{1/x}$ example, $\theta=\pm \pi/2$ are the two Stokes lines - -Everything in this answer comes from - -La classification des connexions irrégulières à une variable, by Malgrange. http://www.numdam.org/item/CIF_1982__17__A1_0/ -Twisted wild character varieties, by Boalch and Yamakawa. https://arxiv.org/abs/1512.08091 - -The definition of a Stokes structure on a sheaf is $4.1$ of the first reference (it's the same as I've written above), how to give a Stokes structure in the ODE case is the top of page $7$. A Riemann Hilbert correspondence (which justifies the above definition of Stokes data) is theorem $4.2$.<|endoftext|> -TITLE: "Basic" loops on standardly embedded surfaces -QUESTION [5 upvotes]: Take a genus $g$ surface $S$ standardly embedded in $\mathbb{R}^3$, by which I mean it is unknotted. Surface $S$ bounds a volume $V$ that deformation retracts on a standardly embedded planar graph $G$ with $\beta_1 = g$, and that only has degree $3$ vertices. -Among the loops on $S$ that are null homotopic in $V$, there is a subset that are boundaries of embedded disks in $V$ that intersect $G$ exactly once for some choice of $G$ as above. -Do these loops (or perhaps close variants) have a name? Do they have an alternate definition? - -REPLY [7 votes]: They are often called meridians of $G$. Note that there are many graphs $G$ to which $V$ deformation retracts (most nonplanar); if you are not particular about which graph $G$ then they are called meridians of $V$. $V$ is called a handlebody and $G$ is a spine of the handlebody. See, for example, Scharlemann's "Refilling meridians in a genus 2 handlebody complement" arXiv:math/0603705<|endoftext|> -TITLE: Connections between two constructions of infinite dimensional Gaussian measures -QUESTION [5 upvotes]: Let me discuss two possible constructions of Gaussian measures on infinite dimensional spaces. Consider the Hilbert space $l^{2}(\mathbb{Z}^{d}) := \{\psi: \mathbb{Z}^{d}\to \mathbb{R}: \hspace{0.1cm} \sum_{x\in \mathbb{Z}^{d}}|\psi(x)|^{2}<\infty\}$ with inner product $\langle \psi, \varphi\rangle_{l^{2}}:= \sum_{x\in \mathbb{Z}^{d}}\overline{\psi(x)}\varphi(x)$. We can introduce in $l^{2}(\mathbb{Z}^{d})$ the discrete Laplacian as the linear operator: -$$(\Delta \psi)(x) := \sum_{k=1}^{d}[-2\psi(x)+\psi(x+e_{k})+\psi(x-e_{k})]$$ -where $\{e_{1},...,e_{d}\}$ is the canonical basis of $\mathbb{R}^{d}$. Because $(-\Delta+m^{2})$ has a resolvent for every $m\in \mathbb{R}$, we can consider its inverse $(-\Delta+m^{2})^{-1}$. It's integral Kernel or Green's function $G(x,y)$ is given by: -\begin{eqnarray} -G(x,y) = \frac{1}{(2\pi)^{d}}\int_{[-\pi,\pi]^{d}}d^{d}p \frac{1}{\lambda_{p}+m^{2}}e^{ip\cdot(x-y)} \tag{1}\label{1} -\end{eqnarray} -where $p\cdot (x-y) = \sum_{i=1}^{d}p_{i}(x_{i}-y_{i})$ and $\lambda_{p} :=2\sum_{i=1}^{d}(1-\cos p_{i})$ is the eigenvalue of $-\Delta$ associated to its eigenvector $e^{ip\cdot x}$. -[First Approach] If $m \in \mathbb{Z}$, let $s_{m} :=\{\phi\in \mathbb{R}^{\mathbb{N}}: \hspace{0.1cm} \sum_{n=1}^{\infty}n^{2m}|\phi_{n}|^{2} \equiv ||\phi||_{m}^{2}<+\infty\}$, $s:=\bigcap_{m\in \mathbb{Z}}s_{m}$ and $s':=\bigcup_{m\in \mathbb{Z}}s_{m}$. Note that $s$ is a Fréchet space when its topology is given by the family of semi-norms $||\cdot||_{m}$ and $s'$ is the dual space of $s$ if $l_{\psi}$ is a continuous linear map on $s$ with $l_{\psi}(\phi) =( \psi,\phi) := \sum_{n=1}^{\infty}\psi_{n}\phi_{n}$. Let $C=(C_{xy})_{x,y \in \mathbb{Z}^{d}}$ be an 'infinite matrix' with entries $C_{xy}:= G(x,y)$. We can consider $C_{xy}$ to be a matrix $C=(C_{ij})_{i,j \in \mathbb{N}}$ by enumerating $\mathbb{Z}^{d}$. Now, let us define the bilinear map: -\begin{eqnarray} -s\times s \ni (\phi, \psi) \mapsto \sum_{n=1}^{\infty}\phi_{i}C_{ij}\psi_{j} \equiv (\phi, C\psi) \tag{2}\label{2} -\end{eqnarray} -Thus, $\phi \mapsto (\phi, C\phi)$ is a quadratic form and we can define: -$$W_{C}:=e^{-\frac{1}{2}(\phi,C\phi)}$$ -Using Minlos' Theorem for $s$, there exists a Gaussian measure $d\mu_{C}$ on $s'$ (or $\mathbb{R}^{\mathbb{Z}^{d}})$ satisfying: -\begin{eqnarray} -W_{C}(\psi) = \int_{s'}e^{i(\psi,\phi)}d\mu_{C}(\phi) \tag{3}\label{3} -\end{eqnarray} -[Second Approach] For each finite $\Lambda \subset \mathbb{Z}^{d}$, set $C_{\Lambda}$ to be the matrix $C_{\Lambda} =(C_{xy})_{x,y \in \Lambda}$ where $C_{xy}$ are defined as before. Then, these matrices $C_{\Lambda}$ are all positive-definite, so that they define Gaussian measures $\mu_{\Lambda}$ on $\mathbb{R}^{\Lambda}$. Besides, these are compatible in the sense that if $\Lambda \subset \Lambda'$ are both finite and $E$ is a Borel set in $\mathbb{R}^{\Lambda}$ then $\mu_{\Lambda}(E) = \mu_{\Lambda'}(E\times \mathbb{R}^{\Lambda'\setminus\Lambda})$. By Kolmogorov's Extension Theorem, there exists a Gaussian measure $\nu_{C}$ with covariance $C$ on $l^{2}(\mathbb{Z}^{d})$ which is compatible with $\mu_{\Lambda}$ for every finite $\Lambda$. -Now, It seems that these two constructions occur when the so-called thermodynamics limit is taken in QFT and Statistical Mechanics. Both Gaussian measures $\mu_{C}$ and $\nu_{C}$ are measures on $\mathbb{R}^{\mathbb{Z}^{d}}\cong \mathbb{R}^{\mathbb{N}}$. I don't know if this is true but I'd expect these two constructions to be equivalent in some sense, but it is not obvious to me that they are. For instance, the first construction provides a Gaussian measure on $s'$ and the second one on $l^{2}(\mathbb{Z}^{d})$. Are there any relation between these two measures? Are they equal? Maybe the Fourier transform of $\nu_{C}$ would give $W_{C}$, proving these two are the same. Anyway, I'm very lost here and any help would be appreciated. - -REPLY [3 votes]: The source of the confusion is not saying explicitly what are the sets and $\sigma$-algebras the measures are supposed to be on. For example, a sentence like ''By Kolmogorov's Extension Theorem, there exists a Gaussian measure $\nu_C$ with covariance $C$ on $l^2(\mathbb{Z}^d)$ which is compatible with $\mu_\Lambda$ for every finite $\mu_\Lambda$.'' is asking for trouble because it seems to say the measure $\nu_C$ is on the set $l^2(\mathbb{Z}^d)$, which is false. -Let's go back to basics. A measurable space $(\Omega,\mathcal{F})$ is a set $\Omega$ equipped with a $\sigma$-algebra $\mathcal{F}$. A measure $\mu$ on the measurable space $(\Omega,\mathcal{F})$ is a map from $\mathcal{F}$ to $[0,\infty]$ satisfying the usual axioms. From now on I will only talk about probability measures. -For best behavior, the $\Omega$ should be a (nice) topological space and $\mathcal{F}$ should be the Borel $\sigma$-algebra for that topology. Suppose one has two topological spaces $X,Y$ and a continuous injective map $\tau:X\rightarrow Y$. Then if $\mu$ is a measure on $(X,\mathcal{B}_X)$ where $\mathcal{B}_X$ is the Borel $\sigma$-algebra of $X$, then one can construct the direct image/push forward measure $\tau_{\ast}\mu$ on $(Y,\mathcal{B}_Y)$ by letting -$$ -\forall B\in\mathcal{B}_{Y},\ -(\tau_{\ast}\mu)(B):=\mu(\tau^{-1}(B))\ . -$$ -This is well defined because a continuous map like $\tau$ is also $(\mathcal{B}_X,\mathcal{B}_Y)$-measurable. Technically speaking $\mu$ and $\tau_{\ast}\mu$ are different measures because they are on different spaces. However, one could argue that they are morally the same. For example, one might be given the measure $\tau_{\ast}\mu$ without knwing that it is of that form, and only later realize that it is and thus lives on the smaller set $\tau(X)$ inside $Y$. -The first construction: -Let $s(\mathbb{Z}^d)$ be the subset of $\mathbb{R}^{\mathbb{Z}^d}$ made of multi-sequences of fast decay $f=(f_x)_{x\in\mathbb{Z}^d}$, i.e., the ones for which -$$ -\forall k\in\mathbb{N}, ||f||_k:=\sup_{x\in\mathbb{Z}^d}\langle x\rangle^k|f_x|\ <\infty -$$ -where $\langle x\rangle=\sqrt{1+x_1^2+\cdots+x_d^2}$. -Put on the vector space $s(\mathbb{Z}^d)$ the locally convex topology defined by the collection of seminorms $||\cdot||_k$, $k\ge 0$. -The strong dual can be concretely realized as the space $s'(\mathbb{Z}^d)$ of multi-sequences of temperate growth. Namely, $s'(\mathbb{Z}^d)$ is the subset of $\mathbb{R}^{\mathbb{Z}^d}$ made of discrete fields $\phi=(\phi_x)_{x\in\mathbb{Z}^d}$ such that -$$ -\exists k\in\mathbb{N},\exists K\ge 0,\forall x\in\mathbb{Z}^d,\ |\phi_x|\le K\langle x\rangle^k\ . -$$ -The vector space $s'(\mathbb{Z}^d)$ is given the locally convex topology generated by the seminorms -$||\phi||_{\rho}=\sum_{x\in\mathbb{Z}^d}\rho_x\ |\phi_x|$ -where $\rho$ ranges over elements of $s(\mathbb{Z}^d)$ with nonnegative values. -The measure $\mu_C$ obtained via the Bochner-Minlos Theorem is a measure on $X=s'(\mathbb{Z}^d)$ with its Borel $\sigma$-algebra $\mathcal{B}_X$. -The second construction: -Let $s_0(\mathbb{Z}^d)$ be the subset of $\mathbb{R}^{\mathbb{Z}^d}$ made of multi-sequences of finite support $f=(f_x)_{x\in\mathbb{Z}^d}$, i.e., the ones for which -$f_x=0$ outside some finite set $\Lambda\subset\mathbb{Z}^d$. Put on the vector space $s_0(\mathbb{Z}^d)$ the finest locally convex topology. Namely, this is the locally convex topology -generated by the collection of all seminorms on $s_0(\mathbb{Z}^d)$ . -Note that $s_0(\mathbb{Z}^d)\simeq \oplus_{x\in\mathbb{Z}^d}\mathbb{R}$. -Let $s'_0(\mathbb{Z}^d)$ be the strong topological dual realized concretely as $\mathbb{R}^{\mathbb{Z}^d}$. One can also define the topology by the seminorms $||\phi||_{\rho}=\sum_{x\in\mathbb{Z}^d}\rho_x\ |\phi_x|$ -where $\rho$ ranges over elements of $s_0(\mathbb{Z}^d)$ with nonnegative values. -However, this is the same as the product topology for $s'_0(\mathbb{Z}^d)=\prod_{x\in\mathbb{Z}^d}\mathbb{R}$. -The measure $\nu_C$ constructed via the Daniell-Kolmogorov Extension Theorem is a measure on $Y=s'_0(\mathbb{Z}^d)=\mathbb{R}^{\mathbb{Z}^d}$ with its Borel $\sigma$-algebra for the product topology a.k.a strong dual topology. -The precise relation between the two measures: -We simply have $\nu_C=\tau_{\ast}\mu_C$ where $\tau$ is the continuous canonical injection due to $X=s'(\mathbb{Z}^d)$ being a subset of $Y=\mathbb{R}^{\mathbb{Z}^d}$.<|endoftext|> -TITLE: What are the occurrences of stacks outside algebraic geometry, differential geometry, and general topology? -QUESTION [20 upvotes]: What are the occurrences of the notion of a stack outside algebraic geometry, differential geometry, and general topology? - -In most of the references, the introduction of the notion of a stack takes the following steps: - -Fix a category $\mathcal{C}$. -Define the notion of category fibered in groupoids/ fibered category over $\mathcal{C}$; which is simply a functor $\mathcal{D}\rightarrow \mathcal{C}$ satisfying certain conditions. -Fix a Grothendieck topology on $\mathcal{C}$; this associates to each object $U$ of $\mathcal{C}$, a collection $\mathcal{J}_U$ (that is a collection of collections of arrows whose target is $U$) that are required to satisfy certain conditions. -To each object $U$ of $\mathcal{C}$ and a cover $\{U_\alpha\rightarrow U\}$, after fixing a cleavage on the fibered category $(\mathcal{D}, \pi, \mathcal{C})$, one associates what is called a descent category of $U$ with respect to the cover $\{U_\alpha\rightarrow U\}$, usually denoted by $\mathcal{D}(\{U_\alpha\rightarrow U\})$. It is then observed that there is an obvious way to produce a functor $\mathcal{D}(U)\rightarrow \mathcal{D}(\{U_\alpha\rightarrow U\})$, where $\mathcal{D}(U)$ is the "fiber category" of $U$. -A category fibered in groupoids $\mathcal{D}\rightarrow \mathcal{C}$ is then called a $\mathcal{J}$-stack (or simply a stack), if, for each object $U$ of $\mathcal{C}$ and for each cover $\{U_\alpha\rightarrow U\}$, the functor $\mathcal{D}(U)\rightarrow \mathcal{D}(\{U_\alpha\rightarrow U\})$ is an equivalence of categories. - -None of the above 5 steps has anything to do with the set up of algebraic geometry. But, immediately after defining the notion of a stack, we typically restrict ourselves to one of the following categories, with an appropriate Grothendieck topology: - -The category $\text{Sch}/S$ of schemes over a scheme $S$. -The category of manifolds $\text{Man}$. -The category of topological spaces $\text{Top}$. - -Frequency of occurrence of stacks over above categories is in the decreasing order of magnitude. Unfortunately, I myself have seen exactly four research articles (Noohi - Foundations of topological stacks I; Carchedi - Categorical properties of topological and differentiable stacks; Noohi - Homotopy types of topological stacks; Metzler - Topological and smooth stacks) talking about stacks over the category of topological spaces. -So, the following question arises: - -What are the occurrences of the notion of a stack outside of the three areas listed above? - -REPLY [8 votes]: Mike Shulman has stack semantics, an application of stacks to logic. This is basically sheaf semantics, a now standard application to logic of sheaves (far from their own origin in geometry), except that sheaf semantics isn't quite powerful enough to capture unbounded quantification in the way that Mike needs in order to do what he wanted to do with set theory (which is what he was doing when he came up with stack semantics). -This is a fairly low-powered application of stacks, as sheaves are almost but not quite sufficient. But simply adopting this approach makes some things easier to talk about, even when one could do them in the old (sheaves-only) way. And if you want to apply this kind of logic to category theory itself instead of to set theory, then the stacks are really necessary.<|endoftext|> -TITLE: Subgroup $\mathrm{E}_6$ generated by $\mathrm{Spin_7}$ and $\mathrm{SL}_3$ -QUESTION [18 upvotes]: Let $\mathbb{O}$ be the octonion algebra (say over $\mathbb{R}$) and let $J_{3}(\mathbb{O})$ be the set of $3 \times 3$ hermitian matrices with octonion coefficients, that is: -$$ J_3(\mathbb{O}) = \left\{ \begin{pmatrix} \lambda_1 & a & b \\ \overline{a} & \lambda_2 & c \\ \overline{b} & \overline{c} & \lambda_3 \end{pmatrix}, \ \ \lambda_i \in \mathbb{R}, \ a,b,c \in \mathbb{O} \right\}$$ -The group $\mathrm{E}_6$ is the group of linear automorphisms of $J_{3}(\mathbb{O})$ which preserve the cubic form: -$$\lambda_1 \lambda_2 \lambda_3 + 2 \mathrm{Re}(a\overline{b}c) - \lambda_2 N(b)^2 - \lambda_3 N(a)^2 - \lambda_1 N(c)^2,$$ where $N$ is the norm over $\mathbb{O}$. -There are many interesting subgroups of $\mathrm{E}_6$ related to this description. $\mathrm{SL}_3(\mathbb{R})$ is one of them. The action of $\mathrm{SL}_3(\mathbb{R})$ on $J_3(\mathbb{O})$ is given by : -$$ \forall g \in \mathrm{SL}_3(\mathbb{R}), \forall A \in J_{3}(\mathbb{O}), \ g\cdot A = g A\,^{t}\! g,$$ -where $^{t}\!g$ is the transpose of $g$. -The group $\mathrm{Spin_8}$ can also be seen as a subgroup of $\mathrm{E}_6$ with the action: -\begin{align*} &\forall (g_1,g_2,g_3) \in \mathrm{Spin}_8,\quad \forall A = \begin{pmatrix} \lambda_1 & a & b \\ \overline{a} & \lambda_2 & c \\ \overline{b} & \overline{c} & \lambda_3 \end{pmatrix} \in J_{3}(\mathbb{O}), \\ &g\cdot A = \begin{pmatrix} \lambda_1 & g_1(a) & g_2(b) \\ \overline{g_1(a)} & \lambda_2 & g_3(c) \\ \overline{g_2(b)} & \overline{g_3(c)} & \lambda_3 \end{pmatrix}, -\end{align*} -where we identify $\mathrm{Spin_8}$ with $\{(g_1,g_2,g_3) \in \mathrm{SO}_8^3, \ \forall (x,y) \in \mathbb{O}, \ g_3(xy) = g_1(x)g_2(y) \}$. -It is well-known (see Harvey's Spinor and calibrations for instance) that the subgroup of $E_6$ generated by $\mathrm{SO_3}$ and $\mathrm{Spin}_8$ is $\mathrm{F}_4$. I think it is equally well-known (I don't have a reference at hand, but it seems to be an easy corollary of the previous statement) that $\mathrm{E}_6$ itself is generated by $\mathrm{SL}_3$ and $\mathrm{Spin_8}$. -Question : Is there an explicit description the subgroup of $\mathrm{E}_6$ (resp. $\mathrm{F}_4$) generated by $\mathrm{SL}_{3}$ and $\mathrm{Spin}_7$ (resp. $\mathrm{SO}_3$ and $\mathrm{Spin}_7$), where $\mathrm{Spin_7}$ is seen in $\mathrm{Spin}_8$ as $\{(g_1,g_2,g_3) \in \mathrm{Spin}_8, \ g_1(1) = 1\}$? - -REPLY [6 votes]: N.B.: I am revising my response for clarity. (The actual answer to the question asked by the OP is still the same, but I think that this re-organization, particularly at the end, makes the structure of the argument for the answer more clear. I was inspired to do this because some people had some difficulty following the original.) I should also say that the main idea is essentially the one that Theo Johnson-Freyd proposed in his first comment on the question. -I'll use the more usual notation -$$ -J_3(\mathbb{O}) = \left\{\ \left.\begin{pmatrix} \lambda_1 & a_3 & {\overline{a_2}} \\ \overline{a_3} & \lambda_2 & a_1 \\ a_2 & \overline{a_1} & \lambda_3 \end{pmatrix}\ \right| \ \ \lambda_i \in \mathbb{R}, \ a_i \in \mathbb{O} \right\} -\tag 1 -$$ -and the cubic form given by -$$ -C = \lambda_1\lambda_2\lambda_3 + 2\,\mathrm{Re}(a_1a_2a_3) - - \lambda_1\,a_1\overline{a_1} - \lambda_2\,a_2\overline{a_2} - - \lambda_3\,a_3\overline{a_3}\,. -$$ -Then $\mathrm{E}_6\subset\mathrm{GL}\bigl(J_3(\mathbb{O})\bigr)\simeq \mathrm{GL}(27,\mathbb{R})$ is the group of linear transformations of $J_3(\mathbb{O})$ that preserve the cubic form $C$ and $\mathrm{F}_4\subset\mathrm{E}_6$ is the subgroup that also fixes $I_3\in J_3(\mathbb{O})$. (Explicitly, $\mathrm{F}_4$ is a maximal compact in this noncompact real form $\mathrm{E}_6^{(-26)}$ -of $\mathrm{E}_6$.) -The subgroup $\mathrm{Spin}(8)\subset{\mathrm{SO}(8)}^3$ is defined as the set of triples $g = (g_1,g_2,g_3)$ that satisfy -$$ -\mathrm{Re}\bigl(g_1(a_1)g_2(a_2)g_3(a_3)\bigr) = \mathrm{Re}(a_1a_2a_3) -$$ -for all $a_i\in\mathbb{O}$. Let $K_i\subset\mathrm{Spin}(8)$ for $1\le i\le 3$ be the subgroup that satisfies $g_i(\mathbf{1}) = \mathbf{1}$ (where $\mathbf{1}\in\mathbb{O}$ is the multiplicative identity). Each of the $K_i$ is isomorphic to $\mathrm{Spin}(7)$, any two of them generate $\mathrm{Spin}(8)$, and the intersection of any two of them is equal to the intersection of all three of them, which is a group isomorphic to $\mathrm{G}_2$, diagonally embedded in ${\mathrm{SO}(8)}^3$ as the automorphism group of the octonions. -As has already been observed, $\mathrm{SL}(3,\mathbb{R})$ acts on $J_3(\mathbb{O})$ preserving $C$ via $a\cdot A = a\,A\,^{t}a$ (usual matrix multiplication), where $a\in\mathrm{SL}(3,\mathbb{R})$ and $A\in J_3(\mathbb{O})$ are arbitrary. This is a faithful action, so, in this way, $\mathrm{SL}(3,\mathbb{R})$ will be regarded as a subgroup of $\mathrm{E}_6$. -Meanwhile, by its very definition, $g = (g_1,g_2,g_3)\in\mathrm{Spin}(8)$ acts on $A\in J_3(\mathbb{O})$ via -$$ -g\cdot \begin{pmatrix} \lambda_1 & a_3 & \overline{a_2} \\ \overline{a_3} & \lambda_2 & a_1 \\ a_2 & \overline{a_1} & \lambda_3 \end{pmatrix} -= \begin{pmatrix} \lambda_1 & g_3(a_3) & {\overline{g_2(a_2)}} \\ \overline{g_3(a_3)} & \lambda_2 & g_1(a_1) \\ g_2(a_2) & \overline{g_1(a_1)} & \lambda_3 \end{pmatrix}\tag 2 -$$ -and this faithful action preserves $C$ as well, so $\mathrm{Spin}(8)$ will also be regarded as a subgroup of $\mathrm{E}_6$. -Now, as mentioned, $\mathrm{SO}(3)$ and $\mathrm{Spin}(8)$ together generate $\mathrm{F}_4\subset \mathrm{E}_6$. Consequently (since there is no connected Lie group that lies properly between $\mathrm{F}_4$ and $\mathrm{E}_6$), it follows easily that $\mathrm{SL}(3,\mathbb{R})$ and $\mathrm{Spin(8)}$ together generate $\mathrm{E}_6$. -We want to show that $\mathrm{SO}(3)$ and $K_1\simeq\mathrm{Spin}(7)$ also suffice to generate $\mathrm{F}_4$ while $\mathrm{SL}(3,\mathbb{R})$ and $K_1\simeq\mathrm{Spin}(7)$ suffice to generate $\mathrm{E}_6$. -To do this, let $h\in\mathrm{SO}(3)\subset\mathrm{SL}(3,\mathbb{R})$ be -$$ -h = \begin{pmatrix}0&-1&0\\-1&0&0\\0&0&-1\end{pmatrix} = h^{-1} = {}^th. -$$ -Then we have -$$ -h\cdot\begin{pmatrix} \lambda_1 & a_3 & \overline{a_2} \\ \overline{a_3} & \lambda_2 & a_1 \\ a_2 & \overline{a_1} & \lambda_3 \end{pmatrix} -= \begin{pmatrix} \lambda_2 & \overline{a_3} & a_1 \\ a_3 & \lambda_1 & \overline{a_2} \\ \overline{a_1} & a_2 & \lambda_3 \end{pmatrix}. -$$ -Consequently, for $g = (g_1,g_2,g_3)\in \mathrm{Spin}(8)$, computation yields -$$ -h(g_1,g_2,g_3)h = \bigl(\ cg_2c,\ cg_1c,\ cg_3c\ \bigr)\in\mathrm{Spin}(8), -\tag 3 -$$ -where $c:\mathbb{O}\to\mathbb{O}$ is conjugation, i.e., $c(a) = \overline{a}$. -(Thus, conjugation by $h$ gives an involution of $\mathrm{Spin}(8)$ that, together with the order $3$ homomorphism $k(g_1,g_2,g_3) = (g_2,g_3,g_1)$, generates a group of automorphisms of $\mathrm{Spin}(8)$ isomorphic to $S_3$ that maps isomorphically onto $\mathrm{Out}\bigl(\mathrm{Spin}(8)\bigr)$. I imagine that this is what Theo Johnson-Freyd had in mind with his initial comment on this question.) -Note that $g_i(\mathbf{1}) =\mathbf{1}$ implies that $cg_ic = g_i$. Consequently, from the above formula $(3)$, it follows that if $g\in K_1$, then $hgh\in K_2$. -Thus the subgroup of $\mathrm{E}_6$ generated by $\mathrm{SO}(3)$ and $K_1 \simeq\mathrm{Spin}(7)$ contains $K_2$ and, hence, $\mathrm{Spin}(8)$ (since $K_1$ and $K_2$ generate $\mathrm{Spin}(8)$). Thus, this group is $\mathrm{F}_4$. Similarly, the subgroup of $\mathrm{E}_6$ generated by $\mathrm{SL}(3,\mathbb{R})$ and $K_1 \simeq\mathrm{Spin}(7)$ contains $K_2$ -and, hence, $\mathrm{Spin}(8)$. Thus, this group is $\mathrm{E}_6$, as desired. -Similar arguments (using similar choices of $h$) suffice to show that, for any of $i= 1$, $2$, or $3$, the subgroup of $\mathrm{E}_6$ generated by $\mathrm{SO}(3)$ and $K_i$ is $\mathrm{F}_4$, while $\mathrm{SL}(3,\mathbb{R})$ and $K_i$ generate $\mathrm{E}_6$.<|endoftext|> -TITLE: What is known about ordinary character values at involutions? -QUESTION [15 upvotes]: Let $G$ be a finite group and let $\chi$ be the character of an irreducible complex representation $\rho$ of $G$ on $V$. -Let $x$ be an involution in $G$. -I'd like to ask the following -Question 1: - -What is known about $\chi(x)?$ - -1a) Are there criteria when $\chi(x)$ is positive / negative / zero ? -Of course, $1_{\text{Aut}(V)}=\rho(x^2)=\rho(x)^2$, such that the only possible eigenvalues of $\rho(x)$ are $\pm 1$. -Moreover, there is an article written by P.X. Gallagher with the title "Character values at involutions" (DOI: https://doi.org/10.1090/S0002-9939-1994-1185260-1) dealing with the case that $\int_G\chi_1\chi_2\chi_3 \neq 0$, where the integral is in the sense of the Haar measure. -EDIT: -The main parts of Gallagher's results are the following ones: -For an involution $\sigma$ of a finite group $G$ and an irreducible complex representation $R$ of $G$, denote by $q$ the proportion of $-1$'s among the eigenvalues of $R(\sigma)$. Then: -$(*)$ $\frac{1}{h}\leq q \leq 1-\frac{1}{h}$, unless $q = 0$ or $1$, -where $h$ is the index of the centralizer $C$ of $\sigma$. -Moreover, if $\int_G\chi_1\chi_2\chi_3 \neq 0$, then $(*)$ is refined to prove that the proportions of $-1$'s among the eigenvalues of $\rho_1, \rho_2$ and $\rho_3$ (i.e., the corresponding representations) at $\sigma$ are the sides of a triangle on a sphere of circumference 2. -$\ $ -1b) When does $\int_G\chi_1\chi_2\chi_3 \neq 0$ happen (necessary / sufficient criteria)? -1c) When does $\int_G\chi_1\chi_2\chi_3 = 0$ happen (necessary / sufficient criteria)? -1d) Are there results apart from Gallagher's result? -1e) Can one deduce additional information, if all considered characters lie in the same 2-block? -Thanks for the help. - -REPLY [3 votes]: Suppose that $\chi$ is an irreducible character of a finite group $G$ and $t$ is an involution in $G$. It is well known that $\chi(1)\equiv\chi(t)$ (mod $2$). Stephen Gagola and Sidney Garrison showed (J. Algebra 74 (1982) 20--51) that if $\chi$ is a faithful orthogonal character of $G$, and $\chi(1)-\chi(t)\equiv4$ (mod $8$), then $G$ has a non-trivial double cover. Moreover if $t$ lies in the commutator subgroup of $G$, then $H^2(G,{\mathbb C}^\times)$ has even order. They also have results related to the restriction of a real-valued character to a Klein-four subgroup of $G$. They used these results to verify that $M_{22}$ has a four-fold cover.<|endoftext|> -TITLE: Asymptotics of sum involving Stirling numbers -QUESTION [7 upvotes]: I've encountered the following sum: -$$ -s_n = \sum_{j=1}^n {n \brace j}(\alpha n)_j \beta^j. -$$ -Here $\alpha$ and $\beta$ are positive constants, $(\alpha n)_j$ is a falling power, and ${n \brace j}$ is the Stirling number of second kind. My computations suggests that $s_n$ grows like $n^n C^n$ (where the constant $C$ depends on $\alpha$ and $\beta$); specifically, $\sqrt[n]{(s_n/n^n)}$ seems to converge to a limit as $n \rightarrow \infty$. -Are there good approaches to figuring out these kinds of limits? A similar situation was discussed in this question, but the sum there had $(\alpha)_j$ rather than $(\alpha n)_j$; the extra dependence on $n$ is causing me headaches! - -REPLY [4 votes]: The sum in question can be written (using Latin letters instead of Greek) as -$$ S_n(a,b):= \sum_{k=0}^n {n \atopwithdelims \{ \} k} k! \binom{na}{k} b^k $$ -where the round brackets denotes an ordinary binomial. It will be shown as $n \to \infty$ -$$ (1)\quad S_n(a,b) \sim \frac{n!}{2}\,\exp{\Big(n\big( h(u_0) + \frac{h''(u_0)}{2}\,u_0^2\big)\Big)} \, - \text{erfc}\big(\sqrt{\frac{n}{2} h''(u_0)} \,\,u_0 \big) $$ -where -$$h(u) = \frac{\log(1+b(e^u-1))^a}{u}, \quad u_0=\frac{1}{a} + \text{W}\big(\frac{1-b}{a\,b}\,e^{-1/a}\big) $$ -and $W()$ is the Lambert W function. -This follows from an asymptotic analysis of an alternative form, -$$(2) \quad S_n(a,b)=n!\,[u^n] (1+b\,(e^u-1))^{n \,a}$$ -where $[u^n]$ is the 'coefficient of' operator. Note that when $b=1,$ the sum can be solved exactly, and gives the form the OP suspects. (This summation is known.) To derive the alternative form, note that sum is the Hadamard product of -$$ \omega_n(z) = \sum_{k=0}^\infty {n \atopwithdelims \{ \} k} k! z^k \quad \text{& } \quad (1+z)^{na}=\sum_{k=0}^\infty \binom{na}{k} z^k $$ -The expression for $\omega_n$ can be written in terrms of Eulerian polynomials, but I prefer the polylogarithm at negative argument, -$$ \omega_n(t) = \frac{\text{Li}_{-n}(t/(1+t))}{1+t} $$ -Taking the Hadamard product, -$$S_n(a,b) = \frac{1}{2 \pi \,i} \oint \frac{dt}{t(1+t)} \text{Li}_{-n}(\frac{t}{1+t}) \big(1+\frac{b}{t}\big)^{na} \, dt$$ -where the integration path surrounds the origin. -Use $\text{Li}_{-n}(t/(1+t))=(-1)^{n+1}\text{Li}_{-n}(1+1/t).$ Let $t \to 1/t$ in the integral and check residue at $\infty.$ Then -$$S_n(a,b) = \frac{(-1)^n}{2\pi i}\oint \frac{ \text{Li}_{-n}(1+t)}{1+t} \big(1+b\,t\big)^{n\,a} \, dt$$ -Now use the known generating function -$$\sum_{n=0}^\infty \frac{u^n}{n!} (-1)^n \frac{ \text{Li}_{-n}(1+t)}{1+t} = -\frac{1}{e^u-1-t}$$ Putting this equation in the penultimate and using Cauchy's theorem results in equation (2). -For the asymptotic analysis, write (2) as a countour integral -$$S_n(a,b)=\frac{n!}{2 \pi \,i} \oint \Big( \frac{(1+b(e^u-1))^a}{u} \Big)^n \frac{du}{u} $$ -Classic saddle point analysis tells us to write this as -$$S_n(a,b)=\frac{n!}{2 \pi \,i} \oint \exp{\big(n h(u)\big)} \frac{du}{u} $$ -where $h$ has been given in (1), then expand $h$ in a power series about $u_0,$ which has also been found explicitly and given in (1). Run the contour through $u_0$ and open the loop so that it becomes a vertical line. Then, so long as certain conditions are satisfied (I have not checked them, other than $h''(u_0)>0$, a necessity) then -$$ S_n(a,b) \sim \exp{(n\ h(u_0))} \frac{n!}{2 \pi} \int_{-\infty}^\infty \frac{dt}{u_0+i\,t} \exp{\big(-\frac{n}{2}\ h''(u_0) t^2 \big)} $$ -The integral is solvable in terms of the complementary error function. For $n$ sufficiently large, it will simplify in the asymptotic limit to an exponential, as long as $h''(u_0)$ isn't too small. A potential problem with this analysis is that if $h''(u_0)$ does become small, and much smaller than $h^{(3)}(u_0)$ then a quadratic expansion of $h$ is not sufficient. -For examples of the accuracy of the approximation, a table is shown. -$$ -\begin{array}{c|lcr} -n & a & b & \text{true} & \text{asym} & \text{% err}\\ -\\ -\hline -20 & 1/2 & 1/2 & 2.7717\ 10^{17} & 2.6867\ 10^{17} & 3.07\\ -{} & 1/2 & 3/2 & 3.7421\ 10^{21} &3.5745\ 10^{21} & 4.48 \\ -{} & 3/2 & 1/2 & 2.1498\ 10^{25} & 2.0862\ 10^{25} &2.96\\ -{} & 3/2 & 3/2 & 1.6663\ 10^{23} & 1.5884\ 10^{23} & 4.68\\ -200 & 1/2 & 1/2 & 6.634\ 10^{374} & 6.617\ 10^{374} & 0.34\\ -{} & 1/2 & 3/2 & 3.336\ 10^{415} & 3.319\ 10^{415} & 0.51\\ -{} & 3/2 & 1/2 & 6.158\ 10^{453} & 6.138\ 10^{453} & 0.33\\ -{} & 3/2 & 3/2 & 8.720\ 10^{521} & 8.673\ 10^{521}& 0.53 -\end{array} -$$ -It's really surprising that the approximation for a problem with 2 parameters is so easily characterized. - -Added 6/3/2020 -There is a much simpler derivation of (2) that bypasses the need for the Hadamard product and the polylogs with negative index. Here it is: -It is known that -$$ {n \atopwithdelims \{ \} k} k! = n![u^n](e^u-1)^k $$ -Then -$$ \sum_{k=0}^n {n \atopwithdelims \{ \} k} k! \binom{na}{k} b^k =$$ -$$ n! [u^n] \sum_{k=0}^\infty \binom{na}{k} b^k (e^u-1)^k = -n! [u^n] \big(1+b(e^u-1) \big)^{n \ a} $$ -by the binomial theorem.<|endoftext|> -TITLE: What are some examples of theorem requiring highly subtle hypothesis? -QUESTION [39 upvotes]: I would like you to expose and explain briefly some examples of theorems having some hypothesis that are (as far as we know) actually necessary in their proofs but whose uses in the arguments are extremely subtle and difficult to note at a first sight. I am looking for hypothesis or conditions that appear to be almost absent from the proof but which are actually hidden behind some really abstract or technical argument. It would be even more interesting if this unnoticed hypothesis was not noted at first but later it had to be added in another paper or publication not because the proof of the theorem were wrong but because the author did not notice that this or that condition was actually playing a role behind the scene and needed to be added. And, finally, an extra point if this hidden hypothesis led to some important development or advance in the area around the theorem in the sense that it opened new questions or new paths of research. This question might be related with this other but notice that it is not the same as I am speaking about subtleties in proof that were not exactly incorrect but incomplete in the sense of not mentioning that some object or result had to be use maybe in a highly tangential way. -In order to put some order in the possible answers and make this post useful for other people I would like you to give references and to at least explain the subtleties that helps the hypothesis to hide at a first sight, expose how they relate with the actual proof or method of proving, and tell the main steps that were made by the community until this hidden condition was found, i.e., you can write in fact a short history about the evolution of our understanding of the subtleties and nuances surrounding the result you want to mention. -A very well known and classic example of this phenomenon is the full theory of classical greek geometry that although correctly developed in the famous work of Euclides was later found to be incompletely axiomatized as there were some axioms that Euclides uses but he did not mention as such mainly because these manipulations are so highly intuitive that were not easy to recognize that they were being used in an argument. Happily, a better understanding of these axioms and their internal respective logical relations through a period of long study and research lasting for millennia led to the realization that these axioms were not explicitly mention but necessary and to the development of new kinds of geometry and different geometrical worlds. -Maybe this one is (because of being the most classic and expanded through so many centuries and pages of research) the most known, important and famous example of the phenomena I am looking for. However, I am also interested in other small and more humble examples of this phenomena appearing and happening in some more recent papers, theorems, lemmas and results in general. -Note: I vote for doing this community wiki as it seems that this is the best way of dealing with this kind of questions. - -REPLY [24 votes]: This is not a perfect example because the subtle hypotheses in question were not "unnoticed"; nevertheless I think it fulfills several of your other criteria. Let us define the "Strong Fubini theorem" to be the following statement: - -If $f:\mathbb{R}^2 \to \mathbb{R}$ is nonnegative and the iterated integrals $\iint f\,dx\,dy$ and $\iint f\,dy\,dx$ exist, then they are equal. - -The Strong Fubini theorem looks innocent enough, but without any measurability hypotheses, it is independent of ZFC. For example, Sierpinski showed that Strong Fubini is false if the continuum hypothesis holds. -In the other direction, a paper by Joe Shipman investigates a variety of interesting hypotheses that imply Strong Fubini, e.g., RVM ("the continuum is real-valued measurable"), which is equiconsistent with the existence of a measurable cardinal. -Here's another one: Let $\kappa$ denote the minimum cardinality of a nonmeasurable set, and let $\lambda$ denote the cardinality of the smallest union of measure-zero sets which covers $\mathbb{R}$. Then the assertion that $\kappa < \lambda$ implies Strong Fubini. - -REPLY [24 votes]: Theorem. Assuming the axiom of choice, the countable union of countable sets is countable. - -Proof. Let $\{A_n\mid n\in\Bbb N\}$ be a family of countable sets, and so we can write $A_n$ as $\{a_{n,m}\mid m\in\Bbb N\}$. -Let $A$ be the union, and define $f(a) = 2^n3^m$ such that $n$ is the least such that $a\in A_n$, and $a=a_{n,m}$. Easily, this is an injection so the union is countable. - -The trained eye, of course will notice the use of the axiom of choice immediately. We choose an enumeration of each $A_n$. But this is very subtle and usually people will not notice that at first. -And of course, this use of choice is necessary. Indeed, it is consistent that the real numbers are a countable union of countable sets! (Still uncountable, though.)<|endoftext|> -TITLE: The normalizer of block diagonal matrices -QUESTION [5 upvotes]: Let $G=\mathrm U_n$ or $\mathrm{GL}_n(\mathbf C)$ and $H$ the subgroup of block diagonal matrices respecting a partition $n=n_1+\dots+n_k$. Is the normalizer $N=N_G(H)$ computed anywhere in the literature? -I guess, but haven’t proved, that it is generated by $H$ and the permutations (“transpositions”) exchanging the partition’s same-length segments ($\smash{n_i=n_j}$, if any). I also suspect this may be discussed in these papers, to which I don’t have access: - -Koĭbaev, V. A., On subgroups of the full linear groups containing a group of elementary block diagonal matrices. ZBL0521.20027. (1982, translation 1983; other translation?) -Borevich, Z. I.; Vavilov, N. A., Ordering of subgroups, containing a group of block-diagonal matrices, in the general linear group over a ring. ZBL0512.20031. (1982; translation?) -Vavilov, N. A., Subgroups of the general linear group over a semi-local ring containing the group of block-diagonal matrices. ZBL0509.20035. (1983; ever translated?) - -REPLY [3 votes]: By request, from my comment: Your guess is correct. Because there are clearly elements in the normaliser arbitrarily permuting same-sized blocks, if you've got an element of the normaliser, you may assume that it fixes each block. But then it must induce an inner isomorphism on each block, and so you can further assume it fixes each block pointwise. But this forces it to be block scalar. -I'm not sure where this would be written down, but someone must have worked it out as part of an example in some notes. If you're willing to combine some theorems, then it may help to think of your block diagonal matrices are examples of Levi subgroups. Let $G$ be an appropriate general linear or unitary group, $M$ a block diagonal subgroup of the sort you describe, and $A$ the subgroup of diagonal matrices. If $g$ belongs to $\operatorname N_G(M)$, then $g A g^{-1}$ is (the group of rational points of) a split maximal torus in $M$, hence is conjugate in $M$ to $A$ (that's one of the theorems; p. 387 of Murnaghan); so, modulo $M$, it suffices to ask which elements of $\operatorname N_G(A)$ preserve $M$. This is an easy computation, since $\operatorname N_G(A)/A$ is a symmetric group (that's another theorem; Example 8.1 of Murnaghan). If this is an argument that it would be satisfactory to cite, I could easily dig up some references; here I've indicated where you can look in the notes of Murnaghan on linear algebraic groups in CMP 4 (MSN), which just state the results but were the first place to look that occurred to me.<|endoftext|> -TITLE: Why do people study Weyl asymptotics and partial-spectral-projections? -QUESTION [5 upvotes]: The major focus of the research that my advisor has me doing centers around the idea of asymptotic behavior of partial-spectral-projections on compact manifolds. In a few sentences, here is the context for the research: - - -$(M,g)$ is a compact Riemannian manifold without boundary, and $-\Delta_g$ is the (positive) Laplace-Beltrami operator of the metric $g$. -The operator $\sqrt{-\Delta_g}$ is defined in the usual way its collection of $L^2$-normalized eigenfunctions is denoted by $\{e_j(x)\}_{j=0}^{\infty}$, with eigenvalues $0 = \lambda_0 < \lambda_1 \leq \lambda_2 \leq \cdots \to \infty$. -For a fixed $\lambda > 0$, we then define the partial-projection operator $$ \sum_{j=0}^{\infty}\langle f,e_j\rangle e_j(x) = f(x) \mapsto \sum_{\{j \,:\, \lambda_j \in [\lambda, \lambda+1)\}} \langle f,e_j \rangle e_j(x) $$ as the projection of $f(x)$ onto the direct sum of eigenspaces which have eigenvalues in the unit-interval $[\lambda, \lambda+1)$. -We then denote the Schwartz kernel of the corresponding integral operator as $K(x,y;\lambda)$, where $$ f(x) \mapsto \int_{M} K(x,y;\lambda)f(y) \,dV_g(y) $$ agrees with the partial-sum definition above. -The goal of our research is to then analyze the big-oh behavior of this Schwartz-kernel as $\lambda \to \infty$. Usually this is formulated as $$ \sup_{x,y \in M}\big| K(x,y;\lambda) - F(x,y;\lambda) \big| = O(\lambda^{n-1}), $$ where the term $F(x,y;\lambda)$ comes from some parametrix approximation or something. - - -At this point I'm a little embarrassed to say admit that while I can do the mathematical research needed, I am unsure as to why people actually care about such a specific kind of linear operator? -I understand that the Weyl law is an old result in functional and harmonic analysis, but sadly I'm not sure why this specific problem is useful in the larger field of research. I've tried asking this of my advisor before, but he has not offered me all that much in the way of an answer. Also, while reading through the literature of similar problems to my own, I find many references to the myriad of results and slightly-different hypotheses, but still an answer of WHY? eludes me. -Specifically, why does everyone also study these partial-projections onto unit-length interval? What would be different if we projected only an interval of length 2? Or length $L$? Or onto a compact set of some fixed, finite measure? -Any insight into these kinds of problems, and their important to the mathematical body at-large would be much appreciated. Thank you in advanced, as always. - -REPLY [2 votes]: The unit length hypothesis at here is not important, and very crude estimates are available using Sobolev embedding only. The main issue is that studying the spectrum on the manifold itself is not enough for recovering underlying topological/geometric information of the manifold. This is a subtle topic even for 2 dimensional surfaces, where a lot of work has been done. -For very recent work, check some papers by Sogge and Xi: -https://arxiv.org/abs/1711.04707 -I would suggest that instead of working through the detailed estimates (on a sphere, on a torus, on negatively curved manifold, etc), think about some other ways to understand the spectrum of the Laplacian on the manifold. For example, a compact Riemann surface of genus $g\ge 2$ can be realized as the quotient of the upper half plane $\mathcal{H}/\Gamma$. There is a lot of interesting work that can be done to understand the relationship between the group action and the spectrum. The interplay between the algebraic nature of the surfaces and flexibility of the analysis tools made the subject really interesting. -A survey paper by Sanark may be a good start: -http://web.math.princeton.edu/facultypapers/sarnak/baltimore.pdf -For 3-manifolds this becomes deep and is related to heat kernels in geometric analysis. The subject is related to Ricci flow and there are plenty written up online already.<|endoftext|> -TITLE: Relations between quantum groups at roots of unity, modular representation theory, and physics -QUESTION [12 upvotes]: I understand that quantum groups at roots of unity are related to physics because they are used in the construction of Reshetikhin-Turaev invariants, conjectured by Witten. Are there other relations of quantum groups at roots of unity to physics? Also, modular representation theory of Lie algebras is related to quantum groups at roots of unity via Andersen-Jantzen-Soergel. Modular representation theory is a very active area of research (cf. work of Lusztig, Bezrukavnikov, Williamson, and others), and I am wondering if there are relations between results/questions in this area and physics. - -REPLY [2 votes]: The area seems to be very broad, let me just give some remarks, which are somewhat close to me. -Many interesting conformal field theories are "rational", in some simple cases it means that some parameter like a central charge is rational/integer $k$. -By some reasons people consider expressions like $(P)\exp( 2\pi i/k J(x) )$, where $J(x)$ are some generators of symmetries of field theories—currents. -And they appeared to be related to quantum groups. So if $k$ is an integer, then $\exp(2\pi i /k)$ is a root of unity. So the point is that roots of unity are related to "rationality" of some field theories. -The oversimplified example is just to consider canonical commutation relations: $[X,Y] = 2\pi i/k$, which after exponentiation gives a quantum-group-style relation: $\exp(X)\exp(Y) = q\exp(Y)\exp(X)$, for $q=\exp(2\pi i/k)$—so $q$ will be a root of unity for $k$ integral. -(Although this is oversimplified, from some very high-level point of view the whole story is about it.) -Probably the most famous example is the Kazhdan–Lusztig equivalence between the category of certain - -integrable representations of the Kac–Moody algebra at a negative level -and the category of (algebraic) representations of the "big" (a.k.a. Lusztig's) quantum group. - -So here you see that we need "negative level", i.e. central charge $k$ is a negative integer, and what you get is a quantum group at $q = \exp(2\pi i/k)$—a root of unity. (I might forget to shift $k$ by dual Coxeter number.) -I guess mathematical proof does not use explicitly calculations like I mentioned above—just take "currents" $J(x)$—and show that $P\exp( J(x)/k)$ generate quantum group, but that's what some physicists were doing. -To put that idea in the right framework - let us think of the famous Drinfeld–Kohno theorem which says that the monodromy of the representation of the Knizhnik–Zamolodchikov equation is given by the corresponding quantum group. -Again you can see that integer values of $k$ would correspond to root of unity for $q$, by the trivial reason that monodromy locally is given by exponent. -In some sense that statement is closely related to Kazhdan–Lusztig theorem—the KZ-equation is given by "currents" $J(x)$ in tensor product of evaluation modules, -one considers its Pexp (i.e. monodromy) and gets the quantum group. -Another example: in -Integrable Structure of Conformal Field Theory III. The Yang-Baxter Relation, -V. V. Bazhanov, S. L. Lukyanov, A. B. Zamolodchikov explicitly construct quantum-group-like relations from the conformal field theory operators. -For certain special values of parameters one can get quantum groups at roots of unity. -If I remember correctly they exploit that in some papers.<|endoftext|> -TITLE: R-matrices and symmetric fusion categories -QUESTION [8 upvotes]: Given a $\mathbb{C}$-linear braided fusion category $\mathcal{C}$ containing a fusion rule of the form e.g. -\begin{equation}X\otimes Y\cong A\oplus B \oplus C\end{equation} -(where $A,B, C, X$ and $Y$ are all simple objects with $A, B, C$ non-isomorphic), we can write the $R$-matrix $R^{XY}=\text{diag}(R^{XY}_{A}, R^{XY}_{B}, R^{XY}_{C})$. My intuition has always been that these scalars for two distinct objects cannot be the same (much like eigenvalues for distinct eigenspaces should not be the same). Is this misguided? That is, -(Q) Can we have scalars $R^{XY}_{A}=R^{XY}_{B}$ for $A\not\cong B$? -If $\mathcal{C}$ is symmetric, then I believe that diagonal matrix $R^{XY}$ can only have $\pm1$s along its diagonal. However, if the answer to (Q) is no, then this would mean that $\mathcal{C}$ can only contain fusion rules of the form $X\otimes Y\cong pA\oplus qB$ and $X\otimes Y\cong pA$ (where $p$ and $q$ are positive integers). This seems too strong. -Thanks! - -REPLY [3 votes]: Basically any representations of any groups will give you counterexamples in the symmetric case. The simplest case, where X and Y are the defining representation of SU(2), already works. For non-symmetric categories you'd expect that "generically" they'll look different, e.g. if you look at the category of representations of a quantum group and vary q, then for all but finitely many q the eigenvalues will be different. But there's plenty of special q where you'll get repeats, and not just symmetric cases. For example, take SU(2) at a level which has a $D_{2n}$ de-equivariantization, then you have a transparent invertible object and if you take the tensor square of the "middle" object it will have both the trivial and the transparent as summands with the same eigenvalue.<|endoftext|> -TITLE: Is a polytope that has in-spheres for faces of all dimensions already regular? -QUESTION [9 upvotes]: Let $P\subset\Bbb R^d$ be a convex polytope (convex hull of finitely many points). -A $k$-in-sphere of $P$ is a sphere centered at the origin to which each $k$-face of $P$ is tangent. So a 0-in-sphere contains all the vertices and is actually a circumsphere, and a $(d-1)$-in-sphere is completely contained in $P$. -$\qquad\qquad\qquad\qquad\qquad$ - -Question: If $P$ has $k$-in-spheres for all $k\in\{0,...,d-1\}$, is $P$ a regular polytope? - -By definition, all these spheres are centered at the origin, hence are concentric. -The answer to the question is Yes for polygons. For $d\ge 3$ note that this property of $P$ is inherited by its faces, and it follows that all 2-faces of $P$ are regular polygons and all edges are of the same length. - -REPLY [8 votes]: This is true in all dimensions, and can be proved by induction (on $d$) applied to the following (slightly stronger) hypothesis: -Theorem: If $P$ is a convex $d$-polytope with $k$-in-spheres for all $k \in [0, d-1]$, then: - -$P$ is regular. -$P$ is determined (up to an element of the orthogonal group $O(d)$) by that $d$-tuple $(r_0, r_1, \dots, r_{d-1})$ of $k$-in-radii. -$P$ is determined completely if additionally a facet (codimension-1 face) $Q$ of $P$ is specified. - -Proof: If the polytope $P$ has squared $k$-in-radii $(r_0^2, r_1^2, \dots, r_{d-1}^2)$, then every facet of $P$ has squared $k$-in-radii $(r_0^2 - r_{d-1}^2, r_1^2 - r_{d-1}^2, \dots, r_{d-2}^2 - r_{d-1}^2)$. By the first two parts of the inductive hypothesis, all facets of $P$ are therefore regular and congruent to each other (being determined by these $k$-in-radii). -Now, given a facet $Q$ of $P$ and a facet $R$ of $Q$, let $\Pi$ be the hyperplane through the origin which contains $R$. Let $Q'$ be the other facet of $P$ which contains $R$. Because the $k$-in-spheres of $Q'$ are the reflections (in $\Pi$) of the $k$-in-spheres of $Q$, and they share a common facet $R$, it follows (from the third part of the inductive hypothesis) that $Q'$ is the reflection of $Q$ through the hyperplane $\Pi$. -As the boundary $\partial P$ (i.e. the union of all facets) is homeomorphic to $S^{d-1}$, we can reach any facet $Q_1$ from any facet $Q_0$ by a 'path' of 'adjacent' (i.e. sharing a common subfacet) facets. Consequently, we can transform any facet into any other facet by a sequence of reflections in hyperplanes through the origin. As each facet is flag-transitive, it therefore follows that $P$ is flag-transitive (i.e. regular) as desired. -Moreover, this reflection procedure of building $P$ from a single facet $Q$ establishes the third part of the theorem. -This leaves the second part of the theorem. Suppose $P$ and $P'$ are two polytopes sharing the same set of $k$-in-spheres. Let $Q$ be an arbitrary facet of $P$, and $Q'$ be an arbitrary facet of $P'$. By the inductive hypothesis, $Q$ and $Q'$ are congruent; let $f$ be an isometry of the ambient space which maps $Q$ to $Q'$. The origin is either mapped to itself or (if we chose the 'wrong' isometry) to $2v$, where $v$ is the centroid of $Q$; we can if necessary reflect again in the hyperplane containing $Q$ to ensure the origin is preserved by $f$. Consequently, $f$ is an element of the orthogonal group $O(d)$ which maps $Q$ to $Q'$. By the third part of the theorem (which we've already proved), $f$ must map $P$ to $P'$, establishing the second part of the theorem.<|endoftext|> -TITLE: Intuition behind Gubinelli derivative -QUESTION [5 upvotes]: I apologise for the confusion of the following sentences. I'm lazy to give more information about Rough path theory as Is a fairly broad subject. -On page 14 of "A Course on Rough Paths - With an Introduction to Regularity Structures" by Peter K. Friz & Martin Hairer has written: -For $\alpha \in (1/ 3; 1 /2]$, define the space of $\alpha$-Hölder rough paths (over V ), -in symbols $\mathcal C^{\alpha} ([0,T]; V )$, as those pairs $(X; \mathbb X) =: \mathbf{X}$ such that -$$ -||X||_{\alpha}:= \sup_{ -s\neq t \in [0;T ]} -\frac{|X_{s,t}|}{|t-s|^{\alpha}} - < \infty , \quad ||X||_{2\alpha}:=\sup_{ -s\neq t \in [0;T ]} -\frac{|\mathbb X_{s,t}|}{|t-s|^{\alpha}} - < \infty , -$$ -and such that the algebraic Chen relation ( is satisfed. -And on page 56 it hase written: - Given a path $X \in \mathcal C^{\alpha}([0, T ]; V )$, we say that $Y \in \mathcal C^{\alpha}([0, T ]; \hat{W} )$, is controlled by $X$ if there exists $Y' \in \mathcal C^{\alpha}([0, T ]; \mathcal L(V , \hat{W})$, so that the remainder term $R^Y$ given implicitly through the relation $$ Y_{s,t }= Y_{s0} X_{s,t} + R_{s,t}^Y , $$ satisfes $||R^Y||_{ 2 \alpha}< 1$. This defines the space of controlled rough paths, $(Y, Y') \in \mathcal D_X^{2α}([0, T ]; \hat{W })$: Although $Y'$ is not, in general, uniquely determined from Y. We call any such $Y'$ the Gubinelli derivative of $Y$ (with respect to $X$). Here, $R_{s,t}^Y$ takes values in $\hat{W}$, and the norm $|| \cdot||$. -Question : What is the intuition behind this idea of the Gubinelli derivative? -Any help is appreciated with thanks in advance. - -REPLY [5 votes]: In a way it is very much like a usual derivative. Recall first that for a regular function $Y$, its derivative $Y'_s$ at a point $s$ is the (unique) number such that -$$ -Y_{t,s}=Y'_s(t-s)+ R_{s,t}, -$$ -where $R_{s,t}\to0$ faster than linearly. If $Y$ is twice differentiable, then $R_{s,t}\lesssim |t-s|^2$. That is, as a function of $t$, $Y_t$ "looks like" the linear function $Y_s+Y'_s(t-s)$, in the neighborhood of $s$. -Now simply replace the linear function by $X$. So we impose -$$ -Y_{t,s}=Y'_sX_{t,s}+R_{s,t} -$$ -with the remainder $R_{s,t}\to0$ faster than the first term, that is, faster than $|t-s|^\alpha$ (The condition $R_{s,t}\lesssim|t-s|^{2\alpha}$ from Friz-Hairer corresponds to the twice differentiable scenario in the previous case). Then as a function of $t$, $Y_t$ "looks like" the path $Y_s+Y'_sX_{s,t}$. This is great news for integration: we can of course integrate $Y_s$ against $dX_t$ (since as a function of $t$ it is just constant), and we can also integrate $X_{s,t}$ against $dX_t$ (by the definition of a rough path). -Actually, I wouldn't focus so much on assigning a meaning to $Y'$ itself, but rather focus on what the existence of a $Y'$ means for $Y$.<|endoftext|> -TITLE: Non-representability of Weil Restriction -QUESTION [5 upvotes]: Let $f \colon R \to S$ be a ring homomorphism and $X$ an $S$-scheme. We define the functor $R_{S/R}(X)$ on $R$-algebras $T$ by -$$ -R_{S/R}(X)(T) = X(T \otimes_R S). -$$ -If $S/R$ is a field extension (more generally, finite and locally free), and $X$ is quasi-projective, then this functor is known to be representable. -Is there an example where this is not representable? Can one find an example that is smooth and finite type over $S$? What about if $X$ is a simple non-separated scheme, such as the affine line with two origins? - -REPLY [8 votes]: Let $R =\mathbb R$, $S = \mathbb C$ (any quadratic field extension works) and let $X$ be the doubled line. -We have a map of functions from the Weil restriction of $X$ to the Weil restriction of $\mathbb A^1$, which is $\mathbb A^2$. So if the representing scheme exists then it must map to $\mathbb A^2$. -Now take $T = \mathbb C[x,y]$ so $T \otimes_R S = \mathbb C[x,y] \times \mathbb C[x,y]$. Consider the map from $T \otimes_R S$ to $X$ where the coordinate on the affine line is $ (x+iy, x-iy)$ but on the first component we map to the affine line with the first origin and on the second we map to the affine line with the second origin. -This defines a map from $T$ to the Weil restriction. If the Weil restriction is a scheme then there must be an open affine $\operatorname{Spec} T'$ in $\operatorname{Spec} T$, that maps to an open affine $\operatorname{Spec} U$ in the Weil restriction, which maps to the Weil restriction, which maps to $\mathbb A^2$. Dualizing we have maps of rings $\mathbb R[x,y] \to \mathbb C[x,y] = T \to T'$ and $\mathbb R[x,y] \to U \to T'$ forming a commutative diagram. -The image of $U$ inside $T'$ either is contained in $\mathbb R(x,y)$ or not. In either case we will derive a contradiction. -If the image is contained in $\mathbb R[x,y]$, then because $\operatorname{Spec} - T'$ contains the origin, the image must be a subring of $\mathbb R(x,y)$ that the map $\mathbb R[x,y] \to \mathbb C$ sending $x$,$y$ to $0$ factors through. Thus it must be $\mathbb R[x,y]$ adjoining some rational functions that are well-defined at $0$. So in fact we have a factorization $\mathbb R[x,y] \to U \to \mathbb R \to \mathbb C$ sending $x,y$ to $0$. But this implies that the $T$-point of the Weil restriction we wrote down earlier, restricted to the origin, descends to $\mathbb R$, which would make it Galois-invariant. But it is not Galois-invariant, because the two components, exchanged by Galois, map to the two different origins. -If the image is not contained in $\mathbb R$, then it contains some rational function $f$ in $\mathbb C(x,y)$. Choose some $u,v \in \mathbb R$, not both $0$, such that the point $(u,v) \in \mathbb A^2$ lies in $\operatorname{Spec} T'$, and for which $f(u,v)$ is well-defined but not an element of $\mathbb R$. This is easy because these are all generic conditions. For such an $x,y$, the induced map $\mathbb R[x,y] \to U \to T' \to \mathbb C$ sending $x$ to $u$ and $y$ to $v$, sends $f$ to $f(x,y)\notin \mathbb R$ and thus has image $\mathbb C$. -Now if the map from the ring of functions of any affine neighborhood of a $\mathbb C$-point of a scheme to $\mathbb C$ has image $\mathbb C$, that point does not descend to $\mathbb R$ (because that can be checked affine-locally). However, our chosen point $(u,v)$ does descend to $\mathbb R$, because $u, v \in \mathbb R$ and $u,v$ are not both $0$ so $u+iv, u-iv \neq 0$, and thus we lie in the affine line with doubled origin. So this contradicts our conclusion that the image of $U$ in $\mathbb C$ is $\mathbb C$.<|endoftext|> -TITLE: Jacobi forms and Kato's modular units -QUESTION [5 upvotes]: this is pretty much just a silly literature question; apologies in advance. Kato uses the following theta function (or slight variants thereof) in his construction of his Euler system: -$$\Theta(\tau, z) = q^{1/12}(e^{\pi iz} - e^{-\pi iz}) \prod_{n\ge 1} (1-q^n e^{2\pi iz})(1-q^ne^{-2\pi i z})$$ -where $q=e^{2\pi i \tau}$. This is the theta function I like as someone who leans heavily towards the algebraic side of things; it's essentially the de Rham realization of a polylogarithm class in the $H^{1,1}$ motivic cohomology of some universal elliptic curve. However, I can't quite figure out the relation to some classical formulas for the Jacobi theta function. It's been stated in several places (for example, Scholl's notes) that the above theta function is essentially the same as the classical "half-integral weight Jacobi form," which I guess is the series -$$ \vartheta(\tau, z)=\sum_{n\in \mathbb{Z}} e^{\pi i n^2 \tau + 2\pi i n z}= \sum_{n\in \mathbb{Z}} q^{n^2/2} e^{2\pi i n z}$$ -which has the similar-looking product expansion -$$\vartheta(\tau, z)= \prod_{n=1}^\infty (1-q^n)(1-e^{2\pi i z}q^n)(1-e^{-2\pi iz}q^n)$$ -(In particular, Scholl says it essentially coincides with the Jacobi theta series $\vartheta_1$, which didn't prove so helpful since I've found that naming conventions for these several classical Jacobi forms are really confusing and inconsistent in the literature. But all of them have very similar forms and are related by a few simple transformations, from what I can tell. It seems they're basically the orbit of the Jacobi form I wrote down under translation by the half-integral lattice, I gather this has something to do with transformations under the action of the metaplectic cover but don't fully understand the significance.) -These almost look same, but I can't quite figure how to get from one to the other. The denominator in the leading power of $q$ bothers me since it implies a monodromy obstruction in Kato's case. Most notably, the latter function is a triple product whereas the former isn't; the Jacobi triple product identity is my understanding of how to expand the classical Jacobi-type products, so the fact that Kato's theta function is missing one of the "triple product" ingredients throws me off, and I can't work out how to relate them. -Is this discussed somewhere in the literature? I'd really love to be able to clarify the relationship here. - -REPLY [4 votes]: The triple product in the Jacobi theta function $\vartheta(\tau,z)$ can be rewritten -\begin{equation*} -\prod_{n \geq 1} (1-q^n) \prod_{n \geq 0} (1-q^n e^{2\pi i(z+\frac{\tau}{2}+\frac12)}) \prod_{n \geq 1} (1-q^n e^{-2\pi i(z+\frac{\tau}{2}+\frac12)}) -\end{equation*} -($2\pi i n$ should be replaced by $2\pi iz$ in your equation). -On the other hand, the theta function used by Kato is essentially the Weierstrass $\sigma$ function (see for example Silverman, Advanced topics in the arithmetic of elliptic curves, I.6.4). This is not surprising as the $\sigma$ function can be used to construct elliptic functions with prescribed divisors (op. cit., I.5.5). -If you work things out, you will see that the discrepancy is essentially the cube of the infinite product $\prod_{n \geq 1} (1-q^n)$, which brings into play $\Delta^{1/8}$, explaining the factor $q^{1/8}$. -Note that Kato really uses the function $\Theta(\tau,z)^{c^2} \Theta(\tau,cz)^{-1}$, which is associated to the divisor $c^2(0) - \sum_{x \in E[c]} (x)$ on the universal elliptic curve. Since this divisor has degree 0, the term $\prod (1-q^n)$ will cancel out.<|endoftext|> -TITLE: A monotone countably cofinal function from $\omega^\omega$ to $\omega^{\omega_1}$ -QUESTION [11 upvotes]: For a set $X$ we endow the set $\omega^X$ of all functions from $X$ to $\omega$ with the natural partial order $\le$ defined by $f\le g$ iff $f(x)\le g(x)$ for all $x\in X$. -A function $f:\omega^\omega\to\omega^X$ is called monotone if for any $\alpha\le\beta$ in $\omega^\omega$ we have $f(\alpha)\le f(\beta)$ in $\omega^X$. - -Question. Is there a monotone function $f:\omega^\omega\to\omega^{\omega_1}$ such that for every countable set $A\subset\omega_1$ and every $\alpha\in\omega^A$ there exists $\beta\in\omega^\omega$ such that $\alpha\le f(\beta){\restriction}_A$? - -Remark. By Proposition 2.4.1(2) in this paper, for every monotone function $f:\omega^\omega\to\omega^{\omega_1}$ there exists $\alpha\in\omega^{\omega_1}$ such that for every $\beta\in\omega^\omega$ we have $\alpha\not\le f(\beta)$. -PS. I learned this question from Jerzy Kąkol who arrived to it studying $\mathfrak G$-bases in locally convex spaces. - -REPLY [5 votes]: The answer is no, if you require that the monotone function $F \, \colon \, \omega^\omega \rightarrow \omega^{\omega_1}$ is total. I will use fairly standard notation, i.e. $f,g \in \omega^\omega$, $\alpha, \beta \in \omega_1$ and $k,m,n \in \omega$. -The proof works as follows: Towards a contradiction assume that such a function $F$ exists. -First we will construct a sequence $(a_n)_{n \in \omega} \in \omega^\omega$ and find an ordinal $\beta < \omega_1$ such that $$\forall n \in \omega \,\, \exists f \in \prod_{m \leq n} a_m \times \prod_{m > n} \omega \, \colon \, F(f)(\beta) \geq n \, ,$$ where $\prod_{m \leq n} a_m \times \prod_{m > n} \omega =\{ f \in \omega^\omega \, \colon \, \forall m \leq n \,\, f(m) < a_m \}$. The idea behind this is that we can require a bound for finitely many values of an input function $f$ and still make $F(f)(\beta)$ arbitrarily large. -In the second step we construct a sequence $(b_n)_{n \in \omega} \in \omega^\omega$ such that $(b_n)_{n \in \omega} \geq (a_n)_{n \in \omega}$, and a sequence $(f_n)_{n \in \omega}$ such that $$\forall n \in \omega \, \colon \, f_n \in \omega^\omega \, \land \,(b_m)_{m \in \omega} \geq f_n \, \land \, F(f_n)(\beta) \geq n \,.$$ But this leads to a contradiction, since the monotonicity of $F$ implies that $\forall n \in \omega \, \colon \, F((b_m)_{m \in \omega})(\beta) \geq n$. -First step: -We will construct $(a_n)_{n\in \omega}$ by induction. For the base case $n=0$ we claim that $$\exists a_0 \in \omega \,\exists \alpha_0 \in \omega_1 \, \forall A \in [\omega_1 \setminus\alpha_0]^{\leq \aleph_0} \, \colon \, F[\prod_{m = 0} a_0 \times \prod_{m>0} \omega] \,\text{is cofinal in} \, \omega^A\, .$$ This means that already $F\restriction \prod_{m = 0} a_0 \times \prod_{m>0} \omega$ dominates $\omega^A$ for every countable $A \subseteq \omega_1$ above $\alpha_0$. -Towards a contradiction assume that the claim is wrong. Therefore, we can construct a sequence $(A_n)_{n \in \omega}$ such that $A_n \in [\omega_1]^{\leq \aleph_0}$ and $\sup A_n < \min A_{n+1}$, and a sequence of functions $(f_n)_{n \in \omega}$ such that $f_n \in \omega^{A_n}$ and $F\restriction \prod_{m = 0} n \times \prod_{m>0} \omega$ does not dominate $f_n$. But if we define $B:= \bigcup_{n \in \omega} A_n \in [\omega_1]^{\leq \aleph_0}$ and $f:= \bigcup_{n \in \omega} f_n \in \omega^B$, we reach a contradiction, since noting that $\omega^\omega = \bigcup_{n \in \omega} \prod_{m = 0} n \times \prod_{m>0} \omega$, there cannot exist a $g \in \omega^\omega$ such that $F(g) \restriction B \geq f$. -Assume inductively that $a_0,..,a_n$ and increasing $\alpha_0,...,\alpha_n$ have already been defined, such that $\forall A \in [\omega_1 \setminus\alpha_n]^{\leq \aleph_0} \, \colon \, F[\prod_{m \leq n} a_m \times \prod_{m>0} \omega] \,\text{is cofinal in} \, \omega^A$. With a similar argument as before, and again noting that $$\prod_{m \leq n} a_m \times \prod_{m>n} \omega = \bigcup_{k \in \omega} \prod_{m \leq n} a_m \times \prod_{m=n+1} k \times \prod_{m>n+1} \omega$$ we can show that $$\exists a_{n+1} \in \omega \, \exists \alpha_{n+1} \in \omega_1 \setminus \alpha_n \, \forall A \in [\omega_1 \setminus\alpha_{n+1}]^{\leq \aleph_0} \, \colon$$ $$F[\prod_{m \leq n} a_m \times \prod_{m=n+1} a_{n+1} \times \prod_{m>n+1} \omega] \,\text{is cofinal in} \, \omega^A \, .$$ So we can find $a_{n+1}$ and $\alpha_{n+1}$ as required. -Now let $\beta > \sup_{n \in \omega} \alpha_n$. It follows from our construction of the $(a_n)_{n \in \omega}$ that $$\forall n \in \omega \,\, \exists f \in \prod_{m \leq n} a_m \times \prod_{m > n} \omega \, \colon \, F(f)(\beta) \geq n \, .$$ -Second step: -Again, we will construct $(b_n)_{n \in \omega}$ and $(f_n)_{n \in \omega}$ by induction. The base case $n=0$ is quite easy: Set $b_0=a_0$ and pick any $f_0 \in \prod_{m = 0} b_0 \times \prod_{m>0} \omega$. Then $F(f_0)(\beta)\geq 0$ trivially holds. -Assume inductively that $b_0,...,b_n$ and $f_0,...,f_n$ have already been constructed such that $$\forall m \leq n \, \colon \, a_m \leq b_m \, \land \, f_m \in \prod_{m \leq n} b_m \times \prod_{m>n} \omega \land \, F(f_m)(\beta) \geq m \, .$$ -By using what we have shown in the first step, we can now find $f_{n+1} \in \prod_{m \leq n} b_m \times \prod_{m>n} \omega$ such that $F(f_{n+1})(\beta) \geq n+1$. We set $b_{n+1}:= \max( \max_{m \leq n+1} f_m(n+1), a_{n+1})+1$. It follows that $$\forall m \leq n+1 \, \colon \, a_m \leq b_m \, \land \, f_m \in \prod_{m \leq n+1} b_m \times \prod_{m>n+1} \omega \land \, F(f_m)(\beta) \geq m \, .$$ -But now we have reached a contradiction, since $\forall n \in \omega \, \colon \, (b_m)_{m \in \omega} \geq f_n$, and therfore the monotonicity (and totality) of $F$ implies that $\forall n \in \omega \, \colon \, F((b_m)_{m \in \omega})(\beta) \geq n$.<|endoftext|> -TITLE: Is there a notion of polynomial ring in "one half variable"? -QUESTION [50 upvotes]: Let $C$ be the category of commutative rings. - -Is there a functor $F :C \to C$ such that $F(F(R)) \cong R[X]$ for every commutative ring $R$ ? - -(Here, we may assume those isomorphisms to be natural in $R$, if needed). -I tried to see what $F(\mathbb Z)$ or $F(k)$ (for a field $k$) should be, but I cannot come up with a contradiction to disprove the existence of $F$. On the other hand, I tried to build such an $F$, without success. I already asked this question (mostly out of curiosity) on MSE last year, but no answer was found. -Some comments however suggest that the existence might involve the axiom of choice, or that requiring $F$ to preserve limits and filtered colimits could be useful. - -REPLY [18 votes]: I don't know how to answer the question as stated, but I believe that by strengthening the question a bit we can see that there does not exist a "reasonable" notion of "polyonomial ring in one-half variable" (unless perhaps we start thinking about enlarging our category or something). That is, let me address the following: - -Modified question: Let $k$ be a commutative ring. Does there exist a functor $F: CAlg_k \to CAlg_k$ and a natural transformation $\iota: Id \Rightarrow F$ such that $F(F(A))$ is naturally isomorphic to $A[x]$ and such that the composite $A \xrightarrow{\iota_A} F(A) \xrightarrow{\iota_{F(A)}} F(F(A)) = A[x]$ is identified under this isomorphism with the usual map $A \to A[x]$? - -It seems to me that this is an extremely minimial, reasonable additional property to ask of a construction deserving the name "polynomial algebra in one half variable". It would be nice to eventually dispense with it, but I hope to shed some light on the original question by adding the assumption of the existence of $\iota$. - -Answer to Modified Question: No, there is no such functor $F$ and natural transformation $\iota$, at least when $k$ is a field. For if there were, then since the composite $F(k) \to k[x] \to F(k)[x]$ has a retract given by evaluation at zero, it would be the case that $F(k)$ is a retract of $k[x]$. There are no retracts of $k[x]$ besides $F(k) = k$ or $F(k) = k[x]$, neither of which is compatible with $F^2(k) \cong k[x]$ and $F^4(k) \cong k[x,y]$. - -EDIT: I was suddenly seized with doubt, so here's a proof of the fact I just used: -Fact: Let $k$ be a field. Let $k \subseteq R \subseteq k[x]$ be a $k$-algebra which is a retract of the $k$-algebra $k[x]$. Then either $R = k$ or $R = k[x]$. -Proof: Let $\pi: k[x] \to R$ be the retraction map, and let $p = \pi(x) \in R \subseteq k[x]$ be the image of $x \in k[x]$ under $\pi$. Since $k[x]$ is a PID, the kernel of $\pi$ is a principal ideal, of the form $(f(x))$ for some $f(x) \in k[x]$. Thus we have $f(p) = 0$. By the following lemma, this implies either that $p \in k$ (in which case $R = k$) or else that $f(x) = 0$ (in which case $R = k[x]$). -Lemma: Let $k$ be a commutative ring, and let $p(x) \in k[x]$ be a polynomial. Suppose that the leading coefficient of $p(x)$ is a non-zero-divisor in $k$. Suppose that $p(x)$ is algebraic over $k$, i.e. that $f(p(x)) = 0$ for some $f(y) \in k[y] \setminus \{0\}$. Then $p(x) \in k$ is a degree-zero polynomial. -Proof: Suppose for contradiction that $\deg p(x) \geq 1$. Then the leading coefficient of $f(y)$ is $a b^n$ where $a$ is the leading coefficient of $f(y)$ and $b$ is the leading coefficient of $p(x)$ (and $n = \deg f$). This leading coefficient vanishes by hypothesis, so that $b$ is a zero-divisor in $k$, contrary to hypothesis. Therefore $\deg p(x) = 0$ as claimed.<|endoftext|> -TITLE: Terminology introduced in recent years with more than one meaning -QUESTION [5 upvotes]: Suppose a term(inology) is recently (in last 20 years) introduced in research mathematics. -It might happen that some one who wish to use it, in the same area of research, for different purposes or see from different point of view realize that, some condition needs to be added or removed for their pov/purpose but still calling by the same name. This creates a slight confusion. - -What are some term(inology) introduced recently (in last 20 years) which have more than one possible meaning because of different point of view or different purpose? - -REPLY [3 votes]: My vote is for the phrase "normal Cayley graph". -Recall that a Cayley graph Cay($G$,$C$) is obtained from a group $G$ and a subset of its elements $C \subseteq G$. The vertex set of Cay($G$,$C$) is $G$ itself, and for each $g \in G$ and $c \in C$ there is an edge from $g$ to $gc$. -Some of my colleagues and co-authors say that Cay($G$,$C$) is a normal Cayley graph if $G$ is a normal subgroup of Aut(Cay($G$,$C$)). -Another set of colleagues and co-authors say that Cay($G$,$C$) is a normal Cayley graph if $C$ is closed under conjugation, (so that $C$ is like a normal subset of $G$). -The first usage involves looking outside $G$ while the second usage involves looking inside $G$.<|endoftext|> -TITLE: The metric difficulty of unknotting unknots -QUESTION [5 upvotes]: Despite the catastrophe for the world and the many victims, at least the lockdown is favorable to two activities: Mathematics and Gardening. The difficulty of handling hedge trimmer wires and garden hoses calls to some mathematical questions relative to the (un)knots in $R^3$, -which are very natural and which I could not solve, nor find references. -Analogous questions actually hold for the Jordan curves in $R^2$. - Any help is welcome. -For $n=2$ or $3$, consider the embeddings of the circle, $S^1\hookrightarrow R^n$, and their isotopies (1-parameter families of embeddings). By the "unknot" in $R^3$, one means the unit circle in the $xy$-plane. Let us define the width of any isotopy $h_t:S^1\hookrightarrow R^n$ ($t\in I=[0,1]$) as the maximum over all $s\in S^1$ of the length of the path $t\mapsto h_t(s)$ in $R^n$. -1) Is there a finite upper bound $A$ such that every smooth Jordan curve $f:S^1\hookrightarrow R^2$ which is contained in the unit disk, is isotopic to the unit circle through an isotopy of width $\le A$? -2) Is there a finite upper bound $B$ such that every smooth knot $f:S^1\hookrightarrow R^3$ which is contained in the unit sphere and isotopic to the unknot, is isotopic to the unknot through an isotopy of width $\le B$? -3) Is there a continuous Jordan curve $f:S^1\hookrightarrow R^2$ which is not isotopic to the unit circle through any $C^0$ isotopy of finite width? -4) Is there a continuous knot $f:S^1\hookrightarrow R^3$ which is $C^0$-isotopic to the unknot, but not through any $C^0$ isotopy of finite width? - -REPLY [3 votes]: The answer is yes for all I think, provided there is a finite width isotopy, as you can just scale the curve to make it arbitrary close to the origin, do the isotopy to a tiny circle there, and expand it back. So the infimum of $A$ or $B$ should be $2$. -Now for 1) if you work in an annulus instead of a disk the answer is no, as you can take a spiral living near the unit circle and consider the boundary of a small neighborhood of it. To isotope it to a circle within the annulus, you'll have to move the curve arbitrary far away depending on the number of turns of the spiral. Clearly that example also works for the other questions. -Another possible formulation would be to use Moebius invariant metrics on the space of curves (which make the scaling trick useless). Then the answer would be no to all I guess.<|endoftext|> -TITLE: Sperner's Lemma implies Tucker's Lemma - simple combinatorial proof -QUESTION [17 upvotes]: Sperner’s Lemma is often called the "combinatorial analog" of Brouwer’s Fixed Point Theorem, and similarly Tucker’s Lemma is often called the combinatorial analog of Borsuk–Ulam’s Theorem. -We can fairly directly show Borsuk–Ulam implies Brouwer, but it seems no direct combinatorial proof is known between Tucker's and Sperner's Lemma. (see a related discussion from 2013/2014 with links to good articles at Sperner's lemma and Tucker's lemma.) - -To my surprise, I find that Sperner's Lemma directly implies Tucker's Lemma in two dimensions. My question: are there any recent results about such a direct combinatorial link for arbitrary dimension? -Edit: As a side comment, there is a striking consequence from Sperner $\Rightarrow$ Tucker: - It is well-known that Tucker $\Rightarrow$ Borsuk-Ulam $\Rightarrow$ Brouwer $\Rightarrow$ Sperner. - So Sperner $\Rightarrow$ Tucker could establish a meaningful scope of equivalence for all these results. - -For clarification, I am adding a 2-dimensional example, and a proof why Tucker’s Lemma follows directly from Sperner’s Lemma. (This example only shows the boundary labelling, not the triangulation and the inside vertices). - -Take a triangulated polygon with vertices labelled -2, -1, 1, or 2, and antipodally symmetric labelling on its boundary, satisfying the conditions of Tucker’s Lemma. -Color the boundary labels such that they meet the conditions of Sperner’s Lemma, like in the example, i.e. $1\mapsto \text{orange}$; $2\mapsto \text{blue}$; $-1, -2\mapsto \text{green}$. Assign the colors to all the Tucker-labelled vertices inside the polygon in the same way -Edit: In the two-dimensional case, such a valid Sperner labelling always exists. Please see the proof in the answer below. -Here is why this Sperner color labelling directly implies Tucker's Lemma: -Because of the valid Sperner coloring of the boundary, a 3-colored Sperner triangle must exist. But this 3-colored triangle either has a complementary green–orange edge $(-1,1)$ or a complementary green–blue edge $(-2,2)$. In other words, the existence of the complementary edge follows directly from Sperner’s Lemma, proving Tucker’s Lemma. -In the two-dimensional case, the Sperner color labelling is always compatible with the Tucker labelling, hence my question about any recent results or ideas in this direction for arbitrary dimensions. -(for a related question see this post Structure of boundary labelling in Sperner‘s Lemma) - -REPLY [3 votes]: Just to close the loop on this. I now found a simple proof that an antipodally symmetric labelling of the boundary always has a valid Sperner coloring (in two dimensions). This means Tucker’s Lemma actually follows directly from Sperner’s Lemma with a combinatorial argument. This is quite a surprise to me, because in the literature, Tucker’s Lemma/Borsuk-Ulam Theorem is generally considered stronger, in the sense that it implies Sperner’s Lemma/Brouwer’s Fixed Point Theorem. -Here is the proof about the compatible labelling, a proof by induction over antipodally symmetric pairs of boundary vertices. It assumes the coloring from the OP questions above. -For the remainder of the proof, we exclude all the cases with a complementary edge on the boundary, as there is nothing more to prove (complementary edge exists). -In the diagrams, the lines do not indicate the triangulation; the lines just indicate the antipodally symmetric pair of vertices. -Induction Base Case($2n=4$): This case obviously allows for a valid Sperner coloring, i.e. a Sperner triangle and hence a complementary edge exists inside the triangulated polygon. - -Induction Step from $2n$ to $2n+2$: -Assume that the boundary has a valid Sperner coloring for its $2n$ antipodally symmetric labelled vertices. Valid Sperner coloring means that the number of color changes on the boundary is uneven (i.e. uneven number of boundary edges with endpoints of different color). -Now we include another pair of antipodally symmetric pair of boundary vertices, to arrive at $2n+2$ vertices with valid Sperner coloring. When including a new pair, we have to insert it in-between two existing pairs of vertices. There are just four different cases to consider: - - -For case A, because we don’t allow for complementary edges on the boundary, we can only include an antipodally symmetric labelled pair of vertices with 1 or 2 on one side, as in the diagram. But this does not add any color change to the boundary, i.e. the boundary just keeps its valid Sperner coloring. An analoguous argument works for case B. - -For case C, because we don’t allow for complementary edges on the boundary, we can only include an antipodally symmetric labelled pair of vertices with 1, 2, or -2 on one side, as in the diagram. But if it is 1, it does not not add any color change to the boundary. And if it is 2 or -2, it just adds an even number of color changes to the boundary. In both cases, the number of color changes remains uneven, i.e. the Sperner coloring of the boundary remains valid.An analoguous argument works for case D. - -Conclusion of Induction: -So we have shown that a valid Sperner coloring for $2n$ antipodally symmetric labelled vertices implies that the Sperner coloring is also valid for $2n+2$ antipodally symmetric labelled vertices vertices. So starting from $2n=4$, it is valid for all $2n$, which concludes the proof.<|endoftext|> -TITLE: An unpublished note by Bloch-Kato on p-divisible groups and Dieudonné crystals -QUESTION [6 upvotes]: I wonder if anyone could find the following unpublished paper of Bloch-Kato: -Spencer Bloch and Kazuya Kato, $p$-divisible groups and Dieudonné crystals, unpublished. -A similar question is here while both links in the question are failed right now. -An unpublished note by Spencer Bloch and Kazuya Kato - -REPLY [8 votes]: The unpublished note was no longer on the Web (also archive.org did not have it cached). I asked professor Moonen for a copy to share with you, here it is: p-divisible groups and Dieudonné crystals by Bloch and Kato.<|endoftext|> -TITLE: Reformulation - Construction of thermodynamic limit for GFF -QUESTION [6 upvotes]: I've posted a question about the thermodynamic limit for Gaussian Free Fields (GFF) a couple days ago and I haven't got any answers yet but I kept thinking about it and I thought it would be better to reformulate my question and exclude the previous one, since now I can pose it in a more concrete way. The problem is basically give mathematical meaning to the infinite volume Gaussian measure associated to the Hamiltonian of the discrete GFF. In what follows, I will formulate the problem and then state the question. -A (lattice) field is a function $\phi: \Lambda_{L} \to \mathbb{R}$, where $\Lambda_{L} := \mathbb{Z}^{d}/L\mathbb{Z}^{d}$. Thus, the space of all fields is simply $\mathbb{R}^{\Lambda_{L}}$. The discrete Laplacian is the linear operator $\Delta_{L}:\mathbb{R}^{\Lambda_{L}}\to \mathbb{R}^{\Lambda_{L}}$ defined by: -\begin{eqnarray} -(\Delta_{L}\phi)(x) := \sum_{k=1}^{d}[-2\phi(x)+\phi(x+e_{k})+\phi(x-e_{k})] \tag{1}\label{1} -\end{eqnarray} -If $\langle \cdot, \cdot \rangle_{L}$ denotes the usual inner product on $\mathbb{R}^{\Lambda_{L}}$, we can prove that $\langle \phi, (-\Delta_{L}+m^{2})\phi\rangle_{L} > 0$ if $\langle \phi,\phi\rangle_{L}> 0$ and $m \neq 0$. Thus, $-\Delta_{L}+m^{2}$ defines a positive-definite linear operator on $\mathbb{R}^{\Lambda_{L}}$. We can extend these ideas to $\mathbb{R}^{\mathbb{Z}^{d}}$ as follows. A field $\phi: \mathbb{Z}^{d}\to \mathbb{R}$ is called $L$-periodic if $\phi(x+Ly) = \phi(x)$ for every $y \in \mathbb{Z}^{d}$. Let $\mathcal{F}_{per}$ be the set of all $L$-periodic fields, so that $\mathcal{F}_{per} \subset \mathbb{R}^{\mathbb{Z}^{d}}$. Now, using the same expression in (\ref{1}) we can define 'infinite volume' Laplacians $\Delta_{per}$ and $\Delta$ on $\mathcal{F}_{per}$ and $l^{2}(\mathbb{Z}^{d}):=\{\phi:\mathbb{R}^{d}\to \mathbb{R}:\hspace{0.1cm} \sum_{x \in \mathbb{Z}^{d}}|\psi(x)|^{2}<\infty\}$, respectivelly. In addition, if $\phi \in \mathcal{F}_{per}$, its restriction $\phi|_{\Lambda_{L}}$ can be viewed as an element of $\Lambda_{L}$, and the action of $\Lambda_{per}$ to $\phi|_{\Lambda_{L}}$ is equivalent to the action of $\Delta_{L}$ to $\phi|_{\Lambda_{L}}$. -The Hamiltonian for the GFF in the lattice $\Lambda_{L}$ is given by: -\begin{eqnarray} -H_{\Lambda_{L}}(\phi) = \frac{1}{2}\langle \phi, (-\Delta_{L}+m^{2})\phi\rangle_{\Lambda} \tag{2}\label{2} -\end{eqnarray} -The first step is to define finite volume measures on $\mathbb{R}^{\mathbb{Z}^{d}}$. For each finite $\Lambda \subset \mathbb{Z}^{d}$, let $C_{\Lambda} =(C_{xy})_{x,y \in \Lambda}$ be the matrix with entries $C_{xy} := (-\Delta_{per}+m^{2})_{xy}$, where $(-\Delta_{per}+m^{2})_{xy}$ is the Kernel of $-\Delta_{per}+m^{2}$ on $\mathbb{R}^{\mathbb{Z}^{d}}$. Because the Kernel of $-\Delta_{per}+m^{2}$ is the same as $-\Delta_{L}+m^{2}$, each $C_{\Lambda}$ is a positive-definite matrix and, thus, define a Gaussian measure $\mu_{\Lambda}$ on $\mathbb{R}^{\Lambda}$. Because this family of Gaussian measures $\mu_{\Lambda}$ is consistent, we can use Kolmogorov's Extension Theorem to obtain a Gaussian measure $\mu$ on $\mathbb{R}^{\mathbb{Z}^{d}}$ (with the product $\sigma$-algebra). Moreover, we can also obtain a family $\{f_{\alpha}\}_{\alpha \in \mathbb{Z}^{d}}$ of random variables such that $\mu_{\Lambda}$ is the joint probability distribution of $\{f_{\alpha}\}_{\alpha \in \Lambda}$. As it turns out, it is possible to prove that these random variables are given by $f_{\alpha}(\phi) = \phi(\alpha)$, $\alpha \in \mathbb{Z}^{d}$. In summary, if $A$ is a Borel set in $\mathbb{R}^{\Lambda}$, we must have: -\begin{eqnarray} -\mu_{\Lambda_{L}}(A) = \frac{1}{Z}\int_{A}e^{-\frac{1}{2}\langle \phi, (-\Delta_{L}+m^{2})\phi\rangle_{L}}d\nu_{L}(\phi) = \mu(A\times \mathbb{R}^{\mathbb{Z}^{d}\setminus \Lambda}) \tag{3}\label{3} -\end{eqnarray} -with $\nu_{L}$ being the Lebesgue measure on $\mathbb{R}^{\Lambda}$. The Gaussian measure $\mu$ is our a priori measure on $\mathbb{R}^{\mathbb{Z}^{d}}$ and, by (\ref{3}), it can be interpreted as a finite volume over $\mathbb{R}^{\Lambda}$. -Now, let $G(x,y)$ the Green function of $-\Delta+m^{2}$ in $l^{2}(\mathbb{Z}^{d})$. If $s_{m}:=\{\psi \in \mathbb{R}^{\mathbb{N}}:\hspace{0.1cm} \sum_{n=1}^{\infty}n^{2m}|\psi_{n}|^{2}\equiv ||\psi||_{m}^{2}\infty\}$, define $s:=\bigcup_{m\in \mathbb{Z}}$ and $s':=\bigcap_{m\in \mathbb{Z}}s_{m}$. Let u $K=(K_{xy})_{x,y \in \mathbb{Z}^{d}}$ be an 'infinite matrix' given by $K_{xy}:= G(x,y)$. If we order $\mathbb{Z}^{d}$, we can consider $K$ to be an 'infinite matrix' $K = (K_{ij})_{i,j \in \mathbb{N}}$. Now, define the following map: -\begin{eqnarray} -s \times s \ni (\psi,\varphi) \mapsto (\psi,K\varphi):= \sum_{i,j=1}^{\infty}\psi_{i}K_{ij}\varphi_{j}\tag{4}\label{4} -\end{eqnarray} -Let $W(\phi):= e^{-\frac{1}{2}(\phi,K\phi)}$. It is possible to prove that $W$ is a positive-definite function on $s$, so that, by Minlos Theorem, there exists a Gaussian measure $\tilde{\mu}_{K}$ on $s'$ such that $W$ is the Fourier transform of $\tilde{\mu}_{K}$. -[Question] I would like to establish a connection between $\mu$ and $\tilde{\mu}_{K}$ (where, here, $\mu$ is the restriction of $\mu$ to $s'\subset \mathbb{R}^{\mathbb{Z}^{d}}$ with its natural $\sigma$-algebra). It seems to me that $\tilde{\mu}_{K}$ is the infinite volume measure of $\mu$, in the sense that when we take $L\to \infty$ one should obtain $\tilde{\mu}_{K}$. In other words, $\tilde{\mu}_{K}$ is the infinite volume Gibbs measure obtained by taking the thermodynamic limit of the measures $\mu_{L}$. But, if I'm not mistaken, to prove that $\tilde{\mu}_{K}$ is the infinite volume Gibbs measure, I should prove that: -\begin{eqnarray} -\lim_{L\to \infty}\int f(\phi)d\mu_{L}(\phi) = \int f(\phi)d\tilde{\mu}_{K}(\phi) \tag{5}\label{5} -\end{eqnarray} -i.e. I should prove that $\mu$ converges weakly to $\tilde{\mu}_{K}$. And I don't know how to prove it. -Note: The above setup is a result of some of my thoughts about the problem during the last few days. I've been using a lot of different references and each one work the problem in a different way, with different notations and different objectives, so I'm trying to put it all together in one big picture. It is possible that my conclusions are not entirely correct or I can be going in the wrong direction, idk. But any help would be appreciated. -Note 2: I think it could be easier to prove a more particular limit such as $\lim_{L\to \infty}\int\phi(x)\phi(y)d\mu_{L}(\phi) = \int \phi(x)\phi(y)d\tilde{\mu}_{K}(\phi)$ and this would be enough to establish the existence of infinite volume correlation functions, which is one of the most important quantities in statistical mechanics. However, I don't think I could conclude that $\tilde{\mu}_{K}$ is the associate infinite volume Gibbs measure for the system just from this limit. Don't I need to prove it for a general $f$ as above? - -REPLY [3 votes]: For $x\in\mathbb{Z}^d$ I will denote by $\bar{x}$ the corresponding equivalence class in the discrete finite torus $\Lambda_{L}=\mathbb{Z}^d/L\mathbb{Z}^d$. -I will view a field $\phi\in\mathbb{R}^{\Lambda_L}$ as a column vector with components $\phi(\bar{x})$ indexed by $\bar{x}\in\Lambda_L$. -The discrete Laplacian $\Delta_L$ then acts by -$$ -(\Delta_L\phi)(\bar{x})=\sum_{j=1}^{d}\left[ --2\phi(\bar{x})+\phi(\overline{x+e_j})+\phi(\overline{x-e_j}) -\right]\ . -$$ -Now take the column vectors -$$ -u_k(\bar{x})=\frac{1}{L^{\frac{d}{2}}} e^{\frac{2i\pi k\cdot x}{L}} -$$ -for $k=(k_1,\ldots,k_d)\in\{0,1,\ldots,L-1\}^d$. -They give an orthonormal basis in $\mathbb{C}^{\Lambda_L}$ which diagonalizes the Laplacian matrix. -Let $C_L=(-\Delta_L+m^2{\rm I})^{-1}$ and denote its matrix elements by $C_L(\bar{x},\bar{y})$. -We then have, for all $x,y\in\mathbb{Z}^d$, -$$ -C_L(\bar{x},\bar{y})= -\frac{1}{L^d}\sum_{k\in\{0,1,\ldots,L-1\}^d} -\frac{e^{\frac{2i\pi k\cdot(x-y)}{L}}}{m^2+ -2\sum_{j=1}^{d}\left[1-\cos\left(\frac{2\pi k_j}{L}\right)\right]} -=:G_L(x,y) -$$ -where we used the formula to define $G_L$ on $\mathbb{Z}^d\times\mathbb{Z}^d$. -Because we assumed $m>0$, we have the trivial uniform bound in $L$ saying -$$ -|G_L(x,y)|\le \frac{1}{m^2}\ . -$$ -Now let $\nu_L$ denote the centered Gaussian probability measure on $\mathbb{R}^{\Lambda_L}$ with covariance matrix $C_L$. We also define an injective continuous linear map -$$ -\tau_L:\mathbb{R}^{\Lambda_L}\longrightarrow s'(\mathbb{Z}^d) -$$ -which sends $\phi\in\mathbb{R}^{\Lambda_L}$ to $\psi\in s'(\mathbb{Z}^d)$ defined by -$\psi(x)=\phi(\bar{x})$ for all $x\in\mathbb{Z}^d$. -Of course $\mathbb{R}^{\Lambda_L}$ has its usual finite-dimensional space topology, whereas -$s'(\mathbb{Z}^d)$ is given the strong topology and the resulting Borel $\sigma$-algebra. -As I explained in my answer to the previous MO question we can use such a map to push forward probability measures. We thus go ahead and define $\mu_L=(\tau_L)_{\ast}\nu_L$ which is a Borel probability measure on $s'(\mathbb{Z}^d)$. -Now we switch gears and consider the Green's function $G_{\infty}(x,y)$ for $-\Delta+m^2$ on $\mathbb{Z}^d$. More explicitly, -$$ -G_{\infty}(x,y)=\frac{1}{(2\pi)^d} -\int_{[0,2\pi]^d}d^d\xi\ -\frac{e^{i\xi\cdot(x-y)}}{m^2+ -2\sum_{j=1}^{d}\left(1-\cos\xi_j\right)}\ . -$$ -The function -$$ -\begin{array}{crcl} -W_{\infty}: & s(\mathbb{Z}^d) & \longrightarrow & \mathbb{C} \\ - & f & \longmapsto & \exp\left( - -\frac{1}{2}\sum_{x,y\in\mathbb{Z}^d} f(x)\ G_{\infty}(x,y)\ f(y) - \right) -\end{array} -$$ -satisfies all the hypotheses of the Bochner-Minlos Theorem for $s'(\mathbb{Z}^d)$. Therefore it is the characteristic function of a Gaussian Borel probability measure $\mu_{\infty}$ on $s'(\mathbb{Z}^d)$. -Finally after all these preliminaries we can state the main result which the OP asked for. -Theorem: When $L\rightarrow\infty$, the measure $\mu_L$ converges weakly to $\mu_{\infty}$. -The proof uses the Lévy Continuity Theorem for $s'(\mathbb{Z}^d)$ which is due to Xavier Fernique. One only has to prove that for all discrete test function $f\in s(\mathbb{Z}^d)$, -$$ -\lim\limits_{L\rightarrow \infty} W_L(f)\ =\ W_{\infty}(f) -$$ -where $W_L$ is the characteristic function of the measure $\mu_L$. -By definition, we have -$$ -W_L(f)=\int_{s'(\mathbb{Z}^d)} \exp\left[ -i\sum_{x\in\mathbb{Z}^d}f(x)\psi(x) -\right]\ d[(\tau_L)_{\ast}\nu_L](\psi)\ . -$$ -By the abstract change of variable theorem, -$$ -W_L(f)=\int_{\mathbb{R}^{\Lambda_L}} \exp\left[ -i\sum_{x\in\mathbb{Z}^d}f(x)\phi(\bar{x}) -\right]\ d\nu_L(\phi) -$$ -$$ -=\exp\left( --\frac{1}{2}\sum_{\bar{x},\bar{y}\in\Lambda_L} -\tilde{f}(\bar{x})\ C_L(\bar{x},\bar{y})\ \tilde{f}(\bar{y}) -\right) -$$ -where we introduced the notation $\tilde{f}(\bar{x})=\sum_{z\in\mathbb{Z}^d}f(x+Lz)$. -Hence -$$ -W_L(f)= \exp\left( - -\frac{1}{2}\sum_{x,y\in\mathbb{Z}^d} f(x)\ G_{L}(x,y)\ f(y) - \right)\ . -$$ -Since the function -$$ -\xi\longmapsto \frac{1}{(2\pi)^d} -\frac{e^{i\xi\cdot(x-y)}}{m^2+ -2\sum_{j=1}^{d}\left(1-\cos\xi_j\right)} -$$ -is continuous on the compact $[0,2\pi]^d$ and therefore uniformly continuous, we have that, for all fixed $x,y\in\mathbb{Z}^d$, the Riemann sums $G_L(x,y)$ converge to the integral $G_{\infty}(x,y)$. -Because of our previous uniform bound on $G_L(x,y)$ and the fast decay of $f$, we can apply the discrete Dominated Convergence Theorem in order to deduce -$$ -\lim\limits_{L\rightarrow\infty} -\sum_{x,y\in\mathbb{Z}^d} f(x)\ G_{L}(x,y)\ f(y)\ =\ -\sum_{x,y\in\mathbb{Z}^d} f(x)\ G_{\infty}(x,y)\ f(y)\ . -$$ -As a result $\lim_{L\rightarrow \infty}W_L(f)=W_{\infty}(f)$ and we are done. -Note that we proved weak convergence which as usual means -$$ -\lim\limits_{L\rightarrow \infty} -\int_{s'(\mathbb{Z}^d)}F(\psi)\ d\mu_L(\psi)\ =\ -\int_{s'(\mathbb{Z}^d)}F(\psi)\ d\mu_{\infty}(\psi) -$$ -for all bounded continuous functions $F$ on $s'(\mathbb{Z}^d)$. -One also has convergence for correlation functions or moments because of the Isserlis-Wick Theorem relating higher moments to the second moment and the previous argument where we explicitly treated the convergence for second moments. -Finally, note that the extension map $\tau_L$ used here is the periodization map, but there are lots of other choices which work equally well. A good exercise, is to construct the massive free field in the continuum, i.e., in $\mathscr{S}'(\mathbb{R}^d)$, as the weak limit of suitably rescaled lattice fields on $\mathbb{Z}^d$ with a mass adjusted as a function of the (rescaled) lattice spacing.<|endoftext|> -TITLE: What clues originally hinted at stability phenomena in algebraic topology? -QUESTION [21 upvotes]: If you didn't know anything about stabilization phenomena in algebraic topology and were trying to discover/prove theorems about the homotopy theory of spaces, what clues would point you toward results such as Freudenthal suspension or the existence of stable homotopy groups of spheres? -References suggest that Freudenthal originally stated his result in this 1938 Paper, although I'm unable to find an English translation. This was published only a few short years after the discovery of the Hopf fibration, so I find it pretty surprising that not only would there have been clear notions of $\pi_{\geq 2}$ at the time, but also enough evidence to suggest looking for things like the suspension map or stable homotopy groups. -Analogous stabilization phenomena do seem to occur elsewhere in mathematics: for instance, vector bundles that become isomorphic after taking Whitney sums with trivial bundles. From there, it may not be that much of a leap to suppose that something similar might work for fibrations. -However, it also seems that Freudenthal's paper predated results like this, and so historically, perhaps the flow of ideas was the other way around. What other results might have motivated his suspension theorem? Or in retrospect, what are some signs that such a thing might have worked and been useful? - -REPLY [5 votes]: $\newcommand{\R}{\mathbb R}\newcommand{\inj}{\hookrightarrow}$This will be an anachronistic answer, because it was discovered a bit later, but: Whitney proved that every -$n$-manifold embeds in $\R^N$ for $N$ large enough, and Wu showed in 1958 that for $N\ge 2n+2$, all such embeddings -are isotopic. This leads to a few interesting stability phenomena: most notably, that every manifold has -canonically the data of the isomorphism type of the normal bundle to the embedding $M\inj\R^N$, up to direct sums -with trivial bundles. (And this leads to stable vector bundles, another stabilization in algebraic -topology…) -How might this have led to Freudenthal's theorem? One upshot is that is bordism groups of immersions stabilize, and -in high codimension are just abstract bordism groups. Thom's work on bordisms showed that bordism groups of -immersions are homotopy groups of certain spaces, called Thom spaces, and the Thom space for $n$-manifolds in -$\R^{N+1}$ is the suspension of the Thom space for $n$-manifolds in $\R^N$. So there are two different reasons -these homotopy groups stabilize (the numbers don't quite match: Freudenthal's theorem is sharper). But in some -alternate history, where Whitney and Wu's work was earlier, one could imagine people asking, “so the homotopy -groups of Thom spaces stabilize, what about everything else?” -(If you modified this by asking for $M\inj\R^N$ to be equipped with a trivialization of its normal bundle, then the -Thom space is a sphere, so this provides another description/proof of the stable homotopy groups of the spheres.)<|endoftext|> -TITLE: A polytope with congruent facets and an insphere that is not facet-transitive? -QUESTION [6 upvotes]: Is there a $d$-dimensional convex polytope (convex hull of finitely many points, not contained in a proper subspace), with $d\ge 4$ and the following properties? - -All facets are congruent, -it has an insphere (a sphere to which each facet is tangent to), and -it is not facet-transitive. - -In 3-dimensional space there is an example with the "memorable" name Pseudo-deltoidal icositetrahedron, depicted below. -I believe its the only such polyhedron. -I am not aware of any higher dimensional examples. - -REPLY [4 votes]: First, a simple remark: If a polytope with congruent facets is inscribed in a sphere, then it is circumscribed about a sphere as well, and the two spheres are concentric. -Next, there is a series of examples described and pictured in my old question -Can the sphere be partitioned into small congruent cells? . Each of these examples is what you want in $R^3$. If you begin with any one such example and place it on a great 2-sphere of the 3-sphere in $R^4$ (say, the "equator"), then suspend it from the poles, you will get an example answering your question. The construction generalizes inductively to all higher dimensions.<|endoftext|> -TITLE: Proving certain inequality related to Primes -QUESTION [5 upvotes]: I was reading the following paper. But I can't understand why the last line concerning $\frac{2}{\pi}$ is true. The proof is a work of Sylvester. -I would be happy if someone helps me in understanding why the last inequality follows. -Thanks in advance. - -REPLY [7 votes]: As Fedor Petrov says, this looks incorrect. The presence of -$$2/\pi = \prod_{n}\left(1-\frac{1}{(2n)^2}\right)$$ -(known as the Wallis product) makes me think Sylvester is mistakenly comparing it to this somehow, with the missing implicit step being 'every prime is at least some even number', so $M\geq W$, since if $2n\leq p$ then we can replace the $(1-1/(2n)^2)$ factor with $(1-1/p^2)$ and only increase the product. -The error is, of course, that 'the greatest even number $\leq p$' is not unique for each $p$ - one gets the same factor appearing for both $2$ and $3$! If one corrects this by omitting $3$ then you get -$$ M > (1-1/9)\cdot\frac{2}{\pi} = \frac{16}{9\pi}.$$ -I don't know if this correction is enough to salvage the remainder of Sylvester's argument. (Note that this is now consistent with the fact that $\prod_p (1-1/p^2)=6/\pi^2$). - -REPLY [6 votes]: I also do not understand. The infinite product of $(1-1/p^2)$ equals $6/\pi^2<2/\pi$.<|endoftext|> -TITLE: quick question about renorming quasi-Banach spaces into p-Banach spaces -QUESTION [5 upvotes]: I have a quick question which is probably supposed to be obvious, but for some reason I just don't see it: How does one re-norm a quasi-Banach space to produce a $p$-Banach space ($0 -TITLE: Understanding Sylvester' s $1871$ paper of primes in arithmetic progression of the forms $4n+3$ and $6n+5$ -QUESTION [6 upvotes]: The following is the proof of infinitude of primes in arithmetic progression of the form $4n+3$ and $ 6n+5$ done by Sylvester in $1871$ in his paper "On the theorem that an arithmetical progression which contains more than one, contains infinite number of primes number." The screenshot is from the book /note "The collected mathematical papers Of James Joseph Sylvester". -I having difficulty in understanding the proof in the case $4n+3$ . -I would be highly grateful if someone helps me in understanding the proof for the case $4n+3.$ -This question has been asked in the following link -Any help would be appreciated. Thanks in advance. - -REPLY [5 votes]: I think the 'identical equation' Sylvester has in mind is -$$\sum_q \mu(q) \frac{x^q}{1-x^{2q}} = x+x^5+x^{13}+x^{17}+x^{25}+x^{29}+\cdots $$ -where the left-hand sum is over all natural numbers $q$ divisible only by primes of the form $4s+3$ and the right hand side is the sum of all powers $x^r$ where $r$ is divisible only by primes of the form $4s+1$. (Sylvester specifies no repeated prime factors for $q$ on the left-hand side, but since I'm using $\mu$, any such summand is killed by $\mu(q) = 0$; note the first summand is for $q=1$.) -Proof. The left coefficient of $x^n$ in the left-hand side is $\sum_{q} \mu(q)$ where the sum is over all square-free $q$ divisible only by primes of the form $4s+3$ such that $n/q$ is odd. It is therefore zero for even $n$. If $n$ is odd let $n = Np_1\ldots p_t$ where $p_i \equiv 3$ mod $4$ for each $i$ and no such prime divides $N$. The sum is then -$$\sum_{q \mid p_1\ldots p_t} \mu(q) = \begin{cases} 1 & \text{if $t=0$} \\ 0 & \text{otherwise.} \end{cases} $$ -Hence the coefficient of $x^n$ is $0$ unless $n$ is divisible only by primes of the form $4s+1$, in which case it is $1$. $\Box$ -The rest of Sylvester's argument seems clear enough to me: if there are only finitely many primes of the form $4s+3$ then the left-hand side is a finite sum, and well-defined when $x=i$ since $i^{2q} = (-1)^q = -1$ as $q$ is odd, and so $1-x^{2q} = 2$. But then, by hypothesis (and infinitely many primes), there are infinitely many primes of the form $4s+1$, making the right-hand side infinite when $x=i$. -There is of course an easier argument, as Euclid take the product of finitely many primes of the form $4s+3$ including the prime $3$, multiply by $4$ and subtract $1$; the result is then divisible by another prime still of the form $4s+3$. -Since I had to make an edit anyway, I'll add that almost the same argument works for primes of the form $6s+1$ and $6s+5$; using the latter instead of primes of the form $4s+3$ to define the left-hand side, the right-hand side is the sum of all powers $x^r$ where $r$ is divisible only by the prime $3$ or primes of the form $6s+1$. But again one can show there are infinitely many primes of the form $6s+5$ by a variation on Euclid's argument. -One feature of interest to me is that Sylvester's argument uses Lambert series rather than the Dirichlet series ubiquitous in analytic number theory.<|endoftext|> -TITLE: What is the Schur multiplier of the Mathieu group $M_{10}$ -QUESTION [9 upvotes]: It is well known that the automorphism group of the alternating group $A_6$ is $P\Gamma L_2(9)$. There are three different index $2$ subgroups of $P\Gamma L_2(9)$, namely the symmetric group $S_6$, the projective general linear group $PGL_2(9)$, and the Mathieu group $M_{10}$. By checking the ATLAS (http://brauer.maths.qmul.ac.uk/Atlas/v3/), the Schur multipliers of those groups $A_6,S_6, PGL_2(9)$ are the cyclic group $Z_6$. What about the Mathieu group $M_{10}$? -Also, I don't know why, in the following book, Page 302, Table 4.1, the Schur multiplier of the group ${\sf C}_2(2)$ is listed to be the cyclic group $Z_2$? -Gorenstein, Daniel, Finite simple groups. An introduction to their classification, Moskva: Mir. 352 p. R. 2.50 (1985). ZBL0672.20010. - -REPLY [10 votes]: $H_2(M_{10},\mathbb Z)\cong H^2(M_{10},\mathbb C^\times)\cong H^3(M_{10},\mathbb Z) = \oplus_{ p | 720} H^3(M_{10},\mathbb Z)_{(p)}$ with $p\in\lbrace 2,3,5\rbrace$. A $p$-primary component is isomorphic to the set of $M_{10}$-invariant elements of $H^3(\text{Syl}_p(M_{10}))$. We can check that $M_{10}$ has semi-dihedral Sylow 2-subgroups and cyclic Sylow 5-subgroups (Wikipedia), both of whose Schur multiplier is trivial. So we're left with the 3-primary component. We can check that $M_{10}$ has elementary abelian Sylow 3-subgroups (Wikipedia), all isomorphic to $\mathbb Z_3^2$, and thus $H^3(M_{10})_{(3)}\cong H^3(\mathbb Z_3^2)^{H/\mathbb Z^2_3}$ by Swan's theorem, where $H$ is the normalizer of $\mathbb Z_3^2\subset M_{10}$. This invariant subgroup is the full group $H^3(\mathbb Z_3^2)\cong\mathbb Z_3$ (see below), so $H_2(M_{10},\mathbb Z)\cong\mathbb Z_3$. -To see that $H^3(\mathbb Z_3^2)^{H/\mathbb Z_3^2}\cong H^3(\mathbb Z_3^2)$, we first note that our $H$ is a maximal subgroup of order 72 so we may as well take it to be the Mathieu group $M_9=\mathbb Z_3^2\rtimes Q_8$ sitting naturally inside $M_{10}$, where $Q_8$ acts as the faithful two-dimensional irreducible representation over $\mathbb Z_3$. So we need to check that $H^3(\mathbb Z_3^2)^{Q_8}\cong H^3(\mathbb Z_3^2)$. For the standard generators $\{I,J\}$ of $Q_8$ and basis $\{a,b\}$ of $\mathbb Z_3^2$ we have $I(a)=a+b$, $I(b)= a-b$, $J(a)=-a+b$, and $J(b)=a+b$. Noting that $H^\ast(\mathbb Z_3^2,\mathbb Z_3) = \mathbb Z_3[x,y]\otimes\Lambda(u,v)$ with $|x|=|y|=2$ and $|u|=|v|=1$, the element $uv$ is $Q_8$-invariant. The induced map $\delta:H^2(\mathbb Z_3^2,\mathbb Z_3)\to H^3(\mathbb Z_3^2,\mathbb Z)$ under the short exact sequence $\mathbb Z\hookrightarrow\mathbb Z\twoheadrightarrow\mathbb Z_3$ is surjective and $Q_8$-equivariant, with image $\langle\delta(uv)\rangle$, and so $H^3(\mathbb Z_3^2)^{Q_8}\cong H^3(\mathbb Z_3^2)$. -We've essentially shown that this agrees with the Schur multiplier of $M_9$ by the same technique (note that the Schur multiplier of $Q_8$ is trivial), $H_2(M_9)\cong H^3(\mathbb Z_3^2)^{Q_8}\cong\mathbb Z_3$.<|endoftext|> -TITLE: A question about Schwartz-type functions used in analytic number theory -QUESTION [5 upvotes]: In analytic number theory we like to weigh our counting functions with a smooth function $f$, so that we may apply Poisson's summation formula and take advantage of Fourier transforms. Typically the weight function $f$ will be a Schwartz type function with the following properties: -1) $f(x) \geq 0 $ for all $x \in \mathbb{R}$; -2) $f(x) = 1$ for $x \in [-X,X]$ say; -3) $f(x) = 0$ for $|x| > X + Y$; and -4) $f^{(j)}(x) \ll_j Y^{-j}$ for $j \geq 0$. -In most applications the dependence on $j$ in the last condition does not matter, since $j$ would be bounded. However, in a problem I am considering it might be worthwhile to make $j$ a (slow growing) function of $X$ so it then becomes relevant to know how the bound depends on $j$. Is it possible to give an explicit example of a function $f$ satisfying the above properties for which the dependence can be made explicit? - -REPLY [6 votes]: The answer to your question is yes, and it is a pretty well-understood topic. -First of all $X$ is more or less irrelevant for the bounds in 4) so let us take $X = Y$, say, for convenience. -Second, we can always scale everything down by $Y$. So without loss of generality $Y = 1$. -Put $g = f'$. Let us also for simplicity assume that $f$ is symmetric so it is enough to study $g$ on $[1, 2]$. The problem therefore is reduced to the following: for which sequences $t_j$ we can find a function $g$ such that $g = 0, x\notin [1, 2]$, $\int g = 1$ and $|g^{j}(x)|\le C t_j$ for some constant $C$. -The answer to this question in more or less full generality is given by the Denjoy-Carleman Theorem: if the sequence $M_j = \frac{t_j}{j!}$ is logarithmically convex (i.e. $\frac{M_{j+1}}{M_j}$ is increasing in $j$) then such a function exists if and only if $\sum_j \frac{1}{jM_j^{1/j}} < \infty$. For example there exists a function $f$ such that -\begin{equation}\label{bound} -|f^{(j)}(x)| \le CY^{-j}j^{(1+\varepsilon)j} -\end{equation} -for any fixed $\varepsilon > 0$ (this is related to the so-called Gevrey classes). -Actually, since you mentioned Fourier transform, let me write about a different result which is more directly applicable to this type of problems: Beurling-Malliavin multiplier theorem. It reads as follows: -Let $w:\mathbb{R}\to \mathbb{R}$ be a nonnegative Lipshitz function (this is a small technical condition). Then there exists a nonzero compactly supported function $f$ with $|\hat{f}(\xi)| \le e^{-w(x)}$ if and only if integral -$$\int_\mathbb{R} \frac{w(x)}{x^2 + 1}dx$$ -is convergent. Moreover, support of the function can be arbitrary small. -Lastly, if you want an explicit function $g$ (and therefore $f$), satisfying the above bound, you can take -$$g(x) = e^{-(1-x)^{-m}}e^{-(x+1)^{-m}}\chi_{(-1, 1)},$$ -see e.g. this paper, section 3.1.<|endoftext|> -TITLE: Does the plane clustered to minimize sum distances^2 to clusters centers ( inertia / "K-means") produce hexagonal clusters / hexagonal lattice? -QUESTION [7 upvotes]: "K-means" is the most simple and famous clustering algorithm, which has numerous applications. -For a given as an input number of clusters it segments set of points in R^n to that given number of clusters. -It minimizes the so-called "inertia" i.e. sum distances^2 to clusters centers = $\sum_{i ~ - ~ cluster~ number} \sum_{X - points~ in ~i-th ~ cluster} |X_{in ~ i-th ~ cluster} - center_{i-th~ cluster} |^2 $ -By some reasons let me ask, what happens for the plane i.e. there is no any natural clusters, but still we can pose minimization task and it will produce something. Let us look on the example: - -So, most clusters look like hexagons. Especially the most central one which is colored in red. -Well, boundary spoils things, also may be not enough sample size/iteration number - simulation is not a perfect thing - but I made many and pictures are similar... -Hexagonal lattice appears in many somewhat related topics, so it might be that some reason exists. -Question 0 What is known on "inertia" minimization on the plane/torus ? (torus - to avoid boundary effects.) (Any references/ideas are welcome). Do hexagons arise as generic clusters ? -Question 1 Consider a torus of sizes R1,R2 , consider the number of clusters to be mn , -is it true that hexagonal lattice will provide the global minima for "inertia" ? (At least for consistent values of R1, R2, m,n (R1=am, R2 = a*n) ). -(Instead of finite number of points we can consider the continuous case and substitute summation over points by the integral. Or we can sample large enough uniform datacloud - as done in simulation). - -Let me mention beautiful survey by Henry Cohn at ICM2010, where lots of optimization problems of somewhat related spirit are discussed and which sound simple, but remain unsolved for years (see also MO78900). That question is not discussed there unfortunately. -The Python code for the simulation above. One can use colab.research.google.com - to run it - no need to install anything - can use google's powers for free. -from sklearn.cluster import KMeans -import numpy as np -import matplotlib.pyplot as plt -from scipy.spatial.distance import cdist -import time -#import datetime - -t0 = time.time() -N = 10**5 # Number of uniformly scattered point -d = 2 # dimension of space -X = np.random.rand(N,d) # Generate random uniform N poins on [0,1]^d -n_clusters = 225 # Number of clusters for Kmeans -clustering = KMeans(n_clusters=n_clusters, - init='k-means++', n_init=10, max_iter=600, # max_iter increased twice from default - tol=0.0001, random_state=None, algorithm= 'full' ).fit(X) # Run K-means with default params - # https://scikit-learn.org/stable/modules/generated/sklearn.cluster.KMeans.html#sklearn.cluster.KMeans - -print(time.time() - t0, ' secs passed' ) # Print time passed - -cluster_centers_ = clustering.cluster_centers_ # -predicted_clusters = clustering.labels_ # - -#################################################################### -# Choose the most central classter - hope boundary effect on it would be negligble -central_point = 0.5 * np.ones(d) # Choose central pint -idx_most_central_cluster = np.argmin( cdist( central_point.reshape(1,-1), cluster_centers_ ) ) # Find cluster most close to central point -coords_most_central_cluster_center = cluster_centers_[idx_most_central_cluster,: ] -mask = idx_most_central_cluster == predicted_clusters -Xm = X[mask,: ] # Select only points from the most central cluster - -####################################################################### -# Plotting -fig = plt.figure( figsize= (20,10 ) ) # 20 - horizontal size, 6 - vertical size -plt.scatter( X[:,0], X[:,1], c = predicted_clusters ) # scatter plot all the points colored according to different clusters -plt.scatter( cluster_centers_[:,0], cluster_centers_[:,1], c = 'blue' ) # Mark centers of the clusters -plt.scatter( Xm[:,0], Xm[:,1], c = 'red' ) # Color by red most central cluster -plt.title('n_sample = '+str(N) + ' n_cluster = ' + str(n_clusters)) -plt.show() - -REPLY [3 votes]: The answer is yes, at least in the limiting case where the number of points tends to infinity. -Specifically, this is known as the quantizer problem (see Chapter 2 of Sphere Packings, Lattices and Groups by Conway and Sloane). The two-dimensional version of the problem was solved by Fejes Tóth, who showed that the hexagonal lattice is optimal. -László Fejes Tóth, 1959: Sur la représentation d'une population infinie par un nombre fini d'éléments -The way that the quantizer problem is formalised in Sphere Packings, Lattices and Groups is to take a large compact ball $B \subsetneq \mathbb{R}^n$ and ask for the limit (as $M \rightarrow \infty$) of the infimum (over all arrangements of $M$ points in the ball) of the normalised mean squared error from a uniform random point in the ball to the closest of the $M$ points: -$$ \dfrac{1}{n} \dfrac{\frac{1}{M} \sum\limits_{i=1}^{M} \int\limits_{V(P_i)} \lVert x - P_i \rVert^2 \; dx}{\left( \frac{1}{M} \sum\limits_{i=1}^{M} \textrm{Vol}(V(P_i)) \right)^{1 + \frac{2}{n}}} $$ -Here, $V(P_i) \subseteq B$ is the Voronoi cell of $P_i$. The connection with $k$-means (where $k = M$ and the ambient dimension is $n$) is that the minimiser of this expression must have each $P_i$ be the centroid of its Voronoi cell $V(P_i)$, and therefore the optimal solution is a fixed point of the $k$-means iteration. The complicated normalisation is to ensure that the limit is sensible (e.g. not $0$ or $\infty$). -For $n = 2$, the limit as $M \rightarrow \infty$ of the infimum of the above expression is $\frac{5}{36 \sqrt{3}} \approxeq 0.0801875$, and is the same as the limit as $M \rightarrow \infty$ of the expression where the points are centred at the vertices of a hexagonal lattice (scaled to have exactly $M$ points inside $B$). -For $n = 3$, the best lattice is the body-centred cubic lattice, but there are more efficient nonlattice arrangements and the quantizer problem is unsolved. -In higher dimensions the problem is unsolved.<|endoftext|> -TITLE: When (why) did we allow manifolds to be non-Hausdorff and/or non-second countable? -QUESTION [17 upvotes]: I was reading David Carchedi's answer for a question on Grothendieck topology for a non-small category. It "reads" like people "choose" if they allow manifolds to be Hausdorff and/or second countable. When I came across the notion of smooth manifolds for the first time, by definition, smooth manifolds are Hausdorff and second countable. - -When (why) did we allow manifolds to be non-Hausdorff and/or non-second countable? - -I have observed this when I am reading about stacks. -Is it because of the constructions when we do in the set up of manifolds (stacks) which resulted in spaces that are "same as" manifolds but are not Hausdorff or not second countable? -Is it because of the influence of algebraic geometry? - -REPLY [8 votes]: If you're asking a historical question, it is probably because of Whitney, who set out a clear account of differentiable manifolds and proved the embedding theorem, that the Hausdorff and second countability assumptions have been regarded as "standard." Nevertheless, violations of these assumptions had been contemplated prior to Whitney's work, notably the non-second-countable Prüfer surface, which dates back to the 1920s. -In addition to the examples mentioned by others, note that people studying general relativity have sometimes considered non-Hausdorff spacetime manifolds (see here for a non-paywalled version). The consensus, however, seems to be that these are mathematical curiosities that are not physically significant. Luc and Placek amusingly quote Penrose as saying, "I must … return firmly to sanity by repeating three times: spacetime is a Hausdorff differentiable manifold; spacetime is a Hausdorff …"<|endoftext|> -TITLE: Smooth Morse function from Forman's discrete Morse function -QUESTION [9 upvotes]: Let $M$ be a smooth manifold and $K$ a triangulation of $M$, so $K$ is a regular CW-complex and in particular a simplicial complex. Assume that $M$ is compact so $K$ is finite. Let $f\colon K \to \mathbb{R}$ be a discrete Morse function (in the sense of Forman). Is is possible to define a smooth Morse function $f'\colon M \to \mathbb{R}$ with the same critical points as $f$ (and satisfying a correspondence between the indexes of the critical points)? Is it possible to do it in "an algorithmic way" (I mean that the proof is constructive)? -As far as I know, the converse was addressed by Gallais and Benedetti, am I right? -I apologize in advance if the questions are to vague or the answers are well-known. Thanks in advance for your time. - -REPLY [9 votes]: You can do the next best think. To a Forman-Morse function $f$ one can associate a flow on the manifold whose stationary points are precisely the barycenters of the faces of your simplicial decomposition. The Conley index of the barycenter of a critical face has the homotopy type of a sphere of the dimension of that face. The Conley index of the barycenter of a non-critical face is homotopically trivial. -Additionally, one can construct a continuous function $\tilde{f}$ on the manifold that decreases along the trajectories of this flow and whose value at a barycenter is equal to the value of $f$ on the corresponding face. As Mike Miller correctly pointed out, a critical face is filled out by the trajectories exiting the barycenter. -For details see Chapter 11 of this paper. The faces of the barycentric subdivision of your simplicial complex are invariant sets of this flow, and on such a face the flow is depicted in Figure 2, p.16 of the above paper. -It took me a while to realize that in Morse theory the gradient flow associated to a Morse function is more important than the function itself. The function plays a sort of accounting role and the Morse condition restricts the nature of the stationary points of the gradient flows. -Remark A while ago I asked this question on MathOverflow that is related to the abundance of discrete Morse functions. They are extremely rare as opposed to the usual smooth Morse functions that are generic.<|endoftext|> -TITLE: Prove that a given distribution is tempered -QUESTION [6 upvotes]: Suppose I have a distribution $E$ such that $\phi \ast E$ is square-integrable for all $\phi \in C_c^\infty \left( \mathbb{R}^d \right)$. Is it possible to prove that $E$ is tempered? It seems plausible to me, but I only get so far: -For brevity, define -\begin{equation} -G_\phi = \phi \ast E -\end{equation} -for any $\phi \in C_c^\infty$. -Now, because convolution is commutative, for all $\phi, \psi \in C_c^\infty$ we have -\begin{equation} -\mathcal{F} \left( \phi \ast \psi \ast E \right) -= \tilde{G}_{\phi \ast \psi} -= \left( 2 \pi \right)^{d/2} \tilde{\phi} \cdot \tilde{G_\psi} -= \left( 2 \pi \right)^{d/2} \tilde{\psi} \cdot \tilde{G_\phi} -\end{equation} -where the tildes denote Fourier transformed quantities. -Then, we get -\begin{equation} -\frac{\tilde{G}_\phi}{\tilde{\phi}} = \frac{\tilde{G}_\psi}{\tilde{\psi}} =: F -\end{equation} -which is of course what we expect, since we would like to interpret $F$ as the Fourier transform of our distribution $E$. -We can now deduce that $F \in L^1_{\mathrm{loc}} \left( \mathbb{R}^d \right) \cap L^2_{\mathrm{loc}} \left( \mathbb{R}^d \right)$ by -\begin{equation} -\left \Vert F \right \Vert_{L^1 \left( K \right)} -\le \left \Vert \frac{1}{\tilde{\phi}} \tilde{\phi} F \right \Vert_{L^1 \left( K \right)} -\le \left \Vert \frac{1}{\tilde{\phi}} \right \Vert_{L^2 \left( K \right)} \left \Vert \tilde{\phi} F \right \Vert_{L^2 \left( K \right)} -< \infty -\end{equation} -and -\begin{equation} -\left \Vert F \right \Vert_{L^2 \left( K \right)} -\le \left \Vert \frac{1}{\tilde{\phi}} \tilde{\phi} F \right \Vert_{L^1 \left( K \right)} -\le \sqrt{\left \Vert \frac{1}{\left \vert \tilde{\phi} \right \vert^2} \right \Vert_{L^\infty \left( K \right)}} \left \Vert \tilde{\phi} F \right \Vert_{L^2 \left( K \right)} -< \infty -\end{equation} -for any compact $K$ by using some $\phi \in C_c^\infty \left( \mathbb{R}^d \right)$ with strictly positive Fourier transform (These do exist). -I seem to be unable to get any further though. -It is now evident, that $F$ is in fact a distribution (as it is locally integrable) and it remains to show that we can approximate any Schwartz function under the integral in a (Schwartz)-continuous way. - -REPLY [4 votes]: Edit: even if another answer has been accepted, I edited mine in order to correct (hopefully) the issues raised in the comments, and - -possibly list more easily readable references to Łojasiewicz's solution of the division problem, and -prove that the tempered distribution $S$, found by using Łojasiewicz's division theorem and such that $\phi\ast S=\phi\ast E$, is equal to $E$. -prove a stronger property that the one required by the asker: namely, if $\phi\ast E\in\mathscr{S}^\prime$ for a single function $\phi\in C^\infty_c(\Bbb R^n)$, then $E\in\mathscr{S}^\prime$. - - -The result can be proved by using Stanisław Łojasiewicz's solution of the division problem in $\mathscr{S}^\prime(\Bbb R^n)$ (see [2] and [3] or [4] pp. 99-101 or [6] chapter VI, §VI.1): the equation -$$ -\Phi S=T\label{div}\tag{DIV} -$$ -has a tempered distribution solution $S\in\mathscr{S}^\prime(\Bbb R^n)$ for every non-null real analytic function $\Phi\in \mathscr{A}(\Bbb R^n)$ and every datum $T\in\mathscr{S}^\prime(\Bbb R^n)$. Indeed, since -$$ -G_\phi=\phi \ast E\in L^2(\Bbb R^n)\qquad \forall \phi\in C_c^\infty(\Bbb R^n), -\label{1}\tag{1} -$$ we have also that $G_\phi\in\mathscr{S}^\prime(\Bbb R^n)$ as a distribution, and thus $\hat{G}_\phi\in\mathscr{S}^\prime(\Bbb R^n)$ by the isomorphism theorem for the Fourier transform in $\mathscr{S}^\prime$ (see [1], chapter VII, §7.1, theorem 7.1.10, p. 164). Then we can choose a test function $\phi\not\equiv 0$ and, by using the division theorem, find a tempered distribution $S$ such that -$$ -\hat{\phi}\hat{S}=\hat{G}_\phi\label{2}\tag{2} -$$ -since -$$ -\phi\in C_c^\infty(\Bbb R^n) \implies\phi\in \mathscr{E}^\prime(\Bbb R^n) -$$ as a distribution, and therefore $\hat{\phi}\in \mathscr{A}(\Bbb R^n)$ i.e. $\hat{\phi}$ is a complex valued real analytic function (see, for example, [1], chapter VII, §7.1, theorem 7.1.14 pp. 165-166). Now applying the inverse Fourier transform to both sides of equation \eqref{2} and considering equation \eqref{1} we have -$$ -\phi\ast S=\phi\ast E\iff \phi\ast(S-E)=(S-E)\ast\phi =0 \label{3}\tag{3} -$$ -Lemma. Equation \eqref{3} implies $S=E$. -Proof for a given $\phi\in C^\infty_c(\Bbb R^n)$, consider the following convolution equation: $\DeclareMathOperator{\invs}{\small{inv}}$ -$$ -\phi\ast\psi(x)=-\varphi(-x)=-\varphi\circ\invs(x) \quad \forall \varphi(x)\in C^\infty_c\Bbb R^N \label{4}\tag{4} -$$ -where $\Bbb R^n \ni x\mapsto \invs(x)=-x$ is the point reflection map. Again by the division theorem, this equation is solvable and its solution, i.e. -$$ -\psi(x)= -\mathscr{F}^{-1} \left[\hat{\phi}^{-1}\right] \ast \varphi\circ\invs(x) -$$ -apart from being tempered as a distribution, is a $C^\infty$ function since it is equal to the convolution of a tempered distribution whit a compactly supported and $C^\infty$-smooth function. Now define $\eta_r\in C_c^\infty(\Bbb R^n)$, $r>0$, as -$$ -\eta_r(x) = -\begin{cases} -1 & |x|r+1, -\end{cases} -$$ -Then the family -$$ -\{\psi_r(x)\}_{r>0}=\{\eta_r(x)\cdot\psi(x)\}_{r>0} -$$ is a family of compactly supported $C^\infty$ functions converging to $\psi$. -Now consider the structure of the convolution on the left side of \eqref{3}: we have that -$$ -(S-E)\ast \phi = \big\langle S-E ,\phi(x-\cdot)\big\rangle -$$ and thus -$$ -\begin{split} -\langle (S-E)\ast \phi, \varphi\rangle & = \int_{\Bbb R^n}\langle S-E ,\phi(x-\cdot)\rangle\varphi(x)\mathrm{d} x\\ -& = \left\langle S-E,\int_{\Bbb R^n}\phi(x-y)\varphi(x)\mathrm{d} x\right\rangle\\ -& = \big\langle E- S,\phi\ast\varphi\circ\invs\big\rangle -\end{split}\label{5}\tag{5} -$$ -so, again considering the relation \eqref{3} and the definition of the family of compactly supported functions $\{\psi_r(y)\}_{r>0}$ we have -$$ -\begin{split} -\lim_{r\to+\infty}\big\langle (S-E)\ast \phi,\psi_r\big\rangle=\langle S-E,\varphi\rangle=0\quad \forall \varphi\in C_c^\infty(\Bbb R^n) -\end{split} -$$ -and thus $E-S=0\iff E=S\;\blacksquare$. -Finally, the above lemma implies $E=S\iff E\in\mathscr{S}^\prime$. -Notes. - -The expositions for Łojasiewicz's solution of the division problem, apart from the original works [2] and [3] are the books by Bernhard Malgrange [4] and Jean-Claude Tougeron [5] (the latter two deal with the work of Malgrange which generalizes Łojasiewicz's solution to systems and even to $C^\infty$ division in some special cases): however, neither of them is particularly readable to the accustomed to the "ordinary" theory of distributions and its application, since the techniques used are more from the theory of analytic sets (varieties) and from the (however related) theory of ideals of smooth functions than from functional analysis. Nevertheless I like the work of Malgrange [4], partly because of its generality and partly because of its newly improved digital version produced by the Tata Institute of Fundalmental research. However, as stated above, they are not easy read. -Jochen Bruning and iolo point out that the solution of \eqref{4} is unique if we chose $\phi$ in such a way that $\hat{\phi}(\xi)>0$ for all $\xi\in\Bbb R^n$: this is alway possible by the Paley-Wiener theorem. -The relation \eqref{5} can also be proved directly by using the standard definition of convolution of distributions: however, using the fact that $\varphi\in C^\infty_c$ simplifies bit the formal development. - -References -[1] Lars Hörmander (1990), The analysis of linear partial differential operators I, Grundlehren der Mathematischen Wissenschaft, 256 (2nd ed.), Berlin-Heidelberg-New York: Springer-Verlag, ISBN 0-387-52343-X/ 3-540-52343-X, MR1065136, Zbl 0712.35001. -[2] Stanisław Łojasiewicz (1959), "Sur le problème de la division" (French), -Studia Mathematica 18, 87-136, DOI: 10.4064/sm-18-1-87-136, MR0107168, Zbl 0115.10203. -[3] Stanisław Łojasiewicz, Sur le problème de la division, (French), Rozprawy Matematyczne 22, pp. 57 (1961), MR0126072, Zbl 0096.32102. -[4] Bernard Malgrange, Ideals of differentiable functions, (English) Studies in Mathematics. Tata Institute of Fundamental Research 3. London: Oxford University Press, pp. 106 (1966), MR0212575, Zbl 0177.17902. -[5] Jean-Claude Tougeron, Ideaux de fonctions différentiables (French) Ergebnisse der Mathematik und ihrer Grenzgebiete. Band 71. Berlin-Heidelberg-New York: Springer-Verlag. pp. VII+219 (1972), MR0440598, Zbl 0251.58001.<|endoftext|> -TITLE: Is there a Riemannian metric on the configuration space of $n$ distinct points with "nice" geodesics? -QUESTION [8 upvotes]: Let $C_n = C_n(\mathbb{R}^3)$ denote the configuration space of $n$ distinct points in $\mathbb{R}^3$. Is there a Riemannian metric $g$ on $C_n$ such that given any two configurations in $C_n$, there is a unique geodesic joining them? -In addition, it would be nice if $g$ was also geodesically complete, and if $g$ came from natural considerations in Physics (for instance if it is the kinetic term of some naturally occuring Lagrangian etc.). -Edit 1: I accepted Andy Putman's answer below, because it does answer negatively my question (thank you!). However, could someone please indicate whether or not there exists a complete Riemannian metric $g$ on $C_n$? Is it more appropriate to create another post perhaps? I just found out that Nomizu and Ozeki proved that any connected smooth (second countable) manifold admits a complete Riemannian metric. This is nice. However, is there a known explicit such complete Riemannian metric $g$ on $C_n$? If two of the points say are going towards each other and seem about to collide, there has to be a repulsive force that forbids collision (in physical terms). - -REPLY [2 votes]: This is just a long comment, and a pretty speculative one at that. However it might perhaps be of interest to you since: - -there is a natural connection to physics, -the construction only works in three dimensions, -the construction is equivariant with respect to the action of the permutations. - -I have in mind the (conjectured) map described by Atiyah in [1] which maps configurations of points to the complex flag manifold: -$$ - C_n(\mathbb{R}^3) \to U(n) / T^n. -$$ -Since the flag manifold is homogeneous, this map would provide a metric on $C_n(\mathbb{R}^3)$ if we could spot a natural metric on the fibres. I don't know if this is possible but for $n=2$, the fibres of the map are pairs of distinct points defining the same direction (first point looking at second point) and so are naturally parameterised by their midpoint $m$ and distance apart $t$. It's a bit of a stretch but if we give this fibre the metric of $dm^2 + (dt/t)^2$ then we get something you might regard as "nice". -[1] Atiyah, M., "Configurations of Points", R. Soc. Lond. Philos. Trans. Ser. A Math. Phys. Eng. Sci. 359 (2001), no. 1784, 1375-1387.<|endoftext|> -TITLE: Are two metrics with the same Levi-Civita connection and the same volume form identical? -QUESTION [7 upvotes]: Given a (smooth, orientable) n-dimensional manifold $M$ with two (pseudo-)Riemannian metrics $g_{1}$ and $g_{2}$ of the same signature that induce the same Levi-Civita connection and satisfy $\sqrt{|\det{g_1}|}\mathrm{d}x^{1}\wedge...\wedge\mathrm{d}x^{n} = \sqrt{|\det{g_2}|}\mathrm{d}x^{1}\wedge...\wedge\mathrm{d}x^{n}$, -is $g_{1}=g_{2}$? - -REPLY [4 votes]: Consider a Riemannian manifold admitting a parallel (with respect to the Levi-Civita connection $\nabla^g$) symmetric bilinear form $\beta$ which is not a multiple of the metric $g$. Then, for an open set of constants $(a,b)\in\mathbb R^2$ $$g_{a,b}:=a g+b\beta$$ is a Riemannian metric, i.e., positive definite. The Levi-Civita connection of $g_{a,b}$ is $\nabla^g$. Moreover, the volume form of $g_{a,b}$ is a constant multiple (depending on $a,b$) of the volume form of $g$, as the volume forms $vol_{a,b}$ are parallel with respect to $\nabla^g$. Clearly, at $(a,b)=(1,0)$ we have $g_{a,b}=g$ and $$\frac{\partial vol_{a,b}}{\partial a}\neq0.$$ Hence, you always find a curve of Riemannian metrics with the same Levi-Civita connection and the same volume form under the above condition. Of course, explicit examples are provided by the answers of Ben and Liviu, but also by product metrics. Analogous arguments apply to the pseudo-Riemann case. -By taking the difference between two different metrics with the same Levi-Civita connection and the same volume form one immediately sees that the space of parallel symmetric bilinear forms must be at least two-dimensional.<|endoftext|> -TITLE: Cohomology of derived tensor product of complexes and Künneth spectral sequence -QUESTION [5 upvotes]: Let $R$ be any commutative ring, let $V^\bullet$ and $W^\bullet$ be (co)chain complexes of $R$-modules, indexed cohomologically. We can also assume that they have both cohomology in nonpositive degrees. Using K-flat resolutions we can define the derived tensor product: -$$ -V^\bullet \overset{\mathbb L}{\otimes}_R W^\bullet. -$$ -I'm looking for assumptions on $V^\bullet$ or $W^\bullet$ that ensure that the following "Künneth formula" holds: -$$ -H^*(V^\bullet \overset{\mathbb L}{\otimes}_R W^\bullet) \cong H^*(V^\bullet) \otimes_R H^*(W^\bullet), -$$ -where the right hand side is the tensor product of graded $R$-modules. Searching the literature, it seems that there should be some spectral sequence involving Tor of the cohomologies, as mentioned for instance in the nLab entry. - -Is there a more precise reference for such a result? -Looking at these Künneth formulas and the spectral sequence, I suspect that the following is true: $H^*(V^\bullet \overset{\mathbb L}{\otimes}_R W^\bullet) \cong H^*(V^\bullet) \otimes_R H^*(W^\bullet)$ holds if $V^\bullet$ or $W^\bullet$ has flat cohomologies, namely $H^k(V^\bullet)$ (say) is a flat $R$-module for all $k$. Is it correct? - -A little more wildly, could one expect that without assumptions one has -$$ -H^*(V^\bullet \overset{\mathbb L}{\otimes}_R W^\bullet) \cong H^*(V^\bullet) \overset{\mathbb L}{\otimes}_R H^*(W^\bullet), -$$ -so taking K-flat resolutions of the cohomology graded $R$-module if necessary? - -REPLY [4 votes]: This isn't a complete answer, but here are some thoughts. I'm sceptical that it is often true. -If $R$ is semisimple then the result holds, because every module over a semisimple ring is projective. -We may as well take $V$ and $W$ to be complexes of projectives, and then ask when is $H(V\otimes W)\cong H(V)\otimes H(W)$? If $R$ is a field this is the usual Kuenneth theorem. -When $V$ has flat cohomology the spectral sequence is not hard to obtain: let $V' \to V$ and $W' \to W$ be flat resolutions; because both $V$ and $W$ are cohomologically bounded above we can take $V'$ and $W'$ to be bounded above. The direct sum total complex of the double complex $V' \otimes W'$ computes the Tor groups. Taking cohomology in the vertical direction and using flatness of $W'$ tells us that the $E_1$ page of the associated spectral sequence is $H(V)\otimes W'$. Now taking cohomology in the horizontal direction and using flatness of $H(V)$ tells us that the $E_2$ page is $H(V)\otimes H(W)\Rightarrow \mathrm{Tor}(V,W)$, as desired. -The same argument when $V$ does not necessarily have flat cohomology gives a spectral sequence with $E_2$ page $\mathrm{Tor}(HV,W)\Rightarrow \mathrm{Tor}(V,W)$. -In general, Tor spectral sequences tend to look like this (as in e.g. https://stacks.math.columbia.edu/tag/061Y) - a spectral sequence of the form $H(V)\otimes H(W)\Rightarrow \mathrm{Tor}(V,W)$ will not exist without this flatness assumption on cohomology (just think about when $V$ and $W$ are genuine $R$-modules!). Of course, when $V$ has flat cohomology then the tensor product $H(V)\otimes H(W)$ is already the derived tensor product. -However, this doesn't answer your main question in the flat cohomology case: you have a spectral sequence, and you are interested in knowing when it collapses (or more generally when the $E_2$ page equals the $E_\infty$ page). -As for your wild question, when $V$ and $W$ are genuine modules, you are asking that $V\otimes^\mathbb{L}W$ be a formal complex. This is true in certain situations: for example the HKR theorem is true on the level of cochains and gives a quasi-isomorphism between the Hochschild complex and the graded module of polyvector fields. Or if $R$ is a ring, $r_1,\ldots, r_n$ is a (finite) regular sequence, $K$ the Koszul complex, and $M$ an $R$-module annihilated by all of the $r_i$ then you can easily check that $R/(r_1,\ldots, r_n)\otimes^\mathbb{L}_R M \simeq K\otimes_R M$ is formal. In general it won't be true, but I can't think of an example off the top of my head.<|endoftext|> -TITLE: Proof of certain $q$-identity for $q$-Catalan numbers -QUESTION [17 upvotes]: Let us use the standard notation for $q$-integers, $q$-binomials, -and the $q$-analog -$$ -\operatorname{Cat}_q(n) := \frac{1}{[n+1]_q} \left[\matrix{2n \\ n}\right]_q. -$$ -I want to prove that for all integers $n\geq 0$, we have -\begin{equation} -\operatorname{Cat}_q(n+2) -= -\sum_{0\leq j,k \leq n} -q^{k(k+2) + j(n+2)} -\left[\matrix{n \\ 2k}\right]_q -\operatorname{Cat}_q(k) -\frac{[n+4]_q}{[k+2]_q} -\left[\matrix{n-2k \\ j}\right]_q. -\end{equation} -I have tried quite a bit, but not succeeded. Using $q$-hypergeometric series, -this is equivalent with proving -$$ -\sum_{\substack{k\geq 0 \\ j \geq 0}} -q^{k(k+2)+j(n+2)} -\frac{ -(q;q)_{n+4} -}{ -(q^{n+3};q)_{n+2} (q;q)_{j} -} -\frac{ -(q^{n-2k+1};q)_{2k} -(q^{n-2k-j+1};q)_{j} -}{ -(q;q)_{k} -(q;q)_{k+2} -} -=1 -$$ -which I have also not managed to prove. -I believe that some WZ-method could solve this easily, -but a human-friendly proof would be preferrable. -Note that the identity above is very similar to a -theorem by Andrews (see reference below). It states that -\begin{equation} -\operatorname{Cat}_q(n+1) -= -\sum_{k \geq 0} -q^{k(k+2)} -\left[\matrix{n \\ 2k}\right]_q -\operatorname{Cat}_q(k) -\frac{(-q^{k+2};q)_{n-k}}{(-q;q)_k}. -\end{equation} -UPDATE: I have managed to find a more -general conjecture, which would imply the one above. -It states that for integer $n \geq 0$, and general $a,c$, -we have -$$ - \sum_{s} - \frac{ (-a q^n)^{s} -q^{-\binom{s}{2}} -(q^{-n};q)_{s} }{ (q;q)_{s} } -{}_{2}\phi_{1}(cq^{s-1}/a,q^{-s};c;q,q) -= -\frac{ (ac ;q)_{n} }{ (c;q)_{n} }. -$$ -I use This book as my main reference for notation and identities. -Andrews, George E., (q)-Catalan identities, Alladi, Krishnaswami (ed.) et al., The legacy of Alladi Ramakrishnan in the mathematical sciences. New York, NY: Springer (ISBN 978-1-4419-6262-1/hbk; 978-1-4419-6263-8/ebook). 183-190 (2010). ZBL1322.11018. - -REPLY [5 votes]: I managed to solve the problem, -in the last general conjecture, one can apply the $q$-Chu-Vandermonde theorem. After some simplification, -the resulting expression can be expressed as a ${}_2\phi_1$ q-hypergeometric series, where one again can apply the $q$-Chu-Vandermonde theorem. -Skipping lots of details the proof still requires a few pages - it can now be found in this paper (Thm. 57)<|endoftext|> -TITLE: Is there a physical reason that fields in QFT are globally defined? -QUESTION [12 upvotes]: I have been trying to read a physics textbook on Quantum Field theory. There seems to me to be a bit of a disconnect in most texts I have looked at between quantum mechanics and quantum field theory, in the passage from multiparticle wave functions to fields. I'm curious if there is a physical reason for this. I'm asking this here because the question has a math physics flavor. -First, let me sketch out my simplistic understanding of field theory from the Hamiltonian point of view. I will ignore relativistic invariance (though if I understand correctly, with a bit of work it can be recovered in this picture). Let $\alpha$ be a bosonic particle and let $V = V_\alpha$ be a space of wavefunctions of a single $\alpha$ particle. For concreteness, let's assume $V = L^2_\mathbb{C}(\mathbb{R}^3),$ corresponding to a scalar field. Note that I don't really care about the details of $V$: any space will do (including a finite-dimensional one for bound particles). The one assumption I will make on $V$ is that we have fixed a real subspace $V_\mathbb{R}\subset V$ compatible with the Hermitian structure. -Then single-particle quantum mechanics says that a wavefunction $\psi\in V$ evolves according to the Schroedinger equation, $\dot{\psi} = -i H_\alpha\psi,$ for $H_\alpha$ the single-particle Hamiltonian. Similarly, for any $n$, there is a non-interacting hamiltonian $H_{\alpha, n} : = \text{Symmetrize}(H\otimes 1\otimes \dots \otimes 1)$ on the bosonic $n$-particle space $S^n(V)$. -Now my understanding is that field theory arises as soon as we perturb the collection of $n$-particle Hamiltonians $$\oplus H_{\alpha, n}\in \prod_n\operatorname{End}(S^n(V))$$ by an interacting term $H_{mix}$ that mixes particle numbers. The new Hamiltonian will now "create" and "annihilate" particles, and its time evolution will now be a time-dependent unitary automorphism $U_t$ of the space of power series $$\mathfrak{F}_{formal} = \widehat{S}^*(V) : = \prod_n S^n(V).$$ (I don't want to be too particular about the analysis here: in particular, perhaps I need to assume that $U_t$ has some decent convergence properties.) -Now $\mathfrak{F}_{formal}$ is of course the space of power series in a neighborhood of $0$ on the affine space $V_\mathbb{R}$ of fields. (Well, technically on its dual, but it has a Hilbert metric.) So the passage from a single bosonic particle state to a superposition of all its multiparticle states moves the quantum state space from $V$ to power series on $V$. (BTW, I'm surprised to have never encountered this point of view written down in a textbook: instead I sort of pieced it together from how physicists talk. Is this a standard, or even a correct point of view?) -Now my question is why field theory doesn't stop at power series. When mathematicians or physicists talk about the Hamiltonian formulation of field theory, the manifold of fields includes all $C^\infty$ (or something) global functions on $V_\mathbb{R}.$ Is this distinction important, and does it come from some specific physical context where one can measure the difference, or is it just an artifact of playing fast and loose with the analysis as physicists are wont to do? - -REPLY [9 votes]: You are describing what is commonly known as second quantization (as you probably already realize). In a nutshell, the main mathematical statement behind second quantization is the following: the algebra $\mathcal{A}_{particles}$ generated by creation/annihilation operators acting on the Fock space $\mathfrak{F}_{particles}$ is isomorphic to the algebra $\mathcal{P}_{field}$ of quantized polynomial observables on the classical phase space of some free field (roughly, a collection of infinitely many harmonic oscillators), and more over the representation of $\mathcal{P}_{field}$ on the Hilbert space $\mathfrak{H}_{field}$ of oscillator states is isomorphic to the representation of $\mathcal{A}_{particles}$ on $\mathfrak{F}_{particles}$. Now, what this "free field" is is determined by the structure of the single particle state space $V$ and Hamiltonian $H$. This field may have a spacetime interpretation, depending on what that structure is, but it may also not. You did not get to the equivalence itself in your description, but it is implicit in the construction of the Fock space. -Historically, this is how the focus shifted from particles to fields. The dictionary could be expanded further and include interactions. When applied to the example of QED (quantum electrodynamics), where initially electrons were quantized as particles and photons as fields, second quantization shows that equally well both electrons and photons can be quantized on equal footing from the start (as fields). One can carry this dictionary quite far, without deciding that either particles or fields are a preferred description. But eventually, the balance shifted towards fields: only those particles and interactions that correspond to local relativistic fields seem to be observed in nature, non-unitarily equivalent representations of $\mathcal{P}_{field}$ require drastic changes to the particle picture, non-perturbative (in the fields) phenomena are not (at least not easily) covered by the dictionary and yet there are good reasons to consider them (phase transitions). -The point of the historical summary is that, given the shift in focus from particles to fields, the mathematical questions change. Namely, one is not concerned a priori with a particle Fock space, rather one is concerned with the quantization of a field theory as an infinite dimensional classical system (either via Hamiltonian or path integral methods). Now, one no longer has a reason to restrict the polynomial field observables $\mathcal{P}_{field}$, other than convenience or technical necessity. It becomes a mathematical challenge to describe as large an algebra of observables $\mathcal{A}_{field} \supset \mathcal{P}_{field}$ as is reasonable. One can think of $\mathcal{A}_{field}$ as a quantization of $C^\infty(V_\mathbb{R})$, which you were wondering about, for a corresponding reasonable interpretation of $C^\infty$ on an infinite dimensional space. If such a quantization of a field system succeeds, and a particle description is possible and desired, then on can simply restrict this quantization to the polynomial observables $\mathcal{P}_{field}$ and use the second quantization dictionary. -P.S.: If you insist on describing Fock space as $\mathfrak{F}_{formal}$ as formal power series on $V$, then it is not a Hilbert space. If you would like Fock space to be a Hilbert space, you should restrict to those series that have finite norm, thus defining $\mathfrak{F}_{particles} = \bigoplus_n S^n(V)$ by the usual direct sum of Hilbert spaces. The connection of this description of Fock space to power series (or at least to polynomials) on $V$ is well-known, though it need not be mentioned in textbooks unless it's needed for some specific remarks. But the connection is mentioned already on Wikipedia.<|endoftext|> -TITLE: Pairwise combinations of distinct elements -QUESTION [6 upvotes]: Consider a set of four elements -$$ -Y^0 = \{ y_1, y_2, y_3, y_4 \} -$$ -Let $Y^1$ be the set that includes all pairwise combinations of distinct elements of $Y^0$ -$$ -Y^1 = \{ y^1_1, \dots, y^1_6 \} := \{ \{y_1, y_2\}, \{y_1, y_3\}, \{y_1, y_4\}, \{y_2, y_3\}, \{y_2, y_4\}, \{y_3, y_4\} \} -$$ -By construction, each element of $Y^1$ contains each element of $Y^0$ at most once. -Now let's build $Y^2$ as the pairwise combinations of distinct elements of $Y^1$ -\begin{align} -Y^2 &= \{ y^2_1 \dots, y^2_{15} \} := \\ -& := \{ \ \{\{y_1, y_2\}, \{y_1, y_3\}\}, \ \{\{y_1, y_2\}, \{y_1, y_4\}\}, \ \{\{y_1, y_2\}, \{y_2, y_3\}\}, \dots, \ \{\{y_2, y_4\}, \{y_3, y_4\}\} \} -\end{align} -12 elements of $Y^2$ are such that one element of $Y_0$ appears twice. 3 elements of $Y_2$ contain only distinct elements of $Y^0$. -At each step $N$ we build $Y^N$ as the set of pairwise combinations of distinct elements of $Y^{N-1}$. -Question: is there a way to know, given $N$, how many elements of $X_N$ include how many repetitions of elements of $X_0$? The information I am looking for is something like "$\ell$ elements of $X_N$ are such that they include an element of $X_0$ three times, another distinct element of $X_0$ two times and a third distinct element of $X_0$ one time" and so on, I am not interested in which specific elements of $X_0$ is repeated. -Addendum: I am adding a picture that should describe the generation of the nested sets better. - -REPLY [2 votes]: Define the signature of an element $t\in Y^N$ as a monomial $s_t(z_1,z_2,z_3,z_4):=z_1^{k_1}z_2^{k_2}z_3^{k_3}z_4^{k_4}$ where $k_i$ is the number of occurrences of $y_i$ in $t$. It is clear that $k_1+k_2+k_3+k_4=2^N$. Let $$S_N(z_1,z_2,z_3,z_4) := \sum_{t\in Y^N} s_t(z_1,z_2,z_3,z_4).$$ -In particular, $S_N(1,1,1,1)=|Y^N|$ with numerical values listed in OEIS A086714. -From the definition of $Y^N$, it follows that -$$S_{N+1}(z_1,z_2,z_3,z_4) = \frac{S_N(z_1,z_2,z_3,z_4)^2-S_N(z_1^2,z_2^2,z_3^2,z_4^2)}2.$$ -In particular, we have -$$S_0(z_1,z_2,z_3,z_4) = z_1+z_2+z_3+z_4,$$ -$$S_1(z_1,z_2,z_3,z_4) = z_1z_2+z_1z_3+z_1z_4+z_2z_3+z_2z_4+z_3z_4,$$ -$$S_2(z_1,z_2,z_3,z_4) = z_1^2(z_2z_3+z_2z_4+z_3z_4)+z_2^2(z_1z_3+z_1z_4+z_3z_4) + z_3^2(z_1z_2+z_2z_4+z_1z_4)+z_4^2(z_2z_3+z_1z_2+z_1z_3)+3z_1z_2z_3z_4.$$ -There may exist a nice representation in terms of symmetric polynomials. - -For example, in terms of monomial symmetric polynomials, we have: -$S_0 = m_{(1,0,0,0)}$, $S_1 = m_{(1,1,0,0)}$, $S_2=m_{(2,1,1,0)}+3m_{(1,1,1,1)}$, $S_3=m_{(4,2,1,1)}+2m_{(3,3,1,1)}+m_{(3,3,2,0)}+5m_{(3,2,2,1)}+9m_{(2,2,2,2)}$, etc. -Here is a sample SageMath code: -m = SymmetricFunctions(QQ).monomial() -S = m[1] -for i in range(5): - print i,":",S - S = (S^2 - sum( t[1]*m[vector(t[0])*2] for t in S ))/2 - S = sum( t[1]*m[t[0]] for t in S if len(t[0])<=4 ) - -producing such representation for first few $N$: -0 : m[1] -1 : m[1, 1] -2 : 3*m[1, 1, 1, 1] + m[2, 1, 1] -3 : 9*m[2, 2, 2, 2] + 5*m[3, 2, 2, 1] + 2*m[3, 3, 1, 1] + m[3, 3, 2] + m[4, 2, 1, 1] -4 : 210*m[4, 4, 4, 4] + 141*m[5, 4, 4, 3] + 92*m[5, 5, 3, 3] + 59*m[5, 5, 4, 2] + 15*m[5, 5, 5, 1] + 59*m[6, 4, 3, 3] + 35*m[6, 4, 4, 2] + 22*m[6, 5, 3, 2] + 8*m[6, 5, 4, 1] + m[6, 5, 5] + 3*m[6, 6, 2, 2] + 2*m[6, 6, 3, 1] + 15*m[7, 3, 3, 3] + 8*m[7, 4, 3, 2] + 2*m[7, 4, 4, 1] + 2*m[7, 5, 2, 2] + m[7, 5, 3, 1] + m[8, 3, 3, 2]<|endoftext|> -TITLE: Orthogonal Cauchy-like matrix -QUESTION [5 upvotes]: Given a $n \times n$ real Cauchy like matrix $C$, i.e., for real vectors $r$, $s$, $x$, $y$ -$$ -C_{ij} = \frac{r_i s_j}{ x_i - y_j} -$$ - -Can a Cauchy-like $C$ be orthogonal, i.e., $C C^T = I$ for $n > 2$? - -There exists such an orthogonal $C$ for $n = 2$ , $x = [1,0.4]$, $y = [6.25,0.625]$, $r = [-1.8114, 1.4811]$, and $s = [2.3367, -0.1225]$ with -$$ -C = \begin{bmatrix} -0.8062 & 0.5916 \\ --0.5916 & 0.8062 -\end{bmatrix} -$$ - -REPLY [4 votes]: Please give a look at this short note.<|endoftext|> -TITLE: Sheaves in combinatorics and discrete geometry -QUESTION [14 upvotes]: I am looking for examples for the application of sheaves, sheaf-like constructions or the (co)homology of sheaves to problems in combinatorics and discrete geometry. -For example given a poset $(P,\leq)$ one can look at the topology given by declaring that -the open sets are order filters $U \subseteq P$, i.e. if $x \in U$ and $x \leq y$ then $y \in U$. -Now, any functor $\mathcal{F}$ from $P$ to some category, e.g. $\mathcal{F}:P \to \mathbf{Ab}$ to the category of abelian groups, -gives a sheaf (e.g. of abelian groups) on the topological space described before. This is called a sheaf on $P$. -I am aware of the following applications of sheaves on posets: - -K. Baclawski used sheaves on posets in Whitney numbers of geometric lattices, in particular sheaf cohomology on posets to answer a question of G.-C. Rota: the Whitney numbers of the first kind of a geometric lattice are the Betti numbers of some homology theory on posets, namely the cohomology of suited sheaves on posets. -S. Yuzvinsky used sheaf cohomology in Cohomology of local sheaves on arrangement lattices to give an interesting characterization of freeness of hyperplane arrangements. -In the subsequent paper The first two obstructions to the freeness of arrangements he used this characterization to prove a conjecture of Falk and Randell that free arrangements are formal. - -I suppose there should be plenty more examples and I am looking forward to your answers. Thanks! - -REPLY [2 votes]: One from mathematical physics. -The enummeration of plane partitions in the image of the moment map of a toric variety compute the Donaldson-Thomas theory of the toric variety $\mathcal{X}$ by identifying plane partitions with monomial ideals of the structural sheaf of $\mathcal{X}$, namely ideal sheaves of $\mathcal{X}$. -The procedure can be generalized by giving a planar bipartite graph and constructing with it the image of the moment map of a toric variety (see Quantum Calabi-Yau and Classical Crystals). In that context the enummeration of perfect matchings on the graph is the same problem as the ennumeration of plane partitions in the image of the moment map of the aforementioned toric variety.<|endoftext|> -TITLE: Why do bees create hexagonal cells ? (Mathematical reasons) -QUESTION [52 upvotes]: Question 0 Are there any mathematical phenomena which are related to the form of honeycomb cells? -Question 1 Maybe hexagonal lattices satisfy certain optimality condition(s) which are related to it? -The reason to ask - some considerations with the famous "K-means" clustering algorithm on the plane. It also tends to produce something similar to hexagons, moreover, maybe, ruling out technicalities, hexagonal lattice is optimal for K-means functional, that is MO362135 question. -Question 2 Can it also be related to bee's construction? - -Googling gives lots of sources on the question. But many of them are focused on non-mathematical sides of the question - how are bees being able to produce such quite precise forms of hexagons? Why is it useful for them? Etc. -Let me quote the relatively recent paper from Nature 2016, -"The hexagonal shape of the honeycomb cells depends on the construction behavior of bees", -Francesco Nazzi: - -Abstract. The hexagonal shape of the honey bee cells has attracted the - attention of humans for centuries. It is now accepted that bees build - cylindrical cells that later transform into hexagonal prisms through a - process that it is still debated. The early explanations involving the - geometers’ skills of bees have been abandoned in favor of new - hypotheses involving the action of physical forces, but recent data - suggest that mechanical shaping by bees plays a role. However, the - observed geometry can arise only if isodiametric cells are previously - arranged in a way that each one is surrounded by six other similar - cells; here I suggest that this is a consequence of the building - program adopted by bees and propose a possible behavioral rule - ultimately accounting for the hexagonal shape of bee cells. - -REPLY [11 votes]: Isn't it just the 2d sphere packing? If one assumes that the larvae needs a disc of fixed radius to grow up to an adult form and that the bees want to have as many cells as possible then the hexagonal lattice is the optimal one.<|endoftext|> -TITLE: Is the class of power-associative binars finitely axiomatizable? -QUESTION [5 upvotes]: A binar is simply a set $S$ equipped with a single binary operation $*$. A power-associative binar is a binar where the subalgebra generated by a single element is associative. Equivalently, they can be axiomatized with the infinite set of equations, $\{(x*x)*x=x*(x*x), ... \}$. Is there some finite set of axioms in the signature $*$ that can axiomatize power-associativity? - -REPLY [6 votes]: The question has been answered, but I will add some remarks -about magma/groupoid/binar. This is in response to some -of the comments on this page: - -What you can currently read on the English Wikipedia: -The term groupoid was introduced in 1927 by Heinrich Brandt -describing his Brandt groupoid (translated from the German Gruppoid). -The term was then appropriated by B. A. Hausmann and Øystein Ore (1937) -in the sense (of a set with a binary operation) used in this article. -… -The term magma was used by Serre [Lie Algebras and Lie Groups, 1965]. -It also appears in Bourbaki's Éléments de mathématique, Algèbre, -chapitres 1 à 3, 1970. -What you can currently read on the French Wikipedia: -L'ancienne appellation « groupoïde de Ore », introduite par Bernard Hausmann et Øystein Ore en 1937 et parfois utilisée jusque dans les années 1960, est aujourd'hui à éviter, l'usage du terme groupoïde étant aujourd'hui réservé à la théorie des catégories, où il signifie autre chose. -[The old name groupoïde de Ore, introduced by Bernard Hausmann and Øystein Ore in 1937 and sometimes used until the 1960s, is to be avoided today, the use of the term groupoid being today reserved for category theory, where it means something else.] -What is currently not on either Wikipedia: -A. (magma) -Peter Shor of MIT has speculated that magma -might have been introduced -as a pun on Ore's name. -To distinguish a groupoid in the sense of Ore -from a groupoid in the sense of -Brandt, the phrase groupoid of Ore was used first, -then shortened to -magma, which in geology means a pile of molten ore. -B. (binar) -In June 1993, there was a conference at MSRI on -Universal Algebra and Category Theory -organized by Ralph McKenzie and Saunders Mac Lane. -At this conference, there was a discussion session where -several topics were discussed, including whether there -could be general agreement on the future -use of the word groupoid. -At this discussion session, George Bergman of -Berkeley proposed using binar to mean an algebraic structure -with a single binary operation. (He also proposed -unar for an algebraic structure with a -single unary operation.) It seemed to me that -Bergman thought this up on the spot, so it is plausible -that he is the source of binar.<|endoftext|> -TITLE: Examples of non-zero negative Steenrod operations -QUESTION [5 upvotes]: In JP May's paper A general algebraic approach to Steenrod operations, Steenrod operations are constructed in wide generality. In this context, it is not necessarily true that negative Steenrod operations are zero. However, I could not find any examples (not in May's paper, not even elsewhere), where one actually proves in some specific situation that some negative Steenrod operation is non-zero. It appears that sheaves of differential graded algebras $A$ (not concentrated in degree $0$) on a topological space $X$ should provide a counterexample, but I would like to see a specific example of $X$ and $A$, possibly with a proof. - -REPLY [8 votes]: Notice the switch in grading from homology to cohomology on May's page 182: $P^s(x) = P_{-s}(x)$. Operations that raise degree when graded homologically lower degree when graded cohomologically. A large part of the motivation for the paper was to give a common framework for the Dyer-Lashof operations in the homology of iterated loop spaces and the Steenrod operations in the cohomology of spaces. But if one grades these the same way, either homologically or cohomologically, one has positive operations and the other has negative operations. The grading May found most sensible for the Steenrod operations on the cohomology of cocommutative Hopf algebras (p. 226) also gives examples. There are a number of more recent papers that make serious use of the large Steenrod algebra (sometimes called the Kudo-Araki-May algebra) constructed using all operations.<|endoftext|> -TITLE: Is every integer $\ge 312$ the sum of two integers with triangular divisors? -QUESTION [5 upvotes]: We say that a natural number $n$ has triangular divisors if it has at least one triplet of divisors $n = d_1d_2d_3$, $1 \le d_1 \le d_2 \le d_3$, such that $d_1,d_2$ and $d_3$ form the sides of a triangle (non degenerate) -E.g.: $60$ has triangular divisors because $60 = 3.4.5$ and $3,4,5$ form a triangle. Note that another triplet of divisors of $60 = 1.4.15$ does not form a triangle but because of the triplet $3,4,5$ the number $60$ qualifies a number with triangular divisors. On the other the number $10$ does not have any triplet of triangular divisors. -I found the following conjectures experimentally. Can they be proved or disproved? - -Weak conjecture: Every integer $\ge 8$ which has triangular divisors can be written as the sum of two integers both of which has triangular divisors. -Strong conjecture: Every integer $\ge 8$ except $11, 14, 15,23, 38, 47, 55, 71, 103, 113$ and $311$ can be written as the sum of two integers both of which has triangular divisors. - -Note: This question was posted in MSE 3 month ago year. It got some upvotes but to answers. Hence posting in MO. -Related question: How many numbers $\le x$ can be factorized into three numbers which form the sides of a triangle? - -REPLY [4 votes]: Not a complete answer, but it made sense I think to summarize the comment thread above: -Theorem: Every sufficiently large positive integer is expressible as a sum of three numbers which have triangular divisors. -To see this, note that every perfect square, $n^2$, has triangular divisors $(1,n,n)$. -Also, if $m$ has triangular divisors, so does $mk^3$ for any $k$. This is because we can just take our three divisors $d_1$, $d_2$, and $d_3$ and scale them up to be $kd_1$, $kd_2$, $kd_3$. -So any number of the form $n^2k^3$ has triangular divisors. But these are precisely the powerful numbers, numbers in which all prime factors are raised to at least the second power. (Note these are sometimes called squarefull numbers.) -To prove the theorem we then use the theorem of Heath-Brown that every sufficiently large positive integer is expressible as the sum of three powerful numbers.<|endoftext|> -TITLE: Further developments of Cartier–Gabriel–Kostant–Milnor–Moore Structure Theorem for cocommutative Hopf algebras -QUESTION [5 upvotes]: A very well-known theorem in Hopf algebra theory (see, for example, Lorenz - A tour of representation theory or the EGNO book (Etingof, Gelaki, Nikshych, and Ostrik - Tensor categories)) states that if $H$ is a cocommutative Hopf algebra over an algebraically closed field $\Bbbk$ of characteristic zero then $$H \cong U(P(H)) \mathbin\# \Bbbk G(H)$$ as Hopf algebras, where $P(H)$ is the space of primitive elements of $H$ and $G(H)$ the group of group-like elements. As far as I know, this theorem is attributed differently to many people, but I believe the list in the title should be comprehensive of all of them. For example, I know that the Milnor–Moore part of the theorem is the one stating that an irreducible cocommutative Hopf algebra $H'$ over a field of characteristic zero satisfies $H' \cong U(P(H'))$. -I know that Nichols proved, in The Kostant structure theorem for $K/k$ Hopf algebras, an analogue of the full CGKMM theorem for Hopf algebroids of the form $(K,H)$, where $K$ is a field extension of $\Bbbk$, that Moerdijk and Mrčun proved, in On the universal enveloping algebra of a Lie algebroid, an analogue of Milnor–Moore for bialgebroids and that, later, Kališnik and Mrčun proved, in A Cartier–Gabriel–Kostant structure theorem for Hopf algebroids, an analogue of the full Cartier–Gabriel–Kostant–Milnor–Moore for Hopf algebroids over the algebra $\mathcal{C}^{\infty}_c(\mathcal{M})$ of smooth functions with compact support over a smooth real manifold $\mathcal{M}$. -Is anybody aware of some further developments/extensions of this result to other classes of Hopf algebras (for example, weak Hopf algebras, general Hopf algebroids, (co)quasi-Hopf algebras, etc.)? -Otherwise, is there anybody aware of counter-examples that suggest this could not be generalized further in some direction? -Edit: Just to add some further motivation to this quest. Let $G$ be a linear algebraic group over an algebraically closed field $\Bbbk$ of characteristic zero. The coordinate algebra $\mathscr{O}(G)$ is a commutative Hopf algebra. If we consider the subspace $\mathscr{O}(G)^\circ$ of $\mathsf{Hom}_\Bbbk\left(\mathscr{O}(G),\Bbbk\right)$ formed by all those linear functionals which vanish on a finite-codimensional ideal, this is a cocommutative Hopf algebra. By the CGKMM Theorem we know that -$$\mathscr{O}(G)^\circ \cong U\left(P\left(\mathscr{O}(G)^\circ\right)\right) ~\#~ \Bbbk G\left(\mathscr{O}(G)^\circ\right).$$ -It happens that $P\left(\mathscr{O}(G)^\circ\right) \cong \mathfrak{g}$, the tangent Lie algebra to $G$ at the neutral element, and $G\left(\mathscr{O}(G)^\circ\right) \cong \mathsf{Alg}_\Bbbk\left(\mathscr{O}(G),\Bbbk\right) \cong G$, so that -$$\mathscr{O}(G)^\circ \cong U\left(\mathfrak{g}\right) ~\#~ \Bbbk G,$$ -where on the right-hand side $G$ is considered as a discrete group and $U(\mathfrak{g}) \cong \mathsf{Dist}_G$, the hyperalgebra of distributions on $G$. -Is there anybody aware of some further results like this one for groupoids? - -REPLY [2 votes]: The super version of the theorem, refers to "hopf superalgebras", or "$\mathbb{Z}_2$-graded hopf algebras" or "hopf algebras in the braided monoidal category of $\mathbb{CZ}_2$-modules": -Let $\mathcal{H}$ be a super-cocommutative hopf superalgebra over an algebraically closed field $k$ of char zero. Then we have the hopf superalgebra isomorphism: -$$ -\mathcal{H}\cong k[G(\mathcal{H})]\ltimes_{\pi} U\big(P(\mathcal{H})\big) -$$ -where, $k[G(\mathcal{H})]$ is the group algebra of the the group $G(\mathcal{H})$ of the grouplikes of $\mathcal{H}$, $U\big(P(\mathcal{H})\big)$ is the universal enveloping algebra of the lie superalgebra $P(\mathcal{H})$ of the primitive elements of $\mathcal{H}$ and the smash product $\ltimes_{\pi}$ is with respect to the representation of $G(\mathcal{H})$ on $P(\mathcal{H})$ determined by: $\pi:G\to Aut(P)$, $\pi(g)x=gxg^{-1}$, for all $g\in G$, $x\in P$. -This has been an old result, first shown by Kostant, at: -B.Kostant, "Graded manifolds, graded Lie theory and prequantization", Differential Geometrical Methods in Mathematical Physics. Lecture Notes in Mathematics, vol 570, p. 177-306, (1977)) -P.S.: What i do not know, is whether there is a corresponding version of the theorem for Hopf algebras in braided monoidal categories. I am not aware of some reference in that direction (although, i think the generalization should not be that difficult to prove). The only related reference i know of is Braided bialgebras of Hecke-type, A. Ardizzoni, C.Menini, D. Stefan, Journal of Algebra 321 (2009) 847–865, where a result -analogous to the Milnor-Moore part of the theorem- is proved for connected, braided bialgebras which are infinitesimally cocommutative (see theorem 5.5 and the resulting discussion). -Edit: another possible line of generalization of the theorem , has to do with the class of quasitriangular hopf algebras. These, generalise the cocommutative hopf algebras, so it is natural to consider the possibility of a "quasitriangular version" of the theorem. See for example: Classification of quasitriangular Hopf algebras<|endoftext|> -TITLE: Distributivity of ! over? -QUESTION [8 upvotes]: Has anyone studied a variant of linear logic, or of its semantic counterpart (exponential modalities on linearly distributive categories / $\ast$-autonomous categories / polycategories) for which there is a distributive law -$$!? \to ?!$$ -Presumably it would need to interact sensibly with the (co)monoidal structure of ! and ?. I haven't thought about whether there is a nice sequent calculus or resource interpretation. I ask for two reasons: - -Such a distributive law should imply that there is a "mixed Kleisli category" whose morphisms from $A$ to $B$ are the morphisms $!A \to ?B$. This would be more symmetrical and pleasing than the usual translation of classical logic involving morphisms $!?A \to ?B$. -In $\rm Chu(Cat,Set)$ I believe there is such a distributive law, or at least a pseudo-distributive law, and a morphism $!A\to ?B$ should be precisely a profunctor from $A^+$ to $B^-$. So this would give a nice way of recovering $\rm Prof$ from $\rm Chu(Cat,Set)$. - -REPLY [2 votes]: I would tend to say "no". -However, besides my comment above, which is not very pertinent, let me mention the paper Combining effects and coeffects via grading, by Marco Gaboardi, Shin-ya Katsumata, Dominic Orchard, Flavien Breuvart and Tarmo Uustalu. They consider "graded" monads and comonads, which include usual monads and comonads as the special case in which the grading is trivial (I don't know who introduced these first, I learned of graded comonads from this note by Paul-André Melliès). The programming language underlying their work is linear and their comonads are graded in a semiring (rather than just a monoid) so that they are a generalization of the exponential modality $!(-)$ of linear logic (the additive structure of the semiring grades weakening/contraction, i.e. the monoidal structure of the comonad, while the multiplicative structure of the semiring grades the actual comonad structure, i.e., the counit and the comultiplication). They then study graded versions of the usual distributive law between monads and comonads in order to account for the simultaneous presence of "quantitative" effects and coeffects in programming languages (e.g., not just tell whether a program may raise an exception but tell, if possible, how many exceptions it will raise, or whatever. This is what the increased expressiveness given by the grading is meant to be used for). -So, forgetting the grading, i.e., if we grade everything with the trivial semiring and the trivial monoid, we are close to what you describe, but not quite: while the trivially graded version of their comonad is, indeed, the $!(-)$ modality of linear logic, the $?(-)$ modality does not fit into their axiomatization, because the trivially graded version of their monad is necessarily strong, and $?(-)$ is not a strong monad. So, strictly speaking, this work says nothing about the distributive law you are looking at. -I skimmed through the references given by Gaboardi et al. in relation with distributive laws and none of them seems to mention linear logic. This supports my belief that no one has ever introduced/studied the variant of linear logic you mention... but of course I can't be sure!<|endoftext|> -TITLE: $x^3+x^2y^2+y^3=7$, and solvable families of Diophantine equations -QUESTION [5 upvotes]: (a) Do there exist integers $x$ and $y$ such that $x^3+x^2y^2+y^3=7$ ? -(b) Is this equation belongs to some family $F$ of equations for which there is a known algorithms for testing if they have an integer solution? For example, such algorithms are known for quadratic equations (in $n$ variables) and for cubic equations in 2 variables, but this equation has degree 4. -(c) Reference request: can you recommend a good book/survey/website which would help to determine if a given (reasonably simple) Diophantine equation belong to some solved case, or is new? - -REPLY [15 votes]: (a) No. There are no integer solutions. The curve $C$ you give has genus $3$ and it has an obvious automorphism $\phi(x,y) = (y,x)$. The quotient curve is an elliptic curve. In particular, if you let $X = -(x+y)$ and $Y = xy$, then the equation becomes $E : Y^{2} + 3XY = X^{3} + 7$. So any integer point on your curve gives rise to an integer point on $E$. As you note, cubic equations in two variables have algorithms for finding their integer points. This elliptic curve has rank $1$ and its integral points are $(-3,4)$, $(-3,5)$, $(186,-2831)$ and $(186,2273)$. None of these lead to integer points on $C$. -(b) There are not many general families of such curves. For curves of genus $0$, there are algorithms for finding their integer points. For curves of genus $1$, the existing procedures rely on being able to compute the Mordell-Weil group and it is currently open whether there is an algorithm to do so. One other family that might be worth mentioned is the family of Thue equations, those of the form $F(x,y) = k$ where $k$ is a constant and $F$ is a homogeneous polynomial in two variables. -(c) For a survey of what number theorists can and cannot handle algorithmically, you might consult Henri Cohen's book "Number Theory, Volume I: Tools and Diophantine Equations." Chapter 6 in that book gives a nice survey of simple Diophantine equations together with some techniques that might suffice to handle those. The class of equations that can be definitely be handled in a systematic or algorithmic way is quite small. (If I had been unable to compute the Mordell-Weil group of $E$ above, the software would not have been able to provably find all the integral points.)<|endoftext|> -TITLE: Categorification of probability theory: what does a "probability sheaf" tell us (if anything) about probability theory? -QUESTION [14 upvotes]: Disclaimer: I only have a superficial knowledge of what category theory and related subjects are concerned with. -So, my understanding is that category theory and related fields of higher mathematics are meant to (A) better organise knowledge within a fields of mathematics (B) build powerful bridges between otherwise dislocated branches of mathematics (C) elevate concepts to a higher level of generality / abstraction so as to expose them to a more unified treatment (which can give one a better understanding of problems / questions in these fields, etc.). -I found Simpson - Probability sheaves and the Giry monad, in which a sheaf for probability theory was constructed. In my layman's quick reading, the manuscript doesn't seem to (1) tell anything new / non-trivial about probability theory (2) help in anyway organize concepts already present within probability theory. So my question is -Question - -How does the construction of a sheaf for probability theory (see paper cited above) help probability theory? -Has any such attempts (to "categorify" probability theory) been made before? - -Thanks in advance for any enlightenment! - -REPLY [4 votes]: Coincidentally, a day after you asked this question, Alex Simpson gave a nice talk (video, slides) where he gave a synthetic formulation of probability theory. In this formulation, random variables are a primitive notion, not maps from a sample space to a measurable space. Hence there's no need to keep track (or even mention) sample spaces at all. That's basically how probability theory was done informally, long before it was encoded in set theory. Several prominent mathematicians (Rota, Tao, Mumford) had suggested that such a reformulation of probability theory would be desirable. -I'd consider this an application of "categorical" probability theory, since I suspect that Simpson arrived at this axiomatisation via the categorical sheaf model he had constructed earlier.<|endoftext|> -TITLE: Jordan algebra identities -QUESTION [7 upvotes]: A Jordan algebra is a vector space with a commutative bilinear operation $\circ$ obeying an identity that's often written as -$$ (x \circ y) \circ (x \circ x) = x \circ (y \circ (x \circ x)) . $$ -I find this identity rather obscure. If we write $x^2 = x \circ x$ and use $L_a$ to stand for left multiplication by $a$, we can rewrite it in a more appealing form: -$$ L_{x^2} L_x = L_x L_{x^2} .$$ -However, I'd be even happier if this were a special case of a more general identity -$$ L_{x^m} L_{x^n} = L_{x^n} L_{x^m} \qquad (\ast) $$ -holding for all $n, m \in \mathbb{N}$. -This more general identity parses in any Jordan algebra, because any Jordan algebra is power-associative: expressions like $x \circ \cdots \circ x$ are independent of how you parenthesize them, so $x^n$ is well-defined. But is this more general identity $(\ast)$ true in every Jordan algebra? - -REPLY [9 votes]: This identity (*) is, indeed, true, and is, in fact, a step in one of the standard ways to prove that Jordan algebras are power-associative: see McCrimmon's 2004 book A Taste of Jordan Algebras, exercise 5.2.2A (question (2)) on page 201. -Edit: another reference, which has the better taste of not being an exercise and of being earlier in a book: Jacobson, Structure and Representations of Jordan Algebras (1968), page 35, just above formula (56). (I also just realized that the fact is mentioned in the Wikipedia article on Jordan algebras, with that reference.) -For completeness of MathOverflow, I might as well copy the essence of the argument: first prove the identity -$$ -L_{(b\circ d)\circ c} = L_{b\circ d}L_c + L_{c\circ d}L_b + L_{b\circ c}L_d - L_b L_c L_d - L_d L_c L_b -$$ -by linearizing the Jordan identity (here we use the fact that the characteristic is not two). Apply this to $b=a^k$ and $c=d=a$, giving -$$ -L_{a^{k+2}} = 2L_{a^{k+1}}L_a + L_{a^2}L_{a^k} - L_{a^k} L_a^2 - L_a^2 L_{a^k} -$$ -From this it follows by induction that all $L_{a^k}$ belong to the algebra generated by $L_a$ and $L_{a^2}$ and, since these commute, they all commute.<|endoftext|> -TITLE: On a property possibly separating countable and not countable cardinals -QUESTION [7 upvotes]: The following question is inspired by Function with vector space , which has been closed for unknown reason and which may have a wellknown answer. Is the following true? -Let $X$ be an uncountable set. Then there is a function $f \colon X \times X \to \mathbb{N}$ such that for any function $g \colon X \to \mathbb{N}$ there is $(x,y) \in X^2$ with $f(x,y) > g(x) + g(y)$. -It is easy to show that this is false if $X$ is countable. - -REPLY [8 votes]: Yes, such a function $f:X\times X\to\mathbb N$ exists if $X$ is uncountable. It will suffice to prove it for $X=\omega_1$. The following proof is based on the same idea as a comment by Ashutosh but uses elementary set theory instead of the result of Todorcevic. -For each ordinal $\alpha\in\omega_1$ choose an injective map $\psi_\alpha:\alpha\to\mathbb N$. Define a function $f:\omega_1\times\omega_1\to\mathbb N$ so that $f(\alpha,\beta)=\psi_\alpha(\beta)$ when $\beta\lt\alpha$. Now consider any function $g:\omega_1\to\mathbb N$. Then $g$ is bounded on some uncountable set $Y\subseteq\omega_1$; say $g(\xi)\le n\in\mathbb N$ for all $\xi\in Y$. Let $Z\subset Y$ be a set of order type $\omega+1$ and let $\alpha=\max Z$. Since $\{f(\alpha,\beta):\beta\in Z\cap\alpha\}=\{\psi_\alpha(\beta):\beta\in Z\cap\alpha\}$ is an infinite subset of $\mathbb N$, we can choose $\beta\in Z\cap\alpha$ with $f(\alpha,\beta)\gt2n\ge g(\alpha)+g(\beta)$. -More generally, if $|X|\gt\aleph_\alpha$, then there is a function $f:X\times X\to\omega_\alpha$ such that, for any function $g:X\to\omega_\alpha$, there are $x,y\in X$ with $f(x,y)\gt g(x)+g(y)$.<|endoftext|> -TITLE: Can triangulations (or some related combinatorial structure) distinguish smooth structures on $RP^4$? -QUESTION [11 upvotes]: There are exotic versions of $RP^4$, constructed by Cappell-Shaneson, which are homeomorphic but not diffeomorphic to the standard $RP^4$. One way to distinguish them is via the $\eta$ invariant of $Pin^+$ Dirac operators on them, c.f. the article "Exotic structures on 4-manifolds detected by spectral invariants" by Stolz, Invent. math. 94, 147-162 (1988) (pdf here). -I was wondering if there was a known combinatorial way to distinguish the smooth structures, e.g. in the following senses: - -Can one construct triangulations of $RP^4$ (e.g. via Morse theory) that must 'correspond' to one of the smooth structures? -If a triangulation by itself can't distinguish smooth structures, is there some additional combinatorial data that one can put on top of the triangulation to distinguish them, like the branching structure on the triangulation? - -The motivation for this question is based on some papers (https://arxiv.org/abs/1610.07628, https://arxiv.org/abs/1810.05833) that construct topological invariants via state-sums on triangulations (generalizing the Crane-Yetter sum) that speculate whether exotic structures can be detected via the state-sum. So it's natural to ask whether such manifolds can even be distinguished combinatorially. And something like this could seem plausible because in 4 dimensions, every manifold is smooth iff it is triangulable. -(If low-brow answers exist, that would be nice since I don't know much about exotic manifolds.) - -REPLY [3 votes]: It is very hard to construct state-sum invariants that distinguish smooth structures in dimension 4, for this simple but crucial fact that is worth mentioning: if $M$ and $N$ are homeomorphic smooth 4-manifolds, it is often the case (I don't remember what condition is needed here) that $M \#_h( S^2 \times S^2)$ and $N\#_h (S^2 \times S^2)$ are diffeomorphic for some $h$. Therefore any combinatorial invariant where the value on $M$ may be deduced from that on $M \# (S^2 \times S^2)$ will not work. So for instance if your invariant is multiplicative on connected sums, it should vanish on $S^2 \times S^2$. -The most famous state-sum invariant in dimension 3 is the Turaev-Viro one, and it is multiplicative on connected sums and is almost never zero.<|endoftext|> -TITLE: A class of unipotent group actions -QUESTION [6 upvotes]: Consider algebraic actions of unipotent groups $G$ on affine spaces $X=\mathbb{C}^n$. I am looking for a condition that would guarantee that the quotient $X/G$ exists and is also an affine space. For instance: -Suppose $X$ is itself isomorphic to a unipotent group and $G$ acts via compositions of left translations with group automorphisms. Suppose the action is such that every point of $X$ has trivial stabilizer. Is it true that $X/G$ is isomorphic to affine space? -UPDATE: It turns out, by replacing $X/G$ with $G\backslash X\rtimes A/A$ where $A$ is the image of $G$ in the group of automorphisms of $X$, the question is reduced to the corresponding question for double cosets, which is again a special case of the original question. So here is an equivalent question: -Suppose a unipotent group $G$ contains unipotent subgroups $G_1, G_2$ such that $G_1\cap x G_2 x^{-1} = \{e\}$ for all $x\in G$. Is the double coset space $G_1 \backslash G / G_2$ isomorphic to affine space? - -REPLY [2 votes]: The answer is "no". In this paper: Winkelmann, J. On free holomorphic ℂ-actions on ℂn and homogeneous stein manifolds. Math. Ann. 286, 593–612 (1990) a free affine linear action of $G_a\times G_a$ on $\mathbb{C}^6$ is given in such a way that the quotient is not an affine variety. So for $X=G_a^6$, $G=G_a^2$, we obtain a counter-example. -By the way, in op. cit. this action is reduced to a triangular algebraic action of $G_a$ on $\mathbb{C}^5$. There is also an example there of a free triangular algebraic action of $G_a$ on $\mathbb{C}^4$ with non-Hausdorff quotient. It is not hard to check that the class of triangular algebraic actions coincides with the class of actions of $G_1$ on $G/G_2$ where $G_1, G_2$ range over unipotent subgroups of unipotent groups $G$.<|endoftext|> -TITLE: Morphism with connected fibers induce surjection on fundamental groups? -QUESTION [8 upvotes]: Let $X,Y$ be path-connected finite CW complexes with base points $x_0,y_0$, let $f\colon X\to Y$ be a surjective continuous map, such that for every $y\in Y$, the fiber $f^{-1}(y)$ is path connected. In this case, is the induced map $$f_*\colon\pi_1(X,x_0)\to\pi_1(Y,y_0)$$ on topological fundamental groups necessarily surjective? -(If this is not true in general, will this be true in the case when $Y$ is a complex algebraic manifold, $X\subset Y\times\mathbb{P}^n$ a quasi-projective variety, and $f=\mathrm{pr}_1$?) -[I think one sufficient condition is that $f$ satisfies the "arc-lifting property": Sufficiently short arc $(-\epsilon,\epsilon)$ centered at any $y\in Y$ can be lifted to an arc in $X$. For then we can cover a path in $X$ by finitely many arcs in $X$, and join the arcs by path in the fibers. But I am not sure if this is always doable?] - -REPLY [6 votes]: Assuming that your map $f\colon X \rightarrow Y$ is a map of CW complexes, the answer is yes. -In fact, you can get away with quite a bit less. Assume that $X$ and $Y$ are arbitrary CW complexes equipped with basepoints $x_0 \in X^{(0)}$ and $y_0 \in Y^{(0)}$ and that $f\colon (X,x_0) \rightarrow (Y,y_0)$ is a map of CW complexes. Furthermore, assume that for all vertices $v \in Y^{(0)}$, the preimage $f^{-1}(v)$ is connected and that for all $1$-simplices $e$ of $Y$, there is some $1$-simplex $E$ of $X$ that is taken to $e$ by $f$. Then I claim that $f_{\ast}\colon \pi_1(X,x_0) \rightarrow \pi_1(Y,y_0)$ is surjective. -There is probably some fancy way of seeing this, but here's a down-to-earth argument. Every element of $\pi_1(Y,y_0)$ can be represented by an edge path in the $1$-skeleton. Let $e_1,e_2,\ldots,e_k$ be the edges traversed by that edge path. The lift to $X$ is then as follows: - -Start at $x_0$. -There is some edge $E_1$ of $X$ projecting to $e_1$; move in the fiber $f^{-1}(y_0)$ to the starting point of $E_1$ and then go across $E_1$. -Letting $y_1$ be the ending point of $e_1$, there again exists some edge $E_2$ of $X$ projecting to $e_2$. Move in the fiber $f^{-1}(y_1)$ to the starting point of $E_2$ and then go across $E_2$. -etc. -At the end of this process, you'll end up at a point of $f^{-1}(y_0)$. Move in the fiber $f^{-1}(x_0)$ back to $x_0$, closing up the loop. - -REPLY [4 votes]: This answer is a complement to Andy's one. If $X$ and $Y$ are complex algebraic varieties then you have the following fact (see more generally Kollár "Shafarevich maps and automorphic forms" Proposition 2.10.2): -If $X$, $Y$ are irreducible algebraic varieties with $Y$ normal and $f:X\to Y$ is a dominant morphism such that the geometric generic fiber is connected then $f_*:\pi_1(X)\to \pi_1(Y)$ is surjective. -In case $Y$ is not normal then the above fact is not true. Take $Y=$ nodal cubic, $X=$ normalization of $Y$ minus one of the two preimages of the node. This situation is realized topologically as follows: it's the map from the sphere minus North Pole to the sphere with North Pole and South Pole identified. (Deleting one of the two points is not necessary to provide a counterexample to the above fact but it provides a counterexample with connected fibers, which is the case you are interested in).<|endoftext|> -TITLE: Structure constants for the double coset algebra of a Young subgroup -QUESTION [7 upvotes]: Fix a Young subgroup $H_\lambda \subseteq \mathcal S_n$, where $\lambda \vdash n$ is a partition of $n$ with $k$ blocks. Inside the group algebra $\mathbb C[\mathcal S_n]$, consider the idempotent -$$\varepsilon = \frac{1}{|H_\lambda|}\sum_{h \in H_\lambda} h.$$ -The double cosets $H_\lambda \backslash \mathcal S_n / H_\lambda$ are indexed by matrices $k \times k$ with non-negative entries and row (resp. column) sums given by the blocks of $\lambda$ (see Richard Stanley's answer to Number of double cosets of a Young subgroup). For a representative $\alpha_x \in \mathcal S_n$ of a given coset, define the element -$$\sigma_x = \varepsilon \alpha_x \varepsilon,$$ -which is an element of the corresponding Hecke algebra; above, $x$ is a matrix with non-negative entries with row and column sums given by $\lambda$. -I would like to know how to find the structure constants of the Hecke algebra, i.e. the coefficients $\gamma_{x,y}^z$ such that -$$\sigma_x \cdot \sigma_y = \sum_z \gamma_{x,y}^z \sigma_z.$$ -Example -In the case $\mathcal S_p \times \mathcal S_q \subseteq \mathcal S_{n=p+q}$, the double cosets can be indexed by integers $0 \leq x \leq \min(p,q)$, where $x$ represents the number of elements of $[1,p]$ mapped to $[p+1,p+q]$. Using basic combinatorics, I can show that -$$\sigma_x \sigma_1 = \frac{1}{pq} \left( (p-x)(q-x) \sigma_{x+1} + x^2 \sigma_{x-1} + (x(p-x)+x(q-x))\sigma_x \right).$$ - -REPLY [5 votes]: There is a combinatorial rule for the structure constants. It appears in Section 2 of https://arxiv.org/abs/1104.1959, although it is not immediately clear that it is indeed what you are looking for. -Suppose that $V$ is a vector space of dimension at least $l(\lambda)$. Then in the setting of Schur-Weyl duality, there is a simultaneous action of $GL(V)$ and $S_n$ on $V^{\otimes n}$. We note that the $\lambda$-weight space of $V^{\otimes n}$ is precisely the permutation module $M^\lambda$ induced from the trivial module of $H_\lambda$. -As a representation of $\mathbb{C}S_n$, $\mathbb{C}S_n \varepsilon$ is $M^\lambda$. So the Hecke algebra you are implicitly considering is -$$ -\mathrm{End}_{\mathbb{C}S_n}(M^\lambda). -$$ -This is related to $\mathrm{End}_{\mathbb{C}S_n}(V^{\otimes n})$, which is why the Schur algebra (appearing in the paper) is relevant here. -However, the most convenient basis is given by -$$ -\xi_x = \frac{1}{|H_\lambda|}\sum_{g \in H_\lambda \alpha_x H_\lambda} g, -$$ -which is to say, double-coset sums divided by $|H_\lambda|$, which is a different normalisation to your elements $\sigma_x$. Explicitly, -$$ -\sigma_x = \frac{|H_\lambda \cap \alpha_x H_\lambda \alpha_x^{-1}|}{|H_\lambda|}\xi_x, -$$ -and it is a standard lemma that if $\alpha_x$ is a minimal-length double-coset representative, $H_\lambda \cap \alpha_x H_\lambda \alpha_x^{-1}$ is also a Young subgroup, so its size can easily be written down. -Note that the double cosets $H_\lambda \alpha_x H_\lambda$ are indexed by square matrices of size $l(\lambda)$ whose row and column sums are $\lambda$. The $(i,j)$-th entry counts how many elements of $\{1,2,\ldots,n\}$ permuted by $S_{\lambda_j}$ are mapped by some (and therefore any) element of the double coset to elements permuted by $S_{\lambda_i}$. -So we let the indexing variables $x,y,z$ be matrices of that form, and write them with subscripts (e.g. $x_{ij}$) to refer to the entries of the corresponding matrices. -To calculate the coefficient of $\xi_z$ in $\xi_x \xi_y$, we consider "cubic matrices of size $l(\lambda)$" (i.e. $l(\lambda) \times l(\lambda) \times l(\lambda)$ 3-tensors) $A_{ijk}$ with entries in $\mathbb{Z}_{\geq 0}$. We require that -$$ -\sum_i A_{ijk} = x_{jk} -$$ -$$ -\sum_j A_{ijk} = z_{ik} -$$ -$$ -\sum_k A_{ijk} = y_{ij} -$$ -Then the coefficient is -$$ -\sum_A \prod_{i,k}\frac{(\sum_j A_{ijk})!}{\prod_j A_{ijk}!}, -$$ -subject to the conditions on $A$ already mentioned. -Let's check this for your example. We take the matrix $x$ to be -$$ -\begin{bmatrix} - p-x & x \\ - x & q-x -\end{bmatrix}. -$$ -(Here there is an conflict of notation between the label of the double coset, and the number of elements it "mixes" between $S_{p}$ and $S_q$.) We take the matrix $y$ to be -$$ -\begin{bmatrix} - p-1 & 1 \\ - 1 & q-1 -\end{bmatrix}. -$$ -So we are going to calculate $\xi_x \xi_y$. Typsetting a 3-tensor $A_{ijk}$ is tricky, so let's consider the "slices" $A_{i1k}$ and $A_{i2k}$ separately. The row-sums of $A_{i1k}$ are $(p-x,x)$ and the column sums are $(p-1,1)$. There are two possibilities for $A_{i1l}$: -$$ -\begin{bmatrix} - p-x & 0 \\ - x-1 & 1 -\end{bmatrix}, -\begin{bmatrix} - p-x-1 & 1 \\ - x & 0 -\end{bmatrix}. -$$ -Similarly, for $A_{i2k}$ we have row-sums $(x,q-x)$ and column sums $(1,q-1)$, so we get -$$ -\begin{bmatrix} - 1 & x-1 \\ - 0 & q-x -\end{bmatrix}, -\begin{bmatrix} - 0 & x \\ - 1 & q-x-1 -\end{bmatrix}. -$$ -So there are four ways of combining these into the full tensor $A_{ijk}$. For each of these, we get a multiple of $\xi_z$, where $z = A_{i1k}+A_{12k}$, and the scalar multiple is a product of (very simple) binomial coefficients. -Choosing the first option for each of $A_{i1k}$ and $A_{i2k}$ gives $z$ with off-diagonal entries $x-1$, and the scalar multiple is $(p-x+1)(q-x+1)$. -Choosing the last of each gives $z$ with off-diagonal entries $x+1$, and the scalar multiple is $(x+1)^2$. -Each of the other two choices gives $z$ with off-diagonal entries $x$, and the scalar multiples are $x(p-x)$ and $x(q-x)$. -To recover your equation, we just need to know that if $z$ has off-diagonal entries equal to $r$, then -$$ -\sigma_r = \frac{1}{{p \choose r}{q \choose r}} \xi_r, -$$ -so for example, you can immediately multiply your equation by $pq$ to turn $\sigma_1$ into $\xi_y$ (and clear a denominator on the right). -I have left out some details to keep this post from being egregiously long, but I would be happy to explain further or give references if it would be helpful. -What is described in this post is just the tip of the iceberg, for example it's possible to interpolate these structure constants to define certain "permutation module Deligne Categories", such as in https://arxiv.org/abs/1909.04100.<|endoftext|> -TITLE: Chromatic number of distance graphs over the integers -QUESTION [5 upvotes]: Let $D\subseteq\mathbb{N}^+$, and consider the graph $G_D$ with vertices set $\mathbb{N}$ and edges set $\{(x,y)\in\mathbb{N}\times\mathbb{N}\;s.t.\;|x-y|\in D\}$. I expect that if $D$ is dense enough in $\mathbb{N}^+$, then the chromatic number of $G_D$ is large. As Wojowu pointed out in the comments, positive density does not guarantee infinite chromatic number. Hence, one can ask the following question: -if $D$ has density (say, e.g., lower asymptotic density) one in $\mathbb{N}^+$, is it true that the chromatic number of $G_D$ is infinite? -Thank you for any suggestion. - -REPLY [4 votes]: Assume that the chromatic number is $k$. Among the numbers $1,2,\dots, N$ there are at least $N/k$ numbers of the same color, say $a_1,\dots, a_t$. Then $D$ does not contain at least $t-1\geq N/k-1$ numbers not exceeding $N-1$, namely $a_i-a_1$, $i\geq2$. Thus the density of $D$ is at most $1-1/k$. Surely, this estimate is tight... - -REPLY [4 votes]: I'll show that if $G_D$ has chromatic number $k$ then $D$ has upper Banach density at most $(k-1)/k$. -So suppose $G_D$ has chromatic number $k$. Let $\mathbb{N}$ be partitioned into $P_1,\ldots,P_k$, where each $P_i$ is independent with respect to $G_D$. Without loss of generality, $P:=P_1$ has upper Banach density at least $1/k$. Let $Q=\{|x-y|:x,y\in P\text{ are distinct}\}$. Then $Q\subseteq \mathbb{N}^+\backslash D$ since $P$ is $G_D$-independent. We claim that $Q$ has lower Banach density at least $1/k$, which implies the desired result for $D$. -(The proof of the claim is an adaptation of Ruzsa's Covering Lemma and/or the well-known fact that if a set $A$ of integers has positive upper Banach density then $A-A$ is syndetic.) -Call a set $X\subset\mathbb{N}$ $P$-separating if $(x+P)\cap (y+P)=\emptyset$ for all distinct $x,y\in X$. Since $P$ has upper Banach density at least $1/k$, it follows that any $P$-separating subset of $\mathbb{N}$ has size at most $k$. So we may choose a $P$-separating set $X$ of maximal size. Now fix $a\in\mathbb{N}^+$ such that $a>\max X$. By maximality, there is some $x\in X$ such that $(a+P)\cap (x+P)\neq\emptyset$. So there are $p,q\in P$ such that $a+p=x+q$. Since $a>x$ it follows that $a\in x+Q$. -Altogether, we have shown that $X+Q$ is cofinite in $\mathbb{N}^+$. Since $|X|\leq k$, it follows that $Q$ has lower Banach density at least $1/k$. - -Remark. The proof actually shows that if $G_D$ has chromatic number $k$ then there are $k$ translates of the complement of $D$ whose union is cofinite in $\mathbb{N}^+$, which I suppose is stronger than saying $D$ has upper Banach density at most $(k-1)/k$.<|endoftext|> -TITLE: Identities of finite inverse semigroups -QUESTION [6 upvotes]: An inverse semigroup is an algebra with two operations: binary $\cdot$ and unary $^{-1}$ such that $\cdot$ is associative and $xx^{-1}x=x, xx^{-1}yy^{-1}=yy^{-1}xx^{-1}$. The Brandt semigroup with 1, $B_2^1$, is the inverse semigroup of $2\times 2$-matrices consisting of 0, I, and the four matrix units $e_{i,j}$, $i,j=1,2$ where $e_{i,j}$ is the matrix with $(i,j)$-entry 1 and other entries 0, $e_{i,j}^{-1}=e_{j,i}$. It is known (Kleiman) that the identities of $B_2^1$ are not finitely based. -Question. Is it known that the identities of any finite inverse semigroup containing $B_2^1$ as an inverse subsemigroup are not finitely based? - -REPLY [4 votes]: I believe this is a well known open question. It has this property as a semigroup but it is not clear as an inverse semigroup. Mark Sapir showed it is contained in a finitely based locally finite variety of inverse semigroups. Your question is problem 3.10.13 in his book Combinatorial algebra: syntax and semantics.<|endoftext|> -TITLE: Conditions under which $\lim_{s\to1^+}\sum_{n=1}^{\infty}\frac{a_n}{n^s}=\sum_{n=1}^{\infty}\frac{a_n}{n}$ -QUESTION [9 upvotes]: I was working with some Dirichlet series and I realized that I have never seen any general conditions under which -\begin{equation} -\sum_{n=1}^{\infty}\frac{a_n}{n}=\lim_{s\to1^+}\sum_{n=1}^{\infty}\frac{a_n}{n^s}\label{1}\tag{1} -\end{equation} -holds. This is obviously not true in a general case since if so there would be a very simple proof of the PNT from just applying this to $a_n=\mu(n)$. My question is: under what conditions does \eqref{1} hold? -I can show that if $\sum_{n=1}^{\infty}\frac{a_n}{n}$ converges then \eqref{1} must hold using a very simple proof, but I can't find any broader statements. The ideal condition that I would like to show is that if the partial sums $\sum_{n=1}^{N}\frac{a_n}{n}$ are bounded then \eqref{1} must hold. I don't know how I would go about proving this though, and any insights on this general area would be greatly appreciated. - -REPLY [2 votes]: In D. J. Newman's paper -A simple analytic proof of the prime number theorem -published in The American Mathematical Monthly -87 (1980) 693-696, Newman proved a result of this type which may also be useful to you (this is a minor variant of the criterion given by KConrad in his answer). -It says the following: -Let $|a_n| \leq 1$ and suppose that the Dirichlet series -$$\sum_{n = 1}^\infty \frac{a_n}{n^s}$$ -admits a holomorphic continuation to the line $\mathrm{re}\, s = 1$. Then -$$\sum_{n = 1}^\infty \frac{a_n}{n^s}$$ -converges for all $\mathrm{re}\, s \geq 1$.<|endoftext|> -TITLE: A variation of the law of large numbers for random points in a square -QUESTION [29 upvotes]: I uniformly mark $n^2$ points in $[0,1]^2$. Then I want to draw $cn$ vertical lines and $cn$ horizontal lines such that in each small rectangle there is at most one marked point. Surely, for a given constant c it is not always possible. -But it seems that for $c=100$ when n tends to infinity, the probability that such a cut exists should tend to one, as a variation of the law of the large numbers. -Do you have any idea how to prove this rigorously? - -REPLY [12 votes]: Interestingly, if we allow the lines to have arbitrary directions, it still requires roughly n^{4/3} (up to a log correction) lines to separate all the points. -https://www.cambridge.org/core/journals/proceedings-of-the-london-mathematical-society/article/economical-covers-with-geometric-applications/486374A93F4351DF26C155F6C3FE35AE<|endoftext|> -TITLE: A conjecture of De Giorgi on weighted Sobolev spaces -QUESTION [12 upvotes]: Let $\mu$ be a probability measure on $\mathbb{R}^d$ which is absolutely continuous with respect to the Lebesgue measure with density $\rho$. Assume that, for all $t>0$, -\begin{align*} -\exp \left(t \rho\right), \exp \left(t \rho^{-1}\right) \in L^1_{loc}. -\end{align*} -Then, a conjecture of De Giorgi asserts that the Meyers-Serrin theorem holds for the weighted Sobolev spaces associated with $\mu$, namely $H=W$. -The only reference I know speaking of this conjecture is this paper -https://iopscience.iop.org/article/10.1070/SM1998v189n08ABEH000344/meta -Thus, I would like to know if there have been recent advances in proving or disproving this conjecture. -Thanks in advance. - -REPLY [5 votes]: I did some diggings and some readings and found out that the conjecture has been solved here -https://link.springer.com/article/10.1134/S1064562413060173 -and extended recently to a wider context in -https://www.degruyter.com/view/journals/crll/2019/746/article-p39.xml<|endoftext|> -TITLE: Does a flat compactification always exist? -QUESTION [8 upvotes]: Let $\pi:X\to S$ be a separated flat morphism of finite type of Noetherian schemes. Does $\pi$ necessarily factor as an open immersion followed by a proper flat morphism? The analogue of this question with the word "flat" replaced by "smooth" has a negative answer (consider an elliptic curve over $\mathbb{Q}_p$ that has bad reduction). - -REPLY [4 votes]: This already fails if $S$ is regular of dimension $3$ and $\pi$ is quasi-finite. Indeed, let $X$ be a normal affine variety over $\mathbf C$ of dimension $3$ with an isolated non-Cohen–Macaulay singularity (e.g. an affine cone over a smooth projective surface $Y$ with $H^1(Y,\mathcal O_Y) \neq 0$). By Noether normalisation, there exists a finite surjection -$$\pi \colon X \to S = \mathbf A^3,$$ -without loss of generality taking the isolated singularity $x_0 \in X$ to the origin $0 \in \mathbf A^3$. Let $U = X \setminus x_0$, which is smooth by assumption, so $\pi|_U$ is flat by miracle flatness. Now I claim that $\pi|_U \colon U \to \mathbf A^3$ does not have a flat compactification. -Indeed, suppose $U \hookrightarrow X' \stackrel{\pi'}\to S$ is a factorisation into an open immersion and a proper flat morphism. Because $\pi'$ is flat and generically finite, it is quasi-finite, hence finite since it is proper. Since $\pi'$ is finite flat and $S$ is regular, we conclude that $X'$ is Cohen–Macaulay. -Let $\bar U \subseteq X'$ be the scheme-theoretic closure of $U$, and let $V = S \setminus 0$. Since $\pi^{-1}(V) \subseteq U$, we conclude that $\bar U \setminus U$ is supported on $\pi'^{-1}(0)$, in particular has dimension $0$. By Hartshorne's connectedness theorem, this implies that there are no other components in $X'$ (otherwise two components would meet only in a $0$-dimensional set), i.e. $\bar U = X'$ set-theoretically. Since $X'$ is generically reduced and Cohen–Macaulay, it is reduced, so $\bar U = X'$ scheme-theoretically. -In particular, $X' \setminus U$ is $0$-dimensional, so $X'$ is regular in codimension $1$ since the same holds for $U$. Since $X'$ is Cohen–Macaulay, this forces $X'$ normal, so it equals the normalisation of $S$ in $K(U)$, which is $X$. But $X$ is not Cohen–Macaulay, contradicting flatness of $\pi'$. $\square$<|endoftext|> -TITLE: "Holomorphic line bundle" + "algebraic after finite cover" implies "algebraic"? -QUESTION [6 upvotes]: Let $X$, $Y$ be complex affine algebraic manifolds (closed submanifolds of $\mathbb{C}^n$), let $f\colon Y\to X$ be a finite covering. Let $\mathcal{L}$ be a holomorphic line bundle on $X$. Suppose $f^*\mathcal{L}$ is an algebraic line bundle. Is $\mathcal{L}$ necessarily algebraic? -(This is true when $f$ admits a section. In general $f_*\mathcal{O}_X$ is locally free, we can recover $\mathcal{L}$ Zariski locally (by the projection formula $f_*\mathcal{L}\cong f_*\mathcal{O}_Y\otimes\mathcal{L}$, then read the factors over $U_i$ over which $f_*\mathcal{O}_{Y}$ is locally free, using $f_*\mathcal{O}_Y|_{f^{-1}(U_i)}\cong\mathcal{O}_{U_i}^{\oplus d}$), but I am not sure if the transitions can be chosen algebraically.) - -REPLY [5 votes]: Yes. This follows from P. Deligne "Equations differentielles..." LNM 163, Proposition II 2.22, since the notion of moderate growth at infinity will coincide for $X$ and $Y$. This works more generally for any coherent sheaf.<|endoftext|> -TITLE: Request for examples: verifying vs understanding proofs -QUESTION [54 upvotes]: My colleague and I are researchers in philosophy of mathematical practice and are working on developing an account of mathematical understanding. We have often seen it remarked that there is an important difference between merely verifying that a proof is correct and really understanding it. Bourbaki put it as follows: - -[E]very mathematician knows that a proof has not really been “understood” if one has done nothing more than verifying step by step the correctness of the deductions of which it is composed, and has not tried to gain a clear insight into the ideas which have led to the construction of this particular chain of deductions in preference to every other one. -[Bourbaki, ‘The Architecture of Mathematics’, 1950, p.223] - -We are interested in examples which, from the perspective of a professional mathematician, illustrate this phenomenon. If you have ever experienced this difference between simply verifying a proof and understanding it, we would be interested to know which proof(s) and why you did not understand it (them) in the first place. We are especially interested in proofs that are no longer than a couple of pages in length. We would also be very grateful if you could provide some references to the proof(s) in question. -We are sorry if this isn’t the appropriate place to post this, but we were hoping that professional mathematicians on MathOverflow could provide some examples that would help with our research. - -REPLY [4 votes]: A good proof is a proof that makes us wiser. If the heart of the proof is a voluminious search or a long string of identities, it is probably a bad proof. If something is so isolated that it is sufficient to get the result popped up on the screen or a computer, then it is probably not worth doing. Wisdom lives in connections. If I have to calculate the first 20 digits of π by hand I certainly become wiser afterwards because I see that these formulas for π that I knew take too much time to produce 20 digits. I will probably devise some algorithms which minimize my effort. But when I get two millions of digits of π from the computer using somebody else's library program I remain as stupid as I was before. - -Yuri I. MANIN, Good Proofs are Proofs that Make us Wiser<|endoftext|> -TITLE: Any real algebraic variety is diffeomorphic to a real algebraic variety defined over $\mathbb{Q}$ -QUESTION [17 upvotes]: Given a smooth proper real algebraic variety can you find a smooth proper real algebraic variety defined over $\mathbb{Q}$ that is diffeomorphic to it? - -REPLY [9 votes]: Edoardo Ballico and Alberto Tognoli proved in their paper "Algebraic models defined over $\mathbb{Q}$ of differential manifolds" (Geom. dedicata 42, 155-161, 1992) that every compact differential manifold is diffeomorphic to the real points of a regular affine variety defined over $\mathbb{Q}$. -For non-smooth algebraic varieties there are obstructions to descend from $\mathbb{R}$ to $\mathbb{Q}$, there is a recent paper by Adam Parusinski and Guillaume Rond "Algebraic varieties are homeomorphic to varieties defined over number fields" arXiv 1810.00808 on this subject. -Edit : let me recall some basic facts about real algebraic sets (I refer to "Real algebraic geometry" Bochnak, Coste and Roy). - -A complete nonsingular affine real algebraic variety is projective (see BCR 3.4 p.74-75) -An algebraic subset of $\mathbb{R}^n$ is complete if and only if it is closed and bounded (3.4.9 and 3.4.10).<|endoftext|> -TITLE: A constant bizarrely related to the Fibonacci Numbers -QUESTION [9 upvotes]: For roughly the past month, I have been studying denesting radicals. For example: the expression $\sqrt[3]{\sqrt[3]2-1}$ is a radical expression that contains another radical expression, so this radical is nested. Is there a way to express this with radicals that are not (or not as) nested? Writing it in this way is referred to as denesting, and indeed, there is one such way. Ramanujan mysteriously found that $$\sqrt[3]{\sqrt[3]2-1}=\sqrt[3]{\frac 19}-\sqrt[3]{\frac 29}+\sqrt[3]{\frac 49}.$$ I found that this is also equal to $\sqrt{\sqrt[3]{\frac 43}-\sqrt[3]{\frac 13}}$, which does not denest it, but it does write the expression under a radical of a coprime degree, which fascinates me. -I have found an abundance of results, one of which fascinates me the most, and is on the constant, $$1-\sqrt[3]{\frac 12}+\sqrt[3]{\frac 14}.$$ This constant is equal to all of the expressions below. Note that the degree of the radicals are, lo and behold, the Fibonacci numbers! $$\sqrt{\frac 32\bigg(\sqrt[3]2-\frac 1{\sqrt[3]2}\bigg)}$$ $$\sqrt[3]{\frac{3^2}{2^2}\big(\sqrt[3]2-1\big)}$$ $$\sqrt[5]{\frac{3^3}{2^3}\bigg(\frac 3{\sqrt[3]2}-\sqrt[3]2-1\bigg)}$$ $$\sqrt[8]{\frac{3^5}{2^5}\bigg(4-\frac{5}{\sqrt[3]2}\bigg)}$$ $$\sqrt[13]{\frac{3^8}{3^8}\bigg(1+\frac{17}{\sqrt[3]2}-\frac{23}{\sqrt[3]4}\bigg)}$$ and, slightly breaking the pattern, $$\sqrt[21]{\frac{3^{14}}{2^{14}}\big(41-59\sqrt[3]2+21\sqrt[3]4\big)}$$ and presumably, this list goes on forever (but the numbers start becoming pretty big). -So... what on earth is going on here? It appears that for some $n$th Fibonacci number $F_n$, this is equal to (for at least most of the time), $$\sqrt[F_n]{\frac{3^{F_{n-1}}}{2^{F_{n-1}}}\big(a+b\sqrt[3]2+c\sqrt[3]4\big)}$$ for some $\{a, b, c\}\subset \mathbb R$, with a minimal polynomial of $4x^3-12x^2+18x-9$. -Can anybody explain these wild affairs? - -Don't know of any other appropriate tags - -REPLY [6 votes]: First of all we get $A=1-\sqrt[3]{\frac{1}{2}}+\sqrt[3]{\frac{1}{4}}=\frac{3\sqrt[3]{2}}{2(\sqrt[3]{2}+1)}$......(1) -And from here you can proof by induction... -$A^{F_n}=A^{F_{n-1}}.A^{F_{n-2}}$ -If $A^{F_{n-1}}=(\frac{3}{2})^{F_{n-2}}(a_{n-1}+b_{n-1}\sqrt[3]{2}+c_{n-1}\sqrt[3]{4})$ -and -$A^{F_{n-2}}=(\frac{3}{2})^{F_{n-3}}(a_{n-2}+b_{n-2}\sqrt[3]{2}+c_{n-2}\sqrt[3]{4})$ -From (1) we get, $A^{F_2}, A^{F_3}$, so we shall get the rest by induction. -Where, $F_2=2, F_3=3$. -$\frac{3}{2}(\frac{\sqrt[3]{2}}{\sqrt[3]{2}+1})^2=\sqrt[3]{2}-\sqrt[3]{\frac{1}{2}}$ -And, -$\frac{3}{2}(\frac{\sqrt[3]{2}}{\sqrt[3]{2}+1})^3=\sqrt[3]{2}-1$<|endoftext|> -TITLE: Left Kan extension that preserves colimit -QUESTION [5 upvotes]: I'd be very happy if the question When do Kan extensions preserve limits/colimits? has been fully answered. But it seems not. -I have a more specific question though. Let $C$ be a site (essentially small to be safe) equipped with some subcanonical Grothendieck topology, $PSh$ be its category of presheaves. Then it's well-known that for any functor $C \to B$ with $B$ cocomplete, the left Kan extension $PSh \to B$ along the Yoneda embedding $C \to PSh$ necessarily preserves colimit. Is it true if I replace $PSh$ by the category of sheaves $Sh$? In other words, does the left Kan extension along $C \xrightarrow{Yoneda} PSh \xrightarrow{sheafification} Sh$ preserve colimit? If in general not, is there any sufficient condition we can impose on the functor $C \to B$? - -REPLY [6 votes]: Usually it does not. We can take $C \to B$ to be the universal example of a functor to a cocomplete category, namely the Yoneda embedding $C \to PSh$. Then the left Kan extension of $y : C \to PSh$ along $j : C \to PSh \to Sh$ is the inclusion $Sh \to PSh$, and this usually does not preserve colimits. The calculation of this left Kan extension can be verified using the coend formula: -$$ -(Lan_j y)(X) = \int^{c : C} Sh(jc, X) \otimes yc -= \int^{c : C} X(c) \otimes yc = X -$$ -where the second equality used the fact that the topology is subcanonical.<|endoftext|> -TITLE: "Real algebraic varieties" vs finite type separated reduced $\mathbb{R}$-schemes with dense $\mathbb{R}$-points -QUESTION [17 upvotes]: This question is partly motivated by a few comments here. Let me denote by $R$ the (real-closed) field of real numbers $\mathbb{R}$; everything is probably the same over an arbitrary real-closed field. -When one has a polynomial subset $V$ of $R^n$, the following two are equally sensible ways of putting a structure sheaf on $V$: - -One is by considering regular functions in the sense of usual scheme theory: in this case the global regular functions are $R$-polynomials in $n$ variables modulo the ideal of those polynomials vanishing on $V$. If, more precisely, we call $X$ the corresponding $R$-scheme (with all of its non-$R$-points too, which by the way are reconstructible from the set $V\subseteq R^n$) and $O_X$ its structure sheaf, then $X(R)=V\subseteq R^n$ and $O_X(X)\simeq R[x_1,\ldots, x_n]/I_X$. - -The other way is by declaring that a regular function is a ratio of polynomials with nonvanishing denominator. We will call such functions $R$-regular, and $R_V$ the resulting structure sheaf. We call $(V,R_V)$ an $R$-algebraic variety. This definition seems to be standard in real algebraic geometry, see e.g. Bochnak-Coste-Roy - Real algebraic geometry (Section 3.2). I think it doesn't change much if we consider the topological space $X$ of the scheme in point 1) instead, endowed with the sheaf $R_X$ that sends an open set $U$ to the rational functions on $U\subseteq X$ that are regular at each point of $U\cap X(R)$. - - -The resulting structure sheaves are not the same. For example, consider the real line: the function $\frac{1}{1+x^2}$ is an $R$-regular function which is not (scheme-theoretically) regular. -Likewise, one can define abstract $R$-algebraic varieties, and $R$-regular maps thereof. -The curious thing is that every projective $R$-algebraic variety is $R$-biregularly isomorphic to an affine one. Indeed, the set theoretic map (example 1.5 in Ottaviani - Real algebraic geometry. A few basics or theorem 3.4.4 in BCR) -$$\mathbb{P}^n(R)\to \operatorname{Sym}^2(R^{n+1})\;\;,\quad (x_0:\ldots : x_n)\mapsto \frac{x_ix_j}{\sum_{h=1}^n x_h^2}$$ -is an $R$-regular embedding. This does not correspond to an everywhere-defined morphism of schemes, as is immediately seen by looking at any component of the map in a standard affine chart of $\mathbb{P}^n$. - -Are there non-(quasi-)affine abstract $R$-algebraic varieties at all? - -Edit: I think the "quasi" in "quasi-affine" may be pleonastic: I haven't checked the details but a quasi-affine $R$-algebraic variety should very often be affine. Indeed, if $X=W\smallsetminus Y$, $Y\subset W \subseteq R^n$ with $W$ affine and $Y$ closed (maybe with some assumptions on $Y$), the real blowup $\operatorname{Bl}_Y W$ is closed in some $\mathbb{P}^{m}\times W$ and the latter is affine; but now the "missing" set $E$ has become a divisor: $X\simeq (\operatorname{Bl}_Y W)\smallsetminus E$, and affine minus a divisor is still affine. - -The above example (the one of projective space embedding in an affine space) shows that the category $\text{$R$-Var}$ of $R$-algebraic varieties is not a full subcategory of schemes over $\operatorname{Spec}(R)$. On the other hand, I think the category $\operatorname{Sch}'_R$ of finite type separated reduced schemes over $\operatorname{Spec}(R)$ is a full subcategory of $\text{$R$-Var}$. [Edit: following the comment of Julian Rosen, we probably also want to require the schemes in $\operatorname{Sch}'_R$ to have dense $R$-points] - -Are there two non-isomorphic schemes in $\operatorname{Sch}'_R$ that become isomorphic in $\text{$R$-Var}$? - -Edit: even before posting, I found example 3.2.8 in BCR. There is also proposition 3.5.2 in BCR, the $R$-biregular isomorphism between the circle $x^2+y^2=1$ and $\mathbb{P}^1_R$. And between the "quadric" sphere and the "Riemann" sphere (i.e. complex projective line thought of as a real algebraic variety). - - -In which other ways does $\text{$R$-Var}$ deviate from $\operatorname{Sch}'_R$? - -Note: I'm not asking how real algebraic geometry deviates from complex algebraic geometry (which is surely addressed in a preexisting MO question). - -Edit: (added following question) - -For non real-closed fields, or fields of positive characteristic, do people consider varieties in the sense of 1) or in the sense of 2)? - -For example, should $1/(1+x^2)$ be a regular function on the line over $\mathbb{F}_7$? (It's a well defined function on a finite field, so there will be a polynomial realizing its values set theoretically, but should it be enough?) -- Or, should 1/(x^2-3) be a regular function on the line over $\mathbb{Q}(\sqrt{2})$? - -REPLY [3 votes]: As for your first question, concerning nonaffine R-varieties as you call them, yes, there are nonaffine R-varieties. However, they are considered pathological. Example 12.1.5 on page 301 -of Bochnak-Coste-Roy, Real algebraic geometry, constructs an R-line bundle over $\mathbf R^2$ whose total space is not affine. In fact, it is not -affine since it does not have any separated complexification. Note that the R-variety itself, however, is separated! -The essential point here is that the set of real points of an irreducible affine scheme over $\mathbf R$ can be reducible. In the aforementioned -example, the irreducible scheme in question is the one defined by the irreducible polynomial $$p=x^2(x-1)^2+y^2\in\mathbf R[x,y,z].$$ The set of -real points in $\mathbf R^3$ defined by $p$ is the disjoint union of the affine lines $$L_0=\{(0,0)\}\times\mathbf R\ \mathrm{and}\ -L_1=\{(1,0)\}\times\mathbf R. $$ This is clearly a reducible subset of $\mathbf R^3$. The separated R-variety that does not have a separated -complexification is the one obtained by gluing the open subsets $$ U_0=\mathbf R^3\setminus L_0\ \mathrm{and}\ U_1=\mathbf R^3\setminus L_1 $$ -along the open subsets $$ U_{01}=U_0\cap U_1\subseteq U_0\ \mathrm{and}\ U_{10}=U_0\cap U_1\subseteq U_1 $$ via the regular isomorphism $$ -\phi_{10}\colon U_{01}\rightarrow U_{10} $$ defined by $$ \phi_{10}(x,y,z)=(x,y,pz). $$ Note that this is indeed a regular isomorphism -since the map $\phi_{01}=\phi_{10}^{-1}$ is the regular map $$ \phi_{01}\colon U_{10}\rightarrow U_{01} $$ defined by $$ -\phi_{01}(x,y,z)=(x,y,\tfrac{z}{p}). $$ -Now, it is easy to see that the R-variety $U$ one obtains is separated, as defined in the -founding paper of the whole theory: Faisceaux algébriques cohérents by Jean-Pierre Serre. Indeed, one easily checks that the diagonal in $U\times -U$ is closed. However, if one wants to construct a real scheme $X$ whose set of real points coincides with $U$, then, inevitably, $X$ will not be -separated. Indeed, the polynomial $p$ defines a nonclosed point $x_0$ in any scheme-wise thickening $X_0$ of $U_0$ since $p$ has zeros in $U_0$, -and similarly it defines a non closed point $x_1$ of any scheme-wise thickening $X_1$ of $U_1$. The gluing morphisms $\phi_{01}$ and $\phi_{10}$ will -extend to open subsets $X_{01}$ of $X_0$ and $X_{10}$ of $X_1$, but they won't contain $x_0$ and $x_1$, respectively. This is because the -polynomial $p$ vanishes at $x_0$. As a result, any scheme-wise thickening of $U$ will be nonseparated! -As for your second question, if I understand correctly, you are asking whether the functor -$$ -F\colon Sch_R'\rightarrow R-Var -$$ -defined by $F(X)=X(\mathbf R)$ is an equivalence onto a full subcategory, where $Sch_R'$ is the category of finite type separated reduced schemes -over $Spec(\mathbf R)$ having dense sets of real points. This is an equivalence onto a full subcategory, its image category, if you localize $Sch_R'$ with respect to inlcusions of open subsets containing all real points: any -morphism of $R$-varieties wil extend to a morphims defined on some open subset containing the real points. Uniqueness is implied by density of -real points and separation. -As for your third question, I can't think of other differences between $R$-varieties and schemes over $\mathbf R$ that differ essentially from -phenomena already present in the example above. -As for your final question about varieties in the sense of $R$-varieties over other fields, Serre certainly did define them in the paper -I mentioned above. I'm not sure whether that has had a follow-up for other fields than real or algebraically closed fields.<|endoftext|> -TITLE: Why the name O for category O? -QUESTION [12 upvotes]: What is the motivation behind naming the category O appearing in the theory of Lie algebras? Does O stand for something? -Here is a question Why the BGG category O? that further confuses me. It seems like there is a notion of when a category is O, is it? - -REPLY [9 votes]: From [Humphreys: Representations of semisimple Lie algebras in the BGG category O], notes for Chapter 1: - -The letter chosen to label the category is the first letter of a - Russian word meaning “basic” - -which is основной.<|endoftext|> -TITLE: Properties of measures that are not countably additive but have countably additive null ideals -QUESTION [5 upvotes]: This is a very naive question, maybe more of a reference request than anything else. -Let $(X, \mathcal X)$ be a measurable space. If $m$ is a real-valued function on $\mathcal X$, we say that $m$ has a countably additive null ideal if $m(\cup_{n=1}^\infty A_n) = 0$ whenever $A_n \in \mathcal X$ and $m(A_n)=0$ for all $n$. -Of course if $m$ is a countably additive measure, then $m$ has a countably additive null ideal. -If $m$ is a merely finitely additive probability measure (i.e. finitely but not countably additive and such that $m(X)=1$) it may or may not have a countably additive null ideal. In a typical example of a merely finitely probability, the null ideal is not countably additive: extend the natural density function to a probability measure $m$ on $(\mathbb N, 2^{\mathbb N})$ by means of a Banach limit and then $m\{n\}=0$ for all $n$ while $m(\mathbb N)=1$. - -I am wondering what can be said about merely finitely additive probabilities with countably additive null ideals. What's a typical example of such a probability? "How similar" are such probabilities to countably additive probabilities, i.e. what properties of countably additive probabilities do such probabilities preserve? Any other interesting results about merely finitely additive probabilities with countably additive null ideals are welcome. - -REPLY [2 votes]: $\newcommand{\N}{\mathbb{N}}\newcommand{\R}{\mathbb{R}}$There are examples on $\R$ with the Borel $\sigma$-algebra $\mathcal{B}$. We take the null ideal to be the meagre Borel sets $\mathcal{M}$ (the $\sigma$-ideal in the Borel sets generated by closed sets with empty interior). -The regular open sets of $\R$ form a complete Boolean algebra $\mathcal{RO}$, and the mapping from $\mathcal{RO} \rightarrow \mathcal{B}/\mathcal{M}$ formed by mapping a regular open set to the equivalence class of Borel sets differing from it by a meagre set is an isomorphism (this uses the Baire category theorem - see for example Fremlin's Measure Theory 514I). What we shall do is define a finitely additive measure $\mu$ on $\mathcal{RO}$ for which the only null element is $\emptyset$. Under the isomorphism above, this defines a finitely-additive Borel probability measure on $\R$ whose null ideal is $\mathcal{M}$. -Let $(U_i)_{i \in \N}$ be a countable base of regular open sets for $\R$ (e.g. open intervals with rational endpoints). By the ultrafilter lemma, for each $i \in \N$, there exists an ultrafilter on $\mathcal{RO}$ containing $U_i$, which defines a finitely-additive measure $\mu_i : \mathcal{RO} \rightarrow [0,1]$ taking only the values $0$ and $1$ and such that $\mu_i(U_i) = 1$. -We then define $\mu : \mathcal{RO} \rightarrow [0,1]$ by $\mu(U) = \sum_{i=1}^\infty 2^{-i} \mu_i(U)$. It is easy to verify that this is a finitely-additive probability measure. Also, for any non-empty regular open $U$ there exists some $i \in \N$ such that $U_i \subseteq U$, and therefore -$$ -\mu(U) \geq \mu(U_i) \geq 2^{-i}\mu_i(U_i) = 2^{-i} > 0. -$$ -So the only $\mu$-null regular open set is $\emptyset$. -The measure $\mu$ is not countably additive because on Polish spaces without isolated points there are no countably-additive Borel probability measures that vanish on meagre sets.<|endoftext|> -TITLE: Two definitions of power operations --- how do they relate? -QUESTION [6 upvotes]: Below are two different stories about power operations for $\mathbb{E}_\infty$-ring spectra, and I am struggling to see how they relate. In the following we let $R$ be an $\mathbb{E}_\infty$-ring spectrum. - -$R$ admits natural maps $R^{\wedge n}_{h\Sigma_n} \to R$. If $X$ is now a space, we may apply $(\,\cdot\,)^{\wedge n}_{h\Sigma_n}$ to a $\Sigma^\infty_+ X \to R$ representing an element of $R^0(X)$, which we then compose with the multiplications $R^{\wedge n}_{h\Sigma_n} \to R$ to obtain a map -$$\mathbb{P}_n \colon R^0(X) \to R^0(X^{\times n}_{h\Sigma_n})$$ -called the $n$-th total power operation of $R$ --- a multiplicative but non-additive map. -There is a category $\mathsf{CAlg}(R)$ of $R$-algebras, and it admits a forgetful functor $U \colon \mathsf{CAlg}(R) \to \mathsf{Sp}$. One then defines a spectrum of power operations on $R$ to be the endomorphism spectrum $\operatorname{Map}(U,U)$. - - -Question. Are these two approaches in any way related? - -At first glance it appears not so, but the forgetful functor $U$ admits a left adjoint $F$ sending a spectrum $Y$ to $R \wedge \bigoplus_n Y^{\wedge n}_{h\Sigma_n}$ --- a formula vaguely similar to what we see in the first definition. I guess if you apply the second definition to the $R$-algebra $\operatorname{Map}(\Sigma^\infty_+ X,R)$ and play around with the adjunction you can make the comparison precise, but I'm struggling with the details. - -REPLY [8 votes]: The first observation is that $U$ is representable in $CAlg(R)$ by $F(S)$ (with $F$ as in your question): $$Map_{CAlg(R)}(F(S),A) \simeq Map_{SMod}(S,A) \simeq U(A).$$ -By some appropriately flowery version of Yoneda's Lemma, it follows that -$$ Hom(U,U) \simeq Map_{CAlg(R)}(F(S),F(S)) \simeq Map_{SMod}(S,F(S)) \simeq R \wedge \bigvee_n B\Sigma_{n+}.$$ -Applying $\pi_0$ to this, is easy to see that the operation corresponding to $1 \in R_0(B\Sigma_n)$ will be precisely the classic $n$th power operation, when applied to the $R$--algebra $A = Map(\Sigma^{\infty}_+X,R)$. -Added later by request ... -Suppose given $f \in \pi_0(F(R))$. Tracing through my equivalences, the associated operation $\theta_f: \pi_0(A) \rightarrow \pi_0(A)$, for $A$ a commutative $R$--algebra, is as follows. -Firstly $f$ can regarded as an $R$-module map $f:R \rightarrow F(R)$. Similarly, $x \in \pi_0(A)$ can be regarded as an $R$-module map $x:R \rightarrow A$. Then $\theta_f(x)$ is the composite -$$ R \rightarrow F(R) \rightarrow F(A) \rightarrow A,$$ -where the first map is $f$, the next is $F(x)$ and the the last -is the structure map for the $R$--algebra: the wedge over $n$ of the maps $A^{\wedge n}_{h \Sigma_n} \rightarrow A$. -For the $n$th power operation, recall that $\displaystyle F(R) = R \wedge \bigvee_m B\Sigma_{m+}$, and let $f_n$ be the composite -$$S^0 \rightarrow B\Sigma_{n+} \rightarrow R \wedge B\Sigma_{n+}\hookrightarrow F(R).$$ -Then $\theta_{f_n}$ will be the $n$th power operation. You can now specialize to $A = Map(X,R)$ if you want.<|endoftext|> -TITLE: A diffeomorphism of complex surfaces mapping subvarieties to subvarieties -QUESTION [10 upvotes]: Let $X$ and $Y$ be smooth projective complex surfaces. If a diffeomorphism from $X$ to $Y$ maps subvarieties to subvarieties does it have to be holomorphic or antiholomorphic? Can we at least verify this for K3 surfaces? - -REPLY [10 votes]: I am posting this answer because the following linear algebra proposition is too long for a comment. -Lemma. Let $A$, respectively $B$, be an invertible $\mathbb{R}$-linear operator on $\mathbb{C}^2$ that is $\mathbb{C}$-linear, resp. $\mathbb{C}$-conjugate linear. -(1). For every $1$-dimensional $\mathbb{C}$-linear subspace $V\subset \mathbb{C}^2$ such that $A(V)$ equals $B(V)$, there exists a generator $\mathbf{v}\in V$ such that $B(\mathbf{v})=r\cdot A(\mathbf{v})$ for a positive real $r$. -(2). For every ordered pair $(\mathbf{v},\mathbf{w})$ of $\mathbb{C}$-linearly independent elements of $V$ -with $B(\mathbf{v})=r\cdot A(\mathbf{v})$ and $B(\mathbf{w})=s\cdot A(\mathbf{w})$ for positive reals $r,s$, for every ordered pair $(\lambda,\mu)\in \mathbb{C}^2$, the elements $A(\lambda\cdot \mathbf{v} +\mu\cdot \mathbf{w})$ and $B(\lambda\cdot \mathbf{v} +\mu\cdot \mathbf{w})$ are $\mathbb{C}$-linearly independent if and only if $r\overline{\lambda}\mu$ equals $s\lambda \overline{\mu}$. -Proof. (1). For any generator $\mathbf{u}\in V$, there exists $\rho \in \mathbb{C}^\times$ such that $B(\mathbf{u})$ equals $\rho\cdot A(\mathbf{u})$. For every complex unit $\zeta$, also $B(\overline{\zeta}\cdot \mathbf{u})$ equals $\zeta^2 \rho A(\overline{\zeta}\cdot \mathbf{u})$. For an appropriate choice of $\zeta$, the complex number $\zeta^2 \rho$ is a positive real number. -(2). This is a straightforward computation. QED -Proposition. Let $A$, respectively $B$, be a nonzero $\mathbb{R}$-linear operator on $\mathbb{C}^2$ that is $\mathbb{C}$-linear, resp. $\mathbb{C}$-conjugate linear. If $A(\mathbf{u})$ and $B(\mathbf{u})$ are $\mathbb{C}$-linearly dependent for every $\mathbf{u}\in \mathbb{C}^2$, then $A(\mathbb{C}^2)$ and $B(\mathbb{C}^2)$ are equal and have complex dimension $1$. -Proof. By the lemma, if $A$ and $B$ are both invertible, and if $V,W$ are distinct $1$-dimensional $\mathbb{C}$-linear subspaces of $\mathbb{C}^2$ with $A(V)=B(V)$ and $A(W)=B(W)$, then there exist generators $\mathbf{v}\in V$ and $\mathbf{w}\in W$ and positive reals $r,s$ such that $A(\lambda\cdot \mathbf{v} + \mu\cdot \mathbf{w})$ and $B(\lambda\cdot \mathbf{v} + \mu\cdot \mathbf{w})$ are $\mathbb{C}$-linearly dependent if and only if $r\overline{\lambda}\mu$ equals $s\lambda \overline{\mu}$. In particular, for $\lambda = 1$ and $\mu=i$, this does not hold, thus the vectors are not $\mathbb{C}$-linearly dependent (they are $\mathbb{C}$-linearly independent). Therefore, for $A$, $B$ as in the statement, at least one of $A$ and $B$ is not invertible. -Thus, there exists a nonzero vector $v\in \mathbb{C}^2$ such that $A(v)=0$ or $B(v)=0$, i.e., either $A(\mathbb{C}^2)$ or $B(\mathbb{C}^2)$ is a $1$-dimensional complex linear subspace of $\mathbb{C}^2$. By the hypothesis on $A$ and $B$, then both $A(\mathbb{C}^2)$ and $B(\mathbb{C}^2)$ are equal to this $1$-dimensional complex linear subspace. QED -Corollary. Let $A$, respectively $B$, be a nonzero $\mathbb{R}$-linear transformation from $\mathbb{C}^n$ to $\mathbb{C}^m$ that is $\mathbb{C}$-linear, resp. $\mathbb{C}$-conjugate linear. If the $\mathbb{R}$-linear transformation $D=A+B$ sends $1$-dimensional $\mathbb{C}$-linear subspaces of $\mathbb{C}^n$ to $\mathbb{C}$-linear subspaces of $\mathbb{C}^m$, then $A(\mathbb{C}^n)$ and $B(\mathbb{C}^n)$ are equal and have $\mathbb{C}$-dimension equal to $1$. -Proof. For the restrictions of $A$ and $B$ to a general $2$-dimensional $\mathbb{C}$-linear subspace of $\mathbb{C}^n$ (considered as an $\mathbb{R}$-linear transformation to the image $2$-dimensional $\mathbb{C}$-linear subspace of $\mathbb{C}^m$), the hypotheses of the proposition hold. In particular, the restrictions of $A$ and $B$ have rank $1$ with equal image. Therefore $A(\mathbb{C}^n)$ and $B(\mathbb{C}^n)$ are equal and have $\mathbb{C}$-dimension equal to $1$. QED -Theorem. Let $X$ be the complex manifold underlying a proper, smooth, connected $\mathbb{C}$-scheme. Let $Y$ be a complex analytic space. Let $f:X\to Y$ be a differentiable function whose $\mathbb{R}$-rank is $>2$ on a connected, dense open subset of $X$. If $f$ maps complex algebraic subvarieties of $X$ to complex analytic subvarieties of $Y$, then either $f$ is holomorphic or anti-holomorphic. -Proof. For every point $p$ in the connected, dense open subset, the derivative map $D=d_p f$ from $T_p X$ to $T_{f(p)}Y$ has rank $>2$. Of course $D$ equals $A+B$ for $A=(D - J_{Y,f(p)}\circ D \circ J_{X,p})/2$ and $B=(D+J_{Y,f(p)}\circ D\circ J_{X,p})/2$, and $A$, respectively $B$, is $\mathbb{C}$-linear, resp. $\mathbb{C}$-conjugate linear. -By the hypothesis on $X$, every $1$-dimensional $\mathbb{C}$-linear subspace of $T_p X$ equals the tangent space to a closed algebraic curve $Z\subset X$ that is smooth at $p$. By the hypothesis, $f(Z)$ is complex analytic, so that $d_p f$ maps this $1$-dimensional $\mathbb{C}$-linear subspace of $T_pX$ to a $\mathbb{C}$-linear subspace of $T_{f(p)} Y$. By the corollary, either $d_p f$ is $\mathbb{C}$-linear or $\mathbb{C}$-conjugate linear, i.e., either $A$ or $B$ is zero. -Since both $A$ and $B$ vary continuously in $p$, either $A$ equals $0$ for all $p$ in the dense open subset of $X$, or $B$ equals $0$ for all $p$ in the dense open subset of $X$. Since the open subset is dense, again by continuity, either $d_p f$ is $\mathbb{C}$-linear for all $p\in X$ or $d_pf$ is $\mathbb{C}$-conjugate linear for all $p\in X$. In other words, either $f$ is holomorphic or anti-holomorphic. QED -Remark. Obviously this is false without the hypothesis that the $\mathbb{R}$-rank of $f$ is $>2$. Indeed, post-compose any nonconstant holomorphic function from $X$ to a Riemann surface $Y$ with a general diffeomorphism of $Y$ to obtain a counterexample.<|endoftext|> -TITLE: Why is the Vandermonde determinant harmonic? -QUESTION [43 upvotes]: It can be checked that the Vandermonde determinant defined as -$$V(\alpha_1, \cdots, \alpha_n) = \prod_{1 \le i < j \le n}(\alpha_i-\alpha_j) $$ -is a harmonic function, that is $\Delta V = 0$ where $\Delta$ is the Laplace operator. Is there a deeper or more intuitive reason why this fact should hold? The straightforward proof of just computing the derivatives and checking doesn't provide any insights. - -REPLY [2 votes]: The other answers already answered the question. I just wanted to point out that this type of question is also part of the Dyson Brownian motion topic. You can see that from the answers already stated, for example Carlo Beenakker. If you look up Dyson Brownian motion you can get to this old blog post of Tao -https://terrytao.wordpress.com/2010/01/18/254a-notes-3b-brownian-motion-and-dyson-brownian-motion/ -Around equation (9) he starts to address the type of algebra you mention. In your notation, his equation (11) says -$$ -\frac{\partial}{\partial \alpha_i} V\, =\, \sum_{\substack{j\\j\neq i}} \frac{V}{\alpha_i-\alpha_j}\, . -$$ -Applying that again and the derivative of $1/x$ you get -$$ -\frac{\partial^2}{\partial \alpha_i^2} V\, =\, \sum_{\substack{j,k\\|\{i,j,k\}|=3}} \frac{V}{(\alpha_i-\alpha_j)(\alpha_i-\alpha_k)}\, . -$$ -Then see his Exercise 25 and the simple algebraic fact that facilitates it. Of course this might not be the deeper reason you want. And then you have all the other deeper connections mentioned by the other folks who already gave you answers.<|endoftext|> -TITLE: Like Bradley’s conjecture (Four incenters lie on a circle) -QUESTION [6 upvotes]: Please don't close this question. Because there is simple configuration with 57 vote up, and don't close. Why you vote up that question and You vote to close this question? -A problem I posed at here since 2014 but no solution: -Let $ABCD$ be a bicentric quadrilateral, $O$ is center of circle $(ABCD)$. Then Incenter of four triangles $OAB,OBC,OCD,ODA$ lie on a circle. - - - -My question: Could You give a your solution for problem above. - - -The problem like Bradley’s conjecture. You can see Bradley’s conjecture at here and page 73, here - -REPLY [4 votes]: Without loss of generality assume that $O=(0,0)$ and the ABCD circle has radius 1. -Then -$$\begin{cases} A=(\cos \alpha,\sin\alpha),\\ -B = (\cos\beta,\sin\beta),\\ -C = (\cos\gamma,\sin\gamma),\\ -D = (\cos\delta,\sin\delta),\\ -AB = 2\sin\frac{\alpha-\beta}2,\\ -BC = 2\sin\frac{\beta-\gamma}2,\\ -CD = 2\sin\frac{\gamma-\delta}2,\\ -AD = 2\sin\frac{\alpha-\delta}2,\\ -\end{cases} -$$ -where we assume $2\pi\geq \alpha\geq \beta\geq \gamma\geq\delta\geq 0$, and by Pitot theorem: -$$(\star)\qquad\sin\frac{\alpha-\beta}2 + \sin\frac{\gamma-\delta}2 = \sin\frac{\beta-\gamma}2 + \sin\frac{\alpha-\delta}2.$$ -For the incircles coordinates we have -$$\begin{cases} -E = \tan(\frac{\pi}4-\frac{\alpha-\beta}4)\left(\cos\frac{\alpha+\beta}2,\sin\frac{\alpha+\beta}2 \right),\\ -F = \tan(\frac{\pi}4-\frac{\beta-\gamma}4)\left(\cos\frac{\beta+\gamma}2,\sin\frac{\beta+\gamma}2 \right),\\ -G = \tan(\frac{\pi}4-\frac{\gamma-\delta}4)\left(\cos\frac{\gamma+\delta}2,\sin\frac{\gamma+\delta}2 \right),\\ -H = \tan(\frac{\pi}4-\frac{\alpha-\delta}4)\left(\cos\frac{\delta+\alpha}2,\sin\frac{\delta+\alpha}2 \right) -\end{cases} -$$ -The points $E,F,G,H$ are concyclic iff -$$\det \begin{bmatrix} -OE^2 & E_x & E_y & 1\\ -OF^2 & F_x & F_y & 1\\ -OG^2 & G_x & G_y & 1\\ -OH^2 & H_x & H_y & 1 -\end{bmatrix} -= 0. -$$ -In our case it takes the form: -$$\det \begin{bmatrix} -\tan(\frac{\pi}4-\frac{\alpha-\beta}4) & \cos\frac{\alpha+\beta}2 & \sin\frac{\alpha+\beta}2 & \cot(\frac{\pi}4-\frac{\alpha-\beta}4)\\ -\tan(\frac{\pi}4-\frac{\beta-\gamma}4) & \cos\frac{\beta+\gamma}2 & \sin\frac{\beta+\gamma}2 & \cot(\frac{\pi}4-\frac{\beta-\alpha}4)\\ -\tan(\frac{\pi}4-\frac{\gamma-\delta}4) & \cos\frac{\gamma+\delta}2 & \sin\frac{\gamma+\delta}2 & \cot(\frac{\pi}4-\frac{\gamma-\delta}4)\\ -\tan(\frac{\pi}4-\frac{\alpha-\delta}4) & \cos\frac{\delta+\alpha}2 & \sin\frac{\delta+\alpha}2 & \cot(\frac{\pi}4-\frac{\alpha-\delta}4) -\end{bmatrix} -= 0, -$$ -which can be verified routinely under the condition $(\star)$. - -To verify the identity we can express everything in terms of $X:=e^{I\frac{\alpha-\beta}2}$, $Y:=e^{I\frac{\beta-\gamma}2}$, $Z:=e^{I\frac{\gamma-\delta}2}$, and $T:=e^{I\frac{\delta}2}$. In particular, we have $\sin\frac{\alpha-\beta}2 = \frac{X-X^{-1}}{2I}$, $\tan(\frac{\pi}4-\frac{\alpha-\beta}4) = \frac{(X-I)I}{X+I}$, $\cos\frac{\alpha+\beta}2 = \frac{X(YZT)^2 + X^{-1}(YZT)^{-2}}2$, and so on. -The following SageMath code verifies that the determinant as a rational function over variables $X,Y,Z,T$ reduces to $0$ w.r.t. the polynomial ideal defined by $(\star)$: -def row(t,z): - return [ (t-I)/(t+I)*I, (z + 1/z)/2, (z - 1/z)/2/I, (t+I)/(t-I)/I ] - -R. = PolynomialRing(QQ[I]) -J = ideal( numerator( (X - 1/X) + (Z - 1/Z) - (Y - 1/Y) - (X*Y*Z - 1/(X*Y*Z)) ) ) -M = matrix(Frac(R), 4, 4, [row(X,X*(Y*Z*T)^2), row(Y,Y*(Z*T)^2), row(Z,Z*T^2), row(X*Y*Z,X*Y*Z*T^2)] ) -print( J.reduce( numerator(det(M)) ) ) - -The code prints "0" (run it online), thus establishing the determinant identity.<|endoftext|> -TITLE: On the growth and bounds for a certain sequence of integers known as Bogotá numbers -QUESTION [6 upvotes]: A Bogotá number is a non-negative integer equal to some smaller number, or itself, times its digital product, i.e. the product of its digits. For example, 138 is a Bogotá number because 138 = 23 x (2 x 3). -Bogotá numbers up to 1000 are 0, 1, 4, 9, 11, 16, 24, 25, 36, 39, 42, 49, 56, 64, 75, 81, 88, 93, 96, 111, 119, 138, 144, 164, 171, 192, 224, 242, 250, 255, 297, 312, 336, 339, 366, 378, 393, 408, 422, 448, 456, 488, 497, 516, 520, 522, 525, 564, 575, 648, 696, 704, 738, 744, 755, 777, 792, 795, 819, 848, 884, 900, 912, 933, 944, 966, 992. -It has been shown that the natural density of Bogotá numbers is 0: https://math.stackexchange.com/questions/3713294/on-the-density-of-a-certain-sequence-of-integers. -The number of Bogotá numbers, B (n), less than or equal to n, for n = 10^ 0, 1, 2, 3..., 9 is 2, 4, 19, 67, 280, 1166, 4777, 19899, 82278, and 340649 as calculated by Freddy Barrera. -Crude estimates and bounds for the value of B (n) are not difficult to come by. How precise can we get? - -REPLY [3 votes]: Fix an integer $b > 1$ and let $p_b(n)$ denote the product of the base-$b$ digits of the integer $n$. Then let $\mathcal{B}_b$ be the set of numbers of the form $p_b(n)n$, for some integer $n \geq 0$, and put $\mathcal{B}_b(x) := \mathcal{B}_b \cap [0, x]$ for all $x > 0$. -The OP is asking for bounds for $\mathcal{B}_b(x)$ (in the special case $b=10$). -I guess that using the same methods of [1], one should be able to prove an upper of the form $x^{c_b + o(1)}$ for some constant $c_b$ depending on $b$. The details are many so I refer directly to the paper (the version on arXiv): - -For a parameter $\alpha > 0$, one splits $\mathcal{B}_b$ into two subsets: $\mathcal{B}_b^\prime$, consisting of the numbers with $p_b(n) > x^\alpha$, and $\mathcal{B}_b^{\prime\prime}$, the remaining numbers with $p_b(n) \leq x^\alpha$. -An upper bound for $\mathcal{B}_b^\prime(x)$ is given considering that each of its elements has a $b$-smooth divisor $> x^\alpha$ (pag. 3, (6) and the three equations before) -An upper bound for $\mathcal{B}_b^{\prime\prime}(x)$ is given by estimating a sum of multinomial coefficients (pag. 4-5, (11) and the two equations before). - -[1] C. Sanna, On numbers divisible by the product of their nonzero base b digits, Quaestiones Mathematicae (in press) https://arxiv.org/abs/1809.05463<|endoftext|> -TITLE: The Calkin representation for Banach spaces -QUESTION [6 upvotes]: Let $X$ be an infinite dimensional Banach space. Let $\Lambda_{0}$ be the set of all finite dimensional subspaces of $X$ directed by the inclusion $\subseteq$. For each $\alpha\in \Lambda_{0}$, let $I_{\alpha}:=\{\beta\in\Lambda_{0}:\alpha\subseteq \beta\}$. Then $\{I_{\alpha}:\alpha\in \Lambda_{0}\}$ is a filter basis and hence is contained in some ultrafilter $\mathcal{U}$. -For an infinite dimensional Banach space $Y$, let $(Y^{*})_{\mathcal{U}}$ be the ultrapower of $Y^{*}$ with respect to $\mathcal{U}$. Let $\widehat{Y}$ be the subspace of $(Y^{*})_{\mathcal{U}}$ defined by $$\widehat{Y}:=\{(y^{*}_{\alpha})_{\mathcal{U}}\in (Y^{*})_{\mathcal{U}}:w^{*}-\lim_{\mathcal{U}}y^{*}_{\alpha}=0\}.$$ For an operator $T:Y\rightarrow X$, we define $\widehat{T}:\widehat{X}\rightarrow \widehat{Y}$ by $\widehat{T}((x^{*}_{\alpha})_{\mathcal{U}})=(T^{*}x^{*}_{\alpha})_{\mathcal{U}}.$ It is easy to see that $\widehat{T}=0$ if $T$ is compact. -Question 1. Is $T$ compact if $\widehat{T}=0$? -Question 2. Let $K$ be a compact, convex and balanced subset of $B_{X}$ and let $\epsilon>0$. We set $A:=K+\epsilon B_{X}$ and define the gauge of $A$ by $$\|x\|_{A}:=\inf\{t>0:x\in tA\}, \quad x\in X.$$ -It is easy to see that $$\epsilon\|x\|_{A}\leq \|x\|\leq (1+\epsilon)\|x\|_{A}, \quad x\in X.$$ We set $Y:=(X,\|\cdot\|_{A})$ and let $j:Y\rightarrow X$ be the formal identity. Is there a constant $C$ such that $\|\widehat{j}\|\leq C\cdot \epsilon$? -Thanks! - -REPLY [3 votes]: The answer to question 2 is yes. -The proof you already know since you proved that $T$ compact implies $\hat{T}$ is zero. (For someone who has not thought about this, it is immediate from the elementary fact that a bounded net in $X^*$ that converges to zero weak$^*$ must converge uniformly to zero on compact subsets of $X$.) So if -$(x^*_\alpha)_\alpha$ is in $\widehat{X}$ -with $\sup \|x_\alpha^\alpha \| \le 1$ and -$x^*_\alpha \to 0$ weak$^*$, then -$x^*_\alpha \to 0$ -uniformly on $K$. Since the unit ball of $Y$ is contained in -$K+\epsilon B_{X}$, -it follows that -$\|\hat{j}\| \le \epsilon$, -so $C$ can be one.<|endoftext|> -TITLE: Is there a condensation from $\aleph_1^{\aleph_0}$ onto a metrizable compact space? -QUESTION [9 upvotes]: Is there a condensation (continuous bijective mapping) from $D^{\aleph_0}$ onto a metrizable compact space ? -$D$ - discrete space of cardinality $\aleph_1$. -CH implies it is a positive answer. In general, I don’t know the answer. - -REPLY [11 votes]: If $|D| < \aleph_\omega$, then there is a condensation from $D^\omega$ onto $\omega^\omega$ (the Baire space) if and only if there is a partition of $\omega^\omega$ into exactly $|D|$ Borel sets. -As far as I know, this theorem was first proved by me and Arnie Miller in - -"Partitions of $2^\omega$ and completely ultrametrizable spaces," Topology and its Applications 184 (2015), pp. 61-71. - -See Theorem 3.9. I still don't know what happens for $|D| \geq \aleph_\omega$. -It is a theorem of Hausdorff that $\omega^\omega$ can be partitioned into $\aleph_1$ Borel sets, regardless of whether $\mathsf{CH}$ holds or not. This, together with the theorem quoted above, provides a positive answer to your question. -It is consistent with any allowed value of $\mathfrak{c}$ that there is a partition of $\omega^\omega$ into $\kappa$ Borel sets for every $\kappa \leq \mathfrak{c}$. (This is Theorem 3.11 in the linked paper.) It is also consistent with any allowed value of $\mathfrak{c}$ that the only partitions of $\omega^\omega$ into Borel sets have size $\leq \aleph_0$, $\aleph_1$, and $\mathfrak{c}$ (see Corollary 3.16 in the linked paper), or size $\leq \aleph_0$, $\aleph_1$, $\kappa$, and $\mathfrak{c}$ for any particular $\kappa$ between $\aleph_1$ and $\mathfrak{c}$ (see Proposition 3.17 and the comments following). Generally, though, it's not known how to get some prescribed list of sizes and not others.<|endoftext|> -TITLE: Who first characterized the real numbers as the unique complete ordered field? -QUESTION [61 upvotes]: Nearly every mathematician nowadays is familiar with the fact that -there is up to isomorphism only one complete ordered field, the -real numbers. -Theorem. Any two complete ordered fields are isomorphic. -Proof. $\newcommand\Q{\mathbb{Q}}\newcommand\R{\mathbb{R}}$Let us observe first that every complete ordered field $R$ is -Archimedean, which means that there is no number in $R$ that is -larger than every finite sum $1+1+\cdots+1$. If there were such a -number, then by completeness, there would have to be a least such -upper bound $b$ to these sums; but $b-1$ would also be an upper -bound, which is a contradiction. So every complete ordered field is -Archimedean. -Suppose now that we have two complete ordered fields, $\R_0$ and -$\R_1$. We form their respective prime subfields, that is, their -copies of the rational numbers $\Q_0$ and $\Q_1$, by computing -inside them all the finite quotients -$\pm(1+1+\cdots+1)/(1+\cdots+1)$. This fractional representation -itself provides an isomorphism of $\Q_0$ with $\Q_1$, indicated -below with blue dots and arrows: - -Next, by the Archimedean property, every number $x\in\R_0$ -determines a cut in $\Q_0$, indicated in yellow, and since $\R_1$ -is complete, there is a counterpart $\bar x\in\R_1$ filling the -corresponding cut in $\Q_1$, indicated in violet. Thus, we have -defined a map $\pi:x\mapsto\bar x$ from $\R_0$ to $\R_1$. This map -is surjective, since every $y\in\R_1$ determines a cut in $\Q_1$, -and by the completeness of $\R_0$, there is an $x\in\R_0$ filling -the corresponding cut. Finally, the map $\pi$ is a field -isomorphism since it is the continuous extension to $\R_0$ of the -isomorphism of $\Q_0$ with $\Q_1$. $\Box$ -My expectation is that this theorem is familiar to almost every -contemporary mathematician, and I furthermore find this theorem -central to contemporary mathematical views on the philosophy of -structuralism in mathematics. The view is that we are entitled to -refer to the real numbers because we have a categorical -characterization of them in the theorem. We needn't point to some -canonical structure, like a canonical meter-bar held in some -special case deep in Paris, but rather, we can describe the -features that make the real numbers what they are: they are a -complete ordered field. -Question. Who first proved or even stated this theorem? -It seems that Hilbert would be a natural candidate, and I would -welcome evidence in favor of that. It seems however that Hilbert -provided axioms for the real field that it was an Archimedean -complete ordered field, which is strangely redundant, and it isn't -clear to me whether he actually had the categoricity result. -Did Dedekind know it? Or someone else? Please provide evidence; it -would be very welcome. - -REPLY [38 votes]: Joel, I believe this was first explicitly stated and proved by E.V. Huntington in his classic paper: COMPLETE SETS OF POSTULATES FOR THE THEORY OF REAL QUANTITIES, Trans. Am. Math. Soc. vol. 4, No. 3 (1903), pp. 358-370. See Theorem II', p. 368. -Edit (June 14, 2020): It is perhaps worth adding that in 1904, the year following the publication of Huntington's paper, O. Veblen published his paper A System of Axioms for Geometry, Trans. Am. Math. Soc. vol. 5, no. 3, pp. 343-384, in which he introduced the idea of a categorical system of axioms. He illustrated his conception with Huntington's above mentioned characterization of the reals (pp. 347-348). No doubt, this is mentioned in the paper referred to below by Ali Enayat.<|endoftext|> -TITLE: Are linear continuous mappings open on totally bounded sets? -QUESTION [5 upvotes]: Let $X$ and $Y$ be locally convex spaces, and $\varphi: X\to Y$ a linear continuous mapping. Suppose first that $S$ is a compact set in $X$. Then $\varphi$, being considered as a mapping from $S$ to $\varphi(S)$, -$$ -\varphi\Big|_S:S\to \varphi(S) -$$ -is open in the sense that for any open set $U$ in $S$ (with respect to the topology induced from $X$) its image $\varphi(U)$ is an open set in $\varphi(S)$ (with respect to the topology induced from $Y$). -This is strange, I was sure that the same must be true if $S$ is not necessarily compact, but just totally bounded (since we can consider the extension of $\varphi$ to the completions of the spaces $X$ and $Y$), but recently I understood unexpectedly that I can't write an accurate proof. Does this mean that there is a counterexample? -So my question: - -Is $\varphi\Big|_S:S\to \varphi(S)$ an open mapping for each totally bounded set $S$ in $X$? - -REPLY [7 votes]: Bounded sets, e.g., unit balls in normed spaces, are always totally bounded for weak topologies. So you only have to find such a ball with two incompatible weak topologies, which is easy to do—say the ball of a suitable dual space with the weak and the weak $\ast$ topology.<|endoftext|> -TITLE: What is the best way to partition the $4$-subsets of $\{1,2,3,\dots,n\}$? -QUESTION [5 upvotes]: Also asked on MSE: What is the best way to partition the $4$-subsets of $\{1,2,3,\dots,n\}$?. -Consider the set $X = \{1,2,3,\dots,n\}$. Define the collection of all $4$-subsets of $X$ by $$\mathcal A=\{Y\subset X: Y\text{ contains exactly $4$ elements}.\}$$ -I want to partition $\mathcal A$ into groups $A_1,A_2,\dots, A_m\subset \mathcal A$ (each of them is a collection of $4$-subsets of $X$) such that $\bigcup_{i=1}^m A_i=\mathcal A$ and such that the intersection of any two distinct $4$-subsets in each $A_k$ has cardinality at most $1$, i.e. such that for all $i\in\{1,\dots,m\}$ and $Y_1, Y_2\in A_i$, we have $$Y_1\neq Y_2 \implies \lvert Y_1\cap Y_2\rvert \le 1.$$ -My question: What can be said about the smallest $m$ (depending on $n$) such that such a partition exists? - -My thoughts: I was expecting that each $A_i$ can contain "roughly" $\frac n4$ elements, so we would have $$m(n)=\Theta\left(\frac{\binom n4}{\frac n4}\right)=\Theta(n^3).$$ In particular, we would have $m(n)\le c n^3$ for some constant $c\in\mathbb R$. -However, I am neither sure if this is correct, nor how to formalize this. - -REPLY [7 votes]: This problem can be reformulated in terms of graph coloring: -Let the graph $G=(V,E)$, $V=\mathcal A$, $(x,y)\in E \leftrightarrow x \cap y \geq 2$ -Then a partition of $\mathcal{A}$ into groups $A_1,A_2,…,A_m$ corresponds to a $m$-coloring of $G$. -The graph $G$ has degree no more than $6{n\choose 2}$, so $G$ can be colored in no more than $6{n\choose 2}+1$ colors by Brooks' theorem. -A lower bound of $\chi$ is presented in Max Alekseyev's comment, which can be interpreted as a clique of $G$.<|endoftext|> -TITLE: For which ordinals do we have $V_\alpha = L_\alpha$? -QUESTION [6 upvotes]: Some elements of $L$ become constructible only in levels higher than its rank level. So I ask: -Let $V$ be such that $V = L$. -For which ordinals $\alpha$ do we have $V_\alpha = L_\alpha$? -Indeed, we have this for $\omega$ and if $\alpha$ is the class of all ordinals. But do we have this for other types of ordinals? Can we, for instance, have $V_\alpha = L_\alpha$ for a singular cardinal? - -REPLY [16 votes]: Yes; in fact the first $\alpha>\omega$ with $V_\alpha=L_\alpha$ has cofinality $\omega$. To obtain this $\alpha$, define $f:Ord\to Ord$ by $f(\xi)=$ the smallest $\eta$ such that $V_\xi\subseteq L_\eta$. Such an $\eta$ exists because of the assumption that $V=L$. Write $f^n$ for the $n$-fold iterate of $f$. Then $\sup_{n\in\omega}f^n(\omega+1)$ is the desired $\alpha$. -More generally, the ordinals you asked about are exactly the fixed points of $f$. Since $f$ is a normal function (increasing and continuous), its fixed points constitute a closed unbounded class of ordinals.<|endoftext|> -TITLE: A natural $\mathbb Q\times \mathbb P$ subset of $\mathbb R$? -QUESTION [9 upvotes]: I would like a simple description of a dense subset of $\mathbb R$ which is homeomorphic to $\mathbb Q\times \mathbb P$. Preferably the description will be of an algebraic nature, and perhaps the set will even be a subgroup of $\mathbb R$. Here $\mathbb R$ is all real numbers, $\mathbb Q$ is all rational numbers, and $\mathbb P$ is all irrational numbers. -The following characterization by Jan van Mill may be useful: $\mathbb Q\times \mathbb P$ is the unique zero-dimensional separable metrizable space which is strongly $\sigma$-complete, nowhere complete, and nowhere $\sigma$-compact. - -REPLY [4 votes]: I made a comment earlier, but let me try converting it to an answer. It's similar in flavor to Ivan's. -By a back and forth argument, all countable dense subsets of $\mathbb{R}$ are homeomorphic to $\mathbb{Q}$, which allows one to replace $\mathbb{Q}$ with the space of quadratic irrationals, which I'll denote by $Q'$. The (regular) continued fraction expansions of elements of $Q'$ are precisely infinite continued fractions that are eventually periodic. By taking continued fractions, we have a homeomorphism $\mathbb{Z} \times \mathbb{N}^\mathbb{N} \cong \mathbb{P}$ defined by -$$(a_0, a_1, a_2, \ldots) \mapsto a_0 + \frac1{a_1 + \frac1{a_2 + \ldots}}$$ -and from there we easily get a homeomorphism $\mathbb{N}^\mathbb{N} \cong \mathbb{P}$. Since $\mathbb{N}^\mathbb{N}$ is homeomorphic to its square via the interleaving map -$$((a_0, a_2, \ldots), (a_1, a_3, \ldots)) \mapsto (a_0, a_1, a_2, a_3, \ldots)$$ -we get a homeomorphism $\mathbb{P} \times \mathbb{P} \to \mathbb{P}$ by interleaving continued fractions. The subset $Q' \times \mathbb{P}$ maps homeomorphically onto its image under this map, and this image is of course dense (it contains for example the dense set $Q'$ of numbers with eventually periodic cf's).<|endoftext|> -TITLE: How to understand the "boundary" of subscheme, as defined in "An elementary characterisation of Krull dimension" -QUESTION [5 upvotes]: In An elementary characterisation of Krull dimension and A short proof for the Krull dimension of a polynomial ring, Coquand, Lombardi, and Roy give an elementary characterization of Krull dimension, which inductively makes use of one of two notions of the "boundary" of a subvariety, given as follows: -Let $R$ be a commutative ring, and $x\in R$. -\begin{align*} -& \operatorname{upper boundary} R^{\{x\}} \mathrel{:=} R/I^{\{x\}}, -&& I^{\{x\}} \mathrel{:=} xR + (\sqrt{0}:x) \\ -& \operatorname{lower boundary} R_{\{x\}} \mathrel{:=} S_{\{x\}}^{-1}R, -&& S_{\{x\}} \mathrel{:=} x^{\mathbb{N}}(1+xR) -\end{align*} -where $(\sqrt{0}:x)$ is the ideal quotient of the nilradical, and $x^{\mathbb{N}}(1+xR) = \{x^n(1+rx) \mathrel\vert \text{$n\in\mathbb{N}$, $r\in R$}\}$. -Upon inspection, $\mathrm{Spec}(R^{\{x\}})$ is $V(x) \cap \overline{\mathrm{Spec}R\setminus V(x)}$, and $\mathrm{Spec}(R_{\{x\}})$ is a localization (not quite open) that is disjoint from the locus $V(x)$. Also, both are trivial exactly when $x\in R^\times \cup \sqrt{0}$. -However, I do not have good intuition for these subschemes. - -How to think about these boundary schemes? Do they represent anything in particular? - -Do these constructions appear anywhere else in the literature? I have not been able to find anything. - -Are they commutative, in that $R^{\{x\}\{y\}} = R^{\{y\}\{x\}}$ and $R_{\{x\}\{y\}} = R_{\{y\}\{x\}}$? - - -I suspect they are commutative, but am unable to prove it, and I have reservations stemming from the fact that permutations of a regular sequence are not necessarily regular. - -Are these very natural constructions? I.e. would it be worth studying them in more detail, in specific cases, or are they primarily instrumental in the characterization of Krull dimension? - -I am willing to restrict to cases where $R$ is integral and Noetherian or has a finitely generated function field. -It seems best to consider first the $R^{\{x_0\}...\{x_k\}}$, where $x_0, ..., x_k$ form a regular sequence, but I was not able to get much further with this assumption. - -REPLY [3 votes]: This is also not really an answer, but it explains an alternative version of the characterisation that I find easier to work with. Let $P_d(R)$ be the polynomial ring over $R$ in variables $x_0,\dotsc,x_d$, and order the monomials lexicographically. Say that $f\in P_d(R)$ is comonic if the lowest monomial has coefficient one. Say that an $R$-algebra homomorphism $\phi\colon P_d(R)\to R$ is thin if the kernel contains a comonic polynomial. After a little translation, the results of Coquand and Lombardi say that $R$ has dimension $\leq d$ iff every such homomorphism is thin.<|endoftext|> -TITLE: Hamming distance to primes -QUESTION [14 upvotes]: There is a positive density of odd numbers which are of the form $2^n+p$ (due to Romanoff), and a positive density which are not of this form (due to van der Corput and Erdos, see this paper for a review and some results on the density). So, for some but not almost all odd numbers, we can get to a prime by subtracting a power of two. -I'm curious about a related question: given an odd integer $m$, is there always a prime number with Hamming distance 1 to $m$? For example, $127 = 1111111_2$ is not of the form $2^n+p$, but it has Hamming distance 1 to a prime, since $383 = 101111111_2$ is prime. -A related question, which implies the first: given an odd integer $m$, does the set $\{m+2^n\mid n\in \mathbb{N}\}$ contain infinitely many primes (or at least one for which $2^n>m$, so that this corresponds to flipping a bit in $m$)? - -REPLY [24 votes]: See OEIS sequences A067760 and A076336. If $n$ is a dual Sierpiński number, there is no $k$ such that $n+2^k$ is prime. There is no prime with Hamming distance $1$ to the Sierpiński number $2131099$, and this may be the least positive integer with this property.<|endoftext|> -TITLE: Every mathematician has only a few tricks -QUESTION [156 upvotes]: In Gian-Carlo Rota's "Ten lessons I wish I had been taught" he has a section, "Every mathematician has only a few tricks", where he asserts that even mathematicians like Hilbert have only a few tricks which they use over and over again. -Assuming Rota is correct, what are the few tricks that mathematicians use repeatedly? - -REPLY [2 votes]: Taylor expansion. Much of the classical theory of statistics (and some of its modern extensions) revolves around performing a second-order Taylor expansion of the likelihood.<|endoftext|> -TITLE: Players alternate moving a $\{\swarrow,\uparrow,\rightarrow\}$ piece on a chessboard -QUESTION [27 upvotes]: Edit $4.$ $-$ Proposing to reopen the question (the related competition should be over by now). -Edit $3.$ $-$ I have just found out that the linked competition (see the "Edit $1$.") is still ongoing. Please close and hide this question and its counterpart (see the "Remark."), until the competition is over. -Edit $2$. $-$ I've added strategies for solving some simple squares (tiles). -Edit $1$. $-$ This problem is equivalent to the problem $20.\space b)$ from Завдання ХХIIІ ТЮМ (2020 р.) which talks about the divisors of $10^{2k}$ where legal moves are $\{\div 10, \times 2,\times 5\}$ to move from the previous divisor to the next one, and you can't revisit a divisor. But, it appears they did not publish the solutions. $-$ Thank you Witold for the reference. -Remark. This problem is equivalent to the even case of the exponent $n=2k$ of the following problem: Winning strategy in a game with the positive divisors of $10^n$ from MSE. (This is a partial cross-post.) - - -The $\{\swarrow,\uparrow,\rightarrow\}$ piece game - -Consider an odd sized chessboard $(2k+1)\times(2k+1)$ with WLOG bottom -left corner square at $(0,0)$ and top right corner square at -$(2k,2k)$. A piece whose allowed moves are $\{(-1,-1),(0,+1),(+1,0)\}$ -is placed on one of the squares. -Two players then alternate moving the piece, such that the piece never -stands on the same square twice. (The piece can't revisit the -squares.) The player that can't move the piece, loses. (The player -that last moved, wins.) -On which starting squares will the first player have a winning -strategy? - -I have brute-forced the game for boards of sizes $(2k+1)=3,5,7,9,11$ with C++ (run it on repl.it). If the second player has a winning strategy, the square is colored blue. Otherwise, the first player has a winning strategy and the square is colored green. - -It seems that if the piece starts on a square $(x,y)$ where $x,y$ are both even, then the second player can always force a win. Can we prove this? -Otherwise, if $x,y$ are not both even, then it seems the first/green player can force a win, unless the square is one of the "exceptions". (Additional blue squares.) -WLOG due to symmetry we list only the exceptions below the main diagonal: - -If $n\le2$ there are no exceptions. -If $n=3$, the exceptions below the diagonal are $(4,3),(6,3)$. -If $n=4$, the exceptions below the diagonal are $(8,3),(5,4),(8,5)$. -If $n=5$, the exceptions below the diagonal are $(6,1),(7,4),(6,5),(10,5),(7,6)$. - -But what will be the exceptions for $n\ge 6$? I do not see a pattern. - - -I do not know how to solve all squares in general, but I can solve specific examples. -Let $n=2k$ and consider some $(n+1)\times(n+1)$ chessboard. -$i)$ Solving the corners -It is easy to show that if the piece starts on any of the four corners $(0,0),(n,0),(0,n),(n,n)$ then the second player has a winning strategy. - - -Starting on $(0,0)$, the second player can keep returning to the main diagonal until the first player can no longer move. ("The main diagonal climber".) - -Starting on $(n,n)$, the first player is forced to move diagonally. Then, the second player can keep forcing the diagonal move by repeating either the $(+1,0)$ or the $(0,+1)$ move until the first player can no longer move. ("The wall crawler".) - -Starting on $(0,n)$ we have a forced $(+1,0)$ move for the first player. This can be forced again and again, by responding with the $(-1,-1)$ move, until we reach $(0,0)$. Now the moves to $(1,0),(2,0)$ are forced. Finally, the second player can keep returning to the corresponding diagonal until the piece reaches $(n,n)$ where we apply "The wall crawler". - -Strategy for starting on $(n,0)$ is symmetric to the previous corner $(0,n)$. - - -$ii)$ Extending the solutions around the $(0,0)$ corner -It can be shown that $(0,0),(2,0),(0,2),(2,2)$ are always a win for the second player, and that $(1,0),(0,1),(1,1),(2,1),(1,2)$ are always a win for the first player. - - -When starting on $(0,2)$ or $(2,0)$, the second player can stay on the diagonal to force the piece to reach $(n,n)$ where they apply "The wall crawler" to beat the first player. Consequently, if we start on $(0,1)$ or $(1,0)$ then the first player can move to $(0,2)$ or $(2,0)$ respectively, and apply the same strategy to beat the second player. - -When starting on $(2,2)$, the first player must use the diagonal move. (Otherwise, the second player can apply "The main diagonal climber".) But, then the second player moves to $(0,0)$ and forces the first player to move to either $(1,0)$ or $(0,1)$. This is losing for the first player because then the second player moves to $(2,0)$ or $(0,2)$ respectively, and applies the previous winning strategy. - -Consequently (to the previous strategy), we have that $(1,1)$ is a win for the first player. - -Consequently (to the previous bullet points), we have that the $(1,2)$ and $(2,1)$ are wins for the first player if they decide to move to the $(2,2)$ square. - - -It looks like extending these strategies to $(4,4)$ is not as simple as the previous extension to $(2,2)$, because the $(3,4),(4,3)$ squares depend on $(n+1)$ (the chessboard dimension) itself. These two squares appear to be a win for the first player, unless the dimension is $(2k+1)=7$ where they are a win for the second player. (According to my brute force solutions.) -That is, the general strategy needs to account for the observed "exceptions". -Remark. Although the question is about the odd sizes of the chessboard, the strategies mentioned above work for both the odd and the even dimensions of the chessboard. If we want to extend these strategies further (to more squares), we do need to also account for the parity of the dimension. (According to my brute force solutions.) - -REPLY [3 votes]: Problems of this contest are given to the school students for long-term work, and work is in the progress. The final stage of the competition is planned on October 2020. Please, remove the solution of this problem from the site (or hide until November 2020).<|endoftext|> -TITLE: Universal property of induced representation -QUESTION [6 upvotes]: Let $H$ be a closed subgroup of the compact Lie group $G$. Let $E$ be a continuous representation of $H$. In the book "Representations of compact Lie groups" by Bröcker and Dieck the induced representation of $E$ is defined as the vector space $iE$ of all continuous functions $f:G\to E$ satisfying $f(g\cdot h)=h^{-1}f(g)$ for all $g\in G$ and $h\in H$. They show that as in the finite case this construction satisfies the Frobenius reciprocity theorem. -Now I wonder whether this construction also satisfies the universal property that we know from the case of finite groups (or, more generally, finite index), i.e., my question is whether the following is true: - -There exists an $H$-linear map $j:E\to iE$ such that for all -$H$-linear maps $g:E\to E'$ to a $G$-module $E'$ there is a unique -map $G$-linear map $g':iE\to E'$ such that $g'\circ j=g$. - -Moreover, is $g'$ continuous if $g$ is? If the answer is "No", is there a better notion of induced representation that makes this true? Or does it help when we restrict to unitary representations? - -REPLY [2 votes]: Your question can be rephrased as "When is the induction the same as coinduction?" This has appeared on MathOverflow before and fancy answer to your question can be found here: When are induction and coinduction of representations of Lie groups isomorphic? When they are compact? Semisimple? -See also Induction and Coinduction of Representations -A direct elementary proof could be perhaps gleaned from https://math.stackexchange.com/questions/225730/left-adjoint-and-right-adjoint-nakayama-isomorphism/226493#226493 as it mentions averaging over group which works equally well for compact Lie groups as it does for finite groups.<|endoftext|> -TITLE: Why does this construction give a (homotopy-invariant) suspension (resp. homotopy cofiber) in an arbitrary pointed model category? -QUESTION [6 upvotes]: In their text Foundations of Stable Homotopy Theory, Barnes and Roitzheim define the suspension of a cofibrant object X of a pointed model category to be the pushout of the diagram $*\leftarrow X\coprod X\to Cyl(X)$, where the second map is the structure map of the cylinder object. By contrast, there is a more manifestly homotopy-invariant definition of suspension given in e.g. Dwyer and Spalinski, which is the homotopy pushout of the diagram $*\leftarrow X\to *$. It is not clear to me why these definitions agree; if we don't assume properness, I don't even see why the first is homotopy-invariant! (If we assume the model category is proper, then the first diagram's pushout is equal to its homotopy pushout.) There is a similar issue with the cofiber, which they define for a cofibration of cofibrant objects $f:A\to X$ as the pushout of $*\leftarrow A\to X$: again, it is not clear why this is homotopy-invariant (with respect to maps between such $f$ in the comma category) unless the model category is proper. Can we drop the properness assumption and still get homotopy colimits or at least homotopy invariance? Even if so, why are the definitions of suspension equivalent? - -REPLY [7 votes]: An argument showing that the two models of suspension are equivalent will probably be based on something like the following: -Assertion: Suppose we are given a commutative diagram of the form -$\require{AMScd}$ -\begin{CD} -\ast @<<< C @= C \\ -@VVV @VVV @VV V \\ -Y @<<< A @>g>> X \\ -@| @VVV @VVV\\ -Y @<<< A/C @>>h > X/C -\end{CD} -in which the vertical directions form cofibration sequences (when I write $A/C$, I mean $A \amalg_C \ast$, where $\ast$ is the zero object), and the maps $g$ and $h$ are cofibrations. -Then the map of pushouts -$$ -Y \cup_A X \to Y \cup_{A/C} X/C -$$ -is a weak equivalence, or better still, it is an isomorphism. -It seems to me that this is true by the assumption of properness, since we have a cofibration sequence given by the pushouts -$$ -\ast\cup_C C \to Y \cup_A X \to Y \cup_{A/C} X/C -$$ -in which the first term is isomorphic to $\ast$. -Let's call the first suspension $SX$ and the second one $\Sigma X$. -Given the assertion, we can show that the two models for suspension are weakly equivalent as follows: -Apply the assertion to the diagram -\begin{CD} -\ast @<<< \ast\amalg X @= X \\ -@VVV @VVV @VVV \\ -\ast @<<< X\amalg X @>g >> \text{Cyl}(X) \\ -@| @VVV @VVV\\ - \ast @<<< X @>>h > CX -\end{CD} -(where $CX = \text{Cyl}(X)/X$) -to get that the map -$$ -SX\to \Sigma X -$$ -is a weak equivalence.<|endoftext|> -TITLE: Conic sections are to cones as quadric surfaces are to what? -QUESTION [5 upvotes]: I just started my Calculus 3 class, which deals mainly with multi-variable Calculus. We went over the different types of 3-D surfaces, mainly Quadric Surfaces (ellipsoids, paraboloids, hyperboloids, etc.) which according to the textbook are "3D analogs of conic sections". I know for conic sections you slice cones with a plane to get ellipses, parabolas, hyperbolas, etc, so I was wondering what is the equivalent for the sliced cone when dealing with these kind of surfaces? How is such a body and the operation of "slicing" it described? And can this be generalized for higher dimensions? -I didn't quite know how to word the problem in a sentence or what field of math it's part of to do a quick google search so if anyone could point me in the right direction I would appreciate it. -Thanks - -REPLY [6 votes]: The thing that makes quadric surfaces "3D analogs of conic sections" is just that they are defined by a single equation of degree 2. It's not a particularly helpful characterization though, I would say. It strikes me more as something a pedagogue would say in a (poor) attempt to relate a new concept to one already known. -[One could see quadric surfaces as "slices" of a certain geometric object, analogously to conic sections, but only if you are only interested in them up to isomorphism (as algebraic varieties). Then a quadric surface can be regarded as a hyperplane section of the image $V \subset \mathbb{P}^9$ of the Veronese embedding $\mathbb{P}^3 \to \mathbb{P}^9$. Note that this is only partially analogous to the representation of a conic section (considered as a plane curve) as the intersection of cone and plane, since the conic section is related to the aforementioned intersection by a projective transformation, which is a lot stronger than saying they are isomorphic as algebraic varieties.]<|endoftext|> -TITLE: Intrinsic characterisation of a class of rings -QUESTION [5 upvotes]: This may be well known, but I was unable to find an answer browsing literature. Let us temporarily call a commutative (unital) ring $R$ an O-ring if there exists an integer $n \ge 1$, a local field of zero characteristic (that is, a finite extension of $ \mathbb{Q}_p$ for some prime $p$) with ring of integers $ \mathcal{O}$ and the unique maximal ideal $\mathfrak{p}$ such that $R$ is isomorphic (as a ring) to $ \mathcal{O}/\mathfrak{p}^n$. Now, it is clear that an O-ring is a finite local ring. It is also easy to see that not all local rings arise in this way. My question is: is there a purely ring-theoretic way of characterising O-rings, without making any reference to local fields at all? I would appreciate any reference to the literature as well. - -REPLY [6 votes]: The following criterion came up when I was writing this answer (but I did not end up using it there): - -Lemma. Let $R$ be a commutative ring. Then $R$ is of the form $\mathcal O_K/\mathfrak p^n$ for a finite extension $\mathbf Q_p \subseteq K$ and $n \in \mathbf Z_{>0}$ if and only if $R$ is finite, local, and $\dim_{R/\mathfrak m} \mathfrak m/\mathfrak m^2 \leq 1$. - -Proof. Clearly any $R$ of the form $\mathcal O_K/\mathfrak p^n$ is finite, local, and has $\dim_{R/\mathfrak m}\mathfrak m/\mathfrak m^2 \leq 1$ (with equality if and only if $n > 1$). Conversely, suppose $R$ is finite, local, and has $\dim_{R/\mathfrak m} \mathfrak m/\mathfrak m^2 \leq 1$. Write $k = R/\mathfrak m$, and set $p = \operatorname{char} k$ and $q = |k|$, so that $k = \mathbf F_q$ with $q = p^r$ for some $r \in \mathbf Z_{>0}$. Write $\mathbf Z_q = W(\mathbf F_q)$ for the Witt vectors (the unique unramified extension of $\mathbf Z_p$ of degree $r$), which is a Cohen ring for $k$. -If $t \in \mathfrak m$ is a generator, then (the proof of) the Cohen structure theorem (Tag 032A) constructs a surjection -$$\phi \colon \mathbf Z_q[[t]] \to R$$ -taking $t$ to $t$. Let $n = \operatorname{length}(R)$, so that $R \supsetneq \mathfrak m \supsetneq \ldots \supsetneq \mathfrak m^n = 0$, where $\mathfrak m^i$ is generated by $t^i$ for all $i$. Let $e \in \{1,\ldots,n\}$ be the integer such that $(p) = \mathfrak m^e$. Then there exists $u \in \mathbf Z_q^\times$ such that $\phi(up) = \phi(t^e)$, i.e. $t^e-up \in \ker\phi$. Thus, $\phi$ factors through -$$\mathbf Z_q[[t]] \twoheadrightarrow \mathbf Z_q\big[\sqrt[e\ \ ]{up}\big] \twoheadrightarrow R,$$ -which realises $R$ as $\mathcal O_K/\mathfrak p^n$ where $K = \mathbf Q_q\big(\sqrt[e\ \ ]{up}\big)$ (and $n = \operatorname{length}(R)$ as above). $\square$ -Remark. So in fact, it sufficies to take $K$ of the form $\mathbf Q_q\big(\sqrt[e\ \ ]{up}\big)$.<|endoftext|> -TITLE: Why did mathematical notation stay so hard to read? -QUESTION [5 upvotes]: One of the first things you learn in a programming 101 course is to write readable code, and to name your variables properly. This notion has seemingly never translated into mathematics. Everywhere you look, there are one letter constants, variables and functions, and an abundance of hard to remember symbols for operators, crammed together into tight, linearly laid out expressions. Characters are often borrowed from Greek, being short of Latin ones. -As soon as you get into college-level mathematics, any non-trivial mathematical expression starts to look like signal noise that programmers would instantly ridicule if it were a programming language. -Was there ever a movement to make mathematics more readable? Is being so succinct really worth it? Does it get better with enough experience? Would using memorable names for mathematical symbols and operators have any downside aside from length? Would a neatly indented, airier layout/syntax for expressions? -I'm not trying to stir up an argument, I'm genuinely curious, having been frustrated by this for a long time, and I'd be keen to hear what actual mathematicians think about this. - -REPLY [5 votes]: Mathematics uses very few variable names in any proof compared to the number of variable names occuring in typical programming languages. Variable names survive only for short passages, except for a small (less than a dozen) global variables. The names are subject to a host of conventions (for example, $\varepsilon$ is a small number, $N$ is a large number) which are not found in programming languages. More distinct symbols are available, as we are not constrained to use ASCII. Try to rewrite a serious piece of mathematics in the style of a programming language, and you will quickly see that it is unreadable.<|endoftext|> -TITLE: Tight sequence of measures -QUESTION [5 upvotes]: This is probably a very easy question for experts in probability or measure theory. -I have a sequence of finite measures $\mu_{n}$ on a non-compact metric space $X$ such that $\mu_{n}$ converges to $\mu$ in the following sense: -$$ \int_{X}fd\mu_{n} \to \int_{X}fd\mu \ \ \ \ \ \text{ for all f continuous with compact support} -$$ -I would like to say that $\mu_{n}(X)\to \mu(X)$. -I know this is false in general, but I have the additional condition that for every $\epsilon>0$ there is $n_{0}\in \mathbb{N}$ and $K\subset X$ compact such that $\mu_{n}(K^{c})\leq \epsilon$ for every $n\geq n_{0}$. This looks very similar to the definition of tight sequence (which guarantees the result I would like). Is this equivalent? -Additional assumptions: X is Polish and locally compact, precisely it is a closed surface with some finitely many points removed. All measures $\mu_{n}$ and $\mu$ are area measures of Riemannian metrics (with singularities at the points removed) on X. - -REPLY [4 votes]: This is really just a comment to add context to your question and the reactions but it will be too long so I am disguising it as an answer. Stripping away all trappings the relevant fact is that an equicontinuous set $A$ in the dual of a locally convex space $E$ is relatively compact for the corresponding weak topology and so that latter coincides on $A$ with the weak topology induced by any dense subspace $E_0$ of $E$. In your case, $E$ is the space of bounded, continuous functions on a locally compact space, provided with the so-called strict topology which was introduced by R.C. Buck in the 50‘s. This has the following three relevant properties which set up the connection to your question. -The dual space is the space of finite tight (or Radon) measures; -A family of measures is equicontinuous if and only if it is bounded and uniformly tight; -The space $E_0$ of continuous functions with compact supports is dense. -This shows that many of the assumptions given above are irrelevant. One can even replace local compactness by tcomplete regularity. The former condition is required to ensure the denseness of $E_0$. In the general case, one can use any subalgebra of $E$ which separates points and is such that for each element of the underlying space there is a function in $E_0$ which does not vanish there.<|endoftext|> -TITLE: Is this formal noncommutative power series identity known? -QUESTION [29 upvotes]: I recently discovered the following cute formal noncommutative power series identity: if $(x_i)_{i \in I}$ is some finite collection of noncommuting variables, then the formal power series -$$ 1 + \sum_{m=1}^\infty \sum_i x_i^m = 1 + \sum_i x_i+ \sum_i x_i^2 + \sum_i x_i^3 + \dotsb$$ -is the reciprocal of the formal power series -\begin{multline*} -\sum_{k=0}^\infty (-1)^k \sum_{i_1 \neq \dotsb \neq i_k} x_{i_1} \dotsb x_{i_k} \\ -= 1 - \sum_i x_i + \sum_{i \neq j} x_i x_j - \sum_{i \neq j \neq k} x_i x_j x_k + \dotsb -\end{multline*} -where summation indices are understood to range in $I$ if not otherwise specified. (Note that we do not require the $i_1,\dots,i_k$ to all be distinct from each other; it is only consecutive indices $i_j, i_{j+1}$ that are required to be distinct. So this isn't just the Newton identities relating power sums with elementary symmetric polynomials, though it seems to be a cousin of these identities.) -For instance, if $\lvert I\rvert=n$ and $x_i=x$, this identity amounts (after summing the geometric series) to the (formal) assertion -$$ \left(1 + \frac{nx}{1-x}\right)^{-1} = 1 - \frac{nx}{1+(n-1)x}$$ -which follows from high school algebra. -Once written down, the general identity is not hard to prove: multiply the two power series together and observe that every non-constant term with a coefficient of $+1$ is cancelled by a term with a coefficient of $-1$ and vice versa. But I am certain that an identity this basic must already be in either the enumerative combinatorics or the physics literature (EDIT: it is very implicitly in the free probability literature, which is how I discovered it in the first place, but to my knowledge it is not explicitly stated there). Does it have a name, and where is it used? Presumably there is also some natural categorification (or at least a bijective or probabilistic proof). - -REPLY [12 votes]: OK, here's an excessively long expansion of my comment. -According to Goulden and Jackson's Enumerative Combinatorics (p. 76), the commutative version of the original formula is due to MacMahon, though they only refer to his book Combinatory Analysis and don't give a more specific reference. I was not able to find this formula in MacMahon's book, but on pages 99–100 of Volume I (Section III, Chapter III) MacMahon gives the related generating function (in modern notation) -$$\frac{1}{1-e_2 -2e_3 -3e_4-\cdots},$$ -for counting derangements of a multiset. Here $e_i$ is the $i$th elementary symmetric function. (MacMahon uses $p_i$ for the elementary symmetric function, which is the modern notation for the power sum symmetric function.) It's not hard to show that (with $p_i$ the power sum symmetric function $x_1^i+x_2^i+\cdots$) we have -$$ -\frac{1}{1-p_1+p_2-\cdots} = \frac{1+e_1+e_2+\cdots}{1-e_2 -2e_3 -3e_4-\cdots}. -$$ -A combinatorial interpretation of the connection between these two generating functions has been given by J. Dollhopf, I. Goulden, and C. Greene, Words avoiding a reflexive acyclic relation, Electronic J. Combin. 11, no. 2, 2004–2006. -Words with adjacent letters different were called waves by L. Carlitz, who gave the (commutative) generating function for them in Enumeration of sequences by rises and falls: a refinement of the Simon Newcomb problem, Duke Math. J. 39 (1972), 267–280. This is probably the first appearance of the generating function, unless it's hiding somewhere in MacMahon. (Carlitz actually solved the more general problem of counting words by rises, falls, and levels.) Nowadays words with adjacent letters different are usually called Smirnov words or Smirnov sequences. This term was introduced by Goulden and Jackson; apparently Smirnov suggested the problem of counting these words though it's not clear that he did anything to solve the problem. According to the review in Mathematical Reviews of O. V. Sarmanov and V. K. Zaharov, A combinatorial problem of N. V. Smirnov (Russian), -Dokl. Akad. Nauk SSSR 176 (1967) 530–532 (I didn't look up the actual paper), -“The late N. V. Smirnov posed informally the following problem from the theory of order statistics: Given $n$ objects of $s+1$ distinct types (with $r_i$ objects of type $i$, $r_1+\cdots+r_{s+1}=n$), find the number of ways these objects may be arranged in a chain, so that adjacent objects are always of distinct types.” -When considered as compositions, i.e., when the entries are added together, Smirnov words are often called Carlitz compositions, as they were studied from this point of view by L. Carlitz, Restricted compositions, -Fibonacci Quart. 14 (1976), no. 3, 254–264. -The generalization that Darij describes in his fourth comment was first proved, as far as I know (though stated in a weaker commutative form) by Ralph Fröberg, Determination of a class of Poincaré series, Math. Scand. 37 (1975), 29–39 (page 35). It was proved (independently) shortly thereafter in L. Carlitz, R. Scoville, and T. Vaughan, Enumeration of pairs of sequences by rises, falls, and levels, Manuscripta Math. 19 (1976), 211–243 (Theorem 7.3). Their statement of the theorem doesn't seem to use noncommuting variables, though their proof contains a formula—equation (7.7)—which is essentially the noncommutative version. (I am not sure that this really makes any difference.) Just to be clear, I'll restate the theorem here, more or less, though not exactly, the way Carlitz, Scoville and Vaughan state it, with some comments in brackets. -Let $S$ be a finite set of objects and let $A$ and $B$ be complementary subsets of $S\times S$. Let $F_A$ be the generating function for all paths [today we would call them words, or possibly sequences] which avoid relations from $A$. [This is referring to a partition of $A$ which is related to their applications of the theorem, but is not really relevant to the theorem.] More specifically, define -$$F_A = 1+\sum s_{i_1}+\sum s_{i_1}s_{i_2}+\sum s_{i_1}s_{i_2}s_{i_3}+\cdots,$$ -where, for example, the last sum is taken over all $i_1,i_2,i_3$ such that $s_{i_1} \mathrel{B}s_{i_2}$ and $s_{i_2}\mathrel{B} s_{i_3}$. (We use lower-case $s_i$'s for both the members of the set $S$ and for the indeterminates [which are presumably commuting] in the enumeration.) -We also introduce -$$\tilde F_B = 1-\sum s_i +\sum s_{i_1}s_{i_2}-\cdots$$ -where the signs alternate and the relations must be from $A$ instead of $B$. -7.3 THEOREM. The functions $F_A$ and $\tilde F_B$ are related by $F_A\cdot \tilde F_B = 1$. -Both Fröberg and Carlitz–Scoville–Vaughhan prove this by showing that all the terms in $F_A\cdot \tilde F_B$ except 1 cancel in pairs. However there is another way to prove it: expand $\tilde F_B^{-1}$ as -$\sum_{k=0}^\infty (1-\tilde F_B)^k$ and use inclusion-exclusion. -Carlitz, Scoville, and Vaughan then apply the theorem to counting Smirnov words. -The Carlitz–Scoville–Vaughan theorem is one of my favorite formulas in enumerative combinatorics, and my 1977 Ph.D. thesis has many applications of it. The slides from a talk I gave about this theorem can be found here.<|endoftext|> -TITLE: What is the groupoid cardinality of the category of vector spaces over a finite field? -QUESTION [19 upvotes]: For any groupoid, it's groupoid cardinality is the sum of the reciprocals of the automorphism groups over the isomorphism classes. Let us consider the category of vector spaces over a finite field $\mathbb F_q$ with only invertible morphisms allowed. -Then, the size of the automorphism groups are $g_n= \prod_{i=1}^n(q^n-q^{i-1}) = q^{n\choose 2}\prod_{i=1}^n(q^i-1)$ so it's groupoid cardinality is the following infinite sum: -$$\sum_{n \geq 1}\frac{1}{g_n}.$$ -Is there a closed form expression for it? Note that as $q\to 1$ in an appropriate sense, the general linear group is supposed to converge to the symmetric group and the groupoid of vector spaces should converge to the groupoid of finite sets. -In this sense, the above infinite sum is an analog of the usual one for $e$ and can perhaps reasonably be called a q-analog of $e$. -Alternatively, in the case of abelian, (even semi-simple) categories, is there a better replacement for the groupoid cardinality that I should be using? - -REPLY [16 votes]: Upon substituting $x=\frac{1}{q}$ we obtain -$$\sum_{n\geq 0}\frac{1}{|\mathrm{GL}_n(\mathbb F_q)|}=\sum_{n\geq 0}\frac{x^{n^2}}{(1-x)(1-x^2)\cdots(1-x^n)}$$ -and this evaluates to the product $\prod_{i\geq 1}\frac{1}{(1-x^{5i-4})(1-x^{5i-1})}$ by the first Rogers-Ramanujan identity. -You can also interpret the second Rogers-Ramanujan identity as giving a groupoid cardinality for the category of affine spaces over $\mathbb F_q$. where the sum becomes $\sum_{n\geq 0}\frac{1}{|\mathrm{AGL}_n(\mathbb F_q)|}$. (Note that I wrote these sums to contain a term for $n=0$, corresponding to the zero dimensional vector/affine space.) -I don't think that there is a good explanation for this evaluation that avoids proving the R-R identities first. In particular it is unclear how to directly relate the exponents of the infinite product side with the category of vector/affine spaces.<|endoftext|> -TITLE: How to pronounce "⊗¬-category"? -QUESTION [19 upvotes]: In his PhD thesis, Categorical Structure of Continuation Passing Style, Thielecke studies the ⊗¬-categories, which are premonoidal categories with structure (namely, a functor ¬ which is adjoint to itself), in order to model the semantics of continuation-passing style languages. -I've been studying his work for some time now, but I'm not really familiar to literature about category theory. So my question is, and it might sound silly: how should I pronounce "⊗¬-category"? In fact, I could rephrase this: if I were to write about those categories in a paper, how should I refer to them? Should one write it in full, or simply just refer to them as ⊗¬-categories all the time and leave the pronunciation as an open problem? - -REPLY [17 votes]: I shared an office with Hayo Thielecke in the late 1990s. -The pronunciation he used was "Tensor-NOT-category".<|endoftext|> -TITLE: Explanations simple enough that non-mathematical audiences can understand -QUESTION [6 upvotes]: The following (debatable) quote is attributed to Einstein: -"You do not really understand something unless you can explain it to your grandmother." -I feel very enlightened when there is a simple explanation of an important idea in mathematics. Below are some of my favorite ones. -My Question: Are there other explanations like this? - -(Credit to my analysis professor, many years ago): The geometric series $\sum_{n=1}^{\infty} \frac{1}{2^n} = 1$ can be explained as follows: take a disc. Cut it in half. Now take half of the disc, and cut that in half. Repeat this process. Then we have $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} +...$ disc. But we started out with a whole disc, so the total is a single disc! - -Definitions, etc. - -By "explanation," I am asking for a proof or a heuristic argument which is simple (in the sense defined below). - -Ideally, the fact is a central part of some subfield of mathematics. (For instance, the geometric series is certainly an important example in calculus since it is the main idea behind many comparison tests. It is also the most basic example of a series that can be computed explicitly (besides perhaps telescoping series). ) The facts themselves must also be simple. - -By "simple", I mean you can explain it to someone without any mathematical background (say a child under 10 years old). In particular, words like "derivative," "group," and "Riemann curvature tensor," are considered to be "too hard," but expressions like "speed/velocity," "symmetry" and "how much a surface/curve curves" are acceptable. (In this regard, words from elementary physics (e.g. Newtonian mechanics, electromagnetism, wave mechanics) are great, but quantum mechanics and relativity are too hard. Notions from middle/high school (e.g. Euclidean geometry) are great too.) - -Simple pictures are okay too, although the picture cannot be too complicated. (Again, the main criteria here is that your average non-mathematical audience can understand it.) - -REPLY [7 votes]: Pythagoras theorem. -Albert Einstein wrote about two pivotal moments in his childhood. The first involved a compass that his father showed him when he was four or five. The second involved his early exposure to Euclidean plane geometry. He was impressed by the idea that a mathematical assertion could “be proved with such certainty that any doubt appeared to be out of the question”. -Steven Strogatz discusses a breathtakingly simple proof of the Pythagorean theorem whose provenance is traced to Einstein as a child. "Though we cannot be sure the following proof is Einstein’s, anyone who knows his work will recognize the lion by his claw." -Einstein's first proof. -The proof relies on the insight that a right triangle can be decomposed into two smaller copies of itself. That’s a peculiarity of right triangles. If you try instead, for example, to decompose an equilateral triangle into two smaller equilateral triangles, you’ll find that you can’t. So Einstein’s proof reveals why the Pythagorean theorem applies only to right triangles: they’re the only kind made up of smaller copies of themselves.<|endoftext|> -TITLE: Historical origin of the empty set -QUESTION [5 upvotes]: The question is in the title: - -Who first claimed the existence / necessity of the empty set ? When did this happen ? - -Of course I know that the notation $\emptyset$ goes back to André Weil, and that the first axiom of ZF is that there exists an empty set, but I ask whether this is an older concept. - -REPLY [13 votes]: "It can be justifiably argued that Boole had inventented the empty set" [in The Mathematical Analysis of Logic (1847)]. -The Empty Set, the Singleton, and the Ordered Pair, Akihiro Kanamori (2003).<|endoftext|> -TITLE: What is the significance of Friedlander-Iwaniec and related theorems? -QUESTION [11 upvotes]: On p.177 of Number Theory Revealed: A Masterclass by Andrew Granville, the author states that "One can ask for prime values of polynomials in two or more variables." (though he later mentions Landau's $n^2+1$ conjecture so I'm not sure why single-variable polynomials are omitted). For example, there are some classical results, like Fermat's Christmas theorem and Euler's $6n+1$ theorem. More recent results include the Friedlander-Iwaniec theorem, which states that there are infinitely many prime numbers of the form $a^2+b^4$ for integers $a$ an $b,$ and the result of Heath-Brown on the infinitude of primes of the form $x^3+2y^3$ for integers $x$ and $y.$ Granville's book mentions other recent advances by Maynard and collaborators of Heath-Brown and Iwaniec. -Initially, I wondered why the Friedlander-Iwaniec theorem is considered to be a breakthrough, so I asked Why does the infinitude of primes of a certain form matter? on math.stackexchange, but it was closed for being "opinion-based". Granville's book cleared up a part of this question for me because it states that the Friedlander-Iwaniec theorem was the first example in which the polynomial is "sparse" in taking on integer values, and that later results of the same type based on norm forms were inspired by Friedlander-Iwaniec (although I don't know to what extent the proofs of the former are related to the proof of the latter). -My question is this: is the area of exploration quoted at the beginning of this post an island in that it is considered to be an inherently interesting question requiring no further justification, or are there external motivations for pursuing it? Some satisfactory motivations (this list is by no means exhaustive) might be that other problems reduce to it, that the techniques developed in the course of solving these problems are widely applicable elsewhere, real world applications (e.g., cryptography), or maybe these are instances of much more general problems for which "Mathematics is not yet ripe" in the words of Erdős. - -REPLY [10 votes]: Suppose you suspect that among some subset $\mathcal{S}$ of the natural numbers, there ought to be infinitely many primes. How would you confirm that this is indeed the case? -Clearly, the "bigger" $\mathcal{S}$ is (measured in terms of natural density), the easier the question should be. The biggest set is of course $\mathbb{N}$ itself, and it was a highly non-trivial theorem of Euclid that confirmed this. What about the next most obvious "big" sets, the arithmetic progressions? Indeed, consider the set $\mathcal{S}(a; q)$ by -$$\displaystyle \mathcal{S}(a; q) = \{qx + a : x \in \mathbb{N}\}.$$ -In the special case $q = 4, a = 3$ one can use Euclid's argument again to show that there are infinitely many primes in the set (but far from the correct density). A much more difficult argument shows that the case $q = 4, a = 1$ also gives infinitely many primes. But we expect that there are infinitely many primes in $\mathcal{S}(a; q)$ whenever $\gcd(a,q) = 1$. This is not known until the work of Dirichlet, almost 2000 years after Euclid! -The most extensive generalization of the theorem of Dirichlet known is the Chebotarev density theorem, which shows that in any number field there are infinitely many primes which split completely (over simplifying here). -So far, all of the sets considered have log density one. If we put $\mathcal{S}(X) = \# \{n \leq X : n \in \mathcal{S} \}$ then the log density is the infimum among all non-negative real numbers $\delta$ such that $\mathcal{S}(X) < X^{\delta}$ for all $X$ sufficiently large. Log density one sets can still be very very hard to count primes in: indeed, we expect the set of primes $p$ such that $p + 2$ is also prime to be infinite, and this set (if infinite) is expected to have log density one. -Thus, producing an infinite set of primes which have log density less than one is a formidable task. Both the Friedlander-Iwaniec theorem and Heath-Brown's theorem are examples where this can be achieved. These should be considered highly exceptional in the sense that similar, perhaps even easier looking polynomials are not amenable to their methods! Perhaps the most infamous is the polynomial $F(x,y) = 4x^3 - y^2$, which counts discriminants of elliptic curves in short Weierstrass form.<|endoftext|> -TITLE: Algebraic geometry additionally equipped with field automorphism operation -QUESTION [6 upvotes]: I am looking for some facts on theory, which is essentially algebraic geometry but with field automorphisms added as 'basic' operations. (Precisely, I mean universal algebraic geometry for (universal) algebra being a field $\mathbb{F}$ and operations being polynomials and field automorphisms) I am also interested in computational aspects. -I am mainly interested in case of field of multivariate rational functions, i.e. $\mathbb{F} = K(x,y)$. -An algebraic set in such setting could be $$\{(f,g) \in (K(x,y))^2 \mid x^2-y^2-f \cdot g(x=\frac{x+y}{2}, y = \frac{x-y}{2}) = 0\}.$$ and ideals are additionally closed under field automorphisms. -I would be grateful for reference to any source that considers theory like this, also the computational aspects. - -REPLY [10 votes]: You should look into difference algebra, which is exactly this setting. (It's called difference algebra because the most classical version looked at the automorphism on polynomials that sends $f(x)$ to $f(x+1)$, which can be used to state the theory of difference equations.) -The hottest area in this is the study of the model theory of fields with automorphisms. Zoe Chatzidakis has some lecture notes on the subject.<|endoftext|> -TITLE: To what extent is a vector bundle on a manifold with boundary determined by its restriction to the interior? -QUESTION [5 upvotes]: Let $M$ be a manifold with boundary $\partial M$ and interior $M_0$. Let $E\rightarrow M_0$ be a fixed vector bundle. How many extensions of $E$ to a vector bundle $E'\rightarrow M$ are there, up to isomorphism? In terms of the bundles monoid: the restriction of $E'$ to $M_0$ gives a monoid morphism $\mathrm{Vec}_k(M)\rightarrow \mathrm{Vec}_k(M_0)$. Is it surjective/injective? -Intuitively, the bundle $E'|_{\partial M}$ is "the limit" of $E$ at $\partial M$, and therefore should be fixed up to isomorphism. -And perhaps in the same vein, the inclusion $\iota : M_0 \rightarrow M$ induces $\iota_*:\pi_1(M_0)\rightarrow \pi_1(M)$. Is this map surjective/injective? Can the bijectivity be deduced from a tubular neighborhood of $\partial M$ in $M$? -Counter-examples are appreciated. - -REPLY [9 votes]: As I indicated in my comment, the inclusion $\iota : M_0 \to M$ is a homotopy equivalence. This can be shown using the fact that the boundary $\partial M$ has a collar neighbourhood; it then boils down to showing the inclusion $(0, 1) \hookrightarrow [0, 1)$ is a homotopy equivalence. Actually, one needs to show that there is a homotopy inverse $j : [0, 1) \to (0, 1)$ to $i$ such that $i\circ j$ and $j\circ i$ are homotopic to identity maps relative to $[\frac{1}{2}, 1)$. This is not difficult, see this answer for some details. -On any paracompact space $X$, there is a natural bijection between isomorphism classes of real vector bundles on $X$ of rank $r$ and $[X, BO(r)]$, the set of homotopy classes of maps $X \to BO(r)$; see section $1.2$ of Hatcher's Vector Bundles and K-Theory for example. In particular, given a map $f : X \to Y$, we get an induced map $f^* : [Y, BO(r)] \to [X, BO(r)]$ which corresponds to pulling back a vector bundle by $f$. The analogous statement is true for complex vector bundles too, one just replaces $BO(r)$ with $BU(r)$. -In the case that $f$ is a homotopy equivalence, then $f^*$ is a bijection: if $g$ is the homotopy inverse of $f$, then $g^*$ is the inverse of $f^*$. In particular, for the homotopy equivalence $\iota : M_0 \to M$, we see that there is a bijection between isomorphism classes of real/complex rank $r$ bundles on $M$ and $M_0$ given by $E \mapsto \iota^*E = E|_{M_0}$. -Finally, as $\iota : M_0 \to M$ is a homotopy equivalence, the induced map $\iota_* : \pi_1(M_0) \to \pi_1(M)$ is an isomorphism. - -As Ben McKay indicates in the comment below, the above does not deal with smooth bundles but topological bundles. The statement for smooth bundles is also true, but requires a bit more work. The key is that every real rank $r$ vector bundle on a smooth manifold $M$ has a classifying map $M \to \operatorname{Gr}_r(\mathbb{R}^N)$ which is unique up to homotopy where $N = r + \dim M + 1$; this is Theorem 3.3.4 of Hirsch's Differential Topology. It follows that isomorphism classes of topological real rank $r$ vector bundles on $M$ are in bijection with $[M, \operatorname{Gr}_r(\mathbb{R}^N)]$; that is, the inclusion $\operatorname{Gr}_r(\mathbb{R}^N) \hookrightarrow \operatorname{Gr}_r(\mathbb{R}^{\infty})$ induces a bijection $[M, \operatorname{Gr}_r(\mathbb{R}^N)] \to [M, \operatorname{Gr}_r(\mathbb{R}^{\infty})] = [M, BO(r)]$. -If the classifying map of a bundle is smooth, then the bundle itself is smooth (the pullback of a smooth bundle by a smooth map is smooth). As every continuous map between smooth manifolds is homotopic to a smooth one, every topological vector bundle on $M$ is isomorphic to a smooth one. Moreover, two smooth maps are homotopic if and only if they are smoothly homotopic which implies that every topological vector bundle is isomorphic to a unique smooth vector bundle up to smooth isomorphism. It follows that isomorphism classes of smooth real rank $r$ vector bundles on $M$ are in bijection with $[M, \operatorname{Gr}_r(\mathbb{R}^N)]$. -Now we can argue as before to deduce that $\iota^*$ induces a bijection between the set of isomorphism classes of smooth real rank $r$ bundles on $M$ and $M_0$. Again, the statement is also true for smooth complex bundles.<|endoftext|> -TITLE: "Gray code" for building teams -QUESTION [7 upvotes]: Motivation. In a team of $n$ people, we had the task to build subteams of a fixed size $k -TITLE: Papers with a large number of coauthors -QUESTION [13 upvotes]: I recently submitted a paper to the preprint arXiv, which was rejected because we didn't list all of the authors on the first page. We chose to follow the polymath model, using a generic name for our group, with a footnote linking to a place with all the actual contributors. -I was surprised that our paper was rejected, as there are many arXiv papers by "D.H.J. Polymath". Does anyone know if this is a recent change? -More importantly, what should be the practice for papers involving a huge number of coauthors in a collaborative setting? Is there a way to petition the arXiv for exceptions in such cases? - -REPLY [5 votes]: This is a matter of opinion, but whatever the policy is, it should be consistently applied. If D.H.J. Polymath is allowed to post papers without listing the members, then other collaborations should be allowed to do so as well. D.H.J. Polymath shouldn't get special treatment just because it is famous. -I'm gradually becoming convinced that the mathematical community as a whole needs to pay more attention to arXiv policies. Seemingly small questions of policy can have a huge impact on mathematical research, simply because such a huge fraction of all mathematical research gets posted to the arXiv—far more than is submitted to any single mathematics journal. There is a myth that there are no barriers to publishing on the arXiv, but this is certainly not true. Figuring out the right policies is not a trivial problem; the founders of viXra stated some legitimate concerns about the arXiv, but I believe that viXra has failed to solve the problems it was intended to solve. Perhaps the American Mathematical Society and other professional societies should investigate some of the concerns and complaints about the arXiv that are being raised by mathematical researchers, and try to change some things about the arXiv if doing so would serve the community better.<|endoftext|> -TITLE: McKean-Singer formula in Heat Kernels and Dirac Operators book -QUESTION [5 upvotes]: I'm reading "Heat Kernels and Dirac Operators" by Berline, Getzler and Vergne. The setting is: $E \to M$ is a $\mathbb{Z}_2$-graded vector bundle on a compact Riemannian manifold $M$ -and $D : \Gamma(M, E) \to \Gamma(M, E) $ is a self-adjoint Dirac operator (especially $D^2$ is Laplace). We denote -by $D^{\pm}$ the restrictions of $D$ to $\Gamma(M, E^{\pm})$. I have some troubles to understand the last part in the second variant of the proof of Theorem 3.50 McKean-Singer: - -It is clamed that the operator $d(e^{-tD^2})/dt$ has a smooth kernel equal to $-D^2e^{-tD^2}$. This don't make any sense to me. Recall if $P: E\to E$ is an integral operator and $s \in E_x, x \in M $ then it's kernel $K: M \times M \to \mathbb{R}$ is defined by equation -$$ Ps_{\vert x}= \int_{\{x \} \times M} K(x,y) s_{\vert y} dy$$ -Especially, $K$ maps from $M \times M$, while $-D^2e^{-tD^2}$ (for fixed $t$) maps from $E$. Any idea how $-D^2e^{-tD^2}$ can be interpreted as an integral kernel? - -REPLY [6 votes]: The assertion is supposed to be that $d(e^{-tD^2})/dt$ has the same smooth kernel as $-D^2 e^{-tD^2}$, i.e. they are the same operator. This is because $e^{-tD^2}$ is the solution operator to the heat equation: $\frac{du}{dt} = -D^2 u$. I'd be comfortable saying this is the definition of $e^{-tD^2}$, but you might prefer to construct $e^{-tD^2}$ using the pseudodifferential calculus in which case this fact would follow from the spectral theorem.<|endoftext|> -TITLE: Canon in algebraic combinatorics and how to study -QUESTION [5 upvotes]: 1) In subjects such as algebraic geometry, algebraic topology there is a very basic standard canonical syllabus of things one learns in order to get to reading research papers. -Is there a similar canon in algebraic combinatorics? (e.g., does someone working in matroids have knowledge of symmetric functions and vice versa?) -2) I want to know how much of EC 1,2 does a regular algebraic combinatorics researcher know? Do I try to solve the vast breadth of problems (at least the ones with difficulty level less equal 3- let's say) in those two books? How about attempting at reading and solving Bourbaki's Lie Groups and Lie Algebras chapters 4-6? This seems like the most read book of Bourbaki, and a treasure trove of Coxeter Group-Root System material. How do I go about studying Macdonald's Symmetric Functions and Hall Polynomials? I mention these books because they appear to be listed as a reference in many of the papers I see. But these are enormous, and reading and solving problems from cover to cover is probably impractical (Is it?). -I want to know how people tackle these kind of classic references. As well as how to practically study algebraic combinatorics. -3) Can someone point out if there is a list of topics-books/notes/videos (similar to https://math.stackexchange.com/questions/1454339/undergrad-level-combinatorics-texts-easier-than-stanleys-enumerative-combinator but with topics such as matroid theory, Coxeter groups, crystal bases included)? - -REPLY [6 votes]: I would suggest that much of what I said in my answer to another MO question applies here, in spades. To a first approximation, the canon is the empty set. Start with a problem, and learn what you need as you go along. -See also How to escape the inclination to be a universalist or: How to learn to stop worrying and do some research.<|endoftext|> -TITLE: Emergence of the orthogonal group -QUESTION [16 upvotes]: Do we know what mathematician first considered, and perhaps named, what we call the group $\mathrm O(n)$, or $\mathrm{SO}(n)$, for some $n>3$? -I mean it specifically as group (not Lie algebra) acting on Euclidean $n$-space. For $n=3$ Jordan (1868) seems a definite upper bound, but for higher $n$ it seems not clear to me that even Cartan (1894) thought in those terms, describing as he does $\mathsf B_l$ and $\mathsf D_l$ as “projective groups of a nondegenerate surface of second order in spaces of $2l$ and $2l-1$ dimensions.” Also please disregard any implicit occurrence of $\mathrm{SO}(4)$ in quaternion theory. - -REPLY [19 votes]: Your quote about Cartan thinking of $B_n$ and $D_n$ as 'projective groups..." is actually Cartan describing the lowest dimensional homogeneous space of these groups (except, of course, for a few exceptional cases such as $D_2$, which is not simple, and therefore should be left out of the description). -If you go just a little bit further in Cartan's 1894 Thesis, to Chapitre VIII, Section 9, you'll see that Cartan describes linear representations as well. For example, of $B_\ell$, he writes "C'est le plus grand groupe linéare et homogéne de l'espace à $2\ell{+}1$ dimensions qui laisse invariante la forme quadratique -$$ -{x_0}^2 + 2x_1x_{1'} +2x_2x_{2'} + \cdots + 2x_\ell x_{\ell'}" -$$ -with a similar description for $D_\ell$. -In fact, he gives the lowest dimensional representation of each of the simple groups over $\mathbb{C}$, including the exceptional ones and, except for $\mathrm{E}_8$, he explicitly describes the equations that define the representation. For example, he writes down an explicit homogeneous cubic in 27 variables and states that $\mathrm{E}_6$ is the the subgroup of $\mathrm{GL}(27,\mathbb{C})$ that preserves this cubic form. -For the summary theorem on the linear representations, see Chapitre VIII, Section 10, where he lists each of the lowest representations and notes the various low dimensional exceptional isomorphisms as well. -Remark 1: Cartan continues to refer to groups of type $B$ and $D$ merely as "the largest groups preserving a quadratic form in $n$ variables" or similar terms for a long time. Even in his papers of 1913–1915 classifying the real forms of the complex simple Lie groups, he uses such terminology, though he clearly finds the special case of the compact real forms of special interest. -The first place that Cartan actually refers to 'orthogonal groups' that I can recall are in his 1926–27 papers on the classification of Riemannian symmetric spaces. There, he begins referring to any subgroup of $\mathrm{GL}(n,\mathbb{R})$ that preserves a quadratic form as 'an orthogonal group' and then, later, finally refers to the maximal group that preserves a positive definite quadratic form as 'the orthogonal group'. I don't recall when or whether he used any notation such as $\mathrm{O}(n)$ or $\mathrm{SO}(n)$. -Whether the term 'orthogonal group' was original to him, I can't say. By that time, of course, Weyl had already started his research on compact Lie groups, and it may be that Weyl had already used the term 'orthogonal group' well before Cartan. -Remark 2: Euler's article (Problema algebraicum ob affectiones prorus singulares memorabile, Novi commentarii academiae scientiarum Petropolitanae 15 (1770) 1771, 75–106) discusses the problem of parametrizing the solutions of the equation $A^TA = I_n$ where $A$ is an $n$-by-$n$ matrix for $n=3$, $4$, and $5$, particularly the rational solutions. He does not use the terminology 'orthogonal' or 'group'. Nevertheless his article does contain some remarkable formulae that clearly anticipate the development of the algebra of quaternions. -For example, identifying $\mathbb{R}^4$ with the quaternions $\mathbb{H}$ in the usual way, it is a now-standard fact that every special orthogonal linear transformation $M$ of $\mathbb{R}^4=\mathbb{H}$ can be written, using quaternion multiplication, in the form $M(X) = A\,X\,\bar B$ where $A$ and $B$ are unit quaternions and $X\in\mathbb{H}$. (This is now the usual way that the double cover $\mathrm{Spin}(3)\times\mathrm{Spin}(3)\to\mathrm{SO}(4)$ is introduced.) Meanwhile conjugation $c:\mathbb{H}\to\mathbb{H}$ is orthogonal but has determinant $-1$, so every element of the non-identity component of $\mathrm{O}(4)$ can be written as $$M'(X) = Ac(X)\bar B = A\,\bar X\, \bar B = A\overline{BX} = Ac(BX).$$ Remarkably, Euler gives this formula for parametrizing $\mathrm{O}(4)$ in the form of the product of matrices $L_A\,c\,L_B$ (where $L_P$ denotes left multiplication by the quaternion $P$), many years before the 'official' discovery of quaternions. - -REPLY [13 votes]: There may be an earlier source, but Adolf Hurwitz 1897 is one upper bound: -A. Hurwitz, Über die Erzeugung der Invarianten durch Integration, Nachr. Ges. Wiss. Göttingen (1897), 71–90. - -Hurwitz’s paper introduced and -developed the notion of an invariant measure for the matrix groups -SO(N) and U(N). He also specified a calculus from which the explicit -form of these measures could be computed in terms of an appropriate -parametrisation — Hurwitz chose to use Euler angles. This enabled him -to define and compute invariant group integrals over SO(N) and U(N). - -source: A. Hurwitz and the origins of random matrix theory in mathematics<|endoftext|> -TITLE: Complex structures on Hermitian symmetric space -QUESTION [9 upvotes]: Let $(M_1,g_1,J_1)$ and $(M_2,g_2,J_2)$ be two simply-connected Hermitian symmetric spaces, which are isometric as two Riemannian manifolds. -Can we find an isometry $\varphi:M_1 \to M_2$ such that -$$ -\varphi^* J_2=J_1? -$$ - -REPLY [8 votes]: The answer is 'yes, we can'. -Since we are in the simply-connected case, by the deRham Theorem, we can assume that $(M_i,g_i)$ for $i=1,2$ are isometric to -$$ -(\mathbb{C}^m,h_0)\times (N_1,h_1)\times\cdots \times (N_k,h_k) -$$ -where $h_0$ is the standard flat metric on $\mathbb{R}^{2m}$ and, for $1\le \ell\le k$, $(N_\ell,h_\ell)$ is an irreducible symmetric space that has a $h_\ell$-parallel, $h_\ell$-orthogonal complex structure $A_\ell$. (I can't use $J_\ell$ because $J_1$ and $J_2$ have already been taken. In fact, because the holonomy group of $h_\ell$ acts irreducibly on the tangent space at any point with commuting ring isomorphic to $\mathbb{C}$, the only $h_\ell$-parallel, $h_\ell$-orthogonal complex structures on $N_\ell$ are $A_\ell$ and $-A_\ell$. It follows that we can assume that -$$ -(M_1,g_1,J_1) = (\mathbb{R}^{2m},h_0,A_0)\times (N_1,h_1,A_1)\times\cdots \times (N_k,h_k,A_k), -$$ -as Hermitian symmetric spaces while -$$ -(M_2,g_2,J_2) = (\mathbb{R}^{2m},h_0,B_0)\times (N_1,h_1,\epsilon_1A_1)\times\cdots \times (N_k,h_k,\epsilon_kA_k), -$$ -for some choice of $\epsilon_i = \pm 1$ while $A_0$ and $B_0$ are orthgonal complex structures on $\mathbb{R}^{2m}$ (not necessarily inducing the same orientation on $\mathbb{R}^{2m}$, of course). -The two flat factors are clearly isometric as Hermitian symmetric spaces, so the only question is whether there is an isometry $c_\ell:(N_\ell,h_\ell)\to (N_\ell,h_\ell)$ that satisfies $c_\ell^*A_\ell = -A_\ell$. -This may not be immediately obvious from the definitions, but it can be proved abstractly from the standard symmetric representation as $G/U$ or simply checked case by case using Cartan's classification of the irreducible Hermitian symmetric spaces. -For the cases AIII (the complex Grassmannians and their duals, including the projective spaces) and BDI (the complex quadrics and their duals) there is an obvious anti-holomorphic isometry. -For the case DIII, the set of positively oriented orthogonal complex structures on $\mathbb{R}^{2n}$ and its noncompact dual, an antiholomorphic involution (in the compact case) is conjugation with an orientation-reversing isometry. -For the case CI, the set of complex Lagrangian subspaces of $\mathbb{C}^{2n} = \mathbb{H}^n$ (and its dual), an antiholomorphic involution (in the compact case) is simply taking the orthogonal complex Lagrangian subspace. -In the first exceptional case EIII, one can rely on the fact that $\mathrm{E}_6\subset\mathrm{SU}(27)$ is defined as the stabilizer of a cubic form $C$ with real coefficients (as per Cartan), so complex conjugation preserves the singular locus of the cone $C=0$, and EIII is just the projectivization of this singular locus (a complex manifold of dimension $16$), which is invariant under the conjugation. -The second exceptional case, EVII (compact type), a complex manifold of dimension $27$, has a similar description as a projectivized orbit of a singular locus of the quartic form Q with real coefficients on $\mathbb{C}^{56}$ stabilized by $\mathrm{E}_7\subset\mathrm{Sp}(28)\subset\mathrm{SU}(56)$. See Cartan's paper on the classification of the real forms for details.<|endoftext|> -TITLE: What is the Borel complexity of this set? -QUESTION [6 upvotes]: Problem. What is the Borel complexity of the set -$$c(\mathbb Q)=\{(x_n)_{n\in\omega}\in\mathbb R^\omega:\exists\lim_{n\to\infty}x_n\in\mathbb Q\}$$ -in the countable product of lines $\mathbb R^\omega$? -Remark. It is easy to see that $c(\mathbb Q)$ is an $F_{\sigma\delta\sigma}$-set in $\mathbb R^\omega$. -Question. Is $c(\mathbb Q)$ of type $F_{\sigma\delta}$ in $\mathbb R^\omega$? - -REPLY [7 votes]: The argument is some standard recursion theoretic trickery (I'm educated enough to recognize that, but not educated enough to know what this is called). - -Theorem. Every $\Sigma^0_3$ subset of Cantor space continuously reduces to your set. - -It follows from this that if your set were $F_{\sigma \delta}$ then every $\Sigma^0_3$ subset of Cantor space would be $\Pi^0_3$, a contradiction with the properness of the Borel hierarchy. I think. -Proof. In front of you stand $\omega$ many creatures. Every second, some subset of these creatures blinks. You have set up video surveillance and have every blink of every creature on tape. By watching these tapes in a dovetailing fashion, you eventually find out about every blink. The event that one of these creatures blinks infinitely many times models (= is an intuitive interpretation of) a general $\Sigma^0_3$ subset of Cantor space: such a set is of the form $\bigcup_n \bigcap_m \bigcup_k C_{n,m,k}$ where $C_{n,m,k}$ is clopen, and a point being in $C_{n,m,k}$ means the time between the $(m-1)$th and the $m$th blink of creature $n$ is $k$ seconds. -Finding a continuous function means we keep watching the tapes, and keep outputting real numbers (actually it's enough to output approximations to the real numbers, but I'll just output exact numbers directly). -We'll keep track of the following data: An irrational number $r \in (0,1)$, and rational numbers $1 = q_0 > q_1 > q_2 > q_3 > \cdots > r$, such that $[r, q_i]$ is disjoint from $\frac{1}{i} \mathbb{Z}$. Note that for any $r \in (0,1)$ we can find such numbers. These should also be indexed by the current time, but I'll keep that implicit. -Now, keep watching the tapes, and do as follows - -every second, output $r$ (i.e. $x_t$ will be the value $r$ output at time $t$) -if creature $i$ blinks, increase $r$ to some irrational number between $(r + q_i)/2$ and $q_i$, and pick new values for the numbers $q_j$ for $j > i$ if needed - -I claim that the resulting sequence $(x_t)_t$ is in $c(\mathbb{Q})$ if and only if some creature blinks infinitely many times. Namely, suppose that the $i$th creature blinks infinitely many times, but no creature with smaller index does. Then after some time $t$, no creature $j < i$ blinks. At that point, we have picked some value $q_i \in \mathbb{Q}$, and since the $i$th creature keeps blinking, we eventually increase $r$ to $q_i$ by the second rule, thus $(x_t)_t \in c(\mathbb{Q})$. -Suppose then that after some time, no creature ever blinks again. Then from that point on, we keep outputting $r$, which is irrational, so $(x_t)_t \notin c(\mathbb{Q})$. -Suppose then that no creature blinks infinitely many times, but there are infinitely many blinks altogether. Since the sequence is monotone increasing and bounded from above (by $q_0 = 1$), there is some limit $\lim_t x_t = r'$. Suppose it is rational, say $r' = n/i$. Consider the last time the $i$th creature blinks. At this time, the value $r$ satisfies $r < q_i$, and $[r, q_i]$ is disjoint from $\frac{1}{i} \mathbb{Z}$. But the way the process is defined, $r$ never increases past $q_i$ unless the $i$th creature blinks, so $r' \in [r, q_i]$ cannot be equal to $n/i \in \frac{1}{i} \mathbb{Z}$. Square. -I think the actual complexity of your set is hinted by the fact it is an intersection of a $F_{\sigma \delta}$ and a $G_{\delta \sigma}$ set, but that's more fine-grained than I'm willing to touch right now, since I'm not exactly a boldface expert. -edit -Scratch that, I will touch it a little. - -Theorem. Every set that is an intersection of a $\Sigma^0_3$ and a $\Pi^0_3$ subset of Cantor space continuously reduces to your set. - -I imagine this pinpoints the exact complexity, since as mentioned your set is an intersection of a $F_{\sigma \delta}$ and a $G_{\delta \sigma}$ set. -Proof. We now have a $\Sigma^0_3$ set $A$ and a $\Pi^0_3$ set $B$, and are interested in reducing $A \cap B$ to your set $c(\mathbb{Q})$. For $A$ we have the set of $\omega$ many magnificent creatures from the previous proof, which occasionally blink. For $B$ we have another $\omega$ many cute creatures, which occationally wag their tails. We interpret $A \cap B$ as the event that some magnificent creature blinks infinitely many times, and no cute creature wags its tail infinitely many times. -In the even coordinates of the output, keep writing (the current value of) $r$ according to the protocol of the previous proof. In the odd coordinates, by default also write $r$, but write $r+\frac{1}{2^i}$ whenever the $i$th creature wags its tail. -If the point of Cantor space is in $B$, then eventually the first $n$ cute creatures stop wagging their tails, and thus the value eventually stays in $[r-\frac{1}{2^n}, r+\frac{1}{2^n}]$ where $r$ is where the monotone sequence converges. If it is additionally in $A$, then $r \in \mathbb{Q}$ so $(x_t)_t \in c(\mathbb{Q})$. -If on the other hand the point is not in $B$ then there is no convergence because we keep seeing consecutive values $(r, r+\frac{1}{2^i})$ for any $i$ that never stops wagging its tail, so $(x_t)_t \notin c(\mathbb{Q})$. If the point is in $B$ but not in $A$, then we have convergence to some element of $\mathbb{R} \setminus \mathbb{Q}$, and again $(x_t)_t \notin c(\mathbb{Q})$. Square.$\;\;\;\;\;$<|endoftext|> -TITLE: is there a ‘nice’ lattice on the set of unlabelled graphs with $n$ vertices? -QUESTION [10 upvotes]: It is easy to endow the set of vertex-labelled graphs with $n$ vertices with a lattice structure: take the union and the intersection of the edge set as meet and join respectively. -However, I wonder whether there is a ‘nice’ lattice structure on unlabelled graphs, i.e., graphs up to isomorphism. -I'd also be happy with a lattice structure on a (reasonably) large subset, if this makes things any easier, e.g., connected or planar graphs on $n$ vertices. -To clarify: a very non-‘nice’ way to make the set of unlabelled graphs on $n$ vertices into a lattice is to pick an arbitrary total order. -An idea for a slightly nicer way might be to set $G < H$ if and only if $H$ has more edges than $G$, however this does not produce a lattice. - -REPLY [2 votes]: I guess this class is way too small, but since the questions is still unanswered, I will give you an example I like. -An EFG (Edge Firing Game) is defined from an undirected graph $G$ with a distinguished vertex called the sink. It starts from an orientation of $G$ and generates a set $L$ of orientations as follows. The initial orientation is in $L$, and if an orientation is in $L$ and has a vertex (other than the sink) with only incoming edges, then the orientation obtained by reversing all these edges also is in $L$. -                       -This operation defines a relation between the orientations in $L$, see above example (the sink is the black square); it is the covering relation of a distributive lattice. This was shown by James Propp in 1993 together with many other interesting results on lattices and orientations. -Interestingly, this game actually encodes all distributive lattices: each of them may be obtained from an EFG, which we have shown in 2001.<|endoftext|> -TITLE: Why aren't $B_n$ and $C_n$ the other way around? -QUESTION [5 upvotes]: In the classification of complex simple Lie algebras/groups, I have always been vaguely puzzled why $B_n$ and $C_n$ are labeled the way they are. I always instinctively want to put the special orthogonal groups together, and so I am tempted to use the letter $B$ for what is standardly called $C$, and vice versa. Looking at the Dynkin diagrams of affine Weyl groups reinforces this instinct of mine, because the vertex of degree 3 makes $\tilde D_n$ look more like $\tilde B_n$ than $\tilde C_n$, at least in my eyes. -Is there some intuitive reason for the standard notation? Or is just a historical accident with no particular rhyme or reason behind it? - -REPLY [7 votes]: "Historical convention" (going back to Lie?) is probably the correct explanation, but note that under what I would call the "standard combinatorial folding procedure" as described by Stembridge in Folding by automorphisms, we produce the Type $B_n$ root system from the Type $A_{2n-1}$ root system, and the Type $C_{n}$ root system from the Type $D_{n+1}$ root system. Though note, as discussed in this prior MO question, that there are two "dual" folding procedures which both arise in Lie theory.<|endoftext|> -TITLE: Surprising properties of closed planar curves -QUESTION [25 upvotes]: In https://arxiv.org/abs/2002.05422 I proved with elementary topological methods that a smooth planar curve with total turning number a non-zero integer multiple of $2\pi$ (the tangent fully turns a non-zero number of times) can always be split into 3 arcs rearrangeable to a closed smooth curve. Here comes one example of what I called the 2-cut theorem. - -I am now writing the introduction to my thesis and I would like to cite more examples of counterintuitive properties of planar closed curves like, I think, the one above. The inscribed square problem, asking whether every Jordan curve admits an inscribed square (pic from Jordan curves admitting only acyclic inscriptions of squares), alredy came to my mind. - -Such a property is still a conjecture for the general case, but proofs have been provided for several special cases and the easier inscribed rectangle problem can be solved with a beautiful topological argument (3Blue1Brown made a very nice video about that https://www.youtube.com/watch?v=AmgkSdhK4K8&t=169s). -My question: what are other surprising/counterintuitive properties of closed planar curves you are aware of? -Note: this question was originally posted on MathStackExchange since I thought it was suitable for a more general audience. Apparently, it requires a slightly broader knowledge of literature. - -REPLY [3 votes]: How about the ``Oval's Problem" of Benguria and Loss. This is a (somewhat surprisingly) open problem connecting spectral theory with plane geometry. The conjecture is that for any simple closed curve $\sigma$ of length $2\pi$ and any periodic function $f:[0,2\pi]\to \mathbb{R}$ one has -$$ -\int_0^{2\pi} |f'(s)|^2+\kappa(s)^2 |f(s)|^2 ds \geq \gamma \int_{0}^{2\pi} |f(s)|^2 ds -$$ -for $\gamma=1$. Here $s$ is the arclength parameter and $\kappa(s)$ is the geodesic curvature. In other words, the lowest eigenvalue of $-\frac{d^2}{ds^2}+\kappa(s)^2$ is bounded from below by $1$ on a closed curve. This lowest bound is achieved by the circle, but is actually achieved by a whole family of "ovals" that deteriorate to a multiplicity two line segment. Interestingly, this problem is related the sharp Lieb-Thirring inequality which is a purely spectral theoretic problem (this was Benguria and Loss's motivation). It is also related to minimal surface theory. -In their paper, Benguria and Loss show that this inequality holds with $\gamma=\frac{1}{2}$. This is actually, sharp if one expands the set of curves to include those that total turning number 1 so one has to use the curve is closed in some way (and not just that $\int_0^{2\pi} k(s)ds=2\pi$. Various other authors have worked on parts of this problem: Burchard and Thomas showed the ovals where local minima of the lowest eigenvalue (so the problem is solved near curves in the family, Linde showed closed convex curves have $\gamma\geq 0.6085$, Denzler showed there is a closed convex curve that minimized the value of $\gamma$ and Bernstein and Mettler discussed the symmetry of the problem (related to projective geometry) and showed some weaker geometric inequalities held for closed convex curves (but not for curves of turning number 1).<|endoftext|> -TITLE: Invertible matrix and 0-1 vector -QUESTION [6 upvotes]: Let $n\ge 1$ and $A$ be any $n\times n$ real invertible matrix. Can we always find a 0-1 vector $b\in\{0,1\}^n$ such that each entry of $Ab$ is nonzero? For example, if $A$ is the identity matrix, then we can (only) take $b$ to be the all-one vector. -Edit: The answer given by Pat Devlin is beautiful. From his construction, we know that there exists a parallelepipedon whose each vertex lies on some coordinate plane. Indeed, for $n=3$ from the matrix -$$A=\left(\begin{matrix}1&0&0\\ -0&1&-1\\ -1&-1&-1 -\end{matrix}\right)$$ -we can obtain the desired parallelepipedon using the columns of $A$. His solution suggests that the invertibility assumption on $A$ seems inappropriate. -Can we put a rectangular solid (particularly, a cube) in $R^3$ such that each vertex lies on some some coordinate plane. In higher dimension, if the column vectors of $A$ are orthogonal with each other (i.e.,$A^T A$ is a nonsingular diagonal matrix), can we always obtain a desired 0-1 vector $b$ such that each entry of $Ab$ is nonzero? -What is a appropriate assumption on $A$ for a positive solution? Does orthogonal matrices suffice? - -REPLY [10 votes]: The answer for $n \leq 2$ is yes, and for $n \geq 3$, it's no. (The $n=1$ case is left to the reader) -The case $n=2$: If $v_i$ are the rows of the matrix, then each equation $v_1 \cdot x = 0$ and $v_2 \cdot x = 0$ describes a co-dimension $1$ space. And the four corners of the square $\{0,1\}^2$ cannot be covered by two lines passing through the origin. -The case $n \geq 3$: For $n \geq 3$, we can cover all the vertices of the cube with three independent hyperplanes through the origin. Namely... -$$x_1 = 0$$ -$$x_2 -x_3 = 0$$ -$$x_1 - x_2 - x_3=0$$ -It's easy to see these three conditions describe a co-dimension three space [i.e., the conditions are independent]. To see they cover the vertices of the cube, note that failure of the first and second equations would imply $x_1 =1$ and $x_2 \neq x_3$. Since each variable is $0,1$, this would force $x_2 + x_3 = 1$, so the third equation will be satisfied. -The $n\geq 3$ case phrased as a matrix: -Make the first row $[1, 0, 0, 0, 0, \ldots]$. Then the second row $[0, 1, -1, 0, 0, \ldots]$. Then make the third row $[1, -1, -1, 0, 0, 0, \ldots]$. These three rows are independent (yay!), so just fill up the rest of the matrix with more linearly independent rows until you have a basis (i.e., an invertible matrix). Then see above argument why $A \vec{x}$ will have at least one of the first three coordinates equal to $0$. -Literature: -Also, there are some really cool results about covering the cube with hyperplanes. See for instance this article about covering all but one of the vertices and also this article about covering the vertices in a way where each equation is needed, and every variable actually shows up. -Edit 1: I changed the equations to handle the $n=3$ as well. -Edit 2 (this part added to address orthogonal case): As per comment, perhaps you’d like the matrix to be orthogonal as well. That’s also fine (at least for $n \geq 4$. I haven’t thought about $n=3$ in this case). -Take the first three rows to be $[1, -1, 0, 0, 0, 0, \ldots]$ and $[0,0,1,-1,0,0,\ldots]$ and $[1,1,-1,-1,0,0,\ldots]$ (and renormalize these so that they each have length 1). Then extend this to an orthonormal basis and fill the rest of the matrix with those rows. This works in the same ways as the above example. (This was actually the first construction I gave, but then I swapped it out for the above, which only needed $n=3$.) ((I notice that I don’t enjoy thinking about the orthogonal $n=3$ case. It might be that you can’t make any orthogonal matrix to do it. On the other hand, it could be that I’m bad at visualizing rotations of the cube that put all its corners onto three orthogonal planes...)) -Edit 3: You seem to want this to be true for really special matrices. I don’t think that’s really going to happen. The cube fits inside a pretty darn small space (union of three planes), so this seems to give us a lot of wiggle room for all sorts of stuff. :-/<|endoftext|> -TITLE: GCH implies acceptability -QUESTION [5 upvotes]: I have been studying the concept of acceptability, particularly in its relation to GCH. -There are many versions of it in the sources I have found, with some slight variations, and some of them are claimed to be equivalent to GCH, some of them are claimed not to be equivalent to GCH. For example, Welch, in A Condensed History of Condensation, claims that GCH is equivalent to acceptability but not to weak acceptability. Friedman and Holy, in A Quasi-Lower Bound on the Consistency Strength of PFA, define a slightly different form of weak acceptability and prove that it is equivalent to GCH. Schindler and Zeman, in Fine Structure, define a fine-structural version of acceptability and say that it can be seen as a stronger form of GCH. -I am interested in any further references providing details and clarifying the relation between acceptability and GCH. -I am particularly interested in knowing whether the fine-structural version given in Schindler and Zeman is implied by GCH in some interesting class of models. More specifically, in the models $\textbf{L}[A]$, does GCH imply that for some appropriate choice of $A$ the $J$-structures $J^A_{\alpha}$ are acceptable for some unbounded class of $\alpha$'s in the sense given in Fine Structure Theory? -EDIT -A more basic question may be helpful. In which precise sense, if any, is fine-structural acceptability (defined for the $J$-hierarchy) equivalent to acceptability defined directly for the $L$-hierarchy? - -REPLY [4 votes]: Another related reference, is the following “An Abstract Condensation Property -” by David Richard Law. -Here is the abstract of it: -Let $A = (A, \dotsc)$ be a relational structure. Say that $A$ has condensation if there is an -$F : A^{< \omega} → A$ such that for every partial order $P$, it is forced by $P$ that substructures of $A$ which are closed under $F$ are isomorphic to elements of the ground model. Condensation holds if every structure in $V$, the universe of sets, has condensation. This property, isolated by Woodin, captures part of the content of the condensation lemmas for $L$, $K$ and other "$L$-like" models. We present a variety of results having to do with condensation in this abstract sense. Section 1 establishes the absoluteness of condensation and some of its consequences. In particular, we show that if condensation holds in $M$, then $M \models \mathrm{GCH}$ and there are no measurable cardinals or precipitous ideals in $M$. The results of this section are due to Woodin. Section 2 contains a proof that condensation implies $\Diamond_κ(E)$ for $\kappa$ regular and $E \subseteq \kappa$ stationary. This is the main result of this thesis. The argument provides a new proof of the key lemma giving GCH. Section 2 also contains some information about the relationship between condensation and strengthenings of diamond. Section 3 contains partial results having to do with forcing "$\operatorname{Cond}(A)$", some further discussion of the relation between condensation and combinatorial principles which hold in $L$, and an argument that $\operatorname{Cond}(G)$ fails in $V[G]$, where $G$ is generic for the partial order adding $\omega_2$ cohen subsets of $\omega_1$.<|endoftext|> -TITLE: Can all proper sublattices of $\mathbb{Z}^n$ be generated cyclically? -QUESTION [5 upvotes]: Let $\Lambda \subset \mathbb{Z}^n$ be a proper sublattice (so that $\Lambda \ne \mathbb{Z}^n$). We say that $\Lambda$ is cyclically generated if there exists a matrix $M \in \text{GL}_n(\mathbb{Z})$ and an element $\mathbf{u} \in \mathbb{Z}^n$ such that $\Lambda$ is equal the $\mathbb{Z}$-span of $\{\mathbf{u}, M \mathbf{u}, M^2 \mathbf{u}, \cdots \}$. Is it true that all proper sublattices are cyclically generated? If so, how would one prove this, and if not, what's a counterexample? - -REPLY [8 votes]: Some standard latticework allows us to write $\Lambda = \text{im}(A)$ for some $n$ by $n$ matrix $A$ (with nonzero determinant). Using Smith normal form, there are some $U, V \in GL_n(\mathbb{Z})$ such that $UAV = D$, where $D$ is a diagonal matrix with diagonal elements $d_1 | d_2 | \dots | d_n$. -Write $\vec{e}_i$ for each of the standard basis vectors. Then let $\vec{u} := U^{-1} d_1 \vec{e}_1$, and let $M := U^{-1}BU$, where $B$ is the bidiagonal matrix with all $1$s on the diagonal, and $\frac{d_{i + 1}}{d_i}$ on the subdiagonal. Note that $det(B) = 1$, and so $B \in GL_n(\mathbb{Z})$; correspondingly, so is $M$. I claim that $(\vec{u}, M)$ is a cyclic pair for $\Lambda$. -Proof: Write $\vec{v}_i := U^{-1} d_i \vec{e}_i$ for each $i$. Note that $\vec{u} = \vec{v}_1$. Then each $\vec{v}_i \in \text{im}(U^{-1}D) = \text{im}(U^{-1}DV^{-1}) = \text{im}(M) = \Lambda$; further, it's not hard to see that they generate $\Lambda$ (by the same reasoning taken backwards). -We can check that $M \vec{v}_i = U^{-1}BU U^{-1} d_i \vec{e}_i = U^{-1} B d_i \vec{e}_i = U^{-1} (d_i \vec{e}_i + d_{i + 1} \vec{e}_{i + 1}) = \vec{v}_i + \vec{v}_{i + 1}$ for $1 \leq i < n$. Noting the base case that $\vec{v}_1 = \vec{u}$, we can use induction to see that $\vec{v}_{i + 1} = M \vec{v}_i - \vec{v}_i \in \text{span}(\{M^j \vec{u}\}_{j = 0}^i)$. As the $\{\vec{v}_i\}$ are a generating set, $\text{span}(\{M^j \vec{u}\}_{j = 0}^{n - 1}) = \Lambda$, and we are done.<|endoftext|> -TITLE: Is there a theory in a finite language that is computably axiomatizable but not by a finite number of axiom schemas? -QUESTION [9 upvotes]: I was told to ask this question on mathoverflow. I asked on math stack exchange whether there is a computably axiomatizable theory that can't be axiomatized by a finite number of axiom schemas. I got an answer, but it was a theory in an infinite language. Now, I am asking whether there is an example in a finite language. - -EDIT by non-OP: this is the above-mentioned MSE question, and this answer gives the definition of "scheme" being used. - -REPLY [8 votes]: This answers complements Fedor Pakhamov's, who provided an example of a computable theory that is not axiomatizable by finitely many schemas. -Following up on the comment by Andreas Blass to the question: Vaught proved that if a theory $T$ is computable and has "a modicum of coding", then $T$ is axiomatizable by a scheme. Vaught's result was improved by Albert Visser, in the paper below, where "the modicum of coding" used by Vaught is reduced to the modest demand that $T$ interprets a non-surjective unordered pairing, where pairing need not be functional. -A. Visser, Vaught's theorem on axiomatizability by a scheme, The Bulletin of Symbolic Logic, vol. 18 (2012), pp. 382-402. -A preprint of Visser's paper can be found here.<|endoftext|> -TITLE: Number of permutations with longest increasing subsequences of length at most $n$ -QUESTION [10 upvotes]: Is there a known expression for, or a nontrivial upper bound on, the number of permutations in $S_k$ with longest increasing subsequence of length at most $n$? -Let $l(\sigma)$ denote the length of the longest increasing subsequence of a permutation $\sigma\in S_k$. It seems like a lot is known about $l(\sigma)$ for a random permutation (and its asymptotic scaling), but are there upper bounds on the number of permutations in $\sigma\in S_k$ with $l(\sigma)\leq n$. -Motivation/context for this question: the moments of traces of random unitaries. It is known that $\int dU |{\rm tr}(U)|^{2k} = k!$ for $k\leq n$, where we integrate over the unitary group $U(n)$ with respect to the Haar measure. More generally, for any $k$ and $n$ one may write the expression as [1] -$$ -\int dU |{\rm tr}(U)|^{2k} = \sum_{\lambda \vdash k,~\ell(\lambda)\leq n} \chi_\lambda(\mathbb{I})^2\,, -$$ -summing over integer partitions $\lambda$ of $k$ with length at most $n$, and where $\chi_\lambda(\mathbb{I})$ is the identity character with respect to $\lambda$. The RHS is then counting the number of pairs of Young tableaux with width $\leq n$, which is equivalent to counting the number of permutations in $S_k$ with no increasing subsequences longer than $n$. -I'm essentially interested in upper bounds on this quantity which are tighter than the trivial bound of $k!$. -[1] E. Rains, "Increasing Subsequences and the Classical Groups," Electron. J. Comb. 5 (1998) R12. http://eudml.org/doc/119270. - -REPLY [2 votes]: This relates to the Stanley-Wilf Conjecture (now a theorem). More generally you can consider $S_k(\sigma)$ the number of permutations of $k$ elements which do not contain the pattern given by the permutation $\sigma$. Here you are looking at the particular case $S_k(12\cdots(n+1))=:u_n(k)$. Exhaustive references on the subject are the books "Combinatorics of Permutations" by Bóna and "Patterns in Permutations and Words" by Kitaev. Theorem 4.10 in Bóna's book gives a very elementary combinatorial proof for the bound -$$ -u_n(k)\le n^{2k}\ . -$$ -A similar bound was conjectured by Arratia for any pattern $\sigma$ of length $n+1$ but this is known to fail for $\sigma=1324$. -Note that -the bound is trivial from the Haar integral formula because $U$ has eigenvalues of modulus one and so $|{\rm tr} (U)|\le n$. -Also, the numbers form a supermultiplicative sequence by a result of Arratia (same article as above). -The supermultiplicative property also follows from the Haar integral: the $S_k(12\cdots(n+1))$ sequence in $k$ being a Stieltjes moment sequence is log-convex. -I first thought that this fact (Feteke's Subadditive Lemma) combined with Regev's asymptotic formula might give a better exponential upper bound (rather than asymptotic). However one ends up with the same upper bound. -That's because Regev's formula gives, after computing a Selberg integral, -$$ -u_n(k)\sim -1!2!\cdots(n-1)!\ (2\pi)^{-\frac{n-1}{2}} -\ 2^{-\frac{n^2-1}{2}}\ n^{\frac{n^2}{2}}\ \frac{n^{2k}}{k^{\frac{n^2-1}{2}}} -$$ -when $k\rightarrow\infty$ (I took the formula from Stanley's ICM survey). -So the correct exponential growth of $n^{2k}$ is already in the trivial bound.<|endoftext|> -TITLE: What were Ramanujan's standard tricks/approaches to solving problems? -QUESTION [10 upvotes]: While trying to formulate an answer to this question, I realized I really have no idea how Ramanujan came up with his formulas. Bruce Berndt has a number of great expository articles, e.g., this one, but I couldn't discern how Ramanujan approached problems. There are famous stories about how solutions seemed to just pop into his head, e.g., when he quickly solved a tricky problem and was asked how, he responded: - -It was clear that the solution should obviously be a continued fraction; I then thought, which continued fraction? And the answer came to my mind. - -According to Wikipedia, Hardy said Ramanujan's results were "arrived at by a process of mingled argument, intuition, and induction, of which he was entirely unable to give any coherent account." In the same article, Ramanujan is quoted as saying "An equation for me has no meaning unless it expresses a thought of God" and crediting his mathematical abilities to his family goddess Namagiri Thayar. -Previous MathOverflow questions have asked how he came up with specific results, and in this mathoverflow question, Tim Chow said "Ramanujan is legendary for having an extraordinary, uncanny intuition, and it is natural to try to understand this intuition better." - -Question: Now that so many of Ramanujan's formulas have been verified, that his notebooks have been carefully studied, and that his results have been understood as a part of a larger theory, has anyone discerned a pattern or a set of standard tricks/approaches that might have been underlying how he came up with his results? - -What I've read from Berndt suggests that Ramanujan's work focused heavily on continued fractions, partition functions, asymptotic formulas, modular forms, zeta functions, $q$-series, Eisenstein series, and mock theta functions. I'd be happy for an answer in any of these individual areas. Berndt suggests Ramanujan worked on slate, and erased his work when finished, recording only the final formulas he discovered, so we can perhaps deduce that Ramanujan had some fairly compact way to do his work. -Side note: while it's fun to have stories of mathematicians so brilliant that no one can understand them, I don't think this is the right point of view if we want to make the field welcoming to newcomers. It's also not very satisfying from the point of view of really understanding what's going on in a field. I hope that one day the mathematical community will understand everything Ramanujan did, and now, 100 years after his death, I'm hoping there has been some progress on this goal. - -REPLY [8 votes]: This is an exposition of my comments via actual examples. I will present a few of the tricks which Ramanujan heavily used (all of these are algebraic manipulation and do not involve anything high brow). -Partial Fractions -Often Ramanujan used to derive partial fractions for many functions (usually made of circular/hyperbolic functions). He never mentioned explicitly the technique used but it appears it was based on analysis of the poles of the function. However this did not involve complex analysis and instead it was an extension of the method used for typical rational functions in such a way to avoid common pitfalls. Partial fractions were then heavily used to obtain many series by comparing coefficients. In particular his formula related to $\zeta(2n+1)$ is derived in this manner (see this thread). Another application of the technique is described here. -Multisection of Series -This involves splitting a power series into multiple series by grouping terms with powers modulo a given number $n$. -Ramanujan used the technique in a different manner by trying to analyze power series for $f(x^{1/n})$ and collecting terms which contain fractional powers of $x$. Using this approach he proved many properties of Rogers Ramanujan continued fraction and also obtained generating functions of $p(5n+4),p(7n+5)$. A nice application of this technique is presented here. -Simplification of algebraic expressions -If $x, y$ are two numbers connected by algebraic equation (in theory) of the form $P(x, y) =0$ where $P$ is a polynomial in $x, y$ with integer coefficients then Ramanujan would often try to guess simple functions like $u=f(x), v=g(y) $ so that the relation between $x, y$ could be transformed into a visually simple form as $F(u, v) =0$ where $F$ need not be polynomial but rather general algebraic function. -Here it appears that he worked by trial and error and put a lot of effort to simplify the form of the algebraic relation. This is clearly seen when one compares Ramanujan's class invariants with the corresponding ones given by Weber. His modular equations are also in much simpler form compared to those given by others. -In this connection it should also be noted that Ramanujan had discovered many algebraic identities which helped him to denest radicals. I don't think there was a technique involved here. The identities were developed in pursuit of specific goals like expressing a number as sum of two cubes in two different ways or in another case for finding simple expressions for singular moduli. Also he wasn't aware of any Galois theory and probably he did not need it. I guess he used his time and skill to figure out these by trial and error (discarding quickly anything which did not seem to meet the desired goal). - -Note: Some of the examples presented above are available on Math.SE and I will add links to them after some time.<|endoftext|> -TITLE: Seeking a combinatorial proof for a binomial identity -QUESTION [8 upvotes]: Let $n\geq m\geq0$ be two integers. The below binomial identity is provable by other means: -$$\sum_{j=0}^m(-1)^j\binom{n+1}j2^{m-j} -=\sum_{j=0}^m(-1)^j\binom{n-m+j}j.$$ - -QUESTION. Can you provide a combinatorial proof for the above identity? I would be thrilled to see as many as possible. - -POSTSCRIPT. I enjoyed the two solutions by Ira & Fedor. Still, more alternating proofs are welcome. - -REPLY [3 votes]: I think I can, if you permit me to multiply it by $2^{n+1-m}$. Then we want to prove $$P:=\sum_{j=0}^m(-1)^j\binom{n+1}j2^{n+1-j} -=2^{n+1-m}\sum_{j=0}^m(-1)^j\binom{n-m+j}j=:Q.$$ -Denote $X=\{1,2,\ldots,n+1\}$, then -$$ -P=\sum_{B\subset A\subset X,|B|\leqslant m} (-1)^{|B|}. -$$ -Fix $A$, denote $a=\max(A)$, and partition possible $B$'s onto pairs of the form $\{C,C\sqcup a\}$, where $C\subset A\setminus \{a\}$. All $B$'s are partitioned onto pairs except those for which $|B|=m$ and $a\notin B$. The sum in each pair 0, therefore -$$ -P=1+(-1)^m|B\subset A\subset X,|B|=m,\max(A)\notin B|. -$$ -Extra 1 comes from the case $B=A=\emptyset$, for which $\max(A)$ does not exist. -Now about $Q$. Consider $B\subset X$, $|B|=m$, and denote by $m-j+1$ the minimal element of $\overline{B}:=X\setminus B$. For fixed $j$, there exist exactly ${n-m+j\choose j}$ such sets $B$. Each of them has $2^{n+1-m}$ oversets $A$. Therefore -$$ -Q=\sum_{B\subset A\subset X,|B|=m} (-1)^{\min(\overline{B})+m+1}. -$$ -Consider the "dominos" $\{1,2\}$, $\{3,4\}$, $\ldots$, and take the first domino which is not contained in $B$. If it contains exactly 1 element from $B$, we may switch this element to the other element of the same domino, and $\min(\overline{B})$ changes its parity. This cancellation in the sum for $Q$ lefts only those $B$'s for which FNFDE (the first not-full domino is empty). Therefore -$$ -Q=(-1)^m|B\subset A\subset X,|B|=m,FNFDE|. -$$ -So $P=Q$ reduces to -$$ -(-1)^m+|B\subset A\subset X,|B|=m,\max(A)\notin B|=|B\subset A\subset X,|B|=m,FNFDE|. -$$ -Subtracting the common part, we should prove that -$$ -(-1)^m+|B\subset A\subset X,|B|=m,\max(A)\notin B,\,\text{not}\, FNFDE|=\\ -|B\subset A\subset X,|B|=m,\max(A)\in B,FNFDE|. -$$ -Fix the first not full domino $\{s,s+1\}$ and $a=\max(A)$. If $a\leqslant s+1$, there is unique possibility which gives $(-1)^m$. Otherwise, if we fix also $B_0:=B\setminus \{s,s+1,a\}$ (it is some set of size $m-1$), and $A_0:=A\setminus \{s,s+1\}$ such that $B_0\subset A_0$, there exist exactly 4 ways to complete the choice of the pair $(B_0,A_0)$ to $(B,A)$ both for the condition $\{\max(A)\notin B,\,\text{not}\, FNFDE\}$ (choose which of $\{s,s+1\}$ belongs to $B$ where another guy from the domino belongs to $A$); and for the condition $\{\max(A)\in B,FNFDE\}$ (choose which of $s,s+1$ belongs to $a$). This proves the result.<|endoftext|> -TITLE: Does bounded Zermelo construct any cumulative hierarchy? -QUESTION [10 upvotes]: ZF is sufficient to construct the von Neumann hierarchy, and prove that every set appears at some stage $V_\alpha$. This is the basis for Scott's trick, for instance. But how much of ZF is needed? Is bounded Zermelo/Mac Lane set theory enough, no Choice assumed? I know Foundation is necessary, and I'm not getting rid of that. I've seen something called the "rank axiom" in discussion of second-order version of original Zermelo (these notes), but I'm sure people have finely calibrated what precisely is needed. -To be honest, all I really want is an ordinal-valued rank function such that sets of rank at most $\alpha$ form a set, for all $\alpha$, and all sets have a rank. So if the von Neumann hierarchy doesn't work, I'm happy to work with something else (and for 'ordinals', I don't need von Neumann ordinals). - -REPLY [8 votes]: One can directly assume in Zermelo that every set belongs to a rank. This does not add any strength at all. But notice that Zermelo may not prove the existence of very many von Neumann ordinals: there may be well-orderings which do not have a von Neumann ordinal as order type. A natural model of this theory is the union -of the V^{omega + n}'s for n a natural number. Notice that in this structure the axioms of Zermelo hold, every object belongs to a V_alpha, but omega+omega does not exist. There are well-orderings with order types far higher than omega+omega (and in this context the Scott representation of the ordinals is available). -There are a couple of additional remarks. It is interesting to note -that the assertion that every set has a rank adds no strength at all to -Zermelo set theory (or to Zermelo set theory with bounded separation) but that adding this assertion to KP, a theory much weaker than Zermelo, gives a theory much stronger than Zermelo. The reason is that KP has a lot of replacement. -The natural way to describe the rank function in "Zermelo with ranks" is probably to use the Scott ordinals as values of the rank function but note that the rank function is not necessarily onto the ordinals. In the absence of replacement, the von Neumann notion of ordinal simply isn't the right notion of ordinal number.<|endoftext|> -TITLE: Direct image of the structure sheaf by an endomorphism of $\mathbb{P}^2$ -QUESTION [14 upvotes]: Let us take 3 quadratic forms on $\mathbb{P}^2$ with no common zero; they define a map $\pi : \mathbb{P}^2\rightarrow \mathbb{P}^2$ of degree 4. It is not difficult to see that $\pi _*\mathscr{O}_{\mathbb{P}^2}\cong \mathscr{O}_{\mathbb{P}^2}\oplus \mathscr{O}_{\mathbb{P}^2}(-1)^3$. Does anyone know how to write down the algebra stucture of $\pi _*\mathscr{O}_{\mathbb{P}^2}$ in terms of this decomposition? - -REPLY [2 votes]: In general, if $f \colon X \to Y$ is a quadruple cover of algebraic varieties, the structure of the algebra $f_* \mathcal{O}_X$ is studied in the paper -D. Hahn, R. Miranda: Quadruple Covers in Algebraic Geometry, Journal of Algebraic Geometry 8 (1999). -You can download the paper from the webpage of the second author.<|endoftext|> -TITLE: Summing infinitely many infinitesimally small variables makes sense in algebra -QUESTION [46 upvotes]: There is an identity $e^x=\lim_{n\to \infty} (1+x/n)^n$, and I always thought it is a purely analytic statement. But then I discovered its curious interpretation in pure algebra: -Consider the ring of formal infinite sums of monomials in infinitely many variables $\varepsilon_1, \varepsilon_2,\ldots$ satisfying $\varepsilon_i^2=0$. -$$ -R=\mathbb{Q}[\![\varepsilon_1, \varepsilon_2, \ldots]\!]/(\varepsilon_i^2: i=1,2,\ldots). -$$ -Then the sum $x=\sum_{i=1}^\infty \varepsilon_i$ makes sense and is not infinitesimally small, in fact we have -$$ -x^n = n! \sum_{1\leq i_11$, where $p_n$ is the power sum symmetric function $x_1^n+x_2^n+\cdots$. Then $\ex$ is the restriction to symmetric functions of the homomorphism on all formal power series in $x_1,x_2,\dots$ that takes each $x_i^2$ to 0 (where $t$ is the image of $p_1$). It has the property that for any symmetric function $f$, -$$\ex(f) = \sum_{n=0}^\infty [x_1x_2\cdots x_n] f \frac{t^n}{n!},$$ -where $[x_1x_2\cdots x_n] f$ denotes the coefficient of $x_1x_2\cdots x_n$ in $f$. In particular, $\ex(h_n) = \ex(e_n) = t^n/n!$ where $h_n$ and $e_n$ are the complete and elementary symmetric functions. -This idea is well known and is very useful in enumerative combinatorics. It allows one to derive exponential generating functions for objects with distinct labels (e.g., permutations or standard Young tableaux) from symmetric function generating functions for objects with repeated labels (e.g., words or semistandard tableaux). There are related homomorphisms that preserve more information; see, for example, section 7.8 of Stanley.<|endoftext|> -TITLE: Upper bounds for number of integral points on short Weierstraß elliptic curve? -QUESTION [6 upvotes]: I hope this question is ok, to post it here, otherwise I will move it to MSE: -Let $y^2 = x^3+ax+b$ be an elliptic curve $E$ where $a,b \in \mathbb{Z}$. -Let $M_x = \max_{x}|x|$ and $M_y = \max_{y}|y|$ where the maximum runs through the integral points $P=(x,y)$ of $E$. If there are no integral points set $M_x = M_y = 0$. Let $N(E)$ be the number of integral points. I have done some small experiments in Sagemath for $a=-n^2,b=n^2, 1 \le n \le 100$ and there seems to hold this inequality: -$$ N(E) < 2 \log( (M_x+1)(M_y+1)+1)$$ - -Question: - -Is there any reason or heuristic this could be true or are there elliptic curves where this is not true? -Here is a picture: -On the x-axis are the number of integral points, on the y-axis is the upper bound above: - -More background why I am interested in this: -Let $R = \operatorname{rad}(\gcd(a,b))$ -Then for each prime $p | \gcd(a,b)$ we have -$$y^2 \equiv x^3 \mod(p)$$ -The number of such solutions $N_p$ is $\ge p$ as $(x,y) = (t^2,t^3)$ for each $t \in \mathbb{F}_p$ are such solutions. My conjecture, which I think can be proven since $\mathbb{F}^*_p$ is cyclic, is that $N_p = p$. -From this it follows that for each integral point $(x,y)$ on $E$, we can find $t,k,l \in \mathbb{Z}$ such that (if $R=1$ set $t=0$): -$$x = t^2+k R$$ -$$y=t^3+lR$$ -In fact, using the Chinese Remainder Theorem, given an integral point $(x,y)$ on $E$, we can compute: -$$\sqrt{x} \equiv t_i \mod(p_i)$$ -$$(y)^{\frac{1}{3}} \equiv t_i \mod(p_i)$$ -for $i=1,\cdots,r$ ,where $R = p_1\cdots p_r$. And we can find $0 \le t \le R-1$. -From this we get upper bounds for $|x|,|y|$: -$$|x| \le R(R+|k|)$$ -and using the elliptic equation we get: -$$|y| \le \sqrt{R^3(R+|k|)^3 + |a|R(R+|k|)+|b|}$$ -Now comes the cheating: -Let $M_k := \max_{k} |k|$ and set -$$B_x := R(R+M_k)$$ -$$B_y := \sqrt{R^3(R+M_k)^3 + |a|R(R+M_k)+|b|}$$ -With the notation of above it is $M_x \le B_x$ and $M_y \le B_y$. - -Question: -Is there any way that we can upper bound $M_k$, if this is not asked too much? - -Thanks for your help! - -REPLY [8 votes]: An LMFDB search for curves with many integer points turns up the curve -20888a1: -$y^2 = x^3 - 52 x + 100$ which has $52$ integral points, -a bit more than your conjectured bound of about $47.052$ -using $(M_x, M_y) = (12214, 1349854)$. -It does seem to be true that curves with many integer points tend to have -a few large ones, but I don't know of any quantitative statement -in this direction. -There are known upper bounds on the size of an integral point on -an elliptic curve, but they are usually quite large -(though still useful for provably listing all integer points).<|endoftext|> -TITLE: Results in “generalised smooth spaces” that did not hold in the case of smooth manifolds -QUESTION [8 upvotes]: Consider the category of smooth manifolds $\text{Man}$. I quote from n-lab page: - -Manifolds are fantastic spaces. It’s a pity that there aren’t more of them. - -I understand that this category $\text{Man}$ is not well behaved in more than one sense or do not have enough objects, for it to be - -closed under pullback, -to have mapping space, an appropriate smooth structure on $\text{Map}(X,Y)$ for manifolds $X$ and $Y$. - -Then, people added more spaces to the category of manifolds, in an attempt to make sure the resulting category has (some) of the nice properties which the category $\text{Man}$ did not had. Some examples are - -Chen spaces (On the proof of "Mapping space is a Chen space"), -Differentiable spaces (I saw the first in the paper, section $2.7$) which are sheaves over the category $\text{Man}$ that are Differentiable stacks over the category $\text{Man}$ (recall that, any manifold is a sheaf over the category $\text{Man}$ that are Differentiable stacks over the category $\text{Man}$). -Frölicher spaces. These are introduced to have a Cartesian closed category (please correct me if I have misunderstood something). - -Question : Are there any (What are the) results that hold in these generalised spaces whose counterparts does not hold true in the set up of smooth manifolds? -There is one result (Lemma $2.35$ in above paper) I am aware of that holds true for Differentiable spaces but there is no appropriate counterpart for smooth manifolds. -Sub questions : - -It looks like diffeological spaces are introduced not to “enrich” (not sure if it is correct word) the category of manifolds, but actually to study sheaves on the category of manifolds. Is that correct? I am not sure to what extent this question is making sense, so feel free to ask for more clarification or ignore it. -I also observe similarity with the notion of “Algebriac spaces”. Those were also (roughly) defined (similar to Differentiable spaces) as sheaves of particular kind (over some appropriate site). I think there are more than a handful of results that holds true in Algebriac spaces but not in the category $\text{Sch}/S$. You can also add them, but I am not sure if I can appreciate them enough. - -REPLY [5 votes]: There are many such results. -Consider some smooth manifolds M and N. -The internal hom Hom(M,N) is a sheaf on smooth manifolds. -We can compute its tangent bundle, -and it turns out that the tangent space at some point f in Hom(M,N), -i.e., f:M→N is a smooth map, equals the vector space -of smooth sections of the vector bundle f*TN. -This is the expected result, but the setting of sheaves allows us to make it completely rigorous and precise with minimal technicalities. -Now take M=N and consider the open subobject of Hom(M,M) consisting of diffeomorphisms. -This is a group object (i.e., an infinite-dimensional Lie group) -and its Lie algebra is precisely the Lie algebra of vector fields on M. -Differential k-forms form a sheaf Ω^k on smooth manifolds. -In particular, morphisms Hom(M,N)→Ω^k are differential k-forms on the infinite-dimensional space of smooth maps M→N. -We also immediately obtain the de Rham complex on Hom(M,N) -in the same manner, and it satisfies the expected properties. -Liekwise, we have a sheaf of groupoids B_∇(G) of principal G-bundles -with connection. -Maps Hom(M,N)→B_∇(G) are principal G-bundles with connection -over the infinite-dimensional space of smooth maps M→N. -Hopkins and Freed compute the de Rham complex of B_∇(G), -and it turns out to be the vector space of invariant polynomials -on the Lie algebra of G. -This means, for instance, that you can immediately start -computing Chern–Weil forms of principal G-bundles with connection -on Hom(M,N), for example. -Now, we can also take G to be any group object in sheaves, -such as, for example, the group Diff(M) of diffeomorphisms -of M considered above. -This immediately allows us to consider principal G-bundles -with connection for such groups. -Other objects that can be encoded in this setting -include the (higher) sheaves of bundle (n-1)-gerbes with connection -and structure abelian Lie group A, denoted by B_∇^n(A). -Morphisms M→B_∇^n(A) are precisely bundle (n-1)-gerbes with connection over M. -Now you can talk about bundle (n-1)-gerbes with connection over Hom(M,N). -The Cheeger–Simons differential refinement of the Chern character -in this language is a morphism B_∇(G)→B_∇^n(A), etc. -So in particular, not only de Rham cohomology, but also differential cohomology make sense in this framework.<|endoftext|> -TITLE: Example of a projective module with non-superfluous radical -QUESTION [5 upvotes]: Let $R$ be a ring with unit. A submodule $N$ of an $R$-module $M$ is called superfluous if the only sumbodule $T$ of $M$ for which $N+T = M$ is $M$ itself. -It is shown, for example, in -[1] F. W._Anderson, K. R. Fuller "Rings and Categories of Modules" (1974) -that if every submodule of $M$ is contained in a maximal submodule, then the radical of $M$ is superfluous (Proposition 9.18). This, in particular, implies that for every finitely generated module $M$ its radical is superfluous. In exercise 9.2. it is explained that divisible abelian groups coincide with their radicals, and therefore their radicals are not superfluous. Divisible abelian groups are not projective objects. -I was curious if it is possible to construct a projective module with non-superfluous radical. - -Question: is there an example of a ring $R$ and a projective $R$-module -$P$ such that the radical $JP$ of $P$ is not superfluous? - -The existence of such module (or, at least, that its non-existence is non-obvious) is somehow hinted by the formulation of Corollary 17.12 in [1]: - -Let $J = J(R)$. If $P$ is a projective left $R$-module such that $JP$ -is superfluous in $P$ (e.g., if ${}_RP$ is finitely generated), then -$J(End({}_RP)) = Hom_R(P,JP)$ and $End({}_RP)/J(End_RP) \cong - End({}_RP/JP)$. - -REPLY [3 votes]: According to Proposition 17.10 in the Anderson-Fuller book (I am using the 1992 second edition; don't know if the first 1974 edition is any different), for any projective module $P$ over any (unital associative) ring $R$, the radical of $P$ is computable as $Rad\,P=JP$, where $J$ is the Jacobson radical of the ring $R$ (just as you say). -Let $p$ be a prime number. Consider the commutative ring $R=\mathbb Z_{(p)}$, that is, the localization of the ring of integers $\mathbb Z$ at the prime ideal $(p)\subset\mathbb Z$. Alternatively, one can consider the ring of $p$-adic integers $R=\mathbb Z_p$, that is the completion of the local ring $\mathbb Z_{(p)}$ at its maximal ideal. In both cases, $J(R)=pR$ is the unique maximal ideal of $R$. -Consider the free $R$-module $F$ with a countable set of generators, $F=R^{(\omega)}$. Let us show that $JF$ is not superfluous in $F$. For this purpose, we will construct a proper submodule $T\subset F$ such that $JF+T=F$. -Consider the $R$-module $Q=R[p^{-1}]$. In other words, $Q$ is just the ring of fractions of the local domain $R$. The $R$-module $Q$ is generated by the sequence of elements $1$, $p^{-1}$, $p^{-2}$, $\dots$; so $Q$ is a countably generated $R$-module. Hence $Q$ is a quotient $R$-module of the $R$-module $F$. -Denote by $T\subset F$ a submodule such that $F/T\cong Q$. So we have a short exact sequence of $R$-modules $0\to T\to F\to Q\to 0$. We want to check that $T+JF=F$. -Indeed, we have $JF=pF$, since $J=pR$. The desired equation $T+pF=F$ is equivalent to $p(F/T)=F/T$. Now $F/T\cong Q$ and we have $pQ=Q$ by construction. -In fact, as it is clear now, any discrete valuation ring can be used in the role of $R$ in this construction (with a prime number $p$ replaced by any uniformizing element).<|endoftext|> -TITLE: Maximality principle in symmetric extensions -QUESTION [5 upvotes]: Let $M$ be a ctm and $P\in M$ a forcing order. -In regular forcing extensions, we have the following well-known Principle: -$$p\Vdash_{M,P}\exists x\phi[x]\;\Longrightarrow\;\exists\sigma\in M^P\;p\Vdash_{M,P}\phi[\sigma]$$ -(where $M^P$ is the class of $P$-names in $M$). -Given an automorphism $f$ of $P$, we can turn $f$ into a function mapping $P$-names to $P$-names, given by -$$\overline{f}(\tau)=\{(\overline{f}(\sigma),f(p))\;|\;(\sigma,p)\in\tau\}$$ -Given a subgroup $H$ of $Aut(P)$, we say that $\tau$ is $H$-invariant iff $\overline{f}(\tau)=\tau$ for all $f\in H$. -We can also fix a filter $\mathcal{F}$ on the set of subgroups of $Aut(P)$ and consider the so-called $\mathcal{F}$-symmetric extension of $M$ given by the evaluation only of those names $\tau$ that are hereditarily $\mathcal{F}$-symmetric, i.e. for some $H\in \mathcal{F}$, $\tau$ is $H$-invariant and for all $(\sigma,p)\in\tau$, $\sigma$ is hereditarily $\mathcal{F}$-symmetric. It can be shown that the resulting model always satisfies ZF, but not always AC. -My question now is whether or not the following principle can be shown to hold (in $M$): Given $p\in P$ such that $p\Vdash_{M,P}^{\mathcal{F}}\exists x\phi[x]$ (i.e. for any $P$-generic filter $G$ containing $p$, there is a hereditarily $\mathcal{F}$-symmetric name $\tau$ such that $M[G]\models\phi[\tau]$), does there necessarily exist a hereditarily $\mathcal{F}$-symmetric name $\sigma$ such that $p\Vdash\phi[\sigma]$? -As far as i can tell, the proof for the usual case can not be modified to prove the symmetric case, since we "stitch together" witnesses along a maximal antichain, which might not be $H$-invariant for any $H\in\mathcal{F}$. - -REPLY [4 votes]: No. Not even remotely. -Consider the Cohen model, i.e. add $\omega$ Cohen reals, permute them amongst themselves, and take finite supports. Let $\dot a_n$ be the canonical name of a Cohen real, and let $\dot A$ be the name of the set of Cohen reals. -$$1\Vdash\exists x(x\in\dot A\land\check 0\in x)$$ -But there is no symmetric name that we can instantiate. If $\dot x$ would be such name, then it is easy to see that its value has to be one of finitely many reals, but then just take a condition which decides that $\check 0\notin\dot a_n$ for any of those finitely many reals.<|endoftext|> -TITLE: On permanents and determinants of finite groups -QUESTION [22 upvotes]: $\DeclareMathOperator\perm{perm}$Let $G$ be a finite group. Define the determinant $\det(G)$ of $G$ as the determinant of the character table of $G$ over $\mathbb{C}$ and define the permanent $\perm(G)$ of $G$ as the permanent of the character table of $G$ over $\mathbb{C}$. Note that due to the properties of the determinant and the permanent, this definition just depends on $G$ and not on the ordering of the conjugacy classes etc. -I'm not experienced with character theory but did some experiments with GAP on this and found nothing related in the literature, which motivates the following questions (sorry, in case they are trivial). Of course finite groups are dangerous and it is tested just for all finite groups of order at most $n \leq 30$ and some other cases, which might not be too good evidence for a question on finite groups. - -Question 1: Are $\perm(G)$ and $\det(G)^2$ always integers? - -I was able to prove this for cyclic groups. Since the character table of the direct product of groups is given by their Kroenecker product, one can conclude that $\det(G)^2$ is also an integer for all abelian groups $G$. Maybe there is a formula for the permanent of the Kroenecker product of matrices to conclduge that $\perm(G)$ is also an integer for all abelian groups or even better a more direct proof that question 1 is true at least for abelian groups. Note that $\det(G)$ is in general not an integer, even for cyclic groups. -Now call a finite group permfect in case $\perm(G)=0$. - -Question 2: Is it true that all finite groups of order $n$ are permfect if and only if $n=4r+2$ for some $r \geq 2$? - -Being permfect could be seen as having a high symmetry. It seems symmetric groups are permfect and for the alternating groups I only found $A_6$ to be permfect yet. - -REPLY [4 votes]: For the sake of completeness, here is the answer to Question 1, part of which -is missing from the other answers: - -Proposition 1. Let $G$ be a finite group. Consider the representations of -$G$ over $\mathbb{C}$. Let $\det G$ denote the determinant of the character -table of $G$. (Note that this is only defined up to sign, since the order of -the rows and of the columns of the character table can be chosen arbitrarily.) -Let $\operatorname*{perm}G$ denote the permanent of the character table of -$G$. Then, $\left(\det G\right)^2$ and $\operatorname*{perm}G$ are integers. - -To prove this, we need the following lemmas: - -Lemma 2. Let $G$ be a finite group. Then, there is a finite Galois field -extension $\mathbb{F}$ of $\mathbb{Q}$ such that all irreducible -representations of $G$ are defined over $\mathbb{F}$. - -Proof of Lemma 2. There is a finite Galois field extension $\mathbb{K}$ of -$\mathbb{Q}$ such that all irreducible representations of $G$ are defined over -$\mathbb{K}$. Indeed, this is known as a splitting field of $G$; its -existence is part of Theorem 9.2.6 in Peter Webb, A Course in Finite Group -Representation Theory, 2016. -Consider this field extension $\mathbb{K}$. Let $\mathbb{F}$ be the Galois -closure of $\mathbb{K}$ over $\mathbb{Q}$ (or any other finite field extension -of $\mathbb{Q}$ that is Galois over $\mathbb{Q}$ and contains $\mathbb{K}$ as -a subfield). Then, all irreducible representations of $G$ are defined over -$\mathbb{F}$ (since they are defined over $\mathbb{K}$, but $\mathbb{F}$ -contains $\mathbb{K}$ as a subfield). This proves Lemma 2. $\blacksquare$ - -Lemma 3. Let $G$ be a finite group. Let $\mathbb{F}$ be a field extension -of $\mathbb{Q}$ such that all irreducible representations of $G$ -are defined over $\mathbb{F}$. Let $\chi:G\rightarrow\mathbb{F}$ be an -irreducible character of $G$. Let $\gamma:\mathbb{F}\rightarrow\mathbb{F}$ be -a $\mathbb{Q}$-algebra automorphism of $\mathbb{F}$. Then, $\gamma\circ -\chi:G\rightarrow\mathbb{F}$ is an irreducible character of $G$. - -Proof of Lemma 3. This is a totally straightforward "isomorphisms preserve -all relative properties of objects they are applied to" argument, but for the -sake of completeness, let me spell it out (at least to some level of detail): -The map $\chi$ is an irreducible character of $G$, and thus is the character -of an irreducible representation $\rho$ of $G$. Consider this $\rho$, and WLOG -assume that $\rho$ is a representation over $\mathbb{F}$. (This can be assumed -since all irreducible representations of $G$ are defined over $\mathbb{F}$.) -Thus, $\rho$ is a group homomorphism from $G$ to $\operatorname*{GL} -\nolimits_{n}\left( \mathbb{F}\right) $ for some $n\geq1$. Consider this $n$. -The $\mathbb{Q}$-algebra automorphism $\gamma:\mathbb{F}\rightarrow\mathbb{F}$ -induces a group automorphism $\widetilde{\gamma}:\operatorname*{GL} -\nolimits_{n}\left( \mathbb{F}\right) \rightarrow\operatorname*{GL} -\nolimits_{n}\left( \mathbb{F}\right) $ that transforms each matrix in -$\operatorname*{GL}\nolimits_{n}\left( \mathbb{F}\right) $ by applying -$\gamma$ to each entry of the matrix. The composition $\widetilde{\gamma} -\circ\rho:G\rightarrow\operatorname*{GL}\nolimits_{n}\left( \mathbb{F} -\right) $ is a group homomorphism (since $\widetilde{\gamma}$ and $\rho$ are -group homomorphisms), and thus is a representation of $G$. Moreover, the -character of this representation $\widetilde{\gamma}\circ\rho$ is $\gamma -\circ\chi$ (since $\operatorname*{Tr}\left( \widetilde{\gamma}\left( -A\right) \right) =\gamma\left( \operatorname*{Tr}A\right) $ for any matrix -$A\in\operatorname*{GL}\nolimits_{n}\left( \mathbb{F}\right) $). We shall -now show that this representation $\widetilde{\gamma}\circ\rho$ is irreducible. -Indeed, let $U$ be a subrepresentation of $\widetilde{\gamma}\circ\rho$ -- -that is, an $\mathbb{F}$-vector subspace of $\mathbb{F}^{n}$ that is invariant -under the action of $\widetilde{\gamma}\circ\rho$. Consider the $\mathbb{Q} -$-module isomorphism $\overline{\gamma}:\mathbb{F}^{n}\rightarrow -\mathbb{F}^{n}$ that applies $\gamma$ to each coordinate of the vector. Since -$\gamma$ is a $\mathbb{Q}$-algebra homomorphism, we can easily see that -$\left( \widetilde{\gamma}\left( A\right) \right) \left( \overline -{\gamma}\left( v\right) \right) =\overline{\gamma}\left( Av\right) $ for -each $A\in\operatorname*{GL}\nolimits_{n}\left( \mathbb{F}\right) $ and each -$v\in\mathbb{F}^{n}$. Thus, we can easily see that $\overline{\gamma} -^{-1}\left( U\right) $ is an $\mathbb{F}$-vector subspace of $\mathbb{F} -^{n}$ that is invariant under the action of $\rho$ (since $U$ is an -$\mathbb{F}$-vector subspace of $\mathbb{F}^{n}$ that is invariant under the -action of $\widetilde{\gamma}\circ\rho$). In other words, $\overline{\gamma -}^{-1}\left( U\right) $ is a subrepresentation of $\rho$. Since $\rho$ is -irreducible, this entails that either $\overline{\gamma}^{-1}\left( U\right) -=0$ or $\overline{\gamma}^{-1}\left( U\right) =\mathbb{F}^{n}$. Since -$\overline{\gamma}$ is an isomorphism, we thus conclude that either $U=0$ or -$U=\mathbb{F}^{n}$. -Forget that we fixed $U$. We thus have shown that if $U$ is a -subrepresentation of $\widetilde{\gamma}\circ\rho$, then either $U=0$ or -$U=\mathbb{F}^{n}$. In other words, the representation $\widetilde{\gamma -}\circ\rho$ is irreducible (since its dimension is $n\geq1$). Thus, its -character is an irreducible character of $G$. In other words, $\gamma\circ -\chi$ is an irreducible character of $G$ (since $\gamma\circ\chi$ is the -character of $\widetilde{\gamma}\circ\rho$). This proves Lemma 3. -$\blacksquare$ -Proof of Proposition 1. Lemma 2 shows that there is a finite Galois field -extension $\mathbb{F}$ of $\mathbb{Q}$ such that all irreducible -representations of $G$ are defined over $\mathbb{F}$. Consider this -$\mathbb{F}$. Let $\Gamma$ be the Galois group $\operatorname*{Gal}\left( -\mathbb{F}/\mathbb{Q}\right) $ (which consists of all $\mathbb{Q}$-algebra -automorphisms of $\mathbb{F}$). The Fundamental Theorem of Galois Theory shows -that the invariant ring $\mathbb{F}^{\Gamma}$ is $\mathbb{Q}$. -Let $\chi_{1},\chi_{2},\ldots,\chi_{r}$ be all irreducible characters of $G$ -(listed without repetition). Note that these characters are maps from $G$ to -$\mathbb{F}$ (since all irreducible representations of $G$ are defined over -$\mathbb{F}$). -Let $c_{1},c_{2},\ldots,c_{r}$ be the conjugacy classes of $G$ (listed without repetition). -Let $\operatorname*{per}A$ denote the permanent of any square matrix $A$. -Let $C$ be the matrix $\left( \chi_{i}\left( c_{j}\right) \right) _{1\leq -i\leq r,\ 1\leq j\leq r}\in\mathbb{F}^{r\times r}$. This matrix $C$ is the -character table of $G$ (for the ordering of characters given by $\chi_{1} -,\chi_{2},\ldots,\chi_{r}$ and the ordering of conjugacy classes given by -$c_{1},c_{2},\ldots,c_{r}$). Thus, the definition of $\operatorname*{perm}G$ -shows that $\operatorname*{perm}G$ is the permanent of $C$. In other words, -$\operatorname*{perm}G=\operatorname*{per}C$. -Let $\gamma\in\Gamma$. Thus, $\gamma$ is a $\mathbb{Q}$-algebra automorphism -of $\mathbb{F}$ (since $\gamma\in\Gamma=\operatorname*{Gal}\left( -\mathbb{F}/\mathbb{Q}\right) $). We shall show that $\gamma\left( -\operatorname*{perm}G\right) =\operatorname*{perm}G$. -The $\mathbb{Q}$-algebra automorphism $\gamma:\mathbb{F}\rightarrow\mathbb{F}$ -induces a $\mathbb{Q}$-algebra automorphism $\widetilde{\gamma}:\mathbb{F}^{r\times -r}\rightarrow\mathbb{F}^{r\times r}$ that transforms each matrix in -$\mathbb{F}^{r\times r}$ by applying $\gamma$ to each entry of the matrix. -We define a map $f:\left\{ 1,2,\ldots,r\right\} \rightarrow\left\{ -1,2,\ldots,r\right\} $ as follows: -Let $i\in\left\{ 1,2,\ldots,r\right\} $. Then, $\chi_{i}:G\rightarrow -\mathbb{F}$ is an irreducible character of $G$. Thus, Lemma 3 (applied to -$\chi=\chi_{i}$) shows that $\gamma\circ\chi_{i}:G\rightarrow\mathbb{F}$ is an -irreducible character of $G$. Hence, $\gamma\circ\chi_{i}=\chi_{j}$ for some -$j\in\left\{ 1,2,\ldots,r\right\} $ (since $\chi_{1},\chi_{2},\ldots -,\chi_{r}$ are all irreducible characters of $G$). This $j$ is uniquely -defined. We define $f\left( i\right) $ to be $j$. -Thus, we have defined a map $f:\left\{ 1,2,\ldots,r\right\} \rightarrow -\left\{ 1,2,\ldots,r\right\} $ with the property that -\begin{equation} -\gamma\circ\chi_{i}=\chi_{f\left( i\right) }\qquad\text{for each } -i\in\left\{ 1,2,\ldots,r\right\} . -\label{eq.darij1.1} -\tag{1} -\end{equation} -If two distinct elements $i$ and $j$ of $\left\{ 1,2,\ldots,r\right\} $ -would satisfy $f\left( i\right) =f\left( j\right) $, then they would -satisfy $\gamma\circ\chi_{i}=\gamma\circ\chi_{j}$ (by \eqref{eq.darij1.1} -) and therefore $\chi_{i}=\chi_{j}$ (since $\gamma$ is invertible), which -would contradict the fact that $\chi_{1},\chi_{2},\ldots,\chi_{r}$ are -distinct. Thus, two distinct elements $i$ and $j$ of $\left\{ 1,2,\ldots -,r\right\} $ always satisfy $f\left( i\right) \neq f\left( j\right) $. In -other words, the map $f$ is injective. Hence, $f$ is a permutation (since $f$ -is an injective map from $\left\{ 1,2,\ldots,r\right\} $ to $\left\{ -1,2,\ldots,r\right\} $). -Now, from $C=\left( \chi_{i}\left( c_{j}\right) \right) _{1\leq i\leq -r,\ 1\leq j\leq r}$, we obtain -\begin{align*} -\widetilde{\gamma}\left( C\right) =\left( \gamma\left( \chi_{i}\left( -c_{j}\right) \right) \right) _{1\leq i\leq r,\ 1\leq j\leq r}=\left( -\chi_{f\left( i\right) }\left( c_{j}\right) \right) _{1\leq i\leq -r,\ 1\leq j\leq r}, -\end{align*} -since each $i,j\in\left\{ 1,2,\ldots,r\right\} $ satisfy -\begin{align*} -\gamma\left( \chi_{i}\left( c_{j}\right) \right) =\left( \gamma\circ -\chi_{i}\right) \left( c_{j}\right) =\chi_{f\left( i\right) }\left( -c_{j}\right) \qquad\left( \text{by \eqref{eq.darij1.1}}\right) . -\end{align*} -Thus, the matrix $\widetilde{\gamma}\left( C\right) $ is obtained from the -matrix $\left( \chi_{i}\left( c_{j}\right) \right) _{1\leq i\leq r,\ 1\leq -j\leq r}$ by permuting the rows (since $f:\left\{ 1,2,\ldots,r\right\} -\rightarrow\left\{ 1,2,\ldots,r\right\} $ is a permutation). In other words, -the matrix $\widetilde{\gamma}\left( C\right) $ is obtained from the matrix -$C$ by permuting the rows (since $C=\left( \chi_{i}\left( c_{j}\right) -\right) _{1\leq i\leq r,\ 1\leq j\leq r}$). Hence, $\operatorname*{per} -\left( \widetilde{\gamma}\left( C\right) \right) =\operatorname*{per}C$ -(since the permanent of a matrix does not change when its rows are permuted). -But the definition of $\widetilde{\gamma}$ yields that $\operatorname*{per} -\left( \widetilde{\gamma}\left( C\right) \right) =\gamma\left( -\operatorname*{per}C\right) $ (since $\gamma$ is a $\mathbb{Q}$-algebra -homomorphism). Hence, $\gamma\left( \operatorname*{per}C\right) -=\operatorname*{per}\left( \widetilde{\gamma}\left( C\right) \right) -=\operatorname*{per}C$. In view of $\operatorname*{perm}G=\operatorname*{per} -C$, this rewrites as $\gamma\left( \operatorname*{perm}G\right) -=\operatorname*{perm}G$. -Forget that we fixed $\gamma$. We thus have shown that $\gamma\left( -\operatorname*{perm}G\right) =\operatorname*{perm}G$ for each $\gamma -\in\Gamma$. In other words, $\operatorname*{perm}G$ belongs to the invariant -ring $\mathbb{F}^{\Gamma}$. In other words, $\operatorname*{perm}G$ belongs to -$\mathbb{Q}$ (since the invariant ring $\mathbb{F}^{\Gamma}$ is $\mathbb{Q}$). -But all values of the characters $\chi_{1},\chi_{2},\ldots,\chi_{r}$ are sums -of roots of unity (because they are traces of matrices $A\in\operatorname*{GL} -\nolimits_{n}\left( \mathbb{F}\right) $ that satisfy $A^{\left\vert -G\right\vert }=I_{n}$, and the eigenvalues of such a matrix are roots of -unity), and thus are algebraic integers. Hence, all entries of the matrix $C$ -are algebraic integers (since all these entries are values of the characters -$\chi_{1},\chi_{2},\ldots,\chi_{r}$). Thus, the permanent $\operatorname*{per} -C$ of this matrix $C$ is an algebraic integer (since the algebraic integers -form a ring). In other words, $\operatorname*{perm}G$ is an algebraic integer -(since $\operatorname*{perm}G=\operatorname*{per}C$). Hence, -$\operatorname*{perm}G$ is an algebraic integer in $\mathbb{Q}$ (since -$\operatorname*{perm}G$ belongs to $\mathbb{Q}$). Since the only algebraic -integers in $\mathbb{Q}$ are integers (because the ring $\mathbb{Q}$ is -integrally closed), this entails that $\operatorname*{perm}G$ is an integer. -A similar argument shows that $\left( \det G\right) ^{2}$ is an integer. -(Here we need to use the fact that the square of the determinant of a matrix -does not change when its rows are permuted. This is because the determinant -gets multiplied by a power of $-1$.) Thus, Proposition 1 is proved. -$\blacksquare$<|endoftext|> -TITLE: Strange and non-strange prime numbers, are there infinitely many of them? -QUESTION [8 upvotes]: Definition. A prime number $p$ is called strange if there exists $k>1$ such that each prime divisior of $p^k-1$ divides $p-1$. -Example 3. The prime number $p=3$ is strange as $3^2-1=8$ has the same prime divisiors as $3-1=2$. -Example 5. The prime number $p=5$ is not strange, since for every $k>1$ the number $5^k-1$ is not a power of $2$ (by the Mihailescu Theorem). By the same reason the prime number $p=17$ is not strange. -Example 7. The prime number $p=7$ is strange since $7^2-1=48$ has the same prime divisors as $7-1=6$. -Example 31. The prime number $p=31$ is strange because $31^2-1=2^6\times 3\times 5$ has the prime divisors as $31-1=2\times 3\times 5$. -Using the small Fermat Theorem, it is possible to prove the following characterization - -Theorem. A prime number $p$ is not strange if and only if for every prime divisor $q$ of $p-1$ the number $p^q-1$ has a prime divisor that does not divide $p-1$. - -This theorem implies that the prime numbers $11,13,19,23,29,37,41,43,47,53,61,67,71,73,79,83,89,97$ are not strange. -Therefore, among prime numbers $<100$ only 3,7, 31 are strange. All these numbers are Mersenne numbers. In his comment Yaakov Baruch observed that each Mersenne number is strange. So, we can ask - -Question 1. Is each strange prime number Mersenne prime? - - -Question 2. Is the set of non-strange numbers infinite? - - -Question 3. Is it true that for any number $x$ and prime numbers $p_1,\dots,p_n$ that not divide $x$, the arithmetic progression $x+p_1\dots p_n\mathbb Z$ contains a non-strange prime number? - -REPLY [15 votes]: Zsigmondy's Theorem shows that the only strange primes are the Mersenne primes. Indeed, it shows that for any $n\geq 2$ the number $p^n-1$ has a prime factor not dividing $p^k-1$ for any $k -TITLE: De Rham's theorem for top-forms in manifolds with boundary -QUESTION [6 upvotes]: In page 79 of Bott-Tu, "Differential Forms in Algebraic Topology", they define the relative de Rham theory as follows: -Let $f:S\to M$ be a smooth map. Define the complex $\Omega^*(f)$ by -$$\Omega^k(f):=\Omega^k(M)\oplus\Omega^{k-1}(S)$$ $$\underline{\mathrm{d}}(\alpha,\beta)=(\mathrm{d}\alpha,f^*\alpha-\mathrm{d}\beta)$$ -It is easy to prove that $\underline{\mathrm{d}}^2=0$ which allows us to define the cohomology $H^*(f)$. As a particular case, one can consider a submanifold $\imath:N\hookrightarrow M$ and define -$$\Omega^*(M,N):=\Omega^*(\imath)$$ - -My interest lies in the case when $N=\partial M$ and $M$ compact, where one can also define the integral of top forms as -$$\int_{(M,\partial M)}(\alpha,\beta):=\int_M\alpha-\int_{\partial M}\beta$$ -It is easy to check, using Stoke's theorem, that -$$\int_{(M,\partial M)}\underline{\mathrm{d}}(\alpha,\beta)=0$$ -Thus, we have a well defined map -$$\tag{1}\label{one}\int_{(M,\partial M)}:H^n(M,\partial M)\to\mathbb{R}$$ -If $\partial M=\varnothing$, then $H^n(M,\partial M)=H^n(M)$ and the previous integral is the standard one. The de Rham's theorem for top-forms then tells us that if $M$ has no boundary -$$\tag{2}\label{two}\int_M:H^n(M)\to\mathbb{R}\quad \text{ is an isomorphism}$$ -However, with boundary we have: - 1. It is surjective (applying \eqref{two} over the boundary and using elements of the form $(0,\beta)$). - 2. Its kernel is isomorphic to $H^n(M)$. Sketch of the proof: for every $[\alpha]\in H^n(M)$, build an element $[(\alpha,\beta)]$ such that $\int_{(M,\partial M)}(\alpha,\beta)=0$ using de Rham's theorem over the boundary. This map is well defined. -I have a heuristic argument to show that $H^n(M)$ is always zero: given $\alpha\in\Omega^n(M)$, take the double of $M$ along the boundary $\partial M$ and extend to some $\widetilde{\alpha}\in\Omega^n(M\sqcup_{\partial M}M)$ such that its integral is zero (using a tubular neighborhood over $\partial M$). Then using \eqref{two} (the double has no boundary) shows that $\widetilde{\alpha}$ is exact and, therefore, its pullback to $M$, which is $\alpha$, is also exact. -This seems a very strong result that I haven't found anywhere, while the proof seems very simple, thus I doubt if there are obstructions to the extension that invalidate the proof. - -So the questions I have in mind (all of them are almost the same question) are: - -Is $H^n(M)=0$ if $M$ is compact with boundary? -Is there a useful characterization of $H^n(M,\partial M)$ that can be used in this context? -Is there a de Rham's theorem like \eqref{two} for manifolds with boundary (with no prescribed boundary conditions)? -Is there a de Rham's theorem like \eqref{two} for relative cohomology? -If $H^n(M)\neq 0$, is there another a map $G:H^n(M,\partial M)\to \mathbb{R}$ such that $(\int_{(M,\partial M)},G)\to\mathbb{R}^2$ is an isomorphism? - -REPLY [4 votes]: It is indeed true that $H^n(M)=0$ if $M$ is a compact manifold with boundary. In particular, $H^n(M,\partial M)\cong\mathbb{R}$ by Lefschetz duality (as Chris Gerig mentioned) and the integral (1) is an isomorphism. -The only reference I have found that states this results is: -Differential forms: theory and practice. Steven Weintraub. Academic Press (Elsevier) 2014. - -Theorem 8.3.10 for compact manifolds with boundary. -Theorem 8.4.8 for non-compact manifold with boundary using compact support forms.<|endoftext|> -TITLE: Short exact sequences every mathematician should know -QUESTION [139 upvotes]: I'd like to have a big-list of "great" short exact sequences that capture some vital phenomena. I'm learning module theory, so I'd like to get a good stock of examples to think about. An elementary example I have in mind is the SES: -$$ -0 \rightarrow I \cap J \rightarrow I \oplus J \rightarrow I + J \rightarrow 0 -$$ -from which one can recover the rank-nullity theorem for vector spaces and the Chinese remainder theorem. -I'm wondering what other 'bang-for-buck' short exact sequences exist which satisfy one of the criteria: - -They portray some deep relationship between the objects in the sequence that is non-obvious, or -They describe an interesting relationship that is obvious, but is of important consequence. - -REPLY [4 votes]: The defining short exact sequence for Milnor's $K_2(R)$ ($R$ any ring) is -$$0\rightarrow K_2(R)\rightarrow St(R)\rightarrow E(R)\rightarrow 0$$ -where $St(R)$ and $E(R)$ are the Steinberg and elementary groups. If we cheat a little on the definition of "short" this extends to -$$0\rightarrow K_2(R)\rightarrow St(R)\rightarrow GL(R)\rightarrow K_1(R)\rightarrow 0$$<|endoftext|> -TITLE: Three integral expressions for integer values of $\zeta(s)$. Could these be further reduced to known integrals? -QUESTION [5 upvotes]: In this MSE-question I've asked about three, similarly shaped, integrals for integer vales of $\zeta(s)$ that I found numerically: -$$\zeta \left( 3 \right) =\frac12{\int_{0}^{1} \frac{1}{x}\big(\zeta(2)-{\it Li_2} \left(1-x\right)\big) \,{\rm d}x} \tag{1}$$ -$$\zeta \left( 4 \right) =\frac{4}{5}{\int_{0}^{1} \frac{1}{x}\big(\zeta(3)-{\it Li_3} \left(1-x\right)\big) \,{\rm d}x} \tag{2}$$ -$$\zeta \left( 5 \right) =\frac12{\int_{0}^{1} \frac{1}{x}\big(\zeta(2)-{\it Li_2} \left(1-x\right)\big)^2 \,{\rm d}x} \tag{3}$$ -ADDED: found one more: -$$\zeta \left( 3 \right) = \frac32 - \frac14{\int_{0}^{1} \big(\zeta(2)-{\it Li_2} \left(1-x\right)\big)^2 \,{\rm d}x} \tag{4}$$ -where ${\it Li_n}(z)$ is the PolyLogarithm. -I have not found any similar expressions at other integer values. -The answer to the MSE-question helped reducing the integral for $\zeta(3)$ to a known integral, however still curious whether the other two could be reduced to something known as well. - -REPLY [6 votes]: These identities, and many more, follow from a theorem in Integrals of polylogarithmic functions, recurrence relations, and associated Euler sums, - -For example, - - - -There are also variations with $\log x$ factor, such as<|endoftext|> -TITLE: When do algebraic closures exist constructively? -QUESTION [6 upvotes]: The field of algebraic numbers exists constructively, since we can represent a number by an irreducible polynomial plus an estimate in rational coordinates that separates it from any other root. -More generally, if we have a countably enumerated field with decidable arithmetic, it seems like we can construct the algebraic closure by picking a countable ordering of the irreducible polynomials, then defining an ordering of the roots of each polynomial that respects the orderings chosen for all previous polynomials. -Questions: - -Is it correct that something like this construction works for any constructive countable field? -Is there a natural broader class of field for which the algebraic closure constructively exists? - -REPLY [4 votes]: This is proved in Theorem VI.3.5 of "A Course in Constructive Algebra" by Ray Mines, Fred Richman, and Wim Ruitenburg. -I'm not aware of any generalizations of this sort.<|endoftext|> -TITLE: Lower bound for diagonal Ramsey numbers —- reference request -QUESTION [5 upvotes]: Using the first moment method, in 1947 Erd\H{o}s gave a lower bound on the diagonal Ramsey numbers $R(k,k)$: -$$ -R(k,k) \geq (1+o(1))\frac{k}{e\sqrt{2}} 2^{k/2}. -$$ -In 1975 Spenser used the Lov\’asz Local Lemma to improve this by a factor of $2$, to $(1+o(1))(k\sqrt{2}/e)2^{k/2}$. -In between these two lower bounds, there is the one you get via the alteration or deletion method: -$$ -R(k,k) \geq (1+o(1))\frac{k}{e} 2^{k/2}. -$$ -I’m trying to discover who first noticed this last bound, and when? Was it before or after Spencer’s improvement? I’ve seen the bound in numerous sets of notes online (and learned it as a graduate student), but with no attribution. Maybe it’s just ``folklore’’. -(I’m leading a summer reading group in Ramsey Theory, and I plan on Tuesday to tell them the history of upper and lower bounds on diagonal Ramsey numbers.) - -REPLY [2 votes]: On p.54 of "Recent Developments in Graphy Ramsey Theory" by Conlon, Fox, Sudakov which appeared in the monograph Surveys in Combinatorics 2015, Czumaj et. al. (eds), is the footnote below: - -Though we do not know of an explicit reference, a simple application of the deletion method which improves Erdos' bound by a factor of $\sqrt{2}$ was surely known before Spencer's work. - -Sounds like it may well be folklore.<|endoftext|> -TITLE: Examples of metric spaces with measurable midpoints -QUESTION [11 upvotes]: Given a (separable complete) metric space $X=(X,d)$, let us say $X$ has the measurable (resp. continuous) midpoint property if there exists a measurable (resp. continuous) mapping $m:X \times X \to X$ such that $d(x,m(x,x')) = d(x',m(x,x')) = d(x,x') / 2$ for all $x,x' \in X$. -It seems to be known (e.g see section 6 of this paper) that continuous midpoint spaces (i.e Polish spaces with the continuous midpoint property) include: - -Hilbert spaces. -Closed convex subsets of Banach spaces. -Hyperconvex spaces. -CAT(0) spaces. - -Hopefully, the collection of measurable midpoint spaces contains much more general examples (for the above list is quite restrictive). - -Question. What are some examples of measurable midpoint spaces ? - -REPLY [2 votes]: We are ultimately looking at the complexity of the multivalued function $\mathrm{MidPoint}_\mathbf{X} : \mathbf{X} \times \mathbf{X} \rightrightarrows \mathbf{X}$ assigning some midpoint to the points here. This includes what choice functions there are, but need not be limited to it. The framework to study the complexity of such operations is Weihrauch reducibility. -Just by definition of the midpoint, it follows that the map from a pair of points to the closed set of midpoints (equipped with the lower Vietoris topology) is continuous. This tells us that $\mathrm{MidPoint}_\mathbf{X} \leq_{\mathrm{W}} \mathrm{C}_\mathbf{X}$, where $\mathrm{C}_\mathbf{X}$ is closed choice on $\mathbf{X}$, which maps non-empty closed sets to some element. With $\mathrm{UC}_\mathbf{X}$ I denote the restriction of $\mathrm{C}_\mathbf{X}$ to singletons. -Everything we need about closed choice for this is found here. -Since we are assuming $\mathbf{X}$ to be Polish1, we immediately get the following: - -If $\mathbf{X}$ is sigma-compact and midpoints are unique, then $\mathrm{MidPoint}_\mathbf{X} \leq_{\mathrm{W}} \mathrm{C}_\mathbb{N}$. This implies that the midpoint map is piecewise continuous, ie that there is a countable cover of $\mathbf{X} \times \mathbf{X}$ by closed sets, such that on each piece the map is continuous. -If $\mathbf{X}$ is sigma-compact, then $\mathrm{MidPoint}_\mathbf{X} \leq_{\mathrm{W}} \mathrm{C}_\mathbb{R}$. This guarantees that there is a Baire class 1 selection function for midpoints, but we get more [Baire class 1 is equivalent to "preimages of opens are $\Sigma^0_2$, so this is much simpler than Borel measurable]. For example, there always is a midpoint which is low (in the computability-theoretic sense) relative to the space. -If midpoints are unique, then $\mathrm{MidPoint}_\mathbf{X} \leq_{\mathrm{W}} \mathrm{UC}_{\mathbb{N}^\mathbb{N}}$. Since the domain of $\mathrm{MidPoint}_\mathbf{X}$ is a Polish space, this already implies that $\mathrm{MidPoint}_\mathbf{X}$ is Borel measurable. -Without any restrictions, we just get that $\mathrm{MidPoint}_\mathbf{X} \leq_{\mathrm{W}} \mathrm{C}_{\mathbb{N}^\mathbb{N}}$. This doesnt rule out that we could avoid Borel measurable selection functions for the midpoint, but any construction would need to be very weird. The best starting point I can think of is using diagonally non-arithmetic functions. - -1 We don't need that the metric defining our midpoints is complete, we just need some equivalent complete metric around.<|endoftext|> -TITLE: A modern reference to the Zsigmondy Theorem -QUESTION [5 upvotes]: I need to cite the classical Zsigmondy Theorem, which was proved in 1892. -Is there any modern reference to this theorem? -I mean some standard textbook in Number Theory containing this theorem together with the proof. - -REPLY [7 votes]: Indeed, it is very difficult to find Zsigmondy's theorem with a proof in a book. However, it is proved in Appendix B to Chapter 30 in -Berkovich, Ya. G.; Zhmudʹ, E. M. Characters of finite groups. Part 2. Translated from the Russian manuscript by P. Shumyatsky [P. V. Shumyatskiĭ], V. Zobina and Berkovich. Translations of Mathematical Monographs, 181. American Mathematical Society, Providence, RI, 1999.<|endoftext|> -TITLE: Topological properties inherited by the Hausdorff metric space -QUESTION [5 upvotes]: Let $(X,d)$ be a metric space and $(K_X , h_d)$ be the associated metric space of nonempty compact subsets of $X$ with the Hausdorff metric. It is well known that $K_X$ inherits certain topological (and analytic) properties from $X$. For example, if $X$ is compact, then so is $K_X$; and if $X$ is complete, then so is $K_X$. -Is there a reference which further explores the properties that $K_X$ inherits from $X$? In particular, if $X$ is locally compact then is $K_X$ also? - -REPLY [4 votes]: Michael, Ernest. “Topologies on spaces of subsets.” Transactions of the American Mathematical Society 71 (1951): 152-182. - -The above paper shows in section $4$ that many properties of $X$, including being locally compact, are inherited by $K(X)$ (the latter space is called $\mathcal C(X)$ in the paper, while the author uses $2^X$ to refer to the hyperspace of closed sets of $X$, for which a more modern notation is $F(X)$). -In my comment I said that being zero-dimensional is also preserved but not shown in the paper, I now noticed that it is in fact shown there, together with more connectedness properties. The relationship between $\dim X$ and $\dim K(X)$ is much harder in general and has been widely studied, there is a whole chapter concerning this topic in the book Hyperspaces by Illanes and Nadler.<|endoftext|> -TITLE: Constructing exotic $\mathbb{R}^4$'s using vector fields on $\mathbb{R}^5$ -QUESTION [9 upvotes]: I was reading a paper of Arnol'd ("Topological Properties of Eigenoscillations -in Mathematical Physics") where he gives the following claim (hopefully I am stating it correctly). -One way to produce smooth 4-dimensional manifolds is to take some smooth, non-vanishing vector field $v$ on $\mathbb{R}^5$. The flow of this vector field defines a smooth $\mathbb{R}$-action on $\mathbb{R}^5$, and then we can just take the quotient of $\mathbb{R}^5$ by this action to produce some smooth 4-manifold $M$. -Arnol'd makes the interesting claim that, given any exotic $\mathbb{R}^4$ we know about, it can be produced by a careful choice of this vector field). -Could anyone shed some light on the details of this construction? - -REPLY [14 votes]: $Exotic(\mathbb R^4) \times \mathbb R$ is diffeomorphic to $\mathbb R^5$ as we know that there exits unique smooth structure on $\mathbb R^5$ (proved by Stalling A reference for smooth structures on R^n). Now there exists a nice $\mathbb R$ action on $Exotic(\mathbb R^4)\times \mathbb R$, i.e, translation along the $\mathbb R$ axis. And push-forward of this action will generate a smooth action on $\mathbb R^5$ which is the one you are looking for.<|endoftext|> -TITLE: When is a distribution having a finite support actually zero? -QUESTION [5 upvotes]: Let $D$ be a differential operator with smooth coefficients in $\mathbb{R}^n$. Suppose $E$ is a bounded open set in $\mathbb{R}^n$. Suppose $u$ is a function that is smooth up to the boundary of $E$. If $D(u\cdot\chi_{E})$, as a distribution in $\mathbb{R}^n$, has a finite support, is it true that $D(u\cdot\chi_{E})$ is actually zero? -The answer is no in the case $n=1$. An example is $D=d/dx$, $E=(0,1)$ and $u(x)=1$ for all $x$. In this case, the support of $D(u\cdot\chi_{(0,1)})$ is $\{0,1\}$. I hope that the answer is yes if $n\geq 2$. If the answer is yes, what weaker conditions in u are actually needed to guarantee that the answer is yes? Any suggestion for appropriate references would be greatly appreciated. - -REPLY [8 votes]: Let $E$ be the square $(0,1)^2$ in $R^2$, $D=\partial_x\partial_y$ and $u=1$. The support of $D(\chi_E u)$ is the set of corners of $E$.<|endoftext|> -TITLE: References for quivers and derived categories of coherent sheaves for a string theory student -QUESTION [8 upvotes]: I'm a student mostly from physics knowledge hoping to learn about the math involved the string theory paper Topological Quiver Matrix Models and Quantum Foam. -Context: The topological string theory partition function can be understood as computing the Donaldson-Thomas invariants of the variety over which the topological string is defined via the Donaldson-Thomas/Gromov-Witten correspondence. That's ok to me, the problem is that I've started to find fascinating, and apparently isolated examples of connections between quiver representations and the derived category of coherent sheaves of some varieties (especially toric cases) that I dont' fully understand. -The prototypical example of the aforementioned connections, is the well known realization of the moduli space of stable, rank $r$ (and $c_{2}=n$) and torsion-free sheaves on $\mathbb{P}^{2}$ as the $\mathcal{M}(n,r)$ quiver variety of the Jordan quiver under the Geiseker stability condition; the first data is precisely what the Donaldson-thomas theory of $\mathbb{P}^{2}$ actually computes. Another example is the Nakamura computation of the $G$-equivariant Hilbert scheme of points over $\mathbb{C}^{3}/G$ where $G$ is a finite $SL(3,\mathbb{C})$ subgroup as described in the page 14 of the paper "The McKay correspondence" using McKay quivers. This latter fact was used in Crystals and Black holes to enummerate tautological sheaves over a crepant resolution of $\mathbb{C}^{3}/G$ to compute the topological string partition function. -My problem: I supspect that the connections are not accidental but I'm uncapable to see what's the precise relation between moduli problems of quiver representations and those ones of sheaves, or where to start to investigate. -My background: I've sudied algebraic geometry from the first four chapters (Varieties,Schemes,Cohomology and Curves) in Hartshorne's textbook, I'm also familiar with the identification between the derived bounded category of coherent sheaves and the D-branes of the topological string B-model. -My weakness: I know very little about representation theory of quivers. -Questions: In Topological Quiver Matrix Models and Quantum Foam - is apparently assumed that we can associate to a given toric variety a quiver whose derived category of representations is isomorphic to the derived bounded category of coherent sheaves of the given toric scheme. -1.-Does anyone know a gentile reference to learn about the mathematical details of how this can be explicity achieved? -2.- What could be a good reference to start learning about quivers focused to understand the papers Topological Quiver Matrix Models and Quantum Foam and Crystal Melting and Black holes given my prior knowledge and physics orientation. -Any comment or reading suggestion is very welcome. - -REPLY [4 votes]: First, that review is somewhat depressing in that it's been over ten years since people figured out how to write down explicit boundary conditions in the B-model for objects in the derived category, but it's still talking about 'tachyon condensation' and locally-free resolutions, which do not always exist. I'm partial to the discussion in my old paper, but see also Kapustin et al and Herbst et al. -For what it's worth, the main bolded statement in the review is wrong -- the D-branes in the B-model do not need to be stable. Stability depends on the Kahler information of the target and has to do with the physical D-branes, not the topological ones. -Anyways, to answer your actual question, when you have an equivalence of categories between the derived category of coherent sheaves on noncompact CY and the derived category of representations of a quiver algebra, you often get that a component of the moduli space of representations with a specific dimension vector is the original noncompact CY. Physically, you can think of this as the moduli space of D0-branes being the cone itself. These D0-branes naturally correspond to representations of the quiver with a fixed dimension vector, and you can find the cone in the moduli stack pretty easily. With a little more work, you can get the GIT quotient too. You can see this in my two papers with Nick Proudfoot 1 and 2. There have been generalizations of this work, but I don't know if it has been proven for all the toric stuff (I've been away from this for a while). I'd start by looking at the work of Alastair Craw. -With respect to quivers, I was going to recommend Harm Derksen's lecture notes for a good introduction, but it looks like he took them down at some point. Sorry I don't have any good recommendation there.<|endoftext|> -TITLE: Subdivision of closed homology manifold reference request -QUESTION [6 upvotes]: I am interested in the barycentric subdivision of closed homology manifolds. -Definition A (finite) simplicial complex $K$ is a closed homology manifold of dimension $n$ if for every $k$-simplex, its link has the homology of $\mathbb{S}^{n-k-1}$. -I wonder whether the barycentric subdivision of a closed homology manifold remains a closed homology manifold and if case it does not, if there are some extra (minimal) conditions I can add to guarantee that. -Do you know of a reference where this or related questions are addressed? -(I have not found an answer in Rourke and Sanderson's "Introduction to Piecewise linear topology", which was my first try.) -Thanks in advance for your time - -REPLY [4 votes]: Using excision, and the decomposition of the star of a simplex $\sigma$ as $\sigma * \mathrm{Lk}(\sigma)$, you can easily show that your definition is equivalent to asking that -$$H_i(|K|, |K| \setminus \{x\} ;\mathbb{Z}) = \begin{cases}\mathbb{Z} & \text{ if $i=n$}\\ -0 & \text{ else}.\end{cases}, \text{ for every $x \in |K|$}.$$ -But this (re)definition of homology manifold only depends on the topological space $|K|$, which is unchanged under barycentric subdivision.<|endoftext|> -TITLE: Fiedler vector, what else? -QUESTION [5 upvotes]: In the spectral analysis of a graph with 1 connected component, the first non-trivial eigenvector (corresponding to the non-zero smallest eigenvalue) is also called the Fiedler vector. This vector is useful in graph partitioning because it minimizes the distance between the connected vertices in the original graph. In other ways, it can creates partitions in which nodes within the same partitions have minimal distance/high similarity and nodes between partitions have minimum edges connecting them. -Now, my question is, as soon as we go to higher eigenvalues, what do eigenvectors mean? Do they represent other partitions of the same graph? Should they be taken into consideration? - -REPLY [4 votes]: Yes. See e.g. the paper Multi-way spectral partitioning and higher-order Cheeger inequalities by Lee, Oveis-Gharan and Trevisan. They show how the first $k$ eigenvectors can be used to find a useful $k$-way partitioning. - -REPLY [3 votes]: The Fiedler vector refers to the second smallest eigenvalue, here is a study of -The third smallest eigenvalue of the Laplacian matrix (2001). - -The relationship between the third smallest eigenvalue of the -Laplacian matrix and the graph structure is explored. For a tree the -complete description of the eigenvector corresponding to this -eigenvalue is given and some results about the multiplicity of this -eigenvalue are given.<|endoftext|> -TITLE: What is the role of topology on infinite dimensional exterior algebras? -QUESTION [6 upvotes]: Wedge products and exterior powers are discussed in W. Greub's book Multilinear algebra as follows. -Definition: Let $E$ be an arbitrary vector space and $p \ge 2$. Then a vector space $\bigwedge^{p}E$ together with a skew-symmetric $p$-linear map $\bigwedge^{p}: E\times \cdots \times E \to \bigwedge^{p}E$ is called a $p$-th exterior power of $E$ if the following conditions are satisfied: -(1) The vectors $\bigwedge^{p}(x_{1},\dotsc,x_{p})\mathrel{:=} x_{1}\wedge \dotsb \wedge x_{p}$ generate $\bigwedge^{p}E$. -(2) If $\psi$ is any skew-symmetric $p$ linear mapping of $\overbrace{E\times \dotsb \times E}^{\text{$p$ times}}$ into an arbitrary vector space $F$, then there exists a linear map $f\colon \bigwedge^{p}E \to F$ such that $\psi = f\circ \bigwedge^{p}$. -Now, we set: -\begin{equation} -\bigwedge E \mathrel{:=} \bigoplus_{n=0}^{\infty} \bigwedge^{p}E, \tag{1}\label{1} -\end{equation} -where $\bigwedge^{0}E \mathrel{:=} \mathbb{C}$ and $\bigwedge^{1}E \mathrel{:=} E$. -Identifying each $\bigwedge^{p}E$ with its image under the canonical injection $i_{p}\colon\bigwedge^{p}E \to \bigwedge E$, we can write $\bigwedge E = \sum_{p=0}^{\infty}\bigwedge^{p}E$. In other words, elements of $\bigwedge E$ can be thought as sequences $(v_{0},v_{1},\dotsc)$ where $v_{p} \in \bigwedge^{p}E$ for each $p\in \mathbb{N}$. Furthermore, there is a uniquely determined multiplication on $\bigwedge E$ such that the following rules hold: -\begin{gather*} -(x_{1}\wedge \cdots \wedge x_{p})(x_{p+1}\wedge \dotsb \wedge x_{p+q}) = x_{1}\wedge \cdots \wedge x_{p+q} \\ -1(x_{1}\wedge \cdots \wedge x_{p}) = (x_{1}\wedge \dotsb \wedge x_{p})1 = x_{1}\wedge \cdots \wedge x_{p}. -\end{gather*} -This turns $\bigwedge E$ into an algebra, which is called exterior (or Grassmann) algebra. -Note that Greub's construction considers arbitrary vector spaces, so that, in particular, we can take $E$ to be infinite dimensional. -Grassmann algebras are used by physicists to study fermionic systems. While searching for some material on Grassmann algebras of infinite dimensional vector spaces, I found lecture notes Fermionic functional integrals and the renormalization group by Feldman, Knörrer and Trubowitz, which has an appendix (page 75) on this topic. Their construction seems interesting, but I'm having trouble trying to relate it with Greub's construction. -The first part of their notes discusses Grassmann algebras of finite dimensional vector spaces. Then, the cited appendix starts with the statement that in order to further generalize it to infinite dimensional vector spaces we need to add a topology on these spaces. This seem not to be necessary in the general case, since Greub's construction does not consider topological vector spaces. However, I think they might have physical motivations in which the addition of a topology might be important. Their construction is as follows. -Let $I$ be a countable set. The Grassmann algebra will be generated by vector from the vector space: -$$E\mathrel{:=} \ell^{1}(I)\mathrel{:=}\{\alpha\colon I \to \mathbb{C}\mathrel: \sum_{i\in I}\lvert a_{i}\rvert < +\infty\}.$$ -$E$ is a Banach space with norm $\|\alpha\| \mathrel{:=}\sum_{i\in I}\lvert a_{i}\rvert$. Let $\mathcal{J}$ be the set of all finite subsets of $I$, including the empty set. Take -$$\mathcal{U}(I) = \ell^{1}(\mathcal{J}) \mathrel{:=}\{\alpha\colon \mathcal{J} \to \mathbb{C}\mathrel: \sum_{I\in \mathcal{J}}\lvert a_{I}\rvert<+\infty\}$$ -where $a_{I} \mathrel{:=} a_{i_{1}}\dotsb a_{i_{p}}$, $I=\{i_{1},...,i_{p}\}$. Then $\mathcal{U}(I)$ is a Banach space with norm $\|\alpha\| = \sum_{I\in \mathcal{J}}\lvert a_{I}\rvert$ and, when equipped with the product: -$$(\alpha \beta)_{I} \mathrel{:=}\sum_{J\subset I} \operatorname{sign}(J, I\setminus J)\alpha_{J}\beta_{I\setminus J},$$ -it becomes an algebra which is called the Grassmann algebra. -With all this being said, let me get to the questions. -Feldman, Knörrer and Trubowitz's construction might not be the most general construction there is (I don't know actually, but I think it's not as I justified before). However, I'd expect their construction to be at least a particular case of Greub's general construction. However, I don't seem to be able to relate these two since the definition of $\mathcal{U}(I)$ strongly depends on its topology. So is the second construction a particular case of the first one? If not, why not? Does it have to do with the hypothesis of $E$ to be a topological vector space? Does the topology on $E$ change the definitions of objects used on Greub's construction? -NOTE: When I ask "does the topology on $E$ changes the definitions of objects on Greub's constructions?", I mean the following. If $E$ is a vector space, $\bigoplus_{n=0}^{\infty}E$ is the space of all sequences $x=(x_{0},x_{1},\dotsc)$, $x_{i} \in E$, with all but finitely many nonzero entries. If $E= \mathcal{H}$ is a Hibert space, on the other hand, $\bigoplus_{n=0}^{\infty}\mathcal{H}$ is the space of sequences with $\|x\|^{2}:=\sum_{n=0}^{\infty}\| x_{i}\|^{2}_{\mathcal{H}}<+\infty$. Thus, although $\mathcal{H}$ is itself a vector space, the norm on $\mathcal{H}$ allows us to define the direct sum in alternative way. In other words, the topology on $\mathcal{H}$ makes the difference when we define direct sums. Maybe the use of Banach spaces by Feldman, Knörrer and Trubowitz implies some modifications like this, say, to define the direct sum (\ref{1}) in an alternative way, so these two constructions might be isomorphic or something like this. -ADDED: Does anyone know this particular construction from Feldman, Trubowitz and Knörrer? Any references on this approach would be really appreciated! - -REPLY [12 votes]: Focusing on exterior powers here is a distraction. The main problem already appears when considering the tensor algebra $T(E)=\oplus_{n\ge 0}E^{\otimes n}$. Once the issue is understood for the tensor algebra, figuring out what to do for the exterior or symmetric algebras (e.g., Fermion or Boson Fock spaces) is trivial, because we are in characteristic zero. In positive characteristic, this becomes subtle as can be seen for example in the recent work "Koszul modules and Green’s conjecture" by Aprodu et al. where a positive characteristic Hermite Reciprocity map is constructed. -Given a vector space $E$, the first step is to consider tensor products like $E^{\otimes n}$. This can be done algebraically as in the mentioned book by Greub. However, when $E$ is an infinite dimensional TVS (topological vector space) the resulting algebraic tensor product $E\otimes\cdots\otimes E$ is a rather unsuitable object for the purposes of analysis. One typically needs to enlarge this space using a completion procedure (topology is essential for that), and one then obtains a topological tensor product $E\widehat{\otimes}\cdots\widehat{\otimes}E$. The caveat is: even when working with Banach spaces, there are lots of ways of doing that. This was Alexander Grothendieck's Ph.D. thesis work. He considered a dozen or so inequivalent reasonable definitions for these completions/versions of the tensor product which depend on the topological structure. In other words, in the course of his explorations Grothendieck found Hell. Luckily, he kept exploring and he eventually also found Paradise: the class of nuclear spaces for which all these different constructions become the same and therefore acquire a cananical feel to them. -Likewise, for the sum $\oplus_{n\ge 0}$ one typically starts with the algebraic direct sum (only finite sums allowed, i.e., we look at almost finite sequences where after a while all the terms are zero) and one then enlarges the space by taking a completion. -The construction by Feldman, Knörrer and Trubowitz is an explicit way (just a choice that works for their purposes) of doing a succession of algebraic constructions followed by topological completions, as explained above. -Now one might think that the algebraic construction as in Greub's book is more general/powerful/etc. than the topological procedure. This is a misconception. For infinite dimensional spaces that are not too big, one could in fact argue the opposite is true. -Take for example the simplest infinite dimensional space: $E=\oplus_{n\ge 0}\mathbb{R}$ which can be viewed as the space of almost finite sequences of real numbers, or the space of polynomials in one variable with real coefficients. Then $T(E)$ constructed algebraically à la Greub is a particular case of the topological completion construction. Indeed, equip $E$ with the locally convex topology defined by the set of all seminorms on $E$. This is also called the finest locally convex topology. With this topology, the space is nuclear in the sense of Grothendieck's general definition (but not nuclear in the sense of the more restrictive definition used by the Russian school around Gel'fand et al., namely, the notion of countably Hilbert nuclear spaces). -So that is a good sign: pretty much any reasonable completion will give you the same $E\widehat{\otimes}\cdots\widehat{\otimes}E$ which will also coincide with the algebraic tensor product (without hats). Finally for the sum one has several possible choices, but one of them will give the algebraic construction. Let us say that a seminorm on the algebraic direct sum $T(E)$ is admissible if and only if it restricts to a continuous seminorm on each summand. Take the locally convex topology on $T(E)$ defined by the set of all admissible seminorms. Take the completion. This will give nothing new. Note that all seminorms are admissible for the case $E=\oplus_{n\ge 0}\mathbb{R}$ but I wanted to introduce a more general construction which can be applied for example to $E=\mathscr{S}(\mathbb{R})$, the Schwartz space of rapidly decaying smooth functions. -Then the $T(E)$ will be isomorphic as a TVS to $\mathscr{D}(\mathbb{R})$, the space of compactly supported smooth functions. -Moral(s) of the story: -For infinite dimensional spaces ordinary bases (Hamel bases) are no good. You need Schauder bases which allow infinite linear combinations. -You will need to base your construction on topology. Even when topology seems to be absent, and one uses purely algebraic direct sums and tensor products, topology is still there hiding behind the scenes as in the $E=\oplus_{n\ge 0}\mathbb{R}$ example. -Recommended reading: -The excellent vignette "Schwartz kernel theorems, tensor products, nuclearity" by Paul Garrett. - -July 2020 Edit: -Let me give more details on the relation between the above general methodology and the particular FKT construction. -First some notation: I will write $\mathbb{N}=\{0,1,2,\ldots\}$, and I will denote the set functions from the set $X$ to the set $Y$ by $\mathscr{F}(X,Y)$. -We start from the $\ell^1$ space $E$ defined as the set of functions $f\in\mathscr{F}(\mathbb{N},\mathbb{C})$ such that -$$ -||f||_E:=\sum_{i\in\mathbb{N}}|f(i)| -$$ -is finite. -The first step is to understand the algebraic tensor product $E\otimes E$. The general construction proceeds via the free vector space with basis indexed by symbols $f\otimes g$ with $f,g\in E$ and quotienting by relations $(f_1+f_2)\otimes g-f_1\otimes g-f_2\otimes g$ etc. Another equally uninspiring construction is to take an uncountable Hamel basis $(e_i)_{i\in I}$, for $E$, produced by the Axiom of Choice, and realize $E\otimes E$ as the subset of $\mathscr{F}(I\times I,\mathbb{C})$ made of functions of finite support (equal to zero except for finitely many elements of $I\times I$). The proper definition is as a solution to a universal problem: $E\otimes E$ together with a bilinear map $\otimes:E\times E\rightarrow E\otimes E$ must be such that for every vector space $V$ and bilinear map $B:E\times E\rightarrow V$, there should exist a unique linear map $\varphi:E\otimes E\rightarrow V$ such that $B=\varphi\circ\otimes$. One can construct such a space more concretely as follows. -Let $E_2$ be the subset of $\mathscr{F}(\mathbb{N}^2,\mathbb{C})$ made of functions $h:(i,j)\mapsto h(i,j)$ which are finite sums of functions of the form $f\otimes g$ with $f,g\in E$. Here $f\otimes g$ is the function $\mathbb{N}^2\rightarrow\mathbb{C}$ defined by -$$ -(f\otimes g)(i,j)=f(i)g(j) -$$ -for all $i,j\in \mathbb{N}$. Note that the definition I just gave also provides us with a bilinear map $\otimes:E\times E\rightarrow E_2$. -Proposition 1: The algebraic tensor product of $E$ with itself can be identified with $E_2$. -The proof relies on the following lemmas. -Lemma 1: For $p,q\ge 1$, suppose $e_1,\ldots,e_p$ are linearly independent elements in $E$ and suppose $f_1,\ldots,f_q$ are also linearly independent elements in $E$. Then the $pq$ elements $e_a\otimes f_b$ are linearly independent in $E_2$. -Proof: Suppose $\sum_{a,b}\lambda_{a,b}e_a\otimes f_b=0$ in $E_2$. Then $\forall i,j\in\mathbb{N}$, -$$ -\sum_{a,b}\lambda_{a,b}e_a(i) f_b(j)=0\ . -$$ -If one fixes $j$, then one has an equality about functions of $i$ holding identically. -The linear independence of the $e$'s implies that for all $a$, -$$ -\sum_{b}\lambda_{a,b}f_b(j)=0\ . -$$ -Since this holds for all $j$, and since the $f$'s are linearly independent, we get $\lambda_{a,b}=0$ for all $b$. But $a$ was arbitrary too, so $\forall a,b$, $\lambda_{a,b}=0$ and we are done. -Lemma 2: Let $B$ be a bilinear map from $E\times E$ into some vector space $V$. -Suppose $g_k,h_k$, $1\le k\le n$ are elements of $E$ satisfying -$$ -\sum_{k}g_k\otimes h_k=0 -$$ -in $E_2$, i.e., as functions on $\mathbb{N}^2$. Then -$$ -\sum_k B(g_k,h_k)=0 -$$ -in $V$. -Proof: This is trivial if all the $g$'s are zero or if all the $h$'s are zero. So pick a basis $e_1,\ldots,e_p$ of the linear span of the $g$'s and pick a basis $f_1,\ldots,f_q$ of the linear span of the $h$'s (no Axiom of Choice needed). We then have decompositions of the form -$$ -g_k=\sum_a \alpha_{k,a}e_a -$$ -and -$$ -h_k=\sum_b \beta_{k,b} f_b -$$ -for suitable scalars $\alpha$, $\beta$. -By hypothesis -$$ -\sum_{k,a,b}\alpha_{k,a}\beta_{k,b}\ e_a\otimes f_b=0 -$$ -and so $\sum_k \alpha_{k,a}\beta_{k,b}=0$ for all $a,b$, by Lemma 1. -Hence -$$ -\sum_k B(g_k,h_k)=\sum_{a,b}\left(\sum_k \alpha_{k,a}\beta_{k,b}\right) B(e_a,f_b)=0\ . -$$ -Now the proof of Proposition 1 is easy. The construction of the linear map $\varphi$ proceeds as follows. -For $v=\sum_{k}g_k\otimes h_k$ in $E_2$, we let $\varphi(v)=\sum_k B(g_k,h_k)$. -This is a consistent definition because if $v$ admits another representation $v=\sum_{\ell}r_{\ell}\otimes s_{\ell}$, then -$$ -\sum_k g_k\otimes h_k\ +\ \sum_{\ell}(-r_{\ell})\otimes s_{\ell}=0 -$$ -and Lemma 2 implies -$$ -\sum_k B(g_k,h_k)=\sum_{\ell} B(r_{\ell},s_{\ell})\ . -$$ -The other verifications that $E_2$ with $\otimes$ solve the universal problem for the algebraic tensor product pose no problem. -The second step is to construct a topological completion $\widehat{E}_2$ for $E_2$. I will use the projective tensor product construction $E\ \widehat{\otimes}_{\pi}E$. -For $h\in E_2$, I will use the $l^1$ norm -$$ -||h||_2=\sum_{(i,j)\in\mathbb{N}^2}|h(i,j)|\ . -$$ -I will also use the seminorm -$$ -||h||_{\pi}=\inf\ \sum_k ||g_k||_E\times||h_k||_E -$$ -where the infimum is over all finite decompositions $h=\sum_k g_k\otimes h_k$. -The projective tensor product is the completion with respect to $||\cdot||_{\pi}$. -The $||\cdot||_1$ is an example of cross norm, i.e., it satisfies -$||f\otimes g||_2=||f||_E\times||g||_E$. Moreover, one has the following easy result. -Proposition 2: For all $h\in E_2$, we have $||h||_2=||h||_{\pi}$. -For the proof use the cross norm property and triangle inequality for $\le$, and for the reverse inequality, approximate $h$ by the truncation where $h(i,j)$ is replaced by zero unless $i,j\le N$. -Now it is clear that the abstract topological tensor product $\widehat{E}_2$ is nothing but the familiar $\ell^1$ space of functions on $\mathbb{N}^2$. -Likewise (but with heavier notations) one can construct $\widehat{E}_n=E\ \widehat{\otimes}_{\pi}\cdots\widehat{\otimes}_{\pi}E$, $n$ times, -with the corresponding $\ell^1$ norm -$$ -||h||_n=\sum_{(i_1,\ldots,i_n)\in\mathbb{N}^n}|h(i_1,\ldots,i_n)|\ . -$$ -The topological exterior power $\widehat{E}_{n,{\rm Fermi}}$ can be identified with the closed subspace of antisymmetric functions inside $\widehat{E}_n$, namely functions $h:\mathbb{N}^n\rightarrow\mathbb{C}$ which satisfy -$$ -h(i_{\sigma(1)},\ldots,i_{\sigma(n)})=\varepsilon(\sigma)\ h(i_1,\ldots,i_n) -$$ -for all $(i_1,\ldots,i_n)\in\mathbb{N}^n$ and all permutations $\sigma$. -We will equip the space with restriction of the norm $||\cdot||_n$. -Now consider the algebraic direct sum $W=\oplus_{n\ge 0}\widehat{E}_{n,{\rm Fermi}}$. -Given (for the moment unspecified) positive weights $w_n$, let us define the norm -$$ -||H||_{\rm Big}=\sum_{n\ge 0}w_n||h_n||_n -$$ -where $H$ is an element of $W$ seen as an almost finite sequence $(h_0,h_1,\ldots)$ -of functions in $\widehat{E}_{0,{\rm Fermi}},\widehat{E}_{1,{\rm Fermi}},\ldots$ -Clearly the completion $\widehat{W}$ is obtained by removing the almost finite restriction but still requiring convergence of the sum defining $||\cdot||_{\rm Big}$. -Finally, to make contact with FKT, to $H=(h_0,h_1,\ldots)\in\widehat{W}$ -we associate the set function $\alpha:\mathcal{J}\rightarrow\mathbb{C}$ where $\mathcal{J}$ is the set of finite subsets of $\mathbb{N}$ (including the empty set), as follows. -For $I=\{i_1,\ldots,i_n\}\in\mathcal{J}$ with $i_1<\cdots -TITLE: Should I expect functions to have analytic continuations? -QUESTION [6 upvotes]: I spend lots of time working with Dirichlet series with bounded coefficients, and I often need to find whether or not they have analytic continuations to the full complex plane. When proving that some mathematical object has some property, I like to know whether I'm working to prove that the object I'm looking has some strange property or if I'm working to prove that it's normal and the numbers aren't just conspiring against me. -For example, when trying to prove whether a number is irrational or not, I know that $100\%$ of numbers are irrational and so I'm trying to show that I didn't happen to pick one of those $0\%$ of numbers. -Sadly, I have no such intuition for analytic continuation. I think that my guess would be that either $100\%$ of $0\%$ of Dirichlet series have analytic continuations, but I could be wrong. To make my question more concrete, - -If $\{a_n\}$ is a sequence of complex numbers chosen uniformly randomly in the unit disk, and $F(s)=\sum_{n=1}^{\infty}\frac{a_n}{n^s}$ is it's Dirichlet series, what is the probability that $F(s)$ has an analytic past the the line $\Re(s)=1$ (not necessarily to the entire plane). - -Answers to variants of this question are also greatly appreciated, like if $a_n$ is chosen uniformly randomly of $[0,1]$ or if we are looking for continuations to the entire complex plane. -EDIT: If it's too complicated to analyse analytic continuation, what about meromorphic ones? Should I expect functions to have meromorphic continuations to $\Re(s)=1$? - -REPLY [9 votes]: I don't think it is reasonable to use "random" Dirichlet series as a guide if you are working with examples that are expected to have some actual structure to them (like most Dirichlet series that arise in practice in number theory). If you are working with Dirichlet series for reasons unrelated to number theory, then perhaps your question is reasonable. What are some reasons you are looking at Dirichlet series with bounded coefficients? -Let's describe a probabilistic model for Dirichlet $L$-functions and see what probability theory predicts about them. -The coefficients of a Dirichlet $L$-function are roots of unity (or $0$), which are on the unit circle, so should we consider a random -Dirichlet $L$-function to be $\sum z_n/n^s$ where $\{z_n\}$ is a sequence chosen independently and uniformly on -the unit circle? That doesn't reflect the multiplicativity of the coefficients of a Dirichlet $L$-function, so we will use a random Euler product, as follows: define a "random" Dirichlet $L$-function to be $L(s) = \prod_p 1/(1 - z_p/p^s)$ where $z_p$ for each prime $p$ is -chosen from a uniform distribution on the unit circle. -For a random number $z = \cos \theta + i\sin \theta$ on the unit circle, -its real and imaginary parts have average value 0 ($\int_0^{2\pi} \cos \theta \,d\theta/2\pi = 0$ and -$\int_0^{2\pi} \sin \theta \,d\theta/2\pi = 0$) and variance 1/2 ($\int_0^{2\pi} \cos^2 \theta \,d\theta/2\pi = 1/2$ and -$\int_0^{2\pi} \sin^2 \theta \,d\theta/2\pi = 1/2$). Note we are not computing -the variance of $z^2$ on the unit circle, which would be 0: $z^2$ is not $\cos^2\theta + i\sin^2\theta$! -The product $L(s)$ converges absolutely and uniformly on compact -subsets of ${\rm Re}(s) > 1$. What happens if $0 < {\rm Re}(s) \leq 1$? -When ${\rm Re}(s) > 1$, a logarithm of $L(s)$ is -$$ -\sum_{p^k} \frac{z_p^k}{kp^{ks}} = \sum_{p} \frac{z_p}{p^s} + \sum_{\substack{p^k \\ k \geq 2}} \frac{z_p^k}{kp^{ks}}, -$$ -where the sum involving $k \geq 2$ is absolutely convergent if ${\rm Re}(s) > 1/2$ since $|z_p^k/kp^{ks}| = 1/kp^{k\sigma}$. The series over primes is an integral: -$$ -\sum_p \frac{z_p}{p^s} = s\int_1^\infty \frac{Z(x)}{x^{s+1}}\,dx, -$$ -where $Z(x) = \sum_{p \leq x} z_p = \sum_{n \leq \pi(x)} z_{p_n}$. Applying the law of the iterated logarithm to the real and imaginary parts of $z_p$ (or $2z_p$ to make the variance 1), $|Z(x)| = O(\sqrt{\pi(x)\log\log \pi(x)})$ for almost all sequences $\{z_p\}$, which makes the integral above absolutely convergent for ${\rm Re}(s) > 1/2$ since $\pi(x)\log\log \pi(x) \sim (x/\log x)\log \log x$. -Therefore for almost all sequences $\{z_p\}$, -$\sum z_p/p^s$ converges for ${\rm Re}(s) > 1/2$, so a random Dirichlet $L$-function as defined here is almost certain to have an analytic continuation from ${\rm Re}(s) > 1$ to ${\rm Re}(s) > 1/2$ (as the exponential of -the analytic continuation of its logarithm) and no zeros -with ${\rm Re}(s) > 1/2$. -There is something inconsistent between random Dirichlet $L$-functions as defined above and actual Dirichlet $L$-functions that go by this name in the "real world" of number theory: for the sequences with $z_p \in \{\pm 1\}$ (a random quadratic Dirichlet $L$-function), with probability 1 the function $L(s)$ does not have an analytic continuation to ${\rm Re}(s) > 1/2 - \delta$ for $\delta > 0$ by Theorem 2 p. 550 of Queffélec's paper on random Euler products; see -"Propriétés presque sûres et quasi-sûres des séries de Dirichlet et des produits d'Euler" Canad. J. Math 32 (1980), 531-558. (I am not aware of a treatment of this issue for random Dirichlet $L$-functions with non-real $z_p$. -The treatment of random Euler products $\prod_p 1/(1 - z_p/p^s)$ for $z_p \in S^1$ in Kowalski's course notes https://people.math.ethz.ch/~kowalski/probabilistic-number-theory.pdf focuses on ${\rm Re}(s) > 1/2$.) Actual -Dirichlet $L$-functions in number theory are not random objects, but highly structured ones, and being able to extend them beyond ${\rm Re}(s) > 1/2$ is a significant feature that you can't predict by only thinking of a "random" $L$-function in a probabilistic sense.<|endoftext|> -TITLE: Inequality for trace of a symmetric product? -QUESTION [10 upvotes]: Let $A$ be a real, positive-definite, symmetric operator on an $n$-dimensional space $V$. Write $\odot^k A$ for the action of $A$ on the symmetric power $\odot^k V$. Let $v_1,\dotsc,v_n$ be a basis for $V$ (an orthonormal basis, if you wish). Write $\alpha_i$ for $\langle v_i, A v_i\rangle$. -Is it the case that $\mathrm{Tr} \odot^k A\geq \sum_{i_1\leq i_2\leq \dotsc \leq i_k} \alpha_{i_1} \dotsb \alpha_{i_k}$ for $k\geq 1$ arbitrary? Or for $k$ even? -Note the answer is yes (a) for $k=2$, (b) when $v_1,\dotsb,v_n$ are eigenvectors of $A$. - -REPLY [2 votes]: After applying an orthogonal transformation, we may and will assume that $v_1, \dots, v_n$ is the canonical basis, so $ \langle Av_i, v_i \rangle= a_{ii}$. So, the question boils down to showing that -$$ \sum a_{11}^{m_1} \cdots a_{nn}^{m_n} \le \textrm{tr } (\textrm{Sym} ^k A)= \sum \lambda_{1}^{m_1} \lambda_2^{ m_1} \cdots \lambda_n^{m_n} . $$ -Both sums run on the set of vectors $(m_1, \dots, m_n)$ of non-negative integers summing up to $k$. Write $F( x_1, \dots, x_n)= \sum x_1^{m_1} \cdots x_n^{m_n}$ with the same condition. Note that $F$ takes is constant set of vertices of the polytope $P$ whose extreme points are all permutations of $(\lambda_1, \dots, \lambda_n)$. By Schur-Horn inequality, the point $(a_1, \dots, a_n)$ belongs to $P$. So, to prove the inequality, it suffices to show that $F$ is convex on $P$. Now, since $A$ is positive, we have $x_i \ge 0$ on $P$, so to show that $F$ is convex, it suffices to show that it is convex in $x_i \in [0, \infty)$ when the other variables are fixed and non-negative. But this is clear, since it is a polynomial with non-negative coefficients, so its second derivative is non-negative. -UPDATE: MTyson pointed out below that my proof of convexity is not correct. At the moment I don't see how to fix it.<|endoftext|> -TITLE: Chromatic t-structures? -QUESTION [7 upvotes]: Questions: Fix a prime $p$ and $n \in \mathbb N_{\geq 1}$. - -Does the category $Sp_{K(n)}$ of $K(n)$-local spectra admit a nontrivial $t$-structure? - -By "nontrivial", I simply mean that $\{0\} \subsetneq Sp_{K(n),\geq 0} \subsetneq Sp_{K(n)}$. - -Does $Sp_{K(n)}$ admit a nontrivial monoidal $t$-structure? - -"Monoidal" means that (1) $\mathbb S_{K(n)} \in Sp_{K(n),\geq 0}$ (where $\mathbb S_{K(n)}$ is the $K(n)$-local sphere) and (2) $Sp_{K(n),\geq 0}$ is closed under the $K(n)$-local smash product. (Evidently I am using homological, rather than cohomological, indexing.) -As usual, the corresponding $T(n)$-local questions are also interesting, though presumably harder. For context, I'd also be interested in hearing about the $E(n)$-local or $T(0) \vee \dots \vee T(n)$-local versions of these questions. -In the above, feel free to interpret "the category $Sp_{K(n)}$" as either "the triangulated category $Sp_{K(n)}$" or as "the stable $\infty$-category $Sp_{K(n)}$" -- whichever is most comfortable. - -There's an easier question which has a negative answer: for $n \in \mathbb N_{\geq 1}$, the category $Mod_{K(n)}$ of $K(n)$-module spectra does not admit a nontrivial $t$-structure. For every object of $Mod_{K(n)}$ is a coproduct of shifts of $K(n)$. So if $0 \neq X \in Mod_{K(n),\geq 0}$, then there is a retract $\Sigma^k K(n)$ of $X$ which is a shift of $K(n)$, so that $\Sigma^k K(n) \in Mod_{K(n),\geq 0}$. Then because $K(n)$ is periodic, every object $Y \in Mod_{K(n)}$ is a coproduct of nonnegative shifts of $\Sigma^k K(n) \in Mod_{K(n),\geq 0}$, and so $Y \in Mod_{K(n),\geq 0}$. -But of course, the category $Sp_{K(n)}$ is much more complicated than the category $Mod_{K(n)}$. -When $n = 0$ (so that $K(n) = H \mathbb Q$) or $n = \infty$ (so that $K(n) = H \mathbb F_p$), $Mod_{K(n)}$ does admit a monoidal $t$-structure given by usual connectivity, and $Sp_{K(n)}$ inherits a monoidal $t$-structure by pullback along the free functor $Sp_{K(n)} \to Mod_{K(n)}$ (which is an equivalence for $n = 0$, of course). I don't think these cases shed much light on the case $n \in \mathbb N_{\geq 1}$, though. - -REPLY [6 votes]: To expand on Tim's answer, the arguments generalize to show that $Sp_{K(n)}$ admits no non-trivial t-structures in general. -The crucial ingredient is that $Sp_{K(n)}$ has no non-trivial localising or colocalising subcategories, see 7.5 in Hovey-Strickland. Thus, to finish the argument it is enough to show that any subcategory $C \subseteq Sp_{K(n)}$ which is closed under limits is in fact colocalising, ie. it's also closed under suspension. -If $X$ is any $K(n)$-local spectrum, then it is well-known (see 7.10 in the aforementioned book) that it can be written as a limit $X \simeq lim \ X \wedge F_{i}$ of its smash products with type n generalized Moore spectra. If $X \in C$, then some desuspension of $X \wedge F_{i}$ is contained in $C$ as well, but as Tim observed these spectra are periodic and so $\Sigma^{n} X \wedge F_{i} \in C$ for all $n \in \mathbb{Z}$. It follows that $\Sigma^{n} X \in C$, ending the argument.<|endoftext|> -TITLE: Involution action on Brauer group of an abelian variety -QUESTION [6 upvotes]: Let $k$ be an algebraically closed field of characteristic $p>2$, let $A/k$ be an abelian variety. Let $\iota\colon A\to A, a\mapsto -a$ be the natural involution. Let $x\in\mathrm{Br}(A)[p]$ be a Brauer class on $A$ of order $p$. Is it necessarily true that $\iota^*x=x$? - -REPLY [4 votes]: $\newcommand{\bG}{\mathbb{G}}$Let $X$ be any smooth scheme over an algebraically closed field $k$ of characteristic $p$. From the short exact sequence $0\to\mu_p\to \bG_m\to\bG_m\to 0$ of sheaves on the flat site of $X$ we get $0\to (\mathrm{Pic}\,X)/p\to H^2_{fl}(X,\mu_p)\to H^2_{fl}(X,\bG_m)[p]\to 0$. -Let now $X$ be an abelian variety. Ogus showed in Proposition 1.2 of "Supersingular K3 crystals" that under certain assumptions on a smooth proper variety (that are satisfied for abelian varieties) there is a canonical inclusion $$H^2_{fl}(X,\mu_p)\hookrightarrow H^2_{\mathrm{dR}}(X/k) $$ -The map is induced by the identification $H^2_{fl}(X,\mu_p)\simeq H^1_{et}(X,\bG_m/(\bG_m)^p)$ and the logarithmic derivative $\bG_m/(\bG_m)^p\xrightarrow{dlog}Z^1\to \Omega^{\bullet}[1]$. Since there is a canonical isomorphism $H^2_{\mathrm{dR}}(X/k)\simeq\Lambda^2 H^1_{\mathrm{dR}}(X/k)$ the multiplication by an integer $[n]:X\to X$ induces multiplication by $n^2$ on $H^2_{\mathrm{dR}}(X/k)$. In particular, the involution $\iota$ acts by identity and hence it does so on $H^2_{fl}(X,\mu_p)$ and $H^2_{fl}(X,\bG_m)[p]=(\mathrm{Br}\,X)[p]$ where the last identification comes from Gabber's theorem.<|endoftext|> -TITLE: Books/Lecture notes which contrast Risch algorithm with basic standard procedure of finding an antiderivative -QUESTION [7 upvotes]: I vaguely remember a book/some lecture notes which introduce integration algorithms such as Risch algorithm by first giving a list of quasi-algorithmic way of evaluating symbolic integrals. (For example, when integrating a rational trignometric function, use one of the substitutions $u=\sin \theta$, $u=\cos \theta, u=\tan \frac12\theta$ and so on; when dealing with a rational function, express it in partial fractions.) Then it goes on to develop the theory of differential algebra. -I could not find the text now. After searching for quite a few sources, I have not found any books which contrast those easy procedures with Risch algorithm. -Where could I find a book which is similar to the one I describe above? - -REPLY [4 votes]: I'm not sure if this is exactly what you're looking for, but my go-to volume for these kinds of question is Symbolic Integration I by Manuel Bronstein. Risch's original treatment is sketchy in many places, and Bronstein did a lot of work to flesh out the details and actually implement Risch's methods. From the Foreword by B. F. Caviness: - -With the advent of general computer algebra systems, some kind of symbolic integration facility was implemented in most. These integration capabilities opened the eyes of many early users of symbolic mathematical computation to the amazing potential of this form of computation. But yet none of the systems had a complete implementation of the full algorithm that Risch had announced in barest outline in 1970. There were a number of reasons for this. First and foremost, no one had worked out the many aspects of the problem that Risch's announcement left incomplete. - Starting with his Ph.D. dissertation and continuing in a series of beautiful and important papers, Bronstein set out to fill in the missing components of Risch's 1970 announcement. Meanwhile working at the IBM T. J. Watson Research Center, he carried out an almost complete implementation of the integration algorithm for elementary functions. It is the most complete implementation of symbolic integration algorithms to date. - -The book is a very nice blend of practical algorithms and general theory. -As the title indicates, Bronstein planned to write at least one more volume, -about the integration of algebraic functions. Sadly, he passed away before he could complete this task. But I think you'll find a lot of relevant information in this first volume.<|endoftext|> -TITLE: Stably trivial non-trivial vector bundles -QUESTION [7 upvotes]: I have two related questions. Can there be a stably trivial non-trivial holomorphic vector bundle over a closed complex manifold? Can there be a stably trivial non-trivial algebraic vector bundle over a smooth proper variety (of arbitary characteristic)? - -REPLY [10 votes]: Assume $E \oplus \mathcal{O} \cong \mathcal{O}^{\oplus n}$. Then, of course, $E \cong \mathcal{O}^{\oplus n}/\mathcal{O}$. On the other hand -$$ -Hom(\mathcal{O}, \mathcal{O}^{\oplus n}) \cong \Gamma(X, \mathcal{O})^{\oplus n}. -$$ -If $X$ is proper, connected and reduced, then $\Gamma(X, \mathcal{O}) = \Bbbk$ (the base field), hence any non-zero morphism $\mathcal{O} \to \mathcal{O}^{\oplus n}$ is given by a non-zero $n$-tuple of elements of the field, therefore any such morphism is isomorphic to the embedding of the first direct summand, hence the quotient is isomorphic to $\mathcal{O}^{\oplus (n - 1)}$. Thus, $E$ is trivial.<|endoftext|> -TITLE: Functions $f \geq 0$ on $\mathbb{R}$ which are of the form $f = |g|^2$ for some entire function $g$ -QUESTION [7 upvotes]: I think the answer to this question must be well known. Is it possible to characterize those functions $f \colon \mathbb{R} \to \mathbb{R}_+$ which are of the form $f(x) = |g(x)|^2, x \in \mathbb{R},$ for some entire function $g \colon \mathbb{C} \to \mathbb{C}$. As a simple counterexample let $f(x) = e^{-1/x^2}$. -Edit: Alexandre Eremenkos answer allows to reformulate the question. -Which entire functions $f \colon \mathbb{C} \to \mathbb{C}$ are nonnegative on $\mathbb{R}$? -In his proof he has (essentially) given a characterization with the help of the Weierstrass factorization theorem. Are there other (more direct) characterizations? I know, it's vague. - -REPLY [16 votes]: These $f$ are exactly those non-negative functions on the real line which are entire (=represented by their Taylor series on the whole real line). For example, $f(x)=(\arctan x)^2$ is not in your class since the Taylor series at $0$ has finite radius of convergence. Neither $f(x)=e^{-1/x^2}$ is -in your class since the Taylor series at zero does not converge to the function). -Proof. Suppose that $g$ is an entire function. -Define $g^*(z)=\overline{g(\overline{z})}$ which is also entire. Then -on the real line $f(z)=|g(z)|^2=g(z)g^*(z)$, so your function $f(x)$ is non-negative on the real line and entire (as a product of entire functions). -Conversely. Let $f$ be an entire function which is non-negative on the real line. -Then all real roots are of even multiplicities, and the rest are symmetric with respect to the real line. Let $X$ be the divisor in the plane which consists of those roots which lie in -the open upper half-plane with their multiplicities, -and real roots with half of their -multiplicities. We have the Weierstrass factorization $f=P e^h$ -where $P$ is the canonical product, and $h$ is entire, both $P$ and $h$ real on the real line. Let $P_1$ be the canonical product over $X$, -then $P=P_1P_1^*$, and set $g=P_1e^{h/2}$. Then on the real line -$$|g(x)|^2=|P_1(x)|^2|e^{h(x)}|=P(x)e^{h(x)}=f(x).$$ -Remark. If $f$ has infinitely many non-real zeros, then there are infinitely many different $g$'s which give such a representation: the zeros can be split between $P_1$ and $P_1^*$ in many ways: if $Y$ is the divisor of zeros of $f$, then any $X$ such that $Y=X+\overline{X}$ will do the job. -Remark 2. How to determine that a function of a real variable is in fact entire. A criterion is that $|f^{(n)}(x)|^{1/n}/n\to 0$ uniformly on compact subsets of the real line. -This follows from the Taylor formula with remainder combined with Stirling's formula.<|endoftext|> -TITLE: Generalization of Weak Nullstellensatz? -QUESTION [9 upvotes]: I believe the following is standard, namely when $k = \bar{k}$ is algebraically closed there is a bijection between points and maximal ideals -\begin{eqnarray*} -k^n &\longrightarrow& \operatorname{Specm}(k[X_1, \ldots, X_n]) \\ -x &\longrightarrow& \ker(\operatorname{ev}_x) -\end{eqnarray*} -where surjectivity follows from Zariski's Lemma. It seems like the following should also be true, by essentially the same argument, but I couldn't find a reference. For $k$ not algebraically closed there is a bijection -\begin{eqnarray*} -\bar{k}^n / \operatorname{Aut}(\bar{k} / k) &\longrightarrow \operatorname{Specm}(k[X_1, \ldots, X_n]) -\end{eqnarray*} -Is there a canonical reference for this ? - -REPLY [3 votes]: Or see Proposition 2.4.6 in Bjorn Poonen's book Rational Points on Varieties (link). This is almost exactly the result you conjectured, just a bit more general: - -Let $X$ be a $k$-variety. Then the map -$$\left\{\text{$\operatorname{Gal}_k$-orbits in $X(\overline{k})$}\right\}\rightarrow \left\{ \text{closed points of $X$} \right\}$$ -given by mapping the orbit of $f \colon \operatorname{Spec} \overline{k} \to X$ to $f(\operatorname{Spec} \overline{k} )$ is a bijection.<|endoftext|> -TITLE: Group structure for distributive lattices -QUESTION [6 upvotes]: On the (finite) Boolean lattice there is a group structure given by the symmetric difference and this group is an elementary abelian 2-group. - -Question: Does there exist a natural group structure on general (finite) distributive lattices? - -Other examples of a group structures for a given lattice would also be interesting. - -REPLY [4 votes]: This question has already been answered, but I will add a slightly different answer. -First, the question and YCor's answer both seem to assume that by a natural group structure we mean that the group is a reduct of the original structure. This means that the group operations are given as words in the original structure, as symmetric difference is given by the Boolean word $x\oplus y = (x\wedge \neg y)\vee (\neg x\vee y)$. I also make this assumption. -Any compatible relation on a structure is also a compatible relation of any reduct. (A relation $R\subseteq A^n$ is compatible on an algebraic structure $A$ if $R$ is a subalgebra of $A^n$.) -YCor's answer uses automorphisms to answer the question. The graph of an automorphism of $A$ is a compatible binary relation on $A$, so any automorphism of a distributive lattice will also be an automorphism of any reduct. Then YCor exhibits a nonidentity automorphism of a distributive lattice that fixes more than half the points, which is something that can't happen in a group. -But to me the more obvious relation to consider is the order relation $\leq$. Every nontrivial distributive lattice has a nondiscrete compatible order, while no nontrivial group has a nondiscrete compatible order. (Even nontrivial ordered groups do not have nondiscrete partial orders that are compatible with all group operations. Reason: $a\leq b$ for compatible $\leq$ implies $a^{-1}\leq b^{-1}$, which implies $a\cdot \underline{a^{-1}}\cdot b\leq a\cdot \underline{b^{-1}}\cdot b$, which implies $b\leq a$.)<|endoftext|> -TITLE: 3-colored triangulations of the sphere $S^2$, and Sperner's Lemma -QUESTION [6 upvotes]: I noticed something about colored triangulations of the topological sphere $S^2$ and have a question about this. - -Observation. If you triangulate the sphere $S^2$ and color the vertices with three colors: then the number of 3-colored triangles is always even (or zero). In particular, there is no coloring with exactly one 3-colored triangle. - -For a proof, view $S^2$ as two triangulated disks with matching coloring of the boundaries that are glued together. As their boundaries have the same number of color changes, we know from Sperner’s Lemma that their triangulations have the same number (mod 2) of 3-colored triangles. So the total number of 3-colored triangles is even or zero. -As an interesting corollary, we get the characterization: A triangulated sphere has zero 3-colored triangles iff all cycles of the triangulation have an even number of color changes. -I looked at the torus, the Klein bottle, and the projective plane, and I find that the observation is also true for them. -Edit: Just for contrast, adding an example below of a "soap bubble" surface, where the two soap bubbles share a common disk. This surface allows for triangulations with even and odd numbers of 3-colored triangles (but like the other surfaces I looked at, cannot have just one). - -Question. I wonder whether this also follows from more general theorems about triangulations of surfaces, or about maximal planar graphs? I have consulted algebraic topology and graph theory texts, but could not find any results in that direction. Would you have a suggestion where else to look, or maybe a reference for that? - -REPLY [3 votes]: Just to close the loop on this: the double-counting argument in the answer of user Matt allows for a nice visual proof of the (2-dim.) Lemma of Sperner. Just want to capture it here, as it connects nicely with the triangulation of the sphere / the maximal planar graph in my OP question. - -Start with a triangulated polygon in the plane, and label each vertex with one of 3 colors. The example just shows the boundary of such a triangulated, 3-colored polygon. Claim (Sperner’s Lemma): If the boundary has an odd number of color changes, then a 3-colored triangle exists in the polygon triangulation. In fact, more generally, an odd number of such 3-colored triangles exists. -Proof: Go to 3-dimensional space, and build a “tent” over the polygon like in the diagram: add a colored vertex, and add the edges between this additional vertex and the boundary vertices of the polygon. This way, we have effectively created a triangulation of the topological sphere $S^2$. -If the boundary of the polygon has an odd number of color changes, this gives an odd number of 3-colored triangles in the “tent” over the polygon. But from the double-counting argument in user Matt’s answer, we know an even number of 3-colored Sperner triangles must exist. Hence the polygon at the bottom must have an odd number of 3-colored triangles (at least one) in its triangulation, which completes the proof.<|endoftext|> -TITLE: When can an $\mathfrak{S}_n$-equivariant map be extended to an $\textrm{O}(n)$-equivariant map? -QUESTION [6 upvotes]: The symmetric group $\mathfrak{S}_n$ can be regarded as a subgroup of the orthogonal group $\textrm{O}(n)$ via the permutation matrices. Let $V$ be a finite dimensional $\textrm{O}(n)$-module and $\varphi: \mathbb{R}^n\to V$ an $\mathfrak{S}_n$-equivariant linear map where $\mathfrak{S}_n$ acts on $\mathbb{R}^n$ in the obvious way. Finally, let $d:\mathbb{R}^n\to\textrm{Sym}_2(\mathbb{R}^n)$ be the map that sends a vector to the corresponding diagonal matrix. -Are there criteria on $\varphi$ that ensure the existence of an $\textrm{O}(n)$-equivariant linear map $\Phi: \textrm{Sym}_2(\mathbb{R}^n)\to V$ such that $\Phi\circ d=\varphi$? -Note that such a map, if it exists, is unique since every real symmetric matrix is diagonalizable with orthogonal matrices. -Edit: This is to show that the map suggested by Aurel is in general not linear. Let $V$ be the representation of $\textrm{O}(n)$ where rotation about angle $t$ acts by $\begin{pmatrix}\cos(6t)&-\sin(6t)\\ \sin(6t)& \cos(6t)\end{pmatrix}$ and reflection at the $x$-axes by $\begin{pmatrix}1&0\\ 0& -1\end{pmatrix}$. Then one can check that the map $\varphi:\mathbb{R}^2\to V$ with $\varphi\binom{1}{1}=0$ and $\varphi\binom{1}{-1}=\binom{1}{0}$ satisfies the conditions described by Aurel. However, $V$ is not part of the decompostion of $\textrm{Sym}_2\mathbb{R}^2$ into irreducibles. - -REPLY [4 votes]: If I have understood the problem correctly, the map $\Phi$ deterimines $\varphi = \Phi \circ d$, so the question amounts to classifying possible compositions $\Phi \circ d$, where $d$ is the "diagonal" map, and $\Phi$ is $O(n)$ equivariant. Clearly the image of $\varphi$ must be contained in the image of $\Phi$ which is isomorphic to a quotient of $\mathrm{Sym}_2(\mathbb{R}^n)$. -Let's begin by noting that as a representation of $\mathfrak{S}_n$, we have the decomposition $\mathbb{R}^n = \mathbf{1}_{\mathfrak{S}_n} \oplus U$, where $\mathbf{1}_{\mathfrak{S}_n}$ is the trivial representation (spanned by the "all ones" vector), and $U$ is the (irreducible) standard representation (consisting of "mean zero" vectors). -Similarly, as a representation of $O(n)$, the representation $\mathrm{Sym}_2(\mathbb{R}^n)$ (which we are interpreting as symmetric $n \times n$ matrices under conjugation) decomposes as $\mathbf{1}_{O(n)} \oplus W$, where $\mathbf{1}_{O(n)}$ is the trivial representation (spanned by the identity matrix), and $W$ is an irreducible representation which consists of symmetric matrices of trace zero. -It is not difficult to see that the map $d: \mathbf{1}_{\mathfrak{S}_n} \oplus U \to \mathbf{1}_{O(n)} \oplus W$ is of the form $f \oplus g$, with $f : \mathbf{1}_{\mathfrak{S}_n} \to \mathbf{1}_{O(n)}$ (it sends the all-ones vector to the identity matrix) and $g: U \to W$. -Now, $\Phi$ must map $\mathbf{1}_{O(n)}$ to an invariant vector (possibly zero), and it must map $W$ to either a copy of $W$ or zero. So, we must give an "intrinsic" description of the image of $g$, i.e. traceless diagonal matrices, inside $W$. -Note that $O(n)$ contains not only $\mathfrak{S}_n$, but the larger hyperoctahedral group $\mathfrak{H}_n = C_2 \wr \mathfrak{S}_n = C_2^n \rtimes \mathfrak{S}_n$, via signed permutation matrices (permutation matrices, but the nonzero entries can be $\pm 1$). The hyperoctahedral group contains a subgroup $C_2^n$ consisting of diagonal matrices with entries $\pm 1$. It is not difficult to check that a matrix that commutes with every element of $C_2^n$ (viewed as a diagonal matrix) must itself be diagonal, and conversely it is clear that every diagonal matrix commutes with $C_2^n$. Phrasing this in terms of the module structure (commuting with = fixed under conjugation by), the diagonal matrices are fixed by the action of $C_2^n$. -We are now in a position to give the characterisation. A map $\varphi: \mathbb{R}^n \to V$ may be written in the form $\Phi \circ d$ if and only if all of the following conditions are satisfied. - -1. The $O(n)$-module generated by the image of $\varphi$ is a quotient of $\mathrm{Sym}_2(\mathbb{R}^n)$. - - -2. The image of the "all ones" vector is $O(n)$ invariant (possibly zero). - - -3. The image of $\varphi$ is fixed pointwise by $C_2^n$ (and therefore has a $\mathfrak{H}_n$ action that factors through $\mathfrak{S}_n$). - -We've demonstrated that these conditions are necessary, so let's check that they are also sufficient. Note that by (2), given $\varphi$ as above, the restriction $\mathbf{1}_{\mathfrak{S}_n} \to V$ determines the restriction $\Phi: \mathbf{1}_{O(n)} \to V$. So it suffices to construct the restriction $\Phi: W \to V$. By (1), we are guaranteed to find such a map whose image agrees with the $O(n)$-module generated by $\varphi(U)$; we check that that (3) implies that $\varphi(U)$ coincides with the image of the (traceless) diagonal matrices in $W$ (because these form an irreducible representation of $\mathfrak{S}_n$, rescaling $\Phi$ is all that is needed to guarantee pointwise agreement of $\varphi$ and $\Phi \circ d$). -As a representation of $\mathfrak{H}_n$, $\mathrm{Sym}_2(\mathbb{R}^n)$ can be decomposed as follows. Recognise that -$$ -\mathbb{R}^n = \mathrm{Ind}_{\mathfrak{H}_1 \otimes \mathfrak{H}_{n-1}}^{\mathfrak{H}_n}(\varepsilon \boxtimes \mathbf{1}_{\mathfrak{H}_{n-1}}), -$$ -where $\varepsilon$ is the sign character of $\mathfrak{H}_1 = C_2$. Now, using a little Mackey theory (to write down the tensor square, and then extract they symmetric part), we see that -$$ -\mathrm{Sym}_2(\mathbb{R}^n) = -\mathrm{Ind}_{\mathfrak{H}_1 \otimes \mathfrak{H}_{n-1}}^{\mathfrak{H}_n}(\mathrm{Sym}_2(\varepsilon) \boxtimes \mathbf{1}_{\mathfrak{H}_{n-1}}) -\oplus -\mathrm{Ind}_{\mathfrak{H}_2 \otimes \mathfrak{H}_{n-2}}^{\mathfrak{H}_n}(\varepsilon \otimes \varepsilon \boxtimes \mathbf{1}_{\mathfrak{H}_{n-1}}) -$$ -where $\varepsilon \otimes \varepsilon = \varepsilon \otimes \varepsilon \otimes \mathbf{1}_{\mathfrak{S}_2}$ is a one-dimensional representation of $\mathfrak{H}_2$ where a signed matrix acts by $(-1)^m$, where $m$ is the number of $-1$'s in the signed matrix. Enthusiasts of wreath products will recognise both summands as being irreducible representations of the hyperoctahedral group. The key point here is that $\mathrm{Sym}_2(\varepsilon) = \mathbf{1}_{C_2}$, so the first summand simply becomes -$$ -\mathrm{Ind}_{\mathfrak{H}_1 \otimes \mathfrak{H}_{n-1}}^{\mathfrak{H}_n}(\mathbf{1}_{\mathfrak{H}_1} \boxtimes \mathbf{1}_{\mathfrak{H}_{n-1}}), -$$ -which is simply $\mathbb{R}^n$, viewed as a $\mathfrak{H}_n$ module that factors through the action of $\mathfrak{S}_n$. In particular, this summand corresponds to diagonal matrices in $\mathrm{Sym}_2(\mathbb{R}^n)$. (The other summand does not factor through the $\mathfrak{S}_n$ action.) -One final comment: If $n \geq 4$, as a representation of $\mathfrak{S}_n$, -$$ -\mathrm{Sym}_2(\mathbb{R}^n) = S^{(n)} \oplus S^{(n)} \oplus S^{(n-1,1)} \oplus S^{(n-1,1)} \oplus S^{(n-2,2)} -$$ -where $S^\lambda$ is a Specht module (irreducible representation) indexed by $\lambda$. The key point here is that $U = S^{(n-1,1)}$ appears with multiplicity 2. This means that trace zero symmetric matrices are not the unique subspace of $\mathrm{Sym}_2(\mathbb{R}^n)$ isomorphic to $U$. This means that condition (3) is not automatic. In particular, Nate's suggestion of understanding the interaction with the Casimir element will help pin down condition (1) (it helps identify the ambient $O(n)$ representation), but some extra information will be required to detect condition (3).<|endoftext|> -TITLE: When is $-1$ in the image of a field norm? -QUESTION [6 upvotes]: Let $p >0$ be an odd prime and let $\mathbb{K} = \mathbb{Q}(\zeta) \subseteq \mathbb{C}$ with $\zeta$ a primitive $p$th root of unity. There is a unique subfield $\mathbb{Q} \subseteq \mathbb{F} \subseteq \mathbb{K}$ satisfying $[\mathbb{F}:\mathbb{Q}]=2$. Specifically $\mathbb{F} = \mathbb{Q}(\alpha)$ where $\alpha^2 = (-1)^{\frac{p-1}{2}}p$. -If $p \equiv -1 \pmod{4}$ then $[\mathbb{K}:\mathbb{F}] = (p-1)/2$ is odd so $\mathrm{N}_{\mathbb{K/F}}(-1) = (-1)^{[\mathbb{K}:\mathbb{F}]} = -1$, where $\mathrm{N}_{\mathbb{K/F}} : \mathbb{K} \to \mathbb{F}$ is the field norm. - -When $p \equiv 1 \pmod{4}$ why is $-1$ not in the image of $\mathrm{N}_{\mathbb{K/F}}$? - -This should be very elementary and for reasons coming from representation theory of finite groups I know this statement is true. However I'd like a straightforward number theory argument for this. -My basic idea was the following. We have $[\mathbb{K}:\mathbb{Q}] = p-1$ and by assumption $4 \mid p-1$ so there exists a unique subfield $\mathbb{Q} \subseteq\mathbb{E} \subseteq \mathbb{K}$ with $[\mathbb{E}:\mathbb{Q}] = 4$. By transitivity of norms, if $-1$ is in the image of $\mathrm{N}_{\mathbb{K/F}}$ then it's also in the image of $\mathrm{N}_{\mathbb{E/F}}$. Hence, it suffices to show it's not in the image of $\mathrm{N}_{\mathbb{E/F}}$. Here is where I got a bit stuck as I wasn't sure what the field $\mathbb{E}$ is exactly. One can get a basis by taking Galois sums of $\zeta$. There might also be a way to use the discriminent. -I have the same question in the local case. So assume $\mathbb{K} = \mathbb{Q}_{\ell}(\zeta)$ with $\ell > 0$ a prime and $\zeta \in \overline{\mathbb{Q}}_{\ell}$ a primitive $p$th root of unity. Looking in Serre's Local Fields the Galois group $\mathrm{Gal}(\mathbb{K}/\mathbb{Q}_{\ell})$ should still be cyclic. So assume $2$ divides $[\mathbb{K}:\mathbb{Q}_{\ell}]$ then there is a unique subfield $\mathbb{Q}_{\ell}\subseteq\mathbb{F} \subseteq \mathbb{K}$ with $[\mathbb{F}:\mathbb{Q}_{\ell}]=2$. The following should be true: - -If $\ell \neq p$ then $-1$ is in the image of the norm map $\mathrm{N}_{\mathbb{K/F}}$. If $\ell = p$ then $-1$ is in the image of the norm map $\mathrm{N}_{\mathbb{K/F}}$ if and only if $p \equiv -1 \pmod{4}$. - -By a block theory argument from finite groups I know the statement when $\ell \neq p$ is true. Hasse's Norm Theorem would then imply that the $\ell = p$ case agrees with the global case. However this all feels far too overblown. There should be an elementary number theory argument for all of this. -Apologies in advance if this is all too elementary. I am sure these answers and arguments are well known. - -REPLY [4 votes]: Global question: If $p = 1 \bmod 4$, then the element $-1 \in F^\times$ is not a norm from $K^\times$, because $F^\times$ is totally real and $K^\times$ is totally complex; thus $-1$ is not a local norm at either of the infinite places of $F$, and hence cannot be a global norm either. -Local question, $\ell \ne p$: If $\ell \ne p$ then the fields $K$ and $F$ you define are both unramified extensions of $\mathbf{Q}_\ell$. Hence $N_{K/F}(O_K^\times) = O_F^\times$, and in particular contains $-1$. (This surjectivity result for the norm map on units follows from the analogous result for the residue fields, which is easy to prove by hand, see slide 6 of these lecture notes by Garrett.) -Local question, $\ell = p$: As pointed out by KConrad in the comments, if you take $\omega$ a $(p-1)$-st root of unity in $\mathbf{Q}_p^\times \subseteq F^\times$, then the composite $F^\times \hookrightarrow K^\times \xrightarrow{N_{K/F}} F^\times$ is raising to the $(p-1)/2$-th power; so its image contains $\omega^{(p-1)/2}$ for every $(p-1)$-st root of unity, i.e. it always contains $-1$. -(You were correct in deducing from Hasse's theorem that if $-1$ is not a norm globally then it had to fail to be a norm locally at some place. However, the bad place is not $p$ but $\infty$.)<|endoftext|> -TITLE: Can a continuous real-valued function on a large product space depend on uncountably many coordinates? -QUESTION [14 upvotes]: Is there a reasonably well-behaved topological space $X$ (ideally Polish), a set $\kappa$, and a continuous function $g: X^\kappa\to\mathbb{R}$ that depends on uncountable many coordinates? -If $X$ is a compact Hausdorff space, the answer is known to be no. To see this, note that the family of all continuous functions depending on only finitely many coordinates satisfies the conditions for the Stone-Weierstrass theorem and is, therefore, uniformly dense. The argument can be found in textbooks. - -REPLY [7 votes]: Bockstein's theorem -Bockstein, M., Un théorème de séparabilité pour les produits topologiques, Fundam. Math. 35, 242-246 (1948). ZBL0032.19103. -This is the case of a product $\prod_{t \in T} X_t$ where all factors are second-countable. I that case any continuous function -$\prod_{t \in T} X_t \to \mathbb R$ depends on countably many coordinates. -PLUG... See Theorem 2.1 in -Edgar, G. A., Measurability in a Banach space, Indiana Univ. Math. J. 26, 663-677 (1977). ZBL0361.46017. -where the special case $X = \mathbb R$ is done. That is, a continuous function $\mathbb R^T \to \mathbb R$ depends on only countably many coordinates.<|endoftext|> -TITLE: If $M$ and $N$ are closed and $M\times S^1$ is diffeomorphic to $N\times S^1$, is it true that $M$ and $N$ are diffeomorphic? -QUESTION [17 upvotes]: If $M$ and $N$ are closed smooth manifolds, and $M\times S^1$ is diffeomorphic to $N\times S^1$, is it true that $M$ and $N$ are diffeomorphic? - -REPLY [23 votes]: The manifolds $M$ and $N$ may not even be homotopy equivalent! -In Compact Flat Riemannian Manifolds: I, Charlap showed that there are two closed flat manifolds $M$ and $N$ of the same dimension which are not homotopy equivalent (this is equivalent to $\pi_1(M) \not\cong \pi_1(N)$ as $M$ and $N$ are aspherical), such that $M\times S^1$ and $N\times S^1$ are diffeomorphic (which is equivalent to $\pi_1(M\times S^1) \cong \pi_1(N\times S^1)$ as closed flat manifolds are determined up to diffeomorphism by their fundamental group). -For a more explicit example, see this excellent answer by George Lowther.<|endoftext|> -TITLE: Proving equivalence of two definitions of a convex-type Hamming distance -QUESTION [5 upvotes]: Update: If somebody can answer my question there, then I will be able to fully answer my question here. -Consider $n\in\mathbb N$ and a non-empty set $M\subset\{0,1\}^n$. I have the following conjecture: -Conjecture. It is true that $$\sup_{\alpha\in[0,1]^n, \lVert \alpha\rVert_2=1}\min_{m\in M} \langle \alpha, m\rangle = \min_{\beta\in[0,1]^M, \sum_{m\in M} \beta_m = 1} \left\lVert\sum_{m\in M}\beta_m m\right\rVert_2.$$ -Here, $\beta_m m$ is just the scalar multiplication of the number $\beta_m$ with $m\in M\subset\{0,1\}^n$. Also, $\lVert \cdot\rVert_2$ is the usual euclidean norm and $\langle\cdot,\cdot\rangle$ is the usual euclidean inner product. (And note, of course, that $[0,1]^M$ is the set of all functions $\beta: M\to[0,1]$ where I will write $\beta_m$ for $\beta(m)$.) - -For instance, it is true if $M=\{m\}$, i.e. if $M$ only contains one element. In that case, the left-hand side equals $$\sup_{\alpha\in[0,1]^n, \lVert \alpha\rVert_2 =1}\langle \alpha,m\rangle.$$ -By Cauchy-Schwarz, we know that $\langle\alpha, m\rangle\le\lVert \alpha\rVert_2\lVert m\rVert_2=\lVert m\rVert_2$ and we have equality if and only if $\alpha=\frac{m}{\lVert m\rVert_2}$. Hence the left-hand side equals $\lVert m\rVert_2$. -The right-hand side is, as we must have $\beta=1$, $\lVert m\rVert_2$. - -If $M=\{m_1, m_2\}$, then we would have to prove -$$\sup_{\alpha\in[0,1]^n, \lVert \alpha\rVert_2 = 1} \min(\langle \alpha, m_1\rangle, \langle\alpha, m_2\rangle) = \min_{\beta\in[0,1]} \lVert \beta\, m_1+(1-\beta)\, m_2\rVert_2.$$ -This is already not obvious to me. However, for example with $M=\{(1,0),(0,1)\}$, both sides can be computed to equal $\frac1{\sqrt 2}$. - -Note: This conjecture is a Lemma that I would need to prove the equivalence of different definitions of convex distance that I found in the context of Talagrand's concentration inequality. - -Another example: Consider $n=4$ and (with a slight abuse of notation) $M=\{m_1,m_2,m_3\}=\{(1,1,0,0),(0,1,1,0),(0,1,1,1)\}$. -The right-hand side is $$\min_{(\beta_1,\beta_2,\beta_3)\in[0,1]^3, \beta_1+\beta_2+\beta_3=1} \lVert (\beta_1,\beta_1+\beta_2+\beta_3,\beta_2+\beta_3,\beta_3)\rVert_2.$$ -It is not too hard to see that the minimizer is $\beta=(1/2,1/2,0)$ for which we have $$\lVert (\beta_1,\beta_1+\beta_2+\beta_3,\beta_2+\beta_3,\beta_3)\rVert_2=\lVert (1/2,1,1/2,0)\rVert_2=\sqrt{\frac32}.$$ -The left-hand side is $$\sup_{(\alpha_1,\alpha_2,\alpha_3,\alpha_4)\in[0,1]^4, \alpha_1^2+\alpha_2^2+\alpha_3^2+\alpha_4^2=1} \min(\alpha_1+\alpha_2,\alpha_2+\alpha_3,\alpha_2+\alpha_3)=\sup_{(\alpha_1,\alpha_2,\alpha_3,\alpha_4)\in[0,1]^4, \alpha_1^2+\alpha_2^2+\alpha_3^2+\alpha_4^2=1}\min(\alpha_1+\alpha_2,\alpha_2+\alpha_3).$$ -The supremum occurs only if $\alpha_1+\alpha_2=\alpha_2+\alpha_3$, which happens for $\alpha=\left(\sqrt{\frac16},\sqrt{\frac23},\sqrt{\frac16},0\right)$. -For that $\alpha$, we have $$\alpha_1+\alpha_2=\alpha_2+\alpha_3=\sqrt{\frac32}$$ and so we indeed have equality of both sides. - -REPLY [3 votes]: Answer inspired by this great answer to a very related question by Paata Ivanishvili. -The right-hand side is equal to $\min_{m\in\operatorname{conv}(M)} \lVert m\rVert_2$, where $\operatorname{conv}(M)$ is the convex hull of $M$ in $\mathbb R^n$. Hence we have: -\begin{equation} -\begin{split} -\min_{m\in\mathrm{conv}(M)} \|m\|_{2}&=\min_{m \in \operatorname{conv} (M)} \max_{\|\alpha\|_{2}\leq 1} \langle \alpha, m\rangle -\\ &\overset{(*)}= \max_{\|\alpha\|_{2}\leq 1} \min_{m \in \operatorname{conv}(M)}\langle \alpha, m\rangle -\\ &\overset{(**)}= \max_{\substack{\alpha \in [0, \infty[^{n}\\\|\alpha\|_{2}\leq 1}} \min_{m \in M}\langle \alpha, m\rangle \\ -&=\max_{\alpha\in[0,1]^n \\ \lVert\alpha\rVert_2=1}\min_{m \in M}\langle \alpha, m\rangle. -\end{split} -\end{equation} -(*): Here, a minimax Theorem was used. -(**): Here, the Bauer maximum principle was used (as the minimum of $m\mapsto\langle\alpha, m\rangle$ will be attained on the extremal points of $\operatorname{conv}(M)$, and they are contained in $M$.) Notice that it is also used that $M\subset\mathbb R_+^n$ in order to restrict the $\alpha$'s to $[0,\infty[^n$. -Note. This works for any compact $M\subset\mathbb R_+^n$, not just $M\subset\{0,1\}^n$.<|endoftext|> -TITLE: Approximation of smooth diffeomorphisms by polynomial diffeomorphisms? -QUESTION [17 upvotes]: Is it possible to (locally) approximate an arbitrary smooth diffeomorphism by a polynomial diffeomorphism? -More precisely: Let $f:\mathbb{R}^d\rightarrow\mathbb{R}^d$ be a smooth diffeomorphism for $d>1$. For $U\subset\mathbb{R}^d$ bounded and open and $\varepsilon>0$, is there a diffeomorphism $p=(p_1, \cdots, p_d) : U\rightarrow\mathbb{R}^d$ (with inverse $q:=p^{-1} : p(U)\rightarrow U$) such that both - -$\|f - p\|_{\infty;\,U}:=\sup_{x\in U}|f(x) - p(x)| < \varepsilon$, $\ \textbf{and}$ -each component of $p$ and of $q=(q_1,\cdots,q_d)$is a polynomial, i.e. $p_i, q_i\in\mathbb{R}[x_1, \ldots, x_d]$ for each $i=1, \ldots, d$? - -Clearly, by Stone-Weierstrass there is a polynomial map $p : \mathbb{R}^d\rightarrow\mathbb{R}^d$ with $\|f - p\|_{\infty;\,U} < \varepsilon$ and such that $q:=(\left.p\right|_U)^{-1}$ exists; in general, however, this $q$ will not be a polynomial map. -Do you have any ideas/references under which conditions on $f$ an approximation of the above kind can be guaranteed nonetheless? -$\textbf{Note:}$ This is a crosspost from https://math.stackexchange.com/questions/3689873/approximation-of-smooth-diffeomorphisms-by-polynomial-diffeomorphisms - -REPLY [3 votes]: An illustration for one of the examples in the answer by Robert Bryant. -It is supposed to convey the feeling of something extremely rigid, unyielding and inflexible. -Image of the square $[-1,1]\times[-1,1]$ under the map $(x,y)\mapsto(x-y^2-2x^2y-x^4,y+x^2)$ (composite of $(x,y)\mapsto(x-y^2,y)$ with $(x,y)\mapsto(x,y+x^2)$).<|endoftext|> -TITLE: A differential inequality involving gradient and laplacian -QUESTION [5 upvotes]: Let $V:\mathbb{R}^{n}\to\mathbb{R}$ smooth, such that $\lim_{|x|\to\infty}V(x)=+\infty$. -What are conditions on $V$ that guarantee the existence of a function $U:\mathbb{R}^{n}\to\mathbb{R}$ such that for all $\alpha>0$ -$$\liminf_{|x|\to\infty} (\nabla V(x) \cdot \nabla U(x) - \alpha\, \Delta U(x)) > 0 \ ?$$ -Of course choosing $U=V$ does the job if $\nabla V$ does not vanish at infinity and we assume $\limsup_{|x|\to\infty}\Delta V(x)\leq0$. This condition seems too restrictive to me, any other idea? -Edit: I've realized that what I actually need is the following: given any $\alpha>0$ prove the existence of a function $U$ (possibly $U=U(\alpha,x)$) satisfying the inequality (under suitable hypothesis on $V$). The fact that $U$ may depend on $\alpha$ should help. - -REPLY [2 votes]: Suppose you know that $\Delta V=o(|\nabla V|^2)$ as $|x|\to\infty$. This implies that for any $\theta\in(0,1)$ you have eventually the inequality $\Delta V\le \theta|\nabla V|^2$. Then for any eventually increasing function $\phi$, if you take $U=\phi(V)$ your expression is bounded below by -$$ -|\nabla V|^2((1-\alpha\theta)\phi'-\alpha \phi''). -$$ -Then you can play with the choice of $\phi$. For instance, if $\phi''=o(\phi')$ and $\phi'>c>0$ you are done.<|endoftext|> -TITLE: Degree of secant varieties of Veronese varieties -QUESTION [9 upvotes]: Consider the degree two Veronese embedding $\mathbb{P}^n\rightarrow\mathbb{P}^N$ and let $V^n_{2}\subset\mathbb{P}^N$ be the corresponding Veronese variety. -Let $Sec_k(V^n_{2})\subseteq\mathbb{P}^N$ be the $k$-secant variety of $V_{2}^{n}$. This is the closure of the union of all $(k-1)$-planes spanned by $k$ independent points on $V_2^n$. -Is there a closed formula for the degree of $Sec_k(V^n_{2})$? -For instance if $k = 1$ we have that $Sec_1(V_2^n) = V_2^n$ has degree $2^n$, while for $k = n$ we have that $Sec_n(V_2^n)\subset\mathbb{P}^N$ is a hypersurface of degree $n+1$. What about the degree of $Sec_k(V^n_{2})\subseteq\mathbb{P}^N$ for $1 < k < n$? - -REPLY [12 votes]: The secant variety $Sec_k(V^n_2)$ is the variety parametrizing $(n+1)\times (n+1)$ symmetric matrices modulo scalar of rank at most $k$ that is of corank at least $n+1-k$. -Then by Proposition 12(b) in -J. Harris; L. W. Tu, On symmetric and skew-symmetric determinantal varieties, Topology 23 (1984), no. 1, 71–84. -the degree of $Sec_k(V^n_2)$ is given by -$$\deg(Sec_k(V^n_2)) = \prod_{i=0}^{n-k}\frac{\binom{n+1+i}{n+1-k-i}}{\binom{2i+1}{i}}$$ -In particular, for $k = n$ you get $n+1$, and for $k = 1$ you get $2^n$.<|endoftext|> -TITLE: Groupoid completion of a topological category vs its homotopy category? -QUESTION [7 upvotes]: Given a category $\mathcal{C}$ enriched in spaces, we can take the nerve (a simplicial space) and then geometric realization to get a space $B\mathcal{C}$. If we view spaces as $\infty$-groupoids, then this process should be thought of as a ($\infty$-)groupoidification. -We can also consider the homotopy category $h\mathcal{C}$, which has the same objects as $\mathcal{C}$ but where the morphisms from x to y are given by $\pi_0 \mathcal{C}(x,y)$. This is an ordinary category and we can take the nerve and geometrically realize it to get the classifying space $Bh\mathcal{C}$. -In general the spaces $B\mathcal{C}$ and $Bh\mathcal{C}$ will be very different, but they might agree on some low dimensional homotopy groups. -Fix an object $x \in \mathcal{C}$. Is it true that $\pi_1(B\mathcal{C}, x)$ is isomorphic to $\pi_1( Bh\mathcal{C}, x)$? If not, what is a good counter example? Are there conditions under which these will be isomorphic? For example I am interested in the case where $\mathcal{C}$ is symmetric monoidal and $x$ is the unit object. -Note that we can view a set as a discrete topological space and so $h\mathcal{C}$ is also a (discrete) topological category. There is a functor $\mathcal{C} \to h\mathcal{C}$, and so there is a natural comparison map $\pi_1(B\mathcal{C}, x)\to\pi_1( Bh\mathcal{C}, x)$. - -REPLY [4 votes]: The $\pi_0$ and $\pi_1$ are the same. The former is obvious since, taking homotopy categories and groupoidifying do not affect connected components. -The fundamental group of an infinity category $S$ by van Kampen has a generator and relation description in terms of the 1 and 2 simplices. In particular, it has generators given by strings of 1-simplices and formal inverses that start and end at $*$ subject to the relation that we can exchange homotopic simplices, and that $ee^{-1}=e^{-1}e=Id$. -This group is the same as the group where we pick a single representative 1-simplex in each homotopy class and add in all the relations involving only these representatives. -Again by van Kampen, this group is exactly the fundamental group of the realization of $Ho(S)$, since we have just named each path component of the morphism space. -Then we simply transfer this back to topologically enriched categories and we are done. -This is the best one can hope for in general, since if $S$ is a Kan complex, its homotopy category is a groupoid and the realization of this is a 1-type (in particular the map you describe is the Postnikov approximation map).<|endoftext|> -TITLE: Which stable homotopy groups are represented by parallelizable manifolds? -QUESTION [29 upvotes]: The Pontryagin-Thom construction allows one to identify the stable homotopy groups of spheres with bordism classes of stably normally framed manifolds. A stable framing of the stable normal bundle induces a stable framing of the stable tangent bundle. -This means that a framed manifold (one whose tangent bundle is trivial, e.g. a Lie group) represents an element of the stable homotopy groups of spheres. -So some elements are represented by honestly framed manifolds (as opposed to stably framed). -What is known about such elements? Is every element of the stable homotopy groups of spheres represented by an honestly framed manifold (i.e. with a trivial tangent bundle)? - -REPLY [13 votes]: Repeating the first part of Oscar's answer and elaborating on comments by Chris and Panagiotis, here is a down-to-earth argument in all cases: -The cases $n=1,3,7$ are fine, since the stable stems are in these degrees generated by $S^1$, $S^3$, $S^7$ with the unstable framing induced by the multiplication in the unit complex numbers, quaternions, or octonions. -In the other cases, we use that the obstruction to destabilising a given stable framing $F$ of an oriented closed manifold $M^n$ lies in $H^n(M,\pi_n(SO/SO(d))$, which is isomorphic (in a preferred way) to $\mathbb{Z}$ if $n$ is even and to $\mathbb{Z}/2$ if $n$ is odd. It is not too hard to see that, with respect to this isomorphism, the obstruction is given by the semi-characterstic: half the Euler characteristic for $n=2d$ and $\sum_{i=0}^d\mathrm{dim}(H_i(M,\mathbb{Z}/2))\text{ mod }(2)$ for $n=2d+1$ and $n\neq1,3,7$. In particular, the obstruction to destabilising is independent of $F$ which is somewhat surprising. -Originally this was proved by to Bredon and Kosinksi [1] who used a more geometric description of this obstruction: it is the degree (mod $2$ if $n$ is odd) of the Gauss map $M\rightarrow{S^n}$ induced by the stable framing $TM\oplus \varepsilon\cong \varepsilon^{n+1}$ (take the image of the canonical vector field in the trivial line bundle and normalize). -Now observe that, as Oscar explained, by doing a couple of trivial surgeries in a ball corresponding to taking connected sums with $S^1\times S^{n-1}$ or $S^2\times S^{n-2}$ and extending the stable framing, any stably framed bordism class in even dimensions contains a representative with trivial Euler-characteristic. The same works with the semi-characteristic in odd dimensions (here at most one surgery is necessary), so by the discussion above every stably framed bordism class has a representative whose stable framing can be destabilised. -[1] G.E. Bredon and A. Kosinski, Vector fields on $\pi$-manifolds. Annals of Math. 84, 85– 90 (1960). - -REPLY [7 votes]: $k \cdot[\mathrm{point}]\in \pi_0^s$ is represented by an honestly framed 0-manifold if and only if $k \geq 0$.<|endoftext|> -TITLE: On reflection properties of convex regions -QUESTION [6 upvotes]: It is well known that any ray of light passing thru a focus of an ellipse will pass thru the other focus after a single reflection from the ellipse boundary. If $A$ and $B$ are the foci of an ellipse, this property of rays holds both ways (those passing thru $A$ meet at $B$ and vice versa). - -Is there a closed convex region $C$ with the property: there exists a pair of points $A$ and $B$ within $C$ such that all rays thru $A$ will reflect once on $C$ and pass thru $B$ but not all rays thru $B$ will pass thru $A$ after one reflection from $C$? - -Is there a closed convex region $C$ such that: there is a pair of points $A$ and $B$ in the interior such that all rays thru $A$ pass thru $B$ after exactly 2 reflections from $C$? It is sufficient for the ray going directly from $A$ to $B$ to get reflected exactly twice from somewhere on $C$ and then pass thru $B$. - - -Note 1: Question 2 can have 'one-way' (convergence only of rays thru $A$ at $B$) and 'both-ways' variants. -Note 2: If not explicit constructions, even existence/non-existence arguments could be sought as answers to these questions. -One can also ask if relaxing convexity has any implications. - -REPLY [2 votes]: The answer to question 1) is no: -Take any ray R passing through B. Since C is convex and B lies in the interior of C, R will intersect C at a point O. Now the ray $\overrightarrow {AO}$ lies in the interior of C and its reflected ray passes through B. But this means we have a path of light $\overrightarrow {AO},\overrightarrow {OB}$. Since C is smooth the tangent at O is unique. Hence we can reverse this to give a light path $\overrightarrow {BO},\overrightarrow {OA}$ and hence the ray R from B will reflect off C at O and pass through A. -Since R was arbitrary, any ray passing through B will reflect off C and pass through A. -For question 2) a good example is to take two parabolas facing each other with a common axis, where the parabolas intersect at points A and A': - -Rays through the focus of either parabola, not passing through A or A' or hitting the other parabola first will be reflected twice and then pass through the other focus. If we move the foci apart we can get an arbitrarily large percentage of directions to work. -Thus the answer to 2) is yes if we allow an arbitrarily small percentage of rays from each focus to be excluded. This is the 2-way version. -Note that we can prove that such any example has to be 2-way in the same way as we did for question 1).<|endoftext|> -TITLE: How to apply Hahn-Banach to the convex hull? -QUESTION [5 upvotes]: I am trying to understand the proof of Lemma 4.1.2 in Michel Talagrand's publication from 1995 on concentration inequalities (see below for the precise question statement): - - - - - -A bit of context: Talagrand fixes a point $x\in X$ (he uses the notation $X=\Omega$) and a subset $A\subset X$, where $X=X_1\times X_2\times\dots X_n$ is the product space of arbitrary non-empty sets $X_1,\dots, X_n$. The $\alpha_i$ and $t$ are all supposed to be positive real numbers. He defines $A_t^c$ as follows: - - - - - -My question. I understand why (4.1.4) implies (4.1.5). However, Talagrand says that "the converse follows from the Hahn-Banach theorem". How does it follow from the Hahn-Banach theorem? - -Note: The problem can be slightly reformulated by saying that we want to prove that for $t>0$ and all $M\subset\{0,1\}^n$, we have that whenever $$\text{for all }\alpha\in]0,\infty[^n, \text{ there exists a } m\in M \text{ such that } \langle \alpha, m \rangle \le t\lVert \alpha\rVert_2,$$ then -$$\min_{m \in \text{Convex hull of } M} \lVert m \rVert_2 \le t.$$ -In fact, if somebody can show this, then I will be able to prove the conjecture formulated by me yesterday. - -REPLY [6 votes]: To solve the problem you mention at the end you can argue in this way: -$$ -\min_{m \in \mathrm{conv} (M)} \|m\|_{2}=\min_{m \in \mathrm{conv} (M)} \max_{\|\alpha\|_{2}\leq 1} \langle \alpha, m\rangle = \max_{\|\alpha\|_{2}\leq 1} \min_{m \in \mathrm{Conv}(M)}\langle \alpha, m\rangle \leq \max_{\|\alpha\|_{2}\leq 1, \alpha \in [0, \infty)^{n}} \min_{m \in M}\langle \alpha, m\rangle \leq t -$$ -The only nontrivial observation was used is min-max theorem, which says that if $X, Y$ are convex compact sets, $f(x,y)$ continuous, convex in $x$ and concave in $y$ then $\min_{x \in X} \max_{y \in Y} f(x,y) = \max_{y\in Y} \min_{x\in X} f(x,y)$. Choose $f(x,y)=\langle x, y\rangle$<|endoftext|> -TITLE: On Glaeser's result for the square-root of a smooth non-negative function -QUESTION [6 upvotes]: One of the results due to Georges Glaeser is the following: there exists a non-negative $C^\infty$ function $f$ on the real line, flat at its zeroes, such that $\sqrt{f}$ is not $C^2$. On the other hand, $\sqrt{f}$ is $C^1$ for any such $f$. -Question 1: is it possible to find $f$ as above such that $\sqrt{f}$ is not twice differentiable at a point? In Glaeser's counterexample mentioned above, $\sqrt{f}$ is twice differentiable with an unbounded second derivative. -Question 2: is it possible to find $f$ as above such that there is no function $g$, $C^2$ on the real line such that $f=g^2$. Here $g$ is allowed to take negative values, which is not the case of $\sqrt f$. -Question 3: is it possible to find $f$ as above such that there is no $C^\infty$ function $g:\mathbb R\longrightarrow \mathbb C$ such that $f=\vert g\vert^2$. - -REPLY [2 votes]: Question 1: In - -Alekseevsky, Dmitri; Kriegl, Andreas; Michor, Peter W.; Losik, Mark -Choosing roots of polynomials smoothly. (English summary) -Israel J. Math. 105 (1998), 203–233. - -it is shown that such an $f$ always has a twice differentiable square root. But this square root is not necessarily positive. -Question 2: -$$ -f(t) = \sin^2(1/t)e^{-1/t} + e^{-2/t} \text{ for }t>0,\quad f(t) = -0\text{ for }t\le0. -$$ -This is a sum of two non-negative $C^\infty$ functions each of which -has a $C^\infty$ square root. -But the second derivative of the square -root of $f$ is not continuous at the origin. This is also a counter example to question 3. - -J.-M. Bony, F. Broglia, F. Colombini and L. Pernazza, Nonnegative. functions as squares or sums of squares, J. Funct. Anal. 232(2006), 137–147. - -have shown that twice differentible is best possible; it cannot be improved to $C^{1,\alpha}$ for any continuity module $\alpha$. -See also - -Bony, Jean-Michel; Colombini, Ferruccio; Pernazza, Ludovico -On the differentiability class of the admissible square roots of regular nonnegative functions. (English summary) Phase space analysis of partial differential equations, 45–53, -Progr. Nonlinear Differential Equations Appl., 69, Birkhäuser Boston, Boston, MA, 2006.<|endoftext|> -TITLE: Are categories special, foundationally? -QUESTION [8 upvotes]: Some folks over at nLab want to use categories as a foundation for all of mathematics, I'm guessing as an alternative to sets. Sets work fine, and so do categories, so I have started wondering what other kinds of objects everything could be founded on. The Wikipedia page for categories has a nice table of different group-like structures. In addition to the group-like structures, there are also a lot of other structures that come up in abstract algebra, like rings, vector spaces and modules. Could there be an v-Lab where everyone gets together to phrase everything in math in terms of vector spaces, or is there something special about categories? Is there some kind of property or requirement that an abstract object must have in order to work as a foundational concept? -There is a lot of theory about how powerful each different kind of logic is (for example, some theorems can be proven in ZFC but not in constructive mathematics), so I am curious if there is a similar multiverse for foundational definitions. -Here is a closely related question, but for groups. It is encouraging to see that for groups the answer is "yes." This makes it even more interesting to ask, what are the requirements for foundational objects in general? - -REPLY [15 votes]: The term "foundations of mathematics" is all well and good when one has fixed a foundation to work with. But when you start trying to compare different foundations of mathematics, you quickly realize that the term "foundations of mathematics" requires a great deal of unpacking. I recall once reading a perspicuous article by Penelope Maddy [1] identifying on the order of a dozen different roles we implicitly expect a "foundations of mathematics" to play. This is a testament to the resounding successes of the most widely-used foundations of math, namely ZFC layered on top of (classical, finitary) first-order logic, and variations thereof. -Anyway, if the question is "What are the requirements for foundational objects in general?", then I'm not sure I can do better than pointing you to Maddy. I don't think the question has a simple answer. -Maybe I'll also point out that there are many different ways one might "use category theory as a foundations of mathematics". - -One way would be to use Lawvere's ETCC (plus additional axioms as needed) in a similar way to the way ZFC (plus additional axioms as needed) is used by most mathematicians. That is, we write down a particular first-order theory (either ZFC or ETCC) and then just encode everything we want to do inside of it. This is perfectly doable, but I think most would agree it is awkward and doesn't really have any tangible benefits. - -Another way would be to use ETCS, or perhaps the theory of an elementary topos with natural number object, or something like this, instead of ZFC. Again, this is perfectly doable. I think there are some who feel that this approach has merit. - -Another way would be to back up, and decide that one isn't interested in just one foundational theory, but in comparing many different ones (note that even the usual foundations are open-ended in this way -- we always allow ourselves to extend ZFC by large cardinal hypotheses and whatnot when necessary). We might find a privileged role for category theory in deciding what is and isn't a "foundations of math", or in comparing different foundations of math. For instance, we might want to compare two different foundations by comparing what each of them says about "the category of sets". - -We might also just use category theory as a tool in the foundations of mathematics. For instance, type theoretic foundations are often compared to classical foundations by building models of the former in the latter, and what a "model" is is often expressed in terms of some kind of category. - -One might stick with ZFC or whatever, but emphasize that category theory plays a privileged role in comparing different types of mathematical objects, based on the empirical fact that mathematical objects often organize themselves naturally into categories, and such categories can be compared via functors, etc. - -As an example of (3), different notions of constructivism can be fruitfully compared in terms of what properties they say the category of sets has (things like being an "exact category", a "pretopos", or whatnot). It's probably due to my ignorance, but I'm not aware of similarly enlightening concepts being more naturally formulated in terms of membership-based set theory. Moreover, new constructivist theories can be formulated and motivated starting from such categorical considerations. - -In another direction, support for certain large cardinal principles, such as measurable cardinals or Vopenka's principle, can be given based on categorical formulations of these principles, and categorical considerations about these formulations. Conceivably, one might imagine formulating new large cardinal principles based primarily on categorical considerations, although I'm not sure this has been done much in practice, unless you count Vopenka's Principle or Weak Vopenka's Principle. I actually find it very suggestive that the strongest large cardinal principles (Reinhardt and Berkeley cardinals) are inconsistent with ZFC, so even classically most naturally retreats to ZF to study them. Although I don't have any evidence for the following scenario, I wouldn't be surprised if even stronger large cardinal principles might one day be found which are inconsistent with ZF, and instead most naturally studied in the context of topos theory or something like that. - -And so forth. - - -[1] I'm not sure, but it may have been What do we want a Foundation to do? (non-paywalled link). Full citation: Maddy P. (2019) What Do We Want a Foundation to Do?. In: Centrone S., Kant D., Sarikaya D. (eds) Reflections on the Foundations of Mathematics. Synthese Library (Studies in Epistemology, Logic, Methodology, and Philosophy of Science), vol 407. Springer, Cham<|endoftext|> -TITLE: On a quasi-separated assumption in a lemma for the homotopy exact sequence of the etale fundamental group -QUESTION [7 upvotes]: Background: -I've seen two versions of the homotopy exact sequence for etale fundamental groups. One from Stacks: - -Stacks 0BTX: Let $k$ be a field with algebraic closure $\overline{k}$. Let $X$ be a quasi-compact and quasi-separated scheme over $k$. If the base change $X_{\overline{k}}$ is connected, then there is a short exact sequence $$ 1\to \pi_1(X_{\overline{k}}) \to \pi_1(X)\to \pi_1(\operatorname{Spec} k) \to 1$$ of profinite topological groups. - -And one from SGA: - -SGA I, Theoreme 6.1 of chapter IX: Suppose $S$ is the spectrum of an Artinian ring $A$ with residue field $k$, $\overline{k}$ an algebraic closure of $k$, $X$ a $S$-scheme, $X_0=X\times_A k$, $\overline{X}_0=X\times_A \overline{k}$, $\overline{a}$ a geometric point of $\overline{X}$, $a$ the image in $X$, and $b$ the image in $S$. We suppose that $X_0$ is quasi-compact and geometrically connected over $k$ (N.B. if $X$ is proper over $S$, this means that $H^0(X_0,\mathcal{O}_{X_0})$ is an artinian local ring whose residue field is radicial over $k$). Then the canonical sequence of homomorphisms $$1\to \pi_1(\overline{X}_0,\overline{a})\to \pi_1(X,a)\to \pi_1(S,b) \to 1$$ is exact, and we have $$\pi_1(S,b)\stackrel{\sim}{\leftarrow}\pi_1(k,\overline{k})= Gal(\overline{k},k).$$ - -A key step in both of these proofs is the following lemma: - -Lemma: Let $X$ be quasi-compact and geometrically connected. If we have a finite etale cover $\overline{Y}$ of $\overline{X}=X\times_k \operatorname{Spec}\overline{k}$, then it comes from a finite etale cover of $X\times_k \operatorname{Spec} K$, where $k\subset K$ is a finite extension. - -Stacks adds the assumption that $X$ is quasi-separated, and SGA omits this assumption. In the case when $X$ is assumed quasi-separated, I think I understand how to show this lemma and I can even write down a recipe for producing the extension we need: using the fact that $X$ is quasi-compact and quasi-separated in combination with the definition of etale morphisms as locally of finite presentation, we can pick a finite affine open cover $U_i=\operatorname{Spec} A_i$ of $X$ and get a finite affine open cover $\overline{U_i}=U_i\times_k \operatorname{Spec} \overline{k}$ of $\overline{Y}$ by spectra of rings of the form $(A_i\otimes_k \overline{k})[x_1,\cdots,x_n]/(f_1,\cdots,f_m)$ which gives us a finite list of coefficients from $\overline{k}$ needed to define the $\overline{U_i}$. Covering $U_i\cap U_j$ with a finite number of open affines $U_{ijk}$ since $X$ is quasi-separated, we see that there's a finite number of coefficients from $\overline{k}$ necessary to define the maps $\overline{U_{ijk}}\to \overline{U_i}$ and $\overline{U_{ijk}}\to \overline{U_j}$, and we can apply the same trick to get a finite list of coefficients of $\overline{k}$ needed to define the data we use to patch together the $\overline{U_i}$ into $\overline{Y}$. We end up with a finite list of elements of $\overline{k}$ which are enough to specify all the data needed to put together $\overline{Y}$, and we can define our cover over a finite extension of $k$ containing all these elements. -Question: How can I prove the lemma when $X$ is not quasi-separated? SGA leaves the proof of the lemma to the reader, and it appears to me that my strategy fails without the quasi-separated hypothesis (some $U_i\cap U_j$ could fail to be quasi-compact and then I would have to deal with a potentially infinite list of elements of $\overline{k}$). Stacks' copy of the lemma seems to rely on quasi-separatedness in an essential way, and I don't see how to remove it. - -REPLY [9 votes]: This is more a comment than an answer: a few years back, in 2011, while working with some friends on SGA1, we also found out that we could not prove this statement without the hypothesis that $X$ is quasi-separated. Our question: Is this hypothesis simply missing in SGA1 ? reached Michel Raynaud and his answer was reported to be something like: Probably, but this is not very interesting.<|endoftext|> -TITLE: How to determine the coefficient of the main term of $S_{k}(x)$? -QUESTION [5 upvotes]: Let $k\geqslant 2$ be an integer, suppose that $p_1,p_2,\dotsc,p_k$ are primes not exceeding $x$. Write -$$ S_{k}(x) = \sum_{p_1 \leqslant x} \dotsb \sum_{p_k \leqslant x} \frac{1}{p_1+\dotsb +p_k}. $$ -By AM-GM inequality, $p_{1}+\dotsb + p_{k} \geqslant k \sqrt[k]{p_{1}\dotsm p_{k}}$, we have -$$ S_{k}(x) \leqslant \frac{1}{k} \sum_{p_{1}\leqslant x}\dotsb \sum_{p_{k} \leqslant x} - \frac{1}{\sqrt[k]{p_{1}\dotsm p_{k}}} = \frac{1}{k} \left( \sum_{p \leqslant x} p^{-\frac{1}{k}} \right)^{k}. $$ -By Prime Number Theorem and summation by parts we see that -$$ \sum_{p \leqslant x} p^{-\frac{1}{k}} = \mathrm{Li}\big( x^{1-\frac{1}{k}} \big) + O \left( x^{1-\frac{1}{k}}\mathrm{e}^{-c\sqrt{\log x}} \right), $$ -Here $\mathrm{Li}(x)$ is the logarithmic integral, and $\mathrm{Li}(x)\sim x/\log x$. Hence -$$ S_{k}(x) \leqslant \left( \frac{k^{k-1}}{(k-1)^{k}} +o(1) \right) \frac{x^{k-1}}{\log^{k} x}. $$ -On the other hand, $p_{1}+\dotsb +p_{k} \leqslant kx$, we have -$$ S_{k}(x) \geqslant \frac{1}{kx} \sum_{p_{1} \leqslant x} \dotsb \sum_{p_{k} \leqslant x} 1 = \frac{1}{kx} \left( \sum_{p \leqslant x} 1 \right)^{k} = \frac{\pi^{k}(x)}{kx} = \frac{(1+o(1))}{k} \frac{x^{k-1}}{ \log^{k} x}. $$ -My question is how to determine the coefficient of the main term of $S_{k}(x)$? Thanks! - -REPLY [2 votes]: Thank you, Mr. Petrov, but you made a little mistake. -A detailed calculation of $c$ is as follows: -Write $g(x)=(1-\mathrm{e}^{-x})^k= \sum\limits_{j=0}^{k} \binom{k}{j} (-1)^{j} \mathrm{e}^{-jx}$, integrating by parts we get -\begin{align} -\int_{0}^{\infty} g(x) x^{-k} \,\mathrm{d} x & = \int_{0}^{\infty} g(x) \,\mathrm{d} \left( \frac{x^{-k+1}}{-k+1} \right) \nonumber \\ - & = \left. \frac{g(x)}{(-k+1)x^{k-1}} \right|_{0}^{\infty} - + \frac{1}{k-1} \int_{0}^{\infty} \frac{g'(x)}{x^{k-1}} \mathrm{d} x, -\end{align} -since $\lim\limits_{x\to 0} \dfrac{g(x)}{x^{k-1}} = \lim\limits_{x\to +\infty} \dfrac{g(x)}{x^{k-1}} = 0$, so that -\begin{align*} - \frac{1}{k-1} \int_{0}^{\infty} \frac{g'(x)}{x^{k-1}} \mathrm{d} x - & = \frac{1}{k-1} \int_{0}^{\infty} g'(x) \, \mathrm{d} \left( \frac{x^{-k+2}}{-k+2} \right) \\ - & = - \left. \frac{g'(x)}{(k-1)(k-2)x^{k-2}} \right|_{0}^{\infty} + \frac{1}{(k-1)(k-2)} \int_{0}^{\infty} \frac{g''(x)}{x^{k-2}} \mathrm{d} x, -\end{align*} -where $g'(x)=k(1-\mathrm{e}^{-x})^{k-1}\cdot \mathrm{e}^{-x}$ and $\lim\limits_{x\to 0} \dfrac{-g'(x)}{(k-1)(k-2)x^{k-2}}= \lim\limits_{x\to +\infty} \dfrac{-g'(x)}{(k-1)(k-2)x^{k-2}}=0$. Hence, integrating by parts $k-1$ times gives -\begin{align} - \int_{0}^{\infty} - \frac{\sum\limits_{j=0}^{k} \binom{k}{j} (-1)^{j}\mathrm{e}^{-jx}}{x^k} \, \mathrm{d} x - & =\frac{1}{(k-1)!}\int_{0}^{\infty} \frac{\sum\limits_{j=0}^{k} \binom{k}{j} (-1)^j(-j)^{k-1} \mathrm{e}^{-jx}}{x} \,\mathrm{d} x \nonumber \\ -& =\frac{1}{(k-1)!}\int_{0}^{\infty} \sum\limits_{j=1}^{k} \binom{k}{j} (-1)^{k+j-1}j^{k-1} \frac{\mathrm{e}^{-jx}}{x} \, \mathrm{d} x. \quad (\ast) -\end{align} -Notice that $(-1)^{k+j-1}=(-1)^{k+j+1}=-(-1)^{k-j}$, and consider the Stirling number of the second kind, we get -\begin{align} -\frac{1}{(k-1)!} \sum_{j=1}^{k} (-1)^{k+j-1} \binom{k}{j} j^{k-1} - & = -k \cdot \frac{1}{k!} \sum_{j=1}^{k} (-1)^{k-j} \binom{k}{j} j^{k-1} \\ - & = -k\cdot S(k-1,k)=0. -\end{align} -Set $\displaystyle a_{j} = \frac{(-1)^{k+j-1}j^{k-1}}{(k-1)!} \binom{k}{j}$, then $\sum\limits_{j=1}^{k} a_{j}=0$. -Using the Frullani's integral formula $\int_{0}^{\infty} \frac{\mathrm{e}^{-jx}- \mathrm{e}^{-Ax}}{x} \mathrm{d} x = \log A - \log j$ with $0 -TITLE: Is it consistent to have a function that is sensitive to subset relation from the power set of a set to that set? -QUESTION [9 upvotes]: Is it consistent with $ZF$ to have a set $S$ and a function $F: P(S) \to S$ such that: - -$\forall X,Y \in P(S): X \subsetneq Y \implies F(X) \neq F(Y)$ - -REPLY [8 votes]: Here's an argument that doesn't use ordinals, as an alternative to the nice proof described by Andrés and Andreas. -Take $F: P(S) \to S$ satisfying your hypothesis. Define a function $\Phi: P(S) \to P(S)$ by -$$ -\Phi(X) = \{F(Y): Y \subseteq X\}. -$$ -Then $\Phi$ is an order-preserving map from the complete lattice $P(S)$ to itself, so there is a least $X \in P(S)$ such that $\Phi(X) \subseteq X$. (That such an $X$ exists is part of the proof of the Knaster-Tarski fixed point theorem, and is in any case easy: put $X = \bigcap\{ Y \in P(S): \Phi(Y) \subseteq Y\}$, then use the fact that $\Phi$ is order-preserving to show that $\Phi(X) \subseteq X$.) -Now: - -$F(X) \in \Phi(X)$ by definition of $\Phi$, so $F(X) \in X$, so if we write $X' = X \setminus\{F(X)\}$ then $X' \subsetneqq X$. - -$\Phi(X') \subseteq \Phi(X) \subseteq X$, so $\Phi(X') \subseteq X$. - -Any subset $Y$ of $X'$ is a proper subset of $X$, so your hypothesis on $F$ gives $F(Y) \neq F(X)$. Hence $F(X) \not\in \Phi(X')$. - - -Combining the three bullet points, we have a proper subset $X'$ of $X$ satisfying $\Phi(X') \subseteq X'$. This contradicts the minimality of $X$.<|endoftext|> -TITLE: Are "large enough" finite etale covers arithmetic? -QUESTION [9 upvotes]: Let $X$ be a variety over a number field $K$. Then it is known that for any topological covering $X' \to X(\mathbb{C})$, the topological space $X'$ can be given the structure of a $\overline{K}$-variety in such a way so that the morphism $f: X' \to X$ inducing the topological map is a finite etale morphism over $\overline{K}$. However, the variety $X'$ and the morphism $f$ may not descend to $K$. -My question is as follows: does there always exist a further finite etale covering $f' : X'' \to X'$ such that the composition $X'' \to X$ may be defined over $K$? -EDIT: Just to be clear, I'd like all the covers involved to be geometrically connected to avoid trivial solutions. - -REPLY [9 votes]: Adding on Will's and Sasha's answers, the condition of having a rational point, or at least a "1-truncated homotopy fixed point" for the action is necessary. For example, let $C_2$ act on the circle $S^1$ by half rotation. The covers of $S^1$ are the standard n-fold ones, and we can ask what it takes to lift the action of $C_2$ to the cover, so that it is "defined over $BC_2$". In particular, we need to lift that half-circle rotation to the n-fold cover, for which the options are $1/2n + k/n$ rounds rotation. For this to be an involution, we need that applying it twice gives the identity, i.e. that $1/n +2k/n$ is an integer. If $n$ is even, this is impossible, and so the double cover of this action on $S^1$ has no cover definable over $BC_2$. To turn this topological picture to arithmetic, take $K=\mathbb{R}$ and let complex conjugation act on $\mathbb{C}^\times$ by $z\mapsto -1/\bar{z}$ (which is a form of the multiplicative group with no rational points). The action on the unit circle is then half rotation, so the Galois story realized to the topological one up to profinite completion. -I would add that what happens topologically is that if we have a fixed point, we can use it to define a "connected" compositum of pointed covers, by taking the component of the tuple of base points lifts. This is what missing in this example esssentially, even though up to isomorphism all covers are actually "the same".<|endoftext|> -TITLE: "Well-known fact" that every irreducible 3-manifold with non-empty boundary has an incompressible surface -QUESTION [6 upvotes]: I have seen in several sources that this results holds, however none of them included the proof. Does anyone know where I can find one? -Also, it would be great if someone could provide me with a counterexample, where irreducibility of the manifold matters. -Thank you. - -REPLY [9 votes]: Counterexample: the 3-ball $B^3$. It is irreducible because it is a compact submanifold of $\mathbb{R}^3$ with connected boundary. It has non-empty boundary and $\partial B$ is simply connected (and incompressible surfaces with boundary must send their boundary to curves which do not bound disks in $\partial B$). Hence if there is an incompressible surface it must be closed. Non orientable closed surfaces do not embed in $B^3$. Moreover, $B^3$ is simply connected and -recall that by Dehn's lemma an orientable, closed surface with $\chi(S)\leq 0$ is incompressible if the inclusion is injective at the level of fundamental group. -However the following is true: - -If $M$ is compact with non-empty boundary, oriented , irreducible, and $\partial$-irreducible* then either $M=B^3$ or $M$ contains an incompressible -and $\partial$-incompressible surface. - -This result relies on the fact that (under the above assumptions), given a class in $H_2(M,\partial M;\mathbb{Z})$ you can represent it by disjoint union of incompressible and and $\partial$-incompressible surfaces. -* -You can find all the details for example in Bruno Martelli. An Introduction to Geometric Topology. https://arxiv.org/pdf/1610.02592.pdf -Proposition 9.4.3 and Corollary 9.4.5. -*Meaning that there are no essential disks, in other words $M$ is not obtained by joining two 3-manifolds with a 1-handle. - -REPLY [5 votes]: The proof might be too long for this fact. However, here is one reference -Algorithmic Topology and Classification of 3-Manifolds by Sergei Matveev in the series Algorithms and computations in Mathematics, Volume 9, 2003, Springer-Verlag. -You may start reading from page 167.<|endoftext|> -TITLE: When does an open manifold admit two linearly independent vector fields? -QUESTION [11 upvotes]: $\DeclareMathOperator{\span}{span}$ -$\DeclareMathOperator{\co}{H}$ -$\newcommand{\kk}{\mathbb{F}}$ -$\newcommand{\qq}{\mathbb{Q}}$ -$\newcommand{\zz}{\mathbb{Z}}$ -$\newcommand{\rr}{\mathbb{R}}$ -$\newcommand{\semi}{\hat{\chi}_2}$ -$\newcommand{\ori}[1]{\textbf{(O$_{\pmb{#1}}$)}}$ -$\newcommand{\nori}[1]{\textbf{(NO$_{\pmb{#1}}$)}}$ -$\newcommand{\rarr}{\rightarrow}$ -Let $M$ be a smooth connected $d$-manifold (without boundary), and write $\span(M)$ for the maximum number of linearly independent vector fields on $M$. By Poincaré-Hopf, we know that if $M$ is closed, then $\span(M) \geq 1$ if and only if $\chi(M) = 0$. It is also known that every open $M$ satisfies $\span(M) \geq 1$. I am interested in the characterization of the condition $\span(M) \geq 2$. This has been achieved for closed manifolds with works of several people (over the years ~1965-90), as I will explain. -Notation for closed $M$: Given a field $\kk$, write $b_j(M;\kk) := \dim_\kk{\co_j(M;\kk)}$. Write $$\semi(M;\kk) := \left( \sum_{j \geq 0} b_{2j}(M;\kk) \!\!\!\mod{\!2}\right) \in \kk_2$$ -for the Kervaire semi-characteristic over $\kk$ (it is always a mod-2 number). For each $0 \leq j \leq d$, write $w_j(M) \in \co^j(M;\kk_2)$ for the $j$-th Stiefel-Whitney class. Write $[M] \in \co_d(M;\kk_2)$ for the $\kk_2$-fundamental class and $$\langle-,[M] \rangle \colon \co^k(M;\kk_2) \rarr \co_{d-k}(M;\kk_2)$$ the associated PD isomorphism. If $M$ is oriented and $d \equiv 0 \!\pmod{4}$, let $\sigma(M)$ denote its signature. -For the characterization of closed $M$ with $\span(M) \geq 2$, first note that when $d=2$ we only have the $2$-torus. Assuming $d \geq 3$, then $M$ has $\span(M) \geq 2$ if and only if in addition to $\chi(M) = 0$, it satisfies one of the following conditions: -$\ori{0}$ : $M$ is orientable, $d \equiv 0\!\pmod{4}$, and $\sigma(M) \equiv 0\!\pmod{4}$. -$\ori{1}$ : $M$ is orientable, $d \equiv 1\!\pmod{4}$, $w_{d-1}(M) = 0$, and $\semi(M;\rr) = 0$ -$\ori{2,3}$ : $M$ is orientable and $d \equiv 2,3 \!\pmod{4}$. -$\nori{0,2}$ : $M$ is non-orientable, $d$ is even, and writing $\zz_{w_1(M)}$ for the orientation sheaf, the twisted Bockstein $$\beta^{*}\colon \co^{d-2}(M;\kk_2) \rightarrow \co^{d-1}(M;\zz_{w_1(M)})$$ sends $w_{d-2}(M)$ to $0$. -$\nori{1}$ : $M$ is non-orientable, $d \equiv 1\!\pmod{4}$, $w_1(M)^2 = 0 = w_{d-1}(M)$, and $$\semi(M;\kk_2) = \langle w_2(M)w_{d-2}(M), [M] \rangle \in \kk_2 \, .$$ -$\nori{3}$ : $M$ is non-orientable, $d \equiv 3\!\pmod{4}$, and $w_1(M)^2 = 0$. -$\nori{1,3}$ : $M$ is non-orientable, $d$ is odd, $w_1(M)^2 \neq 0$, and $w_{d-1}(M) = 0$. -[I can give precise references with a bit more work, but for now I will note that $\ori{0}$ is due to Frank and independently Atiyah for $d > 4$ and due to Randall when $d=4$, $\ori{1}$ is due to Atiyah, $\ori{2,3}$ is due to E. Thomas, $\nori{0,2}$ is due to Pollina, $\nori{1}$ and $\nori{3}$ are due to Randall, $\nori{1,3}$ is due to Mello.] -For open manifolds, I could only find the following: Non-orientable surfaces necessarily have $\span = 1$, and open orientable manifolds of dimension 2,3 are parallelizable. -Can we similarly characterize open $d$-manifolds $M$ with $\span(M) \geq 2$? For instance, can the method in this answer be adapted to show that $\span(M) \geq 2$ whenever $M$ is orientable and $d \equiv 3\!\pmod{4}$? - -REPLY [9 votes]: Throughout we assume $d>4$ and $d$ odd. Denote by $V_{d,2}$ the Stiefel-manifold of orthonormal $2$-frames in $\mathbb R^d$. Since $V_{d,2}$ is $(d-3)$-connected there is a $2$-field over the $(d-2)$-skeleton of $M$. -The first obstruction to extend this $2$-field over the $(d-1)$-skeleton lies in $H^{d-1}(M;\pi_{d-2}V_{d,2}) -=H^{d-1}(M;\mathbb Z_2)$ and is given by $w_{d-1}(M)$. Suppose this class vanishes and consider an extension of the $2$-field over the $(d-1)$-skeleton. But since $M$ is open, there are no $n$-cells for $n>d-1$. Hence the only obstruction to extend a $2$-field from the $(d-2)$-skeleton to the whole open manifold is the Stiefel-Whitney class $w_{d-1}(M)$. -All other obstructions in the theorems you mentioned, are coming from the existence of a $d$-cell of a $d$-dimensional closed manifold.<|endoftext|> -TITLE: Approximating power series coefficients --- Why does a clearly illegitimate method (sometimes) work so well? -QUESTION [9 upvotes]: For reasons that don't matter here, -I want to estimate the power series coefficients -$t_{ij}$ for the rational function -$$T(x,y)= {(1+x)(1+y)\over 1- x y(2+x+y+x y)}=\sum_{i,j} t_{ij}x^iy^j$$ -Using a method that I cannot justify, I get -highly accurate estimates when $i=j$ and highly inaccurate estimates when -$|i-j|$ strays at all far from zero. -My questions are: - -Q1) Why does my apparently illegitimate method work so well when $i=j$? - - -Q2) Why does the answer to Q1) not apply when $i\neq j$ ? - -(Of course, once the answer to Q1) is known, the answer to Q2) might be -self-evident.) - -I'll first present the method, then explain why I think it shouldn't work, -then present the evidence that it works anyway when $i=j$, and then present -the evidence that it rapidly goes haywire when $i\neq j$. -The Apparently Illegitimate Method: -Note that $t_{ij}=t_{ji}$, so we can limit ourselves to estimating -$t_{j+k,j}$ for $k\ge 0$. -I) Define -$$T_k(y)=\sum_jt_{k+j,j}y^j$$ -For example, a residue calculation gives -$$ T_0(y)= {1-y-\sqrt{1-4y+2y^2+y^4}\over y\sqrt{1-4y+2y^2+y^4}} $$ -It turns out that all of the $T_k$ share a branch point at $\zeta\approx .2956$ and are analytic in the disc $r<\zeta$. -II) Write -$$L_k=\lim_{y\mapsto \zeta} T_k(y)\sqrt{y-\zeta}$$. -Discover that $L_0\approx 1.44641$ and $L_k=L_0/\zeta^{k/2}$. -III) Approximate -$$T_k(y)\approx L_k/\sqrt{y-\zeta}$$ -IV) Expand the right hand side in a power series around $y=0$ and equate -coefficients to get -$$t_{ij}\approx \pm{L_0\over\sqrt{\zeta}}\pmatrix{-1/2\cr j\cr}\zeta^{-(i+j)/2} -\approx \pm 2.66036 \pmatrix{-1/2\cr j\cr}\zeta^{-(i+j)/2}\qquad(E1)$$ -Remarks: - -Obviously one could try to improve this approximation -at Step III by using more terms in the power series for $T_k$ at $y=\zeta$. -This doesn't seem to help, except when $k=0$, in which case the original approximation is already quite good. - -For $k\ge 2$, $T_k(y)$ has a zero of -order $k-1$ at the origin. Thus one could modify this method by approximating -$T_k(y)/(y^{k-1})$ instead of $T_k(y)$ -This yields -$$t_{ij}\approx \pm{2.66036}\pmatrix{-1/2\cr 1-i+2j}\zeta^{-(i+j)/2}\qquad(E2)$$ -(E2) is (much) better than (E1) in the range $i\ge 2j+1$, where it gets -exactly the correct value, namely zero. Otherwise, it seems neither systematically better nor worse. - - -Why Nothing Like This Should Work: The expansion of $T_k(y)$ at -$\zeta$ contains nonzero terms of the form -$A_{i,j}(\zeta-y)^j$ for all positive integers $j$. (I'm writing $i=j+k$ to -match up with the earlier indexing.) The truncation at Step III throws all -these terms away. Therefore the expansion around the origin in Step IV -ignores (among other things) the contribution of $A_{ij}$ to the estimate -for $t_{ij}$. So unless we can control the sizes of the $A_{ij}$, we -have absolutely no control over the quality of the estimate. -And in fact, even when $k=0$, the $A_{j,j}$ are not small. -For example, $t_{8,8}=8323$ and my estimate for $t_{8,8}$ is a -respectable $8962.52$. But $A_{8,8}$, which should have contributed to that -estimate and got truncated away, is equal to $58035$. It seems remarkable -that I can throw away multiple terms of that size and have the effects nearly cancel. -I'd like a conceptual explanation for this. -But When $i=j$, It Works Anyway: - -and these get even better if you truncate just slightly farther out. -Why any explanation can't be too general: - -REPLY [9 votes]: The paper 'A New Method for Computing Asymptotics of Diagonal Coefficients of Multivariate Generating Functions' by A. Raichev and M. Wilson has the precise machinery that can solve this problem. Get a copy and these brief notes correspond to their symbols, for the diagonal case -$$ f_{n,n} = [x^n \, y^n ] \frac{I(x)}{J(x)} = [x^n \, y^n ] \frac{(1+x)(1+y)}{1-xy(2+x+y+xy)}.$$ -Solve the simultaneous system for zeros of the denominator $J$; in Mathematica, -Solve[ { x D[J,x] == y D[J,y], J==0 },{x,y} ]. -The proper solution must have both $x$ and $y$ positive. That set is -$$ \mathbf c=(\rho,\rho),\,\rho=(\tau-2/\tau-1)/3, \, \tau=(17+3\sqrt{33})^{1/3} \approx 0.543689.$$ -This solution set, with identical $c_1 = c_2$, falls under the purview of a simplified calculation, in which it can be shown -$$ f_{n,n} \sim \rho^{-2n} \frac{1.5009481}{\sqrt{n}}.$$ -(The true amplitude can be written in terms of $\rho$, and as many decimal places as wanted are possible, but I'm not going to bother to typeset it.) -For comparison: - -$n=40$, $\text{true}=3.4601\times 10^{20}$, $\text{asym} = 3.5261\times 10^{20}$ , $\text{absolute % err} = 1.91\%$. - -$n=200$, $\text{true}=7.6554\times 10^{104}$, $\text{asym} = 7.6847\times 10^{104}$ , $\text{absolute % err} = 0.38\%$. - - -For the non-diagonal case, you will be looking at $f_{an,bn}$. The machinery should work, though it is more complicated. You'll get roots that depend on $(a,b)$ and have to solve a complicated determinant to get the amplitude, also dependent on $(a,b)$. The question is, why does the non-diagonal case deviate rapidly from the diagonal? Is there a way to understand this qualitatively? I believe the answer lies in the modified form -$$ f_{an,bn} \sim c_1(a,b)^{-a n} c_2(a,b)^{-b n} \cdot \operatorname{amp}(a,b)/\sqrt{n}.$$ -The amplitude will vary only like a polynomial upon changing $(a,b)$, but the first two factors have an exponential dependence.<|endoftext|> -TITLE: Is there discrete Morse theory on acyclic categories? -QUESTION [7 upvotes]: Forman introduced discrete Morse theory on finite regular cell complexes. Minian introduced a version of discrete Morse theory for posets which generalizes Forman's original Morse theory https://arxiv.org/abs/1007.1930. So I thought the following problem: -Is there a version of discrete Morse theory for acyclic categories which generalizes Minian's Morse theory? -A desire theorem (It is imaginary and is not logical and currect.) is that there is a "function (functor)" from an acyclic category such that the classifying space of the given acyclic category is homotopy equivalent to a CW complex such that the number of cells = the number of "critical objects". -An acyclic category is a small category in which only the identity morphisms have inverses and any morphism from an object to itself is an identity morphism. A poset $P$ is an acyclic category. - -REPLY [5 votes]: Recently, it appeared in ArXiv (https://arxiv.org/abs/2107.06202) a paper which may be related to this question. The title is: - -Morse theory for loop-free categories - -Note that loop-free categories is the same as acyclic categories. The abstract is: - -We extend discrete Morse-Bott theory to the setting of loop-free (or -acyclic) categories. First of all, we state a homological version of -Quillen's Theorem A in this context and introduce the notion of -cellular categories. Second, we present a notion of vector field for -loop-free categories. Third, we prove a homological collapsing theorem -in the absence of critical objects in order to obtain the Morse -inequalities. Examples are provided through the exposition. This -answers partially a question by T. John: whether there is a Morse -theory for loop-free (or acyclic) categories?<|endoftext|> -TITLE: Why can't we embed Tarski's truth in PA? -QUESTION [9 upvotes]: I recently learned that ZFC can prove $Con(PA)$ because it can give a model of PA, but I'm not given the technical details. (My teacher thinks it is too obvious to even mention.) -What plagues me is that my naïve intuition tells me that the modeling procedures can be imitated in PA, exactly in the same way. -Here is my attempt: -Let $eval_F$ and $eval_T$ be evaluation functions for Formulas and Terms. Let $e$ denote any variable assignment. -We can define these functions recursively, exploiting Tarski's lemma. Explicitly, -\begin{align} -eval_T(\ulcorner v_i\urcorner,e) &= e[i] \\ -eval_T(\ulcorner o\urcorner,e) &= 0 \\ -eval_T(\ulcorner s\tau \urcorner,e) &=eval_T(\ulcorner\tau\urcorner,e)+1 \\ -eval_T(\ulcorner \tau_1 + \tau_2 \urcorner,e) &=eval_T(\ulcorner\tau_1\urcorner,e)+eval_T(\ulcorner\tau_2\urcorner,e) \\ -eval_T(\ulcorner \tau_1 \cdot \tau_2 \urcorner,e) &=eval_T(\ulcorner\tau_1\urcorner,e) \cdot eval_T(\ulcorner\tau_2\urcorner,e) -\end{align} -and -\begin{align} -eval_F(\ulcorner \bot \urcorner,e) &=0 \\ -eval_F(\ulcorner \tau_1 = \tau_2 \urcorner,e) &= \chi_=(eval_T(\ulcorner\tau_1\urcorner,e),eval_T(\ulcorner\tau_2\urcorner,e)) \\ -eval_F(\ulcorner \Phi\to\Psi \urcorner,e) &= \mathrm{sgn}((1-eval_F(\ulcorner \Phi \urcorner,e))+eval_F(\ulcorner \Psi \urcorner,e)) \\ -eval_F(\ulcorner \forall v_i.\Phi \urcorner,e) &=\begin{cases} - 1 & (\forall n.eval_F (\ulcorner\Phi\urcorner,e[i\mapsto n]) = 1) \\ - 0 & (\mathrm{otherwise})\\ -\end{cases} -\end{align} -Then $eval_T$ and $eval_F$ are $\Sigma_1^0$- and $\Sigma_2^0$-defined function respectively. Although $eval_F$ is not decidable, at least we know that $eval_F$ is total over coded PA-formulas and the value is either $0$ or $1$. -If we show that every axiom in PA evaluates to $1$ and the inference rules are truth-preserving, then we can show the soundness of the model: -$$ -\forall \phi:\mathrm{Form}. (Provable(\phi) \to \forall e. (eval_F(\phi,e)=1)) -$$ -If so, we can conclude $Con(PA)$, which is $\neg Provable(\ulcorner \bot \urcorner)$, because $\bot$ evaluates to $0$. -Of course, this violates Gödel's second incompleteness theorem, so I must be wrong somewhere - but I couldn't find where. -I am now suspecting three possibilities: - -We cannot in fact well-define $eval_T$ and $eval_F$ in PA. -We cannot prove that $eval_F$ models the axioms of PA. -The inference rules does not preserve truth generated by $eval_F$. - -I want to know where my argument fails. Thanks in advance. -P.S. -The most suspicious one for me is the second one, especially induction scheme. -Nonetheless I am convinced that induction scheme is provably evaluated to 1, since it reduces to -\begin{align} -\forall \phi:\mathrm{Form}.\forall e. &\forall i. \bigl( -eval_F(\phi,e[i\mapsto 0])=1 \to \\ -&\forall n. (eval_F(\phi,e[i\mapsto n])=1 \to eval_F(\phi,e[i\mapsto n+1])=1) \to \\ -&\forall n. (eval_F(\phi,e[i\mapsto n])=1 )\bigr) -\end{align} -which is an instance of induction scheme of outer PA. - -REPLY [18 votes]: First-order logic does not provide for definitions of functions by recursion. For example, the transitive closure of a binary relation $R$, though definable from $R$ by recursion, is not in general first-order definable from $R$. -Peano Arithmetic, though formulated in first-order logic, does have enough axioms to support some definitions by recursion. Specifically, thanks to the availability in PA of sequence coding, a recursive definition can be reformulated as an explicit definition provided each step in the recursion depends on only finitely many previous results. For example, the natural definition of the factorial function ($0!=1$ and $(n+1)!=n!\cdot (n+1)$) can be rewritten as "$x!=y$ iff there is a sequence $a_0,a_1,\dots,a_x$ of length $x+1$ with $a_0=1$, $a_{n+1}=a_n\cdot(n+1)$ for all $n -TITLE: Rational homotopy invariance of algebraic $K$-theory -QUESTION [12 upvotes]: Suppose that $R\to S$ is a 1-connected morphism of connective structured ring spectra that induces an isomorphism on rational homotopy groups. Is the induced map of (Waldhausen) K-theory spectra -$$ -K(R) \to K(S) -$$ -also an equivalence on rational homotopy? -(The case of the map of group rings $S^0[G] \to \Bbb Z[G]$ was answered in the affirmative in one of Waldhausen's early papers.) -I am looking for a solid reference (assuming the result to be true; I believe it is). - -REPLY [12 votes]: The theorem can be found in more general form in Land, Tamme On the K-theory of pullbacks, Lemma 2.4.<|endoftext|> -TITLE: Directed graph minor theorems -QUESTION [5 upvotes]: In proving the graph minor theorem, Robertson and Seymour proved a stronger statement, namely that the directed graph minor theorem is true, using the definition - -A directed graph is a minor of another if the first can be obtained from a subgraph of the second by contracting edges. - -Since then, many more notions of a “directed graph minor” have arisen in various context (see here and here for examples and references). One of the goals in defining these notions is that the definition Robertson and Seymour gave doesn’t obviously capture the notion that if $G$ is a minor of $H$, then $G$ should be “simpler” than $H$. -I’m looking for references to proofs of the graph minor theorem for these more restrictive definitions of “graph minor.” I’m particularly interested in the following definition (Johnson et al., 2001): - -A graph $G’$ is a butterfly minor of a directed graph $G$ if $G’$ can be obtained from $G$ by a sequence of the following local operations: - -Deleting an edge (a, b); -Contracting an edge (a, b) where b has indegree 1; -Contracting an edge (a, b) where a has outdegree 1. - - -But other definitions would also interest me, especially if they’re amenable to analyzing computational graphs of circuits or ML devices (neural networks, Bayesian networks) - -REPLY [3 votes]: So directed graphs are not well-quasi-ordered by butterfly minors; see the intro of [BPP]. Furthermore, there are reasons to think that many of the FPT results for graph minors may not hold in the directed setting (ie [PW]). -Yet, perhaps surprisingly, it may still be possible to get a structure theorem for butterfly minors! There is an ongoing project to do this; I believe the most recent paper is "The Directed Flat Wall Theorem" by Giannopoulou, Kawarabayashi, Kreutzer, and Kwon. If you Google the authors you should be able to find more information about the project. This is a big task and the results so far are exciting. -Also notable, there is a relationship between "matching-minors" of (undirected) bipartite graphs and butterfly minors; see [HRW]. Furthermore, Johnson's thesis "Eulerian digraph immersion" is on a different kind of directed minor for 4-regular, directed, Eulerian graphs (same Johnson as mentioned in the question). -I'm not really sure why any of these definitions would be useful for studying graphs of circuits or graphs in machine learning, but I sure hope they are! It's nice to see interest from this direction, and I hope something in the post is helpful.<|endoftext|> -TITLE: Minimum cardinality of a cofinal collection of countable subsets of a set -QUESTION [5 upvotes]: Setup - -Let $X$ be a set of cardinality $\kappa\geq \aleph_0$. -Edit: -Based on Todd Eisworth's suggestion: -What is the minimum cardinality of a collection $\hat{X}$ of countable subsets of $X$ such that every countable subset $A \subseteq X$ is contained in some element of $\hat{X}$? - -REPLY [4 votes]: As it is stated in the comments, the question is about the cofinality of $([\lambda]^{\aleph_0}, \subseteq)$. -The following definition is due to Shelah: -$ -cov(\lambda, \mu, \theta, \sigma)=min\{|P|: P$ is a family of subsets of $\lambda$ each of size $< \mu$ such that for every $a \subseteq \lambda, |a|<\theta$, for some $\alpha < \sigma$ and $A_i \in P,$ for $i<\alpha,$ we have $a \subseteq \bigcup_{i<\alpha}A_i \}$. -It is easily sean that -1-$cf([\lambda]^{\aleph_0}, \subseteq)=cov(\lambda, \aleph_1, \aleph_1, 2)$. -By a theorem of Shelah (see [Sh: 355], $\aleph _{\omega +1}$ has a Jonsson Algebra, see also Analytic guides and updates for cardinal arithmetic page 23) -2-$\lambda^{\aleph_0}=cov(\lambda, \aleph_1, \aleph_1, 2)+ 2^{\aleph_0}$. -Note that in particular, if $\lambda^{\aleph_0} > 2^{\aleph_0}$, then -$cf([\lambda]^{\aleph_0}, \subseteq)=\lambda^{\aleph_0}$.<|endoftext|> -TITLE: Good overviews on $\phi^{4}$-field theory? -QUESTION [14 upvotes]: I'm looking for nice overviews on $\phi^{4}$-field theory from the mathematical-physics point of view. To be a little more specific, here are some topics I'd like to read about: -(1) What are the motivations, both from the physics and mathematical point of view, to study $\phi^{4}$-theories? -(2) What has been (mathematically) accomplished so far? What are the most important open questions nowadays? -(3) What are the tools used to (rigorously) study these class of models? Renormalization group, cluster expansions, etc. -My motivation for this question is a rather simple one: I'm teaching myself some field theory but it is really hard to find these discussions on books. In general, books are more interested in solving problems and sometimes I find myself studying some models that I don't know anything about. Also, I'm primarily interested in statistical mechanics, so this helps to narrow the question a bit more. Connections with statistical mechanics and QFT are welcome too, but I don't want a purely QFT reference (I'm sufficiently lost in my own area of interest, after all). Thanks in advance! - -REPLY [2 votes]: In terms of physical motivation of $\phi^4$, if I remember rightly the references in the introduction to this paper might have some pointers. - -Dashen R and Neuberger H 1983 How to get an Upper Bound on the Higgs Mass, Physical Review Letters 50, 24 - -Reference [7] of that article points to rigorous proofs of results which almost show the triviality of $\phi^4$ in four dimensions.<|endoftext|> -TITLE: Trace inequality under consideration of definiteness -QUESTION [5 upvotes]: Let $G \in \mathbb{R}^{3 \times 3}$ a symmetric, but indefinite matrix and $U \in \mathbb{R}^{3\times 3}$ a symmetric and positive definite matrix. I would like to prove the inequality -$$ \text{Tr} \left( G^2 \right) \leq \text{Tr} \left( GUGU^{-1} \right). $$ -If $U$ and $G$ commute, both sides of the inequality are obviously equal. However for more general cases, -I have tried to rearrange the inequality to -$$ \text{Tr}(\underbrace{[UG-GU]}_{\text{skew-symmetric}}\ GU^{-1}) \leq 0 $$ -and then using the Cauchy-Schwarz inequality. Unfortunately, I have not found a solution yet. - -REPLY [7 votes]: Write -$$G=\left( -\begin{array}{ccc} - a & b & c \\ - b & d & e \\ - c & e & f \\ -\end{array} -\right) -$$ -and, without loss of generality, -$$ -U=\left( -\begin{array}{ccc} - 1 & 0 & 0 \\ - 0 & u & 0 \\ - 0 & 0 & v \\ -\end{array} -\right),$$ -where $u,v>0$. Then -$$\text{Tr}(GUGU^{-1})-\text{Tr}(G^2) -=\frac{b^2 (u-1)^2 v+c^2 u (v-1)^2+e^2 (u-v)^2}{u v},$$ -which is manifestly $\ge0$, as desired.<|endoftext|> -TITLE: On the sum of the subgroup orders of a finite group -QUESTION [13 upvotes]: Let $G$ be a finite group. Consider the function providing the sum of the subgroups orders -$$\sigma(G) = \sum_{H \le G} |H|.$$ -Note that if $C_n$ is cyclic of order $n$ then $\sigma(C_n) = \sigma(n)$, with $\sigma$ the usual divisor function. Consider the functions $$\sigma_{-}(n)= \min_{|G|=n} \sigma(G), \ \ \ \ \ \ \sigma_{+}(n)= \max_{|G|=n} \sigma(G). $$ -This post is about a characterization of the extremizers, i.e. the finite groups $G$ such that $\sigma(G) = \sigma_{\pm}(|G|)$. -$$\begin{array}{c|c} n&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15 \newline \hline -\sigma_{-}(n)&1&3&4&7&6&12&8&15&13&18&12&28&14&24&24\newline \hline -\sigma(n)&1&3&4&7&6&12&8&15&13&18&12&28&14&24&24 \newline \hline -\sigma_{+}(n)&1&3&4&11&6&16&8&51&22&26&12&60&14&36&24 \end{array}$$ -We can observe in above table that $\sigma_{-}(n) = \sigma(n)$, and it holds for all $n < 256=2^8$ (by checking on GAP). -Question 1: What are the finite groups $G$ such that $\sigma(G) = \sigma_{-}(|G|)$? Exactly the cyclic groups? - -Next, consider the prime factorization of $n$ $$n=\prod_{i=1}^r p_i^{n_i},$$ then, candidates which come in mind for $\sigma(G) = \sigma_{+}(|G|)$ are the product of prime order cyclic groups: $$G = \prod_{i=1}^r C_{p_i}^{n_i}.$$ -It works often but not always, as $\sigma(S_3) = \sigma_{+}(|S_3|) = \sigma_{+}(6) = 16$ whereas $\sigma(C_2 \times C_3) = 12$; but $S_3 = C_3⋊C_2$, moreover, for $n \le 60$, all the models I found are semi-direct product of prime order cyclic groups. -Question 2: What are the finite groups $G$ such that $\sigma(G) = \sigma_{+}(|G|)$? -Are there semi-direct product of prime order cyclic groups? Or at least supersolvable? - -REPLY [4 votes]: Partial answer: - -The answer to Q1 is "yes" if one restricts $G$ to supersolvable groups or more generally groups that satisfy the converse of Lagrange's theorem. - -If $G$ has the property that for every divisor of the group order there is at least one subgroup of that order, then in particular $\sigma(G)\geq\sigma(|G|)$. Furthermore, if $\sigma(G)=\sigma(|G|)=\sigma(\mathbb{Z}/|G|)$ holds, then there is exactly one subgroup of every order. In particular, all sylow groups are normal so that the group is the direct product of its sylows and inside every sylow subgroup, there is exactly one maximal subgroup so that it is cyclic. - -Another observation: $\sigma$ is "multiplicative" for groups similar to how $\sigma$ is multiplicative for groups, i.e. -$gcd(|G|,|H|)=1 \implies \sigma(G\times H)=\sigma(G)\cdot\sigma(H)$ -This follows from the fact that in these situations, every subgroup of $G\times H$ is uniquely decomposable as $G_0\times H_0$ with $G_0\leq G$ and $H_0\leq H$. -More generally, one can observe that $\sigma(G\rtimes H) = \sum_{G_0\leq G} |G_0| \cdot\sigma(N_H(G_0))$ holds for $gcd(|G|,|H|)=1$, because in this case every subgroup uniquely decomposes as $G_0\rtimes H_0$ by the Schur-Zassenhaus theorem. The right hand side is less than or equal to $\sigma(G)\cdot\sigma(H)$ with equality iff all subgroups of $G$ are $H$-invariant.<|endoftext|> -TITLE: Pell equation and quadratic residues -QUESTION [7 upvotes]: We say an integer $k$ is Pell if there exist some integers $p,q$ such that -$$ -p^2k-q^2=1 -$$ -In studying a physics system we ended up with two weaker notions of Pell: - -We say an integer $k$ is pre-Pell if there exist some integers $p,q$ such that -$$ -pk-q^2=1 -$$ - -We say an integer $k$ is weakly Pell if there exists some Pell integer $k'$ such that the product $kk'$ is also Pell. - - -The physics of the problem led us to conjecture that these two notions are in fact equivalent: -Conjecture: $k$ is pre-Pell if and only if it is weakly Pell. -One of the directions is obvious. We have very superficial knowledge of number theory so the only proof we could come up with was using some very strong conjectures (namely, one of the Hardy-Littlewood conjectures), but I think it is safe to assume that a much simpler, unconditional proof should exist. I would appreciate it if someone could sketch it here, or point me in the right direction. - -REPLY [5 votes]: The answer is yes. As was observed, the condition of being pre-Pell is simply the stipulation that $k$ is a sum of two squares: that is, if $p | k$ and $p \equiv 3 \pmod{4}$ then $p$ must divide $k$ with even multiplicity. If we assume $k$ is square-free, then it is divisible only by $2$ or primes of congruent to $1 \pmod{4}$. -In fact, if $k$ is pre-Pell then all of its odd prime divisors must be congruent to $1 \pmod{4}$. To see this, suppose $x^2 + 1$ is divisible by $p^2$ for some $p \equiv 3 \pmod{4}$. Then in particular it must be divisible by $p$, so the congruence $x^2 + 1 \equiv 0 \pmod{p}$ must be soluble. But this is not the case: $-1$ is a square mod $p$ if and only if $p = 2$ or $p \equiv 1 \pmod{4}$. -We now work on the equivalences. Suppose that $k$ is such that there exists $k^\prime$ such that the negative Pell equation is soluble for both $k^\prime$ and $kk^\prime$. Being a sum of two squares is a necessary condition for negative Pell to be soluble, hence the latter condition implies that $k = (kk^\prime)/k^\prime$ must also be a sum of two squares. To see this, suppose that $kk^\prime$ has a prime divisor $p$ congruent to $3 \pmod{4}$ (otherwise it is obvious that $k$ is a sum of two squares). Suppose that $p^{2m} || kk^\prime$ and $p^{2n} || k^\prime$. Then $p^{2m-2n} || k$, hence $p$ divides $k$ to even multiplicity. Therefore, $k$ is pre-Pell as desired. -The converse is much harder. Suppose that $k$ is a sum of two squares. We can use the fact that for any prime $p \equiv 1 \pmod{4}$, the negative Pell equation $x^2 - py^2 = -1$ is soluble. Thus we may reduce the question to the following: does there exist a prime $p \equiv 1 \pmod{4}$ such that $x^2 - kpy^2 = -1$ is soluble? -Fortunately this can be done via governing fields: that is, for each $k$ there exists a number field $F_k$ such that the 4-rank of the class group of the field $\mathbb{Q}(\sqrt{kp})$ is determined by the splitting behaviour of $p$ in $F_k$. In particular, if the 4-rank of the class group of $\mathbb{Q}(\sqrt{kp})$ is zero, then negative Pell is soluble. We are then done by Chebotarev's density theorem guaranteeing the existence of such a prime (in fact, infinitely many).<|endoftext|> -TITLE: Possible limit involving the gamma function -QUESTION [12 upvotes]: Does $$\lim_{n \to \infty} \int_{0}^{1} \Gamma(x)^{n/(n+1)}dx - n$$ exist? -Here's some background. The integral -$$\int_{0}^{1} \Gamma(x) dx$$ -diverges rather slowly. Inserting the exponent $n/(n+1)$ perhaps leads to a nice surprise---that the floor of resulting integral appears to be $n$. For example, for $n = 100$, the integral has a value of $100.759456...$ - -REPLY [14 votes]: $\newcommand\Ga\Gamma$ -Note that $\Ga(x)=\Ga(1+x)/x$ for $x>0$ and -$-n=1-\int_0^1 x^{-n/(n+1)}\,dx$ for $n>0$. -So, the limit in question is -$$1+\lim_n J_n,$$ -where -$$J_n:=\int_0^1 x^{1/(n+1)}f_n(x)\,dx,$$ -$$f_n(x):=g(x)-h_n(x),$$ -$$g(x):=\frac{\Ga(1+x)-1}x,\quad h_n(x):=\Ga(1+x)\frac{\Ga(1+x)^{-1/(n+1)}-1}x.$$ -Letting $c$ stand for any expressions bounded uniformly over all $x\in(0,1)$ and all $n\ge1$, we have $\Ga(1+x)=1+cx$ and $\Ga(1+x)^{-1/(n+1)}=1+cx/n$, so that $h_n(x)=c/n$ and hence $\int_0^1 x^{1/(n+1)}h_n(x)\,dx\to0$. Thus, the limit in question is -$$1+\int_0^1 g(x)\,dx=0.75330\ldots. $$ - -As seen from the proof, the rate of convergence here is $O(1/n)$. So, the limit value $0.75330\ldots$ is in agreement with the value of the integral you computed for $n=100$.<|endoftext|> -TITLE: Which is the more popular approach to forcing in the literature? -QUESTION [11 upvotes]: There are some interesting questions and answers on the site discussing the different approaches to forcing in set theory, and I understand that the two most important are the ones using countable transitive models (ctm) and Boolean valued ones (bvm), respectively. -My question is primarily about which of those two approaches appears in the literature more often, especially in research articles. I know that it might be difficult to answer, but perhaps an educated guess by the users of the site will be enough for me. -The context of this question is that my team is working on a formal verification of forcing using the ctm approach, and it is important for us to be able (to the extent possible) to represent the actual practice of the subject. It is to be noted that a full formalization of the bvm approach was recently completed by Han and van Doorn. - -REPLY [11 votes]: There are two types of "working with forcing": - -We can develop the theory of forcing, e.g. iterations, where working with canonical forcing notions is somewhat preferable, so dealing with complete Boolean algebras is somehow the most natural approach, and by extension with Boolean-valued models (well, sometimes). -For example, talking about homogeneity conditions is easy when you have them. But maybe you have a rigid partial order which is forcing equivalent to adding a Cohen real (e.g. construct a tree where each node has a unique number of successors). Or perhaps you want to iterate forcings, but the standard definition of iteration given in terms of partial orders is a pre-ordered set. Being able to forego all of that and just find an invariant is great. - -We can use the theory of forcing, e.g. proving various consistency results. In this case it is almost exclusively done with partial orders, and indeed with pre-orders, where we simply ignore all of the forcing theoretic issues that make the "formally correct statements" just a huge pain in the lower-lower-back to state. -I think this is best reflected in Jech's "Set Theory" book (3rd edition, for those keeping track). The basic theory of forcing is developed with Boolean algebras and Boolean-valued models. When forcing is actually used, Jech quickly reverts back to partial orders and pre-orders, instead. - - -Now you can also talk about forcing with topological spaces, forcing with sheaves (or shivs), etc. This is not very common in set theoretic papers in the last few decades. I won't comment on other subjects, as I'm not an expert. -Finally, a word about the foundations of forcing. When one learns about forcing, it is often confusing. The generic object is seemingly black magic, and what's going on with those Cohen reals encoded in limit steps? And what is this "arbitrarily large, but finite fragment of $\sf ZFC$" that Kunen keeps talking about? -Well, the reality is that we can develop forcing in a lot of different ways: - -Just force over countable transitive models of $\sf ZFC$. That's the simplest, most straightforward way to do it. But this requires us to assume more in terms of consistency. - -Just force over countable models of $\sf ZFC$. Oh, but then it gets ugly when talking about things like ordinals and whatnot, because these models are not necessarily well-founded. Also, this requires more consistency, although significantly less than before. - -Use reflection to argue that we can find countable transitive models of any large enough fragment of $\sf ZFC$, force over those, and use a meta-theoretic argument to conclude the proof. - -Use Boolean-valued models to develop forcing as a proper class and argue with Boolean-valued truth that certain statement are consistent. But that's kind of yuck in most cases. - -Instead of Boolean-valued models, define an "internal ultrapower" of the universe by extending the filter base that is the dense open sets to a "generic" filter, and use this model, where the forcing theorem and truth lemma still hold, to finish your argument. In some sense this is a neater version of Boolean-valued models, but in another sense it is quite the opposite. - -Use Feferman's theory, where we add a constant symbol, postulate that it is a countable transitive elementary submodel of the universe, then force over that model. No additional consistency is needed, as Feferman's theory is finitely consistent (assuming $\sf ZFC$ is), so it is given to us. But it sort of makes this specified model somehow... on a pedestal. Also without further assumptions (which are tantamount to (1) with more power) the models of Feferman's theory are ill-founded, which is yuck to think about from a meta-theoretic point of view. - -Use other tricks and machinery to encode forcing and just work syntactically in a theory as weak as $\sf PRA$. (Shudder here.) - - -But what do people actually end up doing (once they grok forcing)? Well. We force over the universe. We just ignore all of it and force over the universe. Because at the end of the day, the goal is to use forcing, and as all of these approaches lead us to the same way, and we anyway define everything and work internally to whatever set-sized model we may have used, we might as well force over the universe. Simply rub your hands, and a generic appears! Magic!<|endoftext|> -TITLE: Characteristic polynomial of the Gcd matrix -QUESTION [19 upvotes]: Let $A_n$ be the $n \times n$-matrix with entries $Gcd(i,j)$ and $f_n$ the characteristic polynomial of $A_n$. - -Question: Is $f_n$ irreducible over $\mathbb{Q}$ for all $n$ except $n=8$? - -This is true for $n \leq 60$. - -REPLY [10 votes]: Edit. See the calculation modulo $2$ below. -This is not an answer, but it is too long for a comment. The matrix $A_n$ can be decomposed as $A_n=D_n^T\Phi_nD_n$ where the entries of $D_n$ (D for divisors, not for diagonal) are either $1$ if $i|j$, or $0$ otherwise, and $\Phi_n={\rm diag}(\phi(1),\ldots,\phi(n))$, $\phi$ being the Euler's totient function. -Notice that the matrix $D_n$ is the $n\times n$ upper-left block of an infinite upper triangular matrix $D$. Its diagonal is made of $1$s, so that $\det D_n=1$. Thus -$$f_n(X)=\det(XD_n^{-T}D_n^{-1}-\Phi_n).$$ -The matrix $D_n^{-T}D_n^{-1}$ is the $n\times n$ upper-left block of $D^{-T}D^{-1}$. The entries of $D^{-1}$ are known to be either $\mu(j/i)$ if $i|j$, or $0$ otherwise. Therefore -$$(D_n^{-T}D_n^{-1})_{ij}=\sum_{k|i,j}\mu\left(\frac ik\right)\mu\left(\frac jk\right).$$ -For instance this is zero when $p^2|j$ but $p$ does not divide $i$. Also the diagonal entry corresponding to $j=i$ equals $2^r$, where $r$ is the number of distinct prime factors of $i$. A closed formula is that the $(i,j)$-entry equals -$$\mu\left(\frac{{\rm lcm}(i,j)}{{\rm gcd}(i,j)}\right)2^{P(i,j)},$$ -where $P(i,j)$ is the number of distinct prime factors of ${\rm gcd}(i,j)$ for which the valuations agree: -$$v_p(i)=v_p(j)\ge1.$$ -Calculation modulo 2. Since $\phi(j)$ is even for every $j\ge3$, the matrix $\Phi_n$ reduces to ${\rm diag}(1,1,0,\ldots,0)$. Thus $f_n(X)=X^n-aX^{n-1}+bX^{n-2}$ where -$$a=(D_n^{-T}D_n^{-1})_{11}+(D_n^{-T}D_n^{-1})_{22}$$ -and $b$ is the minor of $D_n^{-T}D_n^{-1}$ when deleting the two first columns and rows. Because $\det(D_n^{-T}D_n^{-1})=1$, this gives -$$a=(D_nD_n^T)_{11}+(D_nD_n^T)_{22},\qquad b=(D_nD_n^T)\binom{12}{12},$$ -the latter being a $2\times2$ minor. These quantities involve only the two first rows $(1,\ldots,1)$ and $(0101\ldots)$ of $D_n$. We obtain -$$a=n^2+\lfloor\frac n2\rfloor^2=n+\lfloor\frac n2\rfloor,\qquad b=n\lfloor\frac n2\rfloor-\lfloor\frac n2\rfloor^2=(n+1)\lfloor\frac n2\rfloor.$$ -Hence (mod $2$) -$$f_n(X)=\left\{\begin{array}{lcr} X^n, & & n=0,3, \\ -X^{n-1}(X+1), & & n=1 \\ X^{n-2}(X^2+X+1), & & n=2. \end{array}\right.$$<|endoftext|> -TITLE: A variant on Wieferich primes -QUESTION [7 upvotes]: Recall that a Wieferich prime is a prime number $p$ such that -$2^{p-1} \equiv 1 \bmod p^2.$ -It is not known whether there are infinitely many Wieferich primes, nor whether there are infinitely many non-Wieferich primes. In fact there are only $2$ known Wieferich primes. -I'm interested in a slightly different condition which I'm hoping is easier to handle. Namely, I can replace the exponent $p-1$ by the order of $2$ in $(\mathbb{Z}/p\mathbb{Z})^\times$. Moreover, I just want this power to hit the identity with odd $p$-adic valuation. Specifically: - -Are there infinitely many primes $p$ such that $v_p(2^{\mathrm{ord}_p(2)}-1)$ is odd? - -Here $\mathrm{ord}_p(n)$ denotes the order of $n$ in $(\mathbb{Z}/p\mathbb{Z})^\times$ (which divides $p-1$ by FLT), and $v_p$ is the $p$-adic valuation. -Note that the existence of infinitely many non-Wieferich primes would provide a positive answer to my question (since here $v_p(2^{p-1}-1) = 1$). -Ideally I'd also like to know that there collection of such primes has positive density, rather than just being infinite. - -REPLY [11 votes]: Felipe refers to my first ever paper (mathscinet.ams.org/mathscinet-getitem?mr=789713) from 1985 ! However I have a more recent paper that gives a better result along the lines asked for (mathscinet.ams.org/mathscinet-getitem?mr=2997580) which shows that every $2^n-1$, with $n\ne1$ or $6$, has a primitive prime factor that divides it to an odd power.<|endoftext|> -TITLE: Abundancy index and non-solvable finite groups -QUESTION [8 upvotes]: Let $\sigma$ be the sum-of-divisors function. A number $n$ is called abundant if $\sigma(n)>2n$. Note that the natural density of the abundant numbers is about $25 \%$. The abundancy index of $n$ is $\sigma(n)/n$. The following picture displays the abundancy index for the $10000$ first orders of non-solvable groups (see A056866). - -Observe that for $G$ non-solvable with $|G| \le 446040$ then $|G|$ is abundant, with minimal abundancy index $\frac{910}{333} \simeq 2.73$. -Question 1: Are the non-solvable groups of abundant order? -Note that the number of integers $n \le 446040$ with $\sigma(n)/n \ge 910/333$ is exactly $19591$, so of density less than $5 \%$ with more than half of them being the order of a non-solvable group. Among those which are not the order of a non-solvable group, the maximal abundancy index is $512/143 \simeq 3.58$, realized by $n=270270$, whereas there are exactly $896$ numbers $n \le 446040$ with $\sigma(n)/n > 512/143$, which then are all the order of a non-solvable group. -Question 2: Is a number of abundancy index greater than $512/143$ the order of a non-solvable group? -Weaker version 1: Is there $\alpha >3$ such that a number of abundancy index greater than $\alpha$ must be the order of a non-solvable group? -Weaker version 2: Is there $\beta < 1$ such that a number $n$ of abundancy index greater than $\beta e^{\gamma} \log \log n$ must be the order of a non-solvable group? -Recall that $\limsup \frac{\sigma(n)}{n \log \log n} = e^{\gamma}$ with $\gamma$ the Euler-Mascheroni constant. -Finally, there are non-solvable finite groups $G$ with $|G| \gg 446040$ and abundancy index less than $\frac{910}{333}$. The non-abelian simple groups $G$ with $|G|=n \le 749186071932$ and $\sigma(|H|)/|H|>\sigma(n)/n$ for all non-abelian simple groups $H$ of order less than $n$ are exactly -the 39 the simple groups $\mathrm{PSL}(2,p)$ with $p$ prime in {5, 37, 107, 157, 173, 277, 283, 317, 563, 653, 787, 907, 1237, 1283, 1307, 1523, 1867, 2083, 2693, 2803, 3413, 3643, 3677, 4253, 4363, 4723, 5443, 5717, 6197, 6547, 6653, 8563, 8573, 9067, 9187, 9403, 9643, 10733, 11443}. Let $n_p:=|\mathrm{PSL}(2,p)| = p(p^2-1)/2$. -It follows that for $G$ non-abelian simple with $|G| \le 749186071932$ then $|G|$ is abundant, with minimal abundancy index $$\sigma(n_{11443})/n_{11443} = 50966496/21821801 \simeq 2.33.$$ -The following picture displays the adundancy index of $n_p$ for $p$ prime and $5 \le p \le 10^6$. - -The minimal is $579859520520/248508481289 \simeq 2.3333 \simeq 7/3$, given by $p=997013$. -Question 3: Is it true that $\inf_{p \ge 5, \text{ prime}} \sigma(n_p)/n_p = 7/3$? -Question 4: Is the abundancy index of the order of a non-solvable group greater than $7/3$? - -Fun fact: the smallest integer $n$ such that there exists two non-isomorphic simple groups of order $n$ is $20160$, whereas the biggest integer that is not the sum of two abundant numbers is $20161$ (see A048242). Any explanation? - -REPLY [4 votes]: As I mentioned in a comment, Question 2 (in its revised form) has a negative answer, because odd natural numbers have unbounded abundancy index, while the Odd Order Theorem implies all groups of odd order are solvable. -Weaker version 2 has a positive answer: If $\beta$ is sufficiently close to 1, then any $n > 1$ whose abundancy index is greater than $\beta e^\gamma \log \log n$ is a multiple of 60, so there is a group of order $n$ that is unsolvable. -As I mentioned in a different comment, Question 3 is true subject to well-known open conjectures, such as [Dickson's conjecture][1]. In particular, it suffices to show that there are infinitely many primes $p$ such that $p-1$ is 4 times a prime and $p+1$ is 6 times a prime. -[1]: https://en.wikipedia.org/wiki/Dickson%27s_conjecture<|endoftext|> -TITLE: A set of questions on continuous Gaussian Free Fields (GFF) -QUESTION [5 upvotes]: As I said in my previous posts, I'm trying to teach myself some rigorous statistical mechanics/statistical field theory and I'm primarily interested in $\varphi^{4}$, but I know that the absense of this term provides important simplifications to the theory and we can give meaning to the theory when this term is not included using functional integrals and Gaussian measures on functional spaces. My intention with this post is to understand the problems involved in the continuous limit of this theory. I know one usually discretizes the theory to define the objects of interest, but I'm trying to understand the origin of these problems starting from the continuous limit. It's very difficult to find such a complete analysis in books or articles, and I usually find myself having to built the whole picture from little pieces of it, so my intention here is to fill the gaps led by this process. -In what follows, I ask 5 questions and try to answer some of them, but I don't know if my answers and my reasoning are correct. I'd appreciate if you could correct me if needed and add more information, if necessary. -First of all, the idea is to give precise meaning to the probability measure: -\begin{eqnarray} -\frac{1}{Z}\exp\bigg{(}-\int_{\mathbb{R}^{d}}\frac{1}{2}\varphi(x)(-\Delta+m^{2})\varphi(x)dx \bigg{)}\mathcal{D}\varphi \tag{1}\label{1} -\end{eqnarray} -Where $\mathcal{D}\varphi$ is a "Lebesgue measure" in the space of fields. Here, the space of fields will be simply $\mathcal{S}'(\mathbb{R}^{d})$. In what follows, $\mathcal{S}'(\mathbb{R}^{d})$ is equipped with the strong topology and its associate Borel $\sigma$-algebra, i.e. the $\sigma$-algebra generated by its open sets. -Question 1: As I said before, I know that it is usual to discretize the theory and define (\ref{1}) by means of thermodynamic + continuous limits. But is it possible to address the problem directly on $\mathbb{R}^{d}$? -My attempted answer: I think that, once you discretized the theory and saw what are the correct limits and objects you need, you can pose the problem directly on $\mathbb{R}^{d}$ at the end of day, but it is by no means obvious, at a first sight, how to properly define (\ref{1}) or even other objects related to it, such as correlations etc. -In what follows, I'll address the problem directly on $\mathbb{R}^{d}$ assuming my answer to the first question is correct and I'm allowed to do it. -Question 2: Is (\ref{1}) a well-defined measure on its own, for all values of $m \ge 0$? How does the ultraviolet divergencies influence the existence of this measure? Does it play any role on its well-definiteness or just on correlation functions? -My attempted answer: I don't think this is well-defined on its own, because I don't think that the "product Lebesgue measure" $\mathcal{D}\varphi$ is well-defined in $\mathcal{S}'(\mathbb{R}^{d})$. However, I know that we can give meaning to (\ref{1}) if we use Minlos-Bochner theorem. -If my answer to question 2 is correct, I must use Minlos-Bochner. Then, (\ref{1}) is the measure $\mu_{G}(\varphi)$ on $\mathcal{S}'(\mathbb{R}^{d})$ induced by $W(f,f):=e^{C(f,f)}$ (using Minlos-Bochner) where: -\begin{eqnarray} -C(f,g):= \frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\frac{\overline{\hat{f}(\xi)}\hat{g}(\xi)}{|\xi|^{2}+m^{2}}d^{d}\xi \tag{2}\label{2} -\end{eqnarray} -Question 3: Intuitivelly, I know that (\ref{2}) is related to (\ref{1}). This is because $\hat{C}(\xi) = 1/(|\xi|^{2}+m^{2})$ is the Fourier transform of the Green's function $G(x)$ of the massive Laplacian $-\Delta+m^{2}$. Informally: Green's functions are inverse operators and, therefore, the measure induced by Minlos-Bochner theorem is a functional analogue of the usual property that Fourier transform of Gaussians are Gaussians. But, apart from the intuition, how can we related (\ref{1}) to $d\mu_{G}$? In other words, does (\ref{1}) have anything to do with the covariance of $d\mu_{G}$? -My attempted answer: I think the only way to realize $d\mu_{G}$ is the corrected Gaussian measure associated to (\ref{1}) (which was not defined as a Gaussian measure in the first place) is by discretizing the space and recovering the theory with thermodynamic + continuous limits. But starting from Minlos-Bochner's theorem, with covariance (\ref{2}), it doesn't seem obvious to me (apart from intuition) that $d\mu_{G}$ has anything to do with (\ref{1}). -Question 4: As I mentioned before, $d\mu_{G}$ is a Gaussian measure on $\mathcal{S}'(\mathbb{R}^{d})$ while (\ref{1}) seems to be just induced by a bilinear form on $\mathcal{S}(\mathbb{R}^{d})\subset \mathcal{S}'(\mathbb{R}^{d})$. Is (\ref{1}) well-defined only as a subset $\mathcal{S}(\mathbb{R}^{d})$ of $\mathcal{S}'(\mathbb{R}^{d})$? Or is it actually a quadratic form on $\mathcal{S}'(\mathbb{R}^{d})$ (in which case I don't seem to understand it correctly)? -Question 5: If I can, in fact, work the theory directly in the infinite/continuous setup, and all the Gaussian measures are properly defined, is it possible to calculate correlations, say, by using properties of Gaussian measures? -Note: I said, right from the beginning, that the space of fields is $\mathcal{S}'(\mathbb{R}^{d})$ but I know it because I already studied some models before and I knew what was the properly funcional space to consider. However, I believe (not sure) that physicists interpret fields as proper functions e.g. on $\mathcal{S}(\mathbb{R}^{d})$ and (\ref{1}) would be something like a quadratic form $\langle \varphi, (-\Delta+m^{2})\varphi\rangle$ on $\mathcal{S}(\mathbb{R}^{d})$. Then, because of Minlos-Bochner theorem, one notices that $\varphi$ must actually be considered as an element of a larger space $\mathcal{S}'(\mathbb{R}^{d})$ in which (\ref{1}) have no meaning unless $\varphi \in \mathcal{S}(\mathbb{R}^{d})$. This is what I think, but I don't know if I'm completely wrong and fields have physical reasons to be tempered distributions right from the beginning. - -REPLY [5 votes]: Essentially, what is asked is the continuation of my previous MO answer -Reformulation - Construction of thermodynamic limit for GFF -and the solution of the exercise I mentioned at the end of that answer. -There, I explained the construction of Gaussian Borel measures $\mu_m$ on the space $s'(\mathbb{Z}^d)$ of temperate multisequences indexed by the unit lattice in $d$ dimensions. -The measure $\mu_m$ is specified by its characteristic function -$$ -p\longmapsto\exp\left(-\frac{1}{2}\sum_{x,y\in\mathbb{Z}^d}p(x)G_m(x,y)p(y)\right) -$$ -for $p=(p(x))_{x\in\mathbb{Z}^d}$ in $s(\mathbb{Z}^d)$, the space of multisequences with fast decay. -The discrete Green's function $G_m(x,y)$ is defined on $\mathbb{Z}^d\times\mathbb{Z}^d$ -by -$$ -G_m(x,y)=\frac{1}{(2\pi)^d}\int_{[0,2\pi]^d}d^d\xi\ -\frac{e^{i\xi\cdot(x-y)}}{m^2+2\sum_{j=1}^{d}(1-\cos \xi_j)}\ . -$$ -Here we will assume $m\ge 0$ for $d\ge 3$, and $m>0$ if $d$ is $1$ or $2$. -For any integer $N\ge 1$, define the discrete sampling map $\theta_N:\mathscr{S}(\mathbb{R}^d)\rightarrow s(\mathbb{Z}^d)$ which sends a Schwartz function $f$ to the multisequence -$$ -\left(f\left(\frac{x}{N}\right)\right)_{x\in\mathbb{Z}^d}\ . -$$ -This map is well defined and linear continuous. -Indeed, -$$ -\langle Nx\rangle^2=1+\sum_{j=1}^{d} (Nx_j)^2\le N^2\langle x\rangle^2 -$$ -because $N\ge 1$. -So -$$ -||\theta_N(f)||_k:= -\sup_{x\in\mathbb{Z}^d} -\langle x\rangle^k \left|f\left(\frac{x}{N}\right)\right| -\le \sup_{z\in\mathbb{R}^d}\langle Nz\rangle^k|f(z)|\ \le N^k\ ||f||_{0,k} -$$ -where we used the standard seminorms -$$ -||f||_{\alpha,k}=\sup_{z\in\mathbb{R}^d}\langle z\rangle^k|\partial^{\alpha}f(z)| -$$ -for Schwartz functions. -Now consider the transpose map $\Theta_N=\theta_N^{\rm T}$ from $s'(\mathbb{Z}^d)$ to -$\mathscr{S}'(\mathbb{R}^d)$. It is defined by -$$ -\langle \Theta_N(\psi),f\rangle=\langle\psi,\theta_N(f)\rangle=\sum_{x\in\mathbb{Z}^d}\psi(x)f\left(\frac{x}{N}\right) -$$ -for all discrete temperate fields $\psi$ and continuum test functions $f$. -Essentially, -$$ -\Theta_N(\psi)=\sum_{x\in\mathbb{Z}^d}\psi(x)\ \delta_{\frac{x}{N}} -$$ -where $\delta_z$ denotes the $d$-dimensional Dirac Delta Function located at the point $z$. -Now $\Theta_N$ is continuous for the strong topologies. -Indeed if $A$ is a bounded subset of Schwartz space -$$ -||\Theta_N(\psi)||_A=\sup_{f\in A}|\langle \Theta_N(\psi),f\rangle|= -\sup_{p\in \theta_N(A)}|\langle \psi,p\rangle| -$$ -and $\theta_N(A)$ is bounded in $s(\mathbb{Z}^d)$ (because a continuous linear map sends bounded sets to bounded sets). -Suppose we are given sequences $m_N$ and $\alpha_N$ dependent on the UV cutoff $N$. -Define the Borel measure -$$ -\nu_N=(\alpha_N\Theta_N)_{\ast}\mu_{m_N} -$$ -on $\mathscr{S}'(\mathbb{R}^d)$. -Its characteristic function is -$$ -W_N(f)=\int_{\mathscr{S}'(\mathbb{R}^d)}d\nu_N(\phi)\ e^{i\langle\phi,f\rangle} -=\int_{s'(\mathbb{Z}^d)}d\mu_{m_N}(\psi)\ e^{i\langle\psi,\alpha_N\theta_N(f)\rangle} -$$ -by the abstract change of variable theorem. -We then get $W_N(f)=\exp\left(-\frac{1}{2}Q_N(f)\right)$ -where -$$ -Q_N(f)=\frac{\alpha_N^2}{(2\pi)^d}\sum_{x,y\in\mathbb{Z}^d} -f\left(\frac{x}{N}\right)f\left(\frac{y}{N}\right) -\int_{[0,2\pi]^d}d^d\xi\ \frac{e^{i\xi\cdot(x-y)}}{m^2+2\sum_{j=1}^{d}(1-\cos \xi_j)} -$$ -$$ -=\frac{N^{2-d}\alpha_N^2}{(2\pi)^d}\sum_{x,y\in\mathbb{Z}^d} -f\left(\frac{x}{N}\right)f\left(\frac{y}{N}\right) -\int_{[-N\pi,N\pi]^d}d^d\zeta\ \frac{e^{i\zeta\cdot(\frac{x}{N}-\frac{y}{N})}}{N^2 m_N^2+2N^2\sum_{j=1}^{d}\left(1-\cos \left(\frac{\zeta_j}{N}\right)\right)} -$$ -after changing $[0,2\pi]^d$ to $[-\pi,\pi]^d$ by periodicity, then -changing variables to $\zeta=N\xi$, and finally some algebraic rearrangement. -Pointwise in $\zeta\in\mathbb{R}^d$, we have -$$ -\lim\limits_{N\rightarrow\infty} -2N^2\sum_{j=1}^{d}\left(1-\cos \left(\frac{\zeta_j}{N}\right)\right) -=\zeta^2 -$$ -and this is why I put an $N^2$ in the denominator. -Finally, we can pick the right choice for the sequences $m_N$ and $\alpha_N$. For a fixed $m\ge 0$ (or strictly positive if $d=1,2$) we let $m_N=\frac{m}{N}$. Now we pick $\alpha_N$ so that -the prefactor $N^{2-d}\alpha_N^2$ becomes the volume element $N^{-2d}$ for a Riemann sum approximation of a double integral on $\mathbb{R}^d\times\mathbb{R}^d$. -Namely, we pick $\alpha_N=N^{-\frac{d}{2}-1}$. -Equivalently, going back to $\alpha_N\Theta_N(\psi)$, that means choosing -$$ -\alpha_N\sum_{x\in\mathbb{Z}^d}\psi(x)\ \delta_{\frac{x}{N}}=\left(\frac{1}{N}\right)^{d-[\phi]} -\sum_{x\in\mathbb{Z}^d}\psi(x)\ \delta_{\frac{x}{N}} -$$ -where $[\phi]=\frac{d-2}{2}$ is the (canonical) scaling dimension of the free field. I wrote the last equation in a way to explicitly display the lattice spacing $\frac{1}{N}$. -Now an excellent exercise, for graduate students in analysis, is to show that -$$ -\lim\limits_{N\rightarrow \infty}Q_N(f)=\frac{1}{(2\pi)^d}\int_{\mathbb{R}^d} -d^d\zeta\ \frac{|\widehat{f}(\zeta)|^2}{\zeta^2+m^2} -$$ -where the Fourier transform is normalized as -$\widehat{f}(\zeta)=\int_{\mathbb{R}^d}d^dx\ e^{-i\zeta\cdot x} f(x)$. -Finally, Fernique's version of the Lévy Continuity Theorem for $\mathscr{S}'(\mathbb{Z}^d)$, shows that the Borel measures $\nu_N$ converge weakly to the one obtained directly in the continuum using the Bochner-Minlos Theorem.<|endoftext|> -TITLE: Category of modules over an Azumaya algebra and the Brauer group -QUESTION [7 upvotes]: Let $k$ be a field, and let $\alpha \in \mathrm{Br}(k)$. Let $A$ be an Azumaya algebra representing $\alpha$. Then the category $A$–$\mathrm{mod}$ depends only on $\alpha$. -I would like to know whether there's a way to describe which $k$-linear categories arise this way. Thus I'd like to know if there's a way to define the Brauer group of a field $k$ as classifying certain kinds of $k$-linear categories. I'd also like to know if there's a good description of the sum of elements of the Brauer group in terms of categories (is it some sort of tensor product of categories?). -The one condition I can come up with is that it should be a semisimple abelian category over $k$ for which the endomorphism algebra of the unit object is $k$. -An even more bold hope is to express the invariant map $\mathrm{Br}(\mathbb{Q}_p) \to \mathbb{Q}/\mathbb{Z}$ in terms of this category. - -REPLY [8 votes]: $k$-linear cocomplete categories admit a "tensor product over $\text{Mod}(k)$" (thinking of them as module categories over $\text{Mod}(k)$) and the only thing you need to know about it to answer this question is that the tensor product of $\text{Mod}(A)$ and $\text{Mod}(B)$ is $\text{Mod}(A \otimes_k B)$. Azumaya algebras then give rise to invertible categories with respect to this tensor product and I believe every invertible such category has this form although I don't know how to prove it. More details about the tensor product in this blog post. -A more explicit characterization isn't hard but also isn't particularly enlightening - over a field, Azumaya algebras are central simple algebras, so you are looking for $k$-linear cocomplete categories where - -there's exactly one isomorphism class of simple object -every object is a direct sum of copies of this simple object, and -the (categorical) center (endomorphisms of the identity functor) is $k$. - -But the invertibility characterization should hold for $k$ any commutative ring, and is in my opinion very conceptually clean; it tells us that the Brauer group is a kind of "higher Picard group" classifying a categorified version of line bundles. Some difficulties here in general with the difference between the Brauer group and the cohomological Brauer group.<|endoftext|> -TITLE: Élie Cartan's paper "Les groupes réels simples, finis et continus" of 1914 -QUESTION [7 upvotes]: Question 1. -Does Élie Cartan's paper -Les groupes réels simples, finis et continus, -Ann. Sci. École Norm. Sup. (3) 31 (1914), 263–355 -contain a classification of $\Bbb C$-linear involutions of simple complex Lie algebras? -Question 2. -If not, what kind of a similar/related classification does it contain? -Question 3. -In what book/paper is this Cartan's paper discussed? -MathSciNet contains only the title, while zbMATH -contains a review: a translation into German of a few lines from the introduction. - -REPLY [9 votes]: The paper and its progeny are discussed at length in Helgason (1978, p. 537): - -In his paper [2] Cartan classifies the simple Lie algebras over R. His method, which required formidable computations, used the signature of the Killing form although it often happens that two nonisomorphic real forms of the same complex algebra have the same signature. Cartan's statement ([2], p. 263): “Les groupes réels d'ordre $r$ qui correspondent à une même type complexe d’ordre $r$ se classent en général complètement d’après leur caractère,” is therefore not to be taken literally; cf. Lardy ([1], p. 195). After noticing the equivalence of problems B and B' (§1) Cartan (in [12]) simplified his original treatment (see also Lardy [1]). Following his general theory [1] of automorphism of complex simple Lie groups, Gantmacher [2] gave a simplified treatment of the real classification. For further developments of this method see Murakami [3], Wallach [2], and Freudenthal and de Vries [1]. While Gantmacher used a Cartan subalgebra $\mathfrak h$ of $\mathfrak g$ whose “toral part” $\mathfrak h\cap\mathfrak k$ is maximal abelian in $\mathfrak k$, Araki develops in [1] a new method using a Cartan subalgebra $\mathfrak h\subset\mathfrak g$ whose “vector part” $\mathfrak h\cap\mathfrak p$ is maximal abelian in $\mathfrak p$. In addition to a solution to problem B' (§1) this method gives valuable information about the restricted roots and their multiplicities (cf. Exercises F). A modification is given by Sugiura [2]. In the present work we use the method of Kac [1] which at the same time gives a rather explicit description of the automorphism $\sigma$ of finite order.<|endoftext|> -TITLE: Sum involving determinants of binomial coefficients, indexed by partitions -QUESTION [5 upvotes]: I would appreciate some help proving a conjecture related to combinatorics and representation theory. -Given an integer partition $\lambda\vdash n$, define a polynomial in $N$ whose roots are the negatives of the contents of the partition, -$$ [N]_\lambda=\prod_{\square \in \lambda}(N+c(\square)).$$ This polynomial is closely related to the value of a Schur function evaluated at the $N\times N$ identity matrix. On the other hand, given $\nu\vdash m$ and $\rho\vdash k$ contained in $\nu$, Jacobi-Trudi applied to a skew-Schur function leads to a determinant of binomial coefficients -$$s_{\nu/\rho}(1_N)=\det_{1\le,i,j\le m}\left({N+\nu_i-i-\rho_j+j-1 \choose \nu_i-i-\rho_j+j}\right).$$ The final ingredient I need for my question is another determinant of binomials, -$$A_{\lambda\rho}=\det_{1\le,i,j\le k}\left({\rho_i-i \choose \lambda_j-j}\right).$$ -Now, in the course of some physics calculation, I arrived at the quantity -$$ E_{\lambda\nu}(N)=\sum_{\lambda\subset\rho\subset\nu} A_{\lambda\rho}s_{\nu/\rho}(1_N).$$ -I thought this was as far as I could push it, but experimentation convinced me that, as a function of $N$, this guy satisfies -$$ E_{\lambda\nu}(N)\propto [N]_{\nu/\lambda}.$$ -It is very surprising to me that this sum should factor like this. -The question is how to prove the above conjecture. -For example, if $\nu=(2,2,1)$ and $\lambda=(1)$, the six terms in the sum are -$$\{\frac{1}{24}N(N^2-1)(5N-6),-\frac{1}{2}N^2(N-1) ,\frac{1}{3}N(N^2-1) ,\frac{1}{2}N(N-1),-N^2,N\}.$$ When all these are added, the result is proportional to $N(N-2)(N^2-1)=[N]_{(2,2,1)/(1)}$. -Actually, I think I know the proportionality constant when $\nu$ and $\lambda$ are both hooks: -$$E_{\lambda\nu}(N)= \frac{1}{(m-n)!}{m-n \choose m-n-\ell(\nu)+\ell(\lambda)}[N]_{\nu/\lambda}.$$ - -REPLY [2 votes]: Following the OP's comment that the factorization can be obtained using my comments above, I repost them here. -The quantity $E_{\lambda\nu}(N)$ can be computed using Cauchy-Binet. Specifically see Lemma 9.1 in Yeliussizov's (nice) paper: https://arxiv.org/pdf/1601.01581.pdf. -For the quantity $A_{\lambda\rho}$, check equation 53 in loc.cit..<|endoftext|> -TITLE: When is the characteristic polynomial of the character table of a cyclic group irreducible? -QUESTION [6 upvotes]: Let $G=C_n$ be the cyclic group and $f_n$ the characteristic polynomial of its character table (over $\mathbb{C}$) in the ordering so that the character table is given by the discrete Fourier transform matrix (https://en.wikipedia.org/wiki/DFT_matrix) without the factor. - -Question 1: For which $n$ is $f_n$ irreducible (over the smallest field extension of $\mathbb{Q}$ containing its coefficients)? - -I did input this into GAP using the command IrrDixonSchneider to obtain the character table (I hope this is does not change much with respect to irreducibility) and tested for irreducibility with GAP. GAP seems to use another numbering for the character table, so lets call $g_n$ the characteristic polynomial according to GAP (maybe someone can clarify the ordering which GAP uses to get the character table of the cyclic group. I try to see what it is but Im not sure at the moment). -For $n \leq 23$ it was true that the $n$ such that $g_n$ is irreducible coincides with the sequence https://oeis.org/A280862 , which are those $n$ such that $a_n \psi_n = \varphi_n$, where $\varphi_n$ is the Euler phi function (https://oeis.org/A000010), $\psi_n$ is the reduced totient function (https://oeis.org/A002322) and $a_n$ is the greatest common divisor of all $(d-1)$'s, where the $d$'s are the positive divisors of $n$ (https://oeis.org/A258409). - -Question 2: Is this true? - - -Question 3: Does being irreducible depend on the ordering used to obtain the character table here (or even for a general group)? - -REPLY [5 votes]: I think a strong pointer to the answer (for the cyclic group of order $n>2 $, where the character table $C$ is chosen so that the $(i,j)$-entry is $\omega^{(i-1)(j-1)}$ for a fixed complex primitive $n$-th root of unity $\omega$) is as follows: -We have $C\overline{C}^{T} = nI,$ but (with this labelling) we note that $C$ is also symmetric. Hence we have $C\overline{C} = nI.$ As noted in my answer to MO363691 (and also previously noted by Denis Serre in his answer to his own question MO78050), we have $\overline{C} = PC$ where $P$ is a permutation matrix of order $2$ (the number of $2$-cycles in the associated permutation is $\frac{n-1}{2}$ if $n$ is odd, and $\frac{n-2}{2}$ if $n$ is even- this is the number of complex conjugate pairs of linear characters of the cyclic group of order $n$ which are not real-valued). -Now we have $CPC = nI$, so that $P = nC^{-2}$ and (since $P^{2} = I$), we have $C^{2} = nP$. Now the eigenvalues of $C$ are all in the set $\{ \sqrt{n},-\sqrt{n}, i\sqrt{ n}, -i\sqrt{n} \}.$ Since $P$ has the eigenvalue $-1$ with multiplicity $\lfloor \frac{n-1}{2}\rfloor$, we see that $\lfloor \frac{n-1}{2}\rfloor$ of the eigenvalues of $C$ are pure imaginary of absolute value $\sqrt{n}$ . If $n$ is odd, then $\frac{n+1}{2}$ of the eigenvalues of $C$ are real of absolute value $\sqrt{n}$, while if $n$ is even, $\frac{n+2}{2}$ of the eigenvalues are real of absolute value $\sqrt{n}$. -Furthermore, as in the answers to MO3639691 and MO78050 that det(C) is real if $n \equiv 1$ or $2$ (mod $4$), and det(C) is pure imaginary if $n \equiv 0$ or $3$ (mod $4$). -Let $F$ denote the field generated by the coefficients of the characteristic polynomial of $C.$ Then the minimum polynomial $p(x)$ of $C$ in $F[x]$ is certainly a divisor of $(x^{2}-n)(x^{2}+n)$ from our knowledge of the eigenvalues of $C$. -Hence any irreducible factor of $p(x)$ has degree $1$ or $2$. Since the characteristic polynomial of $C$ has degree $n >2,$ the characteristic polynomial of $C$ is definitely not irreducible in $F[x].$<|endoftext|> -TITLE: On the density map of the abundancy index -QUESTION [7 upvotes]: Let $σ$ be the sum-of-divisors function. Let $σ(n)/n$ be the abundancy index of $n$. Consider the density map $$f(x) = \lim_{N \to \infty} f_N(x) \ \ \text{ with } \ \ f_N(x) = \frac{1}{N} \#\{ 1 \le n \le N \ | \ \frac{\sigma(n)}{n} < x \}. $$ -In this paper, Deléglise mentioned that Davenport proved that $f$ is continuous, and proved that $0.752 < f(2) < 0.7526$ (bounds improved by Kobayashi in his PhD thesis). -Let $\alpha = f^{-1}(1/2)$ be the median abundancy index, i.e. the number $\alpha$ such that the integers of abundancy index greater than $\alpha$ have natural density exactly $1/2$. -$$\begin{array}{c|c} -N & f_N^{-1}(1/2) \newline \hline -1 &1.00000000000000 \newline \hline -10 &1.50000000000000 \newline \hline -10^2 &1.54838709677419 \newline \hline -10^3 &1.51485148514851 \newline \hline -10^4 &1.52707249923524 \newline \hline -10^5 &1.52501827363944 \newline \hline -10^6 &1.52384533012867 \newline \hline -10^7 &1.52381552194973 \newline \hline -10^8 &1.52381084043829 -\end{array}$$ -The above table suggests that $\alpha \simeq 1.52381$. -Question 1: What is known about the median abundancy index? Is it even mentioned somewhere? It is true that $|\alpha-1.52381|<10^{-5}$? - -Let $(b_n)_{n \ge 1}$ be the sequence of integers such that for all $k |\sigma(b_n)/b_n - \alpha|.$$ This is the lexicographically first sequence of integers whose adundancy index strictly converge to the median adundancy index. Let us call this sequence the buddhist sequence in reference to the Middle Way in buddhism philosophy. Assuming that $|\alpha-1.52381|<10^{-5}$, here are the first terms of this sequence together with the distance of their adundancy index from $1.52381$: $$ \begin{array}{c|c} -n & b_n & |\sigma(b_n)/b_n -1.52381| \newline \hline -1 & 1 & 0.52381000000000 \newline \hline -2 & 2 & 0.02381000000000\newline \hline -3 & 21& 0.00000047619048\newline \hline -4? & 22099389? & 0.0000002693327? -\end{array} $$ -Observe that $b_3=21$, $\sigma(21)/21 = 32/21$ and $|32/21-1.52381|<10^{-6}$, which is statistically unexpectable, as shown if we consider the variation $(b'_n)$ taking $22$ as initial term: -$$ \begin{array}{c|c} -n & b'_n & |\sigma(b'_n)/b'_n -1.52381| \newline \hline -1&22& 0.112553636363636 \newline \hline -3&26& 0.0915746153846153 \newline \hline -4&27& 0.0423285185185187 \newline \hline -5&46& 0.0414073913043478 \newline \hline -6&58& 0.0279141379310344 \newline \hline -7&62& 0.0245770967741934 \newline \hline -8&74& 0.0167305405405405 \newline \hline -9&82& 0.0127753658536585 \newline \hline -10&86& 0.0110737209302325 \newline \hline -11&94& 0.00810489361702116 \newline \hline -12&106& 0.00449188679245283 \newline \hline -13&118& 0.00161372881355915 \newline \hline -14&122& 0.000780163934426037 \newline \hline -15&3249& 0.000659067405355485 \newline \hline -16&14337& 0.000478759154634911 -\end{array} $$ -So there is a very good chance that $\alpha = 32/21$. If so the buddhist sequence ends with its third term and $b_3=21$ should be called the Buddha number. If not, then we know that the set of abundancy indices is dense, so that the buddhist sequence must have a next term $b_4$, but $\sigma(b_3)/b_3$ is already too close to $\alpha$ compared to its above conjectured approximation, so we cannot conjecture the next term. A possible candidate for $b_4$ is mentioned in above table. -Question 2: Does the buddhist sequence end with its third term? If not what are the next terms? -Below are some additional computations with 10 samples of 100001 random integers between $10^{20}$ and $10^{21}$ suggesting that $\alpha = 32/21$ should be correct (sage lists are numbered from 0). -sage: import random -sage: for t in range(10): -....:     L=[] -....:     for i in range(100001): -....:         b=random.randint(10**20,10**21) -....:         q=sum(divisors(b))/b -....:         L.append(q) -....:     L.sort() -....:     print((32/21-L[50000]).n())   --2.01727393333164e-8 -0.00244355476044226 -0.00201824866273585 --0.00130445314014877 --0.000322772616778371 -0.00102756546533326 --6.74774915307343e-10 --1.48849650772673e-19 --0.0000572173485145812 --6.52303473965081e-20 - -Observation: One sample provides a median close to 32/21 with 20 digits, one with 19 digits, one with 10, one with 8, one with 5, one with 4 and four with 3. -How to explain such statistical irregularities? - -A number with abundancy index greater (resp. less) than $2$ is called an abundant (resp. deficient) number, because the sum of its proper divisors (or aliquot sum) exceeds (resp. subceeds) itself. In the same flavour, a number with abundancy index greater (resp. less) than the median abundancy index $\alpha$ could be called an advantaged (resp. disadvantaged) number. -There is a Collatz-like problem (called Calatan-Dickson conjecture) related to the aliquot sum $s$ asking whether all the aliquot sequences $(s^{\circ r}(n))_{r \ge 0}$ are bounded. A value of $\alpha-1 \simeq 0.52381$ suggests heuristically a positive answer to this problem because $\alpha-1$ is the median for $s(n)/n$, although there are serious counter-example candidates like $n=276$ as $s^{\circ 100}(276)>10^{19}$. There are five such candidates less than $1000$ called the Lehmer Five (see this webpage dedicated to recent advances on the aliquot sequence). - -The following picture displays $f_N$ for $N=10^7$ (which should be a good approximation of $f$, according to above table). - -Observe that the function $f$ seems to make a jump around $\alpha$, whereas it is continuous, so it should be non-differentiable there; moreover the phenomenon happens around many other points (with a Cantor set or fractal flavour), which leads to: -Question 3: Is $f$ a Weierstrass function? What is the meaning of these jumps? - -REPLY [8 votes]: For question 1: -Just for fun I calculated bounds on $\alpha$ in 2018, but have not published them. By using the generalized Deleglise method from my thesis, we find -$$ 1.523812 < \alpha < 1.5238175, $$ -so $\alpha\neq 32/21$. -This was found by calculating the density bounds for equally spaced $x$, then narrowing in when we bracket density $1/2$. Here are some relevant bounds: -$$ 0.50003297 \leq f(1.523812) \leq 0.50018578 $$ -$$ 0.4999934 \leq f(1.523813) \leq 0.5001300 $$ -$$ 0.49995299 \leq f(1.523814) \leq 0.5000895 $$ -$$ 0.49991554 \leq f(1.523815) \leq 0.5000560 $$ -$$ 0.4998909 \leq f(1.523816) \leq 0.500012223 $$ -$$ 0.49986562 \leq f(1.523817) \leq 0.500001975 $$ -$$ 0.49985411 \leq f(1.5238175) \leq 0.49981476 $$ -For question 3: -The function $f$ is known to be singular, that is, continuous, non-constant, and differentiable almost everywhere with derivative zero. Thus, it is not a Weierstrass function. -As for the meaning of the jumps, one way of thinking of these is in terms of the series described in my paper "A new series for the density of abundant numbers." The series for $f(x)$ suddenly gains large terms at certain values of $x$, causing the jumps.<|endoftext|> -TITLE: Off-diagonalize a matrix -QUESTION [8 upvotes]: Consider a self-adjoint matrix $M$ that has block form -$$M = \begin{pmatrix} M_{11} & M_{12} \\ M_{12}^* & M_{11} \end{pmatrix}.$$ -I am wondering if there exists any criterion to decide if this matrix can be transformed by some invertible matrix $T$ -such that $$TMT^{-1} = \begin{pmatrix}0 & C \\ C^* & 0 \end{pmatrix}$$ -for some suitable matrix $C?$ -Notice that one restriction that $\begin{pmatrix}0 & C \\ C^* & 0 \end{pmatrix}$ already puts is that the spectrum of $M$ has to be symmetric with respect to zero as conjugation by $$\begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}$$ shows. -As a first step, one might ask when we can achieve a form -$$TMT^{-1} = \begin{pmatrix}0 & C \\ D & 0 \end{pmatrix}$$ -where $C$ and $D$ are arbitrary matrices? - -REPLY [13 votes]: This is a so-called chiral symmetry. The restriction on the symmetry of the spectrum of $M$ is the only restriction you need, you can then bring $M$ to the desired off-diagonal form by a unitary transformation: -$$M=U\begin{pmatrix}\Lambda&0\\ 0&-\Lambda\end{pmatrix}U^\ast\Rightarrow \Omega^\ast U^\ast MU\Omega =\begin{pmatrix}0&\Lambda\\ \Lambda&0\end{pmatrix},$$ -for $\Omega=2^{-1/2}\begin{pmatrix}1&1\\ -1 &1\end{pmatrix}$. -Here $U$ is the unitary matrix of eigenvectors of $M$; the eigenvalues are contained in the diagonal matrix $\Lambda$.<|endoftext|> -TITLE: Are groups with the Haagerup property hyperlinear? -QUESTION [11 upvotes]: In his 2008 paper Hyperlinear and Sofic Groups: A Brief Guide, Pestov asked (Open Question 9.5) whether every group with the Haagerup property is hyperlinear (or sofic). Has this question been answered in the meanwhile? -A short recap of the relevant definitions: A group is hyperlinear (sofic) if it embeds into the metric ultraproduct of unitary groups equipped with the normalized Hilbert-Schmidt distance (symmetric groups with the normalized Hamming distance). -A group has the Haagerup property if there is a sequence of positive definite functions that vanish at infinity and converge pointwise to the constant function $1$. - -REPLY [4 votes]: Thompson's group $F$ has the Haagerup property [1], but it is not known if it is hyperlinear according to Narutaka Ozawa's comment. -[1] Farley. Finiteness and CAT(0) properties of diagram groups. Topology, 2003.<|endoftext|> -TITLE: Continuous version of the fundamental theorem of invariant theory for the orthogonal group -QUESTION [11 upvotes]: A standard result in the invariant theory of the orthogonal group states the following. -Theorem -Let $(E, \langle .,. \rangle)$ be an n-dimensional euclidean vector space, -let $f : E^m \rightarrow {\bf R}$ -a polynomial function satisfying -$f(g(v_1), ... g(v_m)) = f(v_1,...,v_m)$ -for all isometries $g$ of $E$ and $v_1$,..., $v_m \in E$. -Then such a function is a polynomial function -in the quantities $\{\langle{v_i}{v_j}\rangle\}_{i,j = 1...m}$. -Does the theorem holds in the topological setting, -namely when polynomial is replaced by continuous ? -My guess is that it should be true and the proof should be simpler than its algebraic counterpart, maybe a short computation using SVD. -All references I know present the algebraic proof though. Same question in the differential setting. - -REPLY [3 votes]: Reflecting on the answer of Terence Tao, I guess it boils down to -the fact that an injective proper map between locally compact spaces -is a homeomorphism onto its image. Since we are working with -${\bf R}^n$ here, there is a simple characterisation of proper maps -that leads to the following statement. -Let $\Phi : {\bf R}^n \longrightarrow {\bf R}^k$ be a continuous -map satisfying -$$ -\|\Phi(x)\| \longrightarrow \infty \quad when \quad {\|x\| \rightarrow \infty}. -$$ -Let us define the fiber relation on ${\bf R}^n$ by -$x \sim x' \iff \Phi(x) = \Phi(x').$ -Then $({\bf R}^n/\sim)$ is a locally compact metric space and -$\bar{\Phi} : ({\bf R}^n/\sim) \longrightarrow \Phi({\bf R}^n)$ -is a homeomorphism. -The condition on the norm is there to ensure that -for all compact set $K \subset {\bf R}^k$, $\Phi^{-1}(K)$ is -closed and bounded (hence compact). In particular, the fibers -$\Phi^{-1}(\{y\})$ are compact and thus we can define -a distance on the quotient as follows: -$$ -d(\bar{x}, \bar{x}') = d(\Phi^{-1}(\{\bar{\Phi}(\bar{x})\}), \Phi^{-1}(\{\bar{\Phi}(\bar{x}')\})). -$$ -For the problem at hand, we take -$\Phi(v_1,...,v_l) = (\langle v_i, v_j \rangle)$ -and note that the fibers of $\Phi$ are the orbits of the elements -of ${\bf R}^n$ under the action of the orthogonal group. -Then, for any invariant $f$, we have -$$ -f(v_1,...,v_l) = \bar{f}(\bar{\Phi}^{-1}(\bar{\Phi}(\pi(v_1),..., \pi(v_l)))) - = \bar{f} \circ \bar{\Phi}^{-1}(\langle v_i, v_j\rangle). -$$ -The norm condition also ensures that $\Phi({\bf R}^n)$ is closed, -so $\bar{f} \circ \bar{\Phi}^{-1}$ can be extended to all ${\bf R}^k$ if needed.<|endoftext|> -TITLE: Permanent of a Kronecker product of matrices -QUESTION [5 upvotes]: It is well known that $\det(A \otimes B) = \det(A)^m \det(B)^n$ when $A$ and $B$ are square matrices of size $n$ and $m$ where $\otimes$ denotes the Kronecker product. - -Question: Is there a similar formula for the permanent $per(A \otimes B) $ (instead of the determinant)? - -One motivation is to calculate the permanent of the character table of an abelian group since the permanent for cyclic groups is known and the character table of a direct product $G_1 \times G_2$ of groups is given as the Kronecker product of the character table of $G_1$ and $G_2$. - -REPLY [6 votes]: In this paper: -R. A. Brualdi, Permanent of the direct product of matrices, Pacific J. Math. 16 (1966), 471482 -(the Kronecker product is called here 'direct product') it is shown that, if $A$, $B$ are nonnegative matrices with order $m$, $n$, respectively, it holds true that: -$$ \operatorname{per}(A \otimes B) \geq \operatorname{per}(A)^n \operatorname{per}(B)^m $$ -where the equality holds iff $A$ or $B$ has at most one nonzero term in its permanent expression. Moreover, there exists a minimal number, denoted in the paper by $K_{m,n}$, such that: -$$ \operatorname{per}(A \otimes B) \leq K_{m,n} \operatorname{per}(A)^n \operatorname{per}(B)^m $$ -These numbers satisfy the inequality: -$$ K_{m,n} \geq \frac{(mn)!}{(m!)^n (n!)^m} $$ -and it is conjectured that we actually have an equality. -Furthermore, in the paper: -Marvin Marcus, Permanents ot direct products, Proc. Amer. Math. Soc. 17:226-231 -(1966) -it is proven that if $A$, $B$ are positive semidefinite Hermitian square matrices of order $m$, $n$, respectively, then: -$$ \operatorname{per}(A \otimes B) \geq \left (\frac{1}{n!} \right )^m \left (\frac{1}{m!} \right)^n \operatorname{per}(A)^n \operatorname{per}(B)^m $$ -Equality holds iff at least one of $A$, $B$ has a zero row.<|endoftext|> -TITLE: Homotopy group action and equivariant cohomology theories -QUESTION [8 upvotes]: Many of the introductory notes on generalized equivariant cohomology theories assume that one is working over the category of $G$-spaces or $G$-spectra. However, one thing that concerns me is that the action of $G$ is always strict. A $G$-space $X$ is given by a group homomorphism $G\to \text{Aut}(X)$, where $\text{Aut}(-)$ denotes the group of continuous automorphisms. -If instead I want to allow $\sigma:G\times X\to X$ to solve $\sigma(e)\sim \text{id}_X$ and -$$\sigma\circ(\text{id}_G\times \sigma)\sim \sigma\circ (\mu_G\times \text{id}_X)\,, $$ -only up to homotopies ($\mu_G$ here is the multiplication on $G$) and possibly have higher homotopies I need to think about $\infty$-groupoids, as this paper shows that there is an obstruction to strictifying homotopy group actions. - -Is there a well defined notion of equivariant cohomology theories in this setting? - -For an $\infty$-groupoid, one can take its homotopy quotient (the colimit). Can one define the equivariant cohomology as the cohomology of this quotient? - -REPLY [5 votes]: Much has already been said in the other answers and comments, but let me summarize a few points. -One way to obtain from a category a 'homotopy theory' (aka an $\infty$-category) is to specify a notion of weak equivalence. On the category of $G$-spaces (i.e. topological spaces with strict $G$-action), two of the major notions of weak equivalences are the following: - -A map $X \to Y$ of $G$-spaces is a weak equivalence if the underlying map of spaces is a weak homotopy equivalence, or - -a map $X \to Y$ of $G$-spaces is a weak equivalence if the maps $X^H \to Y^H$ are weak homotopy equivalences for all subgroups $H\subset G$. - - -More generally, you could specify a family $\mathcal{F}$ of subgroups of $G$ and you demand that you have a weak equivalence on $H$-fixed points for all $H\in \mathcal{F}$, but let's focus on the two cases above and call them underlying and genuine. -(Edit: Reacting to Denis's comment a clarification: Why should we consider these two kinds of equivalences? Geometrically, $G$-homotopy equivalences (i.e. we have an equivariant homotopy inverse and the homotopies are also equivariant) are maybe the most relevant notion. As in non-equivariant topology, there is a Whitehead theorem showing that genuine weak equivalences between $G$-CW complexes are $G$-homotopy equivalences. Illman's theorem shows that every compact $G$-manifold has the structure of a $G$-CW complex, so one can say that most nice $G$-spaces have the structure of a $G$-CW complex. If we want a Whitehead theorem for underlying equivalences instead, we must demand that the $G$-action is free though. Sometimes we are happy to do this, but often this is too restrictive. The different families $\mathcal{F}$ correspond to allowing different families of isotropy. ) -It is the $\infty$-category associated with the underlying equivalences that can be modelled by homotopy coherent actions. Taking the coherent nerve of the simplicial category of spaces $\mathcal{S}$, we obtain the $\infty$-category of spaces and the $\infty$-category of spaces with homotopy coherent $G$-action is then modelled/defined as simplicial set maps (aka functors) from $BG$ into this coherent nerve. (If we fix $X$, this is the same as simplicial set maps from $BG$ into $B$ of the homotopy automorphisms of $X$.) This $\infty$-category is equivalent to that associated with $G$-spaces and underlying equivalences. (It is nothing special about starting with a group here. We can instead take functors from an arbitrary small category $\mathcal{C}$ into topological spaces and have a similar story using $B\mathcal{C}$. See e.g. Proposition 4.2.4.4 of Higher Topos Theory.) -We cannot, however, recover from the homotopy coherent action the data of the fixed points $X^H$. If we want to model this homotopy-coherently, we need not only $X$ with a homotopy coherent $G$-action, but we also need all spaces of fixed points $X^H$ with their residual actions and all the restriction maps between them. This can be modelled as a functor from the (nerve of the) orbit category $\mathrm{Orb}_G$ of $G$ into $\mathcal{S}$. In the background is Elmendorf's theorem that shows that there is a Quillen equivalence between $G$-spaces with genuine equivalences and functors from $\mathrm{Orb}_G$ to $\mathrm{Top}$ with underlying equivalences (the Quillen equivalence being given by associating to $G/H$ the fixed points $X^H$). Then one can apply e.g. Proposition 4.2.4.4 of HTT again. -As already remarked by others, some equivariant cohomology theories are only sensitive to underlying equivalences (Borel theories), while others are only invariant under genuine equivalences. The latter are actually more frequent (Bredon cohomology, equivariant K-theory, equivariant bordism...). -The story for spectra is a bit more complicated because there are even more types of weak equivalences one can put on, say, orthogonal spectra with a $G$-action. In Shachar's answer, he describes the case corresponding to underlying equivalences. Genuine equivalences (with respect to a complete universe) require more work. For finite groups, one can consider functors from the Burnside category -- this is the perspective of viewing $G$-spectra as spectral Mackey functors. But this is maybe leading too far here.<|endoftext|> -TITLE: The largest group acting on a non-orientable surface of genus 5 -QUESTION [6 upvotes]: Let $N_5$ denote the non-orientable surface of genus 5. -In Conder's database https://www.math.auckland.ac.nz/~conder/BigSurfaceActions-Genus2to101-ByGenus.txt we can see that the biggest finite group $F$ acting on $N_5$ has order 120. Moreover, the quotient has signature $(0; +; [-]; \{(2,4,5)\})$. -Is there a very concrete description of this group $F$? -To be even more concrete. I would like to the length $n$ of the largest chain of subgroups $1=F_0 -TITLE: Is there an orientable prime manifold covered by a non-prime manifold? -QUESTION [13 upvotes]: A manifold is called prime if whenever it is homeomorphic to a connected sum, one of the two factors is homeomorphic to a sphere. - -Is there an example of a finite covering $\pi : N \to M$ of closed orientable manifolds where $M$ is prime and $N$ is not? - -There are no examples in dimensions two or three. If one is willing to forgo the orientability requirement, then there are examples in dimension three. In this paper, Row constructs infinitely many topologically distinct, irreducible (and hence prime), closed 3-manifolds with the property that none of their orientable covering spaces are prime. -There are examples where $N$ is prime and $M$ is not, such as the double covering $\pi : S^1\times S^2 \to \mathbb{RP}^3\#\mathbb{RP}^3$. - -REPLY [12 votes]: There are examples analogous to Row's in dimensions $n>2$ which are orientable when $n$ is even. I'll give a bit of motivation for the example at the end. -Consider the action of the group $G= \mathbb{Z}^n\rtimes \{\pm I\}=\{ x \mapsto \pm x+ m, m\in \mathbb{Z}^n\}$ on $\mathbb{R}^n$. The subgroup $G_{m/2}=\{x,-x+m\}, m\in \mathbb{Z}^n $ is the stabilizer of $m/2\in \frac12\mathbb{Z}^n$. Remove open balls of radius $r<1/4$ about the lattice points $\frac12\mathbb{Z}^n$ to get the simply-connected manifold $V= \mathbb{R}^n -\mathcal{N}_r(\frac12\mathbb{Z}^n)$. When $n$ is even, $V$ admits an orientation which is $G$-invariant. This induces an orientation on $\partial V$. Since $G_{m/2}$ acts as the antipodal map on the sphere of radius $r$ about $m/2$, the quotient $W'=V/G$ will be a manifold with $2^n$ boundary components (corresponding to $\frac12\mathbb{Z}^n/\mathbb{Z}^n \cong (\mathbb{Z}/2\mathbb{Z})^n$) each of which is homeomorphic to $\mathbb{RP}^{n-1}$. The fundamental group of each boundary component will correspond to some $G_{m/2}$ up to conjugacy. $W'$ has a 2-fold cover $V/\mathbb{Z}^n$ which is homeomorphic to $T^n$ punctured at $2^n$ balls. -Take the $2^n$ boundary components of $W'$, and glue them together in pairs, so that when $n$ is even, the induced orientations get reversed, to get a manifold $W$. For concreteness, let's say that we identify the boundary components corresponding to $m/2+\mathbb{Z}^n$ and $m/2+\frac12^n +\mathbb{Z}^n$, inducing a homomorphism $\alpha_m:G_{m/2}\to G_{m/2+\frac12^n}$. In even dimensions, $W$ will be orientable. Since $\pi_1W'= G$, and the subgroup of a boundary component corresponding to the coset $m/2+\mathbb{Z}^n$ will be conjugate to $G_{m/2}$, we see that $\pi_1 W = G \ast_{m\in 0\times\{0,1\}^{n-1}} \alpha_m$ is a multiple HNN extension by the isomorphisms pairing the subgroups. Each HNN extension will introduce a new group element $t_m$ along with a relation of the form $t_mG_{m/2}t_m^{-1}=G_{m/2+\frac12^n}$. So we can give a relative presentation for the fundamental group as -$$\pi_1 W \cong \langle G, t_m | t_mG_{m/2}t_m^{-1}=G_{m/2+\frac12^n}, m\in 0\times\{0,1\}^{n-1}\rangle.$$ Note that there is some choice here of subgroup representative up to conjugacy which does not affect the overall group isomorphism type. -The claim is that $\pi_1 W$ does not split as a free product. This follows from the Kurosh subgroup theorem, and will be proved below. -Now suppose that $W$ is a non-trivial connect sum $W= W_1 \# W_2$. Then $\pi_1(W)=\pi_1(W_1)\ast \pi_1(W_2)$ by the Seifert-van Kampen theorem. Since $\pi_1(W)$ is not a non-trivial free product, that means that $\pi_1(W_1)=1$ (possibly after reindexing). -We need to show that $W_1'=W_1\backslash D^n$ is homeomorphic to the $n$-ball, and hence $W_1$ is the $n$-sphere. $W_1'$ lifts to the double cover of $W$ coming from the homomorphism $\pi_1(W)\to \mathbb{Z}/2\mathbb{Z}$, which is homeomorphic to $T^n \#( S^{n-1}\times S^1)^{\# 2^{n-1}}$ (a non-prime manifold). In turn, $W_1'$ lifts to the universal cover of this manifold which is a submanifold of $\mathbb{R}^n$ (because it is an infinite connect sum of $\mathbb{R}^n$s). Hence $W_1'$ is an $n$-ball by the Schoenflies Theorem, and we see that $W$ is irreducible and orientable when $n>2$ is even. -Now let's see why $\pi_1 W$ is freely indecomposable. Suppose that $\pi_1 W=A\ast B$. Since $ G < \pi_1 W$ is freely indecomposable, by the Kurosh subgroup theorem $G$ is conjugate to a subgroup of $A$ or $B$, let's say $A$. Moreover, the group $H=\pi_1 W/ \ll \mathbb{Z}^n \gg$ obtained by killing $\mathbb{Z}^n$ will be isomorphic to $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}^{\ast 2^{n-1}}$ since $G/\mathbb{Z}^n\cong \mathbb{Z}/2\mathbb{Z}$. Thus we see that the image $\overline{A}$ of $A$ in $H$ will contain $\mathbb{Z}/2\mathbb{Z}$, and hence will be non-trivial. Moreover, the quotient $H$ will split as a free product $\overline{A}\ast B$. However, $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}^{\ast 2^{n-1}}$ is not a free product, since it has a non-trivial center, a contradiction. -Motivation -If a finitely generated group $G$ splits as a free product, then any Cayley graph for $G$ (associated to a finite generating set) has more than one end. If $G< G'$ is finite index, then the Cayley graphs of $G$ and $G'$ are almost equivalent (quasi-isometric), in fact a Cayley graph for $G$ can be obtained from one for $G'$ by collapsing some finite trees equivariantly (this is basically the Reidemeister-Schreier method). -Hence if $G$ has more than one end, so does $G'$. -Now a theorem of Stallings implies that if a group $G'$ has more than one end, then $G'$ is a graph of groups with finite edge groups. Thus, in this example, we found a manifold whose fundamental group is an HNN extension over $\mathbb{Z}/2\mathbb{Z}$ subgroups, but itself is not a free product. But it has an index 2 subgroup that does split as a free product. -If an $n$-manifold $M$ is a connect sum, and $\pi_k(M)=0$ for $k < n-1$, then a similar argument shows that $\pi_1(M)=A\ast B$ is a non-trivial free product. So any manifold finitely covered by such an $M$ will have fundamental group splitting over a finite group. One can probably find many more examples with such properties. I don't know how to find an example which is a connect sum with simply-connected summands, but finitely covers a prime manifold.<|endoftext|> -TITLE: A generalization of strong primes -QUESTION [6 upvotes]: In this post we denote the sequence of prime numbers as $p_k$ for integers $k\geq 1$. I don't know if the following definition is in the literature. -Definition. We define the $\theta$-strong primes, or strong primes at level $\theta$, as the sequence of those prime numbers $p_n$ that satisfy the inequality $$p_n>\theta\, p_{n-1}+(1-\theta)p_{n+1}\tag{1}$$ -for some fixed real number $0<\theta<1$. -Remark. My definition arises as a generalization of the known as strong primes. See the definition in number theory of strong primes from the Wikipedia Strong prime. - -Question. I would like to know, if my question is interesting, if it is possible to prove that there exists some $\hat\theta$ for which the corresponding sequence of prime numbers has finitely/infinitely many terms. Many thanks. - -Here I emphasize that $0<\theta<1$. -I have no intuition if my question is interesting, I don't know what can be a more interesting question at research level about the study of these sequences $(1)$ and their corresponding $\theta$. I did some calculations for $\theta\neq\frac{1}{2}$ a rational number. - -REPLY [5 votes]: Regarding any $\hat\theta$ for which the prime numbers sequence has finitely/infinitely many terms, consider one which has only finitely many terms. There would then exist a prime index $m$ for which all $n \gt m$ gives -$$p_n \le \hat\theta\, p_{n-1} + (1 - \hat\theta)p_{n+1} \tag{1}$$ -Using the standard definition of prime gaps of -$$g_n = p_{n+1} - p_{n} \tag{2}\label{eq2A}$$ -we then have -$$\begin{equation}\begin{aligned} -p_n & \le \hat\theta(p_{n} - g_{n-1}) + (1 - \hat\theta)(p_{n} + g_{n}) \\ -p_n & \le \hat\theta p_{n} - \hat\theta g_{n-1} + (1 - \hat\theta)p_{n} + (1 - \hat\theta)g_{n} \\ -\hat\theta g_{n-1} & \le (1 - \hat\theta)g_{n} \\ -g_n & \ge \left(\frac{\hat\theta}{1 - \hat\theta}\right)g_{n-1} -\end{aligned}\end{equation}\tag{3}\label{eq3A}$$ -If $\hat\theta \gt 0.5$, then $\frac{\hat\theta}{1 - \hat\theta} \gt 1$, so \eqref{eq3A} shows the prime gaps are strictly increasing for all $n \gt m$. However, this contradicts that there are infinitely many prime gaps of at most $246$ (e.g., see Bounded gaps between primes). If $\hat\theta = 0.5$ instead, then $\frac{\hat\theta}{1 - \hat\theta} = 1$, so the prime gaps are non-decreasing, but this is also not possible since prime gaps can become arbitrarily large and, thus, must decrease later to become the smaller prime gaps, e.g., those at most $246$. -This means the original assumption of there being only a finite number of primes in the sequence must be incorrect, i.e., there are infinitely many $\theta$-strong primes for any $\hat\theta \ge 0.5$. -Update: The PDF version of the arXiv article On the ratio of consecutive gaps between primes, near the bottom of page $8$, states about Erdős - -He mentioned 60 years ago [Erd5]: "One would of course conjecture that -$$\underset{n \to \infty}{\lim \inf}\frac{d_{n+1}}{d_n} = 0 \; \text{ and } \; \underset{n \to \infty}{\lim \sup}\frac{d_{n+1}}{d_n} = \infty \tag{2.2}$$ -but these conjectures seem very difficult to prove." Based on a generalization -of the method of Zhang [Zha] the author proved (2.2) in [Pin2] - -where [Pin2] is - -J. Pintz, Polignac numbers, conjectures of Erdős on gaps between primes and the bounded gap conjecture. arXiv: 1305.6289 [math.NT] 27 May 2013. - -i.e., here. -Note their $d_n$ is the same as $g_n$ in \eqref{eq2A}. If the first part of (2.2) is true, then no matter how close $\frac{\hat\theta}{1 - \hat\theta}$ is to $0$, \eqref{eq3A} cannot always being true for all $n \gt m$ for any integer $m$. This means there are infinitely many $\theta$-strong primes for all $0 \lt \hat\theta \lt 1$. -I haven't read the article or references to try to verify the validity of the claim. However, note the article seems to be basically very similar to at least part of the Springer Link book On the Ratio of Consecutive Gaps Between Primes, but I haven't paid to get a chapter or the entire e-book, or a physical copy of the book, to check on this.<|endoftext|> -TITLE: Depth of modules and regular sequences of endomorphisms -QUESTION [9 upvotes]: Let $(R, \mathfrak{m})$ be a Noetherian local ring and $M$ is a finitely generated $R$-module of depth $t$. It is well known that every maximal regular sequence of $M$ has length $t$. Recall that $x_1, \dotsc, x_t \in \mathfrak{m}$ is an $M$-regular sequence if $x_i$ is a non-zero divisor of $M/(x_1, \dotsc, x_{i-1})M$ for all $i = 1, \dotsc, t$, i.e. the multiplicative map $x_i: M/(x_1, \dotsc, x_{i-1})M \to M/(x_1, \dotsc, x_{i-1})M$ is injective. -Now we consider a sequence of endomorphisms instead of multiplications. -Definition. A sequence of endomorphism $\varphi_1, \dotsc, \varphi_t \in \operatorname{End}(M)$ is called a $M$-regular sequence if -(1) For all $i = 1, \dotsc, t$, $\operatorname{Im}(\varphi_i) \subseteq \mathfrak{m}M$. -(2) For all $i =1, \dotsc, t$, $\varphi_i$ induces an injective endomorphism on $M/(\operatorname{Im}(\varphi_1), \ldots, \operatorname{Im}(\varphi_{i-1}))$. -Question 1. Let $(R, \mathfrak{m})$ be a Noetherian local ring and $M$ be a finitely generated $R$-module of depth $t$. Does every maximal $M$-regular sequence of endomorphims of $M$ have length $t$? -Update: Based on Mohan's answer we will assume the our endomorphisms commute. It is natural to ask the following question. -Question 2. Suppose $\varphi_1, \ldots, \varphi_t$ is an $M$-regular sequence of endomorphisms. Is every permutation of $\varphi_1, \ldots, \varphi_t$ an $M$-regular sequence of endomorphisms? - -REPLY [3 votes]: Here I write a sketch of proof which answers both the questions. Consider the $R$ sub-algebra $S$ of $\operatorname{End} M$ generated by the $\phi_i$s. Then, by our assumption, $S$ is commutative, it is a finite type $R$-module and using the assumption $\phi_i(M)\subset\mathfrak{m}M$, it is also local. $M$ is naturally an $S$-module. Under these hypotheses, it is easy to check that $\operatorname{depth}_R M=\operatorname{depth}_S M$ and that you can easily see answers both the questions.<|endoftext|> -TITLE: Consequences of Gromov's Conjecture -QUESTION [18 upvotes]: In Peter Petersen words, Gromov Betti number estimate is considered one of the deepest and most beautiful results in Riemannian geometry; which asserts that - -Theorem (Gromov 1981). There is a constant $C(n)$ such that -any complete manifold $(M, g)$ with $\sec\geq 0$ and for any field $\Bbb F$ of coefficients satisfies -$$\sum_{i=0}^n b_i(M,\Bbb F)\leq C(n).$$ -and conjectured that the best upper bound is $C(n)=2^n$. - -I want to know - -What would be the consequences if Gromov upper bound Conjecture be true? Will the new results be different from the non exact upper bound $C(n)$? - -REPLY [16 votes]: As Igor mentioned knowing the optimal bound is always better than knowing a non-optimal one such as the bound provided by Gromov's proof. It rules out a lot more examples. A proof of the sharp bound would also likely imply a rigidity result that if the sum of the Betti numbers is exactly $2^n$ then $M$ is a torus. This is quite out of reach with the currently known bound. -Conceptually more interesting is the relation of Gromov's conjecture to several other conjectures. -The strongest of these is a conjecture of Bott that a simply connected nonnegatively curved manifold is rationally elliptic. This means that that the total sum of ranks of rational homotopy groups is finite. This holds for homogeneous spaces for example. Rational ellipticity is a very strong condition. -It's been extensively studied in rational homotopy theory. -It in particular implies that the sum of the rational Betti numbers is at most $2^n$. So Bott's conjecture would imply Gromov's conjecture over $\mathbb Q$. -Ellipticity of $M$ also implies that $\chi(M)\ge 0$. So Bott's conjecture also implies Chern's conjecture that a nonnegatively curved manifold must have nonnegative Euler characteristic. Moreover it's known that a rationally elliptic space $M$ has $\chi(M)> 0$ iff all odd Betti numbers of $M$ vanish. This is also conjectured but not known for manifolds of nonnegative curvature. -Lastly, let me mention that Bott's conjecture is much stronger than Gromov's conjecture. For example it's easy to see that connected sum of at least 3 copies of $\mathbb{CP}^n$ is rationally hyperbolic (i.e. not elliptic) so it should not admit nonnegative sectional curvature according to Bott's conjecture. On the other hand Gromov's conjecture does not rule out the connected sum of $k$ copies of $\mathbb{CP}^n$ unless $k> \frac{2^{2n}-2}{n-1}$.<|endoftext|> -TITLE: Isometric embedding of the modular surface -QUESTION [7 upvotes]: Is there an isometric embedding of the modular surface $X(1)=PSL(2,\mathbb{Z})\backslash \,\mathbb{H}$ into the Euclidean 3-space? For all I know this may be an open problem but I am also curious if anyone studied -it numerically or maybe even made a physical model of it. (Which would probably -look a little scary, with two horns (conical points) and a tail (cusp).) -P.S. To make it clear, I mean a $C^\infty$ embedding outside the conical points. - -REPLY [6 votes]: There is no isometric immersion, let alone embedding, of $X(1)$ into Euclidean $3$-space. Here is a sketch of an argument: -First, let $\mathbb{H}\subset\mathbb{C}$ be the upper half plane endowed with the standard metric $(\mathrm{d}x^2+\mathrm{d}y^2)/y^2$ where $z = x+ i\,y$ with $y>0$. A fundamental domain for the action of $\mathrm{PSL}(2,\mathbb{Z})$ on $\mathbb{H}$ is then defined by the inequalities $|z|\ge 1$ and $|x|\le \tfrac12$. Then one identifies $\tfrac12+i\,y$ with $-\tfrac12+i\,y$ and $\cos\theta + i\,\sin\theta$ with $-\cos\theta + i\,\sin\theta$. The 'conical points' are $z_2 \equiv i$ (of order $2$) and $z_3 \equiv \tfrac12 + i\tfrac{\sqrt3}2$ of order $3$, and the 'cusp' point is $z_1 \equiv +i\,\infty$. -Now suppose that a smooth isometric immersion $f:X(1)\setminus\{z_1,z_2,z_3\}\to\mathbb{E}^3$ exists. Fix a point $z\in X(1)$ distinct from the three $z_i$. There will be a hyperbolic disk $D_r(z)$ of some radius $r>0$ about $z$ that does not contain any of the $z_i$. Because the Gauss curvature of $X(1)$ is $K=-1$, the convex hull of the $f$-image of $D_r(z)$ will contain an Euclidean ball of some positive radius $R>0$. -Meanwhile, let $\epsilon>0$ be a very small positive number and consider the subset $M_\epsilon\subset X(1)$ that consists of the $z = x+i\,y$ that satisfy $y\le 1/\epsilon$ and $d(z,z_2)\ge \epsilon$ and $d(z,z_3)\ge \epsilon$, where $d(z,w)$ is the hyperbolic distance between $z$ and $w$. This $M_\epsilon$ is a compact smooth surface whose boundary $\partial M_\epsilon$ consists of three disjoint circles: - -$C_1$ (the points of the form $z = x+i/\epsilon$), which has length $\epsilon$; - -$C_2$ (the points where $d(z,z_2)= \epsilon$), which has length $\pi\sinh\epsilon$, and - -$C_3$ (the points where $d(z,z_3) = \epsilon$), which has length $\tfrac23\pi\sinh\epsilon$. - - -In particular, when $\epsilon>0$ is taken to be sufficiently small, each of these curves has total length less than $4\epsilon$. -Thus, each $f(C_i)$ must therefore lie in an Euclidean ball $B_i$ of radius at most $4\epsilon$. Hence the $f$-image of the boundary $\partial M_\epsilon$ must lie in an infinite 'slab' of thickness at most $4\epsilon$. (Just take a plane that passes through the centers of the three balls $B_i$ of radius $4\epsilon$ and look at the $4\epsilon$-neighborhood of that plane.) -Now, because the Gauss curvature of $M_\epsilon$ is strictly negative, the $f$-image of $M_\epsilon$ must lie within the convex hull of the image of $\partial M_\epsilon$. In particular, it must lie in the infinite slab of thickness at most $4\epsilon$. -However, if we take $\epsilon -TITLE: What is the appropriate notion of weakly equivalent or Morita equivalent categories internal to a category of generalized smooth spaces? -QUESTION [5 upvotes]: Let $G$ and $H$ be Lie groupoids. We know that there are two notions of equivalences of Lie groupoids: - -Strongly equivalent Lie groupoids: (My terminology) - -A homomorphism $\phi:G \rightarrow H$ of Lie groupoids is called a strong equivalence if there is a Lie groupoid homomorphism $\psi:H \rightarrow G$ and natural transformation of Lie groupoid homomorphism $T: \phi \circ \psi \Rightarrow \mathrm{id}_H$ and $S: \psi \circ \phi \rightarrow \mathrm{id}_G$. In this case $G$ and $H$ is said to be strongly equivalent Lie groupoids. - -Weakly Equivalent or Morita Equivalent Lie groupoids : - -A homomorphism $\phi:G \rightarrow H$ of Lie groupoids is called a weak equivalence if it satisfies the following two conditions - -where $H_0$, $H_1$ are object set and morphism set of Lie groupoid H respectively. Similar meaning holds for symbols $G_0$ and $G_1$. Here symbols $s$ and $t$ are source and target maps respectively. The notation $pr_1$ is the projection to the first factor from the fibre product. from t. Here the condition (ES) says about essential surjectivity and the condition (FF) says about full faithfulness. -One says that two Lie Groupoids $G$ and $H$ are weakly equivalent or Morita equivalent if there exist weak equivalences $\phi:P \rightarrow G$ and $\phi':P \rightarrow H$ for a third Lie groupoid $P$. -(According to https://ncatlab.org/nlab/show/Lie+groupoid#2CatOfGrpds one motivation for introducing Morita equivalence is the failure of the axiom of choice in the category of smooth manifolds ) -What I am looking for: -Now let we replace $G$ and $H$ by categories $G'$ and $H'$ which are categories internal to a category of generalized smooth spaces (For example, category of Chen spaces or category of diffeological spaces... etc). For instance, our categories $G'$ , $H'$ can be path groupoids. -Analogous to the case of Lie groupoids I can easily define the notion of Strongly equivalent categories internal to a category of generalized smooth spaces. -Now if I assume that the axiom of choice fails also in the category of generalized smooth spaces then it seems reasonable to introduce a notion of weakly equivalent or some sort of Morita equivalent categories internal to a category of generalized smooth spaces. -But it seems that we cannot directly define the notion of weakly equivalent or Morita equivalent categories internal to a category of Generalized Smooth Spaces in an analogous way as we have done for Lie Groupoids. Precisely in the condition of essential surjectivity (ES) we need a notion of surjective submersion but I don't know the analogue of surjective submersion for generalized smooth spaces -I heard that Morita equivalence of Lie groupoids are actually something called "Anaequivalences" between Lie groupoids.(Though I don't have much idea about anafunctors and anaequivalences). -So my guess is that the appropriate notion of weakly equivalent or Morita equivalent categories internal to a category of generalized smooth spaces has something to do with anaequivalence between categories internal to a category of generalized smooth spaces. Is it correct? -My Question is the following: -What is the appropriate notion of weakly equivalent or Morita equivalent categories internal to a category of generalized smooth spaces? -EDIT: -In the comments section after the answer by David Roberts we also had a discussion on the following two questions: - -Let $F: G \rightarrow H$ be a Lie groupoid Homomorphism such that $F$ is fully faithful and essentially surjective as a functor between the underlying categories. Let us also assume the $G$ and $H$ are not Morita Equivalent. Then what are the properties that Lie groupoids $G$ and $H$ has in common apart from the trivial fact that they have equivalent underlying categories? - -In papers on Higher gauge theory like Principal 2 bundles and their Gauge 2 groups by Christoph Wockel https://arxiv.org/abs/0803.3692 and the paper Higher Gauge theory: 2-connections by Baez and Schreiber https://arxiv.org/abs/hep-th/0412325 why strong equivalence is preferred over weak equivalence in the notion of Local triviality for Principal-2 bundles over a manifold? (Here equivalence means equivalence between categories internal to a category of generalized smooth spaces) - - -My deep apology for asking two sufficiently different (from the original) question in the comments section. -Thank you. - -REPLY [6 votes]: Apologies for the late answer, I wish I'd found this earlier! -My MSc thesis was actually mainly devoted to developing a notion of Morita equivalence for diffeological groupoids! I don't think I'll have much to add to the answers by David Roberts and Joel Villatoro, and diffeological groupoids are clearly more specific than what you're looking for, but for what it's worth I'd like to contribute my two cents. You can find my thesis here: -Diffeology, Groupoids & Morita Equivalence. I also wrote a paper about the main result (a "Morita Theorem"), which is available on the arXiv here: arXiv:2007.09901 (and which has been accepted for publication in the Cahiers)! -The approach I focussed on in my thesis is that of biprincipal bibundles. This is a slightly different point of view than your question, but it turns out that this gives a notion of Morita equivalence that is equivalent (no pun intended) to the definition using weak equivalences (see Section 5.1.3 of the thesis). The paper by Meyer and Zhu, mentioned in one of David Roberts' comments, also uses this point of view. I don't know to what extent their theory can be used for diffeology. I mainly wanted to focus on this part of your question: - -"...I don't know the analogue of surjective submersion for generalized smooth spaces." - -As David Roberts mentions, one sensible option is to replace surjective submersions by subductions. The entire point of my thesis is essentially to show that the Lie groupoid theory still works if you do this. (If you're looking for a reference on subductions, Section 2.6 of my thesis discusses them in some detail.) This also leads us to an important point that Joel Villatoro raises: - -"By the way, for my part I would recommend taking surjective local subductions as your submersion for the diffeological category." - -The motivation for this could be that the local subductions between smooth manifolds are exactly the surjective submersions (this is proved in the Diffeology textbook by Iglesias-Zemmour), hence directly generalising surjective submersions to the diffeological setting (which is what we were after!). However, as just mentioned, choosing subductions still makes everything work. Besides this, here are two more reasons to consider choosing subductions over local subductions: - -There are naturally occurring bundle-like objects in diffeology that are subductions, but not local subductions. The main examples I have in mind are the internal tangent bundles. For example, if you consider diffeological space that is the union of the two coordinate axes in $\mathbb{R}^2$, its internal tangent bundle is 2-dimensional at the origin, but 1-dimensional everywhere else (see Example 3.17 in arXiv:1411.5425 by Christensen and Wu). The internal tangent bundle of this space is then a subduction, but not a local subduction (thanks to an argument by Christensen). If we want to study such objects to occur in our theory of diffeological groupoid bundles and Morita equivalence, we need to allow subductions and not just local subductions. - -The main reason that we assume the source and target maps of a Lie groupoid $G\rightrightarrows G_0$ are surjective submersions is to ensure that the fibred product $G\times_{G_0}G$ of composable arrows is again a smooth manifold. Since the category $\mathbf{Diffeol}$ of diffeological spaces is (co)complete, this assumption becomes redundant. However, the source and target maps of a diffeological groupoid are always subductions! - - -I elaborate on the choice of subductions over local ones in Sections 4.2 and 4.4.3 in my thesis. My answer to your main question in the setting of diffeology is therefore the same as that of David Roberts: an appropriate notion of Morita equivalence for diffeological groupoids is exactly as for Lie groupoids, but with surjective submersion replaced by subduction. -As to a more structured approach to generalising surjective submersions to the setting of generalised smooth spaces (your actual main question): I believe that the subductions in the category $\mathbf{Diffeol}$ of diffeological spaces are exactly the strong epimorphisms, cf. Proposition 37 in arXiv:0807.1704 by Baez and Hoffnung. In a more abstract setting of generalised smooth spaces you could therefore consider trying to use the strong epimorphisms to replace the surjective submersions!<|endoftext|> -TITLE: Is $\beta^{*}(w_{2k-2}) = 0$ for an open orientable $2k$-manifold? -QUESTION [7 upvotes]: This question is motivated by the vector field question I asked recently. Panagiotis Konstantis answered this question for odd manifolds and I am trying to figure out the even case. -Let $M$ be a smooth oriented manifold (without boundary) of even dimension $2k$ with $k \geq 2$. Steenrod showed that the primary obstruction for lifting the tangent bundle $\tau\colon M \rightarrow BO(2k)$ along the fibration $$V_2(\mathbb{R}^{2k}) \rightarrow BO(2k-2) \rightarrow BO(2k)$$ -is $$ \beta^*(w_{2k-2}) \in H^{2k-1}(M; \pi_{2k-2}(V_2(\mathbb{R}^{2k})) = H^{2k-1}(M; \mathbb{Z})\,,$$ -where $\beta^*$ is the Bockstein operator and $w_{2k-2}$ is the $(2k-2)$th Stiefel-Whitney class of $M$. -Now Theorem 2 of Massey's "On the Stiefel-Whitney classes of a manifold II" paper says that this class vanishes when $M$ is closed. Can we say the same for open $M$? If the cohomology class had field coefficients, we could argue the vanishing as in this answer and the comments under it. But here the class is integral. Can we salvage the situation using other properties the problem has (such as $\beta^*(w_{2k-2})$ being 2-torsion)? - -REPLY [2 votes]: Your question is the following: - -Does $W_{2k-1}$ vanish on an (open) orientable $2k$-manifold? - -I don't know the answer in general, only for $k = 1, 2, 3$. - -We always have $W_1 = \beta^*(w_0) = \beta^*(1) = 0$, so the answer is yes for $k = 1$. - -Note that $W_3 = \beta^*(w_2)$ is the obstruction to the existence of a spin$^c$ structure on an orientable manifold. As every orientable four-manifold is spin$^c$ (see this note by Teichner and Vogt), the answer is also yes for $k = 2$. - -In this paper, Aleksandar Milivojevic and I showed that $W_5$ is the primary obstruction to the existence of a spin$^h$ structure on an orientable manifold (Corollary 2.6). As every orientable six-manifold is spin$^h$ (Corollary 3.10), the answer is also yes for $k = 3$.<|endoftext|> -TITLE: Chern number on non-spin manifold -QUESTION [7 upvotes]: Let $M^4$ be an orientable closed 4-manifold and $c_1$ be the first Chern class of a complex line bundle on $M^4$. Let $b$ be the mod 2 reduction of $c_1$, ie $b=c_1$ mod 2. -We have a relation $w_2 b = b^2$, where $w_n$ is the $n^\text{th}$ Stiefel-Whitney class of the tangent bundle of $M^4$. This implies that if $M^4$ is spin, then the Chern number -on $M^4$ must be even, ie $\int_{M^4} c_1^2 =0$ mod 2. -My question is that for any $M^4$ that is not spin, can we always find a complex line bundle on $M^4$, such that the Chern number on $M^4$ is odd, ie $\int_{M^4} c_1^2 =1$ mod 2. - -REPLY [11 votes]: The Enriques algebraic surface has even intersection form (i.e. for any class $\beta \in H^{2}(M,\mathbb{Z})$, $\int_{M^{4}} \beta^2$ is even) but is not spin by Rokhlin's theorem since the signature of the intersection form is $8$. -A simply connected $4$-manifold is spin $\iff$ the intersection form is even (which doesn't apply to the Enriques surface which has $\pi_{1} = \mathbb{Z}_{2}$). - -REPLY [2 votes]: EDIT: This does not work, in general, as explained by Michael Albanese's comment. Thanks! -If $M$ is not spin, then $w_2(M) \neq 0$. But $w_2$ agrees with $v_2$, the second Wu class, which measures whether the intersection form of $M$ is even or odd. Thus, we can find an element $\alpha \in H^2(M;\mathbb Z)$ such that $\alpha^2$ is an odd number times the cohomological fundamental class of $M$. Now represent $\alpha$ by a map $M \to K(\mathbb Z;2) = BU(1)$, i.e., a complex line bundle $E$ on $M$, then $c_1(E) = \alpha$ is as desired.<|endoftext|> -TITLE: Alternating sum over collections closed under containment -QUESTION [9 upvotes]: Let $\mathscr{C}$ be a collection of subsets of a finite set $P$. Assume $\mathscr{C}$ is closed under containment: if $S\subset P$ is in $\mathscr{C}$, then every set $S'\subset P$ containing $S$ is in $\mathscr{C}$. -What can we say about -$$\sum_{S\in \mathscr{C}} (-1)^{|S|},$$ where $|S|$ is the number of elements of $S$? In particular, is the absolute value of this sum bounded -by the number of minimal elements of $\mathscr{C}$, i.e., -$$\left|\sum_{S\in \mathscr{C}} (-1)^{|S|}\right| \leq |\{S\in \mathscr{C}:\not \exists S'\subsetneq S \;\text{s.t.}\; S'\in \mathscr{C}\}|?$$ - -REPLY [8 votes]: There is an interpretation of your inequality using simplicial complex and $h$-vector. Namely, let $\Delta$ be the set of all $P-S$ with $S$ in your set. Then $\Delta$ is a collection closed under inclusion, hence a simplicial complex. -The invariant you are interested in is, up to sign, the alternating sum of $f_i$: the number of $i$-dim faces of $\Delta$. Then, by a well-known formula, it is equal to $|h_d|$, where $h_i$ forms the $h$-vector of $\Delta$, and $d-1$ is the dimension of $\Delta$. -The right hand side of the inequality you are interested in is the number of facets (maximal elements) of $\Delta$. As people have pointed out, in general the inequality you want does not hold. -However, I would like to point out that it is likely to hold under certain extra topological/homological assumptions. For instance, if $\Delta$ is Cohen-Macaulay, then all the $h_i$ are non-negative, and the number of facets is $f_{d-1}=\sum_{i\geq 0} h_i\geq h_d$, which is what you need. In fact, as $h_0=1$ and $h_1=n-d$ where $n=|P|$, you get something a little stronger. -One can prove other inequalities for $h_i$ under weaker conditions. For example, if $\Delta$ satisfies Serre's conditions $(S_{r})$, one still have non-negativity of $h_{\leq r}$, a result first proved by Murai-Terai. I discussed some of them in a recent talk (but it is perhaps a bit algebraic).<|endoftext|> -TITLE: Is $\mathbb{Q}_p \otimes_{\mathbb{Q}}\mathbb{Q}_p $ coherent? -QUESTION [13 upvotes]: Let $\mathbb{Q}_p$ denote the field of fractions of $\mathbb{Z}_p$. By the answers to this quesition the tensor product $\mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}_p$ cannot be a Noetherian ring (alternatively, this follows because the transcendence degree of $\mathbb{Q} \to \mathbb{Q}_p$ is infinite). One could instead hope for the weaker result that this ring is coherent. Is this true? - -REPLY [15 votes]: You can use the following: -Lemma. Let $A = \operatorname{colim}_i A_i$ be a filtered colimit of coherent rings such that $A$ is flat over each $A_i$. Then $A$ is coherent. -For example, this is true if all the transition maps $A_i \to A_j$ are flat. -Proof. Let $I \subseteq A$ be a finitely generated ideal. Then $I = AI_i$ for some finitely generated ideal $I_i \subseteq A_i$ for some $i$. By assumption, $I_i$ is finitely presented as $A_i$-module, i.e. there is an exact sequence -$$A_i^m \to A_i^n \to A_i \to A_i/I_i \to 0.$$ -By flatness of $A_i \to A$, the sequence -$$A^m \to A^n \to A \to A/I \to 0$$ -is exact as well, i.e. $I$ is finitely presented. $\square$ -Example 1. Let $K \subseteq L$ and $K \subseteq M$ be field extensions. Then $A = L \otimes_K M$ is coherent. Indeed, it can be written as a colimit -$$A = \underset{\substack{\longrightarrow \\ K \subseteq L_i \subseteq L \\ K \subseteq M_j \subseteq M}}{\operatorname{colim}} L_i \underset K\otimes M_j,$$ -where the colimit runs over all finitely generated subextensions $K \subseteq L_i \subseteq L$ and $K \subseteq M_j \subseteq M$. Each $L_i \otimes_K M_j$ is Noetherian, so in particular coherent, and the transition maps are flat because both $L_i \to L_{i'}$ and $M_j \to M_{j'}$ are. -Example 2. The algebraic integers $\bar{\mathbf Z}$ are coherent as the colimit of all $\mathcal O_K$ for $\mathbf Q \subseteq K$ finite. The transition maps $\mathcal O_K \to \mathcal O_L$ are flat because $\mathcal O_K$ is a Dedekind domain and $\mathcal O_L$ is torsion-free.<|endoftext|> -TITLE: Oddness of intersection form of surface bundle -QUESTION [5 upvotes]: Let $\Sigma_g$ be a Riemannian surface of genus $g$. Let $M^4$ be a surface bundle: $\Sigma_g \to M^4 \to \Sigma_h$. When $g=1$, $M^4$ is called a torus bundle. -My question: is there a torus bundle whose intersection form contains an odd diagonal element (if we choose a basis and view the intersection form as a matrix)? -If $M^4=\Sigma_1\times \Sigma_h$, then $M^4$ is spin and its intersection form has only even diagonal elements. -More generally: -For a given fiber $\Sigma_g$, is there a $\Sigma_g$-bundle whose intersection form is odd? - -REPLY [5 votes]: Suppose $\pi\colon M \to \Sigma_g$ is an oriented smooth torus bundle. If $w_2(M) = 0$, then also the second Wu class $v_2(M) = 0$ and $M$ has even intersection form (the converse holds if $H_1(M;\mathbb Z)$ has no $2$-torsion, but we do not need this here). I claim that this is always the case in our situation. -Even better, I claim that $M$ is always parallelizable: stably, $TM$ agrees with the vertical tangent bundle $T_{\pi}$, whose classifying map $E \to B\text{SL}_2(\mathbb R)$ can be identified with the map -$$E \xrightarrow{\pi} \Sigma_g \xrightarrow{(1)} B\text{Diff}^+(T^2) \xrightarrow{(2)} B\text{SL}_2(\mathbb Z) \xrightarrow{(3)} B\text{SL}_2(\mathbb R),$$ -where (1) is the classifying map of $\pi$, (2) is induced from applying $\pi_1$, and (3) is induced from extending coefficients. Since $H^2(B\text{SL}_2(\mathbb Z);\mathbb Z) = \mathbb Z/12$ is torsion, the map composition of (1), (2) and (3) is trivial on second cohomology and hence nullhomotopic, as $B\text{SL}_2(\mathbb R) = K(\mathbb Z,2)$. Thus, $T_{\pi}$ is trivial and $M$ is stably parallelizable. Since $\chi(M) = \chi(\Sigma_g)\chi(T^2) = 0$, $M$ is parallelizable. -If the base is not a surface, I believe that it is possible for torus bundles to be non-spin, see Johannes Ebert's thesis (the last pages of chapter 5), although no concrete examples are constructed there. -For higher genus, note that there are examples of surface bundles over surfaces whose total space has signature $4$, in particular, its intersection form cannot possibly be even. -Also, the total space of the (unique!) nontrivial $S^2$-bundle over $S^2$ is diffeomorphic to $\mathbb CP^2 \# \overline{\mathbb CP^2}$, which has odd intersection form.<|endoftext|> -TITLE: Realizing Stiefel-Whitney classes via vector bundles -QUESTION [12 upvotes]: Let $X$ be a CW complex. If $E$ is a vector bundle over $X$, then it's well-known that the Stiefel-Whitney classes $w_j(E) \in H^j(X,\mathbb F_2)$ of $E$ are determined from the classes $w_{2^k}(E)$ (for $2^k \leq j$) via the Wu formula, using the cup product and the action of the Steenrod algebra. - -Question 1: Does the Wu formula imply any further relations? This is a purely algebraic question which I make more precise in (a) and (b) below. -That is, let $H$ be a nonnegatively-graded $\mathbb F_2$-algebra with an unstable action of the Steenrod algebra satisfying the Cartan formula and $Sq^{|x|}(x) = x^2$ for all homogenenous $x \in H$. Let $W$ be the set of sequences $(w_j \in H^j)_{j \in \mathbb N}$ with $w_0 = 1$ and satisfying the Wu formla. -(a) For any sequence $(v_{2^k} \in H^{2^k})_{k \in \mathbb N}$, does there exist $w \in W$ (necessarily unique) with $w_{2^k} = v_{2^k}$ for all $k \in \mathbb N$? -Presumably (i) the Whitney sum formula and (ii) the universal formula for the Stiefel-Whitney classes of a tensor product of vector bundles are compatible with the Wu formula, so that $W$ is a commutative ring using (i) for addition and (ii) for multiplication. -(b) Is $W$ a polynomial algebra on whichever generators from (a) do exist? - -Question 2: What restrictions beyond the Wu formula are there restricting the Stiefel-Whitney classes of a vector bundle $E$ on a CW complex $X$? This is a genuinely topological question. -Over here Mark Grant describes one such restriction, but ideally I'd like a more systematic discussion. - -If it simplifies matters to assume that $X$ is finite, or even a compact manifold, then that's fine. - -REPLY [3 votes]: Here is a start at answering Question 1: there are indeed further relations between the $w_{2^k}$, or at least conditions on the $w_{2^k}$. -For example, consider the case where $H$ has multiplication which is null except for what is implied by the multiplication being unital. (For example, $H$ may be the cohomology of a suspension space.) -In this case, the Wu formula reduces to -$$Sq^i(w_j) = \binom{j-1}{i} w_{i+j}$$ -So if the $w_{2^k}$'s are given, we are forced to define $w_{2^k + j'} = Sq^{j'} w_{2^k}$ for $0 \leq j' \leq 2^k-1$, which gives us the definition of each $w_j$. So now in the case where $j = 2^k+1$ and $1 \leq i \leq 2^k - 1$, the Wu formula stipulates that $Sq^i Sq^1 w_{2^k} = 0$. This is always the case for $i = 1$, but for all other $i$, the relation $Sq^i Sq^1 = 0$ does not hold in the Steenrod algebra, so I believe there are examples of $H$'s and $w_{2^k} \in H^{2^k}$ where this equation does not hold. So this is an example of some kind of further condition which $w_{2^k}$ may be required by the Wu formula to satisfy.<|endoftext|> -TITLE: Is there a collection of evidence and heuristic arguments against the Riemann hypothesis? -QUESTION [7 upvotes]: There is undoubtedly an overwhelming collection of evidence for the Riemann hypothesis. However, is there any evidence against it ? - -REPLY [6 votes]: (note that the original question before being edited out had an argument about why RH is false and the post below was a refutation of that); the Ivic paper linked by @Mayank contains some good arguments and is worth reading for sure -The mistake in the argument above is quite subtle and is easily seen by doing the above with $M^2(x)=x$, namely that while indeed $g(X) \ll _{\varepsilon}X^{2\Theta-1+\varepsilon}\ll X^{\sigma+\Theta-1}$ the $<<_{\epsilon}$ carries through so the inequality $g(x) << X^{\sigma+\Theta-1}$ is not absolute but it in fact is $g(x) <<_{\sigma-\theta} X^{\sigma+\Theta-1}$ so the final result is actually $f(\sigma) <<_{\sigma-\Theta} \frac{2\sigma - 1}{\sigma-\Theta}$ and obviosuly there is no contradiction since the $<<_{\sigma-\Theta}$ can (and actually does at least when $\Theta=1/2$) very well go to infinity too when $\sigma \to \Theta$. -As noted with $M^2(x)=x$ (which satisfies all the necessary stuff) we get $f(\sigma)=1/(2\sigma-1)$ (and indeed we get the required singularity at $\Theta=1/2$). -Then $g(x) =\log x$ and while indeed $\log x << x^{\sigma -1/2}$ the $<<$ depends on $\sigma-1/2$ so there is no contradiction in $(2\sigma-1)f(\sigma) \to 1, \sigma \to 1/2$ and $f(\sigma)=(2\sigma-1)\int_1^{\infty} (\log x) x^{-2\sigma+1}dx <<_{\sigma-1/2}\int_1^{\infty} x^{-\sigma+1/2}dx <<_{\sigma-1/2}1/2$ as the $<<_{\sigma-1/2} \to \infty, \sigma \to 1/2$<|endoftext|> -TITLE: Is identification of double points of an immersion smooth? -QUESTION [5 upvotes]: Let $f:M^m\to N^n$ be a generic map between smooth manifolds $n>m$. Depending on the pair $(m,n)$ generic maps will have a singular set of double points $\Sigma_2\subset M$. -Let $\phi:\Sigma_2\to \Sigma_2$ be the map of sets that sends $x\in \Sigma_2$ to the other point $y\in \Sigma_2$ such that $f(x)=f(y)$. -$\phi$ can be thought as a $\mathbb{Z}/2$-action on $\Sigma_2$. -For the sake of concreteness one can think of a generic immersion of a $M^3\to N^4$, then the set of double points is generically of dimension 2. - - -Is $\Sigma_2$ a smooth submanifold? -Is $\phi$ a smooth or at least a continuous map? In other words how bad is the action? -How does this generalize to triple points and $n$-points? I.e., can we have any kind of group of $n$ elements acting? - -REPLY [3 votes]: I think the answer to the first 2 questions is yes. Most of the details are in the thesis of Ralph Herbert: -Herbert, Ralph J., Multiple points of immersed manifolds, Mem. Am. Math. Soc. 250, 60 p. (1981). ZBL0493.57012 -The important construction here is the $r$-tuple point manifold $\Delta_r(f)$ of the immersion $f:M^m\looparrowright N^n$, which for $r\ge2$ is defined as follows. Let $F(M,r)\subseteq M^{(r)}$ be the ordered configuration space of $r$-tuples $(x_1,\ldots , x_r)$ of points of $M$ such that $x_i\neq x_j$ whenever $i\neq j$; it is an open submanifold of $M^{(r)}$. Now consider the restriction of the $r$-th Cartesian power of $f$ to this configuration space, which will by abuse of notation will be denoted $f^{(r)}:F(M,r) \looparrowright N^{(r)}$. Generically, $f^{(r)}$ is transverse to the thin diagonal $d_r(N)= \{(n,\ldots , n)\}\subseteq N^{(r)}$ (see Herbert; also Gollubitsky and Guillemin, Stable mappings and their singularities, Chapter III, Corollary 3.3). Then $$\Delta_r(f) :=(f^{(r)})^{-1}(d_r(N))= \{(x_1,\ldots , x_r)\in F(M,r) \mid f(x_i)=f(x_j)\mbox{ for all }1\le i,j\le n\} $$ -is a submanifold of $F(M,r)$ of codimension $rn-n$, therefore of dimension $rm-(rn-n)=n-r(n-m)$. If $M$ is a closed manifold, then so is $\Delta_r(f)$ (compactness is not obvious, it uses the fact that $f$ is locally an embedding). Note that $\Delta_r(f)$ carries smooth free actions of the symmetric group $\mathfrak{S}_r$ and the symmetric group $\mathfrak{S}_{r-1}$ which permute the last $r$ and $r-1$ coordinates, respectively. Let $M_r(f):=\Delta_r(f)/\mathfrak{S}_{r-1}$. -Now consider the map -$$ -\mu_r(f): M_r(f)\to M,\qquad (x_1,[x_2,\ldots , x_r])\mapsto x_1 -$$ -given by projection onto the first co-ordinate. This $\mu_r(f)$ can be shown to be an immersion. Its image is the set of $x\in M$ such that $|f^{-1}f(x)|\ge r$. -Now restrict to the case $r=2$. The immersion $\mu_2(f):M_2(f)\looparrowright M$ is not an embedding when $f$ has triple points or higher. But if we remove the points of the domain where $\mu_2(f)$ fails to be injective, we get an injective immersion onto you $\Sigma_2$ (the "pure" double points), which in fact is an embedding. (A perhaps more convincing argument using general position which works for all $\Sigma_r$ is given on page 25 of Herbert.) -Thus $\Sigma_2\subseteq M$ is an embedded submanifold, and the involution $\phi:\Sigma_2\to \Sigma_2$ is smooth as it is conjugate to the restriction of invoution on $M_2(f)=\Delta_2(f)$ which permutes the factors. -For triple points and higher, I don't see any group action on $\Sigma_r$, since when $f^{-1}f(x)=\{x,x_2,\ldots , x_r\}$ there is no natural way to order the set $\{x_2,\ldots, x_r\}$.<|endoftext|> -TITLE: Algebras Morita equivalent with the Weyl Algebra and its smash products with a finite group -QUESTION [6 upvotes]: My question os motivated, naturally, by the problem of classifying symplectic reflection algebras up to Morita equivalence (a classical reference for rational Cherednik algebras is Y. Berest, P. Etingof, V. Ginzburg, "Morita equivalence of Cherednik algebras", MR2034924; the most up do date work in this subject I know of is I. Losev, Derived equivalences for Symplectic reflection algebras, https://arxiv.org/abs/1704.05144); -and also the problem of understading rings of differential operators on irreducilbe affine complex varieties $X$ up to Morita equivalence (a nice discussion of this lovely problem in the intersection of ring theory and algebraic geometry can be found in Y. Berest, G. Wilson, "Differential isomorphism and equivalence of algebraic varieties", MR2079372) -Given that, my questions are: -(Question 1): What are the more general known conditions on a symplectic reflection algebra $H_{1,c}(V,\Gamma)$ that imples it is Morita equivalent to $\mathcal{D}(V) \rtimes \Gamma$? -(Question 2): What are the recent developments made in the study of equivalence of rings of differential operators up to Morita equivalence (and in particular Morita equivalent to the Weyl algebra) since Berest, Wilson [op. cit.]? -(Question 3): Etingof in "Cherednik and Hecke algebras of varieties with a finite group action", MR3734656, introduces more general versions of rational Cherednik algebras and discuss the possibility of extending the results in Y. Berest, O. Chalykh, Quasi-invariants of complex reflection groups ,MR 2801407, in this setting. So, being optmistic, one hipothetically could obtain results similar as those discussed in Berest, Etingof, Ginzburg [op. cit] regarding Morita equivalence of these generalized rational Cherednik algebras with smash products of rings with differential operatos with a finite groups. Has this line of inquiry lead to results relevant to this discussion so far? -(Question 4): This is totally unrelated to the previous questions. It is more of a very open question in ring theory: are there interesting simple Noetherian algebras, coming from another areas than those above, which are Morita equivalent to a Weyl algebra or a smash product of it with a finite group? - -REPLY [2 votes]: I do not know much on recent developments related to the first three questions asked. However, i know of some old results related mainly to the fourth question: -If $A_1$ is the Weyl algebra over an alg closed field of zero char, with the two generators denoted $p,q$ and $I$ is a non-zero, right ideal, then $M_2(End_{A_1}(I))\cong M_2(A_1)$ and $A_1$ is Morita equivalent to $End_{A_1}(I)$. Furthermore, these algebras are not generally isomorphic: Pick for example $I=p^2A_1+(pq+1)A_1$. Its endomorphism ring is isomorphic to $\{x\in Q|xI\subseteq I\}$, where $Q$ is the quotient division ring of $A_1$. This is not isomorphic to $A_1$ but it is Morita equivalent to it. If you are interested in this example, this is presented in An example of a ring Morita equivalent to the Weyl algebra $A_1$, S.P. Smith, J. of Alg, 73, 552 (1981). -Another result which may be of interest -regarding your fourth question- is that: - -If the semigroup $k\Lambda$ has the same quotient field with $k[t]$, then $D(K)$ is Morita equivalent to $A_1$. - -Here $K$ stands for certain subalgebras of $k[t]$ and $D(K)$ for the ring of differential operators on $K$. This is shown in: Some rings of differential operators which are Morita equivalent to $A_1$, Ian Musson, Proc. of the Am. Math. Soc., 98, 1, 1986 -Finally, if you are interested in examples involving smash products with finite group algebras, i do not have some readily available but i think it is natural to look for such in the graded version of Morita equivalence. -I hope these are of some interest to the OP. Sorry in advance if these are too old and you are already aware of them. -P.S.: One more thing which might be of some interest with respect to the second question: The article Rings graded equivalent to the Weyl algebra, J. of Alg., vol. 321, 2, 2009, generalizes some results of Y. Berest, G. Wilson and Stafford, in the setting of graded module categories. (also i think this article is the first -although i am not sure- which introduces the terminology "graded Morita equivalence")<|endoftext|> -TITLE: An explicit negative solution to the Lüroth problem for non-algebraically closed fields -QUESTION [7 upvotes]: Let $\mathsf{k}$ be a field of characteristic $0$, and consider $\mathsf{k}(x,y)$. -If $\mathsf{k}$ is algebraically closed, then every field $L$ such that the inclusion $\mathsf{k} \subset L \subset \mathsf{K}(x,y)$ holds is a purely transcendental extension of the base field (i.e., Castelnuovo's Theorem implies a positive solution to the Lüroth problem in two dimensions). -Now suppose that $\mathsf{k}$ is not algebraically closed. -Question: can we have a finite group $G$ of field automorphisms of $\mathsf{k}(x,y)$, fixing $\mathsf{k}$, such that $\mathsf{k}(x,y)^G$ is not a purely transcendental extension of $\mathsf{k}$? -I am looking for an explicit example of $G$ such that $\mathsf{k}(x,y)^G$ is not rational. - -REPLY [7 votes]: According to the first paragraph in Shafarevich's paper "On Luroth's problem" (found here http://www.math.ens.fr/~benoist/refs/Shafarevich.pdf) the field of rational functions on the surface $z^2+y^2=x^3-x$ over $\mathbb{R}$ is an example of a non-rational field $F$, containing $\mathbb{R}$ of transcendence degree $2$, that embeds in $\mathbb{R}(u,v)$ (fixing $\mathbb{R}$). I don't know whether or not this embedding can be chosen so that $\mathbb{R}(u,v)/F$ is Galois.<|endoftext|> -TITLE: Cusp forms with integer Fourier-coefficients -QUESTION [9 upvotes]: Let $S_k(\Gamma_1(N))(\mathbb{Z})$ be the set of modular forms of weight $k$ and level $N$ with integer Fourier coefficients. Then is true that any cusp form can be written as $\mathbb{Q}$ linear combination of Hecke eigenforms with integer coefficients ? - -REPLY [10 votes]: No. Take $k = 24$ and $N = 1$. Then $\Delta^{2} = q^{2} - 48q^{3} + 1080q^{4} + \cdots \in S_{K}(\Gamma_{1}(N),\mathbb{Z})$. However, if we write $\Delta^{2} = c_{1} f_{1} + c_{2} f_{2}$, where $f_{1}$ and $f_{2}$ are the Hecke eigenforms (with coefficients in $\mathbb{Q}(\sqrt{144169})$) then (if we order $f_{1}$ and $f_{2}$ appropriately), $c_{1} = \frac{1}{24 \sqrt{144169}}$ and $c_{2} = -\frac{1}{24 \sqrt{144169}}$.<|endoftext|> -TITLE: Non-algebraic holomorphic maps between algebraic curves -QUESTION [14 upvotes]: Let $V$ be a connected smooth complex projective curve of negative Euler characteristic. Can there exist a connected smooth complex algebraic curve $U$ such that there is a non-constant holomorphic map $U\to V$ but no non-constant holomorphic map from the compactification of $U$ to $V$? Note that we are not merely asking that the map $U\to V$ does not extend. -EDIT: the question considers the smooth compactification of $U$. - -REPLY [3 votes]: Just turning my comments into an answer: -Following the OP, let $V$ be a smooth projective connected curve with negative Euler characteristic (i.e., genus at least two) over $\mathbb{C}$. Then $V$ is hyperbolic in the sense that Kobayashi's pseudometric is a metric. In particular, by a theorem of Kwack, it is "Borel hyperbolic", i.e., for every reduced finite type scheme $S$ over $\mathbb{C}$, every holomorphic map $S^{an}\to V^{an}$ is algebraic. -This implies that the answer to the OP's question is no. Indeed, let $\varphi:U^{an}\to V^{an}$ be a non-constant holomorphic map with $U$ a smooth curve. Then there is a morphism of varieties $f:U\to V$ such that $f^{an} = \varphi$. Such a morphism extends to a morphism $\overline{U}\to V$ by the valuative criterion for properness.<|endoftext|> -TITLE: Homeomorphisms and "mod finite" -QUESTION [11 upvotes]: Suppose $f:C\to C$ is a homeomorphism, where $C=\{0,1\}^{\mathbb N}$ is Cantor space. -Suppose $f$ preserves $=^*$ (equality on all but finitely many coordinates). Does it follow that $f$ also reflects $=^*$? -That is, if -$$x=^* y \implies f(x)=^* f(y)$$ -does it follow that: -$$x=^* y \iff f(x)=^* f(y)?$$ -Using Axiom of Choice we can show that a bijection of $C$ that preserves $=^*$ need not reflect $=^*$, so the continuity assumptions are not superfluous. - -REPLY [12 votes]: Define $f : C \to C$ by the formula -$$ f(x) = x_0 \cdot (x \oplus \sigma(x)) $$ -where $\cdot$ is word concatenation, $\oplus : C \times C \to C$ is coordinatewise xor, and $\sigma(x)_i = x_{i+1}$ is the shift. -Clearly this map is continuous and preserves $=^*$. It is a bijection because you can deduce the preimage one coordinate at a time, which amounts to summing the prefix, $f^{-1}(x)_i = \bigoplus_{j \leq i} x_i$. By compactness $f$ is a homeomorphism. However, $f(0^\omega) = 0^\omega$ and $f(1^\omega) = 10^\omega$, so $f$ does not respect $=^*$. -The idea is that a surjective non-injective one-dimensional cellular automaton forgets a finite amount of ("global") information on every step. I picked the xor CA $x \mapsto x \oplus \sigma(x)$ and also wrote the single bit that's forgotten (in the form of the first coordinate) to make it a homeomorphism. -This is simple enough that you can randomly stumble upon the formula, but turning xor injective this way is actually a pretty important idea in cellular automata theory. To name just one example, Kari's proof of the undecidability of reversibility of cellular automata on free abelian groups of rank $\geq 2$ (i.e. whether the CA is a homeomorphism on the full shift) uses this trick to turn a cellular automaton injective if a Turing machine halts; you are applying xor on a one-dimensional snake and you cut off its head if the machine halts. - -REPLY [10 votes]: Even with a homeomorphism, preserving $=^*$ does not imply reflection. There might be an easy example, I'm not sure; at any rate it follows from a result in topological dynamics (the "absorption lemma" of Giordano–Putnam–Skau) that such examples exist. -Let me elaborate a bit: by results of Giordano–Putnam–Skau, there exists a minimal homeomorphism of $X=\{0,1\}^{\mathbb N}$ which induces the relation that you denote $=^*$ (and which is often denoted $E_0$); there is an explicit construction (via a Bratteli diagram) in a paper of Clemens. By this, I mean that being in the same $\varphi$-orbit is the same as being $=^*$-equivalent. -Denote such a homeomorphism by $\varphi$, fix some point $x \in X$, and let $E$ be the relation obtained from $=^*$ by splitting the class of $x$ in its positive semi-orbit under $\varphi$, and its negative semiorbit. (So, all classes are the same as $=^*$, except that the class of $x$ has been split in two pieces). Giordano–Putnam–Skau prove that there is a homeomorphism $g$ of $X$ such that for all $x,y \in X$ one has -$$x=^*y \Leftrightarrow g(x) \mathrel E g(y).$$ -Since $E$ is contained in $=^*$, yet not equal to it, such a $g$ preserves $=^*$ without reflecting it. -Edited to add: the book "Cantor minimal systems" by Ian Putnam is a good reference for information and bibliographical data about this area. Clemens's paper can be found here: Generating equivalence relations by homeomorphisms.<|endoftext|> -TITLE: Does $\pi_k(M)\neq 0$ implies $\operatorname{ind}(\gamma) < k$? -QUESTION [6 upvotes]: Cross post from MSE. and sorry if this is an obvious question. -Here is a line of proof of Theorem 1.15 from -Brendle, Simon, Ricci flow and the sphere theorem, Graduate Studies in Mathematics 111. Providence, RI: American Mathematical Society (AMS) (ISBN 978-0-8218-4938-5/hbk). vii, 176 p. (2010). ZBL1196.53001. - -Let us fix two -points $p, q \in M$ such that $d(p, q) = \operatorname{diam}(M, g) > \frac{\pi}{2}$. Since $\pi_k(M)\neq 0$, there -exists a geodesic $\gamma : [0,1] \to M$ such that $\gamma(0) = \gamma(1) = p$ and $\operatorname{ind}(\gamma) < k$. - -Q: Why $\pi_k(M)\neq 0 \implies \operatorname{ind}(\gamma) < k$? Is this a general fact? -Note: $\operatorname{ind}(\gamma):=$ Morse index of $\gamma$. - -REPLY [11 votes]: Very roughly speaking: Geodesics $\gamma:[0,1]\to M$ with $\gamma(0)=\gamma(1)=p$ and $\operatorname{ind}(\gamma) -TITLE: Origin and variations of problem on $4xy-x-y$ being square -QUESTION [5 upvotes]: One of the forms in which the Diophantine equation in question can be found in the literature is this: - -Solve the equation \begin{eqnarray}z^{2} = 4xy-x-y \qquad \qquad (\ast)\end{eqnarray} in positive integers $x, y$, and $z$. - -There are some other variants of it here and there. It is usually given once the instructor has touched upon the Jacobi symbol. -My questions about the problem are the following two: - -Do you happen to know where it was that the problem first appeared? If I understand correctly, the variation in which one is asked to establish that $z^{2} = 4xyt^{u}-t^{v}-y$ does not admit solutions in positive integers might have first appeared in a Russian compilation of problems of olympiad caliber... - -What other notable variations of (*) do you know? - - -Please let me thank you in advance for your learned replies. - -REPLY [8 votes]: I found an 1899 reference in Google Books: Mathematical Questions and Solutions from the "Educational Times" edited by D. Biddle, vol. 70, Hodgson, London, 1899. -On page 73: - -(Professor Crofton, F.R.S.) Show that $4mn-m-n$ cannot be (1) a square; nor (2) a triangular number. - -Brief solutions by H. W. Curjel and Allan Cunningham are given. It looks like the question was posed in the February 1, 1898 issue of The Educational Times and Journal of the College of Preceptors (vol. 51, page 87). - -Here are the solutions, which could help with making variant problems. -Solutions (1) by H. W. Curjel, M.A.; (2) by Lt.-Col. Allan Cunningham, R.E. -(1) If -$$4mn - m - n = x^2 \quad \text{or} \quad \frac{1}{2}\{y(y+1)\},$$ -$$(4m-1)(4n-1) = 4x^2 + 1 \quad \text{or} \quad \frac{1}{2}\{(2y+1)^2+1\},$$ -which is impossible, since all the prime factors of $X^2+1$ other than 2 are of the form $4N+1$. -(2) Here -$$4mn - m - n = \frac{1}{4}(4m-1)(4n-1) - \frac{1}{4}.$$ -(i) Let this $ = z^2$ (if possible) [I think that should be $x^2$]. Then -$$(4m-1)(4n-1) = 1+(2x)^2,$$ -which gives a sum of squares $\{1^2+(2x)^2\} = $ product of two numbers of form $(4m-1)$, which is impossible. -(ii) Let $4mn-m-n = \frac{1}{2}(x^2+x)$, a triangular number (if possible). Therefore -$$ (4m-1)(4n-1) = 2x^2 + 2x + 1 = (x+1)^2 + x^2,$$ -a sum of squares, which is impossible, as before.<|endoftext|> -TITLE: Manifolds with $w_1(TM)\cup w_1(TM)=0$ and $w_2(TM)=0$ but $w_1(TM)\neq 0$ -QUESTION [5 upvotes]: For a generic dimension $d$, is there an nonorientable manifold $M$ (i.e. $w_1(TM)\neq 0$) with vanishing $w_1(TM)\cup w_1(TM)$ and $w_2(TM)$, i.e., -$$w_1(TM)\cup w_1(TM)=0, ~~~~~ w_2(TM)=0, ~~~~~w_1(TM)\neq 0?$$ -Here $w_i(TM)$ is the $i^{\text{th}}$ Stiefel-Whitney class of the tangent bundle of the manifold $M$. For $d=2$, the Klein bottle is an example. -If such manifolds exist, what kind of structure do they carry? For example, if $w_1(TM)=0$, and $w_2(TM)=0$, then the manifold can be equipped with a spin structure, and we say it is a spin manifold. I'd like to see what is the corresponding structure in the above more complicated case. - -REPLY [6 votes]: A smooth manifold $M$ admits a pin$^+$ structure if and only if $w_2(M) = 0$, and a pin$^-$ structure if and only if $w_1(M)^2 + w_2(M) = 0$; see this page for some information on pin structures. The manifolds you are enquiring about satisfy both conditions and hence admit pin$^+$ and pin$^-$ structures. However, as $w_1(M) \neq 0$, they are non-orientable, so they do not admit spin structures. -There exist manifolds of arbitrary dimension which satisfy your requirements. For example, $K\times S^n$ where $K$ is the Klein bottle.<|endoftext|> -TITLE: Equivariant cohomology of a semisimple Lie algebra -QUESTION [5 upvotes]: Suppose $\mathfrak{g}$ is a real Lie algebra integrating to the connected Lie group $G$. One may consider the $G$-equivariant cohomology of $\mathfrak{g}$ ($\mathfrak{g}^*$) where the $G$-action is the adjoint (coadjoint) representation. That is, the cohomology induced by the cochain complex of $G$-invariant differential forms on $\mathfrak{g}$. (Or, more generally, $G$-invariant $V$-valued forms where $V$ is the target of a linear representation of $G$). When $G$ is compact, it follows from a theorem of Chevalley and Eilenberg that this cohomology vanishes in all degrees. I am however interested in the semisimple case. What is known in general about this cohomology and its vanishing? What is known for specific semisimple Lie algebras? Can this cohomology be tied to the standard Lie algebra cohomology? - -REPLY [3 votes]: It vanishes as well. The complex of differential forms on $\mathfrak{g}$ is null homotopic through the standard null-homotopy coming from the linear homotopy to the origin. Since the $G$-action preserve this null-homotopy, it is $G$-equivariantly null and so its $G$-invariants subcomplex is null as well.<|endoftext|> -TITLE: Totally geodesic submanifolds of complex Grassmannians -QUESTION [9 upvotes]: It is known that the connected totally geodesic complex submanifolds of a projective space ${\rm P}V$ equipped with a Fubini-Study metric are precisely the projective subspaces ${\rm P}Z$, where $Z \subseteq V$ is a complex subspace. Direct implication is obvious, and if for the converse we assume that $N\subseteq {\rm P}V$ satisfies the assumptions, we fix $L \in N$ and a basis $H_1,\ldots, H_k$ for $T_LN \subseteq T_L({\rm P}V)\cong {\rm Hom}(L,L^\perp)$, and let $Z = L \oplus \bigoplus_{i=1}^k H_i[L]$. -For Grassmannians, it is again easy to see that every ${\rm Gr}_k(Z)\subseteq {\rm Gr}_k(V)$ is connected totally geodesic and complex, but the same argument for the converse seems to fail. One reason being that if $N\subseteq {\rm Gr}_k(V)$ is a Grassmannian, then $k$ must divide the dimension of $N$. -Looking around I have found papers here and there discussing particular cases where $\dim V = 4$ and $k=2$, but I'd like to know if there's any reference discussing this more general situation. -Thanks. - -REPLY [4 votes]: There is an extensive literature on totally geodesic submanifolds of symmetric spaces. A good place to start to read about this (and references to the preceeding literature) would be, for example, -Bang-yen Chen and Tadashi Nagano, Totally geodesic submanifolds of symmetric spaces, I, Duke Math Journal 44 (1977), 745–755. -They discuss various complex Grassmannians and their totally geodesic submanifolds in particular examples.<|endoftext|> -TITLE: How to learn a continuous function? -QUESTION [8 upvotes]: Let $\Omega \subset \mathbb{R}^m$ be an open subset bounded with a smooth boundary. -Problem : Given any bounded continuous function $f:\Omega\to\mathbb{R}$, can we learn it to a given accuracy $\epsilon$? ($\epsilon>0$). -Definition : What do you mean by learning a function to a given accuracy $\epsilon$? -Using samples of $f$, at sufficiently large but finite number of data points that are drawn randomly(iid) from the set $\Omega$ (under a uniform probability distribution), and using a sufficiently large but finite number of registers whose precision (arithmetic) is sufficiently large but finite (this finite precision is an important condition), should be able to compute a function $F$ with only a finite number of computations (they could be additions, multiplications, and divisions but performed using registers of finite precision) such that $\|f-F\|_{L^\infty(\Omega)} \le \epsilon$. -compute a function $F$ : Given any query point $x$, one should give out $F(x)$. -Conjecture: There exists a method of learning such that one can derive a bound on required precision $p$ that depends only on $\Omega$ and $\epsilon$ and is independent of $f$. -Question: Has anyone formulated this problem before (any reference). Has anyone solved it? If I solve it, what is its market value? (mathematics market) -PS: solving means coming up with a method to learn such functions in the defined way. -(please feel free to tag appropriately) - -REPLY [3 votes]: Counterexample: sin(1/x) over (0,1) -Learning the function near 0 requires infinitely many samples.<|endoftext|> -TITLE: Recover the characteristic of $k$ from the category of $k$-varieties -QUESTION [14 upvotes]: Can you recover the characteristic of a perfect field from the category of smooth projective geometrically connected varieties over it? - -REPLY [12 votes]: Correction. As correctly noted by Remy van Dobben de Bruyn, there is a mistake in Lemma 4. What follows is a corrected argument, with the original (mistaken) post appended below the corrected argument. -Let $k$ be a perfect field. Denote by $\mathbf{V}$ the category of $k$-schemes that are smooth, projective and geometrically connected. -Proposition. A non-final object $Y$ of $\mathbf{V}$ is a curve if and only if the natural action of the symmetric group $\mathfrak{S}_2$ on $Y\times_{\text{Spec}\ k} Y$ admits a categorical quotient in $\mathbf{V}$. -Proof. For a curve, the geometric quotient is smooth, hence it is an object of $\mathbf{V}$. -Let $Y$ be a non-final object that is not a curve. Denote the dimension by $n\geq 2$. In the category of $k$-schemes, there is a categorical quotient $Y_2$ (even a geometric quotient) of the action of $\mathfrak{S}_2$ on $Y\times_{\text{Spec}\ k}Y$, and it is singular along the image of the diagonal, i.e., the embedding dimension is strictly larger than $2n$. -There is a closed immersion of $Y_2$ into a projective space, and this is an object of $\mathbf{V}$. Thus, if there is a categorical quotient $Z$, it factors this closed immersion. In particular, the morphism from $Y_2$ to $Z$ is also a closed immersion of $k$-schemes. -Since the embedding dimension of $Y$ is strictly larger than $2n$, also $Z$ has dimension strictly larger than $2n$. Thus, the closed immersion is not surjective on closed points. For any closed point of $Z$ that is not in the image of $Y_2$, consider the blowing up $\widetilde{Z}$ of the categorical quotient at that closed point. Since the field is perfect, $\widetilde{Z}$ is still a smooth $k$-scheme (this can fail for the blowing up at a closed point with inseparable residue field). -The closed immersion from $Y_2$ to $Z$ factors through the morphism $\widetilde{Z}\to Z$. Of course the identity map on $Z$ does not factor through $\widetilde{Z}$. This contradicts that $Z$ is a categorical quotient. QED -Now we repeat the last part of the argument from the original post. -Lemma. A $k$-curve $Y$ in $\mathbf{V}$ is isomorphic to $\mathbb{P}^1_k$ if and only if every $k$-curve in $\mathbf{V}$ admits a nonconstant morphism to $Y$. -Proof. This follows from the fact that $\mathbb{P}^1_k$ admits no nonconstant morphism to a curve of positive genus, and every $k$-curve of genus $0$ with a $k$-point is isomorphic to $\mathbb{P}^1_k$. QED -Corollary. The field $k$ is uniquely determined by the category $\mathbf{V}$. -Proof. The proposition and the lemma together establish that there is a characterization of the object $\mathbb{P}^1_k$ in $\mathbf{V}$ in purely categorical terms. The automorphism group of $(\mathbb{P}^1_k,0,\infty)$ is $k^\times$. The automorphism group of $(\mathbb{P}^1,\infty)$ is a semidirect product of $k^\times$ and the additive group $k$. With the structure of both the multiplicative group of $k$ and the additive group of $k$, we can recover the field $k$. QED -Original post. There is a mistake in Lemma 4, as pointed out by Remy van Dobben de Bruyn. -I am writing an answer to address the case of a field $k$ that is not necessarily algebraically closed. Denote by $\mathbf{V}$ the category of $k$-schemes that are smooth, projective and geometrically connected. -Lemma 1. The $k$-scheme $\text{Spec}\ k$ is a final object in $\mathbf{V}$. -Proof. In fact this is a final object in the larger category of all $k$-schemes. QED -Definition 2. A morphism in $\mathbf{V}$ is constant if it factors through a morphism from the domain to a final object. -Definition 3. For an object $Z$ of $\mathbf{V}$, an ordered pair $(f,g)$ of nonconstant morphisms, $f:X\to Z$ and $g:Y\to Z$, is cofinite if every ordered pair $(u,v)$ of morphisms $u:W\to X$, $v:W\to Y$ with $f\circ u$ equal to $g\circ v$ is a pair of constant morphisms. -Lemma 4. A non-final object of $\mathbf{V}$ is a $k$-curve if and only if there exists no cofinite pair of nonconstant morphisms to the object. -Proof. This is a straightforward application of Bertini-type theorems. QED -Lemma 5. A $k$-curve $Y$ in $\mathbf{V}$ is isomorphic to $\mathbb{P}^1_k$ if and only if every $k$-curve in $\mathbf{V}$ admits a nonconstant morphism to $Y$. -Proof. This follows from the fact that $\mathbb{P}^1_k$ admits no nonconstant morphism to a curve of positive genus. QED -Proposition 6. The field $k$ is uniquely determined by the category $\mathbf{V}$. -Proof. The lemmas establish that there is a characterization of the object $\mathbb{P}^1_k$ in $\mathbf{V}$ in purely categorical terms. The automorphism group of $(\mathbb{P}^1_k,0,\infty)$ is $k^\times$. The automorphism group of $(\mathbb{P}^1,\infty)$ is a semidirect product of $k^\times$ and the additive group $k$. With the structure of both the multiplicative group of $k$ and the additive group of $k$, we can recover the field $k$. QED<|endoftext|> -TITLE: How many squares can be formed by using n points? -QUESTION [13 upvotes]: How many squares can be formed by using n points on a 3 dimensional space? - -Like using 4 points, there is 1 square be formed -Using 5 points, still 1 square -Using 6 points, 3 squares can be formed - -REPLY [16 votes]: In the plane $n$ points can determine at most $O(n^2)$ squares. This is because any two distinct points can determine up to three squares. -In $R^3$ this argument no longer holds, since two points can form the corners of arbitrarily many squares. As Gerhard points out, $O(n^3)$ is an upper bound (in any dimension) given that three points determine at most one square. -One can do a bit better than this. Using the Szemerédi–Trotter theorem one can show that a set of $n$ points in $R^3$ determines at most $O(n^{7/3})$ right triangles. It follows that $n$ points determine at most $O(n^{7/3})$ squares (since a square will contain a right triangle). On the other hand, it is certainly easy to see that there exists point sets with at least $\Omega(n^2)$ squares. The bound of $O(n^{7/3})$ is known to be sharp for the less constrained problem of counting right triangles. -Update: A result of Sharir, Shefer and Zahl shows that the number of mutually similar triangles in a point set in $R^3$ is at most $O(n^{\frac{15}{7}})$, where $15/7 = 2.142\ldots$, which implies the same bound for the number of squares. -Closing the gap for squares, however, seems an interesting and non-trivial problem. - -REPLY [6 votes]: In k dimensions, take a regular unit k simplex on k points, and copy it an orthogonal distance of 1. This results in k choose 2 unit squares on 2k points. I invite others to count square arrangements in a hypercube. -Combinatorially, there can be no more squares than three sets of an n set. Indeed, since three points of a square determine the fourth, there are at most a fourth as many squares possible as three sets. I imagine Erdos may have an upper bound for planar arrangements, which should be on the order of n^2, since any two points in the plane determine one of three squares containing those two points. -Gerhard "Dots And Spots And Knots..." Paseman, 2020.07.05.<|endoftext|> -TITLE: density of singular K3 surfaces -QUESTION [5 upvotes]: By singular K3 I mean a smooth complex K3 with Néron-Severi rank equal to 20. -Are singular K3 surfaces dense in the moduli space of polarized K3 surfaces? - -REPLY [7 votes]: A reference for the density of singular K3 surfaces in the period domain (see Will Sawin's answer) is -Piatetski-Shapiro, Ilya I.; Shafarevich, I. R., Arithmetic of K3 surfaces, Trudy Mezhdunarod. Konf. Teor. Chisel, Moskva 1971, Tr. Mat. Inst. Steklova 132, 44-54 (1973) ZBL0293.14010. -In particular, this is a screenshot of page 54:<|endoftext|> -TITLE: Does the étale topos determine the Hodge numbers? -QUESTION [5 upvotes]: Does the small étale topos of a smooth proper variety over a perfect field of positive characteristic determine its Hodge numbers? We consider it as a Grothendieck topos over the étale topos of the field. - -REPLY [2 votes]: No. See Proposition 2.14 in Canonical models of surfaces of general type in positive characteristic<|endoftext|> -TITLE: Equivalence generated by Jacobian minors -QUESTION [5 upvotes]: Let $f,g:\mathbb{R}^m \to \mathbb{R}^n$ be two smooth functions and let $k$ be a strictly positive integer. Write $f \sim_k g$ if at each point in the domain, the determinants of all $k \times k$ minors of the Jacobians of $f$ and $g$ coincide. This is clearly an equivalence relation; and more generally, one could let $f$ and $g$ be smooth maps of smooth manifolds $X \to Y$, and ask that for each $k$-form $\omega$ on $Y$ the pullbacks $f^*\omega$ and $g^*\omega$ agree as $k$-forms on $X$. -Have such equivalences been studied and given an alternate characterization? Here's the sort of thing I'm looking for: if $k=1$ and we are mapping between Euclidean spaces, $f$ and $g$ must lie in the same orbit of the translation group acting on the codomain. Perhaps there is a similar Lie-theoretic reformulation of $\sim_k$ for maps $X \to Y$ even when $k > 1$? - -REPLY [4 votes]: First, let me point out that the OP's suggested generalization to arbitrary target manifolds $Y$ of asking that $f^*\omega = g^*\omega$ for all $k$-forms on $Y$, is not equivalent to the question about equality of $k$-by-$k$ minors. -To see this, consider the simplest case, $X = Y=\mathbb{R}^1$ and $k=1$. The general $1$-form is of the form $\omega = h(x)\,\mathrm{d}x$ (where $x:\mathbb{R}\to\mathbb{R}$ is the identity function), and $f^*\omega = g^*\omega$ for all $1$-forms $\omega$ is equivalent to -$$ -h\bigl(f(x)\bigr)f'(x) = h\bigl(g(x)\bigr)g'(x) -$$ -for all functions $h:\mathbb{R}\to\mathbb{R}$, which clearly implies that $f(x) = g(x)$ for all $x\in\mathbb{R}$. Meanwhile, $f^*\mathrm{d}x = g^*\mathrm{d}x$ only implies $f'(x) = g'(x)$, which is the same as requiring that all the $1$-by-$1$ Jacobian minors of $f$ and $g$ are equal, which does imply that they differ by an additive constant. -Instead, I think that the natural generalization is that one endows the $n$-manifold $Y$ with a coframing, i.e., a basis $\omega = (\omega^1,\ldots,\omega^n)$ of the $1$-forms on $Y$ (which, of course, requires that $Y$ be parallelizable, in fact, $\omega:TY\to\mathbb{R}^n$ defines a linear isomorphism $\omega_y:T_yY\to\mathbb{R}^n$ for each $y\in Y$) and then require that -$$ -f^*(\omega^{i_1}\wedge\cdots\wedge\omega^{i_k}) -= g^*(\omega^{i_1}\wedge\cdots\wedge\omega^{i_k}) -$$ -for all $i_1 -TITLE: Name for a matrices having a specific property -QUESTION [6 upvotes]: is there an established name for the property that a square matrix can be made symmetric by permutation of its columns? -Is it possible to recognize those kind of matrices efficiently? - -REPLY [6 votes]: Here is a suggestion (not an answer) for the second question, at least for real matrices $A$: Suppose that there is a permutation matrix $P$ such that $(AP)^{T} = AP.$ -Then $(AP)^{2} = (AP)(AP)^{T} = AA^{T}$, so that $AP$ is a symmetric square root of the positive semidefinite (symmetric) matrix $AA^{T}$. If the non-zero eigenvalues of $AA^{T}$ are all of algebraic multiplicity one, then there are $2^{r}$ real symmetric square roots of $AA^{T}$, where $r$ is the rank of $A$. This is basically a matter of finding an orthonormal basis of (real) eigenvectors for $AA^{T}$. -In view of comments, let me explain further: For exposition's sake, consider the case where $A$ has full rank $n$ and $AA^{T}$ has no non-zero eigenvalue of multiplicity greater than one. After finding an orthonormal basis of (real) eigenvectors for $AA^{T}$, we have an orthogonal real matrix $U$ such that $UAA^{T}U^{T}$ is diagonal. Then $UAA^{T}U^{T}$ has $2^{n}$ symmmetric square roots, all of which are diagonal. If $Q$ is one of these, then $Q^{\prime} = U^{T}QU$ is a symmetric square root of $AA^{T}$, and each symmetric square root of $AA^{T}$ arises in this way. Hence there is such a permutation matrix $P$ with $AP$ symmetric if and only if one (or more) of the $Q^{\prime}$ as above is such that $A^{-1}Q^{\prime}$ is a permutation matrix. -If $A$ has rank $r -TITLE: Motivation for the Jacobian Variety -QUESTION [12 upvotes]: I've been to many talks in Number Theory and for some reason I've yet to fully grasp, we all seem to like Jacobian Varieties a lot. I know that they are Abelian varieties, which give information about their respective curve, but I'm not sure what information exactly. I know of the analytic description of the Jacobian, but I'm still not exactly sure why the Jacobian is so studied. -In his AMS article, What is a motive Barry Mazur seems to suggest that Jacobians encapsulate all cohomology theories. Is this true? How can I see this? - -REPLY [4 votes]: As outlined by the other answers, the Jacobian $J_X$ of a curve $X$ defined over $\mathbb{F}_q$ indeed encapsulates all cohomology information of $X$. In particular one can read the zeta function $\zeta_X$ directly on $J_X$: the numerator of $\zeta_X$ is simply the (reciprocal) polynomial of the Frobenius $\pi_q$ acting on $J_X$. -In particular André Weil's original proof of the Hasse-Weil bound for curves used Jacobians (implicitely). That was a big motivation in his Foundations of algebraic geometry: the algebraic construction of Jacobians over any field. -By the way over $\mathbb{C}$ the Abel-Jacobi map shows that the Jacobian of $X$ is intimately related to the study of abelian integrals. I think historically that was the prime motivation to study Jacobians. A fun fact is that modular functions coming from hyperelliptic integrals can be used to solve algebraic equations. Cf the appendix of Mumford's TATA2.<|endoftext|> -TITLE: Is there a bijective proof of an identity enumerating independent sets in cycles? -QUESTION [10 upvotes]: Let $C_m$ be the cycle with $m$ vertices, defined so that $C_1$ has a self-loop on its unique vertex. Let $p_m$ be the generating function enumerating the number of ways to choose $k$ vertices in $C_m$ so that no two are adjacent. Thus the coefficient of $z^k$ in $p_m(z)$ is the number of independent sets in $C_m$ of size $k$. -For instance, $p_1(z) = 1$, $p_2(z) = 1+2z$, $p_3(z) = 1+3z$, $p_4(z) = 1+4z + 2z^2$, $p_5(z) = 1 + 5z+5z^2$ and $p_6(z) = 1 + 6z + 9z^2 + 2z^3$. Set $p_0 = 2$. -It is not hard to show by algebraic arguments (related to the theory of Chebyshev polynomials) that if $\ell, m \in \mathbb{N}_0$ with $\ell \ge m$ then -$$p_\ell p_m = p_{\ell+m} + (-1)^m z^{m} p_{\ell-m}.$$ -In particular, $p_m^2 = p_{2m} + 2(-1)^m z^{m}$, and so if $k < m$ then the coefficients of $z^k$ in $p_m^2$ and $p_{2m}$ are equal. I would like a bijective proof of this, or ideally, of the more general identity above. - -Is there a bijective proof that if $k < m$ then the number of independent sets of size $k$ in the disjoint union $C_m \sqcup C_m$ is equal to the number of independent sets of size $k$ in $C_{2m}$? - -REPLY [7 votes]: It seems that I’ve seen this question here before, but I am not sure whether it had a bijective answer. Anyway, here you are. -Enumerate the vertices in two copies of $C_m$ as $1,2,\dots,m$ and $1’,2’,\dots,m’$, respectively. Take any independent set of size $k -TITLE: Smooth map homotopic to Lie group homomorphism -QUESTION [11 upvotes]: Let $G$ and $H$ be connected Lie groups. A Lie group homomorphism $\rho:G\to H$ is a smooth map of manifolds which is also a group homomorphism. -Question: Can we find a smooth (or real-analytic) map $f:G\to H$ which is not homotopic to any Lie group homomorphism? -For example, if $G=H=S^1$, it seems the answer is no. For simplicity, we may begin with the same question but assuming some extra conditions, such like (i) $G,H$ are torus, (ii) $G,H$ are compact, etc. - -REPLY [8 votes]: As Igor shows, every endomorphism of a simple Lie group $G$ has degree $\in\{0,\pm 1\}$. -On the other hand, every compact Lie group admits self maps of other degrees. Namely, the $k$-th power map $g\mapsto g^k$ has degree $k^r$, where $r$ is the rank of the group. So, each $k$ with $|k|\geq 2$ gives an example of a smooth map which is not homotopy equivalent to a homomorphism. -One way to compute the degree of the $k$-th power map is as follows. First, we can find an element $g\in G$ which lies in a unique maximal torus $T^r$ and which is also a regular value of the $k$-th power map. The uniqueness of the maximal torus implies that all $k$-th roots of $g$ lie in $T^r$, so this reduces the degree calculation to $T^r$, where it is obvious.<|endoftext|> -TITLE: Is new $n$-conjecture as follows correct? -QUESTION [9 upvotes]: Given a positive integer $P>1$, let its prime factorization be written as$$P=p_1^{a_1}p_2^{a_2}p_3^{a_3}\cdots p_k^{a_k}.$$ -Define the functions $h(P)$ by $h(1)=1$ and $h(P)=\min(a_1, a_2,\ldots,a_k).$ -Is new the $n$-conjecture, formulated as follows, correct? -Conjecture: if ${P_1,P_2,...,P_n}$ are positive integer and pairwise coprime, then, -$$\min\{h(P_1), h(P_2),...,h(P_n), h(P_1+P_2+...+P_n)\} \leq n+1.$$ -I proposed the case $n=2$ two years ago here (Is the conjecture A+B=C following correct?). Now I reformulate that question as follows: - -Let ${P_1,P_2}$ are coprime, then: -$$\min\{h(P_1), h(P_2), h(P_1+P_2)\} \leq 3$$ - -REPLY [9 votes]: Such attempted generalizations of ABC to four or more variables -often fail to specializations of the identity -$$ -(x^2+xy-y^2)^3 + (x^2-xy-y^2)^3 = 2 (x^6 - y^6). -\label{1}\tag{*} -$$ -One can use elliptic curves to make both $x^2 + xy - y^2$ and $x^2 - xy - y^2$ -"powerful" (of the form $A^2 B^3$), which makes each of the four terms -$(x^2+xy-y^2)^3$, $(x^2-xy-y^2)^3$, $2x^6$, $2y^6$ have $h=6$ -but for a stray factor of $2$ which should not matter in the context of -the ABC conjecture. For example, the pairwise prime numbers -$a,b,c,d$ below satisfy $2a^6 + b^6 + 61^9 c^6 = 2d^6$. -Here $d$ is even but $a$ is odd, so $2a^6$ has a "stray factor of $2$", -and the expansion to $a^6 + a^6 + b^6 + 61^9 c^6 = 2d^6$ loses -pairwise coprimality; so either way we don't quite get a counterexample. -Still, this suggests that generalizations of ABC to four or more variables -can run afoul of identities such as \eqref{1}. -(It is "well known" that the Mason-Stothers theorem forbids the disproof -of ABC itself by such an identity.) -a = 1022288301691921314835532892967014277786302791344455107816139963763145069687359424810667270039489345929029301393007247303344511065237 -b = 4005821025365458069945118311017282675402206671149976403498624129498574702167905733126870212117037684063261425637225699359421949547271 -c = 10621830276852061412855232703108032231130723932854745057900981539571749281974534306702514113168069346943754838515856358759614674721 -d = 4676830625123658957500070687744472849236478810555581279857582626862200039312130562436302012081022720213179152015505627679021327325170<|endoftext|> -TITLE: Is the number of solutions of $\phi(x)=n!$ bounded? If yes, what is its bound? -QUESTION [5 upvotes]: Pillai showed in 1929 that the function $A(n)$ giving the number solutions of the equation $\phi(x)=n$ is unbounded in (S. Pillai, On some functions connected with $\varphi(n)$, Bull. Amer. Math. Soc. 35 (1929), 832–836). I'm interested to know about bounds on solutions of $\phi(x)=n!$ which is assigned A055506 in OEIS, where it is claimed that if $\phi(x) = n!$, then $x$ must be a product of primes $p$ such that $p - 1$ divides $n!$. It is unclear to me if this allows me to prove that there are finitely many solutions of the equation $\phi(x) = n!$. Probably an equivalent question is to ask: is the number of solutions of $\phi(x)=n!$ bounded? If yes, what is its bound ? -Related question: -https://math.stackexchange.com/q/3747571/156150 - -REPLY [3 votes]: We have -$$\frac n{\varphi(n)}=\prod_{p\mid n}\bigl(1-p^{-1}\bigr)^{-1} -\le2\prod_{\substack{p\mid n\\p\ge3}}\frac32 -=2\prod_{\substack{p\mid n\\p\ge3}}3^{\log_3(3/2)} -\le2\prod_{\substack{p\mid n\\p\ge3}}p^{\log_3(3/2)} -\le2n^{\log_3(3/2)}$$ -(where $p$ runs over primes), hence -$$\varphi(n)\le m\implies n\le(2m)^{(1-\log_3(3/2))^{-1}}=(2m)^{\log_23}.$$ -Using a larger cut-off $k$ in place of $3$, the same argument gives -$$\varphi(n)\le m\implies n\le(c_km)^{\log_{k-1}k},$$ -where -$$c_k=\prod_{p -TITLE: Who was Bickart? -QUESTION [6 upvotes]: The term "Bickart points" is often used for the foci of the Steiner circumellipse of a triangle. Who was Bickart, and what was the first publication to use the term? - -REPLY [10 votes]: Sur l’hypocycloïde à trois rebroussements (1913) gives an early reference to L. Bickart (Revue de Mathématiques spéciales, 1908). -This reference says M. Bickart, but that "M." stands for "monsieur", the initial is "L." -Here is one paper by L. Bickart on this geometric topic. -I cannot quite recover the circumellipse construction from this paper, but the constructions seem sufficiently similar to support the conclusion that this is the right Bickart.<|endoftext|> -TITLE: A nontrivial principal bundle which satisfies Leray-Hirsch theorem -QUESTION [12 upvotes]: What is an example of a nontrivial principal bundle whose fibre space $G$, total space $P$ and base space $M$ are compact connected manifolds (the fiber $G$ is a compact Lie group) such that $$H^*(P,\mathbb{Q})=H^*(G,\mathbb{Q})\otimes H^*(M,\mathbb{Q})$$ - -REPLY [2 votes]: There is also a low-dimensional example. Consider the canonical map $\mathbb RP^3 \to \mathbb RP^{\infty} \to \mathbb CP^{\infty}$, classifying the unique non-trivial principal $S^1$-bundle with base $\mathbb RP^3$. Its rational Serre spectral sequences collapses. Of course, its integral Serre spectral sequence does not collapse. The total space of this bundle is the 4-manifold $E = S^1 \times_{\mathbb Z/2} S^3$ whose fundamental group is $\mathbb Z$, not $\mathbb Z \times \mathbb Z/2\mathbb Z$.<|endoftext|> -TITLE: Irreducibility of a polynomial when the sum of its coefficients is prime -QUESTION [6 upvotes]: I came up with the following proposition, but don't know how to prove it. -I used Maple to see that it holds when $ a + b + c + d <300 $. -Let $a,b,c$ and $d$ be non-negative integers such that $d\geq1$ and $a+b+c\geq1$. -If $a+b+c+d$ is a prime number other than $2$, the polynomial $ax^3+bx^2+cx+d$ is irreducible over $Z[x]$. - -REPLY [5 votes]: Francesco Polizzi's idea is enough to solve the problem: - -Lemma. Let $f = ax^3 + bx^2 + cx + d \in \mathbf Z_{\geq 0}[x]$ nonconstant with $d > 0$, such that $f(1)$ is a prime $p > 2$. Then $f$ is irreducible in $\mathbf Z[x]$. - -Proof. Suppose $f = gh$ for $g,h \in \mathbf Z[x]$; we must show that $g \in \{\pm 1\}$ or $h \in \{\pm 1\}$. -First assume $\deg g > 0$ and $\deg h > 0$. Since $\deg f \leq 3$, without loss of generality we may assume $\deg g = 1$ and $m = \deg h \in \{1,2\}$; say -\begin{align*} -g = ux + v, & & h = \alpha x^2 + \beta x + \gamma , -\end{align*} -with $u > 0$ and $\alpha > 0$ or $\alpha = 0$ and $\beta > 0$. If $v \leq 0$, then $f$ has a nonnegative real root, which is impossible because $f$ has non-negative coefficients and $f(0) > 0$. Thus, $v > 0$, and therefore $g(1) \geq 2$, forcing $g(1) = p$ and $h(1) = 1$. By symmetry, this rules out the case $\deg h = 1$. Thus, we may assume $\alpha > 0$. The formulas -\begin{align*} -a = \alpha u, & & b = \beta u + \alpha v, & & c = \gamma u + \beta v, & & d = \gamma v -\end{align*} -give -\begin{align*} -\alpha > 0, & & \alpha v \geq -\beta u, & & \gamma u \geq -\beta v, & & \gamma > 0.\label{1}\tag{1} -\end{align*} -We will show that $h(1) \geq 2$, contradicting the earlier assertion that $h(1) = 1$. This is clear if $\beta \geq 0$, since we already have $\alpha > 0$ and $\gamma > 0$. If $\beta < 0$, then \eqref{1} shows -$$\frac{\gamma}{-\beta} \geq \frac{v}{u} \geq \frac{-\beta}{\alpha} > 0,\label{2}\tag{2}$$ -hence -\begin{align*} -h(1) = \alpha + \beta + \gamma &= \left(1 - \left(\frac{-\beta}{\alpha}\right) + \left(\frac{\gamma}{-\beta}\right)\left(\frac{-\beta}{\alpha}\right)\right)\alpha\\ -&\geq \left(1 - \left(\frac{-\beta}{\alpha}\right) + \left(\frac{-\beta}{\alpha}\right)^2 \right)\alpha. -\end{align*} -The function $1-x+x^2$ attains a minimum of $\tfrac{3}{4}$ at $x = \tfrac{1}{2}$, and is bigger than $1$ on $(1,\infty)$. Thus, we only have to consider $x = \tfrac{-\beta}{\alpha} \in (0,1]$. For $x < 1$, we must have $\alpha \geq 2$, so $(1-x+x^2)\alpha \geq \tfrac{3}{2} > 1$. Finally, for $x = 1$, i.e. $\beta = -\alpha$, \eqref{2} shows that $v \geq u$. Since $p = v+u$ is odd, we must have $v > u$, so \eqref{2} gives $\gamma > -\beta$, so $h(1) = \gamma \geq 2$. -Thus, we conclude that $\deg g = 0$ or $\deg h = 0$; say $g = n$ for $n \in \mathbf Z_{>0}$ without loss of generality. If $n > 1$, then $n = p$, which is impossible since $f(0) = d \in \{1, \ldots, p-1\}$ is not divisible by $p$. So we conclude that $g = 1$. $\square$ -Remark. The argument above relies crucially on the assumption $\deg f \leq 3$. More importantly, for higher degree polynomials the same result is false without a lower bound on $p$ that grows at least linearly in $\deg f$: -Example. Let $p$ be an odd prime. Then the polynomial $f = x^{2p-2} + x^{2p-4} + \ldots + x^2 + 1$ has $f(1) = p$, but it factors as -$$\Big(x^{p-1}+x^{p-2}+\ldots+x+1\Big)\Big(x^{p-1}-x^{p-2}+\ldots-x+1\Big).$$ -Indeed, the above reads -$$\zeta_p(x^2) = \zeta_p(x)\zeta_{2p}(x) = \zeta_p(x)\zeta_p(-x),$$ -which is true because both sides are monic and have the same roots in $\mathbf C$ (namely the $2p^\text{th}$ roots of unity except $\pm 1$, all with multiplicity $1$).<|endoftext|> -TITLE: Intuition about ordinal fixed points $\alpha = \aleph_\alpha$ -QUESTION [11 upvotes]: I wanted to ask for your intuition about ordinal fixed points $\alpha = \aleph_\alpha$, where $\aleph_\alpha$ stands for the $\alpha$-th Aleph number in the Aleph sequence of cardinalities. -For background why I am asking this. I was surprised when I first learned $|\mathbb{Q}| = |\mathbb{N}|$ and $|{\cal P}(A)|>|A|$ for all sets $A$, but gained an intuition over time. It took me a bit longer to understand the convergence proof for Goodstein sequences, because initially I did not understand why a strictly decreasing sequence of ordinals is zero after finitely many steps; I had the wrong intuition about well-ordering (I thought "going downstairs" would be symmetric to "going upstairs"). -Now, I am still unable to find the right intuition for ordinal fixed points, especially for the Aleph sequence. I am aware of the fixed-point lemma for normal functions of ordinals from Veblen. But I have not really gained an intuition from knowing the proof. In a sense, I can understand the proof formally only, but not "morally". -In my intuition (which might be wrong), I am starting from -$0\mapsto\aleph_0\\ -1\mapsto\aleph_1\\ -2\mapsto\aleph_2$ -and so on. The difference in size between the ordinal index $\alpha$ and the Aleph number $\aleph_\alpha$ gets enormously big in a very fast way. My intuition is, the index $\alpha$ can never catch up with $\aleph_\alpha$, even when $\alpha$ is a limit ordinal. In my mental picture, a limit ordinal $\alpha$ can be an extremely "high jump", but it can never really catch up with all the extremely high jumps of the Aleph sequence $(\aleph_\alpha)$ which happen in every single step. -Please could you help me find the right intuition, or maybe point me to the error in my current mental picture? I might be overlooking something obvious! - -REPLY [11 votes]: Your intuition is finitary, and therefore wrong. Compare, for example, the two sequences: - -$\alpha_n=n$, and -$\beta_n=2^n$. - -It is easy to see that $\alpha_n<\beta_n$ for all $n$. We even know from elementary calculus that the rate of change between them is growing very fast as well, so there is no possible way for $\alpha_n$ to be equal to $\beta_n$. Game over, then, I suppose. We can all stop reading this and go get a beer. -But wait a minute, you might say, what about their limit? What is $\sup\alpha_n$ and how does it compare to $\sup\beta_n$? Well, both are $\omega$. -See, limit steps are a great opportunity for the slow and steady to catch up with the rapid. As long as the two sequences are continuous and increasing, they will catch up with one another at some limit points. -And so indeed, setting $\alpha_0=\omega$ and $\alpha_{n+1}=\omega_{\alpha_n}$, will have you with $\alpha=\sup\alpha_n=\omega_\alpha$. And therefore $\alpha=\aleph_\alpha$. The key point here is that the sequence is increasing incredibly fast. Enough to catch up at the limit point. And of course we may replace $\alpha_0$ with any ordinal as our starting point.<|endoftext|> -TITLE: Intuition behind stability and instability in model theory -QUESTION [17 upvotes]: In A survey of homogeneous structures by Macpherson (Discrete Mathematics, vol. 311, 2011), a stable or unstable theory is defined as (Definition 3.3.1): - -A complete theory $T$ is unstable if there is a formula $\varphi(\overline{x}, \overline{y})$ (where $\ell(\overline{x}) = r$ and $\ell(\overline{y}) = s)$, some model $M$ of $T$, and $\overline{a_i} \in M^r$ and $\overline{b_i}\in M^s$ (for $i \in \Bbb{N}$) such that for all $i, j \in\Bbb{N}$, -$$M \vDash \varphi(\overline{a_i}, \overline{b_j})\iff i \leqslant j$$ -The theory $T$ is stable otherwise. - -I wonder what is the intuition behind this definition. More specifically, under what situation is it conceived? Why is it called stable theory? Does the name stable related to the notion of stability in geometry or physics? - -REPLY [5 votes]: Here's a topological way to understand stability of a formula. Depending on your background, you might find it intuitive. -Given a formula $\varphi(\bar x,\bar y)$, a model $M$ and a tuple $\bar b$ in $M$ of the same length as $\bar y$, you can define a formula $f_{\varphi,\bar b}\colon M\to \{0,1\}$ by putting $f_{\varphi,\bar b}(\bar a)=1$ if and only if $\varphi(\bar a,\bar b)$ holds. -Now, let us endow $M^{\lvert x\rvert}$ with the weakest topology which makes all these functions continuous. We may attempt to define stability in the following way (which is not yet complete). - -"Definition" (I think this is not equivalent, although I don't have any counterexamples offhand): the formula $\varphi$ is stable if the set $\{f_{\varphi,\bar b}\mid \bar b\in M^{\lvert y\rvert}\}\subseteq C(M^{\lvert x\rvert})$ is relatively weakly compact. - -Unfortunately, the topology on $M^{\lvert \bar x\rvert}$ itself may not be so nice (for starters, it will often not be Hausdorff). Instead, we consider the space $S_{\varphi}(M)$ --- in model theory, this is the space of $\varphi$-types over $M$. Topologically, it is the compactification of $M^{\lvert \bar x\rvert}$ induced by the family $(f_{\varphi,\bar b})_{\bar b\in M}$ (i.e. the closure of the image of $M^{\lvert \bar x\rvert}$ under the map $M\to \{0,1\}^{M^{\lvert y\rvert}}$, where $\bar a\mapsto (f_{\varphi,\bar b}(a))_{\bar b\in M^{\lvert y\rvert}}$). Note that the functions $f_{\varphi,\bar b}$ uniquely extend to continuous functions on $S_{\varphi}(M)$ (namely, the extensions are the projections). -Having this, we may define stability in the following way: - -The formula $\varphi$ is stable when for every $M\models T$, the set $A_{\varphi,M}:=\{f_{\varphi,\bar b}\mid \bar b\in M^{\lvert y\rvert}\}\subseteq C(S_\varphi(M))$ is relatively weakly compact.<|endoftext|> -TITLE: Possible cardinalities of the remainders of compactifications of $\Bbb R$ -QUESTION [10 upvotes]: With the usual topology on $\Bbb R$, a compactification $\mathrm{id}_{\Bbb R}:\Bbb R\to v\Bbb R$ can have a remainder $v\Bbb R \setminus \Bbb R$ of cardinality $1,2, 2^{\aleph_0}=\mathfrak c,$ or $2^{\mathfrak c}.$ The only possibilities less than $\mathfrak c$ are $1,2.$ -Suppose $\mathfrak c^+<2^{\mathfrak c}.$ What possible cardinals between $\mathfrak c$ and $2^{\mathfrak c}$ can be the cardinals of such remainders? -Is there perhaps a Forcing argument that can answer or partly answer this? - -REPLY [8 votes]: Every connected compact Hausdorff space of weight $\aleph_1$ is the remainder $v \mathbb R \setminus \mathbb R$ of some compactification of $\mathbb R$. In particular, $[0,1]^{\aleph_1}$ is the remainder of a compactification of $\mathbb R$, and therefore $\mathbb R$ has a compactification with remainder of cardinality $2^{\aleph_1}$. -Using forcing, one can show that it is consistent to have $\mathfrak{c} < 2^{\aleph_1} < 2^{\mathfrak{c}}$. (For example, Easton's Theorem immediately implies that we may get a model where $2^{\aleph_0} = \aleph_2$, $2^{\aleph_1} = \aleph_3$, and $2^{\aleph_2} = \aleph_4$, although Easton's Theorem is a bit overkill for this.) Thus it is consistent that $\mathbb R$ has a compactification with cardinality in $[\mathfrak{c}^+,2^{\mathfrak{c}})$. -The result about weight-$\aleph_1$ continua is proved by Dow and Hart in this paper. But the special case of $[0,1]^{\aleph_1}$ is actually much easier to prove, using the fact that $[0,1]^{\aleph_1}$ is separable. Let $\{d_1,d_2,d_3,\dots\}$ be a countable dense subset of $[0,1]^{\aleph_1}$. Map $\mathbb R$ into $[0,1] \times [0,1]^{\aleph_1}$ as follows. First map $\mathbb R$ onto the ray $[1,\infty)$, and then map $[1,\infty)$ into $[0,1] \times [0,1]^{\aleph_1}$ by linearly mapping each interval $[n,n+1]$ to the line segment connecting $(\frac{1}{n},d_n)$ to $(\frac{1}{n+1},d_{n+1})$ in $[0,1] \times [0,1]^{\aleph_1}$. This mapping embeds the ray $[1,\infty)$ in $[0,1] \times [0,1]^{\aleph_1}$, and its boundary in this embedding is precisely the set $\{0\} \times [0,1]^{\aleph_1} \approx [0,1]^{\aleph_1}$. -Edit: It is also possible to find a compactification $v \mathbb R$ of $\mathbb R$ such that $|v \mathbb R \setminus \mathbb R|$ has countable cofinality. In fact, I claim that the set -$$T = \{|v\mathbb R \setminus \mathbb R| \,:\, v\mathbb R \text{ is a compactification of } \mathbb R \}$$ -includes all cardinals of the form $2^\kappa$, where $\aleph_0 \leq \kappa \leq \mathfrak{c}$, and all countable limits of such cardinals. So, for example, in a model of set theory where $2^{\aleph_n} = \aleph_{\omega+n+1}$ for all $n$ (which is consistent, by Easton's Theorem), there is a compactification $v \mathbb R$ of $\mathbb R$ such that $|v \mathbb R \setminus \mathbb R| = \aleph_{\omega+\omega}$. - -Lemma: Suppose $X$ is a connected compact Hausdorff space, and $X$ has a dense subspace $D$ that is both separable and path connected. Then there is a compactification of $\mathbb R$ whose remainder is (homeomorphic to) $X$. - -Proof: The main ideas are already present in the third paragraph above. Let $\{d_1,d_2,d_3,\dots\}$ be a countable dense subset of $D$. Map $\mathbb R$ into $[0,1] \times X$ as follows. First map $\mathbb R$ onto the ray $[1,\infty)$, and then map $[1,\infty)$ into $[0,1] \times X$ by linearly mapping each interval $[n,n+1]$ to some path connecting $(\frac{1}{n},d_n)$ to $(\frac{1}{n+1},d_{n+1})$ in $[0,1] \times D$. This mapping embeds the ray $[1,\infty)$ in $[0,1] \times X$, and its boundary in this embedding is precisely the set $\{0\} \times X \approx X$. -My claim above follows almost immediately from this lemma. Each of the spaces $[0,1]^\kappa$, where $\aleph_0 \leq \kappa \leq \mathfrak{c}$, is separable and path connected, and so $|[0,1]^\kappa| = 2^\kappa \in T$ by the lemma. -To get countable limits of such cardinals, fix some infinite cardinals $\kappa_1,\kappa_2,\kappa_3,\dots \leq \mathfrak{c}$. Let $Y$ be the space obtained by gluing the endpoints of an interval to some (any) point of $[0,1]^{\kappa_1}$ and some (any) point of $[0,1]^{\kappa_2}$, gluing the endpoints of another interval to $[0,1]^{\kappa_2}$ and $[0,1]^{\kappa_3}$, gluing the endpoints of another interval to $[0,1]^{\kappa_3}$ and $[0,1]^{\kappa_4}$, and so on. (In other words, $Y$ is obtained by stringing together the $[0,1]^{\kappa_n}$ like beads on a necklace.) Finally, let $X$ be the one point compactification of $Y$. Then $X$ satisfies the hypotheses of the lemma, and $|X| = \sup_n 2^{\kappa_n}$. -Interestingly, this method seems hopeless for getting a compactification of $\mathbb R$ with a remainder of size $\aleph_\omega$. I wonder if this is possible?<|endoftext|> -TITLE: Weighted Co/ends? -QUESTION [5 upvotes]: Recall: Limits -Recall that the limit of a functor $D\colon\mathcal{I}\to\mathcal{C}$ is, if it exists, the pair $(\mathrm{lim}(D),\pi)$ with - -$\lim(D)$ an object of $\mathcal{C}$, and -$\pi\colon\Delta_{\lim(D)}\Rightarrow D$ a cone of $\lim(D)$ over $D$ - -such that the natural transformation -$$\pi_*\colon h_{\lim(D)}\Rightarrow\mathrm{Cones}_{(-)}(D),$$ -is a natural isomorphism, where - -$\mathrm{Cones}_{(-)}(D)\overset{\mathrm{def}}{=}\mathrm{Nat}(\Delta_{(-)},D)$, and -The component at $X\in\mathrm{Obj}(\mathcal{C})$ of $\pi_*$ is the map $(\pi_*)_X \colon \mathrm{Hom}_\mathcal{C}(X,\lim(D))\to \mathrm{Cones}_X(D)$ sending a morphism $f\colon X\to\lim(D)$ to the cone -$$\Delta_X\xrightarrow{\Delta_f}\Delta_{\lim(D)}\to D$$ -of $X$ over $D$. - -Recall: Ends -Now, the end of a functor $D\colon\mathcal{I}^\mathsf{op}\times\mathcal{I}\to\mathcal{C}$ is the representing object of the functor -$$\mathrm{Wedges}_{(-)}(D)\colon\mathcal{C}^\mathsf{op}\to\mathsf{Sets}$$ -with -$$\mathrm{Wedges}_{(-)}(D)\overset{\mathrm{def}}{=}\mathrm{ExtNat}(\overline{\Delta_{(-)}},\overline{D}),$$ -where - -$\overline{D}\colon\mathsf{pt}\times\mathcal{I}^\mathsf{op}\times\mathcal{I}$ is the unique functor restricting to $D$ under the isomorphism $\mathsf{pt}\times\mathcal{I}^\mathsf{op}\times\mathcal{I}\cong\mathcal{I}^\mathsf{op}\times\mathcal{I}$ and similarly for $\overline{\Delta_{(-)}}$, and where -We are now working with extranatural transformations. - -That is, the object $\int_{A\in\mathcal{C}}D^A_A$ of $\mathcal{C}$ such that -$$h_{\int_{A\in\mathcal{C}}D^A_A}\cong\mathrm{Wedges}_{(-)}(D).$$ -Recall: Weighted Limits -We can generalise limits by replacing $\Delta_{(-)}$ with an arbitrary functor $W\colon\mathcal{C}\to\mathsf{Sets}$. This leads to the notion of the weighted limit of $D\colon\mathcal{I}\to\mathcal{C}$ with respect to the weight $W$. This is the object $\lim_W(D)$ of $\mathcal{C}$ for which we have a natural isomorphism -$$h_{\lim_W(D)}(-)\cong\mathrm{Nat}(W,\mathrm{Hom}_\mathcal{C}(-,D)).$$ -Question: Weighted Ends -Just as with weighted limits, we may define the weighted end of a functor $D\colon\mathcal{I}^\mathsf{op}\times\mathcal{I}\to\mathcal{C}$ with respect to a weight $W\colon\mathcal{I}^\mathsf{op}\times\mathcal{I}\to\mathsf{Sets}$ as the object $\int_{A\in\mathcal{C}}^W D^A_A$ of $\mathcal{C}$ (if it exists) such that we have a natural isomorphism -$$h_{\int_{A\in\mathcal{C}}^W(D)}(-)\cong\mathrm{ExtNat}(\overline{W},\overline{\mathrm{Hom}_\mathcal{C}(-,D)}).$$ -(Or rather that a certain natural transformation $W_*$ induced by $W$ is a natural isomorphism. Note that precomposing extranatural transformations with natural transformations gives back an extranatural transformation, so $\mathrm{ExtNat}(\cdots)$ is indeed a functor.) -Now (finally!) for the actual questions: - -This notion seems to be very natural. Has it been considered somewhere in the literature? -Provided that $\mathcal{C}$ has cotensors, we may write any weighted limit on $\mathcal{C}$ as an end. Can we similarly express weighted ends in terms of ends or limits (possibly weighted)? -Are there any natural occuring examples of this notion? -Everything above can be categorified to the setting of bicategories (with pain). Is there anything remarkable about the resulting notion of a "weighted pseudo biend"? - -REPLY [3 votes]: This is a back-of-the-envelope conjecture, but it seems plausible to me that the "weighted coend" of $T : {\cal I}^\text{op}\times {\cal I}\to {\cal V}$ by $W : {\cal I}^\text{op}\times {\cal I}\to {\cal V}$ is no more than the usual coend of the composition $W\diamond T$, regarding both $W,T$ as endo-profunctors of $\cal I$. -Of course, this only works for coends, and $T$ must be valued on the base of enrichment. But hey, that's a start :-) -In this very special case it's very easy to see that the coend of $T$ weighted by $W$ is the same as the coend of $W$ weighted by $T$; thus this seems to behave like a tensor product of functors. -Googling information about "Coends as traces" (there's a wonderful talk by S. Willerton on the topic), and "shadows for bicategories" (see here, slide 37 of 46) might tell you something interesting.<|endoftext|> -TITLE: How to understand adjoint functors? -QUESTION [11 upvotes]: I asked this same question on MathUnderflow two weeks ago but didn't receive any answer. Now that I am thinking more, it feels like the most suitable place for this question is here. -I have a good grasp of all different definitions/interpretations of adjoint functors, but still do not know have to interpret the left or right adjoint of a give functor, when it exist. It would be a easy to explain my question through an example. -For a particular example consider the inclusion of groupoids into small categories: $$\mathcal{F}:\mathcal{Grpd}\hookrightarrow\mathcal{Cat}.$$ This functor has left adjoint $\mathcal{L}$ which freely invert all existing morphisms of a category. Also it has a right adjoint $\mathcal{R}$ which extract the (maximal) subcategory of all isomorphisms, called "core groupoid" of a category. In some scene this bi-adjunction align with the free-forgetful philosophy. Also we can compute these adjoint functors explicitly as pointwise Kan extensions, then $$\mathcal{L}(\mathcal{C})=\text{Ran}_{\mathcal{F}}(\text{id}_{\mathcal{Grpd}})(\mathcal{C})=\lim(\mathcal{C}\downarrow\mathcal{F}\xrightarrow{\Pi_{\mathcal{C}}}\mathcal{Grpd}\xrightarrow{\mathcal{F}}\mathcal{Cat}).$$ But I don't understand how to interpret this "limit" as a localization process and same for the right adjoint $\mathcal{R}.$ Can we derive it from this formula? If not, how can we see it? -In general, If we know some adjoint of a given functor exist, what is the process to understand it's effect/outcome? - -REPLY [6 votes]: Nice question Bumblebee. So, let us start with some "metaphysics of adjointness": - -THE LEFT AND RIGHT ADJOINTS TO A FUNCTOR -$ \mathcal{F}:\mathcal{C}\hookrightarrow\mathcal{D}$ -ARE THE FREE (LEFT) AND CO-FREE (RIGHT) WAYS TO GO BACK FROM $D$ TO -$C$. - -If you choose some easy examples, for instance $C=Top$ and $D=Set$ and the functor is simply the forgetful functor which "forgets" the topological structure, Left and Right start from a given set and endow it with a topology, in the most economic way (trivial topology) or in the most rigid one (discrete) . Same happens if you replace $Top$ with $Groups$ (or any other algebraic category). -Now, not all functors which have adjoints are forgetful functors, so matters are slightly more subtle sometimes, but the "general metaphysics" of adjointness still holds true. -Now the second part of your question, the scary formula for your example: rather that filling this page with calculations, I want to give you the heuristics (so far I have told you what adjoints are, not whether they exist or how they are calculated). -Here, I use the second "metaphysical principle of adjointness", namely this: - -THINK OF CATEGORIES AS GENERALIZED ORDERS, AND OF ADJOINTNESS AS -GENERALIZED GALOIS CONNECTIONS. - -If you look up Galois connections (see here), how they are defined and calculated, you will also understand cats, by generalizing. Same exact story.....<|endoftext|> -TITLE: Probability of sequence of coin flips palindrome -QUESTION [5 upvotes]: I am flipping a coin until the sequence of coin flips is a palindrome.The length is at least 3(flips). What is the mean number of flips I will perform? What is the probability I won't stop? - -REPLY [7 votes]: Let's distinguish some cases. If the first two flips result in HT or TH then with probability 1 you will flip a palindrome. Indeed, this will happen at the time $n$ when the $n$-th flip coincides with the first one. You can easily calculate the expected value of $n$ to be 4. -If the first two flips result in HH or TT then we distinguish two more cases. Either the third flips coincides with the first two or it doesn't. In the former case then you get a palindrome and it happens with probability 1/4. All the intricacy of the problem is in the last case. Let's prove that with positive probability you won't stop (so in the end the expected value is infinity, as already discussed by mike). -WLOG, assume that the first three flips are TTH. The possible palindromes starting with TTH are of the form TTHTT or TTH$w$HTT where $w$ is a palindrome word (possibly empty). If $n=2k$ or $2k+1$ then the probability of flipping a palindrome word of length $n$ is $\frac{1}{2^k}$. Therefore, by a union bound, the probability of flipping a palindrome (given that the sequence starts with TTH) is less than -$$\frac{1}{4} + \left(2\sum_{k=0}^\infty \frac{1}{2^k}\right)\cdot\frac{1}{8}=\frac{3}{4}$$ -All in all, with probability $>\frac{1}{16}$ you will not flip a palindrome (this lower bound is obviously not optimal).<|endoftext|> -TITLE: Is operadic desuspension inverse to operadic suspension? -QUESTION [5 upvotes]: Given a graded vector space $V$ over a field $k$, consider it's suspension $\Sigma V$ such that $(\Sigma V)^i=V^{i-1}$. For an operad of graded vector spaces over a field $\mathcal{O}$, the operadic suspension $\mathfrak{s}\mathcal{O}$ is defined in several different ways depending on the author. Some standard references might be An Alpine Expedition through Algebraic Topology and Operads in Algebra, Topology and Physics. All the definitions I've seen yield isomorphic graded vector spaces, but the operadic structures differ slightly. In the reference above, the operadic structure is not explicitely defined, it is just said to be induced by the one on $\mathcal{O}$, but it seems to be obvious that $\mathfrak{s}^{-1}\mathfrak{s}\mathcal{O}\cong \mathcal{O}\cong \mathfrak{s}\mathfrak{s}^{-1}\mathcal{O}$ as operads (not only as collections of graded vector spaces). -Here I'm interested in the definition given by Benjamin C. Ward in his Thesis (Section 2.1.2), for which I think that property does not hold. -Background definitions -He defines the operadic suspension as -$$\mathfrak{s}\mathcal{O}(n)=\mathcal{O}(n)\otimes\Sigma^{n-1}sign_n$$ -where $sign_n$ is the sign representation of the symmetric group on $n$ letters. The symmetric group action on the graded vector spaces is the obvious diagonal action, and a diagonal operadic composition is given by the following operadic insertion on $\{\Sigma^{n-1}sign_n\}$. We may identify $\Sigma^{n-1}sign_n$ with the exterior power $\bigwedge^n k$, so it is spanned by the element $e_1\wedge\cdots\wedge e_n$. Therefore, define the $i$-th insertion map -$$\circ_i:\Sigma^{n-1}sign_n\otimes\Sigma^{m-1}sign_m\to \Sigma^{n+m-2}sign_{n+m-1}$$ -as the map -$$(e_1\wedge\cdots\wedge e_n)\otimes (e_1\wedge\cdots\wedge e_m)\mapsto (-1)^{(i-1)(m-1)}(e_1\wedge\cdots\wedge e_{n+m-1}).$$ -We may identify the elements of $\mathcal{O}$ with elements of its operadic suspension, so for $a,b\in\mathcal{O}$ we may write $a\tilde{\circ}_i b$ for the insertion in the suspension. We can compute it in terms of $a\circ_i b$ (the insertion in $\mathcal{O}$) in the following way: -$$\tilde{\circ}_i=(\mathcal{O}(n)\otimes\Sigma^{n-1}sign_n)\otimes (\mathcal{O}(m)\otimes\Sigma^{m-1}sign_m)\cong (\mathcal{O}(m)\otimes \mathcal{O}(m))\otimes (\Sigma^{n-1}sign_n\otimes \Sigma^{m-1}sign_m)\to \mathcal{O}(n+m-1)\otimes \Sigma^{n+m-2}sign_{n+m-1}$$ -The Koszul sign rule on the isomorphism produces a sign with exponent $(n-1)\deg(b)$ and then the insertions are performed diagonaly, so after the identification we get -$$a\tilde{\circ}_i b=(-1)^{(n-1)\deg(b)+(i-1)(m-1)}a\circ_i b.$$ -The operadic desuspension $\mathfrak{s}^{-1}\mathcal{O}$ is defined similarly using $\Sigma^{1-n}sign_n$, so the signs are the same. -Problem -I expected $\mathfrak{s}^{-1}\mathfrak{s}\mathcal{O}\cong \mathcal{O}$ as operads, but I think that the insertions are different. If I compute the insertion induced on $\mathfrak{s}^{-1}\mathfrak{s}\mathcal{O}$ in a similar way as above using the isomorphism -$$(\mathcal{O}(n)\otimes\Sigma^{n-1}sign_n\otimes \Sigma^{1-n}sign_n)\otimes (\mathcal{O}(m)\otimes\Sigma^{m-1}sign_m\otimes \Sigma^{1-m}sign_m)\cong (\mathcal{O}(m)\otimes \mathcal{O}(m))\otimes (\Sigma^{n-1}sign_n\otimes \Sigma^{m-1}sign_m)\otimes (\Sigma^{1-n}sign_n\otimes \Sigma^{1-m}sign_m)$$ -Then, the insertion induced on this product is identified with -$$(-1)^{(1-n)(m-1)}a\circ_i b$$ -which is of course not the same as $a\circ_i b$. So, for this new operad created by the suspension and desuspension to be isomorphic to the original one, we must have an automorphism $f$ on $\mathcal{O}$ such that $f(a\circ_i b)=f(a)\circ_i f(b)=(-1)^{(1-n)(m-1)}a\circ_i b$. I think this automorphism must be then of the form $f(a)=(-1)^{\varepsilon(a)}a$, with $\varepsilon(a)=\pm 1$. But this implies that $(-1)^{(n-1)^2}f(a\circ_i a)=(-1)^{2\varepsilon(a)}a\circ_i a=a\circ_i a$, which is not true for all $n$. -Question -Is my conclusion about this suspension true or am I mistaken? I'm not so sure that $f$ really needs to be of that form, but I can't really find an morphism that makes the two structures isomorphic. Is this definition of operadic suspension used by any other author? - -REPLY [6 votes]: What you really need to show is that -$$f(a\circ_ib)=(-1)^{(n-1)(m-1)}f(a)\circ_if(b).$$ -Here, $n$ is the arity of $a$, $m$ is the arity of $b$, and $\circ_i$ is the infinitesimal composition in $\mathcal{O}$ (once you twist the definition of the infinitesimal composition by your sign, you get the usual equation for operad morphisms). You achieve this with -$$f(a)=-(-1)^{\frac{n(n+1)}{2}}a.$$<|endoftext|> -TITLE: Does anyone know a journal like Expositiones Mathematicae, except not by Elsevier? -QUESTION [5 upvotes]: Meaning primarily expository articles aimed at a wide-ranging mathematical audience, at the same technical level. But not by such a predatory publisher. - -REPLY [4 votes]: Some journals publishing only surveys: Bulletin of the American Mathematical Society, Sugaku Expositions, Russian Mathematical Surveys. Some journals also publish surveys besides research articles.<|endoftext|> -TITLE: Ron L. Graham’s lesser known significant contributions -QUESTION [47 upvotes]: Ron L. Graham is sadly no longer with us. -He was very prolific and his work spanned many areas of mathematics including graph theory, computational geometry, Ramsey theory, and quasi-randomness. His long association with Paul Erdős is of course very well known. Graham’s number, and Graham-Rothschild theorem, and the wonderful book Concrete Mathematics are other well known contributions. -However, some of his contributions may not be as widely known, but deserve to be so. This question is to encourage people to comment on such contributions. I am not familiar with his work on scheduling theory, for example. -He was into magic tricks and the mathematics behind them and co-authored a book on this with Persi Diaconis. And he was into juggling, like Claude Shannon. -Edit: Thanks to @LSpice for pointing out the Meta MathOverflow thread here on personal anecdotes. - -REPLY [5 votes]: Corollary (Graham). A rational number $p/q$ can be written as a sum of finitely many distinct reciprocals of integer squares iff $p/q \in [0,-1+\pi^2/6)~ \cup ~[1,\pi^2/6)$. -For the statement of the full theorem from which this follows, see On Finite Sums of Unit Fractions, with Graham as the sole author. Link. -This result is not very significant, but in literature surround unit/Egyptian fraction-esque problems, this result is cited often; if not for its use than for its novelty.<|endoftext|> -TITLE: Does big and nef imply projectivity? -QUESTION [5 upvotes]: Suppose that we have a compact Kaehler manifold $X$ with big and nef canonical class $c_1(K_{X})$, does it imply that $X$ is projective? By the base point free theorem, big and nef implies semi ample but it is for projective algebraic manifolds. So it seems to suggest that big and nef does not necessarily imply projectivity. But I have seen in literature that people claim that big and nef does imply projectivity. - -REPLY [7 votes]: If $X$ has a big line bundle $L$ then for an appropriate natural number $m$, sections of $L^m$ define a meromorphic map $\varphi: X \dashrightarrow \mathbf P^N$ which is bimeromorphic onto its image. Therefore $X$ is bimeromorphic to the projective variety $\overline{\varphi(X)}$, hence it is Moishezon. But Moishezon plus Kähler equals projective.<|endoftext|> -TITLE: Can one associate a "nice" topos to a von Neumann algebra? -QUESTION [13 upvotes]: The question here inspires my present question. -Reyes proves here that the contravariant functor Spec from the category of commutative rings to the category of sets cannot be extended to the category of noncommutative rings in such a way that every noncommutative ring is assigned to a nonempty set. Reyes also proves that it is impossible to suitably extend the Gelfand spectrum functor to the category of noncommutative C* algebras. -If one loosens the demand a Set-valued functor, then there are nice analogues of Gelfand duality. Please allow me to philosophize for a moment (I do so in order for someone to correct my perhaps inaccurate viewpoint). Even in the commutative case the need for "more open sets" in the Zariski topology led to the development of topos theory by Grothendieck in order to support étale cohomology. Toposes extend the notion of locale, which has the noncommutative relative quantale. It is possible to associate quantales to étale groupoids, to which there have been associated homology theories. Simon Henry's work on Boolean toposes (focusing on the von Neumann algebra/measure theoretic setting...his work goes beyond this) uncovers deeper important connections between von Neumann algebras and toposes. -I have begun to wonder if to find a good homology/cohomology theory for von Neumann algebras will require extracting a topos-like geometric object from the projection lattice of the von Neumann algebra and computing some sort of homology/cohomology of that object. I've read in Henry's papers that the kinds of object coming from projection lattices of von Neumann algebras are substantially different from Grothendieck toposes (in some way that I don't know enough about to ask for). The following question is a bit pie-in-the sky, and most likely completely hopeless, but I wonder if there is an "orienting answer": -Question: Is there hope of associating a "nice" topos to a von Neumann algebra? -This question is laughable, but I ask it nevertheless. What I mean by "nice" here is something like "has a computable cohomology of some kind". The philosophy being that toposes may be the right "noncommutative spaces" that may stand in counterpoint to von Neumann algebras. - -REPLY [11 votes]: (I'm going to be a bit informal to be able to go to the point relatively directly, but if you want more details on some specific aspect. I can try to add them) -Toposes are closely related to topological groupoids, in fact, they can be seen as a special type of localic groupoids or localics stack, the "étale-complete localic groupoids". (see the other answer) -So because we know very well how to attach a C* algebra or Von Neuman algebra to a groupoid it is very natural to expect that one can attach C* or Von Neuman algebra to a topos. Maybe not in full generality as topos corresponds to very general topological spaces and C*-algebras are attached to locally compact topological groupoids, but at least for 'nice topos' it should be possible. And also topos corresponds to Groupoid up to morita equivalence only, so the algebra we produce in general is only well defined up to Morita equivalence. -In some sense my work on this topic at the time was an attempt to give a direct description of the C* algebra or Von Neuman algebra one can attach to a topos (without going through groupoids) or to describe some properties of the Von Neuman algebra directly in terme of the topos (for example its modular time evolution). -And in fact it is possible: -To get a von Neuman algebra you should start with a Boolean topos that satisfies some 'measurability' condition, consider an 'internal Hilbert space object' in the topos and look at its algebra of endomorphisms. The construction works better if one assume that the topos $T$ is in addition 'locally separated' and take an Hilbert space of the form $L^2(X)$ for $X$ such that $T/X$ is separated. In this case you get a close connection between what I call measure theory over $T$ and the modular time evolution of the corresponding Von Neuman Algebra. This is essentially what I study in the paper you linked. -For C* algebra things are a bit more complicated, the best construction I could get to is described here. -Now, to go back to your question: can we go the other way and attach a topos to a von Neuman algebra or C algebra ?* -Essentially, no. At least not in a very interesting way if we do not have some additional structures. Of course it is not possible to give a definitive negative answer to this kind of question, so I'll say "probably not". -The problem is better understood in terms of groupoids than in terms of topos: the convolution algebra of a groupoids contains a lot of information on the groupoids, but if you consider it as a mere C*-algebras clearly a lot of information is lost. -For example, let's consider a groupoid $BG$ with only one object $*$ and $Hom(*,*)=G$ a group (Corresponding to the topos $BG$ of sets with a $G$-action). The kind of Von Neuman algebra or C* algebra you will attach to this topos is a Groupe algebra of $G$. Now if $G$ is abelian you will obtain an abelian Von Neuman algebra. But Abelian Von Neuman algebras corresponds to ordinary measurable spaces, so in this case you get two very different types of toposes that corresponds to exactly the same von Neuman algebra (a BG, and a topos of sheaves over a Boolean locale). The isomorphisms between the two Von Neuman algebra you get is induced by a kind of "Fourier transform" whose origin is purely analytic and non-geometric (at least in this picture). -What I read on this type of example is that if you want to construct a topos (or groupoid) out of an algebra you need something more. What this "something more" is can vary a lot, to give two example: - -For C*-algebra the notion of Cartan subalgebra sometime allow to reconstruct a groupoids, I don't know the literature on this topic but these slides will give you an idea. I assume a similar theory for Von Neuman algebra might be possible. - -One expects there will some connection between module for the algebra that one obtains a some kind of bundle of vector space/hilbert space on the topos. These bundle of vector spaces on the topos generally have a "pointwise tensor product". So one expect the Algebra we obtain to have an additional structure that corresponds to this tensor product, i.e. some sort of "generalized bi-algebra structure". One also expect that this tensor product is enough to recover the geometric object (this is very similar to Tanaka theory). I have a draft that I never finished on this topics if want to see a precise statement. - -There are probably other similar story that can be told. - - -So in some sense I see that C*-algebra/Von Neuman algebra attached to a topos as some kind of invariant, like a homology theory. That reveals a lot of important and sometime hidden informations, but definitely not all the informations. -To finish, I would like to comment on the Bohr topos mentioned in the other answer, as it is is the only such construction present in the literature. I want to emphasize that it does not really answer the question in a satisfying way because the "Bohr topos" is not reall a topos, it is only an ordinary topological space. Indeed, because it is a topos of sheaves on a poset, it actually is a topos of sheaves on a locale, and in fact on a topological space due to a compactness argument, so it will never exhibit any "non-commutative" phenomenon. The construction has been formulated in the language of toposes because many people hope it might be possible to modify the construction to actually produce a topos, and maybe it is, but at the present time what is constructed is really just an ordinary topological space.<|endoftext|> -TITLE: Cohomology theories for spaces vs cohomology theories for spectra -QUESTION [9 upvotes]: It is a standard consequence of the Brown Representability Theorem for $\operatorname{Ho}(\operatorname{Top}_*)$ that the category of generalized cohomology theories for spaces (pointed CW complexes, more specifically) is equivalent to the stable homotopy category $\operatorname{SHC}$ (defined as the homotopy category of the stable model structure on sequential spectra) via representability. On the other hand, we can define a cohomology theory for spectra (as in Barnes and Roitzheim's Foundations of Stable Homotopy Theory) to be a contravariant functor $E^*:\operatorname{SHC}^{op}\to\operatorname{Ab}_*$ such that - -Each exact triangle in $\operatorname{SHC}$ gives rise to a long exact sequence in $\operatorname{Ab}$ -$E^n$ preserves products for each $n$ (that is, it sends wedge sums to products of abelian groups) -$E^*$ preserves suspension up to a specified natural isomorphism $E^{n+1}(\Sigma X)\cong E^n(X)$. - -Then it is easy to check that $[,E]$ is always a cohomology theory for spectra. But is every cohomology theory for spectra thus represented? Given such a cohomology theory $E^*$, we obtain an associated cohomology theory for spaces by restricting to suspension spectra. Thus, by the result for spaces mentioned above, the question becomes whether a cohomology theory for spectra is determined by its restriction to spaces. -This is certainly true if we require $E^*$ to preserve sequential homotopy colimits, because any spectrum is weakly equivalent to a CW spectrum with basepoint at its unique $0$-cell, which is a sequential homotopy colimit of homotopy cofibers of coproducts of shifted sphere spectra. But without this requirement, does the result still hold? If so, why? If not, is there a standard counterexample? - -REPLY [14 votes]: The category of cohomology theories on pointed CW-complexes is not equivalent to the stable homotopy category. The latter projects onto the former, and this projection induces a bijection on isomorphism classes, but there is a kernel, containing superphantom maps, see [Christensen, J.Daniel. “Ideals in Triangulated Categories: Phantoms, Ghosts and Skeleta.” Advances in Mathematics 136, no. 2 (June 1998): 284–339. https://doi.org/10.1006/aima.1998.1735.]. -The category of cohomology theories on spectra is equivalent to the stable homotopy category. This is also called Brown representability. You essentially have to show that all cohomology theories are representable, and the rest follows from Yoneda. -By the aforementioned theorems, any cohomology theory for spectra is determined by the induced cohomology theory for spaces up to a non-canonical isomorphism.<|endoftext|> -TITLE: Index and length of closed geodesics -QUESTION [5 upvotes]: Consider the round metric on $S^n$. The geodesics are (multiples of) great circles, and one can verify that this metric is of Morse-Bott type. The Morse indices of the n-covered great circles are (if I recall correctly) $(n-1), 3(n-1), 5(n-1), \dots, (2k-1)(n-1), \dots$. -Observe in particular that the Morse index of geodesics grows linearly with the length. i.e. there is a constant $C>0$ (depending only on the metric) such that for any geodesic $\gamma$, we have $$\operatorname{ind}(\gamma) > C \operatorname{length}(\gamma).$$ -Question: let $g$ be a small, suitably generic perturbation of the round metric with the property that all geodesics are non-degenerate (such a metric is called "bumpy", I believe). Is it still the case that the Morse index of the geodesics grows linearly with the length? -I am also interested in the weaker question: does there exist a bumpy metric on $S^n$ with the property that the Morse index of the geodesics grows linearly with the length? - -REPLY [5 votes]: This follows from Bonnet-Myers. For a metric near the round metric (in the $C^\infty$ topology), the sectional curvature will be pinched below by $k > 0$, where $k\thickapprox 1$. Hence a segment of length $>\pi/\sqrt{k}$ of any geodesic will be unstable by Myers' theorem. Hence the index of a geodesic of length $L$ should be at least $L\sqrt{k}/\pi$: divide it up into segments of length $\geq \pi/\sqrt{k}$ which each have index at least 1. In fact, using a comparison theorem, one could probably do better and get an extra constant factor depending on the dimension (Myers theorem is using the weaker input of a lower bound on Ricci curvature, so a lower bound on sectional curvature will give a higher index estimate). You may be familiar with it, but Milnor's book on Morse theory is a good introduction to these topics (see Part III). For comparison theorems I like Gallot-Hulin-Lafontaine.<|endoftext|> -TITLE: Curvature of complete conformal metrics on the open unit disk -QUESTION [7 upvotes]: Let $D$ be the unit disk in the complex plane, and assume that $g$ is a Riemannian metric on $D$ which is complete and conformal to the standard Euclidean metric. Can it be the case that the Gaussian curvature of $g$ approaches zero as we approach $\partial D$? - -REPLY [6 votes]: Yes. Take the metric with length element $\rho(z)|dz|$ where $\rho(z)=(1-|z|)^{-2}$. -It is complete since $\int^1\rho(t)dt=\infty$, and the curvature -$$-\rho^{-2}\Delta\log\rho=\rho^{-4}({\rho'}^2-\rho\rho'')=-2(1-r)^2\to 0,$$ -where $r=|z|$ and the primes indicate differentiation with respect to $r$.<|endoftext|> -TITLE: Oriented cobordism classes represented by rational homology spheres -QUESTION [18 upvotes]: Any homology sphere is stably parallelizable, hence nullcobordant. However, rational homology spheres need not be nullcobordant, as the example of the Wu manifold shows, which generates $\text{torsion}({\Omega^{\text{SO}}_{5}}) \cong \mathbb Z/2\mathbb Z$. This motivates the following question. -Which classes in $\Omega^{\text{SO}}_{\ast}$ can be represented by rational homology spheres? -Of course, any such class is torsion, as all its composite Pontryagin numbers, as well as its signature, vanish. - -REPLY [11 votes]: The necessary condition pointed out by Jens Reinhold is also sufficient: any torsion class $x = [M] \in \Omega^{SO}_d$ admits a representative where $M$ is a rational homology sphere. -EDIT: This is Theorem 8.3 in $\Lambda$-spheres by Barge, Lannes, Latour, and Vogel. They also calculate the group of rational homology spheres up to rational h-cobordism, and more. I'll leave my argument below: -To prove this, we first dispense with low-dimensional cases: in any dimension $d < 5$ the only torsion class is $0 = [S^d]$. The high dimensional case follows from Claims 1 and 2 below. -I'll write $MX$ for the Thom spectrum of a map $X \to BO$ and $\Omega^X_d \cong \pi_d(MX)$ for the bordism group of smooth $d$-manifolds equipped with $X$-structure. Representatives are smooth closed $d$-manifolds $M$ with some extra structure, which includes a continuous map $f: M \to X$. -Claim 1: if $d \geq 5$ and $X$ is simply connected and rationally $\lfloor d/2 \rfloor$-connected, then any class in $\Omega^X_d$ admits a representative where $M$ is a rational homology sphere. -Claim 2: There exists a simply connected space $X$ such that $\widetilde{H}_*(X;\mathbb{Z}[\frac12]) = 0$, and map $X \to BSO$ such that the image of the induced map $\Omega^X_d = \pi_d(MX) \to \pi_d(MSO) = \Omega_d^{SO}$ is precisely the torsion subgroup, for $d > 0$. -Proof of Claim 1: Starting from an arbitrary class in $\Omega^X_d$ we can use surgery to improve the representative. Since $X$ is simply connected and $d > 3$ we can use connected sum and then surgery on embeddings $S^1 \times D^{d-1} \hookrightarrow M$ to make $M$ simply connected. Slightly better, such surgeries can be used to make the map $M \to X$ be 2-connected, meaning that its homotopy fibers are simply connected. From now on we need not worry about basepoints and will write $\pi_{k+1}(X,M) = \pi_k(\mathrm{hofib}(M \to X))$. These are abelian groups for all $k$. -If there exists a $k < \lfloor d/2\rfloor$ with $\widetilde{H}_k(M;\mathbb{Q}) \neq 0$ we can choose $\lambda \in H_k(M;\mathbb{Q})$ and $\mu \in H_{d-k}(M;\mathbb{Q})$ with intersection number $\lambda \cdot \mu \neq 0$. If $d = 2k$ for even $k$ we can additionally assume $\lambda \cdot \lambda = 0$, since the signature of $M$ vanishes. The rational Hurewicz theorem implies that $\pi_k(M) \otimes \mathbb{Q} \to H_k(M;\mathbb{Q})$ is an isomorphism, and the long exact sequence implies that $\pi_{k+1}(X,M) \otimes \mathbb{Q} \to \pi_k(M)\otimes\mathbb{Q}$ is surjective. After replacing $\lambda$ by a non-zero multiple, we may therefore assume that it admits a lift to $\pi_{k+1}(X,M)$. Such an element can be represented by an embedding $j: S^k \times D^{d-k} \hookrightarrow M$, together with a null homotopy of the composition of $j$ with $M \to X$. In the case $k < d/2$ this follows from Smale-Hirsh theory, in the case $d = 2k$ we must also use $\lambda \cdot \lambda = 0$ to cancel any self-intersections. (Actually there could also be obstructions to this in the case $d=2k$ for odd $k$, but those obstructions vanish after multiplying $\lambda$ by 2.) The embedding and the nullhomotopy gives the necessary data to perform surgery on $M$ and to promote the surgered manifold to a representative for the same class in $\Omega^X_d$. -Performing the surgery gives a new manifold $M'$ where $H_k(M';\mathbb{Q})$ has strictly smaller dimension than $H_k(M;\mathbb{Q})$ and $\widetilde{H}_*(M';\mathbb{Q}) = 0$ for $* < k$. This is seen in the same way as in Kervaire-Milnor. The case $d > 2k+1$ is easy, similar to their Lemma 5.2. In the case $d = 2k+1$ the diagram on page 515 shows that we can kill the homology class $j[S^k]$ and at worst create some new torsion in $H_k(M')$. In the case $d = 2k$ the diagram on page 527 shows that we can kill the homology class $j[S^k]$ and at worst create some new torsion in $H_{k-1}(M')$. -In finitely many steps we arrive at a representative where $\widetilde{H}_k(M;\mathbb{Q}) = 0$ for all $k \leq \lfloor d/2\rfloor$. Poincaré duality then implies that $H_*(M;\mathbb{Q}) \cong H_*(S^d;\mathbb{Q})$. $\Box$. -Proof of Claim 2: Finiteness of the stable homotopy groups of spheres implies that $\pi_d(MX)$ is a torsion group for $d > 0$ for any such $X$. Therefore we can never hit more than the torsion in $\pi_d(MSO)$, all of which is exponent 2 by Wall's theorem. The difficult part is to construct an $X$ where all torsion is hit. -The non-trivial based map $S^1 \to BO$ factors through $\mathbb{R} P^\infty \to BO$, whose image in mod 2 homology generates the Pontryagin ring $H_*(BO;\mathbb{F}_2)$. We can freely extend to double loop maps -$$\Omega^2 S^3 \to \Omega^2 \Sigma^2 \mathbb{R}P^\infty \to BO$$ -where the second map then induces a surjection on mod 2 homology. Both $\Omega^2 \Sigma^2 \mathbb{R}P^\infty$ and $BO$ split as $\mathbb{R} P^\infty$ times their 1-connected cover, so the induced map of 1-connected covers $\tau_{\geq 2}(\Omega^2 \Sigma^2 \mathbb{R}P^\infty) \to \tau_{\geq 2}(BO) = BSO$ also induces a surjection on mod 2 homology. -Now let $X = \tau_{\geq 2}(\Omega^2 \Sigma^2 \mathbb{R}P^\infty)$ with the map to $BSO$ constructed above. Take 1-connected covers of the double loop maps above, Thomify, 2-localize, and use the Hopkins-Mahowald theorem to get maps of $E_2$ ring spectra -$$H \mathbb{Z} _{(2)} \to MX_{(2)} \to MSO_{(2)}.$$ -(See e.g. section 3 of this paper.) -We can view $MX_{(2)} \to MSO_{(2)}$ as a map of $H\mathbb{Z}_{(2)}$-module spectra, and hence $MX/2 \to MSO/2$ as a map of $H\mathbb{F}_2$-module spectra. The induced map $H_*(MX/2;\mathbb{F}_2) \to H_*(MSO/2;\mathbb{F}_2)$ is still surjective (it looks like two copies of $H_*(X;\mathbb{F}_2) \to H_*(BSO;\mathbb{F}_2))$, and inherits the structure of a module map over the mod 2 dual Steenrod algebra $\mathcal{A}^\vee = H_*(H\mathbb{F}_2;\mathbb{F}_2)$. Both modules are free, because any $H\mathbb{F}_2$-module spectrum splits as a wedge of suspensions of $H\mathbb{F}_2$. In fact the Hurewicz homomorphism $\pi_*(MX/2) \to H_*(MX/2;\mathbb{F}_2)$ induces an isomorphism -$$\mathcal{A}^\vee \otimes \pi_*(MX/2) \to H_*(MX/2;\mathbb{F}_2),$$ -and similarly for $MSO$. Therefore the map $\pi_*(MX/2) \to \pi_*(MSO/2)$ may be identified with the map obtained by applying $\mathbb{F}_2 \otimes_{\mathcal{A}^\vee} (-)$ to the map on homology, showing that the induced map $\pi_*(MX/2) \to \pi_*(MSO/2)$ is also surjective. Now any 2-torsion class $x \in \pi_d(MSO)$ comes from $\pi_{d+1}(MSO/2)$, hence from $\pi_{d+1}(MX/2)$ and in particular from $\pi_d(MX)$. $\Box$<|endoftext|> -TITLE: Probability that random high dimensional vectors are all on the convex hull -QUESTION [8 upvotes]: Say I pick $n$ i.i.d. random standard normal points in $\mathbb{R}^d$. Roughly, as long as $n$ is much smaller than exponential in $d$, with high probability all points will be on the convex hull. This is because with high probability they will all be near the radius $\sqrt{d}$ sphere and all almost orthogonal, and thus each point is the furthest in its own direction from the origin. Let $p(n,d)$ be the failure probability that at least one point is in the interior of the convex hull. -Question: What's the best upper bound on $p(n,d)$ as a function of $n$ and $d$? I care most about the regime $d \gg 1$, $n \in O(\operatorname{poly}(d))$. - -REPLY [6 votes]: It's not too bad to see that the probability is at most $2n^2 e^{-d/2e}$. Let $x_1,\ldots,x_n$ be the points. We will use a union bound, so it is sufficient to examine the probability that $x_1$ is in the convex hull of $x_2,\ldots,x_n$. This happens if and only if there are $\lambda_j \in [0,1]$ with $\sum \lambda_j = 1$ and $$x_1 = \sum_{j = 2}^n \lambda_j x_j\,.$$ -Take an inner product with $x_1$ to see that this implies $$\| x_1 \|_2^2 = \sum_{j = 2}^n \lambda_j \langle x_1, x_j \rangle.$$ -Thus $$P(x_1 \in \mathrm{conv}(x_2,\ldots,x_n)) \leq P( \|x_1\|^2 \leq \max_{j \geq 2} |\langle x_1, x_j \rangle|).$$ -If we divide by $\|x_1\|$, the RHS probability bound becomes -$$P\left(\|x_1\| \le \max_{j \ge 2} \left|\left\right|\right).$$ -$\|x_1\|^2 \sim \chi^2_d$ and $\left \sim N(0,1)$, so from $\chi^2_d$ and $N(0,1)$ tail bounds we have -\begin{align*} -P(\|x_1\| \le t\sqrt{d}) &\le \left(t e^{(1-t^2)/2}\right)^d \\ -P\left(\left \ge t\sqrt{d}\right) &\le \frac{\left(e^{-t^2/2}\right)^d}{t\sqrt{2\pi d}} -\end{align*} -for any $t \in (0,1)$. Matching the base of the exponents gives -\begin{align*} -t e^{(1-t^2)/2} &= e^{-t^2/2} \\ -t &= e^{-1/2} \approx 0.606531 -\end{align*} -whence union bounding shows -\begin{align*}P(x_1 \in \mathrm{conv}(x_2,\ldots,x_n)) - &\le (n-1) \left(1 + \frac{1}{\sqrt{2\pi d/e}}\right) e^{-d/2e} \\ - &< 2ne^{-d/2e} -\end{align*} -and so -$$P(\exists~j \text{ s.t. }x_j \in \mathrm{conv}(x_1,\ldots,x_{j-1},x_{j+1},\ldots,x_n)) < 2n^2 e^{-d/2e}.$$ -I do not know if this is optimal, but it's worth noting that it's basically the strategy you suggested. When $n$ is exponentially large in $d$ the probability does not tend to $0$ provided the exponent is big enough, which is where this bound breaks.<|endoftext|> -TITLE: A smooth function $\mathbb{R}\to\mathbb{R}$ agrees with an analytic function on a bounded infinite set -QUESTION [6 upvotes]: Fix a smooth function $f:\mathbb{R}\to\mathbb{R}$. Do there exist real numbers $a0$ such that $f|_S=g|_S$? -If $g$ is only required to be defined on $(a, b)$ the question has a positive answer. In fact, we can take any $(a, b)$ we like and set $g=f(a)+\mathrm{sin}(\frac{1}{x-a})$. -We can not require $g$ to be defined on all of $\mathbb{R}$ since we can take $f(x)=\mathrm{exp}(-\frac{1}{|x|})$ for $x\neq 0$ and $f(0)=0$. Then by the pigeonhole principle $S$ must contain infinitely many positive numbers or infinitely many negative numbers; in either case $g$ does not extend to $\mathbb{R}$. Coincidentally, this shows that an arbitrary $(a, b)$ won't do in the original problem. - -REPLY [3 votes]: The answer is no, not necessarily. -Let $f$ be any smooth function such that the Taylor series of $f$ about any point $p$ has zero radius of convergence; see this MO answer for an explicit example. -Suppose that $g$ is a smooth function such that for some sequence $(x_n)$ convergent to $p$ (and such that $x_n \ne p$) we have $f(x_n) = g(x_n)$. It is rather easy to see that $f^{(k)}(p) = g^{(k)}(p)$ for every $k \geqslant 0$, and hence the Taylor series of $f$ and $g$ about $p$ coincide. In particular, the Taylor series of $g$ about $p$ has zero radius of convergence, and consequently $g$ is not real-analytic at $p$.<|endoftext|> -TITLE: Compact operator without eigenvalues? -QUESTION [5 upvotes]: Consider the operator $M$ on $\ell^2(\mathbb{Z})$ defined by for $u\in \ell^2(\mathbb Z)$ -$$Mu(n)=\frac{1}{\vert n \vert+1}u(n).$$ This is a compact operator! -Then, let $l$ be the left-shift and $r$ the right-shift on $\ell^2(\mathbb Z).$ -We consider the compact operator on $\ell^2(\mathbb Z;\mathbb C^2)$ defined by -$$T:=\begin{pmatrix} 0 & l M \\ rM & 0 \end{pmatrix}$$ -My question is: Even though $T$ is not normal, since $$T^*T= \begin{pmatrix} MlrM & 0 \\ 0 & Mr lM \end{pmatrix}=M^2$$ -whereas $$TT^*= \begin{pmatrix} lM^2 r & 0 \\ 0 & rM^2 l \end{pmatrix}\neq M^2$$ does $T$ have eigenvalues? - -REPLY [9 votes]: Note that $$T^2 = \begin{pmatrix}lMrM&0\\0&rMlM\end{pmatrix},$$ and hence the eigenvectors of $T^2$ are $$v_j = (e_j, 0) , \qquad w_j = (0, e_j),$$ with corresponding eigenvalues $$\lambda_j = \frac{1}{(1 + |j|) (1 + |j+1|)} \, , \qquad \mu_j = \frac{1}{(1 + |j|) (1 + |j-1|)} \, ,$$ respectively. In particular, the eigenspaces of $T^2$ are four-dimensional: for $j \ge 0$, the eigenspace corresponding to $\lambda_j = \mu_{j+1} = \lambda_{-j-1} = \mu_{-j}$ is spanned by $v_j$, $u_{j+1}$, $v_{-j-1}$ and $u_{-j}$. -If $u$ is an eigenvector of $T$, then it is also an eigenvector of $T^2$, and hence it is a linear combination of $v_j$, $u_{j+1}$, $v_{-j-1}$ and $u_{-j}$ for some $j \ge 0$. By inspection, $$ T(a v_j + b u_{j+1} + c v_{-j-1} + d u_{-j}) = \frac{a u_{j+1} + d v_{-j-1}}{1 + |j|} + \frac{b v_j + c u_{-j}}{1 + |j + 1|} $$ corresponds to a simple block $4\times4$ matrix, and it is now an elementary exercise to find the eigenvectors. - -REPLY [4 votes]: Sure, let $\mu_n = \frac{1}{|n|+ 1}$, then for each $n$ the vector $\sqrt{\mu_{n-1}}e_n\oplus \pm\sqrt{\mu_n}e_{n-1}$ is an eigenvector with eigenvalue $\pm\sqrt{\mu_n\mu_{n-1}}$.<|endoftext|> -TITLE: Example of a C*-algebra whose $K_1$ is uncountable -QUESTION [8 upvotes]: We know that if $A$ is a separable $C^{*}$-algebra then $K_1(A)$ is countable. -Can anybody give an example of a C*-algebra for which $K_1(A)$ is uncountable? - -REPLY [9 votes]: Nik's answer nails it but if you prefer something representable on a separable Hilbert space then you may consider the suspension $SM$ of any ${\rm II}_1$-factor $M$. Indeed, as $M$ is tracial, $K_0(M) \cong \mathbb R$ and the suspension simply reverses the $K$-groups. -Actually you may produce further commutative examples that are representable on separable Hilbert spaces: for example $S\ell_\infty$, since $K_0(\ell_\infty)$ comprises $\mathbb{Z}$-valued continuous functions on $\beta \mathbb N$.<|endoftext|> -TITLE: Could computing the next prime in a finite Euler product be made rigorous? -QUESTION [18 upvotes]: It is well known that: -$$\zeta(s):=\prod_{n=1}^{\infty} \frac{1}{1-p_n^{-s}} \qquad \Re(s) \gt 1$$ -with $p_n =$ the $n$-th prime. It also known that: -$$\zeta(2n):= \frac{(-1)^{n+1} B_{2n}(2\pi)^{2n}}{2(2n)!}$$ -where $B_{2n}$ is the $2n$-th Bernoulli number. -Now define the function: -$$f(k,N,x):= \zeta(2k) - \left(\prod_{n=1}^{N} \frac{1}{1-p_n^{-2k}}\right)\cdot \left(\frac{1}{1-x^{-2k}}\right) \qquad \Re(s) \gt 1$$ -where $k, N \in \mathbb{N}$ and $x$ is the unknown next prime ($p_{N+1}$) to be computed. -I found numerically that solving $x$ in $f(k,N,x)=0$ for some $N$, yields an increasingly accurate approximation of $p_{N+1}$ when $k$ increases. For example take the first 6 primes and try to derive the 7th prime (17): - -$f(6, 6, x) = 0 \rightarrow x = 16.64054...$ -$f(12, 6, x) = 0 \rightarrow x = 16.95214...$ -$f(24, 6, x) = 0 \rightarrow x =16.99830...$ - -The key question is how high $k$ needs to go to ensure that $x=p_{N+1}$ after rounding. In the following Maple code, I simply used $k=2N$ and it already correctly generates all 'next' primes up to $N=60$: -Digits:=600 -for N from 1 to 60 do ithprime(N), ithprime(N+1), round(fsolve(f(2*N, N, x), x = 0 .. 300)) end do - -I immediately acknowledge that this is a highly inefficient and impractical algorithm to generate primes. However, is there more to say about the minimally required value of $k$ (maybe as a function of $N$) to ensure that rounding $x$ will correctly yield $p_{N+1}$? - -REPLY [15 votes]: $2k=1+p_N$ works for $N>1$, but $2k\le 0.56 \, p_N$ will fail if -$p_{N+2}=p_{N+1}+2$. -With $q=p_{N+1}$, we have -$$ -\frac{1}{1-q^{-2k}} < \frac{1}{1-x^{-2k}} = \frac{1}{1-q^{-2k}} \prod_{p>q} \frac{1}{1-p^{-2k}} . -$$ -It follows that -$$ -q^{-2k} < x^{-2k} < q^{-2k} + \sum_{j\ge 2} (q+j)^{-2k} < q^{-2k} +\frac{1}{(q+1)^{2k-1}(2k-1)}. -$$ -Taking logarithms, and using $\log(1+y)\le y$, we get -$$ --2k \log q < -2k \log x < -2k\log q + \frac{q+1}{\exp\{(1-o(1)) 2k/q\} (2k-1)}. -$$ -Dividing by $-2k$ and exponentiating, we have -$$ -q > x > q - \frac{q(q+1)}{\exp\{(1-o(1))2k/q\} 2k (2k-1)}. -$$ -We want the last expression to be less than $1/2$. Since $q/p_N \to 1 $ as $N\to \infty$, we need $k\ge (1+o(1)) c \, p_N$, -where $c=0.45...$ is the solution to $e^{2c}4 c^2=2$. -So $2k=1+p_N$ works for large $N$. We can check with a computer that it also works for small $N$. -Similar calculations show that when $p_{N+2}=p_{N+1}+2$, it is necessary that $k>0.28 \, p_N$ when $N$ is large. -Edit: Using the inequality $\log(1+y)\le y$ was somewhat wasteful and not necessary. Also, as the OP points out in the comments, we can use the ceiling function instead of rounding, since $x -TITLE: A series that is rational? -QUESTION [9 upvotes]: Let $k=\mathbb F_q(T)$. Can one prove (or disprove) that the series $\sum_{n\ge0}(1-TX^{q^n})Y^{q^n}\in k[[X,Y]]$ belongs to $k(X,Y)$? At first, it looked like it was simple. But in fact, I have no clue to attack this question. I thought about Dwork-Polya-Bertrandias theorem, but I did not find a several variables version of this theorem. - -REPLY [23 votes]: If you set $T=0$ or $X=0$ then you get the series $\sum_{n\geq 0} Y^{q^n}$. This cannot be rational because a rational power series in one variable that is not a polynomial cannot have arbitrarily long sequences of 0 coefficients (since the coefficients satisfy a linear recurrence relation with constant coefficients).<|endoftext|> -TITLE: Infinity-homotopies -QUESTION [11 upvotes]: Koszul duality for operads allows for straightforward generalizations of $A$-infinity algebras and $A$-infinity morphisms for the so called Koszul operads $\mathcal{O}$, among which we find the associative operad. Good accounts of this can be found in standard references. However, I've been unable to find a generalization of $A$-infinity homotopies between $A$-infinity morphisms. Does anyone know of any reference(s) where this is treated in the greatest possible generality? Ideally, for Koszul operads over an arbitrary commutative ground ring, but anything is welcome. - -REPLY [5 votes]: I don't know if you found an answer since you posted the question, but I will write this just in case: there is a "cute" (easy) definition in case of nonsymmetric operads which generalises the A-infinity story rather trivially (derivation homotopy), while for symmetric operads the definition is more involved. There is a comparison of various possibilities in my paper with Poncin https://link.springer.com/article/10.1007/s10485-015-9407-x , and an explanation that these possibilities actually come from a suitable model category structure in the definitive treatment by Bruno Vallette https://aif.centre-mersenne.org/item/AIF_2020__70_2_683_0/<|endoftext|> -TITLE: Visualize (co)sketeton of a simplicial set (geometrical intuition) -QUESTION [7 upvotes]: I want to understand if there is an intuition approchable with -most possible 'elementary geometrical' knowledge for -$n$-(co)skeleta of simplicial sets? -Formally sketleton & coskeleton functions arise as follows: For $\Delta$ the simplex category write $\Delta_{\leq n}$ for its full subcategory on the objects -$[0],[1],\cdots,[n][0], [1], \cdots, [n]$. -The inclusion $\Delta|_{\leq n} \hookrightarrow \Delta$ -induces a truncation functor -$$\mathrm{tr}_n: \mathit{sSet}= [\Delta^{\mathrm{op}},Set] \to [\Delta_{\leq n}^{\mathrm{op}},\mathit{Set}]$$ -that takes a simplicial set and restricts it -to its degrees $\leq n$. -This functor has a left adjoint, given by left Kan extension -$\mathrm{sk}_n: [\Delta_{\leq n},\mathit{Set}] \to \mathit{SSet}$ called the $n$-skeleton -and a right adjoint, given by right Kan extension -$\mathrm{cosk}_n : [\Delta_{\leq n},Set] \to SSet$ called the $n$-coskeleton. -Now set $F: \Delta^{\mathrm{op}} \to Set, [n] \mapsto X_n$. The picture one conventionally has in mind thinking intuitionally/geometrically about $X$ is -that one thinks $X_n$ as "the set of $n$-simplices/cells of the -"simplicial complex" $X$ (only as geometrical intuition). -How can I think in this naive manner about $\mathrm{sk}_n(X)$ and -$\mathrm{cosk}_n(X)$? -The $\mathrm{sk}_n(X)$ might be considered as a "subcomplex" of $X$ -obtained from $X$ by killing all $m$-simplices with $m > n$. -The way all $\ell$-simplices for $\ell \le n$ are "glued together" -stays the same as for $X$, ie for $\ell$-simplices happens nothing. -If we keep thinking about $X$ as a simplicial complex, -which picture should one have thinking about $\mathrm{cosk}_n(X)$? -How it deviates from original $X$? - -REPLY [8 votes]: For $k \le n$, the $k$-simplices in $\mathrm{cosk}_n(X)$ are the same as in $X$. For larger $k$, there is a unique $k$-simplex for every $n$-skeleton of a $k$-simplex you find in $X$, that is, $(\mathrm{cosk}_n(X))_k \cong \mathrm{Hom}(\mathrm{sk}_n \Delta^k, X)$. -You can also think inductively: again, for $k \le n$ the $k$-simplices in $\mathrm{cosk}_n(X)$ are the same as in $X$; then for each $k > n$ if you already know the simplices of dimension less than $k$ in $\mathrm{cosk}_n(X)$, you get the $k$-simplices by filling in uniquely every empty $k$-simplex you find in $\mathrm{cosk}_n(X)$. That is, for $k>n$, $(\mathrm{cosk}_n(X))_k \cong \mathrm{Hom}(\partial \Delta^k, \mathrm{cosk}_n(X))$.<|endoftext|> -TITLE: Perfect $\mathbb Z_\ell$-modules -QUESTION [6 upvotes]: Disclaimer : I asked this question on Maths.StackExchange 20 days ago (and started a bounty) here but got no answer, so I'm asking it here now (with no modification). -$\newcommand{\l}{\ell} \newcommand{\Z}{\mathbb Z}$ -I'm trying to understand a remark in Weil's conjectures for function fields I, by Gaitsgory-Lurie. -Specifically it's remark 2.3.4.3., which essentially states the following : an object $M$ in $\mathrm{Mod}_{\mathbb Z_\l}$ is perfect if and only if it's $\l$-complete and $M\otimes_{\Z_\l}\Z/\l$ is perfect in $\mathrm{Mod}_{\Z/\l}$ (this latter condition implies that this is also true for $M\otimes_{\Z_\l}\Z/\l^d$, for any $d\geq 0$) -They state it without proof, and it looks like there is no proof elsewhere in the book (as far as I can see, although I haven't checked the whole book). -What is clear to me is the forward direction : indeed $\Z_\l\otimes_{\Z_\l}\Z/\l$ is perfect over $\Z/\l$ and $\Z_\l$ is $\l$-complete, and these conditions are closed under retracts, so the sub-$\infty$-category of those $M$'s that satisfy the latter conditions is a stable subcategory, contains $\Z_\l$ and is closed under retracts so it contains all perfect $\Z_\l$-modules. -The reverse direction, however, does not seem so clear. I think one ought to use the fact that the $\infty$-category of $\l$-complete modules is equivalent to $\underset{d}{\varprojlim} \mathrm{Mod}_{\Z/\l^d}$ but that doesn't seem to be enough since we want $M$ to be compact in the whole $\mathrm{Mod}_{\Z_\l}$, not just in $\l$-complete modules. - -REPLY [4 votes]: Over a PID, you can check perfectness on homology. So the claim is that the homology of an $\ell$-complete $\mathbb{Z}_\ell$-module is finitely generated if and only if it is so mod $\ell$. The homology groups are $\ell$-complete, and so this follows from the long exact sequence of mod $\ell$-reduction.<|endoftext|> -TITLE: Summing over normalized characters of the permutation group -QUESTION [5 upvotes]: Let $\chi_\lambda(\mu)$ be the usual characters of the irreducible representations of the permutation group $S_n$. The normalized character is the quotient $\chi_\lambda(\mu)/f^\lambda$, where $f^\lambda=\chi_\lambda(1)$ is the dimension of the representation. -Can I hope for a nice formula expressing their sum -$$ \sum_{\lambda\vdash n}\frac{\chi_\lambda(\mu)}{f^\lambda},$$ -in terms of the parts of $\mu$? - -REPLY [7 votes]: The quantity you are asking about is in fact a well-known expression: When multiplied by $n!$, it is the number of ordered pairs $\sigma, \tau \in S_{n}$ such that -$[\sigma, \tau] = \mu$, where $[\sigma, \tau] = \sigma^{-1}\tau^{-1}\sigma \tau$ is the commutator of $\sigma$ and $\tau$. However, I do not know how to relate this to the disjoint cycle structure of $\mu$, except to say that this quantity is clearly zero if $\mu$ is an odd permutation. - -REPLY [4 votes]: This is just an observation. I normalize your problem by $n!$, to get rid of denominators. -Let $A_n(\mu) := n! \sum_{\lambda \vdash n} \chi^{\lambda}(\mu)/f^\lambda$. -Define $B_n(x) := n! \sum_{\lambda \vdash n} \frac{p_\lambda(x)}{f^\lambda}$. -Then $A_n(\mu) = \langle B_n(x), s_\mu \rangle$. -That is, $A_n(\mu)$ is the coefficient of $s_\mu$ when expanded in the Schur basis. -The Schur expansion of $B_n(x)$ for $n=1,2,\dotsc$ -are -\begin{array}{l} -s_{1} \\ -4 s_{2} \\ -15 s_{3}+6 s_{21}+9 s_{111} \\ -76 s_{4}+64 s_{22}+44 s_{31}+76 s_{211}+12 s_{1111} \\ -368 s_{5}+628 s_{32}+416 s_{41}+580 s_{221}+792 s_{311}+344 s_{2111}+200 s_{11111} -\end{array} -Perhaps there is some pattern...<|endoftext|> -TITLE: A royal road to Coulomb branches of 3D $\mathcal{N}=4$ gauge theories -QUESTION [17 upvotes]: So, I've been very interested recently with the developements of the (now mathematically precise) theory of Coulomb branches - in particular because of its recent applications on representation theory and symplectic geometry. -They have been considered by mathematical physicists for a time, but without a rigorous definition. The first attempt to give such a definition was made by H. Nakajima, Introduction to a provisional mathematical definition of Coulomb branches of 3-dimensional N=4 gauge theories. -This project was completed by A. Braverma, N. Finkelberg, H. Nakajima in -Towards a mathematical definition of Coulomb -branches of 3-dimensional N=4 gauge theories, II, and their companion papers -Coulomb branches of 3d N=4 quiver gauge theories and slices in the affine Grassmannian. With two appendices by Braverman, Finkelberg, Joel Kamnitzer, Ryosuke Kodera, Nakajima, Ben Webster and Alex Weekes. -Ring objects in the equivariant derived Satake category arising from Coulomb branches. Appendix by Gus Lonergan. -Now, I've become rather convinced that to proceed with my research I need to have a rather firm grasp on the tree papers above by Braverman, Finkelberg and Nakajima. -Now, I unfortunately have no good physics intuition, and I've found that some parts of the above papers are very technical and are motived by a lot of different constructions in geometric representation theory in the range of the past 20 years. The final form of this theory is a remarkable achievement of novel ideas and technical mastery, and I'm feeling a little bit lost on what should be the relevant things to focus on (important previous work and mathematical techiniques and machinery). -Hence, the question: what is the royal road to Coloumb branches? -PS: the two papers I'm trying to fully appreciate are: -J. Kamnitzer, P. Tingley, B. Webster, A. Weekes and O. Yacobi, On category O -for affine Grassmannian slices and categorified tensor products. -and -A. Weekes, Generators of Coulomb branches of quiver gauge theories, arXiv:1903.07734. - -REPLY [7 votes]: I've written a survey paper on this topic: -Symplectic resolutions, symplectic duality, and Coulomb branches. -I begin with symplectic resolutions in general, then discuss symplectic duality, and then the Coulomb branch construction of Braverman-Finkelberg-Nakajima. Throughout, I discuss quiver varieties and (generalized) affine Grassmannian slices. I hope this is helpful!<|endoftext|> -TITLE: Proving a majorization inequality for the singular value of the product of two matrices without using tensor product -QUESTION [5 upvotes]: For any two matrices $\mathbf{A},\mathbf{B} \in \mathbb{C}^{n \times n}$, we know that the following majorization inequality holds -$$ -\tag{1} -\label{grz} -\sigma^{\downarrow}(\mathbf{A}\mathbf{B}) \prec_w \sigma^{\downarrow}(\mathbf{A})\sigma^{\downarrow}(\mathbf{B}), -$$ -where $\sigma^{\downarrow}(\cdot)$ denotes the vector of singular values, ordered in the decreasing order. This is equivalent to the following system of inequalities -$$ -\tag{2} -\label{sysineq} -\sum_{i=1}^k\sigma_i^{\downarrow}(\mathbf{A}\mathbf{B}) \leq \sum_{i=1}^k \sigma_i^{\downarrow}(\mathbf{A})\sigma_i^{\downarrow}(\mathbf{B}), -$$ -for $k=1,\dots,n$. -Proof: -In all the textbooks or papers that I have seen, the proof of this majorization inequality is as follows. By the sub-multiplicativity of the spectral norm, one has -$$ -\sigma_1^{\downarrow}(\mathbf{A}\mathbf{B}) \leq \sigma_1^{\downarrow}(\mathbf{A})\sigma_1^{\downarrow}(\mathbf{B}). -$$ -By employing this inequality to the anti-symmetric tensor powers (i.e. the compound matrices) $\wedge^k(\mathbf{A})$ and $\wedge^k(\mathbf{B})$, we have -$$ -\sigma_1^{\downarrow}\big((\wedge^k \mathbf{A})(\wedge^k \mathbf{B})\big) \leq \sigma_1^{\downarrow}\big(\wedge^k \mathbf{A}\big)\sigma_1^{\downarrow}\big(\wedge^k \mathbf{B}\big), -$$ -for $k=1,\dots,n$. Then using the facts that $\wedge^k(\mathbf{A}\mathbf{B}) = (\wedge^k \mathbf{A})(\wedge^k \mathbf{B})$ and $\sigma_1^{\downarrow}\big(\wedge^k \mathbf{A}\big) = \prod_{i=1}^k \sigma_i^{\downarrow}(\mathbf{A})$, it follows that -$$ -\tag{3} -\label{lwm} -\prod_{i=1}^k\sigma_i^{\downarrow}(\mathbf{A}\mathbf{B}) \leq \prod_{i=1}^k \sigma_i^{\downarrow}(\mathbf{A})\sigma_i^{\downarrow}(\mathbf{B}), -$$ -for $k=1,\dots,n$. Finally, inequality \eqref{grz} follows using the fact that log-weak majorization inequality \eqref{lwm} implies weak majorization inequality \eqref{grz} [Bhatia, Matrix analysis, Example II.3.5 (vi)]. -Question: -Can we prove the majorization inequality \eqref{grz} without resorting to the tensor products and employing no facts about them? -Thanks in advance! -My attempt: -By the maximal characteristic of the singular values, we know that -\begin{equation} - \sigma_i(\mathbf{A}) = \max_{\substack{\|\bf{x}_i\|=\|\bf{y}_i\|=1 \\ \bf{x}_i \bot \text{span}\{\bf{x}_1,\dots, \bf{x}_{i-1}\} \\ \bf{y}_i \bot \text{span}\{\bf{y}_1,\dots, \bf{y}_{i-1}\}}}\big|\langle \mathbf{A}\bf{x}_i,\bf{y}_i \rangle\big|, -\end{equation} -for $i=1,\dots,n$. Using this formula, we can demonstrate that the inequalities \eqref{sysineq} are equivalent to the following system of inequalities: -\begin{equation} -\max_{\substack{\|\bf{x}_i\|=\|\bf{y}_i\|=1, \;i \in [k] \\ \bf{x}_1 \bot \dots \bot \bf{x}_k \\ \bf{y}_1 \bot \dots \bot \bf{y}_k}} \sum_{i=1}^k \big|\langle \mathbf{A}\mathbf{B} \bf{x}_i,\bf{y}_i \rangle\big| \leq \max_{\substack{\|\bf{x}_i\|=\|\hat{\bf{x}}_i\|=1, \;i \in [k] \\ \bf{x}_1 \bot \dots \bot \bf{x}_k \\ \hat{\bf{x}}_1 \bot \dots \bot \hat{\bf{x}}_k}} \max_{\substack{\|\bf{y}_i\|=\|\hat{\bf{y}}_i\|=1, \;i \in [k] \\ \bf{y}_1 \bot \dots \bot \bf{y}_k \\ \hat{\bf{y}}_1 \bot \dots \bot \hat{\bf{y}}_k}} \sum_{i=1}^k\big| \langle \mathbf{B}\bf{x}_i,\hat{\bf{x}}_i \rangle \langle \mathbf{A}\bf{y}_i,\hat{\bf{y}}_i \rangle\big|, -\end{equation} -for $k=1,\dots,n$. All I can show is that for each $i=1,\dots,k$, we have -\begin{equation} -\begin{split} -\big|\langle \mathbf{A}\mathbf{B} \bf{x}_i,\bf{y}_i \rangle\big| &= \big|\langle \mathbf{B} \bf{x}_i, \mathbf{A}^\mathsf{H}\bf{y}_i \rangle\big| \\ -& \leq \|\mathbf{B}\bf{x}_i\| \|\mathbf{A}^\mathsf{H}\bf{y}_i\| \\ -& = \max_{\|\hat{\bf{x}}_i\|=1} \big|\langle \mathbf{B}\bf{x}_i,\hat{\bf{x}}_i \rangle\big| \max_{\|\hat{\bf{y}}_i\|=1} \big|\langle \mathbf{A}^\mathsf{H}\bf{y}_i,\hat{\bf{y}}_i \rangle\big|, -\end{split} -\end{equation} -where $\mathbf{A}^\mathsf{H}$ is the conjugate transpose of $\mathbf{A}$. The inequality and the last equality follow by the Cauchy-Schwarz inequality. Therefore -\begin{equation} -\max_{\substack{\|\bf{x}_i\|=\|\bf{y}_i\|=1 \\ \bf{x}_1 \bot \dots \bot \bf{x}_k \\ \bf{y}_1 \bot \dots \bot \bf{y}_k}} \sum_{i=1}^k \big|\langle \mathbf{A}\mathbf{B} \bf{x}_i,\bf{y}_i \rangle\big| \leq \max_{\substack{\|\bf{x}_i\|=\|\hat{\bf{x}}_i\|=1 \\ \bf{x}_1 \bot \dots \bot \bf{x}_k}} \max_{\substack{\|\bf{y}_i\|=\|\hat{\mathbf{y}}_i\|=1 \\ \mathbf{y}_1 \bot \dots \bot \mathbf{y}_k}} \sum_{i=1}^k\big| \langle B\mathbf{x}_i,\hat{\mathbf{x}}_i \rangle \langle A\hat{\mathbf{y}}_i,\bf{y}_i \rangle\big|. -\end{equation} -However, these inequalities are weaker than what we want. -Bhatia, Rajendra, Matrix analysis, Graduate Texts in Mathematics. 169. New York, NY: Springer. xi, 347 p. (1996). - -REPLY [3 votes]: We prove that -$$\sum_{i=1}^k \sigma^\downarrow_i(AB) - = \sup_{U}|\mathrm{Tr}(UAB)| - \le \sup_{U,V}|\mathrm{Tr}(UAV^*B)| - =\sum_{i=1}^k \sigma^\downarrow_i(A)\sigma^\downarrow_i(B),$$ -where $U$ and $V$ run over all partial isometries (or contractions) of rank (at most) $k$. -The only nontrivial is $\le$ part of the rightmost equality. -For the proof of this, we may assume that $A$ and $B$ are positive. -Then by the Cauchy--Schwarz inequality, $|\mathrm{Tr}(UAV^*B)|$ attains -the supremum $\mathrm{Tr}(UAU^*B)$ at some rank $k$ partial isometry $U$ (and $V=U$). -Let's denote by $\tilde{A}$ (resp.\ $\tilde{B}$) the truncated operator -$UAU^*$ (resp.\ $B$) on $\mathop{\mathrm{ran}} U$. -Then $\tilde{A}$ and $\tilde{B}$ are positive operators of rank at most $k$ -satisfying -$\sigma^\downarrow(\tilde{A})\prec_w\sigma^\downarrow(A)$, -$\sigma^\downarrow(\tilde{B})\prec_w\sigma^\downarrow(B)$, and -$$\mathrm{Tr}(UAU^*B)=\mathrm{Tr}(\tilde{A}\tilde{B}).$$ -For the computation of $\mathrm{Tr}(\tilde{A}\tilde{B})$, -we may assume that $\mathop{\mathrm{ran}} U={\mathbb C}^k$ and $\tilde{A}$ is -the diagonal matrix with entries $\sigma^\downarrow(\tilde{A})$. -Let's denote by $\beta$ the diagonal entries of the positive matrix $\tilde{B}$. -Then it satisfies $\beta^\downarrow\prec\sigma^\downarrow(\tilde{B})$. -Hence in conclusion -$$\sup_{U,V}|\mathrm{Tr}(UAV^*B)| - = \mathrm{Tr}(\tilde{A}\tilde{B}) - = \sum_{i=1}^k\sigma^\downarrow_i(\tilde{A})\beta_i - \le \sum_{i=1}^k\sigma^\downarrow_i(A)\sigma^\downarrow_i(B).$$ -Here, we have used (twice) the following fact. -For any positive eventually-zero sequences $\alpha,\beta,\gamma$ with -$\beta^\downarrow\prec_w\gamma^\downarrow$, one has $\sum_i\alpha^\downarrow_i\beta_i \le \sum_i\alpha^\downarrow_i\gamma^\downarrow_i$, because -$$\sum_i\alpha^\downarrow_i\beta_i - = \sum_i\bigl((\alpha^\downarrow_i-\alpha^\downarrow_{i+1})\sum_{j=1}^i\beta_j\bigr).$$<|endoftext|> -TITLE: Time-saving (technology) tricks for writing papers -QUESTION [86 upvotes]: I have over the years learned some tricks which saves a lot of time, -and I wish I had known them earlier. Some tricks are LaTeX-specific, but other tricks are more general. Let me start with a few examples: - -Use LaTeX macros and definitions for easy reuse. This is particularly useful for when making many similar-looking figures. Another example is to make a macro that includes the $q$ when typing q-binomial coefficients. This ensures consistency. - -In documents with many Tikz figures, compilation time can become quite brutal. However, spreading out all figures in many documents is also inconvenient. Solution: Use one standalone file, where each figure appears as a separate .pdf page. Then include the .pdf pages as figures in the main document. All figures are in one .tex-file, making it easy to reuse macros. I find this trick extremely useful, as it does not lead to duplicate code spread over several files. - -Use bibtex and .bib files. I prefer to use doi2bib to convert doi's to a .bib entry (some light editing might be needed). - -For collaboration, use git. Also, Dropbox or similar for backups. Keeping track of versions has saved me several times. - -Learn Regular expressions, for search-and-replace in .tex files. This is useful for converting hard-coded syntax into macros. - -Get electronic (local) copies of standard references, and make sure to name them in a sane manner. Then it is easy to quickly search for the correct book. These are available when the wifi is down, or while traveling. - -Do file reorganization and cleanup regularly. Get final versions of your published papers, and store in a folder, as you'll need them for job applications. Hunting down (your own!) published papers in pay-walled journals can be surprisingly tedious! - -Take the time to move code snippets from project-specific notebooks, and turn into software packages for easy reuse. Also, it is sometimes worth to spend time optimizing code - waiting for code to run does not seem like a big deal, but I have noticed that small improvements in my work-flow can have big impact. I am much more likely to try out a conjecture if it is easy to run the code. - -REPLY [7 votes]: I have a rather different approach to most posting here. I write my papers in LyX, which uses LaTeX on the backend, but presents a friendlier interface overtop. I feel like I can use less of my brain for writing proper LaTeX, and concentrate more on the mathematics. (Of course, one still needs/wants to remember the basic codes like \alpha, etc, but LyX will automatically handle a fair number of fiddly details.) And that you can immediately see what you are typing (where you are typing it) helps avoid many typos. - -There was some discussion about spellchecking. I'll particularly comment that LyX includes a spellchecker, and it straightforwardly underlines unknown words in red, just like any modern word-processor. - -A drawback of LyX is that negotiation (more than usual) is needed when writing with coauthors. - -On managing bibtex, I like the MathSciNet bibtex, which is generally of a reliable quality. No one has yet noticed, but you don't need a MathSciNet subscription to get at the bibtex: you can use the MRLookup interface, which is available from anywhere. It will only return the top three hits of a search, but the interface is simpler than full MathSciNet, and I prefer to use it even from a university IP when I'm looking for something specific. - -I use with BibDesk (on macOS) to manage my master bib file.<|endoftext|> -TITLE: Is every open topological $d$-manifold homotopy equivalent to a CW-complex of dimension $\leq d-1$? -QUESTION [9 upvotes]: Let $M$ be a connected open topological $d$-manifold (without boundary). -Whitehead showed that if $M$ has a PL structure, there exists a subcomplex of dimension $\leq d-1$ onto which $M$ deformation retracts. -Can we still find a homotopy equivalent CW complex of dimension $\leq d-1$ when $M$ is not PL? - -REPLY [6 votes]: $\DeclareMathOperator{\co}{H}$ -$\DeclareMathOperator{\ch}{C}$ -$\newcommand{\zz}{\mathbb{Z}}$ -$\newcommand{\nn}{\mathbb{N}}$ -$\newcommand{\A}{\mathcal{A}}$ -$\newcommand{\B}{\mathcal{B}}$ -$\DeclareMathOperator{\lf}{lf}$ -Let me put together an answer following the pointers in the comments. By Whitehead's result stated in the question and smoothability in lower dimensions we may assume $d \geq 4$. Write $\pi := \pi_1(M)$ for brevity. -As an ANR, M has the homotopy type of a CW-complex, so by a result of Wall it suffices to show that $\co^{j}(M; \A) = 0$ whenever $j \geq d$ and $\A$ is a $\zz \pi$-module. Writing $w$ for the orientation $\zz \pi$-module, by Poincaré duality we have $$\co^j(M;\A) \cong \co^{\lf}_{d-j}(M; \A \otimes_{\zz} w)\,,$$ -where $\co^{\lf}_{*}$ denotes the locally finite singular homology (sometimes called the Borel-Moore homology). Therefore the only nontrivial thing to check is the vanishing of the 0-th locally finite homology for every $\zz \pi$-module $\B$. Writing $\tilde{M}$ for the universal cover of $M$, this amounts to showing that the first differential -$$\partial_1 \otimes_{\zz\pi} \B \colon \ch^{\lf}_1(\tilde{M}) \otimes_{\zz\pi} \B \rightarrow \ch^{\lf}_0(\tilde{M}) \otimes_{\zz\pi} \B$$ -of the locally finite singular chain complex is surjective, for which $\partial_1$ being surjective before tensoring is enough. We can verify $\partial_1$ is surjective by elementary means: Fix a locally finite singular 0-chain $\sigma$; it is necessarily supported on a discrete subset of $\tilde{M}$. Since $\tilde{M}$ is non-compact (and second-countable), we can find a countably infinite discrete subset $$\{x_n : n \in \nn\} \subseteq\tilde{M}$$ which contains the support of $\sigma$. Thus $\sigma$ is a formal sum of the form $$\sigma = \sum_{n \in \nn}a_n x_n$$ -with $a_n \in \zz$. Now for each $n \in \nn$ we can find a path $\gamma_n : [0,1] \rightarrow M$ connecting $x_{n+1}$ to $x_{n}$ such that the formal sum $$\tau := \sum_{n \in \nn}b_n \gamma_n$$ -with the coefficients $b_n := \sum_{j \leq n} a_j$, is a locally finite $1$-chain with $\partial_1(\tau) = \sigma$.<|endoftext|> -TITLE: The inconsistency of Graham Arithmetics plus $ \forall n, n < g_{64}$ -QUESTION [12 upvotes]: As you all know, Ronald Graham just passed away. He is famous for many fabulous contributions to finite combinatorics, and much much more, but perhaps none of them is as popular as the infamous Graham number (see here): $g_{64}$. -This number is truly huge, though in more recent times it has been dwarfed by several other "finite inaccessible numbers" (allow me to call them that way), such as Friedman's TREE(3) . -Anyway, it is big enough to play the role of "infinite number" in what follows. -Now, start with Robinson's arithmetics $Q$ and "beef it up" so that it contains all the recursive definitions necessary to talk about $g_{64}$ (for instance, all equations to define upper arrow notation). -As a tribute to Graham, let us call this theory GRAHAM. -Define T as GRAHAM + $ \forall n, n < g_{64}$ -Clearly, T is classically inconsistent. But: just like Parikh's feasible arithmetics, there are, as far as I see, no short proofs of its inconsistency without appealing to induction. -Now the question: - -assuming a fixed proof system (say Gentzen), can we give a measure of -the shortest induction-free proof that T is not consistent? - -NOTE: I have mentioned Parikh's aritmetics, because obviously T and Parikh system share some common features. -But in his case, he keeps some induction and augments arithmetics with an additional unary predicate expressing feasibility for which induction does not apply , whereas here there is no such predicate, but on the other hand there is no induction whatsoever. - -REPLY [7 votes]: The length of the least proof of contradiction in $\mathsf{Graham}+\forall n (ns$ to $s$. We put $\mathsf{len}(\tau)=n$ and $\tau_i=(A_i\cap [0,s))\cup\{s\}$. It is easy to see that - -$\tau$ supports the set of axioms of $T$ -$\tau,i\Vdash \varphi$, for any $i\le \mathsf{len}(\tau)$ and axiom $\varphi$ of $T$ -$\mathsf{len}(\tau)>\log_2(\log_2(g_{64}))$. - -Finally assume for a contradiction that $P$ is a proof of the sequent $\lnot \mathsf{Graham},\exists x \lnot(x -TITLE: What is the Zhu algebra of a lattice vertex algebra? -QUESTION [8 upvotes]: Associated to a vertex algebra $V$ is an associative algebra $A(V)$, the Zhu algebra. Its defining property is approximately that the representations of $V$ and of $A(V)$ are the same. -In vertex operator algebras associated to affine and Virasoro algebras, Frenkel and Zhu prove for instance that the Zhu algebra of the affine vertex algebra $V_k(\mathfrak{g})$ is $U(\mathfrak{g})$. -Question: Is the Zhu algebra of the vertex algbera $V_k(L)$ associated to lattice $L$ known? -Writing $\mathfrak{h}=L\otimes_{\mathbf{Z}}\mathbf{C}$, there is a map $V_k(\mathfrak{h})\hookrightarrow V_k(L)$, which gives a map $U(\mathfrak{h})\to A(V_k(L))$. - -REPLY [4 votes]: I'm expanding Reimundo Heluani's link, which gives the answer when $L$ is an even positive definite lattice. Write $\mathfrak{h}=L\otimes_{\mathbf{Z}}\mathbf{C}$. Every $\alpha\in L$ gives two elements, $E_\alpha\in \mathbf{C}[L]$ and $\alpha\in\mathfrak{h}$. -The Zhu algebra is -$$A(V_L)\ =\ U(\mathfrak{h})\otimes\mathbf{C}[L]/\ (\alpha -(\alpha,\alpha)/2)E_\alpha.$$ -The algebra structure on $ U(\mathfrak{h})$ is the usual one, the structure on $\mathbf{C}[L]$ is almost the usual one -$$E_\alpha\cdot E_\beta\ =\ \text{const.}\cdot E_{\alpha+\beta}$$ -(for the constants see equations 2.9 and 2.10 of arxiv.org/abs/q-alg/9605032), which together with -$$[\alpha, E_\beta]\ =\ (\alpha,\beta) E_\beta$$ -give the algebra structure on $U(\mathfrak{h})\otimes\mathbf{C}[L]$. -It is a finitely generated algebra, and contains a copy of $U(\mathfrak{h})$ within it.<|endoftext|> -TITLE: Loop spaces motivation -QUESTION [10 upvotes]: I read that one of the main goals of utilization simplicial methods is to prove that a space is a loop space. On the other hand where lies the main importance to recognize topological spaces as loop spaces? Surely, if a space is a loop space then its connected components obtain a magma structure via concatenation because if $X$ is homotopy equivalent to a loop space $\Omega(Y)$ then $\pi_0(X)= \pi_1(Y)$ and thus there is a map $*: \pi_0(X) \times \pi_0(X) \to \pi_0(X)$. -Is this the only motivation behind the study of loop spaces? Or what other interesting aspects can be gained from studying loop spaces from the viewpoint of homotopy theory? - -REPLY [7 votes]: There are several useful points in the comments, but I want to go beyond them and try to give a more comprehensive answer, so this question doesn't linger unanswered. Some great sources are May's Geometry of Iterated Loop Spaces (GILS) and A Concise Course in Algebraic Topology (CCAT). As the OP points out, if $X$ is a loop space, then concatenation makes $\pi_0(X)$ a group. - -Is this the only motivation behind the study of loop spaces? Or what other interesting aspects can be gained from studying loop spaces from viewpoint of homotopy theory? - -No, there is much more to loop spaces than the observation that concatenation yields a group structure on $\pi_0(X)$. First of all, motivation is easy. Since homotopy theory is intimately tied to paths, loop spaces themselves are fundamental objects, e.g., because they allow you to shift dimension as $\pi_i(\Omega X)\cong \pi_{i+1}(X)$. The operation of taking loops is connected to the suspension operation as follows. Let $X$ and $Y$ denote based spaces, $F(X,Y)$ denote the space of based maps between them (so $\Omega X = F(S^1,X)$), and $\Sigma X = X\wedge S^1$ denote the suspension of $X$. The usual hom-tensor adjunction tells you that $F(\Sigma X,Y) \cong F(X,\Omega Y)$. Taking $\pi_0$, we have $[\Sigma X,Y] \cong [X,\Omega Y]$, and composition of loops turns this set into a group. Hence, $\Omega Y$ is a cogroup object in the homotopy category of pointed spaces, and this is used in Hovey's book to set up the much more general homotopy theory encoded by model categories. -Since the theory of $\Omega$-spectra starts with a sequence of based spaces $T_n$ and weak equivalences $T_n\to \Omega T_{n+1}$, loop spaces are also foundational to stable homotopy theory. They pop up in the long exact sequences induced by fiber and cofiber sequences, that allow us to compute things in stable homotopy theory. They pop up in Postnikov towers and Brown representability (since $K(A,n) = \Omega K(A,n+1)$). One way to prove Bott periodicity is to study the homotopy equivalence of $H$-spaces $\beta: BU \times \mathbb{Z} \to \Omega^2(BU\times \mathbb{Z})$. So, there's plenty to motivate the study of loop spaces. Let's say more about "interesting aspects." -The introduction to Adams' book Infinite Loop Spaces mentions work of Morse and Serre computing the number of geodesics on a Riemannian manifold using loop spaces, work of Serre on $H$-spaces and the Pontryagin product on $H_*(X)$, and the development of the Leray-Serre spectral sequence and its resulting homology calculations. Loop spaces give us more to compute with, and double loop spaces, $n$-fold loop spaces, and infinite loop spaces, give us even more. -As has been pointed out in the comments, the recognition principle says that $n$-fold loop spaces $\Omega^n Y$ are (up to homotopy) the same thing as $E_n$-algebras. For $n=1$, these are the same as $A_\infty$-spaces, as discussed in the preface to GILS. The Moore path space trick Naruki mentioned (parametrizing loops by $[0,t)$) gives a model for a strictly associative and unital topological monoid of loops, and the usual loop space is a deformation-retract, which is one way to understand the $A_\infty$-space structure. It's easy to show that $\Omega Y$ is a grouplike homotopy associative $H$-space, but the $A_\infty$-structure is better. -As pointed out in the link Najib provided, the $i^{th}$ stable homotopy group of $X$ is equal to $\pi_{i+k} \Sigma^{k} X = \pi_i \Omega^k \Sigma^k X$ for sufficiently large $k$, so spaces of the form $\Omega^k \Sigma^k X$ for $1\leq k \leq \infty$ contain a tremendous amount of information about $X$. As pointed out in the preface to GILS, this leads you naturally to the James construction and to Dyer-Lashof operations, which are essential for understanding the algebraic structure of the (co)homology of $X$, for an understanding of power operations, and for computations in the Adams spectral sequence. In GILS, May finds geometric approximations to these spaces, and descriptions of $H_*(\Omega^n \Sigma^n X)$ as functors of $H_*(X)$. The resulting understanding of Dyer-Lashof operations is the foundation upon which much computational work has been done, as wonderfully summarized in an article of Tyler Lawson.<|endoftext|> -TITLE: Tiling with similar tiles -QUESTION [7 upvotes]: Question 1: Is there a polygon $P$ that - -cannot tile the plane - -and - -tiles the plane when copies of $P$ and some other polygon(s) all similar in shape to $P$ but of different size(s) can be used? - -Basically, with copies of $P$ alone we should be able to form a layout with gaps which are all similar in shape to P and of different size(s). -Motivation: In basic tiling, we are constrained to use congruent copies of a candidate polygonal region (congruent up to some isometry) to fill the plane without gaps. Here, we consider relaxing the constraint to allow scaled copies of the candidate region and try to see whether this relaxation can non-trivially improve the chance a candidate region has of being a tile. -Note: Two cases to this question – $P$ is convex and not necessarily so. -Question 2: Is there a $P$ such that it is not a rep-tile but a large copy of $P$ can be tiled by several units all similar to $P$? -Note: By rep-tile, we mean a polygon that can be cut into some finite number of equally scaled down copies of itself. So, the $P$ one is looking for cannot be tiled by any finite number of equally scaled down copies of itself but can be tiled with copies of itself which have been scaled down by factors that are different among themselves. - -REPLY [5 votes]: Here is an answer to Question 2. The following shape (attributed to Karl Scherer on this website) tiles into similar shapes of different sizes. - -Convincing myself that it is not a rep-tile took me several case distinctions - I found it easiest to start with one of the right angles and construct the tiling from there until deriving a contradiction (angles of $\pi/3$ can only be "filled" in one way, right angles and angles of $2\pi/3$ can be "filled" in two different ways).<|endoftext|> -TITLE: fiberwise-quasi-compact implies quasi-compact? -QUESTION [9 upvotes]: Let $f\colon X\to \mathbb{A}^n_{\mathbb{C}}$ be a morphism of $\mathbb{C}$-schemes. Suppose $f$ is (a) separated, (b) flat, (c) locally of finite type, (d) all fibers are quasi-compact, is $X$ necessarily quasi-compact? - -REPLY [11 votes]: Here is a counterexample: -Example. We will define $X$ as a union of affine varieties -$$U_0 \subseteq U_1 \subseteq \ldots$$ -as follows: start with $U_0 = \mathbf A^1 \times (\mathbf A^1 \setminus 0) \subseteq \mathbf A^2 = V_0$ with its natural projection to $\mathbf A^1$, and let $Z_0 = \mathbf A^1 \times 0$ be the complement of $U_0$ in $V_0$. -Choose a sequence of points $x_1,x_2,\ldots$ on $\mathbf A^1$. Define $V_i$ as the blowup of $V_0$ in the points $(x_1,0), \ldots, (x_i,0)$, so we have maps -$$\ldots \to V_i \to V_{i-1} \to \ldots \to V_0.$$ -Let $E_i$ be the exceptional divisor for $V_i \to V_{i-1}$, let $Z_i$ be the strict transform of $Z_0$ in $V_i$, and let $U_i$ be its complement in $V_i$. For each $i$, the centre of the blowup $V_i \to V_{i-1}$ is contained in $Z_{i-1}$, giving an isomorphism -$$V_i\setminus(E_i \cup Z_i) \stackrel\sim\longrightarrow V_{i-1}\setminus Z_{i-1},$$ -hence an open immersion -$$U_{i-1} = V_{i-1}\setminus Z_{i-1} \cong V_i\setminus (E_i \cup Z_i) \hookrightarrow V_i \setminus Z_i = U_i.$$ -Define $X$ as the union. The maps $U_i \to \mathbf A^1$ are compatible, so they give a map $X \to \mathbf A^1$. It is flat since $X$ is integral and dominant over the Dedekind scheme $\mathbf A^1$. It is separated and locally of finite type since $X \to \operatorname{Spec} k$ is. Finally, the fibres are quasi-compact: each step $U_{i-1} \hookrightarrow U_i$ only modifies the fibre over $x_i$. But $X$ itself is not quasi-compact. $\square$<|endoftext|> -TITLE: Computational complexity of proof verification -QUESTION [5 upvotes]: Let $\mathcal{L}$ be a recursive first-order theory, with a deductive system $\Xi$ (for instance, Hilbert-Ackerman proof system). Let $\phi$ be a formula and let $l=(\psi_1, \ldots, \psi_n=\phi)$ be a sequence of formulas. - -Question 1: Suppose we what want to discuss the (asymptotical) computational complexity cost of checking wether $l$ constitutes a proof for the pair $(\mathcal{L}, \Xi)$. What are the relevant numerical parameters, depending on $L$, involved in such a complexity function, and to which complexity class it belongs (P, NP, etc)? - -Question 2: How much the complexity of verifying $l$ is a proof changes if we change the deductive system (Gentzen's style, for instance), or consider a suitable higer-order theory, or etc? - - -I apologize in advance about question 2, I hope it makes sense (albeit it is a somewhat non-rigorous question). -The motivation of these questions are the very famous work of Gödel, On the lenght of proofs, and naturally, P=NP? problem. - -REPLY [10 votes]: Your setup doesn't provide any complexity restrictions on determining whether a formula is an axiom or not, beyond demanding that this is computable. Thus, you won't be able to limit the complexity of proof verification either. -Determining the axioms is going to be the only issue, though. Any reasonable proof system will make verifiying proofs a polynomial-time problem relative to an oracle for the axioms. Typically, quadratic time would do: You read through the proof step by step, check whether the pre-requisites are really there; and if so, if the deduction step matches the rules.<|endoftext|> -TITLE: Is Gauss map of a free boundary convex disk a diffeomorphism? -QUESTION [6 upvotes]: I asked this question on MSE, but obtained no answer. Maybe this is the right place to post it. -Let $D$ be a properly embedded free boundary disk in the closed unit ball $\mathbb{B}^3$ of $\mathbb{R}^3$. This means that $D$ is a smooth disk embedded in this ball, $D \cap \partial \mathbb{B}^3 = \partial D$ and this intersection is orthogonal. By orthogonality here I mean this: if $N$ is a unit normal along $D$ (its Gauss map), then $\langle N(x), x \rangle = 0$ for all $x \in \partial D$. -Assume that $D$ is strictly convex, that is to say, the principal curvatures are positive at each point of $D$ with respect to the fixed unit normal $N: D \to \mathbb{S}^2$. Does it follow that $N$ is a diffeomorphism onto its image? Equivalently, is $N$ injective? -The motivation is the following: if $S$ is a closed and connected surface in $\mathbb{R}^3$ which is also convex, then $N : S \to \mathbb{S}^2$ is a local diffeomorphism, hence a covering map. Since $\mathbb{S}^2$ is simply connected, this implies that $N$ is a global diffeomorphism. -What happens when the surface is a disk? - -REPLY [4 votes]: The answer is yes. To show this one can use the fact that any topological immersion (locally one-to-one continuous map) of an n-dimensional disk into a sphere of the same dimension is an embedding (globally one-to-one) for $n\geq 2$, provided only that the map is one-to-one on the boundary of the disk. A proof may be found in -Gauss map, topology, and convexity of hypersurfaces with nonvanishing curvature, -Topology, 41 (2002) 107-117. -So it remains to show that $N$ is one-to-one on the boundary $\partial D$ of the disk $D$. To see this one can extend $D$ to a complete $\mathcal{C}^1$ convex surface by attaching to $\partial D$ all the rays which are orthogonal to $S^2$ from outside. These rays belong to a convex cone $C$ with apex at the center $o$ of $S^2$. Since $D$ has positive curvature, it follows from basic differential geometry that $\partial D$ has positive geodesic curvature in $S^2$, and hence is strictly convex, which in turn yields that $C$ is strictly convex. So $N$ will be one-to-one along $\partial D$, since $N$ is just the Gauss map of $C\setminus\{o\}$. -Incidentally, it is not necessary to assume that $D$ is convex or even embedded, but it is enough that it have positive curvature and satisfy the free boundary condition; see the following paper with Changwei Xiong -Nonnegatively curved hypersurfaces with free boundary on a sphere, -Calc. Var. Partial Differential Equations, 58 (2019), Art. 94, 20 pp.<|endoftext|> -TITLE: Reference on Fourier analysis on compact groups -QUESTION [5 upvotes]: I am looking for a reference for Fourier analysis on compact (Lie) groups. The kind of theorems I would like the book to cover/do are the Peter-Weyl theorem, define Fourier transforms and use the Peter-Weyl theorem to derive the Plancherel theorem.The Peter-Weyl theorem(s?) can be found in multiple references but most of the books I look into do Fourier analyis on locally compact abelian groups. -Any suggestion is welcome, thanks! I am aware of Sepanski's text but I am looking for an alternate text. - -REPLY [4 votes]: For a reference which 1) has what you need, 2) is short and elementary (only undergrad point set topology needed, e.g., Ascoli/Arzela Theorem) yet includes detailed proofs, 3) is very clear and pedagogical, 4) is free; I think you will have a hard time finding better than the note "Haar Measure" by Joel Feldman, on his UBC webpage.<|endoftext|> -TITLE: A problem in additive combinatorics -QUESTION [11 upvotes]: $\color{red}{\mathrm{Problem:}}$ -$n\geq3$ is a given positive integer, and $a_1 ,a_2, a_3, \ldots ,a_n$ are all given integers that aren't multiples of $n$ and $a_1 + \cdots + a_n$ is also not a multiple of $n$. -Prove there are at least $n$ different $(e_1 ,e_2, \ldots ,e_n ) \in \{0,1\}^n $ such that $n$ divides $e_1 a_1 +\cdots +e_n a_n$ -$\color{red}{\mathrm{My\, Approach:}}$ -We can solve this by induction (not on $n$, as we can see in Thomas Bloom's answer). But I approached in a different way using trigonometric sums. Can we proceed in this way successfully? -$\color{blue}{\text{Reducing modulo $n$ we can assume that $1\leq a_j\leq n-1$}.}$ -Throughout this partial approach, $i$ denotes the imaginary unit, i.e. $\color{blue}{i^2=-1}$. -Let $z=e^{\frac{2\pi i}{n}}$. Then $\frac{1}{n}\sum_{k=0}^{n-1}z^{mk} =1$ if $n\mid m$ and equals $0$ if $n\nmid m$. -Therefore, if $N$ denotes the number of combinations $e_1a_1+e_2a_2+\cdots+e_na_n$ with $(e_1,e_2,\ldots, e_n)\in\{0,1\}^n$ such that $n\mid(e_1a_1+e_2a_2+\cdots+e_na_n)$, then $N$ is equal to the following sum, -$$\sum_{(e_1,e_2,\ldots, e_n)\in\{0,1\}^n}\left(\frac{1}{n}\sum_{j=0}^{n-1}z^{j(e_1a_1+e_2a_2+\cdots+e_na_n)}\right)$$ -By swapping the order of summation we get, -$$N=\frac{1}{n}\sum_{j=0}^{n-1}\prod_{k=1}^{n}(1+z^{ja_k})$$ -Clearly, the problem is equivalent to the following inequality: -$$\left|\sum_{j=0}^{n-1}\prod_{k=1}^{n}(1+z^{ja_k})\right|\geq n^2\tag{1}$$ -This is actually IMO shortlist $1991$ problem $13$. No proofs are available except using induction. So if we can prove inequality $(1)$, it will be a completely new proof! In fact, inequality $(1)$ is itself very interesting. -$\color{red}{\mathrm{One\, more\, idea\, (maybe\, not\, useful):}}$ -Let $\theta_{jk}=\frac{ja_k\pi}{n}$ and $A=\sum_{k=1}^{n}a_k$, then we get, -$$(1+z^{ja_k})=\left(1+\cos\left(\frac{2ja_k\pi}{n}\right)+i\sin\left(\frac{2ja_k\pi}{n}\right)\right)=2\cos(\theta_{jk})(\cos(\theta_{jk})+i\sin(\theta_{jk}))$$ -Therefore, -$$\left|\sum_{j=0}^{n-1}\prod_{k=1}^{n}(1+z^{ja_k})\right|=2^n\left|\sum_{j=0}^{n-1}\prod_{k=1}^{n}\cos(\theta_{jk})e^{i\theta_{jk}}\right|$$ -So we get one more equivalent inequality, -$$\left|\sum_{j=0}^{n-1}\prod_{k=1}^{n}\cos(\theta_{jk})e^{i\theta_{jk}}\right|=\left|\sum_{j=0}^{n-1}e^{i\frac{\pi Aj}{n}}\prod_{k=1}^{n}\cos(\theta_{jk})\right|\geq\frac{n^2}{2^n}\tag{2}$$ -$\color{red}{\text{Remark:}}$ According to the hypothesis of the question, $n\nmid A$. Therefore $e^{i\frac{\pi A}{n}}\neq\pm1$. -Can we prove this inequality? -Any hint or help will be appreciated. Thank you! -It was posted before on Math Stack Exchange - -REPLY [2 votes]: The induction argument suggested by Thomas actually goes back to the last paper of Olson, namely J. E. Olson, A problem of Erdős on abelian groups, Combinatorica 7 (1987), 285–289.<|endoftext|> -TITLE: On hereditarily reflexive Banach spaces -QUESTION [9 upvotes]: It was proved by W.B. Johnson and H.P. Rosenthal [Studia Math. 43 (1972), 77–92] -that every Banach space $X$ with $X^{**}$ separable is hereditarily reflexive: -every infinite dimensional closed subspace of $X$ contains an infinite dimensional -reflexive subspace. -Suppose that $X$ is separable and $X^{**}/X$ reflexive. Is $X$ hereditarily reflexive? -Of course, we would have a positive answer if each infinite dimensional closed subspace of such a space $X$ contains an infinite dimensional subspace $Y$ with $Y^{**}$ separable. - -REPLY [6 votes]: The question has a negative answer: -Following the idea in Bill Johnson's comment, I looked at the work of Argyros. In this paper (see the reference below), there are several examples of hereditarily indecomposable Banach spaces containing no infinite dimensional reflexive subspaces. -One of these spaces $X$ has the additional property that $X^{**}/X$ is isomorphic to $\ell_2(\Gamma)$ with $\Gamma$ uncountable: see Proposition 5.4 in the paper. -[reference] S.A. Argyros, A.D. Arvanitakis, A.G. Tolias. -Saturated extensions, the attractors method and Hereditarily James Tree Spaces. -pp. 1-90 in "Methods in Banach space theory". J.M.F. Castillo and W.B. Johnson eds. -London Math. Soc. Lecture Note Ser. 337, Cambridge University Press, 2006.<|endoftext|> -TITLE: Fixed point stack for a torus action -QUESTION [5 upvotes]: In this paper, M. Romagny defines for an action of a group scheme $G$ on a stack $X$ the fixed point stacks $X^G$ associated -to the group action on a stack and in Theorem 3.3 he proves that if - -the group $G$ is proper and flat of finite representation -$X$ is a Deligne-Mumford stack - -then $X^G$ is algebraic. Later in this note, he proves that condition 2. can be relaxed to $X$ being algebraic with the diagonal being locally of finite presentation. -I am mostly interested in actions by complex tori on algebraic stacks locally of finite type. In this case, one doesn't have the properness from condition 1. Is it still true that the stack of fixed points $X^G$ is algebraic? -I am aware that it is common to take fixed points of Deligne-Mumford stacks as in Graber--Pandharipand with respect to the torus action, but the same approach doesn't seem to work in the completely general case. - -REPLY [4 votes]: I hadn't seen this question until Arkadij contacts me directly. The answer is yes: if $G$ is a group scheme of multiplicative type then the fixed point stack is algebraic. This is now here : https://arxiv.org/abs/2101.02450.<|endoftext|> -TITLE: Arthur's Simple Trace Formula -QUESTION [5 upvotes]: In Deligne–Kazhdan–Vigneras's "Représentations des groupes réductifs sur un corps local," they use the Simple Trace Formula to prove cases of the local Jacquet–Langlands correspondence over nonarchimedean fields. -Let's recall some setup for the Simple Trace Formula. Let $F$ be a global field, let $G$ be a connected reductive group over $F$, write $Z$ for the center of $G$, and let $\omega$ be a unitary character of $Z(F)\backslash Z(\mathbb{A})$. Write $L^2(G,\omega)$ for the space of $L^2$-functions on $G(F)\backslash G(\mathbb{A})$ where $Z(\mathbb{A})$ acts via $\omega$, and write $L^2_0(G,\omega)$ for the subspace of cusp forms. -In what follows, $v$ will range over places of $F$. Let $f=\prod_vf_v$ be a function on $G(\mathbb{A})$ on which $Z(\mathbb{A})$ acts via $\omega^{-1}$ such that every $f_v$ is a smooth function on $G(F_v)$ with compact support mod $Z(F_v)$. Such $f$ naturally yield linear operators $\rho_0(f)$ on $L^2_0(G,\omega)$. Under extra assumptions on $f$, the Simple Trace Formula says that -$$\operatorname{tr}\rho_0(f) = \sum_\gamma\operatorname{vol}(Z(\mathbb{A})G_\gamma(F)\backslash G_\gamma(\mathbb{A}))\int_{G_\gamma(\mathbb{A})\backslash G(\mathbb{A})}dg\,f(g^{-1}\gamma g),$$ -where $\gamma$ ranges over elliptic regular conjugacy classes in $Z(F)\backslash G(F)$, and $G_\gamma$ denotes the centralizer of $\gamma$ in $G$. -Usually, the extra assumptions on $f$ are that $f_v$ is supercuspidal at one place $v$ and $f_{v'}$ is supported on the elliptic regular elements of $G(F_{v'})$ at one place $v'$. However, in part e. of the introduction of "Représentations des groupes réductifs sur un corps local," Deligne–Kazhdan–Vigneras say that Arthur announced the Simple Trace Formula also holds when $f_v$ and $f_w$ are supercuspidal at two places $v$ and $w$. -Question: Does the Simple Trace Formula indeed hold in this case? If so, where could I find a proof? Thank you in advance! - -REPLY [5 votes]: Yes, I have not known Deligne, Kazhdan and Vigneras to lie. A sketch of the proof, at least with the key details for GL(2), is given in Lecture V of - -Steve Gelbart, Lectures on the Arthur--Selberg Trace Formula - -Added remarks: In that Lecture, Gelbart addresses both two kinds of simple traces formulas. The one you are asking about is essentially Prop 2.1. While Gelbart states the hypotheses in terms of vanishing of hyperbolic orbital integrals at two places, in the proof he explains how this is related to vanishing of unipotent orbital integrals, which is the condition to be a supercusp form at 2 places. As I remember, while technically he doesn't state that being a supercusp form at 2 places suffices, you should be able to work this out from what he does, at least in the case of GL(2). You'll probably need to look at papers of Arthur, Deligne, Kazhdan, Vigneras, etc for details general G. For me personally, reading Rogawski was also useful for understanding these things.<|endoftext|> -TITLE: Galois embedding question for dihedral groups -QUESTION [9 upvotes]: Let $D_n$ be the dihedral group of order $2n$. Then all the quotients of $D_n$ are dihedral as well, and of the form $D_k$ with $k \mid n$. So for a field $K/\mathbb{Q}$ with $\operatorname{Gal}(K/\mathbb{Q}) \cong D_n$, there exists, for any $k \mid n$, a subfield $F \subseteq K$ with $\operatorname{Gal}(F/\mathbb{Q}) \cong D_k$. -My question is about the reverse question. Given a number field $F/\mathbb{Q}$ with $\operatorname{Gal}(F/\mathbb{Q}) \cong D_k$, is there a field $K \supset F$ such that $\operatorname{Gal}(K/\mathbb{Q}) \cong D_n$ for any $n$ a multiple of $k$? -I've been told that this is called the "Galois embedding problem" and is not true for many types of groups. I was wondering if anyone could point me in the right direction for what is known about this in the dihedral case. -Thanks, -MC - -REPLY [8 votes]: The answer is "no", in general, since there may be local obstructions. Suppose, for example, that $k$ and $n$ are odd prime powers, and let $L/\mathbb{Q}$ be the unique intermediate quadratic in $F$. A necessary condition for the existence of $K$ is that the cyclic degree $k$ extension $F/L$ embeds into a cyclic degree $n$ extension $K/L$. Every prime $\mathfrak{p}$ of $L$ that is totally ramified in $F$ would have to be totally ramified in $K$, so if the residue characteristic of $\mathfrak{p}$ is coprime to $n$, then you need $n$ to divide the order of the multiplicative group of the residue field of $\mathfrak{p}$. This is a genuine restriction. For concreteness, take $k=3$, $n=9$. Then there are infinitely many $D_3$ extensions of $\mathbb{Q}$ in which $7$ is split in the quadratic and ramified in the cubic, but none of them embed inside a $D_{9}$ extension, because $7$ cannot be totally tamely ramified of degree $9$. -You can upgrade such local conditions to also ensure that the bottom cyclic group of order $2$ normalises the top group and acts on it by $-1$, so that the whole extension is dihedral. If all such local conditions are satisfied, then you can prove with class field theory that the sought-for embedding always exists. In particular if $k$ and $n/k$ are coprime, then the embedding will always exist. See [1, Section 3.1] for some examples of how such constructions work. -[1] A. Bartel, Large Selmer groups over number fields, Math. Proc. Cambridge Philos. Soc. 148 no. 1 (2010), pp. 73–86.<|endoftext|> -TITLE: Binomial theorem for content polynomials of partitions -QUESTION [6 upvotes]: Let $\lambda$ be a partition, represented by a usual Young diagram in which $1\le i\le \ell(\lambda)$ labels the rows and, for each $i$, $1\le j\le \lambda_i$ labels the columns. For each box $\square$ in the diagram, $c(\square)=j-i$ is its content. The polynomial -$$ P_\lambda(x)=\prod_{\square\in\lambda}(x+c(\square))=\prod_{(i,j)\in\lambda}(x+j-i)$$ -is the content polynomial. -I would like to know some family of coefficients $b^\lambda_{\mu\nu}$, as a function of three partitions, such that -$$P_\lambda(x+y)=\sum_{\mu,\nu\subset \lambda}b^\lambda_{\mu\nu}P_\mu(x)P_\nu(y).$$ -(Notice that content polynomials are not linearly independent, so this equation alone is not enough to uniquely determine the coefficients. I ask for some family) - -REPLY [10 votes]: If you lift this to the level of symmetric functions then the structure constants are uniquely determined. Suppose $x$ denotes a set of $m$ variables and $y$ denotes a set of $n$ variables. Then you can start with the identity of schur polynomials -$$s_{\lambda}(x,y)=\sum_{\mu,\nu}c_{\mu,\nu}^{\lambda}s_{\mu}(x)s_{\nu}(y)$$ -where $c_{\mu,\nu}^{\lambda}$ are the Littlewood-Richardson coefficients. Setting all the variables to $1$ gives -$$P_{\lambda}(m+n)=\sum_{\mu,\nu}b_{\mu,\nu}^{\lambda}P_{\mu}(m)P_{\nu}(n)$$ -where $$b_{\mu,\nu}^{\lambda}=c_{\mu,\nu}^{\lambda}\frac{\prod_{\square\in\lambda}h_{\square}}{\prod_{\square\in\mu}h_{\square}\prod_{\square\in\nu}h_{\square}}$$ -and the $h_{\square}$ represent the hook lengths of each partition.<|endoftext|> -TITLE: Looking for sufficient conditions for positive Fourier transforms -QUESTION [12 upvotes]: I am looking for some sufficient conditions for an even, continuous, nonnegative, non-increasing, non-convex function to be non-negative definite. In other words -$$ -\int_0^\infty f(x)\cos(x\omega) \, dx\ge 0, \quad \omega \in \mathbb{R}. -$$ -The function $f(x)=\exp(-x^{\alpha})[x^{\alpha} \log(x) + \delta/\alpha ]$, $\alpha\in(1,2)$, $\delta\ge 2$. Note that for $\alpha\in(0,1)$, $f(x)$ is convex. -I have tried complex detour numerically (as in Tuck, E. O.: On positivity of Fourier transforms, Bull. Austral. Math. Soc., 74 (2006) 133– -138) – did not work out. It seems the imaginary part always increases around $0$. -I have tried Polya type conditions () - no luck either. -For $\alpha\in(0,1)$ everything works out as the function is convex. -Thank you in advance for any hints or references! - -REPLY [2 votes]: How about the following reference T. Gneiting, Criteria of Po´lya type for radial positive definite functions, Proc. Am. Math. Soc. 129 (2001), 2309–2318. I have plotted the results for 6th derivative all numeric but it seems working,<|endoftext|> -TITLE: A $q$-analogue of a characterization of polynomials by binomial coefficients -QUESTION [5 upvotes]: Considering the binomial coefficient $\binom{x}{m}$ as a polynomial in $x$, the span of $\binom{x}{0}, \binom{x}{1}, \ldots, \binom{x}{d}$ is exactly the polynomials of degree $\le d$. A closely related characterization is that this subspace is the kernel of $\Delta^{d+1}$, where $\Delta : \mathbb{C}[x] \rightarrow \mathbb{C}[x]$ is the difference operator defined by $(\Delta P)(x) = P(x) - P(x-1)$. Roughly stated, my question asks for a $q$-analogue of either of these results. -To make this more concrete, use the standard notation, so $[n]_q = 1+q+\cdots + q^{n-1}$, $[n]_q! = [n]_q[n-1]_q\ldots [1]_q$, and $\binom{n}{d}_q = \frac{[n]_q!}{[d]_q![n-d]_q!}$. Fix $N \in \mathbb{N}$ and for each $m \in \mathbb{N}_0$ let -$$u^{(m)}_q = \bigl( \binom{0}{m}_q, \binom{1}{m}_q, \ldots, \binom{N-1}{m}_q \bigr) \in \mathbb{C}[q]^{N}$$ -By the first paragraph, for $P \in \mathbb{C}[x]$, we have -$$\bigl( P(0), P(1), \ldots, P(N-1) \bigr) \in \bigl\langle u^{(0)}_1, u^{(1)}_1, \ldots, u^{(d)}_1 \bigr\rangle$$ -if and only if $\mathrm{deg} P \le d$. In a current research problem, it's useful that the same holds for $( P(N-1), \ldots, P(1), P(0))$; in fact the evaluation points $0$, $1, \ldots, N-1$ can be varied by an arbitrary affine transformation. - -Is there an analogous characterization of the span $\langle u_q^{(0)}, u_q^{(1)}, \ldots, u_q^{(d)} \rangle$? - -As a follow up, what transformations preserve this space? In particular, is it invariant under a $q$-analogue of the affine transformation just mentioned? Despite some thought I have not found any reasonable answer to these questions, but I find it hard to believe that there is nothing to be said. - -REPLY [3 votes]: We have -$$\binom{n}d_q=\frac{(q^n-1)\ldots(q^n-q^{d-1})}{(q^d-1)\ldots(q^d-q^{d-1})}=f_d(q^n)=g_d([n]_q) -$$ -where $f_d$ and $g_d$ are polynomials of degree $d$ (depending on $q$ of course). Therefore $$\bigl( P([0]_q), P([1]_q), \ldots, P([N-1]_q) \bigr) \in \bigl\langle u^{(0)}_q, u^{(1)}_q, \ldots, u^{(d)}_q \bigr\rangle$$ -if and only if $\deg P\leqslant d$.<|endoftext|> -TITLE: What are some examples of proving that a thing exists by proving that the set of such things has positive measure? -QUESTION [42 upvotes]: Suppose we want to prove that among some collection of things, at least one -of them has some desirable property. Sometimes the easiest strategy is to -equip the collection of all things with a measure, then show that the set -of things with the desired property has positive measure. Examples of this strategy -appear in many parts of mathematics. - -What is your favourite example of a proof of this type? - -Here are some examples: - -The probabilistic method in combinatorics As I understand it, a -typical pattern of argument is as follows. We have a set $X$ and want to -show that at least one element of $X$ has property $P$. We choose some -function $f: X \to \{0, 1, \ldots\}$ such that $f(x) = 0$ iff $x$ satisfies -$P$, and we choose a probability measure on $X$. Then we show that -with respect to that measure, $\mathbb{E}(f) < 1$. It follows that -$f^{-1}\{0\}$ has positive measure, and is therefore nonempty. - -Real analysis One example is Banach's -proof -that any measurable function $f: \mathbb{R} \to \mathbb{R}$ satisfying -Cauchy's functional equation $f(x + y) = f(x) + f(y)$ is linear. -Sketch: it's enough to show that $f$ is continuous at $0$, since then -it follows from additivity that $f$ is continuous everywhere, which makes -it easy. To show continuity at $0$, let $\varepsilon > 0$. An -argument using Lusin's theorem shows that for all sufficiently small -$x$, the set $\{y: |f(x + y) - f(y)| < \varepsilon\}$ has positive -Lebesgue measure. In particular, it's nonempty, and additivity then -gives $|f(x)| < \varepsilon$. -Another example is the existence of real numbers that are -normal (i.e. normal to every base). -It was shown that almost all real numbers have this property -well before any specific number was shown to be normal. - -Set theory Here I take ultrafilters to be the notion of measure, an -ultrafilter on a set $X$ being a finitely additive $\{0, 1\}$-valued -probability measure defined on the full $\sigma$-algebra $P(X)$. Some -existence proofs work by proving that the subset of elements with the -desired property has measure $1$ in the ultrafilter, and is therefore nonempty. -One example is a proof that for every measurable cardinal -$\kappa$, there exists some inaccessible cardinal strictly smaller than -it. Sketch: take a $\kappa$-complete ultrafilter on $\kappa$. Make an inspired choice of function $\kappa \to \{\text{cardinals } < -\kappa \}$. Push the ultrafilter forwards along this function to give -an ultrafilter on $\{\text{cardinals } < \kappa\}$. Then prove that the set -of inaccessible cardinals $< \kappa$ belongs to that ultrafilter ("has -measure $1$") and conclude that, in particular, it's nonempty. -(Although it has a similar flavour, I would not include in this list the cardinal arithmetic proof of the -existence of transcendental real numbers, for two reasons. First, -there's no measure in sight. Second -- contrary to -popular belief -- this argument leads to an explicit construction -of a transcendental number, whereas the other arguments on this list -do not explicitly construct a thing with the desired properties.) - - -(Mathematicians being mathematicians, someone will probably observe that -any existence proof can be presented as a proof in which the set of things -with the required property has positive measure. Once you've got a thing -with the property, just take the Dirac delta on it. But obviously I'm -after less trivial examples.) -PS I'm aware of the earlier question On proving that a certain set is -not empty by proving that it is actually -large. That has some good -answers, a couple of which could also be answers to my question. But my -question is specifically focused on positive measure, and excludes -things like the transcendental number argument or the Baire category -theorem discussed there. - -REPLY [7 votes]: A very famous and important theorem in the theory of metric embeddings is known as "Assouad's Embedding Theorem". It concerns doubling metric spaces: metric spaces for which there is a constant $D$ such that every ball can be covered by $D$ balls of half the radius. - -Theorem (Assouad, 1983): For every $\epsilon\in (0,1)$ and $D>0$, there are constants $L$ and $N$ such that if $(X,d)$ is doubling with constant $D$, then the metric space $(X,d^\epsilon)$ admits an $L$-bi-Lipschitz embedding into $\mathbb{R}^N$. - -This theorem is widely used throughout metric geometry and analysis on metric spaces. (See, e.g., here or here.) -An $L$-bi-Lipschitz embedding is simply an embedding that preserves all distances up to factor $L$. It's easy to see that the doubling condition is necessary for this theorem to hold. Moreover, there are known doubling metric spaces (the Heisenberg group for one) that are doubling but do not admit a bi-Lipschitz embedding into any Euclidean space, so one cannot allow $\epsilon=1$ in Assouad's theorem. -This means, of course, that the constants $L$ and $N$ must blow up as $\epsilon\rightarrow 1$, and this is reflected Assouad's proof. -Except, that's not quite true. In a really surprising construction, Naor and Neiman showed in 2012 that the dimension $N$ in Assouad's theorem can be chosen independent of the ``snowflake'' parameter $\epsilon$ as $\epsilon\rightarrow 1$. (The distortion $L$ must necessarily blow up in general.) In other words, one need not use too many dimensions for the embedding, no matter how close $\epsilon$ gets to $1$. I believe this shocked many people. -The construction of Naor and Neiman is probabilistic: they construct a random Lipschitz map from $(X,d^\epsilon)$ into $\mathbb{R}^N$, and show that it is bi-Lipschitz with positive probability. The proof is also a nice application to geometry of the Lovasz Local Lemma. -Assouad's paper: http://www.numdam.org/article/BSMF_1983__111__429_0.pdf -Naor-Neiman's paper: https://www.cs.bgu.ac.il/~neimano/Naor-Neiman.pdf<|endoftext|> -TITLE: Is the Hilbert cube the countable union of punctiform spaces? -QUESTION [5 upvotes]: Recall that a (separable) metric space is called punctiform, if all its compact subspaces are zero-dimensional. While "natural" spaces would seem to be punctiform if they already themselves zero-dimensional, there are even infinite dimensional punctiform spaces. The constructions I have seen however are still yielding spaces that "feel sparse" to me. -The Hilbert cube $[0,1]^\omega$ is large in the sense that it is not a countable union of zero-dimensional spaces. What I am now wondering is whether we can write the Hilbert cube as a countable union of punctiform spaces. Note that I do not want to impose any complexity constraints on the pieces. -If the answer should be "yes", I'd be very interested in understanding the structure of the punctiform spaces involved. -Had the answer been "no", this would have answered an open problem in computability theory, see Question 5 on Page 99 (v1) here: https://arxiv.org/abs/1904.04107 - -REPLY [5 votes]: The Hilbert cube can be written as the union of two punctiform spaces. Just take any Bernstein set $X\subset[0,1]^\omega$ and observe that compact subsets in $X$ and $Y=[0,1]^\omega\setminus X$ are at most countable. So, $X$ and $Y$ are punctiform spaces and $X\cup Y=[0,1]^\omega$. -A less trivial fact says that the Hilbert cube cannot be written as the countable union of hereditarily disconnected sets; for the proof of this fact, see Main Lemma in this paper. A topological space $X$ is called hereditarily disconnected if all connected subspaces in $X$ are singletons.<|endoftext|> -TITLE: Closure of the product of subfunctors -QUESTION [5 upvotes]: Background: - -Let $X: \textbf{CRing} \to \textbf{Set}$ be a presheaf on the category of affine schemes and $Z \subseteq X$ a subfunctor. One defines $Z$ to be closed if for every ring $A$ and every morphism $f: \text{Hom}(A , -) \to X$ the inverse image $f^{-1}(Z)$ is of the form $R \mapsto \{ \varphi : A \to R | \varphi(I) = 0 \}$ for some ideal $I \subseteq A$. -The intersection of subfunctors is defined naively, as is the closure (denoted by $\overline{Z}$) of a subfunctor $Z \subseteq X$ (it is the intersection of all closed subfunctors of $X$ containing $Z$). -If $Y$ is another presheaf, the product of $X$ and $Y$ is also defined naively. - -Context: In section 1.14 of Jens Jantzen's great book "Representations of Algebraic Groups", the following is stated: If $X$ and $Y$ are presheaves which are schemes over a noetherian ring $k$ and $Z \subseteq X$ is a subscheme, and if $Z, X$ are algebraic and $Y$ is flat, then $\overline{Z \times Y} = \overline{Z} \times Y$. For the proof, he references Demazure-Gabriel I, section 2, 4.14 (although in my copy of Bell's translation this reference unfortunately doesn't exist). -Actual Question: -Is this true for general presheaves? I.e. if $X$ and $Y$ are presheaves and $Z \subseteq X$ is a subfunctor, is it true that $\overline{Z \times Y} = \overline{Z} \times Y$? I worry that it isn't because of the conditions in Jantzen stated above, but I haven't been able to decide either way. (Also side question: does anyone know the correct reference in the translation?) - -REPLY [4 votes]: This is not true even for affine schemes. Let $k = \mathbb{Z}$, let $X = \operatorname{Spec} \mathbb{Z}$, let $Y = \operatorname{Spec} \mathbb{F}_p$, and let $Z \cong \operatorname{Spec} \mathbb{Z} [ p^{-1}]$. The closure of $Z$ in $X$ is $X$ itself, but $Z \times Y \cong \operatorname{Spec} \{ 0 \}$, which is already closed in $X \times Y$. Of course, $Y$ is not flat.<|endoftext|> -TITLE: Smallest family of subsets that generates the discrete topology -QUESTION [12 upvotes]: If $X$ is a finite set, what is the smallest (in cardinality) family of open subsets $\mathcal U\subseteq 2^X$ such that $\mathcal U$ generates the discrete topology, i.e. if $\mathcal U\subseteq \tau\subseteq 2^X$ and $\tau$ is a topology, then $\tau=2^X$? - -REPLY [17 votes]: Let $\mathcal{U}=\{A_1,\ldots,A_k\}$. Then for any element $x\in X$ there should exist a set $I(x)\subset \{1,\ldots,k\}$ such that $\cap_{i\in I(x)} A_i=\{x\}$. Note that $I(x)$ is not contained in $I(y)$ for $x\ne y$. Therefore $|X|\leqslant \binom{k}{\lfloor k/2\rfloor}$ by Sperner's theorem. On the other hand, if $|X|\leqslant \binom{k}{\lfloor k/2\rfloor}$, we may construct an injection $f$ from $X$ to $\lfloor k/2\rfloor$-subsets of $\{1,\ldots,k\}$ and define $A_i=\{x:i\in f(x)\}$. Then $\cap_{i\in f(x)} A_i=\{x\}$. -So the answer is the minimal $k$ for which $|X|\leqslant \binom{k}{\lfloor k/2\rfloor}$. - -REPLY [9 votes]: Here is a construction that is within a constant factor of optimal. -You can find such an $\cal U$ containing $2 \lceil \log_2 |X|\rceil $ sets: identify $X$ with a subset of $\{0,1\}^{\lceil \log_2 |X|\rceil}$ and take $U_i$ to be the set of all elements whose $i$-th coordinate is $0$, and $V_i$ to be the set of all elements whose $i$-th coordinate is $1$. Then each singleton is an intersection of appropriate sets $U_i$ and $V_i$, so the generated topology includes all singletons and thus is discrete. -On the other hand, we cannot do much better than that: if $\cal U$ contains fewer than $\log_2 |X|$ sets, then there are two points contained in exactly the same sets of $\cal U$ (and clearly these points cannot be distinguished by the resulting topology). To find such a pair, let $U_1,U_2,\dots$ be an enumeration of the elements of $\cal U$. Let $X_1 = X$ and inductively let $X_{i+1}$ be the larger set of $X_i \cap U_i$ and $X_i \setminus U_i$. Note that in every step we keep at least half of the elements, hence the last $X_i$ contains at least two elements.<|endoftext|> -TITLE: On folding a polygonal sheet -QUESTION [5 upvotes]: Consider a polygonal sheet $P$ of area $A$ with $N$ vertices (it material is not stretchable or tearable). Let $n$ be a positive integer >=2. -Question: Let $P$ lie on a flat plane. We need to fold up $P$ so that it now occupies only an area $A/n$ of the plane. It is also needed that the folding is as uniform as possible - ie the number of layers of the material above any given point should be as close to $n$ as possible. We need an algorithm that does it and an estimate of its complexity. -Example: If $P$ is a rectangle of area $A$ and $n$ is an integer, it is easy to see that we can fold it to an area $A/n$ such that it is exactly $n$ layers thick throughout - the 'creases' could simply be $n-1$ equally spaced parallel lines. It appears that no other shape of $P$ has this property of 'perfectly uniform foldability'. Which is the shape(s) of $P$ that causes greatest variation in the number of layers for a given $n$? -Further possibilities: One can further ask: Minimize the perimeter of the area $A/n$ region that is covered by the folded polygon. Alternatively, We could require $P$ to be folded as uniformly as possible so that it can be packed into a rectangular or square box of some specified dimensions - and area not necessarily equal to $A/n$ with n an integer. - -REPLY [2 votes]: For every $k\ge 2$ there is a non-rectangular shape allowing a uniform folding for all $n$ that are multiplies of $k$: - -The reason is that you can uniformly fold it into a rectangle with $k$ layers. - -If you are looking for convex shapes, then $k=2$ above is convex. -Here is a non-rectangular convex shape admitting a uniform folding with three layers. -If you want to make it rectangular you will get six layers, but then you can proceed in all multiples of 6. - -More generally, every regular $n$-gon admits a $2n$-layer folding. And it can be further made into a rectangular $4n$-layer folding (and then every multiple thereof). - -The differently colored creases in the pentagon are to be understood as alternatingly folded upwards and downwards. -Or even better, for every $n$ there is a non-rectangular convex shape admitting an $n$-layered folding, or a $2n$-layerd rectangular folding (and then every multiple thereof). - -So an interesting question might be whether for every $n$ there is a non-rectangular convex shape admitting a rectangular $n$-layer folding.<|endoftext|> -TITLE: Name for a class of almost symplectic manifolds -QUESTION [6 upvotes]: A $2n$-dimensional manifold $M$ is said to be almost symplectic if it possesses a non-degenerate two-form $\omega \in \Omega^2(M)$. Equivalently, an almost symplectic structure is a $G$-subbundle $P \subset F(M)$ of the frame bundle where $G < GL(2n,\mathbb{R})$ is isomorphic to the symplectic group $Sp(2n,\mathbb{R})$. -The intrinsic torsion of such a $G$-structure is captured by the three-form $d\omega \in \Omega^3(M)$. The bundle $\wedge^3 T^*M$ breaks up into the Whitney sum of two $G$-stable sub-bundles corresponding to the $\omega$-traceless 3-forms and their $\omega$-perpendicular complement. This therefore gives rise to four types of almost symplectic manifolds: - -symplectic, where $d\omega = 0$ -locally conformal symplectic, where $d\omega = \omega \wedge \varphi$ for some one-form $\varphi$ which is closed and hence $\varphi = df$ locally, allowing us to construct a local symplectic form $e^{-f}\omega$. -name?, where the volume form $\omega^n$ is left invariant by the hamiltonian vector fields $X_f = \omega^\sharp(df)$ -generic, where $d\omega$ is none of the above. - -My question is whether there is an accepted name for the third type. I would also appreciate a link to where this classification was made explicit for the first time. -Thank you. -Edit As Robert Bryant pointed out below, the condition name? is actually $d\omega^{n-1} = 0$. I will leave the question unmodified, except for this. - -REPLY [4 votes]: I'm a bit confused by your question, because I believe that, if one defines an $\omega$-Hamiltonian vector field to be a vector field of the form $X_f = \omega^\#(\mathrm{d}f)$ where $f$ is a (smooth) function on $M$, then $\omega^{n}$ is always invariant under the flow of $X_f$. -To see this, recall that, when $n>1$, if $\omega$ is a non-degenerate $2$-form on $M^{2n}$, its exterior derivative can be written uniquely in the form -$$ -\mathrm{d}\omega = \phi\wedge\omega + \psi -$$ -where $\phi$ is a $1$-form and $\psi\in\Omega^3(M)$ is $\omega$-primitive, i.e., $\omega^{n-2}\wedge\psi = 0$. -Meanwhile, by Cartan's formula for the Lie derivative with respect to $X_f$, we have, since $\iota(X_f)\omega = -\mathrm{d}f $ (where $\iota(X)$ denotes interior product with $X$), -$$ -\begin{align} -\mathcal{L}_{X_f}\omega^n &= n\,\omega^{n-1}\wedge \mathcal{L}_{X_f}\omega -= n\,\omega^{n-1}\wedge\bigl(\iota(X_f)(\mathrm{d}\omega) + \mathrm{d}(\iota(X_f)\omega)\bigr)\\ -&=n\,\omega^{n-1}\wedge\bigl(\iota(X_f)(\phi\wedge\omega + \psi) + \mathrm{d}(-\mathrm{d}f))\bigr)\\ -&=n\,\omega^{n-1}\wedge\bigl(\phi(X_f)\wedge\omega + \phi\wedge\mathrm{d}f+ \iota(X_f)\,\psi)\bigr)\\ -&= n\,\omega^{n-1}\wedge\bigl(\iota(X_f)\,\psi\bigr) =0 -\end{align} -$$ -since $\omega^{n-2}\wedge\psi=0$ implies -$$ -0 = \iota(X_f)(\omega^{n-1}\wedge\psi) = (n{-}1)\,\omega^{n-2}\wedge(-\mathrm{d}f)\wedge\psi + \omega^{n-1}\wedge\bigl(\iota(X_f)\,\psi\bigr). -$$ -Instead of the 'Hamiltonian flow invariance' criterion you propose for name? (by which, I think, you are trying to capture the condition $\phi=0$), you should instead just require that $\omega^{n-1}$ be closed. (This condition is sometimes known as 'balanced' in the literature.) -By the way, your second type "locally conformally symplectic" is only apt if $n>2$. When $n=2$, you do not automatically get that $\phi$ is closed from the condition $\mathrm{d}\omega = \phi\wedge\omega$. (In fact, it is generically not true, even though $\psi$ does vanish identically when $n=2$.)<|endoftext|> -TITLE: Rectangles in rectangles and $(b^2-a^2)^2\le (ax-by)^2+(bx-ay)^2$ -QUESTION [6 upvotes]: When does an $a\times b$ rectangle fit inside an $x\times y$ rectangle? I have an algebraic condition which I can diagram geometrically, and I'd like a good geometric argument. -Assume $0 -TITLE: A global mathematics library -QUESTION [23 upvotes]: Just for personal interest, I am not (yet) professionally involved in it. My question is about the state of arts in digitalization of mathematics and to what extent it is possible and reasonable. -There are different levels of digitalizations: - -OCR scan all historical mathematical texts -organize metadata of references and authors (e.g. as graphs) -extract mathematical objects (like theorems, definitions, etc) -extract proofs and ideas -formalize mathematics such that it can be completely checked by theorem provers - -The main effort I found is -https://imkt.org/ -Steps 3/4 and 5 may be of independent interest and should be understood more parallel than chronological. Point 5 is more interesting in having (error-free) formalized math. It should also be allowed to choose different foundations of mathematics and the possibility to switch in between them. Point 3/4 is more interesting for a researcher that wants all references for a definition, a theorem, a keyword. It would be a wonderful source for doing data analysis of mathematical knowledge (historical, social, semantical, etc). In contrast to 5 it can contain errors and speculations. The main interest is in identifying and referencing mathematical objects over all the produced texts in history of math. -My question is: -The goal of https://imkt.org/ is huge but when looking at its first projects it looks (sry) a little bit disappointing. The main focus (also of other literature that I scanned through) lies on connecting the existing databases and languages, i.e. of theorem provers, computer algebra systems (and maybe wikis). I understand that different applications in math demand different systems (e.g. integer series http://oeis.org/ should also be part of it?) Can or shouldn't there be one system that contains everything that can be accessed (and is stored not just referenced!) through the same system? Are my dreams of such a system over the top? - -One of the largest issues is the copyright of the big publishers. -More and more is going in direction open math. Until then it is -unclear to what extent a library can be complete (gaps are somehow -missing the point of this system). - -The other issue is the efficiency in producing the content extraction -and to advance it by advertising the advantages of such a library to -mathematicians such that it will move itself at some point. - - -There have been many efforts in the past that were abandoned again or here for years (like Mizar) but far from being known and used in daily mathematics. - -REPLY [3 votes]: There's a lot of work lately on formalizing large parts of classical mathematics using proof assistants, see for example the Lean prover and the accompanying library mathlib. You can see an overview of what has been formalized here; it includes a lot of classical abstract algebra and real analysis, although there's also a lot of undergraduate math they haven't formalized yet which they list here. If you think that's all "trivial", there's ongoing work to formalize the proof of the independence of the continuum hypothesis, the proof that it's possible to evert the sphere, etc. -An interesting facet of this project is that all the proofs are kept under a version control system and you can see the entire history of every proof. It's common for the initial proof of some result to be very long and verbose, only for people to find ways to shorten it or make it more elegant later -- for example, von Neumann's proof of the Radon-Nikodym theorem. Where before you might have had to trace through the citations of loads of historical papers that your library might not even have, now you can see exactly how this process played out by just doing git blame.<|endoftext|> -TITLE: Examples of $\aleph_0$-categorical nonhomogeneous structures -QUESTION [10 upvotes]: Macpherson in a survey of homogeneous structures, states that there are many $\aleph_0$-categorical structures which are not homogeneous. I'd like to know more of such examples. -Edit: The homogeneity mentioned here is the ultrahomogeneity that is defined as every isomorphism between two finite substructures of a structure $M$ can be extended to an automorphism of $M$. There is another homogeneity known as $ℵ_0$-homogeneity that is defined as two $n$-tuples with the same type in $M^n$ must be on the same orbit of $\rm Aut$$(M)$. -$\aleph_0$-categorical structures need not be ultrahomogeneous, but are always $\aleph_0$-homogeneous. So $\aleph_0$-homogeneous is a weaker notation than ultrahomogeneous in general. These two types of homogeneity become equivalent if and only if $\rm Th$$(M)$ has quantifier elimination. - -REPLY [12 votes]: Here is my favorite example. - -Theorem. Fix $n\geq 1$. Then there is a unique (up to isometry) countable metric space $(M_n,d)$ satisfying the following properties: - -$d(x,y)\in\{0,1,2,\ldots,n\}$ for all $x,y\in M_n$ -Any finite metric space with distances in $\{0,1,2,\ldots,n\}$ embeds as a subspace of $M_n$. -Any partial isometry between two finite subspaces of $M_n$ extends to a total isometry of $M_n$. - - -I don't really know who to attribute this to (perhaps Casanovas & Wagner, or Delhomme, Laflamme, Pouzet, Sauer). The point is that the class of finite metric spaces with distances in $\{0,1,\ldots,n\}$ is a Fraisse class in an appropriate relational language, and so $M_n$ is the Fraisse limit. In particular, for all $k\leq n$, add a binary relation $d_k(x,y)$ interpreted as "$d(x,y)\leq k$". In this language, $M_n$ is homogeneous. But.... -View $M_n$ as a graph under just the relation $d_1(x,y)$. Then $M_n$ is still $\aleph_0$-categorical because the metric is "definable" from the graph language using existential quantifiers. (Specifically, the defining properties of $M_n$ force the metric $d$ to coincide with the "path metric" given by distance $1$. E.g, $d(x,y)\leq 2$ iff $\exists z(d_1(x,z)\wedge d_1(z,y))$.) However, in the graph language, $M_n$ is homogeneous if and only if $n=1$ or $n=2$. Indeed, if $n\geq 3$ then we can find points $a,b,c,d\in M_n$ such that $d(a,b)=2$ and $d(c,d)=3$. In the graph language, the finite substructures $\{a,b\}$ and $\{c,d\}$ are isomorphic. But an automorphism of $M_n$ has to respect the metric, and so there is no automorphism sending $(a,b)$ to $(c,d)$. -The point, of course, is that by looking only at the distance $1$ relation, we lose quantifier elimination. -Note, on the other hand, $M_1$ is a countably infinite complete graph and $M_2$ is the countable Rado graph, both of which are homogeneous as graphs. -By the way, $M_n$ is an example of what is called a metrically homogeneous graph. For more on these, see Cherlin's work on the classification program for these graphs.<|endoftext|> -TITLE: Extending a holomorphic vector bundle: a reference request -QUESTION [9 upvotes]: Let $Y$ be a complex manifold, $X\subset Y$ a compact submanifold, and $E\to X$ a holomorphic vector bundle. Can $E$ be extended -to a bundle over an open neighborhood of $X$ in $Y$? (Four years ago I have asked this question on MO Extending the tangent bundle of a submanifold for the case $E=T_X$.) -After tinkering with this problem for a while I found a necessary condition, there is an invariant in $H^2(X, \mathcal{N}_{X/Y}^*\otimes End(E))$ which must be zero for an extension to be possible. So far, so good. -Now I think about writing it down and submitting somewhere (assuming it is a new result). But, in a decent paper there are supposed to be references to known results in the same direction, right? And this is what the real problem is: damned if I have a clue where to look! It is all miles away from areas I am familiar with (mostly differential geometry), and this far I could not find anything remotely relevant. So, it would be nice if someone helps me with this. - -REPLY [11 votes]: It seems to me that your result follows by Proposition 1.1 of the paper -P. A. Griffiths: The extension problem in complex analysis. II: Embeddings with positive normal bundle, Am. J. Math. 88, 366-446 (1966). ZBL0147.07502, -that can be freely downloaded here. The statement is the following: - -Proposition 1.1 (Griffiths 1966). If $\alpha$ is a holomorphic vector bundle $\mathbf{E} \to X$, then $$\omega(\alpha_{\mu-1}) \in H^2(X,\,\Omega(\mathrm{Hom}(\mathbf{E}, \, \mathbf{E})(\mu))).$$ - -Here $\omega(\alpha_{\mu-1})$ is the obstruction to extending $\mathbf{E}$ to the $\mu$th infinitesimal neighborhood $X_{\mu}$ of $X$ in $Y$, provided that you already have an extension $\alpha_{\mu-1}$ to $X_{\mu-1}$, and $$H^2(X,\,\Omega(\mathrm{Hom}(\mathbf{E}, \, \mathbf{E})(\mu))= H^2(X, \mathrm{End}(\mathbf{E}) \otimes \mathrm{Sym}^{\mu}(N_{X/Y}^*)).$$ -In order to have an extension of $\mathbf{E}$ to a genuine analytic neighborhood of $X$ in $Y$, all these obstruction classes must vanish. In fact, if I understand correctly, you only provided the obstruction class for the extension of $\mathbf{E}$ to the first infinitesimal neighborhood $X \subset X_1$ of $X$ in $Y$.<|endoftext|> -TITLE: An abstract characterization of line integrals -QUESTION [6 upvotes]: Let $M$ be a smooth manifold (endowed with a Riemann structure, if useful). If $\omega \in \Omega^1 (M)$ is a smooth $1$-form and $c : [0,1] \to M$ is a smooth curve, one defines the line integral of $\omega$ along $c$ as -$$I(\omega, c) = \int _0 ^1 \omega_{c(t)} (\dot c (t)) \ \mathrm d t \ .$$ -This is clear, but this is just a formula that does not give any insight into the innards of the concept. - -Is is possible to define the concept of line integral by purely abstract properties? - -To give two analogies, the algebraic tensor product is defined by some universal property, and then it is shown that it exists and is essentially unique. Similarly, the Haar measure on locally-compact groups is defined as a regular, positive measure, that is invariant under (left) translations, and then is shown to exist and be essentially unique. Do you know of any similar approach for line integrals? - -To clarify: if $\mathcal C$ is the space of smooth curves in $M$, I am attempting to understand line integration as a map $I : \Omega ^1 (M) \times \mathcal C \to \mathbb R$ uniquely determined by some properties: what are these properties? For sure, linearity in the first argument is among them. What else is needed? - -REPLY [2 votes]: I'll suggest here another possible characterisation, expanding on a suggestion of the OP in one of the comments. Again, this is an assertion that certain known properties of line integration characterise it uniquely; this doesn't provide a "new" construction of line integration. Unlike my previous answer, here all the action takes place on the one manifold $M$. -To avoid various technicalities, I'm going to redefine $\mathcal C_M$ to be the set of immersed paths, i.e. smooth paths $c\colon[0,1]\to M$ such that $\dot c(t)\neq0$ for all $t\in[0,1]$. I think this restriction could probably be removed with enough effort. -Theorem: -For any manifold $M$, line integration is the unique function $I\colon\Omega^1(M)\times\mathcal C_M\to\mathbb R$ satisfying the following properties: - -(additivity in the path) Suppose that $c_1$ and $c_2$ are two immersed paths that are composable, i.e. all the derivatives $c_1^{(i)}(1)=c_2^{(i)}(0)$. Then $I(\omega,c_1c_2)=I(\omega,c_1)+I(\omega,c_2)$ for all $\omega\in\Omega^1(M)$. Here $c_1c_2$ denotes the composite path, defined by$$c_1c_2(t)=\begin{cases}c_1(2t)&0\leq t\leq1/2\\c_2(2t-1)&1/2\leq t\leq1.\end{cases}$$ -(additivity in the $1$-form) We have $I(\omega_1+\omega_2,c)=I(\omega_1,c)+I(\omega_2,c)$ for all $\omega_1,\omega_2\in\Omega^1(M)$ and all $c\in\mathcal C_M$. -(locality) If $\omega\in\Omega^1(M)$ satisfies $\omega_{c(t)}(\dot c(t))=0$ for all $t\in[0,1]$, then $I(\omega,c)=0$. -(exact forms) If $f\colon M\to\mathbb R$ is smooth, then $I(\mathrm df,c)=f(c(1))-f(c(0))$. - - -The proof uses two lemmas. -Lemma 1: Let $c$ be an immersed path. Then there is a non-negative integer $N$ such that for all $0\leq k<2^N$, the restriction $c|_{[2^{-N}k,2^{-N}(k+1)]}$ of $c$ to the interval $[2^{-N}k,2^{-N}(k+1)]$ is an embedding. -Proof (outline): This follows from the standard fact that an immersion is locally an embedding (see e.g. this MO question), and that $[0,1]$ is compact. -Lemma 2: Let $c$ be an embedded path in $M$ and $\omega\in\Omega^1(M)$. Then there exists a smooth function $f\colon M\to\mathbb R$ such that $\omega_{c(t)}(\dot c(t))=\mathrm df_{c(t)}(\dot c(t))$ for all $00$. This is possible by Borel's Lemma, which says that we can choose smooth maps $(-\epsilon,0]\to M$ and $[1,1+\epsilon)\to M$ having the same higher-order derivatives at $0$ and $1$ as $c$, respectively. -Decreasing $\epsilon$ if necessary, we may even assume that $c\colon(-\epsilon,1+\epsilon)\hookrightarrow M$ is an embedding. The tubular neighbourhood theorem then implies that the embedding $c$ extends to an embedding $\tilde c\colon(-\epsilon,1+\epsilon)\times (-1,1)^{d-1}\hookrightarrow M$, where $d=\dim(M)$. In other words, we have $c(t)=\tilde c(t,0,\dots,0)$ for all $t$. -We now extend $f_0$ as follows. By Borel's Lemma again, we may extend $f_0$ to a smooth function $f_0\colon(-\epsilon,1+\epsilon)\to\mathbb R$, and then extend this again to a smooth function $f_0\colon(-\epsilon,1+\epsilon)\times(-1,1)^{d-1}\to\mathbb R$. Multiplying by an appropriate bump function if necessary, we may assume that $f_0$ vanishes outside $(-\frac12\epsilon,1+\frac12\epsilon)\times(-\frac12,\frac12)^{d-1}$. -We've now constructed an extension $f=f_0\circ\tilde c^{-1}$ of $f$ on the open neighbourhood $\mathrm{im}(\tilde c)$ of the image of $c$. Moreover, we've ensured that this extension has compact support (it vanishes outside a compact subspace), so we can extend $f$ to all of $M$ by specifying that it is $0$ outside $\mathrm{im}(\tilde c)$. This yields the desired $f$. This proves Lemma 2. - -Proof of Theorem: We show unicity. Let $I$ and $I'$ be two functions $\Omega^1(M)\times\mathcal C_M\to\mathbb R$ which satisfy the given conditions. We need to show that $I(\omega,c)=I'(\omega,c)$ for all $\omega\in\Omega^1(M)$ and all immersed paths $c$. -To do this, suppose first that $c$ is embedded. By Lemma 2 we can choose a smooth map $f\colon M\to\mathbb R$ such that $\omega_{c(t)}(\dot c(t))=\mathrm df_{c(t)}(\dot c(t))$ for all $c\in[0,1]$. Using additivity in the $1$-form, locality and the condition about exact forms, we find that $I(\omega,c)=I(\mathrm df,c)=f(1)-f(0)$. Since the exact same argument applies to $I'$, we have $I(\omega,c)=I'(\omega,c)$. -Now let us deal with the general case. By Lemma 1 we can choose a non-negative integer $N$ such that $c|_{[2^{-N}k,2^{-N}(k+1)]}$ is an embedded path for all $0\leq k<2^N$. A repeated application of the additivity property implies that $I(\omega,c)=\sum_{k=0}^{2^N-1}I(\omega,c|_{[2^{-N}k,2^{-N}(k+1)]})$ and similarly for $I'$. Since we already know that $I$ and $I'$ agree on embedded paths, we obtain that $I(\omega,c)=I'(\omega,c)$, as desired. This concludes the proof. - -Remark: -If one only cares about the integrals of closed 1-forms, then this whole setup can be significantly simplified: one doesn't need to restrict to immersed paths, and one can replace the locality condition above with the more natural condition: - -(locality') If $\omega$ vanishes on an open neighbourhood of the image of $c$, then $I(\omega,c)=0$.<|endoftext|> -TITLE: Examples of amenable Banach algebras which have non-amenable subalgebra -QUESTION [5 upvotes]: I am looking for examples of amenable Banach algebras which have non-amenable subalgebra -I know - -1: Each amenable Banach algebra has a bounded approximate identity - - -2: If $I$ be a closed ideal in an amenable Banach algebra, then - - -$I$ amenable if and only if $I$ has a bounded approximate identity - -Does anybody have an example? -Thank you for any suggestions! - -REPLY [12 votes]: Mateusz's answer mentions lots of good mathematics but I feel obliged to point out that the fundamental example which answers your original question in the negative is $M_2({\bf C})$. (Banach algebras behave very differently from ${\rm C}^*$-algebras and $L^1$-group algebras.) -The point is that the algebra -$$ -{\bf C}[x] / (x^2) \cong -\left\{ -\begin{pmatrix} a & b \cr 0 & a \end{pmatrix} -\colon a,b \in {\bf C} \right\} -\subset M_2({\bf C}) -$$ -is not amenable, and should not be "generalized amenable" in any sensible version of "generalized amenability". -Non-amenability can be seen in many ways, but the most direct one -- if you use the definition of amenability in terms of derivations -- is to note that the mapping -$$ -\begin{pmatrix} a & b \cr 0 & a \end{pmatrix} \to b -$$ -is a non-zero derivation. In some sense this is the philosophical idea behind derivations on associative algebras, they are the 1st-order terms that arise when one perturbs a homomorphism.<|endoftext|> -TITLE: "Intersection number" of a cardinal -QUESTION [6 upvotes]: Let $\kappa$ be an infinite cardinal. We call a cardinal $\lambda \leq 2^\kappa$ intersecting if there is ${\cal C}\subseteq {\cal P}(\kappa)$ such that - -for every $A\in {\cal C}$ we have $|A|=\kappa$, -$|A_0\cap A_1|<\lambda$ whenever $A_0\neq A_1\in {\cal C}$, and -$|{\cal C}| > \kappa$. - -We denote the smallest intersecting cardinal of $\kappa$ by $i(\kappa)$. For instance we have $i(\aleph_0) = \aleph_0$ (also see the concept of an almost disjoint family). By the comments of users bof and Alessandro Codenotti, we always have $i(\kappa) \leq \kappa$ for any infinite cardinal $\kappa$. -Question. If $\kappa$ is an infinite cardinal, is there a cardinal $\alpha\geq\kappa$ with $i(\alpha) < \alpha$? - -REPLY [7 votes]: Since this question is still unanswered I thought I might write down some of what you can get out of Baumgartner's paper. -In Baumgartner's notation (see the beginning of section 2), $A(\kappa,\lambda,\mu,\nu)$ means that there exists a family of sets $F$ such that - -$F\subseteq P(\kappa)$, -$|F| = \lambda$, -$|X| = \mu$ for all $X\in F$, and -$|X\cap Y| < \nu$ for all $X,Y\in F$ with $X\neq Y$. - -Hence the connection is that $\lambda$ is intersecting (in your notation) if and only if $A(\kappa,\kappa^+,\kappa,\lambda)$ holds. -In Theorem 3.4(a) Baumgartner proves that, assuming GCH, for any cardinals $\nu \le \mu \le \kappa$, $A(\kappa,\kappa^+,\mu,\nu)$ holds if and only if $\mu = \nu$ and $cf(\mu) = cf(\kappa)$. Since we're only interested in the case where $\mu = \kappa$, this implies that, under GCH, $i(\kappa) = \kappa$ for all $\kappa$. Note that this conclusion already follows from bof's comments. -The other side is partly covered by Theorem 6.1, which says: assuming GCH holds in $V$, for any cardinals $\nu \le \kappa \le \lambda$ such that $\nu$ is regular, there is a forcing extension $V[G]$ which preserves the cofinalities (hence cardinals) of $V$, in which $A(\kappa,\lambda,\kappa,\nu)$ is true. Hence you can make $i(\kappa) = \omega$ true for any particular $\kappa$, starting from a model of GCH. -It remains to show the consistency of the statement in your question, i.e. for all $\kappa$ there is some $\alpha \ge \kappa$ such that $i(\alpha) < \alpha$. Maybe someone who knows about class forcing can step in.<|endoftext|> -TITLE: Publishing undergraduate research -QUESTION [10 upvotes]: Sorry if something like this has already been asked, I searched but I couldn't find anything similar to my question. -I'm a senior undergraduate and currently doing my senior thesis. My senior thesis is not original work, however it's quite demanding and I'm learning a lot of high level topics. I have been lurking around arxiv and started reading "Solved and unsolved problems in Number theory" by Daniel Shanks. My plan is to work on some open problems and play around with them so that I can try to get a publication before I graduate. My main reason for trying to get a publication is to increase my chances to get into a good graduate program (my GPA is not that great and I don't have the money to apply to many programs, so unless I publish something I'll probably only apply to safety schools). -With that being said if I were to do original work, how would I go about publishing? I might end up modifying a problem too much and proving something that might not be interesting, so I feel it'll get rejected from a journal for not being profound. I will also attack problems with all I know, so I might also end up using some heavy tools that aren't part of an undergraduate curriculum so I don't if i would send them to an undergraduate research journal. Maybe I could just upload on dropbox or arxiv, but then it's not a publication. -I have thought about asking my advisors about this, but I'll rather not since I'm aware I'm probably being overly ambitious and should probably focus on my thesis instead. Which I can agree with, hence I'll probably play around with problems on the weekends only or once a week. I'm also aware I might end up not publishing anything all, however in my mind unless I give it a shot I won't know. Either way I'll have fun and end up learning a lot about research so I don't see a downside. -(In case my background is relevant, my senior thesis is about perfectoid spaces. I've already taken a graduate course on commutative algebra, have taken a basic course on p-adic analysis,started learning about point free topology, already know the basics of category theory, still learning more about algebraic geometry, will learn about adic spaces soon/already know a bit about krull valuations, learning about homological algebra through weibel's book, started reading szamuely's galois theory book, will have to learn about etale cohomology soon, will also learn some things from almost mathematics, etc.) - -REPLY [5 votes]: I'm adding this answer in response to Annie Lee's question, because it's too long to fit in a comment. -Publishing as a grad student should definitely be done in consultation with an advisor. Unlike publishing as an undergraduate, a grad student's first papers serve as their introduction to the experts in their field, and largely determine whether they get a good postdoc. For this reason, it's very wise to have papers on arxiv before hitting the job market, but in many fields it's ok if these papers haven't been published yet (as long as the advisor's letter certifies that the work is solid). It's important not to put out anything shoddy, because your reputation really matters at that stage of your career. There's also the danger of getting scooped, e.g., if your PhD thesis has two parts, and you post the first part to arxiv a full year before the second, someone else might come along and do the second part before you can. An advisor will help you determine how likely this is to happen, and the pros and cons of advertising your work before it's fully finished. That's an important conversation anyway, so you can determine how much to say at conferences. An advisor can also talk to other experts in the field to encourage them not to try to prove what you are trying to prove in your thesis. -All that said, I think it's very valuable for grad students to have an early introduction to the world of publishing, as I previously wrote here. If you've never submitted a paper, responded to a referee report, etc. this can be anxiety inducing to do during your first job, while trying to juggle a million other things. Also, having an actual publication shows postdocs that you have what it takes to see a project through to completion. I was fortunate to write a paper as a second year grad student, in a totally different area than my main research (homotopy theory), and this gave me both confidence to get me through the hard times during my PhD research, and experience in choosing a journal, choosing an editor, when to keep polishing and when to submit, etc. Pro tip: the time to submit is almost always sooner than you think, because some degree of polishing can be done after the referee report, and the referee will always have suggestions for things to change (so, sending something too perfect can lead the referee to suggest big changes instead of obvious small changes). -In order to help give grad students this important experience with early publications, I joined the editorial board of the Graduate Journal of Mathematics, as I wrote about here. If you work out some interesting result or new example, and publishing it separately won't harm your PhD thesis, please feel free to submit to our journal! I think it's the only one of its kind, unlike the plethora of undergrad journals I mentioned in my other answer. To submit, you just pick an editor on the editorial board and email the pdf.<|endoftext|> -TITLE: How does the number of trees on $n$ vertices *up to isomorphism* grow as $n \to \infty$? -QUESTION [7 upvotes]: It is well known that the number of labelled trees on $n$ vertices is equal to $n^{n-2}$. - -We do not expect any such exact formula for the number of isomorphism types of trees on $n$ vertices. But what are the sharpest asymptotics, or best upper and lower bounds known, as $n \to \infty$? -Has anyone studied the number of homeomorphism types of trees on $n$ vertices? Again, I don't expect an exact answer, and am mostly interested in the asymptotics as $n \to \infty$. - -REPLY [16 votes]: For Q1 the answer is known to be $\sim C_1C_2^n n^{-5/2} $ for $C_1\approx 0.5349496061...$ and $C_2\approx 2.9955765856...$. This can be found in Flajolet and Sedgewick's "Analytic Combinatorics" (see p.481) with the main ingredients being singularity analysis and the relation -$$I(z)=H(z)-\frac{1}{2}\left(H(z)^2-H(z^2)\right)$$ -where $I$ is the generating function for unrooted unlabelled trees and $H$ is the generating function for rooted unlabelled trees. This is all originally from Otter's paper -"The Number of Trees" from Annals of Mathematics, Vol 49, no.3 pp. 583-599. -For Q2, homeomorphism classes of trees on $n$ vertices correspond to homeomorphically irreducible trees (also sometimes called series-reduced tress, or topological trees, see OEIS) of at most $n$ vertices. This enumeration problem appears in the movie Good Will Hunting and there is even a numberphile video about it. -These were enumerated by Harary and Prins in - -Frank Harary and Geert Prins, "The number of homeomorphically irreducible trees and other species" Acta Math., 101 (1959), 141-162 - -and for the asymptotics see - -F. Harary, R. W. Robinson and A. J. Schwenk, Twenty-step algorithm for determining the asymptotic number of trees of various species, J. Austral. Math. Soc., Series A, 20 (1975), 483-503 - -I also believe that both Q1 and Q2 are exposited in the book "Graphical Enumeration" by Harary and Palmer, but I'll have to check.<|endoftext|> -TITLE: Locally minimal simplicial categories -QUESTION [7 upvotes]: Given a (fibrant) simplicially enriched category $\mathcal{C}$, I'm interested in the possibility of replacing it with a weakly equivalent one (in Bergner model structure) such that all the mapping spaces are minimal Kan complexes, that I read about for example in $\textit{Higher Topos Theory}$, section 2.3.3. -The naive idea, that is, individually replacing every mapping space with a minimal model, is bound not to work, because the construction of minimal models for a Kan complex is overtly not functorial, since it relies on an inductive choice of representatives for homotopy classes of simplices in each dimension. -Is there a way to obviate this issue and somehow choose such representatives in a way that is compatible with compositions in $\mathcal{C}$? I expect the construction won't give a functor $s \mathcal{Cat} \to s \mathcal{Cat}$ even in case of a positive answer, but I would be quite content with performing this on a single given simplicially enriched category. -In case of a negative answer, is there any other way that I can obtain a simplicially enriched category with the desired property? Feel free to add any conditions you like on $\mathcal{C}$ if it helps. Thanks! - -REPLY [4 votes]: It's not possible in general to ensure that all the hom-spaces in a simplicial category are minimally fibrant. Here's a counterexample inspired by Isbell. -Consider $Set$ with its cartesian monoidal structure (the same approach, mutatis mutandis, will work with the cocartesian monoidal structure). Let $C \subset Set$ be the subcategory with two objects, $\{\emptyset\}$, $\mathbb N$, and with morphisms the bijections. Then after making a choice of bijection $\mathbb N \cong \mathbb N \times \mathbb N$, the groupoid $C$ inherits a monoidal structure equivalent to the restriction of the cartesian monoidal structure on $Set$. As Zhen Lin shows at the linked argument of Isbell, there is no strict monoidal structure on $C$ equivalent to this monoidal structure. -(Of course, in line with the strictification theorem, the monoidal structure can be strictified after passing to an equivalent groupoid which is not skeletal; the point is that the flexibility gained by having extra isomorphic copies of each object is essential; this monoidal category cannot be simultaneously skeletized and strictified.) -Let $BC$ be the 1-object bicategory (with groupoidal homs) associated to $C$, and let $bC$ be any simplicial category corresponding to $BC$. Suppose that $D$ were a simplicial category with minimally-fibrant hom-spaces equivalent to $bC$. By passing to a skeleton, we may assume that $D$ has one object. -Now, we know the homotopy type of the hom-space of $D$ -- it is the classifying space of $C$ -- which is a minimal Kan complex! By uniquness of minimal models, this implies that the hom-space of $D$ is in fact isomorphic to the classifying space of $C$. Therefore the composition on $D$ -- which is strictly associative and unital -- restricts to a strict monoidal structure on $D$ equivalent to the one we've chosen. This contradicts Isbell's theorem that no such strictification exists. -EDIT: I think Isbell's argument uses some non-invertible morphisms, so doesn't quite apply to $C$ as above. Here is an alternative argument to Isbell's in the cocartesian variation. That is, let $C' \subset Set$ be the subcategory with two objects $\emptyset, \mathbb N$, and morphisms the bijections. The cocartesian monoidal structure on $Set$ restricts to a monoidal structure on $C'$ after we choose a bijection $\mathbb N \cong \mathbb N \amalg \mathbb N$. Without loss of generality, this bijection corresponds to the partition of $\mathbb N$ into the even numbers and odd numbers. Such a monoidal structure cannot be strictified. To see this, observe that if $f: \mathbb N \to \mathbb N$ is any nontrivial bijection, then $(1 \otimes f) \otimes 1$ moves some number which is congruent to 2 mod 4, and fixes any number which is not 2 mod 4, whereas $1 \otimes (f \otimes 1)$ moves some number which is congruent to 1 mod 4, and fixes any number which is not 1 mod 4. Therefore these two bijections are distinct, contradicting the strictness of the monoidal structure.<|endoftext|> -TITLE: Closed walks on an $n$-cube and alternating permutations -QUESTION [10 upvotes]: Let $w(n,l)$ denote the number of closed walks of length $2l$ from a given vertex of the $n$-cube. Then, it is well-known that -$$\cosh^n(x)=\sum_{l=0}^{\infty}\frac{w(n,l)}{(2l)!}x^{2l}.$$ -Differentiating both sides, we get -$$n \cdot \cosh^{n-1}(x)\cdot \sinh(x) = \displaystyle\sum_{l=1}^{\infty}\frac{w(n,l)}{(2l-1)!}x^{2l-1}.$$ By the Cauchy product of the Maclaurin series of $n\cosh^{n-1}(x)$ and $\sinh(x)$ and comparing coefficients of the LHS and RHS, we get the recursion -$$w(n,l)=n\sum_{k=1}^{l}\binom{2l-1}{2k-1}w(n-1,l-k).$$ -The above recursion has the following simple combinatorial interpretation. Let us count the total number of closed walks of length $2l$ on the $n$-cube. W.L.O.G, let the initial step be along the 1st dimension. Then, out of the remaining $2l-1$ steps, choose $2k-1$ more places to step back and forth the "1st" dimension. Note that there is exactly one way for this once the $2k-1$ places are chosen. For the remaining $2l-2k$ steps, we take steps in every dimension except the 1st, resulting in $w(n-1,l-k)$ ways. As $k$ is the number of times we walk back and forth the 1st dimension, we sum $k$ from 1 to $l$ ($k>0$ as the initial step is along the 1st dimension). Finally, as the initial step can be taken in $n$ dimensions, we multiply by $n$ and get the above recursion. -My question is the following. To obtain the above recursion, we considered the Cauchy product of the Maclaurin series of $n\cdot \cosh^{n-1}(x)$ and $\sinh(x)$. This, however, is equivalent to the Cauchy product of the Maclaurin series of $n \cdot \cosh^n(x)$ and $\tanh(x),$ which by the same method gives -$$w(n,l)=n\sum_{k=1}^{l}(-1)^{k+1}\binom{2l-1}{2k-1}A(2k-1)w(n,l-k),$$ -in which the "tangent numbers" $A(2k-1)=T_k$ count the number of alternating permutations of $2k-1$ elements (note how the dimension of $w$ is unchanged). I was wondering if a combinatorial interpretation of the above was possible, in a similar fashion to the first recursion. The $(-1)^{k+1}$ term hints inclusion-exclusion, but I'm unable to come up with a satisfactory explanation. -The following post on $w(n,l)$ focuses on a closed-form expression, without mention of recursive formulae. -Number of closed walks on an $n$-cube - -REPLY [3 votes]: This is a kind of inclusion-exclusion related to the identity -$$ -\sum_{k=1}^m (-1)^{k+1} \binom{2m-1}{2k-1}A(2k-1)=1 \quad\quad(1) -$$ -for all $m=1,2,\ldots$. -For a route on the $n$-cube with first step being vertical we label other $2k-1$ vertical steps, take a weight $(-1)^{k+1}A(2k-1)$ for such a configuration and sum up. For given $k$, you may choose $2k-1$ places of vertical steps, after removing them and the first step you get a route of length $2(l-k)$. So the sum of weights of all configurations is -$$\sum_{k=1}^{l}(-1)^{k+1}\binom{2l-1}{2k-1}A(2k-1)w(n,l-k).$$ -On the other hand, the sum of weights of all configurations for a fixed route equals 1 due to (1). Thus the result. -You may ask how to prove (1) сombinatorially. This is most probably known, but for any sake here is a short proof. -Consider such configurations: -(i) $(x_1,\ldots,x_{2m-1})$ is a permutation of $1,\ldots,2m-1$ and $k\in \{1,\ldots,m\}$; -(ii) $2k-1$ first terms $x_1,\ldots,x_{2k-1}$ are labelled and form an alternating permutation: $x_1x_3<\ldots >x_{2k-1}$; -(iii) other terms are decreasing: $x_{2k}>x_{2k+1}>\ldots>x_{2m-1}$. -Define the weight of such configuration as $(-1)^{k+1}$. The sum of all weights is LHS of (1) (we start with fixing $k$, next fixing the set $\{x_1,\ldots,x_{2k-1}\}$, next fix an alternating permutation on this set). On the other hand, any permutation except $\pi=(2m-1,2m-2,\ldots,1)$ is counted twice with opposite weights, and $\pi$ is counted once with weight 1. - -REPLY [2 votes]: Equation (1) from the above answer can also be viewed as the case in which $n=1$ for $w(n,l).$ This is simply because the number of closed walks of length $2l$ on a one-dimensional cube is always 1 regardless of $n$.<|endoftext|> -TITLE: Bounded non-symmetric domains covering a compact manifold -QUESTION [6 upvotes]: This question is somewhat related to this other question of mine. -I was wondering which are the known examples of bounded domains $\Omega$ in $\mathbb C^n$ admitting a compact free quotient. -By a theorem of Siegel, such a domain must be holomorphically convex. Moreover, if the boundary is sufficiently regular, say $C^2$ (even if, by a recent theorem of A. Zimmer, $C^{1,1}$ suffices), by the classical theorem of Wong-Rosay, then $\Omega$ must be biholomorphic to the unit ball. -Of course all bounded symmetric domains give such examples, by a classical theorem of E. Borel. But I am interested in more "exotic" examples, specifically non symmetric examples. -The only I am aware of live in $\mathbb C^2$ and are the universal covers of Kodaira fibrations (see this question for more details). - -Is it possibile for instance to construct higher dimensional analogous of the universal cover of a Kodaira fibration? - -For dimension $n\ge 4$, there exists (many in fact) homogeneous bounded domains in $\mathbb C^n$ which are non symmetric. In 1979 J. E. D’Atri proved that there exists bounded homogeneous non-symmetric domains, for $n\ge 6$, whose Bergman metric has positive holomorphic sectional curvature somewhere. -Unfortunately, they can never cover a compact manifold. Indeed, it was shown by J. Hano in 1957 that if a homogeneous bounded domain covers a manifold of finite volume, i.e. if its automorphism group admits a lattice so that it's unimodular, than the domain is in fact symmetric. -Remark that I am really looking for compact, discrete, free quotients. -Thank you very much in advance. - -REPLY [6 votes]: A silly generalization would be a direct product of Koadaira's surface and, say, a Riemann surface. A better construction is below. -I will use -Griffiths, Phillip A., Complex-analytic properties of certain Zariski open sets on algebraic varieties, Ann. Math. (2) 94, 21-51 (1971). ZBL0221.14008. -specifically, Lemma 6.2: Suppose that $U\to S$ is a (nonsingular) holomorphic family of compact Riemann surfaces such that $S$ is uniformed by a bounded contractible domain of holomorphic in ${\mathbb C}^n$. Then the same holds for $U$. -Given this, one inducts: Take a compact Kodaira surface $K\subset {\mathcal M}_g$, let $\xi: {\mathcal M}_{g,1}\to {\mathcal M}_{g}$ be the "universal curve." Then the pull-back of $\xi$ to $K$ is a holomorphic family of genus $g$ Riemann surfaces $U\to K$ as in Griffiths lemma. Hence, $U$ is compact and is again uniformed by a bounded domain in ${\mathbb C}^3$. To continue, one needs a trick since $U$ lies in ${\mathcal M}_{g,1}$ and the universal curve over that will have noncompact fibers. However, one can regard a puncture on a genus $g$ surface as an orbifold cone-point of order $2$, hence, ${\mathcal M}_{g,1}$ (as an orbifold) holomorphically embeds in some ${\mathcal M}_{h}$. To prove this, take a 2-dimensional oriented compact connected orbifold ${\mathcal O}$ of genus $g$ with one cone point of order $2$. It admits a finite manifold-covering $S_h\to {\mathcal O}$. Hence, the moduli space of ${\mathcal O}$ embeds holomorphically (as an orbifold) in ${\mathcal M}_h$. -Thus, $U$ is embedded in ${\mathcal M}_{h}$ and, so we can continue. -Edit. I do not know how to prove that in general these domains are non-symmetric (in dimension 2 this is understood). But, in all the examples I am aware of, the compact complex manifolds given by this construction are non-rigid and, hence, cannot be locally symmetric (except for trivial families which I ignore).<|endoftext|> -TITLE: Residues in several complex variables -QUESTION [33 upvotes]: I am trying to educate myself about the basics of the theory of residues in several complex variables. As is usually written in the introduction in the textbooks on the topic, the situation is much harder when we pass from one variable to several variables. -So for $n=1$ we have: - -For a holomorphic $f$ with an isolated singularity at point $a$, the residue of $f$ at $a$ is defined as -$$res_a f = \frac{1}{2\pi i} \int_{\sigma} f dz$$for a small loop $\sigma$ around $a$. - -For $n>1$ we have: - -(Shabat, vol. II) For a meromorphic $f$ defined on $D \subset \mathbb{C}^n$ with the indeterminacy locus $P \subset D$, choose a basis $\sigma_{\alpha}$ of $H_1(D \setminus P, \mathbb{Z})$ and define the residue of $f$ with respect to $\sigma_{\alpha}$ to be $$res_{\sigma_{\alpha}} f=\frac{1}{(2\pi i)^n} \int_{\sigma_{\alpha}} f dz$$ - -(Griffith-Harris, Chapter 5) Let $U$ be a ball $\{z\in \mathbb{C}^n \ | \ ||z||< \varepsilon\}$ and $f_1,...,f_n \in \mathcal{O}(\bar{U})$ be holomorphic functions with an isolated common zero at the origin. Take $\omega=\frac{g(z) dz_1 \wedge ... \wedge dz_n}{f_1(z)...f_n(z)}$ and $\Gamma=\{z \ : \ |f(z_i)|=\varepsilon_i\}$. The (Grothendieck) residue is given by $$Res_{ \{0\}} \omega=\frac{1}{(2 \pi i)^n} \int_{\Gamma} \omega .$$It can further be viewed as a homomorphism $$\mathcal{O}_0/(f_1,...,f_n) \to \mathbb{C}$$ - -In the "General theory of higher-dimensional residues", Dolbeault discusses residue-homomorphism, homological residues, cohomological residues, residue-currents, etc. - - -So since there are so many various things called residue, my question is - -What structure are all these things trying to capture, so that we call -all these various things "residue"? - -In Chapter 3, Griffiths and Harris outline a general principle when discussing distributions and currents: -$$(*) \quad D T_{\psi} - T_{D \psi} = \text{"residue"},$$where $T_{\psi}$ is the current $T_{\psi}(\phi)=\int_{\mathbb{R}^n} \psi \wedge \phi$ (this discussion takes plane on $\mathbb{R}^n$). They illustrate that by applying this principle to the Cauchy kernel $\psi=\frac{dz}{2 \pi i z}$: -$$\phi(0)=\frac{1}{2 \pi i} \int_{\mathbb{C}} \frac{\partial \phi(z)}{\partial \bar{z}} \frac{dz \wedge d \bar{z}}{z} \ \iff \bar{\partial}(T_{\psi})=\delta_{0}.$$ -This is a nice example, but later on when they discuss the Grothendieck residue (2) in Chapter 5 they do not explain how it fits into the philosophy $(*)$. I also do not see how (0), (1) and (3) fit into this philosophy. So maybe one can explain how $(*)$ might be a potential answer to the question I ask. - -REPLY [3 votes]: I believe residue currents encompass most definitions of residues in several complex variables. Residue currents as developed in the 20th century are discussed for example in the survey "Residue currents" by Tsikh and Yger. Given a tuple $(f_1,\dots,f_p)$ defining a complete intersection, i.e., such that $\{ f_1 = \dots = f_p = 0 \}$ has codimension $p$, there is an associated residue current $\mu^f$, as first defined by Coleff and Herrera. Their definition is by taking limits of integrals similar to in 2), but where $g$ in the definition of $\omega$ should be a test form and you should consider limits where $\epsilon_1,\dots,\epsilon_p$ tend to $0$ in an appropriate way. -Just as the Grothendieck residue, these residue currents can also be defined with the help of Bochner-Martinelli forms, as was first done by Passare, Tsikh and Yger. In fact, if $B_f$ is the Bochner-Martinelli form of $f$, then the action of $\mu^f$ on a test form $\varphi$ is given by -$\lim_{\epsilon \to 0} \int_{\{|f|=\epsilon\}} B_f \wedge \varphi$. -In the absolute case in 2), i.e., when $p=n$, and you take a cut-off function $\chi$ with compact support that is $\equiv 1$ at the origin, then $\chi \omega$ is a test-form, and the action of $\mu^f$ on $\chi \omega$ equals the Grothendieck residue. With the help of the representation in terms of Bochner-Martinelli forms, it is immediate that $\mu^f$ acting on $\chi \omega$ equals the Grothendieck residue of $\omega$. -I would believe that also the case 1) should be possible to represent by residue currents, by taking a holomorphic function $g$ whose zero set contains the indeterminacy locus $P$ and letting $\mu^g$ act on an appropriate form, but I'm not familiar enough with the definition of Shabat to describe this. -More recently, there are also residue currents defined more generally for coherent sheaves by Andersson and Wulcan, "Residue currents with prescribed annihilator ideals", not just complete intersections as considered above. - -Regarding how $(*)$ fits into this picture, I don't know if this has been explicitly elaborated on earlier, but at least it is discussed in "Direct images of semi-meromorphic currents" by Andersson and Wulcan. -A semi-meromorphic form $\psi$ is a form that is locally a smooth form times a meromorphic form, and one might identify the form with its corresponding principal value current. An almost semi-meromorphic form is the push-forward of a semi-meromorphic form under a modification. The Bochner-Martinelli form $B_f$ is an example of an almost semi-meromorphic form. (When $p=1$, it is indeed even meromorphic.) -If $\psi$ is an almost semi-meromorphic form on $X$ that is smooth outside a subvariety $Z$, then $\bar\partial\psi$ is a smooth form on $X \setminus Z$, and it turns out that $\bar\partial\psi|_{X\setminus Z}$ has a principal value extension to $X$ that is again an almost semi-meromorphic form. In this way, there is a $\bar\partial$-operator acting on almost semi-meromorphic currents. Andersson and Wulcan define the residue of $\psi$ as the current $R(\psi)=\bar\partial T_\psi - T_{{\bar\partial} \psi}$, see section 4.4 of their paper. The residue is thus the difference between this $\bar\partial$-operator on almost semi-meromorphic forms and the $\bar\partial$-operator acting in the sense of currents. As is basically detailed in their Example 4.18, the current $\mu^f$ is then in fact the residue of the Bochner-Martinelli form $B_f$.<|endoftext|> -TITLE: What kinds of papers does the Indiana University Mathematics Journal publish? -QUESTION [10 upvotes]: I am wondering if anyone can provide information on the Indiana University Mathematics Journal, as I have been able to find very little (ie aims and scope) on their website. I have the following list of questions. - -What kinds of papers do they "like" to publish? Should submitted papers be solving open problems, introducing entirely new concepts, etc.? -Do they put more weight on certain fields over others? -Roughly, what are their review times? - -In general, I suppose my main question is, "If I have a new paper, how can I tell that it is a good fit for the journal"? This question may be impossible to answer, but I would greatly appreciate some intuition. - -REPLY [20 votes]: The key portion of the FAQ reads: - -The initial review is handled by the Managing Editor and by members of the Editorial Board and/or other departmental reviewers, depending on the area of expertise. This initial review is usually completed within weeks and, for most manuscripts, this is when it is determined whether or not a manuscript might eventually be accepted for publication. A paper deemed a candidate for publication will be sent to an external referee, whose report weighs heavily on the formal editorial acceptance or rejection of the manuscript. - -The essential thing here is that the initial quick opinion and suggestion of a referee is done by members of the IU math department. So you should submit papers which are close enough to an IU faculty member's interests that they'll be familiar with the topic and able to suggest an appropriate referee. So all of mathematics is ok subject-wise (since we're a broad department), but within a given topic the closer it is to IU expertise the better the fit is. -The other two important parts in terms of your questions about turnaround is that the initial quick opinion step takes only a few weeks, and that there's typically a single referee which means it's faster than journals with two referees. Part of keeping the first stage fast is that sometimes papers are rejected simply because no appropriate referee was identified within two weeks, so please don't take it personally if this happens, the point is to keep the turnaround quick so you can submit elsewhere without losing time. -(Full disclosure: I am a member of the IU math department and do regularly review for them.)<|endoftext|> -TITLE: Group cohomology of Q/Z -QUESTION [5 upvotes]: What is the group cohomology $H^{d}(\mathbb{Q}/\mathbb{Z}, \mathbb{Z})$ with trivial action? -Can it be computed succinctly using the short exact sequence $0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0$? - -REPLY [9 votes]: This can be computed using standard tools of algebraic topology. The idea is first to compute the homology groups $H_*({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ and then use the universal coefficient theorem to get the cohomology groups $H^*({\mathbb Q}/{\mathbb Z},{\mathbb Z})$. -The group ${\mathbb Q}/{\mathbb Z}$ is the union of an infinite increasing sequence of finite cyclic subgroups $G_n$. For example one can take $G_n$ to be cyclic of order $n!$. An Eilenberg-MacLane space $K({\mathbb Q}/{\mathbb Z},1)$ can be constructed as an increasing union of $K(G_n,1)$'s since homotopy groups commute with direct limits. Homology groups also commute with direct limits so $H_i({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ is the direct limit of the groups $H_i(G_n,{\mathbb Z})$. The latter is $G_n$ for odd $i$ and $0$ for even $i>0$. Thus $H_i({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ is ${\mathbb Q}/{\mathbb Z}$ for odd $i$ and $0$ for even $i>0$. -To apply the universal coefficient theorem we have $\operatorname{Hom}({\mathbb Q}/{\mathbb Z},{\mathbb Z})=0$ and hence $H^i({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ is $\operatorname{Ext}({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ for even $i>0$ and $0$ for odd $i$. -Computing $\operatorname{Ext}({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ takes a little work so let me just refer to my algebraic topology book where this is done in Section 3F, specifically on page 318 a few lines above Proposition 3F.12. The answer is that $\operatorname{Ext}({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ is the direct product (not direct sum) of the additive groups of $p$-adic integers for all primes $p$. -Comparing this calculation with Robert Kropholler's answer, his group $A$ is the product of the groups of $p$-adic integers and his map ${\mathbb Z}\to A$ is the obvious diagonal map to this product. -In particular the cohomology groups of ${\mathbb Q}/{\mathbb Z}$ are uncountable in each even positive dimension. This is an instance of the general fact that for an abelian group $A$ which is not finitely generated either $\operatorname{Hom}(A,{\mathbb Z})$ or $\operatorname{Ext}(A,{\mathbb Z})$ is uncountable, which is the Proposition 3F.12 mentioned above.<|endoftext|> -TITLE: Why study infinite loop spaces? -QUESTION [10 upvotes]: What makes an infinite loop space an interesting object of study for homotopy theorists? The reason I ask this question is that I found a lot of results treating the question of whether a given space is an infinite loop space. So it seems that the property of a space being homotopy equivalent to an infinite loop space opens totally new possibilities and techniques to study the space. I would glad if anybody could take some time to give an overview about the most well-known directions one can study a space recognized as infinite loop space. - -REPLY [3 votes]: There is so much much more. For one historical starting point among many, you see that many spaces of interest are infinite loop spaces and that tells you how to calculate things about them. For just one example, to add to David's parenthetical clause, almost everything we know about characteristic classes for topological bundles comes from the infinite loop structure of BTop. This is very concrete and calculational and tells us geometrically about topological cobordism. At another extreme, knowing that algebraic K-theory is given by E infty ring spectra is the starting point for derived algebraic geometry. I could go on for pages and pages. The emerging equivariant story is even richer and promises far more to come.<|endoftext|> -TITLE: Are any interesting classes of polynomial sequences besides Sheffer sequences groups under umbral composition? -QUESTION [8 upvotes]: This question on math.stackexchange.com has 35 views, three up-votes, and not a word from anybody, so I'm posting it here. -Let us understand the term polynomial sequence to mean a sequence $(p_n(x))_{n=0}^\infty$ in which the degree of $p_n(x)$ is $n.$ -The umbral composition $((p_n\circ q)(x))_{n=0}^\infty$ (not $((p_n\circ q_n)(x))_{n=0}^\infty$) of two polynomial sequences $(p_n(x))_{n=0}^\infty$ and $(q_n(x))_{n=0}^\infty,$ where for every $n$ we have $p_n(x) = \sum_{k=0}^n p_{nk} x^k,$ is given by -$$ -(p_n\circ q)(x) = \sum_{k=0}^n p_{nk} q_k(x). -$$ -An Appell sequence is a polynomial sequence $(p_n(x))_{n=0}^\infty$ for which $p\,'_n(x) = np_{n-1}(x)$ for $n\ge1.$ -A sequence of binomial type is a polynomial sequence $(p_n(x))_{n=0}^\infty$ for which $$ p_n(x+y) = \sum_{k=0}^n \binom n k p_k(x) p_{n-k}(y) $$ -for $n\ge0.$ -A Sheffer sequence is a polynomial sequence $(p_n(x))_{n=0}^\infty$ for which the linear operator from polynomials to polynomials that is characterized by $p_n(x) \mapsto np_{n-1}(x)$ is shift-equivariant. A shift is a mapping from polynomials to polynomials that has the form $p(x) \mapsto p(x+c),$ where every term gets expanded via the binomial theorem. -At least since around 1970, it has been known that - -Every Appell sequence and every sequence of binomial type is a Sheffer sequence. -The set of Sheffer sequences is a group under umbral composition. -The set of Appell sequences is an abelian group under umbral composition. -The set of sequences of binomial type is a non-abelian group under umbral composition. -The group of Sheffer sequences is a semi-direct product of those other two groups. -For every sequence $a_0, a_1, a_2, \ldots$ of scalars there is a unique Appel sequence $(p_n(x))_{n=0}^\infty$ for which $p_n(0) = a_n$ for $n\ge0.$ -For every sequence $c_1, c_2, c_3, \ldots$ of scalars there is a unique sequence $(p_n(x))_{n=0}^\infty$ of binomial type for which $p\,'_n(0) = c_n$ for $n\ge1.$ This can be proved by induction on $n.$ (And in every case $p_0(0)=1$ and $p_n(0)=0$ for $n\ge 1.$) - -So my question is whether Sheffer sequences exhaust the list of interesting classes of polynomial sequences that are groups under this operation? Are there any others of interest? - -REPLY [8 votes]: Another equivalent characterization of Sheffer sequences is that they fit into a generating function of the form -$$\sum_{n=0}^{\infty}\frac{p_n(x)}{n!}t^n=f(t)e^{xg(t)}.$$ -Most of the results on Sheffer sequences apply to a more general setting where we work with a function $\Psi(x)=\sum_{n\geq 0}x^n/c_n$ and define $\Psi$-Sheffer sequences as those which satisfy a generating function of the form -$$\sum_{n=0}^{\infty}\frac{p_n(x)}{c_n}t^n=f(t)\Psi(xg(t)).$$ -These $\Psi$-Sheffer sequences also form a group under Umbral composition and this group is also a semidirect product of its $\Psi$-Appell subgroup and $\Psi$-binomial type subgroup. It should be noted that abstractly these groups are all isomorphic no matter the choice of $\Psi$: Let $A$ be the group of invertible power series $\mathbb C[[x]]^{\times}$ under multiplication and let $B$ be the (nonabelian) group $x\mathbb C[[x]]^{\times}$ under composition. We can let $B$ act on $A$ by composition and the result is that the group of $\Psi$-Sheffer sequences is isomorphic to the semidirect product of $B$ and $A$. -The details and proofs can be found in Steven Roman's papers "The Theory of the Umbral Calculus I-III", where he gives lots of examples of families of special polynomials that can be treated by this new umbral setting: Chebyshev, Jacobi, Gegenbauer etc. For a treatment that is a more modern you can see S. Zemel "Generalized Riordan groups and operators on polynomials".<|endoftext|> -TITLE: A subcontinuous function, which is not continuous -QUESTION [5 upvotes]: Let $E$ be a Banach space and $T: E\rightarrow E$ be a mapping. $T$ is said to be subcontinuous if for any sequences $(u_n)_{n\in\mathbb{N}}$ in $E$ that converge strongly to $u$ the sequence $(T(u_n))_{n\in\mathbb{N}}$ converges weakly to $T(u)$. - -I am looking for a subcontinuous function which is not continuous. I thought I can find something in $l^2$, but I didn't. -I already posted it on math.stackexchange, but to no avail -Thank you - -REPLY [10 votes]: Let $(e_n)$ be the standard orthonormal basis of $\ell^2$: recall that, as a sequence, $(e_n)$ converges weakly to $0$. Now define a map $f\colon\mathbb{R} \to \ell^2$ by $f(\frac{1}{n})=e_n$ and $f(t)=0$ if $t\leq 0$, and interpolating linearly between $\frac{1}{n}$ and $\frac{1}{n+1}$: this is continuous at every point except at $0$ where it is subcontinuous (because any sequence of the form $t_k\, e_{n_k} + (1-t_k)\, e_{n_k+1}$ with $0\leq t_k\leq 1$ and $n_k\to+\infty$, converges weakly to $0$ when $k\to+\infty$). If you want a subcontinuous function $\ell^2\to\ell^2$, just right-compose $f$ with a nonzero continuous linear map $\ell^2\to\mathbb{R}$, e.g., $e_0^*$.<|endoftext|> -TITLE: Positivity of $ \int_{-\infty}^{\infty} \left\{{2^{1/\beta-1/2} \over v}\right\}^{it} { \Gamma\{(it+1)/\beta\}\over \Gamma\{(it+1)/2\} }dt$ -QUESTION [6 upvotes]: I have the following function -$$ -\int_{-\infty}^{\infty} \left\{{2^{1/\beta-1/2} \over v}\right\}^{it} -{ \Gamma\{(it+1)/\beta\}\over \Gamma\{(it+1)/2\} }dt -$$ -where $1<\beta<2$, $v>0$. Need to show it is positive. -The inverse Mellin transform of -$$ - \left\{2^{1/\beta-1/2} \right\}^{it} -{ \Gamma\{(it+1)/\beta\}\over \Gamma\{(it+1)/2\} } -$$ -is -$$ -{C \over v}\int_{-\infty}^{\infty} \left\{{2^{1/\beta-1/2} \over v}\right\}^{it} -{ \Gamma\{(it+1)/\beta\}\over \Gamma\{(it+1)/2\} }dt -$$ - -REPLY [11 votes]: $\newcommand\Ga\Gamma -\newcommand{\R}{\mathbb{R}} -\newcommand{\de}{\delta} -\newcommand{\ga}{\gamma} -\newcommand{\Si}{\Sigma}$ -We have to show that for $a:=-\ln(2^{1/b-1/2}/v)\in\R$ and $b:=\beta\in(1,2)$, -\begin{equation*} -I(a):=\int_{-\infty}^{\infty} e^{-iat}R(t)\,dt>0, \tag{1} -\end{equation*} -where -\begin{equation*} - R(t):=\frac{\Ga\big((1+it)/b\big)}{\Ga\big((1+it)/2\big)}. \tag{2} -\end{equation*} -The key is Euler's product formula -\begin{equation*} - \Ga(z)=\frac1z\,\prod_{j=1}^\infty\frac{(1+1/j)^z}{1+z/j} -\end{equation*} -for $z\in\mathbb C\setminus\{0,-1,-2,\dots\}$, which yields -\begin{equation*} - \frac{\Ga(s+it)}{\Ga(s)}=\prod_{j=1}^\infty(1+1/j)^{it} - \Big/\prod_{j=0}^\infty\Big(1+\frac{it}{j+s}\Big); \tag{3} -\end{equation*} -here and in what follows, $s$ is any positive real number and $t$ is any real number. -Based on (3), it is easy to obtain - -Lemma 1: $\ln|\Ga(s+it)|\sim-\pi|t|/2$ as $|t|\to\infty$. - -The proof of Lemma 1 will be given at the end of this answer. -It also follows from (3) that -\begin{equation*} - R(t)=c\prod_{j=1}^\infty(1+1/j)^{iht}f_j(t), \tag{4} -\end{equation*} -where $c:=\Ga(1/b)/\Ga(1/2)>0$, -\begin{equation} - h:=\frac1b-\frac12=\frac{2-b}{2b}, -\end{equation} -and -\begin{equation} - f_j(t):=\frac{1+it/(1+2j)}{1+it/(1+bj)}, -\end{equation} -so that $f_j$ is the characteristic function (c.f.) of a random variable (r.v.) $X_j\sim p_j\de_0+(1-p_j)Exp(-1/(1+bj))$, where in turn $p_j:=(1+bj)/(1+2j)\in(0,1)$, $\de_0$ is the Dirac distribution supported on the set $\{0\}$, and $Exp(-1/(1+bj))$ is the exponential distribution with mean $-1/(1+bj)$, supported on the interval $(-\infty,0]$. Here and in what follows, $j$ is any natural number. Note that $EX_j=-\frac{hj}{(j+1/b)(j+1/2)}$ and $Var\,X_j\le1/(bj)^2\le1/j^2$. So, the series -\begin{equation} - \sum_{j=1}^\infty(X_j-EX_j)=:S -\end{equation} -converges almost surely. Hence, by (4) -\begin{equation*} - R(t)=ce^{ihc_1t}f_S(t), -\end{equation*} -where $f_S$ is the c.f. of the r.v. $S$ and -\begin{equation} - c_1:=\sum_{j=1}^\infty\Big(\ln(1+1/j)+EX_j\Big) \\ - =\sum_{j=1}^\infty\Big(\ln(1+1/j)-\frac{j}{(j+1/b)(j+1/2)}\Big)\in\R -\end{equation} -(in fact, $c_1=(\ga b-2 b+b \ln4+2 \psi\left(1+1/b\right))/(2-b)$, where $\ga=0.577\dots$ is the Euler constant and $\psi:=\Ga'/\Ga$; however, the actual value of $c_1$ does not matter here). -So, $R$ is the c.f. of the r.v. $T:=hc_1+S$. Also, by Lemma 1, $R$ is in $L^1$. It now follows that the function $I$, defined by (1), is $2\pi$ times the density of the r.v. $T$. Thus, $I(a)\ge0$ for all real $a$, as desired. -It remains to provide -Proof of Lemma 1: By (3), -\begin{equation*} - \frac{|\Ga(s+it)|}{\Ga(s)}=\prod_{j=0}^\infty\frac{j+s}{|j+s+it|} - =\exp\{-\Si_{s,t}/2\}, -\end{equation*} -where -\begin{equation} - \Si_{s,t}:=\sum_{j=0}^\infty\ln\Big(1+\frac{t^2}{(j+s)^2}\Big). -\end{equation} -Since $\ln\big(1+\frac{t^2}{(j+s)^2}\big)$ is nonincreasing in $j$, we have -\begin{equation} - J_{s,t}\le\Si_{s,t}\le J_{s,t}+\ln\big(1+\frac{t^2}{s^2}\big), -\end{equation} -where -\begin{equation} - J_{s,t}:=\int_0^\infty\ln\big(1+\frac{t^2}{(x+s)^2}\big)\,dx\sim\pi|t| -\end{equation} -as $|t|\to\infty$, which completes the proof of Lemma 1 and the entire answer. -(In fact, integrating by parts, for $t\ne0$ we find -\begin{equation} - J_{s,t}=\pi|t|-s \ln \left(s^2+t^2\right)-2 t \arctan(s/t)+2 s \ln s\sim\pi|t|.) -\end{equation} -The proof of Lemma 1 and the entire answer are now complete.<|endoftext|> -TITLE: When size matters in category theory for the working mathematician -QUESTION [58 upvotes]: I think a related question might be this (Set-Theoretic Issues/Categories). -There are many ways in which you can avoid set theoretical paradoxes in dealing with category theory (see for instance Shulman - Set theory for category theory). -Some important results in category theory assume some kind of ‘smallness’ of your category in practice. A very much used result in homological algebra is the Freyd–Mitchell embedding theorem: - -Every small abelian category admits an fully faithful exact embedding in a category $\text{$R$-mod}$ for a suitable ring $R$. - -Now, in everyday usage of this result, the restriction that the category is small is not important: for instance, if you want to do diagram chasing in a diagram on any category, you can always restrict your attention to the abelian subcategory generated by the objects and maps on the diagram, and the category will be small. -I am wondering: -What are results of category theory, commonly used in mathematical practice, in which considerations of size are crucial? -Shulman in [op. cit.] gives what I think is an example, the Freyd Special Adjoint Functor Theorem: a functor from a complete, locally small, and well-powered category with a cogenerating set to a -locally small category has a left adjoint if and only if it preserves small limits. -I would find interesting to see some discussion on this topic. - -REPLY [25 votes]: The other answers are good, but I would like to point out that Ivan's "uncheatable" lemma can in fact be cheated. The proof of that lemma (due to Freyd) makes inescapable use of classical logic, and in constructive mathematics it is possible to have a non-poset that is complete for the size of its own set of objects (a complete small category). It is even possible to have a category "of sets" with this property (e.g. those called "modest sets" in realizability). Then all Kan extensions into such a category exist, all sheafifications of modest presheaves exist, and presumably all Bousfield localizations of modest spectra exist (although the latter may get you into HoTT water when you try to do it constructively). -About the only thing the category of modest sets lacks is a subobject classifier (it is locally cartesian closed). So these days I prefer the following argument as the "least cheatable" (calling something "uncheatable" sounds like a challenge) manifestation of size issues in category theory. -Lemma 1: Any endofunctor of a complete small category has a fixed point. -Proof: If $C$ is complete-small, so is the category of $F$-algebras for any endofunctor $F:C\to C$. But any complete small category has an initial object (by essentially the same argument that any complete meet-semilattice also has all joins), and an initial $F$-algebra is a fixed point of $F$ (by Lambek's lemma). -Lemma 2: If $C$ is an elementary topos, the double powerset functor $X \mapsto \Omega^{\Omega^X}$ has no fixed point. -Proof: By Cantor's diagonalization argument. -Thus, no elementary topos can have all limits of the size of its collection of objects.<|endoftext|> -TITLE: Is a direct sum of flabby sheaves flabby? -QUESTION [17 upvotes]: Consider a family of flabby (= flasque) sheaves $(\mathcal F_i)_{i\in I}$ of abelian groups on the topological space $X$. -My question : is their direct sum sheaf $\mathcal F=\oplus _{i\in I} \mathcal F_i$ also flabby? -Here is the difficulty: -Given an open subset $U\subset X$ a section $s\in \Gamma(U,\mathcal F)$ consists in a collection of sections $s_i\in \Gamma(U,\mathcal F_i)$ subject to the condition that for any $x\in U$ there exists a neighbourhood $x\in V\subset U$ on which almost all $s_i\vert V \in \Gamma(V,\mathcal F_i)$ are zero. -Now, every $s_i$ certainly extends to a section $S_i\in \Gamma(X,\mathcal F_i)$ by the flabbiness of $\mathcal F_i$. -The problem is that I see no reason why the collection $(S_i)_{i\in I}$ should be a section in $\Gamma(X,\oplus _{i\in I} \mathcal F_i)$, since I see no reason why every point in $X$ should have a neighbourhood $W$ on which almost all the restrictions $S_i\vert W$ are zero. -Of course any direct sum of flabby sheaves is flabby if the space $X$ is noetherian, since in that case we have $\Gamma(U,\mathcal F) =\oplus_{i\in I} \Gamma(U,\mathcal F_i)$ for all open subsets $U\subset X$. -I have only seen the fact that direct sums of flabby sheaves are flabby (correctly) used on noetherian spaces, actually schemes, so that my question originates just from idle curiosity... - -REPLY [21 votes]: No, a direct sum of flabby sheaves need not be flabby. -Take $X=\{1,1/2,1/3,1/4,\dots\}\cup\{0\}$ with the subspace topology from $\mathbb R$, and let $\mathcal F$ be the sheaf whose sections over an open $U\subseteq X$ are the functions $U\to\mathbb F_2$ (not necessarily continuous). This is a flabby sheaf. I claim that the infinite direct sum $\mathcal F^{\oplus\mathbb N}$ of countably many copies of $\mathcal F$ is not flabby. -To see this, let $U=X\setminus\{0\}$, and for $i\in\mathbb N$ let $s_i\colon U\to\mathbb F_2$ denote the function sending $1/i$ to $1$ and all other elements of $U$ to $0$. Thus each $s_i$ is a section of $\mathcal F$ over $U$. Observe that $s=(s_i)_{i\in\mathbb N}\in\Gamma(U,\mathcal F^{\oplus\mathbb N})$, since locally on $U$ all but finitely many of the sections $s_i$ are equal to zero (the topology on $U$ is discrete). -I claim that this section $s$ doesn't extend to a section of $\mathcal F^{\oplus\mathbb N}$ over all of $X$. Indeed, if $s$ extended to a section $\tilde s=(\tilde s_i)_{i\in\mathbb N}$, then there would be a neighbourhood of $0$ in $X$ on which all but finitely many of the $\tilde s_i$ were equal to $0$. But this would imply that $\tilde s_i(1/i)=s_i(1/i)=0$ for all sufficiently large $i$, which is impossible. Thus $s$ does not extend.<|endoftext|> -TITLE: Determining the kernel of the localization map when defining the localization by generators and relations à la Serre -QUESTION [7 upvotes]: All rings considered will be commutative and unitary. Let $A$ be a ring, $S \subseteq A$ a multiplicatively closed subset. The localization $\lambda_S : A \longrightarrow A[S^{-1}]$ can be characterized as a ring homomorphism $\lambda : A \longrightarrow B$ with the following three properties: - -(LC1) $\lambda$ localizes $S$, i.e. $\lambda(s)$ is invertible in $B$ for all $s \in S$; - -(LC2) for every $b \in B$ there is $s \in S$ such that $s b \in \text{im $\lambda$}$; - -(LC3) $\ker \lambda = \{a \in A \,| \,\exists s \in S: sa = 0\}$. - - -One way to achieve this is to define the localization by means of generators and relations: take an indeterminate $T_s$ for each $s \in S$, form the polynomial ring $A[T] = A[T_s|s \in S]$ over $A$ in these indeterminates and quotient out the ideal generated by the $sT_s - 1$ , $s \in S$, thus defining the localization $A[S^{-1}]$: -\begin{equation*} - A[S^{-1}] := A[T_s|s \in S]\,/\,(sT_s-1|s \in S). - \end{equation*} -The structure map $\lambda_S : A \longrightarrow A[S^{-1}]$ then comes along as the composite -\begin{equation*} - A \longrightarrow A[T_s|s \in S] \longrightarrow A[S^{-1}]. - \end{equation*} -See [1], pp. I-7-8. The question is how to verify properties (LC1-3) for this construction. In fact, (LC1-2) are straightforward, but (LC3) seems hard. It is known to be true, since it holds in the other widespread model of localization, given by $\mu_S : A \longrightarrow S^{-1}A$ with -\begin{equation*} - S^{-1}A := A \times S / \sim, - \end{equation*} -where $\sim$ denotes the equivalence relation -\begin{equation*} - (a,s) \sim (b,t) :\iff \exists u \in S:\, u(ta-sb) = 0, - \end{equation*} -and -\begin{equation*} - \mu_S(a) := a/1, - \end{equation*} -where, for $(a,s) \in A \times S$, $a/s$ denotes its equivalence class in $S^{-1}A$. Here, $(LC3)$ is trivial for $\mu_S$, holding by construction. Since both $\lambda_S$ and $\mu_S$ are universal among the ring homomorphisms localizing $S$, it holds for $\lambda_S$, too. But to show this directly for $\lambda_S$ using its definition, is surprisingly difficult: if $\lambda_S(a) = 0$, this means there are $s_1, \dots s_n \in S$ and polynomials $p_1(T), \dots, p_n(T) \in A[T]$ such that ($T_i:=T_{s_i}$) -\begin{equation*} - a = \sum_{i=1}^n p_i(T)(s_iT_i - 1). - \end{equation*} -From this I can conclude -\begin{equation*} - a = -\sum_{i=1}^n a_i \quad,\quad a_i := p_i(0) - \end{equation*} -but this is, for the time being, the end of the flagpole. In the best -of all possible worlds, I would have $p_i(T) = a_i$; this would give -$a_is_i = 0$ for $i=1, \dots, n$, and so $sa = 0$ with $s := s_1 \cdots s_n$, but I see no reason for that. -So does somebody know what is needed to make progress towards (LC3)? -[1] Serre, J.-P., -Algèbre locale - Multiplicités -(Lecture Notes in Mathematics 11). Springer 1965 - -REPLY [2 votes]: The proof of (LC3), in the given setting, is surprisingly difficult, or, at least, elaborate. Let $a \in A$ with $\lambda_S(a) = [a] = 0$ in $A[S^{-1}]$, -i.e. one has -\begin{equation} -\tag{1} - a \in (sT_s-1\,|\,s \in S). -\end{equation} -To show is that -\begin{equation} -\tag{2} - sa = 0 -\end{equation} -for some $s \in S$. Because of (1), there are elements $s_1, \dots s_n \in S$ and polynomials $p_1(T), \dots,p_m(T) \in A[T]$ such that -\begin{equation} - a = \sum_{i=1}^n p_i(T) (s_iT_i - 1) \quad - \text{in $A[T]$},\quad,\quad T_i := T_{s_i}. - \end{equation} -As a first reduction, we may assume $p_i(T) = p_i(T_1, \dots,T_n)$ for all $i$, so that -\begin{equation} -\tag{3} - a = \sum_{i=1}^n p_i(T_1, \dots, T_n) (s_iT_i - 1) - \quad \text{in $A[T]$}. -\end{equation} -Namely, let $T' \subseteq T$ be those indeterminates which either equal some $T_i$, or which appear in at least one $p_i(T)$, $i = 1, \dots, n$, so that we may write $T' = \{T_1, \dots, T_n, T_{n+1}, \dots, T_q\}$. -By eventually introducing dummy terms with coefficient 0, we may assume $p_i(T) = p_i(T') = p_i(T_1, \dots, T_q)$, so that $$a = \sum_{i=1}^n p_i(T_1, \dots, T_q) (s_iT_i - 1)$$. Putting $p_i(T):=0$ for $i=n+1, \dots, q$ then gives -\begin{equation*} - a = \sum_{i=1}^q p_i(T_1, \dots, T_q) (s_iT_i - 1) - \quad \text{in $A[T]$}, -\end{equation*} -which upon renaming $q$ by $n$ gives (3). -To prove that $sa = 0$ for some $s \in S$ we proceed by induction on $n$. For $n = 1$ we start with -\begin{equation*} - a = p(T_s) (sT_s - 1) \quad \text{in $A[T]$} -\end{equation*} -for some indeterminate $T_s \in X$. We abbreviate notation by writing $u := T_s$, so that we have the equation -\begin{equation} - a = p(u) (su - 1) \quad \text{in $A[T]$}. -\end{equation} -Let $p(u) = \sum_{k=0}^d a_k u^k$; then -\begin{equation*} -\begin{split} - p(u) (su - 1) - &= \sum_{k=0}^d sa_k u^{k+1}-\sum_{k=0}^d a_ku^k\\ - &= \sum_{k=1}^{d+1} sa_{k-1} u^k - - \sum_{k=0}^d a_k u^k\\ - &= sa_du^d + \sum_{k=1}^d(sa_{k-1}-a_k) u^k-a_0\\ - &= a, -\end{split} -\end{equation*} -so that -\begin{equation*} - a_0=-a \quad,\quad a_k=sa_{k-1}\,,\,k=1,\dots, d-1 - \quad,\quad sa_d = 0, -\end{equation*} -hence -\begin{equation*} - a_k = -s^ka \,,\, k=0, \dots, d \quad,\quad - sa_d = 0 , -\end{equation*} -so that -\begin{equation*} - s^{d+1}a = -sa_d = 0, -\end{equation*} -as was to be shown. This establishes the base clause of the induction. -We now assume that $n \ge 1$, and that, with $k < n$, -\begin{equation*} - a = \sum_{i=1}^k p_i(T_1,\dots,T_n)(s_iT_i-1) - \quad \text{in $A[T]$} -\end{equation*} -implies that $sa = 0$ for some $s \in S$, and we want to show that the same is true for $k = n$. So we assume, with a given ring $A$, that $a \in \ker \lambda_S$ and (2) holds. We put $A' := A[T_n]/(s_nT_n - 1)$. The projection $A \longrightarrow A'$ then realizes(!) the localization $$\lambda_{S'} : A \longrightarrow A[S'^{-1}]$$ with $S' := \{s_n\}$; in particular, $A'= A[S'^{-1}]$. The canonical map -\begin{equation*} - A[T_n] \longrightarrow A[T] \longrightarrow - A[S^{-1}] -\end{equation*} -induces, by passing to the quotient, $$A'= A[S'^{-1}] \longrightarrow A[S^{-1}] = (A[S'^{-1}])[S^{-1}]$$, which realizes the localization -\begin{equation*} - \lambda_S' : A[S'^{-1}] \longrightarrow (A[S'^{-1}]) - [S^{-1}]. -\end{equation*} -The localization map $\lambda_S : A \longrightarrow A[S^{-1}]$ then factors as the composite of localizations -\begin{equation*} - A \longrightarrow A' \longrightarrow A[S^{-1}] = - A \longrightarrow A[S'^{-1}] \longrightarrow - (A[S'^{-1}])[S^{-1}]. -\end{equation*} -Let $\overline{a} \in A' = A[S'^{-1}]$ be the image of $a \in A$ under $A \longrightarrow A'$. Then $\lambda_S(a) = \lambda_S'(\overline{a}) = 0$. and so, by (3), -\begin{equation*} - \overline{a} = \sum_{i=1}^{n-1} - \overline{p_i}(T_1, \dots, T_{n-1}) (s_iT_i - 1) - \quad \text{in $A'[T]$} -\end{equation*} -with $\overline{p_i}(T_1, \dots, T_{n-1}) = p_i(T_1,\dots, T_{n-1},1/s_n)$, $i=1, \dots, n-1$, since $s_nT_n - 1 = 0$ in $A' = A[S'^{-1}]$. Therefore, by the induction hypothesis, $s\overline{a} = \overline{sa} = 0$ for some $s \in S$. Thus $sa \in \ker \lambda_{S'}$, and so, by the base clause $n=1$ applied to $\lambda_{S'}$, -\begin{equation*} - s_n^{d+1}(sa) = (s_n^{d+1}s)a = 0, -\end{equation*} -which finishes the proof. As a byproduct of the proof we obtain that $s$ in (2) -may be chosen as a product of the $s_i$'s (with repeated factors), i.e. as -an element of the multiplicative closure of $\{s_1, \dots, s_n\}$.<|endoftext|> -TITLE: Invariant neighborhood in a G-CW complex -QUESTION [6 upvotes]: Let $G$ be a discrete group and $X$ be a $G$-CW complex. For any $x\in X$ and open neighborhood $U$ of $x$, I am interested in the question that whether we can find a $G_x$-invariant open neighborhood $V$ of $x$ that is contained in $U$. This is not true in general, for example, if $X$ is the cone over $\mathbb R$ with the $G=\mathbb Z$ action by translation, then the cone point $x_0$ has a neighborhood which does not contain any $G_{x_0}$-invariant open neighborhood. But I am wondering whether the following is true: -Any $G$-CW complex $X$ is $G$-homotopy equivalent to a $G$-CW complex $Y$ so that any open neighborhood $U$ of $y\in Y$ contains a $G_y$-invariant open neighborhood $V$. -As a special case, I am wondering whether the following is true: -Let $\mathcal F$ be a family of subgroups of $G$ which is closed under conjugation and finite intersection. Can we always find a model $E_{\mathcal F}G$ for the universal $G$-CW complex relative to $\mathcal F$ so that any open neighborhood of $x\in E_{\mathcal F}G$ contains a $G_x$-invariant open neighborhood? -Thank you! - -REPLY [4 votes]: I will call the property of a $G$-CW-complex that inside every neighborhood of a point one can find a $G$-invariant neighborhood property $A$. -As in your example, a graph where an edge stabilizer has infinite index in one of the adjacent vertex stabilizers does not have property $A$. -Further $G$-subcomplexes of $G$-CW-complexes with property $A$ have property $A$; given an open set in the subcomplex, extend it to an open set of the whole complex, choose that invariant neighborhood there and intersect back. -Now look at the free group in $a,b$ with the family of subgroups $\mathcal{F}$ containing the trivial group and all conjugates of $\langle a \rangle$ and $\langle b \rangle$. One model for $E_FG$ is given by the Bass-Serre tree, which does not have property $A$. But why can't there be a better model? -Then its one-skeleton would also have property $A$. But by the defining property of $E_FG$ we can find a point $p_a$ with $\langle a \rangle$ is contained in the stabilizer of $p_a$. Since $\langle a \rangle$ is maximal in $F$, it actually follows that $\langle a \rangle$ is the stabilizer of $p_a$. Choose $p_b$ analogously. -Since the one-skeleton is connected, we can choose a finite path from $p_a$ to $p_b$. we want to show that on that path there is some edge, whose stabilizer group has infinite index in the stabilizer of one adjacent vertex. If this was not the case, then $\langle a \rangle$ and $\langle b \rangle$ would be comensurable in the free group, which they are not.<|endoftext|> -TITLE: Algebraicity of a ratio of values of the Gamma function -QUESTION [9 upvotes]: The following ratio: -$$\frac{\Gamma(2/5)^3}{\pi\Gamma(1/5)}$$ -has kept appearing in my research, and the only thing I know about its value is that it is $\cong 0.7567213$, whence the following two questions: -Is the value of this ratio an algebraic number? -What is the exact value of this ratio? - -REPLY [8 votes]: This number is expected to be transcendental. This answer gives a conceptual framework for studying the algebraicity of such $\Gamma$ ratios, and in fact a completely explicit criterion (which is only conjectural, when it comes to establishing transcendence). -Your number is equal to -\begin{equation*} -\frac{\Gamma(2/5)^3}{\Gamma(1/5)^2 \Gamma(4/5)} -\end{equation*} -up to multiplication by an algebraic number. This ratio is described by the vector of exponents $(-2,3,0,-1)$ and the criterion there is not met for $u=2$. In fancy terms, the Bernoulli distribution $B_1$ does not vanish on this vector.<|endoftext|> -TITLE: Classification of subgroups of finitely generated abelian groups -QUESTION [10 upvotes]: A finitely generated abelian group $A$ is isomorphic to a direct sum of cyclic groups. I am interested in an extension of this result on couples of abelian groups $(A,B),$ where $B$ is a subgroup of $A.$ Consider the category of such couples $(A,B),$ where morphism $f:(A,B)\to (A',B')$ is a homomorphism $f:A\to A'$ such that $f(B)\subseteq B'.$ A couple $(A,B)$ is called cyclic if $A$ and $B$ are cyclic groups. -Question 1: Is it true that any couple of finitely generated abelian groups $(A,B)$ is isomorphic to a direct sum of cyclic couples? -If $A$ is free, it follows from the Smith normal form theorem. If there was a version of the Smith normal form theorem for arbitrary homomorphisms of finitely generated abelian groups, then, I believe, this result would follow. -Question 2: Is there a version of the Smith normal form theorem for arbitrary homomorphisms of finitely generated abelian groups? - -REPLY [3 votes]: For finite abelian groups see Sapir, M. V. -Varieties with a finite number of subquasivarieties. -Sibirsk. Mat. Zh. 22 (1981), no. 6, 168–187, 226.<|endoftext|> -TITLE: An enumeration problem for Dyck paths from homological algebra -QUESTION [10 upvotes]: In their article "On n-Gorenstein rings and Auslander rings of low injective dimension" Fuller and Iwanaga gave a homological characterisation of 2-Gorenstein Nakayama algebras with global dimension at most three, see theorem 3.16. there. Now Nakayama algebras (we always assume they are acyclic) are in natural bijection to Dyck paths. Call a Dyck path nice in case the corresponding Nakayama algebra is 2-Gorenstein with global dimension at most 3, see below for an elementary combinatoria description. I noticed with the computer that nice Dyck paths seem to be enumerated by $2^{n-2}$ (thats why I call them nice) and the subclass of nice Dyck paths with global dimension at most two by the Fibonacci numbers. This leads to the following question: - -Question 1: Is there is a bijective proof mapping nice Dyck paths to some known/nice combinatorial objects? - -Furthermore, to every nice Dyck path there is associated a canonical bijection and I wonder what this bijection is (there is a motivation to call this bijection homological rowmotion as it generalises the classical rowmotion from certain posets to more general combinatorial objects such as certain Dyck paths). - -Question 2: What is the associated bijection to a nice Dyck path? - -I currently have no elementary description so question 2, is more of a guess from the data what it might be. -An $n$-Kupisch series (which we can identify with a Dyck path via its area sequence) is a list of $n$ numbers $c:=[c_1,c_2,...,c_n]$ with $c_n=1$, $c_i \ge 2$ for $i \neq n$ and $c_i-1 \leq c_{i+1}$ for all $i=1,...,n-1$ and setting $c_0:=c_n$. -The number of such $n$-Kupisch series is equal to $C_{n-1}$ (Catalan numbers). -Here are some examples of the nice Dyck paths for small $n$ together with the bijection on $\{1,..,n\}$. -$n=2$: - [ [ 2, 1 ], [ [ 1, 2 ], [ 2, 1 ] ] ] - -$n=3$: - [ [ 2, 2, 1 ], [ [ 1, 3 ], [ 2, 1 ], [ 3, 2 ] ] ], - - [ [ 3, 2, 1 ], [ [ 1, 2 ], [ 2, 3 ], [ 3, 1 ] ] ] - -n=4: - [ [ 2, 2, 2, 1 ], [ [ 1, 4 ], [ 2, 1 ], [ 3, 2 ], [ 4, 3 ] ] ], - - [ [ 3, 2, 2, 1 ], [ [ 1, 2 ], [ 2, 4 ], [ 3, 1 ], [ 4, 3 ] ] ], - - [ [ 2, 3, 2, 1 ], [ [ 1, 3 ], [ 2, 1 ], [ 3, 4 ], [ 4, 2 ] ] ], - - [ [ 4, 3, 2, 1 ], [ [ 1, 2 ], [ 2, 3 ], [ 3, 4 ], [ 4, 1 ] ] - -n=5: - [ [ [ 3, 2, 2, 2, 1 ], [ [ 1, 2 ], [ 2, 5 ], [ 3, 1 ], [ 4, 3 ], [ 5, 4 ] ] ], - - [ [ 2, 3, 2, 2, 1 ], [ [ 1, 3 ], [ 2, 1 ], [ 3, 5 ], [ 4, 2 ], [ 5, 4 ] ] ], - - [ [ 4, 3, 2, 2, 1 ], [ [ 1, 2 ], [ 2, 3 ], [ 3, 5 ], [ 4, 1 ], [ 5, 4 ] ] ], - - [ [ 2, 2, 3, 2, 1 ], [ [ 1, 4 ], [ 2, 1 ], [ 3, 2 ], [ 4, 5 ], [ 5, 3 ] ] ], - - [ [ 3, 2, 3, 2, 1 ], [ [ 1, 2 ], [ 2, 4 ], [ 3, 1 ], [ 4, 5 ], [ 5, 3 ] ] ], - - [ [ 3, 3, 3, 2, 1 ], [ [ 1, 5 ], [ 2, 4 ], [ 3, 1 ], [ 4, 2 ], [ 5, 3 ] ] ], - - [ [ 2, 4, 3, 2, 1 ], [ [ 1, 3 ], [ 2, 1 ], [ 3, 4 ], [ 4, 5 ], [ 5, 2 ] ] ], - - [ [ 5, 4, 3, 2, 1 ], [ [ 1, 2 ], [ 2, 3 ], [ 3, 4 ], [ 4, 5 ], [ 5, 1 ] ] ] - -n=6: - [ [ 2, 3, 2, 2, 2, 1 ], [ [ 1, 3 ], [ 2, 1 ], [ 3, 6 ], [ 4, 2 ], [ 5, 4 ], [ 6, 5 ] ] ], - - [ [ 4, 3, 2, 2, 2, 1 ], [ [ 1, 2 ], [ 2, 3 ], [ 3, 6 ], [ 4, 1 ], [ 5, 4 ], [ 6, 5 ] ] ], - - [ [ 2, 2, 3, 2, 2, 1 ], [ [ 1, 4 ], [ 2, 1 ], [ 3, 2 ], [ 4, 6 ], [ 5, 3 ], [ 6, 5 ] ] ], - - [ [ 3, 2, 3, 2, 2, 1 ], [ [ 1, 2 ], [ 2, 4 ], [ 3, 1 ], [ 4, 6 ], [ 5, 3 ], [ 6, 5 ] ] ], - - [ [ 2, 4, 3, 2, 2, 1 ], [ [ 1, 3 ], [ 2, 1 ], [ 3, 4 ], [ 4, 6 ], [ 5, 2 ], [ 6, 5 ] ] ], - - [ [ 5, 4, 3, 2, 2, 1 ], [ [ 1, 2 ], [ 2, 3 ], [ 3, 4 ], [ 4, 6 ], [ 5, 1 ], [ 6, 5 ] ] ], - - [ [ 3, 2, 2, 3, 2, 1 ], [ [ 1, 2 ], [ 2, 5 ], [ 3, 1 ], [ 4, 3 ], [ 5, 6 ], [ 6, 4 ] ] ], - - [ [ 2, 3, 2, 3, 2, 1 ], [ [ 1, 3 ], [ 2, 1 ], [ 3, 5 ], [ 4, 2 ], [ 5, 6 ], [ 6, 4 ] ] ], - - [ [ 4, 3, 2, 3, 2, 1 ], [ [ 1, 2 ], [ 2, 3 ], [ 3, 5 ], [ 4, 1 ], [ 5, 6 ], [ 6, 4 ] ] ], - - [ [ 3, 3, 3, 3, 2, 1 ], [ [ 1, 5 ], [ 2, 6 ], [ 3, 1 ], [ 4, 2 ], [ 5, 3 ], [ 6, 4 ] ] ], - - [ [ 4, 3, 3, 3, 2, 1 ], [ [ 1, 2 ], [ 2, 6 ], [ 3, 5 ], [ 4, 1 ], [ 5, 3 ], [ 6, 4 ] ] ], - - [ [ 2, 2, 4, 3, 2, 1 ], [ [ 1, 4 ], [ 2, 1 ], [ 3, 2 ], [ 4, 5 ], [ 5, 6 ], [ 6, 3 ] ] ], - - [ [ 3, 2, 4, 3, 2, 1 ], [ [ 1, 2 ], [ 2, 4 ], [ 3, 1 ], [ 4, 5 ], [ 5, 6 ], [ 6, 3 ] ] ], - - [ [ 3, 3, 4, 3, 2, 1 ], [ [ 1, 5 ], [ 2, 4 ], [ 3, 1 ], [ 4, 2 ], [ 5, 6 ], [ 6, 3 ] ] ], - - [ [ 2, 5, 4, 3, 2, 1 ], [ [ 1, 3 ], [ 2, 1 ], [ 3, 4 ], [ 4, 5 ], [ 5, 6 ], [ 6, 2 ] ] ], - - [ [ 6, 5, 4, 3, 2, 1 ], [ [ 1, 2 ], [ 2, 3 ], [ 3, 4 ], [ 4, 5 ], [ 5, 6 ], [ 6, 1 ] ] ] ] - -In the following I give the elemenatary combinatorial description of nice Dyck paths. Sadly, it is quite complicated at the moment despite the possibly very nice enumeration. -I found the following combinatorial characterisation of those Dyck paths (compare with Combinatorics problem related to Motzkin numbers with prize money I ): -The CoKupisch series $d$ of $c$ is defined as $d=[d_1,d_2,...,d_n]$ with $d_i:= \min \{k | k \geq c_{i-k} \} $ and $d_1=1$. One can show that the $d_i$ are a permutation of the $c_i$. A number $a \in \{1,...,n \}$ is a descent if $a=1$ or $c_a >c_{a-1}$. Define a corresponding set, indexed by descents: $X_1 := \{1,2,...,c_1-1 \}$, and $X_a := \{ c_{a-1}, c_{a-1}+1 ,..., c_a -1 \}$ for descents $a > 1$. -A $n$-Kupisch series is called $2-$Gorenstein if it satisfies the following condition: - - -condition: for each descent $a$, and each $b \in X_a$: -either $c_{a+b} \geq c_{a+b-1}$ or $d_{a+b-1} = d_{a+b + c_{a+b}-1} - c_{a+b}$ is satisfied. - - -Now an $n$-Kupisch path is nice if and only if it is 2-Gorenstein and it has global dimension at most 3. Sadly there is no nice formal description of global dimension at most 3 but it can be pictured in a nice way in a Dyck path. -Call an $i$ with $1 \leq i \leq n-1$ good in case one of the following three conditions hold: - -$c_{i+1}=c_i -1$ (equivalent to the simple module $S_i$ having projective dimension one) - -($c_{i+1}>c_i-1 $ and $c_{i+c_i}=c_{i+1}-c_i+1$) (equivalent to $S_i$ having projective dimension two) - -($c_{i+1}>c_i-1 $ and $c_{i+c_i}>c_{i+1}-c_i+1$ and $c_{i+c_{i+1}+1}=c_{i+c_i}-c_{i+1}+c_i-1$) (equivalent to $S_i$ having projective dimension three) - - -Now the 2. condition is: - - -condition: Every $i$ with $1 \leq i \leq n-1$ is good. - - -So an n-Kupisch series (=Dyck path) is nice if and only if it satisfies condition 1. and 2. - -REPLY [4 votes]: This is a conjectural answer. -Let $w = 0\dots01$ be a binary word of length $n$. Then $\phi(w)$ is the Dyck path $U^{(n+1)/2} (UD)^{(n-1)/2} D^{(n+1)/2}$ if $n$ is odd and $U^{n/2} (UD)^{n/2} D^{n/2}$ if $n$ is even. -Let $w = 0^{n_1} 1 0^{n_2} 1 \dots 0^{n_k} 1$ be any binary word ending with a $1$. Then $\phi(w) = \phi(0^{n_1} 1) \phi(0^{n_2} 1)\dots \phi(0^{n_k} 1)$. -Finally, to obtain the nice Dyck path, apply the Lalanne-Kreweras involution https://www.findstat.org//Mp00120.<|endoftext|> -TITLE: Disjoint sets with twice ratio -QUESTION [10 upvotes]: Given are a positive integer $n$ and positive real numbers $a_1,\dots,a_n,b_1,\dots,b_n$. A subset $S\subseteq N=\{1,\dots,n\}$ is called $a$-good if $$\sum_{i\in S}a_i\geq \frac{1}{2}\left(\sum_{i\in N\backslash S}a_i-\min_{i\in N\backslash S}a_i\right),$$ -and $b$-good if $$\sum_{i\in S}b_i\geq 2\left(\sum_{i\in N\backslash S}b_i-\min_{i\in N\backslash S}b_i\right).$$ -Are there always two disjoint subsets, one $a$-good and the other $b$-good? -If $a_i = b_i$ for all $i$, this is true by a greedy algorithm. - -REPLY [4 votes]: An affirmative answer to your question would follow from a famous conjecture about envy-free allocations. It is also known that if you replace the '2' in $b$-goodness with '$\frac 12$', then it is true. -A good paper to read (first) on the topic is Almost Envy-Freeness with General Valuations. -An allocation of some items among some players is envy-free up to any good (EFX) if no player i would envy any other player j IF an arbitrary item of j was to be removed. -(Note that players can value different items differently, but let's suppose that their valuation function is additive.) -The conjecture is that an EFX allocation always exists. -This would imply a positive answer to your question as follows. -Let player A evaluate $n$ items as $(a_i)$, while players B and B' as $(b_i)$. -Take an EFX allocation among these 3 players, then merge B and B'. -An easy calculation shows that the conditions of your question will be satisfied.<|endoftext|> -TITLE: When is a bi-Lipschitz homeomorphism smoothable? -QUESTION [15 upvotes]: Suppose I have a smooth Riemannian manifold $X$ with induced distance function $d$, and a bi-Lipschitz (with respect to $d$) homeomorphism -$$\phi: X \to X.$$ -Under what circumstances could $\phi$ be smoothable to a diffeomorphism? By "smoothable" in this case I mean "homotopic to a diffeomorphism through bi-Lipschitz homeomorphisms" (this might not be standard, I suppose). Are there any clear obstructions? - -REPLY [3 votes]: There is some interest in a related question in non-linear elasticity, specifically people there would consider a function "smoothable" if there is a close-by (in some norm applying both to function and its inverse) diffeomorphism. -In 2D there are some results with regards to this, see e.g. Smooth approximation of bi-Lipschitz orientation-preserving homeomorphisms by Danieri & Pratelli, who prove that there is a close diffeomorphism in some bi-Sobolev norm for domains in the plane (which if I am not mistaken should imply the same result at least for compact manifolds). But the proof uses a bi-Lipschitz extension theorem, so I am not sure if one can construct homotopies from that result easily and there seems to be no way to extend this to higher dimensions.<|endoftext|> -TITLE: Does Morita theory hint higher modules for noncommutative ring? -QUESTION [11 upvotes]: Two possibly noncommutative rings are called Morita equivalent if their left-module categories are equivalent. In the commutative case, Morita equivalence is nothing more than ring isomorphism. Otherwise, there are many known examples where this does not hold. -That means modules alone are not enough to characterize the ring. Are there notions of higher modules or higher structures, such that the corresponding higher Morita equivalence is nothing more than ring isomorphism? -In short, can you find a better notion of modules that faithfully capture their underlying ring? - -REPLY [3 votes]: Although I think my answer "pointed categories" is an important one, there is another way that the question could be interpreted: What is an interesting class of rings which are recoverable up to isomorphism from their categories of modules? And perhaps, if we are not trying to be too universal or functorial, we won't mind if the isomorphism is not unique. -For this, I strongly recommend Morita's original paper introducing his equivalences. (Kiiti Morita, Duality for modules and its applications to the theory of rings with minimum condition, Sci. Rep. Tokyo Kyoiku Daigaku, Sect. A 6, 83-142 (1958). You can find a PDF by googling.) Among the many things in that paper is a theorem saying that rings satisfying a natural minimality condition called "basic" are Morita equivalent iff they are isomorphic (commutative rings, I believe, satisfy this condition) and that every finite-dimensional ring is Morita equivalent to one satisfying this minimality condition. See my answer to a related question.<|endoftext|> -TITLE: What are the benefits of writing vector inner products as $\langle u, v\rangle$ as opposed to $u^T v$? -QUESTION [103 upvotes]: In a lot of computational math, operations research, such as algorithm design for optimization problems and the like, authors like to use $$\langle \cdot, \cdot \rangle$$ as opposed to $$(\cdot)^T (\cdot)$$ -Even when the space is clearly Euclidean and the operation is clearly the dot product. What is the benefit or advantage for doing so? Is it so that the notations generalize nicely to other spaces? -Update: Thank you for all the great answers! Will take a while to process... - -REPLY [7 votes]: Lots of great answers so far, but I'll add another (hopefully at least good) answer: the notation $v^T u$ makes it somewhat difficult to speak of collections of bilinear pairings depending on a parameter. Typical examples: - -"Let $\langle \cdot, \cdot \rangle_i$ be a finite set of inner products on a vector space $V$" -"Let $\langle \cdot, \cdot \rangle_p$, $p \in M$, be a Riemannian metric on a manifold $M$" -"Let $\langle \cdot, \cdot \rangle_t$ be a continuously varying family of inner products on a Hilbert space $H$" - -These are all difficult to express using the transpose notation. The closest you can get is to write, for instance $v^T A_i u$ where $A_i$ is a family matrices, but particularly when one is speaking of continuously varying families of inner products you run into all sorts of difficult issues with coordinate systems, and it becomes very difficult to keep things straight.<|endoftext|> -TITLE: Reference to a covering theorem by Ketonen -QUESTION [5 upvotes]: I am reading a paper by Goldberg, and he uses a theorem by Ketonen, which is highlighted in red below: - -Do you know where can I find the theorem and it's proof? -Link to the article: https://arxiv.org/abs/2002.07299 -Thank you. - -REPLY [4 votes]: Sorry I didn't give a reference. The theorem first appeared in spirit in Ketonen's paper "Strong compactness and other cardinal sins," Theorem 1.3. -The precise result I needed is proved in my thesis as Theorem 7.2.12, which states that if $j : V \to M$ is an elementary embedding, $\lambda$ is a regular uncountable cardinal, and $\delta$ is an $M$-cardinal, then $\text{cf}^M(\sup j[\lambda])\leq \delta$ if and only if $j$ is "$(\lambda,\delta)$-tight," which just means $j[\lambda]$ is contained in a set in $M$ of $M$-cardinality at most $\delta$. -Lemma 7.2.7 states that an ultrapower embedding $j : V\to M$ is $(\lambda,\delta)$-tight if and only if $M$ has the "$({\leq}\lambda,{\leq}\delta)$-cover property," which just means that every subset of $M$ of cardinality at most $\lambda$ is contained in an element of $M$ that has $M$-cardinality at most $\delta$.<|endoftext|> -TITLE: Closed form solution for $XAX^{T}=B$ -QUESTION [7 upvotes]: Let $d \times d$ matrices $A, B$ be positive definite. Is there a closed form solution for the following quadratic equation in $X$? -$$X A X^{T} = B$$ -Thank you. - -REPLY [15 votes]: $B^{-1/2}XAX^TB^{-1/2}=I$, so $B^{-1/2}XA^{1/2}=Q$ must be orthogonal. On the other hand, for any orthogonal $Q$, it is simple to verify that $X = B^{1/2}QA^{-1/2}$ solves the equation, so this is a complete parametrization of the solutions. -Here $A^{1/2}$ is the symmetric square root of $A$ (if you prefer you can work with the Cholesky factor and obtain similar results).<|endoftext|> -TITLE: Specific notions of forcing from the point of view of category theory -QUESTION [12 upvotes]: I'm trying to learn about the topos of sheaves and the double negation topology to try to go through the independence of CH from a categorical perspective. I'm curious in general about what the standard set theoretic notions of forcing would look like in category theory. -For example: what corresponds to Prikry forcing in the topos of sheaves point of view? Is there a categorical interpretation of the Levy-Solovay theorem about large cardinals being unaffected by small forcing? Is this known or has it been investigated somewhere? Thanks in advance for any references or answers/comments. - -REPLY [14 votes]: The topos version of forcing with a poset $P$ regards $P$ as a category, forms the topos of presheaves on it, and then passes to the subtopos of double-negation sheaves. The presheaf topos amounts to strong forcing (which obeys only intuitionistic logic) and the sheaf subtopos amounts to the much more widely used weak forcing. This was, at least for the Cohen models violating CH, in the early work of Lawvere and Tierney, presented in Lawvere's paper "Quantifiers and sheaves" (Proceedings of the 1970 ICM in Nice. Big pdf (see p329), individual article). If I remember correctly, Marta Bunge worked out the topos version of the forcing that adjoins a Suslin tree (JPAA 1974, doi:10.1016/0022-4049(74)90020-6, see also the erratum). -More information about these topics is in work of John Bell (connecting it with Boolean-valued models, if I remember correctly) and in Peter Freyd's paper "All topoi are localic" (JPAA 1987, doi:10.1016/0022-4049(87)90042-9). Andre Scedrov and I included much of this material in the early, background sections of "Freyd's models for the axiom of choice" (Memoirs A.M.S. 404 (1989) --- this was before "404" became a widely known synonym for "file not found").<|endoftext|> -TITLE: Intersection form of surface bundle over surface -QUESTION [7 upvotes]: Let $\Sigma_g$ be a Riemannian surface of genus $g$. Let $M^4$ be a surface bundle over surface: $\Sigma_g \to M^4 \to \Sigma_h$. $\Sigma_g$ is the fiber and $\Sigma_h$ is the base space. -My question: is there a surface bundle over surface $M^4$ such that it has a 2-cocycle $c \in H^2(M^4;Z)$ satisfying -(1) $\int_{M^4} c^2 =\pm 1$, and -(2) $\int_{\Sigma_g} c =0$ -Also: How to construct surface bundles with known odd intersection form? This may help to answer the first question. -See a related question: -Oddness of intersection form of surface bundle -A further question: is there a surface bundle over surface $M^4$ such that it has a 2-cocycle $c \in H^2(M^4;Z)$ satisfying -(1) $\int_{M^4} c^2 =\pm 1$, -(2) $\int_{\Sigma_g} c =0$, and (3) $c = w_2$ mod 2. -What is the signature and $g$ for such surface bundle? -Note that the condition (1) implies that $M^4$ is not spin and $w_2$ is non-trivial. - -REPLY [9 votes]: Yes, such a thing exists, but I don't know an explicit example. -To see that it exists, it is clearest to me to consider the universal situation. For any $k \in \mathbb{Z}$ there is a space $\mathcal{S}_g(k)$ which classifies oriented surface bundles -$$\Sigma_g \to E \overset{\pi}\to B$$ -equipped with a class $c \in H^2(E; \mathbb{Z})$ such that $\int_{\Sigma_g} c = k$. Associated to such a family there are characteristic classes -$$\kappa_{i,j} = \int_\pi e(T_\pi E)^{i+1} \cdot c^j \in H^{2(i+j)}(B;\mathbb{Z}),$$ -where $T_\pi E$ denotes the tangent bundle of $E$ along the fibres of $\pi$, and $e(T_\pi E)$ denotes its Euler class. (The classes $\kappa_{i,0}$ are the usual Miller--Morita--Mumford classes $\kappa_i$.) -In - -J. Ebert and O. Randal-Williams, Stable cohomology of the universal Picard varieties and the extended mapping class group. Doc. Math. 17 (2012), 417–450. - -Johannes Ebert and I studied, among other things, the low-dimensional integral cohomology of $\mathcal{S}_g(k)$, and showed that as long as $g$ is large enough (I think $g \geq 6$ will do) one has -$$H^1(\mathcal{S}_g(k);\mathbb{Z})=0 \quad\quad H^2(\mathcal{S}_g(k);\mathbb{Z})\cong\mathbb{Z}^3$$ -where the isomorphism in the second case is given by a basis of cohomology classes $\lambda, \kappa_{0,1}, \zeta$, where the outer two are related to the $\kappa_{i,j}$ by the identities -$$12 \lambda = \kappa_{1,0} \quad\quad 2\zeta = \kappa_{0,1} - \kappa_{-1,2}.$$ -In particular, applying this with $k=0$ and using that every second homology class is represented by a map from an oriented surface, it follows that there is a surface bundle -$$\Sigma_g \to E \overset{\pi}\to \Sigma_h$$ -for some $h$ (which is uncontrollable using this method) with a class $c \in H^2(E;\mathbb{Z})$ satisfying $\int_{\Sigma_g}c = 0$, and having -$$\int_{\Sigma_h}\lambda=\text{whatever you like} \quad\quad \int_{\Sigma_h}\kappa_{0,1}=1 \quad\quad \int_{\Sigma_h}\zeta = 0$$ -and hence having -$$\int_E c^2 = \int_{\Sigma_h} \int_\pi c^2 = \int_{\Sigma_h} \kappa_{-1,2} = \int_{\Sigma_h} \kappa_{0,1} = 1.$$<|endoftext|> -TITLE: Proper morphisms with geometrically reduced and connected fibers -QUESTION [5 upvotes]: Let $f: X \to S$ be a proper morphism ($S$ locally noetherian), and $X \to S' \to S$ its Stein factorisation. By Zariski's Main Theorem the number of geometric connected components of the fibers of $f$ can be read from the cardinal of the fibers of the finite $S' \to S$. In particular if all fibers of $f$ are geometrically connected, then $S' \to S$ is radicial. -I expect that if furthermore the fibers of $f$ are geometrically reduced (and $f$ is surjective and $S$ reduced in order to remove trivial counterexamples), then $S'=S$ that is $f$ is an $\mathcal{O}$-morphism (viz. $f_*\mathcal{O}_X = \mathcal{O}_S$). Strangely I only find this fact when $f$ is furthermore assumed to be flat, for instance: https://stacks.math.columbia.edu/tag/0E0L. -Here is an outline of a demonstration (suggested by a friend): we want to show that $S' \to S$ is an isomorphism. Since it is a surjection by assumption on $f$, it suffices to show that it is an immersion. By our assumptions on $f$, $S' \to S$ has geometrically connected and reduced fibers. We way assume that $S=\textrm{Spec} A$ and $S'=\textrm{Spec} B$, with $A \to B$ finite. Let $C$ be the cokernel of $A \to B$ (seen as a $A$-module). If $p$ is a prime ideal in $A$, $B \otimes_A \overline{k}(p) = \overline{k}(p)$ (since it is connected and reduced over $\overline{k}(p)$), so $C \otimes_A \overline{k}(p)=0$, so $C=0$. -Is the above proof indeed correct? Does the hypotheses already imply that $f$ is flat? Is there a reference to this result somewhere in the literature, presumably in EGA? - -REPLY [6 votes]: Here is a standard example. Take $\mathbb{P}^1\subset\mathbb{P}^3$ of large degree and let $S$ be the cone, with the vertex $p$, only singular point. Let $f:X\to S$ the blow up of $p$. One can check that $X$ is smooth and thus the Stein factorization $S'$ is the normalization of $S$. The fiber over $p$ in $X$ is smooth irreducible (scheme-theoretically), but the fiber in $S'$ is not reduced.<|endoftext|> -TITLE: Equivalence of families indexes of Fredholm operators -QUESTION [5 upvotes]: Let $F=F(H,H)$ be the space of bounded Fredholm operators in a Hilbert space $H$ with topology inherited from the norm operator topology, and let $X$ be a compact topological space. -For a continuous map $T\colon X\to F$, there exists a closed subspace $W\subseteq H$ with $\dim H/W<\infty$ such that $W\cap\ker T_x=0$ for all $x\in X$ and $H/T(W) =\bigcup_{x\in X} H/T_x(W)$ is a vector bundle over $X$ (See appendix of K-Theory, Anderson & Atiyah). -Then one can show that -$$\mbox{Ind}_1(T) = [X\times H/W] - [H/T(W)] \in K(X)$$ -does not depend on $W$. -On the other hand, there exists a finite dimensional subspace $V\subseteq H$ such that $V+T_x(H) = H$ for all $x\in X$, so we can define $T^V\colon X\to F(H\oplus V, H)$ by $T^V_x(u,v) = T_x u + v$. Then $T^V_x$ is surjective and $\dim\ker T^V_x$ is constant on $x$. Thus $\ker T^V = \bigcup_{x\in X} \ker T_x$ is also a vector bundle over $X$. One can show that -$$ \mbox{Ind}_2(T) = [\ker T^V] - [X\times V] \in K(X)$$ -does not depend on $V$. -These index maps are called the family index of families of Fredholm operators in $H$, and it made me suspect that they are equal. - -Question: Is it true that $$[X\times H/W] - [H/T(W)] = [\ker T^V] - [X\times V] \qquad (1)$$ in $K(X)$ ? -Is there any reference that proves the equivalence of these indexes? - - -Edit: We can shrink $W$ or augment $V$ in order to have $\dim H/W = \dim V$. Say $H/W \cong V \cong \mathbb{C}^N$, so that $X\times H/W \cong X\times V \cong X\times\mathbb{C}^N$, and therefore -$$\mbox{Ind}_1(T) = [X\times\mathbb{C}^N] - [H/T(W)]\ ,$$ -$$\mbox{Ind}_2(T) = [\ker T^V] - [X\times\mathbb{C}^N]\ .$$ -Equation $(1)$ becomes -$$[X\times\mathbb{C}^N] - [H/T(W)] = [\ker T^V] - [X\times\mathbb{C}^N]$$ -and it holds iff there exists $k\geq0$ such that -$$\ker T^V \oplus H/T(W) \oplus (X\times\mathbb{C}^k) \cong X\times\mathbb{C}^{2N+k}$$ -But why does there exists such $k$? - -REPLY [2 votes]: Notice that $V^\perp\cap\ker T_x^*=0$ for every $x$, because for every $w\in V^\perp\cap\ker T_x^*$, $u\in H$, and $v\in V$, one has -$$\langle w,T_x(u)+v \rangle = \langle w,T_x(u) \rangle = 0.$$ -By composing two isomorphisms $\ker P_{V^\perp}T \ni u \mapsto u\oplus T(-u) \in \ker T^V$ and $\ker P_{V^\perp}T\cong H/T^*(V^\perp)$, one sees $\mathrm{Ind}_2(T)=-\mathrm{Ind}_1(T^*)$. -Thus the quality of two indices follows from the fact that self-adjoint Fredholm operator -$\left[\begin{smallmatrix} & T^*\\ T & \end{smallmatrix}\right]$ has index zero in either definition.<|endoftext|> -TITLE: Size of the category of cohomology theories -QUESTION [5 upvotes]: I'd like to understand the structure of the functor category Coh whose objects are cohomology functors from a category of Spaces to the category of graded commutative rings GCR. Spaces could be any of the familiar geometric categories: topological spaces, manifolds, algebraic varieties, schemes, etc. -My first question is about just the size of Coh. For any choice of Spaces there are several well known cohomology functors (singular, de Rham, etale, ...) and new ones keep cropping up (syntomic, prismatic) - and they all agree too, on suitable subcategories of Spaces after suitable extensions of scalars, hinting at a fundamental core to it all (e.g., motives) - but I don't know how many more there can be. Is there a systematic way to enumerate or even construct them all? The latter is extremely unlikely because construction of any one we have has been a highly creative and painstaking task, but can we at least know how many in some sense are still out there? Is it even a discrete set, or can we in fact "deform" cohomology theories in families in certain settings? -Same question of size applies to the sets of natural transformations among cohomology theories, i.e., the Hom sets in Coh. Apart from the standard comparison isomorphisms what do we know about other natural transformations, even just for two well known cohomologies, for example, Betti and de Rham? -Sorry if the scope of the question is too broad and I should have made it more manageable by fixing a particular category of spaces and coefficient system for their cohomology. Please feel free to pick a setting that makes for a satisfactory answer. - -REPLY [4 votes]: It is perhaps more natural to study homology theories (or cohomology theories) up to `Bousfield equivalence', where two theories are equivalent if they send the same maps to isomorphisms. (So, for example, classical cohomology with coefficients in a field $k$ are sorted by the characteristic of $k$.) This has been studied, with much sophistication, beginning with papers by Bousfield around 1970, who wrote about the lattice of such localization functors, and gave examples of both orderly and unusual behavior. -I would suggest looking up his papers, and also those of Mark Hovey (e.g. Cohomological Bousfield classes. J. Pure Appl. Algebra 103 (1995), no. 1, 45–59, or Hovey, Mark; Palmieri, John H. The structure of the Bousfield lattice. Homotopy invariant algebraic structures (Baltimore, MD, 1998), 175–196, Contemp. Math., 239, Amer. Math. Soc., Providence, RI, 1999.) A paper that shows that versions of your question can be proved to be undecidable is Casacuberta, Carles; Scevenels, Dirk; Smith, Jeffrey H. Implications of large-cardinal principles in homotopical localization. Adv. Math. 197 (2005), no. 1, 120–139. Have fun exploring this! -[I should add that, when restricted to the category of finite CW complexes, the Bousfield classes are known: they are detected by the sequence of Morava K-theories associated to each ordinary prime.]<|endoftext|> -TITLE: Smooth projective surface with geometrically integral reduction -QUESTION [6 upvotes]: Let $S$ be a geometrically connected smooth projective surface over $\mathbb{Q}_p$. Can it be put in a proper flat $\mathbb{Z}_p$-scheme with a geometrically integral special fiber? - -REPLY [5 votes]: That is not true. There are probably shorter answers than the following. Let $K$ be a field, and denote a separable closure by $K^{\text{sep}}$. Let $n>1$ be an integer. -Definition. A Severi-Brauer variety over $K$ of relative dimension $n-1$ is a proper, smooth $K$-scheme whose base change to $K^{\text{sep}}$ is isomorphic to projective space of dimension $n-1$ over $K^{\text{sep}}$. -There is a natural bijection between the set of $K$-isomorphism classes of Severi-Brauer varieties over $K$ of relative dimension $n-1$ and the subset of $\text{Br}(K)[n]$ of those ed. $n$-torsion elements in the Brauer group of $K$ whose order index divides $n$. In particular, the identity element in this subset corresponds to the isomorphism class of projective space, i.e., the isomorphism class of any Severi-Brauer variety over $K$ of relative dimension $n-1$ that has a $K$-rational point. -For a finite, separable field extension $L/K$ of degree $d$, there are restriction and corestriction group homomorphisms, -$$ -\text{Br}(K)\to \text{Br}(L), \ \ \text{Br}(L)\to \text{Br}(K), -$$ -whose composition is the "multiplication by $d$" map. -Finally, for every finite extension of $\mathbb{Q}_p$, local class field theory gives a natural isomorphism, -$$ -\text{inv}_K:\text{Br}(K) \to \mathbb{Q}/\mathbb{Z}, -$$ -ed. and the index of every element equals the order of that element. -Now let $K$ be a finite extension of $\mathbb{Q}_p$, and let $X_K$ be a Severi-Brauer variety over $K$ of relative dimension $n-1$ whose image in $\text{Br}(K)[n]$ is a generator for this cyclic group of order $n$. By the restriction-corestriction homomorphisms, for a finite field extension $L/K$, the base change of $X_K$ over $L$ has an $L$-rational point only if the degree $d$ of the field extension is divisible by $n$. -On the other hand, if $X_K$ has a proper, flat model over the ring of integers of $K$ whose special fiber has nonempty smooth locus an irreducible component that is geometrically integral, then by the Lang-Weil estimates together with Hensel's Lemma, for every sufficiently large integer $d$, there is an unramifield field extension $L/K$ of degree $d$ such that the base change has an $L$-rational point. Therefore, there is no proper, flat model of $X_K$ over the ring of integers of $K$. -In particular, for the integer $n=3$, there exists a Severi-Brauer scheme over $\mathbb{Q}_p$ of relative dimension $n-1=2$ that has no proper, flat model over $\mathbb{Z}_p$ (ed. . . . with an irreducible component that is geometrically integral!). - -REPLY [4 votes]: Let $X$ be an integral regular scheme which is proper over $\mathbb{Z}_p$. Assume that the special fibre $X_{\mathbb{F}_p}$ is irreducible and let $k$ be the algebraic closure of $\mathbb{F}_p$ in the function field of $X_{\mathbb{F}_p}$. Then $k$ is a birational invariant of the generic fibre $X_{\mathbb{Q}}$. (The field $k$ is something like the field of definition of the geometric irreducible components of $X_{\mathbb{F}_p}$, with $k = \mathbb{F}_p$ if and only if $X_{\mathbb{F}_p}$ is geometrically irreducible). -This is a special case of a more general result given in [1, Cor. 2.3] (which also makes precise what I mean by "birational invariant of the generic fibre", also allows the special fibre to be reducible, and applies to general dvrs). -So consider the $\mathbb{Z}_p$-scheme -$$ x^2 - ay^2 = p z^2$$ -where $a \in \mathbb{Z}_p^\times$ is a non-square (model of a plane conic). This is easily check to be integral regular and proper. But the special fibre is -$$x^2 - ay^2 \equiv 0 \bmod p$$ -which has $k = \mathbb{F}_p(\sqrt{a})$ in the above notation. Therefore, by the above result, there is no regular integral proper model of the generic fibre with geometrically integral special fibre. -For a counter-example involving surfaces, just take the fibre product of the above scheme with $\mathbb{P}^1_{\mathbb{Z}_p}$. -Your question allowed arbitrary (not necessarily regular) models. If you are given a non-regular model $Y$ for the above conic, then you can perform resolution of singularities to obtain a regular model $\widetilde{Y}$ (since $\dim Y \leq 3$). If $Y$ had a geometrically integral special fibre, then the special fibre of $\widetilde{Y}$ would have an irreducible component which is geometrically integral, which is not allowed by the more general [1, Cor. 2.3]. -[1] Skorobogatov - Descent on toric fibrations<|endoftext|> -TITLE: What subsystem of second-order arithmetic is needed for the recursion theorem? -QUESTION [8 upvotes]: In its simplest version, the recursion theorem states that for any $m\in\mathbb{N}$ and any function $g:\mathbb{N}\rightarrow\mathbb{N}$, there exists a function $f:\mathbb{N}\rightarrow\mathbb{N}$ such that $f(0)=m$ and $f(n+1) = g(f(n))$. There are many more complicated versions, with multiple variables and parameters and course-of-values recursion and so on. But that’s the gist of it. -Now the recursion theorem, no matter which version of it you take, is a statement in the language of second-order arithmetic. And I'm pretty sure that it’s provable in $Z_2$, i.e. full second-order arithmetic. But my question is, what is the weakest subsystem of second-order arithmetic capable of proving it? -Do different versions of the recursion theorem require different subsystems to prove it? - -REPLY [11 votes]: As Wojowu already pointed out $\mathsf{RCA}_0$ proves recursion theorem. You could find a proof in Simpson's book [1], Section II.3. -In fact primitive recursion theorem is equivalent to $\Sigma^0_1\textsf{-Ind}$ over $\mathsf{RCA}_0^{\star}$. Here $\mathsf{RCA}_0^{\star}$ is $\mathsf{EA}+\Delta^0_1\text{-}\mathsf{CA}+\Delta^0_0\text{-}\mathsf{Ind}$ and $\mathsf{RCA}_0=\mathsf{RCA}_0^{\star}+\Sigma^0_1\text{-}\mathsf{Ind}$. So we need to prove in $\mathsf{RCA}_0^{\star}+\mathsf{PrimRec}$ any given instance -$$\exists y\;\varphi(0,y)\land \forall x\;(\exists y\;\varphi(x,y)\to \exists y\;\varphi(x+1,y))\to \forall x\;\exists y\varphi(x,y)$$ -of $\Sigma^0_1\textsf{-Ind}$, where $\varphi$ is $\Delta^0_0$. -Indeed let us reason in $\mathsf{RCA}_0^{\star}+\mathsf{PrimRec}$. We assume $\exists y\;\varphi(0,y)\land \forall x\;(\exists y\;\varphi(x,y)\to \exists y\;\varphi(x+1,y))$ and claim $\forall x\;\exists y\varphi(x,y)$ and claim that $\forall x\exists y\varphi(x,y)$. Using $\Delta^0_1\textsf{-CA}$ and premise of induction we form the following function $g(x)$: -$$g(x)=\begin{cases}\langle y_0,\ldots,y_{n-1},\min\{y_{n}\mid \varphi(n,y_{n})\}\rangle &\text{, if $x=\langle y_0,\ldots,y_{n-1}\rangle$ and}\\ & \text{$\;\;\;$ $\varphi(i,y_i)$, for all $i< n$}\\ 0&\text{, otherwise}\end{cases}$$ -We applying primitive recursion to $g$ and put $f(0)=\langle \rangle$. The resulting $f$ maps $n$ to a sequence $\langle y_0,\ldots,y_{n-1}\rangle$ such that $\varphi(0,y_0),\ldots,\varphi(n-1,y_{n-1})$. Note that the latter fact could be verified by $\Delta^0_0\textsf{-Ind}$. Thus we prove the instance of $\Sigma^0_1\textsf{-Ind}$. -[1] Simpson, S. G. (2009). Subsystems of second order arithmetic (Vol. 1). Cambridge University Press.<|endoftext|> -TITLE: There is a 3-connected 5-regular simple $n$-vertex planar graph iff $n$ satisfies....? -QUESTION [5 upvotes]: Is there any characterization on the set of integers $n$ such that there is a 3-connected 5-regular simple $n$-vertex planar graph? - -REPLY [9 votes]: There is a 3-connected 5-regular simple $n$-vertex planar graph if and only if $n=12$ or $n \ge 16$ is even. See Recursive generation of 5-regular graphs by Mahdieh Hasheminezhad, Brendan D. McKay, Tristan Reeves in WALCOM: Algorithms and Computation, eds. Das and Uehara, Lecture Notes in Computer Science, vol 5431, Springer 2009. The number of such graphs is given in OEIS A308489. -They use a set of 7 graphs that are irreducible under a system of expansions & reductions and, as is common for contemporary graph theory, computer assistance. E.g., "The program completed execution in 21 seconds. In total, 39621 induced subgraphs were found..." - -REPLY [8 votes]: There are no such graphs when $n$ is odd, by the handshaking lemma. -Conversely, for all even $n \geq 224$, we claim such a graph exists. -In particular, given two planar 5-regular graphs $G$, $H$ each drawn on the surface of a sphere, we can define the 'connected sum' of the graphs as follows: - -remove a small disk (containing one vertex) from the sphere on which $G$ is drawn; -remove a small disk (containing one vertex) from the sphere on which $H$ is drawn; -combine the two resulting hemispheres at their equator. - -The resulting graph (which may depend on the chosen vertices) has $|G| + |H| - 2$ vertices, and inherits the planarity, 5-regularity, and 3-connectedness of $G$ and $H$. -Now, given an even integer $n \geq 224$, we can find integers $i, j \geq 0$ such that $n = 2 + 10i + 58j$. Then we can construct an $n$-vertex graph with the desired properties by taking the connected sum of $i$ copies of the icosahedron and $j$ copies of the snub dodecahedron. - -This leaves finitely many values of $n$ to check, namely the even numbers between 14 and 222.<|endoftext|> -TITLE: Random products of $SL(2,R)$ matrices and Furstenberg's theorem -QUESTION [5 upvotes]: I was recently learning Furstenberg's theorem on random products of $SL(2,R)$ matrices, and came across with a simple example that confused me: -Considering random products of two matrices $A=\begin{pmatrix} -2 &0\\ -0 &1/2 -\end{pmatrix}$ and $B=\begin{pmatrix} -0 &1\\ --1 &0 -\end{pmatrix}$ with probability $1/2$ and $1/2$, it is mentioned in literature that the Lyapunov exponent for this random product is $0$. See, e.g., page 29 in Damanik's survey article: https://arxiv.org/abs/1410.2445 -I understand that if we focus on a specific matrix element of the product, it has no well-defined limit. However, if we study its norm, say the Frobenius norm $||A||=\sqrt{\sum_{i,j}|a_{ij}|^2}$. I found that by taking average, this norm indeed grows exponentially, and the Lyapunov exponent is positive. -So my question is, why it is said "the lyapunov exponent for this example is $0$" in literature? Thanks! - -REPLY [6 votes]: Because its Lyapunov exponent is zero, but that is not what you are computing. -Instead of looking at random walks on $SL(2, \mathbb{R})$, let me focus on random walks on $\mathbb{R}_+^*$, as there is the same issue. Let $(X_n)$ be i.i.d. in $\mathbb{R}_+^*$, and to make things simple, assume that there are only finitely many values. Let $P_n := X_n \ldots X_1$. -The Lyapunov exponent of this random walk is the real $\Lambda$ such that -$$\lim_{n \to + \infty} \frac{\ln (P_n)}{n} = \Lambda.$$ -By the law of large numbers, $\Lambda = \mathbb{E} (\ln(X_1))$. For instance, if $X_1$ takes values $2$ and $1/2$ each with probability $1/2$, the Lyapunov exponent is $0$: the Markov chain $(P_n)$ will oscillate between very large and very low values. -However, if you compute the expectation of the norm, a short computation gets you $\mathbb{E} (P_n) = (5/4)^n$, which grows exponentially fast. But that does not mean that the Lyapunov exponent is $\ln (5/4)$. The issue is merely that the exponential does not commute with the expectation: -$$1 = e^{\mathbb{E}(\ln(P_n))} \leq \mathbb{E} (e^{\ln(P_n)}) = (5/4)^n.$$ -To get back to a general random walk, and very roughyl, we have $\ln(P_n) \simeq \mathcal{N} (n\mu, n\sigma^2)$. The Lyapunov exponent is the constant $\mu$. However, -$$\mathbb{E} (P_n) \simeq \mathbb{E} (e^{\mathcal{N} (n\mu, n\sigma^2)}) = \mathbb{E} (e^{n(\mu+\frac{\sigma^2}{2})}),$$ -so taking the norm as you get gives you an error coming from the diffusion of the random walk (well, in practice, the exact value of $\sigma^2/2$ for this error is wrong, but I don't think the heuristics is too bad at this level).<|endoftext|> -TITLE: Euler characteristic with compact support of spaces of Euclidean lattices -QUESTION [9 upvotes]: Has the Euler characteristic with compact support of $\mathrm{SL}_n(\mathbb R)/\mathrm{SL}_n(\mathbb Z)$ been computed ? References? Thanks. - -REPLY [2 votes]: I think that the euler characteristic is 0 for the following reasons. -Firstly, the space $SL_N(\mathbb{R})$ is a bundle over the symmetric space $SO(N,\mathbb{R})\backslash SL_N(\mathbb{R}) = SP(n,\mathbb{R})=X$, the space of symmetric positive-definite real matrices of determinant 1. For a discussion of this symmetric space, see e.g. Bridson-Haefliger II.10. Then $SL_N(\mathbb{R})/SL_N(\mathbb{Z})$ is a bundle over $X/SL_N(\mathbb{Z})$ with fiber $SO(N,\mathbb{R})$. Note that this is an orbifold bundle, but that by passing to a torsion-free subgroup, one can assume that it is a manifold (and since you're interested in euler characteristic, this just multiplies by the index). -Now the space $X/SL_N(\mathbb{Z})$ admits a bordification by Borel-Serre. Hence $SL_N(\mathbb{R})/SL_N(\mathbb{Z})$ has a bordification by an $SO(N,\mathbb{R})$-bundle over the Borel-Serre bordification. Hence it is the interior of a manifold with boundary $M$. In this case, $H^*_c(SL_N(\mathbb{R})/SL_N(\mathbb{Z}))\cong H^*(M,\partial M)$. Then by Lefschetz duality, $\chi(H^*_c(M,\partial M))=\chi(M)$. -But since $M$ is a bundle with fiber $SO(N,\mathbb{R})$, and $\chi(SO(N,\mathbb{R}))=0$ (any Lie group has a nowhere vanishing vector field), we have $\chi(M)=\chi(SO(N,\mathbb{R}))\times \chi(X/SL_N(\mathbb{Z})) =0$, since the euler characteristic of bundles is the product of the euler characteristic of the base and the fiber.<|endoftext|> -TITLE: Hodge numbers rule out good reduction -QUESTION [9 upvotes]: A theorem of Fontaine says that if a geometrically connected smooth proper variety $X$ over $\mathbb{Q}$ has good reduction everywhere then $h^{i, j}(X)=0$ for $i\neq j$, $i+j\leq 3$. -This means that the variety's Hodge numbers can "disqualify" it from having good reduction everywhere. For example, there do exist varieties whose non-zero Hodge numbers are $h^{0, 0}=h^{1, 0}=h^{0, 1}=h^{1, 1}=1$ but none of them have good reduction everywhere (in this case it was known before Fontaine but it is just an example). -Can this happen locally? We consider the geometrically connected smooth proper varieties over $\mathbb{Q}_p$ whose Hodge numbers are equal to some fixed values. Can they all have bad reduction (assuming at least one such variety exists)? - -REPLY [8 votes]: Because the inverse Hodge problem is a very difficult question, I think it's unlikely that there will be an answer for this question for any given set of Hodge numbers. For most given Hodge diamonds, we don't even know whether it can be realised by any smooth projective variety (or Kähler manifold); and when it can, we know little about the geometry of these varieties. -However, Kotschick and Schreieder proved that the Hodge ring of varieties in characteristic $0$ is generated by $\mathbf P^1$, $\mathbf P^2$, and any elliptic curve $E$. Since we can choose these with good reduction, this implies that there are no linear relations (like $h^{i,j}(X) = 0$ for a given $(i,j) \in \{0,\ldots,n\}^2$) among the Hodge numbers of smooth projective varieties with good reduction.<|endoftext|> -TITLE: Analyzing the decay rate of Taylor series coefficients when high-order derivatives are intractable -QUESTION [5 upvotes]: This could be a soft question. I am trying to show that the $n$-th Taylor series coefficient of a function is $O(n^{-5/2})$. However, because the function is a function composition of another function with itself, it seems intractable to compute high-order derivatives. I was wondering if there are methods that can bound the asymptotic decay rate of Taylor series coefficients without obtaining the exact coefficients. For example, can complex analysis help here? -Thank you so much! - -The function that I am trying to analyze is $f(x)=g(g(x))$, where $g(x) = \frac{1}{\pi}\left( x\cdot (\pi-\arccos(x)) + \sqrt{1-x^2} \right)$. I conjecture that its $n$-th Taylor coefficient about $x=0$ is $O(n^{-5/2})$. I have shown that the $n$-th Taylor coefficient of $$g(x)= \frac{1}{\pi} + \frac{x}{2} + \sum_{n=1}^\infty \frac{(2n-3)!!}{(2n-1)n!2^n \pi} x^{2n}$$ is $O(n^{-5/2})$. - -REPLY [9 votes]: Complex analysis can help. The rate of Taylor coefficients is determined by: -a) the radius of convergence, which is equal to the radius of the largest disk $|z|1$ and let -\begin{equation*} - D_r:=\{z\in R\colon|z|\pi/4\}. -\end{equation*} -The main difficulty is to show that $g(D_r)\subseteq R$ for some $r>1$. -First here, take any real $t>0$. Then there is some real $u_t>0$ such that for all complex $z$ with $|z|\le1$ and $|\arg z|\ge t$ we have $|\frac1\pi+\frac z2|\le\frac1\pi+\frac12-u_t$, whence -\begin{equation*} - |g(z)|\le\Big|\frac1\pi+\frac z2\Big|+ \sum_{n=1}^\infty \frac{(2n-3)!!}{(2n-1)n!2^n \pi}|z|^{2n} - \le\frac1\pi+\frac12-u_t+ \sum_{n=1}^\infty \frac{(2n-3)!!}{(2n-1)n!2^n \pi} - =g(1)-u_t=1-u_t. -\end{equation*} -Since $g$ is analytic on $R$, it is uniformly continuous on any compact subset of $R$. So, there is some real $r_t>1$ such that $|g(z)|<1$ for all complex $z\ne1$ with $|z|\le r_t$ and $|\arg z|\ge t$. -Also, it follows that $|g(z)|<1$ for all complex $z$ with $|z|<1$. -Thus, to prove that $g(D_r)\subseteq R$ for some $r>1$, it suffices to show that -$\Im g(z)\ne0$ for all $z\ne1$ with $|z|\ge1$ and $|\arg(z-1)|>\pi/4$ that are close enough to $1$. -To see this, note that $h(u)\sim\frac1{\pi\sqrt2\,\sqrt{1-u}}$ as $u\to1$. Then (0) yields -\begin{equation*} -g(z)=z+c_1\cdot(z-1)^{3/2}, \tag{1} -\end{equation*} -where $c_1=c_1(z)$ converges to a nonzero complex number as $z\to1$. So, the conclusion that -$\Im g(z)\ne0$ for all $z\ne1$ with $|z|\ge1$ and $|\arg(z-1)|>\pi/4$ that are close enough to $1$ follows, which does show that $g(D_r)\subseteq R$ for some $r>1$. -So, $f=g\circ g$ is analytic on $D_r$ for such an $r$. -Moreover, (1) implies -\begin{equation*} - f(z)=g(g(z))=g(z+c_1\cdot(z-1)^{3/2})\\ - =z+c_1\cdot(z-1)^{3/2}+\hat c_1\cdot(z-1+c_1\cdot(z-1)^{3/2})^{3/2} \\ - =z+\tilde c_1\cdot(z-1)^{3/2}, \tag{2} -\end{equation*} -where $\hat c_1=\hat c_1(z)\sim c_1(z)$ and $\tilde c_1=\tilde c_1(z)\sim2c_1(z)$ as $z\to1$. -Similarly, since $h(u)\sim\frac1{\pi\sqrt2\,\sqrt{1+u}}$ as $u\to-1$, the Taylor formula for $g(z)$ at $z=-1$ (together with the observation $g(-1)=g'(-1)=0$) yields $g(z)=c_{-1}\cdot(z+1)^{3/2}$, where $c_{-1}=c_{-1}(z)$ converges to a complex number as $z\to-1$. Therefore and because $g$ is analytic at $0$, we have -\begin{equation*} - f(z)=g(g(z))=c_0+\hat c_0\cdot g(z) - =c_0+\tilde c_0\cdot(z+1)^{3/2}, \tag{3} -\end{equation*} -where $c_0:=g(0)$, $\hat c_0=\hat c_0(z)=O(1)$, and $\tilde c_0=\tilde c_0(z)=O(1)$ as $z\to-1$. -Now we are finally in a position to use (with thanks to Alexandre Eremenko) Theorem VI.5 (with $\alpha=-3/2$, $\beta=0$, $\rho=1$, $r=2$, $\zeta_1=1$, $\zeta_2=-1$, $\mathbf D=D_r$, $\sigma_1(z)=z$, $\sigma_2(z)=c_0$), which yields the $n$th coefficient for $f$: -\begin{equation*} - [z^n]f(z)=O(n^{\alpha-1})=O(n^{-5/2}), -\end{equation*} -as you conjectured. - -For an illustration, here is the set $\{g(z)\colon z\in R,|z|<2\}$:<|endoftext|> -TITLE: Continuity of eigenvectors -QUESTION [6 upvotes]: Let $\mathbb{C} \ni z \mapsto M(z)$ be a square matrix depending holomorphically on a parameter $z$ with the property that $\operatorname{dim}\ker(M(z)))=1$ for $z $ away from a discrete set $D \subset \mathbb{C}$ and $\operatorname{dim}\ker(M(z)))\ge 1$ for $z \in D.$ -I ask: Is it always possible to choose a continuous vector $\mathbb{C} \ni z \mapsto v(z)$ such that $M(z)v(z)=0?$ - -REPLY [7 votes]: Yes. Let the size of your matrix be $n$. -Your condition implies that there is an $n-1\times n-1$ submatrix whose determinant is not identically equal to $0$. -Assume without loss of generality that this is the submatrix formed by the first $n-1$ rows and columns. Then we can set $u_n=1$ and find a vector $u(z)$ such that -$M(z)u(z)=0$ by solving the system of $n-1$ linear equations in $n-1$ variables. This requires only arithmetic operations -on the entries of $M$, so $u(z)$ will be meromorphic on $\mathbb{C}$. Let $D$ be the divisor of poles of $u$. According to a theorem of Weierstrass there is an entire function $f$ having zeros at $D$. Then $v(z)=f(z)u(z)$ -is a solution to your problem, which is not only continuous but holomorphic.<|endoftext|> -TITLE: Is there any relationship between Szemerédi's theorem and Sunflower conjecture? -QUESTION [6 upvotes]: I have observed some similar things between a reformulation of the Sunflower conjecture (see also conjecture 1.3 in Improved bounds for the sunflower lemma) and Szemerédi's theorem such that for Szemerédi's theorem we have an often-used equivalent finitary version which states that: - -Theorem: for every positive integer $k$ and real number $\delta\in (0,1]$ there exists a positive integer $N=N(\delta,k)$ such that every subset of $\{1,2,\cdots,N\}$ of size at least $δN$ contains an arithmetic progression of length $k$ say $k\text{-arithmetic progression}$ - -The Sunflower conjecture states that : - -Conjecture: Let $r ≥ 3$. There exists $c = c(r)$ such that any $w$-set system $F$ of size $|F| ≥ c^w$ contains an $r$-sunflower - -We see that both of them investigate about existence of such constant with such size bounds for which we have $k\text{-arithmetic progression}$ or $r$-sunflower. Also, another thing that attracted my attention is that for Erdős conjecture on arithmetic progressions, Erdős and Turán made in 1936 the weaker conjecture that any set of integers with positive natural density contains infinitely many 3-term arithmetic progressions. This was proven by Klaus Roth in 1952, and generalized to arbitrarily long arithmetic progressions by Szemerédi in 1975 in what is now known as Szemerédi's theorem. In the opposite direction, Kostochka proved that any $w$-set system of size -$$ -|F| \geq - cw! \cdot {(\log \log \log w/ \log \log w)}^{w} -$$ -must contain a $3$-sunflower for some absolute constant $c$. Now for quantitative bounds of $r_k(N)$(the size of the largest subset of $\{1, 2, \ldots, N \}$ without an arithmetic progression of length $k$) it is an open problem to determine its exact growth rate. In the same time it is an open problem to determine the exact rate growth or exact bounds for The sunflower (the sunflower lemma via Shannon entropy). My question here is: - -Question: According to similar things which I have cited above between the Sunflower conjecture and Szemerédi's theorem, is there any non-trivial relationship between them? And in which context we can consider Sunflower coincide with arithmetic progression? - -REPLY [7 votes]: I do not know of a direct connection to Roth or Szemerédi over the integers. However, the paper - -N. Alon, A. Shpilka and C. Umans, On Sunflowers and Matrix Multiplication, 2012 IEEE 27th Conference on Computational Complexity, Porto, 2012, pp. 214-223, doi:10.1109/CCC.2012.26 (author pdf) - -shows that a proof of the Sunflower Conjecture would imply a proof of the Erdos-Szemeredi sunflower theorem, which also follows from a bound of $(3-\delta)^n$ for the capset problem, a strong form of Roth over $\mathbb{F}_3^n$ (which is already known due to Croot-Lev-Pach). See this 2016 blog post by Gil Kalai for more discussions along this line. -It is also noteworthy that the Erdős-Szemerédi sunflower conjecture (which has been proved and follows from the capset problem) also implies that if $|S|=C\log(n)$ is a subset of $[n]$ for a large constant $C$, then there are three disjoints $X, Y, Z$ whose subset sums are identical, and thus the sums of the subsets $X, X \cup Y, X \cup Y \cup Z$ are in arithmetic progression; see - -P. Erdős, A. Sárközy, Arithmetic progressions in subset sums, Discrete Mathematics 102 Issue 3 (1992) pp 249–264, doi:10.1016/0012-365X(92)90119-Z -(Core pdf).<|endoftext|> -TITLE: Simple examples of equivariant cobordism -QUESTION [10 upvotes]: Let $Y$ be an oriented 3-manifold with a free action by a finite group $G$. If I understand correctly, there exists a multiple of $NY$ of $Y$ and an oriented manifold $X$ such that $\partial X = NY$ and $G$ extends to a free action on $X$. (That is, the equivariant oriented cobordism group is finite. Here, I believe $NY$ should be interpreted as $N$ disjoint copies of $Y$ - note that $N$ is nonzero.) I am trying to understand some very simple examples of this. For instance, if $Y = S^3$ and $G$ is a cyclic group (so that the quotient is a lens space), what is the manifold $X$? -EDIT: For a concrete mention of this claim, see the bottom of the first page of https://www.maths.ed.ac.uk/~v1ranick/papers/aps002.pdf -I realize that the claim is from equivariant bordism theory (as mentioned in one of the comments) but I am not very familiar with this, so I just gave the place where I first saw it. - -REPLY [2 votes]: For the concrete case of the cyclic group $C_p$ acting linearly on $S^3$, there's a very explicit construction. Call $\omega = e^{2\pi i/p}$. Fix an integer $q$ coprime with $p$, and let us look at the action $\lambda_q$ on $\mathbb{C}^2$ generated by the diagonal matrix with entries $\omega, \omega^q$. $\lambda_q$ restricts to an action (which I will still call $\lambda_q$) on $S^3$ (as the unit sphere in $\mathbb C^2$ whose quotient is $L(p,q)$. -Take $M = V(x_0^p + \dots + x_3^p)$, the Fermat hypersurface of degree $d$ in $\mathbb{CP}^3$. On $\mathbb{CP}^3$ we define two actions $\psi$ and $\phi$ of $C_p$. Calling $g$ the generator of $C_p$, the two actions are given by: -$\phi_g(x_0:x_1:x_2:x_3) = (\omega x_0 : \omega^{-q}x_1 : x_2 : x_3)$, and $\psi_g(x_0:x_1:x_2:x_3) = (x_0 : x_1 : \omega x_2 : x_3)$. -I claim that: - -Both actions preserve $M$. -The fixed point set of the action $\phi$ on $M$ is the set of points $(0:0:1:\omega^a)$ as $a$ varies, and the action is semi-free (i.e. there are no new fixed points appearing when you take powers of $g$). -$\psi_g$ cyclically permutes the fixed points of $\phi_g$ on $M$. -$\phi_g$ and $\psi_g$ commute. -The linearised action of $\phi_g$ on $M$ at $(0:0:1:1)$ is $\lambda_{-q}$. - -I will not justify these points (the only one that maybe requires some care is the last one), and I will just take them for granted. Now take a small $\phi_g$-invariant ball $B$ in $M$ centred at $(0:0:1:1)$ and remove all its $\psi_g$-orbit in $M$, to get $M_0$. The action of $\phi$ on $M_0$ is free (because $\phi_{g^k}$ has the same fixed points as $\phi_g$ for each $0 < k< p$) and it extends the linear action $\lambda_q$ on $pS^3 = \sqcup_k \psi_{g^k} \partial B$. (Note that there is a $-q$ in an exponent of $\omega$ when defining $\phi$: this is because the boundary of $M_0$ is the boundary of $B$ with its orientation reversed.)<|endoftext|> -TITLE: Slodowy slice intersecting a given orbit "minimally"? -QUESTION [8 upvotes]: Let $\mathfrak{g}$ be a complex semisimple Lie algebra. Is it true that for any $X\in\mathfrak{g}$, there exists an $\mathfrak{sl}_2$-triple $(e,h,f)$ in $\mathfrak{g}$ such that - -We have $X\in e+Z_{\mathfrak{g}}(f)$, and -$\dim Z_{\mathfrak{g}}(X)=\dim Z_{\mathfrak{g}}(e)$? - -($Z_{\mathfrak{g}}(-)$ always the centralizer in $\mathfrak{g}$. One can deduce from the above conditions that the conjugacy orbit of $X$ must then meet the slice transversally, like in classical Kostant section situation.) -When $X$ is regular this is the well-known result about Kostant section. When $\mathfrak{g}=\mathfrak{gl}_n$ this is also true and can be deduced from the rational form of $X$. It might be that this question can be answered by a simple reference, but thanks a lot in advance in all cases! - -REPLY [3 votes]: Woooo, I think I finally find a proof!$ -\newcommand{\Lie}{\operatorname{Lie}} -\newcommand{\Lg}{\mathfrak{g}} -\newcommand{\Ll}{\mathfrak{l}} -\newcommand{\Ad}{\operatorname{Ad}}$ -Let $G=\operatorname{Aut}(\Lg)^\circ$ be a corresponding complex Lie group. Write $X=X_s+X_n$ the Jordan decomposition into semisimple and nilpotent part. The centralizer $L:=Z_G(X_s)$ is a Levi subgroup. Denote by $\Ll=\Lie L$ and $Z(\Ll)=\Lie Z(L)$ the center of $\Ll$. Write $\mathcal{O}:=Z(\Ll)\cdot\Ad(L)X_n$, a locally closed subvariety in $\Ll$. Note that $\mathcal{O}\cong Z(\Ll)\times\Ad(L)X_n$ has an obvious map to $Z(\Ll)$. -Pick a parabolic subgroup $P$ containing $L$ as its Levi subgroup. Let $U$ be the unipotent radical of $P$ so that $P=LU$. There exists a nilpotent $G$-orbit in $\Lg$ that intersects $X_n+\Lie U$ at a dense open subset of $X_n+\Lie U$. Let $e$ be any element in the intersection. We have to prove the two asserted property in the question, i.e. - -$\Ad(G)X$ meets the Slodowy slice $e+Z_{\Lg}(f)$ and -$\dim Z_G(X)=\dim Z_G(e)$. - -We have $Z_G(X)=Z_L(X_n)$ and also $\dim Z_G(e)=\dim Z_L(X_n)$ [LS79, Theorem 1.3]. This proves (2). -Consider the “generalized Grothendieck-Springer alternation” given by -$$ -\tilde{\mathfrak{g}}_X:=\{(g,\gamma)\in (P\backslash G)\times\mathfrak{g}\;|\;\Ad(g)\gamma\in\Lie P\text{ is such that its image under}\Lie P\twoheadrightarrow\Lie L\text{ lies in }\mathcal{O}\}. -$$ -It is the usual Grothendieck–Springer alternation if $X$ is regular semisimple, hence the name. There is a natural smooth map $\mu:\tilde{\mathfrak{g}}_X\rightarrow Z(\Ll)$ via $\mathcal{O}\twoheadrightarrow Z(\Ll)$ and another natural $G$-equivariant map $\pi:\tilde{\Lg}_X\rightarrow\Lg$ sending $(g,\gamma)\mapsto\gamma$. For $\zeta\in Z(\Ll)$ that are $(G/L)$-regular (this means $Z_G(\zeta)=L$), we have $\pi(\mu^{-1}(\zeta))=\Ad(G)(\zeta+X_n)$. For the central fiber we have instead $\pi(\mu^{-1}(0))\supset\Ad(G)e$. -Since $e+Z_{\Lg}(f)$ meets $\Ad(G)e$ transversally, there exists a neighborhood (analytically or algebraically, whichever) $U$ of $e$ in $\Lg$ such that for any orbit $\Ad(G)Y$ we have $\Ad(G)Y\cap U\not=\emptyset\implies\Ad(G)Y\cap (e+Z_{\Lg}(f))\ne\emptyset$. The preimage $\pi^{-1}(U)$ intersects $\mu^{-1}(0)$. Thanks to smoothness of $\mu$, there exists a neighborhood $U_{Z(\Ll)}$ of $0\in Z(\Ll)$ such that $\pi^{-1}(U)$ intersects $\mu^{-1}(\zeta)$ for any $\zeta\in U_{Z(\Ll)}$. For $\zeta\in U_{Z(\Ll)}\cap Z(\Ll)^{\text{$(G/L)-reg$}}$ this means $U$ intersects $\Ad(G)(\zeta+X_n)$, and thus the asserted property (1) is true for ($X$ replaced by) $\zeta+X_n$. -Since nilpotent orbits are stable under scalar scaling, that (1) is true for $\zeta+X_n$ implies that it is true for $c(\zeta+X_n)$ for all $c\in\mathbb{C}^{\times}$ and $\zeta\in U_{Z(\Ll)}\cap Z(\Ll)^{\text{$(G/L)-reg$}}$, and thus also for $c\zeta+X_n$ since $c(\zeta+X_n)$ and $c\zeta+X_n$ are also conjugate. Since such $c\zeta$ covers all of $Z(\Ll)^{\text{$(G/L)-reg$}}$, the assertion for $X=X_s+X_n$ is proved. -[LS79] Lusztig, G.; Spaltenstein, N., Induced unipotent classes, J. Lond. Math. Soc., II. Ser. 19, 41-52 (1979). ZBL0407.20035. -Sadly, I can't recall why I needed this cute property ….<|endoftext|> -TITLE: Reference for graduate-level text or monograph with focus on "the continuum" -QUESTION [7 upvotes]: I always had the dream to design a course for my graduate students like "mathematical models of the continuum". This course should cover history of real numbers, the Measure Problem, the Banach-Tarski Paradox, but also Baire/Cantor/Polish spaces, and it should look at infinite games and determinacy as well. -Obviously I am conscious about the text books and articles covering all these concepts. However, I seem to be unable to find a rigorours, graduate-level monograph that captures all of this in one, which I could potentially use as a text. -Would someone be able to point me to a good reference for that? - -REPLY [6 votes]: A great textbook for your course would be "The Structure of the Real Line" by Lev Bukovský. It covers all of the topics you mentioned, except for the Banach-Tarski Paradox, and provides all necessary topological and measure-theoretic background.<|endoftext|> -TITLE: What is known about the "unitary group" of a rigged Hilbert space? -QUESTION [18 upvotes]: Suppose that $(E,H)$ is a rigged (infinite dimensional, separable) Hilbert space, i.e. $H$ is a Hilbert space, and $E$ is a Fréchet space, equipped with a continuous linear injection $E \rightarrow H$ with dense image. -Define now the group $U(E,H)$ to consist of those invertible, continuous, linear transformations on $E$, which extend to unitary transformations of $H$. - -What is known about the group $U(E,H)$? - -For example: - -Does the map $U(E,H) \rightarrow U(H)$ have dense image? -Is it a Fréchet Lie group? If so: - -What is its Lie algebra? Is it just the continuous linear skew-adjoint operators on $E$? -Is the map $U(E,H) \times E \rightarrow E$ smooth? -Do the smooth vectors of the representation $U(E,H) \times H \rightarrow H$ consist exactly of $E$? - - -Is $U(E,H)$ contractible? - -I am most interested in the case that $E$ is a nuclear Fréchet space. -Perhaps a good first step towards a Fréchet Lie group structure on $U(E,H)$ would be to equip the space of skew operators on $E$ with the structure of Fréchet (Lie) algebra. -It's not so obvious to me how this should go. Is there anything known in this direction? - -REPLY [4 votes]: These are just some thoughts on you question, so not an answer but too long for a comment. I waited a few days before posting this since, rather than addressing your post directly, I am going to look at a variant. This is usually not the done thing but my excuse is that I hope that it will contain material which could be of interest to you. -First of all, as has already been pointed out, your questions only make sense if you specify the relevant topologies. Experience suggests that the norm is not suitable for such questions. However, there are available a range of well-behaved and well-studied weaker ones which might do very well. -My main thesis is that most of the rigged Hilbert spaces which arise in mathematical physics have a special form and it is worth while starting there. The setting is an unbounded self adjoint operator $T$ on a Hilbert space. $E$ is then the intersection of the domains of definition of its powers. This is a Fréchet space with its natural structure, even Fréchet nuclear (Pietsch) under conditions on the spectral behaviour of $T$ which are satisfied by many of the classical operators (Laplace, Laplace-Beltrami, Schrödinger). -My main suggestion is that one study the space of unitary operators on the Hilbert space which commute with $T$. Such operators then map $E$ continuously into itself. -This space can be explicitly calculated for most of the situations mentioned above and this may provide some insight into your original question. -As an example, for the case of the standard one-dimensional Schrödinger operator the Fréchet space is that of the rapidly decreasing smooth functions and the corresponding operator space is the countable product of circle groups, regarded as multipliers on the coefficients of the Hermite expansion. As a compact group it won’t be dense anywhere in a non-trivial way but it will, presumably, be an infinite dimensional Lie group. -Similar remarks apply to the other classical cases where the spectral behaviour of the operator is known.<|endoftext|> -TITLE: Values of a pair of determinants -QUESTION [6 upvotes]: Let $\mathbf{x} = (x_0, x_1, x_2), \mathbf{y} = (y_0, y_1, y_2)$ be vectors over a field $\mathbb{F}$ of characteristic zero. Define the function -$\displaystyle S(\mathbf{x}, \mathbf{y}) = x_2 (y_0^2 - 2 y_1 y_2) + x_1 (2 y_2^2 - y_0 y_1) + x_0 (y_1^2 - y_0 y_2) = \begin{vmatrix} x_2 & x_1 & x_0 \\ y_2 & y_1 & y_0 \\ y_1 & y_0 & 2 y_2 \end{vmatrix}$ -and $T(\mathbf{x}, \mathbf{y}) = S(\mathbf{y}, \mathbf{x})$. -Curiously, I found that for fixed $(s,t) \in \mathbb{F}^2$ the set of solutions to $s = S(\mathbf{x}, \mathbf{y}), t = T(\mathbf{x}, \mathbf{y})$ is stable under the map -$\mathbf{x} \mapsto \begin{bmatrix} x_0 + 2 x_1 + 2 x_2 \\ x_0 + x_1 + 2 x_2 \\ x_0 + x_1 + x_2 \end{bmatrix}, \mathbf{y} \mapsto \begin{bmatrix} y_0 + 2 y_1 + 2 y_2 \\ y_0 + y_1 + 2 y_2 \\ y_0 + y_1 + y_2 \end{bmatrix}.$ -Moreover, the matrix defining this linear map which is -$M = \begin{bmatrix} 1 & 2 & 2 \\ 1 & 1 & 2 \\ 1 & 1 & 1 \end{bmatrix}$ -has determinant one. -Thus, if we define the group $\mathcal{G} \subset \text{GL}_3(\mathbb{F})$ to be the set of $3 \times 3$ matrices $A$ over $\mathbb{F}$ such that $S(\mathbf{x}, \mathbf{y}) = S(A \mathbf{x}, A \mathbf{y})$ for all $\mathbf{x}, \mathbf{y} \in \mathbb{F}^3$, then we have shown that $M \in \mathcal{G}$. Further, $M$ has infinite order so $\mathcal{G}$ contains infinitely many elements. -Is it possible to determine $\mathcal{G}$ in a reasonable way? - -REPLY [3 votes]: First, by transposing and interchanging two rows $S({\bf x},{\bf y}) = -\left|\begin{matrix} x_0 & y_0 & 2y_2 \\ x_1 & y_1 & y_0 \\ x_2 & y_2 & y_1 \end{matrix} \right|$. Notice also that if $T = \left[\begin{matrix}0&0&2 \\ 1&0 &0 \\ 0& 1& 0\end{matrix} \right]$ then $T{\bf y} = \left[\begin{matrix} 2y_2 \\ y_0 \\ y_1\end{matrix}\right]$. Hence, $S({\bf x},{\bf y}) = -\left| {\bf x}\ \ {\bf y}\ \ T{\bf y} \right|$ -Suppose $A\in \mathcal G$ then for all ${\bf x},{\bf y} \in \mathbb F^3$ -$$ -S({\bf x},{\bf y}) = S(A{\bf x}, A{\bf y}) = -\left|{\bf x}\ \ {\bf y} \ \ |A|A^{-1}TA{\bf y}\right|. -$$ -Now when ${\bf x} = e_3$ then this gives -$$ -y_0^2 - 2y_1y_2 = y_0\langle|A|A^{-1}TA{\bf y}, e_2\rangle - y_1\langle|A|A^{-1}TA{\bf y}, e_1\rangle. -$$ -Subsequently, when ${\bf y} = e_1$ this gives $1 = \langle|A|A^{-1}TAe_1, e_2\rangle$, and when ${\bf y} = e_2$ this gives $0 = \langle|A|A^{-1}Tae_2, e_1\rangle$. Moreover, when ${\bf y} = e_1+e_3$ then after some computation $$\langle|A|A^{-1}TAe_1, e_1\rangle = \langle|A|A^{-1}TAe_3, e_3\rangle$$ -When ${\bf x} = e_2$ this gives -$$ -2y_2^2 - y_0y_1 = y_2\langle|A|A^{-1}TA{\bf y}, e_1\rangle - y_0\langle|A|A^{-1}TA{\bf y}, e_3\rangle, -$$ -${\bf y} = e_1$ gives $0 = \langle|A|A^{-1}TAe_1, e_3\rangle$, and ${\bf y} = e_3$ gives $2 = \langle|A|A^{-1}Tae_3, e_1\rangle$. -Moreover, when ${\bf y} = e_1+e_2$ then $$\langle|A|A^{-1}TAe_1, e_1\rangle = \langle|A|A^{-1}TAe_2, e_2\rangle$$ -When ${\bf x} = e_1$ this gives -$$ -y_1^2 - y_0y_2 = y_1\langle|A|A^{-1}TA{\bf y}, e_3\rangle - y_2\langle|A|A^{-1}TA{\bf y}, e_2\rangle -$$ -${\bf y}=e_2$ gives $1 = \langle|A|A^{-1}TAe_2, e_3\rangle$ and ${\bf y} = e_3$ gives $0 = \langle|A|A^{-1}TAe_3, e_2\rangle$ -If we define $\lambda := \langle|A|A^{-1}TAe_1, e_1\rangle$ then the above argument concludes -$$ -|A|A^{-1}TA = \left[\begin{matrix}\lambda&0&2 \\ 1&\lambda &0 \\ 0& 1& \lambda\end{matrix} \right] = \lambda I + T -$$ -and so -$$ -TA = \frac{1}{|A|}\lambda A + \frac{1}{|A|}AT. -$$ -One won't get anymore out of these equations. Furthermore, any $A$ satisfying this equation gives -$$ -S(A{\bf x}, A{\bf y}) = -|{\bf x} \ \ {\bf y} \ \ \lambda{\bf y} + T{\bf y}| = S({\bf x}, {\bf y}) -$$ -by column replacement. Your $M$ corresponds to the case where $\lambda = 0$, that is commutes with $T$, and $|A| = 1$. However, there may be more matrices in $\mathcal G$. -Therefore, -$$ -\mathcal G = \left\{ A\in {\rm GL}_3(\mathbb F) : TA = \frac{1}{|A|}\lambda A + \frac{1}{|A|}AT, \lambda\in \mathbb F \right\} -$$<|endoftext|> -TITLE: What subsystem of third order arithmetic proves the real numbers are Dedekind complete? -QUESTION [7 upvotes]: Reverse mathematics is mainly about subsystems of second-order arithmetic, but in recent years it’s expanded to cover subsystems of third-order arithmetic as well. Now the fact that the real numbers are Dedekind complete is (encodable as) a statement in the language of third order arithmetic. And I think it’s probably provable using full third order arithmetic. -But my question is, what is the weakest subsystem of third-order arithmetic capable of proving that the real numbers are Dedekind complete? -By the way, the fact that the real numbers form a real closed field is provable even in $RCA_0$, so my question is really about the interpretability of the second-order theory of real numbers. - -REPLY [2 votes]: The answer to your question (unsurprisingly) depends on the formalisation of "being a subset of $\mathbb{R}$". Alex Kreuzer [1] has used characteristic functions to represent subsets of Cantor space $2^\mathbb{N}$. Dag Normann and I have adopted this formalism in e.g. [2, 3] for subsets of $\mathbb{R}$, as it yields nice results that generalise the notion of open/closed set from Reverse Mathematics. -Using the "sets as characteristic functions" formalism, Kohlenbach'ssystem RCA$_0^\omega$ from [0] plus Every bounded subset of $\mathbb{R}$ has a surpremum -is a conservative extension of WKL$_0$. One uses the intuitionistic fan functional from [0] to establish this. -References -[0] Kohlenbach, U., Higher order reverse mathematics, Reverse mathematics 2001, Lect. Notes Log., vol. 21, ASL, 2005, pp. 281–295. -[1] Kreuzer, A., Measure theory and higher order arithmetic. Proc. Amer. Math. Soc. 143 (2015), no. 12, 5411–5425. -[2] Normann D. and Sanders S., Open sets in Reverse Mathematics and Computability Theory, Journal of Logic and Computability 30 (2020), no. 8, pp. 40. -[3]____, On the uncountability of R, Submitted, arxiv: https://arxiv.org/abs/2007.07560 (2020), pp. 29.<|endoftext|> -TITLE: Why did Robertson and Seymour call their breakthrough result a "red herring"? -QUESTION [26 upvotes]: One of the major results in graph theory is the graph structure theorem from Robertson and Seymour -https://en.wikipedia.org/wiki/Graph_structure_theorem. It gives a deep and fundamental connection between the theory of graph minors and topological embeddings, and is frequently applied for algorithms. -I was working with this results for years, and now heard someone saying that Robertson and Seymour called this result a "red herring". Is this true, and why would they possibly call such a breakthrough result a red herring? -(Edit: my question refers to the 2003 graph structure theorem, not to the graph minor theorem which they established later) - -REPLY [5 votes]: If you are interested in tangles in the sense of Robertson and Seymour, this is just to provide some perspective on it. I am working on this for my Ph.D. project and I thought maybe it is considered helpful if I share this high-level, intuitive perspective here (it is not a detailed definition): -The best and shortest description, I think, is given in the following quote. It mentions the proof of the graph minor theorem, but in fact it is the same concept of tangles as in the proof of the graph structure theorem: -Quote: „Originally devised by Robertson and Seymour as a technical device for their proof of the graph minor theorem, tangles have turned out to be much more fundamental than this: they define a new paradigm for identifying highly connected parts in a graph. -Unlike earlier attempts at defining such substructures—in terms of, say, highly connected subgraphs, minors, or topological minors—tangles do not attempt to pin down this substructure in terms of vertices, edges, or connecting paths, but seek to capture it indirectly by orienting all the low-order separations of the graph towards it. -In short, we no longer ask what exactly the highly connected region is, but only where it is. For many applications, this is exactly what matters. -Moreover, this more abstract notion of high local connectivity can easily be transported to contexts outside graph theory. This, in turn, makes graph minor theory applicable beyond graph theory itself in a new way, via tangles.“ End of Quote. -This quote is from Reinhard Diestel, in the preface to the 5th edition of his Graph Theory book.<|endoftext|> -TITLE: Is $\arcsin(1/4) / \pi$ irrational? -QUESTION [6 upvotes]: Is $\arcsin(1/4) / \pi$ rational? An approximation given by a calculator seem to suggest that it isn't, but I found no proof. Thanks in advance! - -REPLY [14 votes]: This is a partial case of the classical result. -https://en.wikipedia.org/wiki/Niven%27s_theorem<|endoftext|> -TITLE: Tensor product of unit and co-unit in a closed compact category -QUESTION [6 upvotes]: Consider a compact closed category, i.e., a symmetric monoidal category with a unit $\eta$ and co-unit $\epsilon$. It seems natural to demand that the tensor product of two units (for different objects) is a unit again (for the tensor product of the objects). That is, we have an equality between the morphisms -$$ -1\xrightarrow{\simeq} -1\otimes 1\xrightarrow{\eta_A\otimes \eta_B} -(A^*\otimes A)\otimes (B^*\otimes B) \xrightarrow{\simeq} -(A^*\otimes B^*)\otimes (A\otimes B) -$$ -and -$$ -1\xrightarrow{\eta_{A\otimes B}} (A^*\otimes B^*)\otimes (A\otimes B) -$$ -$\simeq$ stands for some combination of unitors, associators and braidings, I can spell it out if you insist. The analogous equation should hold for the co-unit. -Can this equation be derived from the axioms for compact closed categories? Or is it somehow trivially implied from the way closed compact categories are defined? If not, is there a name for this property, or is it equivalent to some other known property? -Are there closed compact categories for which this equation doesn't hold? I know that it holds for the compact closed categories of finite vectorspaces, and sets and relations. If I'm trying to define a unit and co-unit for the symmetric monoidal category of super vector spaces, imposing the above equation seems to prevent me from doing that (however, I'm also not sure if super vector spaces can be extended to a compact closed category). - -REPLY [5 votes]: The definition of compact closed category merely says that for every object $A$ there exists an object $A^\star$ and a unit $\eta_A$ and counit $\varepsilon_A$ satisfying the snake equations. That is, it is a property, not structure. Since this data is unique up to isomorphism, you can choose different units and counits for different objects as you please, as long as they satisfy the snake equation. If you choose $\eta_A$ for $A$ and $\eta_B$ for $B$, then their combination $1 \to (A^* \otimes B^*) \otimes (A \otimes B)$ will satisfy the snake equation for $A \otimes B$, and hence you could use it for $\eta_{A \otimes B}$. See chapter 3 of Heunen&Vicary.<|endoftext|> -TITLE: Potential p-norm on tuples of operators -QUESTION [6 upvotes]: Consider $\left[\begin{matrix}A \\ B\end{matrix}\right] \in B(H)^2$. One can define -$$ -\left\|\left[\begin{matrix}A \\ B\end{matrix}\right]\right\|_p = \| |A|^p + |B|^p\|^{1/p}. -$$ -Q: Is this a norm? -Consider the matrices $C = \left[\begin{matrix}1 & 0 \\ 0 & 0\end{matrix}\right]$ and $D = \left[\begin{matrix}0 & 1 \\ 0 & 0\end{matrix}\right]$. Then -$$ -\left\|\left[\begin{matrix}C+D \\ D+C\end{matrix}\right]\right\|_p -= 2^{1/p}||C+D\| = 2^{1/p+1/2} -$$ -while -$$ -\left\|\left[\begin{matrix}C \\ D\end{matrix}\right]\right\|_p + \left\|\left[\begin{matrix}D \\ C\end{matrix}\right]\right\|_p -= 2\|C|^p + |D|^p\|^{1/p} = 2\|I\|^{1/p} = 2. -$$ -Therefore, when $1\leq p< 2$ then -$$ -\left\|\left[\begin{matrix}C+D \\ D+C\end{matrix}\right]\right\|_p > \left\|\left[\begin{matrix}C \\ D\end{matrix}\right]\right\|_p + \left\|\left[\begin{matrix}D \\ C\end{matrix}\right]\right\|_p -$$ -and so $\|\cdot\|_p$ is not a norm. -The $p=2$ case does give a norm as $$ -\| |A|^2 + |B|^2\|^{1/2} = \left\| [A^* B^*]\left[\begin{matrix}A \\ B\end{matrix}\right] \right\|^{1/2} = \left\|\left[\begin{matrix}A & 0 \\ B & 0\end{matrix}\right] \right\| -$$ -which allows one to use the triangle inequality for $M_2(B(H))$. -My question is what happens for all of the other cases, $p>2$? - -REPLY [7 votes]: No, the expression -$$ -\left\|\left[\begin{matrix}A \\ B\end{matrix}\right]\right\|_p = \| |A|^p + |B|^p\|^{1/p}. -$$ -is not a norm for any $20$ (large). Then a small computation gives that $\|A_1+B\| = s + \langle A_1 u,u\rangle + O(1/s)$, so making $s \to +\infty$, we obtain that 2. implies that $\langle A_1 u,u\rangle \leq \langle A_2 u,u\rangle$. The lemma is proven. -Assume for a contradiction that your expression was a norm. Then for any $C$ of norm $<1$, we can write $C$ as the average of two unitaries (see here), and therefore (for arbitrary $A,B$) we can write $\left[\begin{matrix} CA \\ B\end{matrix}\right]$ as a convex combination of elements of the form $\left[\begin{matrix}UA \\ B\end{matrix}\right]$ for unitares $U$, which all have the same "norm" $\| |A|^p + |B|^p\|^{1/p}$, and therefore we would have -$$\| |CA|^p+|B|^p\| ^{1/p}\leq \| |A|^p + |B|^p\|^{1/p}.$$ -So by the Lemma we would have -$$ |CA|^p \leq |A|^p$$ -for any $A$ and any $C$ of norm $\leq 1$. Note that every operator $\leq |A|^2=A^* A$ can be written as $|CA|^2$ for some $C$ of norm $\leq 1$. So we have reached the desired conclusion that the map $t\mapsto t^{p/2}$ is operator monotone.<|endoftext|> -TITLE: In infinite dimensions, is it possible that convergence of distances to a sequence always implies convergence of that sequence? -QUESTION [15 upvotes]: This is a cross-posted on MSE here. -Let $(X,d)$ be a metric space. Say that $x_n\in X$ is a P-sequence if $\lim_{n\rightarrow\infty}d(x_n,y)$ converges for every $y\in X.$ Say that $(X,d)$ is P-complete if every P-sequence converges. Problem 1133 of the College Mathematics Journal (proposed by Kirk Madsen, solved by Eugene Herman) asks you to prove that $$\text{compact}\Longrightarrow\text{P-complete}\Longrightarrow\text{complete}$$ and that none of these implications go both ways. The implications follow by showing that $$\text{sequence}\Longleftarrow\text{P-sequence}\Longleftarrow\text{Cauchy sequence},$$ since a P-sequence (and thus a Cauchy sequence) converges iff it has a convergent subsequence. To give counterexamples to the converses, there are several possible directions. My question specifically involves normed vector spaces (although it is overkill for the original problem). -For any $n\geq 0$, any norm on $\mathbb R^n$ induces a P-complete metric. This distinguishes compactness and P-completeness, since $\mathbb R^n$ obviously isn't compact when $n>0$. To differentiate P-completeness and completeness, we can note that a Hilbert space is P-complete iff it is finite-dimensional (otherwise, we take a non-repeating sequence of vectors from an orthonormal basis and get a P-sequence that doesn't converge). I wonder if other infinite-dimensional normed spaces (necessarily Banach) might be P-complete. But my knowledge of Banach spaces is very limited, so I don't have much intuition about what examples to try. Also, the property of P-completeness (unlike compactness and completeness) is not closed-hereditary, so we can't just try an something by embedding it in a larger example. -Question: What is an example of an infinite dimensional, P-complete Banach space? -Examples I've tried: - -$\ell^p$ spaces for all $1\leq p< \infty$. They are not P-complete, since the sequence $e_n=(0,\dotsc,0,1,0,\dotsc)$ is a P-sequence but not Cauchy. As was pointed out to me in the comments, $\ell^\infty$ is not P-complete, but you need a different sequence as a counterexample. -$C(X)$ for $X$ compact Hausdorff, first-countable and infinite. There must be an accumulation point $p\in X$. We can take a sequence of bump functions $f_k$ converging (pointwise) to the characteristic function $\chi_p$. For any $g\in C(X)$, we have $\lim d(g,f_k)=\lVert g-\chi_p\rVert_\infty$. Thus $(f_k)$ is a P-sequence that does not converge (uniformly), because the pointwise limit is discontinuous. - -REPLY [10 votes]: That every Banach space is contained in a $P$-complete Banach space follows immediately from the following -Theorem. -Let $X$ be a Banach space. Then there exists a Banach space $Y$ containing $X$ in which no separated sequence is a $P$-sequence. -Modulo "abstract nonsense", which I will explain later, the theorem follows from the following proposition, which comes from Christian Remling's remark that the unit vector basis $(e_n)$ of $c_0$ is not a $P$-sequence in $\ell_\infty$. -Proposition. Suppose that $(x_n)$ is a normalized basic sequence in a Banach space $X$. Then there is an isometric embedding $S$ from $X$ into $X \oplus_\infty \ell_\infty$ such that no subsequence of $(Sx_n)$ is a $P$-sequence. -Proof: Since $(x_n)$ is normalized and basic and $\ell_\infty$ is $1$-injective, there is $\alpha >0$ and a contraction $T: X \to \ell_\infty$ such that for all $n$, $Tx_n = \alpha e_n$. Define $S$ from $X$ into $X \oplus_\infty \ell_\infty$ by -$Sx := (x,Tx)$. Since $T$ is a contraction, $S$ is an isometric embedding. We show that $(Sx_n)$ does not contain a $P$-convergent subsequence; this is basically Christian's comment. Let $A$ be any infinite set of natural numbers and take an infinite subset $B$ of $A$ so that $A\setminus B$ is also infinite. Then the distance from $Sx_n$ to $-1_B$ is $1+\alpha$ if $n$ is in $B$ and one otherwise, so $(x_n)_{n\in A}$ is not a $P$-sequence. -Now comes the soft souping up. By iterating the Proposition transfinitely, we get for any Banach space $X$ a superspace $Z$ such that no normalized basic sequence in $X$ is a $P$-sequence in $Z$. Iterate this $\omega_1$ times to get an increasing transfinite sequence $X_\lambda$, $\lambda < \omega_1$, of Banach spaces with $X_1 = X$ so that no normalized basic sequence in $X_\lambda$ is a $P$-sequence in $X_{\lambda+1}$. Let $Y$ be the union of $X_\lambda$ over $\lambda < \omega_1$. Every sequence in $Y$ is in some $X_\lambda$, hence no normalized basic sequence in $Y$ is a $P$-sequence. This property carries over to the completion of $Y$ by the principle of small perturbations. -Now suppose that $Y$ is a Banach space in which no normalized basic sequence is a $P$-sequence. We claim that also no separated sequence in $Y$ is a $P$-sequence. Certainly no non norm null basic sequence in $Y$ is a $P$-sequence, and $P$-sequences are bounded, so it is enough to consider a general separated sequence $(x_n)$ that is bounded and bounded away from zero. If the sequence has a basic subsequence, we are done. But it is known (and contained, for example, in the book of Albiac and Kalton), that if such an $(x_n)$ has no basic subsequence then it has a subsequence that converges weakly, so without loss of generality we can assume that $x_n - x$ converges weakly to zero but is bounded and bounded away from zero. But then $x_n - x$ has a basic subsequence, hence $x_n - x$ cannot have a $P$-subsequence, whence neither can $x_n$. -EDIT 7/27/20: -The reduction of the problem to the theorem above is a consequence of things proved, but perhaps not always explicitly stated, in any course that contains an introduction to metric spaces: -Theorem. Let $M$ be a metric spaces. Then one and only one of the following is true. -A. $M$ is totally bounded. -B. $M$ contains a separated sequence. -A corollary is that every sequence in a metric space either contains a Cauchy subsequence or a separated subsequence.<|endoftext|> -TITLE: Calculating $n$-dimensional hypervolumes ($n \sim 50$), for example -QUESTION [5 upvotes]: I have a question regarding efficient and possibly simple algorithms for computing volumes of $n$-dimensional polytopes. -The polytope of concern isn't arbitrary: it is obtained by applying a linear transformation to a unit hypercube, whose associated matrix has the form $I - L$, where $I$ is identity matrix and $L$ is an arbitrary (in reality sparse) matrix with zeros on its diagonal. -The main difficulty of the problem is that I do not need to calculate the whole volume of the polytope, but only the volume of the part lying in the first hyperoctant (i.e. whose point have all the coordinates positive). So what I really need is an algorithm to compute quickly and efficiently the volume of the intersection of a polytope and the first hyperoctant. -If it helps to know the context of the question, I am doing research in the field of complex networks, where $L$ is the adjacency matrix of a $E-R$ network. The hypervolume which I am concerned with would be a sort of a measure of resiliency of the network. - -REPLY [4 votes]: I'm not an expert but I think I can at least point you at the right direction. -Exact computation of volume is #P-Hard when given only vertices or only halfspaces. You can find a relevant discussion in MO979. Some software for exact computation: cdd, lrs, vinci and Qhull. -The dimension of your polytope is not very large and its H-representation (representing polytope by bounding halfspaces) requires $2n$ halfspaces for the cube after a linear transformation and another $n$ for intersecting the first hyperoctant/nonnegative orthant. The low number of halfspaces is a positive as it is much easier to compute volume when there are few halfspaces. -In the last 30 years, there has been progress in algorithms for efficiently sampling polytopes. For introduction, see Geometric Random Walks: A Survey by Santosh Vempala and How to compute the volume in high dimension? by Miklós Simonovits. In [1], Lee and Vempala give an algorithm to compute volume of a polytope up to $\epsilon$ multiplicative error in $O(mn^{2/3}\epsilon^{-2}\cdot mn^{w-1})$ where $n$ is number of dimension and $m$ is number of halfspaces/linear inequalities and logarithmic factors are suppressed. So for polytopes likes yours it should run in $O(n^{w+1+2/3}\epsilon^{-2})$. -Some software for volume approximation: VolEsti (implements Hamiltonian Monte Carlo), polytopewalk (Vaidya walk and John walk), polyrun (Hit-and-Run, BallWalk, SphereWalk, GridWalk), PolytopeSamplerMatlab (Hamiltonian Monte Carlo) of those only VolEsti seems as mature as cdd, you can find comparsion of VolEsti to mc-stan HMC here. Out of all software mentioned I'll guess VolEsti is your best choice. -Also, are you trying to optimize the volume of the polytope subject on some constraints on $L$? -[1]: Convergence Rate of Riemannian Hamiltonian Monte Carlo and Faster Polytope Volume Computation by Yin Tat Lee, Santosh S. Vempala, https://arxiv.org/pdf/1710.06261.pdf<|endoftext|> -TITLE: Is there an algebraic version of Darboux's theorem? -QUESTION [11 upvotes]: Let $M$ denote a smooth manifold, and $\omega \in \Omega^2(M, \mathbb{R})$ a symplectic form. The classical version of Darboux's theorem states that for any $x \in M$, there exists an open neighborhood $U$ of $x$ together with local coordinates $p_1,\dots, p_n, q_1, \dots, q_n$ on $U$ such that -\begin{equation} -\omega \vert_U = dp_1 \wedge dq_1 + \dots + dp_n \wedge dq_n. \tag{1} -\end{equation} -I learned from this question that the theorem also holds in the complex analytic context, which leads to my question. All of the ingredients in Darboux's theorem make sense algebraically. Specifically, let $X$ denote a smooth variety over an algebraically closed field $k$ of characteristic $0$. A symplectic form on $X$ is a closed 2-form $\omega \in H^0(X, \Omega^2_{X/k})$ inducing a non-degenerate pairing on the tangent space to each closed point of $X$. For example, we have the 'standard' symplectic form on $\mathbb{A}^{2n}_k = \text{Spec} \, k[p_1,\dots, p_n, q_1, \dots, q_n]$ given by (1). -My question, vaguely stated, is whether or not there is a version of Darboux's theorem for such symplectic forms. -For example, since $X$ is smooth, there exists an open neighborhood $U$ of $x$, and an étale $k$-morphism $f: U \rightarrow \mathbb{A}^n_k$, where $n = \text{dim}(X)$. Perhaps such a result would state that $n = 2m$ is even, and that $f$ can be chosen so that $\omega$ is the pullback of the standard form on $\mathbb{A}^{2m}_k$ under $f$. - -REPLY [17 votes]: In fact, the opposite is true. For $X$ a smooth proper variety over $\mathbb C$ and $\omega$ a nonzero symplectic form on $X$. Then there is no nonempty Zariski open set on which $\omega$ is the pullback (under any map) of the standard symplectic form. The same should work for etale open sets, and even for smooth morphisms. -The reason is that the standard symplectic form is exact (i.e. is the derivative of a 1-form), so its cohomology class vanishes - we can just use ordinary de Rham cohomology for this. Thus its pullback under any map has vanishing cohomology class. -But it is not possible for a nonzero holomorphic $2$-form on a smooth projective variety $X$ to have zero cohomology class when restricted to a nonempty open set $U$. If it did, then the associated class in $H^2(X, \mathbb C)$ would lie in the image of the natural map from the $H^2$ of $X$ supported in $X \setminus U$. But this cohomology group is generated by the classes of the $\dim X-1$-dimensional irreducible components of $X \setminus U$, which are sent to their divisor classes. So the associated class in $H^2(X, \mathbb C)$ would have to be a linear combination of divisor classes. -But divisor classes are sent by Hodge theory to $(1,1)$-forms, while holomorphic $2$-forms are $(0,2)$-forms, so they can never be equal.<|endoftext|> -TITLE: The strength of "There are no $\Pi^1_1$-pseudofinite sets" -QUESTION [12 upvotes]: For $\Gamma$ a set of second-order sentences in the empty language, say that a set $X$ is $\Gamma$-pseudofinite if $X$ is infinite but for every sentence $\varphi\in\Gamma$ which is satisfied in every finite pure set we have $X\models\varphi$. For example, $\mathsf{ZF}$ proves that the sentence "I can be linearly ordered, and every linear ordering of me is discrete" is true of exactly the finite sets, and so $\Sigma^1_1\wedge\Pi^1_1$-pseudofinite sets do not exist; in the other direction, $\mathsf{ZF}$ proves that $\omega$ is $\Sigma^1_1$-pseudofinite. -The interesting case is $\Pi^1_1$. While $\mathsf{ZFC}$ proves that there are no $\Pi^1_1$-pseudofinite sets (consider "Every linear ordering of me is discrete"), James Hanson showed in $\mathsf{ZF}$ that amorphous sets are $\Pi^1_1$-pseudofinite. My question is whether amorphousness is more-or-less the only way we get $\Pi^1_1$-pseudofinite sets: - -Over $\mathsf{ZF}$, does "There are no amorphous sets" imply "There are no $\Pi^1_1$-pseudofinite sets?" - -Note that this is a bit weaker than asking whether every $\Pi^1_1$-pseudofinite set is amorphous. FWIW I think the answer to that question is negative (I suspect e.g. that the union of two $\Pi^1_1$-pseudofinite sets is $\Pi^1_1$-pseudofinite). - -REPLY [6 votes]: My "hunch" in the comments to the question appears to be correct! This model comes from Howard, Paul E.; Yorke, Mary F., Definitions of finite, Fundam. Math. 133, No. 3, 169-177 (1989). ZBL0704.03033. The paper has a few confusing typos and, in particular, the proof of Theorem~15 appears insufficient, so I'm sketching the argument in some detail, with another proof of that theorem. - -$\newcommand{\supp}{\operatorname{supp}\nolimits} -\newcommand{\fix}{\operatorname{fix}\nolimits}$Fix a ground model $\mathfrak{M}$ of ZFA+AC where the set $U$ of atoms is countably infinite and fix a dense linear ordering ${<}$ of $U$ without endpoints. -Let $\mathcal{P}$ be the lattice of finite interval partitions of $U$, i.e., patitions of $U$ into finitely many blocks where each block is an interval of any shape. -This is a lattice under refinement $P \leq Q$ iff every block of $P$ is contained in a block of $Q$. -The meet $P \sqcap Q$ consists of all nonempty intersections of a block from $P$ and a block from $Q$. -The join $P \sqcup Q$ is more complicated: the blocks of $P \sqcup Q$ are maximal unions of the form $B_1\cup B_2 \cup \cdots \cup B_k$ where $B_1,B_2,\ldots,B_k \in P \cup Q$ and $B_1 \cap B_2, B_2 \cap B_3, \ldots, B_{k-1} \cap B_k$ are all nonempty. -Given an interval partition $P$, we write ${\sim_P}$ for the associated equivalence relation: $x \sim_P y$ iff $x$ and $y$ belong to the same block of $P$. -Let $G$ be the group of permutations $\pi$ of $U$ with finite support $\supp\pi = \{ x \in U : \pi(x) \neq x \}$. -Given an interval partition $P$ of $U$, let $$G_P = \{ \pi \in G : (\forall B \in P)(\pi(B) = B) \}.$$ -Note the following facts: - -$G_{P \sqcap Q} = G_P \cap G_Q$. -If $\{x\}$ is a block of $P$ for each $x \in \supp\pi$ then $\pi G_P\pi^{-1} = G_P$. -$G_{P \sqcup Q}$ is the subgroup generated by $G_P \cup G_Q$. - -It follows that these subgroups generate a normal filter $\mathcal{F}$ of subgroups of $G$. -Let $\mathfrak{N}$ be the symmetric submodel associated to $\mathcal{F}$: -$$\mathfrak{N} = \{ X \in \mathfrak{M} : \fix(X) \in \mathcal{F} \land X \subseteq \mathfrak{N} \}.$$ -Note that by 3, for every $X \in \mathfrak{N}$ there is a coarsest interval partition $\supp(X)$ such that $G_P \subseteq \fix(X)$, namely $$\supp(X) = \bigsqcup \{ P : G_P \subseteq \operatorname{fix}(X) \}.$$ -Lemma. For any set $X$ in $\mathfrak{N}$, if $\pi \in \fix(X)$ then for every $x_0 \in U$, $\pi(x_0)$ is not in a block of $\supp(X)$ which is adjacent to that of $x_0$. -Proof. Suppose, for the sake of contradiction, that $A,B$ are adjacent blocks of $\supp(X)$ and $x_0 \in A$, $\pi(x_0) \in B$ for some $x_0$. -We will show that for any $a \in A$ and $b \in B$ the transposition $(a,b)$ fixes $X$. Note that at least one of $A$ or $B$ must be infinite. Let's assume that $B$ is infinite, the other case is symmetric. - -Suppose $a = x_0$ and $b \notin \supp\pi$. -Then $(a,b) = (x_0,b) = \pi^{-1}(\pi(x_0),b)\pi$. -Suppose $a = x_0$ and $b \in \supp\pi$. -Then pick $b' \in B \setminus \supp\pi$ and note that $(a,b) = (a,b')(b,b')(a,b')$. -Suppose $a \neq x_0$. -Then $(a,b) = (a,x_0)(x_0,b)(a,x_0)$ - -It follows that any permutation of $A \cup B$ fixes $X$, which contradicts the fact that $\supp(X)$ is the coarsest partition such that $G_{\supp(X)} \subseteq \fix(X)$. -Claim 1 (Howard & Yorke, Theorem 15). $\mathfrak{N}$ contains no amorphous sets. -Proof. Suppose $X \in \mathfrak{N}$ is infinite. If $G_{\supp(X)}$ fixes $X$ pointwise, then $X$ is wellorderable and therefore not amorphous. -Pick $x_0 \in X$ such that $P_0 = \supp(x_0)\sqcap\supp(X)$ properly refines $\supp(X)$. -Let $A,B$ be two adjacent blocks of $P_0$ which belong to the same block of $\supp(X)$. -Suppose $A$ has a right endpoint $a$; the case where $B$ has a left endpoint is symmetric. -Let $P_1$ be obtained from $P_0$ by replacing $A$ with $A\setminus\{a\}$ and $B$ with $B\cup\{a\}$. -Note that for $\phi,\psi \in G_{P_1}$, $\phi(x_0) = \psi(x_0) \iff \phi(a) = \psi(a)$. -Fix $b \in B$ such that $B\cap(-\infty,b)$ and $B\cap[b,+\infty)$ are both infinite. -Let -$$X_0 = \{ \pi(x_0) : \pi \in G_{P_1}, \pi(a) < b \}$$ -and -$$X_1 = \{ \pi(x_0) : \pi \in G_{P_1}, \pi(a) \geq b \}.$$ -These are two disjoint infinite subsets of $X$. -Moreover, $X_1, X_2 \in \mathfrak{N}$ since they are both fixed by $G_Q$ where $Q$ is a refinement of $P_0, P_1$, and $\{(-\infty,b),[b,+\infty)\}$. -Therefore $X$ is not amorphous. -Claim 2. $U$ is $\Pi^1_1$-pseudofinite in $\mathfrak{N}$. -Sketch. -Suppose, for the sake of contradiction, that $$(\forall Y \subseteq X^n, Z \subseteq X^m,\ldots )\phi(X,Y,Z,\ldots)$$ is a $\Pi^1_1$ statement which is true of every finite set $X$ but false for $X = U$. -Let $Y \subseteq U^n, Z \subseteq U^m,\ldots$ be sets in $\mathfrak{N}$ such that $\lnot\phi(U,Y,Z,\ldots)$. -Let $P = \supp(Y)\sqcap\supp(Z)\sqcap\cdots$ -Note that there are only finitely many possibilities for $Y, Z, \ldots$ -For example, when $n=1$ then $Y$ must be a union of some of the intervals from $P$. -When $n=2$, $Y$ must be a boolean combination of cartesian products of two intervals from $P$ and the diagonal set $\{(x,x) : x \in U\}$. And so on... -By an EF-style argument, if $V \subseteq U$ is a finite set such that that contains all the endpoints of intervals from $P$ each of the sets $P \cap B$ is sufficiently large when $B$ is an infinite interval from $P$, then $\phi(U,Y,Z,\ldots)$ is equivalent to $\phi(V, Y \cap V^n, Z \cap V^m,\ldots)$. It follows that $\lnot\phi(V, Y \cap V^n, Z \cap V^m, \ldots)$ for some finite set $V \subseteq U$, but this contradicts the assumption. - -Since the ultimate goal is to obtain a more concrete understanding of what $\Pi^1_1$-pseudofinite means, I will propose an alternate conjecture. -Recall Tarski's notion of II-finite: every chain of subsets of $X$ has a maximal element. This is equivalent to the $\Pi^1_1$ statement: every total preordering of $X$ has a maximal element. So every $\Pi^1_1$-pseudofinite set is II-finite. It seems that the converse might be true but I will only propose the following: -Conjecture. There is no infinite $\Pi^1_1$-pseudofinite set if and only if there is no infinite II-finite set.<|endoftext|> -TITLE: If $A$ is a cofibrant commutative dg-algebra over a commutative ring of characteristic $0$, then its underlying chain complex is cofibrant -QUESTION [6 upvotes]: Let $R$ be a commutative ring with characteristic $0$, namely it contains the field of rational numbers. Higher Algebra Proposition 7.1.4.10 tells that the category of commutative $R$-dg-algebras $\mathrm{CAlg^{dg}}(R)$ has a model structure induced from the projective model structure on chain complexes $\mathrm{Ch}(R)$, where weak equivalences are quasi-isomorphisms and fibrations are surjections (so, a morphism of commutative dg-algebras is a fibration or a weak equivalence if its underlying morphism of chain complexes is such). -In the proof of the following Proposition 7.1.4.11, an unproved claim (a condition in 4.5.4.7) is implicitly used, namely: - -The forgetful functor $\mathrm{CAlg^{dg}}(R) \to \mathrm{Ch}(R)$ preserves fibrant-cofibrant objects. - -Now, every object is fibrant with respect to the considered model structures, so this claim boils down to check that if $A$ is a cofibrant object in $\mathrm{CAlg^{dg}}(R)$, then its underlying chain complex is cofibrant with respect to the projective model structure on $\mathrm{Ch}(R)$. -How can I prove this? If $R$ were a field then it would be very easy, because every chain complex over a field is cofibrant. I feel that I should somehow use that $R$ has characteristic $0$ (it contains $\mathbb Q$), but can't precisely figure out how. - -REPLY [5 votes]: Cofibrant CDGAs are retracts of cellular ones. A cellular cofibrant CDGA is a free commutative graded algebra on a (possibly transfinite) sequence of generators $x_1,x_2,\dots$ such that $d(x_i)$ only depends on previous generators. A linear basis is given by monomials $x_{i_1}^{n_{i_1}}\cdots x_{i_r}^{n_{i_r}}$ such that $i_j -TITLE: Should cohomology of $\mathbb{C} P^\infty$ be a polynomial ring or a power series ring? -QUESTION [9 upvotes]: Some people define total cohomology of a space $X$ to be $\bigoplus_{i \geq 0} H^i(X)$, which would make $H^*(\mathbb{C} P^\infty)$ a polynomial ring in one generator of degree 2. -However, it seems like thinking of $H^*(\mathbb{C} P^\infty)$ as a power series ring is more natural for several reasons. For one thing, if cohomology is like the dual of homology, then the dual to an infinite direct sum is a direct product. Many algebraic formulas are also simplified if one allows for the entire power series ring rather than the polynomial ring. -Question: Are there compelling reasons to define total cohomology as $\bigoplus_i H^i$ or as $\prod_i H^i$? - -Addendum: -The question itself is quite concrete, but there are other reasons I am contemplating this, so perhaps I should list them. - -If I think of $H^*\mathbb{C}P^\infty$ as somehow Koszul dual to a circle, this question might be closer to whether one should think of (this kind of) Koszul duality as always happening in a filtered/pro setting. If there are strong views/philosophies on viewing infinite projective space as an instance of Koszul duality, or on whether Koszul duality should always ask for filtration (e.g., adic-near-a-point) structures, do share. - -One can think of $\mathbb{C}P^\infty$ as a space in its own right, or as a filtered diagram of spaces. This changes, for example, what kind of condensed set I think of $\mathbb{C}P^\infty$ as. Accordingly, the cohomology of the condensed set obtained as an ind-object of $\mathbb{C}P^n$ should look more pro-y (and hence look more like a power series), while the cohomology of the condensed set called "what does $\mathbb{C}P^\infty$ represent as a space" feels more like a polynomial ring. - -REPLY [13 votes]: In my opinion the most natural statement is that $H^*(\mathbb{CP}^\infty)$ is a graded power series ring. That is we can write -$$H^*(\mathbb{CP}^\infty)=\lim_n \mathbb{Z}[x]/x^n$$ -where the limit is taken in the category of graded rings and $x$ has degree 2. Note that in this particular case (where the ring of coefficients is concentrated in degree 0) it coincides with the graded polynomial ring. This has the advantage that the formula works for all complex-oriented cohomology theories, e.g. for complex K-theory: -$$KU^*(\mathbb{CP}^\infty)=\lim_n KU^*[x]/x^n$$ -in which case it does not coincide with the graded polynomial ring! -Whether to present graded rings as direct sums or direct products is largely a matter of personal preference, although the direct sum option has always seemed the most natural to me. - -One quick comment about one of your addenda: you can always think of homotopy types as the ind-category of finite homotopy types (precisely, this is true at the level of the ∞-category of spaces), so every cohomology ring $E^*X$ has a natural enhancement to a pro-(graded ring) (it's not quite true that $E^*X$ is always the limit of this pro-ring, because of the possible presence of $\lim^1$-terms, but let's ignore this for the moment). Under this correspondence $H^*\mathbb{CP}^\infty=\mathbb{Z}[[x]]$ seen as a pro-(graded ring) in the canonical way. This is the start of the connections between homotopy theory and formal geometry, which has been extremely fruitful. This book is an excellent source if you want to learn more about this.<|endoftext|> -TITLE: Appearance of proof relevance in "ordinary mathematics?" -QUESTION [14 upvotes]: I've been wondering recently what—if any—applications proof theory has to ordinary mathematics (by which I mean algebra, analysis, topology, and so on). In particular, I'd be fascinated to see a proof of a theorem about ordinary mathematical objects that relies on proof relevance. I'm afraid I'm unable to make the question more concrete, but what I would hope to see is something that "smells like" the following: - -The proposition $P$ is simply connected, in the sense of HoTT. It follows that the ring $R$ is nilpotent-free. - -Note that I don't care at all what is proved, or about what kind of "ordinary" object; only that it's a readily-understood "non-foundational" statement that isn't completely trivial. The proof-theoretic ingredients should also not be too trivial: in particular, requiring "$P$ is inhabited" is not very unusual at all! Can you cook up anything like this? Has anything like it ever been used in a serious way? - -REPLY [9 votes]: Proof mining (which even has a short Wikipedia article!), and area in large part developed by Kohlenbach, is mentioned briefly in the comments, and I thought it deserves a bigger mention. Roughly speaking, proof mining is the idea that from non-constructive proofs of existence one can often extract effective bounds. For example, from the standard proof of the irrationality of $\sqrt{2}$ it is not hard to show that for any irreducible rational $\frac{a}{b} > 1$ we must have that $|\frac{a}{b} - \sqrt{2}| > \frac{1}{2b^2}$ (try it yourself!). This ties in with the irrationality measure of a real number. -Especially in its applications to functional analysis, such as in [Kohlenbach, U.; Leuştean, L.; Nicolae, A.; Quantitative results on Fejér monotone sequences. Commun. Contemp. Math. 20 (2018), no. 2], proof mining really shines through as a powerful set of techniques. -I attended a summer school in 2016 in which Kohlenbach presented these slides -- they are a gold mine of information, but can be rather dense at times. However, they provide an excellent overview for many important concepts in the area (such as Herbrand normal forms) and highlight many applications. A good introductory text is also this text by Kohlenbach and Oliva.<|endoftext|> -TITLE: Is there an abstract logic that defines the mantle? -QUESTION [7 upvotes]: It is a known result by Scott and Myhill that the second-order version of $L$ yields $\mathrm{HOD}$. -Recently, Kennedy, Magidor, and Väänänen (Inner models from extended logics: Part I and II) investigated inner models given by logics with generalized quantifiers, which yields a logic intermediate between first-order and second-order logic. It motivates the following question: - -Is there a logic that produces the mantle? - -(Here the choice of the mantle is somewhat arbitrary; we may replace it by 'generic mantle', 'symmetric mantle' or whatever. I will focus on the mantle in this question, but I welcome discussing other cases.) -Of course, the answer is trivial if we assume, like $V=L$ or $V=L[G]$ for some $L$-generic $G$. I want to ask the existence of logic which defines the mantle uniform to models of ZFC. - -Is there a (ZFC-definable) abstract logic $\mathcal{L}$ such that the inner model given by $\mathcal{L}$ is (ZFC-provably) the mantle? -(Under model-theoretic terms, is there $\mathcal{L}$ such that for any model $M$ of $\mathsf{ZFC}$, the inner model given by $\mathcal{L}$ is the mantle of $M$?) - -Here are some of my rough thoughts: - -Sublogics of higher-order logics are not the candidate for $\mathcal{L}$: the corresponding inner models of higher-order logics are $\mathrm{HOD}$ (if my reasoning is correct), so the sublogics yield a submodel of $\mathrm{HOD}$. However, $\mathrm{HOD}$ need not be the mantle. (Theorem 70 of Fuchs, Hamkins, and Reitz (Set-theoretic geology).) - -We can rule out $\mathcal{L}_{\kappa\kappa}$, which yields Chang model. The inner model given by $\mathcal{L}_{\kappa\kappa}$ is the least transitive model of ZF that contains all ordinals and is closed under $<\kappa$-sequences (Theorem II of Chang's Sets constructible using $L_{\kappa\kappa}$.) However, the mantle need not be closed under $<\kappa$-sequences. (A generic extension of $L$ would be an example.) - - -I would appreciate any comments or answers. - -REPLY [2 votes]: Combining Goldberg's comment and Hamkins' answer seems to work. Especially, for any inner model $M$ of ZF, we have an abstract logic $\mathcal{L}$ whose corresponding inner model $L^\mathcal{L}$ is $M$. -Consider the sublogic of $\mathcal{L}_{\infty,\omega}$ such that infinite conjunction and disjunctions are only allowed to set of formulas in $M$. In fact, $\mathcal{L}=\mathcal{L}_{\infty,\omega}^M$. -Define $\psi_A$ for $A\in M$ as Hamkins defined: to repeat the definition, -$$\psi_A(x):= \bigvee_{a\in A} (\forall v : v\in u\leftrightarrow \psi_a(u)).$$ -Then $\psi_A(x)$ is a member of $M$ by induction on $A\in M$. -We can see that if $A\in M$, $A\subseteq V_\alpha^M$ then $$A=\{u\in V^M_\alpha \mid V^M_\alpha\models \psi_A(u)\}.$$ -Hence the $\alpha$th hierarchy $L_\alpha^\mathcal{L}$ contains $V^M_\alpha$ (It can be shown by induction on $\alpha$.) Therefore $M\subseteq L^\mathcal{L}$. -On the other hand, an inductive argument shows that the $\alpha$th hierarchy $L^\mathcal{L}_\alpha$ is a member of $M$ (we need the absoluteness of the satisfaction relation for $\mathcal{L}$ between $M$ and $V$), so $L^\mathcal{L}\subseteq M$.<|endoftext|> -TITLE: A graph similar to the Bruhat graph, what is it called? -QUESTION [5 upvotes]: The weak Bruhat graph (or 1-skeleton of the permutohedron) $B_n$ can be constructed as follows: - -the vertices of $B_n$ are the permutations of the tuple $(1,...,n)$, two are joined by an edge, if they can be transformed into each other by a transposition of two neighboring numbers, e.g. - -$$(5,3,\underline{\smash{4,1}},2) \quad\mapsto \quad (5,3,\underline{\smash{1,4}},2).$$ -Now, consider the graph $X_n$, basically defined in the same way, but also allowing the "cyclic transposition": - -the vertices of $X_n$ are the permutations of the tuple $(1,...,n)$, two are joined by an edge, if they can be transformed into each other by a transposition of two neighboring numbers, or the first and the last number, i.e. - -$$(\underline 5,3,4,1,\underline 2) \quad\mapsto \quad (\underline 2,3,4,1,\underline 5).$$ - -Question: Is there a name for the graph $X_n$ in the literature, and what is it used for? - -$B_n$ is a spanning subgraph of $X_n$. -In fact, $B_n$ can be obtained from $X_n$ by deleting a matching. -In contrast to the Bruhat graph, $X_n$ is always arc-transitive. -It has $n!$ vertices and degree $n$ (if $n\ge 3$). For example, $X_3=K_{3,3}$. - -REPLY [6 votes]: A Markov chain on the symmetric group with this transition graph (but with directed edges and weights) was investigated by Lam and Williams. This has since received considerable attention, and has been connected to "TASEP on a ring" if you are looking for search words (it doesn't appear that the graph itself has a name in this context). -I should point out that one should not give this graph the tempting names "circular Bruhat graph" or "cyclic Bruhat graph" as both of these have been used to refer to the graph underlying Postnikov's circular Bruhat order, which is different.<|endoftext|> -TITLE: What are all invariant polynomials on the space of algebraic curvature tensors? -QUESTION [7 upvotes]: Let $V = (\mathbb{R}^n, g)$, where $g$ is the Euclidean inner product on $V$. Denote by $G$ the orthogonal group $O(V) = O(n)$ and by $\mathfrak{g}$ the Lie algebra of $G$. -Let $W \subset \Lambda^2V^* \odot \Lambda^2V^*$ be the subset satisfying the algebraic Bianchi identity. More precisely, let $R(v_1,v_2,v_3,v_4)$ denote an element of $\Lambda^2V^* \odot \Lambda^2V^*$. Thus $R$ is skew-symmetric in $v_1$ and $v_2$ and it is also skew-symmetric in $v_3$ and $v_4$. Moreover -$$ R(v_3,v_4,v_1,v_2) = R(v_1,v_2,v_3,v_4). $$ -Then $R \in W$ if and only if, in addition to the conditions above, $R$ also satisfies the following identity (known as the algebraic Bianchi identity): -$$ R(v_1,v_2,v_3,v_4) + R(v_2,v_3,v_1,v_4) + R(v_3,v_1,v_2,v_4) = 0. $$ -Now my question can be formulated. What is an explicit description of the ring $\mathbb{C}^G[W]$ of $G$-invariant polynomials on $W$ (with $W$ being the space of algebraic curvature tensors, if I may call it so) and $G$ acting on $W$ by restricting its natural action on $\Lambda^2V^* \odot \Lambda^2V^*$. -Also, if one fixes a degree $d > 0$, what is an explicit description of the space of $G$-invariant homogeneous polynomials in $W$ of degree $d$? -I was thinking at first about the Chern-Weil homomorphism, but I think this only gives a proper subspace of $G$-invariant polynomials on $W$, and not all of them (I am not a 100% sure). This has probably been studied in the literature. I don't have access to MathSciNet anymore though (due to some budget cuts by my University). -Edit 1: I notice some overlap with the post Invariant polynomials in curvature tensor vs. characteristic classes, but the posts are sufficiently different. - -REPLY [7 votes]: I think this is unlikely to have a very nice answer. When $n=2$ and $n=3$, the answer is simple, but, already for $n=4$, it's not likely to be easy to give a set of generators and relations for the $\mathrm{O}(n)$-invariant polynomials on the vector space $\mathcal{R}_n$ of algebraic curvature tensors in dimension $n$. (I'm avoiding the OP's notation of $W$ for this space because it doesn't explicitly reference the dimension $n$ and I don't want to confuse it with the space of Weyl curvature tensors.) -Since $\mathcal{R}_n$ has dimension $\tfrac1{12}n^2(n^2-1)$ and since, for $n>2$ the generic element of $\mathcal{R}_n$ has only a finite stabilizer in $\mathrm{O}(n)$, the dimension of the ring of $\mathrm{O}(n)$-invariant polynomials on $\mathcal{R}_n$ will be -$$ -\frac1{12}n^2(n^2-1) - \frac12n(n-1) = \frac1{12}(n+3)n(n-1)(n-2), -$$ -so there will always be at least that many independent generators and, when $n>3$, many more, plus a bunch of relations, since the quotient space will not be 'smooth' near the origin. -Once one gets above the low degrees when $n>3$, to compute the dimensions of the graded pieces of this ring will be complicated (essentially, one is asking for the Hilbert series of the ring of invariants). (However, the dimension of the grade 1 piece is 1, and the dimension of the grade $2$ piece is $2$ for $n=3$ and $3$ for $n>3$. If one were using $\mathrm{SO}(4)$ for $n=4$, the dimension of the grade $2$ piece would be $4$.) -I imagine that the answers for $n=4$ are known (though I don't know them) since it is, in principle, just a representation-theoretic computation. - -REPLY [5 votes]: I am not sure that this has a "nice" answer. Your question can be reformulated as follows. Let $\mathcal{A}_n$ be the space of algebraic curvature tensors on $\mathbb{R}^n$. A homogenous polynomial $P$ on $\mathcal{A}_n$ is the same as an element of $S^k\mathcal{A}_n$, the $k$-th symmetric tensor power of $\mathcal{A}_n$. Now if $H_k$ is the space of homogeneous polynomials of degree $k$ on $\mathcal{A}_n$, then $H_k \subset S^k \mathcal{A}_n$ is a subrepresentation of $G$. -In other words, a recipe to obtain an answer to your question for specific $k$, $n$, is the following. Decompose the $G$-representation $S^k \mathcal{A}_n$ into irreducible $G$-representations and count the number of trivial representations among those. This can be done for low $k$, $n$ using software such as LiE. -Note that as a $G$-representation, the space $\mathcal{A}_n$ splits into the direct sum of three irreducible representations: -$$ \mathcal{A}_n = \mathbb{R} \oplus S^2_0(\mathbb{R}^n) \oplus \mathcal{W},$$ -where $\mathcal{W}$ is the space of Weyl curvature tensors (i.e. those curvature tensors that are additionally entirely trace-free). -A quick check on LiE shows that there are plenty such polynomials: For example, looking for polynomials that depend on the Weyl part alone and $n$ large, there is one of degree 2 and four of degree 3. I doubt there is a good general answer.<|endoftext|> -TITLE: Path integral as quantum mechanics on the tangent bundle -QUESTION [6 upvotes]: Let $X$ be a configuration space, a finite-dimensional manifold. By "quantum mechanics on $X$" I mean a linear evolution equation on complex-valued functions on $X$, determined by a Hamiltonian $H\in \text{End} [L^2(X,\mathbb{C})]$, with endomorphisms defined in an appropriate densely defined sense. (I am not requiring that $H$ is Hermitian, at least for now.) -Now solving this evolution equation is equivalent to writing down the time-$T$ evolution matrix $U_T= e^{-iTH},$ with "matrix coefficients" $U_T(x,y) : = \langle y|U_T|x\rangle.$ Here $x, y\in X$ points and $|x\rangle$ the delta-function on $X$ (this is not in $L^2,$ but $U_T(x,y)$ can be made sense of as a function of $x, y$ in a suitable distributional sense). -The path integral formalism gives (at least in theory) another way of computing $\langle y|U_T|x\rangle$ (for Hermitian $H$). Namely, the matrix coefficient $\langle y|U_T|x\rangle$ can be written as a limit of integrals $$\int dx_1dx_2\dots dx_N \langle y| U_{\epsilon} |x_N\rangle \langle x_N|U_{\epsilon}|x_{N-1}\rangle \langle x_{N-1}|U_{\epsilon}|x_{N-2}\rangle\cdots \langle x_2|U_\epsilon|x_1\rangle\langle x_1|U_\epsilon|x\rangle$$ -where $\epsilon = T/N$ and $N$ goes to $\infty$, with the integrand a multiplicative function in the consecutive pairs $(x_k, x_{k+1})$. Taking a continuous limit, this "morally" reduces to an integral $$\int D\gamma \exp \big( i S(\gamma)\big)$$ over paths $\gamma:[0,T]\to X$ from $x$ to $y$, with the "action" $S(\gamma) = \int_0^T dt L\big(\gamma(t), \dot{\gamma}(t)\big),$ and $L$ the Lagrangian, a functional on $TX$. Here $\gamma$ is the continuous limit of the sequence $x_0, x_1,\dots, x_N$ and the form of the action encodes its multiplicative structure and its dependence only on "neighboring pairs" (corresponding to the data $\gamma(t), \dot{\gamma}(t)$). -In practive it is difficult to make sense of the path integral: rigorous definitions use the Wiener measure (associated to some metric on $X$), corresponding to the Brownian random process, and concentrated on nowhere differentiable paths, which means that the action functional cannot be treated as a function. On the other hand, there are "bounded-energy" probability measures on paths which give Brownian motion in a limit but are more nicely behaved: only concentrated on differentiable (or piecewise differentiable) paths, and well-behaved with respect to the $C^1$ topology on paths (where the action functional is explicitly defined and continuous). One such approximation (if I understand correctly) is to consider piecewise-linear paths with direction changing according to a Poisson process. Another is to consider everywhere differentiable paths with derivative undergoing a more continuous random process. -Now here's my question. It should be possible to write down a (potentially non-Hermitian) Hamiltonian $H'$ on the tangent space $TX$ with the property that its evolution operator $U_T'$ has matrix coefficients $\langle x', v' | U_T'| x, v\rangle$ which compute the expectation of $\exp\big( -i S(\gamma)\big)$ in one of these bounded-energy measures over differentiable (or piecewise differentiable) paths that start at $x$ with derivative $v$ and end at $x'$ with derivative $v'.$ Indeed, all that's needed is to incorporate the Lagrangian, the standard dynamic relationship between $x$ and $v$ and some kind of a stochastic term on each tangent fiber. -This seems to me like a promising and straightforward way to replace the analytic difficulties involved in the path integral with ordinary quantum mechanics (i.e. exponentiating operators). If this works, someone must have tried it. Are there problems with this approach or references where it is done? - -REPLY [4 votes]: Path integrals over simultaneous position/velocity, or more commonly position/momentum, degrees of freedom are known as phase space path integrals. I don't know very much about the rigorous construction of path integral measures, and even less so about their phase space version. However, there does appear to be at least one classic work on the subject by Berezin, who discussed in particular the regularity of the paths on which the path integral measure is concentrated. Then there were also some follow-ups by Daubechies and Klauder. Perhaps these works could point you in the right direction. - - -Berezin, F. A., Feynman path integrals in a phase space, Sov. Phys. Usp. 23, 763 (1980). - -Klauder, John R.; Daubechies, Ingrid, Quantum mechanical path integrals with Wiener measures for all polynomials Hamiltonians, Phys. Rev. Lett. 52, No. 14, 1161-1164 (1984). ZBL0979.81518. - -Daubechies, Ingrid; Klauder, John R., Quantum-mechanical path integrals with Wiener measure for all polynomials Hamiltonians. II, J. Math. Phys. 26, No. 9, 2239-2256 (1985). ZBL0979.81517.<|endoftext|> -TITLE: Relating smooth concordance and homology cobordism via integral surgeries -QUESTION [12 upvotes]: Let $K_0$ and $ K_1$ be knots in $S^3$. They are called smoothly concordant if there is a smoothly properly embedded cylinder $S^1 \times [0,1]$ in $S^3 \times [0,1]$ such that $\partial (S^1 \times [0,1]) = -(K_0) \cup K_1$. -Let $Y_0$ and $ Y_1$ be integral homology spheres, i.e., $H_*(Y_i; \mathbb Z) = H_*(S^3; \mathbb Z)$. They are called homology cobordant if there exists a smooth compact oriented $4$-manifold such that $\partial X = -(Y_0) \cup Y_1$ and $H_*(X,Y_i; \mathbb Z)=0$ for $i=0,1$. -I cannot explicitly figure out but I made some progress. How can we concretely prove that the following well-known theorem: Let $S_n^3(K)$ denotes $3$-manifold obtained by the $n$-surgery on the knot $K$ in $S^3$. -Theorem: If $K_0$ is smoothly concordant to $K_1$ in $S^3$, then for all $n$, $S_n^3(K_0)$ is homology cobordant to $S_n^3(K_1)$. -Addition: Can we use this theorem to obtain "strong" obstructions for knots being smoothly concordant? - -REPLY [3 votes]: I am not quite sure it is a “strong” obstruction but it is “nice” at least to me: -Observation: The left-handed trefoil and the right-handed trefoil are not smoothly concordant in $S^3$. -Let $K_0$ and $K_1$ respectively denote the left-handed trefoil and right-handed trefoil. Assume that $K_0$ and $K_1$ are smoothly concordant in $S^3$. Then by theorem, we know that $S^3_{-1}(K_0)$ and $S^3_{-1}(K_1)$ are homology cobordant. -Observe that $S^3_{-1}(K_0)$ is the Brieskorn sphere $\Sigma(2,3,5)$ while $S^3_{-1}(K_1)$ is the Brieskorn sphere $\Sigma(2,3,7)$. This can be done by Kirby calculus. For example, see Chapter 3 in Saveliev's book. -But Fintushel-Stern $R$-invariants of $\Sigma(2,3,5)$ and $\Sigma(2,3,7)$ are not same and Fintushel-Stern $R$-invariant provides a homology cobordism invariant. Hence we have reached a contradiction. It is worthy to note that this invariant can be easily computed due to Neumann-Zagier’s shortcut. -This conclusion also can be derived Ozsváth-Szabó $d$-invariant because $d(\Sigma(2,3,5))=-2$ and $d(\Sigma(2,3,7))=0$, see the example section in their paper. As Golla emphasized, this obstruction also comes from Frøyshov's $h$-invariant. -Further note: Let $\Theta^3_\mathbb Z$ denote integral homology cobordism group. It is the set of integral homology spheres modulo smooth homology cobordism. Then $d$- and $h$-invariants provide the following surjective group homomorphisms: $$d: \Theta^3_\mathbb Z \to 2 \mathbb Z,\ \ \ \ \ \ \ \ h: \Theta^3_\mathbb Z \to \mathbb Z.$$<|endoftext|> -TITLE: Are there first-order statements that second order PA proves that first order PA does not? -QUESTION [5 upvotes]: Are there first-order statements that second order PA proves that first order PA does not? Is this known one way or the other? Could you share an example? (edit: to clarify, by 'second order PA' I don't mean any first order theory) - -REPLY [8 votes]: This question more or less reduces to asking for natural arithmetic statements unprovable in PA. A very interesting but lesser-known example that I learned about recently concerns fusible numbers. The term "fusible" may be unfamiliar but it arises as a natural generalization of a well-known puzzle involving measuring lengths of time by burning fuses (finite pieces of string) of different lengths. The puzzle is an old one and was not specifically constructed to produce unprovable statements, so the unprovable statements involving fusible numbers are arguably the most "natural" PA-unprovable statements around.<|endoftext|> -TITLE: Identity involving the probability that a random walk stays below a curve -QUESTION [13 upvotes]: I'm looking for a direct proof of the following identity: -Let $W_n$ be a simple random walk with $W_0=0$. For all $x>0$ we have -$$ -\lim _{N\to \infty} \sqrt{N} \cdot \mathbb P \Big( \forall n \le N , \ W_n \le x\sqrt{N} -x\sqrt{N-n} \Big) -=\frac{e^{-\frac{x^2}{2}}}{\int _x^\infty e^{-\frac{y^2}{2}}dy}. -$$ -I have an indirect proof that follows from a specific model that I've been working on related to DLA. -It's not hard to show that the probability in the left hand side decays like $C/\sqrt{N}$ (because one can show that it behaves like the probability that a random walk stays negative for $N$ steps). This is the reason for the factor $\sqrt{N}$ inside the limit. - -REPLY [5 votes]: Example 2 of arXiv:0704.2826 considers the analogous problem for the continuous-time random walk, in the more general case that the curve has the form $g(t)=a+b\sqrt{T-t}$ with $a+b\sqrt T\geq 0$. The random walk starting at the origin stays below that curve for all $t -TITLE: Is there a minimal inner model for determinacy? -QUESTION [14 upvotes]: Assume $\sf ZF+AD$. Is there some inner model $M$ containing all the ordinals such that $M\models\sf ZF+AD$ as well? -What if we require $\omega_1$ and/or $\omega_2$ to be computed correctly? -Can we say anything about these models (e.g. $M\models V=L(\Bbb R)$)? -Is it at all consistent? -There's no real reason to expect a minimal model, of course, since Woodin cardinals are involved. But $\sf AD$ is also a very strong axiom and might have unexpected consquences. - -REPLY [17 votes]: Assuming AD, Woodin showed that no inner model that is missing a real correctly computes $\omega_1$. (This almost follows from Theorem 9 of Velickovic-Woodin's "Complexity of the set of reals of inner models of set theory.") -Therefore if you take an inner model $M$ of AD + $V = L(\mathbb R)$ whose $\omega_1$ is as small as possible, this model has no proper inner model of AD. -(This argument shows that inclusion is a wellfounded partial order of the inner models of $\text{AD}+ V = L(\mathbb R)$...) -If one generically wellorders the reals of $M$ without adding reals, one obtains a model of ZFC in which there is a minimum inner model of $\text{AD}$. Update: this might be true, but as Dmytro Taranovsky points out in the comments, it is far from clear. -In general, however, there need not be a minimum inner model of $\text{AD}$. Assume there is a model of $\text{AD}$ containing only countably many reals. (This hypothesis is justified in the following paragraph.) Let $M$ be a minimal model of $\text{AD}$ with this property. Then there is a real $g$ that is $M$-generic for Cohen forcing. Let $N = L(\mathbb R)^{V[g]}$. By another theorem of Woodin, there is an elementary embedding $j : M\to N$. (There are more details in Kechris-Woodin's "Generic codes for uncountable ordinals.") Therefore $N$ is a minimal model of $\text{AD}$, since this is first order expressible, and $N$ is obviously distinct from $M$. -Of course, after forcing with $\text{Col}(\omega,\mathbb R)$ over a model of $\text{AD}^{L(\mathbb R)}$, there is an inner model of $\text{AD}$ that contains countably many reals, so the hypothesis of the above paragraph is consistent. In fact, the hypothesis is true: using $\mathbb R^\#$, one can build an inner model containing only countably many reals that is elementarily embeddable into $L(\mathbb R)$. (Take a countable elementary substructure of the mouse $\mathbb R^\#$ and then iterate away the top measure.) Therefore large cardinals imply that there is no minimum inner model of $\text{AD}$.<|endoftext|> -TITLE: Reconstruct a variety from its crystalline topos -QUESTION [16 upvotes]: Let $k$ be a perfect field of positive characteristic. Let $X$ be a smooth projective geometrically connected $k$-scheme with a $k$-point. -Can we reconstruct $X$ from its small crystalline topos $((X/W(k))_{\mathrm{cris}}, \mathcal{O}_{X/W(k)})$ considered with the structure morphism to $((\mathrm{Spec}\:k/W(k))_{\mathrm{cris}}, \mathcal{O}_{\mathrm{Spec}\:k/W(k)})$? -Can we at least find the Hodge numbers? - -REPLY [3 votes]: Edit: This answer is probably wrong, sorry. The issue is indicated [in bold] below. - -Yes, we can reconstruct $X$ (as a $k$-scheme). -It would be kind of trivial if you had asked for the small Zariski topos instead of the small crystalline topos since the small Zariski topos $\mathrm{Sh}(X)$ is just the space $X$ viewed as a topos (generalized space). So the question is how to reconstruct the small Zariski topos (including the structure sheaf) from the small crystalline topos. -Claim. Let $X$ be a scheme over an arbitrary base scheme $S$. Then $\mathrm{Sh}(X)$ (or rather $X$ itself, as a locale) is the localic reflection of $(X / S)_{\mathrm{cris}}$. -Proof. The opens of the localic reflection are given by the subterminal objects of the topos in question, i.e. by the subsheaves $\mathcal{F}$ of the terminal sheaf on the small crystalline site of $X$ over $S$. Whenever $\mathcal{F}(U \hookrightarrow T)$ is inhabited for an $S$-PD-thickening $T$ of an open subscheme $U$ of $X$, then also $\mathcal{F}(U \rightarrow U) =: \mathcal{F}(U)$ is inhabited, since there is a morphism from $(U \rightarrow U)$ to $(U \hookrightarrow T)$. -[But this morphism does not cover $(U \hookrightarrow T)$ and there is in general no morphism in the opposite direction. So $\mathcal{F}(U)$ inhabited probably doesn't imply $\mathcal{F}(U \hookrightarrow T)$ inhabited.] -Also, for a cover $U_i$ of $U$, the sheaf condition for $\mathcal{F}$ says that $\mathcal{F}(U)$ is inhabited if the $\mathcal{F}(U_i)$ are. In summary, a subsheaf of the terminal sheaf is precisely given by an open of $X$. (There is an isomorphism of frames.) $\blacksquare$ -Since $\mathcal{O}_{X/S}(U \rightarrow U) = \mathcal{O}_X(U)$ we also have the structure sheaf. -Note that we don't even need the structure morphism you specified, only the one to $\mathrm{Sh}(\mathrm{Spec}\:k)$.<|endoftext|> -TITLE: Schur Weyl duality for the supergroup $\text{GL}(m|n)$ -QUESTION [9 upvotes]: Let $G$ be the supergroup $\text{GL}(m|n)$. It has a tautological representation $V= \mathbb{C}^{m|n}$. -For every natural number $d$ we have a natural map $$\Phi_d:\mathbb{C} S_d\to \text{End}_G(V^{\otimes d})$$ -where $\sigma\in S_d$ is sent to the linear transformation given by tensor permuting $V^{\otimes d}$ according to $\sigma$. To what extent does Schur-Weyl duality generalise from the case $n=0$ to the case of super vector space? That is: -Question 1: Is $\Phi_d$ surjective? -Question 2: What is the kernel of $\Phi_d$ ? Can it also be described using some combinatorial condition similar to the case $n=0$? - -REPLY [6 votes]: Schur Weyl duality holds in the super case, as well. There is the double centralizer property, thus a positive answer to Q1, and also a characterization of the kernel as those ideals of $\mathbb C[S_d]$ which correspond to partitions that don't fit inside the (m,n)-hook. -See the paper "Hook Young diagrams with applications to combinatorics and to representations of Lie superalgebras" by Berele and Regev. For a more recent textbook treatment this is also done nicely in chapter 11 of Musson's book "Lie Superalgebras and Enveloping Algebras". - -REPLY [4 votes]: This is a result of Sergeev, but I can only find the article in Russian at the moment. If I remember right, the map is surjective, and the kernel can be worked out from the fact that a Schur functor applied to the standar representation of $\mathfrak{gl}(m|n)$ vanishes iff the Young diagram contains the box at position (m+1,n+1), i.e. if they do not fit in an "(m,n) hook." For example, see the discussion in these slides of Serganova. Note that the classical case the diagrams fitting in an (m,0)-hook are exactly the ones with m or fewer rows, while in the entirely odd case they need to have n or fewer columns.<|endoftext|> -TITLE: Is it true that $\{x^3-2x+y^3-2y+z^3-2z: x,y,z\in\mathbb Z\}=\mathbb Z$? -QUESTION [6 upvotes]: A well known conjecture states that -$$\{x^3+y^3+z^3:\ x,y,z\in\mathbb Z\}=\{m\in\mathbb Z:\ m\not\equiv\pm4\pmod 9\}.$$ -For $m=33,\, 42$ an integer solution to the equation $x^3+y^3+z^3=m$ was only found last year. -In 2017, Tyrell asked whether -$$\left\{\frac{x(x+1)(x+2)}6+\frac{y(y+1)(y+2)}6+\frac{z(z+1)(z+2)}6:\ x,y,z\in\mathbb Z\right\}=\mathbb Z,$$ see the question with the website http://math.stackexchange.com/questions/2472205. Few weeks ago Alkan (cf. Numbers of the form $x^2(x-1) + y^2(y-1) + z^2(z-1)$ with $x,y,z\in\mathbb Z$) conjectured that -$$\left\{\frac{x^2(x-1)}2+\frac{y^2(y-1)}2+\frac{z^2(z-1)}2:\ x,y,z\in\mathbb Z\right\}=\mathbb Z.$$ -I think it's interesting to find a cubic polynomial $P(x)$ with integer coefficients such that -$$\{P(x)+P(y)+P(z):\ x,y,z\in\mathbb Z\}=\mathbb Z.$$ -This led me to pose the following conjecture. -Conjecture. Each $m\in\mathbb Z$ can be written as a sum of three numbers of the form $x^3-2x\ (x\in\mathbb Z)$. In other words, we have -$$\{x^3-2x+y^3-2y+z^3-2z: x,y,z\in\mathbb Z\}=\mathbb Z.$$ -As $P(x)=x^3-2x$ is an odd function, the conjecture can be reduced to the case $m\in\mathbb N=\{0,1,2,\ldots\}$. Via computation I found that those natural numbers $n\le1000$ not in the set -$$\{x^3-2x+y^3-2y+z^3-2z:\ x,y,z\in\{-1000,\ldots,1000\}\}$$ -are -\begin{gather}70,\ 75,\ 83,\ 86,\ 139,\ 185,\ 198,\ 237,\ 253,\ 262,\ 275, -\ 305,\ 338,\ 355,\ 362, -\\397, 414,\ 415,\ 422,\ 426,\ 457,\ 458,\ 509,\ 535,\ 558,\ 562,\ 564,\ 580,\ 583, -\\ 593, \ 613,\ 614,\ 635,\ 642,\ 673,\ 677,\ 684, \ 693,\ 697,\ 722,\ 735,\ 779,\ 782, -\\ 790,\ 791,\ 793,\ 807,\ 818,\ 850,\ 851,\ 870,\ 888,\ 898,\ 908,\ 943,\ 957. -\end{gather} -Let $S$ denote the set of these numbers. -QUESTION. Can we find an explicit solution of the equation -$$n=x^3-2x+y^3-2y+z^3-2z\ \ (x,y,z\in\mathbb Z)$$ for each $n\in S$? - -REPLY [11 votes]: 0 = P(7) + P(10) + P(-11) - = P(3250) + P(2293) + P(-3593) - = P(6266) + P(13243) + P(-13695) - = P(11700) + P(13277) + P(-15797) - = P(37555) + P(131381) + P(-132396) - = P(747511) + P(1059490) + P(-1171307) - = P(5529835) + P(22681597) + P(-22790636) - = P(8042677) + P(13682243) + P(-14552100) - = P(14270088) + P(39054467) + P(-39679475) - = P(29292092) + P(81358953) + P(-82605425) - = P(42588445) + P(291524359) + P(-291827018) - = P(56973565) + P(71715599) + P(-82119294) - = P(35977605) + P(866776048) + P(-866796709) - = P(143141833) + P(102053460) + P(-158684449) - = P(784428376) + P(3091918585) + P(-3108657737) - = P(129810373) + P(136917575) + P(-168147294) - - Though I expanded the search range to 10^10, no solution for n=558 was found. - On the other hand, there are many solutions for n=1 below. - - 1 = P(1439) + P(2554) + P(-2698) - = P(-1506) + P(-2432) + P(2611) - = P(-5214) + P(-11006) + P(11383) - = P(-8516) + P(-17400) + P(18055) - = P(13952) + P(70243) + P(-70426) - = P(18457) + P(10233) + P(-19451) - = P(18949) + P(56163) + P(-56873) - = P(21394) + P(107636) + P(-107917) - = P(21599) + P(61917) + P(-62781) - = P(75215) + P(256620) + P(-258756) - = P(132479) + P(517316) + P(-520196) - = P(525599) + P(2589115) + P(-2596315) - = P(697638) + P(803074) + P(-950033) - = P(-140064) + P(-198656) + P(219583) - = P(-198846) + P(-913333) + P(916464) - = P(-257810) + P(-1509380) + P(1511883) - = P(-617569) + P(-1930917) + P(1951749) - = P(-887510) + P(-1092290) + P(1260399) - = P(-931224) + P(-1288696) + P(1433823) - = P(1384739) + P(2458622) + P(-2597096) - = P(1602719) + P(9519294) + P(-9534414) - = P(4092479) + P(28437689) + P(-28465913) - = P(4875121) + P(2381859) + P(-5057717) - = P(9192959) + P(73135432) + P(-73183816) - = P(-1288696) + P(-931224) + P(1433823) - = P(-6063625) + P(-20241211) + P(20420995) - = P(-6919820) + P(-21816096) + P(22045735) - = P(-8121991) + P(-32025689) + P(32198879) - = P(18740159) + P(167927031) + P(-168004791) - = P(24544311) + P(124666228) + P(-124982552) - = P(62900639) + P(689911189) + P(-690085429) - = P(96931304) + P(198453683) + P(-205880474) - = P(-11745176) + P(-17900062) + P(19447931) - = P(-20241211) + P(-6063625) + P(20420995) - = P(-24301082) + P(-68349676) + P(69358667) - = P(-41154429) + P(-47640292) + P(56234034) - = P(-42083576) + P(-117387233) + P(119163144) - = P(-95843081) + P(-181052723) + P(189595899) - = P(106254719) + P(1271978124) + P(-1272225228) - = P(123437629) + P(177749151) + P(-195715037) - = P(-119444557) + P(-275690964) + P(282970698) - = P(-120282709) + P(-113449262) + P(147367664) - = P(-169017105) + P(-182314167) + P(221641415) - = P(-181052723) + P(-95843081) + P(189595899) - = P(-190571214) + P(-1169296181) + P(1170981088) - = P(-1129360025) + P(-3749403040) + P(3783251348) - - Solution for n=338 was found using LLL algorithm for X^3+Y^3=1. - 338= P(109043424)+ P(223729659)+ P(-232050701) - Only n=558 remains. - - Added new solutions. - 422= P(31441077)+ P(52488141)+ P(-56007428) - 426= P(-11575473)+ P(-42374626)+ P(42660619) - 509= P(4620839)+ P(7911642)+ P(-8405584) - 583= P(-2697799)+ P(-3187069)+ P(3732685) - - - Added new solutions. - 185=P(-14114372)+ P(-283189)+ P(14114410) - 198=P(-142960)+ P(-613349)+ P(615927) - 614=P(-412307)+ P(-16619)+ P(412316) - 793=P(-296708)+ P(-387970)+ P(438851) - - - - 262 = P(10239) + P(5400) + P(-10717) - 275 = P(38314) + P(4857) + P(-38340) - 305 = P(8535) + P(5187) + P(-9131) - 355 = P(2568) + P(982) + P(-2615) - 362 = P(6547) + P(636) + P(-6549) - 397 = P(-2029) + P(-973) + P(2101) - 414 = P(1059) + P(576) + P(-1113) - 457 = P(-7709) + P(-6134) + P(8832) - 535 = P(-11999) + P(-2241) + P(12025) - 562 = P(-3435) + P(-862) + P(3453) - 564 = P(-848) + P(-751) + P(1011) - 580 = P(-2295) + P(-825) + P(2330) - 593 = P(1563) + P(458) + P(-1576) - 613 = P(18873) + P(1623) + P(-18877) - 635 = P(10566) + P(9745) + P(-12816) - 642 = P(-5020) + P(-3871) + P(5693) - 673 = P(4487) + P(566) + P(-4490) - 677 = P(5967) + P(1087) + P(-5979) - 684 = P(4316) + P(2750) + P(-4660) - 693 = P(3575) + P(702) + P(-3584) - 697 = P(-17181) + P(-2952) + P(17210) - 722 = P(-1051) + P(-311) + P(1060) - 735 = P(1934) + P(1460) + P(-2179) - 779 = P(3781) + P(1593) + P(-3873) - 790 = P(-152491) + P(-8563) + P(152500) - 791 = P(11265) + P(8599) + P(-12735) - 818 = P(2003) + P(874) + P(-2057) - 850 = P(9047) + P(1510) + P(-9061) - 851 = P(1105) + P(264) + P(-1110) - 870 = P(6390) + P(1917) + P(-6447) - 888 = P(3928) + P(1444) + P(-3992) - 898 = P(1709) + P(929) + P(-1796) - 908 = P(4950) + P(4172) + P(-5788) - 943 = P(-5848) + P(-3743) + P(6320) - 957 = P(-4297) + P(-3091) + P(4775)<|endoftext|> -TITLE: Is $\sum_{k=1}^{n}\frac{(n-1)!}{(k-1)!}$ composite for $n\geq 4$? -QUESTION [12 upvotes]: Define $a_n$ as follows: -$$ -a_1=1,\ \ a_{n+1}=na_n+1\ -$$ -At this time, the sequence $a_n$ is as follows: -$$ -a_n=\sum_{k=1}^{n}\frac{(n-1)!}{(k-1)!} -$$ -I made some discoveries about this sequence. -The first:$$a_k\equiv 0\pmod{m}\Rightarrow a_{k+Nm}\equiv 0\pmod{m}~~~~\forall k,m,N\in\mathbb{N}$$ -The second:$$ n\geq 4\,\Rightarrow\,a_n ~\mathrm{is~composite} $$ -I was able to prove the first, but not the second. My expectation is that the second is correct, but I'm not sure it can be proved. My friend used computer and check $a_n$ is composite for $4\leq n\leq 48$. After $a_{49}$, it is too large number to check on his computer. Please let me know if you come up with a proof method. Any help is welcome! -(I am a Japanese college student. I'm sorry for my poor English.) - -REPLY [8 votes]: $a_n$ is composite for $4 \le n \le 2016$. -$a_{2017}$ appears to be prime (it passes a strong pseudoprime test). I have not tried to certify that it is prime (this would take a while as the number has 5789 digits).<|endoftext|> -TITLE: Grothendieck derivators vs $\infty$-categories -QUESTION [20 upvotes]: I have some questions on derivators and $(\infty,1)$-categories, -I would be grateful if someone could help me. - -Is there some problems that $(\infty,1)$-categories/derivators can resolve but derivators/$(\infty,1)$-categories cannot resolve? - -Why do so many people prefer $(\infty,1)$-categories than Grothendieck derivators? - -Is there a good place to learn about $(\infty,1)$-categories than Grothendieck derivators but with a historical and comparing point of view? - - -Thank you in advance!! - -REPLY [21 votes]: The short answer is that $(\infty,1)$-categories are the "real" object of interest. Derivators are a tool for working with them that is sometimes (for some people) easier to use, but doesn't remember all the information and hence is not always applicable. -When you work with $(\infty,1)$-categories, you have to deal explicitly with higher-dimensional coherence all the time. Everything is determined only up to equivalence. This can be kind of a pain, so it's convenient to have 1-categorical structures that neverthless carry $\infty$-categorical information. -The classical kind of 1-categorical structure used for this is a Quillen model category. This carries "too much" information, in that two objects can be equivalent in an $(\infty,1)$-category but not isomorphic in the model category it presents, and similarly two model categories can present equivalent $(\infty,1)$-categories but not be equivalent categories. Thus a model category needs a notion of "weak equivalence" between its objects, and similarly we have a notion of "Quillen equivalence" between model categories. But a model category contains all the information of an $(\infty,1)$-category, and so we can work with all the higher coherences as necessary. -A derivator is sort of a "dual" to a model category: it carries "too little" information. An advantage is that it is not subject to the issues of weak equivalence: two objects of an $(\infty,1)$-category are equivalent if and only if they are isomorphic in the corresponding derivator, and similarly two (locally presentable) $(\infty,1)$-categories are equivalent if and only if their corresponding derivators are equivalent (in a 1-categorical sense: derivators are the objects of a 2-category just like 1-categories are, and here we mean equivalence internal to that 2-category). -But a derivator doesn't have enough information to do everything we might want to with higher coherences. It's essentially an enhancement of the homotopy 1-category (the quotient by homotopies) that remembers the notion of homotopy-coherent diagram, and therefore also of homotopy limit and colimit (and homotopy Kan extension). This is sufficient for a surprising amount of homotopy coherence, at least when working only within a single $(\infty,1)$-category. But there are some things it can't do (or not very well), notably those that involve diagrams of $(\infty,1)$-categories and things like functor $(\infty,1)$-categories. -So the answer to your first question is yes, there are things you can do with $(\infty,1)$-categories but not with derivators; but no, anything you can do with a derivator can also be done with an $(\infty,1)$-category. And I think this mostly answers your second question as well: given that derivators are not good enough for everything, even if they make certain things easier, it's understandable that many people prefer not to learn two different languages and stick to $(\infty,1)$-categories even if they happen to be doing something that might be a bit easier with derivators. (On the other hand, I personally feel that it is easier to be sloppy with $(\infty,1)$-categories --- partly because being precise is so much work --- and easier to be precise with derivators, which is one reason that I still use the latter sometimes.) -Historically, derivators were introduced (by Heller, Grothendieck, and Franke, fairly independently) before practical notions of $(\infty,1)$-category were available. So at the time they were the only way of getting rid of the "weak equivalences"; but even with that advantage they never really caught on. I'm not quite sure why not; perhaps the requisite 2-category theory was also offputting at the time? -As for references, I don't have any suggestions other than those on the nLab page. I particularly like the expository aspect of Moritz Groth's work. You may also be interested in this blog post of mine.<|endoftext|> -TITLE: NP-hardness of a sequence problem -QUESTION [5 upvotes]: Given $n$ binary sequences $s_i$ ($1\le i\le n$) with common period $T$. Let $s_i^{t_i}$ denote the sequence obtained by cyclically shifting $s_i$ for $t_i$ bits. The $n$ sequences form a good system if under any combination of $\{t_i\}_{i=1}^n$, for each sequence $s_i$ there always exists $\tau_i$ such that $s_i^{t_i}(\tau_i)=1$ and $s_j^{t_j}(\tau_j)=0$ for $j\ne i$. For example, $s_1=1010$ and $s_2=1100$ is a good system, while $s_1=0001$ and $s_2=1000$ is not a good system. -Is the problem of deciding whether a system is good NP-hard? -The background of the problem is below. We want to design a code for each of the $n$ users. User $i$ with code $s_i$ transmits its packet in slot $t$ if $s_i(t)=1$. We want to check whether a set of codes can ensure that even the users are not time-synchronized, each of them can successfully transmit a packet under any clock drift among users. If two or more users transmit at the same slot, none of them succeeds. - -REPLY [4 votes]: I assume you mean "there always exist $\tau_i$ such that $s_i^{t_i}(\tau_i) = 1$ and $s_j^{t_j}(\tau_i) = 0$ for $i \neq j$", i.e. you want that no matter how the sequences are shifted, each sequence has at least one bit which is zero in the other shifted sequences, and that's the slot when it manages to send its packet in your application. -(What you have written currently is "there always exist $\tau_i$ such that $s_i^{t_i}(\tau_i) = 1$ and $s_j^{t_j}(\tau_j) = 0$ for $i \neq j$". If they are chosen separately for each $i$ this just means each of the $s_i$ contain both $1$ and $0$. If they are chosen once and for all, this is impossible unless $n = 1$.) -Your problem as I interpret it is clearly in co-NP, as you check that all ($\forall$) shifts satisfy a (polynomial-time checkable) constraint, so it is probably not NP-hard, as that would collapse the polynomial hierarchy. I'll complement your problem and sketch a proof of NP-hardness of the resulting problem, meaning your problem is co-NP-complete. -Notation: On the set $X = \{0,1\}^{\mathbb{Z}_T}$ we have the shift action of $\mathbb{Z}_T = \mathbb{Z}/T\mathbb{Z}$ by $\sigma(s)_i = s_{i+1}, \sigma : X \to X$. For $s, s' \in X$ define $(s \cup s')_i = \max(s_i, s'_i)$. Write $s \leq s'$ for $\forall i: s_i \leq s'_i$. -The complemented problem: Consider a set of sequences $S = (s_i)_i$, $s_i \in X$. We say $i$ is a bad index for $S$ if $s_i \leq \bigcup_{j \neq i} \sigma^{t_j}(s_j)$ for some $t_j \in \mathbb{Z}_T$. We say $S$ is bad if there exists a bad index. Clearly $S$ is bad if and only if it is not good. The problem we prove NP-complete is identifying bad sets of sequences. -First, we will make sure $i = 1$ is the only possible bad index, i.e. $s_1$ is the only sequence that could possibly be the union of the others. For this, we will put an arithmetic progression $a_i$ in $s_i$, $i > 1$. This progression should be longer than $n$ and such that any other $s_j$ covers at most one element of it. I'll write some formulas for completeness. -Pick some $M$ (a parameter for future purposes). If $a_i$ is the sequence with support $\{kM(n^2+i) \;|\; k = 0,1,...,n+1\}$, then any shift of $a_i$ covers at most one position of any other $a_{i'}$: if $kM(n^2+i) = k'M(n^2+i')$, $k, k' \in \{1, ..., n+1\}$, $i, i' \in \{2, ..., n\}$ and $i' > i$, then $k/k' = (n^2+i')/(n^2+i) \in (1, \frac{n^2+n}{n^2+2}] \subset (1, \frac{n+1}{n})$, but clearly $k/k' > 1 \implies k/k' \geq (n+1)/n$. Now just include $a_i \leq s_i$ for each $i \geq 2$, and make sure that all other remaining things we include in the sequences $s_i$ fit within a single interval of length $Mn^2$ which is sufficiently far from $0$ (pick e.g. $T = 100 M n^3$ and there's plenty of space left, since the total length of $a_i$ is less than $2Mn^3$). -Now, consider a SAT instance with $n-1$ variables and clauses, $x_i, \phi_i, i \in \{2,...,n\}$. To reduce SAT, we want $\exists$ to have to make a binary choice for each $i > 1$, which will represent a choice between $x_i$ and $\neg x_i$. Pick arithmetic progressions $b_i$ similarly as we did with $a_i$ (but on a smaller scale; pick a suitable $M$ so we can do all that follows in an interval of length $Mn^2$ as we promised ourselves in the previous paragraph). The sequence $s_1$ contains one copy of $b_i$ while $s_i$ contains two copies of $b_i$ at distance $h$ from each other. If $\exists$ is to win, the copy of $b_i$ in $s_1$ has to be covered by one of the copies in $s_i$ (note that as long as $b_i$ fits in an interval of length $Mn^2$, the existing $a_j$-bits in the $s_j$ are not helpful for covering it). -Now, we can add for each clause of the SAT instance a single bit in $s_1$. These bits are in arithmetic progression with distance $2h$ between them. Depending on whether $x_j$ or $\neg x_j$ appears in the clause (or neither), we put a $1$ in the position in $s_j$ such that the correct clause bit is covered. (The bits coming from the choice $x_i = \top$ do not touch any clause bits if we choose the $x_i = \bot$ alignment for $s_i$, since this gives only a displacement of $h$; and vice versa for the $x_i = \top$ alignment.)<|endoftext|> -TITLE: Enriched vs ordinary filtered colimits -QUESTION [19 upvotes]: Filtered categories can be defined as those categories $\mathbf{C}$ such that $\mathbf{C}$-indexed colimits in $\mathrm{Set}$ commute with finite limits. -Similarly, for categories enriched in $\mathbf{V}$ (where the appropriate notion of colimits is colimits weighted by enriched presheaves) one can define a presheaf $W \colon \mathbf{C}^{\mathrm{op}} \rightarrow \mathbf{V}$ to be ($\kappa$-)flat if $W$-weighted colimits in $\mathbf{V}$ commute with finite ($\kappa$-small) limits in $\mathbf{V}$ (for some regular cardinal $\kappa$). Borceux, Quinteiro, and Rosický take this as a starting point to develop a theory of accessible and presentable $\mathbf{V}$-categories in their paper "A theory of enriched sketches". -BQR show that in some ways flat weighted colimits are closely related to ordinary (conical) filtered colimits. For example, they show that if $\mathbf{C}$ has finite ($\kappa$-small) weighted limits, then a presheaf on $\mathbf{C}$ is ($\kappa$-)flat if and only if it is a ($\kappa$-)filtered ordinary colimit of representable presheaves. However, they give a counterexample that shows this need not be true for arbitrary $\mathbf{C}$ - but in this example it is still true that flat presheaves are filtered colimits of absolute colimits of representables. -Question 1: A $\kappa$-filtered ordinary colimit of absolute colimits of representables is always a $\kappa$-flat presheaf. Is anything further known (or expected) about the other direction, i.e. whether every $\kappa$-flat presheaf can be decomposed as such a colimit (or some variant involving two cardinals)? -Let me add a second closely related question that indicates why one might care about the first one. BQR prove that if $\mathbf{M}$ is a presentable $\mathbf{V}$-category then its underlying ordinary category is also presentable. -Question 2: Suppose $\mathbf{M}$ is a cocomplete $\mathbf{V}$-category whose underlying category is presentable. Does this imply that $\mathbf{M}$ is a presentable $\mathbf{V}$-category? -(This would be the case if the two classes of presheaves in the first question coincide.) - -REPLY [5 votes]: We deal with Question 1 in this paper, joint with Steve Lack. -In general it is not true that every $\kappa$-flat presheaf lies in the closure of the resperesntables under $\kappa$-filtered colimits and absolute colimits. That's the case for example when $\mathcal V=\mathbf{Set}^G$, for a non-trivial finite group $G$ (see Section 5 of the paper). -However, there are many bases of enrichments for which the equality holds. In fact, whenever $\mathcal V$ is locally $\kappa$-presentable as a closed category and $\mathcal V(I,-)$ is (weakly) cocontinuous and (weakly) strong monoidal, then every $\kappa$-flat presheaf is a $\kappa$-filtered colimit of representables. (In this case absolute colimits reduce to the usual splittings of idempotents.) Therefore the existence and preservation of $\kappa$-flat colimits is equivalent to that of $\kappa$-filtered colimits in any $\mathcal V$-category (for $\mathcal V$ as above). Examples of such bases are the cartesian closed categories $\mathbf{Set}$ of sets, $\mathbf{Cat}$ of small categories, $\mathbf{SSet}$ of simplicial sets, $\mathbf{2}$ of the free-living arrow, $\mathbf{Pos}$ of partially ordered set, etc. -We also investigate the case when $\kappa$-flat colimits are generated by $\kappa$-filtered colimits plus other absolute colimits (such as finite direct sums and copowers by dualizable objects). Examples of $\mathcal V$ for this setting include the monoidal categories $\mathbf{CMon}$ of commutative monoids, $\mathbf{Ab}$ of abelian groups, $R\textit{-}\mathbf{Mod}$ of $R$-modules for a commutativering $R$, and $\mathbf{GAb}$ of graded abelian groups.<|endoftext|> -TITLE: Conjecture about minimal number of edge crossings in complete bipartite graphs -QUESTION [5 upvotes]: I am interested in the status of the conjecture about the minimum number of edge crossings $cr(K_{m,n})$ in a drawing of the complete bipartite graph $K_{m,n}$. -The Wikipedia article https://en.wikipedia.org/wiki/Tur%C3%A1n%27s_brick_factory_problem led me to study the original papers of Zarankiewicz (On a problem of P. Turan concerning graphs) from 1954 and of Urbanik (Solution du problème posé par P. Turán) from 1955. -I wondered whether someone could tell me whether an asymptotic approach has been successfully attempted (letting $n\to\infty$). If so, I would be very interested in any references for that. - -REPLY [7 votes]: The Electronic Journal of Combinatorics has many Dynamic Surveys one of which is The Graph Crossing Number and its Variants: A Survey by Schaefer which first appeared in 2013 and has been updated as recently as Feb 14, 2020. From the bottom of page 40 onto page 41 you will find this conjecture for complete bipartite graphs discussed (with many references). As far as I can tell from the survey the conjecture is open (both for exact values and asymptotically). -One paper you might be interested in is Zarankiewiczʼs Conjecture is finite for each fixed $m$ by Christiana, Richter, and Salazar. This paper shows that for each $m$ if the conjecture holds up for all $n$ up to some very large $N(m)$ (which is an explicit value), then the conjecture is true for $K_{n,m}$ with any $n$. -This survey also references another survey Turán’s Brick Factory Problem: The Status of the Conjectures of Zarankiewicz and Hill by Székely. (I haven't be able to access this survey so I don't exactly what's inside.)<|endoftext|> -TITLE: Proof of a 'well-known' result of Shimura on periods of modular forms -QUESTION [8 upvotes]: It is often noted in the literature that there are certain complex periods that allow one to normalize the modular symbol associated to a modular form in such a way that its coefficients are algebraic. This process of normalization by complex periods is regularly attributed to Shimura, though I can't seem to find a concrete reference explaining this result. -More precisely, let $ -\Gamma=\Gamma_0(N)$ and fix an eigenform $f\in S_k(\Gamma)$. The modular symbol $\xi_f\in \operatorname{Hom}_{\Gamma}(\operatorname{Div}^0(\mathbb{P}^1(\mathbb{Q})),V_{k-2}(\mathbb{C}))$, where $V_{k-2}(\mathbb{C})$ is the space of homogeneous polynomials with complex coefficients of degree $k-2$, attached to $f$ is defined by -$$ -\xi_f(\{r\}-\{s\})=2\pi i \int_s^r f(z)(zX+Y)^{k-2}dz. -$$ -One can expand this into a homogeneous polynomial $\sum_{j=0}^{k-2} c_jX^jY^{k-2-j}$ where $c_j=\binom{k-2}{j}2\pi i \int_s^rf(z)z^jdz$. -The matrix $\begin{pmatrix} -1 &0\\ 0&1\end{pmatrix}$ normalizes $\Gamma$, -so the modular symbols come equipped with an involution, and hence there is a unique eigenspace decomposition $\xi_f=\xi_f^++\xi_f^-$, with $\xi^\pm$ in the $\pm 1$-eigenspace. -The following theorem is stated in the literature (see, for example, [Greenberg-Stevens, 3.5.4], [Bertolini-Darmon,1.1], or [Pollack-Weston,page 7]). -Theorem. There exists complex numbers $\Omega_f^\pm$ such that $\xi_f^\pm/\Omega_f^\pm$ takes values in $V_{k-2}(K_f)$, where $K_f$ is the number field generated by the Fourier coefficients of $f$. -Greenberg-Stevens cite this 1977 paper of Shimura, Pollack-Weston cite Shimura's book on automorphic functions, and the Bertolini-Darmon does not give a reference. I could not find anything helpful in Shimura's automorphic function book, but I think theorem 1 from the 1977 paper is probably what we want. For simplicity, I state it below in the case where $f$ has rational coefficients. -Theorem. (Shimura, Theorem 1) Fix a primitive Dirichlet character $\chi$. There exist complex numbers $u_f^\pm$ such that -$$ -\frac{L(f_\chi,j)}{u_f^\epsilon\tau(\chi)(2\pi i)^j}\in K_fK_\chi -$$ -where $0< j< k$, $\epsilon$ is the sign of $\chi(-1)(-1)^j$, $\tau(\chi)$ is the classical Gauss sum, and $L(f_\chi,s)=\sum\chi(n)a_nn^{-s}$ is the $L$-function of $f$ twisted by $\chi$. -In fact, Shimura gives precise (though rather noncanonical) descriptions of these periods $u_f^\pm$: they are essentially the value of the $L$-function at $k-1$. - -I would like to know how the first theorem stated above follows from this theorem 1 of Shimura. - -It seems like a nontrivial exercise, or perhaps I am just having some trouble connecting the dots. I would also be content to see a reference which outlines a proof of the first theorem above. -My thoughts are roughly the following. With the notation as above, let $m$ be the conductor of $\chi$. I know that (see [Mazur-Tate-Teitelbaum, 8.6], for example) one has the following connection between coefficients of the modular symbols and special values of $L$-functions -$$ -\frac{j!}{(-2\pi i)^{j+1}}\frac{m^{j+1}}{\tau(\bar \chi)}L(f_{\bar\chi},j+1) =\sum_{a\in (\mathbb{Z}/m\mathbb{Z})^\times}\chi(a)\int_{-a/m}^{i\infty}f(z)(mz+a)^j dz, -$$ -for $0\leq j \leq k-2$. This tells us, for example, that certain weighted sums of the coefficients of $\xi_f(\{\infty\}-\{-a/m\})$ can be scaled to be algebraic. Even more, after writing down the symbols $\xi_f^\pm$, I can find periods $\Omega_f^\pm$ such that, roughly speaking, $$ -\frac{1}{\Omega_f^\pm}\sum\chi(a)(\text{$j$th coefficient of $\xi_f^\pm(\{\infty\}-\{a/m\}$}) ) -$$ -is algebraic, but again, this only tells me that (a) certain weighted sums of the coefficients are algebraic, and (b) only gives information about the modular symbol evaluated at $\{\infty\}-\{a/m\}$, which as far as I can tell, is not the generality needed for the first theorem above. -(I posted this question on MSE a few days ago, but did not have much luck there. I hope re-posting it here is not too much of a faux pas.) - -REPLY [2 votes]: You can find a proof of this theorem (the first in the OP) in Section 5.3 of the following paper by Pasol and Popa: https://arxiv.org/abs/1202.5802 -The idea is to use the action of Hecke operators. More precisely, the map $f \mapsto \xi_f^{\pm}$ is Hecke-equivariant, the Hecke operators preserve the rational structures of both sides, and the eigenspaces are 1-dimensional. -This theorem could also, in principle, be deduced from Shimura's theorem (Theorem 1 in the OP), but the proof I have in mind would be very technical. The idea is to start from the formula expressing the values $L(f,\chi,j+1)$ in terms of modular symbols and then take the inverse Fourier transform. But there are many technical problems due to the fact that the Dirichlet characters are not necessarily primitive, and Shimura's formula is a priori only for primitive characters. Nevertheless, in the case of weight 2, Merel has proved a completely general formula expressing modular symbols in terms of twisted $L$-values, see the article Symboles de Manin et valeurs de fonctions $L$.<|endoftext|> -TITLE: Characterization of metrics such that the volume of balls doesn't depend on their centers? -QUESTION [6 upvotes]: Given a finite metric space $(X,d)$, -when does it hold that for all $y\in X$ and $r>0$, $\#B(y,r)$ does not depend on $y$? -Here $ B(y,r):=\{x\in X: d(x,y)\le r\} $ denotes a ball of radius $r$ centered at $y$, -and $\#A$ denotes the cardinality of $A$. -An example that does not satisfy the above property is $Z$-distance on $\{0,1\}^n$. -It is defined as $ d_Z(x,y) := \max\{\#(\mathrm{supp}(x)\setminus\mathrm{supp}(y)),\#(\mathrm{supp}(y)\setminus\mathrm{supp}(x))\} $ for $x,y\in\{0,1\}^n$. -Here $ \mathrm{supp}(x):=\{i\in[n]:x_i\ne0\} $ denotes the support (set of nonzero locations) of a vector $x\in\{0,1\}^n$. -(The notation $Z$-distance is perhaps nonstandard. My motivation comes from coding theory, in particular error-correcting codes for $Z$-channels which only zero out bits but do not flip zeros to ones. One can check that $Z$-distance is indeed a metric.) -All normed spaces satisfy the above property. -It is easy to see that the above property is equivalent to the condition: for all $y,z\in X$ and $r>0$, $ \#(B(y,r)\setminus B(z,r)) = \#(B(z,r)\setminus B(y,r)) $. -However, I couldn't further simply it. -My question is: is there, or is it possible to have, a characterization of metrics satisfying the above property? - -REPLY [9 votes]: I do not know the answer in general, but in the case in which $X$ is a subset of $\mathbb{R}^n$ there are some known remarkable results: - -If $X\subset\mathbb{R}^2$ is bounded, then $X$ consists of vertices of -a regular $n$-gon or $X$ is a union of the vertices of two regular -$n$-gons having having the same center and radius. - - -If $X\subset\mathbb{R}^n$ is bounded, then $X$ is contained in a sphere. - -You can find proofs of these and other results in -B Kirchheim, D. Preiss, -Uniformly distributed measures in Euclidean spaces. -Math. Scand. -90 (2002), no. 1, 152–160. -The proofs are surprisingly difficult and deep. In view of these results I would expect that it is difficult to find a general and satisfactory answer.<|endoftext|> -TITLE: Compactly supported chern character -QUESTION [5 upvotes]: It is a standard result that for a CW complex $X$, the chern character -$$\text{ch}: K^*(X)\otimes_{\mathbb{Z}} \mathbb{Q}\to H^*(X,\mathbb{Q})$$ -induces an isomorphism. Suppose now that $X$ is an open manifold and consider the chern character with compact support -$$\text{ch}_{\text{cs}}: K^*_{\text{cs}}(X)\otimes_{\mathbb{Z}} \mathbb{Q}\to H^{*}_{\text{cs}}(X,\mathbb{Q})$$ -Is it still true that it is an isomorphism? It seems to be so for the case of $\text{Tot}(E\to S)$ where $S$ is a compact space and $E$ is a vector bundle. I suppose this is well known, but I couldn't find a reference. - -REPLY [6 votes]: Yes, this is true. For any generalized cohomology theory $E$, the compactly supported $E$-cohomology of a space $X$ is -$$E_{\mathit{cs}}^*(X) := \varinjlim\limits_{K\subseteq X:\text{ $K$ compact}} E^*(X, X\setminus K).$$ -The Chern character is a natural isomorphism of cohomology theories, so is compatible with the limit above, hence induces an isomorphism on compactly supported rational $K$-theory. -One reference for this definition of compactly supported generalized cohomology is Ranicki-Roe, “Surgery for Amateurs”, Remark 2.1.<|endoftext|> -TITLE: Ramsey's number R(4,4) with arithmetic progressions -QUESTION [6 upvotes]: Can 17 positive integers in arithmetic progression be found such that that no four of them have, pairwise, a common divisor greater than 1, but, likewise, no four of them are, pairwise, relatively prime? -Because R(4,4)=18, 18 such numbers are impossible. -At https://puzzling.stackexchange.com/questions/100391/seventeen-positive-integers/100477#100477 it has been shown that 17 numbers with the required property, but not necessarily in arithmetic progression, can be easily found based on Paley's graph of order 17. The question arises: could those 17 numbers be in arithmetic progression? - -REPLY [5 votes]: No, there cannot be 17 such numbers in arithmetic progression (and there cannot be 5 such numbers with the corresponding property for triples). -Suppose we have such an arithmetic progression of length $k$, say $x,x+d,\ldots,x+(k-1)d$. I claim that if a prime $p$ divides any two of them then either it divides all of them (which cannot be the case), or else $p -TITLE: How many squares can be formed by $n$ points in general position in the plane? -QUESTION [5 upvotes]: [This is much in the spirit (but different from) the questions from different posters: How many squares can be formed by using n points? and How many squares can be formed by using n points: revisited?] - - -Let $A$ be a set of $n$ points in the plane in general position. By general position we mean that no $3$ points are co-linear. What is the maximum number of squares that can be formed with vertices in $A$? - -I note that there are trivial upper and lower bounds for this problem: -[Trivial Upper Bound] Given $n$ arbitrary points in the plane, noting that any two points determine at most $3$ squares it follows that there are at most $O(n^2)$ squares with vertices in $A$. -[Trivial Lower Bound] Place four points at the corner of a square, and repeat taking care to avoid all lines generated by pairs of points already placed in the plane until we've placed $n$ points. This clearly gives a lower bound of $\Omega(n)$. -I can improve the implied constant in both the upper and lower bound by being a bit more clever. The problem, however, is to - -Improve (asymptotically) on either the upper or lower bound just given. - -REPLY [6 votes]: We can get a lower bound on the order of $n \log n$. -I'll describe how to arrange $4^n$ points in general position to get $n 4^{n-1}$ squares. -The arrangement is described recursively. For the base case $n=1$, we have $4^1 = 4$ points, and you can probably guess how we should arrange them to get $1 \cdot 4^{1-1} = 1$ squares. Now suppose we have an arrangement $A$ of $4^{n-1}$ points, in general position, giving us a total of $(n-1)4^{n-2}$ squares. Take $4$ copies of $A$ (a total of $4^n$ points). Place the $4$ copies of $A$ at the $4$ corners of a "large" square, and then rotate each copy of $A$ by a "random" angle $\theta$ (the same angle for each of our $4$ copies of $A$). This gives us our new arrangement of points. -If the square mentioned above is large enough, then no points from $3$ distinct copies of $A$ can lie on a line. And it is not hard to show that, with probability $1$, a randomly chosen $\theta$ will have the property that no two points from a given copy of $A$ will lie on a common line with a different copy of $A$. So for a "large" square and a "random" angle, we get a set of $4^n$ points in general position. -In each small copy of $A$, we get $(n-1)4^{n-2}$ small squares, for a total of $4(n-1)4^{n-2} = (n-1)4^{n-1}$ small squares in our new arrangement. In addition to these, we get $|A| = 4^{n-1}$ additional large squares, by connecting the $4$ corresponding points in each of our $4$ copies of $A$. This gives a total of $n4^{n-1}$ squares, as promised.<|endoftext|> -TITLE: Learning from unsuccessful attempts at the Poincaré conjecture -QUESTION [8 upvotes]: This is reposted from MSE, but perhaps it is more appropriate to post it here. Let me know if I'm wrong. -Among the many unsuccessful attempts at solving the Poincaré conjecture, I'm wondering if there was an approach that went along the lines of showing that a closed simply connected $3$-manifold could be endowed with a Lie group structure. This might not be enough to solve it, since there are other compact $3$-dimensional Lie groups (if I'm not wrong), but it's tempting to think about since the only spheres which are Lie groups are of dimension $0, 1$ or $3$. This makes it feel like there is this special additional structure that can be used for the case $n=3$ as opposed to solving the generalized Poincaré conjecture, where no higher dimensional spheres are Lie groups. -So I'm wondering if someone tried this. If they did, what hurdle/subtlety did they find in the course of the attempt that prevented a solution? Did we learn something interesting from it? - -REPLY [15 votes]: Thurston approaches 3-manifolds by cutting them up along various surfaces (one first cuts along spheres [Kneser-Milnor] and then along tori [Jaco-Shalen-Johannson]) into pieces which each admit a locally homogeneous geometric structure, modelled on a homogeneous space with an invariant Riemannian metric. A compact, simply connected manifold with such a structure is homogeneous. But a Lie group is homogeneous under its own left translations (or right translations), with many invariant Riemannian metrics (given by left, or right, translation of any positive definite inner product on any one tangent space. So Thurston's programme contains your programme as a special case. -Thanks to Allen Hatcher for correcting my description of the decomposition.<|endoftext|> -TITLE: Does Rademacher's convergent series for p(n) define an analytic function? -QUESTION [22 upvotes]: Let $p(n)$ be the number of partitions of $n\geq 0$. We can let $n$ be -any complex number in Rademacher's convergent infinite series for -$p(n)$. (See e.g. equation (24) here.) -For what $n$ does it converge? Does it define an analytic function for -such $n$? If so, what are the properties of this analytic function -(singularities, branch points, domain of existence, etc.)? This question -asked for an analytic continuation of $p(n)$, so I am suggesting a -possible answer. - -REPLY [20 votes]: Not a direct answer to the question, but a brief numerical exploration of this function. -First, a trivial observation: we can write either $e^{\pi i x}$ or $\cos(\pi x)$ in the formula for the exponential sum $A_k(n)$ in the Rademacher series. This makes no difference at integer $n$, but gives different generalizations for noninteger $n$. The cosine version has the nice property of being real-valued on the real line and conjugate symmetric. -Here is a plot of the cosine-extended $p(n)$ on the real line: - -The exponential version (real and imaginary parts): - -Either version of the function seems to have simple zeros at the negative integers $-1, -2, -3, ...$. (This is quite nice if correct, because it matches the obvious combinatorial interpretation $p(-n) = 0$.) The cosine version has additional zeros on the negative real line (the first near $-0.93$). -At half-integers, it appears that all terms in the cosine version of the Rademacher series except the first term vanish, and so one has a trivial closed-form evaluation of $p(k+\tfrac{1}{2})$, $k \in \mathbb{Z}$. This is not the case for the exponential version. -Taking this leading term as a cue for the asymptotics on the real line, the origin is a turning point between exponential growth to the right and $\text{oscillation} \cdot O(n^{-1})$ behavior to the left. -Viewed in the imaginary direction, the exponential version appears to grow exponentially as $n \to +i \infty$ but remains small as $n \to -i \infty$. Plot of the real and imaginary parts of $p(i x)$: - -The cosine version looks like the upper-half-plane exponential version: - -There are additional zeros in complex plane. Since the Rademacher series converges slowly when the imaginary part of $n$ is large, it's a bit difficult to explore these zeros numerically. Here is a low-resolution plot of the exponential version of $p(z)$ on $z \in [-4,4] + [-2,2] i$: - -And the cosine version: - -The slow convergence also makes it difficult to search numerically for other potential closed forms (it is expensive to get more than ~6 digits). A faster algorithm for computing $p(n)$ to high precision near the origin would be very exciting. -Unoptimized Python implementation that I used to create these plots: https://gist.github.com/fredrik-johansson/7c2711887811ef9f2d7038b8451a4e63<|endoftext|> -TITLE: Presentations of $\mathbf{PGL}_3(\mathbb{F}_2)$ by three involutions -QUESTION [5 upvotes]: I am searching for (two) presentations of the group $\mathbf{PGL}_3(\mathbb{F}_2)$ for which the generators are involutions $a, b, c$, and such that the following relations are present [among extra relations, of course], in two separate cases: -REP. 1: $(ab)^4 = 1, (bc)^4 = 1, (ac)^2 = 1$; -REP. 2: $(ab)^3 = (bc)^3 = (ac)^3 = 1$. -Can anyone help me out here ? (Thanks !!) - -REPLY [4 votes]: There is a pretty direct argument here. Note first that $G = {\rm PGL}(3,2) = {\rm SL}(3,2)$ is a finite simple group of order 168. Suppose it were so generated. -Note next that $a$ inverts both $ab$ and $ac$, so normalizes $\langle ab, ac \rangle$. -Now $bc = (ca)(ba)^{-1} \in \langle ab,ac \rangle$, so that $a$ normalizes $\langle ab,bc,ac \rangle.$ The reason I leave the redundant generator visible is that by symmetry, it follows that $b$ and $c$ both normalize $\langle ab,bc,ac \rangle.$ -Thus $\langle ab,ac,bc \rangle$ is normal in $\langle a,b,c \rangle = G.$ -Hence $\langle ab, bc, ac \rangle$ is either trivial or all of ${\rm PGL}(3,2)$, by simplicity. -Triviality of the subgroup $\langle ab,bc,ac \rangle$ yields $a = b = c$, so we have a contradiction. -Suppose then that $\langle ab,bc,ac \rangle = G.$ I give the complete argument in the case of REP2: let $x = ab, y = bc, z = ac.$ Then $x^{3} = y^{3} = z^{3}= 1$, by assumption, and $xy = z.$ The following argument was known to W. Burnside. -Now $\langle x^{-1}y, yx^{-1} \rangle$ is an Abelian normal subgroup of $G$, because $(xyx)(yxy) = 1$, as $xy$ has order $3$, yielding $yxy = (xyx)^{-1} = (x^{-1}y)(yx^{-1})$, while also $(yx^{-1})(x^{-1}y) = yxy$ because $x$ and $y$ have order 3. -Thus $x^{-1}y$ and $yx^{-1}$ commute. But $x(x^{-1}y)x^{-1} = yx^{-1},$ so that $x$ normalizes $\langle x^{-1}y, yx^{-1} \rangle$. Since $x^{-1}y$ certainly normalizes $\langle x^{-1}y, yx^{-1} \rangle,$ we see that $\langle x^{-1}y, yx^{-1} \rangle \lhd \langle x,y \rangle = G$, a contradiction. -There is a similar well-known argument to exclude the case of REP1, since then $\langle ab,ac,bc \rangle$ is a "$(2,4,4)$"-group -Later edit: Instead of using a generators and relations type argument, it is also possible to see via a character calculation that if $u,v,w$ are elements of order $4,4$ and $2$ respectively in $G$ with $uv = w$, then $\langle u,v \rangle$ is a cyclic $2$-group. -For $G$ has 6 complex irreducible characters of degrees $1,3,3,6,7,8$. Let $u$ be an element of order $4$ of $G$ and $w$ be an involution (there is a single class of elements of order $4$ and a single class of involutions in $G$). Only the irreducible characters $\chi$ of degrees $1,3,3$ and $7$ have $\chi(u)\chi(w) \neq 0$, and we find from the well-known character formula that the number of times $w$ is expressible as a product of two conjugates of $u$ in $G$ is $\frac{168}{16} [ 1 -\frac{1}{3}-\frac{1}{3} - \frac{1}{7}] = 2$. On the other hand, if we take $D = C_{G}(w)$ to be a Sylow $2$-subgroup in which $w$ is central, then $w$ is already expressible as $x^{2}$ and as $y^{2}$, where $x$ and $y$ are the two elements of order $4$ in $D$. -Hence whenever $w$ is expressible as a product of two elements of order $4$ in $G$, we see that these two elements of order $4$ are equal. -But rep 1 gives elements $ab, bc$ of order $4$ with $(ab)(bc) = ac$ of order $2$, and, as we have seen, we should have $\langle ab,bc \rangle = G$, a contradiction.<|endoftext|> -TITLE: Issue UPDATE: in graph theory, different definitions of edge crossing numbers - impact on applications? -QUESTION [183 upvotes]: QUICK FINAL UPDATE: Just wanted to thank you MO users for all your support. Special thanks for the fast answers, I've accepted first one, appreciated the clarity it gave me. I've updated my torus algorithm with ${\rm cr}(G)$. Works fine on my full test set, i.e. evidence for ${\rm cr}(G)={\rm pcr}(G)$ on torus. More on this later, will test sharper bound from last answer as well. I'm going to submit in time! Thanks again MO users for all your help! - -Original post: -I apologize if „crisis“ is too strong a word, but I am in a mode of panic, if that's the right word: In two weeks, I should be submitting my Ph.D. Thesis, but I have just received bad news, or I should say information that makes me very concerned. It is really an emergency situation: -My thesis is in computer science, algorithms related to graph drawings on the sphere and the torus. One of the cornerstone mathematical results I am relying on is the graph edge crossing lemma (or edge crossing inequality). It gives a lower bound for the minimum number of edge crossings ${\rm cr}(G)$ for any drawing of the graph $G$ with $n$ vertices and $e$ edges -$${\rm cr}(G)\geq \frac{e^3}{64n^2}$$ -for $e>4n$. - -PROBLEM: I am reading in the article of Pach and Tóth that there is a possibility that mathematics papers on crossing numbers operate with different definitions. There is the crossing number ${\rm cr}(G)$ (minimum of edge crossings in a drawing of $G$), but also the pair crossing number -${\rm pcr}(G)$, the minimum number of edge pairs crossing in a drawing of $G$. -I double-checked my algorithms and, based on this definition, I clearly apply the pair crossing number ${\rm pcr}(G)$ - - -CRITICAL QUESTION: Can you confirm to me that the edge crossing lemma remains valid on the sphere and the torus also for the pair crossing number ${\rm pcr}(G)$? - -Reference: János Pach and Géza Tóth. Which crossing number is it anyway? J. Combin. Theory Ser. B, 80(2): 225–246, 2000. -And Wikipedia article as a starting point https://en.wikipedia.org/wiki/Crossing_number_inequality - -REPLY [28 votes]: @user161819 I wanted to make a comment but it got too long, so putting it as an answer. But please take it just as a comment for later, once everything is finished: - -If I understand your comment to my answer correctly, you are aiming to change your algorithm for the torus so it works with ${\rm cr}(G)$. I think the whole MO community is keeping their fingers crossed, wishing you that you can successfully complete everything in time! - -Looking at the far horizon, I wanted to make a suggestion to you. Once you have changed your torus algorithm and completed your thesis, you will have effectively two algorithms in your hands for the torus: The old one based on ${\rm pcr}(G)$ and the new one based on ${\rm cr}(G)$. I am saying the obvious here, keep both of them, they can really be fruitful for future research. -(1) Obviously, your two algorithms could support research on the big open question whether ${\rm pcr}(G)\stackrel{\rm ?}{=}{\rm cr}(G)$ or not. They -could produce experimental evidence, ideas, and insights for a -future proof of equality, or an actual counterexample. (Again, I am -saying the obvious here.) -(2) To really pressure-test ${\rm pcr}(G)\stackrel{\rm ?}{=}{\rm cr}(G)$ on the torus, it would be interesting to also try the -best known to date lower bound for ${\rm cr}(G)$ -$$\frac{1}{29}\frac{e^3}{n^2}$$ for graphs with $e>7n$. This lower bound is from -Eyal Ackerman (2019): "On topological graphs with at most four -crossings per edge", Computational Geometry, 85: 101574, 31, -doi:10.1016/j.comgeo.2019.101574 (probably you are aware of it from -the Wikipedia article that you quoted). -I think your question and this whole topic are really important. László Székely calls it one of the "foundational problems" and devotes a whole section to it in his article Turán’s Brick Factory Problem: The Status of the Conjectures of Zarankiewicz and Hill. In: R. Gera et al. (eds.)(2016): Graph Theory—favorite conjectures and open problems. 1.) - -For now, fingers crossed that you can complete your thesis in time!<|endoftext|> -TITLE: Number of Fuchsian groups with same trace field -QUESTION [6 upvotes]: Let $\Gamma,\Sigma\subset \mathrm{SL}_2({\mathbb R})$ be cocompact arithmetic subgroups. They are called commensurable in the wider sense, if there exists -$g\in \mathrm{SL}_2({\mathbb R})$, such that the intersection of $\Gamma$ and $g\Sigma g^{-1}$ has finite index in both. -The trace field of $\Gamma$, denoted ${\mathbb Q}(\mathrm{tr}\,\Gamma)$ is the field extension of $\mathbb Q$ generated by all traces of elements of $\Gamma$. -Next let $\Gamma^{(2)}$ be the subgroup of $\Gamma$ generated by all squares $\gamma^2$ with $\gamma\in\Gamma$. The invariant trace field is defined as $I(\Gamma)={\mathbb Q}(\mathrm{tr}\,\Gamma^{(2)})$. -The (invariant) trace field is a number field and commensurable groups have the same invariant trace field. -My question is this: -For a given number field $K$, is it true that there is only a finite number of commensurability classes $[\Gamma]$ with $K=I(\Gamma)$? - -REPLY [5 votes]: No, this follows from a result of Bogwang Jeon. He showed that given a number field $K$ and quaternion algebra $A$ over $K$ with $A\otimes_K \mathbb{R} \cong M_2(\mathbb{R})$, one can find a fuchsian surface of genus $g$ having $K$ as its invariant trace field and $A$ as the invariant quaternion algebra. Now one observes that there are infinitely many isomorphism classes of quaternion algebras over $K$ which split over $\mathbb{R}$. See e.g. Theorem 7.3.6 of MacLachlan-Reid, which states that two quaternion algebras over $K$ are isomorphic iff they have the same ramification set of places, and that any admissible ramification set is realized by a quaternion algebra. Hence there are infinitely many quaternion algebras from the infinitude of prime ideals in the ring of integers of a number field (corresponding to the non-Archimedean places). -Since the invariant quaternion algebra and trace field are commensurability invariant (Theorem 3.3.4 and Corollary 3.3.5 of MacLachlan-Reid) , this implies the infinitude of incommensurable fuchsian groups with a given invariant trace field.<|endoftext|> -TITLE: Cohomological behavior of the embedding $Gr(3,5)\to Gr(3,6)$ -QUESTION [6 upvotes]: The following question is particularly interesting for me: -Does the natural map $Gr(3,5)\to Gr(3,6)$ induce a surjection -$$H^4(Gr(3,6),\mathbb{Z})\to H^4(Gr(3,5),\mathbb{Z})?$$ -Here $Gr(k,n)$ means the real grassmannian of rank $k$. - -REPLY [5 votes]: Let me give more details on Nanjun Yang's answer (which I think is correct). There is no need for any results concerning Chow-Witt rings of Grassmannians, the surjectivity can be deduced from the results of Casian and Kodama linked in Danny Ruberman's answer to this MO-question (or other sources of specific knowledge about the structure of the cohomology of the real Grassmannians). -Here's the relevant facts on the cohomology of the real Grassmannians ${\rm Gr}_k(\mathbb{R}^n)$. The non-torsion classes are generated by Pontryagin classes of the tautological sub- and quotient bundle, plus an additional class in degree $n-1$ in the case both $k$ and $n-k$ are odd (not a characteristic class but related to an Euler class on some other Grassmannian). All torsion is 2-torsion, and it's given by the image of the integral Bockstein from the mod 2 cohomology (i.e. generated by Bocksteins of Stiefel-Whitney classes). Relations between these generators all come in some way from the Whitney sum formula. -In the specific case we have ${\rm H}^4({\rm Gr}(3,5),\mathbb{Z})=\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$. A generator of the $\mathbb{Z}$-summand is the first Pontryagin class of either of the tautological bundles. A generator for the 2-torsion summand is the Bockstein of ${\rm w}_1^3$ (with ${\rm w}_1$ the first Stiefel-Whitney class of either of the tautological bundles). In Young diagram terms, this corresponds to the partition $(2,1,1)$. This can be explicitly found (for ${\rm Gr}(2,5)={\rm Gr}(3,5)$ in the paper of Casian and Kodama. -On the other hand we have ${\rm H}^4({\rm Gr}(3,6),\mathbb{Z})=\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}^{\oplus 2}$, generated again by the Pontryagin class and classes corresponding to the partitions $(3,1)$ and $(2,1,1)$. The restriction morphism induced from the inclusion ${\rm Gr}(3,5)\hookrightarrow{\rm Gr}(3,6)$ maps characteristic classes of the tautological bundles on ${\rm Gr}(3,6)$ to the corresponding characteristic classes of the tautological bundles on ${\rm Gr}(3,5)$. Because Pontryagin and Stiefel-Whitney classes are stable, we get the required surjectivity. -The general moral here is that we get surjectivity whenever the cohomology group of the smaller Grassmannian is generated by stable characteristic classes. The only problematic class is the non-characteristic class in degree $n-1$ (corresponding to the maximal hook partition). That only exists on ${\rm Gr}(k,n)$ whenever $k$ and $n-k$ are both odd. That is exactly precluded by the conditions in Nanjun Yang's answer.<|endoftext|> -TITLE: Proving Hall's marriage theorem using Sperner's lemma -QUESTION [11 upvotes]: In the paper Hall's theorem for hypergraphs (Aharoni and Haxell, 2000), -the authors prove a theorem on the existence of perfect matchings in bipartite hypergraphs, using Sperner's lemma. At the last page (6), they say that "we have here a topological proof of Hall's theorem" (for bipartite graphs). I thought it should be easy to write this proof explicitly, since a simple bipartite graph is just a bipartite hypergraph in which each hyperedge is of size 2. But there is a problem: during the proof, the authors assume that the sets of neighbors (i.e., the sets $N(x)$ for each vertex $x\in X$, where $X$ is one part of the hypergraph) are pairwise-disjoint. For a hypergraph, this assumption is without loss of generality, since we can add dummy vertices to edges, and this does not affect the theorem conditions or concludion. But in a graph, we cannot add vertices to edges. -So my question is: is there an explicitly-written proof of Hall's marriage theorem, using Sperner's lemma (or a similar topological theorem)? - -REPLY [6 votes]: In addition to Carlo Beenakker's answer that gives Hall via Sperner directly, I think you can also get it by applying Hall's Theorem for Hypergraphs as follows. Let $G$ be a bipartite graph with partite sets $X, Y$, and write $V(X) = \{x_1, \ldots, x_n\}$. For each $i$, define a $1$-uniform hypergraph $H_i$ with vertex set $N(x_i)$ and edge set $\{ \{y\} : y \in N(x_i) \}$. -Letting $\mathcal{A} = \{H_1, \ldots, H_n\}$, we see that $\mathcal{A}$ has a disjoint set of representatives if and only if $G$ has a matching that saturates $X$. Now the condition of Aharoni and Haxell's Theorem 1.1 applied to the family $\mathcal{A}$ is clearly equivalent to Hall's Condition on $G$. Since classical Hall's Theorem falls out of the hypergraph version so quickly, I think it's fair to think of the topological proof of the hypergraph version as also being a topological proof of Hall's theorem.<|endoftext|> -TITLE: What is the probability that a random chord in a sphere touches opposite hemispheres? -QUESTION [5 upvotes]: (edited) Consider the unit sphere $\mathbb{S}^2\subset \mathbb{R}^3$, and its upper $(z>0)$ and lower $(z<0)$ hemispheres. -Draw two independent, uniformly distributed points $X,Y$ on $\mathbb{S}^2$. -Given $\theta\in[0,\pi/2]$, what is the probability that $X$ and $Y$ belong to different hemispheres among the above two, conditioning by the event that the chord $[X,Y]$ makes an angle $\theta$ with the $z$-axis? -Numerically I find that this probability is $\cos{\theta}$. -Could anyone help me justify this $\cos{\theta}$? - -REPLY [7 votes]: This is not a true "no pen or paper" solution requested by fedja, but at least it avoids integrals. :-) -Let $X$ and $Y$ be independent random vectors on the unit sphere. Write $E = (X - Y) / |X - Y|$ for the unit vector parallel to the chord $XY$, and $Z = \tfrac{1}{2} (X + Y)$. - -Claim: Conditionally on $E = e$, the projection $Z = \tfrac{1}{2}(X + Y)$ of the chord $XY$ onto its perpendicular bisector plane $\pi_e = \{v : v \perp e\}$ (the entire chord projects onto a single point) is uniformly distributed over the unit disk in $\pi_e$. -Given the above claim, the proof is straightforward. Indeed: given any unit vector $e$ such that $\theta = \arcsin |e \cdot (0,0,1)|$ as in the question, and conditionally on $E = e$, $X$ and $Y$ belong to different hemispheres if and only if $Z$ belongs to an ellipse, which is the projection of the equator of the unit sphere onto $\pi_e$. This ellipse has semi-axes $1$ and $\cos \theta$ (this becomes pretty clear if one draws a picture). The area of this ellipse is equal to $\cos \theta$ times the area of the unit disk, and the desired result follows. Thus, it remains to prove the claim. - -Proof of the claim: Since the random variable $X \cdot Y$ is uniformly distributed over $[-1, 1]$ (Archimedes's theorem!), the random variable -$$ \|Z\|^2 = \|\tfrac{1}{2} (X + Y)\|^2 = \tfrac{1}{2} (1 - X \cdot Y) $$ -is uniformly distributed over $[0, 1]$. By rotational symmetry, $\|Z\|^2$ and $E$ are independent. It follows that conditionally on $E = e$, $\|Z\|^2$ is uniformly distributed over $[0, 1]$. Again by symmetry, the conditional distribution of $Z$ (given $E = e$) is invariant under rotations of $\pi_e$, and so it follows that this conditional distribution is uniform over the unit disk on $\pi_e$, as desired.<|endoftext|> -TITLE: Optimizing the gradient norm on the unit sphere -QUESTION [5 upvotes]: Let $ \Bbb S^{d-1}=\{(x_1,\cdots ,x_d): x_1^2+ \cdots +x_d^2=1\}\subset \Bbb R^d$ be the unit -sphere. Let $\nabla u= (\partial_{x_1}u,\cdots, \partial_{x_d}u)$ be the gradient of a function $u\in C_c^\infty(\Omega)$ with $\Omega \subset \Bbb R^d$ open. For $e\in \Bbb S^{d-1}$, we write $$\nabla u(x)\cdot e = \partial_{x_1}u(x) e_1+\cdots+\partial_{x_d}u(x) e_d.$$ -Then, for $p\geq 1$, what is the following quantity? -\begin{split} -I(u) =\sup_{e\in \mathbb{S}^{d-1}}\int_\Omega |\nabla u(x)\cdot e|^pdx -\end{split} -I have tried to use the fact that for all $z\in \Bbb R^d$ we have, $$|z|=\sup_{e\in \mathbb{S}^{d-1}}|z\cdot e|$$ And only got the upper estimate $$I(u)\leq \int_\Omega |\nabla u(x)|^pdx$$ -In fact I am expecting an equality here. - -REPLY [2 votes]: LONG COMMENT: My collaborators and I have in fact studied the function -$$ x\in \mathbb{R}^d \mapsto \left(\int_{\mathbb{R}^d} |x\cdot\nabla f|^p\,dx\right)^{1/p} $$ -quite extensively, for example in this paper. Here are a few highlights: - -It defines a norm on $\mathbb{R}^d$. The unit ball of this norm is called the $L^p$ polar projection body of the function $f$. The unit ball of the dual norm is called the $L^p$ projection body. This generalizes the definition of the $L^p$ projection and polar projection bodies of a convex body, which were studied here. -If we denote the norm by $\|\cdot\|_{f,p}$, then the sup of $\|e\|_{f,p}$ over all unit vectors $e$ is half the diameter of the polar projection body, where the diameter is defined to be the maximum distance between two parallel supporting hyperplanes. -Also, note that the quantity -$$ \int_{S^{d-1}} \|e\|_{f,p}^{-n}\,de = \int_{S^d}\left(\int_{\mathbb{R}^d} |e\cdot\nabla f|^p\,dx\right)^{-n/p}\,de $$ -is the volume of the polar projection body. Although this does not define a norm on any function space, we proved that it satisfies a sharp affine Sobolev inequality. We call the inequality affine, because this quantity is invariant under the action of $SL(d)$ acting on $\mathbb{R}^d$. It is also not hard to show that this inequality implies the standard sharp Sobolev inequality on $\mathbb{R}^d$ that uses the standard Euclidean norm. - -A paper that discuss this function, when $f$ is a probability density function can be found here: -E. Lutwak, D. Yang, G. Zhang. Moment-entropy inequalities, Annals of Probability 32 (2004) 757-774.<|endoftext|> -TITLE: Ramification and reduction -QUESTION [6 upvotes]: Let $K$ a local field ($K$ finit extension of $\mathbb{Q}_p$), $\mathcal{O}_K$ the integer of $K$ and $k$ the residue field of $\mathcal{O}_K$. -Let $\psi:\mathbb{P}^1_K\to\mathbb{P}^1_K$ a finit separable morphism, $\widetilde{\psi}=\Psi:\mathbb{P}^1_{\mathcal{O}_K}\to\mathbb{P}^1_{\mathcal{O}_K}$ a model of $\psi$ that is $\Psi$ is the extension of scalar of $\psi$ ie $\Psi=\psi\times_{\mathcal{O}_K}\text{Id}_K$. -$$ \require{AMScd} -\begin{CD} -\mathbb{P}^1_K @>{\psi}>> \mathbb{P}^1_K\\ -@VV{\alpha}V @VVV \\ -\mathbb{P}^1_{\mathcal{O}_K} @>{\Psi}>> \mathbb{P}^1_{\mathcal{O}_K} -\end{CD} $$ -Let $\overline{\Psi}=\Psi\times_{\mathcal{O}_K}\text{Id}_k$ the reduction of $\Psi$. -$$ \require{AMScd} -\begin{CD} -\mathbb{P}^1_k @>{\overline{\Psi}}>> \mathbb{P}^1_k\\ -@VV{i}V @VVV \\ -\mathbb{P}^1_{\mathcal{O}_K} @>{\Psi}>> \mathbb{P}^1_{\mathcal{O}_K} -\end{CD} $$ -If the branching points (ie ramification points) $P_1,\ldots,P_n$ of $\psi$ are $K$-rationnals, as $\mathbb{P}^1_{\mathcal{O}_K}(\mathcal{O}_K)=\mathbb{P}^1_K(K)$ (by mutliplication of denominators) one can take their reductions $\overline{P_1},\ldots,\overline{P_n}\in\mathbb{P}^1_k(k)$. -Question: I'd like to prove that if the ramification indices of the $P_i$ are resp. $e_i$, they are the same for the $\overline{P_i}$ and if there are ``coalescence'' then the ramification indices of the resulting ramification point $\overline{Q}$ is the sum of the indices $e_i$ for which $\overline{P_i}=\overline{Q}$. I don't have the beginning of an explanation of that, if it's true... -I guess that we shouldn't have wild ramification so the sums of $e_i$ of point that collapse in the same point shouldn't be nul in $k$. -I guess that a general reference for that is SGA1 (Exposé X) but for the moment it's to difficult for me... If someone has a simpler reference for my specific case I'l take it! Thanks! -If you find this question to easy for mathoverflow feel free to answer here in mathstackexchange and tell me in a comment. - -REPLY [6 votes]: In your setting, you can just do everything concretely using the derivative. -The correct statement is for $\overline{Q}$ in $\mathbb P^1_k$, -$$e(\overline{Q}) + \operatorname{swan}(\overline{Q}) = 1 + \sum_{\substack{ i \in \{1,\dots n \} \\ \overline{P}_i = \overline{Q} }} (e_i - 1).$$ -This is under your assumptions except that we need to assume $\overline{\Psi}$ is separable. -To prove this, first we can assume by a change of variables that $\overline{\Psi}(\overline{Q}) \neq \infty$. Then express $\widetilde{\psi}$ as a rational function $f$ in $\mathbb Z_p[X]$, without a pole at $\overline{Q}$, and thus without a pole at any of the $P_i$ that reduce to $Q$. Now consider its derivative $\frac{df}{dx}$. -In characteristic zero, this function vanishes exactly at the ramification points $P_1,\dots, P_n$, and its order of vanishing at $P_i$ is $e_i-1$. -In characteristic $p$, its order of vanishing at a point is $e$ plus the Swan conductor minus $1$. -Now we just need to know that the order of vanishing of $\frac{df}{dx}$ at $\overline{Q}$ is the sum of its orders of vanishings at $P_i$ for all the $P_i$ that reduce to $Q$. This follows from factoring the numerator of $\frac{df}{dx}$ into linear factors, and noting that the order of vanishing is the number of linear factors that vanish at a point. -We can't rule out wild ramification here, as the example $x^p-x$ (for $K = \mathbb Q ( p^{1/(p-1)})$) shows. In that case we have one point $\infty$ with $e=p$ and $p-1$ points (the $p-1$st roots of $p^{-1}$) with $e=2$, that all reduce to $\infty$, and in the reduction, $\infty$ has $e=p$ and $\operatorname{swan}=p-1$.<|endoftext|> -TITLE: Groupoids as models of symmetric simplicial sets -QUESTION [6 upvotes]: In the Elephant, Peter Johnston remarks that internal categories may be regarded as simplicial objects that “preserve all limits that happen to exist in $\Delta^{op}$“ (I guess you might call this a flat functor). This is because the join in $\Delta$ is a limit. -Does a similar statement exist for the symmetric simplicial set/groupoid correspondence? It’s clear that the nerve of a groupoid yields a symmetric simplicial set, the question is then whether or not they correspond to flat functors on the site for symmetric simplicial sets. - -REPLY [7 votes]: You can definitely characterize groupoids as presheaves on $Fin_+$ preserving some colimtis (i.e. sending some colimits in $Fin_+$ to limits in Set). In fact Groupoids are the presheaf on $Fin_+$ that preserve the colimits comming from $\Delta$. -However, the Category $Fin_+$ has much more colimits than $\Delta$, in fact it all non-empty finite colimits. Groupoids do not preserves most of these colimits. -For example, in $Fin_+$ you have $\{1\} \coprod \{1\} \simeq \{1,2\}$ , but given a a groupoid $G$, $G(\{1\})$ is the set of objects, and $G(\{1,2\})$ is the set of arrow, so the isomorphisms $G(\{1,2\}) \simeq G(\{1\}) \times G(\{1\})$ means that your groupoid has a unique arrow between any two objects. -In fact, I claim that the colimits in $Fin_+$ that exists are exactly the finite non-empty colimits, and the category of presheaf preserving them is equivalent to the category of sets. In terms of the usual description of groupoids as presheaf on $Fin_+$, these corresponds to anti-discrete groupoids.<|endoftext|> -TITLE: Reference request for the explicit formula for $\sum_{n\leq x} \Lambda(n)n^{-s}$ -QUESTION [6 upvotes]: Denote by $\Lambda(n)$ th e von Mangoldt function, which is equal to $\log p$ if $p\geq 2$ is a prime, and $0$ otherwise. Let $\rho$ denote a complex zero of the Riemann $\zeta$-function. If i recall well, i once heard sometime ago that -$$\sum_{n\leq x} \Lambda(n)n^{-s} = -\frac{\zeta'}{\zeta}(s) + \frac{x^{1-s}}{1-s} - \sum_{|Im \rho| \leq x} \frac{x^{\rho-s}}{\rho-s} + O(\log^{2}x)$$ for $s\neq 1, s\neq \rho$ and $s\neq -2k, k\in \mathbb{N}$. -Does anyone have a reference for this result ? - -REPLY [5 votes]: I don't know a reference off-hand, but here is a sketch of the proof (the details need to be checked carefully, and I have not done it). One can start from -$$\sum_{n\leq x}\frac{\Lambda(n)}{n^s}=\frac{1}{2\pi i}\int_{(\sigma)}-\frac{\zeta'(z)}{\zeta(z)}\cdot\frac{x^{z-s}}{z-s}\,dz,\qquad\sigma>\max(1,\mathrm{Re}\,s),$$ -which is a variant of Theorem 5.1 in Montgomery-Vaughan: Multiplicative number theory I, and can be proved in the same way. As in the theorem, the RHS is understood as a Cauchy principal value, while in case of $x\in\mathbb{N}$ the term corresponding to $n=x$ in the LHS is counted with weight $1/2$. -The integrand is meromorphic with simple poles at $z=s$, $z=1$, and $z=\rho$. The corresponding residues are $-\zeta'(s)/\zeta(s)$, $x^{1-s}/(1-s)$, and $-m_\rho\cdot x^{\rho-s}/(\rho-s)$, where $m_\rho$ is the multiplicity of $\rho$. So one can derive the OP's display by performing the following steps: - -Truncate the integral on the RHS to $|\mathrm{Im}\,z|\leq x$ and estimate the error introduced. Perturb $x$ slightly if it is very close to some $\mathrm{Im}\,\rho$. -Extend the truncated contour (which is a vertical line segment) to a rectangle containing the points $s=1$ and $s=0$, hence all the $\rho$'s with $|\mathrm{Im}\,\rho|\leq x$. By the Residue Theorem, the integral weighted by $1/(2\pi i)$ equals the sum of corresponding residues listed above. -Estimate the contribution of the horizontal line segments of the rectangular contour, as well as of the vertical line segment to the left of $s=0$. -The LHS equals the sum of residues listed in item 2, up to the error terms listed in items 1 and 3. - -P.S. See also Terry Tao's valuable remarks below the original post.<|endoftext|> -TITLE: Rank of sum of two matrices -QUESTION [7 upvotes]: Given matrices $A, B \in \Bbb R^{3 \times 3}$ whose ranks satisfy $\mbox{rank} (A), \mbox{rank} (B) \geq 2$, I would like to prove that for large (or small) enough scalar $\alpha \in \mathbb{R} \setminus \{0\}$ the following does hold. -$$\mbox{rank} (A+\alpha B) \geq 2$$ -This seems to be true by hand waving argument, but I would like to find some short proof or reference that formally proves it. Thanks. - -REPLY [12 votes]: Since rank$(A)\geq 2$, the matrix $A$ has a $2\times 2$ submatrix $a$ with nonzero determinant; the determinant is a continuous function of the matrix elements, so adding a sufficiently small perturbation $\alpha B$ to $A$ will leave $\det a\neq 0$ and hence the rank of $A+\alpha B$ remains $\geq 2$. -If instead of small $\alpha$ you wish to take large $\alpha$, define $B'=\alpha B$, $A'=\alpha A$ and work with $B'+(1/\alpha)A'$, where $B'$ has rank $\geq 2$.<|endoftext|> -TITLE: Why is a dynamical system not a dynamic system? -QUESTION [12 upvotes]: This is a research question in the history of math, I suppose. -As a non-native english speaker I became used to mathematical expressions like 'dynamical' and 'tangential'. When using them in daily conversation as substitutes for 'dynamic' and 'tangent' I got frowned upon by native english speakers who claimed to have never heard of these words before. -Some references suggest -https://www.merriam-webster.com/dictionary/dynamical -https://www.merriam-webster.com/dictionary/tangential -that indeed they can almost mean the same as 'dynamic' and 'tangent' but for some reason nobody seems to use these words that way. -So whereas in English: -A dynamic person-a dynamical system -the adjective is different -it is in French -une personne dynamique- un système dynamique -and in German -Eine dynamische Person - ein dynamisches System -the adjective is the same. -I am wondering when (and maybe also why) these expressions started deviating in English. - -REPLY [11 votes]: Thing is, they do not mean the same thing. At least, not in theory. -Dynamic the adjective means "exhibiting continual change". -Dynamics the noun means "the study of forces and their relation to motion". -Dynamical the adjective means "relating to the study of dynamics." -A "dynamic" system is a system exhibiting continual change. A "dynamical" system is a system relating to the study of dynamics. (Since OP is Chinese, this is also why DS is 動力系統 and not 不定系統.) -Similarly, -Tangent the adjective means the geometric notion of touching but not intersecting. -Tangent the noun refers first to the geometric construct of the line tangent to a shape, and then also to the idea of "objects that can be split off without making a turn", whence the idea of "going on a tangent" when you derail the discussion with something related but not directly relevant. (You won't be going on a tangent if you change the topic or the direction of discussion abruptly.) -Tangential the adjective refers to the quality of "tangent". Hence you make a "tangential remark" while you "go on a tangent". Hence you look for "tangent lines" while compute "tangential forces". (The force itself is not tangent, but it is directly along the line that is tangent to the object.) - -This, of course, gets muddled by the fact that English is perfectly happy with attributive nouns. English being remarkably loosey goosey about grammar for a Western language, understanding in theory why things are the way they are is probably much less useful than accepting their use as a convention. (After all, what is language but a convention to enable communication?) - -Homework exercise: discuss transverse versus transversal<|endoftext|> -TITLE: Is a function of several variables convex near a local minimum when the derivatives are non-degenerate? -QUESTION [7 upvotes]: This is a cross-post. -Let $U \subseteq \mathbb R^n$ be an open subset, and let $f:U \to \mathbb R$ be smooth. Suppose that $x \in U$ is a strict local minimum point of $f$. -Let $df^k(x):(\mathbb R^n)^k \to \mathbb R$ be its $k$ "derivative", i.e. the symmetric multilinear map defined by setting -$df^k(x)(e_{i_1},\dots,e_{i_k})=\partial_{i_1} \dots \partial_{i_k}f(x)$. -Assume that $df^j(x) \neq 0$ for some natural $j$. Let $k$ be the minimal such that $df^k(x) \neq 0$. Since $x$ is a local minimum, $k$ must be even. -Suppose now that $df^k(x)$ is non-degenerate, i.e. $df^k(x)(h,\dots,h) \neq 0$ for any non-zero $h \in \mathbb R^n$. (Since $x$ is a minimum, this is equivalent to $df^k(x)$ being positive-definite, i.e. $df^k(x)(h,\dots,h) > 0$ for any non-zero $h \in \mathbb R^n$). - -Question: Is $f$ is strictly convex in some neighbourhood of $x$? - -In the one-dimensional case, when $f$ is a map $\mathbb R \to \mathbb R$, the answer is positive: -We have $f^k(x)>0$, and the Taylor expansion of $f''$ near $x$ is -$$ -f''(y) = {1 \over (k-2)!} f^{(k)}(x)(y - x)^{k-2} + O((y - x)^{k-1}). -$$ -Thus, $f''(y)>0$ for $y \ne x$ sufficiently close to $x$, so $f$ is strictly convex around $x$. - -Returning back to the high-dimensional case, if $k>2$, we have $\text{Hess}f(x)=df^2(x)=0$, and I guess that we should somehow prove that $\text{Hess}f(y)$ becomes positive-definite for $y$ sufficiently close to $x$. -Perhaps we need to understand the Taylor's expansion of $\text{Hess}f$ around $x$, similarly to the one-dimensional case, but I am not sure how to do that. -Is there a nice way? - -Comment: -It is certainly not enough to assume that $df^k(x)$ is non-zero. Indeed, consider $ f(x,y) = x^2 y^2 + x^8 + y^8$. -$f$ has a strict global minimum at $(0,0)$. -$$\det(\text{Hess}f(x,y))=3136 x^6 y^6 + 112 x^8 + 112 y^8 - 12 x^2 y^2,$$ which is negative when $x=y$ is small and nonzero. Thus, $f$ is not convex at a neighbourhood of zero. -Note that $\text{Hess}f(0,0)=0$; The first non-zero derivative at $(0,0)$ is the fourth-order derivative $df^4(0)$. It is degenerate, however, since $df^4(0)(h^1e_1+h^2e_2,h^1e_1+h^2e_2)=4(h^1)^2(h^2)^2$ vanishes when either $h_i$ is zero. -So, non-vanishing of some derivatives does not ensure convexity. - -REPLY [12 votes]: Let -$$\begin{aligned} f(x,y) & = x^4 - x^2 y^2 + y^4 \\ & = \tfrac{1}{2} x^4 + \tfrac{1}{2} y^4 + \tfrac{1}{2} (x^2 - y^2)^2 . \end{aligned}$$ -Then $f$ is a strictly positive (except at the origin, of course) homogeneous polynomial of degree $4$, and hence $d^j f(\vec 0) = 0$ for $j < 4$ and $d^4 f(\vec 0) > 0$ (indeed: $d^4 f(\vec 0)(\vec h, \vec h, \vec h, \vec h) = 4! f(\vec h) > 0$ whenever $\vec h \ne \vec 0$). On the other hand, $$\partial_{xx} f(0,y) = -2 y^2 < 0$$ whenever $y \ne 0$, and so $f$ is not convex near $0$.<|endoftext|> -TITLE: Concrete examples of statements not provable in PRA + $\epsilon_0$-induction that are provable in PA? -QUESTION [10 upvotes]: It is well-known that $\mathbf{PRA}$ plus $\epsilon_0$-induction on bounded formulas cannot prove all $\mathbf{PA}$ theorems (essentially because $I\Sigma_1$ plus $\epsilon_0$-induction on bounded formulas is finitely axiomatizable while the latter isn't). Are there any concrete examples known (preferable "natural")? -(This is an exact copy of my question on MSE, but I expect I won't get any answer there, so I hope I'm allowed to cross-post it here) - -REPLY [8 votes]: One example is $I\Sigma_2$ (or rather, the conjunction of its finite axiomatization). -Notice that the theory $I\Sigma_1+\epsilon_0$-induction for bounded formulas is axiomatized by $\Pi_3$ sentences: in particular, the $\epsilon_0$-induction schema can be written in prenex form as -$$\forall u\,\forall x\,\exists y\,\forall z\,\bigl[\bigl(\neg\theta(u,y)\land(z\prec y\to\theta(u,z))\bigr)\lor\theta(u,x)\bigr],$$ -where $\theta\in\Delta_0$ and $\prec$ is the order representing $\epsilon_0$. By a theorem of Leivant [1], $I\Sigma_2$ is not provable from any set of $\Pi_3$ (or even $\Sigma_4$) sentences consistent with $Q$. More generally, for $n\ge1$, $I\Sigma_n$ is not provable from any set of $\Sigma_{n+2}$ sentences consistent with $Q$. -The reason is that $I\Sigma_n$ proves the uniform $\Sigma_{n+1}$-reflection schema for $Q$: -$$\mathrm\forall x\,\bigl(\mathrm{Pr}_Q(\ulcorner\phi(\dot x)\urcorner)\to\phi(x)\bigr),\qquad\phi\in\Sigma_{n+1},$$ -where $\mathrm{Pr}_Q$ denotes the formalized provability predicate for $Q$. This schema is equivalent to the single sentence -$$\def\rfn{\mathrm{RFN}}\rfn_Q(\Sigma_{n+1})=\forall x\,\bigl(\mathrm{Pr}_Q(x)\to\mathrm{Tr}_{n+1}(x)\bigr),$$ -where $\mathrm{Tr}_{n+1}$ is the truth definition for $\Sigma_{n+1}$-formulas (defined so that it is vacuously true for non-$\Sigma_{n+1}$-formulas). Moreover, it is easy to see that -$$\rfn_Q(\Sigma_{n+1})\equiv\rfn_Q(\Pi_{n+2}).$$ -Now, if we assume for contradiction that $\rfn_Q(\Pi_{n+2})$ is provable from a consistent extension of $Q$ by $\Sigma_{n+2}$ sentences, there is a single $\Sigma_{n+2}$ sentence $\phi$ consistent with $Q$ such that -$$Q+\phi\vdash\rfn_Q(\Pi_{n+2})\vdash\mathrm{Pr}_Q(\ulcorner\neg\phi\urcorner)\to\neg\phi,$$ -that is, -$$Q+\phi\vdash\mathrm{Con}_{Q+\phi},$$ -contradicting the second incompleteness theorem. -Reference: -[1] Daniel Leivant: The optimality of induction as an axiomatization of arithmetic, Journal of Symbolic Logic 48 (1983), no. 1, pp. 182–184, doi: 10.2307/2273332.<|endoftext|> -TITLE: List chromatic index of a particular graph -QUESTION [6 upvotes]: Consider the graph $G$ of order $n$ consisting of two disjoint cliques of even order $\frac{n}{2}=p+1$ (where $p$ is odd prime) joined by a bipartite graph (that is, deleting the edges of the two disjoint cliques from $G$ leaves a bipartite graph) of maximum degree $p$. Then, does the graph have list chromatic index $\le 2p+1$? The bipartite graph is also quite specific, in that it has one vertex in each partite set of degree exactly equal to $0,1,2,\dotsc,p$. -My view is that, by Schauz - Proof of the list edge coloring conjecture for complete graphs of prime degree paper, we have that the disjoint cliques are chromatic edge-choosable. In addition, the edges joining the two cliques is a bipartite graph, which is again chromatic edge-choosable by the Galvin's theorem. Thus, it makes me think the above question has a positive answer. By the way, the graph has chromatic index equal to $2p$, that is the graph is of class $1$. Any hints? - -REPLY [5 votes]: Greedy coloring works here to show $2p$-choosability, I believe, and the hypothesis that $p$ is prime doesn't appear to be necessary. Write the cliques as $A = \{a_1, \ldots, a_{p+1}\}$ and $B = \{b_1, \ldots, b_{p+1}\}$, taking the notation so that $a_i$ has exactly $i-1$ neighbors in $B$ and vice versa. -First color the edges in the bigraph between $A$ and $B$; observe that each such edge is adjacent (in $L(G)$) to at most $2p-1$ previously colored edges when it is processed, thus has a color available. (Alternatively, just use Galvin's theorem for this part; then these edges only need to have lists of size $p$.) -Then color the edges $a_ia_j$ within $A$, ordering the edges so that $i + j$ is non-increasing. Observe that an edge $a_ia_j$ with $i \leq j$ has, within the clique $A$, exactly $p+1-j$ previously-colored adjacent edges at its $a_i$-endpoint and $(p+1)-i-1 = p-i$ previously-colored adjacent edges at its $a_j$-endpoint, for a total of $$2p+1-(i+j)$$ previously-colored adjacent edges within $A$. Furthermore, $a_ia_j$ has exactly $$(i-1) + (j-1) = i+j-2$$ previously-colored adjacent edges going to $B$. Thus, each edge $a_ia_j$ within $A$ is adjacent to exactly $2p-1$ previously-colored edges when it is processed, and therefore has a color available. Coloring $B$ the same way finishes the proof.<|endoftext|> -TITLE: A Rademacher ‘root 7’ anti-concentration inequality -QUESTION [23 upvotes]: Let $r_1,r_2,r_3,\dotsc$ be an IID sequence of Rademacher random variables, so that $\mathbb P(r_n=\pm1)=1/2$, and $a_1,a_2,\dotsc$ be a real sequence with $\sum_na_n^2=1$. For $S=\sum_na_nr_n$, does the following inequality always hold? -$$ -\mathbb P\left(\lvert S\rvert\ge1/\sqrt7\right)\ge1/2.\tag{*}\label{star} -$$ -Is this a known result, or conjecture? -Stated in another way, for each finite sequence $a_1,a_2,\dotsc,a_N$ with $\sum_na_n^2=1$, if we look at all $2^N$ sums $\pm a_1\pm a_2\pm a_3\pm\dotsb\pm a_N$, do we always have at least as many with absolute value at least $1/\sqrt7$ as with absolute value strictly less than $1/\sqrt7$? -Background -The statement is very similar to a conjecture of Tomaszewski, that at least as many of the $2^N$ sums have absolute value less than or equal to 1 as have value greater than one. This conjecture is equivalent to the concentration inequality $\mathbb P(\lvert S\rvert\le1)\ge1/2$, and is still unproven, other than possibly in a (not currently peer reviewed, as far as I know) paper Keller and Klein - Proof of Tomaszewski's conjecture on randomly signed sums posted on the arXiv recently. -We can also consider more general anti-concentration inequalities, which bound $\mathbb P(\lvert S\rvert \ge x)$ from below. For $x=0$ the lower bound is trivially 1 and, for $x > 1$, it is $0$. For $0 < x < 1$, then, as in a previous answer of mine, the Paley–Zygmund inequality can be used to obtain -$$ -\mathbb P(\lvert S\rvert \ge x) \ge (1-x^2)^2/3, -$$ -but this is far from optimal. Also, for $x=1$, there does exist a strictly positive anti-concentration bound, as shown in the answers to another question of mine, An $L^0$ Khintchine inequality. It was shown by Oleszkiewicz, in 1996, that a (non-optimal) lower bound of $1/10$ applies when $x=1$ (in On the Stein property of Rademacher sequences) and also conjectured that the optimal bound is $7/32$, but this is still open as far as I am aware. -Running a Monte Carlo simulation to randomly pick the values of $a$ and compute the probability suggests that the optimal lower bound for $\mathbb P(\lvert S\rvert\ge x)$ is piecewise constant in $x$, so that there is only a small set of $x$ values at which the bound changes (specifically, $x=0,1/\sqrt7,1/\sqrt5,1/\sqrt3,2/\sqrt6,1$). As mentioned, for $x=1$ it is an open conjecture, so my question here is regarding the smallest non-trivial value for $x$. -Note that \eqref{star} is best possible, in the sense that it does not hold if either of the inequalities are strict and, consequently, does not hold if either the $1/\sqrt7$ inside the probability or the $1/2$ outside is increased. Considering $a_n=1/\sqrt7$ for $n\le7$, we obtain $\mathbb P(\lvert S\rvert > 1/\sqrt7)=29/64$, and considering $a_n=1/\sqrt2$ for $n\le 2$ gives -$$ -\mathbb P(\lvert S\rvert\ge1/\sqrt7)=\mathbb P(\lvert S\rvert > 0)=1/2. -$$ -This example also shows that for any $x \gt 0$, the anti-concentration bound can never be better than 1/2, so \eqref{star} is also optimal in this sense, and it would immediately follow that we have optimal inequalities -$$ -\mathbb P(\lvert S\rvert\ge x)\ge1/2 \tag{**}\label{starstar} -$$ -for all $0 \lt x \le 1/\sqrt7$. In fact, it is not difficult to prove \eqref{starstar} for $x\le0.693/\sqrt7$ as follows: Let $\lVert a\rVert_\infty=\max_n\lvert a_n\rvert$. We split into two cases, - -$\lVert a\rVert_\infty\ge x$. In this case, choose $n$ such that $\lvert a_n\rvert\ge x$. Flipping the sign of $r_n$ does not affect the distribution of $S$, but if $\lvert S\rvert < x$ then it changes its value such that $\lvert S\rvert\ge x$, so that $\mathbb P(\lvert S\rvert\ge x)\ge\mathbb P(\lvert S\rvert \lt x)$, giving the result. -$\lVert a\rVert_\infty < x$. In this case, if we let $\Phi(x)$ be the standard normal cumulatiive probability function, the Berry–Esseen theorem gives -$$ -\mathbb P(\lvert S\rvert\ge x)\ge 2\Phi(-x)-2C\lVert a\rVert_\infty \gt 2\Phi(-x)-2Cx -$$ -for a global constant $C$. We can use $C=0.56$ (as stated in the Wikipedia page, this was proven by Shevstova in 2010, in An improvement of convergence rate estimates in the Lyapunov theorem). Evaluating the right hand side with $x=0.693/\sqrt7$ gives a value greater than 1/2. QED - -Finally, I mention that I have confirmed \eqref{star} numerically on a dense grid of values for $a$ and, by bounding the interpolation errors, should in principle lead to a (rather unsatisfactory) proof. - -REPLY [6 votes]: Addressed in Theorem 1.3 in Dvořák and Klein - Probability mass of Rademacher sums beyond one standard deviation (not yet peer reviewed). It describes a computer program that verifies $\Pr[\lvert S\rvert \geq 1/\sqrt{7} - \epsilon] \geq 1/2$, with concrete $\epsilon > 0$. Giving it more time (polynomial in $1/\epsilon$), it empirically seems we may take $\epsilon\to 0$. -Apologizes for the unsatisfactory answer :)<|endoftext|> -TITLE: Resolving singularities in one fell swoop -QUESTION [8 upvotes]: From what I understand, resolution of singularities (in characteristic 0) is proved and implemented inductively. You repeatedly blow up your variety along subsets of your singular locus in a way that decreases the "severity" of your singular locus. -At the end of "On the problem of resolution of singularities in positive characteristic" (link), Hauser says: - -"From Hironaka’s theorem it follows (at least in characteristic zero) that there does exist another ideal structure on the singular locus of a variety so that the induced blowup with this center resolves the singularities in one single stroke. Formidable!" - -So in contrast to the usual inductive way of resolving singularities, there is a way of resolving singularities "all at once." -Question: What does Hauser mean by "another ideal structure" on the singular locus? -The following was my initial thought. If $k$ is a field of characteristic 0 and $X$ is a variety over $k$ with singular locus $Z$, we want some ideal sheaf $\mathcal{I}$ such that $\mathcal{O}_X/\mathcal{I}$ has support $Z$, and such that blowing up $X$ at $\mathcal{I}$ is smooth. Actually constructing $\mathcal{I}$ is presumably difficult, since it should compensate for any difficulties that arise from resolving $Z$. -However, blow-ups are determined by their center, so this initial thought isn't right. For example, there are isolated singular points that require multiple blow-ups to be resolved. How could such a singular point ever be resolved by a single blow-up? Perhaps there's a larger center containing the singular locus that one should blow up? -Edit: As Donu points out, if a singular locus can be resolved, it can be resolved by a single blow-up for entirely formal reasons. So the locus cut out by an ideal that resolves our singular locus in one fell swoop should contain, but need not be equal to, our singular locus. This answers the previous paragraph. -The question then becomes whether the singular locus determines the "resolving ideal" in any tractable way. - -REPLY [7 votes]: The reasons for this are rather formal. Any projective birational morphism of varieties is the blow up of some ideal, see chap II, theorem 7.17, of Hartshorne. So although the statement you quote sounds striking, I don't think it is a terribly useful. -Let me add a few more remarks, even though it won't answer your modified question. Blowing up a complicated ideal or closed subscheme would hard understand geometrically. What makes Hironaka, and various other resolution proofs, useful is not just that the resolution exists, but that it can be achieved by blowing up along a succession of smooth centres.<|endoftext|> -TITLE: Weight enumerator classifiers -QUESTION [8 upvotes]: Let $f(x,y)$ be a polynomial with integer coefficients. What conditions guarantee that this is the weight enumerator of a binary linear code of size $n$ and dimension $k$? -I’m almost certain that the answer to this question is unknown...so instead i’ll settle for anything that is conjectural. -There’s a list of necessary conditions: - -$f$ must be homogeneous of degree $n$ with non-negative coefficients. - -The $x^n$ coefficient has to be $1$ since the zero vector is the unique weight $0$ vector. - -The $y^n$ coefficient has to be $0$ or $1$ since the all $1$’s vector either belongs to the code or doesn’t. - -The sum of the coefficients has to be $2^k$ since every vector has a unique weight and so is counted exactly once by some coefficient. - -The MacWilliams transform ($g(x,y) = \frac{1}{2^k}f(x+y,x-y)$) has to have all of the above properties but with coefficient sum $2^{n-k}$ since if $f$ corresponds to a code then $g$ would correspond to the dual code. - - -Are there any more necessary conditions missing? - -REPLY [4 votes]: This answer has grown so split it into sections. So, let me summarize. -First in Section 1 we give a counterexample showing more conditions are needed to guarantee we have the weight enumerator of a linear code. In Section 2 we show that the conditions in the question do classify weight enumerators for constant weight linear codes. Finally in Section 3 we give some additional necessary conditions (but not a complete solution). -Section 1. A "MacWilliams coincidence" example: -Let $f(x,y) = \sum_{j=0}^n a_j x^{n-j}y^j$. One other simple condition not in the question is $a_j \leq \binom{n}{j}$. But we also need some lower bounds of the coefficients $a_j$. -Consider the nonlinear code -\begin{align*} - 00000000 & & 11111111 \\ - 11000000 & & 00111111 \\ - 10100000 & & 01011111 \\ - 10010000 & & 01101111 \\ - 10001000 & & 01110111 \\ - 10000100 & & 01111011 \\ - 10000010 & & 01111101 \\ - 10000001 & & 01111110 -\end{align*} -which is binary of length $8$ with $16$ elements. -It's weight enumerator is -$$ f(x,y) = x^8 + 7x^6y^2 + 7x^2y^6 + y^8$$ -and -$$ \frac{1}{16}f(x+y, x-y) = f(x,y).$$ -This example is in "The MacWilliams identities for nonlinear codes" by MacWilliams, Sloane, and Goethals which can be viewed on Sloane's webpage. -We see $f$ satisfies all the conditions we have listed so far. -However, $f$ cannot be the weight enumerator of any linear code. -Indeed if it were for some linear code then $u+v$ must have weight $2$ for any two codewords $u$ and $v$ of weight $2$. -This means for any two weight $2$ codewords $u$ and $v$ they have exactly one coordinate in common. -We may assume the first two weight $2$ codewords are $11000000$ and $10100000$. -If the next weight $2$ codeword is $10010000$, then $11110000$ would be a codeword. -Alternatively the next weight $2$ codeword could be $01100000$, but then we cannot complete to seven codewords of weight $2$ without forcing a codeword on weight $4$. -In additional to the conditions in the question we need some lower bound on coefficients. We see here that the seven codewords of weight $2$ force the existence of a weight $4$ codeword. (I guess the above shows if $a_2 > 3$, then $a_4 > 0$.) -Section 2. MacWilliams is enough for constant weight: -A constant weight linear is a linear code where all codewords have the same weight (with the exception of the all zero codeword). -It is known that any $k$-dimensional constant weight binary linear code has generator matrix whose columns consist (with possible replication allowed) of all nonzero binary words of length $k$ along with zero columns. -For example, the matrix -$$ -\begin{bmatrix} -1 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 0 & 1 \\ -0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 1 -\end{bmatrix}$$ -generates $2$-dimension constant weight of weight $6$. -So, the weight generator of a constant weight linear code looks like -$$f(x,y) = x^n + (2^{k}-1)x^{n-m2^{k-1}}y^{m2^{k-1}}$$ -for some $k$, $m$, and $n$. -It turns out that this is the only possible way to satisfy our conditions with only $x^n$ and one other monomial. -Assume $f(x,y) = x^n + (2^{k}-1)x^{n-a}y^a$ satisfies MacWilliams in the sense that -$$\frac{1}{2^k}f(x+y,x-y) = \frac{1}{2^k} \left( (x+y)^n + (2^k - 1)(x+y)^{n-a}(x-y)^a\right)$$ -has integer coefficients. -This would mean that -$$\frac{1}{2^k} \left( (x+y)^n - (x+y)^{n-a}(x-y)^a\right)$$ -has integer coefficients. -The coefficient of $x^{n-1}y$ in -$$(x+y)^n - (x+y)^{n-a}(x-y)^a$$ -is $2a$. -Hence, $2^k$ divides $2a$ and so $a = m2^{k-1}$. -Section 3. some (easy) lower bounds: -Here are two fairly simple conditions based off of $a_1$ and $a_{n-1}$ to get lower bounds on the coefficients. -Let $[n] = \{1,2,\dots, n\}$ and let $\binom{A}{j}$ denote all $j$-element subsets of a set $A$. We let $\{e_i : i \in [n]\}$ denote the standard basis. For any $B \subseteq [n]$ set $e_B = \sum_{i \in B} e_i$. Assume $f(x,y)$ is the weight enumerator of a linear binary code $C$. Then there is some $A \subseteq [n]$ with $|A| = a_1$ such that $e_i \in C$ if and only if $i \in A$. It follows that $e_B \in C$ for $B \subseteq A$. So, $e_B$ is a codeword of weight $j$ for each $B \in \binom{A}{j}$. Hence, we get the condition -$$\binom{a_1}{j} \leq a_j$$ -for each $1 \leq j \leq n$. -Also, we have $A \subseteq [n]$ with $|A| = a_{n-1}$ such that $e_{[n]} + e_i \in C$ if and only if $i \in A$. Now for any $B \subseteq A$ we have -$$\sum_{i \in B} (e_{[n]} + e_i) = |B|e_{[n]} + e_B$$ -which is either $e_B$ or $e_{[n]} + e_B$ depending on the parity of $|B|$. -Hence, we have -$$\binom{a_{n-1}}{2j} \leq a_{2j}$$ -and -$$\binom{a_{n-1}}{2j+1} \leq a_{n-2j-1}.$$ -Ideally this sort of idea could be continued in a more complicated way with $a_2, a_3, \dots$ for codes with larger minimum distance. But this at least cuts down the candidate space some from the conditions in the original question. There are certainly more conditions. As we saw earlier $a_2 > 3$ implies $a_4 > 0$. It would be nice to find some rule saying $a_2 > m$ implies $a_4 > \phi(m)$ for some $\phi$.<|endoftext|> -TITLE: Deep applications of the Pettis integral? -QUESTION [9 upvotes]: In the Notes section of chapter 2 of Diestel and Uhl's Vector Measures they make the comment: -"Presently the Pettis integral has very few applications. But our prediction is that when (and if) the general Pettis integral is understood it will pay off in deep applications." ( Google books link ) -That was back in $1977$ and when I did a little searching to find out how well that prediction had fared I found little: wikipedia provides a cursory description and Encyclopedia of Mathematics is arguably better but terser. This question asks why (Dunford-)Pettis integrals are useful, but I would not say it covers deep applications per se. -So: are there deep applications to the Pettis integral, and if so, what are they? - -REPLY [10 votes]: Diestel and Uhl commented on this in Measure theory and its applications:<|endoftext|> -TITLE: Is there an $\mathbb{R}$-valued cohomology theory for varieties over $\mathbb{F}_p$? -QUESTION [12 upvotes]: If $E$ is a supersingular elliptic curve over $\mathbb{F}_{p^m}$ with $m\geq 2$ its endomorphism ring is a maximal order in a quaternion algebra ramified at $p$ and $\infty$ so there can't be a Weil cohomology with coefficients in $\mathbb{Q}_p$ or $\mathbb{R}$. -For varieties over $\mathbb{F}_p$ there is a $\mathbb{Q}_p$-valued cohomology theory (the crystalline cohomology). -Is there an $\mathbb{R}$-valued cohomology theory as well? - -REPLY [16 votes]: Let me advertise a conjectural positive answer to a slightly different question. This actually works not only over $\mathbb F_p$ but over its algebraic closure $\overline{\mathbb F}_p$. -Consider the following $\mathbb R$-linear category, sometimes called the category of real isocrystals $\mathrm{Isoc}_{\mathbb R}$. Objects are finite-dimensional $\mathbb C$-vector spaces $V$ together with a ($\mathbb C$-linear) grading $V=\bigoplus_{i\in \mathbb Z} V_i$, and a $\mathbb C$-antilinear graded isomorphism $\alpha: V\to V$ (i.e., $\alpha(xv)=\overline{x}\alpha(v)$ for $x\in \mathbb C$ and $v\in V$, and $\alpha(V_i)=V_i$) such that $\alpha^2|_{V_i} = (-1)^i$. (Thus, $\alpha$ induces a real structure on the even part of $V$, and a quaternionic structure on the odd part of $V$.) - -Conjecture (See Conjecture 9.5). There is a Weil cohomology theory for varieties over $\overline{\mathbb F}_p$ with values in $\mathrm{Isoc}_{\mathbb R}$. - -The grading should correspond to the weight decomposition (always split in this case, as all motives over $\overline{\mathbb F}_p$ are pure). In particular, we see that for motives concentrated in even weights, a fibre functor with $\mathbb R$-coefficients ought to exist, even for varieties over $\overline{\mathbb F}_p$, refining the previous answer. -The conjecture is known to follow from the Tate conjecture. However, I'd believe there should be a direct way to construct it (like etale and crystalline cohomology), but I don't have any insight into how. -Why is this an analogue of isocrystals? Kottwitz has constructed for any local or global field $F$ an $F$-linear Tannakian category, that for nonarchimedean local fields reduces to isocrystals, and for $\mathbb R$ gives the above category. For function fields, it gives a category of isoshtukas; for number fields, a linear-algebraic description is unknown. Conjecturally, a Weil cohomology theory should even exist with values in Kottwitz' category for $F=\mathbb Q$. The latter Weil cohomology theory should even induce a fully faithful functor from motives over $\overline{\mathbb F}_p$ into Kottwitz' category for $F=\mathbb Q$; this is closely related to the Langlands--Rapoport conjecture.<|endoftext|> -TITLE: Homotopy equivalent Postnikov sections but not homotopy equivalent -QUESTION [19 upvotes]: Two pointed, connected CW complexes with the same homotopy groups need not be homotopy equivalent (Are there two non-homotopy equivalent spaces with equal homotopy groups?). Moreover, having the same homotopy and homology groups is also not enough (Spaces with same homotopy and homology groups that are not homotopy equivalent?). -Question: Suppose that two pointed, connected CW complexes have homotopy equivalent $n$-Postnikov sections for every $n \geq 1$. Are the spaces homotopy equivalent? -I expect the answer to the question to be negative, but I'm having a hard time finding a counterexample. So far, I know that such a counterexample would have both spaces have the same homology and homotopy groups, and that both spaces would need to have nontrivial homology and homotopy groups in arbitrarily high dimension. - -REPLY [24 votes]: This is a pretty well-known phenomenon, linked with phantom maps. -One of the first existence results was Brayton Gray's paper - -Spaces of the same $n$-type, for all $n$, Topology -5 (1966) 241--243 - -Clarence Wilkerson classified the spaces of the same $n$-type for all $n$ in - -Classification of spaces of the same $n$-type for all $n$, Proc. Amer. -Math. Soc., -60 (1976) 279--285 (1977)<|endoftext|> -TITLE: Finding solutions of the differential equation $x\frac{d}{d x}\left(x\frac{d y}{d x}\right)=4(y-x^{2}y+y^{3})$ -QUESTION [7 upvotes]: In my research I have come across the following non-linear differential equation: -$$x\frac{d}{d x}\left(x\frac{d y}{d x}\right)=4(y-x^{2}y+y^{3})$$ -I want to find the general solution of this equation for $x\geq0$, but didn't manage to do it. -I have found some facts about the solutions by investigating the asymptotics: - -For $x\ll1$, the solutions go like $y=Cx^2$ for some constant $C$, or diverge. -For $x\gg1$, the solutions go like $y=x$, or go to $0$ (probably after oscillating, found by numerics), or diverge. - -In particular, I am interested in a one of the solutions. By looking upon the numerics, I believe that there is only one solution that goes like $y=x$ for $x\gg1$, and it does not diverge at $x=0$. I didn't manage to prove this by now. -I am interested in finding this solution analytically, or if impossible, just find appropriate $C$ analytically. -Thanks! - -REPLY [3 votes]: Is this too naive? To obtain the large-$x$ behavior, I take a series expansion of $y(x)$ in powers of $1/x$, -$$y(x)=c x + a_0 + a_1/x + a_2/x^2+ a_3/x^3 +\cdots,$$ -substitute into -$$R(x)=x\frac{d}{d x}\left(x\frac{d y}{d x}\right)-4(y-x^{2}y+y^{3}),$$ -expand $R(x)$ in powers of $1/x$ and demand that the leading order terms vanish. It follows that $c=1$, $a_{n}=0$ for any even $n$, while for odd $n$ I find -$$a_1=-\frac{3}{8},\;\;a_3=-\frac{9}{128},a_5=-\frac{99}{1024},\;\;a_7=-\frac{11637}{32768},\;\;a_9=-\frac{627669}{262144},$$ -so coefficients of the form $a_{2n+1}=-b_n/2^{2n+1}$ with integer $b_n$.<|endoftext|> -TITLE: Restriction of holomorphic functions on $G$-invariant subspace -QUESTION [6 upvotes]: Let $X$ be a complex manifold with a holomorphic action of a complex reductive group $G$. Let $Y \subset X$ be a $G$-invariant reduced complex analytic subspace. Is the restriction -$$ -\mathcal{O}_X^G \to \mathcal{O}_Y^G -$$ -surjective (on stalks)? Here $\mathcal{O}_X^G$ is the sheaf of $G$-invariant holomorphic functions on $X$ and similarly for $Y$. - -REPLY [2 votes]: After writing an answer in terms of global sections, then realising the original question was in terms of stalks, I realised that I don't understand what "the sheaf of $G$-invariant holomorphic functions on $X$" means: what is $\mathscr O_X^G(U)$, if $U$ is a non-$G$-stable open subset of $X$? I wonder if you might mean the sheaf of holomorphic functions on the quotient $X/G$, which is defined by putting $\mathscr O_{X/G}(\overline U_X) = \mathscr O_X(U_X)^G$ for any open subset $\overline U_X$ of $X/G$ with pullback $U_X \subseteq X$. I'll answer the question that way, but, since the technique I describe is very robust, hopefully I can adapt it to whatever the correct interpretation is. -Fix a point $\overline y \in Y/G$, and a germ $f$ at $\overline y$ (that is, an equivalence class of $G$-invariant holomorphic functions on $Y$ defined on the pullback of an open neighbourhood in $Y/G$ of $\overline y$). -Fix an open subset $\overline U_X$ of $X/G$ containing $\overline y$ that is small enough that $f$ is defined on $\overline U_X \cap Y/G$ and that, if we write $U_X$ for the pullback to $X$ of $\overline U_X$, then $f$ extends to a holomorphic function on $U_X$. Then consider the space $\mathscr V$ of functions on $X$ whose restriction to $Y$ is proportional to $f$. This is a $G$-representation, hence also a $\mathfrak g$-representation. -Let $K$ be the compact form of $G$. Since $K$ is compact, the representation $\mathscr V$ has a non-$0$, $K$-fixed vector $F$. The span of $F$ is annihilated by $\mathfrak k$, hence by $\mathfrak g = \mathfrak k \otimes_{\mathbb R} \mathbb C$; so $F$ itself is $G$-fixed.<|endoftext|> -TITLE: Interpretations of modal logic where $\Box$ means "valid" -QUESTION [8 upvotes]: Consider the propositional modal language in one propositional letter, $p$. -Recall that a pointed Kripke frame is a Kripke frame $(W,R)$ with a designated world $w_0\in W$, and a sentence is valid in a pointed Kripke frame iff it is true at $w_0$ for every interpretation of the propositional letters as subsets of $W$. -I'm wondering if it's possible to find a finite model in which $\Box$ means "valid". More precisely, is it possible to find a finite transitive reflexive pointed Kripke model $(W,R,w_0, [[\cdot]])$ such that -$w_0 \Vdash \Box A$ if and only if $A$ is valid in $(W,R,w_0)$? -Certainly it can be done in an infinite frame. For instance, over the infinite tree that has omega many daughters at any node, you can make each sentence satisfiable in the frame true at one of the daughters of the base node. And this even works with infinitely many propositional letters. -(For context: I got thinking about this question after coming back to this earlier question about logical interpretations of $\Box$.) - -REPLY [10 votes]: $\def\R{\mathrel R}$No, this is not possible. -Recall that the depth of a point $x$ in a transitive frame $(W,R)$ is the maximal length $d$ of a strictly increasing chain starting at $x$, i.e., $x_1,\dots,x_d$ such that $x_d=x$ and $x_{i+1}\R x_i$, $x_i\not\R x_{i+1}$. -There are formulas in one variable that are satisfiable only in frames of depth $\ge d$ (cf. Thm. 12.21 in Chagrov&Zakharyaschev, Modal logic). Moreover, it is possible to define them in such a way that when satisfied in a model of depth exactly $d$, they force a particular value for $p$ in all points in the root cluster; we will obtain a contradiction from this. An explicit construction of such formulas follows below. -Consider the formulas -$$\begin{align} -\theta_1(p)&=\Box p,\\ -\theta_{i+1}(p)&=p^{i+1}\land\Diamond\theta_i(p)\land\Box\Bigl(p^{i+1}\lor\bigvee_{j\le i}\theta_j(p)\Bigr), -\end{align}$$ -where -$$p^i=\begin{cases}\phantom{\neg}p&\text{if $i$ is odd,}\\\neg p&\text{otherwise.}\end{cases}$$ -We will use the property that these formulas are pairwise contradictory; moreover, the following formulas are valid: -$$\theta_j\to\Box\neg\theta_i,\qquad j -TITLE: pursuit-evasion based on Schroeder's upper bound for graphs of genus $g$ -QUESTION [6 upvotes]: I am following Schroeder's work on pursuit-evasion games on graphs (often called "cops and robbers"). In his 2001 publication ("The copnumber of a graph is bounded by $\lfloor 3/2 {\ \rm genus}(G)+3\rfloor$". In: Categorical perspectives (Kent, OH, 1998). Trends in Mathematics, pp. 243-263. Birkhäuser, Boston 2001) he derived an upper bound for the cop number $c(G)$ that depends on the genus $g$ of the surface on which the graph $G$ can be embedded: $c(G)\leq \lfloor 3g/2 +3\rfloor$. -My most recent reference for this result is "Topological directions in Cops and Robbers" from 2018, Anthony Bonato and Bojan Mohar, arXiv:1709.09050v2 . -This gives $c(G)\leq 4$ if $G$ can be embedded on a torus. Now, I have worked extensively to come up with an example of a graph $G$ that actually hits this bound, i.e. I have searched for $G$ with $c(G)=4$, but with no success. So I am starting to see strong evidence for the conjecture $c(G)\leq 3$ if $G$ can be embedded on a torus. Question: Is someone aware of a more recent reference for this conjecture? It appears lower than any other bound I have seen in the literature so far (N.B. I would also be interested in references beyond torus embeddings) - -REPLY [7 votes]: What you conjecture has been conjectured (more or less explicitly) a few times before. In the paper by Bonato and Mohar that you reference, it is dubbed the Andreae-Schroeder conjecture. -I recently proved that it is true, i.e. the cop-number of toroidal graphs is at most 3, see this ArXiv preprint. See also this preprint, where a general bound $c(G) \leq \frac{4g}{3} + \frac {10}3$ is proved.<|endoftext|> -TITLE: Conjecture of van der Holst and Pendavingh related to bound for Colin de Verdière invariant -QUESTION [9 upvotes]: In their 2009 paper (“On a graph property generalizing planarity -and flatness”. In: Combinatorica 29.3 (May 2009), pp. 337–361. issn: 1439-6912. -doi: 10.1007/s00493-009-2219-6.), van der Holst and Pendavingh defined a new minor monotone -graph invariant $\sigma(G)$ for a graph $G$: the minimal integer $k$ such that every CW-complex whose 1-skeleton is $G$ admits a certain even mapping into $\mathbb R^k$. -They were able to prove $\mu(G)\leq\sigma(G)+2$, where $\mu(G)$ is the Colin de Verdière number of $G$ which is minor monotone as well (Colin de Verdière. “Sur un nouvel invariant des graphes et un critère de planaritè”. In: Journal of Combinatorial Theory, Series B 50.1 (1990), pp. 11–21. -issn: 0095-8956. doi: 10.1016/0095-8956(90)90093-F.) -My main interest is in the conjecture of van der Holst and Pendavingh in that paper. They conjectured that actually $\mu(G)\leq\sigma(G)$ might hold. Question: What is known about the status of this conjecture? (I have difficulties tracing it as their new invariant $\sigma(G)$ does not seem to have a commonly agreed name yet). - -REPLY [7 votes]: I would like to add one important aspect: it was known that $\mu(G)$ and $\sigma(G)$ can deviate by a large amount for larger values $k$. Now we have the proof of the improved (sharper) bound $\mu(G)\leq\sigma(G)$, but even though this is an improvement, Kaluza and Tancer also showed that a large gap exists already for small values of $k$: They showed there is a graph $G$ such that $\mu(G)\leq7$ and $\sigma(G)\geq8$ ("Even maps, the Colin de Verdière number, and representations of graphs" on arxiv. Here is the link https://arxiv.org/pdf/1907.05055.pdf). -Now, a suspension of $G$ (adding a new vertex to $G$ and connecting it to all vertices of $G$) increases both $\mu(G)$ and $\sigma(G)$ by exactly one (unless $G$ is the complement of $K_2$). Therefore $\mu(G)\leq7$ and $\sigma(G)\geq8$ implies that for every $k \in\mathbb N$, $k \geq 7$, there must exist a graph $G_k$ with $\mu(G)\leq k$ and $\sigma(G)\geq k+1$, i.e. strict inequality for all large values of $k$. Finally, the authors also show that the gap between $\mu(G)$ and $\sigma(G)$ is asymptotically large.<|endoftext|> -TITLE: Primality test for $N=2^a3^b+1$ -QUESTION [7 upvotes]: Can you prove or disprove the following claim: - -Let $N=2^a3^b+1$ , $a>0 , b>0$ . If there exists an integer $c$ such that $$c^{(N-1)/3}-c^{(N-1)/6} \equiv -1 \pmod{N}$$ then $N$ is a prime. - -You can run this test here. I have verified this claim for all composite $N$ up to $2^{100} \cdot 3^{100}+1$ with $2 \le c \le 100$ , and for all prime $N$ from this list. - -REPLY [15 votes]: Yes. Obviously this $c$ and $N$ are coprime. We get $c^{(N-1)/2}+1=(c^{(N-1)/6}+1)(c^{(N-1)/3}-c^{(N-1)/6}+1)$ is divisible by $N$. Therefore $c^{N-1}-1$ is divisible by $N$, and $N-1$ is divisible by $k:={\rm {ord}}(c)$, where ${\rm ord}(x)$ denotes the multiplicative order of $x$ modulo $N$. But $(N-1)/2$ is not divisible by $k$, since $c^{(N-1)/2}\equiv -1\pmod N$. Assume that $(N-1)/3$ is divisible by $k$; then $c^{(N-1)/3}\equiv 1 \pmod N$, $c^{(N-1)/6}\equiv c^{(N-1)/3}+1\equiv 2 \pmod N$ and $$1\equiv c^{(N-1)/3}= (c^{(N-1)/6})^2\equiv 4\pmod N,$$ -a contradiction. So neither $(N-1)/2$ nor $(N-1)/3$ is divisible by $k$; thus, $k=N-1$. But $k$ must divide $\varphi(N)$, so $N-1\leqslant \varphi(N)$ and $N$ is prime.<|endoftext|> -TITLE: Irreducible representation of $S_n$: contained in tensor powers of the standard representation? -QUESTION [13 upvotes]: Let $S_n$ be the permutation group and $V = \operatorname{Fun}(X,\mathbb{k})$ functions from $X=\{1,\dotsc,n\}$ to some field $\mathbb{k}$. How can I prove that every irreducible representation of $S_n$ occurs in $V^{\otimes m}$ for integer $m$ large enough? -This is a standard fact when $n!\neq 0$ in $\mathbb{k}$. - -REPLY [10 votes]: Benjamin Steinberg answered the question, but I wanted to unwind his idea into an explicit formula. -Let $V$ be the representation with basis $e_1,\dotsc, e_n$, where a permutation $\sigma$ acts by sending $e_i$ to $e_{\sigma(i)}$. -Let $W$ be an irreducible representation of $S_n$. -Fix a linear form $l$ on $W$. We can map $W$ to $V^{\otimes n}$ by sending $w \in W$ to -$$f(w) = \sum_{ \sigma \in S_n} l ( \sigma^{-1} (w))\cdot e_{\sigma (1) }\otimes e_{\sigma(2)} \otimes \dotsm\otimes e_{ \sigma(n) }.$$ -Then we have -\begin{align*} -\sigma' (f(w)) & {}= \sum_{ \sigma \in S_n} l ( \sigma^{-1} (w))\cdot \sigma'( e_{\sigma (1) }\otimes e_{\sigma(2)} \otimes \dotsm e_{ \sigma(n) } ) \\ -& {}= \sum_{ \sigma \in S_n} l ( \sigma^{-1} (w))\cdot e_{ \sigma'(\sigma(1))} \otimes e_{ \sigma'(\sigma(2))} \otimes \dotsm \otimes e_{ \sigma'(\sigma(n))} \\ -& {}= \sum_{ \sigma \in S_n} l ( (\sigma' \circ \sigma)^{-1}( \sigma'(w)))\cdot e_{ \sigma'(\sigma(1))} \otimes e_{ \sigma'(\sigma(2))} \otimes \dotsm \otimes e_{ \sigma'(\sigma(n))} \\ -& {}= \sum_{ \sigma \in S_n} l ( \sigma^{-1} (\sigma' (w) ))\cdot e_{\sigma (1) }\otimes e_{\sigma(2)} \otimes \dotsm\otimes e_{ \sigma(n) } = f(\sigma' (w)) -\end{align*} -using the change of variables $\sigma \mapsto \sigma' \circ \sigma$ in the last line. So $f$ is a homomorphism, and because $W$ is irreducible, and $f$ is nontrivial (as long as $l$ is a nontrivial linear form, since all the terms in the sum give different basis vectors in the tensor product), this map is the inclusion of a subrepresentation, as desired. -One can get embeddings into $V^{\otimes m}$ for higher values of $m$ by just putting repetitions in the sequence of basis vectors being tensored, or for $n-1$ by removing the last term, but one can't go lower than $n-1$, because of the sign representation (except maybe in characteristic $2$, I guess).<|endoftext|> -TITLE: Sylow subgroups of abelian profinite groups -QUESTION [5 upvotes]: If $G$ is a finite abelian group, then we have a decomposition -$$G\cong \prod_{p} G(p)$$ -where $G(p)$ is the $p$-Sylow subgroup of $G$. This product makes sense as for all but finitely many primes $p$, we have $G_p=\{0\}$. This is proven by showing that the cardinality of $G$ and $\prod_{p} G(p)$ agree. If we now assume that $P$ is a profinite abelian group, there still exists the notion of a $p$-Sylow subgroup $P(p)$ which is now a pro-$p$-group. I'm curious if there exists an isomorphism -$$P\cong \prod_{p} P(p).$$ - -REPLY [10 votes]: This is Proposition 2.3.8 of Ribes and Zaleskii - Profinite groups (second edition). (I originally gave references specifically for the finer structure of profinite Abelian groups, but assuming finite generation, in Section 4.3 of the same book.)<|endoftext|> -TITLE: Survey of recent developments of the Gelfand-Kirillov dimension -QUESTION [7 upvotes]: It is almost two decades since the now classical books by McConnell and Robinson's - -[ Noncommutative Noetherian rings. With the cooperation of L. W. Small. Revised edition. Graduate Studies in Mathematics, 30. American Mathematical Society, Providence, RI, 2001 ], - -and Krause and Lenagan's - -[ Growth of algebras and Gelfand-Kirillov dimension. Revised edition. Graduate Studies in Mathematics, 22. American Mathematical Society, Providence, RI, 2000. ], - -which are were (and still are in my opinion), the standard references on almost everything related to the Gelfand-Kirillov dimension, appeared. -Time has passed, and a lot of new work on this dimensional invariant has been done. -I am looking for references, surveys and pherhaps lecture notes on the Gelfand-Kirillov dimension which covers relevant developments regarding this invariant in the last 20 years. -Regarding its computational aspects, one has for instance - -J. Bueso, J. Gomés-Torrecillas, A. Verschoren, [ Algorithmic methods in non-commutative algebra. -Applications to quantum groups. Mathematical Modelling: Theory and Applications, 17. Kluwer Academic Publishers, Dordrecht, 2003 ], - -but it does not cover all aspects of recent developments. - -REPLY [4 votes]: This list is certainly far from being complete, but it contains some important results obtained in the last 20 years. -The following thesis discusses some recent results obtained by Bell (see Section 5): -Michelle Roshan Marie Ashburner (2008). A Survey of the Classification of Division Algebras over Fields. Master Thesis, University of Waterloo -This is a survey on GK dimension of graded PI-algebras: -L. Centrone, On some recent results about the graded Gelfand-Kirillov dimension of graded -PI-algebras, Serdica Math. J. 38(1-3) (2012), 43-68. -Centrone also wrote other papers where he proved some interesting results on GK dimension. For instance, he wrote: -L. Centrone, The graded Gelfand-Kirillov dimension of verbally prime algebras, Linear Multilinear Algebra 59(12) (2011), 1433-1450. -and -L. Centrone, A note on graded Gelfand-Kirillov dimension of graded algebras, J. Algebra -Appl. 10(5) (2011), 865-889. -For some results on Hopf algebras with finite GK dimension, see: -Zhang, G. (2013). Hopf algebras of finite Gelfand-Kirillov dimension. PhD Thesis, University of Washington -To conclude, GK dimension has been recently extended to algebras over commutative domains by Zhang and Bell. Now, GK dimensions can be studied on many new structures. In the following paper, GK is studied for skew PBW extensions -Reyes, A.: Gelfand–Kirillov dimension of skew PBW extensions. Rev. Col. Mat. 47(1), 95–111 (2013) -while in this one it has been studied for rings: -Lezama, O., Venegas, H. Gelfand–Kirillov dimension for rings. São Paulo J. Math. Sci. 14, 207–222 (2020). -I'm not aware of any survey discussing all these new developments.<|endoftext|> -TITLE: Higher order generalization of Cauchy-Schwarz? -QUESTION [11 upvotes]: Is there a generalization of the Cauchy-Schwarz inequality along the following lines? Let $V$ be an inner product space (for simplicity of notation, let us work over the real numbers). Let $v_1, \ldots, v_n$ be in $V$. Let $G$ denote the Gram matrix of the $v_i$, namely, $G$ consists of all possible $(v_i, v_j)$, as $i,j = 1, \ldots, n$, where $(-,-)$ is the inner product in $V$. The usual Cauchy-Schwarz inequality, with $n=2$, can be written as follows, to get rid of square roots: -$$ \det(G) = (v_1,v_1)(v_2,v_2) - (v_1,v_2)^2 \geq 0, $$ -with equality iff $v_1$ and $v_2$ both belong to some $1$-dimensional subspace of $V$. So in this case, for $n=2$, the LHS is a homogeneous polynomial in $G$ of degree $2$, and equality is achieved iff $v_1$ and $v_2$ both belong to some $1$-dimensional subspace. -For the general $n$ case, is there a higher degree homogeneous polynomial in $G$ which is non-negative for any $v_1, \ldots, v_n$ in $V$, and which vanishes iff the $v_i$, for $i = 1,\ldots, n$ all lie in some $1$-dimensional subspace of $V$? -(I suspect there may be such a polynomial of degree $2 \lfloor \frac{n(n+1)}{4} \rfloor$. So for instance, if $n=2$, the expected degree is $2$. If $n=3$, the expected degree is $6$, and so on.) - -REPLY [16 votes]: Yes, because the OP stated that the ground field is $\mathbb{R}$, one can simply take the octic polynomial -$$ -Q(v_1,v_2,\ldots,v_n) = \sum_{1\le i < j\le n} \bigl((v_i,v_i)(v_j,v_j)-(v_i,v_j)^2\bigr)^2, -$$ -which will do the trick.<|endoftext|> -TITLE: Literature request: Schatten class difference of semigroups -QUESTION [9 upvotes]: Let $\mathcal{H}$ be a Hilbert space and $A,B$ two operators on it (not necessarily self-adjoint) such that $A, A+B$ are generators of strongly continuous one parameter semigroups $e^{-tA},e^{-t(A+B)}$, $t>0$. I would like to ask for some literature on the consequences of the condition -\begin{equation} -e^{-t(A+B)}-e^{-tA} \in S_{p}(\mathcal{H}) -\end{equation} -where $p\geq 1$ and $S_{p}(\mathcal{H})$ is the p-Schatten class of operators on $\mathcal{H}$. In particular, I would like results crucially using that this difference is Hilbert-Schmidt, for example, and not just that it is compact as a result. - -REPLY [8 votes]: The problem is discussed in a more general setting (operator ideals in Banach spaces) for the so-called analytic semigroups (parabolic problems) in -Blunck, S.; Weis, L., Operator theoretic properties of differences of semigroups in terms of their generators, Arch. Math. 79, No. 2, 109-118 (2002). ZBL1006.47036. -The paper seems to be freely accessible. The idea is, if the differences (of some fractional powers) of the resolvents belong to the ideal and have a nice asymptotics, then the differences of the semigroups belong to the same ideal and have a nice decay. -The application they give is $e^{t(\Delta - V)} - e^{t\Delta}$ on the whole space, where the semigroups are not even compact, but the difference belongs to the Schatten classe you wish for.<|endoftext|> -TITLE: Axiomatic definition of quantum groups -QUESTION [11 upvotes]: This is a question I've discussed with a lot of mathematicians, and have read some mathematical texts about, and watched some conference talks about: what is, axiomatically, a quantum group? -There are many classes of noncommutative algebras that everybody agrees is a quantum group (or quantum algebra): quantizations of certain coordinate rings, quantizations of enveloping algebras, quantizations of semisimple algebraic groups, multiparameter quantizations of the Weyl algebra, etc; but what is the state of the art of attempts to give an axiomatic definition for this class of algebras? -A related MO question is What is quantum algebra?. A nice and leisure discussion, albeit not axiomatic, is Shahn Majid's 'What Is... a Quantum Group' (here). - -REPLY [6 votes]: I would say that if you are looking for a concrete definition then it's better to adopt the Tannakian point of view and to focus on the category of representations of the quantum group rather than on the algebra itself. So take as your fundamental object a tensor category (a special type of rigid abelian monoidal category - see here for details). A "quantum group" is then some way of realising the category as a category of representations or corepresentations. There can be different algebras which do this job, and they can come in different flavours, such as Hopf algebras or compact quantum groups in the sense of Woronowicz. This allows one to view the various quantum groups floating around as tools to study the category itself, removing the need for any axiomatic definition. -If the object is really deserving of the name quantum group, then the tensor category should be braided, as is the case for quasi-triangular Hopf algebras and their category of modules. (see the comment of Sam Hopkins above.)<|endoftext|> -TITLE: Non-abelian Ext functor and non-abelian $H^2$ -QUESTION [6 upvotes]: Let $G$ be a group and -$$0\rightarrow K\rightarrow M\rightarrow N\rightarrow 0$$ -a short exact sequence of groups. Now these are abelian groups, if I want to show that $\text{Hom}(G,M)\rightarrow \text{Hom}(G,N)$ is surjective, I would show that $\text{Ext}^1(G,K)=0$. However, if I'm studying the same question for non-abelian groups, then I do not have the tool of derived categories at my disposal. Can this be overcome with (non-abelian) cohomology? - -REPLY [4 votes]: EDITED, taking into account the comments of Donu Arapura. -As JLA wrote, a homomorphism $f\colon G\to N$ gives an extension -\begin{equation}\label{e:E} -1\to K\to E\to G\to 1.\tag{E} -\end{equation} -This extension defines a homomorphism -$$b\colon G\to \operatorname{Out} K$$ -called the band (lien, kernel) of \eqref{e:E}. -By definition, $H^2(G,K,b)$ is the set of isomorphism classes of extensions \eqref{e:E} bound by $b$. -A cohomology class $\eta(E)\in H^2(G,K,b)$ is called neutral if the extension \eqref{e:E} splits, that is, there exists a homomorphism $G\to E$ such that the composite homomorphism $G\to E\to G$ is the identity automorphism of $G$. In this case we obtain an action $\varphi$ of $G$ on the normal subgroup $K$ of $E$, and we obtain an isomorphism $E\overset{\sim}{\to}K\rtimes_\varphi G$ with the semidirect product. -There may be more that one neutral class in $H^2(G,K,b)$: they correspond to semidirect products with different actions $\varphi$ of $G$ on $K$. I have read that there may be no neutral elements, but I don't know examples. (In the Galois cohomology setting, for a connected reductive group, by Douai's theorem there always exists a neutral element in nonabelian $H^2$; see [2], Proposition 3.1). -If $K$ is abelian, then $\operatorname{Out} K = \operatorname{Aut} K$, so $b$ is just an action of $G$ on $K$, and $H^2(G,K,b)$ is the usual abelian group cohomology $H^2(G,K)$, where $G$ acts on $K$ via $b$. -The set $H^2(G,K,b)$ can be described in terms of cocycles. See Section 1.14 in Springer [1]. -The band $b$ defines an action of $G$ on the center $Z=Z(K)$, and we may consider the usual (abelian) group cohomology $H^2(G,Z)$. From the cocyclic description of $H^2(G,K,b)$ it is clear that $H^2(G,Z)$ naturally acts on $H^2(G,K,b)$. -Moreover, if the set $H^2(G,K,b)$ is nonempty, then $H^2(G,Z)$ acts on it simply transitively; see Mac Lane, Homology, Theorem IV.8.8. The set $H^2(G,K,b)$ is nonempty if and only if a certain obstruction $\operatorname{Obs}(G,K,b)\in H^3(G,Z)$ vanishes; see Mac Lane, Theorem IV.8.7. -Note that we should not think that $H^2(G,K,b)$ "equals" $H^2(G,Z)$. First, $H^2(G,K,b)$ does not have a distinguished unit element. Secondly, $H^2(G,K,b)$ has a distinguished subset $N^2(G,K,b)$ of neutral elements. This is important because in many applications one uses nonabelian $H^2$ in order to determine whether a given extension \eqref{e:E} is split or not. -As far as I know, nonabelian $H^2$ is mostly used in the Galois cohomology setting. -Namely, if $k$ is an algebraic closure of a field $k_0$ of characteristic 0, $G=\operatorname{Gal}(k/k_0)$, -and $Y$ is a quasi-projective $k$-variety with additional structure (say, an algebraic group or a homogeneous space) such that for any $\sigma\in G=\operatorname{Gal}(k/k_0)$ there exists an isomorphism $\alpha\colon\sigma Y\overset{\sim}{\to}Y$, then it defines an extension -$$1\to \operatorname{Aut} Y\to E\to G\to 1,$$ -where $E$ is the set of such pairs $(\alpha,\sigma)$ -with a suitably defined composition law. -We obtain the cohomology class $\eta(Y)\in H^2(k_0,\operatorname{Aut} Y,b)$ -of this extension for a suitable band $b$. -The variety $Y$ (with additional structure) admits a $k_0$-model if and only if $\eta(Y)$ is neutral, that is, the extension splits; -see this question. -For nonabelian $H^2$ in Galois cohomology see: -[1] T. A. Springer, Non-abelian $H^2$ in Galois cohomology, in: Algebraic Groups and Discontinuous -Subgroups, Proc. Sympos. Pure Math. 9, Amer. Math. Soc., Providence, 1966, 164-182. -[2] M. Borovoi, Abelianization of the second nonabelian Galois cohomology. Duke Math. J. 72 (1993), 217-239. -[3] Flicker, Scheiderer, Sujatha, Grothendieck's theorem on non-abelian $H^2$ and local–global principles. -J. Amer. Math. Soc. 11 (1998), no. 3, 731–750. -See also newer papers (they refer to these three), and this preprint.<|endoftext|> -TITLE: Mori's cone theorem -QUESTION [6 upvotes]: I need the proof (reference) of Mori’s theorem about this implication : -Let $X$ be a projective complex manifold. If $X$ contains no rational curves, then $K_K$ is nef. - -REPLY [6 votes]: A more precise form of this result is originally due to Miyaoka and Mori, see Theorem 3.6 page 67 in -Olivier Debarre: Higher-dimensional algebraic geometry, Universitext. New York, NY: Springer. xiii, 233 p. (2001). ZBL0978.14001.<|endoftext|> -TITLE: Comparing generic versions of $\mathbb{R}$ -QUESTION [12 upvotes]: This question was previously asked and bountied at MSE, unsuccessfully. - -I'm currently interested in the behavior of cardinalities in generic extensions of models of $\mathsf{ZF+\neg AC}$, and especially of $\mathsf{ZF+AD}$. The following seems both natural and much simpler than various other questions I've asked on the topic: - -Suppose $V$ is a transitive model of $\mathsf{ZF+AD}$. Let $c,d$ be mutually Cohen generic over $V$; is there in $V[c,d]$ a bijection between $\mathbb{R}^{V[c]}$ and $\mathbb{R}^{V[d]}$? - -"Obviously" the answer should be yes, but I don't see how to prove that. We have a set $\mathcal{X}\in V$ of names for reals such that every real in a Cohen extension is named by some element of $\mathcal{X}$, and each Cohen real $a$ induces an equivalence relation $\sim_a$ on $\mathcal{X}$ as $\nu\sim_a\mu\leftrightarrow\nu[a]=\mu[a]$. So really this question is asking for a bijection in $V[c,d]$ between $\mathcal{X}/\sim_c$ and $\mathcal{X}/\sim_d$. Intuitively this should exist since Cohen forcing is as homogeneous as one could hope; however, in the absence of choice in $V[c,d]$ I don't actually see how to build one. -I would also be very interested in a partial negative answer for $\mathsf{ZF}$-models, but the determinacy case is really my main point of focus. - -To preempt one natural attempt, note that Cohen forcing kills determinacy so we can't use determinacy in $V[c,d]$ even though we have it in $V$. While at first glance this might appear to contradict (say) the generic absoluteness of the theory of $L(\mathbb{R})$ given large cardinals, there is no discrepancy since $(L(\mathbb{R}))^V[G]\not=(L(\mathbb{R}))^{V[G]}$ in general. - -REPLY [4 votes]: The answer seems to be no. Moreover: Suppose that every set of reals has the property of Baire. Let $\mathbb{C}$ be Cohen forcing and let $P$ be any wellorderable partial order. If $(c,d)$ is generic for $\mathbb{C} \times P$, then there is no injection in $V[c,d]$ from the Cantor space of $V[c]$ to any set in $V[d]$. -Here's a proof, which is a variation on the standard argument showing that if all sets of reals have the property of Baire, then there is no function choosing between complementary $\mathbb{E}_{0}$ degrees. We will call members of the Cantor space reals. Suppose that $\tau$ is a $\mathbb{C} \times P$ name for a function mapping the reals of $V[g_{left}]$ into $V[g_{right}]$, where $g_{left}$ and $g_{right}$ are the $\mathbb{C} \times P$-names for the left and right coordinates of the generic. We will find a condition forcing the realization of $\tau$ not to be injective. -Let $N$ be the set of nice Cohen names $\sigma$ for elements of the Cantor space which are also functions with domain $\mathbb{C}$, where each value $\sigma(p)$ has the form $\check{t}$, for $t$ an element of $\mathbb{C}$ of the same length as $p$. That is, $N$ is the set of nice $\mathbb{C}$-names for reals such that each condition of $\mathbb{C}$ decides the initial segment of the realization of the name up to the length of the condition (and maybe more). -Let $\langle q_n : n \in \omega \rangle$ be the natural enumeration of $\mathbb{C}$, where shorter sequences are listed before longer ones, and sequences of the same length are listed in lexicographic order (any enumeration in odertype $\omega$ would work). Given $p \in \mathbb{C}$, let $n_p$ be such that $p = q_{n_{p}}$. There is a natural bijection $b$ between the Cantor space and $N$, where we let $x$ in the Cantor space correspond to the set of pairs $(p, \check{t})$, where $t = \langle x(n_{p \upharpoonright i}) : i \leq |p| \rangle$. -Applying the assumption that every subset of the Cantor space has the property of Baire, and the Kuratowski-Ulam theorem to deal with the wellorderable poset $P$, we get $s$ in $\mathbb{C}$ and $(p_0, p_1) \in \mathbb{C} \times P$ such that, for comeagerly many $x$ extending $s$, there is some $\mathbb{C}$-name $\rho$ with $(p_0,p_1)$ forcing $(b(x)_{g_{left}}, \rho_{g_{right}}) \in \tau$. That is, $(p_0,p_1)$ forces some $P$-name $\rho$ to represent via $g_{right}$ the $\tau$-value for the $g_{left}$-realization of $b(x)$. However, we can find $x$ and $x'$ extending s and in this comeager set such that $x$ and $x'$ disagree at exactly one value $n^*$ with $q_{n^*}$ compatible with $p_0$, and with this $q_{n^*}$ a proper extension of $p_0$. So $p_0$ doesn't decide whether or not the realizations of $b(x)$ and $b(x')$ will be the same. -Moreover, there exist $P$-names $\rho$ and $\rho'$ such that $(p_0,p_1)$ forces both -$(b(x)_{g_{left}}, \rho_{g_{right}}) \in \tau$ -and -$(b(x')_{g_{left}}, \rho'_{g_{right}}) \in \tau$. -Let $p'_1$ be a strengthening of $p_1$ deciding whether or not $\rho$ and $\rho'$ have the same realization. -Since there are extensions of $p_0$ forcing that the realizations of $b(x)$ and $b(x')$ will be the same, and $\tau$ is a name for a function, it must be that $p'_1$ forces the realizations of $\rho$ and $\rho'$ to be the same. The condition $q_{n^*}$ however extends $p_0$ and forces that the realizations of $b(x)$ and $b(x')$ will be different, and therefore $(q_{n^{*}}, p'_1)$ forces that the realization of $\tau$ will not be an injection.<|endoftext|> -TITLE: When annihilator of ideal and ideal is co maximal -QUESTION [6 upvotes]: Let $R$ be a commutative ring with identity. It is not always true that ideal $J$ and annihilator of ideal $ann(J)$ are co maximal (ex: integral domain) - -Is there a sufficient (necessary) condition( or ring) under which this happen $J$ and $ann(J)$ are comaximal? - -REPLY [11 votes]: The necessary and sufficient condition is that $J$ be generated by an idempotent. -A. Assume $J=(e)$ is generated by an idempotent $e$ ($e^2=e$). Then the annihilator of $J$ contains $1-e$, so the sum of $J$ and its annihilator contains $e + (1-e)=1$. This shows that $J$ and its annihilator are comaximal. -B. Now suppose that the annihilator $K$ of $J$ is comaximal with $J$. That is, $J+K=R$. Then there exist $j\in J$ and $k\in K$ such that $j+k=1$. Multiplying through by $j$ one learns $j^2+jk = j^2 = j$, so $j\in J$ is an idempotent. Similarly, $k\in K$ is an idempotent. Since $J$ annihilates $K$, we have $kJ=0$, hence for $x\in J$ we have $x = x-0=x-kx = (1-k)x = jx$. We now have that $j\in J$ (i.e. $(j)\subseteq J$) and, for all $x\in J$ we have $x = jx$ (i.e. $J\subseteq (j)$), so we get that $J = (j)$. This shows that $J$ is generated by an idempotent. -August 3 Edit: -In response to the question below, -I wonder is this is a standard result (it seems to be a standard one)? -the answer is Yes, essentially. It is standard that the following are equivalent for unital rings: -(1) $J$ and $K$ are complementary ideals of $R$. $(J+K=R$, $J\cap K=(0))$ -(2) $J$ is generated by a central idempotent $(J=(e))$ and $K$ -is generated by the complementary idempotent $(K=(1-e))$. -(3) $R$ factors as $R\cong J\times K\cong R/ J\times R/K$. -Where does your condition fit into this picture? (Condition = $J$ is comaximal with its annihilator.) -It is a well-known fact that if $J$ and $K$ are comaximal, then their intersection equals their product. Therefore they are disjoint iff they annihilate each other. In particular, if $J$ annihilates $K$ and is comaximal with $K$, then $J$ is a complement to $K$. -Here is how you prove the well-known fact that if $J+K=R$, then -$J\cap K=JK$. It uses the fact that $JK\subseteq J\cap K$ for any two ideals. -$$ -\underline{J\cap K}=R(J\cap K)=(J+K)(J\cap K)=J(J\cap K)+K(J\cap K)\subseteq \underline{JK}\subseteq \underline{J\cap K}. -$$<|endoftext|> -TITLE: Algebraic graph invariant $\mu(G)$ which links Four-Color-Theorem with Schrödinger operators: further topological characterizations of graphs? -QUESTION [25 upvotes]: 30 years ago, Yves Colin de Verdière introduced the algebraic graph invariant $\mu(G)$ for any undirected graph $G$, see [1]. It was motivated by the study of the second eigenvalue of certain Schrödinger operators [2,3]. It is defined in purely algebraic terms as the maximum corank in a set of generalized Laplacian matrices of $G$. -It turned out to be very powerful concept, linking algebraic with topological graph theory (and, by conjecture, with graph coloring). For example, - -Four Color Theorem: Colin de Verdière conjectures $\chi(G)\leq\mu(G)+1$ where $\chi(G)$ is the chromatic number of $G$, see [4]. If true, this would prove the Four Color Theorem. - -Graph minor monotone: The property $\mu(G)\leq k$ is closed under taking graph minors of $G$, meaning $\mu(g)\leq \mu(G)$ if $g$ is a minor of $G$, see [1]. So, by the Robertson–Seymour Graph Minor Theorem, the property $\mu(G)\leq k$ can be characterized by a finite number of excluded graph minors. - -Embeddability: $\mu(G)$ characterizes this topological property for several families of graphs: embeddable in a line $(\mu\leq1)$, outerplanar $(\mu\leq2)$, planar $(\mu\leq3)$, or linklessly i.e. flat embeddable in ${\mathbb R}^3$ $(\mu\leq4)$, see [1,2]. - -Embeddings in more general surfaces: If $G$ embeds in the real projective plane or in the Klein bottle, then $\mu\leq5$. If it embeds in the torus, $\mu\leq6$. If it embeds in a surface $S$ with negative Euler characteristic $\psi$, then $\mu\leq 4−2\psi$, see [4] - - -Now, I have two questions, the first one being the main one: - -MAIN QUESTION: Is someone aware of further embeddability characterizations based on $\mu(G)$ beyond the results in bullet point No. 3? In No. 3, we have full characterizations, while the results in No. 4 are just implications for $\mu(G)$ in case $G$ can be embedded, i.e. just in one direction. - - -Quoting [3]: "The parameter was motivated by the study of the maximum -multiplicity of the second eigenvalue of certain Schrödinger operators. These operators -are defined on Riemann surfaces. It turned out that in this study one can approximate -the surface by a sufficiently densely embedded graph $G$, in such a way that $\mu(G)$ is the -maximum multiplicity of the second eigenvalue of the operator, or a lower bound to it." -SECOND QUESTION: So it appears $\mu(G)$ was developed to resolve a problem in Schrödinger operator theory. I wondered when/how the idea emerged to study $\mu(G)$ as a graph invariant in its own right? I looked at [1] and [CV 1] but could not find an answer. - -References -[1] Yves Colin de Verdiere (1990): Sur un nouvel invariant des graphes et un critère de planarité, J. Combin. Th. (B) 50, 11–21. -[2] L. Lovasz & A. Schrijver (1998): A Borsuk theorem for antipodal links and a spectral characterization of linklessly embeddable graphs, Proc. Amer. Math. Soc. 126, 1275–1285. -[3] H. van der Holst, L. Lovasz & A. Schrijver (1999): The Colin de Verdière graph parameter, pp. 29– -85 in: Graph Theory and Combinatorial Biology (L. Lovasz et al., eds.), János Bolyai Math. -Soc., Budapest. -[4] Andries E. Brouwer, Willem H. Haemers (2011): Spectra of graphs, Springer Monograph. -Earlier work that Colin de Verdière cites in his article [1]: -[CV 1] Y. COLIN DE VERDIÈRE, Spectres de variétés riemanniennes et spectres de graphes, Proc. Intern. Congress of Math., Berkeley 1986, 522-530. -[CV 2] Y. COLIN DE VERDIÈRE, Sur la multiplicité de la premiere valeur propre non nulle du laplacien, Comment. Math. Helv. 61 (1986), 254-270. -[CV 3] Y. COLIN DE VERDIÈRE, Sur une hypothèse de transversalité d’Arnold, Comment. Math. Helv. 63 (1988). 184-193. -[CV 4] Y. COLIN DE VERDIÈRE, Constructions de laplaciens dont une partie finie du spectre est donné, Ann. Sci. École Norm. Sup. 20 (1987), 599-615. -https://en.wikipedia.org/wiki/Colin_de_Verdi%C3%A8re_graph_invariant - -REPLY [6 votes]: Embeddability in any surface but the sphere (or plane) can probably not be characterized via the Colin de Verdière number. -Suppose that $K_n$ is the largest complete graph that embedds into a surface $S$. -This shows that the best we can hope for is "$G$ embedds in $S$ $\Leftrightarrow$ $\mu(G)\le\mu(K_n)= n-1$". -The following is still a bit hand-wavy (maybe someone can help): -I can imagine, that a disjoint union of sufficiently many $K_n$ can no longer be embedded into $S$ (except if $S$ is a sphere/plane). My intuition is that any additional $K_n$ must embedd in one of the regions given by the embedding of the previous $K_n$, and this region is probably "of a lesser genus" (if the genus is not already 0). -For example, this is true for $S$ being the projective plane: $K_5$ embedds in $\Bbb R P^2$, but $K_5+K_5$ does not (see here). -Also, a claim in this question seems to support this in the orientable case. -But we also have $\mu(K_n+\cdots +K_n)=\mu(K_n)=n-1$ (see [1]), contradicting the desired characterization. - -[1] van der Holst, Lovász, Schrijver: -"The Colin de Verdière graph parameter", Theorem 2.5<|endoftext|> -TITLE: Smooth functions that resemble random walks -QUESTION [11 upvotes]: If the Riemann hypothesis holds, then the Mertens function $M(n)\equiv\sum_{x\leq n} \mu(n)$ behaves much like a 1D random walk. This includes the statements that - -$M(n)$ changes sign infinitely often -$M(n)=O(\sqrt{n})$ (ignoring subleading logarithmic corrections). - -It is also believed that 3) $\mu(n)=M(n)-M(n-1)$ "looks random". This seems to be a topic of current research, but is sometimes phrased as the "Mobius randomness law" (Eq 5 here), which says that for functions of low complexity $\xi (n)$ -$\sum_{n\leq N} \xi(n) \mu(n) = o(\sum_{n\leq N} |\xi(n)|)$ -Some weaker analogue of this conjecture is proved in the linked note. -Now, the Mertens function can be extended to the reals through an integral expression -$$M(x) = \int^{c+\mathrm{i}\infty}_{c-\mathrm{i}\infty } \frac{ds}{2\pi \mathrm{i}} \, \frac{x^s}{s \zeta(s)}\,\,\,\,\, (*)$$ -My question is: Does anyone know of a function $f:\mathbb{R}\rightarrow \mathbb{R}$, which is a deterministic$\dagger$ combination of known analytic functions (e.g., Eq. $(*)$) and which can also be proved to obey conditions 1., 2., and 3. above? Here 3. would mean that $f(n)-f(n-1)$ "looks random" in some sense, e.g., the sense described above. Perhaps there are many examples; if so, what's the simplest? I don't particularly care whether $f(n)$ takes integer values; I just want it to look like a random walk. -**EDITS/ I've updated the wording of this question. There is an obvious set of examples if I merely insist on 1. and 2. I should have emphasized the need for $f(n)-f(n-1)$ to "look random". -$\dagger=$ I want $f$ to be expressible as a deterministic combination of known functions; so I won't accept e.g., a fourier series with randomly chosen coefficients (see Carlo's answer below). Morally, I'm interested in the appearance of randomness from seemingly deterministic expressions (e.g., $(*)$). - -REPLY [3 votes]: Smooth random functions, random ODEs, and Gaussian processes (2018) describes an approach that takes a finite Fourier series on the interval $(0,1)$ with randomly chosen coefficients. The integral of this function approaches Brownian motion in the limit that the number $M$ of Fourier coefficients tends to infinity. -The plot shows three such functions, for $M=1/\lambda=5,25,$ and $125$. - -For $M=1000$ the curve is a Brownian path within plotting accuracy, the plot below shows 10 realizations.<|endoftext|> -TITLE: Does $\forall x \forall y\ (x \in y) \lor \lnot (x \in y)$ imply excluded middle? -QUESTION [10 upvotes]: Suppose that we take constructive set theory and add the axiom $\forall x \forall y\ (x \in y) \lor \lnot (x \in y)$. Does this imply excluded middle, or are there still some formulas $\varphi$ for which $\varphi \lor \lnot \varphi$ isn't provable using this new axiom? - -REPLY [16 votes]: It depends how much separation is available. If you can construct the set $\{ z \in \{ \emptyset \} \;|\; \varphi \}$ then you can show $\varphi \vee \neg \varphi$. So for theories with full separation, like IZF, you can derive excluded middle, whereas for CZF where you only have separation for bounded formulas, you can only get excluded middle for bounded formulas. -Edit: See Rathjen, Indefiniteness in semi-intuitionistic set theories: On a conjecture of Feferman for a set theory with bounded excluded middle, but in which $\mathbf{CH} \vee \neg \mathbf{CH}$ is unprovable.<|endoftext|> -TITLE: Quiver and relations of Schur algebras -QUESTION [5 upvotes]: Assume that the Schur algebra $S(n,r)$ with $n \geq r$ is not representation-finite. - -Question: For which $n$, $r$ is the quiver and relations of the blocks of $S(n, r)$ explicitly known? - -I just found the cases $n=r=4$ and $n=r=5$ in Xi - On representation types of $q$-Schur algebras. - -REPLY [5 votes]: Erdmann's article Schur algebras of finite type shows that $S(n,r)$ has finite representation type in prime characteristic $p$ if and only if $n=2$ and $r < p^2$ or $n \ge 3$ and $r \le 2p$ or $p=2$, $n=2$ and $r=5$ or $7$. In these cases the quiver and relations for (the basic algebra Morita equivalent to) each block are found explicitly. Quivers and relations are also found for some blocks of Schur algebras of infinite representation type. For example, Proposition 5.2 gives a basic algebra Morita equivalent to the principal block of $S(2,p^2)$ or $S(2,p^2+1)$ when $p > 2$. There are further results in Section 5 on blocks of $S(3,r)$ of infinite type. -More recently Doty, Erdmann, Martin and Nakano have classified all the tame Schur algebras. As one would expect, their paper gives some information about the Ext quivers. For example, see page 153 for the quiver for the basic algebra of $S(2,6)$ in characteristic $2$: it has wild representation type. -I can't find any explicit results in these papers for $n \ge r$. Since the authors' main interest is in the finite/tame/wild distinction, they concentrate on the blocks at the `threshold', where typically $n < r$. The corollary on page 143 of the second paper outlines a method for embedding the module category of $S(n,d)$ in the module category of $S(n',d)$, whenever $n' \ge d$.<|endoftext|> -TITLE: Torsion & Artin groups -QUESTION [11 upvotes]: I have seen it conjectured several times that Artin groups are torsion-free. This is a very basic question one could ask about these groups. Intuitively, to me it seems like it must be true however it seems impossible to prove. I am curious if anyone is actually working on this / what kind of methods people may have tried to prove such a thing. - -REPLY [10 votes]: This is a consequence of the $K(\pi,1)$ conjecture, stating that there is an explicit $K(\pi,1)$ for Artin groups which is the complement of a complex hyperplane arrangement whose real locus are the hyperplanes of the reflections in the associated Coxeter group quotient by the action of the Coxeter group. See Conjecture 2.2 in Paris' survey Lectures on Artin groups and the $K(\pi, 1)$ conjecture for the statement and the discussion after for known cases. Since this is a finite-dimensional space (whose complex dimension is bounded by that of the number of generators), this implies that there cannot be any torsion. Recently Paolini and Salvetti have announced a Proof of the $K(\pi, 1)$ conjecture for affine Artin groups. See also the introduction of their paper for a summary of previous results and McCammond's survey The mysterious geometry of Artin groups: the affine case of the torsion-free problem was solved by McCammond and Sulway in Artin groups of Euclidean type.<|endoftext|> -TITLE: Homology of the étale homotopy type -QUESTION [5 upvotes]: $\DeclareMathOperator\Et{Et}$Let $X$ be a scheme and denote by $\Et(X)$ the associated étale homotopy type. Then by the work of Artin–Mazur, we know that for an abelian group $A$, we have -$$H^n(\Et(X),A)=H^n_{\text{ét}}(X,A)$$ -and -$$\pi^1(\Et(X))\cong \pi^1_{\text{alg}}(X).$$ -Therefore I wonder what $H_n(\Et(X),A)$ is? Naïvely I would hope that it is $H_{n,\text{ét}}(X,A)$, however, it seems that no survey or reference mentions this. - -REPLY [11 votes]: I'm sure there are easier and better ways to think about this, but here's how I like to think about it. -Work on the big pro-etale site on all schemes, which maps to the pro-etale site of a point, $\pi: \mathrm{Sch}_{\mathrm{proet}}\to \ast_{\mathrm{proet}}$. Also, let's consider (hypercomplete) sheaves of anima. Then $\pi^\ast$ has a left adjoint $\pi_\natural$, and (for our given scheme $X$) $\pi_\natural(X)$ is a condensed anima that "is" the etale homotopy type of $X$. Concretely, if $X_\bullet\to X$ is a proetale hypercover by w-contractible $X_\bullet$, then $\pi_\natural(X)$ is represented by the simplicial extremally disconnected profinite set $\pi_0(X_\bullet)$. By adjunction, there is a natural map $X\to \pi^\ast \pi_\natural (X)$. -Now giving a $\mathbb Q_\ell$-local system $\mathbb L$ (same for other coefficient rings) is the same as giving a map $X\to \pi^\ast B\mathrm{GL}_n(\mathbb Q_\ell)$, i.e. equivalently a map $\pi_\natural(X)\to B\mathrm{GL}_n(\mathbb Q_\ell)$, i.e. a $\mathbb Q_\ell$-local system on the condensed anima $\pi_\natural(X)$. -Now what is homology of $X$ with coefficients in $\mathbb L$? One definition uses the formalism of solid $\mathbb Q_\ell$-sheaves $D_\blacksquare(X,\mathbb Q_\ell)$ (no reference for them in the case of schemes yet, sorry!). In that setting, pullback $D_\blacksquare(\ast,\mathbb Q_\ell)\to D_\blacksquare(X,\mathbb Q_\ell)$ has a left adjoint, which takes $\mathbb L$ to the homology of $X$ with coefficients in $\mathbb L$. Here $D_\blacksquare(\ast,\mathbb Q_\ell)$ is the "usual" derived category of solid $\mathbb Q_\ell$-modules. -Working with the condensed anima $\pi_\natural(X)$ instead, one can apply a similar procedure, and the two answers will agree, so indeed the homology of $X$ with coefficients in $\mathbb L$ agrees with the homology of $\pi_\natural(X)$ with coefficients in $\mathbb L$. -In practice, if $X$ is sufficiently nice, then these homology groups will be finite-dimensional over $\mathbb Q_\ell$ and dual to cohomology, so this would indeed follow from the usual statements about cohomology. For general $X$, this statement about homology is however slightly finer (as cohomology is always the dual of homology, but not conversely). -Edit: Here is a way to phrase the answer so that it does not involve a reference to $D_\blacksquare(X,\mathbb Q_\ell)$. One can instead use the full derived category of pro-etale $\mathbb Q_\ell$-sheaves $D(X_{\mathrm{proet}},\mathbb Q_\ell)$; then, just like for sheaves of anima, the pullback along $f: X_{\mathrm{proet}}\to \ast_{\mathrm{proet}}$ has a left adjoint $f_\natural$, and $f_\natural \mathbb L\in D(\ast_{\mathrm{proet}},\mathbb Q_\ell)$ is a complex of condensed $\mathbb Q_\ell$-vector spaces that can be considered as the homology of $\mathbb L$. Again, a similar construction can be done for $\pi_\natural(X)$, and these two notions of homology agree. -On the other hand, I expect that it is extremely difficult to compute this notion of homology even for $X=\mathbb P^1_k$ for $k$ an algebraically closed field. However, passing to the solidification, one can compute it in practice, and I guess it usually agrees with the homology of the Artin--Mazur pro-(homotopy type).<|endoftext|> -TITLE: Updates on a least prime factor conjecture by Erdos -QUESTION [5 upvotes]: In the 1993 article "Estimates of the Least Prime Factor of a Binomial Coefficient," Erdos et al. conjectured that -$$\operatorname{lpf} {N \choose k} \leq \max(N/k,13)$$ -With finitely many exceptional $(N,k)$. Here, $\operatorname{lpf}(x)$ denotes the smallest prime factor of $x$. -I am posting here to ask whether any progress has been made toward this conjecture. - -REPLY [8 votes]: The authors of the paper you mention, Erdos, Lacampagne, and Selfridge, define $p(m)$ to be the least prime divisor of $m$ and concern themselves what can be said about $p(\binom{n}{k}).$ I suspect that Selfridge wrote the article. It has his style of saying a lot in a succinct way which is puzzling but solvable with some thought on the part of the reader. The conjecture stated in the abstract is $$p(\binom{n}{k}) \leq \max(\frac{n}{k},29).$$ That shouldn't be thought of as their big conjecture but rather a terse and amusing way of capturing some of the the main points. -The short answer to your question is that they did a lot of computation, made some observations that had theoretical backing and strong computational support. No-one, as far as I know, has challenged or refuted them and perhaps it isn't especially attractive to try further computation. Or perhaps it is, but not to report "I didn't find anything else either." -Aside: They are perhaps more interested in the growth rate of $g(k),$ the minimal $n>k$ with $p(\binom{n}{k})>k.$ I feel compelled to quote a small stretch of the article: - -That is a whole mess of conjectures, but not snappy enough for an abstract. That is the subject of section 1 of the paper. They and others explored $g(k)$ up to about $k=140$ and with more powerful computers the results were later extended to about $k=200.$ The current record lower bound is $$g(k) \geq exp(c(\log^3k/\log \log k)^{1/2}).$$ -Getting back to the conjecture you ask about, the first puzzle is - -The stated conjecture is clearly false. $\binom{n}{n-1}=n$ is prime when $n$ is. I can imagine Selfridge saying "Well of course we don't mean that." And if you read in further, the investigation is only for $k\frac{n}{k}?$ After all, there is a $0 \leq j \frac{n}{k}.$ -They say that they wrote a program to find all cases of $p(\binom{n}{k})=p>\frac{N}{k}$ with $p>5$ and $k \leq 12000.$ It must not have been entirely exhaustive because they say that there was only one output other than the twelve mentioned for $331 5.$ Anyone familiar with Pascal's Triangle $\mod 2$ will realize that for every $k>2$ there are lots of cases of $p(\binom{n}{k})=3$ with $2k -TITLE: Proof of Rashevskii-Chow theorem -QUESTION [5 upvotes]: I'm looking for a good quotation and comprehensive explaination of the theorem of Chow-Rashewski. -I'm writing my thesis on sub-Riemannian Geometry and a special control problem. Therefore I want to state the theorem of Chow–Rashewski in its sub-Riemannian version and prove it: -Let $M$ be a connected manifold and $\Delta$ a distribution on $M$ that is bracket generation then there is for every $p,q\in M$ a curve that is almost everywhere horizontal that connects $p$ with $q$. -And I also have an additional question sometimes it says that this curve has to be horizontal almost everywhere and sometimes it says everywhere, why? -Thank you all for your explainations so far. I was working now on the book of Agrachev and I understand everthing apart of the connection between the bracket generating condition and the differentials of the functions $\phi_{i}$ could once again please someone help me to understand this last but central step? - -REPLY [4 votes]: As a reference, in addition to the classical ones cited above, I can recommend the following: -Agrachev, Andrei; Barilari, Davide; Boscain, Ugo, A comprehensive introduction to sub-Riemannian geometry., ZBL07073879. -The proof of the Chow-Rashewski theorem is in Section 3.2. An electronic version of the book is also freely available online (https://www.imj-prg.fr/~davide.barilari/ABB-v2.pdf) -The idea is of course the same as the one in the proof given above by Piotr Hajlasz, but I think that the presentation in the book is more geometric and concise. -Concerning your last question (everywhere vs almost everywhere). Horizontal curves might not be differentiable at certain points (e.g. think at a curve with a corner). In order to define a length, the tangent vector of an horizontal curve $\gamma:[0,1]\to M$ should be defined almost everywhere on $[0,1$]. There are then several regularity classes of curves which one might use (all used in the literature): - -$\gamma \in W^{1,1}$ that is absolutely continuous curves (the largest class one can think of) -$\gamma \in W^{1,2}$ that is absolutely continuous curves whose tangent vector is $L^2$ (slightly smaller, but natural in view of minimization of the energy functional, and furthermore the space of "admissible velocities" is Hilbert) -$\gamma \in W^{1,\infty}$ that is curves that are locally Lipschitz in charts (as I comment blow, also this class is natural as one can always reduce to this case when dealing with the length-minimization problem) - -in any case, of course, the tangent vector, which is defined almost everywhere, is required to belong to the sub-Riemannian distribution. The proof of the Chow-Rashevskii theorem shows that connectivity is achieved by horizontal curves that are concatenation of a finite number of smooth curves, which belongs to all the classes above (so the choice of regularity class above is irrelevant). -It turns out that also the sub-Riemannian distance (defined as the infimum of the length of horizontal curves between two points) does not depend on the choice of the regularity class. This is due to the fact that, within a given regularity class ($W^{1,1}$, $W^{1,2}$ or $W^{1,\infty}$) one can always reparametrize the curve, without changing its length, in such a way that the reparametrized curve has constant speed. This is proved in Section 3.6 of the book by Agrachev, Barilari and Boscain. \ No newline at end of file